Algebra
BOARD ANSWER PAPER : MARCH 2015 ALGEBRA Q.1. Attempt any five of the following sub-questions:
i.
The given sequence is 1, 4, 7, 10, …
Here, t1 = 1, t2 = 4, t3 = 7, t4 = 10
t2 t1 = 4 1 = 3
t3 t2 = 7 – 4 = 3
t4 t3 = 10 7 = 3
The difference between two consecutive terms is constant.
The given sequence is an A.P.
ii.
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
iii.
By adding given equations, we get
12 x + 13 y = 29
13 x + 12 y = 21
25x + 25y = 50
25 (x + y) = 50
50
x+y=
25
x+y=2
iv.
Given, Sn =
[½]
[½]
[1]
[½]
[½]
n
n 1
For n = 10
10
10 1
10
S10 =
11
S10 =
v.
vi.
[½]
[½]
The given equation is
x2 + 3x – 4 = 0
Substituting x = 1 in L.H.S of the above equation, we get
L.H.S. = (1)2 + 3 (1) – 4
=1+3–4
=0
L.H.S = R.H.S
1 is the root of the given quadratic equation.
[½]
The given equation is
x+y=5
Substituting x = 3 in above equation, we get
3+y=5
y=2
[½]
[½]
Q.2. Attempt any four of the following sub-questions:
i.
x 2 – 7x + 12 = 0
x 2 – 4x – 3x + 12 = 0
x (x – 4) – 3 (x – 4) = 0
(x – 4) (x – 3) = 0
x–4=0
or
x–3=0
x=4
or
x=3
[½]
[½]
[½]
[½]
[½]
1
Board Answer Paper: March 2015 ii.
iii.
iv.
v.
vi.
The given A.P. is 4, 9, 14, ….
Here, a = 4, d = 9 – 4 = 5, n = 10
Since, tn = a + (n – 1)d
t10 = 4 + (10 – 1)5
= 4 + 45
t10 = 49
[½]
[½]
[½]
[½]
The given equation is 5x + ay = 19
The point (x, y) = (2, 3) lies on the graph of the equation, hence it statifies the equation.
Substituting x = 2 and y = 3 in the given equation, we get
5 (2) + a (3) = 19
10 + 3a = 19
3a = 9
a=3
When a die is thrown,
S = {1, 2, 3, 4, 5, 6}
A = Event of getting an odd number
A = {1, 3, 5}
[½]
[½]
[1]
[1]
[1]
The inter-relation between the measures of central tendency is
Mean Mode = 3(Mean Median)
Mode = Mean 3(Mean Median)
Mode = 101 – 3(101 – 100)
Mode = 101 – 3(1)
Mode = 98
x = 1 is the root of the given equation
kx2 – 7x + 5 = 0
It satifies the given equation
k(1)2 – 7(1) + 5 = 0
k–7+5=0
k–2=0
k=2
[½]
[½]
[½]
[½]
[½]
[½]
[1]
Q.3. Attempt any three of the following sub-questions:
i.
Measures of central
Area (in
Crop
hectares)
angle ()
2 2
Jowar
40
Wheat
60
Sugarcane
50
Vegetables
30
Total:
180
40
360 = 80
180
60
360 = 120
180
50
360 = 100
180
30
360 = 60
180
360
[1½]
Algebra
The pie diagram representation is as follows:
Wheat
Jowar
120
80
[1½]
60
100
Vegetables
Sugarcane
ii.
When two coins are tossed,
S = {HH, HT, TH, TT}
[½]
n(S) = 4
[½]
Let A be the event that at the most one tail turns up
A = {HH, HT, TH}
[½]
n(A) = 3
[½]
P(A) =
n(A)
n(S)
[½]
P(A) =
3
4
[½]
iii.
The given simultaneous equations are
x+y=7
….(i)
and x – y = 5
….(ii)
From equation (i), y = 7 – x
x
0
1
–1
y
7
6
8
(x, y)
(0, 7)
(1, 6)
(–1, 8)
From equation (ii), y = x – 5
x
0
1
–1
y
–5
–4
–6
(x, y)
(0, –5)
(1, – 4)
[½]
(–1, – 6)
3
Board Answer Paper: March 2015 Scale : (On both axes
1 cm = 1 unit)
Y
8
(1, 8)
7
(0, 7)
(1, 6)
6
5
4
3
2
1
(6, 1)
[2]
X
6
5
4
2
3
1
0
1
2
1
3
4
5
6
X
2
3
4
5
(1,6)
(1,4)
(0,5)
6
Y
The two lines intersect at point (6, 1)
(6, 1) is the solution of given simultaneous equations.
Solution set = {(6, 1)}
iv.
The number of seats arranged row wise is as follows:
20, 22, 24, ….
[½]
[½]
This sequence is an A.P. with
a = 20, d= 22 – 20 = 2, n = 22
[½]
Now, tn = a + (n – 1) d
[½]
t22 = 20 + (22 – 1) 2
[½]
= 20 + 21 2
= 62
4 4
The number of seats in the twenty second row is 62.
[1]
Algebra
v.
x2 + 11x + 24 = 0
x2 + 11x = – 24
….(i)
1
[½]
2
Now, third term = coefficient of x
2
1
[½]
2
= 11
2
=
Adding
121
4
[½]
121
on both sides of (i), we get
4
x2 + 11x +
121
121
= – 24 +
4
4
2
11
25
x =
4
2
[½]
Taking square root on both sides, we get
x+
11
5
=
2
2
x=
11 5
2
2
x=
11 5
2
2
or
x=
11 5
2
2
x=
6
2
or
x=
16
2
x=–3
or
x = –8
– 3 and – 8 are the roots of the given quadratic equation.
[1]
Q.4. Attempt any two of the following sub-questions:
i.
The sample space for two digit numbers without repeating the digits is
S = {10, 12, 13, 14, 15, 20, 21, 23, 24, 25, 30, 31, 32, 34, 35, 40, 41, 42, 43, 45, 50,
51, 52, 53, 54}
[1]
P is the event that the number so formed is even.
P = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52 , 54}
[1]
Q is the event that the number so formed is greater than 50.
Q = {51, 52, 53, 54}
[1]
R is the event that the number so formed is divisible by 3.
R = {12, 15, 21, 24, 30, 42, 45, 51, 54}
[1]
5
Board Answer Paper: March 2015 ii.
Age (in years)
Class interval
10 20
20 30
30 40
40 50
50 60
60 70
Total:
No. of patients
Frequency fi
60
42
55 f
70
53
20
fi = 300
Cumulative frequency
(less than type)
60
102 c.f.
157
227
280
300
[1]
Here, fi = N = 300
300
N
=
= 150
2
2
[½]
Cumulative frequency (less than type) which is just greater than (or equal) to 150 is 157.
Median class is 30 40.
[½]
Now, L = 30, f = 55, c.f. = 102, h = 10.
[½]
N
h
Median = L + c.f.
2
f
[½]
= 30 + (150 102)
10
= 30 + 48 0.1818
55
= 30 + 8.73 = 38.73
The median age of a patient is 38.73 years.
[1]
iii.
It is given that
+=5
3 + 3 = 35
Now, ( + )3 = 3 + 32 + 32 + 3
( + )3 = 3 + 3 + 3( + )
(5)3 = 35 + 3(5)
125 = 35 + 15
125 – 35 = 15
15 = 90
[1]
[½]
[½]
=
= 6
the required quadratic equation is
x2 ( + )x + = 0
x2 5x + 6 = 0
i.e.
90
15
[1]
[½]
[½]
Q.5. Attempt any two of the following sub-questions:
i.
The instalments are in A.P.
[½]
Here, S10 = 4000 + 500 = 4500
[½]
Also, n = 10, d = 10
[½]
Now, Sn =
6 6
n
[2a + (n 1)d]
2
[½]
Algebra
10
[2a + (10 1)(10)]
2
S10 =
4500 = 5 [2a + 9 (10)]
4500 = 2a 90
5
900 + 90 = 2a
990 = 2a
[½]
990
2
a = 495
Also,tn = a + (n 1)d
a=
[1]
[½]
t10 = 495 + (10 1)(10)
= 495 + 9 (10)
[½]
= 495 90
t10 = 405
The first instalment is ` 495 and the last instalment is ` 405.
ii.
Let there be x number of tickets each of ` 20 and y number of tickets each of ` 40.
According to the first condition,
x + y = 35
… (i)
According to the second condition,
20x + 40y = 900
[½]
20(x + 2y) = 900
x + 2y =
900
20
x + 2y = 45
… (ii)
[½]
[½]
[½]
[1]
Subtracting equation (i) from (ii)
x + 2y = 45
x + y = 35
() ()
()
y = 10
Substituting y = 10 in equation (i), we get
x + 10 = 35
[1]
x = 35 10
[1]
There were 25 tickets of ` 20 each and 10 tickets of ` 40 each sold.
x = 25
[½]
iii.
Speed (in km/hr)
No. of Students
2030
6
3040
80
4050
156
5060
98
6070
60
7
Board Answer Paper: March 2015 Y
[5]
Scale: On X-axis: 1 cm = 10 km/hr
On Y-axis: 1 cm = 10 students
160
150
140
Histogram
130
120
110
100
No. of Students
90
80
Frequency
Curve
70
60
50
40
30
20
10
X
0
8 8
Y
20
30
40
50
60
Speed (in km/hr)
70
X
Algebra
Q.1.
BOARD ANSWER PAPER : JULY 2015 ALGEBRA Attempt any five of the following sub-questions:
i.
Given, tn = n + 2
For n = 1, t1 = 1 + 2 = 3
For n = 2, t2 = 2 + 2 = 4
ii.
iii.
iv.
v.
2.
3y2 = 10y + 7
3y2 10y 7 = 0 is in the standard form.
4 3
2 7
[½]
= 28 6 = 22
[½]
When two coins are tossed,
S = {HH, HT, TH, TT}
The given sequence is 1, 3, 6, 10, …
Here, t1 = 1, t2 = 3, t3 = 6, t4 = 10
t2 t1 = 3 1 = 2
t3 t2 = 6 3 = 3
t2 t1 t3 t2
The difference between two consecutive terms is not constant.
The given sequence is not an A.P.
Let the length of the rectangle be x cm and its breadth be y cm.
According to the given condition,
2(x + y) = 36
36
x+y=
2
x + y = 18 is the required equation.
iii.
The given A.P. is 7, 13, 19, 25, …
Here, a = 7, d = 13 – 7 = 6, n = 18
Since, tn = a + (n – 1)d
t18 = 7 + (18 – 1)6
= 7 + 102
t18 = 109
When a die is thrown,
S = {1, 2, 3, 4, 5, 6}
P is the event that the number is odd.
P = {1, 3, 5}
iv.
Here, Dx = 18, Dy = 15 and D = 3
Using Cramer’s rule, we get
x=
Dx
D
[1]
[½]
[½]
[½]
x + y = 18
Attempt any four of the following subquestions:
i.
x = 4 is the root of the given equation x2 7x + k = 0
It satisfies the given equation
(4)2 7(4) + k = 0
16 28 + k = 0
–12 + k = 0
k = 12
[1]
= (4 7) (2 3)
vi.
ii.
[½]
[½]
[½]
[½]
[½]
[1]
[½]
[½]
[½]
[½]
[1]
[1]
[½]
1
Board Answer Paper: July 2015
18
3
x=
x=6
[½]
and y =
[½]
D
15
3
y=
y=5
x = 6 and y = 5
v.
Let = 5, = 7
then, + = 5 + 7 = 12
and = 5 (7) = 35
The required quadratic equation is
x2 – ( + ) x + = 0
x2 – 12x + 35 = 0
vi.
3.
Dy
The inter-relation between the measures of central tendency is
Mean – Mode = 3 (Mean – Median)
Mean – 180 = 3 (Mean – 156)
Mean – 180 = 3 Mean – 468
468 – 180 = 3 Mean – Mean
288 = 2 Mean
Mean =
Mean = 144
[½]
[½]
[½]
[½]
[½]
[½]
288
2
[1]
Attempt any three of the following subquestions :
i.
The given equation is 2x 2 + 5x + 2 = 0
Comparing it with ax 2 + bx + c = 0, we get
a = 2, b = 5, c = 2
[½]
2
[1]
b b 4ac
2a
[½]
=
5 (5) 2 4(2)(2)
2(2)
[½]
=
5 25 16
4
x =
5 9
5 3
=
4
4
5 3 2
1
5 3 8
x =
=
=
or x =
=
= 2
4
4
2
4
4
1
, 2 are the roots of the given equation.
2
=
ii.
iii.
2 2
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24,
25, 26, 27, 28, 29, 30}
n(S) = 30
A is the event that ticket drawn bears a number which is a perfect square.
A = {1, 4, 9, 16, 25}
n(A) = 5
Given, t18 = 52 and t39 = 148
Now, tn = a + (n 1)d
t18 = a + (18 1)d
52 = a + 17d
[1]
[1]
[½]
[1]
[½]
Algebra
a + 17d = 52
... (i)
Also, t39 = a + (39 1)d
148 = a + 38d
a + 38d = 148
... (ii)
Adding (i) and (ii), we get
a + 17d = 52
a + 38d = 148
2a + 55d = 200
… (iii)
n
Also, Sn = [2a + (n – 1)d]
2
56
[2a + (56 – 1)d]
S56 =
2
= 28 [2a + 55d]
= 28 (200)
…[From (iii)]
S56 = 5600
Sum of the first 56 terms of the A.P. is 5600.
iv.
[½]
[½]
[1]
[1]
The given equation is
3x – y = –6
3x + 6 = y
y = 3x + 6
x
y
(x, y)
0
6
(0, 6)
–1
3
(1, 3)
1
9
(1, 9)
[½]
Y
Scale : (On both axes
1 cm = 1unit)
(1, 9)
9
8
7
6
(0,6 )
5
4
(–1, 3)
3
2
1
(2, 0)
X
6
5
4
3
2
3x – y = –6
1
0
1
2
3
4
5
6
X
1
[1½]
The points of intersection of graph with X-axis is (–2, 0) and with Y-axis is (0, 6).
[1]
3
Board Answer Paper: July 2015
v.
Part of the
day
Percentage of
electricity
used
Morning
30
Afternoon
40
Evening
20
Night
10
Total:
100
Measure of central
angle ()
30
100
40
100
20
100
10
100
360 = 108
360 = 144
360 = 72
360 = 36
360
[1½]
The pie diagram representation is as follows:
Morning
Afternoon
108
144
36 Night
[1½]
72
Evening
4.
Attempt any two of the following subquestions:
i.
A card is drawn from a pack of 52 cards
n(S) = 52
a.
Let A be the event of getting a king card.
The pack of cards consists of 4 king cards.
n(A) = 4
4
1
n(A)
=
=
P(A) =
52 13
n(S)
b.
Let B be the event of getting a face card.
There are 12 face cards in a pack of cards.
n(B) = 12
[1]
3
n(B) 12
P(B) =
=
=
52 13
n(S)
ii.
3x4 13x2 + 10 = 0
Let x2 = m,
The given quadratic equation becomes
3m2 13m + 10 = 0
3m2 3m 10m + 10 = 0
3m(m 1) 10(m 1) = 0
(m 1)(3m 10) = 0
m 1 = 0 or
3m 10 = 0
m=1
m=
10
3
or
x2 =
10
3
But x = m
4 4
x2 = 1
[1]
[1]
or
2
[1]
[½]
[½]
[1]
[1]
Algebra
x = 1
The solution set is 1, 1,
or
x=
10
3
10
10
,
3
3
[1]
iii.
Bowling speed
(km/hr)
Class interval
85 100
100 115
115 130
130 145
Here, the maximum frequency is 11.
Modal class is 100 115.
Now, L = 100, fm = 11, f1 = 9, f2 = 8, h = 15
Mode = L +
[1]
[1]
f m f1
h
2 f m f1 f 2
11 9
= 100 +
15
2 11 9 8
= 100 +
5.
No. of players
Frequency
fi
9 f1
11 fm
8 f2
5
[½]
[½]
2
15 = 100 + 6 = 106
5
The modal bowling speed of a player is 106 km/hr.
Attempt any two of the following subquestions:
i.
Let there be x number of rows and y number of students in each row
total number of students = xy
According to the first condition,
(y 3)(x + 10) = xy
xy 3x + 10y 30 = xy
3x + 10y 30 = xy xy
3x + 10y = 0 + 30
3x + 10y = 30
3x 10y = 30
… (i)
According to the second condition,
(y + 5)(x 10) = xy
xy + 5x 10y 50 = xy
5x 10y 50 = xy xy
5x 10y = 50 + 0
5x 10y = 50
… (ii)
Subtracting equation (i) from (ii), we get
5x 10y = 50
3x 10y = 30
() (+)
(+)
2x
x=
[1]
[½]
[1]
[1]
= 80
80
2
x = 40
[1]
Substituting the value of x in equation (i), we get
3(40) 10y = 30
120 10y = 30
10y = 30 120
10y = 150
5
Board Answer Paper: July 2015
10y = 150
y = 15
The total number of students = xy = 40 15 = 600.
The total number of students participating in the drill is 600.
[1]
ii.
[½]
Let the temperatures from Monday to Friday in A.P. be a 2d, a d, a, a + d, a + 2d.
According to the first condition,
a 2d + a d + a = 0
3a 3d = 0
ad =0
a=d
According to the second condition,
a + d + a + 2d = 15
2a + 3d = 15
2a + 3a = 15
…[ d = a]
5a = 15
a=3
d=3
150
10
y=
[½]
[1]
[1]
[½]
…[ d = a]
a 2d = 3 2 3 = 3
ad=33=0
a=3
a+d=3+3=6
a + 2d = 3 + 2 3 = 9
The temperatures of each of the five days are 3 C, 0 C, 3 C, 6 C and 9 C
respectively.
Thus,
[½]
[½]
[½]
[½]
iii.
House Rent (in ` per month)
Number of families
400-600
200
800-1000
300
1000-1200
50
Scale: On X-axis: 2 cm = ` 200
On Y-axis: 1 cm = 50 Families
Y
Number of families
600-800
240
300
250
200
150
100
50
X
0
6 6
Y
400
600
800
1000
House rent per month in (`)
1200
X
[5]
Algebra
BOARD ANSWER PAPER : MARCH 2016 ALGEBRA Q.1. Attempt any five of the following sub-questions:
i.
Given, tn = 3n – 4
For n = 1, t1 = 3(1) – 4 = –1
For n = 2, t2 = 3(2) – 4 = 6 – 4 = 2
ii.
iii.
iv.
