Question

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

Answer 1

$\huge\tt{Required\:Answer:}$

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:

★ 20 paisa coins = 20 ★

★ 25 paisa coins = 30 ★

Step-by-step explanation:

Given:

• Type of coins lady has 20 paisa and 25 paisa coins
• Number of coins she has 50
• Total amount of money she has is Rs 11.50

To Find:

• Each type of coins lady she have

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

★ Since, all the terms are in Paisa . Therefore, changing Rs 11.50 into Paise, we have (11)100 + 50 = 1150 paisa ★

According to the question:

$\small\implies{\sf }$ 20x + 25y = 1150

$\small\implies{\sf }$ 4x + 5y = 230............(1) [ Divide both sides by 5]

† She has 50 coins in total †

∴ x + y = 50..........(2)

→4x + 4y = 200 [Multiply both sides by 4]

★ Now, Subtracting equation (2)*4 from (1)

$\small\implies{\sf }$ 4x + 5y = 230

$\small\implies{\sf }$ 4x + 4y = 200

$\small\implies{\sf }$ y = 30

★ Substitute the value of y in equation (2) ★

$\small\implies{\sf }$ x + y = 50

$\small\implies{\sf }$ x + 30 = 50

$\small\implies{\sf }$ x = 50–30

$\small\implies{\sf }$ x = 20

Hence, Number of 20 paisa coins lady has = x = 20 coins

Number of 25 paisa coins lady has = y = 30

★ Verification ★

→ 20 paisa coins + 25 paisa coins = 50

→ x + y = 50

→ 20 + 30 = 50

LHS = RHS