Students who want to get complete knowledge on Simultaneous Linear Equations Word problems can check this Worksheet on Problems on Simultaneous Linear Equations. Our System of Linear Equations Word Problems Worksheet is helpful to improve your preparation levels. It contains a number of examples of Simultaneous Linear Equations. So, practice all the questions from the Simultaneous Linear Equations Worksheet and develop your skills. Also, have a look at Worksheet on Simultaneous Linear Equations and prepare well for the exam.
1. The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number.
Solution:
Let the two-digit number be xy.
According to the data provided in the question,
The sum of the digits of a two-digit number is 7.
x + y = 7 —– (i)
If the numbers are reversed, then the number is increased by 27.
The sm of two digits of the number as can be written as 10x + y
If the numbers are reversed, then the number can be written as 10y + x
10y + x = 10x + y + 27
10x + y + 27 – x – 10y = 0
9x – 9y = -27
9(x – y) = -27
x – y = -3 —– (ii)
Adding equations (i) & (ii)
x + y + x – y = 7 + (-3)
2x = 7 – 3
2x = 4
x = 4/2
x = 2
Substituting x = 2 in equation (i)
2 + y = 7
y = 7 – 2
y = 5
Therefore, the required 2 digit number is 25.
2. Mahesh bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487. For the second time, he bought 6 bushes and 2 trees and totaled $232. The bills do not list the per-item price. What were the costs of one bush and of one tree?
Solution:
Let the cost of each bush be x, one tree is y.
As per the question,
The first condition is he bought 13 bushes and 4 trees from the nursery for the first time and its total cost was $487
13x + 4y = 487 —- (i)
The second condition is he bought 6 bushes and 2 trees and totaled $232
6x + 2y = 232 —– (ii)
Multiply the second equation by 2.
2(6x + 2y = 232)
12x + 4y = 464 —– (iii)
Subtract equation (iii) from equation (i)
12x + 4y – (13x + 4y) = 464 – 487
12x + 4y – 13x – 4y = -23
-x = -23
x = 23
Putting x = 23 in equation (i)
13(23) + 4y = 487
299 + 4y = 487
4y = 487 – 299
4y = 188
y = 188/4
y = 47
So, costs of one bush is $23, and of one tree is $47.
3. The first number is six times the second number. The difference between the numbers is 60. Find those two numbers?
Solution:
Let the first number be x, the second number be y.
As per the given data,
The first condition is the first number is six times the second number.
x = 6y —- (i)
The second condition is the difference between the numbers is 60.
x – y = 60 —– (ii)
Put equation (i) in equation (ii)
6y – y = 60
5y = 60
y = 60/6
y = 10
Substituting y = 10 in equation (i)
x = 6 x 10
x = 60
Therefore, the two numbers are 60, 10.
4. One number is three times the other number. The sum of two numbers is 24. Find the two numbers?
Solution:
Assume that two numbers are x, y.
Given that,
One number is three times the other number.
x = 3y —– (i)
The sum of two numbers is 24.
x + y = 24 —– (ii)
Substituting x = 3y in equation (ii)
3y + y = 24
4y = 24
y = 24/4
y = 6
Putting y = 6 in equation (ii)
x + 6 = 24
x = 24 – 6
x = 18
so, the two numbers are 18, 6.
5. It takes 3 hours for a boat to travel 27 miles upstream. The same boat can travel 30 miles downstream in 2 hours. Find the speeds of the boat and the current?
Solution:
Let x be the speed of the boat (without current), let y be the speed of the current.
According to the given problem,
The boat can travel 27 miles upstream in 3 hours
Upstream speed = (Upstream distance)/(Upstream time)
Upstream speed = 27/3 = 9 miles/hour
The boat can travel 30 miles downstream in 2 hours.