Given equation is 2x2 – x – 3 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = –1, c = –3
Let = –2 and = –3
the required quadratic equation is x2 – ( + ) x + = 0
i.e., x2 – (–2 – 3) x + (–2) (–3) = 0
i.e., x2 + 5x + 6 = 0
4 2
= (4 1) – (–2 3)
3 1
=4+6
= 10
v.
vi.
2.
S = {Sunday, Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday}
20 30 50
= 25
2
2
30 40 70
= 35
Class mark of 30 – 40 =
2
2
Class mark of 20 – 30 =
Attempt any four of the following subquestions:
i.
Given, a = 11, d = –2
t1 = a = 11
t2 = t1 + d = 11 + (–2) = 9
t3 = t2 + d = 9 + (–2) = 7
The first three terms of the A.P. are 11, 9 and 7.
ii.
iii.
x2 + 11x + 24 = 0
x2 + 8x + 3x + 24 = 0
x (x + 8) + 3 (x + 8) = 0
(x + 8) (x + 3) = 0
x + 8 = 0 or x + 3 = 0
x = –8 or x = –3
[1]
[½]
[½]
[½]
[½]
[1]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
x 5
= 31
3 4
4(x) – 3 (–5) = 31
4x + 15 = 31
4x = 31 – 15
4x = 16
x=4
iv.
When a die a thrown
S = {1, 2, 3, 4, 5, 6}
n (S) = 6
[½]
[½]
[1]
[1]
1
Board Answer Paper : March 2016
a.
A is the event of getting number divisible by 3
A = {3,6}
n(A) = 2
P(A) =
b.
B is the event of getting number less than 5.
B = {1, 2, 3, 4}
n(B) = 4
P(B) =
n(A) 2 1
n(S) 6 3
[1]
n(B) 4 2
n(S) 6 3
[1]
v.
Number of
words
Class
mark (xi)
600 – 800
800 – 1000
1000 – 1200
1200 – 1400
1400 – 1600
Total:
700
900
1100
1300
1500
Mean x
No. of
candidates
frequency (fi)
14
22
30
18
16
f i= 100
fixi
9800
19800
33000
23400
24000
fixi = 10000
[1]
fi xi 110000
= 1100
fi
100
The mean number of words written are 1100.
[1]
vi.
Subject
Marks
Measure of central
angle()
Marathi
95
95
360 = 95
360
Hindi
90
90
360 = 90
360
English
95
95
360 = 95
360
Mathematics
80
80
360 = 80
360
Total
360
360
[1]
The pie diagram representation is as follows:
Hindi
90
95
English
2 2
Marathi
95
80
Mathematics
[1]
Algebra
3.
Attempt any three of the following subquestions :
i.
The given equation is
6x2 7x 1 = 0
Comparing it with ax2 + bx + c = 0,
we get a = 6, b = 7, c = 1
x =
=
[1]
b b 2 4ac
2a
[½]
(7) (7) 2 4(6)(1)
2(6)
[½]
7 73
7 49 24
=
12
12
x=
7 + 73 7 73
,
are the roots of the given equation.
12
12
[1]
a.
Let the three boys be B1, B2, B3 and the two girls be G1 and G2.
A committee of two is to be formed.
The sample space is
S = {B1B2, B1B3, B1G1, B1G2, B2B3, B2G1, B2G2, B3G1, B3G2, G1G2}
n(S) = 10
A is the event that the committee contains at least one boy.
A={ B1B2, B1B3, B1G1, B1G2, B2B3, B2G1, B2G2, B3G1, B3G2}
n(A) = 9
P(A) =
b.
B is the event that the committee contains one boy and one girl.
B = {B1G1, B1G2, B2G1, B2G2, B3G1, B3G2}
n(B) = 6
P(B) =
ii.
[1]
9
n(A)
=
10
n(S)
[1]
6
3
n(B)
=
=
10
5
n(S)
[1]
iii.
Diameter (in mm)
Class interval
Class mark
(xi)
33 35
36 38
39 41
42 44
45 47
Total:
d =
34
37
40 A
43
46
di = xi A
di = xi 40
6
3
0
3
6
No. of screws
Frequency
(fi)
9
21
30
20
18
fi = 98
fi di
51
=
= 0.52
98
fi
Mean x = A + d
= 40 + 0.52
x = 40.52
The mean diameter of the head of a screw is 40.52 mm.
fidi
54
63
0
60
108
fi di = 51
[1½]
[½]
[1]
3
Board Answer Paper : March 2016
iv.
Marks Scored
0 20
20 40
40 60
60 80
80 100
3
8
15
17
7
Number of students
Y
Scale: On X-axis: 2 cm = 20 marks
On Y-axis: 1 cm = 2 Students
18
16
No. of students
14
12
10
8
6
4
2
X
0
20
60
80
Marks
40
Y
100
X
[3]
v.
4 4
Rainfall (in mm)
Class mark
(xi)
–
20 25
25 30
30 35
35 40
40 45
45 50
–
17.5
22.5
27.5
32.5
37.5
42.5
47.5
52.5
No. of Years
Frequency
(fi)
0
2
5
8
12
10
7
0
[1]
Algebra
Y
Scale: On X-axis: 1 cm = 5 mm
On Y-axis: 1 cm = 2 years
12
No. of Years
10
8
6
4
2
X
0
17.5
Y
4.
22.5
27.5
32.5
37.5 42.5 47.5 52.5
Rainfall (in mm)
Attempt any two of the following subquestions:
i.
Given, t11 = 16, t21 = 29
a.
Since, tn = a + (n 1)d
t11 = a + (11 1)d
16 = a + 10d
a + 10d = 16
Also, t21 = a + (21 1)d
29 = a + 20d
a + 20d = 29
Subtracting (i) from (ii), we get
a + 20d = 29
a + 10d = 16
() () ()
10d = 13
d=
13
10
... (i)
[½]
... (ii)
[½]
13
in (i), we get
10
13
= 16
10
a + 13 = 16
a = 16 13
a=3
13
= 1.3
a = 3 and d =
10
The 1st term is 3 and the common difference is 1.3
b.
Now, tn = a + (n 1)d
t34 = 3 + (34 1)1.3
= 3 + 33 1.3
= 3 + 42.9
t34 = 45.9
The 34th term is 45.9
[3]
[½]
Substituting d =
a + 10
X
[½]
[1]
5
Board Answer Paper : March 2016
c.
Given, tn = 55
Since, tn = a + (n 1)d
55 = 3 + (n 1)1.3
55 3 = (n 1)1.3
52 = (n 1)1.3
52
=n1
1.3
520
=n1
13
40 = n 1
n = 40 + 1 = 41
tn = 55 for n = 41
ii.
The given simultaneous equations are
7
13
= 27
2 x 1 y 2
... (i)
13
7
= 33
2 x 1 y 2
... (ii)
Let
1
1
be p and
be q
y2
2 x 1
equations (i) and (ii) becomes
7p + 13q = 27
13p + 7q = 33
Adding equations (iii) and (iv),
7p + 13q = 27
13p + 7q = 33
20p + 20q = 60
20 (p + q) = 60
p+q=
[1]
... (iii)
... (iv)
60
20
p+q=3
Subtracting equation (iii) from (iv),
13p + 7q = 33
7p + 13q = 27
() () ()
6p – 6q = 6
6(p – q) = 6
p–q=
p–q=1
... (v)
[1]
... (vi)
[1]
6
6
Now, adding equations (v) and (vi),
p+q=3
p–q=1
2p
4
2
p=
p=2
Substituting p = 2 in equation (v), we get
2+q=3
q=3–2
q=1
(p, q) = (2,1)
6 6
=4
[1]
Algebra
Resubstituting the values of p and q, we get
2=
2(2x + 1) = 1
4x + 2 = 1
4x = 1 – 2
4x = –1
x=
(x, y) = , 1
4
Let P(C) be x
P(B) = 2 × P(C)
P(B) = 2x
and P(A) = 2 × P(B)
P(A) = 4x
Given, P(A) + P(B) + P(C) = 1
4x + 2x + x = 1
7x = 1
x=
1
4
= ,
7
7
1
2
P(B) = 2x = 2 × =
7
7
1
P(C) = x =
7
4
2
1
P(A) = , P(B) = , P(C) =
7
7
7
iii.
5.
1
1
and 1 =
y2
2 x 1
1
4
and
and
and
and
1(y + 2) = 1
y+2=1
y=1–2
y = –1
and
y = –1
1
[1]
[½]
… (Given)
[½]
… (Given)
[½]
1
7
[1]
[½]
P(A) = 4x = 4 ×
[½]
[½]
Attempt any two of the following subquestions:
i.
Let the divisor be x
quotient = x
[½]
1
x
and remainder = x =
2
2
[½]
Now, Divisor Quotient + Remainder = Dividend.
[½]
x
x x + = 6123
2
x
x2 + = 6123
2
2x2 + x = 2 6123
2x2 + x = 12246
2x2 + x 12246 = 0
Comparing with ax2 + bx + c = 0, we get
a = 2, b = 1, c = –12246
x=
=
=
b b 2 4ac
2a
[½]
….[Multiplying both sides by 2]
[½]
[½]
1 1 4 2 12246
2 2
1 1 97968
4
7
Board Answer Paper : March 2016
1 97969
4
1 313
=
4
1 313
1 313
or
=
4
4
312
314
or
=
4
4
157
x = 78 or x =
2
157
as a divisor cannot be negative as well as a fraction.
But x
2
=
[½]
x = 78
The divisor is 78.
[½]
ii.
The numbers from 50 to 350 which are divisible by 6 are 54, 60, 66, .... 348
This sequence is an A.P. with
a = 54, d = 60 54 = 6, tn = 348
But tn = a + (n – 1) d
348 = 54 + (n 1) 6
348 – 54 = (n 1) 6
294 = (n – 1) 6
[1]
294
=n–1
6
49 = n – 1
49 + 1 = n
n = 50
iii.
[½]
[½]
[1]
n
Now, Sn = [t1 + tn]
2
50
[54 + 348]
S50 =
2
= 25 (402)
S50 = 10050
Also, t15 = a + (15 – 1) d
= 54 + 14 6 = 54 + 84
t15 = 138
The sum of all numbers from 50 to 350 which are divisible by 6 is 10050 and 15th
term of the A.P. is 138.
Let the digit in the hundreath’s place be x and the digit in the unit place be y.
Digit
Original Number
Reversed Number
8 8
[1]
H
x
y
T
x+y+1
y +x+1
[1]
[1]
[½]
U
y
x
The original number = 100x + 10 (x + y + 1) + y
= 100x + 10x + 10y + 10 + y
= 110x + 11y + 10
According to the first condition,
110x + 11y + 10 = 17 [(x) + (x + y + 1) + y]
110x + 11y + 10 = 17 (2x + 2y + 1)
110x + 11y + 10 = 34x + 34y + 17
110x – 34x + 11y – 34y = 17 – 10
76x 23y = 7
[½]
…. (i)
[1]
Algebra
The reversed number = 100y + 10 (y + x + 1) + x
= 100y + 10y + 10x + 10 + x
= 11x + 110y + 10
According to the second condition.
(110x + 11y + 10) + 198 = 11x + 110y + 10
110x – 11x + 11y – 110y = 10 – 198 – 10
99x – 99y = –198
Dividing both sides, by 99 we get
x – y = –2
Multiply equation (ii) by 23, we get
23x 23y = –46
Subtracting (iii) from (i), we get
76x – 23y = 7
23x – 23y = – 46
()
(+)
(+)
53x
= 53
x=1
Substituting value of x in equation (ii), we get
1 – y = –2
1+2=y
y=3
Required Number = 110x + 11y + 10
= 110(1) + 11(3) + 10
= 110 + 33 + 10 = 153
Required Number is 153
[½]
…. (ii)
[1]
…. (iii)
[½]
[½]
[½]
9
Algebra
BOARD ANSWER PAPER : JULY 2016 ALGEBRA Q.1. Attempt any five of the following sub-questions:
i.
By adding given equations, we get
3x + 2y = 10
2x + 3y = 15
5x + 5y = 25
5(x + y) = 25
25
5
x+y=
x+y=5
ii.
The given A.P. is 3, 5, 7, …
t1 = 3, t2 = 5
d = t2 – t1 = 5 – 3
d=2
iii.
[½]
[½]
[½]
When two coins are tossed,
S = {HH, HT, TH, TT}
[1]
iv.
x (x + 3) = 7
x2 + 3x = 7
x2 + 3x – 7 = 0 is standard form.
[1]
v.
mean x =
=
2.
[½]
fi xi
fi
[½]
100
25
mean x = 4
[½]
vi.
Given equation is 2x2 + 18 = 6x
i.e., 2x2 – 6x + 18 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = – 6, c = 18
[1]
Attempt any four of the following subquestions:
i.
The given A.P. is 1, 7, 13, 19, ...
Here, a = 1, d = 7 1 = 6
Since, tn = a + (n 1)d
t18 = 1 + (18 1)6
= 1 + 17 6
= 1 + 102
t18 = 103
The eighteenth term of the given A.P. is 103.
ii.
Let = 3 and = 8
then, + = 11 and = 24
the required quadratic equation is
x2 – ( + )x + = 0
i.e., x2 – 11x + 24 = 0
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
1
Board Answer Paper: July 2016
iii.
The given simultaneous equations are
4x + 3y = 4
6x + 5y = 8
… (i)
… (ii)
Equations (i) and (ii) are in ax + by = c form.
D=
4 3
= (4 5) (3 6)
6 5
= 20 18 = 2 0
Dx =
4 3
= (4 5) (3 8)
8 5
= 20 24 = 4
Dy =
4 4
= (4 8) (4 6)
6 8
= 32 24 = 8
By Cramer’s rule, we get
Dx
D
4
x=
2
x=
and
y=
and
y=
[1]
Dy
D
8
2
x = 2
and y = 4
x = 2 and y = 4 is the solution of the given simultaneous equations.
iv.
The sample space for two digit numbers without repetition of digit is
S = {10, 12, 13, 14, 20, 21, 23, 24, 30, 31, 32, 34, 40, 41, 42, 43}
n(S) = 16
P is the event that the number so formed is an even.
P = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42}
n(P) = 10
v.
The inter-relation between the measures of central tendency is
Mean Mode = 3(Mean Median)
54.6 54 = 3(54.6 Median)
0.6 = 3 (54.6 – Median)
0.6
= 54.6 Median
0.2 = 54.6 Median
3
Median = 54.4
[1]
[½]
[½]
[½]
[½]
[½]
[½]
Median = 54.6 0.2
[1]
vi.
Subject
2 2
Marks
Measure of central
angle()
Marathi
85
85
360 = 85
360
English
90
90
360 = 90
360
Science
85
85
360 = 85
360
Mathematics
100
100
360 = 100
360
Total:
360
360
[1]
Algebra
The pie diagram representation is as follows:
English
90
85
Science
Marathi
85
100
Mathematics
[1]
Attempt any three of the following subquestions :
i.
Number of weeks
Price of sugar per kg (in `)
Frequency (fi)
Class interval
4
28 30
8
30 32
22
32 34
12
34 36
6
36 38
24
Y
Scale: On X- axis: 1 cm = ` 2
On Y- axis: 1 cm = 2 weeks
22
20
18
Number of weeks
3.
16
14
12
10
8
6
4
X
0
Y
30
28
34
32
38
36
Price of sugar per kg. (in `)
X
[3]
3
Board Answer Paper: July 2016
ii.
a.
The sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
A is the event of getting head on the middle coin.
A = {HHH, HHT, THH, THT}
n(A) = 4
P(A) =
b.
B is the event of getting exactly one tail.
B = {HHT, HTH, THH}
n(B) = 3
P(B) =
4
1
n(A)
=
=
8
2
n(S)
[1]
[1]
3
n(B)
=
8
n(S)
[1]
iii.
No. of patients
Frequency
fi
60
42
55 f
70
53
20
fi = 300
Age (in years)
Class interval
10 20
20 30
30 40
40 50
50 60
60 70
Total:
iv.
60
102 c.f.
157
227
280
300
Here, fi = N = 300
300
N
=
= 150
2
2
Cumulative frequency (less than type) which is just greater than (or equal) to 150 is 157.
Median class is 30 40.
Now, L = 30, f = 55, c.f. = 102, h = 10.
N
h
Median = L + c.f.
2
f
= 30 + (150 102)
Cumulative frequency
(less than type)
[½]
[½]
[½]
[½]
10
= 30 + 48 0.1818
55
= 30 + 8.73 = 38.73
The median age of a patient is 38.73 years.
[1]
The given equation is 7y2 5y 2 = 0
Comparing it with ay 2 + by + c = 0, we get
a = 7, b = 5 , c = 2
[1]
y =
=
2
b b 4ac
2a
[½]
(5) ( 5) 2 4(7)(2)
2(7)
[½]
5 25 56 5 81
59
=
=
14
14
14
59
14
5 9 4 2
=
= 1 or y =
=
=
y=
14
14
14
14
7
2
are the roots of given equation.
1,
7
=
4 4
[1]
Algebra
v.
Rainfall (in cm)
No. of Years
Frequency
(fi)
2
5
8
12
10
7
20 25
25 30
30 35
35 40
40 45
45 50
Y
Scale: On X-axis: 1 cm = 5 cm
On Y-axis: 1 cm = 2 years
12
No. of Years
10
8
6
4
2
X
0
20
Y
4.
25
30
35
45
50
40
Rainfall (in cm)
X
Attempt any two of the following subquestions:
i.
a.
Given, t11 = 16, t21 = 29
Since, tn = a + (n 1)d
t11 = a + (11 1)d
16 = a + 10d
a + 10d = 16
Also, t21 = a + (21 1)d
29 = a + 20d
a + 20d = 29
Subtracting (i) from (ii), we get
a + 20d = 29
a + 10d = 16
() ()
()
... (i)
[½]
... (ii)
[½]
10d = 13
d=
13
10
[1]
5
Board Answer Paper: July 2016
Substituting d =
a + 10
13
= 16
10
a + 13 = 16
a = 16 13
a=3
13
a = 3 and d =
= 1.3
10
b.
The 1st term is 3 and the common difference is 1.3
Now, tn = a + (n 1)d
t34 = 3 + (34 1)1.3
= 3 + 33 1.3
= 3 + 42.9
t34 = 45.9
The 34th term is 45.9
ii.
iii.
