Verbal Reasoning - Arithmetic Reasoning - Discussion
Discussion Forum : Arithmetic Reasoning - Section 2 (Q.No. 1)
1.
A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly ?
Answer: Option
Explanation:
Suppose the boy got x sums right and 2x sums wrong.
Then, x + 2x = 48 
3x = 48 x = 16.
Discussion:
8 comments Page 1 of 1.
Nitika sharma said:
2 years ago
Num of right sums = x.
Num of wrong sums = 2*x.
As per the question;
The sum of all attempts = 48.
x+2x = 48.
3x = 48,
x = 48/3,
x = 16 Answer.
Num of wrong sums = 2*x.
As per the question;
The sum of all attempts = 48.
x+2x = 48.
3x = 48,
x = 48/3,
x = 16 Answer.
(1)
VINOD padahetty said:
4 years ago
X+2x = 48,
3x = 48,
X =48/3,
X = 16.
3x = 48,
X =48/3,
X = 16.
(1)
Himani said:
5 years ago
X + 2X = 48,
3X = 48,
X = 48/3,
X = 16.
3X = 48,
X = 48/3,
X = 16.
(1)
Kalyan said:
9 years ago
The answer is 48/4 = 12 A is right.
Justin Biber said:
9 years ago
@Swaroop told an easy method.
Crammer said:
9 years ago
Let x = sums wrong .
2x = sums right ("twice as many sums wrong as he got right")
x + 2x = 48,
3x = 48,
x = 16,
2x = 32 = sums right.
Shouldn't the answer be 32 instead of 16?
2x = sums right ("twice as many sums wrong as he got right")
x + 2x = 48,
3x = 48,
x = 16,
2x = 32 = sums right.
Shouldn't the answer be 32 instead of 16?
Swaroop said:
10 years ago
d+2d = 48.
3d = 48.
d = 48/3 = 16.
3d = 48.
d = 48/3 = 16.
Shashi said:
1 decade ago
x+2x = 48.
3x = 48.
x = 48/3 = 16.
3x = 48.
x = 48/3 = 16.
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