Page 0
MATHEMATICS TEST PAPER
CLASS : X
SA – II PAPERS OF VARIOUS SCHOOLS
Units
Marks
1. Number Systems
17
2. Algebra
25
3. Geometry
37
4. Co-Ordinate Geometry
06
5. Heron’s Formula
05
Total (Theory)
90
IMPORTANT TIPS:

Prepare a list of all important formulae and concepts for quick revision

Understand the usage of the formulae

If there are any doubts clear your doubts as soon as possible from your teacher

Focus on NCERT books for studying as majority of exam revolves around NCERT books, but practice
questions from other books in addition to NCERT

Complete all NCERT questions, solved examples and exemplar.

Practice a few questions from previous chapters as you study a new chapter. This will make sure you stay
in touch with the concepts.

Practices writing down the steps you take for solving a question as the process also gain marks.

Do not skip any topic, complete the syllabus thoroughly

Feel free to think out your answers on rough paper. It helps in understanding the question quicker and you
can use a column in your answer sheet for rough work during exams.

Pay attention to details, a single misplaced decimal sign or - or + can lead to an incorrect answer in
questions you otherwise know. Always write out questions even if you know them mentally, writing helps
improve your answering style

Practice answering previous year papers and sample papers within the time limits of the official exam.

Definitely practice the last terms question papers

Identify trends in the question paper

Identify your weak areas, where you are unable to answer and where you answer incorrectly

Increase your speed in answering so you are able to complete the exam and have some time for revision
during the exam

Create a study schedule and focus on studying and practicing weak areas and doing a few practice
questions in areas you are comfortable in with theorems and concepts remember to understand the
applications thoroughly.

Keep your formulae/concept list with you and read it at least 3 times a day.

Long answer questions are scoring and generally come from algebra and geometry so prepare accordingly
Page 1
………………………………………………..
SUMMATIVE ASSESSMENT – II
SUBJECT : MATHEMATICS
Max. Marks : 90
SECTION : A
Time : 3 : 00 Hrs.
I.
Question numbers 1 to 4 carry one mark each
1.
A ladder, leaning against a wall, makes an angle of 60º with the horizontal. If the foot of the ladder is 2.5 m away
from the wall, find the length of the ladder.
2.
For what value of k will k+9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?
3.
In see figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and CAB  30, find
PCA.
4.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a
red card nor a queen.
SECTION : B
II.
Question number 5 to 10 carry two marks each.
5.
In See Figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides
AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that. AB + CD = BC + DA.
6.
The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
7.
In See Figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If
OP = 2r, show that OTS  OST  30.
8.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
9.
Let P and Q b the points of trisection of the line segment joining the points A(2, -2) and (B(-7, 4) such that P is
nearer to A. Find the coordinates of P and Q.
10.
Solve for x :
2x  9  x  13
Page 2
SECTION : C
III.
11.
12.
13.
14.
15.
Question numbers 11 to 20 carry three marks each.
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical
22
vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use   )
7
In See figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12cm. BC is joined. Find the area
of the shaded region. (Take   3.14)
A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is
5
completely submerged in water, the water level in the cylindrical vessel rises by 3 cm. Find the diameter of the
9
cylindrical vessel.
In See Figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and
diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the
22
cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq.metre. (Use   )
7
16.
In See Figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm
22
where AOC  40. (Use   )
7
17.
A man standing on the deck of a ship, which is 10m above water level, observes the angle of elevation of the top of
a hill as 60º and the angle of depression of the base of hill as 30º. Find the distance of the hill from the ship and the
height of the hill.
The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing
the digits is 594 less than the original number. Find the number.
18.
19.
If the roots of the quadratic equation (a  b)x 2  (b  c)x  (c  a)  0 are equal, prove that 2a = b + c.
20.
From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card
is drawn at random. Find the probability that drawn card is :
(i)
a black King
(ii) a card of red colour
(iii) a card of black colour
Page 3
SECTION : D
IV.
21.
22.
23.
24.
25.
Question numbers 21 to 31 carry four marks each.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state
government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the
whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with
conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per
sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem?
22
(Use   )
7
In See figure, two equal circles, with centres O and O’, touch each other at X.OO’
produced meets the circle with centre O’ at A. AC is tangent to the circle with centre
DO'
.
O, at the point C. O’D is perpendicular to AC. Find the value of
CO
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the
numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
In See Figure, the vertices of ABC are A(4, 6), B(1, 5) and C(7, 2). A linesegment DE is drawn to intersect the sides AB and AC at D and E respectively
AD AE 1

 . Calculate the area of ADE and compare it with area of
such that
AB AC 3
ABC.
In See Figure, is show a sector OAP of a circle with centre O, containing . AB is perpendicular to the radius



