HCF and LCM – Railway Maths Part III

Total Questions: 52

1. If the sum of the LCM and the HCF of two numbers is 1050 and if their HCF is 630 less than their LCM, then the product of those two numbers is _____. [Group D 16/09/2022 (Morning)]

Correct Answer: (d) 176400
Solution:

Let the L.C.M be x, then H.C.F be x - 630
x + x - 630 = 1050
⇒ 2x = 1050 + 630
⇒ 2x = 1680 ⇒ x = 840
Hence,
L.C.M = 840 and H.C.F = 840 - 630 = 210
So,
product of the number = L.C.M × H.C.F
= 840 × 210 = 176400

2. The highest common factor of a set of coprime numbers is equal to: [Group D 16/09/2022 (Evening)]

Correct Answer: (b) 1
Solution:

Co- prime numbers have always H.C.F 1.

3. The LCM of two numbers is 588. Which of the following CANNOT be their HCF? [Group D 16/09/2022 (Evening)]

Correct Answer: (c) 35
Solution:

588 = 2 × 2 × 3 × 7 × 7
Now, by options 35 cannot be the H.C.F of 588 because 5 is not a factor of 588.

4. Find the smallest natural number that can be divided by the first five 2-digit positive even numbers without leaving a remainder. [Group D 17/09/2022 (Morning)]

Correct Answer: (c) 5040
Solution:

First five 2 digit positive even number
= 10, 12, 14, 16, 18
LCM of 10, 12, 14, 16, 18 = 5040
So, smallest natural number = 5040

5. The traffic signals at three different road crossings change after every 45 seconds, 75 seconds and 90 seconds respectively. If they all change simultaneously at 9:30:00 a.m., then at what time will they next change again simultaneously? [Group D 17/09/2022 (Afternoon)]

Correct Answer: (d) 9 : 37 : 30 a.m.
Solution:

LCM of 45, 75 and 90 = 450
450 second = 7.5 min
So, next change = 9:30 + 7.5 min
= 9 : 37 : 30 am

6. Three numbers are co-prime to one another and the product of the first two and the last two are 432 and 945, respectively. Find the LCM of the three numbers. [Group D 17/09/2022 (Evening)]

Correct Answer: (b) 15120
Solution:

Let the number are x, y and z
According to the question,
xy = 432 ,yz = 945
Middle Number
= HCF of 432 and 945 = 27
First number = 16
Second number = 35
So, LCM of 27, 16 and 35 = 15120

7. The greatest number, which divides 2000 and 2200 to leave 22 and 38 respectively, as remainders, is: [Group D 18/09/2022 (Afternoon)]

Correct Answer: (b) 42 
Solution:

According to question,
To get remainder of 22 and 38
= 2000 - 22 = 1978 and 2200 - 38 = 2162
H.C.F of 1978 and 2162
=2162 - 1978 = 184, Therefore,
highest common factor of 1978, 2162 and 184 = 46

8. The greatest number which on dividing 50, 58 and 69 leaves remainders of 1, 2 and 6, respectively, is: [Group D 19/09/2022 (Morning)]

Correct Answer: (a) 7
Solution:

50 - 1 = 49,  58 - 2 = 56 and 69 - 6 = 63
H.C.F of 49 , 56 and 63 = 7
Hence, 7 is the required number.

9. Find the scale with the greatest possible length which can measure poles of length 3 m 96 cm, 5 m 28 cm, and 7 m 92 cm exact number of times. [Group D 19/09/2022 (Afternoon)]

Correct Answer: (d) 1 m 32 cm
Solution:

3m 96cm , 5m 28cm and 7m 92cm
or 396cm , 528cm and 792cm
H.C.F of 396 , 528 and 792
396 = 2 × 2 × 3 × 3 × 11
528 = 2 × 2 × 2 × 2 × 3 × 11
792 = 2 × 2 × 2 × 3 × 3 × 11
H.C.F = 2 × 2 × 3 × 11 = 132
Hence, the required greatest possible scale length = 1m 32cm

10. The HCF and the LCM of two numbers are 15 and 300, respectively. If the difference of the two numbers is 15, find the sum of those two numbers. [Group D 19/09/2022 (Afternoon)]

Correct Answer: (c) 135
Solution:

Let the two number be  x and y.
Given, H.C.F = 15 and L.C.M = 300
15x - 15y = 15
x - y = ....... (1)
Product  of the number = L.C.M × H.C.F
= 15 × 300 = 4500
(15x) × (15y) = 4500 ⇒ xy = 20...... (2)
By solving eq. (1) and (2) we get,
x = 5 and y = 4
Number = 75 and 60
Hence, the sum of the numbers
= 75 + 60 = 135