Solving word problems on simple 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.

**Example 1 :**

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

**Solution :**

Let "x" be the numerator.

"The denominator of the fraction exceeds the numerator"

From the above information, fraction = x / (x+5) ----------(1)

"If 3 be added to both, the fraction becomes 3/4"

From the above information, we have (x+3) / (x+5+3) = 3/4

(x+3)/(x+8) = 3/4 ---------> 4(x+3) = 3(x+8)

4x + 12 = 3x + 24 ---------> x = 12

(1)--------> x / (x+5) = 12 / (12+5) = 12/27

**Hence, the required fraction is 12/27**

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**Example 2 :**

If thrice of A's age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A's present age.

**Solution :**

Let "x" be A's present age.

A's age 6 years ago = x - 6

Thrice of A's age 6 years ago = 3(x-6) -----------(1)

Twice his present age = 2x ----------(2)

According to the question, (2) - (1) = A's present age

2x - 3(x-6) = x ----------> 2x - 3x + 18 = x -----------> 18 = 2x

18 = 2x -----------> 9 = x

**Hence, A's present age is 9 years **

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**Example 3 :**

A
number consists of two digits. The digit in the tens place is twice
the digit in the units place. If 18 be subtracted from the number, the
digits are reversed. Find the number.

**Solution :**

Let "x" be the digit in units place.

Then the digit in the tens place = 2x

According to the question, (2x)x - 18 = x(2x)

(2x)x - 18 = x(2x) -----> (2x).10 + x.1 - 18 = x.10 + (2x).1

20x + x - 18 = 10x + 2x -------> 21x - 18 = 12x

21x - 18 = 12x ------> 9x = 18 ------> x = 2

So, the digit at the units place = x = 2

and the digit at the tens place = 2x = 2(2) = 4

**Hence the required number is 42**

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**Example 4 :**

For
a certain commodity, the demand equation giving demand "d" in kg, for a
price "p" in dollars per kg. is d = 100(10-p). The supply equation
giving the supply "s" in kg. for a price "p" in dollars per kg is s =
75(p-3). The market price is such at which demand equals supply. Find
the market price.

**Solution :**

Since the market price is such that demand (d) = supply (s), we have

100(10-p) = 75(p-3)

1000 - 100p = 75p - 225

p = 7

**Hence, the market price is $7
**

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**Example 5 :**

The fourth part of a number exceeds the sixth part by 4. Find the number.

**Solution :**

Let "x" be the required number.

Fourth part of the number = x/4

Sixth part of the number = x/6

According to the question, we have x/4 - x/6 = 4

3x/12 - 2x/12 = 4 ------> (3x - 2x) / 12 = 4 -----> x / 12 = 4 ------> x = 48

**Hence, the required number is 48
**

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**Example 6 :**

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

**Solution :**

Let "x" be the length of the rectangle.

Then, width of the rectangle = (2/3)x

Perimeter = 80 cm -----> 2(l + w) = 80 -------> l + w = 40

l + w = 40 -------> x + (2/3)x = 40 -------> (3x+2x) / 3 = 40

(3x+2x) / 3 = 40 ------> 5x = 120 ------> x = 24

So, **length** = x = **24 cm **

and **width** = (2/3)x = (2/3)24 = **16 cm**

Area = l x w = 24x16 = 384 square cm.

**Hence, area of the rectangle is 384 square cm**

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**Example 7 :**

18 is taken away from 8 times of a number is 30. Find the number.

**Solution :**

Let "x" the required number.

According to the question, we have 8x - 18 = 30

8x - 18 = 30 --------> 8x = 48 --------> x = 6

**Hence, the required number is 6**

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**Example 8 :**

In a triangle, the second angle is 5° more than the first angle. And the third angle is three times of the first angle. Find the three angles of the triangle.

**Solution :**

Let "x" be the first angle.

Then the second angle = x + 5° and third angle = 3x

Sum of three angle in any triangle = 180°

x + (x+5) + 3x = 180 ------> 5x + 5 = 180 ------> x = 35

So, the first angle = x = 35°

the second angle = x + 5° = 35 + 5° = 40°

the third angle = 3x = 3(45°) = 135°

**Hence, the three angles of the triangle are 35****°, 40****° and 135****°**

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**Example 9 : **

If a number of which the half is greater than 1/5 th of the number by 15, find the number.