 1
OA and meets OP produced at B. Prove that the perimeter of shaded region is r  tan   sec  
180 

26.
27.
28.
29.
30.
The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of
numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.
A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return
downstream to the same spot. Find the speed of the stream.
Draw an isosceles ABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose
3
sides are of the corresponding sides of ABC.
4
Prove that tangent drawn at any point of a circle is perpendicular to the radius through the point of contact.
As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing
directly towards it, changes from 30º to 60º. Find the distance travelled by the ship during the period of
observation. (Use 3  1.73)
31.
A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more
than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth
of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park.
Page 4
VIDHYA BHAVAN PUBLIC SCHOOL
SUMMATIVE ASSESSMENT – II, 2015 - 2016
SUBJECT : MATHEMATICS
Max. Marks : 90
SECTION : A
Time : 3 : 00 Hrs.
I.
Question numbers 1 to 4 carry one mark each
1.
What constant number must be added or subtracted to 4x 2  12x  8  0 to solve it by method of completing the
square?
A player sitting on the top of a tower of height 20 m observes the angle of depression of a ball lying on the ground
is 60º. Find the distance between the foot of the tower and the ball.
A die is thrown once. Find the probability of getting a number which has at least 2 factors.
A (-2, 1) and E (3, 4) are the extremeties of the median AE of the ABC. Find the coordinates of its centroid.
2.
3.
4.
SECTION : B
II.
Question number 5 to 10 carry two marks each.
5.
If the quadratic equation kx 2  2kx  6  0 has equal roots, then find the value of k.
Find the first term if k  2, 4k  6 and 3k  2 are the three consecutive terms of an A.P.
6.
7.
AB is a chord of a circle with centre O. At B, a tangent PB is drawn such that its length is 24cm. The distance of P
from the centre is 26cm. If the chord AB is of length 16cm, find its distance from the centre.
P
A
8.
B
9.
Draw a line segment AB of length 9 cm. Divide it internally in the ratio 3:2. Measure the two parts.
In 2 concentric circles, prove that all chords of the outer circle which touch the inner circle are of equal length.
10.
Area of a sector of a circle of radius 36 cm is 54 cm 2 . Find the length of the corresponding arc of the sector.
SECTION : C
III.
Question numbers 11 to 20 carry three marks each.
11.
Find the roots of the quadratic equation 2x 2  7x  5 2  0 by factorisation method.
The 8th term of an AP is zero. Prove that its 38th term is triple of its 18th term.
Construct a tangent to a circle of radius 3 cm from a point on the concentric circle of radius 5 cm and measure its
length. Also, verify the length by actual calculation.
The angle of elevation of the sun is 60º. Find the length of the shadow cast by a tree of height 12 m.
Set A = {1, 3, 8} and Set B = {2, 5, 6} are given
Two integers will be randomly selected from the sets above, first integer from Set A and second integer from Set
B.
(a) What is the probability that the sum of the two integers will be odd?
(b) What is the probability that the sum of the two integers will be multiple of 5?
(c) What is the probability that the sum of the two integers will be greater than 11?
12.
13.
14.
15.
Page 5
16.
17.
18.
19.
20.
Show that the points A(0, -1), B(2, 3), C(10, 7) and D(8, 3) taken in order are the vertices of a parallelogram. Do
they make a rectangle?
A(0, 0), B(-8, 2) and C(10, 6) are the three vertices of a triangle ABC. If P is the mid-point of BC, find the area of
ABP.
For an advertisement purpose, a mall in the city is spreading a white light in the night time over a sector of angle
70º to a distance of 10 km. Find the area of the city over which the light is spread.
A model of tank made by a child is in the shape of a cylinder of base diameter 21 cm and length 18 cm fitted with
22
conical ends each of 9 cm height. Find the capacity of this tank. (Use   )
7
A cylindrical bottle with radius 5 cm and height 14 cm has jam packed in it. The jam is spread over bread of
measure 11 cm × 10 cm. How thick can the layer of jam be spread, if the entire jam is to cover 10 bread pieces?
SECTION : D
IV.
Question numbers 21 to 31 carry four marks each.
21.
Find the positive value of k for which x 2  kx  64  0 and x 2  8x  k  0 will have real roots.
If the numbers a, b, c, d and e are in an A.P. then prove that a  4b  6c  4d  e  0
A thief runs away from a police station with a uniform speed of 100 m/min. After one minute a policemen runs
behind the thief to catch him. He goes at a speed of 100 m/min in the 1 st minute and increases his speed 10m each
succeeding minute. After how many minutes, the policeman will catch the thief?
In the given figure, O is the center of the circle. Determine APC, if DA and DC are tangents and <ADC = 50º.
22.
23.
24.
A
D
50
P
O
C
25.
26.
27.
28.
29.
30.
31.
Construct DEF ~ ABC in which AB = 5.2 cm B  45 and BC = 6 cm, using scale factor 1 : 2.
The angle of elevation of the top B of a tower AB from a point X on the ground in 60º. At a point y, 40 m vertically
above X, the angle of elevation of the top is 45º. Find the height of the tower AB and the distance XB.
In a class of 50 students, there are 18 girls and 32 boys. Of these students, 20 are vegetarians and rest non
vegetarians and 12 of the vegetarians are boys. If a student is chosen at random, what is the probability of
(A) choosing a girl?
(B) choosing a non vegetarian student?
(C) choosing a boy student who is non vegetarian?
(D) choosing a girl who is vegetarian?
If the point P (3, 4) is equidistant from the points A (a + b, b – a) and B (a – b, a + b), then prove that 3b – 4a = 0.
A brooch is made with silver wire in the form of a circle with diameter 35 mm for the inauguration of a Social club
in a school to keep environment clear in its neighborhood. The wire is also used in making 5 diameters which
divide the circle into 10 equal sectors. Find
(i)
Total length of the silver wire required
(ii) The area of each sector of the brooch.
(iii) How can you help to keep environment pollution free?
A archery target has three regions formed by three concentric circles as shown in the figure. If the diameters of the
concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas and perimeter of three regions.
A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and
20 cm, respectively. Find the capacity also find the cost of milk which can completely fill the container, at the rate
of Rs. 25 per litre (Use   3.14)
Page 6
……………………………………..
SUMMATIVE ASSESSMENT – II, 2014 - 2015
SUBJECT : MATHEMATICS
Max. Marks : 90
SECTION : A
Time : 3 : 00 Hrs.
I.
Question numbers 1 to 4 carry one mark each
1.
In figure a tower AB is 20 m high and BC, its shadow on the ground, is 20 3 m long. Find the Sun’s altitude.
A
B
C
2.
Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the
dice is 6.
3.
If the quadratic equation px 2  2 5px  15  0 has two equal roots, then find the value of p.
4.
In figure, PQ is a chord of a circle with centre O and PT is a tangent. If QPT  60, find PRQ.
SECTION : B
II.
Question number 5 to 10 carry two marks each.
5.
In an AP, if S5  S7  167 and S10  235, then find the AP, where Sn denotes the sum of its first n terms.
6.
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.
In figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If
PRQ  120, then prove that OR = PR + RQ.
7.
P
O
R
Q
8.
In figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are
respectively of lengths 6 cm and 9 cm. If the area of ABC is 54cm 2 , then find the lengths of sides AB and AC.
Page 7
9.
Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear.
10.
Solve the following quadratic equation for x : x 2  2ax  (4b 2  a 2 )  0
SECTION : C
III.
11.
12.
13.
Question numbers 11 to 20 carry three marks each.
Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed
immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of
diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m and the canvas to
be used costs Rs. 100 per sq. m, find the amount, the association will have to pay. What values are show by these
22
associations? (Use   )
7
A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of
diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.
A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can
have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per 100 sq. cm.
(Use   3 14)
14.
504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter
22
of the sphere and hence find its surface area. (Use   )
7
15.
Solve for x : 3x 2  2 2x  2 3  0
The angle of elevation of an aeroplane from a point A on the ground is 60º. After a flight of 15 seconds, the angle
16.
17.
18.
19.
20.
of elevation changes to 30º. If the aeroplane is flying at a constant height of 1500 3 m, find the speed of the plane
in km/hr.
Find the area of the minor segment of a circle of radius 14 cm, when its central angle is 60º. Also find the area of
22
the corresponding major segment. (Use   )
7
th
rd
The 13 term of an AP is four times its 3 term. If its fifth term is 16, then find the sum of its first ten terms.
2
Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP  AB.
5
A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of
3
2
getting a white ball is
and that of a black bal is , then find the probability of getting a red ball. If the bag
5
10
contains 20 black balls, then find the total number of balls in the bag.
SECTION : D
IV.
21.
22.
23.
Question numbers 21 to 31 carry four marks each.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30º. The angle of
depression of the reflection of the cloud in the lake, at A is 60º. Find the distance of the cloud from A.
A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is
(i)
a card of spade or an ace
(ii) a black king
(iii) neither a jack nor a king
(iv) either a king or a queen
In figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and
QR with centre at O, the intersection of its diagonals. Find the total area of the two flowers beds (shaded parts).
Page 8
24.
25.
26.
27.
28.
29.
30.
31.
From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The
height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into
22
a cylindrical wire of 1.4 cm thickness. Find the length of the wire. (Use   )
7
The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than
the shorter side, then find the lengths of the sides of the field.
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of
the arc.
A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed
which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first
speed of the truck.
An arithmetic progression 5, 12, 19, …. has 50 terms. Find its last term. Hence find the sum of its last 15 terms.
Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ABC  60. Now construct another triangle
5
whose sides are times the corresponding sides of ABC.
7
Find the values of k for which the points A(k +1, 2k), B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
Page 9
………………………………………………..
SUMMATIVE ASSESSMENT – II, 2015 – 2016
SUBJECT : MATHEMATICS
Max. Marks : 90
SECTION : A
Time : 3 : 00 Hrs.
I.
Question numbers 1 to 4 carry one mark each
1.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a
red card nor a queen.
2.
A ladder, leaning against a wall, makes an angle of 60º with the horizontal. If the foot of the ladder is 2.5 m away
from the wall, find the length of the ladder.
3.
In see figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and CAB  30, find
PCA.
4.
For what value of k will k+9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?
SECTION : B
II.
Question number 5 to 10 carry two marks each.
5.
In See Figure, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides
AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that. AB + CD = BC + DA.
6.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
7.
The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
8.
Let P and Q bet the points of trisection of the line segment joining the points A(2, -2) and (B(-7, 4) such that P is
nearer to A. Find the coordinates of P and Q.
9.
In See Figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If
OP = 2r, show that OTS  OST  30.
10.
Solve for x :
6x  7  (2x  7)  0
Page 10
SECTION : C
III.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Question numbers 11 to 20 carry three marks each.
A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical
22
vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (Use   )
7
In See figure, O is the centre of a circle such that diameter AB = 13 cm and AC = 12cm. BC is joined. Find the area
of the shaded region. (Take   3.14)
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
In See Figure, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and
diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the
22
cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq.metre. (Use   )
7
A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is
5
completely submerged in water, the water level in the cylindrical vessel rises by 3 cm. Find the diameter of the
9
cylindrical vessel.
A man standing on the deck of a ship, which is 10m above water level, observes the angle of elevation of the top of
a hill as 60º and the angle of depression of the base of hill as 30º. Find the distance of the hill from the ship and the
height of the hill.
In See Figure, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm
22
where AOC  40. (Use   )
7
There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at
random. Find the probability that the number on the selected card
(i)
is divisible by 9 and is a perfect square
(ii) is a prime number greater than 80.
Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the
squares of the other two by 60. Find the numbers.
The sums of first n terms of three arithmetic progressions are S1, S2 and S3 respectively. The first term of each A.P.
is 1 and their common differences are 1, 2 and 3 respectively. Prove that S1  S3  2S2 .
Page 11
SECTION : D
IV.
21.
22.
23.
Question numbers 21 to 31 carry four marks each.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state
government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the
whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with
conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per
sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem?
22
(Use   )
7
The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of
numbers of houses proceeding the house numbered X is equal to sum of the numbers
of houses following X.
In See Figure, the vertices of ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment
DE is drawn to intersect the sides AB and AC at D and E respectively such that
AD AE 1

 . Calculate the area of ADE and compare it with area of ABC.
AB AC 3
24.
In See figure, two equal circles, with centres O and O’, touch each other at X.OO’
produced meets the circle with centre O’ at A. AC is tangent to the circle with centre
DO'
.
O, at the point C. O’D is perpendicular to AC. Find the value of
CO
25.
A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return
downstream to the same spot. Find the speed of the stream.
In See Figure, is show a sector OAP of a circle with centre O, containing . AB is perpendicular to the radius
26.



 1
OA and meets OP produced at B. Prove that the perimeter of shaded region is r  tan   sec  
180 

27.
28.
29.
30.
31.
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
1
Two pipes running together can fill a tank in 11 minutes. If one pipe takes 5 minutes more than the other to fill
9
the tank separately, find the time in which each pipe would fill the tank separately.
From a point on the ground, the angle of elevation of the top of a tower is observed to be 60º. From a point 40 m
vertically above the first point of observation, the angle of elevation of the top of the tower is 30º. Find the height
of the tower and its horizontal distance from the point of observation.
4
Draw a triangle with sides 5 cm and 7 cm. Then draw another triangle whose sides are
of the corresponding
5
sides of first triangle.
A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from
the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.
Page 12
……………………………………………..
SUMMATIVE ASSESSMENT – II
SUBJECT : MATHEMATICS
Max. Marks : 90
I.
1.
SECTION : A
Time : 3 : 00 Hrs.
Question numbers 1 to 4 carry one mark each
Find the common difference of the A.P.
1 1  2q 1  4q
,
,
2q 2q
2q
2.
In the given figure O is the centre of a circle and AR is the tangent to the circle. If AB is a chord such that
RAB  40, find AOB.
3.
The angle of elevation of the top of a building at a point on the ground 15 m away from the foot of the building is
45º. Find the height of the building.
1
A solid metallic cylinder of base radius 5 cm and height
of 360 cm. It is melted to form one solid sphere. Find
2
the radius of the sphere.
4.
SECTION : B
II.
5.
6.
7.
8.
9.
10.
Question number 5 to 10 carry two marks each.
Find the number of natural numbers from 110 to 990 which are divisible by both 2 and 5.
Solve the quadratic equation for x : 4 2x 2  5x  3 2  0
Cards marked with numbers 2, 3, 4, ……, 71 are placed in a box. One card at random is drawn from the box. Find
the probability that the number one the card drawn is a perfect square.
A bag contains 7 black balls and some green balls. If the probability of drawing a green ball from the bag is thrice
that of a black ball. Find the number of green balls in the bag.
22 

