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All questions of HCF & LCM for RRB NTPC/ASM/CA/TA Exam
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "HCF and LCM" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations. Q. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.- a)4
- b)7
- c)9
- d)13
Correct answer is option 'A'. Can you explain this answer?
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "HCF and LCM" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
Q. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
a)
4
b)
7
c)
9
d)
13
|
Pallabi Deshpande answered |
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
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?
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
|
|
Pallabi Deshpande answered |
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 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?
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
|
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
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18
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?
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
|
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 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?
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
|
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
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:- a)4
- b)5
- c)6
- d)8
Correct answer is option 'A'. Can you explain this answer?
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
a)
4
b)
5
c)
6
d)
8
|
Sameer Rane answered |
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
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?
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
|
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 smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:- a)1008
- b)1015
- c)1022
- d)1032
Correct answer is option 'B'. Can you explain this answer?
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
a)
1008
b)
1015
c)
1022
d)
1032
|
|
Pallabi Deshpande answered |
Required Number = (L.C.M of 12, 16, 18,21,28)+7
= 1008 + 7
= 1015
Product of two co-prime numbers is 117. Then their L.C.M. is- a)9
- b)117
- c)39
- d)13
Correct answer is option 'B'. Can you explain this answer?
Product of two co-prime numbers is 117. Then their L.C.M. is
a)
9
b)
117
c)
39
d)
13
|
Arnav Saini answered |
Understanding Co-prime Numbers
Co-prime numbers are two integers that have no common factors other than 1. This means their greatest common divisor (GCD) is 1.
Product of Co-prime Numbers
When two co-prime numbers are multiplied, their product is equal to the product of their LCM and GCD. Since the GCD of co-prime numbers is 1, we can express this relationship as:
- Product = LCM × GCD
- Given: Product = 117
- GCD = 1 (for co-prime numbers)
Calculating LCM
Using the relationship:
- 117 = LCM × 1
- Therefore, LCM = 117
This indicates that the least common multiple (LCM) of the two co-prime numbers is simply their product, which is 117.
Conclusion
Hence, the LCM of two co-prime numbers whose product is 117 is:
- Correct answer: Option B (117)
This conclusion is valid because, for co-prime numbers, the LCM is always equal to the product of the numbers, confirming that the answer is indeed 117.
Co-prime numbers are two integers that have no common factors other than 1. This means their greatest common divisor (GCD) is 1.
Product of Co-prime Numbers
When two co-prime numbers are multiplied, their product is equal to the product of their LCM and GCD. Since the GCD of co-prime numbers is 1, we can express this relationship as:
- Product = LCM × GCD
- Given: Product = 117
- GCD = 1 (for co-prime numbers)
Calculating LCM
Using the relationship:
- 117 = LCM × 1
- Therefore, LCM = 117
This indicates that the least common multiple (LCM) of the two co-prime numbers is simply their product, which is 117.
Conclusion
Hence, the LCM of two co-prime numbers whose product is 117 is:
- Correct answer: Option B (117)
This conclusion is valid because, for co-prime numbers, the LCM is always equal to the product of the numbers, confirming that the answer is indeed 117.
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?
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
|
Gowri Chakraborty answered |
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 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?
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
|
|
Gowri Chakraborty answered |
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.
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?
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
a)
9000
b)
9400
c)
9600
d)
9800
|
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).
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 product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?- a)160
- b)150
- c)120
- d)140
Correct answer is option 'A'. Can you explain this answer?
The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?
a)
160
b)
150
c)
120
d)
140
|
Nilesh Banerjee answered |
Given Information:
- Product of two numbers = 1280
- H.C.F. of the numbers = 8
Calculating the Numbers:
To find the numbers, we first need to express 1280 as the product of the H.C.F. and the L.C.M. of the numbers.
1280 = H.C.F. * L.C.M.
1280 = 8 * L.C.M.
L.C.M. = 1280 / 8
L.C.M. = 160
Therefore, the L.C.M. of the two numbers is 160.