Since, (3, 1) is the point of intersection of the lines
ax + by = 7 and bx + ay = 5
the point (x, y) = (3, 1) satisfies the two equations.
ax + by = 7
and bx + ay = 5
Putting x = 3 and y = 1 in the above equations, we get
3a + b = 7
and 3b + a = 5
Multiplying equation (iv) by 3, we get
3a + 9b = 15
Subtracting equation (v) from (iii), we get
3a + b = 7
3a + 9b = 15
() () ()
8b = 8
8b = 8
b=1
Putting b = 1 in equation (iv), we get
3(1) + a = 5
3+a=5
a=53
a=2
a = 2 and b = 1.
Let P(C) = x
P(B) = 2 × P(C)
… (Given)
P(B) = 2x
and P(A) = 2 × P(B)
… (Given)
P(A) = 4x
Given, P(A) + P(B) + P(C) = 1
4x + 2x + x = 1
7x = 1
x=
6 6
13
in (i), we get
10
1
7
[1]
[1]
[½]
… (i)
… (ii)
… (iii)
… (iv)
[½]
[½]
… (v)
[½]
[1]
[1]
[½]
[½]
[½]
[1]
Algebra
5.
1
4
= ,
7
7
1
2
P(B) = 2x = 2 × =
7
7
1
P(C) = x =
7
4
2
1
P(A) = , P(B) = , P(C) =
7
7
7
P(A) = 4x = 4 ×
[½]
[½]
[½]
Attempt any two of the following subquestions:
i.
The numbers from 50 to 250 which are divisible by 6 are 54, 60, 66, .... 246
This sequence is an A.P. with
a = 54, d = 60 54 = 6, tn = 246
But tn = a + (n – 1) d
246 = 54 + (n 1) 6
246 – 54 = (n 1) 6
192 = (n – 1) 6
192
=n–1
6
32 = n – 1
32 + 1 = n
n = 33
[1]
[½]
[½]
[1]
n
Now, Sn = [t1 + tn]
2
33
33
[54 + 246] =
(300)
S33 =
2
2
= 33 (150)
ii.
S33 = 4950
Also, t13 = a + (13 – 1) d
= 54 + 12 (6) = 54 + 72
t13 = 126
The sum of all numbers from 50 to 250 which are divisible by 6 is 4950 and 13th term of
the A.P. is 126.
9 x2
[1]
1
1
3 x 20 = 0
2
x
x
Now, x 2
[1]
2
1
1
= x + 2
2
x
x
[½]
The equation becomes
1
2
1
9 x 2 3 x 20 = 0
x
x
1
x
Put x = m
9(m2 + 2) 3m 20 = 0
9m2 + 18 3m 20 = 0
9m2 3m 2 = 0
9m2 + 3m 6m 2 = 0
3m(3m + 1) 2(3m + 1) = 0
(3m + 1)(3m 2) = 0
3m + 1 = 0 or 3m 2 = 0
m=
1
2
or m =
3
3
[½]
…(i)
[½]
[1]
7
Board Answer Paper: July 2016
When m =
x
1
, from equation (i), we get
3
1 1
=
x
3
Multiplying both sides by 3x, we get
3x2 3 = x
3x2 + x 3 = 0
Here, a = 3, b = 1, c = 3
x=
b b 2 4ac 1 (1) 2 4(3)( 3)
=
2(3)
2a
1 1 36
1 37
=
6
6
2
When m = , from equation (i), we get
3
1 2
x =
x 3
[1]
=
Multiplying both sides by 3x, we get
3x2 3 = 2x
3x2 2x 3 = 0
Here, a = 3, b = 2, c = 3
x=
b b 2 4ac
2a
x =
(2) (2) 2 4(3)( 3)
2(3)
=
2 4 36
6
=
2 40
6
=
2(1 10)
2 2 10
1 10
=
=
6
3
6
The solution set is
1 10 1 10 1 37 1 37
,
,
,
3
6
6
3
[½]
iii.
Let the digit in the hundreath’s place be x and the digit in the unit place be y.
[½]
Digit
Original number
Reversed Number
8 8
[1]
H
x
y
T
x+y+1
y +x+1
U
y
x
The original number = 100x + 10 (x + y + 1) + y
= 100x + 10x + 10y + 10 + y
= 110x + 11y + 10
According to the first condition,
110x + 11y + 10 = 17 [(x) + (x + y + 1) + y]
110x + 11y + 10 = 17 (2x + 2y + 1)
110x + 11y + 10 = 34x + 34y + 17
110x – 34x + 11y – 34y = 17 – 10
76x 23y = 7
The reversed number = 100y + 10 (y + x + 1) + x
= 100y + 10y + 10x + 10 + x
= 11x + 110y + 10
[½]
…. (i)
[1]
[½]
Algebra
According to the second condition.
(110x + 11y + 10) + 198 = 11x + 110y + 10
110x – 11x + 11y – 110y = 10 – 198 – 10
99x – 99y = –198
Dividing both sides, by 99 we get
x – y = –2
Multiply equation (ii) by 23, we get
23x 23y = –46
Subtracting (iii) from (i), we get
76x – 23y = 7
23x – 23y = –46
()
(+)
(+)
53x
= 53
x=1
Substituting value of x in equation (ii), we get
1 – y = –2
1+2=y
y=3
Required Number = 110x + 11y + 10
= 110(1) + 11(3) + 10
= 110 + 33 + 10 = 153
Required Number is 153
….(ii)
[1]
….(iii)
[½]
[½]
[½]
9
Geometry
BOARD ANSWER PAPER : MARCH 2015 GEOMETRY
1.
Solve any five sub-questions:
i.
A(ABC)
AB
=
DC
A(DCB)
A(ΔABC)
2
=
3
A(ΔDCB)
ii. ---- [Ratio of the areas of two triangles having equal base is equal
[½]
to the ratio of their corresponding heights.]
[½]
Equation of the line is: y = –2x + 3
comparing it with y = mx + c, we get
[½]
m = 2 and c = 3
slope = 2 and y-intercept = 3
iii.
sin =
opposite side
hypotenuse
sin =
BC
AC
[½]
sin =
1
2
[½]
iv.
Diagonal of a square =
=
[½]
2 (side)
[½]
2 (10)
Diagonal of a square = 10 2 cm
[½]
v.
Volume of a cube = l3
[½]
3
But, volume of cube = 1000 cm
2.
l3 = 1000
l = 10 cm
vi.
If two circles touch internally, then the distance between their centres is difference of their radii.
[½]
the distance between their centres = 5 3 = 2 cm
[½]
----[Taking cube root on both sides]
[½]
Solve any four sub-questions:
i.
sin =
5
13
---- [Given]
We know that, sin2 + cos2 = 1
[½]
2
5
2
+ cos = 1
13
cos2 = 1
cos2 =
144
169
cos =
12
13
[½]
25
169
[½]
----[Taking square root on both sides]
[½]
1
Board Answer Paper: March 2015 ii.
A
D
115
B
C
Draw angle of 115
[½]
Draw arcs
[½]
Obtain point D
[½]
Join BD (angle bisector of ABC)
[½]
iii.
Let C (3, 5) (x1, y1), D (2, 3) (x2, y2)
[½]
slope of line CD =
y2 y1
x2 x1
[½]
=
3 5
2 3
[½]
=
8
5
slope of line CD =
iv.
Radius (r) = 5 cm
8
5
[½]
Length of arc (l) = 10 cm
Area of sector =
=
radius
length of arc
2
[½]
5
10
2
[½]
= 5 5 = 25
Area of sector is 25 cm2.
v.
In PQR,
[1]
Seg RS is the angle bisector of PRQ
2 2
PS PR
SQ QR
6 15
8 QR
QR =
QR = 20
15 8
6
---- [Given]
---- [By property of angle bisector of a triangle]
[½]
[½]
[½]
[½]
Geometry
D
vi.
A
Y
C
B
1
2
1
mAEB =
2
1
mEAD =
2
1
mEAD =
2
mAEB =
X
E
m(arc AYC)
---- [Inscribed angle theorem]
40 = 20
---- (i)
m(arc DXE)
---- [Inscribed angle theorem]
100 = 50
---- (ii)
[½]
---- [Remote interior angle theorem]
---- [From (i) and (ii)]
[½]
---- [D – A B]
[½]
With respect to BAE, EAD is the exterior angle.
EAD = ABE + AEB
50 = ABE + 20
ABE = 30
i.e., DBE = 30
3.
Solve any three sub-questions:
i.
MPN = 40
QMP = 90
QNP = 90
In MQNP,
MPN + QMP + QNP + MQN = 360
[½]
---- [Given]
---- [Tangent is perpendicular to radius]
[1]
---- [Sum of the measures of the angles
of a quadrilateral is 360]
[½]
40 + 90 + 90 + MQN = 360
220 + MQN = 360
MQN = 360 220
MQN = 140
[½]
[1]
ii.
P
M
2.8 cm
C
7 cm
Q
L
Draw the circle of radius 2.8 cm
Plot point L such that ML = 7 cm
Draw perpendicular bisector of seg ML
Draw circle with centre C
Draw tangents LP and LQ
[½]
[½]
[1]
[½]
[½]
3
Board Answer Paper: March 2015 iii.
Let A1 and A2 be the areas of larger triangle and smaller triangle respectively and h1 and h2 be
their corresponding heights.
A1
6
=
5
A2
---- (i) [Given]
h1 = 9
---- (ii) [Given]
A1
h
= 1
A2
h2
[½]
[½]
---- [Ratio of the areas of two triangles having equal base
is equal to the ratio of their corresponding heights.]
[½]
---- [From (i) and (ii)]
[½]
6
5
=
9
h2
h2
=
59
6
h2 =
h2 = 7.5 cm
The corresponding height of the smaller triangle is 7.5 cm.
15
2
iv.
[1]
C
A
45
M
[½]
30 m
B
10 m
D
Let AB and CD represent the height of the two buildings.
AB = 30 m
Let BD represent the width of the road.
BD = 10 m
Draw seg AM seg CD.
CAM is the angle of elevation.
CAM = 45
In ABDM,
B = D = 90
M = 90
---- [Construction]
A = 90
---- [Remaining angle of ABDM]
ABDM is a rectangle
---- [Each angle is 90]
AM = BD = 10 m
---- (i)
[Opposite sides of a rectangle]
DM = AB = 30 m
---- (ii)
In right angled AMC,
tan CAM = tan 45
CM
AM
tan 45 =
1=
CM = 10 m
---- (iii)
CD = CM + DM
---- [CMD]
CD = 10 + 30 = 40 m
---- [From (ii) and (iii)]
The height of the second building is 40 m.
4 4
CM
10
[½]
[½]
[½]
---- [From (i)]
[½]
[½]
Geometry
v.
Given: For the sphere, radius (r) = 4.2 cm
To find: Volume, total surface area
Solution:
4 3
r
3
4 22
= 4.2 4.2 4.2
3 7
[½]
Volume of the sphere =
[½]
= 4 22 1.4 0.6 4.2 = 310.46 cm3
Total surface area of sphere = 4r2
=4
4.
22
4.2 4.2
7
[½]
[½]
[½]
= 4 22 0.6 4.2 = 221.76 cm2
The volume of the sphere is 310.46 cm3 and the surface area of the sphere is 221.76 cm2.
[½]
Solve any two sub-questions:
i.
Given: ABCD is a cyclic quadrilateral.
A
D
O
[½]
B
C
To prove: BAD + BCD = 180, ABC + ADC = 180
Proof:
1
m(arc BCD)
2
1
m BCD = m(arc BAD)
2
mBAD =
[½]
---- (i) [Inscribed angle theorem]
[½]
---- (ii) [Inscribed angle theorem]
[½]
Adding equations (i) and (ii), we get
1
1
m(arc BCD) + m(arc BAD)
2
2
1
mBAD + mBCD = [m(arc BCD) + m(arc BAD)]
2
1
---- [Measure of a circle is 360]
BAD + BCD = 360
2
mBAD + mBCD =
ii.
BAD + BCD = 180
---- (iii)
[½]
In ABCD,
BCD +BAD +ABC +ADC = 360 ---- [Sum of measures of angles of a quadrilateral]
180 + ABC + ADC = 360
---- [From (iii)]
ABC + ADC = 360 180
ABC + ADC = 180
Hence, the opposite angles of a cyclic quadrilateral are supplementary.
L.H.S. = sin6 + cos6
= (sin2 )3 + (cos2 )3
= (sin2 + cos2) [(sin2 )2 sin2 cos2 + (cos2)2]
---- [ a3 + b3 = (a + b) (a2 ab + b2)]
= 1[sin4 sin2 cos2 + cos4 ]
4
2
2
2
---- [ sin2 + cos2 = 1]
2
[½]
4
= sin + 2sin cos 3sin cos + cos
= sin4 + 2sin2 cos2 + cos4 3 sin2 cos2
[½]
[½]
[½]
[½]
[½]
[½]
[½]
5
Board Answer Paper: March 2015 = (sin2 + cos2 )2 3sin2 cos2
= (1)2 3sin2 cos2
= 1 3sin2 cos2
= R.H.S.
sin6 + cos6 = 1 3sin2 cos2
iii.
----[ (a + b)2 = a2 + 2ab + b2] [½]
Given:
[½]
[½]
For the cylindrical part:
20
= 2 cm
10
diameter 2
= = 1 cm
radius (r1) =
2
2
Diameter = 20 mm =
height (h1) = height of test tube height of hemispherical portion
h1 = 15 1 = 14 cm
For the hemispherical portion:
Radius (r2) = 1 cm
To find: Capacity (volume) of the test tube
Solution:
Capacity of test tube
= volume of cylindrical part + volume of hemispherical portion
2
= r12h1 + r23
3
[½]
[½]
[½]
[½]
2
3
= ( 1 1 14) + ( 1 1 1)
[½]
[½]
2
= 14 = 3.14(14 + 0.666)
3
5.
= 3.14 14.666 = 46.05 cm3
Capacity of the test tube is 46.05 cm3.
[1]
Solve any two sub-questions:
i.
E
A
[½]
B
Given:
D
C
In ABC, ray AD bisects BAC
To Prove that:
BD AB
=
DC AC
[½]
Construction: Draw a line parallel to ray AD, passing through point C.
Extend BA to intersect the line at E.
Proof:
In BEC,
seg AD || side EC
---- [By construction]
6 6
BD AB
DC AE
---- (i) [By B.P.T.]
line AD || line EC on transversal BE
BAD AEC
---- (ii) [Corresponding angles]
[½]
[½]
[½]
Geometry
line AD || line EC on transversal AC.
CAD ACE
---- (iii) [Alternate angles]
[½]
Also, BAD CAD
---- (iv) [ Ray AD bisects BAC]
[½]
AEC ACE
---- (v) [From (ii), (iii) and (iv)]
[½]
In AEC,
AEC ACE
---- [From (v)]
AE = AC
---- (vi) [Sides opposite to congruent angles]
[½]
BD AB
=
DC AC
---- [From (i) and (vi)]
[½]
l
ii.
m=2
A (2, 6)
n=3
P (x, y)
[½]
B (3, 4)
Let A (2, 6) (x1, y1), B (3, 4) (x2, y2)
m:n=2:3
Let P divide AB internally in the ratio 2 : 3
By internal division section formula,
[½]
[½]
mx2 nx1 my2 ny1
,
mn
mn
[½]
2(3) 3(2) 2( 4) 3(6)
,
23
23
[½]
P
P
P
P ,
P (0, 2)
6 6 8 18
,
5
5
0 10
5 5
[½]
3
Equation of line with slope (m) =
and passing through the point P (0, 2) (x1, y1) in
2
slope point form y y1 = m(x x1) is
[1]
3
y 2 = (x 0)
2
[½]
2(y 2) = 3x
2y 4 = 3x
3x 2y + 4 = 0
The equation of the line is 3x 2y + 4 = 0.
iii.
RST ~ UAY
S = A = 50
---- [c. a. s. t]
[½]
RS ST RT
UA AY UY
---- (i) [c. s. s. t]
[½]
[½]
Since, corresponding sides are in the ratio 5 : 4
RS ST RT 5
UA AY UY 4
---- [From (i)]
[½]
6
7.5 5
UA AY 4
---- (ii)
[½]
7
Board Answer Paper: March 2015 Consider,
6
5
UA 4
5 UA = 6 4
UA =
UA = 4.8 cm
---- [From (ii)]
6 4
5
[1]
7.5 5
Now, Consider
AY 4
---- [From (ii)]
AY 5 = 7.5 4
AY =
AY = 6 cm
In UAY, UA = 4.8 cm, AY = 6 cm and A = 50
7.5 4
5
[1]
Y
[1]
U
8 8
50
4.8 cm
A
Geometry
BOARD ANSWER PAPER : JULY 2015 GEOMETRY
1.
Solve any five sub-questions:
i.
A ABE
A(BAD)
=
BE
AD
....[Ratio of areas of two triangles having equal base is
equal to the ratio of their corresponding heights.]
ii.
A ABE
A(BAD)
=
6
9
=
2
3
A ΔABE
A(ΔBAD)
[½]
[½]
Diagonal of a square =
=
2 (side)
[½]
2 (16)
Diagonal of a square = 16 2 cm
[½]
iii.
If two circles touch internally, then the distance between their centres is difference of their radii.
[½]
the distance between their centres = 8 3 = 5 cm
[½]
iv.
Given, cos =
But cos 30 =
3
2
3
2
[½]
cos = cos 30
= 30
[½]
v.
Equation of line in slope-intercept form is y = mx + c.
[½]
Here, m = 2 and c = 5
Equation of line is y = 2x + 5.
vi.
Total surface area of cube = 6 l2
[½]
[½]
2
= 6 (9)
2.
Total surface area of cube = 486 cm2
[½]
Solve any four sub-questions:
i.
In ABC,
line l || side BC
AP
AY
=
PB
YC
4
5
=
x
8
x =
x = 10 units
.... [Given]
.... [By B.P.T.]
[½]
[½]
8 5
4
[1]
1
Board Answer Paper : July 2015
ii.
MPN = 40
.... [Given]
QMP = 90
.... [Tangent is perpendicular to radius]
[½]
QNP = 90
In MQNP,
.... [Sum of the measures of the angles of
a quadrilateral is 360]
MPN + QMP + QNP + MQN = 360
40 + 90 + 90 + MQN = 360
220 + MQN = 360
MQN = 360 220
MQN = 140
[½]
[½]
[½]
iii.
3 .5 cm
P
iv.