If the perimeter of a protractor is 36 cm. Find its area  Take   
7 

If the radii of two concentric circles with centre O are 3 cm and 5 cm. AB is a chord of outer circle which touches
the inner circle at P. Find the lengths of AB.
SECTION : C
III.
Question numbers 11 to 30 carry three marks each.
11.
Find five numbers in A.P., whose sum is 12
12.
1
and the ratio of first and last number is 2:3.
2
Find two consecutive odd positive integers sum of whose squares is 1354.
13.
Draw a triangle with side BC = 8 cm, C  30 and A  105. Then construct a triangle whose sides are
2
times
5
the corresponding sides of ABC.
Page 13
14.
If angle between two tangents PT and QT drawn from a point T to a circle of radius ‘r’ and centre ‘O’ is 60º, then
prove that PT =
15.
3r.
The angle of elevation of the top of a building from a point A on the ground is 30º. On moving a distance 30 m
towards the base to a point B, the angle of elevation changes to 45º. Find the height of the building and the distance
of its base from point A. [Take 3  1.732]
16.
17.
Prove that area of a triangle whose vertices are A(x, x – 2), B(x + 23, x + 2) and C(x + 3, x) is independent of x.
A conical tent has six ribs which are equally spaced. Assuming the base of conical tent to be circular of radius 300
18.
2 

cm. Find the area between two consecutive ribs of the tent.  Take   
7 

In the given fig. PQRS is a trapezium right angled at S. Area of trapezium is 29.4 sq. cm. PQ || SR; PQ = 3 cm, SR
22
= 11 cm. If PSX is a quadrant of a circle, Find the shaded area. [Take   ]
7
19.
A farmer connects a pipe of internal diameter 25 cm from a canal into a cubical tank in his field of side 2.5 m. If
20.
22 

water flows through the pipe at the rate of 3.5 km/hr, how much time will it take to fill the tank?  Take   
7 

A square table cloth has 9 circular designs each of diameter 28 cm. Find the area of the table cloth with no design.
22 

 Take   7 


SECTION : D
IV.
21.
22.
23.
Question numbers 21 to 31 carry four marks each.
In a school, the Administration decided of planting trees in and around school to reduce Air Pollution. It was
decided that the number of trees that each section of each class will plant, will be the same as the class in which
they are studying. There are three sections fo each class.
(i)
Find out, how many trees were planted by the students from class I to class XII.
(ii) Also reflect and tell why this decision was taken by the School Administration?
1
 3x  1   x  2 
Find the roots of the following equation : 3 
  2  3x  1   5; x  2, 3
x

2

 

A truck is driven at a speed limit of 70 km/hr on a highway. He has to cover a distance of 420 km. If he increases
his speed by 10 km/hr, he can reach his destination 1 hour earlier. The driver has to drive at a uniform speed. What
was his original speed?
Page 14
24.
25.
26.
27.
Prove that the parallelogram circumscribing a circle is a rhombus.
A path of length 32 m is constructed along a chord PQ of a circular park of radius 20 m. Two more paths are
constructed from an external point T to the park and tangential to it at P and Q. If the cost of digging the path
(taking it to be narrow of negligible width) is Rs. 120 per meter, find the total cost incurred.
A circular target of radius 12 cm is to be hit as a game. The dart board has 3 concentric circles inside the bigger
board of radii 4 cm, 7 cm and 9 cm, dividing the board into 4 regions. If a shot hit the target, find the probability of
hitting each region if you shoot 144 times.
A man standing on a tower of height 12 m observes a cliff. The angle of elevation of the top of the cliff is 30º and
the angle of depression of the base of the cliff is 45º. Find the height of the cliff and its distance from the tower.
(Use 3  1.732)
28.
29.
30.
31.
If the coordinates of the mid-points of the sides of a triangle XYZ are P(2, 1), Q (3, -5) and R(6, 4). Find its
centroid.
The tree vertices of a parallelogram are A(-2, 1), B(4, 1) and C(7, 4). Find the fourth vertex D.
A container open at the top is in the form of a frustum of a cone of height 21 cm with radii of its lower and upper
circular ends as 16 cm and 6 cm respectively. Find the cost of juice which can completely fill the container at the
22 

rate of Rs. 42 per litre.  Use  
7 

A gardener planted flowers on the sides of a square lawn of side 56 m in the form of circular bed of flowers. If the
centre of each flower bed is the point of intersection O of the diagonals of the square lawn. Find the area of flower
22 

beds. (Shaded area)  Use  
7 

Page 15
THE MILLENIUM SCHOOL
SUMMATIVE ASSESSMENT – II
SUBJECT : MATHEMATICS
Max. Marks : 90
I.
1.
2.
3.
4.
SECTION : A
Time : 3 : 00 Hrs.
Question numbers 1 to 4 carry one mark each
a, a + d, a + 2d, ….., l are n terms of an A.P. Find (n – 1)th term of it in terms of ‘l’.
A ladder of length 18 m touches a wall at a height of 9 m. Find the angle made between the ladder and the ground.
What is the sum of probabilities of all the elementary events of an experiment?
The line segment joining the points A(-1, 8) and B gets bisected by origin. Find the coordinates of the point B.
SECTION : B
II.
5.
6.
7.
8.
9.
10.
Question number 5 to 10 carry two marks each.
If a number is added to twice its sq are, the result is 21. Find the number.
Which term of the AP : 3, 15, 27, 39, ….. will be 132 more than its 54th term?
PA and PB are tangents to a circle. MQN is another tangent touching the circle in Q. If PB = 8cm and PM = 5.5cm,
find length of MQ.
Two circles touch internally at P. How many tangents can be drawn to the circles from an external point? What is
the relation between the tangents?
To construed a pair of tangents to a circle which are inclined to each other of angle of 60º. It is required to
construct tangents at the end points of two radii of the circle. What will be the angle between the two radii?
A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the remaining solid.
SECTION : C
III.
Question numbers 11 to 20 carry three marks each.
11.
If the quadratic equation (1  a 2 )b 2 x 2  2abcx  (c2  m 2 )  0 in x has equal roots, prove that c 2  m 2 (1  a 2 ).
12.
How many terms of the AP : -15, -13, -11,… are needed to make the sum – 55?
If PA and PB are two tangents drawn to a circle with centre O such that BPA  120, prove that OP = 2PB.
13.
14.
16.
The angle of elevation of the top of a tower from certain point is 30º. If the observer moves 20 metres towards the
tower, the angle of elevation of the top increases by 15º. Find the height of the tower.
In a hospital there are 200 beds for patients. Of these 120 are occupied by males and remaining by females. 20% of
the males and 40% of the females are suffering from malaria and rest of them from Dengue. If a patient is selected
at random, find the probability that he/she is :
(a) Female patient
(b) Male patient suffering from malaria
(c) Female patient suffering from dengue
If (1, 5), (p, 1) and (4, 11) are collinear, find the value of p.
17.
Find the value of x for which the distance between the points A(x, -3) and B(-4, 2) is 5 5 units.
18.
In summer vacations, a summer camp is organised for students where they stay in conical tents of radius 12 m.
Each tent has 2 m wide path around it for movement. If the land where these tents are erected is a rectangular field
15.
22 

of dimensions 1.4 km × 200 m, how many tents are possible in this field?  Use  
7 

Page 16
19.
Find the volume of an oil container, which is in the shape of a cylinder with hemispherical ends. Total length of oil
20.
22 

container is 23 m and radius is 3 m.  Use  
7 

Find the area of the shaded region in the given figure
SECTION : D
IV.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
Question numbers 21 to 31 carry four marks each.
1
3
 4x  3 
 2x  1 

Express 
  10  4x  3   3,  x  2 , x  4  in standard form and then find its roots by factorisation
2x

1






method.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms
of the AP.
5
4
3
  3, x  0,
Solve for x :
2x  3 x
2
A circle touches the side BC of ABC at P and the sides AB and AC produced at Q and R respectively. If the
sides AB, BC and AC of the ABC are 13 cm, 15 cm, 18 cm respectively, show that AQ = AR = 23 cm.
Construct a ABC whose sides are in the ratio 2 : 3 : 4 and perimeter 12 cm. Now construct A'BC' similar to
2
ABC such that A 'B  AB.
5
At a point on level ground, the angle of elevation of a tower is found to be such that its tangent is 5/12. On walking
192 metres towards the tower, the tangent of the angle of elevation is 3/4. Find the height of the tower.
Surya and Samhita are studying in the same class what is the probability that both will have :
(i)
different birthdays
(ii) Same birthdays
Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4), taken in order, are the vertices of a square. Also, find its area.
A farmer had a square field of side 150 m. He divided this field between his son and daughter such that he gave all
the four corners in the form of four sectors of circles of radius 49 m each to his daughter and an equal amount of
area to his son. The remaining area he left for charitable purpose. Find out the area he left for charity. What values
22
of the farmers are depicted through this action? (Use   )
7
A plot is in the form of a rectangle ABCD having a semicircle on BC. If AB = 60 m and BC = 28 m, find the area
and perimeter of the plot.
A well with 14 m inside diameter is 10 m deep. Earth taken out of it and spread all around to a width of 5m to form
an embankment. Find the height of embankment.
Page 17
ANSWERS : PAPER – 1
1.
Length of ladder = 5m
2.
5.
subjective solved in soln
6.
11.
h  2 cm
12.
16.
Area  410.66cm2
18.
Original number  852
22.
Ratio 
25.
30.
1
3
26.
23.
3.
K = 18
7.
6
13
4.
P (1, 0) Q (4, 2)
9.
8.
diameter  18 cm
13.
Area  36  36 cm 2
17.
Distance of hill from ship = 10 3m , Height of hill = 40 m
14.
1
10
13
(ii)
(iii)
23
23
23
15.
10.
x  8, 20
Total cost = 16,500/-
Cost  1,35,864 / 
21.
19.
20.
(i)
1
2
24.
5
Ar of ADE  sq. unit , Ar of ABC  9  ar. of ADE
6
Probability =
x  35
27.
Speed of stream = 8 km/hr
200
3
31.
Length = 7m, Breath = 4 m
Distance =
PCA  60
28.
Construction
29.
SOLUTIONS
1.
A
5m
cos 60 
2.5

