# Shahid had a certain number of Re 1 coins, Rs 2 coins and Rs 10 coins. If..

Shahid had a certain number of Re 1 coins, Rs 2 coins and Rs 10 coins. If the number of Re 1 coins he had is six times the number of Rs 2 coins Shahid had, and the total worth of his coins is Rs 160, find the maximum number of Rs 10 coins Shahid could have had. 18 Respuestas
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18 Respuestas
• If he had 5, ₹2 coins the value will be ₹10. Then the number of ₹1 coin is 30(5*6) that values ₹30. Then the maximum number of ₹10 coin is 12, that costs ₹120. 10+30+120=160

• 1(6x)+2(x)+10(y) = 160 8x+10y = 160 Min money by ₹1 & ₹2 coins = LCM(8,10). =40 Remaining amount = 160 - 40 = 120. Max number of ₹10 coins = 120 / 10 = 12

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• Take this logically For 1 two rupee coin sum of 1&2 =6+2=8 For 2 two rupee coin sum of 1&2 ruppe coin=12+4=16 For 3 two rupee coin sum of 1&2 ruppe coin=18+6=24 For 4 two rupee coin sum of 1&2 ruppe coin=24+8=32 For 5 two rupee coin sum of 1&2 ruppe coin=30+10=40 NOW the digits are in multiple of 10 and we can add ₹10 coins to make up 160 We need 12 coins So maximum is 12 coins

• Let x , y, z be no of Re1, Rs2 and Rs 10 coins respectively .
equations are:-
X+Y+Z=160;
X=6Y;
Solving these two , we get Z=12

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• X - re 1
Y - rs 2
Z - rs 10. Where X Y S are no. Of coins
So X = 6y, X+2y+10z = 160
From 1st eqn , 2nd eqn becomes 8y + 10 z = 160
The maximum no. For z as a possibility is z = 12 and y = 5. I.e 8(5) + 10(12) = 40+120 = 160 Otras discusiones relacionadas 