Question

two numbers are such that the ratio between them 3:5 if each
is increased by 10 the ratio between the new number so from the is 5:7 find the original number

Answer 1

Solutions :-


Given :

Initial ratio of two numbers = 3 : 5
Ratio increased by 10
Final ratio of two numbers = 5 : 7


Let the the two numbers be 3x and 5x respectively.


A/q

$= > \frac{3x + 10}{5x + 10} = \frac{5}{7} \\ \\ = > 7(3x + 10) = 5(5x + 10) \\ \\ = > 21x + 70 = 25x + 50 \\ \\ = > 21x - 25x = 50 - 70 \\ \\ = > - 4x = - 20 \\ \\ = > x = \frac{ - 20}{ - 4} = 5$

Therefore,
First number = 3x = 3 × 5 = 15
Second number = 5x = 5 × 5 = 25



Hence,
The original number are 15 and 25

Answer 2

let the required no be 3x
then, the other no be 5x
a/q 3x+10/5x+10=5/7
7(3x+10)=5(5x+10)
21x+70=25x+50
21x-25x=50-70
-4x=-20
X=20÷4
X=5
then the no is 3×5=15
5×5=25