**Solution :**

Let "x" be the required number.

Half of the number = (1/2)x

1/5 the of the number = (1/5)x

According to the question, we have (1/2)x - (1/5)x = 15

(1/2)x - (1/5)x = 15 ------> (5x - 2x) / 10 = 15 -------> 3x = 150

3x = 150 ----------> x = 50

**Hence, the required number is 50**

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**Example 10 : **

Three persons A, B and C together have $51. B has $4 less than A. C has got $5 less than A. Find the money that A, B and C have.

**Solution :**

Let "x" be the money had by A, then **A = x**

"B has $4 less than A" ------> **B = x - 4**

"C has got $5 less than A" -------> **C = x - 5 **

According to the question, **A + B + C = 51**

A + B + C = 51-------> x + (x-4) + (x-5) = 51 ---------> 3x + 9 = 51

3x + 9 = 51 -------> 3x = 42 -------> x = 14

Money had by A = x = 14 ---------------------> **A = $14**

Money had by A = x-4 = 14- 4 = 10 --------> **B = $10**

Money had by A = x-5= 14-5 = 9 ------------> **C = $9**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 11 :**

The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is

**Solution :**

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

**Hence the age of the youngest boy is 15 years**

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**Example 12 :**

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

**Solution :**

Sum of the terms in the given ratio = 3+5 = 8

So, no. of boys in the school = 720x(3/8)= 270

No. of girls in the school = 720x(5/8)= 450

Let "x" be the no. of new boys admitted in the school.

No. of new girls admitted = 18 (given)

**After the above new admissions, **

no. of boys in the school = 270+x

no. of girls in the school = 450+18 = 468

The ratio after the new admission is 2 : 3 (given)

So, (270+x) : 468 = 2 : 3

3(270+x) = 468x2 (using cross product rule in proportion)

810 + 3x = 936

3x = 126

x = 42 **Hence the no. of new boys admitted in the school is 42**

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**Example 13 :**

The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly income of the second person.

**Solution :**

**From the given ratio of incomes ( 4 : 5 ), **

Income of the 1st person = 4x

Income of the 2nd person = 5x

**(Expenditure = Income - Savings)**

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

So, (4x - 50) : (5x - 50) = 7 : 9

9(4x - 50) = 7(5x - 50) (using cross product rule in proportion)

36x - 450 = 35x - 350

x = 100

Then, income of the second person = 5x = 5(100) = 500.

**Hence, income of the second person is $500**

Let us look at the next problem on "Word problems on simple equations"

**Example 14 :**

The ratio of the prices of two houses was 16:23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11:20. Find the original price of the first house.

**Solution :**

From the given ratio 16:23,

original price of the 1st house = 16x

original price of the 2nd house = 23x

**After increment in prices, **

price of the 1st house = 16x + 10% of 16x = 16x + 1.6x = 17.6x

price of the 2nd house = 23x+477

**After increment in prices, the ratio of prices becomes 11:20 **

Then we have, 17.6x : (23x + 477) = 11 : 20

20(17.6x) = 11(23x+477) (using cross product rule)

352x = 253x + 5247

99x = 5247

x = 53

Then, original price of the first house = 16x = 16(53) = 848

**Hence, original price of the first house is $848**

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**Example 15 :**

If the angles of a triangle are in the ratio 2:7:11, then find the angles.

**Solution :**

From the ratio 2 : 7 : 11,

the three angles are 2x, 7x, 11x

In any triangle, sum of the angles = 180

So, 2x + 7x + 11x = 180°

20x = 180 -------> x = 9

Then, the first angle = 2x = 2(9) = 18°

The second angle = 7x = 7(9) = 63°

The third angle = 11x = 11(9) 99°

**Hence the angles of the triangle are (18°, 63°, 99°)**

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**Example 16 :**

The ratio of two numbers is 7:10. Their difference is 105. Find the numbers.

**Solution :**

From the ratio 7 : 10,

the numbers are 7x, 10x.

Their difference = 105

10x - 7x = 105 ------> 3x = 105 --------> x = 35

Then the first number = 7x = 7(35) = 245

The second number = 10x = 10(35) = 350

**Hence the numbers are 245 and 350.**

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**Example 17 :**

The ratio between the speeds of two trains is 7:8. If the second train runs 400 kms. in 5 hours, then, find the speed of the first train.