1 2.5

2
 5m
C
60
B
2.5m
2.
2K  1  K  9  2K  7  2K  1
K  10  8  K  18
3.
BD  DC
D
DBC  DCB  30
DCA  ACB  BCD  180
C
DCA  90  30  180
DCA  60
30
A
4.
60
B
No. of Red Card = 26
2 Black Queen Total = 26 + 2 = 28 Cards
52  28  24 cards
24 cards are neither Red nor Queen
Probability 
5.
24 6

52 13
AB  CD  BC  DA
Property = Length of tangents from a point on circle are equal
AB  a  b CD  c  d
BC  b  c
AD  a  d
abcd bcad
 a bcd a bcd
6.
a 4  a  3d  0
P.T  a 25  3a11
a 25  a  24d
a11  a  10d
Page 18
a  24d  3 (a  10d)
a  24d  3a  30d
2a  6d  0
7.
cos  
a  3d  0
already given
T
r
B

2r H
O
  60

P
OTS  30
S
8.
A(3,0) B(6,4) C (1,3)
Applying distance formula
AB  32  42  5
BC  49  1  5 2
AC  42  32  5
AB = AC isosceles triangle
9.
10.
A
(2, 2)
1
1
P
Q
1
B
( 7, 4)
 7  4 4  4 
P
,
3 
 3
 14  2 8  2 
Q
,
3 
 3
P(1,0)
Q (4,2)
2x  9  x  13
2x  9  13  x
Squaring both side
2x  9  169  x 2  26x
x 2  28x  160  0
x 2  8x  20x  160  0
x(x  8)  20 (x  8)  0
(x  8) (x  20)  0
x  8, 20
11.
P(x, y)
A (a  b, b  a)
Given :
PA  PB
B(a  b, a  b)
PA2  PB2 (Squaring & applying distance formula)
 x  (a  b) 
2
  y  (b  a)    x  (a  b)    y  (a  b) 
2
2
2
 x  (a  b)    x  (a  b)    y  (a  b)    y  (b  a) 
 x  a  b  x  a  b x  a  b  x  a  b    y  a  b  y  b  a  y  a  b  y  b  a 
2
2
2
2
2(x  a)x  2b  2(y  b)x  2a
b(x  a)  a(y  b)
bx  ab  ay  ab
bx  ay
12.
Vol. of cone = Vol. of cylinder
Page 19
1
  25  24    100  h
3
25  24
2
3  100
h  2 cm
h
13.
Area of shaded region = r 2  (Area of Right angle)
2
 13  1
     12  5
2 2
36.36 cm 2
14.
Net Vol. of liquid + Vol. of sphere = Vol. of final volume
4
32 

r 2 h  (6)3  r 2  h  
3
9 

I
32 
4

r 2  h    r 2 h  (6)3
9 
3

32

 4
r2  h 
 h   666
9

 3
32
cm
9
h
r
46669
r2 
32  3
r 2  81
r  9cm
diameter  18 cm
15.
Total CSA = CSA of Cylinder + CSA of Cone
3
3
 2   2.1     2.8
2
2

3 22
  7  33 m 2
2 7
Total cost = 33m2  500m2
 16500 / 
16.
Area of Ring = Area of outer – Area of inner
Circle
Circle

22
22
 14  14   7  7
7
7

22
 196  22  21  462cm2
7
Area of sector Ring =
40
40
   14  14 
 77
360
360
1 22
22  7
   147 
9 7
3
 Area of shaded region = Area of Ring – Area of sector ring
 22  21 
22  21
9
8
  22  21  410.66 cm 2
9
Page 20
17.
AC  Hill
ED  Ship
A
ln  ECD
tan 30 
10
1 10

  a  10 3
a
3 a
E
60
a
30
 distance between hill and ship  10 3m
ln ABE
tan 60 
b  30 m
30
D
 height of hill = 30 + 10 = 40 m
18.
10m
10m
b
b
 3
a
10 3
B
C
a
Let the three digit be a – d, a, a + d
a  d  a  a  d  15  3a  15  a  5
100  (a  d)  10a  a  d  594  100(a  d)  10a  a  d
100a  100d  a  d  594  100a  100d  a  d
99d  99d  594
d  3

a d 538

a 5

a d 53 2
Original No  852
19.
(a  b)x 2  (b  c)x  (c  a)  0
Roots are  & given = 
  
2 
(b  c)
ab
  
cb
ab
2 
ca
ab
(  )
ca
ab
2

 (c  b) 
ca

 
a b
 2(a  b) 
(c  b) 2 (a  b)  4(a  b) 2 (c  a)
c2  b 2  2bc  4 (ac  a 2  bc  ab)
b2  c2  2bc  4ac  4a 2  4bc  4ab
4a 2  b2  c2  4ab  2bc  4ac  0
(2a  b  c) 2  0
2a  b  c
20.
J, Q, K of Red are 6 and they are removed
Left out cards  52  6  46
2
1

26 23
(a)
Black King  2  Prof =
(b)
Card of Red colour  20  Prof =
20 10

46 23
(c)
Card of Block colour  26 Prof =
26 13

26 23
Page 21
21.
Total CSA = CSA of cylinder + CSA of cone
The slant height of cone = 3  5m
Total CSA = 2 
22
22
 2.8  3.5   2.8  3.5
7
7
Total CSA = 3 
22
 2.8  3.5
7
 35
2.1
35
cost of one tent  1500
Now cost bear by one school 
50
 22  2  8  3  5  1500
7  50
 1,35,864 / 

2 8
Value  helping and co-operation
22.
ln ADO & ACO
A  A
(common angle)
D  C
(Each 90)
ADO ~ ACO

[AA corollary]
DO' AO' r


CO AO 3r
DO' 1

CO 3
23.
(1,2,3,4) (1,4,9,16)


x
y
Total No. of Pairs of 16
Now x  y less than 16 are 8
and x  y equal to 16 are 2
As in question only less than 16 is asked
So Probability 
24.
8 1

16 2
A (4, 6) B (1, 5) C (7, 2)
Ar of ABC 

1
 x1 (y2  y3 )  x 2 (y3  y1 )  x 3 (y1  y 2 ) 
2
1
 4(5  2)  1 (2  6)  7(6  5) 
2
1
15
 (12  4  7)  sq. unit
2
2
AD AE 1


AB AC 3
By converse of Thales Theorem
DE || BC
Given 

and ADE  ABC[SAS]

ar of ADE  AD 

ar of ABC  AB 
2
Page 22
2
 1  15 5
ar of ADE      sq. unit
2 6
 3
5
ar of ADE  sq. unit
6
ar of ABC  9  ar of ADE
25.
Perimeter of shaded region
Length of sector =

 2 r
360
Length AB  tan  
AB
 AB  r tan 
r
BP  OB  OP
BP  rsec   r
Perimeter 

r  r tan   r sec   r
180
 

Perimeter  r  tan   sec  
1
180 

26.
House no  1, 2, 3, - - - - - - - - - - 49
S49  Sn  Sn
n  No. of the house
S49  2Sn
49  50
n(n  1)
2
2
2
n 2  n  1225  0
On solving we get n = 34.5
 n  35 x  35
Thus house no. 35 is one which satisfy the condition
27.
Speed of motor (SM )  24 km / hr
Speed of stream (SS )  9
Upward stream speed = SM  SS 
32
t1
Downward stream speed = SM  SS 
Given 
....(1)
32
t2
...(2)
t1  t 2  1
t1  t 2  1
32
32

1
24  SS 24  SS
SS2  64SS  576  0
SS (SS  8)  72 (SS  8)  0
SS  8 km / hr

SS  72 km / hr (Rejected)
Speed of stream = 8 km/hr
Page 23
28.
Do it your self
29.
To prove  OB  BP
Proof :  of the two lines OP & OB
O
By fig. we see OP  OB
OP  OA
A
and OA  OB  radius of circle
Thus OB is the shortest of all line draw from O to BP
And hence the shortest line / distance is the  distance
between a point and line.
30.
B
P
ln OAB
tan 60 
O
100
100
 a
m
a
3
100m
ln OAC
tan 30 
b
31.
a  b  100 3
100

100
a  b b  100 3 
3
B
C
A
200
m
3
Area of Rect. Park = 4 + Area of isoc. Triangle park
1
(  3)  4   12 (  3)
2
2
60
30
A
D
 2m Rejected
(  3)m
 9  14  0
E
 7cm, b  4m
12m
 7, 2

 

C
B
ANSWERS : PAPER – 2
1.
No. added / substracted =
5.
K6
10.
Length of arc = 9.4 cm
14.
 4 3m
6.
15.
3
2
a 5
(i)
20
m
3
2.
distance =
7.
Distance = 6 cm
11.
x   2, 

2
2
(ii) (iii)
9
9
9
5
2
3.
Probability =
8.
Construction
1
2
9.
12.
13.
Construction
Area  61.11 km 2
16.
17.
17 sq. unit
18.
22.
19.
Volume  8316 cm3
20.
Thickness  1cm
21.
K = 16
24.
APC  115
25.
Construction
26.
Height = 60  20 3 m XB  40
27.
(i)
30.
Perimeter  1: 2 : 3 , Area 
9
3
2
4
(ii) (iii) (iv)
25
5
5
25
28.
29.
4 
G  ,3 
3 
4.

