Question

To pass an examination, 40% marks are essential. A obtains 10% less than the pass marks and B obtains 11.11% marks less than A. What percent less that the sum of A’s and B’s marks should C obtain to pass the exam ?

Answer 1

Answer:

Percentage less marks obtain by C as sum of marks of A and B is 41.1%

Step-by-step explanation:

Given as :

The percentage of passing marks = 40%

Let The total marks of exam = 100 marks

So, Passing marks = 40 % of 100

Or, Passing marks = $\dfrac{40}{100}$ × 100

i,e Passing marks = 40

Again

marks obtain by A = 10% less than pass marks

i.e marks obtain by A =  40 - 10% of 40

Or, marks obtain by A =40 -  $\dfrac{10}{100}$ × 40

Or, marks obtain by A = 36

Again

Marks obtain by B = 11.11% marks less than A

i.e Marks obtain by B = 36 - 11.11% of 36

or, Marks obtain by B = 36 - $\dfrac{11.11}{100}$ × 36

Or, Marks obtain by B = 32

Let The percentage less marks obtain by C = x %

Or, Sum of marks obtain by A and B = 36 + 32 = 68

∵ passing marks = 40

So, The percentage less marks obtain by C = $\dfrac{68-40}{68}$ × 100

Or,  x = $\dfrac{28}{68}$ × 100

∴   x = 41.1 %

So, The percentage less marks obtain by C = x = 41.1%

Hence, Percentage less marks obtain by C as sum of marks of A and B is 41.1%  Answer