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Lesson Posted on 28/06/2017 Tuition/Class I-V Tuition Tuition/Class VI-VIII Tuition Tuition/Class IX-X Tuition +1 Tuition/Class VI-VIII Tuition/Mathematics less

How To Do Unitary Metho Sums?

Padmini R.

I am an experienced, qualified teacher and tutor with over 15 years of experience in teaching Maths and...

We know, in unitary method, we first find the value of one quantity from the value of the given quantity. Then we use this value to find the value of the required quantity. While working out the problems using unitary method we come across certain variations in which the values of two quantities depend... read more

We know, in unitary method, we first find the value of one quantity from the value of the given quantity. Then we use this value to find the value of the required quantity. While working out the problems using unitary method we come across certain variations in which the values of two quantities depend on each other in such a way that a change in one, results in a corresponding change in the other; then the two quantities are said to be in variation and the two types of variation which occur are called direct and inverse variations.

 Solved examples of mixed problems using unitary method:

1. If 24 painters working for 7 hours a day, for painting a house in 16 days. How many painters are required working for 8 hours a day will finish painting the same house in 12 days?

Solution:

 24 painters working for 7 hours paint a house in 16 days.

1 painter working for 7 hours paints a house in 16 × 24 days.

1 painter working for 1 hour paints a house in 16 × 24 × 7 days.

Let the required number of painters be x, then;

x painters working for 1 hour a day paint the house in (16 × 24 × 7)/x days

x painters working for 8 hours a day paint the house in (16 × 24 × 7)/(x  ×  8) days

But the number of days given = 12

According to the problem;

(16 × 24 × 7)/(x × 8) = 12

2688/8x = 12

8x × 12 = 2688

96x = 2688

x = 2688/96

x = 28

Therefore, 28 painters working for 8 hours a day will finish the same work in 12 days.

 2. 11 potters can make 143 pots in 8 days. How many potters will be required to make 169 pots in 4 days?

Solution:

11 potters can make 143 pots in 8 days.

1 potter can make 143 pots in 8 × 11 days.

1 potter can make 1 pot in (8 × 11)/143 days.

Let the number of potters required be x, then;

 x potters can make 1 pot in (8 × 11)/( 143 × x) days

x potters can make 169 pots in (8 × 11 × 169)/(143 × x ) days

But the number of days given = 4

 According to the problem;

(8 × 11 × 169)/(143 × x ) = 4

14872/143x = 4

572x = 14872

x = 14872/572

x = 26

Therefore, 26 potters are required to make 169 pots in 4 days.

We will learn ‘what direct variation is’ and how to solve different types of problems on some situations of direct variation.

If two quantities are related in such a way that the increase in one quantity results in a corresponding increase in the other and vice versa, then such a variation is called a direct variation.

 If the two quantities are in direct variation then we also say that they are proportional to each other

Suppose, if the two quantities ‘x’ and ‘y’ are in direct variation, then the ratio of any two values of x is equal to the ratio of the corresponding values of y.

i.e., x1x2=y1y2">x1/x2=y1y/2

or, x1y1=x2y2">x1/y1=x2/y2

Some situations of direct variation:

  • More articles, more money required to purchase
  •  Less articles, less money required to purchase.
  • More men at work, more work is done.
  •  Less men at work, less work is done.
  •  More money borrowed, more interest is to be paid.
  • Less money borrowed, less interest to be paid.
  • More speed, more distance covered in fixed time.                
  • Less speed, less distance covered in fixed time.
  •  More working hours, more work will be done.
  • Less working hours, less work will be done.
  • Problems on different situations of direct variation:

3. A motor bike travels 280 km in 40 liters of petrol. How much distance will it cover in 9 liters of petrol?

Solution:

  • This is the situation of direct variation.
  • Less quantity of petrol, less distance covered.
  • In 40 liters of petrol, distance covered = 280 km
  • In 1 liter of petrol, distance covered = 280/40 km
  • In 9 liters of petrol, distance covered = 280/40 × 9 km = 63 km.

We will learn ‘what inverse variation is’ and how to solve different types of problems on some situations of inverse variation.

If two quantities are related in such a way that increase in one quantity causes corresponding decrease in the other quantity and vice versa, then such a variation is called an inverse variation or indirect variation.

If the two quantities are in inverse variation then we say that they are inversely proportional. Suppose, if two quantities x and y vary inversely with each other, then the values of x is equal to the inverse ratio of the corresponding values of y.

Problems on different situations of inverse variation:

1. If 48 men can do a piece of work in 24 days, in how many days will 36 men complete the same work?

Solution: 

This is a situation of indirect variation.

Less men will require more days to complete the work.

48 men can do the work in 24 days

1 man can do the same work in 48 × 24 days

36 men can do the same work in (48 × 24)/36 = 32 days

Therefore, 36 men can do the same work in 32 days.

2. 100 soldiers in a fort had enough food for 20 days. After 2 days, 20 more soldiers join the fort. How long will the remaining food last?

 Solution:

More soldiers, therefore, food lasts for less days.

This is a situation of indirect variation.               

