Question

A bag contains one rupee,fifty paise, twenty paise and twenty five paise coins in the ratio 1:2:4:5. if the total value of coins is Rs.400, then the number of 20 paise coins exceeding those of twenty paise coins is
a.100
b.400
c.200
d.500​

Answer 1

The given proportion is 1 : 2 : 4 : 5.

Let, x be the common multiple.

Then, number of

• 1 rupee coins = x
• 50 paise coins = 2x
• 25 paise coins = 4x
• 20 paise coins = 5x

By the given condition,

(100 * x) + (50 * 2x) + (25 * 4x) + (20 × 5x) = 400 × 100

or, 100x + 100x + 100x + 100x = 40000

or, 400x = 40000

or, x = 100

So the number of 20 paise coins = 5 * 100

= 500

& the number of 25 paise coins = 4 * 100

= 400

Therefore the number of 20 paise coins exceeds the number of 25 paise coins by (500 - 400) = 100.

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Answer 2

Complete question:

A bag contains one rupee,fifty paise, twenty paise and twenty five paise coins in the ratio 1:2:4:5. if the total value of coins is Rs.400, then the number of 20 paise coins exceeding those of 25 paise coins is

a.100

b.400

c.200

d.500​

Answer:

The number of 20 paise coins exceeding those of twenty paise coins is a. 100

Step-by-step explanation:

From question, the ratio of rupees are 1 : 2 : 4 : 5

Let 1 rupee coins = x

Let 50 paise coins = 2x

Let 25 paise coins = 4x

Let 20 paise coins = 5x

From question, the total value of coins is Rs. 400, thus,

(100 × x) + (50 × 2x) + (25 × 4x) + (20 × 5x) = 400 × 100

100x + 100x + 100x + 100x = 40000

400x = 40000

x = 40000/400

∴ x = 100

So, the number of 25 paise coin is:

⇒ 5 × 100 = 500

The number of 20 paise coin is:

⇒ 4 × 100 = 400

The number of 20 paise coins exceeding 25 paise coins is:

∴ 500 - 400 = 100