## ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 1

**Choose the correct answer from the given four options (1-2):**

Question 1.

If the sum of two integers is -21 and one of them is -10 then the other is

(a) -32

(b) 32

(c) -11

(d) 11

Solution:

Let the other integer be x

∴ x + (-10) = -21

x = -21 + 10

x = -11 (c)

Question 2.

The number of natural numbers between the smallest natural number and the greatest 2-digit number is

(a) 90

(b) 97

(c) 98

(d) 99

Solution:

Smallest natural number = 1

Greatest 2-digit number = 99

⇒ So the natural numbers between 1 and 99 are

2, 3, 4, ………….., 98

∴ Total Natural numbers = 98 – 1 = 97

Question 3.

Find the value of 25 × 37 × 8 × 6 by suitable arrangement.

Solution:

25 × 37 × 8 × 6

= 25 × 8 × 37 × 6

= 200 × 222

= 44400

Question 4.

Write four consecutive integers preceding -97.

Solution:

The four consecutive integers preceding

-97 is

= -97 – 1 = -98

= -98 – 1 = -99

= -99 – 1 = -100

= -100 – 1 = -101

Question 5.

Write the greatest and the smallest 4-digit numbers using four different digits with the condition that 5 occurs at ten’s place.

Solution:

Greatest 4-digit number when digit 5 always at tens place = 9857

Smallest 4-digit number when digit 5 is a tens place = 1052

Question 6.

Write all possible natural numbers formed by the digits 7, 0 and 3. Repetition of digits is not allowed.

Solution:

The given digits are 7, 0, 3 and repetition of digits is not allowed.

The one-digit numbers that can be formed are 7 and 3.

We are required to write 2-digit numbers.

Out of the given digits, the possible ways of choosing the two digits are

7, 0, 3, 0, 3, 7

Using the digits 7 and 0, the numbers are 70.

Similarily, Using the digits 3 and 0, the numbers are 30

Using the digits 3 and 7, the numbers are 37 and 73.

Hence, all possible 2-digit numbers are 30, 70, 37, 73

Now, We are required to write 3-digit numbers using the digits 7, 0, 3

and the repetition of the digits is not allowed. Keeping 0 at unit’s place,

The 3-digit number obtained are 370 and 730.

Keeping 3 at unit’s place, the 3-digit number obtained is 703.

Keeping 7 at unit’s place, the 3-digit number obtained is 307.

Hence, all possible 3-digit numbers are 370, 730, 703 and 307.

All possible numbers using the digits 7, 0 and 3 are

3, 7, 73, 37, 70, 30, 703, 307, 730, 370,

Question 7.

Find the value of: -237 – (-328) + (-205) – 76 + 89.

Solution:

-237 – (-328) + (-205) – 76 + 89

= -237 + 328 – 205 -76 + 89

= 91 – 281 + 89

= -190 + 89 = -101

Question 8.

Abhijeet’s school is 3 km 520 m away from his home. One day while returning from his school, just after covering 1 km 370 m distance, he saw a woman who was bleeding, he took her to the nearest hospital which was 2 km 775 m away from that place and got her admitted. He came back to his home which was 4 km 565 m from the hospital.

(i) Find the distance covered by Abhijeet on that day.

(ii) What value of life is depicted by Abhijeet?

Solution:

Distance of Abhijeet’s school from home = 3 km 520 m

One day, Abhijeet covered = 1 km 370 m and saw injured women,

took her to hospital.

A distance of hospital from that place = 2 km 775 m

Then, he came back home and travel distance = 4 km 565 m

∴ Total distance covered = 3 km 520 m

(ii) Social responsibility.

Question 9.

Arrange the following integers in descending order:

-353, 207, -289, 702, -335, 0, -77.

Solution:

Descending order:

-353, 207, -289, 702, -335, 0, -77

Question 10.

Find the smallest five-digit number which is exactly divisible by 254.

Solution:

Smallest 5-digit number = 10000

On dividing 10000 by 254, we get,

Remainder = 94

So, 254 – 94 = 160, should be added to 10000

to get the smallest 5-digit number divisible by 254

∴ Smallest 5-digit number divisible by 254

= 10000 + 160 = 10160