Class 8th Sample Paper Solution

31 Pages·2016·1.56 MB·English

Pioneer’s

Aspire Scholarship Exam
Solution
8th CBSE
Examination Centre: Pioneer Education, Sector – 40-D

General Instructions:
The question paper contains 90 multiple choice questions.
There are two sections in the question paper
Section: A– MATHEMATICS (1 to 45)
Section: B– SCIENCE (46 to 90)
Each right answer carries 4 marks and − 1 for every wrong answer.
The Maximum marks are 360 and Maximum Time 2.00 hrs.
Give your response in the OMR Sheet provided with the Question Paper.
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Section – A {Mathematics}
2/3 4/3
1 1
1. Simplify :
216 27
3 9 4 4
(a) (b) (c) (d)
4 4 9 3
Ans. (c)
Solution:
2/3 4/3
1 1
216 27

2/3
1 1
27 4/3 a m
63 am

1 2 1 62 n
am amn
6 33 4/3 34
6 6 4

3 3 3 3 9
2. Find the value of x, if ( 6)x 2 1.
(a) x = 3 (b) x = 4 (c) x = 2 (d) x = 1
Ans. (c)
Solution:
x 2
( 6)x 2 1 6 1/2 60 [ 60 1]
1
6 2 x 2 60 [Using (am)n = amn]
1
x 2 0 [Using am = an]
2
x 2 0 x 2
Hence, the value of x = 2
3. Find x, if 5x + 5x–1 = 750.
(a) x = 4 (b) x = 3 (c) x = 1 (d) x = 2
Ans. (a)
Solution:
5x 5x 1 750
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5x 5x.5 1 750
1
5x 5x 750
5
5 5x 5x 750 5
5x 5 1 750 5
750 5 53 5 6
5x 5x
6 6
5x 54 x = 4.
4
1
4. Express the answer with positive indices : 3 x4y
3 xy7
y8 x2 x4 y8
(a) (b) (c) (d)
x2 y4 y8 x4
Ans. (d)
Solution :
4
1
3 x4y
3 xy7

4
1
1/3 1
x4y ma am
1/3
xy7
4
x4/3 y1/3 n
= am amn
x1/3 y7/3
4
4 1 1 7
= x3 3 y3 3
4 4
= x1 y 6/3 xy 2
= x 4 y8
y8
=
x4

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5. The difference between compound interest and simple interest at 5% per annum in 2 years is Rs. 30.
Find the sum.
(a) 14, 000 (b) 12, 000 (c) 13000 (d) 11000
Ans. (b)
Solution:
Given R = 5%, T = 2 years, C.I. – S.I. = Rs. 30, P = ?
P R T P 5 2 P
S.I.
100 100 10
2 2
5 21 441P
and A P 1 P
100 20 400
441P
C.I. A P P
400
A.T.Q.,
C.I. – S. I. = Rs. 30
41P P 41P 40P
30 30
400 10 400
P
30 P 400 30 Rs.12,000
400
6. At what rate per cent per annum, compound interest will Rs. 10, 000 amount to Rs. 13, 310 in three
years?
(a) 15% (b) 20% (c) 10% (d) 5%
Ans. (c)
Solution:
Let the rate b R% per annum.
We have, P = Rs. 10, 000, A = Rs 13, 310, and n = 3 years.
3
R
From formula, A = P 1
100
3
R
We get, 13, 310 = 10, 000 1
100
3 3 3
13,310 R 11 R
1 1
10,000 100 10 100
R 11 R 1
1
100 10 100 10
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100
R 10
10
Hence, rate = 10% per annum.
7. By selling 288 hens, Malleshwari lost the S.P. of 12 hens. Find her loss percent.
(a) 2 % (b) 4% (c) 6% (d) 8%
Ans. (b)
Solution:
Let S.P. of 1 hen = Rs. 1
S.P. of 288 hens = Rs. 288 1 = Rs. 288
and Loss = S.P. of 12 hens
= Rs. 12 1 = Rs. 12
C.P. of 288 hens = S.P. + Loss
= Rs. 288 + Rs. 12 = Rs. 300
Loss 12
Therefore, loss% = 100 100 4
C.P. 300
Thus, Malleshwari’s loss is 4%.
8. A table with marked price Rs. 2400 was sold to a customer for Rs. 2200. Find the rate of discount
allowed on the table.
1 1 1 1
(a) 8 %. (b)10 %. (c) 6 %. (d) 4 %.
3 3 3 3
Ans. (a)
Solution:
M.P. = Rs. 2400, S.P. = Rs. 2200
Discount = Rs. 2400 – Rs. 2200 = Rs. 200
discount
Rate of discount = 100%
M.P.
200 1
= 100% 8 .
2400 3
9. Divya bought a set of cosmetic items for Rs. 345 including 15% sales tax and a purse for Rs. 110
including 10% sales tax. What per cent is the sales tax charged on the whole transaction?
(a) 15.7% (b) 17.5% (c) 13.7% (d) 13.9%
Ans. (c)
Solution:
Let the LP of cosmetic items = Rs. x
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Given, Rate of sales Tax = 15%
15
Given, Sales price = 345 = x + x
100
115x
or 345
100
34500
or x Rs.300
115
Let the list price of purse be y.
and Rate of Sales Tax = 10%
10
Given, Sales price = 110 = y + y
100
110
or 110 = y
100
y = Rs. 100
Total sales price of cosmetics and Purse
= 345 + 110 = Rs. 455
and = 300 + 100 = Rs. 400
Hence, Discount = Total sales price – total list price
= Rs. 455 – Rs. 400 = Rs. 55,
i.e., Divya pays Rs. 55 towards the sales tax on purchase of above items.
55
Percentage of sales tax = 100 13.7%
400
10. Price of a DVD player including 12% VAT is 4480. Find the price of DVD player before VAT is added.
(a) 4500 (b) 6000 (c) 5000 (d) 4000
Ans. (d)
Solution :
Let price of DVD player before VAT is added = Rs. x
12
Value added tax (VAT) = x
100
12
Price including VAT = x + x
100
12
x x 4480
100
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112x
4480 x 4000
100
Price before VAT in added = Rs. 4000
11. Find the area of rectangle if its length is (3x + 5) and breadth is (2y + 4).
(a) 6xy + 12x + 10y + 20 (b) 6xy – 12x + 10y + 20
(c) 6xy + 12x + 8y + 20 (d) 6xy + 12x + 8y + 20
Ans. (a)
Solution:
Given Length = 3x + 5 and breadth = 2y + 4
Area = length breadth
Area = (3x + 5) (2y + 4)
= 3x (2y + 4) + 5(2y + 4)
= 6xy + 12x + 10y + 20
1 1 2
12. Find the product : a 2b a2 ab 4b2 .
3 9 3
1 1 1 1
(a) 8b3 a3 (b) a3 8b3 (c) a2 8b2 (d) a3 8b3
27 27 27 27
Ans. (d)
Solution:
1 1 2
a 2b a2 ab 4b2
3 9 3

