# QUESTIONS ON LCM & HCF (PART-I)

## QUESTIONS ON LCM & HCF

#### QUERY 1

The ratio of the sum to the LCM of two natural numbers is 7:12. If their HCF is 4, then the smaller number is:

A) 20
B) 16
C) 12
D) 8

RONNIE BANSAL
The ratio of the sum to the LCM of two natural numbers is 7 : 12 (GIVEN)

Now think 2 such co-prime numbers whose sum is 7 and LCM is 12. They are 3 & 4 of course. But the HCF of the given numbers is 4 (given); so multiply these prime numbers by the HCF i.e. 4 (here) to find the smaller number & the greater number as it has been asked. Hence the smaller number here is 12 (option ‘C’)

#### QUERY 2

A girl gives away each of four girls 1/12, 5/18, 7/30,7/48 of the apples in a basket and has only just enough apples to be able to do so without dividing an apple, the minimum number of apples she had?

A) 533
B) 530
C) 720
D) 500

MAHA GUPTA
We need to find the LCM of the denominators of the fractions. The LCM is 720. So she had 720 apples in all out of which she gives away 60+200+168+105 = 533 (option ‘A’)

NOTE: Some guys think the girl distributed all the apples she had. Nowhere the question says she distributed all the apples she had.

#### QUERY 3

Find the LCM and HCF OF 25/6 and 15/4

A) 3/8
B) 5/12
C) 5/16
D) 4/13

MAHA GUPTA
LCM of fractions = LCM of numerators/HCF of denominators
HCF of fractions = HCF of numerators/LCM of denominators

Therefore the LCM of above = (LCM of 25 and 15)/(HCF of 6 and 4)
= 75/2
And HCF of above = (HCF of 25 and 15)/(LCM of 6 and 4)
= 5/12 (option ‘B’)

#### QUERY 4

Product of two co-prime numbers is 117; then their LCM is?

A) 13
B) 39
C) 117
D) 9

MAHA GUPTA
Product of two numbers = HCF*LCM of both
But the numbers are co-prime, therefore their HCF is 1

Hence from above
117 = 1*LCM
Thus the LCM = 117 (option ‘C’)

#### QUERY 5

The HCF of two numbers is 8 and the sum of numbers is 224. How many such pairs of numbers satisfy this condition?

A) 6
B) 4
C) 8
D) 7

MAHA GUPTA
The HCF of the numbers is 8; means both the numbers are divisible by 8 and both the quotients thus got are co-prime.

Now let those numbers be 8a and 8b
=> 8a + 8b = 224
=> a + b = 224/8 = 28

Now we just need to know the pairs of co-prime numbers whose sum is 28

Obviously they are (1, 27), (3, 25), (5, 23), (9, 19), (11, 17), (13, 15)
Hence the required number = 6 (option ‘A’)

#### QUERY 6

84 Maths books, 90 physics books, and 120 chemistry books of equal thickness have to be stacked topic-wise. How many books will be there in each stack so that each stack will have the same height too?

A) 6
B) 12
C) 21
D) 18

MAHA GUPTA
Find the biggest number that can divide 84, 90 and 120. Such number, of course, is called the HCF (Highest Common Factor)

Now HCF of 84, 90 and 120 = 6 (option ‘A’)

#### QUERY 7

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 am. The time when they will next make a beep together at the earliest, is?

A) 10:28 am
B) 10:30 am
C) 10:31 am
D) 10:15 am

MAHA GUPTA
Obviously both the bells will make a beep together in the time equal to the LCM of the time they make a beep.

Now the LCM of the times they make a beep = LCM of 60 and 62 seconds i.e. 1860 seconds
= 1860/60 minutes = 31 minutes

Thus the time at which they will now make a beep together = 10 am + 31 minutes = 10:31 am (option ‘C’)

#### QUERY 8

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

MAHA GUPTA
Obviously all the bells will toll together in the time equal to the LCM of the time of their intervals.

Now the LCM of the times of their intervals = LCM of 2, 4, 6, 8, 10, 12 i.e. 120 seconds
= 2 minutes

But they will also toll together once in the beginning

Thus the number of times they will toll together in 30 minutes = 30/2 + 1 = 16 (option ‘D’)

#### QUERY 9

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is?

A) 1
B) 2
C) 3
D) 4

MAHA GUPTA
Let the numbers be 13m and 13n.

Then, 13m × 13n = 2028

=> mn = 12.

To find such pairs of numbers as is said in the question, we find the co-primes with the product of the numbers; so , the co-primes with product 12 are (1, 12) and (3, 4).

We see that there are two such pairs, which is our answer (option ‘B’)

#### QUERY 10

Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is?

A) 75
B) 81
C) 85
D) 89

MAHA GUPTA
Let the numbers be a, b, c

So, according to the question
a × b = 551
b × c = 1073

We see ‘b’ is common to both, and as all numbers are co-prime, so ‘b’ is the LCF of both 551 and 1073

Therefore, ‘b’ = HCF of 551 and 1073 = 29

Now as the first product is the product of the first two numbers, hence the first number i.e. ‘a’ = 551/29 = 19;
Similarly the third number i.e. ‘c’ = 1073/29 = 37

So, sum of the three numbers = 19 + 29 + 37 = 85 (option ‘C’)

TRICK
a*bb*c 5511073

=> a*bb*c 19*2929*37

=> a1937

=> a = 19, and c = 37

Now as a*b = 551, therefore ‘b’ = 551/19 = 29

Hence sum of the numbers = 19 + 29 + 37 = 85 (option ‘C’)

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