Question

If the ratio of speeds of a and b is 5:4, then what start can a give to b so that in a 1km race, b must cover 10% less distance than a?​

Answer 1

Given : ratio of speeds of a and b is 5:4,

To find : what start can a give to b so that in a 1km race, b must cover 10% less distance than a

Solution:

Speed of A  =  5S  km/hr

Speed of B = 4S km/hr

A will cover 1 km

b must cover 10% less distance than a    => 1 - (10/100)1  = 0.9 km

Let say  x  km (0 < x < 1) is the start given to B

Minimum Distance remained for  B = 0.9 - x   km  

Time taken by  B = (0.9 - x)/4S  hr

Time taken by  A  = 1/5S  hr

1/5S = (0.9 - x)/4S

=> 4 = 4.5 - 5x

=>  5x = 0.5

=>x = 0.1  km

=> x = 100  m

At least 100 m start to be given to b

He will cover 100 m + 800 m = 900m

while a will cover 1000m  ( 1 km)

0.1 km = 100 m start to be given so that b must cover 10% less distance than a

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Answer 2

Answer:

Ratio of speeds of a and b = 5: 4

Let,

Speed of a = 5 km/h

Speed of b = 4 km/h

A covers 1 km, then b must cover 10% less than distance of a.

$= > 1 - ( \frac{10}{100} )1 = 0.9km$

Let x km is the start given to b.

Distance remained for b = 0.9-x km

Time taken (b) = (0.9-x) / 4s h

Time taken (a) = 1/5s h

Now,

$\frac{1}{5s} = \frac{(0.9 - x)}{4s} \\ => 4 = 4.5 - 5x \\ = > 5x = 0.5 \\ = > x = 0.1km \\ = > x = 100km$

100 m to b, will cover 100m+800m= 900m

a will cover 1000m= 1km.