Please solve question no 27.....

Total coins = 220
Let there be x coins of 1Re., y coins of 50 p, and z coins of 25 paise.

Given 110Rs = 11000 paise && 80Rs = 8000 paise
11000 paise = x*100 + y*50 + z*25...(i)
8000 paise = x*100  + z*50 + y*25....(ii)   [since, number of 25p and 50p coins interchanged]
Subtracting: (i) - (ii)
3000 paise = 25*y - 25* z
3000 = 25* (y-z)
120 = y - z
So the difference between 50 p and 25 p coins = 120 and number of 50 p coins is greater than 25 p coins.
So the number of 50 p coins has to be greater than 120 (then only we will get a difference of 120)

Assume, 160 50 P coins, then there are 40 25 p coins..so the remaining (220-200) 20 coins are of 1 re.
Now check: 1st day: 20*1re + 160*50P + 40*25P = 20+80+10=110Rs.
2nd day: 20*1re + 40*50P + 160*25P = 20+20+40 = 80Rs.
Condition satisfied.
So he had 160 coins of 50 paise and 40 coins of 25 paise.
Therefore, option (c) is the correct answer.

Regards