# Model Question Paper-3 ML Aggarwal Class-7 ICSE Maths

Model Question Paper-3 Solved Class-7 ML Aggarwal ICSE Mathematics . APC Understanding Mathematics for ICSE Class-7 Model Question Paper-3 Solutions based on Chapter-1 to 9. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

## Model Question Paper-3 ML Aggarwal Class-7 ICSE Maths

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### ML Aggarwal Class-7 ICSE Maths** Model Question Paper 3**

( based on chapter-1 to 9)

Time Allowed-1 hour

max mark-25

**General Instructions**

- All questions are compulsory.
- The question paper consists of 29 questions divided into four sections A, B, C and D.
- Section A comprises of 8 questions of 1 mark each.
- Section B comprises of 6 questions of 2 marks each.
- Section C comprises of 10 questions of 4 marks each and
- Section D comprises of 5 questions of 6 marks each.
- Question numbers 1 to 8 in Section A is multiple choice questions where you are to select one correct option out of the given four.

**Section – A**

Choose the correct answer from the given four option (1-2)

Question 1

Question 1.

The number of integers between -16 and 5 is

(a) 19

(b) 20

(c) 21

(d) 22

Answer

Number of integer between -16 and 5 is -15 to 4 = 20 (b)

Question 2.

50 m 5 cm is the same as

(a) 50.5 m

(b) 50.05 m

(c) 50.005 m

(d) 5.05 m

Answer

50 m 5 cm = 50.05 m (b)

Question 3

……………..

Answer

Question 4

The number 5,540,000,000,000 in the scientific notation can be written as:

(a) 554 × 10^{10}

(b) 55.4 × 10^{11}

(c) 5.54 × 10^{12}

(d) 5.54 × 10^{11}

Answer

5,540,000,000,000 = 5.54 × 10^{12} (c)

Question 5.

The number of unlike terms in the expression 5x^{2}y – 2xy^{2} – 2yx^{2} + 3y(xy + y^{2}) + 7 is

(a) 3

(b) 4

(c) 5

(d) 6

Answer

Number of unlike terms

5x^{2}y – 2xy^{2} – 2yx^{2} + 3y(xy + y^{2}) + 7

= 5x^{2}y – 2xy^{2} – 2yx^{2} + 3xy^{2} + 3y^{3} + 7

= 3x^{2}y + xy^{2} + 3y^{3} + 7 = 4 (b)

Question 6

x = -2 is a solution of the equation

(a) 2x + 5 = 9

(b) 3x – 1 = 5

(c) 4x + 3 = 1

(d) 5x + 12 = 2

Answer

5x + 12 = 5 × (-2) + 12 = -10 + 12 = 2 (d)

Question 7

The ratio of the number of girls to the number of boys in a class is 5 : 4. If there are 16 boys in the class, then the number of students in the class is

(a) 20

(b) 32

(c) 36

(d) 45

Answer

**Section – B**

**Question numbers 9 to 14 are of 2 marks each.**

Question 9.

Using suitable properties, evaluate:

238 × (-44) + (-238) × 56.

Answer

238 × (-44) + (-238) × 56

= 238 × (-44) – (238) × 56

= 238 (-44 – 56)

= 238 × (-100)

= -23800

Question 10

State whether each of the following statement is true or false for the sets P and Q where P = {letters of TITLE} and Q = {letters of LITTLE}

(i) P ↔ Q

(ii) P = Q

Solution:

P = {letters of TITLE} = {T, I, L, E}

Q = {letters of LITTLE} = {L, I, T, E}

(i) P ↔ Q – True

(ii) P = Q – True

Question 11

Evaluate: -3 – (-2)

Answer

Question 14

If ₹ 4000 amounts to ₹ 5000 in 2 years, find the rate of simple interest per annum.

Solution:

Principal (P) = ₹ 4000

Amount (A) = ₹ 5000

S.I. = A – P = ₹ 5000 – ₹ 4000 = ₹ 1000

Time (T) = 2 years

**Section – C**

**Question numbers 15 to 24 are of 4 marks each.**

Question 15.

Simplify:

……………….

