WORD PROBLEMS ON SIMPLE EQUATIONS

About "Word problems on simple equations"

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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°

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

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

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

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

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

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

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

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

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

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°)

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

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.

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

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.

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

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

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

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. 

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

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 135

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.

HTML Comment Box is loading comments...

ALGEBRA

Variables and constants

Writing and evaluating expressions

Solving linear equations using elimination method

Solving linear equations using substitution method

Solving linear equations using cross multiplication method

Solving one step equations

Solving quadratic equations by factoring

Solving quadratic equations by quadratic formula

Solving quadratic equations by completing square

Nature of the roots of a quadratic equations

Sum and product of the roots of a quadratic equations 

Algebraic identities

Solving absolute value equations 

Solving Absolute value inequalities

Graphing absolute value equations  

Combining like terms

Square root of polynomials 

HCF and LCM 

Remainder theorem

Synthetic division

Logarithmic problems

Simplifying radical expression

Comparing surds

Simplifying logarithmic expressions

Negative exponents rules

Scientific notations

Exponents and power

COMPETITIVE EXAMS

Quantitative aptitude

Multiplication tricks

APTITUDE TESTS ONLINE

Aptitude test online

ACT MATH ONLINE TEST

Test - I

Test - II

TRANSFORMATIONS OF FUNCTIONS

Horizontal translation

Vertical translation

Reflection through x -axis

Reflection through y -axis

Horizontal expansion and compression

Vertical  expansion and compression

Rotation transformation

Geometry transformation

Translation transformation

Dilation transformation matrix

Transformations using matrices

ORDER OF OPERATIONS

BODMAS Rule

PEMDAS Rule

WORKSHEETS

Converting customary units worksheet

Converting metric units worksheet

Decimal representation worksheets

Double facts worksheets

Missing addend worksheets

Mensuration worksheets

Geometry worksheets

Comparing  rates worksheet

Customary units worksheet

Metric units worksheet

Complementary and supplementary worksheet

Complementary and supplementary word problems worksheet

Area and perimeter worksheets

Sum of the angles in a triangle is 180 degree worksheet

Types of angles worksheet

Properties of parallelogram worksheet

Proving triangle congruence worksheet

Special line segments in triangles worksheet

Proving trigonometric identities worksheet

Properties of triangle worksheet

Estimating percent worksheets

Quadratic equations word problems worksheet

Integers and absolute value worksheets

Decimal place value worksheets

Distributive property of multiplication worksheet - I

Distributive property of multiplication worksheet - II

Writing and evaluating expressions worksheet

Nature of the roots of a quadratic equation worksheets

Determine if the relationship is proportional worksheet

TRIGONOMETRY

SOHCAHTOA

Trigonometric ratio table

Problems on trigonometric ratios

Trigonometric ratios of some specific angles

ASTC formula

All silver tea cups

All students take calculus 

All sin tan cos rule

Trigonometric ratios of some negative angles

Trigonometric ratios of 90 degree minus theta

Trigonometric ratios of 90 degree plus theta

Trigonometric ratios of 180 degree plus theta

Trigonometric ratios of 180 degree minus theta

Trigonometric ratios of 180 degree plus theta

Trigonometric ratios of 270 degree minus theta

Trigonometric ratios of 270 degree plus theta

Trigonometric ratios of angles greater than or equal to 360 degree

Trigonometric ratios of complementary angles

Trigonometric ratios of supplementary angles 

Trigonometric identities 

Problems on trigonometric identities 

Trigonometry heights and distances

Domain and range of trigonometric functions 

Domain and range of inverse  trigonometric functions

Solving word problems in trigonometry

Pythagorean theorem

MENSURATION

Mensuration formulas

Area and perimeter

Volume

GEOMETRY

Types of angles 

Types of triangles

Properties of triangle

Sum of the angle in a triangle is 180 degree

Properties of parallelogram

Construction of triangles - I 

Construction of triangles - II

Construction of triangles - III

Construction of angles - I 

Construction of angles - II

Construction angle bisector

Construction of perpendicular

Construction of perpendicular bisector

Geometry dictionary

Geometry questions 

Angle bisector theorem

Basic proportionality theorem

ANALYTICAL GEOMETRY

Analytical geometry formulas

Distance between two points

Different forms equations of straight lines

Point of intersection

Slope of the line 

Perpendicular distance

Midpoint

Area of triangle

Area of quadrilateral

Parabola

CALCULATORS

Matrix Calculators

Analytical geometry calculators

Statistics calculators

Mensuration calculators

Algebra calculators

Chemistry periodic calculator

MATH FOR KIDS

Missing addend 

Double facts 

Doubles word problems

LIFE MATHEMATICS

Direct proportion and inverse proportion

Constant of proportionality 

Unitary method direct variation

Unitary method inverse variation

Unitary method time and work

SYMMETRY

Order of rotational symmetry

Order of rotational symmetry of a circle

Order of rotational symmetry of a square

Lines of symmetry

CONVERSIONS

Converting metric units

Converting customary units

WORD PROBLEMS

HCF and LCM  word problems

Word problems on simple equations 

Word problems on linear equations 

Word problems on quadratic equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation 

Word problems on unit price

Word problems on unit rate 

Word problems on comparing rates

Converting customary units word problems 

Converting metric units word problems

Word problems on simple interest

Word problems on compound interest

Word problems on types of angles 

Complementary and supplementary angles word problems

Double facts word problems

Trigonometry word problems

Percentage word problems 

Profit and loss word problems 

Markup and markdown word problems 

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

Pythagorean theorem word problems

Percent of a number word problems

Word problems on constant speed

Word problems on average speed 

Word problems on sum of the angles of a triangle is 180 degree

OTHER TOPICS 

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

Time, speed and distance shortcuts

Ratio and proportion shortcuts

Domain and range of rational functions

Domain and range of rational functions with holes

Graphing rational functions

Graphing rational functions with holes

Converting repeating decimals in to fractions

Decimal representation of rational numbers

Finding square root using long division

L.C.M method to solve time and work problems

Translating the word problems in to algebraic expressions

Remainder when 2 power 256 is divided by 17

Remainder when 17 power 23 is divided by 16

Sum of all three digit numbers divisible by 6

Sum of all three digit numbers divisible by 7

Sum of all three digit numbers divisible by 8

Sum of all three digit numbers formed using 1, 3, 4

Sum of all three four digit numbers formed with non zero digits

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6

HTML Comment Box is loading comments...