Conclusion:
The L.C.M. of the numbers will be 160, which matches with option A. So, the correct answer is option A) 160.
- Product of two numbers = 1280
- H.C.F. of the numbers = 8
Calculating the Numbers:
To find the numbers, we first need to express 1280 as the product of the H.C.F. and the L.C.M. of the numbers.
1280 = H.C.F. * L.C.M.
1280 = 8 * L.C.M.
L.C.M. = 1280 / 8
L.C.M. = 160
Therefore, the L.C.M. of the two numbers is 160.
Conclusion:
The L.C.M. of the numbers will be 160, which matches with option A. So, the correct answer is option A) 160.
The product of two numbers is 2028 and their HCF is 13. The number of such pairs is- a)2
- b)1
- c)4
- d)3
Correct answer is option 'A'. Can you explain this answer?
The product of two numbers is 2028 and their HCF is 13. The number of such pairs is
a)
2
b)
1
c)
4
d)
3
|
Target Study Academy answered |
Here, HCF = 13
Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028
The possible pairs are : (1, 12), (3, 4), (2, 6)
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2
Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028
The possible pairs are : (1, 12), (3, 4), (2, 6)
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2
The product of two numbers is 2160 and their HCF is 12. Number of such possible pairs is- a)1
- b)2
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
The product of two numbers is 2160 and their HCF is 12. Number of such possible pairs is
a)
1
b)
2
c)
3
d)
4
|
Ssc Cgl answered |
HCF = 12
Numbers = 12x and 12y
where x and y are prime to each other.
∴ 12x × 12y = 2160
= 15 = 3 × 5, 1 × 15
Possible pairs = (36, 60) and (12, 180)
Hence , Number of such possible pairs is 2.
Numbers = 12x and 12y
where x and y are prime to each other.
∴ 12x × 12y = 2160
= 15 = 3 × 5, 1 × 15
Possible pairs = (36, 60) and (12, 180)
Hence , Number of such possible pairs is 2.
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?
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
|
|
Pallabi Deshpande answered |
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?
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
|
Arya Roy answered |
Let the numbers 13a and 13b.
Then, 13a * 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be co prime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 * 1, 13 * 12) and (13 * 3, 13 * 4).
Clearly, there are 2 such pairs.
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?
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
|
Dhruv Mehra answered |
Required Time = LCM of 252, 308 and 198 seconds = 2772 seconds = 46 minutes 12 seconds.
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?
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
|
|
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
The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?- a)12
- b)8
- c)35
- d)24
Correct answer is option 'C'. Can you explain this answer?
The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?
a)
12
b)
8
c)
35
d)
24
|
Glance Learning Institute answered |
LCM = 2 × 2 × 2 × 3 × 5
Hence, HCF = 4, 8, 12 or 24
According to question
35 cannot be H.C.F. of 120.
Hence, HCF = 4, 8, 12 or 24
According to question
35 cannot be H.C.F. of 120.
The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is- a)3
- b)4
- c)1
- d)2
Correct answer is option 'D'. Can you explain this answer?
The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is
a)
3
b)
4
c)
1
d)
2
|
EduRev SSC CGL answered |
Let the number be 15x and 15y, where x and y are co –prime.
∴ 15x × 15y = 6300
So, two pairs are (7, 4) and (14, 2)
∴ 15x × 15y = 6300
So, two pairs are (7, 4) and (14, 2)
Find the lowest common multiple of 24, 36 and 40.- a)120
- b)240
- c)360
- d)480
Correct answer is option 'C'. Can you explain this answer?
Find the lowest common multiple of 24, 36 and 40.
a)
120
b)
240
c)
360
d)
480
|
|
Aarav Sharma answered |
Solution:
To find the lowest common multiple (LCM) of 24, 36, and 40, we need to follow the following steps:
Step 1: Find the prime factorization of each number.
- Prime factorization of 24: 2 × 2 × 2 × 3
- Prime factorization of 36: 2 × 2 × 3 × 3
- Prime factorization of 40: 2 × 2 × 2 × 5
Step 2: Identify the highest power of each prime factor.
