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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.
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
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
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 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
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
|
|
Ishani Rane answered |
Greatest number of 4−digits is 9999.
Now, 15=3×5
25 = 5×5
40 = 2×2×2×5
and 75 = 3×5×5
L.C.M. of 15,25,40 and 75 is 2×2×2×3×5×5 = 600.
On dividing 9999 by 600, the remainder is 399.
Required number = (9999−399) = 9600.
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
|
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.
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.
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 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.
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