Question

Mother, daughter and an infant combined age is 74, and mother's age is 46 more than daughter and infant. If infant age is 0.4 times of daughter age, then find daughters age.

Answer 1

Answer: 10 yrs

Step-by-step explanation:

Let the daughter's age be x

Therefore, infant's age=4x/10

Therefore, mother's age=46+4x/10+x

B.T.P.

x+4x/10+46+4x/10+x=74

x+x+4x/10+4x/10+46=74

2x+8x/10=74-46

28x/10=28

28x=28*10

x=280/28

x=10

Therefore, daughter's age=10 yrs

Answer 2

Answer:10yrs(dont forget unit)

Step-by-step explanation:

Let mother's age be x years daughter's age be y years

Assuming for better understanding

Infant's age to be z years

So;

x+y+z=74

x=46+y+z

But;

z=0.4(y)

z=0.4y

There fore infants age is 0.4y

Substitute

x+y+0.4y=74

x+1.4y=74........eq 1

x=46+y+0.4y

x=46+1.4y

x-1. 4y=46........eq 2

Solving

x+1.4y=74

x-1.4y=46

(+) (+) (+)

2x+0 =120

2x=120

X= 60

Substituting value of x in eq 2

60-1.4y=46

-1.4y=46-60

-1.4y=-14

y =14/1.4 (cancelling.

negative)

y=10