HCF and LCM – Railway Maths Part II

Total Questions: 50

1. The HCF and LCM of two numbers is 10⁸ and 10¹²·7³ respectively. If one of the numbers is 10¹², then the other number is _____. [NTPC CBT-I 11/02/2021 (Evening)]

Correct Answer: (d) 10⁸ × 7³
Solution:

Product of two number = LCM × HCF
10¹² × 2nd number = 10⁸ × 10¹²7³
⇒ 2nd number = (10⁸ × 10¹²7³) / 10¹² = 10⁸ × 7³

2. If the LCM of p and q is pq , then the numbers of p, q must be: [NTPC CBT-I 12/02/2021 (Morning)]

Correct Answer: (c) prime
Solution:

LCM of prime numbers = product of numbers
So, the LCM of p and q is pq, then the numbers of p, q must be prime.

3. Two positive numbers are in the ratio 5 : 4 and the product of their LCM and HCF is 18000. Find the sum of the two numbers. [NTPC CBT-I 02/03/2021 (Morning)]

Correct Answer: (a) 270
Solution:

Let, two +ve numbers are = 5x and 4x
5x × 4x = 18000
20x² = 18000 ⇒ x² = 900 ⇒ x = 30
Sum of the numbers = 5x + 4x = 9x = 30 × 9 = 270

4. The LCM of (a³ + b³) and (a⁴ − b⁴) is: [NTPC CBT-I 08/03/2021 (Morning)]

Correct Answer: (d) (a³ + b³)(a + b)(a² − b²)
Solution:

The LCM of (a³ + b³) and (a⁴ − b⁴) is
= (a³ + b³)(a² + b²)(a − b)

5. The LCM of two prime numbers x and y (x > y) is 119. The value of 3y − x is: [NTPC CBT-I 14/03/2021 (Morning)]

Correct Answer: (c) 4
Solution:

LCM of two prime numbers = 119
119 = 7 × 17
Both prime numbers are X = 17, Y = 7
Now, 3y − x = 3 × 7 − 17 = 21 − 17 = 4

6. Find the least square number that is exactly divisible by 4, 5, 6, 15 and 18. [NTPC CBT-I 21/03/2021 (Morning)]

Correct Answer: (d) 900
Solution:

LCM of 4, 5, 6, 15 and 18 = 180
Hence, 180 × 1 = 180 (it’s not a perfect square number)
180 = 2² × 3² × 5
For perfect square, we have to multiply with 5
Hence, least perfect square = 180 × 5 = 900

7. What is the smallest sum of money which contains Rs. 2.50, Rs. 20, Rs. 1.20 and Rs. 7.50? [NTPC CBT-I 03/04/2021 (Evening)]

Correct Answer: (b) Rs. 60
Solution:

LCM of (250, 2000, 120 and 750) paise = 6000 paise
6000 paise = 60 Rs.

8. After spending ¼th of pocket money on chocolates and ⅛th on pizza, a girl is left with Rs. 40. How much money did she have at first? [RRB JE 23/05/2019 (Morning)]

Correct Answer: (b) Rs. 64
Solution:

L.C.M. of 4 and 8 = 8
Let total pocket money be 8 units.
Money spent on chocolates = 2 units
Money spent on Pizza = 1 units
Money left with him (5 units) = 40 Rs.
Total pocket money (8 units) = 8 × 8 = 64 Rs.

9. What is the GCD of these polynomials? [RRB JE 23/05/2019 (Evening)]

(x³ + x² + x + 1) and (x³ + 2x² + x + 2)?

Correct Answer: (c) (x² + 1)
Solution:

(x³ + x² + x + 1) and (x³ + 2x² + x + 2)
x³ + x² + x + 1 = x²(x + 1) + (x + 1) = (x + 1)(x² + 1)
x³ + 2x² + x + 2 = x²(x + 2) + (x + 2) = (x + 2)(x² + 1)
Greatest common divisor (GCD) of (x³ + x² + x + 1) and (x³ + 2x² + x + 2) = (x² + 1)

10. LCM of two numbers 'p' and 'q' is 935 and (p > q). What is the maximum sum of the digits of 'q' out of the possible pairs 'p', 'q'? [RRB JE 30/05/2019 (Afternoon)]

Correct Answer: (a) 8
Solution:

According to question,
L.C.M (p, q) = 935, p > q
So, possible pair of p and q are (85, 11), (55, 17), (187, 5)
Therefore, maximum sum of the digit of q = 1 + 7 = 8