23.
time  6 min



3 1 m
(a) 285 mm (b) 96.25 m 2
1
9
:1:  1: 4 : 36
4
4
31.
Cost = 550/-
SOLUTIONS
1.
4x 2  12x  8  0
(2x) 2  2  2x  3  8  1  1  0
Page 24
(2x  3) 2  (1) 2  0
Thus the factor added or substracted is 1.
2.
tan 60 
a
A
20
20
 3
a
a
20
3
20 3

 a
m
3
3
3
20m
60
C
3.
B
a
Die thrown No.  1, 2, 3, 4, 5, 6
No. having atleast two factors  2, 3, 4, 5, 6
Probability =
4.
5
6
2
A(2,1)
1
E(3, 4)
F
 6  2 8 1 
4 
F
,
 F  , 3

3 
 3
4 
5.
Kx 2  2Kx  6  0
 
2K
K
Roots , 
  
2  2
2 
6
K
6

K
[  ]
1
6
 K6
K
 1
6.
4K  6  K  2  3K  2  4K  6
3K  8  K  4
4K  12
7.
First term = 5
K 3
PB2  PC  PA
(24) 2  (26  r) (26  r)
(24) 2  262  r 2
O
A
r  100  r  10 cm
C
P
2
D
ln ODB
OB  10
DB  8
OD  6 cm Pythogarus Theorem
B
8.
9.
ln ONC & OMA
ON  OM (Each radius)
N  M (Each 90)
OA  OC (Radius of outer circle)
ONC  OMA [RHS]
 NC  MA
2NC  2MA
CD  AB
[CPCT]
H.P.
Page 25
10.
Area of Sector 
54 

 r 2
360

   36  36
360
  15

 2r
360
Length Arc of Sector 
15
22
 2   36
360
7
 9.4 cm

11.
2x 2  7x  5 2  0
2x 2  2x  5x  5 2  0
2x(x  2)  5 (x  2)  0
( 2x  5) (x  2)  0
x   2,
12.
x
5
2
a8  a  7d  0  a  7d
...(1)
a 38  3918
a  37d  3 (a  17d)
a  37d  3a  51d
[from (1)]
7d  37d  21d  51d
30d  30d
13.
14.
tan 60 
a
A
12
a
12
12  3
m
3
3
12m
60
a  4 3m
15.
C
a
B
A = {1, 3, 8} B = {2, 5, 6} Total Cases = 9
(a) Sum of two integer odd [Event Odd = Odd]
(8, 5) (1, 2) (1, 6) (3, 2) (3, 6)
5
9
Sum multiple of 5  (3, 2) (8, 2)
Pr of 
(b)
2
9
Sum of integer greater than 11
(8, 5) (8, 6)
Prof 
(c)
Prof 
2
9
Page 26
16.
Using distance formula
C(10,7)
D(8,3)
AB  4  16  20
BC  64  16  80
CD  4  16  20
AD  64  16  80
 opp. side are equal thus it is a parallogram
AC  100  64  164
A(0, 1)
B(2,3)
BD  36  0  6
A  BD
As diagonal are not of equal length thus it doesnot form a Rectangle
17.
18.
19.
P is mid point of BC
Property  Median of  divides the triangle in two part of equal
Area (-8, 2)
1
 ar ABP  ar ABC
2
1 1
   8(6  0)  10(0  2)
2 2
1
68
  48  20 
 17 sq. unit
4
4
ar ABP  17 sq. unit
Area of sector 
B
P
C(10,6)
70 22
  10  10
360 7
 61.11 km 2
Volumes of Tank = Vol. of Cylinder + 2  Vol. of Cone
21 21 21
1
21 21
   18  2       9
2 2 2
3
2 2
21 21
2 

    18   9
2 2
3 

21 21 22
    24
2 2 7
 8316 cm3
Vol. of Tank  
20.
Vol. of jam in cylinder  10 1110  thickness
2
 5  5  14  10  11  10  thickness
7
22  5  5  14
 1cm
thickness 
7  10  11  10
21.
x 2  kx  64  0
D1  K 2  256
x 2  8x  K  0
D2  64  4K
K2  256  0
(K  16)(K  16)  0
64  4K  0
4K  64
K  16
22.
A(0,0)
K = 16
a b c
d
a1  a a 2  b
a a
b a ba  b
9cm
10.5
18cm
10.5
9cm
16
16
e
d  b  a  common difference
Page 27
c  a  2b  2a  2b  a
d  a  3b  3a  3b  2a
e  a  4b  4a  4b  3a
a  4b  bc  4d  e
a  4b  6(2b  a)  4(3b  2a)  4b  3a
a  4b  12b  6a  12b  8a  4b  3a
O H.D.
23.
Distance cover by thief = Distance cover by police
(t  1)
t  100 
 2  100  (t  2)  10 
2
t 100  (t  1)(200  10t  20)
Simplifying we get
t 2  3t  18  0
(t  6) (t  3)  0
t  6 min
t  3 min (Rejected)
24.
ODC  25  ODC  ODA
A
DOC  65 [180  (90  25)]
2DOC  130  AOC
Reflex Angle  230
230
APC 
 115
2
APC  115
D
25
O
P
C
25.
26.
ln BXA
tan 60 
B
b  40
a
b
3a  b  40
D
ln BDC
b
 ba
a
Using (2) in (1)
tan 45 
....(2)
a
C
40m
40m
60
3a  a  40
a
45
....(1)
40
3 1

 20( 3  1)
3 1
3 1
X
a
A
a  20 ( 3  1)  b
Height of tower AB  40  20 3  20
 (60  20 3)m
ln XBA
cos 60 
a
1
a
 
 XB  2a
XB
2 XB
XB  2  20 ( 3  1)  XB  40( 3  1)m
Page 28
27.
50 student
Total  50 student
18 Girls
32 Boys
28.
30 Non Veg
20 Veg
12 Boys
(a)
Choosing a girl 
Pr of 
(b)
Non-veg Student  Prof =
(c)
Boys who is Non-veg  Prof =
(d)
Girl who is Veg  Prof =
8 Girls 20 Boys
10 Girls
18 9

50 25
30 3

50 5
20 2

50 5
8
4

50 25
Given  PA  PB
P(3, 4)
PA 2  PB2
(a  b  3) 2  (b  a  4) 2  (a  b  3) 2  (a  b  4) 2
A
(a  b, b  a)
(a  b  3) 2  (a  b  3) 2  (a  b  4) 2  (b  a  4) 2
B
(a  b,a  b)
 a  b  3  a  b  3 a  b  3  a  b  3  a  b  4  b  a  4a  b  4  b  a  4
(2a  b)(2b)  (2b  8)(2a)
(a  3)b  (b  4)a
ab  3b  ab  4a
29.

4a  3b  3b  4a  0
Diameter = 2r = 35 mm
(a) Total length of silver wire used
360
 36  1 Rectangle
10
 2r  5  2r

22
 35  5  35  285mm
7
Area of Each sector 
(b)
36
36 22 35 35
  
360 7 2 2
 96.25 mm 2
30.
da : d b : dc
1: 2 : 3
Radius =
diameter
 dividing by 2
2
ra : rb : rc
1
3
:1:
2
2
Pa : Pb : PC
2
3
: 2 : 2 
2
2

 : 2 : 3

da : d b : dc
P :P :P
 a b c
1: 2 : 3
1: 2 : 3
Aa : Ab : Ac


: :
4
4
1
9
:1:
4
4
Aa : A b : Ac
1
9
:1:
4
4
Page 29
31.
1
Volume of frustum   h (r12  r1r2  r2 2 )
3
r1  10 cm
r2  20 cm
h  30 cm
1 22
Vol.    30 (100  400  200)
3 7
1 22
Vol.    30  700  22000cm3
3 7
Vol. of bucket  22000 cm3
1 cm3 
1
litre
1000
1
litre
1000
Vol  22 litre
V  22000 
Cost of Milk to fill contains = 22  25
Cost = 550 Rs.
Vol. of bucket = 22 litre
Cost of Milk = 550 Rs.
ANSWERS : PAPER – 3
1.
  30
2.
6.
P4
7.
11.
3,79,500 / 
15.
2
x  6, 
3
18.
S10  175
22.
(i)
P(E) 
1
9
3.
P3
4.
8.
AB = 9 cm AC = 12 cm
12.
5.4 cm
16.
PRQ  120
9.
cos t  33.92 / 
13.
speed  720 km / m
17.
1, 6, 11, 16, 21 …….
5.
2x  y  3  0
14.
10.
x  a  2b
Diameter = 21 cm
Surface area = 1386 m 2
Area of Minor segment =
7
(44  21 3)m 2
3
Area of Major segment  598  21 m 2
19.
P(3,4)

1
11
2
(ii)
(iii)
(iv)
26
13
13
13
P(R) 
23.
Area  504 m 2
25.
 24m b  10m
29.
 348 sum of last 15 term = 4485
26.
3
Total balls  50
10
20.
27.
24.
28.
Speed  60 km / hr
30.
Construction
21.
Distance = 40 m
Length = 158.4 cm
31.
3
K  ,1
2
SOLUTIONS
1.
2.
3.
tan  
20
1
 tan  
20 3
3
  30
Total No. of Cases = 36
Favourable  (1, 6) (6,1) (2, 3) (3, 2)
Pr of 
4 1

