Aptitude - Problems on Ages - Discussion
Discussion Forum : Problems on Ages - General Questions (Q.No. 11)
11.
The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
Answer: Option
Explanation:
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
20x = 80
x = 4.
Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
Discussion:
42 comments Page 5 of 5.
Ronak said:
1 decade ago
The total of ratios is 20 (4+7+9).
The total of ages will become 56 + (8x3) = 56+24 = 80.
Now 20 x 4 = 80,
So the multiple is 4. So the common multiple is 4 for all.
Now we get present ages of all.
4x4 = 16.
7x4 = 28.
9x4 = 36.
The total of ages will become 56 + (8x3) = 56+24 = 80.
Now 20 x 4 = 80,
So the multiple is 4. So the common multiple is 4 for all.
Now we get present ages of all.
4x4 = 16.
7x4 = 28.
9x4 = 36.
Anu said:
1 decade ago
When should we use (+) and (-) symbols under this section means problems based on age.
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