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.
