Math, asked by sayonapurohit, 9 months ago

Sahir runs in a triathlon consisting of three phases in the
following manner. Running 12 km, cycling 24 km and
swimming 5 km. His speeds in the three phases are in the
ratio 2:6:1. He completes the race in n minutes. Later, he
changes his strategy so that the distances he covers in each
phase are constant but his speeds are now in the ratio 3:8:1.
The end result is that he completes the race taking 20 minutes
more than the earlier speed. It is also known that he has not
changed his running speed when he changes his strategy.
17. What is his initial speed while swimming?
(a) 1/2 km/min (b) 0.05 km/min
(C) 0.15 km/min (d) None of these
18. If his speeds are in the ratio 1:3:1, with the run-
ning time remaining unchanged, what is his finishing
time?
(a) 500/3 min
(b) 250/3 min
(c) 200/3 min
(d) 350/3 min
19. What is Sahir's original speed of running?
(a) 9 kmph
(b) 18 kmph
(c) 54 kmph
(d) 12 kmph

Answers

Answered by sahitibyna
4

Step-by-step explanation:

3x is for swimming speed and 6x is for running speed

Attachments:
Answered by Tulsi4890
5

Given:

The distance to be covered via running = 12km

The distance to be covered via cycling = 24km

The distance to be covered via swimming = 5km

The ratio of speeds initially= 2 : 6: 1

The new ratio= 3 : 8 : 1

To Find:

Initial swimming speed

The finishing time if the ratio is 1 : 3 : 1

Initial running speed

Solution:

Let x be the ratio multiplier in the 1st case and y  in the 2nd case.

Hence, in 1st case: Running speed = 2x, Cycling speed = 6x, and Swimming speed = x

Similarly, in the 2nd case: Running speed = 3y, Cycling speed = 8y, and Swimming speed = y

Given that, the running speed is the same.

⇒ 2x = 3y           - (1)

We know that Time = Distance/Speed

Therefore,

Case 1:

n/60 = 12/(2x) + 24/(6x) + 5/x      

=> n/60 = 6/x + 4/x + 5/x

=> n/60 = 15/x

=> n = 900/x            -(2)

Case 2:

(n+20)/60 = 12/(3y) + 24/(8y) + 5/y

=> (n+20)/60 = 4/y + 3/y + 5/y

=> (n+20)/60 = 12/y

=> n+20 = 720/y            -(3)

Substituting y=2x/3 from eq.(1) in eq.(3)

n+20 = 1080/x

or 20 = 1080/x - 900/x

or 20 = 180/x

=> x = 9

=> y = 6

If the speed ratio is 1:3:1, time taken will be: 12/18 + 24/ 54 + 5/18

= 12 + 8 + 5 / 18 hr

= 25 X 60 / 18 = 250 / 3 mins

Hence, the initial speed while swimming = x = 9km/hr = (C) 0.15 km/min

The finishing time = (b) 250 / 3 mins

The original speed of running = 2x = 2 X 9 = (b) 18 kmph

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