Aptitude - Problems on H.C.F and L.C.M

Why Aptitude Problems on H.C.F and L.C.M?

In this section you can learn and practice Aptitude Questions based on "Problems on H.C.F and L.C.M" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) with full confidence.

Where can I get Aptitude Problems on H.C.F and L.C.M questions and answers with explanation?

IndiaBIX provides you lots of fully solved Aptitude (Problems on H.C.F and L.C.M) questions and answers with Explanation. Solved examples with detailed answer description, explanation are given and it would be easy to understand. All students, freshers can download Aptitude Problems on H.C.F and L.C.M quiz questions with answers as PDF files and eBooks.

Where can I get Aptitude Problems on H.C.F and L.C.M Interview Questions and Answers (objective type, multiple choice)?

Here you can find objective type Aptitude Problems on H.C.F and L.C.M questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided.

How to solve Aptitude Problems on H.C.F and L.C.M problems?

You can easily solve all kind of Aptitude questions based on Problems on H.C.F and L.C.M by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Problems on H.C.F and L.C.M problems.

Exercise :: Problems on H.C.F and L.C.M - General Questions

1. 

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A. 4
B. 7
C. 9
D. 13

Answer: Option A

Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

     = H.C.F. of 48, 92 and 140 = 4.


2. 

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276
B. 299
C. 322
D. 345

Answer: Option C

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.


3. 

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4
B. 10
C. 15
D. 16

Answer: Option D

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1 = 16 times.
2


4. 

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A. 4
B. 5
C. 6
D. 8

Answer: Option A

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4


5. 

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000
B. 9400
C. 9600
D. 9800

Answer: Option C

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.




Have a question?

Ask your question now !

Create your own website ?

Web Hosting Services in India