Question

In an examination, A gets 10% marks less than B and B gets 10% marks less than C. If A gets 810 marks, what marks does C get?

(A) 900 (B) 945 (C) 973 (D) 1,000​

Answer 1

Option D

Step-by-step explanation:

Given :-

In an examination, A gets 10% marks less than B and B gets 10% marks less than C. A gets 810 marks.

To find :-

What marks does C get?

(A) 900 (B) 945 (C) 973 (D) 1,000

Solution :-

Let the marks got by C be X

Then, The marks got by B = 10% marks less than C

=> C - 10% of C

=> X -(10% of X)

=> X -(10%×X)

=> X -(10/100)×X

=> X -(1/10)×X

=> X-(X/10)

=> (10X-X)/10

=> 9X/10

Marks got by A = 10% less than B

=> (9X/10) - (10% of 9X/10)

=> (9X/10) -[(10/100)×(9X/10)]

=> (9X/10) -[(1/10)×(9X/10)]

=> (9X/10) - (9X/100)

=> 9X[(1/10)-(1/100)]

=> 9X[(10-1)/100]

=> 9X(9/100)

=> 9X(9/100)

=> (9X×9)/100

=> 81X/100

The marks got by A = 81X/100

According to the given problem

The marks got by A = 810

=> 81X/100 = 810

=> 81X = 810×100

=> 81 X = 81000

=> X = 81000/81

=> X = 1000

The marks of C = 1000

Answer:-

Marks got by C in the examination is 1000

Check:-

Marks got by C = 1000

10% of 1000 = (10%) of 1000

=> (10/100)×1000

=> 10×10

=> 100

10% less than 1000 = 1000-100 = 900

Marks of B = 900

Marks of A = 10% less than B

=> 900- (10% of 900)

=> 900 - (10×900/100)

=> 900- (9000/100)

=> 900-90

=> 810

Marks of A = 810

Verified the given relations in the given problem.