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Age Problem
The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago. Four years ago, Jennifer, who is a year younger than Margaret, was 3/2 times as old as Abigail. The sum of the ages of Jennifer and Margaret two years, hence, will be:
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let the present age of Abigail=a
let the present age of Jennifer=j
let the present age of Margaret=m
The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago
that means ,
(a+j+m)-9=52
=> a+j+m=61------------------(1)
Four years ago, Jennifer, who is a year younger than Margaret,was 3/2 times as old as Abigail
means,
m-j = 1 ----------------------(2)
j - 4 = 3/2(a-4) -----------------------------(3)
from no 3 equation we can get a = (2j + 4)/3
now put the value of a in no 1 equation
and we can get 5j + 3m = 179------------------------(4)
now from no 2 equation m=1+j
put ithis into no 4 equation
and we get j = 22
and so from equation 2 we can get m = 23
so The sum of the ages of Jennifer and Margaret two years, hence, will be= 22+23+4 = 49
- 8 years agoHelpfull: Yes(5) No(0)
- ans=44
solution:
Abigail =a
Jennifer =j
Margaret=m
then
The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago.
(a+j+m)-3=52
present age is a+j+m=55
Four years ago, Jennifer, who is a year younger than Margaret
j-4=m+1 =>j-m=5
was 3/2 times as old as Abigail.
j-4=(3/2)a => -(3/2)a+j=4
after solving a,m,j
a=13
j=23.5
m=18.5
The sum of the ages of Jennifer and Margaret two years, hence, will be: j+m+2
=23.5+18.5+2
=44(ans)
- 8 years agoHelpfull: Yes(2) No(4)
- answer is 49
- 8 years agoHelpfull: Yes(1) No(1)
- DHILEEP BRO, BUT THE ANSWER IS NOT MATCHING AS YOU SOLVED,THE OPTIONS ARE 51,47,48,49.
- 8 years agoHelpfull: Yes(0) No(2)
- 27
x+3/2x+3/2x-1=55
x=8
m+j=27
- 8 years agoHelpfull: Yes(0) No(1)
- ans=44
solution:
Abigail =a
Jennifer =j
Margaret=m
then
The sum of the ages of Abigail, Jennifer and Margaret was 52 three years ago.
(a-3 + j-3 + m-3)=52
present age is a+j+m=61
Four years ago, Jennifer, who is a year younger than Margaret
j-4=(m-4)+1 =>j-m=1
was 3/2 times as old as Abigail.
j-4=(3/2) (a-4) => 3a-2j=4
after solving a,m,j
a=16.5
j=22.75
m=21.75
The sum of the ages of Jennifer and Margaret two years, hence, will be: j+m+2
=22.75+21.75+2
=46.5(ans) - 7 years agoHelpfull: Yes(0) No(1)
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