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

Model Question Paper-2 Class-6 ML Aggarwal ICSE Maths Solutions. APC Understanding Mathematics for ICSE Class-6 Model Question Paper-2 Solutions based on Chapter-4 to 6. Visit official Website for detail information about ICSE Board Class-6 Mathematics.

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

### Model Question Paper 2

( based on chapter-4 to 6)

Time Allowed-1 hour

max mark-25

Note :

(i) All Questions are compulsory

(ii) Questions 1-2 carry 1 mark each, Questions 3-5 carry 2 marks each, Questions 6-8 carry 3 marks each and Questions 9-10 carry 4 marks each.

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

#### Question 1.

Every composite number has at least
(a) 1 factor
(b) 2 factors
(c) 3 factors
(d) 4 factors

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.

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

#### Question 3.

Find the prime factorisation of 1320.

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).

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

Reduce the fraction $\frac{714}{1386}$ in its simplest form.
Solution:

#### Question 6.

Simplify the following:
$2 \frac{3}{14}-3 \frac{5}{6}-\frac{2}{5}+2 \frac{1}{2}$

#### Question 7.

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

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.

#### Question 9.

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

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.?

LCM = 2 × 2× 2 × 3 × 3 × 5 = 360

After = 360 minutes
360 minutes = 6 hours
They will ring together again at 11 a.m.

— End of Model Question Paper-2 Class-6 ML Aggarwal Solutions  :–