Downstream speed = (Downstream distance) / (Downstream time)
Downstream speed = 30/2 = 15 miles/hour
The linear equation for upstream speed is x – y = 9 —— (i)
downstream speed is x + y = 15 ——- (ii)
Adding equations (i) and (ii)
x – y + x + y = 9 + 15
2x = 24
x = 24/2
x = 12
Substituting x = 12 in equation (ii)
12 + y = 15
y = 15 – 12
y = 3
Therefore, the speed of boat is 12 miles/hour, current is 3 miles/hour.
6. The sum of two numbers is 28 and their difference is 4. Find the two numbers?
solution
Let us take the two numbers as x, y.
As per the data given in the question,
The sum of two numbers is 28.
x + y = 28 —— (i)
The difference between the two numbers is 4.
x – y = 4 ——- (ii)
Adding equations (i) and (ii)
x + y + x – y = 28 + 4
2x = 32
x = 32/2
x = 16
Substitute x = 16 in equation (i)
16 + y = 28
y = 28 – 16
y = 12
So, the two numbers are 16, 12.
7. The sum of two numbers is 14. The difference in their squares is 28. Find the numbers?
Solution:
Assume that the two numbers are x and y.
According to the first condition in the question,
The sum of two numbers is 14.
x + y = 14 —— (i)
y = 14 – x
According to the second condition in the question,
x² – y² = 28 —- (ii)
Substitute y = 14 – x in equation (ii)
x² – (14 – x)² = 28
x² – (14² + x² – 28x) = 28
x² – 196 – x² + 28x = 28
28x – 196 = 28
28x = 28 + 196
28x = 224
x = 224/28
x = 4.6
Substitute x = 4.6 in equation (i)
4.6 + y = 14
y = 14 – 4.6
y = 9.4
Therefore, the two numbers are 4.6, 9.4.
8. The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2200 people enter the fair and $5050 is collected. How many children and how many adults attended?
Solution:
Let us take the number of children who attended the fair as x, the number of adults who attended the fair as y.
The total number of people who attended the fair is 2200.
x + y = 2200 —- (i)
x = 2200 – y
The admission fee at a small fair is $1.50 for children and $4.00 for adults and the total amount collected is $5050.
1.5x + 4y = 5050 —— (ii)
Putting x = 2200 – y in the second equation.
1.5(2200 – y) + 4y = 5050
3300 – 1.5y + 4y = 5050
3300 + 2.5y = 5050
2.5y = 5050 – 3300
2.5y = 1750
y = 1750/2.5
y = 700
Substitute y = 700 in equation (i)
x + 700 = 2200
x = 2200 – 700
x = 1500
So, the number of children attended the small fair is 1500, the number of adults attended the fair is 700.
9. Emma has 23 notes of £20 and £5 in her handbag. The amount of money she has in the bag is £340.00. Find the number of notes of each type.
Solution:
Let us take the number of £20 notes as x, £5 notes as y.
The total number of notes in the handbag is 23.
x + y = 23 —- (i)
As per the second condition in the question,
20x + 5y = 340 —– (ii)
Multiply the equation (i) by 5.
5(x + y) = 23 x 5
5x + 5y = 115 —— (iii)
Subtracting equation (i) from equation (ii)
5x + 5y – (20x + 5y) = 115 – 340
5x + 5y – 20x – 5y = -225
-15x = -225
x = 225/15
x = 15
Putting x = 15 in equation (i)
15 + y = 23
y = 23 – 15
y = 8
So, Emma has 15 notes of £20, 8 notes of £5 in her handbag.
10. A family goes to the cinema. 4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50. Calculate the costs of each adult and child ticket.
Solution:
Let us take the cost of the ticket for adults as x, cost of the ticket for the child as y.
According to the question,
4 adults and 2 child tickets cost $47, 1 adult and 3 child tickets cost $25.50.
