In Ruchika's wallet there are Rs. 16, consisting of 10 paise, 20 paise and Re. 1 coins.

Ques - In Ruchika's wallet there are  Rs. 16, consisting of 10 paise, 20 paise and Re. 1 coins. The ratio of number of coins 10 paise coins and 20 paise is 6 : 1. The maximum number of Re. 1 coin is:

(a) 5

(b) 12

(c) 8 

(d) 4

Solution - In Ruchika's wallet there are  Rs. 16.

Firstly,

Change the rupees into paise = 16 × 100

= 1600

consisting of 10 paise, 20 paise and Re. 1 coins.

Let,

10 paise = 10x

20 paise = 20y

Re. 1 = 100z

A/C

10x + 20y + 100z = 1600

The ratio of number of coins 10 paise coins and 20 paise is 6 : 1  (given)

x : y = 6 : 1

x/y = 6/1

x = 6y

10x + 20y + 100z = 1600

10(6y) + 20y + 100z = 1600

60y + 20y + 100z = 1600

80y + 100z = 1600

20 (4y + 5z) = 1600

4y + 5z = 1600/20

4y + 5z = 80

Now,

Take z = 1,2,3,4,5 and check

we get,

z = 4

4y + 5z = 80

4y + 5(4) = 80

4y = 80 - 20

4y = 60

y = 60/4

y = 15

So,

Re. 1 = 100z

= 100(4)

= 400paise

Therefore,

Minimum 4 coins of Re. 1 will be there.

Hence,

The correct answer is option (d) 4.

Thank You