Solving Word Problems Using Linear Equations in One Variable

Algorithm of Solving Word Problems Using Equations

Examples of Problem Solving

Problem 1. One side of a triangle is twice as long as the second side and 3 cm shorter than the third side. Find the sides of the triangle if its perimeter is 43 cm.

Solution:

Denote the side AB=x. Fill in the table:

The perimeter: P = AB+AC+BC = x+2x+(2x+3) = 43

$$5x+3 = 43 \iff 5x = 40 \iff x = 40:5 = 8$$

The lengths:

AB=x=8 cm,AC=2x=16 cm,BC=2x+3=19 cm

Answer: 8 cm,16 cm and 19 cm

Problem 2. At a speed of 70 km/h, a train travels the distance between two stations half an hour faster than at a speed of 60 km/h. Find this distance.

Solution:

Let x be the distance between the stations.

Fill in the table:

As stated in the problem, the difference in the times spent:

$$ \frac{x}{60} - \frac{x}{70} = \frac{1}{2} $$

Solve the equation: $ \frac{x}{60} - \frac{x}{70} = \frac{1}{2} | \times 420 \iff 7x-6x = 210 \iff x = 210 $

The distance between the stations is 210 km.

Answer: 210 km

Problem 3. A team of workers was supposed to produce a certain number of parts in 5 days, but was able to complete the task in 4 days, making 12 more parts every day. How many parts did the team produce?

Solution:

Let x be the total number of the parts produced.

Fill in the table:

As stated in the problem, the difference in the number of parts produced daily:

$$ \frac{x}{4} - \frac{x}{5} = 12 $$

Solve the equation: $ \frac{x}{4} - \frac{x}{5} = 12 | \times 20 \iff 5x-4x = 240 \iff x = 240 $

The team produced 240 parts.

Answer: 240 parts

Problem 4. The sum of two numbers is 90. If the greater of them is divided by the smaller, then the quotient is 3 and the remainder is 6. Find these numbers.

Solution:

Let x be the smaller of the numbers. Then, the greater is 90-x. As stated in the problem: 90-x=3x+6

Solve the equation:

$$ 90-6 = 3x+x \iff 4x = 84 \iff x = 21 $$

The smaller number is x=21, the greater number is 90-x=69.

Answer: 21 and 69

Problem 5. The mother is 37 years old and the daughter is 13 years old. When was (or, will be) the daughter three times younger than her mother? And, twice as young?

Solution:

Let x be the number of next years. Then, the mother will be 37+x years old, and the daughter will be 13+x years old.

The quotient of the ages:

$$ \frac{37+x}{13+x} = 3 \iff 37+x = 3(13+x) \iff 37+x = 39+3x \iff 37-39 = 3x-x \iff $$

$$ \iff 2x = -2 \iff x = -1 $$

The daughter was three times younger than her mother 1 year ago.

The second quotient:

$$ \frac{37+x}{13+x} = 2 \iff 37+x = 2(13+x) \iff 37+x = 26+2x \iff 37-26 = 2x-x \iff $$

$$ \iff x = 11 $$

The daughter will be half her mother's age in 11 years.

Answer: 1 year ago; in 11 years

Problem 6. How old are the father and the son, if a year ago the son was 5 times younger, and next year he will be 4 times younger?

Solution:

Let x be the son’s age this year.

Fill in the table:

For both the father and the son, the difference in times is three years

(x+1)-(x-2) = 3 = 4(x+1)-5(x-2)

Solve the equation:

$$ 4(x+1)-5(x-2) = 3 \iff 4x+4-5x+10 = 3 \iff 4x-5x = 3-14 \iff -x = -11 $$ $$ x = 11 $$

Now, the son is 11 years old.

Next year, the father will be 4(x+1)=4∙12=48 years old. Thus, he is 47 now.

Answer: 11 and 47.

Problem 7. The sum of the digits of a certain two-digit number is 7. After interchanging its digits, you get a two-digit number that is greater by 9 than the given one. Find the given number.

