Arun Sharma Test: HCF & LCM

# Arun Sharma Test: HCF & LCM

Test Description

## 10 Questions MCQ Test CSAT Preparation for UPSC CSE | Arun Sharma Test: HCF & LCM

Arun Sharma Test: HCF & LCM for Quant 2022 is part of CSAT Preparation for UPSC CSE preparation. The Arun Sharma Test: HCF & LCM questions and answers have been prepared according to the Quant exam syllabus.The Arun Sharma Test: HCF & LCM MCQs are made for Quant 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Arun Sharma Test: HCF & LCM below.
Solutions of Arun Sharma Test: HCF & LCM questions in English are available as part of our CSAT Preparation for UPSC CSE for Quant & Arun Sharma Test: HCF & LCM solutions in Hindi for CSAT Preparation for UPSC CSE course. Download more important topics, notes, lectures and mock test series for Quant Exam by signing up for free. Attempt Arun Sharma Test: HCF & LCM | 10 questions in 10 minutes | Mock test for Quant preparation | Free important questions MCQ to study CSAT Preparation for UPSC CSE for Quant Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Arun Sharma Test: HCF & LCM - Question 1

### Find the lowest common multiple of 24, 36 and 40.

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 1 Arun Sharma Test: HCF & LCM - Question 2

### The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 2 Arun Sharma Test: HCF & LCM - Question 3

### Detailed Solution for Arun Sharma Test: HCF & LCM - Question 3 Arun Sharma Test: HCF & LCM - Question 4

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 4 Arun Sharma Test: HCF & LCM - Question 5

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 5

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Arun Sharma Test: HCF & LCM - Question 6

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 6 Arun Sharma Test: HCF & LCM - Question 7

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 7 Arun Sharma Test: HCF & LCM - Question 8

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 8

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Arun Sharma Test: HCF & LCM - Question 9

The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 9

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

Arun Sharma Test: HCF & LCM - Question 10

252 can be expressed as a product of primes as:

Detailed Solution for Arun Sharma Test: HCF & LCM - Question 10

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

## CSAT Preparation for UPSC CSE

72 videos|64 docs|92 tests
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Arun Sharma Test: HCF & LCM Page
In this test you can find the Exam questions for Arun Sharma Test: HCF & LCM solved & explained in the simplest way possible. Besides giving Questions and answers for Arun Sharma Test: HCF & LCM, EduRev gives you an ample number of Online tests for practice

## CSAT Preparation for UPSC CSE

72 videos|64 docs|92 tests