9) Two numbers are such that the ratio
between them is 3:5 if each is increased
by 10 the ratio between the new numbers
So formed is 5:7 find the original
numbers.
Answers
let them be x and y respectively.
then
x/y = 3/5
5x = 3y .............................................[1]
after adding 10
(x+10)/(y+10) = [5/7]
7x+70 = 5y+50
7x +20 = 5y......................................[2]
from eq 1 we get
y = 5x/3
putting this value in 2
we get..
7x -(5x/3)+20 = 0
(21x-5x)/3 +20 = 0
16x/3 +20 = 0
(16x+60)/3 = 0
16x = -60
x = -60/16 = -15/4
y = - 5(15/4)/3 = -75/12 = -25/4
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Answer:
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→ 15 and 25 .
Step-by-step explanation:
Let x and y be the two numbers .
Now,
CASE 1 .
→ Two numbers are such that the ratio between them is 3 : 5.
A/Q,
∵ x : y = 3 : 5
⇒ 5x = 3y .
∵ x = 3y / 5 ........( 1 ).
CASE 2 .
→ If each number in increased by 10, the ratio between the new number so formed is 5 : 7.
A/Q,
∵ ( x + 10 ) : ( y + 10 ) = 5 : 7 .
⇒ 7( x + 10 ) = ( y + 10 ) 5 .
⇒ 7x + 70 = 5y + 50 .
⇒ 7x + 70 - 50 = 5y .
⇒ 7x + 20 = 5y. ........( 2 ).
Put value of 'x' from equation ( 1 ) in ( 2 ) .
⇒ 7× 3y/5 + 20 = 5y .
⇒ ( 21y + 100 ) / 5 = 5y .
⇒ 21y + 100 = 25y .
⇒ 100 = 25y - 21y .
⇒ 100 = 4y .
⇒ 100 / 4 = y .
∴ y = 25 .
Therefore ,
∵ y = 25 ,
Put y = 25 in equation ( 1 ), we get
⇒ x = 3 × 25 / 5
⇒ x = 3 × 5
∴ x = 15
Original numbers are x and y = 15 and 25 .
Hence, it is solved .
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