Correct option is (A) 3

Let Ram has x number of Rs 1 coins, y number of Rs 2 coins and z number of Rs 5 coins.

\(\therefore\) x+y+z = 18    ____________(1)    \((\because\) Ram has total 18 coins)

Their value is Rs 54

\(\therefore\) x + 2y + 5z = 54    ____________(2)

Given that number of Rs. 2 coins is more than of Rs. 5 coins.

\(\therefore\) y > z    ____________(3)

\(\because\) Total number of coins is 18.

\(\therefore0<x\leq18,0<y\leq18\;\&\;0<z\leq18\)    ____________(4)

Subtract equation (1) from (2), we obtain

y + 4z = 36    ____________(5)

By considering inequilities (3) & (4), we can conclude the possible values of y & z.

(i) If z = 1, then y = 32    (From (5))

which is contradictions of inequility (4).

(ii) If z = 2 then y = 28

which is not possible.

(iii) If z = 3 then y = 24 which is not possible.

(iv) If z = 4 then y = 20 which is not possible.

(v) If z = 5 then y = 16 then y+z = 16+5 = 21 but total coins are 18.

Hence, this case is not possible.

(vi) If z = 6 then y = 12 then y+z = 6+12 = 18

Then, x = 0 which is not possible because Ram has Rs 1 coin.

(vii) If z = 7 then y = 36 - 28 = 82    (From (5))

Then x = 18 - y - z

= 18 - 8 - 7

= 18 - 15 = 3

Also x + 2y + 5z = 3+16+35

= 54             (Satisfied)

Hence, Ram has 3 Rs 1 coin, 8 Rs 2 coin and 7 Rs 5 coin.

Hence, number of Rs 1 coin Ram has = 3.

Correct option is A) 3