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Numerical Ability
Percentage
In an examination a candidateobtained 32% marks and failed by 16 marks. Another candidate obtained 36% marks which is 10 marks more than the minimum necessary marks to pass. Find the passing marks?
Read Solution (Total 3)
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- Let us Assume the pass marks be 'X'
Now,
Candidate 1 obtained 32% of marks which is 16 marks less than the pass marks
i.e 32%=X-16----------(1)
similarly,
Candidate 2 obtained 36% of marks which is 10 marks more than pass marks
i.e 36%=X+10---------(2)
Now,(2)-(1) gives
4%=26
=>32%=(32/4)*26=208
Therefore ,Pass marks=208+16=224
similarly if we consider 36%=(36/4)*26=234(10 marks more than the pass marks) - 7 years agoHelpfull: Yes(3) No(0)
- Difference in percentage mark obtained = 4%
Difference between there mark scored = 26
4% of total mark = 26
32% of total mark = 26 * 8 = 208
then pass mark= 208 +16= 222 - 6 years agoHelpfull: Yes(0) No(0)
- let total amount be x
a = 32x/100
b =36x/100
(36x/100)-10 =(32x/100)+16
x =650
a=(32(650)/100)-10)
a =224
similarly b =224
pass mark is 224 - 6 years agoHelpfull: Yes(0) No(0)
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