A. 2km/hr
B. 3km/hr
C. 4 km/hr
D. 5 km/hr
Explanation:
Distance covered upstream in 45/4min =3/4 km
Distance covered upstream in 1 hr =(3/4 × 2/15 × 60) km/hr = 4 km/hr
Distance covered downstream in 15/2 min = ¾ km
Distance covered downstream in 1hr
= (3/4 × 2/15 × 60)km/hr = 6 km/hr
Speed of the man in still water
= ½ (6 + 4)km/hr = 5 km/hr
A. 2
B. 3
C. 4
D. 5
A. Rs.1261
B. Rs.9261
C. Rs.9621
D. Rs.9162
Explanation:
A = 8000(21/20)3 = 9261
[B].
[C].
[D].
Answer: Option B
Explanation:
| Volume of water displaced | = (3 x 2 x 0.01) m3 |
| = 0.06 m3. |
| Mass of man | = Volume of water displaced x Density of water |
| = (0.06 x 1000) kg | |
| = 60 kg. |
[B].
[C].
[D].
Answer: Option A
Explanation:
Let the sum be Rs. x. Then,
| C.I. = | x | 1 + | 4 | 2 | – x | = | 676 | x | – x | = | 51 | x. | ||||||
| 100 | 625 | 625 |
| S.I. = | x x 4 x 2 | = | 2x | . | ||
| 100 | 25 |
| 51x | – | 2x | = 1 | |
| 625 | 25 |
x = 625.
[B].
[C].
[D].
Answer: Option B
Explanation:
Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x.
| Then, | (52)7.5 x (5)2.5 | = 5x |
| (53)1.5 |
| 5(2 x 7.5) x 52.5 | = 5x | |
| 5(3 x 1.5) |
| 515 x 52.5 | = 5x | |
| 54.5 |
5x = 5(15 + 2.5 – 4.5)
5x = 513
x = 13.