Math, asked by AnyDhillon, 7 months ago

The ratio of milk and water in a vessel is 5:3.How much part of a mixture is replaced by 14 1/2% part of water so that the ratio of new mixture becomes 3 : 5?

Answers

Answered by rkcomp31
0

Answer:

Thus 6.348 x or @ 81 % of solution to be replaced with14 1/2 % water

Step-by-step explanation:

Assume milk=5x litres and water is 3x litres  in this mixture-1

Total mixture=5x+3x=8x

Now let  y litre  mixture is removed and y*(14 1/2) =y*29/(2*100)

=29y/200 litre water is added

Now In the Resulting miture

Milk=(8x-y)*5/8=(40x-5y)/8........................(1)

and water= (8x-y)*3/8+29y/200

Water=1/200{ 75(8x-y) +29y}

Water=( 600x-46y)/20 litre............................(2)

As per given new milk : water=3:5

(40x-5y)/8 : (600x-46y)20 :: 3:5

5/8*(40x-5y)=3/20(600x-46y)

(200x-25y)/8=(1800x-138y)/20

4000x-500y=14400x-1104y

10400x=1604y

y=10400x/1604=6.48 x

Thus 6.48 x of 8 x

=6.48*100/8=81 %

Thus 6.348 x or @ 81 % of solution to be replaced with14 1/2 % water

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