Question

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 1

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 
2x=48-x
3x=48
x=16 ans

Answer 2

Total wrong answers = twice of right answers
Let wrong answers be W and right answers be R
W=2R
Total wrong answers+Total right answers = 48
W+R = 48
2R+R = 48 (Since W=2R)
3R = 48
Therefore, R=16