I have a total of ₹300 in coins of denomination ₹1,₹2 and ₹5.The number of ₹2 coins is 3 times the number of ₹5 coins.The total number of coins is 160.How many coins of each denominations are with me?
Let the number of ₹5 coins be x
then the number of ₹2 = 3x
and the number of ₹1 = 160 - x - 3x
= 160 - 4x
Now for the value:
5x + 2(3x) + 1(160 - 4x) = 300
5x + 6x + 160 - 4x = 300
7x = 140
x = 20
You have 20 ₹5 coins
60 of the ₹2 coins and
80 of the ₹1 coins
check:
do the coins total 160 ? YES
value: 5(20) + 2(60) + 1(80) = 300
my answer is correct
To solve this problem, we can use a system of equations:
Let's assume:
Number of ₹1 coins = x
Number of ₹2 coins = y
Number of ₹5 coins = z
We are given the following information:
1. The number of ₹2 coins is 3 times the number of ₹5 coins:
y = 3z
2. The total number of coins is 160:
x + y + z = 160
3. The total value of the coins is ₹300:
1x + 2y + 5z = 300
Now, we will solve this system of equations to find the values of x, y, and z.
From equation 1, we have y = 3z.
Substituting this value in equation 3, we get: 1x + 2(3z) + 5z = 300
Simplifying: x + 2z + 5z = 300
x + 7z = 300
Now, we have two equations:
x + 7z = 300 (equation 4)
x + y + z = 160 (equation 2)
We can solve these two equations simultaneously to find the values of x and z.
From equation 2, we have x = 160 - y.
Substituting this value in equation 4, we get: (160 - y) + 7z = 300
Simplifying: 160 - y + 7z = 300
- y + 7z = 140 (equation 5)
Now, we have two equations:
- y + 7z = 140 (equation 5)
x + 7z = 300 (equation 4)
Let's solve this system of equations to find the values of x and z.
Subtract equation 5 from equation 4 to eliminate x:
(x + 7z) - (- y + 7z) = 300 - 140
Simplifying: x + 7z + y - 7z = 160
Combining like terms: x + y = 160 (equation 6)
Now we have two equations:
x + y = 160 (equation 6)
- y + 7z = 140 (equation 5)
From equation 6, we have x = 160 - y.
Substituting this value in equation 5, we get: - y + 7z = 140
Now, we have two equations:
- y + 7z = 140 (equation 5)
160 - y = 7z (equation 7)
Rearranging equation 7, we get: y = 160 - 7z.
Now we have two equations:
- y + 7z = 140 (equation 5)
y = 160 - 7z (equation 7)
Substituting equation 7 into equation 5, we have: -(160 - 7z) + 7z = 140
Simplifying: -160 + 7z + 7z = 140
14z = 300
Dividing by 14 on both sides, we get: z = 300/14
Simplifying: z = 21.4285714
Since z represents the number of ₹5 coins, we cannot have a fraction of a coin. Therefore, z should be a whole number. But 21.4285714 is not a whole number.
Hence, this problem does not have a feasible solution.