WORD PROBLEMS ON SIMULTANEOUS LINEAR EQUATIONS

About "Word problems on simultaneous linear equations"

Solving word problems on simultaneous linear equations is sometimes a difficult job for some students.

Actually it is not. There is a simple trick behind it.

The picture given below tells us the trick.

Examples

To have better understanding on solving word problems on simultaneous linear equations, let us look at some examples. 

Example 1 :

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction. 

Solution :

Let "x/y" be the required fraction. 

"If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1"

From the above information, we have (x+2) / (y+1) = 1

(x+2) / (y+1) = 1 -----> x+2 = y+1 -----> x - y = -1 ----------(1)

"In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2"

From the above information, we have (x-4) / (y-2) = 1/2

(x-4) / (y-2) = 1/2 -----> 2(x-4) = y-2 -----> 2x - y = 6----------(1)

Solving (1) and (2), we get x = 7 and y = 8

So, x/y = 7/8

Hence, the required fraction is 7/8

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Example 2 :

A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?

Solution :

x ------> no. of adult tickets,         y ------> no. of kids tickets

According to the question, we have x + y = 548 ---------(1)

And also, 10x + 5y = 3750 ---------> 2x + y = 750 --------(2)

Solving (1) & (2), we get x = 202 and y = 346

Hence, the number of adults tickets sold = 202

the number of kids tickets sold = 346 

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Example 3 :

A number consists of three digits of which the middle one is zero  and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297.Find the number. 

Solution :

Let "x0y" be the required three digit number. (As per the given information, middle digit is zero)

"The sum of the other digits is 9" ---------> x + y = 9 --------(1)

"Interchanging the first and third digits" --------> y0x

From the information given the question, we can have

y0x - x0y = 297

(100y + x)  -  (100x + y)  =  297 ---------> 100y + x - 100x -y = 297

-99x + 99y = 297 ----------> -x + y = 3 --------(2)

Solving (1) & (2), we get  x = 3 and y = 6

So, x0y = 306

Hence the required number is 306

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Example 4 :

A manufacturer produces 80 units of a product at $22000 and 125 units at a cost of $28750. Assuming the cost curve to be linear, find the equation of the line and then use it to estimate the cost of 95 units.

Solution :

Since the cost curve is linear, its equation will be y = Ax + B.

(Here y = Total cost, x = no. of units)

80 units at $22000 --------> 22000 = 80A + B -------(1)

125 units at $28750 --------> 28750 = 125A + B -------(2)

Solving (1) and (2), we get A  = 150 and B = 10000

So, the equation of the line is y = 150x + 10000 --------(3)

To find the cost of 95 units, plug x = 95 in (3).

(3) -------> y = 150(95) + 10000

y = 14250 + 10000

y = 24250

Hence, the cost of 95 units is $24250

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Example 5 :

Y is older than X by 7 years. 15 years back X's age was 3/4  of Y's age. Find the present their present ages.

Solution :

X's present age = "x" and Y's present age = "y"

Y is older than X by 7 years --------> y = x + 7 --------(1)

15 years back--------> X's age = x-15 and Y's age = y -15

According to the question, we have (x-15) = (3/4)(y-15)

4(x-15) = 3(y-15) -------> 4x - 60 = 3y - 45 -----> 4x = 3y + 15 ------(2)

Solving (1) and (2), we get x = 36 and y = 43.

Hence, the present ages of "x" and "y" are 36 and 43

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Example 6 :

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers. 

Solution :

Let "x" and "y" be the required two numbers such that x > y.

From the information given in the question, we have

x + y = 16 ----------(1)

and 1/5(x)  =  (1/3)y ---------> 3x = 5y -------> 3x - 5 y = 0 --------(2)

Solving (1) and (2), we get x = 10 and y = 6.

Hence, the two numbers are 10 and 6

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Example 7 :

The wages of 8 men and 6 boys amount to $33. If 4 men earn $4.50 more than 5 boys, determine the wages of each man and boy.

Solution :

Let "x" and "y" be the wages of each man and boy.

From the information given in the question, we have

8x + 6y = 33 ----------(1)

Wages of 4 men = 4x

Wages of 5 boys = 5y

According to the question, we have 4x - 5y = 4.50 -----------(2)

Solving (1) and (2), we have x = 3 and y = 1.5.

Hence, the wages of each man and each boy are $3 and $1.50 respectively

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Example 8 :

A number between 10 and 100 is five times the sum of its digits. If 9 be added to it the digits are reversed. Find the number.

