Answer :
Sure! Let's solve each of the problems step-by-step:
### Problem 1
Mr. Yeboah is 10 years older than his wife Rebecca. Thirty years ago, he was twice as old as his wife.
1. Let Rebecca's current age be [tex]\( R \)[/tex].
2. Therefore, Mr. Yeboah's current age is [tex]\( R + 10 \)[/tex].
Thirty years ago:
- Rebecca's age was [tex]\( R - 30 \)[/tex].
- Mr. Yeboah's age was [tex]\( (R + 10) - 30 \)[/tex].
According to the problem:
[tex]\[ (R + 10) - 30 = 2 \times (R - 30) \][/tex]
Simplify and solve:
- [tex]\( R - 20 = 2R - 60 \)[/tex]
- [tex]\( 60 - 20 = 2R - R \)[/tex]
- [tex]\( 40 = R \)[/tex]
Rebecca is currently 40 years old.
### Problem 2
Thirty balls were removed from a bag. The number of balls left is one-third of what was originally in the bag.
1. Let the original number of balls be [tex]\( B \)[/tex].
2. After removing 30 balls, the balls left are [tex]\( B - 30 \)[/tex].
According to the problem:
[tex]\[ B - 30 = \frac{1}{3}B \][/tex]
Solve for [tex]\( B \)[/tex]:
- [tex]\( 3(B - 30) = B \)[/tex]
- [tex]\( 3B - 90 = B \)[/tex]
- [tex]\( 2B = 90 \)[/tex]
- [tex]\( B = 45 \)[/tex]
The number of balls now in the bag after removing 30 is:
[tex]\[ B - 30 = 45 - 30 = 15 \][/tex]
So, there are 15 balls in the bag now.
### Problem 3
The sum of two integers is 80. One integer is 20 more than three times the other.
1. Let the smaller integer be [tex]\( x \)[/tex].
2. Then, the larger integer is [tex]\( 3x + 20 \)[/tex].
According to the problem:
[tex]\[ x + (3x + 20) = 80 \][/tex]
Simplify and solve:
- [tex]\( 4x + 20 = 80 \)[/tex]
- [tex]\( 4x = 60 \)[/tex]
- [tex]\( x = 15 \)[/tex]
The larger integer is:
[tex]\[ 3x + 20 = 45 + 20 = 65 \][/tex]
The positive difference between the integers:
[tex]\[ 65 - 15 = 50 \][/tex]
The positive difference between the integers is 50.
### Problem 4
A man has 50 coins consisting of nickels and quarters. The total value is [tex]$7.50.
1. Let the number of nickels be \( N \).
2. The number of quarters is \( 50 - N \).
The value equation:
\[ 0.05N + 0.25(50 - N) = 7.5 \]
Solve for \( N \):
- \( 0.05N + 12.5 - 0.25N = 7.5 \)
- \( -0.2N + 12.5 = 7.5 \)
- \( -0.2N = -5 \)
- \( N = 25 \)
The amount in dollars from nickels:
\[ 0.05 \times 25 = 1.25 \]
So, he has $[/tex]1.25 in nickels.
### Problem 5
There are three outlets at the bottom of a tank. Outlet 1 drains in 10 hours, outlet 2 in 20 hours, and outlet 3 in 30 hours.
1. The rate of Outlet 1 is [tex]\( \frac{1}{10} \)[/tex] tanks per hour.
2. The rate of Outlet 2 is [tex]\( \frac{1}{20} \)[/tex] tanks per hour.
3. The rate of Outlet 3 is [tex]\( \frac{1}{30} \)[/tex] tanks per hour.
Combined rate:
[tex]\[ \frac{1}{10} + \frac{1}{20} + \frac{1}{30} = \frac{6 + 3 + 2}{60} = \frac{11}{60} \][/tex]
Time to drain the tank:
[tex]\[ \frac{1}{\frac{11}{60}} = \frac{60}{11} \approx 5.45 \][/tex]
The tank will take approximately 5.45 hours to drain.
### Problem 6
Dilute 200 ml of a 30% sugar solution with 50 ml of water.
1. Initial sugar content is [tex]\( 0.3 \times 200 = 60 \)[/tex].
2. Total volume after adding water is [tex]\( 200 + 50 = 250 \)[/tex].
New concentration:
[tex]\[ \frac{60}{250} = 0.24 \][/tex]
The concentration of the resultant solution is 24%.
### Problem 7
20% of candies in bag A are Tom Toms; the rest are Hacks. Bag B has twice as many candies as bag A with half Tom Toms.
1. Let the candies in bag A be [tex]\( C_A \)[/tex].
2. Candies in bag B is [tex]\( 2C_A \)[/tex].
3. Total candies in bag C is [tex]\( C_A + 2C_A = 3C_A \)[/tex].
Tom Toms in bag A: [tex]\( 0.2 \times C_A \)[/tex].
Tom Toms in bag B: [tex]\( 0.5 \times 2C_A = C_A \)[/tex].
Total Hacks:
[tex]\[ 0.8 \times C_A + C_A = 1.8C_A \][/tex]
Percentage of Hacks in bag C:
[tex]\[ \frac{1.8C_A}{3C_A} \times 100\% = 60\% \][/tex]
The percentage of candies that are Hacks in bag C is 60%.
### Problem 8
Jojo drove from town A to B at 60 km/hr and back at 40 km/hr.
The average speed over the two-way trip is given by the harmonic mean:
[tex]\[ \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = 48 \][/tex]
The average speed over the entire journey is 48 km/hr.
