"Translating word problems into equations" is a much needed stuff to the students who learn solving word problems in math

**Example 1 :**

**Example 2 :**

**Example 3 :**

**Example 4 :**

**Example 5 :**

The age of a father is thrice the sum of the ages of his two sons and 5 years hence his age will be twice the sum of their ages. Find the present age of the father.

**Solution:**

**Step 1:**

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

1. The age of the father is thrice the sum of the ages of his two sons. (At present)

2. After 5 years, his age would be twice the sum of their ages. (After 5 years)

**Step 2:**

**Target of the question:** Present age of the father

**Step 3:**

Introduce required variables for the information given in the question.

Let "x" be the present age of the father.

Let "y" be the sum of present ages of two sons.

Clearly, the value of "x" to be found. Because that is the target of the question.

**Step 4:**

Translate the given information as mathematical equation using "x" and "y".

__First information: __

The age of the father is thrice the sum of the ages of his two sons.

__Translation (i) :__

The Age of the father --------> x

is --------> =

thrice the sum of the ages of his two sons --------> 3y

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

x = 3y ----(1)

__Second Information:__

After 5 years, his age would be twice the sum of their ages.

__Translation (ii) :__

Age of the father after 5 years --------> x + 5

Sum of the ages of his two sons after 5 years --->y+5+5 = y +10

(Here we have added 5 two times.The reason is there are two sons)

Twice the sum of ages of two sons --------> 2(y+10)

would be --------> =

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

x + 5 = 2(y+10) ----(2)

**Step 5:**

Solve equations (1) & (2) :

Plug x = 3y in equation (2) ===> 3y + 5 = 2(y+10)

3y + 5 = 2y + 20

y = 15

Plug y = 15 in equation (1) ===> x = 3 (15)

x = 45

**Therefore, the present age of the father is 45 years.**

**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**

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**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**

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**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**

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**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**

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**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**

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**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**

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**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**

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**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****°**

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**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**

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**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**

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**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**

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**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**

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**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**

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**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**

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**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°)**

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**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.**

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**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.**

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**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**

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**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 example on "Algebra word problems worksheet with answers"

**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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