R
Draw a circle of radius 3.5 cm
[½]
Extend the line passing through R
[½]
Draw the perpendicular at point R
[1]
Y
P
X
O
A
X
Y
The initial arm rotates by 220 in clockwise direction.
The angle is more than 180 and less than 270
2 2
AOP lies between 270and 180.
The terminal arm lies in quadrant II.
[1]
[1]
Geometry
v.
Given, radius (r) = 3 cm
height (h) = 7 cm
Curved surface area of cylinder = 2rh
=2
The curved surface area of the cylinder is 132 cm2.
vi.
Given, radius (r) = 10 cm, central angle () = 72
[1]
r 2
360
[½]
=
72
3.14 10 10
360
[½]
=
1
314
5
Area of sector =
3.
22
3 7 = 132 cm2
7
[1]
Area of sector = 62.8 cm2
[1]
Solve any three sub-questions:
i.
seg AQ is the median on side BC.
BQ = QC =
=
1
BC
2
.... [Q is the midpoint on side BC]
1
10
2
= 5 units
[1]
In ABC,
seg AQ is the median
AB2 + AC2 = 2AQ2 + 2BQ2
2
.... [By Apollonius theorem]
2
122 = 2AQ + 2(5)
122 = 2AQ2 + 50
122 50 = 2AQ2
72 = 2AQ2
AQ2 =
AQ2 = 36
AQ = 6 units
The length of the median on side BC is 6 units.
ii.
Line CM is a tangent at M and line CA is a secant.
CM2 = CA CB
[½]
[½]
72
2
.... [Taking square root on both sides]
.... (i) [Tangent secant property]
[1]
[1]
Line CN is a tangent at N and line CA is a secant.
CN2 = CA CB
.... (ii) [Tangent secant property]
CM2 = CN2
.... [From (i) and (ii)]
CM = CN
.... [Taking square root on both sides]
[1]
[1]
3
Board Answer Paper : July 2015
iii.
T
O
70
60
P
5.4 cm
M
Draw PMT of given measure
Draw
the
perpendicular
bisectors of side PM and side
TM
Draw circumcircle by taking O as
centre
iv.
v.
[1]
[1]
L.H.S. = sec2 + cosec2
=
1
1
cos 2 sin 2
=
sin 2 cos 2
cos 2 sin 2
=
1
cos sin 2
=
1
1
cos 2
sin 2
2
1
1
.... cos
,sin
sec
cosec
[½]
[½]
.... [ sin2 +cos2 = 1]
[1]
[½]
= sec2 ·cosec2
= R.H.S.
[1]
[½]
sec2 + cosec2 = sec2 ·cosec2
Let A (3, 11), B (6, 2), C (k, 4)
Slope of a line =
y2 y1
x2 x1
For line AB:
Let A (3, 11) (x1, y1), B (6, 2) (x2, y2)
Slope of line AB =
2 11
9
9
=
= = 1
6 (3) 6 3 9
.... (i)
[1]
.... (ii)
[1]
For line AC:
Let A (3, 11) (x1, y1), C (k, 4) (x2, y2)
Slope of line AC =
4 4
4 11
7
=
k ( 3) k 3
Geometry
4.
Since, points A, B and C are collinear,
Slope of AB = Slope of AC
1 =
k 3 = 7
k = 7 + 3
k = 4
k=4
The value of k is 4.
7
k3
.... [From (i) and (ii)]
[1]
Solve any two sub-questions:
i.
Given: ABCD is a cyclic quadrilateral.
A
D
[½]
O
B
C
To prove: BAD + BCD = 180, ABC + ADC = 180
[½]
Proof:
mBAD =
1
m(arc BCD)
2
....(i) [Inscribed angle theorem]
[½]
m BCD =
1
m(arc BAD)
2
....(ii) [Inscribed angle theorem]
[½]
Adding equations (i) and (ii), we get
mBAD + mBCD =
1
1
m(arc BCD) + m(arc BAD)
2
2
mBAD + mBCD =
1
[m(arc BCD) + m(arc BAD)]
2
BAD + BCD =
BAD + BCD = 180
1
360
2
[½]
....[Measure of a circle is 360]
....(iii)
[½]
BCD +BAD +ABC +ADC = 360
....[Sum of measures of angles of a quadrilateral]
[½]
180 + ABC + ADC = 360
....[From (iii)]
ABC + ADC = 360 180
ABC + ADC = 180
In ABCD,
Hence, the opposite angles of a cyclic quadrilateral are supplementary.
A
P
ii.
30 45
E
30
[½]
C
24 m
45
B
D
[1]
5
Board Answer Paper : July 2015
Let AB represent the lighthouse and CD be the ship.
CD = 24 m
The distance of the ship from lighthouse is BD
Draw ray AP || seg BD and CE AB
PAC and PAD are the angles of depression
PAC = 30 and PAD = 45
Also, ACE = PAC = 30
....[Alternate angles]
ADB = PAD = 45
In ECDB
B = D = 90
E = 90
....[construction]
C = 90
....[Remaining angle of ECDB]
ECDB is a rectangle
BE = CD = 24 cm
....(i)
EC = BD
....(ii)
In ABD,
tan ADB = tan 45
AB
BD
tan 45 =
1=
AB = BD
In AEC,
tan ACE = tan 30
tan 30 =
1
AE
=
EC
3
1
AB BE
=
EC
3
1
BD 24
=
BD
3
AB
BD
....(iii)
[½]
....[From (i), (ii) and (iii)]
[1]
AE
EC
BD =
3 (BD 24)
BD =
3 BD 24 3
[½]
3 BD BD = 24 3
BD
BD =
=
3 1 = 24 3
24 3
3 1
24 3 ( 3 1)
3 1
3 1
=
24 3 3
2
= 12 3 3
= 36 + 12 3
6 6
= 36 + 12 (1.73)
= 36 + 20.76
BD = 56.76
The distance of the ship from lighthouse is 56.76 m.
[1]
Geometry
iii.
A(4,7)
F
E
B(2, 3)
D
C(0, 1)
Let A (4, 7) (x1, y1); B (2, 3) (x2, y2) and C (0, 1) (x3, y3)
By midpoint formula,
x2 x3 y2 y3 2 0 3 1
,
,
=
2 2
2
2
D (x, y) =
2 4
= , = (1, 2)
2 2
x3 x1 y3 y1 0 4 1 7
,
,
=
2 2
2
2
E (x, y) =
4 8
= , = (2, 4)
2 2
x1 x2 y1 y2 4 2 7 3
,
,
=
2
2
2
2
4 2 10
2 10
, = , = (1, 5)
=
2 2
2 2
F (x, y) =
[1]
For Median AD:
A (4, 7) (x1, y1) and D (1, 2) (x4, y4)
Equation of median AD in two-point form is:
x x1
y y1
x1 x4 y1 y4
x4
y7
=
4 1 7 2
x4
y7
=
4 1
5
x4
y7
=
5
5
x4=y7
xy4+7=0
xy+3=0
For median BE:
B (2, 3) (x2, y2) and E (2, 4) (x5, y5)
Equation of median BE in two-point form is:
....[Multiplying both sides by 5]
[1]
x x2
y y2
x2 x5 y2 y5
x 2
y 3
=
2 2
3 4
x2
y 3
=
4
1
x + 2 = 4 (y 3)
x + 2 = 4y 12
x 4y + 2 + 12 = 0
x 4y + 14 = 0
For median CF:
C (0, 1) (x3, y3) and F (1, 5) (x6, y6)
Equation of median CF in two-point form is:
[1]
x x3
y y3
x3 x6 y3 y6
7
Board Answer Paper : July 2015
5.
x0
y 1
=
0 1
1 5
x
y 1
=
1
4
4x = y 1
4x y + 1= 0
The equations of medians are x y + 3 = 0; x 4y + 14 = 0; 4x y + 1 = 0.
[1]
Solve any two sub-questions:
1
2
i.
Consider, 2ab = 2 A(XYZ) YZ = 2 ( XZ YP) YZ
2ab = XZ YP YZ
Consider, b4 + 4a2 = YZ4 + 4[A(XYZ)]2
1
2
.... (i)
2
[1]
1
4
= YZ4 + 4 XY YZ = YZ4 + 4 XY2 YZ2
2
2
b 4 4a 2
= YZ 2 + XY2
YZ2
In XYZ,
XYZ = 90
YZ 2 + XY2 = XZ2
b4 + 4a2 = YZ 2 XZ2
b 4 4a 2 =
b 4a = YZ XZ
2ab
XZ YP YZ
=
4
2
YZ XZ
b 4a
4
YZ2 XZ2
ii.
In ABC,
A + B + C = 180
A+ 55 + 65 = 180
A + 120 = 180
A = 180 120
A = 60
ABC LMN
A = L = 60
B = M = 55
.... (ii)
[½]
.... (iii) [By Pythagoras theorem]
.... [From (ii) and (iii)]
[½]
[½]
.... (iv)
[1]
4
b + 4a 2
AB
BC
AC
=
=
LM
MN
LN
AC
3
Since,
=
LN
5
AB
BC
3
=
=
LM
MN
5
AB
3
=
LM
5
5.1
3
=
LM
5
3 LM = 5.1 5
LM =
5.1 5
3
[½]
.... [Dividing (i) by (iv)]
2ab
YP =
[1]
.... [Taking square root on both sides]
2
8 8
= YZ4 + XY2 YZ2
b + 4a = YZ [YZ 2 + XY2]
4
.... [Sum of the measures of all angles of a triangle is 180]
[½]
[1]
.... [Given]
....
[c.a.s.t]
....
[½]
.... (i) [c.s.s.t]
[½]
.... (ii) [Given]
.... [From (i) and (ii)]
[½]
Geometry
LM = 8.5 cm
In LMN, LM = 8.5 cm, M = 55, L = 60
[1]
N
[1]
L
60
55
M
8.5 cm
iii. Volume of cylinder = r2h
Now, volume of ink filled in the cylindrical container = 71% of volume of cylindrical
ink container
2
= 71% of r1 h1
71
6 6 14
100
71 3614
cm3 .... (i)
=
100
[½]
[½]
=
Also, the volume of ink filled in the refill = 84 % of volume of the cylindrical refill
= 84% of r22 h 2
84
1 1
12
10 10
100
84 12
cm3
=
10000
[1]
[½]
=
.... (ii)
[1]
Number of refills that can be filled with ink
=
volumeof ink filledin cylindricalcontainer
volumeof ink filledin the refill
71 36 14
100
=
84 12
10000
10000
71 36 14
=
= 71 50
84 12
100
[½]
.... [From (i) and (ii)]
= 3550
The number of refills that can be filled with ink is 3550.
[1]
9
Geometry
BOARD ANSWER PAPER : MARCH 2016 GEOMETRY 1.
Solve any five sub-questions:
i.
DEF ~ MNK
....[Given]
A(DEF)
DE 2
A( MNK) MN 2
....[By theorem on areas of similar triangles]
A(DEF) (2) 2
A( MNK) (5) 2
A(ΔDEF)
4
A(ΔMNK) 25
ii.
In ∆ABC,
A = 30, C = 60, B = 90
[½]
[½]
....[Given]
∆ABC is a 30– 60– 90 traingle
BC =
1
AC
2
BC =
1
16
2
BC = 8
iii.
PQS =
1
m (arc PMQ)
2
....[Tangent secant theorem]
PQS =
1
110
2
....[Given]
PQS = 55
iv.
= –30
....[Given]
cos (–30) = cos 30
....[ cos (–) = cos ]
....[side opposite to 30]
[½]
[½]
cos (–30) =
v.
Slope = tan
[½]
[½]
3
2
= tan 60
Slope =
vi.
By using Euler’s formula,
3
F+V=E+2
6 + V = 10 + 2
V = 12 – 6
V=6
[½]
[½]
[½]
....[ = 60]
[½]
[½]
[½]
1
Board Answer Paper : March 2016
2.
Solve any four sub-questions:
i.
In PRQ
Seg RS is the angle bisector of PRQ
PS
PR
=
SQ
QR
9 18
6 QR
....[Given]
....[By property of angle bisector of a triangle]
[½]
18 6
9
QR =
QR = 12 units
ii.
C
[½]
[½]
B
9
P
A
12
Line AP is a tangent to the circle at A and line PC is a secant.
.... [Tangent secant property]
AP2 = CP BP
(12)2 = CP 9
144 = CP 9
144
CP =
9
CP = 16 units
BP + BC = CP
.... [PBC]
9 + BC = 16
BC = 16 9
BC = 7 units
iii.
[½]
[½]
[½]
[½]
[½]
A
E
O
B
D
C
Draw ABC of given measure
Draw
the
perpendicular
bisectors of side BC and side
2 2
AC
[1]
Draw circumcircle by taking O as
centre
[1]
Geometry
iv.
Y
P
X
O
A
X
[½]
Y
v.
The initial arm rotates by 130 in anti-clockwise direction.
The angle is more than 90 and less than 180
AOP lies between 90and 180.
The terminal arm lies in quadrant II.
radius
length of arc
2
9
= 16
2
[½]
=98
= 72
Area of sector is 72 cm2.
vi.
Radius (r) = 1.4 cm
Surface area of the sphere = 4r2
[½]
[1]
[½]
22
1.4 1.4
=4
7
3.
= 4 22 0.2 1.4
Surface area of the sphere = 24.64 cm2
Solve any three sub-questions:
i.
17 cm
A
11 cm
B
1
OB = BD
2
1
OB = 26
2
[½]
[1]
D
O
C
Let ABCD be the parallelogram and its diagonals AC and BD intersect each other at O.
AB = 11 cm, AD = 17 cm, BD = 26 cm
[1]
Radius (r) = 9 cm
length of arc (l) = 16 cm
Area of sector =
[½]
[½]
.... [Diagonals of a parallelogram bisect each other]
OB = 13 cm
In ABD,
O is the midpoint of diagonal BD
seg AO is the median of ABD
AB2 + AD2 = 2OA2 + 2OB2
(11)2 + (17)2 = 2OA2 + 2(13)2
121 + 289 = 2OA2 + 2 169
410 = 2(OA)2 + 338
410 338 = 2OA2
72 = 2OA2
.... (i)
[½]
.... [Diagonals of a parallelogram bisect each other]
.... [By Apollonius theorem]
[½]
[½]
72
= OA2
2
3
Board Answer Paper : March 2016
OA2 = 36
OA = 6 cm
.... (ii) [Taking square root on both sides]
1
OA = AC
2
1
6 = AC
2
[½]
.... [Diagonals of a parallelogram bisect each other]
.... [From (ii)]
AC = 12 cm
The length of the other diagonal is 12 cm.
[½]
ii.
a.
mPQR =
1
m(arc PCR)
2
mPQR =
1
26
2
mPQR = 13
....(i)
b.
1
mSPQ = m(arc QDS)
2
.... [Inscribed angle theorem]
mSPQ =
mSPQ = 24
SRQ SPQ
.... [Angles subtended by the same arc are congruent]
SRQ = 24
.... (ii)
With respect to ARQ, SRQ is the exterior angle.
SRQ = RAQ + AQR
.... [Remote interior angle theorem]
24 = RAQ + 13
.... [From (i) and (ii)]
RAQ = 24 13
mRAQ = 11
c.
iii.
.... [Inscribed angle theorem]
.... [Given]
[1]
1
48
2
[1]
[1]
A
l
O
K
B
i.
P
4 4
For drawing a circle
of radius 3.5 cm
ii. For drawing chord
BK
iii. For drawing BAK
iv. For drawing BKP
[½]
[½]
[1]
[1]
Geometry
iv.
sec =
2
3
cos =
1
3
= 1 =
sec 2
2
.... [Given]
3
.... cos
1
sec
3
2
cos =
sin2 + cos2 = 1
sin2 +
3
=1
4
3
sin2 = 1
4
43
2
sin =
4
1
2
sin =
4
1
sin =
2
[½]
2
3
= 1
2
sin2 +
But, lies in the quadrant IV.
sin is negative.
sin =
.... [Taking square root on both sides]
1
2
cosec =
1
sin
cosec =
1
= 2
1
2
cosec = 2
Consider,
[1]
[½]
[½]
3
1 cos ec 1 ( 2) 1 2
=
=
=
= 3
1 cos ec 1 ( 2) 1 2 1
v.
1 cosec
= 3
1 cosec
Let P (2, 3) (x1, y1), Q (4, 7) (x2, y2)
The line passes through points P and Q.
Equation of line in two-point form is:
x x1
y y1
=
x1 x2 y1 y2
x2 y 3
=
2 4 37
x2
y 3
=
2
4
4 (x 2) = 2 (y 3)
4x + 8 = 2y + 6
2y = 4x + 6 8
2y = 4x 2
y = 2x 1
.... [Dividing both sides by 2]
y = 2x 1 is the required equation of line which is of the form y = mx + c.
[½]
[½]
[½]
[½]
[½]
[1]
5
Board Answer Paper : March 2016
4.
Solve any two sub-questions:
i.
Given: A circle with centre O, an external point P of the circle. The two tangents through the
point P touch the circle at the points A and B.
To prove: PA = PB
Construction: Draw seg OA, seg OB and seg OP.
[½]
[½]
[½]
A
O
[½]
P
B
ii.
Proof:
PAO = PBO = 90
....[Tangent is perpendicular to radius.]
In the right angled PAO and the right angled PBO,
seg OA seg OB
....[Radii of the same circle]
hypotenuse PO hypotenuse PO
....[Common side]
PAO PBO
....[Hypotenuse side test]
seg PA seg PB
....[c.s.c.t.]
PA = PB
Hence, the lengths of the two tangent segments drawn to a circle from an external point
are equal.
[½]
[½]
[½]
[½]
A
[½]
60
B
30
40 m
C
Let AB represent the height of the tree and BD represent the width of river.
D and C are the initial and final positions of the observer.
DC = 40 m
ADB and ACD are the angles of elevation.
ADB = 60 and ACD = 30
In right angled ABD,
tan 60 =
D
3=
AB =
[½]
AB
BD
3 BD
tan 30 =
6 6
AB
BD
.... (i)
In right angled ABC,
AB
BC
AB
1
=
3 BD DC
AB
1
=
BD
40
3
.... [BDC]
[½]
Geometry
BD 40
=AB
3
3 BD 3 = BD + 40
3BD = BD + 40
3BD BD = 40
2BD = 40
BD =
BD = 20 m
Now, AB =
3 BD =
BD 40
3
.... (ii)
[½]
.... [From (i) and (ii)]
[½]
.... [From (i)]
[½]
[½]
40
2
3 BD = 3 20
AB = 20 3 = 20 1.73
AB = 34.6 m
The height of the tree is 34.6 m and width of the river is 20 m.
iii.