36 9
 
2 5
P
P
  
15
P
[  ]
Page 30
4.
2  2 5
2 
 5
5
15
P
15
 P3
P
QPT  60
OPQ  90  60  30
ln OPQ is an isosceles triangle
POQ  120
Reffix of POQ  240
PRQ 
5.
240
 120
2
5
7
(2a  4d)  (2a  6d)  167
2
2
10a  20d  14a  42d  334
24a  62d  334
12a  31d  167
...(1)
S10  235
10
(2a  9d)  235
2
2a  9d  47
12a  31d  167
12a  54d  282
23d  115
d=5
2a  9  5  47
2a  2
a 1
A.P.  1, 6,11,16, 21........
6.
...(2)  6
...(1)
...(2)
In Right Triangle ABC
A(4,7)
AC  AB  BC [Pythogarus Theo.]
2
2
2
9  16  (P  4) 2  16  (P  7) 2  O
9  P2  16  8P  P2  49  14P
2P2  22P  56  0
P2  11P  28  0
P2  4P  7P  28  0
P(P  4)  7(P  4)  0
P  4, 7
B(P,3)
C(7,3)
P  7  Rejected
P4
Page 31
7.
ln ORP & ORQ
P
P  Q (Each 90)
r
30
OR = OR (Common side)
OP  OQ (radius)
ORP  ORQ
[RHS]
R
r
ln OPR
Q
cos30 
r
2r
 OR 
OR
3
tan 30 
PR
r
PR 
r
3

60
O
PR  RQ
(CPCT)
OR = PR + RQ
2r
r
r
2r 2r




3
3
3
3
3
8.
Ar of ABC  54cm 2
A
Using heron’s formula
a = 15 b = a + 9 c = a + 6
a
a
E
F
15  a  9  a  6 30  2a
S

 15  a
2
2
6
2916  a (15  a)  6  9
B
9
3cm
(54)2  (15  a)(15  a  15)(15  a  a  9)(15  a  a  b)
6
D
9
C
54  54
 a (a  15)
69
a 2  15a  54  0
a 2  3a  18a  54  0
a(a  3)  18 (a  3)  0
a 3
 AB  9cm a  3, 18 a  18 Rejected
AC  12 cm
9.
3 point are collinear then Area = 0
A(x, y) B(5,7) C(4, 5)
x(2)  5(5  y)  4(y  7)  0
2x  25  5y  4y  28  0
2x  4y  5y  3  0
2x  y  3  0
10.
x 2  2ax  (4b 2  a 2 )  0
Solving this by completing the sq. method
x 2  2ax  a 2  a 2  4b2  a 2  0
(x  a) 2  4b 2  0
(x  a  4b) (x  a  2b)  0
x  a  2b
x  a  2b
Page 32
11.
CSA of one tent = CSA of cylinder + CSA of Cone.
Slant height of cone = 3.5 m
 2
 3.5m
2.8m
22 4.2
22 4.2

4 
 3.5
7
2
7
2
CSA of 1 tent = 75.9m2
Cost of one tent = 75.9 100  7590 / 
2.1m
4m
Cost of 100 tent = 7590 100  759000 / 
Association will pay 
12.
759000
 379500 / 
2
2.1m
Volume of hemisphere = 72  Vol. of Cylinder
As 10% is wasted in transfer
90% of Vol. of hemisphere = 72  Vol. of 1bottle
9 2
    18  18  18  72    3  3  h
10 3
9  2  18  18  18
10  3  72  9
h = 5.4 cm
h
13.
Max. diameter of hemisphere is 10 cm
Total surface are of solid =
Total surface Area of
CSA of


area of circle Circle hemisphere
 6 10 10  r 2  2r 2
 600  r 2
 600 
TSA of solid =
4750 2
cm
7
Cost of Painting 
14.
22
4200  550
 25 
7
7
4750 5

Rs.  33.92 Rs.
7
100
Volume of sphere = 504  Vol. of 1cone
4 3
1
35 35
r  504       3
3
3
20 20
r3 
504  7  7  3
4 4 4
r3 
2  2  2  3 3 7  7  3
4 4 4
2  3  7 21

4
2
diameter  21cm
r
r  10.5 cm
Surface area = 4r 2
 4
15.
22 21 21
   1386m 2
7 2 2
3x 2  2 2x  2 3  0
3x 2  3 2x  2x  2 3  0
Page 33
3x(x  6)  2(x  6)  0

3x  2
x
16.
 x  6   0
2
3
x 6
ln CBA
C
1500 3
a
a  1500 m
15 sec D
tan 60 
1500 3
60
ln DAE
tan30 
30
a
A
1500 3
1 1500 3


ab
3 1500  b
B
b
E
1500  b  4500  b  3000m
 speed 
3000
 200m / sec
15
Speed = 200 
17.
18 km
 720 km / hr
5 hr
Area of Minor Area of Area of


Segment
Sector Triangle
OAB is an equilateral triangle
O
60 22
3
  14 14 
14 14
360 7
4
Area of Minor segment 
14

14
60 60
 22
3
 196 


 76 4 
 196
60
A
B
 44  21 3)
21 4


7
44  21 3 m 2
3
Area of Area of Mirror

Circle
Segment
Area of Major segment =

22
7
 14  14  (44  21 3)
7
3

22
7  44
 14  14 
 49 3
7
3
44  7
3
 616  84.87  102.66
 44  14  49 3 
616  17.79  598.21m2
18.
Given: 
a13  4a 3
...(1)
a  (2d  4(a  2d)
a  4d  16
a  12d  4a  8d
4d
 4d  16
3
a 5  16
...(2)
Page 34
4
 16
3
3a  4d
4d 
4d
3
d 3
a
S10  5 (8  9  3)
a4
 5  35  175
19.
12  3
5
a 3
a
3
2
AP 2
AP 2
 

AB 5
PB 3
P
(a, b)
A
(1, 2)
14  6
5
b4
B
(6,7)
b
P(3, 4)
20.
Prof of white ball = P(W) 
Prof of black ball = P(B) 
3
10
2
5
Prof of Red ball = P(R)  ?
As the bag contain only this 3 type of Ball

P(W)  P(B)  P(R)  1
3 2
  P(R)  1
10 5
3
10
No. of black balls = 20
P(R) 
P(B) 
2 20

(Total no. of balls)
5 n
2 20

 n  50 balls
5 n
21.
ln ABC
tan 30 
B
a
b
1 a
  ba 3
3 b
a
A
20m
E
30
60
b
C
20m
D
ln ACF
tan 60 
a  40
 b 3  a  40
b
3a  a  40
a  20m
a  20
F
b  20 3m
AB  a 2  b2  400  1200
AB  40m
22.
Total playing cards are 52
Page 35
(i)
A card of spade or an acr = 13 + 3 = 16
(ii)
16 4

52 13
A black king = 2
(iii)
2
1

52 26
A jack a king = 8 cards
(iv)
44 22 11


52 26 13
Either a king or queen = 8 cards
Pr of 
Pr of 
Pr of 
Pr of 
23.
8
2

52 13
Area of flower bed
= Area of sector (OSAP) – Area of iso. Right triangle

90 22
1
  21 2  21 2   21 2  21 2
360 7
2
R
21 2m
A
B
90
 1 22 1 
 441 2    
4 7 2
 441  2 
42m
S
O
2
 63  4  252 m 2
7
Q
P
 Area of two flower bed = 2  252  504m 2
24.
Left out solid = Vol. of Cylinder = Vol. of sphere
4
   4.2  4.2  10     4.2  4.2  4.2
3
4
2


  4.2   10   4.2 
3


10 cm
28 
2

  4.2    10  
5 

r  42
22
   4.2  4.2 
5
Now the left out cylinder is made in a wire
Which is a cylinder in shape with r  7cm
Left out solid = (7) 2  h
  4.2  4.2 
22
   7   7  h
5
6  6  22
 158  4 cm  h
5
25.
As ABCD is a rectangle

ln ABC we can apply
a  16
Pythogaus Theorem
(a  16)  (a  14)  a
2
2
C
D
a
2
(a  16  a  14)(a  16  a  14)  a 2
(2a  30)  2  a 2
A
a  14
B
Page 36
4a  60  a 2
a 2  4a  60  0
a 2  6a  10a  60  0
a(a  b)  10(a  6)  0
a  10,  6
a  6  Rejected

Sides of Rectangle are
Shorted side = 10 m Longer side = 24 m
26.
ln POA & POB
A
PO  PO (Common side)
r
AO  OB (Radius)
POB  POA
PA  PB
O
P
A  B (Each 90)
r
[RHS Cong.]
B
[CPCT]
27.
28.
Avg. speed 

Total distance
Total time
50  200 350

 70km / hr
5
5
150  200
150
200

5
5  20
Simplifying (1) we get
Avg. speed = 70 =
...(1)
150
200

5
5
5  20
30
40

1
5 5  20
305  600  405  52  205
52  505  600  600  0
5(5  10)  60(5  10)  0
5  60km / hr
29.
a 5
5  10km / hr  Rajected
d7
 a  (50  1)d
 5  49  7  348
Sum of cost 15 term = Sum of 50 term – Sum of first 35 term
Sum of cost 15 term = S50  S35
50
35
 5  348  10  34  7 
2
2
 4485

