# ALGEBRA WORD PROBLEMS

Solving algebra word problems is sometimes a difficult job for some students.

Actually it is not.

We have to apply the following simple tricks.

2. Denote the unknown by "x"

3. If it is needed, use two variables "x" and "y".

4. Form equations with "x" & "y"  and solve for them.

Let us see how these tricks are being applied to solve algebra word problems.

## Examples

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 = 60 -------> x = 20

Money had by A = x = 20 ---------------------> A = \$20

Money had by A = x-4 = 20- 4 = 16 --------> B = \$16

Money had by A = x-5= 20-5 = 15 ------------> C = \$15

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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)

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

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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.

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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

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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 135

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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

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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

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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

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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.

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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

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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.

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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°.

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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

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Example 31 :

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 32 :

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 33 :

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 34 :

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 35 :

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 36 :

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

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

Example 37 :

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

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

Example 38 :

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

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

Example 39 :

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.

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

Example 40 :

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

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

Example 41 :

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.

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

Example 42 :

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.

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

Example 43 :

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

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

Example 44 :

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.

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

Example 45 :

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.

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

Example 46 :

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

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

Problem 47 :

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.

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

Problem 48 :

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.

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

Problem 49 :

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.

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

Problem 50 :

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 "How to solve Algebra word problems.

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