- 2: highest power is 3 (from 24 and 40)
- 3: highest power is 2 (from 24 and 36)
- 5: highest power is 1 (from 40)
Step 3: Multiply the highest powers of each prime factor.
- 2³ × 3² × 5¹ = 360
Therefore, the LCM of 24, 36, and 40 is 360, which is option C.
To find the lowest common multiple (LCM) of 24, 36, and 40, we need to follow the following steps:
Step 1: Find the prime factorization of each number.
- Prime factorization of 24: 2 × 2 × 2 × 3
- Prime factorization of 36: 2 × 2 × 3 × 3
- Prime factorization of 40: 2 × 2 × 2 × 5
Step 2: Identify the highest power of each prime factor.
- 2: highest power is 3 (from 24 and 40)
- 3: highest power is 2 (from 24 and 36)
- 5: highest power is 1 (from 40)
Step 3: Multiply the highest powers of each prime factor.
- 2³ × 3² × 5¹ = 360
Therefore, the LCM of 24, 36, and 40 is 360, which is option C.
The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?- a)8
- b)12
- c)24
- d)35
Correct answer is option 'D'. Can you explain this answer?
The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?
a)
8
b)
12
c)
24
d)
35
|
|
Ssc Cgl answered |
We know that:
LCM is the least common multiple of the given numbers whereas HCF is the highest common factor of those numbers.
Then, LCM is the multiplication of one common factor of the numbers and the other different factors of the numbers.
Write the LCM = 120 into factored form, that is
120 = 2 × 2 × 2 × 3 × 5
= 4(2 × 3 × 5)
⇒ 4 is the common factor of the numbers.
So, the HCF of three numbers is a multiple of 4.
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...
Therefore, 35 is not the multiple of 4, then 35 cannot be their HCF.
LCM is the least common multiple of the given numbers whereas HCF is the highest common factor of those numbers.
Then, LCM is the multiplication of one common factor of the numbers and the other different factors of the numbers.
Write the LCM = 120 into factored form, that is
120 = 2 × 2 × 2 × 3 × 5
= 4(2 × 3 × 5)
⇒ 4 is the common factor of the numbers.
So, the HCF of three numbers is a multiple of 4.
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...
Therefore, 35 is not the multiple of 4, then 35 cannot be their HCF.
The HCF and LCM of two numbers are 13 and 455 respectively. If one of the number lies between 75 and 125, then, that number is :- a)91
- b)78
- c)117
- d)104
Correct answer is option 'A'. Can you explain this answer?
The HCF and LCM of two numbers are 13 and 455 respectively. If one of the number lies between 75 and 125, then, that number is :
a)
91
b)
78
c)
117
d)
104
|
Bayshore Academy answered |
HCF = 13
Let the numbers be 13x and 13y.
Where x and y are co-prime.
∴ LCM = 13 xy
∴ 13 xy = 455
∴ Numbers are 13 × 5 = 65 and 13 × 7 = 91
Let the numbers be 13x and 13y.
Where x and y are co-prime.
∴ LCM = 13 xy
∴ 13 xy = 455
∴ Numbers are 13 × 5 = 65 and 13 × 7 = 91
The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of the two numbers is :- a)276
- b)299
- c)345
- d)322
Correct answer is option 'D'. Can you explain this answer?
The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of the two numbers is :
a)
276
b)
299
c)
345
d)
322
|
|
Ssc Cgl answered |
HCF is 23. So the other two numbers would be,
(23 * 13) and (23 * 14).
Thus Larger Number = 23 * 14 = 322.
(23 * 13) and (23 * 14).
Thus Larger Number = 23 * 14 = 322.
Chapter doubts & questions for HCF & LCM - Mathematics for RRB NTPC / ASM 2025 is part of RRB NTPC/ASM/CA/TA exam preparation. The chapters have been prepared according to the RRB NTPC/ASM/CA/TA exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for RRB NTPC/ASM/CA/TA 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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