**Solution :**

From the given ratio 7 : 8,

Speed of the first train = 7x

Speed of the second train = 8x ----------(1)

Second train runs 400 kms in 5 hours (given)

**[Hint : Speed = Distance / Time]**

So, speed of the second train = 400/5 = 80 kmph -------(2)

From (1) and (2), we get

8x = 80 -------> x = 10

So, speed of the first train = 7x = 7(10) = 70 kmph.

**Hence, the speed of the second train is 70 kmph.**

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**Example 18 :**

Two numbers are respectively 20% and 50% are more than a third number, Find the the ratio of the two numbers.

**Solution :**

Let "x" be the third number.

Then,

the first number = (100+20)% of x = 120% of x = 1.2x

the first number = (100+50)% of x = 150% of x = 1.5x

First no. : second no. = 1.2x = 1.5x

1.2x : 1.5x---------------> 12x : 15x

Dividing by (3x), we get 4 : 5

**Hence, the ratio of two numbers is 4:5**

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**Example 19 :**

If $782 is divided among three persons A, B and C in the ratio 1/2 : 2/3 : 3/4, then find the share of A.

**Solution :**

Given ratio ---> 1/2 : 2/3 : 3/4

First let us convert the terms of the ratio into integers.

L.C.M of denominators (2, 3, 4) = 12

When we multiply each term of the ratio by 12, we get

12x1/2 : 12x2/3 : 12x3/4 ------> 6 : 8 : 9

From the ratio 6 : 8 : 9,

Share of A = 6x

Share of B = 8x

Share of C = 9x

We know that ( A + B + C ) = 782

6x + 8x + 9x = 782 --------> 23x = 782 ------> x = 34

Share of A = 6x = 6(34) = 204

**Hence, the share of A = $ 204. **

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**Example 20 :**

An amount of money is to be divided among P, Q and R in the ratio 3 : 7 : 12. he between the shares of P and Q is $2400. What will be the difference between the shares of Q and R?

**Solution :**

From the given ratio 3 : 7 : 12,

Share of P = 3x

Share of B = 7x

Share of C = 12x

Difference between the shares of P and Q is $ 2400

That is,

Q - P = 2400 -------> 7x - 3x = 2400 -------> 4x = 2400 -------> x = 600

R - Q = 12x - 7x = 5x = 5(600) = 3000

**Hence, the difference between the shares of Q and R is $3000. **

**Let us look at the next problem on "Word problems on simple equations"**

**Example 21 :**

The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest number ?

**Solution :**

Let "x" be the first odd number number.

Then, the five consecutive odd numbers are

x, x+2, x+4, x+6, x+8

Average of five consecutive odd numbers = 61

So, we have

[x + (x+2) + (x+4) + (x+6) + (x+8)] / 5 = 61

[x + (x+2) + (x+4) + (x+6) + (x+8)] = 305

5x + 20 = 305

5x = 285

x = 57

Then,

x+2 = 59, x+4 = 61, x+6 = 63, x+8 = 65

**Hence, the five consecutive odd numbers are **

**57, 59, 61, 63, 65**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 22 :**

The average of five numbers is 27. If one number is excluded, the average becomes 25. Find the excluded number.

**Solution :**

Average of five numbers = 27

Sum of the five numbers / 5 = 27

Sum of the five numbers = 135 ----------(1)

After having excluded one number, there would be four numbers.

Average of four numbers = 25

Sum of the four numbers / 4 = 25

Sum of the four numbers = 100 ----------(1)

(1) - (2) ===>

Excluded no. = (sum of five numbers) - (sum of four numbers)

Excluded number = 135 - 100

Excluded number = 135

**Hence, the excluded number is 13****5**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 23 :**

The average age of 35 students in a class is 16 years. Out of 35 students, the average age of 21 students is 14. What is the average age of remaining 14 students?

**Solution :**

Average of 35 students = 16

Sum of the ages of 35 students / 35 = 16

Sum of the ages of 35 students = 560 ----------> (1)

Average of 21 students = 14

Sum of the ages of 21 students / 21 = 14

Sum of the ages of 21 students = 294 ----------> (1)

(1) - (2) ===>

Sum of the ages of remaining 14 students = (1) - (2)

Sum of the ages of remaining 14 students = 560 - 294

Sum of the ages of remaining 14 students = 266

Average of the remaining 14 students = 266 / 14 = 19

**Hence, the average of the remaining 14 students is 19**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 24 :**

If a number of which the half is greater than one fifth of the number by 15, then find the number.

