Question

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

Answer 1

Answer:

4 children

Step-by-step explanation:

Given :

• The sum of ages of husband and his wife is four times times the sum of ages of their children.
• Four years ago, the ratio of sum of their ages to sum of ages of their children was 18:1
• Two years hence the ratio will be 3:1

To find :

the number of children they have

Solution :

Let the present age of

husband be 'a' years

wife be 'b' years

Also, let the number of children be 'n'

and the sum of the ages of their children be 'c' years

As given,

a + b = 4c ––(1)

Four years ago,

husband's age = a – 4

wife's age = b – 4

Sum of their ages = a – 4 + b – 4

= a + b – 8

= 4c – 8 (eqn. (1))

sum of the ages of children = c – 4n

As given,

(4c – 8)/(c – 4n) = 18/1

4c – 8 = 18(c – 4n)

4c – 8 = 18c – 72n

18c – 4c = 72n – 8

14c = 72n – 8

2(7c) = 2(36n – 4)

7c = 36n – 4 ––(2)

After two years,

husband's age = a + 2

wife's age = b + 2

Sum of their ages = a + 2 + b + 2

= a + b + 4

= 4c + 4 (eqn. (1))

sum of the ages of children = c + 2n

As given,

(4c + 4)/(c + 2n) = 3/1

4c + 4 = 3(c + 2n)

4c + 4 = 3c + 6n

4c – 3c = 6n – 4

c = 6n – 4

Put c = 6n – 4 in equation (2),

7c = 36n – 4

7(6n – 4) = 36n – 4

42n – 28 = 36n – 4

42n – 36n = 28 – 4

6n = 24

n = 24/6

n = 4

Therefore, they have 4 children.

Answer 2

★CORRECT QUESTION:-

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

${\boxed{\underline{\tt{ \orange{Required \: \: answer \: \: is \: \: as \: \: follows:-}}}}}$

4 Children

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★SOLUTION:-

• Let C be the present sum of Chikdren's age.

• Let P be the present sum of Parent's age.

• Let n be the Number of Children

➝According to the question,

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

$\therefore \sf{P = 4C} --------------(1)$

Then,

4 Years ago The ratio of their ages to the ages of

their children was 18:1

So, it will reduce the sum of the parents ages by 8,

The children's age by 4n

$\therefore \sf{P - 8 = 18(C - 4n) }$

$\therefore \sf{P - 8 = 18C - 72n}--------------- (2)$

Now, replace P with 4C ( From (1) )

$: \implies\sf{4C -8=18C-72n}$

Subtract 18C from both the sides and arrange then as:-

$:\implies\sf{-14C +72n =8} --------------- (3)$

after 2 years, the ratio is 3:1.

that will increase the parents age sum by 4, the Children age by 2n

$:\implies\sf{P+4=3(C+2n)}$

$:\implies\sf{P +4=3C +6n} ----------------(4)$

Replace P again with 4C (From (1))

$:\implies\sf{4C +4=3C +6n}$

Subtract 3C from both the sides and rearrange them:-

$:\implies\sf{C = 6n -4} --------------------- (5 )$

In the (3) replace C with (6n-4)

$: \implies \sf{ - 14(6 n - 4) + 72n = 8} \\ : \implies \sf{ - 84n + 56 + 72n = 8} \\ : \implies \sf{ - 12n = 8 - 56} \\ : \implies \sf{n = \frac{ - 48}{ - 12}} \\ \\ { \boxed{ \underline{ \pink{ \huge{ n = 4}}}}}$

So, the Number of Children's = 4