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# HCF and LCM Questions for IIFT

Practice IIFT solved HCF and LCM Questions paper tests, which are the practice question to have a firm grasp on the HCF and LCM topic in the IIFT exam. In this article, one can find the top 20 very Important HCF and LCM Questions for IIFT based on asked questions in previous exam papers. The IIFT question papers contain actual questions asked with detailed solutions. Download these important HCF and LCM Questions for IIFT PDF with detailed solutions. Click on the link given below to download the PDF.

Question 1:Â Let $x$ be the least number divisible by 8, 12, 30, 36 and 45 and $x$ is also perfect square. What is the volue of $x$?

a)Â 3600

b)Â 4900

c)Â 2500

d)Â 4225

Solution:

To find the least divisible number , we have find LCM of the given numbers

8, 12, 30, 36 , 45

LCM =Â $2\times\ 2\times\ 2\times\ 3\times\ 3\times\ 5$

= 360

As we have given, the number is also a perfect square

All the factors should be inÂ  pair, for making that we have to multiply it by 2 and 5 ,

i.e;Â $2\times\ 2\times\ 2\times\ 2\times\ 3\times\ 3\times\ 5\times\ 5$

= 3600

Hence, Option A is correct.

Question 2:Â The proportion among three numbers is 3 : 4 : 5 and their LCM is 1800. The second number is:

a)Â 90

b)Â 120

c)Â 30

d)Â 150

Solution:

Given, LCM = 1800

The proportion among three numbers is 3 : 4 : 5

Let the numbers are 3a, 4a, 5a respectively

LCM of the numbers = 3 x 4 x 5 x a = 60a

$=$>Â  60a = 1800

$=$>Â  a = 30

$\therefore\$Second number = 4a = 4(30) = 120

Hence, the correct answer is Option B

Question 3:Â The sum of two numbers is 1215 and their HCF is 81. If the numbers lie between 500 and 700, then the sum of the reciprocals of the numbers is ………

a)Â $\frac{5}{702}$

b)Â $\frac{5}{378}$

c)Â $\frac{5}{1512}$

d)Â $\frac{5}{1188}$

Solution:

HCF of the two numbersÂ is 81 soÂ two numbers are 81x and 81y.

81(x + y) = 1215

x + y = 15

numbers lie between 500 and 700 so,

For theÂ first number –

81 $\times$ 7 = 567

For the second number-

$81 \times 8$ = 648

sum of both numbers = 567 + 648 = 1215

The sum of the reciprocals of the numbers = $\frac{1}{81x} + \frac{1}{81y}$

= $\frac{81x + 81y}{81x \times 81y} = \frac{1215}{567 \times 648} = \frac{1215}{367416}$

= $\frac{5}{1512}$

Question 4:Â The LCM of165, 176, 385 and 495 is k. When is divided by the HCF of the numbers, the quotientis p. What is the value of p?

a)Â 2520

b)Â 5040

c)Â 6720

d)Â 3360

Solution:

As per the given question,

Numbers areÂ 165, 176, 385 and 495

LCM of the above $=2\times 3\times5\times 7\times 8\times 9\times 11=55440=K$
HCF of the given number =11
Now, LCM is getting divided by HCF$=\dfrac{55440}{11}=5040$

Question 5:Â What is the sum ofthe greatest three digit number and the smallest four digit number such that their HCF is 23?

a)Â 2001

b)Â 2002

c)Â 1984

d)Â 1998

Solution:

GreatestÂ  Three-digit number divisible by 23

then number = $23\times 43$ = 989

Smallest four-digit number divisible by 23

then number = $23\times 44$= 1012

so total = 989 + 1012

= 2001Ans

Question 6:Â The HCF and LCM of two numbers are 8 and 48 respectively. If the ratio of the two numbers is 2 : 3, then the larger of the two numbers is:

a)Â 24

b)Â 18

c)Â 48

d)Â 16

Solution:

Given that,Â  H C F = 8 and LC M = 48

let numbers are 2a and 3aÂ  Where ” a” is a common factor

so we use HCF and LCM formula,

$HCF\times LCM = 2a\times3a$

$\Rightarrow 8\times 48 = 6a^2$

$\Rightarrow a^2 = \dfrac{8 \times48}{6}Â$

$\Rightarrow a^2= 64$

$\Rightarrow a = 8$

then the largest number = 3a = $3\times8$=24Ans

Question 7:Â In finding the HCF of two numbersby division method, the quotients are 1, 8 and 2 respectively, and the last divisor is 105. What is the sum of the numbers?

a)Â 3570

b)Â 3885

c)Â 3780

d)Â 3675

Solution:

Let the two numbers be $a$ and $b$

The table shows that when $b$ is divided by $a$, quotient is 1 and remainder is $c$, which becomes divisor in 2nd step and $a$ becomes dividend, now quotient is 8 and remainder is 105 and in the final step, 105 becomes divisor with $c$ being the dividend, 2 is quotient and remainder is 0.

