at an examination in which full marks were 500. a got 10% less than b, b got 25% more than c and c got 20% less than d. if a got 360marks, what percentage of full marks was obtained by d?
D obtained 80% of the full marks.
Step-by-step solution
Let the marks of A, B, C and D be A,B,C,DA,B,C,DA,B,C,D.
- Full marks = 500
- Given: A=360A=360A=360
1. Relation between A and B
A got 10% less than B.
That means A is 90% of B:
A=0.9BA=0.9BA=0.9B
360=0.9B⇒B=3600.9=400360=0.9B\Rightarrow B=\frac{360}{0.9}=400360=0.9B⇒B=0.9360=400
So, B=400B=400B=400.
2. Relation between B and C
B got 25% more than C.
That means B is 125% of C:
B=1.25CB=1.25CB=1.25C
400=1.25C⇒C=4001.25=320400=1.25C\Rightarrow C=\frac{400}{1.25}=320400=1.25C⇒C=1.25400=320
So, C=320C=320C=320.
3. Relation between C and D
C got 20% less than D.
That means C is 80% of D:
C=0.8DC=0.8DC=0.8D
320=0.8D⇒D=3200.8=400320=0.8D\Rightarrow D=\frac{320}{0.8}=400320=0.8D⇒D=0.8320=400
So, D=400D=400D=400.
4. Percentage of full marks obtained by D
Full marks = 500, D’s marks = 400.
Percentage=400500×100=80%\text{Percentage}=\frac{400}{500}\times 100=80%Percentage=500400×100=80%
So, D obtained 80% of the full marks. TL;DR: Working back from A’s 360 marks through the percentage relations leads to D = 400 marks out of 500, i.e. 80%.
