Question

A lady has only 20-paisa coins and 25-paisa coins in her purse.If she has 50 coins in all totalling Rs.11.50,how many coins of each kind does she have ?

Answer 1

Answer:

Let the number of 20-paisa coins be x and 25-paisa coins be y.

Total number of paise =11.50×100=1150

$\: \: \: \: \: \: \: 20 \times x + 25 \times y = 1150 \\ = >20x + 25y = 1150 \\ = > 5(4x + 5y) = 1150 \\ = > 4x + 5y = \frac{1150}{5} \\ = > 4x + 5y = 230 \: \: \: \: \:-1$

Total number of coins = 50

$\: \: \: \: \: \: \: x + y = 50 \:\:\: -(2)$

After multiplying equation(2)by 4 .

Subtracting (2) from (1) we get :

$\begin{lgathered}\: \: \: \: 4x + 5y = \: \: \: 230 \\ \: \: \: \: 4x + 4y = - 200 \\----------- \\ - \: \: \: \: \: \: - \: \: \: \: \: \: = \: \: \: \: 30 \\ \: \: \: \: \: \: \: \: \: y = 30\end{lgathered}$

Substituting value of y in (2) :

$\:\:\:\:\:\:\:\:x+y = 50$

$=> x+30=50$

$=>x = 50-30$

$=> x = 20$

Total number of 20 paise coins = 20

Total number of 25 paise coins = 30

Answer 2

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Given:

A lady has only 20 paisa coins and 25 paisa coins in her purse.If she has 50 coins in all totalling Rs.11.50.

To find:

The coins of each kind does she have.

Explanation:

Let the number of 20 paisa coins be R &

Let the number of 25 paisa coins be M

We have,

Total coins= Rs.11.50        [Convert into paisa]

We know that 1 rupees= 100paisa

Then, Rs.11.50 rupees= [11.50× 100]paisa

Total coins = 1150paisa

According to the question:

→ R+ M= 50

→ R= 50 -M.................(1)

&

→ 20R+ 25M= 1150................(2)

Putting the value of R in equation (2), we get;

→ 20(50-M) +25M= 1150

→ 1000- 20M +25M =1150

→ 1000+ 5M= 1150

→ 5M= 1150 -1000

→ 5M= 150

→ M= $\frac{\cancel{150}}{\cancel{5}}$

→ M= 30

Now, putting the value of M in equation (1), we get;

→ R= 50 - 30

→ R= 20

Thus,

She have coins of R= 20

She have coins of M= 30.