Solution :

Let "xy" be the required number between 10 and 100. (Two digit number)

"A number between 10 and 100 is five times the sum of its digits"

From the information above, we have

xy = 5(x+y) -------> 10x + y = 5x + 5y ------> 5x - 4y = 0 -------(1)

"If 9 be added to it the digits are reversed"

xy + 9 = yx ---------> 10x + y + 9 = 10y + x -------> 9x - 9y = -9

9x - 9y = -9 ----------> x -  y = -1 ---------(2)

Solving (1) and (2), we get x = 4 and y = 5.

Hence, the required number is 45

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Example 9 :

The age of a man is three times  the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man.

Solution :

Let "x" be the present age of the man and "y" be the sum of the present ages of two sons.

Present age of the man is 3 times the sum of the ages of 2 sons.

So, x = 3y ------(1)

5 years hence, age of the man will be double the sum of the ages of his two sons.

So, x+5 = 2(y+5+5)  (There are two sons, so 5 is added two times)

x+5 = 2(y+10) --------> x = 2y + 20 - 5 -----> x = 2y +15 -------(2)

Solving (1) and (2), we get x = 45

Hence the present age of the man is 45 years.

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Example 10 :

A trader has 100 units of a product. A sells some of the units at $6 per unit and the remaining units at $8 per units. He receives a total of $660 for all 100 units. Find the number units sold in each category.

Solution :

"x" -------> no. of units sold at $6/unit

"y" -------> no. of units sold at $8/unit

According to the question, x + y = 100 ------------(1)

6x + 8y = 660 ------3x + 4y = 330 ----------(2)

Solving (1) and (2), we get x = 70 and y = 30

Hence, the no. of tickets sold at $6 per unit = 70

the no. of tickets sold at $8 per unit = 30

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Example 11 :

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

Solution :

Let "x" and "y" be the two numbers such that x > y  

Given : One number is greater than thrice the other number by 2

So, we have   x  =  3y + 2 ---------(1)

Given : 4 times the smaller number exceeds the greater by 5

So, we have  4y = x + 5 ---------(2)

Plugging  (1) in (2), we get  4y = 3y +2 + 5 -------> 4y = 3y + 7

4y = 3y + 7 -----> y = 7

Plugging y = 7 in (1), we get   x = 3(7) + 2

Therefore x = 23

Hence the two numbers are 23 and 7. 

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Example 12 :

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Solution :

Let "xy" be the required two digit number. 

Given : Two digit number is 7 times the sum of its digits. 

So, we have   xy  =  7(x+y) -------> 10x + y = 7x + 7y

10x + y = 7x + 7y ------> 3x - 6y = 0

x - 2y = 0 --------(1)

Given : The number formed by reversing the digits is 18 less than the given number

So, we have    xy  -  yx  =  18 

(10x + y) - (10y + x) = 18 ------> 10x + y -10y - x  =  18

9x - 9y  =  18 -----------> x - y  =  2 --------(2)

Solving (1) and (2), we get x = 4 and y = 2

xy = 42

Hence the required number is 42.  

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Example 13 :

In a rectangle, if the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm . If the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm . Find the area of the rectangle.

Solution :

Let "x" and "y" be the length and width of the rectangle respectively.

Then, area of the rectangle = xy  

Given : if the length is increased and the width is reduced each by 2 cm then the area is reduced by 28 sq.cm

So, we have  (x + 2)(y -2)  =  xy - 28

xy - 2x + 2y - 4  =  xy - 28 ---------> - 2x + 2y  =  -24

 - x + y = -12 ---------(1)

Given : if the length is reduced by 1 cm and the width increased by 2 cm , then the area increases by 33 sq.cm

So, we have  (x - 1)(y + 2)  =  xy + 33

xy + 2x - y - 2  =  xy + 33 --------->  2x - y  =  35 ---------(2)

Solving (1) and (2), we get x = 23 and y = 11

Area of the rectangle  =  xy  =  (23)(11)  =  253 sq.cm

Hence the area of the rectangle  =  253 sq.cm

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Example 14 :

Sum of incomes of A and B is $2640. If B's income is 20% more than A, find the income of A and B. 

Solution :

Let "x" and "y" be the incomes of A and B respectively.

Then,  x + y  =  2640  --------(1)

Given : B's income is 20% more than A

So, we have     y  =  120% of x 

y  =  1.2x --------(2)

Plugging  y  =  1.2x  in (1) ------> x + 1.2x = 2640

2.2x  =  2640 --------> x  =  1200

Plugging   x  =  1200 in (2) ------> y = 1.2(1200)

y  =  1440

Hence the incomes of A and B are $1200 and $1440. 

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Example 15 :

Sum of the cost price of two products is $50. Sum of the selling price of the same two products is $52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.  

Solution :

Let "x" and "y" be the cost prices of two products. 