### Problem 1
Mr. Yeboah is 10 years older than his wife Rebecca. Thirty years ago, he was twice as old as his wife.
1. Let Rebecca's current age be [tex]\( R \)[/tex].
2. Therefore, Mr. Yeboah's current age is [tex]\( R + 10 \)[/tex].
Thirty years ago:
- Rebecca's age was [tex]\( R - 30 \)[/tex].
- Mr. Yeboah's age was [tex]\( (R + 10) - 30 \)[/tex].
According to the problem:
[tex]\[ (R + 10) - 30 = 2 \times (R - 30) \][/tex]
Simplify and solve:
- [tex]\( R - 20 = 2R - 60 \)[/tex]
- [tex]\( 60 - 20 = 2R - R \)[/tex]
- [tex]\( 40 = R \)[/tex]
Rebecca is currently 40 years old.
### Problem 2
Thirty balls were removed from a bag. The number of balls left is one-third of what was originally in the bag.
1. Let the original number of balls be [tex]\( B \)[/tex].
2. After removing 30 balls, the balls left are [tex]\( B - 30 \)[/tex].
According to the problem:
[tex]\[ B - 30 = \frac{1}{3}B \][/tex]
Solve for [tex]\( B \)[/tex]:
- [tex]\( 3(B - 30) = B \)[/tex]
- [tex]\( 3B - 90 = B \)[/tex]
- [tex]\( 2B = 90 \)[/tex]
- [tex]\( B = 45 \)[/tex]
The number of balls now in the bag after removing 30 is:
[tex]\[ B - 30 = 45 - 30 = 15 \][/tex]
So, there are 15 balls in the bag now.
### Problem 3
The sum of two integers is 80. One integer is 20 more than three times the other.
1. Let the smaller integer be [tex]\( x \)[/tex].
2. Then, the larger integer is [tex]\( 3x + 20 \)[/tex].
According to the problem:
[tex]\[ x + (3x + 20) = 80 \][/tex]
Simplify and solve:
- [tex]\( 4x + 20 = 80 \)[/tex]
- [tex]\( 4x = 60 \)[/tex]
- [tex]\( x = 15 \)[/tex]
The larger integer is:
[tex]\[ 3x + 20 = 45 + 20 = 65 \][/tex]
The positive difference between the integers:
[tex]\[ 65 - 15 = 50 \][/tex]
The positive difference between the integers is 50.
### Problem 4
A man has 50 coins consisting of nickels and quarters. The total value is [tex]$7.50.
1. Let the number of nickels be \( N \).
2. The number of quarters is \( 50 - N \).
The value equation:
\[ 0.05N + 0.25(50 - N) = 7.5 \]
Solve for \( N \):
- \( 0.05N + 12.5 - 0.25N = 7.5 \)
- \( -0.2N + 12.5 = 7.5 \)
- \( -0.2N = -5 \)
- \( N = 25 \)
The amount in dollars from nickels:
\[ 0.05 \times 25 = 1.25 \]
So, he has $[/tex]1.25 in nickels.
### Problem 5
There are three outlets at the bottom of a tank. Outlet 1 drains in 10 hours, outlet 2 in 20 hours, and outlet 3 in 30 hours.
1. The rate of Outlet 1 is [tex]\( \frac{1}{10} \)[/tex] tanks per hour.
2. The rate of Outlet 2 is [tex]\( \frac{1}{20} \)[/tex] tanks per hour.
3. The rate of Outlet 3 is [tex]\( \frac{1}{30} \)[/tex] tanks per hour.
Combined rate:
[tex]\[ \frac{1}{10} + \frac{1}{20} + \frac{1}{30} = \frac{6 + 3 + 2}{60} = \frac{11}{60} \][/tex]
Time to drain the tank:
[tex]\[ \frac{1}{\frac{11}{60}} = \frac{60}{11} \approx 5.45 \][/tex]
The tank will take approximately 5.45 hours to drain.
### Problem 6
Dilute 200 ml of a 30% sugar solution with 50 ml of water.
1. Initial sugar content is [tex]\( 0.3 \times 200 = 60 \)[/tex].
2. Total volume after adding water is [tex]\( 200 + 50 = 250 \)[/tex].
New concentration:
[tex]\[ \frac{60}{250} = 0.24 \][/tex]
The concentration of the resultant solution is 24%.
### Problem 7
20% of candies in bag A are Tom Toms; the rest are Hacks. Bag B has twice as many candies as bag A with half Tom Toms.
1. Let the candies in bag A be [tex]\( C_A \)[/tex].
2. Candies in bag B is [tex]\( 2C_A \)[/tex].
3. Total candies in bag C is [tex]\( C_A + 2C_A = 3C_A \)[/tex].
Tom Toms in bag A: [tex]\( 0.2 \times C_A \)[/tex].
Tom Toms in bag B: [tex]\( 0.5 \times 2C_A = C_A \)[/tex].
Total Hacks:
[tex]\[ 0.8 \times C_A + C_A = 1.8C_A \][/tex]
Percentage of Hacks in bag C:
[tex]\[ \frac{1.8C_A}{3C_A} \times 100\% = 60\% \][/tex]
The percentage of candies that are Hacks in bag C is 60%.
### Problem 8
Jojo drove from town A to B at 60 km/hr and back at 40 km/hr.
The average speed over the two-way trip is given by the harmonic mean:
[tex]\[ \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = 48 \][/tex]
The average speed over the entire journey is 48 km/hr.
