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

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Step-by-step explanation:

THE NUMBER OF EACH COINS SHE HAVES;

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

= > 20x + 25y = 1150. 

Dividing the term by 5 we get,

= > 4x + 5y = 230 ------ (1)

Given that she has 50 coins in total.

= > x + y = 50.   ----- (2)

On solving (1) & (2) * 4, we get

4x + 5y = 230

4x + 4y = 200

-----------------------

 y = 30

Substitute y = 30 in (2), we get

= > x + y = 50

= > x + 30 = 50

= > x=50-30

= > x = 20.

The number of 20 paise coins = 20.

The number of 25 paise coins = 30.

Answer 2

Answer:

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

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

Total number of coins = 50

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

After multiplying equation(2)by 4 .

Subtracting (2) from (1) we get :

$\: \: \: \: 4x + 5y = \: \: \: 230 \\ \: \: \: \: 4x + 4y = - 200 \\----------- \\ - \: \: \: \: \: \: - \: \: \: \: \: \: = \: \: \: \: 30 \\ \: \: \: \: \: \: \: \: \: y = 30$

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