Problem on Ages
Today we're gonna see another 3 problems...
1. The ratio of Kevin's age to Sana's age is 7:3. The product of their ages is 756. The ratio of their ages after 6 years will be:
Solution:
Let their ages be 7x and 3x respectively.
The product of their ages (7x)(3x) = 756
21x2= 756
x2= 756/21 = 36
x = 6
The ratio of their ages after 6 years will be (7x+6) : (3x+6)
48: 24
2: 1
2. The present ages of three persons are in the proportions 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages in years.
Solution:
Let their present ages be 4x, 7x and 9x respectively.
Eight years ago, the sum of their ages = (4x-8) + (7x-8) + (9x-8) = 56
20x-24 = 56
20x = 80
x = 4
Their present ages are 4(4) = 16 years, 7(4) = 28 years and 9(4)= 36 years respectively.
3. The ratio of the ages of a man and his wife is 4:3. After 4 years, this ratio will be 9:7. If at the time of marriage, the ratio was 5:3, then how many years ago were they married?
Solution:
Let their ages are 4x and 3x respectively.
After 4 years, the ratio will be 9:7
So, (4x+4)/(3x+4) = 9/7
7(4x+4) = 9(3x+4)
28x+28 = 27x+36
x = 8
Their ages are 32 and 24 years respectively.
At the time of marriage, the ratio was 5:3
They were married 'a' years ago.
Hence, (32-a)/(24-a) = 5/3
3(32-a) = 5(24-a)
96-3a = 120-5a
2a = 24
a = 12 years.
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