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All questions of HCF & LCM for CDS Exam

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
  • a)
    196
  • b)
    630
  • c)
    1260
  • d)
    2520
Correct answer is option 'B'. Can you explain this answer?

Aditya Kumar answered
L.C.M. of 12, 18, 21 30                 2 | 12  -  18  -  21  -  30
                                         ----------------------------
= 2 x 3 x 2 x 3 x 7 x 5 = 1260.       3 |  6  -   9  -  21  -  15
                                         ----------------------------
Required number = (1260 � 2)            |  2  -   3  -   7  -   5

= 630.

The Greatest Common Divisor of 1.08, 0.36 and 0.9 is:
  • a)
    0.03
  • b)
    0.9
  • c)
    0.18
  • d)
    0.108
Correct answer is option 'C'. Can you explain this answer?

Faizan Khan answered
Given numbers are 1.08 , 0.36 and 0.90
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18            

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
Correct answer is option 'D'. Can you explain this answer?

LCM of 2, 4, 6, 8 10 and 12 is 120.

So, after each 120 seconds, they would toll together.

Hence, in 30 minutes, they would toll 30*60 seconds / 120 seconds = 15 times

But then the question says they commence tolling together. So, they basically also toll at the "beginning" ("0" second).

So, total tolls together = 15+1 = 16

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
  • a)
    1677
  • b)
    1683
  • c)
    2523
  • d)
    3363
Correct answer is option 'B'. Can you explain this answer?

Manoj Ghosh answered
LCM of 5, 6, 7 and 8 = 840.
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 x 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 x 2) + 3 = 1683 which is divisible by 9.
Hence, 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
  • a)
    74
  • b)
    94
  • c)
    184
  • d)
    364
Correct answer is option 'D'. Can you explain this answer?

Ishani Rane answered
7x = 6a+4 = 9b+4 = 15c+4 = 18d+4

7x - 4 = 6a = 9b = 15c = 18d

LCM(6,9,15,18) = 90

7x - 4 = 90y

7x = 90y + 4 = 84y + 6y + 4

7x’ = 6y+4

6x’ + x’ = 6y+4

x’ = 6y’ + 4

y’ = 0 → x’ = 4 → y = 4 → x = 364/7 = 52

Ans: 52*7 = 364

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
  • a)
    9000
  • b)
    9400
  • c)
    9600
  • d)
    9800
Correct answer is option 'C'. Can you explain this answer?

Sagar Sharma answered
To find the greatest number of four digits that is divisible by 15, 25, 40, and 75, we need to find the least common multiple (LCM) of these numbers.

Finding the LCM:
1. Prime factorize each number:
- 15 = 3 × 5
- 25 = 5 × 5
- 40 = 2 × 2 × 2 × 5
- 75 = 3 × 5 × 5

2. Identify the highest power of each prime factor:
- 3 occurs once
- 5 occurs twice
- 2 occurs three times

3. Multiply these prime factors together:
LCM = 3 × 5 × 5 × 2 × 2 × 2 = 600

Finding the greatest number of four digits divisible by 600:
To find the greatest number of four digits divisible by 600, we need to find the largest multiple of 600 that is less than 10,000.

Divide 10,000 by 600:
10,000 ÷ 600 = 16 remainder 400

Subtract the remainder from 10,000:
10,000 - 400 = 9,600

Therefore, the greatest number of four digits divisible by 15, 25, 40, and 75 is 9600 (option C).

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
  • a)
    3
  • b)
    13
  • c)
    23
  • d)
    33
Correct answer is option 'C'. Can you explain this answer?

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

Therefore, L.C.M. of 5, 6, 4 and 3 = 60. Number to be added = (60 - 37) = 23.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
  • a)
    26 minutes and 18 seconds
  • b)
    42 minutes and 36 seconds
  • c)
    45 minutes
  • d)
    46 minutes and 12 seconds
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Given:
- A completes a round in 252 seconds
- B completes a round in 308 seconds
- C completes a round in 198 seconds

To find:
- After what time will they again at the starting point?

Solution:
To find the time at which they will again meet at the starting point, we need to find the least common multiple (LCM) of the given times taken by A, B, and C to complete one round.

Step 1: Prime factorization of the given times:
- 252 = 2 * 2 * 3 * 3 * 7
- 308 = 2 * 2 * 7 * 11
- 198 = 2 * 3 * 3 * 11

Step 2: Finding the LCM:
The LCM is obtained by taking the highest powers of all prime factors involved.
- LCM = 2 * 2 * 3 * 3 * 7 * 11 = 2772 seconds

Step 3: Converting the LCM to minutes and seconds:
- 2772 seconds = 2772 / 60 minutes (quotient) + 2772 % 60 seconds (remainder)
- 2772 seconds = 46 minutes + 12 seconds

Therefore, they will meet again at the starting point after 46 minutes and 12 seconds. Hence, the correct answer is option D.

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
  • a)
    40
  • b)
    80
  • c)
    120
  • d)
    200
Correct answer is option 'A'. Can you explain this answer?

Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'B'. Can you explain this answer?

Aarav Sharma answered
Solution:

Given, the product of two numbers is 2028 and their H.C.F. is 13.
Let the two numbers be 13a and 13b (where a and b are co-primes)
Therefore, 13a × 13b = 2028
=> ab = 12
So, the possible pairs of (a, b) are (1, 12) and (3, 4)
Hence, the possible pairs of numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
Therefore, there are two pairs of numbers whose product is 2028 and H.C.F. is 13.

Therefore, option 'B' is the correct answer.

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
  • a)
    101
  • b)
    107
  • c)
    111
  • d)
    185
Correct answer is option 'C'. Can you explain this answer?

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

 ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

 Greater number = 111.

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:
  • a)
    12
  • b)
    16
  • c)
    24
  • d)
    48
Correct answer is option 'D'. Can you explain this answer?

Dhruv Mehra answered
Let the numbers be  3x and 4x . Then their H.C.F = x. So, x=4

 Therefore, The numbers are 12 and 16

 L.C.M of 12 and 16 = 48

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