2
1 1 1
2
= a 2b a a 2b 2b
3 3 3
3
1 1
= a 2b 3 a3 8b3
3 27
13. Simplify : (4m + 5n)2 + (5m + 4n)2
(a) 41m2 41n2 (b) 41m2 41n2 80mn (c) 16m2 25n2 40mn (d) None of these
Ans. (b)
Solution:
(4m + 5n)2 + (5m + 4n)2
= 16m2 + 25n2 + 40mn + 25m2 + 16n2 + 40mn
= 41m2 + 41n2 + 80mn.
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14. Add : 3(a + b +c) and (a + b) (a – b – c).
(a) a2 + b2 + ac + bc + 3a + 3b + 3c (b) a2 – b2 – ac – bc + 3a + 3b + 3c
(c) a2 – b2 + bc + ac + 3a + 3b + 3c (d) a2 – b2 – ac – bc – 3a – 3b – 3c
Ans. (b)
Solution:
3(a + b +c) and (a + b) (a – b – c)
= 3a + 3b + 3c + a2 – ab – ac + ab – bc – b2 – c2
= 3a + 3b + 3c + a2 – b2 – c2 – ac – bc
15. By what number (–9)–1 should be multiplied so that the product may be equal to (–12)–1?
3 3 4 4
(a) (b) (c) (d)
4 4 3 3
Ans. (a)
Solution:
Product = (–12)–1
Given number = (–9)–1
The required number = [Product] (Given number)
1 1
= 12 9
1
1 1 12

12 9 1
9
1 9 1 9 3
= .
12 1 12 1 4
3
Thus, (–9)–1 should be multiplied by to get (–12)–1 as the product.
4
40
5
16. Simplify the following: [5 1 5 19]2 .
2
40 20
1 1
(a) 240 (b) (c) (d) 220
2 2
Ans. (b)
Solution:
40 40
5 2 5
[5 1 19 ]2 5 20
2 2
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40 40
1 5
= 5 40
40
2
40
1
= 5 40 540
40
2
1
= 5 40 40 [ 1 40 = 1 as 40 is an even number]
240
40 40
1 1 1
= 50 1 .
240 2 2
17. The number 0.000675 can be written as
(a) 675 10–6 (b) 67.5 10–5 (c) 6.75 10–4 (d) All of these
Ans. (d)
Solution:
0.000675
675 675
= 675 10 6 or 67.5 10 5 or 6.75 10 4
1000000 106
18. On a scale map 0.6 cm represents 6.6 km. If the distance between two points on the map is 80.5 cm,
what is the actual distance between these points?
(a) 885.5 km. (b) 885.7 km (c) 887.7 km (d) 887.5 km
Ans. (a)
Solution:
Distance on map 0.6 80.5
Actual distance 6.6 x
Values are directly proportional : Using
a a
1 2 we get
b b
1 2
0.6 80.5

6.6 x
0.6x 80.5 6.6
x = 885.5 km.

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19. 120 men had food provision for 200 days. After 5 days, 30 men died due to an epidemic.
How long will the remaining food last?
(a) 240 days (b) 250 days (c) 260 days (d) 230 days
Ans. (c)
Solution:
Since 30 men die after 5 days. Therefore, the remaining food is sufficient for 120 men for 195 days.
Suppose the remaining food lasts for x days for the remaining 90 men.
Thus, we have the following table.
Number of men 120 90
Number of days 195 x
We note that more men will consume the food in less number of days and less number of men will
consume the food in more number of days. So, it is a case of inverse variation.
Ratio of number of men = Inverse ratio of number of days
120 : 90 = x : 195
120 x

90 195
120 195
x 260
90
Hence, the remaining men will consume the food in 260 days.
20. The price of sugar goes up by 20%. By how much percent must a house wife reduce her consumption so
that the expenditure does not increase?
2 2 2 2
(a) 16 % (b) 16 % (c) 16 % (d) 16 %
5 3 7 9
Ans. (b)
Solution :
Let the consumption of sugar originally be 100 kg and its price be Rs 100.
Then,
New price of 100 kg sugar = Rs 120 [ Price increases by 20%]
Now, Rs 120 can fetch 100 kg sugar
100 250
Rs 100 can fetch = 100 kg sugar = kg sugar
120 3
250 50 2
Reduction in consumption = 100 % % 16 %
3 3 3
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