Question

Juma has 21 coins whose total value is 72. There are twice as many 5 sh.coins as there are 10 sh. coins. The rest are 1sh. coin. Find the number of 10sh. coins that Juma has.​

Answer 1

Explanation:

Let's use algebra to solve this problem.

Let: x = the number of 10 shilling coins

2x = the number of 5 shilling coins (since there are twice as many 5 shilling coins as 10 shilling coins)

Now, we can set up an equation based on the total value of the coins:

10x (value of 10 shilling coins) + 5(2x) (value of 5 shilling coins) + 1(21 - 3x) (value of 1 shilling coins) = 72

Now, let's solve for x:

10x + 10x + 21 - 3x = 72

Combine like terms:

17x + 21 - 3x = 72

Now, subtract 21 from both sides:

17x - 3x = 72 - 21

14x = 51

Now, divide both sides by 14 to solve for x:

x = 51 / 14

Now, calculate x:

x ≈ 3.64

Since you can't have a fraction of a coin, round x to the nearest whole number. So, Juma has 4 ten-shilling coins.

And since there are twice as many 5 shilling coins as 10 shilling coins, he has 2x = 2 * 4 = 8 five-shilling coins.

The remaining coins are 1 shilling coins, so to find the number of 1 shilling coins, subtract the total number of 10 and 5 shilling coins from the total number of coins (21):

Number of 1 shilling coins = 21 - (4 + 8) = 21 - 12 = 9.

So, Juma has 4 ten-shilling coins, 8 five-shilling coins, and 9 one-shilling coins.