Question

there are some coins in a bag. The number of 50 Paisa coins are double than that of one rupees coins and that if 25 paisa coins are nine times that of one rupees coin. If total worth of the bag is Rs 544, Find the number of 25 paisa coins.​

Answer 1

Answer:

Let a and b be number of 25 paise and 50 coins, respectively. Then

Given that:

x+y=140                                       

25x+50y=5000                                                [Rs.50=5000 paise]

x+2y=200                                   

On subtracting equation (1) from equation (2), we get 

y=60

So, x=80

Hence, there are 80 coins of 25 paise and 60 coins of 50 paise in the bag.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• The number of 50 paisa coins are double than that of one rupees coins.

• 25 paisa coins are nine times that of one rupees coin.

• Total worth of the coins in bag is Rs 544.

Let assume that

• Number of one rupee coin = x

So,

• Number of 50 paisa coins be 2x

• Number of 25 paisa coins be 9x.

So, number of Coins in bag are

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{1 \: rupee \: coin \: = \: x} \\ \\ &\sf{50 \: paisa \: coin \: = \: 2x}\\ \\ &\sf{25 \: paisa \: coin \: = \: 9x} \end{cases}\end{gathered}\end{gathered}$

Now,

• Total amount in bag in denominations of 1 rupee coin = 100 × x = 100x paisa

• Total amount in bag in denominations of 50 paisa coin = 50 × 2x = 100x paisa

• Total amount in bag in denominations of 25 paisa coin = 25 × 9x = 225x paisa

Since, It is given that

Total amount in bag = Rs 544 = 54400 paisa

$\rm\implies \:100x + 100x + 225x = 54400$

$\rm\implies \:425x = 54400$

$\bf\implies \:x = 128$

So, Number of coins in bag are

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{1 \: rupee \: coin \: = \: x = 128} \\ \\ &\sf{50 \: paisa \: coin \: = \: 2x = 256}\\ \\ &\sf{25 \: paisa \: coin \: = \: 9x = 1152} \end{cases}\end{gathered}\end{gathered}$

Hence,

• Number of 25 paisa coin in bag is 1152