A baby having born in between, the average age of the family is same today. When was the baby born?
a) 1 year ago
b) 2 years ago
c) 3 years ago
d) just today
e) none of the above
2007-07-01 20:48:26 · 6 answers · asked by BLAZER...! 1 in Science & Mathematics ➔ Mathematics
total ages =85 then
now the total ages=85+(3*5)+x
avg=17=(100+x)/6
hence ,x=2
so it was born 2 yrs ago...
2007-07-01 20:57:40 · answer #1 · answered by Agni n 1 · 0⤊ 0⤋
Correct answer is (b) 2 years ago.
Three years ago the total of the ages must have been 85 (17*5)
There are now 6 members in the family and if the average age is the same then the ages must add up to 102 (6*17)
BUT the ages of the 5 people have increased by 3 years so now the total of the five is 85+5*3=100. Therefore the age of the infant must be 2.
2007-07-02 03:58:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The problem can be solved algebraically by setting up a simple equation.
5 x 17 + 15 + x = 6 x 17
where x is the age of the baby.
100 + x = 102
x = 2
So, the baby is 2 years old or was born 2 years ago.
2007-07-02 04:05:01 · answer #3 · answered by Swamy 7 · 0⤊ 0⤋
3 years ago: average of 5 members' age = 17 years
===> total age = 5 * 17 = 85
add current year: 85 + (3 * 5) = 100 years
this year: average of 6 members' age (incl. infant's) = 17 years
===> total age = 6 * 17 = 102 years
difference = 102 - 100 = 2 years
therefore, the baby was born 2 years ago.
2007-07-02 04:16:55 · answer #4 · answered by jurassicko 4 · 0⤊ 0⤋
c) 2 year ago
average of today is 100/5 = 20
in between 17 & 20 is 18.5
the child was born 1.5 years ago
I choose 2 yrs
2007-07-02 04:04:28 · answer #5 · answered by CPUcate 6 · 0⤊ 0⤋
b) 2 years ago
2007-07-02 03:57:08 · answer #6 · answered by sweet n simple 5 · 0⤊ 0⤋