# HOW TO SOLVE WORD PROBLEMS USING SIMULTANEOUS EQUATIONS

## About "How to solve word problems using simultaneous equations"

"How to solve word problems using simultaneous equations ?" is the question having had by almost all the students who study math in high school level.

There is a simple trick behind it.

The picture given below tells us the trick.

## 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

After having gone through the examples explained above, we hope that students would have understood "How to solve word problems using simultaneous equations".

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

4. Solving Word Problems on Permutations and Combinations

5. Solving Word Problems on HCF and LCM

7. Solving Word Problems on Time and Work

8. Solving Word Problems on Trains

9. Solving Word Problems on Time and Work.

10. Solving Word Problems on Ages.

11.Solving Word Problems on Ratio and Proportion

12.Solving Word Problems on Allegation and Mixtures.

13. Solving Word Problems on Percentage

14. Solving Word Problems on Profit and Loss

15. Solving Word Problems Partnership

16. Solving Word Problems on Simple Interest

17. Solving Word Problems on Compound Interest

18. Solving Word Problems on Calendar

19. Solving Word Problems on Clock

20. Solving Word Problems on Pipes and Cisterns

21. Solving Word Problems on Modular Arithmetic