Question

The sum of ages of husband and his wife is 4 times the sum of ages of their children. 4 years ago the ratio of the sum of their ages to children was 18:1 Two years hence the ratio will be 3:1. How many children do they have?





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Answer 1

Answer:-

They have total 4 children.

$\rule{100}2$

Step-by-step explanation:-

The sum of ages of husband and his wife is 4 times the sum of ages of their children.

Let the age of husband be H, age of wife be W and age of children be C years.

As per the given condition,

Sum of ages of husband and wife i.e. H + W is 4 times the sum of ages of their children.

$\implies\:\sf{H\:+\:W\:=\:4(C)}$

$\implies\:\sf{H\:+\:W\:=\:4C}$ $\sf{....(A)}$

$\implies\:\sf{H\:=\:4C\:-\:W}$ $\sf{....(1)}$

Four years ago the ratio of the sum of their ages to children was 18:1.

Here sum of their ages means sum if ages of husband and wife.

• Four years ago, age of husband was (H - 4) years
• Age of wife was (W - 4) years
• Age of children was (C - 4) years

As per the given condition,

But wait! We have to find the number of children they have.

Let the number of children be C'.

$\implies\:\sf{\dfrac{(H-4)\:+\:(W-4)}{C-4C'}\:=\:\dfrac{18}{1}}$

$\implies\:\sf{H\:+\:W\:-\:8\:=\:18(C-4C')}$

$\implies\:\sf{4C\:-\:W\:+\:W\:-\:8\:=\:18C\:-\:72C'}$ $\sf{[From(1)]}$

$\implies\:\sf{14C\:=\:72C'\:+\:8}$

$\implies\:\sf{14C\:-\:8\:=\:72C'}$

$\implies\:\sf{14C\:=\:72C'\:+\:8}$ $\sf{...(2)}$

Two years hence the ratio of their ages will be 3:1.

• Age of husband after 2 years = (H + 2) years
• Age of wife = (W + 2) years
• Age of children = (C +2) years

$\implies\:\sf{\dfrac{(H+2)(W+2)}{C+2C'}\:=\:\dfrac{3}{1}}$

$\implies\:\sf{\dfrac{H\:+\:W\:+\:4}{C\:+\:2C'}\:=\:\dfrac{3}{1}}$

$\implies\:\sf{H\:+\:W\:+\:4\:=\:3C\:+\:6C'}$

$\implies\:\sf{4C\:+\:4\:=\:3C\:+\:6C'}$ $\sf{{[From(A)]}}$

$\implies\:\sf{C+\:4\:=\:6C'}$

$\implies\:\sf{C\:=\:6C'\:-\:4}$ $\sf{....3}$

Substitute value of equation (3) in (2)

$\implies\:\sf{14(6C'-4)\:=\:72C'\:+\:8}$

$\implies\:\sf{84C'\:-\:56\:=\:72C'\:+\:8}$

$\implies\:\sf{12C'\:=\:48}$

$\implies\:\sf{C'\:=\:4}$

•°• They have four children.

Answer 2

Step-by-step explanation:

QUESTION:-

The sum of ages of husband and his wife is 4 times the sum of ages of their children. 4 years ago the ratio of the sum of their ages to children was 18:1 Two years hence the ratio will be 3:1. How many children do they have?

let

the present age of husband =M

the present age of wife= w

the sum of their children age =s

GIVEN;

THE sum of ages of husband and his wife is 4 time the sum of ages of their children

so,

husband + wife=4* s

Let

the number of children be N

four years ago, the ratio of sum of their ages to the sum of ages of their children was 18:1.

so,

=>M+W-8/S-4N=18/1

=>M+W-8=18{S-4N}

=>M+W-8=18S-72N

here we have to change m+w

=>4S-8=18S-72N

=>4S-18S=-72N+8

=>-14S=-72N+8

=>72N-8=14S

Two years hence the ratio will be 3:1

=>M+W+4/S+2N=3/1

=>M+W+4=3{S+2N} (HERE WE ALSO CHANGE M+W)

=>4S+4=3S+6N

=>4S-3S=6N-4

=>6N-4=S

again we have to change S

so,

=>14S=72N-8

=>14(6N-4)=72N-8

=>84N-56=72N-8

=>84N-72N=-8+56

=>12N=48

=>N=48/12

=>N=4

now putting value of N

S=>6N-4

=>6*4-4

=>20(Here the sum of ages of children

HENCE:-

the sum of age of husband and wife =4s

=4*20==>80years

Number of children=4children