1, 4, 9, 16, 23, 25, 36
Question:
1, 4, 9, 16, 23, 25, 36
[A].
9
23
25
36
1, 4, 9, 16, 23, 25, 36
[A].
[B].
[C].
[D].
Answer: Option B
Explanation:
Each of the numbers except 23, is perfect square.
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(256)0.16 x (256)0.09 = ?
Question:
(256)0.16 x (256)0.09 = ?
[A].4 [B].
16 [C].
64 [D].
256.25 Answer: Option A
Explanation:
(256)0.16 x (256)0.09 = (256)(0.16 + 0.09)
= (256)0.25
= (256)(25/100)
= (256)(1/4)
= (44)(1/4)
= 44(1/4)
= 41
= 4
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Question: A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
[A].Rs. 650 [B].
Rs. 690 [C].
Rs. 698 [D].
Rs. 700 Answer: Option C
Explanation:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?A. Rs.10,840
B. Rs.10,720
C. Rs.10,560
D. Rs.10,280
E. None of these.
Explanation:
Let the rate of interest be R% p.a.
4400{[1 + R/100]2 – 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred’s and ten’s places are respectively:
Question:
476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred’s and ten’s places are respectively:
[A].7 and 4 [B].
7 and 5 [C].
8 and 5 [D].
None of these Answer: Option C
Explanation:
Let the given number be 476 xy 0.
Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.
And, (0 + x + 7) – (y + 6 + 4) = (x – y -3) must be either 0 or 11.
x – y – 3 = 0 y = x – 3
(17 + x + y) = (17 + x + x – 3) = (2x + 14)
x= 2 or x = 8.
x = 8 and y = 5.
A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
Question: A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
[A].Rs. 4991 [B].
Rs. 5991 [C].
Rs. 6001 [D].
Rs. 6991 Answer: Option A
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) – 34009 ]
= Rs. (39000 – 34009)
= Rs. 4991.
Video Explanation: https://youtu.be/tP4hszReksU
1, 8, 27, 64, 125, 216, (….)
Question:
1, 8, 27, 64, 125, 216, (….)
[A].354 [B].
343 [C].
392 [D].
245 Answer: Option B
Explanation:
Numbers are 13, 23, 33, 43, 53, 63.
So, the missing number is 73 = 343.