Question

Two numbers are in the ratio 2:7. if the first number is increased by 4 and the second number is decreased by 1, the resultant numbers are in the ratio 1:3. find the original numbers.​

Answer 1

Two numbers are in the ratio 2:7.

Let us assume that first number is 2M and second number is 7M.

If the first number is increased by 4 and the second number is decreased by 1, the resultant numbers are in the ratio 1:3.

First number = 2M + 4

Second number = 7M - 1

Ratio of first and second number is 1:3.

According to question,

⇒ (2M + 4)/(7M - 1) = 1/3

Cross-multiply them

⇒ 3(2M + 4) = 1(7M - 1)

⇒ 6M + 12 = 7M - 1

⇒ 6M - 7M = - 1 - 12

⇒ - M = - 13

⇒ M = 13

Therefore,

First number = 2M = 2(13) = 26

Second number = 7M = 7(13) = 91

Verification

From above calculations M = 13. Substitute value of M in (2M + 4)/(7M - 1) = 1/3

→ (2*13 + 4)/(7*13 - 1) = 1/3

→ (26 + 4)/(91 - 1) = 1/3

→ 30/90 = 1/3

→ 1/3 = 1/3

Answer 2

$\huge\underline\frak\blue{Given}$

Two numbers are in the ratio 2:7. if the first number is increased by 4 and the second number is decreased by 1, the resultant numbers are in the ratio 1:3.

$\huge\underline\frak\blue{To\:find}$

Find the original numbers.

$\huge\underline\frak\blue{Solution}$

Let the number be x

According to the given condition

$\implies\sf \large\frac{2x+4}{7x-1}=\large\frac{1}{3}$

$\implies\sf 3(2x+4)=7x-1$

$\implies\sf 6x+12=7x-1$

$\implies\sf 12+1=7x-6x$

$\implies\sf x=13$

$\large{\boxed{\bf{x=13}}}$

$\large{\boxed{\bf{First\:no.=2x=2×13=26}}}$

$\large{\boxed{\bf{Second\:no.=7x=7×13=91}}}$

$\huge\underline\frak\blue{Verification}$

Substitute the value of x in LHS

RHS = 1/3

LHS = $\sf \large\frac{2x+4}{7x-1}$

$\implies\sf \frac{2x+4}{7x-1}$

$\implies\sf \frac{2×13+4}{7×13-1}$

$\implies\sf \cancel\frac{30}{90}$

$\implies\sf \frac{1}{3}$

LHS = RHS for the value x = 13