Aptitude - Problems on Ages - Discussion
Discussion Forum : Problems on Ages - General Questions (Q.No. 4)
4.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?
Answer: Option
Explanation:
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, B's age = 2x = 10 years.
Discussion:
79 comments Page 3 of 8.
Bradley S said:
11 months ago
Based on a simplistic alternate method the answer is 9.
As follows.
A + 2 = B.
7 + 2 = 9 .
Then
2B = C
2 *9 = 18
Sums 9 + 18 = 27
which verifies that B = 9.
As follows.
A + 2 = B.
7 + 2 = 9 .
Then
2B = C
2 *9 = 18
Sums 9 + 18 = 27
which verifies that B = 9.
(6)
Nagendra said:
1 decade ago
The population of town was 24000. If males increases by 6% and females by9%, it becomes 25620. Number of males and females are?- Please any one explain me...
Oro Oghenevwegba said:
12 months ago
A + B + C = 27,
A = 2 + B,
B = 2C,
B = x.
Equate the above into A + B + C = 27.
2 + 2C + 2C + C = 27.
2 + 5C = 27.
5C/5 = 25/5.
C = 5.
B = 2C.
B = 2×5 = 10.
A = 2 + B,
B = 2C,
B = x.
Equate the above into A + B + C = 27.
2 + 2C + 2C + C = 27.
2 + 5C = 27.
5C/5 = 25/5.
C = 5.
B = 2C.
B = 2×5 = 10.
(7)
Thiru said:
1 decade ago
Hi,
As per the question a+b+c=27, here a= b as of now reduced two years from b. So here b=a+2 years, c value is twice value of b, so which mean b*2.
As per the question a+b+c=27, here a= b as of now reduced two years from b. So here b=a+2 years, c value is twice value of b, so which mean b*2.
Al1cee said:
1 decade ago
A+B+C = 27.
Let A = B-2.
B = It self.
C = 2B (B is twice older than C).
Substituting the givens A+B+C.
B-2+B+2B = 27.
4B = 27+2.
B = 29/4.
B = 7.
Let A = B-2.
B = It self.
C = 2B (B is twice older than C).
Substituting the givens A+B+C.
B-2+B+2B = 27.
4B = 27+2.
B = 29/4.
B = 7.
Venkata ramana said:
6 years ago
A = B+2
B = 2C
A+B+C = 27,(B+2)+2C+C = 27,B+2+3C = 27, B+3C = 25, 2C+3C = 25,5C = 25, C = 5.
Then B is twice as old as C, B = 2C, B = 2(5) = 10.
B = 2C
A+B+C = 27,(B+2)+2C+C = 27,B+2+3C = 27, B+3C = 25, 2C+3C = 25,5C = 25, C = 5.
Then B is twice as old as C, B = 2C, B = 2(5) = 10.
Wisdom said:
1 decade ago
a=x;
b=x-2;
c=(x-2)/2;
a+b+c=27
x+x-2+[(x-2)/2]=27
[2x+2x-4+x-2]/2=27
5x-6=54
5x=60
x=12
b=x-2;
b=10 we got answer
b=x-2;
c=(x-2)/2;
a+b+c=27
x+x-2+[(x-2)/2]=27
[2x+2x-4+x-2]/2=27
5x-6=54
5x=60
x=12
b=x-2;
b=10 we got answer
Suraj said:
9 years ago
If b = x,
then,
a = 2 + x
c = 2 + 2 + x = 4 + x.
Then,
a + b + c = 27.
(2 + x) + x + (4 + x) = 27.
3x = 27 + 2 + 4.
x = 33/3.
x = 10.
then,
a = 2 + x
c = 2 + 2 + x = 4 + x.
Then,
a + b + c = 27.
(2 + x) + x + (4 + x) = 27.
3x = 27 + 2 + 4.
x = 33/3.
x = 10.
Tanishq said:
5 years ago
a = two years more than b+2.
b = twice on c 2x.
Assume c = x.
b+2+2x+x.
Substitute b.
2x+2+2x+x = 5x.
5x = 27-2.
x = 5.
b = 2*5 = 10.
b = twice on c 2x.
Assume c = x.
b+2+2x+x.
Substitute b.
2x+2+2x+x = 5x.
5x = 27-2.
x = 5.
b = 2*5 = 10.
(3)
Mahesh bist said:
6 years ago
Here, let B= x then A= 2+x then always c= x/2.
According to question,
A+B+C = 27,
(2+x)+x+x/2 = 27,
5x+4 = 27*2,
x = 50/5,
x = 10.
According to question,
A+B+C = 27,
(2+x)+x+x/2 = 27,
5x+4 = 27*2,
x = 50/5,
x = 10.
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