LCM and HCF MCQs (Top 20 breakthrough tips)

Que 1: The sum and the difference of the LCM and HCF of the two numbers are 312 and 264, respectively. Find the numbers if their sum is 168.

Solution: Here the sum of all given pairs is 168. (LCM and HCF MCQs)

Let the HCF of two numbers is H then the numbers are Hx and Hy

LCM=Hxy

Hxy + H= 312 …..(i)

Hxy – H= 264 …..(ii)

Add both equations:

LCM=Hxy=288

Put this value in any of the two equations: 

H=24

Now, Only option (d) satisfies this condition of HCF=24.

Que 2: A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contain the same number of liters of a drink, and doesn’t want to mix any two drinks in a can. What is the least number of cans required?

Solution:

Mazza= 80 ltr (LCM and HCF MCQs)

Pepsi= 144 ltr

Sprite= 368 ltr

The Capacity of each Can= HCF= 16 (Highest Common Factor)

The number of cans required for these three drinks= (5, 9, 23)

Total cans= 5+9+23=37

Option (c) is correct.

Que 3: If the LCM of ‘a’ and ‘b’ is a & LCM of ‘b’ and ‘c’ is b, then what is the LCM of ‘c’ and ‘a’?

Solution: (LCM and HCF MCQs)

c=c

b=cx

a=cxy

LCM of a & c= a (LCM and HCF MCQs)

Option (a) is correct.

Que 4: LCM of two prime numbers x and y (x>y) is 161. The value of 3y-x is:

Solution: x & y are Prime numbers (LCM and HCF MCQs)

161=7×23

x=23 and y=7

So, 3y-x= -2

Option (a) is correct.

Que 5: How many pairs of positive integers x and y exist such that HCF(x,y) +LCM(x,y)= 91?

Solution: HCF(x,y) +LCM(x,y)= 91

x=Ha

y=Hb

Here H is the HCF of two numbers. a & b are Co-Prime numbers. (LCM and HCF MCQs)

H+Hab= 91

H×(ab+1)=91×1 =7×13

If H=1 then ab=90

(a,b)={(1,90), (2,45), (5,18), (9,10)} ⇒ 4 Pairs

If H=7 then ab=12

(a,b)={(1,12), (3,4)} ⇒ 2 Pairs

If H=13 then ab=6

(a,b)={(1,6), (2,3)} ⇒ 2 Pairs

Total Pairs=8

Option (b) is correct. (LCM and HCF MCQs)

Que 6: The HCF and LCM of the two numbers are 11 and 385 respectively. If one number lies between 75 and 125, then that number is

Solution: HCF=H=11

Numbers are Hx and Hy

LCM=Hxy =385

xy=35= 5×7

The required number= 11×7= 77

Option (a) is correct. (LCM and HCF MCQs)

Que 7: Find the least five-digit number, which when divided by 12,18,21, and 28 leaves the same remainder.

Solution:

12=2²×3

18=2×3²

21=3×7

28=2²×7

LCM= 2²×3²×7=252

Now the Required number= 252×k+ R

Here k is a whole number and R is the Remainder. (LCM and HCF MCQs)

The least number of 5 digits of 10,000

10,252 mod 252= 172

It means 10,252- 172=10,080 is divisible by all the given divisors.

Now the value of R should be less than 12.

Option (b) is correct.

Que 8: The product of two co-prime numbers added to their HCF gives us 91. How many such pairs are possible?

Solution: Let the co-prime numbers be x & y.

HCF=1

xy+1=91

xy=90

(x,y)={(1,90), (2,45), (9,10)}

Only 3 such pairs are possible. (LCM and HCF MCQs)

Option (b) is correct.

Que 9: How many pairs of integers (x,y) exist such that the product of x,y, and HCF(x,y)= 1080?

