# QUESTIONS ON LCM & HCF (PART-I)

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## 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*b}⁄_{b*c }= ^{551}⁄_{1073}

=> ^{a*b}⁄_{b*c}_{ }= ^{19*29}⁄_{29*37}

=> ^{a}⁄_{c }= ^{19}⁄_{37}

=> 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’)