Question

a bus conductor gets a total of 220 coins of 25 paise, 50 paise and Rs 1 rupees daily one day he got 110 rupees and next day he got 80 rupees in that number of coins of 25 paise and 50 paise coins are interchange then find the total number of 50 paise coins and 25 paise coins

Answer 1

Answer:

Step-by-step explanation: x+3y=-9

Answer 2

ANSWER:

Bus conductor will have 200 coins of 50 paise and 25 paise after two days.

SOLUTION:

Given, a bus conductor gets a total of 220 coins of 25 paise, 50 paise and Rs 1 rupees daily  

Total coins = 220

Let there be x coins of 1Rs, y coins of 50 p and z coins of 25 paise.

Given 110Rs = 11000 paise & 80Rs = 8000 paise

$11000 \text { paise }=x \times 100+y \times 50+z \times 25$ → (i)

$8000 \text { paise }=x \times 100+z \times 50+y \times 25$ → (ii)  [since, number of 25p and 50p coins interchanged]

Subtracting: (i) - (ii)

3000 paise = 25y - 25z

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 remaining (220-200) 20 coins are of 1 re.

Now check: 1st day: $20 \times 1 \mathrm{re}+160 \times 50 \mathrm{P}+40 \times 25 \mathrm{P}=20rs +80rs +10rs =110 \mathrm{Rs}$

2nd day: $20 \times 1 \mathrm{re}+40 \times 50 \mathrm{P}+160 \times 25 \mathrm{P}=20 \mathrm{rs}+20 \mathrm{rs}+40 \mathrm{rs}=80 \mathrm{Rs}$

Condition satisfied.

So he had 160 coins of 50 paise and 40 coins of 25 paise and on other day he will have 40 coins of 50 paise and 160 coins of 25 paise.

Hence, he will have 200 coins of 50 paise and 25 paise after two days.