Model Question Paper-3 Class-8 ML Aggarwal Solutions

Model Questions Paper-3 Class-8 ML Aggarwal ICSE Mathematics Solutions. APC Understanding Mathematics for ICSE Class-8 Model Question Paper-3 Solutions of Section-A , B, C and D based on Chapter-1 to 9. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.

Model Questions Paper-3 Class-8 ML Aggarwal ICSE Mathematics Solutions

( Based on chapter-1 to 9)

Time Allowed-2 and 1/2 hour

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-: Select Topics :-

Section-A

Section-B

Section-C

Section-D

 


Section-A , Paper-3 Class-8 ML Aggarwal

Questions 1 to 8 are of 1 mark each.
Question 1.
Product of rational number \frac{-4}{7} and its additive inverse is
(a) \frac{16}{49}
(b) \frac{-16}{49}
(c) 1
(d) 0

Answer

Product of rational number \frac{-4}{7} and its
additive inverse – \left(\frac{-4}{7}\right)=\frac{4}{7}
\frac{-4}{7} \times \frac{4}{7}=\frac{-16}{49} (b)

Question 2.
What should be subtracted from \frac{-3}{5} to get \frac{-2}{3}?
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 1
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 2

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Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 3

Question 3.
The value of \left(\frac{2}{3}\right)^{0}+\left(\frac{2}{3}\right)^{-1} is
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 4

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 5

Question 4.
Standard form of 0.00000000234 is
(a) 2.34 × 10-8
(b) 23.4 × 10-10
(c) 2.34 × 10-9
(d) 234 × 10-1

Answer

0.00000000234 = 2.34 × 10-9 (c)

Question 5.
The volume of a cube is 216 m3. Length of its side is
(a) 3 m
(b) 6 m
(c) 9 m
(d) 12 m

Answer

Volume of a cube = 216 m3
∴ Side – \sqrt[3]{216}=\sqrt[3]{6 \times 6 \times 6} = 6 m (b)

Question 6.
When the difference of a 2-digit number ab (a > b) and the number obtained by reversing the digits is divided by 9, then quotient is
(a) a + b
(b) b – a
(c) a – b
(d) none of these

Answer

The difference of the two-digit number ab (a > b)
and the number reversing the digits = \frac{a b-b a}{10} = a – b (c)

Question 7.
Amar buys a toy for ₹ 120 and sells it for ₹ 144, his profit % is
(a) 5%
(b) 10%
(c) 20%
(d) 24%

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Answer

C.P. of a toy = ₹120
S.P. = ₹144
∴ Profit = S.P. – C.P. = ₹144 – ₹120 = ₹24
and Profit % = \frac{24 \times 100}{\text { C.P. }}=\frac{24 \times 100}{120} = 20% (c)

Question 8.
If 567×8 is divisible by 9, then value of x is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

∵ 576×8 is divisible by 9
∴ Sum of digits 8 + x + 7 + 6 + 5 is also divisible by 9
⇒ 26 + x is divisible by 9
⇒ 26 + x = 27
⇒ x = 27 – 26 = 1
∴ x = 1


 Paper-3 Class-8 ML Aggarwal Section-B


Questions 9 to 14 are of 2 marks each.


Question 9.
Ramesh bought a fan for ₹ 1320 including 10% VAT. Find the price of the fan before VAT was added.

Answer

Value of a fan including VAT = ₹1320
VAT = 10%
∴ Price of the fan before adding VAT
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 6

Question 10.
Find A in the addition
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 7

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 8
A + 5 = 2 or 12
∴ A = 12 – 5 = 7

Question 11.
Find the value of x for which \left(-\frac{4}{5}\right)^{2} \times \left(-\frac{5}{4}\right)^{2 x}=\frac{625}{256}.

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 9
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 10

Question 12.
Using appropriate properties of operations of rational numbers, evaluate:
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 11

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 12

Question 13.
What number should be subtracted from the sum of \frac{-5}{12} and \frac{2}{3} to get \frac{5}{4} ?

Answer

ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 13
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 14

Question 14.
If n(ξ) = 40, n(A) = 23, n(B’) = 28 and n(A ∪ B’) = 9, find n(A ∩ B).

Answer

n(ξ) = 40, n(A) = 23, n(B’) = 28
n(A ∪ B’) = 9
n(B) – n(ξ) = -n(B’)
= 40 – 28 = 12
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 15
n(A ∪ B) = n(ξ) – n(A ∪ B)’ = 40 – 9 = 31
∴ n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 23 + 12 – 31 = 35 – 31 = 4


Section-C, Model Question Paper-3 Class-8 ML Aggarwal Solutions


Questions 15 to 24 are of 4 marks each.
Question 15.
In an auditorium, the number of rows is equal to the number of chairs in each row. If there are 5184 chairs in the auditorium, find the number of rows.

Answer

In an auditorium number of rows = number of chairs in a row
Total chairs = 5184
Number of rows = \sqrt{5184} = 72
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 16

Question 16.
Insert 5 rational numbers between \frac{1}{3} and \frac{1}{2}.

Answer

5 rational numbers between \frac{1}{3} and \frac{1}{2}
LCM of 3, 2 = 6
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 17

Question 17.
Find the smallest number by which 10584 should be multiplied so that product is a perfect cube. Also find the cube root of the product.

Answer

10584
Factors of 10584
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 18
= 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7
Grouping the same factors in triplet, 7 × 7 are left unpaired.
∴ In order to complete a perfect cube, we shall multiply by 7.
So product = 10584 × 7 = 74088 and cube root = 2 × 3 × 7 = 42

Question 18.
In a 3-digit number unit’s digit is 2 less than hundred’s digit and hundred’s digit is 3 less than the ten’s digit. If the sum of the original number and number obtained by reversing the digits is 968, find the number.

Answer

In a 3-digit number,
Let tens digit = x
Then hundreds, digit = x – 3
and unit digit = x – 3 – 2 = x – 5
∴ Number = (x – 5) + 10(x) + 100(x – 3)
= x – 5 + 10x + 100x – 300
= 111x – 305
By reversing the digits
Unit digit = x – 3
Ten’s digit = x
and hundred’s digit = x – 5
∴ Number = x – 3 + 10x + 100x – 500 = 111x – 503
According to the condition,
111x – 305 + 111x – 503 = 968
⇒ 222x – 808 = 968
222x = 968 + 808 = 1776
x = \frac{1776}{222} = 8
and hundred’s digit = x – 5
∴ Number = 111 × 8 – 305 = 888 – 305 = 583

Question 19.
Rohit purchased some oranges at 3 for ₹5 and sold them at 5 for ₹12. Thus he gained ₹143 in all. How many oranges did he purchase?

Answer

LCM of 3, 5 = 15
Let Rohit purchased = 15 oranges
∴ C.R of 15 oranges = \frac{15}{3} × 5 = ₹25
and S.R of 15 oranges = \frac{15}{5} × 12 = ₹36
Gain = S.P. – C.P. = ₹36 – ₹25 = ₹11
If gain is ₹11, then number apples = 15
and if gain is ₹143, then number of apples = \frac{15}{11} × 143 = 195

Question 20.
How much per cent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 20% he still earns a profit of 20%?

Answer

Let cost price = ₹100
Profit = 20%
ML Aggarwal Class 8 Solutions for ICSE Maths Model Question Paper 3 19
∴ He should marked price ₹(150 – 100) = 50% above the cost price

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