R. Gupta Solutions for Chapter: L.C.M. and H.C.F., Exercise 1: EXERCISE

Author:R. Gupta

R. Gupta Quantitative Aptitude Solutions for Exercise - R. Gupta Solutions for Chapter: L.C.M. and H.C.F., Exercise 1: EXERCISE

Attempt the practice questions on Chapter 2: L.C.M. and H.C.F., Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. NABARD Officers Assistant Manager (Grade A) & Manager (Grade B) Phase I - Quantitative Aptitude solutions are prepared by Experienced Embibe Experts.

Questions from R. Gupta Solutions for Chapter: L.C.M. and H.C.F., Exercise 1: EXERCISE with Hints & Solutions

HARD
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The product of two numbers is 2160 and their H.C.F. is 12. How many such pairs of numbers can be possibly formed?

HARD
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The sum of two numbers is 216 and their H.C.F. is 27. These numbers are:

HARD
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The H.C.F. and the L.C.M. of two numbers are 50 and 250 respectively. On dividing one of these numbers by 2, 50 is obtained as quotient. The numbers are:

MEDIUM
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The H.C.F. of three numbers is 12. If the three numbers are in the ratio of 1:2:3, then the numbers are:

MEDIUM
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The greatest four-digit number completely divisible by 2, 3, 4 and 5 is:

HARD
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

Three bells ring respectively at an interval of 15 seconds, 20 seconds and 24 seconds. If they ring continuously for 12 minutes then how many times, during this period, will they ring together?

MEDIUM
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

The smallest number, on being successively divided by 5, 6, 8, 9 and 12 leaves 1 as remainder in each case and is completely divisible by 13, will be:

HARD
Andhra Pradesh State Cooperative Bank Assistant
IMPORTANT

If in the process of finding H.C.F. of two numbers by continued division method, 49 is the last divisor and quotients obtained (from the beginning) are 17, 3 and 2 respectively, then the numbers are: