9. Manas has a small box with 740 coins of denominations ₹1, ₹2, and ₹5. If the number of ₹2 coins is 3 times the number of ₹1 coins, and the number of ₹5 coins is three times the number of ₹2 coins, how many coins of each denomination does he have?



Answer :

Sure, let's solve this step-by-step!

We are given a total amount of ₹740 in coins of denominations ₹1, ₹2, and ₹5. We also know the following relationships about the number of coins for each denomination:

1. The number of ₹2 coins is three times the number of ₹1 coins.
2. The number of ₹5 coins is three times the number of ₹2 coins.

Let's denote the number of ₹1 coins as [tex]\( x \)[/tex].

From the first relationship, the number of ₹2 coins will be [tex]\( 3x \)[/tex].

From the second relationship, the number of ₹5 coins will be [tex]\( 3 \times (3x) = 9x \)[/tex].

Now, we can write the total value of the coins in terms of [tex]\( x \)[/tex]:

- The value of the ₹1 coins is [tex]\( 1 \times x = x \)[/tex].
- The value of the ₹2 coins is [tex]\( 2 \times 3x = 6x \)[/tex].
- The value of the ₹5 coins is [tex]\( 5 \times 9x = 45x \)[/tex].

Summing these values should give us the total amount of ₹740:

[tex]\[ x + 6x + 45x = 740 \][/tex]

Combining like terms, we get:

[tex]\[ 52x = 740 \][/tex]

Solving for [tex]\( x \)[/tex]:

[tex]\[ x = \frac{740}{52} = 14 \][/tex]

So, the number of ₹1 coins is [tex]\( x = 14 \)[/tex].

The number of ₹2 coins is [tex]\( 3x = 3 \times 14 = 42 \)[/tex].

The number of ₹5 coins is [tex]\( 9x = 9 \times 14 = 126 \)[/tex].

Therefore, Manas has:
- 14 coins of ₹1 denomination,
- 42 coins of ₹2 denomination,
- 126 coins of ₹5 denomination.