4x + 2y = 47 —— (i)
x + 3y = 25.5 —– (ii)
x = 25.5 – 3y
Putting x = 25.5 – 3y in equation (i)
4(25.5 – 3y) + 2y = 47
102 – 12y + 2y = 47
102 – 10y = 47
102 – 47 = 10y
10y = 55
y = 55/10
y = 5.5
Substitute y = 5.5 in equation (ii)
x + 3(5.5) = 25.5
x + 16.5 = 25.5
x = 25.5 – 16.5
x = 9
Therefore, the ticket cost for adults is $9, the ticket cost for child is $5.5.
11. The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increase by 2 units. If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units. Find the length and breadth of the rectangle.
Solution:
Let the length and breadth of the rectangle is l and b respectively.
The area of a rectangle is lb.
The area of a rectangle gets reduced by 10 square units if its length is reduced by 4 units and breadth is increased by 2 units.
(l – 4)(b + 2) = (lb – 10) sq units
lb + 2l – 4b – 8 = lb – 10
lb – lb – 2l + 4b = 10 – 8
4b – 2l = 2 —- (i)
If we increased the length by 3 units and breadth by 4 units, the area is increased by 96 square units.
(l + 3)(b + 4) = (lb + 96)
lb + 3b + 4l + 12 = lb + 96
lb + 3b + 4l – lb = 96 – 12
3b + 4l = 84 —— (ii)
Multiplying equation (i) by 2
4b – 2l = 2 x 2
8b – 4l = 4 —- (iii)
Adding equation (iii) in equation (ii)
8b – 4l + 3b + 4l = 84 + 4
11b = 88
b = 88/11
b = 8
Substituting b = 11 in equation (i)
4(11) – 2l = 2
44 – 2 = 2l
42 = 2l
l = 42/2
l = 21
So, the length and breadth of the rectangle is 21 units, 8 units.
12. If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12 and if 3, 8 is subtracted from the numerator and denominator it becomes 4/5. Find fractions.
Solution:
Let the numerator of the fraction be x, the denominator of the fraction be y.
As stated in the question,
If 5 is added to the numerator and 9 is subtracted from the denominator it becomes 21/12.
(x + 5)/(y – 9) = 21/12
Cross multiply the fractions.
12(x + 5) = 21(y – 9)
12x + 60 = 21y – 189
12x – 21y + 189 + 60 = 0
12x – 21y + 249 = 0 —- (ii)
If 3, 8 is subtracted from the numerator and denominator it becomes 4/5.
(x – 3)/(y – 8) = 4/5
Cross multiply the fractions.
5(x – 3) = 4(y – 8)
5x – 15 = 4y – 32
5x – 4y = -32 + 15
5x – 4y = -17
5x – 4y + 17 = 0
5x = 4y – 17
x = (4y – 17)/5 — (ii)
substitute equation (ii) value in equation (i)
12[(4y – 17)/5] – 21y + 249 = 0
[(48y – 204)/5] – 21y + 249 = 0
48y – 204 – 105y + 1245 = 0
-57y + 1041 = 0
57y = 1041
y = 1041/57
y = 18.26
Putting y = 18.26 in equation (ii)
x = (4(18.26) – 17)/5
x = (73.052 – 17)/5
x = 56.0526/5
x = 11.210
Therefore, the required fraction is 1121/1826.
13. Chole and Tino have a combined age of 48. Three years ago Chole was double the age Tino is now. Find the present ages of Chole and Tino.
Solution:
Let us take the age of chole and Tino as x, y.
As per the data given in the question,
The combined age of Chole and Tino is 48.
x + y = 48 —- (i)
Three years ago Chole was double the age Tino is now
x – 2y = 3
x = 3 + 2y —– (ii)
Putting x value in equation (i)
3 + 2y + y = 48
3 + 3y = 48
3y = 48 – 3
3y = 45
y = 45/3
y = 15
Substituting y = 15 in equation (ii)
x = 3 + 2(15)
x = 3 + 30
x = 33
So, the present age of Chole is 33 years, Tino is 15 years.