30.
A(K  1, 2K) B(3K, 2K  3) C(5K 1, 5K)
ABC are collinear so area = 0
0
1
 (K  1) (2K  3  5K)  3K (5x  2K)  (5K  1)(2K  2K) 
2
Page 37
O  (K  1)(3  3K)  3K  3K  (5K  1) X  3)
O  3 (1  K 2 )  9K 2  15K  3
O  3  3K2  9K2  15K  3
6K2  15K  9  0
2K2  5K  3  0
2K2  2K  3K  3  0
2K(K  1)  3(K  1)  0
(2K  3) (K  1)  0
3
K  ,1
2
ANSWERS : PAPER – 4
1.
d  1
6.
x
3
4 2
2.
AOB  80
, 2
7.
P(E) 
1
10
9 5 11
2, , , ,3
4 2 4
3.
Height = 15 m
4.
r  15 cm
8.
no. of green ball  21
9.
Area  206  18 cm 2
12.
25 and 27
10.
AB  8 cm
15.
Height of building = 15( 3  1)m , distance from A to foot of building = (45  15 3)m
18.
Area  15.54cm2
19.
23.
Speed  60km / hr
24.
25.
27.
height  18  928m
28.
 11 
G  ,0 
3 
11.
time  5
5
min
11
20.
13.
1512 cm 2
cos t  10, 240 / 
29.
D (1, 4)
5.
14.
Construction
21.
16.
234 trees
26.
0.11, 0.22, 0.22, 0.43
30.
Cost  358  512 / 
n  89
17.
1
10
22.
x  5,
31.
Area  896m2
SOLUTIONS
1.
Common difference
1  2q 1 1  4q 1  2q



2q
2q
2q
q
Common difference are same in an AP
d  a 2  a1 
2.
1  2q 1 1  2q  1


 1
2q
2q
2q
RAB  40
BAD  90  RAB  90  40  50
AOB is an isosceles triangle

AOB  50  50  180
ADB  80
3.
tan 45 
h
 h  15m
15
h
45
15m
Page 38
4.
Volume of Cylinder = Vol. of Sphere
(5) 2  180 
4 3
r
3
r  3 5 5 5 3 3 3
r  15 cm
5.
Divisible by both 285
110  first No. Now L.C.M. of 285 = 10
 common diff. of AP is 10
110, 120, 130,………… 990
990  110  (n  1) 10
99  11  (n  1)
88  1  n
6.
n  89
4 2x 2  5x  3 2  0
4 2x 2  8x  3x  3 2  0
4 2x(x  2)  3(x  2)  0
(x  2) (4 2x  3)  0
x 2
7.
3
4 2
Perfect square from 2 – 71
4, 9, 16, 25, 36, 49, 64
Prob. 
8.
x
7
1

70 10
Black balls = 7
P(B) 
Green balls = n
7
n7
P(G)  3P(B)
P(G) 
n
n7
n
7
 3
n7
n7
n  21Balls
9.
Perimeter of Protector = Perimeter of semicircle
36  r
r
36

Area of Protactor =  

10.
36 36 1
 
  2
36  36  7
 206 18 cm2
22  2
By Pythogarus Theorem in OPB
O
(5) 2  (3)2  PB2
PB  4 cm

AB  8 cm
3
D
5
B
Page 39
11.
Let the five no. in A.P. are
a  2d, a  d, a, a  d, a  2d
a  2d  a  d  a  a  d  a  2d 
5a 

25
2
25
5
 a
2
2
a  2d 2
  3a  bd  2a  4d
a  2d 3
a  10 d 
d
5
 10 d
2
1
4
9 5 11
 Number are 2, , , , 3
4 2 4
12.
Consecutive odd  ve integer be 2n  1, 2n  1
(2n  1) 2  (2n  1) 2  1354
4n 2  1  4n  4n 2  1  4n  1354
8n 2  1352  n 2  169
n  13
 No are 25 and 27
13.
14.
ln POT
tan 30 
P
r
PO
1
r


PT
3 PT
PT  3r
15.
30
O
T
Q
ln DBC
tan 45 
D
h
ha
a
ln DAC
tan 30 
h
h
1
h


30  a
3 30  a
30  a  3h
30  3h  h  h 
30
3 1

 15( 3  1)m
3 1
3 1
45
30
A
30m
B
a
C
Height of building  15( 3  1)m
Distance from A foot of building  30  15 3  15
 45  15 3m
16.
1
 x1 (y2  y3 )  x 2 (y3  y1 )  x 3 (y1  y 2 ) 
2
A(x, x  2) B(x  23, x  2) C(x  3, x)

Page 40

1
 x(x  2  x)  (x  23)(x  x  2)  (x  3)(x  2  x  2) 
2
1
 (2x  2x  46  4x  12)
2
 17 sq. unit Hence proved Area is independent of x.
17.
18.
Area of Trap 
29.4 
ab
h
2
3  11
 h  h  4.2 cm
2
Area of shaded Region = Area of Trap -
Area of Circle
4
22 4.2  4.2

7
4
 29.4  13.86
 29.4 
 15.54 cm 2
19.
Time taken by Pipe to fill tank =
Time taken 
Time taken 
20.
Volume of cubical tank
Area of cross section of pipe  Rate of flow
2.5  2.5  2.5
min
22 25 25 35 1000


 
7 200 200 10 60
60
5
min  5 min
11
11
Area of square = 84  84  7056cm2
One side of square = 3  diameter of 1circle
 3  28  84cm
Area of region with no design = 7056  9  Area of one circle
 7056  9 
22
 14  14
7
 1512cm2
21.
No. of trees each class will plant
 (1  2  3  4...........12)  3
12  (12  1)
 3  234 trees
2
Value  Environmental awareness and social responsibility

22.
 3x  1   x  2 
3
  2
5
 x  2   3x  1 
Taking L.C.M. and cross Multiply
3(3x  1) 2  2(x  2) 2  5(x  2)(3x  1)
3(9x 2  1  6x)  2(x 2  4  4x)  5(3x 2  5x  2)
25x 2  15x 2  26x  25x  5  10  0
10x 2  51x  5  0
Page 41
10x 2  50x  x  5  0
10x(x  5)  1(x  5)  0
(10x  1)(x  5)  0
x
23.
1
,5
10
Let the original speed = S1 km / hr
Time taken = t hr
420  S1t
t
S1  t  (S1  10) (t  1)
...(1)
420
S1
S1  10  10t
S1  10 
...(2)
10  420
S1
S12  10S1  4200
S12  10S1  4200  0
S12  60S1  70S1  4200  0
S1 (S1  60)  70(S1  60)  0
S1  60km / hr
24.
S1  70km / hr Rejected
Given ABCD is a 11gm
c
PB  BQ  a 
QC  RC  b 
 Length of tangent drawn from ext. point are equal
RD  SD  c 
SA  AP  d 
b
R
C
D
b
c
Q
S
As the given ABCD is a 11gm
a d bc
...(1)
[AB  CD Opp. sides equal]
d
a b dc
...(2)
[BC  AD Opp. sides equal]
A
a
B
d
Solving (1) & (2)
a d bc
a bdc
a
P
d  b  b  d  2b  2d  b  d

a d a b
abab
25.
[AB  BC]
H.P.
P
Length PQ = 32 m
Cost of making PQ  32 120
 3840 Rs.
20m
T
16m
O
R
ln Right triangle PRQ
Q
16
4
 sin  
20
5
ln Right triangle OTP
sin  
tan  
PT
PO
....(1)
Page 42
sin  
4
4
by pythogours theorem tan  
5
3
 PT
80

 PT  x  m
3 20
3
P
20
16
Cost of making PT & TQ  PT  TQ ext point tangent are equal
Cost of making PT & TQ  2 
80
 120
3
 6400 Rs.
2D
R
O
T
 Total cost = 3840 + 6400
= 10240 Rs.
X
Q
P
26.
There are 4 circle R – 1, R – 2, R – 3, R – 4 are the 4 regions
Area of 1st circle  A1 
22
 7  7  154m 2
7
Area of 3rd circle = A3 
22
 9  9  254.57m 2
7
Area of 4th circle  A 4 
22
 12  12  452.16m 2
7
Prob. to hit R  1 
27.
R4
R 3
22
 4  4  50.28m 2
7
Area of 2nd circle  A 2 
O
20
R2
R 1
A1
50.28

 0.11
A 4 452.16
Prob. To hit R  1 
A 2  A1 103.72

 0.22
A4
452.16
Prob. To hit R  3 
A3  A 2 100.57

 0.22
A4
452.16
Prob. To hit R  4 
A 4  A3 197.59

 0.43
A4
452.16
ln ECD
tan 45 
A
12
x  12m
x
a
ln AEB
30
E
tan 30 
a
1
a
12

 a 
a 4 3
x
3 12
3
Height of cliff.  12  4 3  12  69  18.928m
B
45x
12m
12m
45
D
x
C
Page 43
28.
G  Centroid
X
Property  Co-ordinate of Centroid of triangle is
same when taken with three coordinates of triangle or
it is the co-ordinate of mid-point of sides

 2  3  6 1 5  4 
 11 
G
,
 G  ,0 

3
3 

3 
G
Y
29.
Given ABCD is a 11gm
Using property of 11gm that
Diagonal bisect each other
Thus using midpoint formula for diagonal AC
Volume of frustum =
Z
C(7, 4)
O
Now again using midpoint formula for diagonal BD
5 4a

 a 1
2
2
D(1,4)
P(2,1)
D(a, b)
 2  7 1 4 
5 5
O 
,
 O , 

2
2 

2 2
30.
Q(3, 5)
R(6, 4)
A(2,1)
B(4,1)
5 b 1

b4
2
2
1
h(r12  r1r2  r2 2 )
3
h = 21 cm r1  6cm r2  16 cm
1 22
Vol. of frustum    21(36  256  96)
3 7
 22  388 cm3 [1000cm 2  1 lit]
22  388  42
Rs.
1000
 358.512 Rs.
Cost of milk  Vol. of frustum  42 
31.
Area of 1 flower bed = Area of sector (OBPA) – Area of Isoc. Right triangle
In a square diagonal bisect each atleast 90
And it the side of square is a then length of diagonal of square is a 2
diagonal = 56 2m
Square side = 56 cm
56 2
 28 2m
2
Area of 1 flower bed = Area of sector – Area of isoc. Right triangle
Radius of circle flower bed =

