HCF and LCM – (Railway Maths Part VIII)

Total Questions: 27

1. Two numbers have their HCF and LCM to be 72 and 1008 respectively. If one of the numbers is 144, then find the other number. [RRB JE 26/06/2019 (Morning) ]

Correct Answer: (b) 504
Solution:

2. Find the LCM and HCF of a(a + b) and b(a + b), where a, b are coprime. [RRB JE 26/06/2019 (Morning) ]

Correct Answer: (a) ab(a + b), (a + b)
Solution:

3. Find the LCM of set of prime numbers a, b and c. [RRB JE 26/06/2019 (Evening) ]

Correct Answer: (a) abc
Solution:

LCM of the prime numbers is the product of those numbers. Therefore the LCM of a,b and c = abc

4. Find the largest number of four digits that is exactly divisible by 27, 18, 15 and 12. [RRB JE 26/06/2019 (Evening) ]

Correct Answer: (d) 9720
Solution:

LCM of 27, 18, 15 and 12 = 540
Largest 4-digit no. = 9999
On dividing 9999 by 540 , remainder will be 279
Therefore , the largest four digit number that is divisible by 27 , 18 , 15 and 12 is → 9999 - 279 = 9720

5. Find the greatest number that will divide the numbers 400, 435 and 541 leaving remainders 9, 10 and 14 respectively. [RRB JE 27/06/2019 (Morning) ]

Correct Answer: (a) 17
Solution:

6. Find the smallest 3 digit number that is exactly divisible by 12, 15 and 20. [RRB JE 27/06/2019 (Morning) ]

Correct Answer: (d) 120
Solution:

LCM of 12, 15, and 20 = 60
According to options smallest 3 digit number is 120 which is divisible by 12 , 15 , 20

7. LCM of 36 and K is 72. Find the possible values of 'K'. [RRB JE 27/06/2019 (Evening) ]

Correct Answer: (b) 8, 24, 72
Solution:

LCM (k , 36) = 72 …(given) So , 72 will be divisible by both k and 36. So , the possible value of k is (8 , 24 , 72)

8. Find the HCF of a³b³c³, a²b²c², abc and: [RRB JE 27/06/2019 (Evening) ]

Correct Answer: (d) abc
Solution:

H.C.F. (a³ b³ c³, a² b² c², abc, a² bc) = abc

9. If two numbers with HCF 9 are in the ratio 5 : 7, what is their difference? [RRB JE 28/06/2019 (Evening) ]

Correct Answer: (c) 18
Solution:

Let A and B be the numbers, and their HCF = 9
According to question,
Ratio of numbers (A : B) = 5 : 7
So, number (A) = 5 × 9 = 45 and number
(B) = 7 × 9 = 63
Required difference = 63 − 45 = 18

10. The sum of two numbers is 35. Their product is 150. Find the HCF of the sum of their reciprocals and the difference of the reciprocals. [RRB JE 28/06/2019 (Evening) ]

Correct Answer: (d) 1/30
Solution:

Let two number be a and (35 − a)
According to the question
a × (35 − a) = 150
a² − 35a + 150 ⇒ a = 30, 5
So, the numbers are 30 and 5
Now sum of their reciprocals
⇒ 1/30 + 1/5 ⇒ 7/30
And the difference of the reciprocals
⇒ 1/5 − 1/30 ⇒ 5/30
Required H.C.F. (7/30, 5/30) = 1/30