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Sagot :
Step by step Explanation:
To solve for the number of each type of coin in the jar, let's define variables for the number of each type of coin:
Let [tex]\( x_1 \), \( x_2 \), \( x_5 \)[/tex], and [tex]\( x_{10} \)[/tex] represent the number of 1p, 2p, 5p, and 10p coins respectively.
From the problem statement, we have the following information:
1. Total number of coins: [tex]\( x_1 + x_2 + x_5 + x_{10} = 20 \)[/tex]
2. Probability of picking each type of coin:
- Probability of picking a 1p coin: [tex]\( \frac{x_1}{20} = \frac{1}{5} \)[/tex]
- Probability of picking a 2p coin: [tex]\( \frac{x_2}{20} = \frac{1}{2} \)[/tex]
Let's solve these equations step by step:
From [tex]\( \frac{x_1}{20} = \frac{1}{5} \)[/tex]:
[tex]\[ x_1 = \frac{1}{5} \times 20 = 4 \][/tex]
From [tex]\( \frac{x_2}{20} = \frac{1}{2} \)[/tex]:
[tex]\[ x_2 = \frac{1}{2} \times 20 = 10 \][/tex]
Now, using the equation [tex]\( x_1 + x_2 + x_5 + x_{10} = 20 \)[/tex], substitute [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ 4 + 10 + x_5 + x_{10} = 20 \][/tex]
[tex]\[ x_5 + x_{10} = 20 - 14 \][/tex]
[tex]\[ x_5 + x_{10} = 6 \][/tex]
We also know the total value of the coins in the jar is 59 pence:
[tex]\[ 1 \cdot x_1 + 2 \cdot x_2 + 5 \cdot x_5 + 10 \cdot x_{10} = 59 \][/tex]
Substitute [tex]\( x_1 = 4 \)[/tex] and [tex]\( x_2 = 10 \)[/tex]:
[tex]\[ 1 \cdot 4 + 2 \cdot 10 + 5 \cdot x_5 + 10 \cdot x_{10} = 59 \][/tex]
[tex]\[ 4 + 20 + 5x_5 + 10x_{10} = 59 \][/tex]
[tex]\[ 24 + 5x_5 + 10x_{10} = 59 \][/tex]
[tex]\[ 5x_5 + 10x_{10} = 35 \][/tex]
[tex]\[ x_5 + 2x_{10} = 7 \][/tex]
Now, solve the system of equations:
1. [tex]\( x_5 + x_{10} = 6 \)[/tex]
2. [tex]\( x_5 + 2x_{10} = 7 \)[/tex]
Subtract the first equation from the second:
[tex]\[ (x_5 + 2x_{10}) - (x_5 + x_{10}) = 7 - 6 \][/tex]
[tex]\[ x_{10} = 1 \][/tex]
Substitute [tex]\( x_{10} = 1 \)[/tex] into [tex]\( x_5 + x_{10} = 6 \)[/tex]:
[tex]\[ x_5 + 1 = 6 \][/tex]
[tex]\[ x_5 = 5 \][/tex]
So, the number of each type of coin in the jar is:
- 1p coins [tex](\( x_1 \))[/tex]: 4
- 2p coins [tex](\( x_2 \))[/tex]: 10
- 5p coins [tex](\( x_5 \))[/tex]: 5
- 10p coins [tex](\( x_{10} \))[/tex]: 1
Therefore, there are 4 coins of 1p, 10 coins of 2p, 5 coins of 5p, and 1 coin of 10p in the jar.
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