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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.

in all the student got 16 sums right and 32 sums wrong
the answer is 16 sums right and 16 * 2 = 32 sums wrong total 16 + 32 = 48


Let the no. of sums he got right be x
ten the no. of sums he got wrong are 48-x
according to the question 
x=16 ans
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Total wrong answers = twice of right answers
Let wrong answers be W and right answers be R
Total wrong answers+Total right answers = 48
W+R = 48
2R+R = 48 (Since W=2R)
3R = 48
Therefore, R=16

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