ITL Public School
Summative Assessment - 2 (2014-15)
Mathematics (Set -B)
Date:
Time: 3 hours
Class:VIII
M.M : 90
General Instructions:
1. Read the question paper carefully and answer legibly.
2. All questions are compulsory.
3. The question paper consists of 31 questions divided into four sections A, B, C and D.
4. Section A comprises of 4 questions of 1 mark each, Section B comprises of 6 questions of 2
marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 11
questions of 4 marks each.
5. Use of calculators is not permitted.
Q.2
Section A
Find the area of a rectangle with sides (in cm) are 2x and 5xy
Area= Length X Breadth =2x X 5xy =10x2y
Express 729 - 3 as a power with the base 9.
1
1
729 – 3=
 33 =9-9
3
729
9
Q.3
Identify the terms and their coefficients for the expression
Q.1
1
1
ab – mn.
12
ab,mn
4
Coefficients= 3,-1
Find the ratio of 200 l to 40 kl
200l 5
Ratio=

40kl k
5:k
1
Terms are =
Q.4
1
Section B
-1
-1
-2
Q.5
Find the value of (3
6 )
1
1
1 1 1
    2  9 
18
2
3 6 3
Q.6
If the weight of 18 sheets of thick paper is 60 g, how many sheets of paper would weigh 3 kg.
Weight of 18 sheets = 60 g
1 kg=1000g
3.5 kg =3.5 × 1000g =3500g
18
Number of sheets in 3500g =  3500  1050
60
Q.7
Q.8
3
Express the numbers appearing in the following statements in standard form:
(a) Size of a plant cell is 0.00001275 m.
(b) Average radius of the Sun is 695000 m.
0.00001275= 1.275 × 10-5
695000=6.95 × 105
Convert the following ratio to percentage – 2: 3: 5
2 + 3 + 5 = 10
2/10 x 100 % = 20 %
3/10 x 100 % = 30 %
5/10 x 100 % = 50 %
2
1+1
2
0.5
1
0.5
2
1
1
2
1
0.5
0.5
Find the sum of ab – bc, bc – ca, ca – ab
Sum= ab-bc+bc-ca+ca-ab = 0
Q.10 The area of a trapezium shaped field is 960 m2. The distance between two parallel sides is 15 m
and one of the parallel side is 40 m. Find the length of the other parallel side
( a  b)
Area of trapezium=
h
2
(a  20)
960=
 15
2
960 × 2 = (a+20)×15
1920
 a  20
15
128  20  a
a  108
Hence the other parallel side is of 108 m.
Section C
Q.11 A road roller takes 900 complete revolutions to move once over to level a road. Find the area of
the road if the diameter of the road roller is 98 cm and its length is 1 m
In one revolution, the roller will cover an area equal to its lateral surface area.
Thus, in 1 revolution, area of the road covered = 2πrh
=2×22/7×49×1= 3.08 m2
In 900 revolutions, area of the road covered
=900× 3.08
= 2772 m2
Q.9
Q.12
2
2
1
0.5
0.5
3
0.5
0.5
0.5
0.5
0.5
0.5
3
If Rahul had Rs. 12000 left after spending 75 % of his money, how much did he have in the
beginning?
Let amount of money Rahul had in the beginning be x
75% of x =12000
75
 x  12000
100
120000
x=
 16000
75
Hence statement.
0.5
0.5
0.5
0.5
0.5
Q.13 Draw the line passing through (3, 6) and (6, 3). Find the coordinates of the points at which this
line meets the x – axis and y – axis.
Plot (4,5) and (5,4) on the graph paper.
Write down all coordinates of the points.
4
Q.14 Factorize: 81 – x
0.5
3
1.5
1.5
3
= 92-(x2)2
= (9+x2)(9-x2)
=(9+x2)(3-x)(3+x)
1.5
1.5
8B 1 of 3
Q.15 Find the factors of 4a2 + 12a + 8.
=4(a2+3a+2)
=4(a+1)(a+2)
Factors are -1 and -2
Q.16 Using algebraic identities, evaluate:
1.05 9.5= (1+0.05)(1+8.5)
(x+a)(x+b) = x2+(a+b)x +ab
1+(0.05+8.5)1+0.05×8.5=9.975
Q.17
3
1
1
0.5
0.5
1.05
Find the value of x for which ( )
5
 
6
4 x 3 4
5
 
6
4 x 1
5
5
 
 
6
6
4x+1=5
4x=4
X= 1
3
1.5
1
0.5
9.5
( )
=( )
1
5
1
5
when the bases are same, we equate the powers
1
Q.18 Ankit took a loan of Rs. 90000 from a bank. If the rate of interest is 10% per annum, find the
amount he would be paying after 1 year and 6 months if the interest is compounded half yearly.
P=Rs 90000, R= 10% p.a T= 1.5 yrs
CI=P 1  R 

