Question

17. The sum of the ages of 3 persons is 150 years. 10 years ago their ages were in the ratio
7:8:9. Their present ages are
(
(b) (40, 60, 50)
(c) (35, 45, 70)
(d) none of these
10
The numbers 14 16 25​

Answer 1

I hope it helps you................

Answer 2

Answer:

$\sf{Their \ present \ ages \ are \ 45 \ years,}$

$\sf{50 \ years \ and \ 55 \ years \ respectively.}$

Given:

$\sf{\leadsto{The \ sum \ of \ the \ ages \ of \ three \ persons}}$

$\sf{is \ 150 \ years.}$

$\sf{\leadsto{10 \ years \ ago \ their \ ages \ were \ in}}$

$\sf{the \ ratio \ of \ 7:8:9.}$

To find:

$\sf{Their \ present \ ages.}$

Solution:

$\sf{Let \ the \ constant \ be \ x \ for \ the}$

$\sf{ratio \ of \ their \ ages \ 10 \ years \ ago.}$

$\sf{\therefore{Their \ ages \ 10 \ years \ ago \ were}}$

$\sf{7x, \ 8x \ and \ 9x \ respectively.}$

$\sf{Hence, \ their \ present \ ages \ are}$

$\sf{(7x+10), \ (8x+10) \ and \ (9x+10) \ respectively.}$

$\sf{According \ to \ the \ given \ condition}$

$\sf{(7x+10)+(8x+10)+(9x+10)=150}$

$\sf{\therefore{24x+30=150}}$

$\sf{\therefore{24x=150-30}}$

$\sf{\therefore{24x=120}}$

$\sf{\therefore{x=\dfrac{120}{24}}}$

$\sf{\therefore{x=5}}$

$\sf{Their \ present \ ages \ are}$

$\sf{\leadsto{7x+10=7(5)+10=35+10=45 \ years,}}$

$\sf{\leadsto{8x+10=8(5)+10=40+10=50 \ years,}}$

$\sf{\leadsto{9x+10=9(5)+10=45+10=55 \ years.}}$

$\sf\purple{\tt{\therefore{Their \ present \ ages \ are \ 45 \ years,}}}$

$\sf\purple{\tt{50 \ years \ and \ 55 \ years \ respectively.}}$