Question

Six students take an exam two times. The increase in average marks of all the students from first test to second test is 30%.The difference of the total marks of first five students in the two tests is 20% of their total marks in second test. If in the first test the marks of the sixth student were mistakenly written 1/9 times more than the original, what is the actual percentage increase in his marks?

A. 57% B. 100% C. 72.22% D. 80 E. None of these​

Answer 1

Answer:

please f.o.l.l.o.w. me

Step-by-step explanation:

let the first test average be x then

total marks of 6 students in the first test is 6x.

the average marks in the second test is x+(x of 30%) =13x/10 .

total marks of students in second test = 13x/10 ×6

=39x/5

difference of 5 students in the two tests=39x/5 ×20%

=39x/25

Actual total marks of 6 students in first test =6x ÷1/9=54x

increase=54x-6x =48x

percentage= 48x ×100/6x = 800%

hence,the actual percentage increase in his marks will be 800%...

may be my answer is correct ...