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03 Dec 2020, 07:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (02:18) correct
22%
(03:08)
wrong
based on 69
sessions
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The sum of A and B's age is 43 years. In 11 years, A's age will be 7/6 times B's age. What is B's present age?
A. 24 years
B. 22 years
C. 20 years
D. 19 years
E. 18 years
A. 24 years
B. 22 years
C. 20 years
D. 19 years
E. 18 years
03 Dec 2020, 07:20
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Bunuel
The sum of A and B's age is 43 years.
Let B = B's PRESENT age
So, 43 - B = A's PRESENT age
In 11 years....
B + 11 = B's age IN 11 YEARS
(43 - B) + 11 = A's age IN 11 YEARS
..., A's age will be 7/6 times B's age.
We can write: (43 - B) + 11 = (7/6)(B + 11)
Simplify the left side: 54 - B = (7/6)(B + 11)
To eliminate the fraction, multiply both sides of the equation by 6 to get: 6(54 - B) = (7)(B + 11)
Expand: 324 - 6B = 7B + 77
Add 6B to both sides of the equation: 324 = 13B + 77
Subtract 77 from both sides: 247 = 13B
Solve: B = 247/13 = 19
Answer: D
General Discussion
06 Dec 2020, 04:29
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Let the age of A be \(a\) and B be \(b\).
therefore, equation 1 is \(a+b=43\) and equation 2 is \(a+11\)=\(\frac{7}{6}(b+11)\).
solving equation 2 further we get \(6a+66=7b+77\),
\(6a=7b+11\),
\(a=\frac{(7b+11)}{6}.\)
Now substituting value of a in equation 1.
\(\frac{(7b+11)}{6}+b=43\).
\(\frac{7b+6b+11}{6}=43.\)
\(7b+6b+11=43*6\).
\(13b+11=258.\)
\(b=\frac{258-11}{13} => \frac{247}{13 }=> 19.\)
Therefore B's present age is 19 and the OA is D.
therefore, equation 1 is \(a+b=43\) and equation 2 is \(a+11\)=\(\frac{7}{6}(b+11)\).
solving equation 2 further we get \(6a+66=7b+77\),
\(6a=7b+11\),
\(a=\frac{(7b+11)}{6}.\)
Now substituting value of a in equation 1.
\(\frac{(7b+11)}{6}+b=43\).
\(\frac{7b+6b+11}{6}=43.\)
\(7b+6b+11=43*6\).
\(13b+11=258.\)
\(b=\frac{258-11}{13} => \frac{247}{13 }=> 19.\)
Therefore B's present age is 19 and the OA is D.