Then,  x + y  =  50  --------(1)

Let us assume thatr "x" is sold at 20% profit

Then, the selling price of "x" = 120% of "x"

selling price of "x" = 1.2x

Let us assume thatr "y" is sold at 20% loss

Then, the selling price of "y" = 80% of "y"

selling price of "x" = 0.8y

Given : Selling price of "x"  +  Selling price of "y"  =  52

1.2x + 0.8y  =  52 -------> 12x + 8y  =  520

3x + 2y  =  130 --------(2)

Solving (1) and (2), we get x  =  30 and y  =  20 

Hence the cost prices of two products are $30 and $20.

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Example 16 :

In a school, there are 880 students in total. If there is 20% more boys than girls, find the number of boys and girls in the school ?   

Solution :

Let "x" and "y" be the no. of boys and girls respectively.

Then,  x  +  y  =  880

Given : There is 20% more boys than girls

Then,  we have  x  =  120% of y  -------> x   =  1.2y  --------(1)

Plugging  x  =  1.2y  in (2) --------> 1.2y  +  y  =  880

2.2y  =  880 ---------->  y  =  400

Plugging  y  =  400  in (1) --------> x    =   1.2(400)

x  =  480

Hence the number of boys  =  480 and

the number of girls  =  400

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Problem 17 :

Mr. Lenin invests some amount in deposit A and some amount in deposit B. The total money invested is $2500. He gets 10% income on deposit A and 20% income on deposit B. If the total income earned be $380, find the amount invested in A and B separately    

Solution :

Let "x" and "y" be amounts invested in A and B respectively. 

Then, we have   x  +  y  =  2500  --------(1)

From the given information, we have

10% of x  +  20% of y  =  380

0.1x + 0.2y  =  380 --------> x + 2y  =  3800 -------(2)

Solving (1) and (2), we get   x =  1200  and  y  =  1300

Hence, the amount deposited in A and B are $1200 and $1300 respectively.  

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Problem 18 :

The sum of two numbers is 209. If one number is 7 less than two times of the other, then find the two numbers.     

Solution :

Let "x" and "y" be the two numbers. 

Given : The sum of two numbers is 209

x  +  y  =  209 -------(1)

Given: One number is 7 less than two times of the other

So, we have    x  =  2y - 7 -------(2)

Plugging (2) in (1) --------> 2y - 7 + y = 209

3y - 7  =  209 -------> 3y  =  216 ------->  y  =  72

Plugging  y  =  72   in (2) --------> x  =  2(72) - 7

x = 144 - 7 --------->  x  =  137

Hence, the two numbers are 137 and 72.  

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Problem 19 :

The perimeter of the rectangle is 158 cm. If the length is 7 more than 3 times the width, find the area of the rectangle. 

Solution :

Let "x" and "y" be the length and width of the rectangle respectively.  

Given : perimeter of the rectangle is 158 cm

Then, we have   2x + 2y  =  158 ------(1) 

Given : The length is 7 more than 3 times the width

So, we have   x  =  3y + 7 --------(2)

Plugging (2) in (1) --------> 2(3y+7) + 2y = 158

6y + 14  + 2y  =  158 --------> 8y  =  144

Therefore  y  =  18

Plugging  y  =  18  in (2) --------> x  =  3(18) + 7

x  =  61

So, length = 61 cm and width = 18 cm

Area of the rectangle  =  length x width 

=  61 x  18 

=  1098  

Hence, the area of the rectangle is  1098 sq.cm. 

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Problem 20 :

A trader gains one third of the cost price as profit on a product and one fourth of the cost price as profit on other product. Total profit earned on these two products is $43.  The sum of the cost prices of two products is $150. Find the cost price of each product. 

Solution :

Let "x" and "y" be the cost prices of the two products.

Then, we have   x + y  =  150 -------(1)

Given : One third of the cost price as profit on a product

So, profit on the first product = (1/3)x  =  x/3

Given : One fourth of the cost price as profit on the other product. 

So, profit on the second product = (1/4)y  =  y/4

Total profit earned on these two products  =  $43

x/3  +  y/4  =  43 -------> (4x + 3y) / 12  = 43

4x + 3y  =  516 --------(2)

Solving (1) and (2), we get x  =  66  and y  =  84

Hence, the cost prices of two products are $66 and $84 

After having gone through the examples explained above, we hope that students would have understood "Word problems on simultaneous linear equations".

Apart from the examples on word problems on simultaneous linear equations, if you want to know more about "Word problems on simultaneous linear  equations, please click here.

Please click the below links to know "How to solve word problems in each of the given topics"

1. Solving Word Problems on Simple Equations

2. Solving Word Problems on Simultaneous Equations

3. Solving Word Problems on Quadratic Equations

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5. Solving Word Problems on HCF and LCM

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17. Solving Word Problems on Compound Interest

18. Solving Word Problems on Calendar

19. Solving Word Problems on Clock

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21. Solving Word Problems on Modular Arithmetic

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