[½]
A(5,4)
[½]
P
B(3,2)
D(x, y) C(1,8)
Let A (5, 4) (x1, y1);
B (3, 2) (x2, y2) and
C (1, 8) (x3, y3).
D (x, y) is the midpoint of BC.
By midpoint formula,
x2 x3 y2 y3
,
2
2
D (x, y) =
3 1 2 8 2 10
,
= ,
= (1, 5)
2 2 2
2
=
[½]
Let D (1, 5) (x4, y4)
For Median AD:
The equation of the median AD in two-point form is:
x x1
y y1
x1 x4 y1 y4
[½]
x5
y4
=
5 1
4 5
[½]
x5
y4
=
5 1
45
x5
y4
=
6
9
x5
y4
=
2
3
.... [Multiplying both the sides by 3]
3 (x 5) = 2 (y 4)
3x 15 = 2y 8
7
Board Answer Paper : March 2016
3x 2y 15 + 8 = 0
3x 2y 7 = 0
[½]
4 8
y1 y3
=
5 1
x1 x3
Slope of line AC =
48
12
=
=3
4
4
=
[½]
The slope of parallel lines are equal.
Slope of line (l) = Slope of line AC = 3
By slope-point form, equation of line l passing through B (3, 2) (x1, y1) and having
slope (m) = 3 is:
[½]
y y1 = m (x x1)
y (2) = 3 [x (3)]
y + 2 = 3 (x + 3)
y + 2 = 3x + 9
3x + 9 y 2 = 0
3x y + 7 = 0.
[½]
The equation of median AD is 3x 2y 7 = 0 and the equation of the line parallel to AC
and passing through point B is 3x y + 7 = 0.
5.
Solve any two sub-questions:
i.
In ABF,
EB = EF = a
.... [Given]
E is midpoint of seg BF.
seg AE is the median
2
2
2
AB + AF = 2AE + 2EB
AB2 + a2 = 2a2 + 2a2
2
2
[½]
2
.... [By Apollonius theorem]
[½]
[½]
2
AB + a = 4a
AB2 = 4a2 a2
AB2 = 3a2
AB = 3 a
.... [Taking square root on both sides]
[1]
In ACE
FC = FE = a
F is the midpoint of seg EC
seg AF is the median
AE2 + AC2 = 2AF2 + 2FC2
8 8
.... [Given]
2
2
2
2
2
[½]
.... [By Apollonius theorem]
2
a + AC = 2a + 2a
2
AC + a = 4a
AC2 = 4a2 a2
AC2 = 3a2
AC =
AB = AC =
3a
[½]
.... [Taking square root on both sides]
3a
[½]
[1]
Geometry
ii.
U
U
[1]
R
S
4.5 cm
H
V
R
Analytical figure
S
4.5 cm
H
V
S1
o
S2
S3
o
S4
S5
X
Draw SHR of given measure
Draw a ray making an acute angle at S
with side SV, mark points S1, S2, …., S5 such that
SS1 = S1S2 = S2S3 = S3S4 = S4S5
Join S3 and H and draw seg S5V parallel
to S3H, where V is the point on extended SH
Draw VU || side HR
SVU is the required triangle similar to SHR.
iii.
For the cylindrical pipe:
diameter = 20 mm
radius =
Rate of flow of water through the pipe = 15 m/minute
= 15 100 cm/minute
= 1500 cm/minute
Water flow’s through a distance (h) of 1500 cm in a minute
diameter 20
= = 10 mm = 1cm
2
2
[1]
[½]
[1]
[1]
[½]
[½]
[½]
9
Board Answer Paper : March 2016
Volume of water flowing through the pipe in 1 minute = r2h
= 1 1 1500
= 1500 cm3
For the conical vessel:
diameter = 40 cm
radius =
diameter 40
= = 20 cm
2
2
[½]
[½]
[½]
depth (h) = 45 cm
volumeof conical vessel
Time taken to fill the conical vessel =
volumeof water flowing through pipe in1minute
1
3
1
3
Volume of conical vessel = r2h = 20 20 45 cm3
[1]
[½]
1
20 20 45
20 20 45
=
= 3
3 1500
1500
6000
=
1500
10
10 = 4 min
Time taken to fill the conical vessel is 4 min.
[1]
Geometry
BOARD ANSWER PAPER : JULY 2016 GEOMETRY Time: 2 Hours
Max. Marks: 40
Note:
i.
Solve All questions. Draw diagrams wherever necessary.
ii.
Use of calculator is not allowed.
iii.
Figures to the right indicate full marks.
iv.
Marks of constructions should be distinct. They should not be rubbed off.
v.
Diagram is essential for writing the proof of the theorem.
Q.P. SET CODE
1.
Solve any five sub-questions:
i.
RP : PK = 3 : 2
A(TRP)
RP
=
A(TPK)
PK
A
----[Given]
---- [Ratio of the areas of two triangles having equal heights
is equal to the ratio of their corresponding bases.]
A(ΔTRP) 3
=
A(ΔTPK) 2
[½]
[½]
ii.
If two circles touch externally, then the distance between their centres is sum of their radii.
[½]
the distance between their centres = 4 + 3 = 7 cm
[½]
iii.
slope = tan
[½]
= tan 45
slope of the line is 1
[½]
iv.
By using Euler’s formula,
F+V=E+2
[½]
12 + V = 30 + 2
V = 32 – 12
V = 20
v.
Diagonal of a square =
[½]
=
2 (side)
2 (8)
Diagonal of a square = 8 2 cm
vi.
[½]
[½]
Y
X
O
A
X
P
Y
The initial arm rotates by 305in anticlockwise direction.
The angle is more than 270 and less than 360.
AOP lies between 270 and 360.
The terminal arm lies in quadrant IV.
[½]
[½]
1
Board Answer Paper : July 2016
2.
Solve any four sub-questions:
i.
M
P
A
B
N
ii.
Draw seg AB of length 9.7 cm
Take point P at a distance 3.5 cm from A
Draw MN AB through point P
[½]
[½]
[1]
The terminal arm passes through (3, 4)
x = 3 and y = 4
r2 = x 2 + y2
[½]
2
2
2
r = (3) + (4)
r2 = 9 + 16
r2 = 25
r=5
---- [Taking square root on both sides]
Let the angle formed be .
By definition of trigonometric ratios in standard position, we get
sin =
y
4
=
r
5
[1]
DMN and AQR are similar.
Reason:
In DMN and AQR,
DMN AQR
DNM ARQ
DMN AQR
iv.
m (arc AKC) + m (arc BMC) = 180
40 + m (arc BMC) = 180
m (arc BMC) = 180 – 40
m (arc BMC) = 140
v.
Slope of line (m) =
iii.
2 2
[1]
---- [Each is 55]
---- [Each is of same measure]
---- [By AA test of similarity]
[1]
[½]
[½]
[1]
6
and the line passes through the point P(0, 6) (x1, y1)
7
Equation of line in slope point form is:
y y1 = m(x x1)
[½]
6
y 6 = (x 0)
7
6
y6= x
7
7(y 6) = 6x
7y 42 = 6x
6x 7y + 42 = 0
The equation of line is 6x 7y + 42 = 0.
[½]
[½]
[1]
Geometry
vi.
radius (r) = 21 cm, central angle () = 60
r2
360
60o 22
21 21
=
360 7
1
= 22 3 21
6
1386
=
6
Area of sector =
3.
Area of sector = 231 cm2
Solve any three sub-questions:
i.
Line CB is a tangent to the circle at B and line AC is a secant
CB2 = CA CD
….(i) [Tangent secant property]
Since, line CB is tangent at B and AB is a diameter
ABC = 90
In right angled ABC
….[By Pythagoras Theorem]
AB2 + BC2 = AC2
2
2
AB + CA CD = AC
….[From (i)]
2
2
(2r) + AC CD = AC
(2r)2 = AC2 – AC CD
4r2 = AC (AC – CD)
4r2 = AC AD
….[A D C]
2
i.e. 4(radius) = AC AD
ii.
PQ
QT
=
PR
TR
3.6
4
=
x
5
3.6 5 = 4 x
3.6 5
=x
4
x = 4.5 cm
PR = 4.5 cm
iii.
---- [By property of angle bisector of a triangle]
[1]
[½]
[½]
[½]
[½]
[1]
[½]
[1]
---- [ PR = x]
Now, QR = QT + TR
---- [QTR]
QR = 4 + 5
QR = 9 cm
Perimeter of PQR = PQ + QR + PR
= 3.6 + 9 + 4.5 = 17.1 cm
The value of x is 4.5 cm and the perimeter of PQR is 17.1 cm.
l:b:h=5:4:2
---- [Given]
Let the common multiple be x
length (l) = 5x, breadth (b) = 4x, height (h) = 2x
Total surface area of cuboid = 2(lb + bh + hl)
But, total surface area of cuboid = 1216 cm2 ---- [Given]
2(lb + bh + hl) = 1216
(lb+ bh + hl) =
[½]
In PQR,
Ray PT is the angle bisector of QPR.
[½]
[½]
[½]
[½]
[½]
[½]
1216
2
3
Board Answer Paper : July 2016
5x 4x + 4x 2x + 2x 5x = 608
20x2 + 8x2 + 10x2 = 608
38x2 = 608
x2 =
x2 = 16
x=4
length = 5x = 5 4 = 20 cm
608
38
---- [Taking square root on both sides]
[1]
breadth = 4x = 4 4 = 16 cm
height = 2x = 2 4 = 8 cm
The dimensions of the cuboid are (20 16 8) cm.
[1]
iv.
R
I
S
v.
M
7 cm
L.H.S. =
1 cos A
=
1 cos A
=
(1 cosA)2
1 cos2 A
=
(1 cosA)2
sin 2 A
1 cosA 1 cos A
1 cos A 1 cos A
4 4
---- [On rationalising the denominator]
[1]
[1]
[1]
[½]
[½]
----[ sin2 = 1 cos2 ]
[½]
=
1 cos A
sin A
[½]
=
1
cos A
sin A sin A
[½]
= cosec A cot A = R.H.S
Draw RST of given measure
Draw the angle bisectors of S
and T
Draw incircle by taking I as centre
T
1 cos A
= cosec A cot A
1 cos A
….[
1
= cosec ]
sin
[½]
Geometry
4.
Solve any two sub-questions:
i.
Given: ABCD is a cyclic quadrilateral.
A
D
[½]
O
B
C
To prove: BAD + BCD = 180, ABC + ADC = 180
Proof:
1
m(arc BCD)
2
1
m BCD = m(arc BAD)
2
mBAD =
[½]
---- (i) [Inscribed angle theorem] [½]
---- (ii) [Inscribed angle theorem] [½]
Adding equations (i) and (ii), we get
1
1
m(arc BCD) + m(arc BAD)
2
2
1
mBAD + mBCD = [m(arc BCD) + m(arc BAD)]
2
1
---- [Measure of a circle is 360]
BAD + BCD = 360
2
mBAD + mBCD =
[½]
BAD + BCD = 180
---- (iii)
In ABCD,
BCD +BAD +ABC +ADC = 360 ---- [Sum of measures of angles of a quadrilateral]
180 + ABC + ADC = 360
---- [From (iii)]
ABC + ADC = 360 180
ABC + ADC = 180
Hence, the opposite angles of a cyclic quadrilateral are supplementary.
[½]
ii.
[½]
[½]
A
[½]
60
30
C
40 m
Let AB represent the height of the tree and BD represent the width of river.
D and C are the initial and final positions of the observer.
DC = 40 m
ADB and ACD are the angles of elevation.
ADB = 60 and ACD = 30
In right angled ABD,
AB
tan 60 =
BD
AB
3 =
BD
B
AB =
3 BD
D
---- (i)
[½]
[½]
5
Board Answer Paper : July 2016
In right angled ABC,
AB
BC
tan 30 =
AB
1
=
3 BD DC
AB
1
=
3 BD 40
BD 40
=AB
3
3 BD =
3 BD 3 = BD + 40
3BD = BD + 40
3BD BD = 40
2BD = 40
BD =
BD = 20 m
BD 40
3
---- (ii)
[½]
---- [From (i) and (ii)]
[½]
40
2
[½]
3 BD = 3 20
Now, AB =
---- [BDC]
---- [From (i)]
[½]
AB = 20 3 = 20 1.73
AB = 34.6 m
The height of the tree is 34.6 m and width of the river is 20 m.
iii.
Given, diameter = 0.9 m
radius (r) =
[½]
diameter 0.9
=
= 0.45 m
2
2
[½]
height (h) = length of roller = 1.8 m
Curved surface area of roller = 2rh
[½]
= 2 3.14 0.45 1.8
= 6.28 0.81 = 5.0868 m2
[1]
Area of ground pressed by roller in 1 revolution = curved surface area of roller
[½]
2
= 5.0868 m
Area of ground pressed by roller in 500 revolutions = 500 5.0868
= 500
=
5.
i.
Given:
In PQR, line l || side QR.
Line l intersects side PQ and side PR in
points M and N respectively, such that
PMQ and PNR.
6 6
50868
10000
50868
= 2543.4 m2.
20
Area of the ground pressed by the roller in 500 revolutions is 2543.4 m2.
Solve any two sub-questions:
[½]
[1]
Geometry
P
N
M
l
[½]
Q
R
PM
PN
=
MQ NR
[½]
Construction: Draw seg QN and seg RM.
[½]
To Prove that:
Proof:
In PMN and QMN, where PMQ,
A(PMN)
PM
=
A(QMN) MQ
---- (i) [Ratio of the areas of two triangles having equal heights
is equal to the ratio of their corresponding bases.]
[1]
In PMN and RMN, where PNR,
A(PMN)
PN
=
A(RMN)
NR
---- (ii) [Ratio of the areas of two triangles having equal heights
is equal to the ratio of their corresponding bases.]
[1]
A(QMN) = A(RMN)
---- (iii) [Areas of two triangles having equal bases and
equal heights are equal.]
[½]
A(PMN)
A(PMN)
=
A(QMN)
A(RMN)
---- (iv) [From (i), (ii) and (iii)]
[½]
PM
PN
=
MQ NR
---- [From (i), (ii) and (iv)]
[½]
ii.
Let A (4, 8) (x1, y1);
B (5, 5) (x2, y2);
C (2, 4) (x3, y3) and
D (1, 7) (x4, y4)
Slope of a line AB =
=
y2 y1
5 8
=
5 4
x2 x1
3
= –3
1
[½]
Slope of line BC =
y3 y 2
4 5 1 1
=
= =
x3 x2
2 5 3 3
---- (ii)
[½]
Slope of line DC =
y 4 y3
7 4 3
=
= = –3
x4 x3
1 2 1
---- (iii)
[½]
Slope of line AD =
y4 y1 7 8
1 1
=
= =
x4 x1 1 4
3 3
---- (iv)
[½]
---- [From (i) and (iii)]
[½]
Slope of line AB = Slope of line DC
---- (i)
line AB || line DC
---- (v)
[½]
Slope of line BC = Slope of line AD
---- [From (ii) and (iv)]
[½]
line BC || line AD
---- (vi)
[½]
ABCD is a parallelogram
---- [From (v) and (vi)]
Points (4, 8), (5, 5), (2, 4) and (1, 7) are the vertices of a parallelogram.
[1]
7
Board Answer Paper : July 2016
iii.
T
E
T
120
A
E
6.3 cm
H
[1]
M
Analytical figure
120
A
M
H
6.3 cm
A1
A2
A3
A4
A5
A6
A7
Draw AMT of given measure
Draw a ray making an acute angle at A
with side AM, mark points A1, A2, …., A7 such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
Join A7 and M and draw seg A5H parallel
to A7M, where H is the point on AM
Draw HE || side MT
AHE is the required triangle similar to AMT.
8 8
X
[1]
[½]
[1]
[1]
[½]
Science and Technology
BOARD ANSWER PAPER : MARCH 2015 SCIENCE AND TECHNOLOGY
SECTION A
1.
(A)
(a)
(b)
Rewrite the following statements with suitable words in the blanks:
i.
The device used for producing electric current is called a generator.
ii.
Stratosphere, the second layer of the atmosphere reaches 48 km above the
earth’s surface.
2.
Column A
eosin
oxidation
2
1
Column B
synthetic indicator
losing hydrogen
(c)
(B)
Rewrite the following statements by selecting the correct options:
i.
(C) When phenolphthalein is added to NaOH, the colour of the solution will become
pink.
ii.
[1]
Rewrite the following table so as to match the second column with the first column: i.
ii.
[1]
CaOCl2
(D)
[1]
[1]
[1] [1]
If the potential difference across the ends of a conductor is 220 V and the
resistance of the conductor is 44 (ohm), then the current flowing through is
5 A.
[1]
iii.
(B)
1 A = 103 mA
[1]
iv.
(B)
The distance between principal focus and optical centre of the lens is focal length.
[1]
v.
(B)
When rays of light are incident on a glass slab, then the incident ray and
emergent ray are parallel to each other.
[1]
Answer any five of the following:
i.
a.
Scattering of light depends upon the particle size and wavelength of the incident light.
b.
In the visible range of light, red light has greater wavelength.
c.
Red light can travel larger distance without getting scattered.
d.
This enables viewer to spot red colour distinctly even from a distance.
Hence, danger signals are red in colour.
[½]
[½]
[½]
[½]
ii.
[2]
CuO + 2HCl CuCl2 + H2O
Copper
chloride
iii.
iv.
Water
Newlands’ Octaves law states that “When the elements are arranged in an increasing order of
their atomic masses, the properties of the eighth element are similar to the first.”
[2]
1 = 3 108 m/s
2 = 1.5 108 m/s
To find:
12
Formula: 12 = 1
2
From Formula,
3108
=
=2
1 2
1.5108
[1]
The refractive index of the medium with respect to air is 2.
[1]
Given:
1
Board Answer Paper : March 2015
v.
a.
b.
c.
d.
Resistances in series
The voltage across each resistor is
different.
The current through each resistor is the
same.
This combination is used to increase
the effective resistance of the circuit.
This combination decreases the current
in the circuit.
Resistances in parallel
The voltage across each resistor is same.
The current in various resistors are
inversely proportional to the resistances.
This combination is used to decrease the
effective resistance of the circuit.
This combination increases the current in
the circuit.
[2]
vi.
A
B1
C
B
F
P
A1
Object (AB) between centre of curvature and focus
Image (A1B1) beyond centre of curvature, real, inverted and magnified.
