Word Problem: Speed, investment & Work Grade 7 worksheet | Live ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

A 545-kilometer, 5-hour trip to Town A was driven at two speeds. The car’s speed is 105 kilometers per hour on the first part of the trip and 115 kilometers per hour for the second part. How long did the car drive at each speed?

#### Solution:
Let:
- \( t_1 \) be the time (in hours) the car drove at 105 km/h.
- \( t_2 \) be the time (in hours) the car drove at 115 km/h.

From the problem, we have two equations:
1. The total time equation:
\[
t_1 + t_2 = 5
\]
2. The total distance equation:
\[
105t_1 + 115t_2 = 545
\]

We can solve these equations simultaneously. First, solve the first equation for \( t_1 \):
\[
t_1 = 5 - t_2
\]

Substitute \( t_1 = 5 - t_2 \) into the second equation:
\[
105(5 - t_2) + 115t_2 = 545
\]

Expand and simplify:
\[
525 - 105t_2 + 115t_2 = 545
\]
\[
525 + 10t_2 = 545
\]

Solve for \( t_2 \):
\[
10t_2 = 545 - 525
\]
\[
10t_2 = 20
\]
\[
t_2 = 2
\]

Now substitute \( t_2 = 2 \) back into \( t_1 = 5 - t_2 \):
\[
t_1 = 5 - 2 = 3
\]

Thus, the car drove:
- 3 hours at 105 km/h,
- 2 hours at 115 km/h.

Answer:
\[
\boxed{3 \text{ hours at } 105 \text{ km/h}, 2 \text{ hours at } 115 \text{ km/h}}
\]

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Problem 2:

Two runners start at the same point and travel in opposite directions. One runner travels 6 kilometers per hour slower than the other. If the two runners are 76 kilometers apart after 2 hours, what is their speed?

#### Solution:
Let:
- \( v \) be the speed of the faster runner (in km/h).
- \( v - 6 \) be the speed of the slower runner (in km/h).

The distance covered by the faster runner in 2 hours is:
\[
2v
\]

The distance covered by the slower runner in 2 hours is:
\[
2(v - 6)
\]

Since they are traveling in opposite directions, the total distance between them after 2 hours is the sum of the distances they each traveled:
\[
2v + 2(v - 6) = 76
\]

Simplify the equation:
\[
2v + 2v - 12 = 76
\]
\[
4v - 12 = 76
\]

Solve for \( v \):
\[
4v = 88
\]
\[
v = 22
\]

Thus, the speed of the faster runner is 22 km/h, and the speed of the slower runner is:
\[
v - 6 = 22 - 6 = 16 \text{ km/h}
\]

Answer:
\[
\boxed{22 \text{ km/h} \text{ and } 16 \text{ km/h}}
\]

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Problem 3:

Matthew received 30,000 as a gift from his parents. He then put part of the money in an account which offers 6% interest and the rest into stocks paying 7%. If the total income is 2,000 baht, how much did he invest in stocks?

#### Solution:
Let:
- \( x \) be the amount invested in the account offering 6% interest.
- \( 30,000 - x \) be the amount invested in stocks offering 7%.

The interest earned from the account is:
\[
0.06x
\]

The interest earned from stocks is:
\[
0.07(30,000 - x)
\]

The total interest earned is given as 2,000 baht:
\[
0.06x + 0.07(30,000 - x) = 2,000
\]

Expand and simplify:
\[
0.06x + 0.07 \cdot 30,000 - 0.07x = 2,000
\]
\[
0.06x + 2,100 - 0.07x = 2,000
\]
\[
-0.01x + 2,100 = 2,000
\]

Solve for \( x \):
\[
-0.01x = 2,000 - 2,100
\]
\[
-0.01x = -100
\]
\[
x = 10,000
\]

Thus, the amount invested in the account is 10,000 baht, and the amount invested in stocks is:
\[
30,000 - x = 30,000 - 10,000 = 20,000 \text{ baht}
\]

Answer:
\[
\boxed{20,000}
\]

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Problem 4:

Ana invested 15,000, some at 8% and the rest at 7%. If she wants to earn 1,110 baht, how much should she invest at each rate?

#### Solution:
Let:
- \( x \) be the amount invested at 8%.
- \( 15,000 - x \) be the amount invested at 7%.

The interest earned from the 8% investment is:
\[
0.08x
\]

The interest earned from the 7% investment is:
\[
0.07(15,000 - x)
\]

The total interest earned is given as 1,110 baht:
\[
0.08x + 0.07(15,000 - x) = 1,110
\]

Expand and simplify:
\[
0.08x + 0.07 \cdot 15,000 - 0.07x = 1,110
\]
\[
0.08x + 1,050 - 0.07x = 1,110
\]
\[
0.01x + 1,050 = 1,110
\]

Solve for \( x \):
\[
0.01x = 1,110 - 1,050
\]
\[
0.01x = 60
\]
\[
x = 6,000
\]

Thus, the amount invested at 8% is 6,000 baht, and the amount invested at 7% is:
\[
15,000 - x = 15,000 - 6,000 = 9,000 \text{ baht}
\]

Answer:
\[
\boxed{6,000 \text{ at } 8\%, 9,000 \text{ at } 7\%}
\]

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Problem 5:

Chris and Ben have to stuff and mail 1000 envelopes for a new marketing campaign. Ben can do the job alone in 6 hours. If Chris helps, they can get the job done in 4 hours. How long would it take Chris to do the job by himself?

#### Solution:
Let:
- \( B \) be Ben's work rate (envelopes per hour). Since Ben can complete the job in 6 hours, his work rate is:
\[
B = \frac{1000}{6} = \frac{500}{3} \text{ envelopes per hour}
\]

- \( C \) be Chris's work rate (envelopes per hour). We need to find \( C \).

When Chris and Ben work together, their combined work rate is:
\[
B + C = \frac{1000}{4} = 250 \text{ envelopes per hour}
\]

Substitute \( B = \frac{500}{3} \) into the equation:
\[
\frac{500}{3} + C = 250
\]

Solve for \( C \):
\[
C = 250 - \frac{500}{3}
\]
\[
C = \frac{750}{3} - \frac{500}{3}
\]
\[
C = \frac{250}{3} \text{ envelopes per hour}
\]

To find how long it takes Chris to do the job alone, use the formula:
\[
\text{Time} = \frac{\text{Total Work}}{\text{Work Rate}}
\]
\[
\text{Time} = \frac{1000}{\frac{250}{3}} = 1000 \cdot \frac{3}{250} = 12 \text{ hours}
\]

Answer:
\[
\boxed{12}
\]

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Final Answers:

1. \(\boxed{3 \text{ hours at } 105 \text{ km/h}, 2 \text{ hours at } 115 \text{ km/h}}\)
2. \(\boxed{22 \text{ km/h} \text{ and } 16 \text{ km/h}}\)
3. \(\boxed{20,000}\)
4. \(\boxed{6,000 \text{ at } 8\%, 9,000 \text{ at } 7\%}\)
5. \(\boxed{12}\)