Thus, $c=105\times2+0=210$

$b=8c+105=8\times210+105=1785$

$a=1\times b+c=1785+210=1995$

Thus, sum of numbers = $a+b=1995+1785=3780$

=> Ans – (C)

Question 8:Â What is the HCF of $\frac{4}{5}, \frac{6}{8}, \frac{8}{25} ?$

a)Â $\frac{1}{5}$

b)Â $\frac{1}{100}$

c)Â $\frac{1}{200}$

d)Â $\frac{1}{50}$

Solution:

HCF of $\frac{4}{5}, \frac{6}{8}, \frac{8}{25}$

= Ratio of HCF (4,6,8) : LCM (5,8,25)

= $2:200=\frac{1}{100}$

=> Ans – (B)

Question 9:Â In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients are 1. 11 and 2, respectively. What is sum of the two numbers?

a)Â 833

b)Â 867

c)Â 816

d)Â 901

Solution:

Last divisor is 17 and quotients is 2 so
Second last divisor = 17 $\times$ 2 = 34
AndÂ quotients is 11 so,
Second last dividend = 34Â $\times$ 11 + 17 = 391
quotients is 1 so,
firstÂ dividend = 391 $\times$ 1 + 34 = 425
Sum of the two numbers =Â 391 +Â 425 = 816

Question 10:Â The LCM of two numbers x and yis 204 times their HCF. If their HCF is 12 and the difference between the numbers is 60, then x + y = ?

a)Â 660

b)Â 426

c)Â 852

d)Â 348

Solution:

HCF = 12
LCM = 204 $\times$ 12
xy =Â 204 $\times$ 12Â $\times$ 12 = 29376
x – y = 60
$(x – y)^2$ = 3600
$x^2 + y^2 -2xy = 3600$
$x^2 + y^2 -2xy = 3600$
$(x + y)^2 -4xy = 3600$
$(x + y)^2 = 3600 + 4 \timesÂ 29376$
$(x + y)^2 = 121104$
x + y = 348

Question 11:Â Let x be the least number which when divided by 15, 18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to makeit a perfect square?

a)Â 39

b)Â 37

c)Â 43

d)Â 36

Solution:

LCM ofÂ 15, 18, 20 and 27 = 540
Number = $\frac{540k +Â 10}{31}$
= $\frac{527k + 13k +Â 10}{31}$
Value of k = 4
So, number =Â 540 $\times$ 4Â + 10 = 2160 +Â 10 = 2170
Square of 47 = 2209
Number added to make it perfect square = 2209 – 2170 =Â 39

Question 12:Â The HCF of two numbers is 21 and their LCM is 221 times the HCF. If one of the numbers lies between 200 and 300, then the sum ofthe digits of the other number is:

a)Â 14

b)Â 17

c)Â 18

d)Â 15

Solution:

HCF = 21
LCM =Â 221 $\times$ 21
Number of productsÂ  = HCF $\times$ LCM
Number of products =Â 221 $\times$ 21Â $\times$ 21
=Â 13 $\times$Â 17 $\times$ 21 $\times$ 21
So 1 numberÂ  =Â 13 $\times$ 21 = 273
Another number =Â 17 $\times$ 21 = 357
Sum of the digits of the other number = 3 +Â 5 +Â 7 = 15

Question 13:Â When 12, 16, 18, 20 and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousandsâ€™ place in x ?

a)Â 5

b)Â 8

c)Â 4

d)Â 3

Solution:

Number = (LCM ofÂ 12, 16, 18, 20 and 25)k + 4
= 3600k +Â 4
The number should be divisible by the 7 so,
Value of K = 5
Number =Â 3600 $\times$ 5Â + 4 = 18000 +Â 4 = 18004
The digit at the thousandsâ€™ place = 8

Question 14:Â The number of pair of positive integers whose sum is 99 and HCF is 9 is:

a)Â 2

b)Â 4

c)Â 3

d)Â 5

Solution:

