Question

33800. How much
e) A's age is 4 times that of B's age. 10 years ago, A's age was 19 times that of B's age. Find the
present ages of A and B.​

Answer 1

Answer:

Age of A = 48 years

Age of B = 12 years

Step-by-step explanation:

given that,

A's age is 4 times that of B's age.

Let the age of B be x

so,

the age of A = 4x

now,

also given that,

10 years ago, A's age was 19 times that of B's age

so,

10 years ago

4x - 10 = 19(x - 10)

4x - 10 = 19x - 190

19x - 4x = 190 - 10

15x = 180

x = 180/15

x = 12

so,

the age of B = 12 years

now,

age of A = 4x

= 4(12)

= 48 years

so,

________________

Age of A = 48 years

Age of B = 12 years

________________

Verification :

Age of A = 4 times the age of B

48 = 4 × 12

48 = 48

__________

10 years ago, A's age was 19 times that of B's age

4x - 10 = 19(x - 10)

4(12) - 10 = 19(12 - 10)

48 - 10 = 19(2)

38 = 38

_____________

hence verified

Answer 2

Answer:-

A's age = 48 years

B's age = 12 years

Given:-

A's age is 4 times that of B's age.

10 years ago, A's was 19 times that of B's age.

To find :-

The present ages of A and B.

Solution :-

Let the present age of A be x years and B be y years.

A/Q

$x = 4y-------eq.1$

A's age 10 years ago = ( x - 10)

B's age 10 years ago = ( y -10)

then,

→$(10-x) = 19(10-y)$

→$10-x = 190 -19y$

→$-x +19y = 190 -10$

→$-x +19 y = 180-----eq.2$

Put the value of x in eq. 2

→$x = 4y$

→$-x +19y = 180$

→$-4y +19y = 180$

→$15y = 180$

→$y = \dfrac{180}{15}$

→$y = 12$

put the value of y in eq.1

x = 4y

x = 4 × 12

x = 48

hence,

A's age present age is 48 years

A's age present age is 48 years B's present age is 12 years.