90 22
1
  28 2  28 2   28 2  28 2
360 7
2
28 2  28 2  11 
 7  1
2



28  28  4
 28  16  448 m 2
7
 Area of two flower bed = 2  448m2
 896m2
Page 44
ANSWERS : PAPER – 5
1.
a n 1   d
7.
MQ  2.5 cm
  30
2.
8.
n  5,11
3.
1
One
4.
B(1, 8)
9.
Angle between radii = 120
13.
14.
height  10( 3  1)m
17.
x = 6, - 14
18.
454 tents
19.
22.
13, 8, 3
23.
x  1, 2
24.
11.
12.
16.
P  1
21.
4 1
x ,
3 8
27.
(i)
31.
height  5.15m
364
1
(ii)
365
365
28.
A  34 sq. unit
29.
No.  3
5.
15.
(i)
n  65th term
10.
Area  1366.35m2
2
3
6
(ii) (iii)
5
25
25
Volume  594m3
25.
6.
Construction
Area  7408 m 2
20.
Area  17.71 m 2
26.
height  180m
30.
Area  1372m2
Perimeter  192m
SOLUTIONS
1.
a n 1  a  (n  1  1)d
 a  (n  1)d
a n 1  a  (n  1)d  d
a n 1   d
2.
ln ABC
sin  
A
9
1
 sin  
18
2
  30
18m
9m
Q
B
C
3.
Let E1E2 E3 ..........En  are all elements events of an event E
P(E1 )  P(E 2 )  P(E3 ).......P(E n )  1
4.
Using mid-point formula
A
1 a
O
 a 1
2
O
5.
8b
 b  8
2
Acc. To given condition
(1,8)
O
(0,0)
B
(a, b)
B (1,  8)
Let the no. be x
x  2x  21  2x  x  21  0
2
2
2x 2  6x  7x  21  0
2x(x  3)  7(x  3)  0
(x  3)(2x  7)  0
x=3,
6.
7
Rejected
2
3, 15, 27, 39…..
a  3 d  12
a n  132  a 54
Page 45
a n  132  3  53 12
3  (n  1)12  132  3  53 12
12(n  1  53)  132
n  54  11
n  65

7.
65th term is the am.
PA = PB
[Length of tangent]
PA = 8cm PM = 5.5 cm
MA = MQ [Length of tangent from ext. point]

AM  PA  PM  8  5  5  2  5cm

N
P
MQ  2  5 cm
B
Q
5.5
M
A
8cm
8.
C1 & C2 are two circles and touch each other internally at P
P
Thus only one tangent can be drawn to circle
C1
C2
9.
A
Now As we know PBOA forms a cyclic quad.
Thus sum of opp. angle is 180 in cyclic Quad.

APB  60
then AOB  120
60
P
x
O
B
10.
Surface area of Remaing solid
= Surface area of cube 
 6  14  14 
Area of base circle CSA of

of cone
cone
22
22
77  77 5
7
7
 1176  154  154 5
 1366.35 m 2
7 5
14
7
11.
(1  a )b x  2abcx  (c  m )  0  (, )
2
2
2
2
2
    given
 
2abc 2
b
(1  a 2 )
  
c 2  m2
b2 (1  a 2 )
Page 46
2 
2abc
2
b (1  a 2 )
2 
c 2  m2
b2 (1  a 2 )
Substituting from (1)

ac
b(1  a 2 )

...(1)

a 2c2  c2  m2 1  a 2
a 2 c2
c 2  m2

b2 (1  a 2 )2 b2 (1  a 2 )

a 2c2  c2  a 2c2  m2  a 2m2
c2  m 2 (1  a 2 ) H.D.
12.
a  15 d  2
n
(30  (n  1)  2)
2
110  n (30  2n  2)
 55 
110  n (2n  32)
55  n 2  16n  n 2  16n  55  0
n 2  5n  11n  55  0
n(n  5)  11(n  5)  0
n  5,11
As it is dec. A.P. thus two times it will take a particular sum value
13.
ln POB it is Right
A
Angled Triangle
cos 60 
PB
PO
P
O
60
1 PB

 PO  2PB
2 PO
120
H.D.
B
14.
ln OAB
O
h
 ha
a
ln OAC
tan 45 
tan 30 
h
h
1
h


a  20
3 a  20
h  20  3h  h 
20
3 1

( 3  1)
3 1
C
h  10( 3  1)m
15.
45
30
 20m 
B
a
A
200 beds
80 Females
120 males
20%
24 Males
Malaria
96 Males
Dengue
40%
32 Females
Malaria
48 Female
Dengue
Page 47
(b)
80 2

200 5
Male suffering from Malaria
(c)
24
3

200 25
Female Patient suffering from Dengue
(a)
Female patient  Pr of 
Pr of 
Pr of 
16.
48
6

200 25
A(1, 5) B (P, 1) C (4, 11) are collinear
 Area of  from ABC are Zero
0
1
1(1  11)  P(11  5)  4(5  1) 
2
O  (10  6P  16)
6  6P  0
17.
P  1
A(x, 3) B(4,2) AB  5 5
AB2  (x  4) 2  (3  2) 2
125  x 2  16  8x  25
x 2  8x  41  125  0
x 2  8x  84  0
x 2  6x  14x  84  0
x(x  6)  14(x  6)  0
x  6,
18.
14
Total No. of tents erected will be
Area of Rectangular Field  n  Area of 1 Circle with radius (14 m)
12
22
1400  200  n   14  14
7
200m
2
1400  200
n
 n  454.54
22  2  14

454 tents can be ereded
19.
1400m
Volume of Tank =
Volume of Cylinder + 2  Vol. of hem.shpere
3m
22
2 22
 3  3  17  2    3  3  3
7
3 7
23m
22
22
 (3)2 17  4 
 9  21
7
7
17m
 66  9  594m3
3m
3m
20.
Area of shaded Region =
Area of
Area of

Rect. Semi  Circle
Page 48
6 4 
24 
21.
22 2  2

7
2
44 168  44

 17.71m 2
7
7
 4x  3 
 2x  1 
 2x  1   10  4x  3   3




(4x  3) 2  10(2x  1) 2  3(2x  1)(4x  3)
24x 2  64x  1  24x 2  6x  9
48x 2  58x  8  0
24x 2  29x  4  0
24x2  32x  3x  4  0
8x(3x  4)  1(3x  4)  0
(8x  1) (3x  4)  0
1 4
x ,
8 3
22.
a 4  a 8  24
a 6  a10  44
a  3d  a  7d  24
a  5d  a  9d  44
2a  10d  24
2a  14d  44
a  5d  12
a  7d  22
...(1)
...(2)
Solving (1) & (2)
a  5d  12
a  7d  22
2d  10
d5
Sub. In (1)
a  25  12  a  13
13,  8,  3
23.
5
4
 3
2x  3 x
5
4  3x

2x  3
x
5x  (4  3x) (2x  3)
5x  8x  12  6x 2  9x
6x 2  6x  12  0
x2  x  2  0
(x  1) (x  2)  0
x  1,  2
Page 49
24.
A
Length of Tangent from ext. point are equal
AB  13
BC  15
AC  18
a  c  13
...(1)
Solving (1) & (2)
a  b  15
...(2)
a  c  13
b  c  18
...(3)
a  b  15
c  b  2

AQ  AR  8cm
C
C
R
Q
b
a
c  b  18
2C = 16
c=8
a 
B
b
P
C
25.
26.
ln ABC
A
b
a
3 h
tan 2   
4 a
h
tan 2 
a
4h
3
h
tan 1 
192  a
5 (192  a)  12h
2
1
ln ABD

5
h

12 192  a
D
192m
C
a
B
960  5a  12h
4h
 12h
3
20
16b
960  12h  h  960 
3
3
h  180m
960  5 
27.
Total 365 days in a year
1
365
364
Prob. Of diff. birthday =
365
Prob. Of same birthday =
28.
ABCD is a square
D( 4, 4)
C( 1, 1)
A(1,7)
B(4, 2)
AB  9  25  34
BC  25  9  34
CD  9  25  34
AD  25  9  34
Thus all four sides are equal Now it can be from by or square.
Thus we find length of diagonal
AC  4  64  68
Page 50
BD  64  4  68
Now length of diagonal are equal thus it is a square.
Area of square = ( 34)2  34 sq. unit
29.
Field Area given to
Daughter 
P 4gm A
B
90 22
  49  49  4
360 7
 7546 sq.m
90
Q
An equal amount of area is given to son
 Total Area given to Son daughter = 15,092m2
D
C
150m
 Area given to Charity
Area of sq. field – Area given to son & daughter
150 150  15,092
22,500 15,092  7408 m2
30.
Area of Plot = Area of Rect. – Area of Semi Circle
D
C
22 14 14

7
2
 1680  308
 60  28 
28m
 1372 m2
Perimeter of Plot = AB + DC + AD + Parameter of Semicircle
A
 60 
B
22
 60  60  28  14
7
 192m
 Vol. of em. bankment
31.
Volume of Well = Vol. of outer cylinder – Vol. of Inner
cylinder
(7)2 10  (12)2  h  (7)2  h
h
49 10  (144  49)h
490
h
95

5m
h  5.15m
10m
7
Page 51
Page 52
Page 53