Question

Ram has 18 coins in the denominations of rs1,rs2, rs5. If their total value isrs54 and the number ofrs2 coins are greater than that ofrs5coins then find the number ofrs1 coins with him

Answer 1

Answer:

Let rs. 1 be x, rs.2 be y, rs. 5 be z.

ATQ,

(x+y+z= 18) first equation

(x+y+2+z x 5= 54). second equation

(y+4z=36)

given, (y > z)

z=7 and y= 8

x= 18-8-7= 18-15= 3

so, x=3

Hence, rs.1 coins are 3, rs.2 coins are 8 and rs.5 coins are 7.

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Answer 2

Step-by-step explanation:

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

∴ x+y+z = 18 ____________(1) (∵ Ram has total 18 coins) Their value is Rs 54 ∴ x + 2y + 5z = 54 ____________(2) Given that number of Rs. 2 coins is more than of Rs. 5 coins.

∴ y > z ____________(3)

∵ Total number of coins is 18.

∴ 0 < x ≤ 18 , 0 < y ≤ 18 & 0 < z ≤ 18. ___________(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.