0.002 x 0.5 = ?
Question:
0.002 x 0.5 = ?
[A].
0.0001
0.001
0.01
0.1
0.002 x 0.5 = ?
[A].
[B].
[C].
[D].
Answer: Option B
Explanation:
2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
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The certain worth of a certain sum due sometime hence is Rs. 1600 and the true discount is Rs. 160. The banker’s gain is:
Question:
The certain worth of a certain sum due sometime hence is Rs. 1600 and the true discount is Rs. 160. The banker’s gain is:
[A].Rs. 20 [B].
Rs. 24 [C].
Rs. 16 [D].
Rs. 12 Answer: Option C
Explanation:
B.G. = (T.D.)2 = Rs. 160 x 160 = Rs. 16. P.W. 1600 3, 5, 11, 14, 17, 21
Question: 3, 5, 11, 14, 17, 21
[A].21 [B].
17 [C].
14 [D].
3 Answer: Option C
Explanation:
Each of the numbers except 14 is an odd number.
The number ’14’ is the only EVEN number.
If the rate of interest be 4% per annum for first year 5% per annum for the second year and 6% per annum from the third year then the compound interest of Rs 10000 for 3 years will be_________?
If the rate of interest be 4% per annum for first year 5% per annum for the second year and 6% per annum from the third year then the compound interest of Rs 10000 for 3 years will be_________?A. Rs 1575.20
B. Rs 1600
C. Rs 1625.80
D. Rs 2000
Explanation:
Amount = Rs[10000 × (1 + 4/100) × (1 +5/100) × (1 × 6/100)]
= Rs(1000 × 26/25 × 21/20 × 53/50)
= Rs (57876/5) = Rs 11575.20
C.I = Rs (11575.20 – 10000) = Rs 1575.20The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:
Question:
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:
[A].Rs. 400 [B].
Rs. 500 [C].
Rs. 600 [D].
Rs. 800 Answer: Option B
Explanation:
Let the sum be Rs. P.
Then, P 1 + 10 2 – P = 525 100 P 11 2 – 1 = 525 10 P = 525 x 100 = 2500. 21 Sum = Rs . 2500.
So, S.I. = Rs. 2500 x 5 x 4 = Rs. 500 100 22, 33, 66, 99, 121, 279, 594
Question:
22, 33, 66, 99, 121, 279, 594
[A].33 [B].
121 [C].
279 [D].
594 Answer: Option C
Explanation:
Each of the number except 279 is a multiple of 11.
A person can swim in still water at 4 km/h. If the speed of water 2 km/h, how many hours will the man take to swim back against the current for 6km?
A person can swim in still water at 4 km/h. If the speed of water 2 km/h, how many hours will the man take to swim back against the current for 6km?A. 3
B. 4
C. 4 (1/2)
D. Insufficient data
Explanation:
M = 4
S = 2
US = 4 – 2 = 2
D = 6
T = 6/2 = 3