Speed upstream = (5 - 1) kmph = 4 kmph.
Let the required distance be x km.
Then, | x | + | x | = 1 |
6 | 4 |
⟹ 2x + 3x = 12
⟹ 5x = 12
⟹ x = 2.4 km.
Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.
and 11¼ minutes as 675 seconds.
Rate upstream = | ❨ | 750 | ❩m/sec | = | 10 | m/sec. |
675 | 9 |
Rate downstream = | ❨ | 750 | ❩m/sec | = | 5 | m/sec. |
450 | 3 |
∴ Rate in still water = | 1 | ❨ | 10 | + | 5 | ❩m/sec |
2 | 9 | 3 |
= | 25 | m/sec |
18 |
= | ❨ | 25 | x | 18 | ❩km/hr |
18 | 5 |
= 5 km/hr.
Less persons, More days (Indirect Proportion)
More working hours per day, Less days (Indirect Proportion)
Persons | 30 | : | 39 | ] | :: 12 : x |
Working hours/day | 6 | : | 5 |
∴ 30 x 6 x x = 39 x 5 x 12
⟹ x = | (39 x 5 x 12) |
(30 x 6) |
⟹ x = 13.
Work done = | 5 |
8 |
Balance work = | ❨ | 1 - | 5 | ❩ | = | 3 |
8 | 8 |
Let the required number of days be x.
Then, | 5 | : | 3 | = :: 10 : x | ⟺ | 5 | x x = | 3 | x 10 |
8 | 8 | 8 | 8 |
⟹ x = | ❨ | 3 | x 10 x | 8 | ❩ |
8 | 5 |
⟹ x = 6.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
∴ 125 : 150 :: 35 : x ⟺ 125 x x = 150 x 35
⟹ x = | 150 x 35 |
125 |
⟹ x = 42.
Rate downstream = | ❨ | 16 | ❩kmph = 8 kmph. |
2 |
Rate upstream = | ❨ | 16 | ❩kmph = 4 kmph. |
4 |
∴ Speed in still water = | 1 | (8 + 4) kmph = 6 kmph. |
2 |
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr.
∴ | 30 | + | 30 | = 4 | 1 |
(15 + x) | (15 - x) | 2 |
⟹ | 900 | = | 9 |
225 - x2 | 2 |
⟹ 9x2 = 225
⟹ x2 = 25
⟹ x = 5 km/hr.
Speed downstream = (10 + x) mph,
Speed upstream = (10 - x) mph.
∴ | 36 | - | 36 | = | 90 |
(10 - x) | (10 + x) | 60 |
⟹ 72x x 60 = 90 (100 - x2)
⟹ x2 + 48x - 100 = 0
⟹ (x+ 50)(x - 2) = 0
⟹ x = 2 mph.
Speed downstream = (x + 3) kmph,
Speed upstream = (x - 3) kmph.
∴ (x + 3) x 1 = (x - 3) x | 3 |
2 |
⟹ 2x + 6 = 3x - 9
⟹ x = 15 kmph.
Distance travelled = | ❨ | 18 x | 12 | ❩km = 3.6 km. |
60 |
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