Question

manav has a small box with rs. 740 in coins of denominator rs 1 ,2 and 4 . if the number of coins of denomination of rs 2 and 3 times the number of coins of rs1 and the number of coins of denominators of rs 5 are twice the number of coins of rs 2 . how many coins of each denominator does manav have?
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Answer 1

Answer:

So the answer is RS: 5/-✌✌✌❤❤

Step-by-step explanation:

Let us assume x,y,z be no of coins of 1, 2 and 5

respectively

y = 3x and z = 2y

Value = x+2y+5z = 740 becomes x+2(3x)+5(2)

(3x) = 740

37x = 740

x = 20: y = 60 and z = 120

There are 20 coins of 1 re, 60 coins of Rs. 2

and 120 coins of Rs. 5/-

Answer 2

Answer:

Let us assume x,y,z be no of coins of 1 , 2 and 5 respectively

y = 3x and z = 2y

Value = x+2y+5z = 740 becomes x+2(3x)+5(2)(3x) = 740

37x = 740

x = 20: y = 60 and z = 120

There are 20 coins of 1 re, 60 coins of Rs. 2 and 120 coins of Rs. 5/-