3.
Answer any five of the following:
i.
The role of citizen in pollution control:
a.
Citizens should plant trees and develop gardens, parks and open grounds in the locality.
b.
Citizens should save fossil fuels and reduce pollution.
c.
They should minimize electricity consumption.
d.
They should use public transport instead of private vehicles.
e.
Citizens should use non-conventional source of energy like solar, wind energy, tidal energy.
f.
They should maintain vehicles in well tuned condition.
g.
Citizens should keep home and public places clean and should keep their own locality
free from pollution.
[Any six points: ½ mark each]
[3]
ii.
[1]
iii.
a.
b.
The band of coloured components of a light beam is called as spectrum.
White light is composed of seven colours. When white light is incident a prism, it splits
up into its constituent colours.
c.
Each colour bends through different angles with respect to incident ray. So the rays of
each colour emerge along different path and become distinct.
Hence, we get a spectrum of seven colours when white light is dispersed by a prism.
a.
b.
iv.
a.
The four most common electrical appliances based on heating effect of electric current
are electric iron, electric heater, electric toaster and electric oven.
Finely heated platinum wire is used in surgery for cutting tissues much more efficiently
than a knife.
Gypsum is obtained on mixing Plaster of Paris with water.
2CaSO4.H2O + 3H2O 2CaSO4.2H2O + Heat.
Plaster of Paris
b.
c.
2 2
[2]
Water
[1]
[1]
[2]
[1]
[1]
Gypsum
Gypsum is used as a raw material in manufacturing cement.
POP is used in making statues and decorating roofs. It is also used in surgical
bandages.
[1]
[1]
Science and Technology
v.
Metals
Mg, Hg
vi.
a.
b.
Non-metals
C, S
12 + 2.5 + 2.5
4
4
c.
Metalloids
Si, As
2
[3]
[1]
=
17
=
[1]
2
2
=
[1]
3
4.
2
Attempt any one of the following:
(A) a.
Short circuiting takes place if a live wire and a neutral wire come in direct contact or
touch each other.
b.
During a short circuit, the resistance of the circuit becomes very small.
c.
A huge amount of current flows through the circuit during a short circuit.
d.
The flow of a large amount of current in the circuit beyond the permissible value (of
current) is called overloading. It may cause fire.
e.
The effects of overloading can be avoided by not connecting many appliances (especially
of high power rating) at a time in the circuit.
(B)
a.
b.
c.
d.
e.
Negative power indicates that 8 students use spectacles having concave lens.
Positive power indicates that 2 students use spectacles having convex lens.
Generally, most of the students use the spectacles having concave lenses of suitable focal
length.
Most of the students suffer from myopia or near-sightedness.
Two possible reasons for myopia:
1.
The ciliary muscles are unable to relax sufficiently.
2.
Increase in the distance between eye lens and retina due to lengthening of eyeball
or curved lens.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[½]
[½]
SECTION B
5.
(A)
(a)
(b)
(B)
6.
Find the correlation in the given pair and rewrite the answer:
i.
Tinning : Tin :: Galvanizing : Zinc.
ii.
Mammals : Reptiles :: Amphibia : Fishes.
[1]
[1]
State True or False:
i.
True
ii.
False
iii. True
[1]
[1]
[1]
Rewrite the following statements by selecting the correct options:
i.
(A) The molecular formula of acetic acid is CH3COOH.
ii.
(C) Carbon dioxide enters into the leaves through tiny pores present on the surface of
the leaf called stomata.
iii. (A) CuSO4 solution is blue in colour.
iv.
(D) Yeast reproduces by budding.
v.
(B) Raisins put in water absorb water by the process of osmosis.
Solve any five of the following:
i.
a.
Common salt is an ionic compound having strong force of attraction between the
oppositely charged Na+ and Cl ions.
b.
So, a large amount of heat energy is required to break these forces of attraction and to
melt or boil the common salt.
Thus, common salt has high melting point and boiling point.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
3
Board Answer Paper : March 2015
ii.
The pancreas with their associated structures:
Gall bladder
Stomach
[2]
Bile duct
Common duct
Pancreatic duct
iii.
a.
b.
iv.
Pancreas
Connecting link between Peripatus and Annelida:
Segmental nephridia, thin cuticle and parapodia like appendages.
Connecting link between Peripatus and Arthropoda:
Trachea and open circulation.
Two plant hormones and their functions:
a.
Gibberellins: Help in growth of stem
b.
Cytokinins: Promote cell division
[1]
[1]
[1]
[1]
v.
i.
ii.
iii.
Toilet soap
High quality of fats and oils are used as
raw materials.
Expensive perfumes are added.
Toilet soaps do not contain free alkalies
which are harmful to skin.
Laundry soap
Cheaper quality of fats and oils are
used as raw materials.
Cheaper perfumes are added.
Laundry soaps contain free alkalies
which contribute towards its cleaning
action.
[Any two points :1 mark each]
vi.
7.
4 4
Objectives of sustainable development are :
a.
Reduce pollution by using eco-friendly technology.
b.
Restrain the use of natural resources to ensure availability for the future generation.
c.
Protection of environment.
d.
Social equality in accessing resources.
e.
Continuous economic growth.
[Any four points :½ mark each]
Answer any five of the following:
i.
Alloy:
a.
An alloy is a homogenous mixture of two or more metals or a metal and a non-metal in
definite proportion.
b.
Alloys do not corrode easily.
Example:
a.
Brass (copper and zinc)
b.
Bronze (copper and tin)
Cells that assist the neuron in their function Neuroglia
The small gap between the consecutive neurons Synapse
Part of the brain that co-ordinates the voluntary functions Forebrain (Cerebrum)
ii.
a.
b.
c.
iii.
Fertilization:
a.
During the process of fertilization, sperms enter through the vaginal passage, travel
upwards and reach the oviduct.
b.
In oviduct, one of the sperm fuses with egg and the fertilization is completed.
Development:
a.
After fertilization, the egg cell (zygote) begins to divide and redivide to form a ball of
cells called a blastocyst.
b.
This blastocyst implants itself in the wall of the uterus.
[2]
[2]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Science and Technology
c.
The embryo develops inside the uterus. It obtains all the nutrients and oxygen from its
mother’s blood through the umbilical cord.
d.
Development of the foetus takes place for nine months.
Birth:
a.
Nine months onwards, the baby is ready to take birth. It begins to move down towards
the vaginal passage.
b.
During birth, the cervix gradually opens and the baby is released through the vagina.
iv.
v.
Vestigial organs:
Vestigial organs are the non-functional organs in some organisms but have essential functions
in other organisms.
Examples of vestigial organs in human beings: Vermiform appendix, wisdom teeth.
Examples of vestigial organs in plants: Scale like leaves on Indian pipe plant, Stamens
which lack anthers in some flowers.
a.
b.
c.
vi.
8.
a.
b.
Recycling is a type of green technology that uses old material to make new products. Many
waste products from the industries such as paper, glass, plastics and metals can be recycled.
Example: Old used tyres are recycled to create play ground flooring in parks to provide
soft surfaces that increase the safety of the children playing there.
Advantages of recycling:
1.
It conserves energy and raw materials.
2.
It saves space used in landfills.
3.
It protects environment by effective handling of waste materials.
4.
It reduces the cost of production.
[Any two points:½ mark each]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
Sexual reproduction gives rise to variation.
Importance of variation in survival of species:
Variations give rise to variety and diversity. They enable organisms to adapt and
survive in the changing environmental conditions. Variations also help to prevent the
complete extinction of plant and animal species.
[1]
[2]
(A)
i.
ii.
iii.
iv.
v.
Anaerobic reaction = CO2 + Ethanol
Reaction in human muscles = Lactic acid
Aerobic respiration = CO2 + H2O
Reaction in plant cells = Starch
Reaction in liver = Glycogen
[1]
[1]
[1]
[1]
[1]
(B)
i.
ii.
The other two names of ethanol: Ethyl alcohol, Spirit.
The structural formula of ethanol:
[1]
H
iii.
iv.
H
H
C
C
OH
H
H
Properties of ethanol:
i.
It is a colourless liquid.
ii.
It is combustible and burns with a blue flame.
Action of phosphorus trichloride on ethanol:
When ethanol reacts with phosphorus trichloride (PCl3), it forms ethyl chloride and
phosphorus acid.
3C2H5OH + PCl3 3C2H5Cl + H3PO3
Ethanol
Phosphorus
trichloride
Ethyl
chloride
Phosphorus
acid
[1]
[1]
[2]
5
Science and Technology
BOARD ANSWER PAPER : JULY 2015 SCIENCE AND TECHNOLOGY SECTION A
1.
(A)
(a)
i.
ii
Fill in the blanks:
Very fine particles mainly scatter blue light.
1 mA = 103 A.
(b)
Match the column ‘A’ with column ‘B’:
i.
ii.
iii.
(B)
i.
ii.
iii.
iv.
v.
2.
2.
3.
1.
Column ‘B’
Scandium
Gallium
Germanium
Choose the correct alternative and rewrite the following:
Mirror used by a dental surgeon is concave.
(D) If the potential difference across it is doubled and the resistance is halved, current
passing through a resistance becomes four times.
(B) If three equal resistances are given, they can be arranged in four combinations.
(B) Inside water, an air bubble behaves always like a concave lens.
(D) A glass slab is placed in the path of convergent light. The point of convergence of light
undergoes lateral shift.
Answer the following questions (any five):
i.
Dobereiner’s triads:
i.
In 1829, Dobereiner classified existing elements in a tabular form by placing three
elements having similar properties in a group called triad.
ii.
In each triad, the elements were placed according to increasing order of their atomic
masses.
iii. The atomic mass of the middle element in each triad was approximately the mean of
the atomic masses of other two elements.
Eg. In the triad of Lithium, Sodium and Potassium, the atomic mass of Sodium (23) is the
mean of atomic masses of Lithium (6.9) and Potassium (39).
ii.
iii.
Column ‘A’
Eka-boron
Eka-Aluminium
Eka-Silicon
When edible oil is stored in an iron or tin container for a long time, it undergoes
oxidation reaction.
b. Due to oxidation reaction, the taste and smell of edible oil changes and it becomes rancid,
if food is cooked in this oil, its taste changes.
Hence, edible oil is not allowed to stand for a long time in an iron or tin container.
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[½]
[½]
[½]
[½]
a.
Power (P) = 750 W
Time (t) = 2 hrs / day
Number of days (n) = 30
To find: Energy consumed = ?
Formula: Energy consumed = P t n
Solution:
From formula,
Energy consumed = 750 2 30
= 45000 W h
= 45 kWh
Energy consumed = 45 kWh
[1]
[1]
Given:
[1]
[1]
1
Board Answer Paper : July 2015
iv.
i.
ii.
iii.
Conductors
Substances having very low resistance
are called conductors.
They allow charges to flow freely.
iv.
They contain large number of free
electrons.
Generally, conductors are metals.
Eg.
Copper, silver, gold, iron etc.
Insulators
Substances
having
infinite
resistance are called insulators.
They do not allow the charges to
flow.
They contain practically no free
electrons.
Generally, insulators are non
metals.
Rubber, plastic, glass, ebonite etc.
[Any two points: 1 mark each] [2]
v.
Uses of sodium carbonate (washing soda):
a.
Washing soda is used in washing clothes as a cleansing agent.
b.
It is very useful in manufacturing detergent powder, paper and glass.
c.
It is also used to refine petroleum.
d.
It plays an important role in making the water soft and potable.
[½]
[½]
[½]
[½]
vi.
Screen
Dispersion of light
3.
Answer the following questions (any five):
i.
a.
Silver chloride (AgCl)
b.
Double displacement reaction.
c.
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Silver nitrate Sodium chloride
ii.
a.
b.
c.
iii.
iv.
2 2
Diagram:
Labels:
[1]
[1]
[1]
[1]
[1]
Silver chloride Sodium nitrate
When two or more substances combine to form a single product, then the chemical
reaction is known as combination reaction.
The chemical reaction which is accompanied by absorption of heat is called
endothermic reaction.
The chemical reaction in which reactants gain oxygen atom or lose hydrogen atom to
form products is known as oxidation reaction.
Concave mirrors are used:
a.
in torches and headlights.
b.
in flood lights.
c.
as reflecting mirrors for projector.
d.
to collect heat radiations in solar devices.
e.
in shaving mirror, dentist’s mirror.
f.
in solar furnaces.
The domestic appliances based on heating effect of electric current are:
a.
Electric iron
b.
Electric heater
c.
Electric toaster
d.
Electric oven
e.
Electric kettle
f.
Electric bulb
[1]
[1]
[1]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
Science and Technology
v.
Formation of rainbow:
a.
The beautiful phenomenon of the rainbow is a combination of different phenomena like
dispersion, refraction and reflection of light.
b.
The rainbow appears in the sky after a rain shower. When sunlight enters the water
droplets present in the atmosphere, water droplets act as small prism.
c.
These drops refract and disperse the incident sunlight.
d.
This dispersed light gets reflected inside the droplet and again is refracted.
As a collective effect of all these phenomena, the seven coloured rainbow is observed as
shown in figure.
Rain drop
Sunlight
Red
a.
b.
c.
[½]
[½]
[½]
Violet
[1]
Formation of rainbow
vi.
[½]
A convex lens forms real, diminished and inverted image when an object is placed
beyond 2F1.
The image is formed between F2 and 2F2.
The formation of image is as shown in the figure.
L
A
[1]
[1]
F2 B1 2F2
B
2F1
F1
O
A1
Object (AB) beyond 2F1
Image (A1B1) between F2 and 2F2, Diminished, real and inverted.
4.
[1]
Answer the following question (any one):
i.
a.
B
C
N
S
A
B1
Iron core
Armature coil
D
Slip rings
R1
R2
B2
Axle
G
Electric AC generator
b.
c.
Principle of electric generator:
Electric generator works on the principle of electromagnetic induction. When the coil
of electric generator rotates in a magnetic field, magnetic field induces a current in this
coil. This induced current then flows into circuit connected to the coil.
Function of slip rings:
The ends of armature coil are connected to two slip rings. Slip rings rotate along with
the armature coil.
[2]
[1]
[1]
3
Board Answer Paper : July 2015
d.
1.
2.
ii.
Any two uses of a generator:
Generators are used in cinema halls, office buildings, hospitals, etc to provide
electricity in case of power failure in the city.
Generators provide backup electrical power to household appliances such as
refrigerators, water heaters, etc.
Sources of noise pollution:
a.
Industrial activity
b.
construction activity
c.
generator sets
d.
loud speakers
e.
public address systems
f.
music systems
g.
vehicular horns
h.
other mechanical devices, etc.
[½]
[Any six sources : ½ mark each] [3]
Impact of noise pollution on human body:
i.
Impact of noise pollution on human beings depends on the noise intensity, frequency
and exposure duration.
ii.
Noise pollution can cause auditory fatigue and deafness.
iii. It can cause communication interference, sleep interference, concentration interference,
ill temper, annoyance, violent behaviour, mental disorientation, bickering and loss of
working efficiency.
iv.
Noise pollution can cause physiological effects such as nausea, fatigue, anxiety, visual
disturbances, insomnia, hypertension, cardio vascular disease.
4 4
[½]
[½]
[½]
[½]
[½]
Science and Technology
BOARD ANSWER PAPER : JULY 2015 SCIENCE AND TECHNOLOGY SECTION B
1.
(A)
(a)
i.
ii.
(b)
(c)
Fill in the blanks:
Chlorine is a greenish coloured poisonous gas.
Compounds with identical molecular formula but different structures, are called
isomers.
Name the following:
Neuroglia
iii.
iv.
v.
2.
[1]
Column ‘I’
Inhibits plant growth
Cytokinins
Cellular respiration
Bile
5.
4.
3.
1.
Column ‘II’
Abscisic acid
Promote cell division
Mitochondria
Breaks large fat globules into smaller ones
Choose the correct alternative and rewrite the following:
Oxygen is released in plants during the process of photosynthesis.
To observe the hydra bud clearly, Raju should see it first under the low power lens and then
under the high power lens in order to see all of the above.
Reaction of iron nails with copper sulphate solution is an example of displacement reaction.
CuSO4 is blue in colour.
Acetic acid turns blue litmus red.
Answer the following questions (any five):
i.
a.
When a piece of calcium is placed in water, initially it sinks in water as its density is
greater than that of water.
b.
Calcium reacts with water less vigorously to form calcium hydroxide and hydrogen
gas.
Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)
Calcium
c.
d.
[1]
Match the following:
i.
ii.
iii.
iv.
(B)
i.
ii.
[1]
Water
Calcium
Hydrogen hydroxide
Since sufficient heat is not evolved during the reaction, hydrogen does not catch fire.
Instead, calcium starts floating because the bubbles of hydrogen gas formed stick to the
surface of calcium metal.
Hence, calcium floats over water during its reaction with water.
[½]
[½]
[½]
[½]
[1]
[1]
[1]
[1]
[1]
[½]
[½]
[½]
[½]
ii.
i.
ii.
iii.
Toilet soap
High quality of fats and oils are used as
raw materials.
Expensive perfumes are added.
Toilet soaps do not contain free
alkalies which are harmful to skin.
Laundry soap
Cheaper quality of fats and oils are
used as raw materials.
Cheaper perfumes are added.
Laundry soaps contain free alkalies
which contribute towards its cleaning
action.
[Any two points: 1 mark each] [2]
iii.
a.
b.
There is no definite excretory system or organs present in plants for the removal of
excretory products. Gaseous excretory materials are eliminated by diffusion.
Many waste products are stored in the vacuoles of leaves, flowers, fruits and bark
which fall off later on. Whereas other waste products are stored as resins and gum in
old xylem.
[½]
[½]
1
Board Answer Paper : July 2015
c.
d.
Plants also excrete some waste substances into the soil around them. In some plants,
waste is in the form of calcium oxalate crystals called raphides.
Rubber, latex, gum, resins and essential oils like eucalyptus or sandalwood oil are plant
wastes useful to human beings.
iv.
[½]
[½]
Dendrites
Cyton (cell body)
Nucleus
Axon
Diagram:
Any two labels:
Neuron
v.
a.
b.
c.
d.
vi.
a.
b.
3.
Regeneration is the process of asexual division in some multicellular organisms,
through which they can reconstruct the entire body from the isolated body cells.
Regeneration is carried out by specialised regenerative cells. These cells proliferate
and make large number of cells.
These later develop into various cell types and tissues and help in the production of
new organisms.
