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C
1) Choose appropriate antonym for the given word: TRANQUILITY
No Explanation found. Add Explanation and get +2 points.
S
2) What will come at the place of question mark ? 6, 13, 28, 59, ?, 249.
Explanation :
First term → 6
Second term → (6*2+1) = 13
Third term → (13*2+2) = 28
Fourth term → (28*2+3) = 59
Fifth term → (59*2+4) = 122
Sixth term → (122*2+5) = 249
So, the required term = 122.
S
3) In the following number series a wrong number is given. Find out the wrong number. 1, 3, 10, 21, 64, 129, 356, 777
Explanation :
The given pattern is,
2nd term → 1*2+1 = 3
3rd term → 3*3+1 = 10
4th term → 10*2+1 = 21
5th term → 21*3+1 = 64
6th term → 64*2+1 = 129
7th term → 129*3+1= 388
So, 7th term 356 is wrong and must be replaced by 388.
S
4) What will come at the place of question mark ? 9, 27, 31, 155, 161, 1127, ?
Explanation :
Given pattern is,
2nd term → 9*3 = 27
3rd term → 27+4 = 31
4th term → 31*5 = 155
5th term → 155+6 = 161.
6th term → 161*7 = 1127
Missing term → 1127+8 = 1135.
R
5) 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:
Explanation :
The HCF of a group of numbers will be always a factor of their LCM.
HCF is the product of all common prime factors using the least power of each common prime factor.
LCM is the product of highest powers of all prime factors.
Clearly, the numbers are (23 x 13) and (23 x 14)
∴ Larger number = (23 x 14) = 322
R
6) 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 ?
Explanation :
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
R
7) 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:
Explanation :
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
R
8) Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:
Explanation :
Let the numbers be 3x, 4x and 5x
Then, their L.C.M. = 60x
So, 60x = 2400 or x = 40
so The numbers are (3 x 40), (4 x 40) and (5 x 40)
Hence, required H.C.F. = 40
R
9) The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
Explanation :
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)
[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 x 1, 13 x 12) and (13 x 3, 13 x 4)
Clearly, there are 2 such pairs.
R
10) The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Explanation :
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
So Required number = (90 x 4) + 4 = 364
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