3
200 
2n
3
=90000 1  5   90000(1.05) 3  104186.25

3
2
100 
Amount=CI-P= Rs14186.25
1
Q.19 Show that: (2ab + 3c)2 – 24abc = (2ab – 3c)2 .
LHS=(2ab + 3c)2 – 24abc = 4a2b2 + 9c2 + 12abc – 24abc
= 4a2b2 + 9c2 - 12abc
RHS= (2ab – 3c)2 = 4a2b2+9c2 – 12abc
LHS =RHS, hence verified.
Q.20 A rectangular piece of paper 11 cm × 4 cm folded without overlapping to make a cylinder of
height 4 cm. Find the volume and LSA of the cylinder.
Height of the cylinder = h = 4 cm
Perimeter of the base of the cylinder = 2πr = 2×22/7×r
44r/7 = 11
. . r = 1.75
Therefore, r =1.75 cm
Volume of the cylinder = V = πr2h
=22/ 7 ×1.75×1.75×4
= 38.5 cm .
Hence the volume of the cylinder is 38.5 cm .
LSA of cylinder=2πr(r + h)
=2π×1.75(1.75+4)
=63.25 cm
3
1
0.5
1
0.5
3
0.5
0.5
0.5
3
3
0.5
0.5
0.5
Section D
2
2
2
Q.21 Factorize: 25a – 4b + 28bc – 49c .
25a2 − 4b2 + 28bc − 49c2 = 25a2 − (4b2 − 28bc + 49c2)
= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]
= (5a)2 − [(2b − 7c)2]
[Using identity (a − b)2 = a2 − 2ab + b2]
= [5a + (2b − 7c)] [5a − (2b − 7c)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (5a + 2b − 7c) (5a − 2b + 7c)
4
1
1
1
1
Q.22 Draw the graph for the following and check whether it is a linear graph.
Side of square
2.5
3
3.5
4.5
(in cm)
Perimeter (in
10
12
14
18
cm)
Drawing the correct x and y axis.
Marking the correct values on the axis.
Checking whether it is linear or not.
Q.23 Plot the points P (2,1) Q(5,1) R(5,4) and S(2,4) on the graph. Join them in order and name the
figure obtained.
Make x and y axis on the graph properly.
Mark all the points correctly.
Join them.
Name the figure.
4
1
2
1
4
1
1
1
1
Q.24 Subtract the sum of x (x + 2y + 3z) and – y (x – 2y + z) from z (– 3x – y + z).
=z (–3x – y + z)-[ x (x + 2y + 3z) – y (x –2 y + z)]
=-3zx-zy+z2-[x2+2xy+3xz-xy+2y2-yz]
= -6zx+z2-x2-y2-xy
4
2
1
1
Q.25
4
Simplify : (a) (4x2 + 9) (4x2 – 9)
(b) (
)
2
2
4
2
2
(a) (4x + 9) (4x – 9) = 16x – 36x +36 x – 81 =16x4-81
2
(
)
2
 5 p   3q 
      2  pq
 3   5 
25 p 2 9q 2


 2 pq
9
25
2+2
8A 2 of 3
Q.26
Factorize the expressions and divide them as directed:
39 y 3  2(25 y 2  49)
39 y[(5 y ) 2  7 2 )]
39y3(50y2 – 98) 26y2(5y + 7) =

26(5 y  7)
26 y 2 (5 y  7)
39 y (5 y  7)(5 y  7)

26(5 y  7)
39 y (5 y  7)
=
26
4
Q.27 A milkman sold two of his buffaloes for Rs.40, 000 each. On one he made a gain of 5 % and on
4
2
1
1
the other a loss of 10 %. Find his overall gain or loss.
S.P. of each buffalo = Rs 40,000
The milkman made a gain of 5% while selling one buffalo.
This means if C.P. is Rs 100, then S.P. is Rs 105.
105
Rs40000
100 = Rs 42000
C.P. of one buffalo =
Also, the second buffalo was sold at a loss of 10%.
This means if C.P. is Rs 100, then S.P. is Rs 90.
100
40000
90 = Rs 44444.44
∴C.P. of other buffalo =
0.5
1
0.5
1
Total C.P. = Rs 42000 + Rs 44444.44 = Rs 86444.44
Total S.P. = Rs 40000 + Rs 40000 = Rs 80000
Loss = Rs 44444.44 − Rs 40000 = Rs 4444.44
Thus, the overall loss of milkman was Rs 4444.44.
0.5
0.5
Q.28 Water is pouring into cuboidal reservoir at the rate of 60 l per minute. If the volume of the
reservoir is 108 m3. Find the number of hours it will take to fill the reservoir.
Volume of cuboidal reservoir = 108 m3 = (108 × 1000) L = 108000 L
It is given that water is being poured at the rate of 60 L per minute.
That is, (60 × 60) L = 3600 L per hour
4
1
2
Required number of hours
= 30 hours
Thus, it will take 30 hours to fill the reservoir.
1
Q.29 Simplify using laws of exponents:
4
,x
9 x
3
=
5 
3 3
10
10  6
 2  10 2  x 6
4
9 x
9 x
 9
9
2
5  2  10
5  200
3
=
0
3
Q.30 The population of a place increased to 1,25,000 in 2006 at a rate of 5% per annum.
(a) Find the population in 2004.
(b) What would be its population in 2008?
(i) It is given that, population in the year 2006 = 1,25,000
2+1
+1
4
1
Therefore,
05
1,25,000 = (Population in 2004)
Population in 2004 1,25,000 
20 20

 113378.685
21 21
Thus, the population in the year 2001 was approximately 113378
20 20
(ii) Population in 2008 = 1,25,000 

 113378
21 21
0.5
1
2
1 
21 21

1,25,0001    1,25,000  
 1,37812.5
20 20
 20 
Thus, the population in the year 2005 would be 24438.
1
Q.31 There are four solid gold cubes of side 1 cm, 2 cm, 3 cm, 4 cm. it was divided among two
brothers Rahul and Hemant such that Rahul got 3 cubes of side 1cm, 2cm and 3 cm and
Hemant got one cube of 4 cm. Is the distribution equal? Comment on your answer by justifying
the answer.
Volume of cube with side 1 cm= side 3= 1 cm3
Volume of cube with side 2 cm= side 3= 8 cm3
Volume of cube with side 3 cm= side 3= 27 cm3
Volume of cube with side 4 cm= side 3= 64 cm3
→Volumes of cubes with side 1,2 and 3cm < Volume of cube with side 4cm
1+8+27<64
36<64
Therefore the distribution is not equal.
4
2
1
1
8B 3 of 3