﻿ Systems of Equations Word Problems | Problems & Solutions Systems of Equations Word Problems

Systems of Equations Word Problems
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1.
A fair ticket for children is $\$$1.50 and for adult is \$$4.50. If in a day 2000 people enter and total collection is$\$$4050. Find how many children and adult enter on that day? a. Adult = 1600, Children = 350 b. Adult = 350, Children = 1650 c. Adult = 400, Children = 1650 d. Adult = 1700, Children = 300 Solution: Let number of adult = x and number of children = y Total number: x + y = 2000.....(1) Total collection: 4.5x + 1.5y = 4050.....(2) Now we have 2 equations and 2 variables From equation (1) y = 2000 - x substitute y value in equation (2) 4.5x + 1.5(2000-x) = 4050 x = 350 substitute x value in equation 350 + y = 2000 y = 2000 - 350 y = 1650 Adult = 350 children = 1650 Correct answer : (2) 2. If the difference between two numbers is 49 and the sum is 123. Find the numbers? a. 86 and 35 b. 80 and 37 c. 86 and 37 d. 80 and 30 Solution: Let the numbers are x and y as given x + y = 123.......(1) x - y = 49......(2) add equation (1) and (2) 2x = 172 x = \frac{172}{2} x = 86 substitute x value in equation (1) 86 + y = 123 y = 123 - 86 y = 37 The numbers are 86 and 37 Correct answer : (3) 3. In a farmhouse total animals are 20, some of them are hens and some are cows. Total legs of these animals are 66. Find how many hens and cows are in the farmhouse? a. Hens = 7, cows = 13 b. Hens = 10, cows = 10 c. Hens = 5, cows = 15 d. Hens = 6, cows = 14 Solution: Let total hen = x total cow = y By given data x + y = 20......(1) as we know each cow have 4 legs and each hen have 2 legs 2x + 4y = 66....(2) multiply equation (1) with -4 and add with equation (2) -2x = -14 x = 7 substitute x value in equation (1) 7 + y = 20 y = 20 - 7 y = 13 hence, total hens = 7 and total cows = 13 Correct answer : (1) 4. Maria bought 5 tee shirts in \$$18. Each half sleeve tee shirt cost is $\$$3 and full tee sleeve shirt cost is \$$4. Find how many each type of tee she bought? a. Half sleeve tee = 1 ,Full sleeve tee = 4 b. Half sleeve tee = 5 ,Full sleeve tee = 0 c. Half sleeve tee = 2 ,Full sleeve tee = 3 d. Half sleeve tee = 3 ,Full sleeve tee = 2 Solution: Let half sleeve tee = x and full sleeve tee = y by given data, x + y = 5....(1) 3x + 4y = 18...(2) multiply by -4 in equation (1) and add with equation (2) x = 2 substitute x value in equation (1) 2 + y = 5 y = 3 hence, half sleeve tee = 2 and full sleeve tee = 3 Correct answer : (3) 5. A group of 100 students went to a trip. They used cars and buses altogether 7 vehicles. If in a car 4 students can sit and in a bus 40 students can sit. Find how much each car and bus they use? a. 2 cars and 5 buses b. 5 cars and 2 buses c. 4 cars and 3 buses d. 6 cars and 1 bus Solution: Let car = a and bus =b as given a + b = 7......(1) 4a + 40b = 100.....(2) multiply by -4 in equation (1) and add with equation (2) 36b = 72 b = 2 substitute b value in equation (1) a + 2 = 7 a = 5 Answer is 5 cars and 2 buses Correct answer : (2) 6. 12 men and 6 women can paint a wall in 12 hours. 6 men and 8 women can paint it in 15 hour. Find the time to paint the same wall by 1 men alone and 1 women alone. a. Men = 210 hours, Women = 200 hours b. Men = 225 hours, Women = 225 hours c. Men = 200 hours, Women = 225 hours d. Men = 225 hours, Women = 200 hours Solution: Let m = the part of the job by 1 men in 1 hour and w = the part of the job by 1 women in 1 hour By given we have, 12(12m + 6w) = 1.....(1) 15(6m + 8w) =1.....(2) 12m + 6w =$\frac{1}{12}$6m + 8w =$\frac{1}{15}$use elimination: -8(12m + 6w) =$\frac{1}{12}$(-8) 6(6m+8w) =$\frac{1}{15}$(6) m =$\frac{1}{225}$substitute m value in equation (1) 12(12($\frac{1}{225}$) + 6w) = 1 w =$\frac{1}{200}$Hence 225 hours to paint the wall by 1 men and 200 hours by a women Correct answer : (4) 7. In a test you have 30 questions worth 100 points. Some questions are 2 points and some are 4 points. Find how many each type of questions are in the test? a. 2 points = 10, 4 points = 20 b. 2 points = 15, 4 points = 15 c. 2 points = 20, 4 points = 10 d. 2 points = 25, 4 points = 5 Solution: Let 2 points questions = x and 4 points questions = y as given x + y = 30...(1) 2x + 4y = 100...(2) Multiply by -4 in equation (1) and add with equation (2) -2x = -20 x = 10 substitute x value in equation(1) 10 + y = 30 y = 30 - 10 y = 20 Answer is 2 points questions = 10 and 4 points questions = 20 Correct answer : (1) 8. Let two angles are supplementary. The measure of one angle is 40 degrees smaller than 3 times the other angle. Find the angles. a. 50 degrees and degrees 130 b. 55 degrees and degrees 125 c. 30 degrees and degrees 150 d. 55 degrees and degrees 125 Solution: Let two angles are x and y By given, x + y = 180...(1) x = 3y - 40....(2) substitute the x value from equation (2) to equation (1) 3y - 40 + y = 180 4y = 180 + 40 4y = 220 y = 55 again substitute y value in equation (2) x = 3(55) - 40 x = 125 Answer is 55 degrees and 125 degrees Correct answer : (2) 9. The equations 6x + 3y = 38 and 2x + 3y = 26 represent the money collected from a fair tickets. if each adult ticket cost is$\$$x and each child ticket cost is \$$y. Find the exact cost for each adult. a. $\$$5 b. \$$6 c.$\$$3 d. \$$2

Solution:

Given x = ticket for each adult
and y = ticket for each child

6x + 3y = 38....(1)
2x + 3y = 26....(2)

Multiply by -1 in equation(1) and add equation (2)
-4x = -12
x = 3

Ticket for each adult = $\$\$3

10.
Three small containers and one large container hold 10 cups of water. 1 large container minus 2 small containers hold 5 cups of water. How many cups of water can each container hold? a. Small container = 2 cup, Large container = 8 cups b. Small container = 0 cup, Large container = 10 cups c. Small container = 1 cup, Large container = 7 cups d. Non of them

Solution:

Let x = small container
and y = large container

3x + y = 10...(1)
y - 2x = 5...(2)

multiply by -1 in equation (2) and add with equation (1)
5x = 5x
x = 1

substitute x value in equation (1)
3 + y = 10
y = 7