The capacity to regenerate is very high among some animals. In Planaria, when the
body is cut into many pieces, each piece develops into a whole new organism.
Organisms which are structurally intermediate between two different groups are called
‘Connecting Links.’
Example: Peripatus is considered as a connecting link between Annelida and
Arthropoda.
Answer the following questions (any five):
i.
Mechanism of Mendel’s monohybrid cross:
a.
Monohybrid cross involves crossing of two plants with one pair of contrasting
characters. Mendel selected dominant red flowered (RR) and recessive white flowered
(rr) pea plants as parents for crossing.
b.
After crossing of the parents, Mendel obtained the F1 generation, which possessed red
coloured flowers, but they were different from red-flowered plants of parental
generation, because these plants outwardly showed the same characteristics (i.e. red
flowers) but were genetically different. This happened because the F1 generation also
contained white recessive coloured factors.
c.
From this, Mendel concluded that red was dominant over white and the F1 generation
was ‘phenotypically’ red and ‘genotypically’ a hybrid of red and white.
Pure red Pure white flowers …P1 generation
RR
rr
R
r
Rr
Plants with red flowers
2 2
[1]
[1]
[½]
[½]
[½]
[½]
[1]
[1]
[½]
[½]
[½]
…Gametes
…F1 generation
[½]
Science and Technology
d.
e.
The F1 generation red flowered plant possess Rr genotype and thus, produces two
types of gametes ‘R’ and ‘r’.
The F1 plants were allowed to self-pollinate to produce second filial generation.
(F2 generation). The F2 generation can be represented in Punnett square as:
R
RR
Homozygous
red
Rr
Heterozygous
red
R
r
f.
r
Rr
Heterozygous red
rr
Homozygous
white
From F2 generation, Mendel observed that the phenotypic ratio of the offsprings was
approximately 3 red (dominant):1 white (recessive), whereas the genotypic ratio was 1
dominant (RR):2 hybrid (Rr):1 recessive (rr), which is based on genetic constitution.
[½]
[½]
ii.
XY
Parents:
Gametes:
Children:
X
XX
[1]
XX
Y
XX
X
X
XY
XY
[1]
Boy
Girl
Sex determination in human beings
iii.
a.
b.
1.
2.
iv.
a.
b.
c.
Seismonastic movement:
The movement of plant part in response to the stimulus of touch is called as
seismonastic movement.
Type of movements in plants:
Growth dependent movement:
e.g. movement of root system of plants in response to the stimulus of water and gravity
i.e. hydrotropic and gravitropic movements respectively.
Growth independent movement:
e.g. Opening of lotus petals in the morning and tuberose at night.
Metal A is more reactive than B as it has to lose only one electron from the outermost
orbit while metal B has to lose two electrons to get stable electronic configuration.
Metal A with electronic configuration of 2, 8, 1 is sodium (Na). Metal B with
electronic configuration of 2, 8, 2 is Magnesium (Mg).
Magnesium reacts with dil. HCl to form magnesium chloride and hydrogen gas.
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
Magnesium
v.
Hydrochloric
acid
Magnesium
chloride
Hydrogen
Allotropes are two or more forms of same element that differ from each other in their
physical properties but have same or similar chemical properties.
b.
Two allotropic forms of carbon: Diamond and graphite
Uses of diamond:
a.
Diamonds are used as precious stones in jewellery.
b.
Black diamonds are used for cutting glass.
[Any one use: ½ mark]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
a.
[1]
[1]
3
Board Answer Paper : July 2015
Uses of graphite:
a.
Graphite is used in making carbon electrodes.
b.
It is used as lubricants and in lead pencils.
[Any one use: ½ mark]
vi.
i.
ii.
iii.
4.
Answer the following question (any one):
i.
a.
The organs of the human male reproductive system:
1.
Testes
2.
Epididymis
3.
Vas deferens
4.
Seminal vesicle
5.
Prostate gland
6.
Penis.
b.
Functions:
1.
Testes: Testes produce sperms. They secrete hormone testosterone, which brings about
changes in boys during puberty.
2.
Epididymis: Sperms are stored and matured in the epididymis.
3.
Vas deferens: It is a passage through which the sperms travel towards the urethra.
4.
Seminal vesicles and prostate gland:
They produce ejaculatory fluid which helps the sperm in transport and provides
nutrition.
5.
Penis:
It delivers the sperms to the site of fertilization.
[Write function of any two organs: 1 mark each]
ii.
4 4
Green plants synthesize their food by the process of photosynthesis.
The factors which take part in this process are carbon dioxide, water, chlorophyll and
sunlight.
Chemical equation of photosynthesis:
Chlorophyll
C6H12O6 + 6O2
6CO2 + 6H2O
Sunlight
The functions of the Maharashtra Pollution Control Board:
i.
MPCB plans comprehensive programmes for the prevention, control or abatement of
pollution.
ii.
MPCB inspects sewage or trade effluent treatment and disposal facilities.
iii. MPCB supports and encourages the developments in the fields of pollution control,
waste recycle reuse, eco-friendly practices, etc.
iv.
It educates and guides the entrepreneurs in improving environment by suggesting
appropriate pollution control technologies and techniques.
v.
It creates public awareness about the clean and healthy environment and considers
the public complaints regarding pollution.
[1]
[1]
[1]
[½]
[½]
[½]
[½]
[½]
[½]
[2]
[1]
[1]
[1]
[1]
[1]
Science and Technology
BOARD ANSWER PAPER : MARCH 2016 SCIENCE AND TECHNOLOGY SECTION A
1.
(A)
(B)
i.
Fill in the blanks:
The modern periodic table consists of seven periods.
[1]
ii.
The formulae of chloride of metal M is MCl2. The metal M belongs to 2nd or II A
[1]
iii.
Corrosion can be prevented by using anti rust solution.
[1]
(b)
i.
Find the odd one out:
Thermometer: Others work on magnetic effect of electric current.
[1]
ii.
Bar magnet: Others are appliances which work on magnetic effect of electric current.
[1]
Choose the correct alternative and rewrite the following:
(C) Reaction of iron nails with copper sulphate solution is an example of Displacement
reaction.
[1]
ii.
(A)
Dilute NaOH can be tested with red litmus paper.
[1]
iii.
(B)
In series combination current remains constant.
[1]
iv.
(D)
An object of 10 cm is placed in front of a plane mirror. The height of image will be
10 cm .
[1]
When a ray of light travels from air to glass slab and strikes the surface of separation at
90, then it passes unbent.
[1]
v.
2.
(a)
i.
(C)
Answer the following questions (any five):
i.
a.
The potential difference between two points is said to be 1 volt if 1 joule of work is
done in moving 1 coulomb of electric charge from one point to another.
1 volt =
b.
ii.
1 joule
1coulomb
Electric power is defined as the rate at which electrical energy is consumed.
The major harm done to human beings by air pollution are:
a.
Short term effects of air pollution on human beings:
1.
Irritation of eyes, nose, mouth and throat.
2.
Respiratory infections such as bronchitis, pneumonia.
3.
Headache, nausea and allergy
4.
Asthma attacks
5.
Reduced lung functioning.
b.
Long term effects of air pollution on human beings:
1.
Chronic pulmonary disease
2.
Cardio vascular disease
3.
Lung cancer
4.
Premature death
[1]
[1]
[1]
[1]
1
Board Answer Paper : March 2016
iii.
Power of lens = + 2 D
P=
1
f
[1]
1
P
1
f = = 0.5 m = 50 cm
2
f=
Focal length of the lens = 50 cm.
[1]
iv.
Applications of Sodium bicarbonate (baking soda):
a.
Sodium bicarbonate is used to prepare light and spongy breads, cakes and dhokalas.
b.
It also helps to reduce acidity in stomach.
c.
It is very useful in preparing CO2 gas and is one of the contents of fire extinguishers.
[Any two: 1 mark each]
[2]
v.
a.
When an object is placed within the focal length of convex lens, one gets a magnified
and erect image of the object.
b.
Thus, the watch repairer can see the minute parts of the watch more clearly with the
help of simple microscope than the naked eye, without any strain on the eye. A
magnification of about 20 times is obtained by simple microscope.
Hence, the watch repairers make use of simple microscope while repairing watches.
[1]
[1]
vi.
NEUTRAL
0
3.
ACIDIC
BASIC
7
pH scale
14
Answer the following questions (any five):
i.
Features of Mendeleev’s periodic table:
a.
The horizontal rows in the periodic table are called periods. There are seven periods
numbered from 1 to 7.
b.
Properties of elements in a particular period show regular gradation from left to right.
c.
Vertical columns in the periodic table are called groups. There are eight groups numbered
from I to VIII. Groups I to VII are further divided into A and B subgroups.
ii.
Redox reactions:
Redox reaction is a chemical reaction in which both oxidation and reduction take place
simultaneously.
Eg.
[2]
[1]
[1]
[1]
[1]
Reduction (loss of oxygen)
BaSO4 + 4C BaS + 4CO
Oxidation (gain of oxygen)
In the above example, BaSO4 is reduced to BaS by removing oxygen atom and carbon is
oxidized to carbon monoxide.
iii.
2 2
Defect of human eye shown in the figure is hypermetropia. In this defect person can
see distant objects clearly but cannot see nearby object clearly.
b.
Reason for hypermetropia:
1.
Weak action of ciliary muscles cause low converging power of eye lens.
2.
The distance between the eye lens and the retina decreases on account of either
shortening of eye ball or flattening of lens. In this case focal length of eye lens is too
long.
Correction:
i.
Hypermetropia can be corrected by using spectacles having convex lenses of suitable
focal length.
[1]
[1]
a.
[1]
[1]
Science and Technology
ii.
The convex lens produces convergence of the light rays passing through it and then
light rays are converged by eye lens. As a result the image of the nearby object is
formed on the retina.
L1
L
Image at retina
O
O
Remedy of hypermetropia
iv.
Refraction of light:
The phenomenon of change in the direction of light when it passes from one transparent
medium to another is called refraction of light.
Relationship between refraction of light and refractive index:
a.
The extent of change in the direction of light ray is different for different media and
depends upon the relative speed of propagation of light in different media. The relative
speed of propagation of light is basically measured with the help of refractive index of
medium.
b.
The refractive index (12) of second medium with respect to the first is given by the
ratio of the magnitude of velocity of light in first medium to that in second medium,
i.e., 12 =
v.
vi.
[1]
[1]
velocityof light in medium first
.
velocity of light in mediumsecond
Thus, the refraction of light depends upon refractive index of the material.
[1]
a.
b.
[1]
The band of coloured components of a light beam is called as spectrum.
White light is composed of seven colours. When white light is incident a prism, it splits
up into its constituent colours.
c.
Each colour bends through different angles with respect to incident ray. So the rays of
each colour emerge along different path and become distinct.
Hence, we get a spectrum of seven colours when white light is dispersed by a prism.
a.
1.
2.
b.
1.
2.
3.
4.
c.
1.
2.
3.
4.
[1]
Use sound proof tiles on walls of classroom or curtains to avoid spread of noise
pollution.
Teacher should engage students in interesting activities during free time due to
which students will not make noise.
Purchase energy efficient products and operate them efficiently.
Use daylighting in your home by using energy efficient windows.
Use solar water heater rather than electric water heater.
Turn off lights, fans, air conditioners, T.V., Computers etc. when not in use.
[Any two points: ½ mark each]
Bursting of fire crackers near silence zones such as near hospitals and schools
must be avoided.
Fire crackers should be avoided.
Fire crackers which do not produce loud sound and cause less pollution must be
preferred.
[Any two points: ½ mark each]
Answer the following question (any one):
i.
Expression for equivalent resistance in series:
+
V
I
C
D
R1
R2
R3
A
+
[1]
[1]
[½]
[½]
[1]
[1]
[5]
+
E
K
Resistors in series combination
[1]
3
Board Answer Paper : March 2016
a.
ii.
Let R1, R2 and R3 be the three resistors connected in series between points C and D as
shown in the circuit diagram.
b.
An electric circuit is completed by connecting an ammeter (A), a voltmeter (V) a plug
key (K) and a battery (E).
c.
Suppose I is the current and V is the P.D. across points C and D.
d.
Suppose V1, V2 and V3 are the P.D.s across resistors R1, R2 and R3 respectively, such
that
….(1)
V = V1 + V2 + V3
e.
If Rs is the equivalent resistance, then using Ohm’s law,
V = IRs and V1 = IR1;
V2 = IR2; V3 = IR3
Substituting the values in equation (1) we get,
IRs = IR1 + IR2 + IR3
IRs = I(R1 + R2 + R3)
….(2)
Rs = R1 + R2 + R3
Equation (2) represents equivalent resistance in series combination of resistors.
For ‘n’ number of resistors connected in series,
Rs = R1 + R2 + R3 + … + Rn
Characteristics of series combination of resistors:
a.
Resistance of the combination of resistors is equal to the sum of the resistances of
individual resistors.
Rs = R1 + R2 + R3 + …+ Rn
b.
The effective resistance in a series combination is greater than the individual
resistances.
c.
Hence, this combination is used to increase resistance of the circuit.
d.
Current is the same in every part of the circuit.
e.
The total voltage across the combination is equal to the sum of the voltage across the
separate resistors.
[Any two characteristics: 1 mark each]
a.
[1]
[2]
C
B
S
N
A
D
R1 R2
B1
[1]
B2
Iron core
Armature coil
Split ring
Axle
Electric motor
b.
c.
4 4
Electric motor works on the principle that a current carrying conductor placed on a
magnetic field experiences a force.
Appliances in which electric motor is used :
1.
Electric fans
2.
Hair dryers
3.
Electric cranes
4.
Rolling mills
[2]
[1]
[½]
[½]
[½]
[½]
Science and Technology
BOARD ANSWER PAPER : MARCH 2016 SCIENCE AND TECHNOLOGY SECTION B
1.
2.
(A)
Answer the following sub-questions :
i.
Liver is the largest gland in the body.
ii.
Vas deferens is a part of male reproductive system, rest all are parts of female reproductive
system.
[1]
iii.
False:
[1]
Aquatic animals breathe at a faster rate than the terrestrial animals.
[1]
iv.
rr : Homozygous : : Rr : Heterozygous
[1]
v.
Main ore of aluminium-Bauxite.
[1]
(B)
Choose the correct alternative and rewrite the following:
i.
Hydrogen is liberated when acetic acid reacts with sodium metal.
[1]
ii.
For binary fission, Amoeba requires one parent cells.
[1]
iii.
A solution of CuSO4 in water is blue in colour.
[1]
iv.
Raisins are formed by drying grapes. The process that takes place during formation of raisins
from grapes is dehydration.
[1]
v.
Ethanoic acid has a pungent odour.
[1]
Answer the following questions (any five):
i.
Voluntary movements
ii.
Involuntary movements
i.
Voluntary
movements
are
the
movements which can be controlled by
our will.
Involuntary movements are
movements
which
cannot
controlled by our will.
the
be
ii.
Voluntary movements are controlled by
motor areas in the fore brain and
cerebellum in the hind brain.
Involuntary movements are controlled
by the mid-brain and the medulla
oblongata of the hind brain.
Eg.
Riding a bicycle, picking up an object
from the floor, etc.
Breathing, sneezing, etc.
[1]
[1]
a.
Cellular DNA is the source of information for synthesizing proteins in the cell.
[½]
b.
DNA possesses a double helical structure, which consists of two strands coiled around
each other.
[½]
A fragment of DNA that provides complete information about one protein is referred to
as “Gene” for that protein.
[½]
During sexual reproduction, each of the parent contributes equally to the DNA of the
progeny.
[½]
c.
d.
1
Board Answer Paper : March 2016
iii.
Stigma
Petal
Style
Carpel
Anther
Ovary
Filament
Stamen
Sepal
iv.
v.
vi.
Longitudinal section of a typical flower
i.
The root system of the plants are specialized for absorbing water and minerals from the
soil.
ii.
The roots respond to the stimuli of gravity and water and grow towards the soil. Thus,
the root system of plants show gravitropic movement.
Hence, the roots of plants grow away from light.
a.
b.
c.
a.
b.
c.
d.
3.
Drinking water sources such as rivers, lakes should be protected from water pollution.
Rain water harvesting plant should be constructed to conserve rain water.
Waste water should be taken to sewage treatment sites where it undergoes various
methods of treatment. Such water can be used for agriculture and other purposes.
[Any two: 1 mark each]
Fossils are the dead remains of plants and animals which existed in the past.
They are formed by various methods.
Sometimes, impressions of plants and animals are formed on mud which get converted
into fossils at later stage.
At other times, plants and animals get burried in the soil and the soft part of the body
gets decayed, while the hard part (bones) remains in the soil in the form of fossils.
Answer the following questions (any five):
i.
Alloy:
a.
An alloy is a homogenous mixture of two or more metals or a metal and a non-metal in
definite proportion.
Eg. a.
Brass (copper and zinc)
b.
Bronze (copper and tin)
In human males, two dissimilar chromosomes are present in the 23rd pair, the longer
‘X’ and the shorter ‘Y’.
b.
In females, the 23rd pair contains two similar ‘X’ chromosomes.
c.
All children inherit one ‘X’ chromosome from their mother. The other chromosome is
inherited from the father.
d.
If the chromosome inherited from the father is ‘X’, then the offspring is a daughter and
if it is ‘Y’, then the offspring is a son.
Thus, the sex of the child is determined by the male sex chromosome.
ii.
[1]
[1]
[2]
[½]
[½]
[½]
[½]
[1]
[1]
[1]
a.
XY
Parents:
Gametes:
Children:
X
XX
[½]
[½]
[½]
[½]
XX
Y
XX
X
X
XY
XY
Girl
Boy
Sex determination in human beings
2 2
[2]
[1]
Science and Technology
iii.
iv.
Neurons are classified, according to their functions, into three groups:
a.
Sensory neurons: They conduct impulses from the sense organs to the brain and spinal
cord.
b.
Motor neurons: They conduct impulses from the brain and spinal cord to the effector
organs like muscles and glands.
c.
Association neurons: They perform integrative functions of the nervous system.
a.
b.
c.
d.
v.
Three ‘R mantra’ is nothing but ‘Reduce, Reuse and Recycle’ approach to eliminate waste
and conserve resources.
‘Reduce’ means using fewer resources in the first place.
Instead of throwing things away, trying to find ways to use them again is ‘Reuse’.
Recycled items are put through a process that makes it possible to create new products
out of the materials from the old ones.
a.
b.
c.