Solution:

Let x be the first digit, that is, the number of tens in the given number.

Fill in the table:

As stated in the problem, the difference between the values

(10(7-x)+x)-(10x+(7-x)) = 9

Solve the equation:

$$ (70-10x+x)-(10x+7-x) = 9 \iff 70-9x-9x-7 = 9 \iff $$ $$ \iff -18x = 9-63 \iff -18x = -54 \iff x = 3 $$

The first digit is x=3, the last digit is 7-x=4.

The given number is 34.

Answer: 34

Problem 8. According to the schedule, the bus should travel from the village to the station at a speed of 32 km/h and arrive at the station half an hour before the train leaves. But due to inclement weather, the bus was traveling slower by 7 km/h, and was late for the train by 12 minutes. What is the distance between the village and the station?

Solution:

Let x be the distance between the village and the station.

Fill in the table:

The difference between the scheduled time and the actual arrival:

30 min+12 min = 42 min = $\frac{42}{60}$ h = 0,7 h

$ \frac{x}{25}- \frac{x}{32} = 0,7 | \times 32 \cdot 25 $

$ 32x-25x = \frac{7}{10} \cdot 32 \cdot 25 = 7 \cdot 16 \cdot 5 $

$ 7x = 7 \cdot 16 \cdot 5 \iff x = 16 \cdot 5 = 80 $

The distance is 80 km.

Answer: 80 km

Problem 9*. If the digit 4 is added on the right and on the left of the two-digit number, then the result will be 54 times greater than the given number. Find the given two-digit number.

Solution:

Let x be the given number.

If we write 4 on the left and on the right, the resulting four-digit number has the following places: the first 4 indicates the thousands place, x stays for the number of tens (that is, the hundreds and tens places), the last 4 is for the ones. The quotient:

$$ \frac{4000+10x+4}{x} = 54 $$

Solve the equation: $ 4004+10x = 54x \iff 4004=44x \iff x = \frac{4004}{44} = \frac{1001}{11} = 91 $

The given number is x=91.

Answer: 91

Problem 10. A certain number of notebooks were purchased for the exam. If you put them in packs of 45 pieces, there will be one extra notebook, but if you put them in packs of 50 pieces, then you’ll lack 4 notebooks in one pack. How many notebooks were bought if there are one more of 45-pcs-packs than 50-pcs-packs?

Solution:

Let n be the number of 50-pcs-packs.

The total numbers are equal:

$$ 45(n+1)+1 = 50n-4 \iff 45n+45+1 = 50n-4 \iff 45n-50n = -4-46 \iff $$

$$ \iff-5n = -50 \iff n = 10 $$

Thus: 50 $\cdot$ 10-4 = 496 notebooks.

Answer: 496 notebooks.

Problem 11*. A company purchased two batches of goods for 125 thousand dollars. The first batch was sold with a profit of 25%, the second - with a profit of 50%. The total profit was 40%. What was the cost of each batch?

Solution:

Let x be the cost of the first batch.

Fill in the table:

The sum is:

1,25x+1,5(125-x) = 1,4 $\cdot$ 125 | $\times$ 4

5x+6(125-x) = 1,4 $\cdot$ 500

5x+750-6x = 700

x = 50

The cost of the first batch is x=50 K$, the second batch 125-x=75 K$

Answer: 50 K$ and 75 K$

Problem 12*. One ingot contains 10% of silver, and the second - 40%. The second ingot is 3 kg heavier than the first. The ingots were fused, and an ingot with 30% of silver was obtained. What is the mass of the resulting ingot?

Solution:

Let x be the mass of the first ingot.

Fill in the table:

The mass of silver in the sum:

0,1x+0,4(x+3) = 0,3(2x+3) |$\times$ 10

x+4(x+3) = 3(2x+3) $\iff$ x+4x+12 = 6x+9 $\iff$ x = 3

The mass of the first ingot is x=3 kg. The mass of the resulting ingot is 2x+3=9 kg.

Answer: 9 kg