# Dedicated for NMAT 2023 # AP GP Progression NMAT 2023

Question 1

How many numbers between 900 and 3000 are exactly divisible by both 12 and 40?

a) 18 b) 17 c) 12 d) 19

Solution :

If a number is divided by 12 and 40 then it must be divided by their LCM which is 120.
Numbers between 900 and 3000 which are divisible by 120 are
960, 1080,…, 2880.
i.e., the first number which is divisible by 120 and greater than 900 is 960.

The number 3000 is also divisible by 120 but we have to find the numbers between 900 and 3000.
So the last number which is divisible by 120 and less than 3000 is 2880.
The numbers are 960, 1080,.., 2880.

This is an A.P where a = 960, d = 120 and t(n) = 2880
We know that the n-th term in A.P = a + (n – 1)d
2880 = 960 + (n – 1)120
2880 = (8 + (n – 1))120
24 = 8 + n – 1
n = 17.
Hence, there are 17 numbers between 900 and 3000 that are exactly divisible by both 12 and 40.

Question 2

Find the number of 2-digit numbers which are exactly divisible by 3.

a) 19 b) 24 c) 30 d) 35

Solution :

The two digit numbers are 10, 11, 12,…,99
The 1st two digit number that is exactly divisible by 3 is 12
and the last two digit number that is exactly divisible by 3 is 99.
Then the two digit numbers that are divisible by 3 are 12,15,…,99.

This is an A.P where a = 12, d = 3 and t(n) = 99
Then t(n) = a + (n-1)d
99 = 12 +(n-1)3
33 = 4 + n – 1
n = 30.
Hence, there are 30 two-digit numbers that are divisible by 3.

Question 3

Find the number of positive integers which are less than 1000 and greater than 100 and are divisible by 2,3 and 4.

a) 75 b) 72 c) 79 d) 80

Solution :

If a number is divided by 2,3 and 4 then it must be divided by their LCM 12 also.
Numbers between 100 and 1000 which are divisible by 12 are
108, 120,…, 996.
i.e., the first number which is divisible by 12 and greater than 100 is 108.
And the last number which is divisible by 12 and less than 1000 is 996.
Then the numbers are 108, 120,…, 996.

This is an A.P where a = 108, d = 12 and t(n) = 996
We know that the n-th term in A.P = a + (n-1)d
996 = 108 + (n-1)12
83 = 9 + n-1
n = 75.
Hence, there are 75 numbers between 100 and 1000 that are exactly divisible by 2,3 and 4.

Question 4

Find the number of 3 digit numbers which are exactly divisible by 4.

a) 222 b) 220 c) 198 d) 225

Solution :

The three digit numbers are 100, 101, 102,…,999
The 1st three digit number that is exactly divisible by 4 is 100
and the last three digit number that is exactly divisible by 4 is 996.
Then the three digit numbers that are divisible by 4 are 100,104,…,996.

This is an A.P where a = 100, d = 4 and t(n) = 996
Then t(n) = a + (n-1)d
996 = 100 +(n-1)4
249 = 25 + n – 1
n = 225.
Hence, there are 225 three digit numbers that are divisible by 4.

#### Popular Post

• ###### Time Speed Distance in NMAT 2023 Lorem ipsum dolor sit amet consectetur adipiscing elit ut elit tellus luctus nec.

## Pricing Plans For Every Need

Lorem ipsum dolor sit amet consectetur adipiscing elit semper dalar elementum tempus hac tellus libero accumsan dolor sit amet consctur. ## 2000/-

Best books for NMAT Exam!

12 Yellow Books

Full Syllabus

Focus on Vocab Grammar

Critical Reasoning

Quant All topics covered

DI & DS intensive prep ## 2500/-

Best Shortcuts for NMAT Exam

Shortcuts for Quant

G Strategy for DI DS

Vocab Grammar G Strategy

Logical Reasoning Shortcuts

Live Classes and Recordings

Weekly Targets ## 2000/-

Best Mocks for NMAT Exam