"How to solve word problems in algebra" is a much needed stuff to the students who would like to learn solving word problems in algebra

**Example 1 :**

**Example 2 :**

**Example 3 :**

**Example 4 :**

**Example 5 :**

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:**

**Step 1 :**

Let us understand the given information. There are two information given in the question.

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

2. If 9 be added to it, the digits are reversed.

**Step 2 :**

**Target of the question:** What is that number?

**Step 3 :**

A number between 10 and 100 is clearly a two digit number.

Introduce required variables for the information given in the question.

Let "x" be the digit at one's place.

Let "y" be the digit at ten's place.

So, the number is "xy" and we have to find "xy".

**Step 4 :**

Any two digit number can be written as given below.

56 = 10x5 + 1x6

(5 is at ten's place and 6 is at one's place)

In the same way we can write our two digit number "xy"

That is, xy = 10x + 1y

Now, translate the given information as mathematical equation using "x" and "y".

**Step 5 :**

__First information:__

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

__Translation:__

A number between 10 and 100 -------> "xy" = 10x + y

is --------> = (equal)

Five times sum of the digits --------> 5(x+y)

Equation related to the first information using "x" and "y" is

10x + y = 5(x+y)

10x + y = 5x + 5y

5x - 4y = 0 ---------(1)

**Step 6 :**

__Second Information:__

If 9 be added to it, the digits are reversed.

__Translation:__

A number between 10 and 100 -------> "xy"

When we add 9 to "xy", the digits are reversed. That is, "xy" will become "yx".

Equations related to the second information using "x" and "y" is

xy + 9 = yx

10x + y + 9 = 10y + x

9x + 9 = 9y

Divide by 9 on both the sides.

x + 1 = y

(or) y = x + 1 ---------(2)

**Step 7 :**

Solve equations (1) & (2) :

Plug y = x + 1 in equation (1) ===> 5x - 4(x+1) = 0

5x - 4x - 4 = 0

x = 4

Plug x = 4 in equation (2) ===> y = 4 + 1

y = 5

So, xy = 45

**Therefore, the required number between 10 and 100 is 45.**

Let us look at some practice questions on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 6 :**

Let us look at the next example on "Algebra word problems worksheet with answers"

**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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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****°**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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 "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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 "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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°)**

Let us look at the next problem on "How to solve word problems in algebra

**Problem 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.**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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.**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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**

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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. **

Let us look at the next problem on "How to solve word problems in algebra"

**Problem 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. **

**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**

**6. Solving Word Problems on Numbers**

**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**

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