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.
Answers
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: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