**Solution :**

Let "x" be the required number.

Half of the numb er = (1/2)x = x/21

One fifth of the number = (1/5)x x/5

**Given :** Half is greater than one fifth of the number by 15

So, we have

x/2 - x/5 = 15

(5x - 2x) / 10 = 15

3x / 10 = 15

3x = 150

x = 50

**Hence, the required number is 50**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 25 :**

A students is asked to divide half of a number by 6 and other half by 4 and then to add the two quantities. Instead of doing so, the student divides the given number by 5. if the answer is 4 short of the correct answer, find the number.

**Solution :**

Let "x" be the required number.

The two halves are x/2 and x/2.

Given : Divide half of a number by 6 and other half by 4 and then to add the two quantities.

[x/2] / 6 + [x/2] / 4 = x/12 + x/8 = 5x / 24 ---------(1)

Given : Divide the given number by 5

x/5 ---------(2)

From the given information,

(2) - (1) = 4

5x / 24 - x/5 = 4

(25x - 24x) / 120 = 4

x / 120 = 4

x = 480

**Hence, the required number is 480**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 26 :**

Two numbers are in the ratio 6 : 7. If the sum of two numbers is 104, find the numbers.

**Solution :**

Two numbers are in the ratio 6 : 7

Then, the two numbers are 6x and 7x

Sum of the two numbers = 104 ---------> 6x + 7x = 104

13x = 104 ------> x = 8

Therefore, the first number = 6x = 6(8) = 48

the second number = 7x = 7(8) = 56

**Hence, the two numbers are 48 and 56. **

**Let us look at the next problem on "Word problems on simple equations"**

**Example 27 :**

The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is

**Solution :**

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

**Hence the age of the youngest boy is 15 years**

**Let us look at the next problem on "Word problems on simple equations"**

**Example 28 :**

In a game, Alex and Jose scored 720 points in total. If Alex scored two times of the points scored by Jose, find the points scored by Alex and Jose.

**Solution :**

Let "x" and "y" be the points scored by Alex and Jose respectively.

Given : Alex and Jose scored 720 points in total.

So, we have x + y = 720 -------(1)

Given : Alex scored two times of the points scored by Jose

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

Plugging (1) in (2), we get 2y + y = 720

3y = 720 --------> y = 240

Plugging y = 240 in (1), we get x + 240 = 720

x = 480

**Hence, the points scored by Alex and Jose are 480 and 240. **

**Let us look at the next problem on "Word problems on simple equations"**

**Example 29 :**

In a triangle, the first angle is 20% more than the third angle. Second angle is 20% less than the third angle. Then find the three angles of the triangle.

**Solution :**

Let "x" be the third angle.

Then the first angle = 120% of x = 1.2x

The second angle = 80% of x = 0.8x

Sum of the three angles in any triangle = 180°

Then, we have x + 1.2x + 0.8x = 180°

3x = 180° --------> x = 60°

Then the first angle = 1.2(60°) = 72°

The second angle = 0.8(60°) = 48°

**Hence, the three angles of the triangle are 72°, 60° and 48°. **

**Let us look at the next problem on "Word problems on simple equations"**

**Example 30 :**

Three persons A, B and C are working in a company. The salary of B is $50 more than 2 times of A's salary. C's salary is 3 times of B's salary. If A, B and C are getting together $3800, find the salary of A, B and C separately.

**Solution :**

Let "x" be the salary of A

Given : The salary of B is $50 more than 2 times of A's salary

Then, the salary of B = 2x + 50

Given : C's salary is 3 times of B's salary.

Then, the salary of C = 3(2x+50) = 6x + 150

Given : A, B and C are getting together $3800

So, we have x + (2x+50) + (6x+150) = 3800

9x + 200 = 3800 ------> 9x = 3600 -------> x = 400

Salary of B = 2(400) + 50 = 850

Salary of C = 3(850) = 2550

**Hence, the salary of A = $400 **

**the salary of B = $850**

**the salary of C = $2550 **

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

Apart from the examples, if you want to know more above word problems on simple equations, please click here.

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