A group of 74 students of Class XI went on picnic. The number of students who went on picnic by each section is as follows: Section A = 15 Section B = 24 Section C = 15 Section D = 20 5 students from Section A spent as much as 4 students from Section B. 12 Section B students spent as much as 9 Section C students and 6 Section C students spent as much as 8 Section D student. If every student of a particular section spent the same amount and all the students spent Rs. 23,370 altogether, find the amount spent by each student of Section D?

A group of 74 students of Class XI went on picnic. The number of students who went on picnic by each section is as follows: Section A = 15 Section B = 24 Section C = 15 Section D = 20 5 students from Section A spent as much as 4 students from Section B. 12 Section B students spent as much as 9 Section C students and 6 Section C students spent as much as 8 Section D student. If every student of a particular section spent the same amount and all the students spent Rs. 23,370 altogether, find the amount spent by each student of Section D? Correct Answer Rs. 307.50

GIVEN:

A group of 74 students of Class XI went on picnic. The number of students who went on picnic by each section is as follows:

Section A = 15

Section B = 24

Section C = 15

Section D = 20

Every student of a particular section spent the same amount

All the students spent Rs 23,370 altogether

So, 15A + 24B + 15C + 20D = 23,370

5 students from Section A spent as much as 4 students from Section B.

So, 5A = 4B       ---- (i)

12 Section B students spent as much as 9 Section C students

12B = 9C       ---- (ii)

6 Section C students spent as much as 8 Section D student

6C = 8D       ---- (iii)

(i) + (ii) + (iii), we get:

5A + 12B + 6C = 4B + 9C + 8D

→ 5A + 8B – 3C – 8D = 0

Multiplying the above equation by 3, we get:

15A + 24B – 9C – 24D = 0

On subtracting ‘15A + 24B – 9C – 24D = 0’ and ‘15A + 24B + 15C + 20D = 23,370’, we get:

-24C – 44D = - 23,370

Or, 24C + 44D = 23,370

Again, 6C = 8D      ---- (iii)

Multiplying the above equation by 4, we get:

24C = 32D

Now, substituting the value of 24C in ‘24C + 44D = 23,370’, we get:

32D + 44D = 23,370

Or, 76D = 23,370

Or, D = 307.5