Answer

Question 16

Vikram’s monthly salary is ₹ 12750. He spends of his salary on food and out of the remaining, he spends on rent and on the education of children. Find

(i) how much he spends on each item?

(ii) how much money is still left with him?

Answer

Monthly salary of Vikram = ₹ 12750

Spent on food = of 12750 = ₹ 2550

Remaining salary = ₹ 12750 – ₹ 2550 = ₹ 10200

Spent on rent = of ₹ 10200 = ₹ 2550

Spent on Education = × ₹ 10200 = ₹ 1700

Amount still left with him = ₹ 10200 – ₹ (2550 + 1700)

= ₹ 10200 – ₹ 4250

= ₹ 5950

Question 17

Insert five rational numbers between and

Answer

Question 18

Afzal can walks 5 km in one hour. How much distance will he cover in 2 hours 40 minutes? What are the health advantages of having a brisk walk?

Answer

In one hour, Afzal can walk = 5 =

** **

Question 19

If a vehicle covers a distance of 57.72 km in 3.7 litres of petrol. How much distance will it cover in one litre of petrol?

Solution:

In 3.7 litres of petrol a vehicle can travel = 57.72 km

It will cover in 1 litre =

= 15.6 km

Answer

Question 22

Solve the equation: 3(2x – 1) – 2(2 – 5x) = 1

Answer

3(2x – 1) – 2(2 – 5x) = 1

⇒ 6x – 3 – 4 + 10x = 1

⇒ 16x – 7 = 1

⇒ 16x = 1 + 7 = 8

⇒ x = =

Question 23

If 74% of the population of a village is illiterate and the number of literate people is 2158, then find the population of the village.

Answer

** Section -D**

**Question numbers 25 to 29 are of 6 marks each.**

Question 25.

If we represent the distance above the ground by a positive rational number and that below the ground by a negative rational number, then answer the following question:

An elevator descends into a mine shaft at the rate of 4 metre per minute. If it begins to descend from 7 metre above the ground, what will be its position after 18 minutes from the ground?

Answer

Speed of elevator into the mine = 4 = m/min

Time taken = 18 m

Total distance covered = × 18 = m = 85.5 m

Distance above the ground = 7.5 m

Distance below the ground = 85.7 – 7.5 = 78 m

It is below the ground.

It will be = -78 m

#### Question 26

In a competition, the question paper consists of 25 questions. 4 marks are awarded for every correct answer, 2 marks are deducted for every incorrect answer and no marks for not attempting a question. If Vaishali scored 58 marks and got 17 correct answers, how many questions she attempted incorrectly? How many questions she did not attempt?

Answer

Total number of questioins = 25

For correct answer = 4 marks per question

Deduction of 2 marks for incorrect answer

No marks for unattemtped question

Vaishali got 58 marks and got 17 correct questions

Let x be the questions of wrong answer

17 × 4 – 2x + (25 – 17 – x) × 0 = 58

68 – 58 = 2x + 0

2x = 10

x = 5

Number of wrong answer questions = 5

and number of questions not attempted = 25 – 17 – 5 = 3

#### Question 27.

Divide ₹ 216000 into two parts such that one-fourth of one part is equal to one-fifth of the other part. Find the two parts.

Answer

Amount = ₹ 216000

Let of one part = of second part = x

First part = 4x

Second part = 5x

4x + 5x = 216000

⇒ 9x = 216000

⇒ x = = 24000

First part = ₹ 24000 × 4 = ₹ 96000

Second part = ₹ 24000 × 5 = ₹ 120000

Question 28.

If a table is sold for ₹ 437 at a loss of 8%, find its cost price. At what price must it be sold to gain 10%?

Answer

S.P. of table = ₹ 437

Loss = 8%

Question 29.

Solve the inequality:

3 – 2x ≥ x – 10, x ∈ N.

Also, represent its solution set on the number line.

Answer

3 – 2x ≥ x – 10, x e N

⇒ -2x – x ≥ -10 – 3

⇒ -3x ≥ -13

⇒ 3x ≤ 13

⇒ x ≤

⇒ x ≤ 4

x = {4, 3, 2, 1}

– : End of** Model Question Paper-3 Solved Class-7 ML Aggarwal **Solutions :–

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