Given, HCF of two numbers = 9

Let the two numbers be $9x$ and $9y$

Sum of the numbers = 99

$=$>Â $9x+9y=99$

$=$>Â $9\left(x+y\right)=99$

$=$>Â $x+y=11$

$=$> The pair of positive integers which satisfies the above condition is (1,10) (2,9) (3,8) (4,7) (5,6)

$=$> The possible pair of numbers are (9,90) (18,81) (27,72) (36,63) (45,54)

$\therefore\$The number of positive pair of integers = 5

Hence, the correct answer is Option D

Question 15:Â The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28.

a)Â 9828

b)Â 9928

c)Â 9882

d)Â 9288

Solution:

LCM ofÂ 12, 18, 21 and 28 = 252

The greatest four digit number = 9999

On dividing 9999 by 252, the remainder is 171

$\therefore\$The greatest four digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 = 9999-171 = 9828

Hence, the correct answer is Option A

Question 16:Â The ratio of two numbers is 3:4 and their LCM is 120. The sum of numbers is:

a)Â 140

b)Â 35

c)Â 70

d)Â 105

Solution:

Let the two numbers be $3x$ and $4x$

$=$> HCF of two numbers = $x$

Given, LCM of two numbers = 120

We know that,Â Product of two numbers is equal to Product of LCM and HCF of two numbers

$=$>Â  $3x\times4x=120\times x$

$=$> Â $x=10$

$\therefore\$Sum of the numbers = $3x+4x=7x=7\left(10\right)=70$

Hence, the correct answer is Option C

Question 17:Â The LCM of two numbers is 12 times their HCF. The sum of the HCF and LCM is 403. If one of the number is 93, then the other is

a)Â 120

b)Â 124

c)Â 112

d)Â 116

Solution:

Given, LCM of two numbers is 12 times their HCF

$=$>Â  LCM = 12 HCF

Sum of HCF and LCM is 403

$=$> HCF + LCM = 403

$=$> HCF + 12 HCF = 403

$=$> 13 HCF = 403

$=$>Â  HCF = 31

$\therefore\$LCM = 12 HCF = 12(31)

$=$>Â  LCM = 372

Let the second number be $x$

Product of numbers = Product of their LCM and HCF

$=$>Â  93 x $x$ = 372 x 31

$=$>Â  $x$ = 124

$\therefore\$The other number is 124

Hence, the correct answer is Option B

Question 18:Â Let x be the greatest number which when divides 6475, 4984 and 4132, the remainder in each case is the same. What is the sum of digits of x a?

a)Â 4

b)Â 7

c)Â 5

d)Â 6

Solution:

Let x be the greatest number which when divides 6475, 4984 and 4132, the remainder in each case is the same,then

6475 – 4984 = 1491

4984 – 4132 = 852

6475 – 4132 = 2343

Required Number = HCF of 1491, 852 and 2343

i.e; 213 is the HCF

Sum of the Digits = 2 +Â 1 + 3 = 6

Hence,Â Option DÂ is correct.

Question 19:Â HCF and LCM of two numbers p and q is A and B respectively, if A + B = p + q, then the value of $A^3 + B^3$ is:

a)Â $p^3 + q^3$

b)Â $q^3$

c)Â $p^3$

d)Â $p^3 – q^3$

Solution:

As we know,

$L.C.M\times\ H.C.F=product\ of\ numbers$

So, $A\times\ B=p\times\ q$……….(i)

But A + B = p + q (given)……..(ii)

As we know,

$A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)$

$A^3+B^3=\left(A+B\right)\left(\left(A+B\right)^2-3AB^{ }\right)$

So,

$A^3+B^3=\left(p+q\right)\left(\left(p+q\right)^2-3pq\right)$

So,

$A^3+B^3=p^3+q^3$

Hence, Option A is correct.

Question 20:Â The largest three digit number that is exactly divisible by 6, 7 and 8 is:

a)Â 999

b)Â 168

c)Â 358

d)Â 840

Solution:

6 = $2\times3\times1$

7 = $7\times1$

8 = $2\times2\times2\times1$

The LCM ofÂ 6, 7 and 8 isÂ 168.

The largest three-digit number is 999.

The required number should be completely divisible byÂ 168 which is theÂ LCM of 6, 7 and 8.

So when we will divide 999 byÂ 168, then the remainder willÂ beÂ 159.

Hence the required number = 999-159

=Â 840