Solution: Let HCF= H

Numbers are Hx and Hy

Product of all three=H³xy=1080 

H³xy=2³×3³×5

If H=1

(x,y)={(2³×3³×5,1), (2³×3³,5), (2³×5,3³), (3³×5,2³)} ⇒ 4 Pairs

If H=2

(x,y)={(3³×5,1), (3³,5)} ⇒ 2 Pairs

If H=3

(x,y)={(2³×5,1), (2³,5)} ⇒ 2 Pairs

If H=2×3=6

(x,y)={(1,5)} ⇒ 1 Pairs

Total Pairs=9 (LCM and HCF MCQs)

Option (c) is correct

Que 10: The greatest number which on dividing 1657 and 2037 leaves the remainder 6 and 5 respectively, is

Solution:

1657-6=1651 and 2037-5=2032

2032-1651=381

381 is divisible by only one option which is 127.

Option (b) is correct.

Que 11: Ram wants to utilize his unused field and plans to plant some trees, he plants 88 Guava, 132 Papaya, and 220 Sugarcane trees in equal rows (in terms of the number of trees). Also, he wants to make distinct rows of trees (i.e. only one type of tree in one row). Calculate the minimum number of rows.

Solution:

The number of trees in a row= HCF(88,132,220)=44

The number of rows of respective trees= (2,3,5)

Total rows= 2+3+5=10 (LCM and HCF MCQs)

Option (b) is correct.

Que 12: Find the least number that should be added to 2014 such that the resulting number can be divided by 3,4,5, and 6 leaving no remainder.

Solution: LCM(3,4,5,6)=60

(2014+60) mod 60= 34

It means 60-34=26 should be added to satisfy the given condition.

Option (d) is correct.

Que 13: The product of two numbers is 2028 and HCF is 13. The number of such pairs is:

Solution: HCF=13

Numbers are 13x and 13y

Product= 13²xy=2028

xy=12 (LCM and HCF MCQs)

(x,y)={(1,12), (3,4)} ⇒ 2 Pairs

Option (b) is correct.

Que 14: Find the least number which when divided by 5 leaves the remainder of 4, when divided by 6 leaves the remainder of 5, when divided by 7 leaves 6, when divided by 8 leaves the remainder of 7, and when divided by 9 leaves the remainder 8?

Solution:

5x+4 ⇒5p-1

6y+5 ⇒6q-1

7z+6 ⇒7r-1

8m+7 ⇒8s-1

9n+8 ⇒9t-1

The required number= LCM(5,6,7,8,9)×k -1

=2520×k-1

For the least value, k=1

The least number= 2520-1= 2519

Option (c) is correct.

Que 15: Find the sum of the digits of the smallest 4-digit number which when divided by 4 leaves remainder 1, when divided by 5 leaves remainder 3, and when divided by 7 leaves remainder 2?

Solution:

Let the number be 7x+2

Here x is a Natural number.

Now 7x+2 mod 5=3 (Given)

2x+2 mod 5=3 

It is possible for x={3, 8, 13, 18,….}  

7x+2 mod 4=1 (Given)

3x+2 mod 4=1

It is possible for x={1, 5, 9, 13……}

So, For x=13 all the conditions are satisfied

7x+2=7×13+2=93

The general formula for such numbers= LCM(7,5,4)×k +93

=140k +93

Here k is a whole number.

Put k=7 to find the smallest four-digit number.

The required number= 140×7 +93=1073

The Sum of digits= 1+0+7+3=11

Option (b) is correct.

Que 16: If N is divided by 3,4,5 and 6 successively it leaves remainders 2,1,1 and 4 respectively.  How many such 4-digit numbers are possible?

Solution: The sequence of divisors: 3, 4, 5, and then 6

The sequence of respective remainders: 2, 1, 1, and 4

Short Trick: From right to left apply this formula 

(Divisor+Remainder)×Previous Divisor= New Divisor

(6+4)×5=50

(50+1)×4=204

(204+1)×3=615

(615+2)=617   …….The End

The least Possible number that satisfies this condition is 617.

The General formula of such numbers= LCM(3,4,5,6)×k +617

=60×k +617

Here k is a Whole number.

For the least four-digit number 

60k +617> 1000

k>6.38

For the largest four-digit Number

60k +617< 9999

k<156.36

It means k={7,8,9……,156}⇒150 numbers

So, a total of 150 such numbers exist.

Option (c) is correct.