1. 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
2. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
A. 276
B. 299
C. 322
D. 345
3. 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
4. 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
5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A. 9000
B. 9400
C. 9600
D. 9800
6. 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
7. 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
8. The G.C.D. of 1.08, 0.36 and 0.9 is:
A. 0.03
B. 0.9
C. 0.18
D. 0.108
9. 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
10. 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
Answers With Explanation
1. (A)
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
2. (C)
Clearly, the numbers are (23 x 13) and (23 x 14)
Larger number = (23 x 14) = 322.
3. (D)
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds (2 minutes).
In 30 minutes, they will toll together (30/2) + 1 = 16 times.
4. (A)
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
5. (C)
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
6. (C)
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.
7. (A)
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.
8. (C)
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
9. (B)
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
10. (D)
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
You might like this:
No comments:
Post a Comment