Metal A with electronic configuration of 2,8,1 is sodium (Na).
Metal B with electronic configuration of 2,8,8,2 is calcium (Ca).
Metal A is more reactive than B as it has to lose only one electron from the outermost
shell while metal B has to lose two electrons, to achieve a stable electronic
configuration.
d.
Both sodium and calcium react with dil. HCl to form corresponding salt and release
hydrogen gas.
2Na(s) + 2HCl(aq) 2NaCl(aq) + H2(g)
Sodium
Ca(s)
Calcium
vi.
4.
Hydrochloric
acid
+
2HCl(aq)
Hydrochloric
acid
Sodium
chloride
Hydrogen
CaCl2(aq) + H2(g)
Calcium
chloride
Hydrogen
Disadvantages of a large family size:
a.
Large families affect both the individual as well as the community life.
b.
Large families may lead to economic pressure and poor housing.
c.
Children may get neglected at home.
d.
The children may also suffer from malnutrition due to insufficient medical care.
e.
There may also be lack of better education for the children.
f.
Large families may lead to poor health of mother in the family.
Answer the following question (any one):
i.
a.
Anaerobic reaction = 5
b.
Reaction in human muscles = 4
c.
Aerobic respiration = 3
d.
Reaction in plant cells = 1
e.
Reaction in liver = 2
ii.
a.
b.
1.
2.
3.
4.
5.
A group of organic compounds containing same functional group, which can be
represented by the same general formula and which, more or less, show similar trends
in their properties is called homologous series.
The important characteristics of homologous series are as follows:
The general formula of all compounds in the series is the same.
They have the same functional group.
Physical properties like melting point, boiling point, density, etc. generally show a
gradual change with increase of molecular formula in the series.
Chemical properties of the members of the series show close resemblance because of
the presence of the same functional group in them.
Consecutive members of the series differ from one another by CH2 group
(methylene group), and their molecular weight differs by 14 units.
[Any four characteristics:1 mark each]
[1]
[1]
[1]
[1]
[1]
[1]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[½]
[1]
[1]
[1]
[1]
[1]
[1]
[4]
3
Science and Technology
BOARD ANSWER PAPER : JULY 2016 SCIENCE AND TECHNOLOGY SECTION A
Q.1. (A)
Answer the following sub-questions:
(a) Fill in the blanks and rewrite the completed statements.
i.
To increase the effective resistance in a circuit the resistors are connected in
series.
ii.
(b)
State whether the following statements are true or false:
i.
True
ii.
(c)
2.
The phenomenon of splitting of light into its component colours is dispersion.
False. The chemical reactions in which heat are liberated is called exothermic
reactions.
Telescope
[1]
[1]
[1]
[1]
[1]
(B)
i.
Rewrite the following statements by selecting the correct options:
(B) When the resistance of a conductor increases, the current decreases.
ii.
(C)
[1]
The phenomenon of change in the direction of light when it passes from one
transparent medium to another is called refraction.
[1]
iii.
(B)
When CO2 is passed through fresh lime water, it turns milky.
[1]
iv.
(A)
A concave mirror is used by a dentist.
[1]
v.
(C)
The equivalent resistance of a parallel combination of two resistors of 30 and 60 is
20
[1]
Attempt any five of the following:
i.
Given:
1 = 3 108 m/s
2 = 1.5 108 m/s
To find:
12 = ?
Formula:
12
=
1
2
[1]
Solution:
From Formula,
12
=
3108
=2
1.5108
The refractive index of the medium with respect to air is 2.
[1]
ii.
a.
b.
c.
[½]
[½]
iii.
Grills of doors and windows are generally made up of iron metal.
Before painting, iron metal is in direct contact with the moisture in atmosphere.
Due to the effect of moist air, a reddish brown coloured substance called rust is formed
on the surface of iron. This damages the grills of doors and windows.
d.
Painting prevents corrosion as it prevents iron metal from coming in contact with
moisture and oxygen from the atmosphere.
Hence, grills of doors and windows are always painted before they are used.
a.
b.
c.
The enrichment of water bodies by inorganic plant nutrients like nitrate, phosphate
occurring either naturally or due to human activity is called eutrophication.
The process of eutrophication takes place due to introduction of nutrients and
chemicals through discharge of domestic sewage, industrial effluents and fertilizers
from agricultural fields.
These promote excessive growth of phytoplankton and algae in water. When algae die,
decomposition of organic substances use oxygen which results in depletion of oxygen
in water.
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Board Answer Paper : July 2016
d.
Bloom of algae blocks penetration of oxygen, light and heat into the water body which
results in death of most of the organisms below the surface of the water.
iv.
Metals:
Sodium
Nonmetals: Carbon, Oxygen
Metalloids: Boron
v.
a.
b.
[½]
[½]
[1]
[½]
Stretch the thumb, forefinger and middle finger of the right hand so that they are
perpendicular to each other.
If the forefinger indicates the direction of magnetic field and the thumb shows the
direction of motion of the conductor, then the middle finger will show the direction of
induced current.
[1]
[1]
vi.
i.
Oxidation reaction
The chemical reaction in which
reactants gain oxygen atom or lose
hydrogen atom to form products is
known as oxidation reaction.
Reducing agent undergoes oxidation.
ii.
Eg. 2Cu(s) + O2(g)
2CuO(s)
3.
Reduction reaction
The chemical reaction in which reactants
lose oxygen atom or gain hydrogen atom
to form products is known as reduction
reaction.
Oxidising agent undergoes reduction.
C(s) + 2H2(g) CH4(g)
[Any two points: 1 mark each]
Attempt any five of the following questions:
i.
a.
The type of lens used in Aniket’s spectacle is concave lens with negative power.
b.
Aniket is suffering from Myopia or near sightedness.
c.
Given: Power of lens = 0.5d.
To find: Focal length (f) = ?
Formula: P =
1
f
[2]
[1]
[1]
[½]
Solution:
From Formula,
f =
ii.
iii.
2 2
1
1
= 2 m = – 200 cm
=
P
0.5
Demerits of Mendeleev’s periodic table:
a.
No fixed position was given to hydrogen because it resembled alkali metals as well as
halogens.
b.
Isotopes of same element have different atomic masses. So each of these isotopes should
be given different position. But isotopes which were chemically similar had to be given
same position.
c.
At certain places, an element of higher atomic mass has been placed before an element
of lower mass.
Eg. Cobalt (Co = 58.93) is placed before nickel (Ni = 58.71).
d.
Some elements placed in the same sub-group had different properties.
Eg. Manganese (Mn) is placed with halogens which totally differ in the properties.
[Any three points : 1 mark each]
Formation of rainbow:
a.
The beautiful phenomenon of the rainbow is a combination of different phenomena like
dispersion, refraction and reflection of light.
b.
The rainbow appears in the sky after a rain shower. When sunlight enters the water
droplets present in the atmosphere, water droplets act as small prism.
c.
These drops refract and disperse the incident sunlight. This dispersed light gets
reflected inside the droplet and again is refracted.
[½]
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Science and Technology
d.
As a collective effect of all these phenomena, the seven coloured rainbow is observed
as shown in figure.
Rain drop
Sunlight
Red
Violet
[1]
Formation of rainbow
iv.
a.
b.
c.
v.
In torches, the source of light is kept at the focus of concave mirror to obtain parallel
beam of light.
[1]
In projector lamps, the source of light is kept at the centre of curvature of concave
mirror, to obtain an image of the same size.
[1]
In flood lights, the source of light is kept just beyond the centre of curvature of concave
mirror to obtain proper light.
[1]
The major harm done to human beings by air pollution are:
i.
Short term effects of air pollution on human beings:
a.
Irritation of eyes, nose, mouth and throat.
b.
Respiratory infections such as bronchitis, pneumonia.
c.
Headache, nausea and allergy
d.
Asthma attacks
e.
Reduced lung functioning.
[Any three short term effects: ½ mark each]
ii.
Long term effects of air pollution on human beings:
a.
Chronic pulmonary disease
b.
Cardio vascular disease
c.
Lung cancer
d.
Premature death
[Any three long term effects: ½ mark each]
vi.
a.
The strength of an acid or a base is measured on a scale that reads from 0 (most acidic)
to 14 (most basic). Such a scale is called pH scale.
[1½]
[1½]
[1]
b.
In pH, p stands for ‘potenz’, which means ‘strength’ in German.
c.
pH helps in measuring hydrogen ion concentration in solutions and the value of pH
indicates the acidic or basic nature of a solution.
[½]
When the pH value is in between 0 to 7, then the solution is acidic in nature, when the
pH value of a solution is 7, then the solution is neutral, and when the pH value lies
between 7 to 14, then the solution is basic in nature.
[½]
d.
NEUTRAL
0
ACIDIC
BASIC
7
pH scale
14
[1]
3
Board Answer Paper : July 2016
4.
Attempt any one of the following:
i.
Expression for equivalent resistance in series:
+
V
I
C
R1
A
+
R2
D
R3
+
[1]
E
K
Resistors in series combination
a.
b.
c.
d.
e.
Let R1, R2 and R3 be the three resistors connected in series between points C and D as
shown in the circuit diagram.
An electric circuit is completed by connecting an ammeter (A), a voltmeter (V) a plug
key (K) and a battery (E).
Suppose I is the current and V is the P.D. across points C and D.
Suppose V1, V2 and V3 are the P.D.s across resistors R1, R2 and R3 respectively, such
that
V = V1 + V2 + V3
….(1)
If Rs is the equivalent resistance, then using Ohm’s law,
V = IRs and V1 = IR1;
V2 = IR2; V3 = IR3
Substituting the values in equation (1) we get,
IRs = IR1 + IR2 + IR3
IRs = I(R1 + R2 + R3)
Rs = R1 + R2 + R3
….(2)
Equation (2) represents equivalent resistance in series combination of resistors.
For ‘n’ number of resistors connected in series,
Rs = R1 + R2 + R3 + … + Rn
If number of resistors are connected in series, then resultant effects are as follows:
a.
Resistance of the combination of resistors is equal to the sum of the resistances of
individual resistors.
Rs = R1 + R2 + R3 + …+ Rn
b.
The effective resistance in a series combination is greater than the individual
resistances. Hence, this combination is used to increase resistance of the circuit.
c.
Current is the same in every part of the circuit.
d.
The total voltage across the combination is equal to the sum of the voltage across the
separate resistors.
ii.
4 4
A.C. generator:
A generator which converts mechanical energy into electrical energy in the form of
alternating current is called A.C. generator.
Construction: The main components of A.C. generator are:
i.
Armature coil
ii.
Strong magnets
iii. Slip rings
iv.
Brushes
a.
Armature coil: A large number of turns of insulated copper wire wound on iron core
in rectangular shape forms an armature coil ABCD as shown in figure.
b.
Strong magnets: The armature coil is placed in between two pole pieces (N and S) of
a strong magnet. This provides a strong magnetic field.
c.
Slip rings: The two ends of the armature coil are connected to two brass rings R1 and
R2. These rings rotate along with the armature coil.
d.
Brushes: Two carbon brushes B1 and B2 are used to press the rings.
[½]
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[½]
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[½]
[½]
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Science and Technology
Working:
a.
When the armature coil ABCD rotates in the magnetic field provided by the strong
magnets, it cuts the magnetic lines of forces.
b.
Thus, the changing magnetic field produces induced current in the coil. The direction of
induced current is determined by the Fleming’s right hand rule.
B
C
N
S
A
B1
[1]
Iron core
Armature coil
D
Slip rings
R1
R2
B2
G
Electric AC generator
Axle
[1]
5
Science and Technology
BOARD ANSWER PAPER : JULY 2016 SCIENCE AND TECHNOLOGY SECTION B
1.
(A)
(a)
i.
Fill in the blanks:
Abscissic acid inhibits growth leading to wilting of leaves.
[1]
ii.
The organic compounds having double or triple bond in them are termed as
unsaturated hydrocarbons.
[1]
Lymph flows in one direction.
[1]
iii.
(b)
H
H
HCCH
H
(c)
[1]
Budding
[1]
(B)
i.
Choose the correct alternative and rewrite the following:
(B) Fermentation is a type of anaerobic respiration.
ii.
(A)
iii.
iv.
v.
2.
H
(A)
(D)
(D)
Some acetic acid is treated with solid NaHCO3, the resulting solution will be
colourless.
[1]
Ramesh observed a slide of Amoeba with elongated nuclei. It would represent binary
fission.
[1]
Twenty dry raisins were soaked in 50 ml of water and kept for one hour at 50oC. All
raisins absorbed water.
[1]
Iron is less reactive than Al.
[1]
Answer the following questions (any five):
i.
a.
Gold and silver are noble metals. They are malleable, ductile and shiny.
b.
They do not undergo corrosion due to attack of moisture, air, acids under normal
conditions.
Hence, gold and silver are used to make jewellery.
ii.
a.
b.
When a piece of calcium is placed in water, initially it sinks in water as its density is
greater than that of water.
Calcium reacts with water less vigorously to form calcium hydroxide and hydrogen
gas.
Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)
Calcium Water
c.
[1]
Calcium
hydroxide
[1]
[1]
[½]
[½]
[½]
Hydrogen
Since sufficient heat is not evolved during the reaction, hydrogen does not catch fire.
Instead, calcium starts floating because the bubbles of hydrogen gas formed stick to the
surface of calcium metal.
Hence, calcium floats over water during its reaction with water.
[½]
1
Board Answer Paper : July 2016
iii.
iv.
Physical properties of ethanol:
a.
Ethanol is a colourless liquid.
b.
It has a pleasant odour.
c.
Its boiling point is 78 C and freezing point is 114 C.
d.
It is combustible and burns with blue flame.
[Any two properties: 1 mark each]
a.
The root system of the plants are specialized for absorbing water and minerals from the
soil.
b.
The roots respond to the stimuli of gravity and water and grow towards the soil.Thus,
the root system of plants show gravitropic movement.
Hence, the roots of plants grow away from light.
v.
a.
b.
In birds and lepidopterans like moths and butterflies, female decide the sex.
In these living organisms, females are heterozygous carrying XY chromosomes
and males are homozygous carrying XX chromosomes.
vi.
Molecular formula : C6H6
Structural formula:
H
[2]
[1]
[1]
[1]
[1]
C
H
C
C
H
H
C
C
H
C
H
3.
Answer the following questions (any five):
i.
Vestigial Structures:
Vestigial structures are the non-functional organs in some organisms but have essential
functions in other organisms.
Examples of vestigial organs in human beings:
a.
Vermiform appendix
b.
Wisdom teeth
c.
Ear muscles
d.
Plica semilunaris
ii.
a.
b.
c.
d.
e.
f.
2 2
Charles Robert Darwin proposed the scientific theory that the branching pattern of
evolution resulted from a process that he called natural selection.
Natural selection is the process of selection of characteristics that contribute to the
fitness of the organism for its survival.
According to Charles Darwin, there exists competition between individuals of the same
type to survive. There are also many differences found in the individuals of the same
species.
Some individuals have certain essential factors which are necessary for them to survive
the struggle for existence. The other organisms, which do not possess these factors and
are not fit enough to survive, and die.
These useful factors appear in the next generation of the species and the struggle for
survival continues. This process goes on for generations, each generation exhibiting
certain variations.
After a long period of time, the species is very different from the original species.
Thus, according to Charles Darwin, the criterion for natural selection is the successful
adaptation for growth and reproduction in the given environment.
[1]
[1]
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[½]
[½]
[½]
[½]
[½]
[½]
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Science and Technology
iii.
Petal
Anther
Ovary
[Diagram: 1mark,]
[Four labels: ½ mark each]
[3]
Reproduction is the fundamental characteristic of living organisms through
which all living organisms produce new individuals of the same species.
It is necessary to maintain the number of individuals of species and to prevent
their extinction.
[1]
Vegetative propagation is the form of asexual reproduction in which new plants
are produced from the vegetative parts, i.e. root, stem, leaves and buds of the
plant.
Plants produced by vegetative propagation take less time to grow and bear
flowers and fruits earlier than those produces from seeds.
[1]
The process of transfer of pollen grains from anther to stigma is called
pollination.
There are two types of pollination: self pollination and cross pollination.
[1]
Sepal
Longitudinal section of flower
iv.
a.
1.
2.
b.
1.
2.
c.
1.
2.
v.
vi.
Classification of the nervous system:
a.
The nervous system of human body can be classified as:
1.
Central Nervous System (CNS)
2.
Peripheral Nervous System (PNS)
3.
Autonomic Nervous System (ANS)
b.
The Central nervous system comprises of the brain and spinal cord and regulates all
activities of the body.
c.
The Peripheral nervous system includes all the nerves forming a network and spreading
throughout the body. Communication between the central nervous system and other
parts of the body is brought about by peripheral nervous system.
d.
The Autonomic nervous system comprises of all the nerves present in the involuntary
organs like heart, stomach, lungs, etc.
a.
b.
c.
Copper is a more reactive element as compared to silver. Thus, copper can displace
silver from its solution.
When copper coin is dipped in silver nitrate solution, a thin layer of silver gets
deposited on the surface of copper coin due to which the coin gets a silver shine.
The balanced chemical equation for this reaction is given by
2AgNO3(aq) + Cu(s) Cu(NO3)2(aq) + 2Ag(s)
Silver
nitrate
4.
Copper
Copper
nitrate
Silver
Answer any one of the following questions:
i.
a.
Recycling is a type of green technology that uses old material to make new products.
Many waste products from the industries such as paper, glass, plastics and metals can
be recycled.
b.
Example of reuse:
Reuse of empty plastic or metal cans to store things.
[1]
[½]
[1]
[½]
[1]
[1]
[1]
[1]
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Board Answer Paper : July 2016
c.
ii.
a.
b.
4 4
Advantages of recycling:
1.
It conserves energy and raw materials.
2.
It saves space used in landfills.
3.
It protects environment by effective handling of waste materials.
4.
It reduces the cost of production.
[Any three: 1 mark each]
[3]
Organs of Human respiratory system:
1.
Pharynx
2.
Larynx
3.
Trachea
4.
Bronchi
5.
Bronchioles
6.
Lungs
[½]
[½]
[½]
[½]
[½]
[½]
Inhalation:
1.
Inhalation is the process by which air from atmosphere is inhaled (or taken into
the body).
2.
When the muscular diaphragm of the body contracts, volume of the thoracic
cavity increases and air pressure inside the cavity decreases.
3.
The air from outside, enters the lungs through the nostrils and the alveolar sacs
are filled with air, rich in oxygen.
[1]
[½]
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