# H.C.F and L.C.M Problems Question & Answers

## HCF and LCM Problems

Quantitative aptitude questions are asked in many competitive exams and placement exam.'HCF and LCM' is a category in Quantitative Aptitude. Quantitative aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS Exam, CSAT, CLAT, Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams, Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

We have a large database of problems on “H C F and L C M” answered with explanation. These will help students who are preparing for all types of competitive examinations.

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

 A) 123 B) 127 C) 235 D) 305

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

Rate this : 2 1

Product of two co-prime numbers is 117. Their L.C.M  should be

 A) 1 B) 117 C) Equal to their H.C.F D) cannot be calculated

Explanation:

H.C.F of co-prime numbers is 1. So, L.C.M = $\inline&space;\fn_jvn&space;\frac{117}{1}$ =117

Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

Rate this : 2 2

The G.C.D of 1.08, 0.36 and 0.9 is

 A) 0.03 B) 0.9 C) 0.18 D) 0.108

Explanation:

Given numbers are 1.08 , 0.36 and 0.90

H.C.F of 108, 36 and 90 is 18                  [ $\inline&space;\fn_jvn&space;\because$ G.C.D is nothing but H.C.F]

$\inline&space;\fn_jvn&space;\therefore$ H.C.F of given numbers = 0.18

Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

Rate this : 1 4

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

 A) -2 B) -1 C) 1 D) 2

Explanation:

H. C. F of two prime numbers is 1.  Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with  product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

$\inline&space;\fn_jvn&space;\therefore$ 3y-x = (3 x 7)-23 = -2

Subject: H.C.F and L.C.M Problems - Quantitative Aptitude - Arithmetic Ability

Rate this : 0 2

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

 A) 77 B) 88 C) 99 D) 110

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  $\inline&space;\fn_jvn&space;\Rightarrow$  ab = 35

Now, co-primes with product  35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is  (55,77)

Hence , required number = 77