Dear Reader, Below are three problems where you will have to apply formulas to arrive at correct answers. For your benefit, we have derived some important formulas. However, in exams, you can remember these formulas to save plenty of time.
Sri Ramanujam arts and science college is situated on the banks of river Cauvery in the famous town of Kumbakonam. During Audi festival, boat rowing competition takes place and youngsters participate in large numbers. In one such occasion, Syed Thamim, rows 28 km with the stream and 16 km against the stream taking 4 hours each time. Can you find the speed in kmph of Syed in still water and the speed of river Cauvery in kmph, at that point of time ?
(a) 6kmph and 2 kmph (b) 6.5kmph and 1.5 kmph (c) 5.5 kmph and 1.5 kmph (d) 7 kmph and 2 kmph
Answer : c) 5.5 kmph and 1.5 kmph
If downstream speed is Sdown and upstream speed is Sup
Then, Speed of Syed in still water can be found using the formula (Sdown + Sup) / 2 ...(1)
Similarly, Speed of Cauvery can be found using (Sdown - Sup) / 2 ...(2)
Now lets solve the problem using data given in question ;
Sdown = Distance Covered when travelling with the stream / Time taken = 28 / 4 =7 kmph
Sup = Distance Covered when travelling against the stream / Time taken =16 / 4 =4 kmph
Using formula 1, Speed of Syed =1/2 (7 + 4) = 5.5 kmph
Using formula 2, Speed of Cauvery = ½ (7 - 4) = 1.5 kmph
So option (c) is correct.
Sneha Rajput is a good swimmer. She has the ability to swim long distances due to constant practice. On Sundays, she swims to a nearby tiny island to pray Ganesha and return. In one such occasion, she swims the distance downstream in 2 hours and returns upstream in 5 hours. If the stream flows at the rate of 3 km per hour find the speed of Sneha, in still water.
(a) 8 kmph (b) 6kmph (c) 7.5 kmph (d) 7 kmph
Answer : d) 7 kmph
Let the distance of island from shore be d Km.
Speed of Sneha in still water = Sdown + Sup / 2 ...(1)
Speed of Stream = Sdown - Sup / 2 ...(2)
Note: Sdown refers to downstream speed while Sup represents upstream speed.
Dividing eq 1 by eq 2 we get
Speed of Sneha / Speed of Stream = (Sdown + Sup) / (Sdown - Sup)
Speed of Sneha = Speed of Stream (Sdown + Sup) / (Sdown - Sup) ...(3)
Note : Remember the above formula so that you can solve the questions of current type easily. This formula can be an effective shortcut that can save valuable time.
From data given in question the values are as follows
Speed of Stream = 3 Kmph
Sdown = d/2 kmph
Sup = d/5 kmph
Substituting the above values in eq 3 we get
Speed of Sneha = 3 (d/2 + d/5)/(d/2 - d/5)
= 3 (1/2 + 1/5)/ (1/2 - 1/5)
= 3 (7/10)/(3/10)
= 7 kmph
Therefore speed of Sneha in still water = 7 kmph
Kashinath, can row 5 kmph in still water.When the rate of flow of the river is 2.0 kmph , it takes Kashinath 1 hour to row to a ghat and return. How far is Kashinath’s starting place from the ghat ?
(a) 2 km (b) 2.1 km (c)2.5km (d) 2.8 km
Answer : b) 2.1 km
Let d be the distance of ghat from the point where Kashinath starts to swim.
Let Sk be the speed of Kashinath in still water
Let Sr be the speed of stream (river)
Then, Sdown = Sk + Sr ...(1)
And Sup = Sk - Sr ...(2)
Time taken to travel downstream to ghat = d / Sdown
Time taken to travel upstream from ghat = d / Sup
Total time taken, T = d / Sdown + d / Sup
Substituting values from eq1 and eq2 in the above equation we get,
Or, T = d/(Sk + Sr) + d/(Sk - Sr)
T = d(Sk - Sr) + d(Sk + Sr) / (Sk + Sr)(Sk - Sr)
T = 2dSk / (Sk2-Sr2)
Or d = T (Sk2-Sr2) / 2Sk ...(3)
Note : Remember the above formula as this can prove to be a shortcut.
Substitute T = 1 hour, Sk = 5 Kmph and Sr = 2 Kmph in equation 3 we get
d = 1(52 - 22) / 2x5 = 21 / 10=2.1 km