Question

Two numbers are in ratio 7:11 . If 7 is added to each of the numbers, the ratio becomes 2:3. Find the numbers.​

Answer 1

Answer:

$let \: the \: numbers \: be \: 7x \: and \: 11x \\ now = \frac{7x + 7}{11x + 7} \\ according \: to \: the \: question \: \\ \frac{7x + 7}{11x + 7} = \frac{2}{3} (cross \: mutiply) \\ 3 \times (7x + 7) = 2 \times (11x + 7) \\ 21x + 21 = 22x + 14 \\ 21 - 14 = 22x - 21x \\ 7 = x \\ the \:first \: number \: is \: 7x + 7 = 7 \times 7 + 7 \\ = 49 + 7 \\ = 56 \\ the \: second \: number \: is \: 11x + 7 = 11 \times 7 + 7 \\ = 77 + 7 \\ = 84$

Answer 2

Answer:

let the numbers be X and y. X=49 and Y=77

Step-by-step explanation:

Let the numerator be X and the denominator be y

Given : X:Y=7:11 = X/Y = 7/11

and X+7/y+7= 2/3

after cross multiplication it becomes

11x=7y which can be written as y=11x/7 -①

and 3x+21=2y+14 -②

let us substitute

① in equation ②

3x+21=2y+14

=2y-3x=21-14

=2(11x/7)-3x=7

=(22x)/7 -3x=11

after taking LCM

(22x-21x)/7=7

X=49

as we found X we can find y by substituting X in equation ①

y=11x/7

=(11*49)/7

y=77