## ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 2

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

Question 1.

Every composite number has atleast

(a) 1 factor

(b) 2 factors

(c) 3 factors

(d) 4 factors

Solution:

3 factors (c)

Question 2.

Which of the following collections form a set?

(a) Collection of 5 odd prime numbers

(b) Collection of 3 most intelligent students of your class

(c) Collection of 4 vowels in English alphabet

(d) Collection of months of a year having less than 31 days.

Solution:

Collection of months of a year having less than 31 days. (d)

Question 3.

Find the prime factorisation of 1320.

Solution:

1320 = 2 × 2 × 2 × 3 × 5 × 11

Question 4.

If A = {x : x is a positive multiple of 3 less than 20} and B = {x : x is an odd prime number less than 20}, then find n(A) + n(B).

Solution:

A = {3, 6, 9, 12, 15, 18}

B = {3, 5, 7, 11, 13, 17, 19}

n(A) = 6

n(B) = 7

∴ n(A) + n(B) = 13

Question 5.

Reduce the fraction \(\frac{714}{1386}\) in its simplest form.

Solution:

Question 6.

Simplify the following:

\(2 \frac{3}{14}-3 \frac{5}{6}-\frac{2}{5}+2 \frac{1}{2}\)

Solution:

Question 7.

A number is divisible by 5 and 8 both. By what other numbers with that number be always divisible?

Solution:

1, 2, 4, 10, 20, 40

Question 8.

Arrange the fractions \(\frac{2}{3}, \frac{7}{9}, \frac{5}{8}, \frac{3}{5}\) in ascending order.

Solution:

Question 9.

Find the smallest number of 5-digits which is divisible by 12, 15 and 18.

Solution:

12, 15 and 18

∴ LCM of given numbers = 2 × 2 × 3 × 3 × 5 = 180

Smallest number of 5-digits = 10000

We divide 10000 by 180 and find the remainder.

According to the given condition,

we need the least number of 5-digits

which is exactly divisible by 12, 15 and 18.

The required number = 10000 + (180 – 100) = 10080

Question 10.

Three bells are ringing continuously at intervals of 30, 40 and 45 minutes respectively. At what time will they ring together if they ring simultaneously at 5 A.M.?

Solution:

LCM = 2 × 2× 2 × 3 × 3 × 5 = 360

After = 360 minutes

360 minutes = 6 hours

They will ring together again at 11 a.m.