Since 20 soldiers join the fort after 2 days, therefore, the remaining food  is sufficient for 100 soldiers and 18 days.

1. 12 typists working for 4 hours to type a book in 18 days. In how many days 4 typists will work for 8 hours to type same book?

Solution:

This is a situation of indirect variation.

12 typists working for 4 hours type a book in 18 days

 

1 typist working for 4 hours types a book in 18 × 12 days.

1 typist working for 1 hour types a book in 18 × 12 × 4 days.

4 typists working for 1 hour type a book in (18 × 12 × 4)/4

4 typists working for 8 hours type a book in (18 × 12 × 4)/(4 × 8) days.

Therefore, 4 typists working for 8 hours type a book in 27 days.

2. 16 men can build a wall in 56 hours. How many men will be required to do the same work in 32 hours?

Solution:

This is a situation of inverse variation

More the number of men, the faster will they build the wall.

In 56 hours, the wall is built by 16 men.

In 1 hour, the wall is built by 16 × 56 men.

In 32 hours, the wall is built by (16 × 56)/32 men

Therefore, in 32 hours, the wall is built by 28 men.

3. If 72 workers can do a piece of work in 40 days, in how many days will 64 workers complete the same work?

Solution:

This is a situation of indirect variation.

Less workers will require more days to complete the work.

72 workers can do the work in 40 days

1 worker can do the same work in 72 × 40 days

64 workers can do the same work in (72 × 40)/64

Therefore, 64 workers can do the same work in 45 days.

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Lesson Posted on 19/06/2017 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

Multiplication of even number with 6

Ajinkya

Let's take an example. 26x6 = 156 26/2=13 13+2=15 Answer = 156 244x6 = 1464 244/2=122 122+24=146 Answer=1464
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Lesson Posted on 18/06/2017 Tuition/Class VI-VIII Tuition Tuition/Class IX-X Tuition Tuition/Class VI-VIII Tuition/Mathematics +1 Tuition/Class IX-X Tuition/Mathematics less

Profit and loss important formulas

Padmini R.

I am an experienced, qualified teacher and tutor with over 15 years of experience in teaching Maths and...

IMPORTANT FACTS Cost Price: The price, at which an article is purchased, is called its cost price, abbreviated as C.P. Selling Price: The price, at which an article is sold, is called itsselling prices, abbreviated as S.P. Profit or Gain: If S.P. is greater than C.P., the seller is said to have a profit... read more

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price:

The price, at which an article is sold, is called itsselling prices, abbreviated as S.P.

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

IMPORTANT FORMULAE

  1. Gain = (S.P.) - (C.P.)

  2. Loss = (C.P.) - (S.P.)

  3. Loss or gain is always reckoned on C.P.

  4. Gain Percentage: (Gain %) =

(Gain/C.P )×100

5.Loss Percentage: (Loss %)=

  1. (Loss/C.P )× 100
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Lesson Posted on 16/06/2017 Tuition/Class IX-X Tuition Tuition/Class VI-VIII Tuition Tuition/Class IX-X Tuition/Mathematics +1 Tuition/Class VI-VIII Tuition/Mathematics less

Linear equation worksheet

Soumi Roy

I am an experienced,qualified tutor with over 5 years of experience in teaching across many boards including...

1) 6r + 7 = 13 + 7r 2) 13 − 4x = 1 − x3) −7x − 3x + 2 = −8x − 8 4) −8 − x = x − 4x5) −14 + 6b + 7 − 2b = 1 + 5b 6) n + 2 = −14 − n7) n − 3n = 14 − 4n 8) 7a − 3 = 3 + 6a9) 5 + 2x = 2x + 6 10) −10 + x... read more

1) 6r + 7 = 13 + 7r

2) 13 − 4x = 1 − x
3) −7x − 3x + 2 = −8x − 8

4) −8 − x = x − 4x
5) −14 + 6b + 7 − 2b = 1 + 5b

6) n + 2 = −14 − n
7) n − 3n = 14 − 4n 8) 7a − 3 = 3 + 6a
9) 5 + 2x = 2x + 6

10) −10 + x + 4 − 5 = 7x − 5
11) −8n + 4(1 + 5n) = −6n − 14

12) −6n − 20 = −2n + 4(1 − 3n)
13) 4n − 40 = 7(−2n + 2)

14) 7(5a − 4) − 1 = 14 − 8a
15) −31 − 4x = −5 − 5(1 + 5x)

16) 38 + 7k = 8(k + 4)
17) 8x + 4(4x − 3) = 4(6x + 4) − 4

18) 3(1 − 3x) = 2(−4x + 7)
19) 4(−8x + 5) = −32x − 26

20) −3(x − 1) + 8(x − 3) = 6x + 7 − 5x

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Lesson Posted on 15/06/2017 Tuition/Class IX-X Tuition Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

How To Solve Exponents?

Padmini R.

I am an experienced, qualified teacher and tutor with over 15 years of experience in teaching Maths and...

1. Simplifying fractional exponents: The base b raised to the power of n/m is equal to: bn/m = (m√b)n = m√(bn) Example: The base 2 raised to the power of 3/2 is equal to 1 divided by the base 2 raised to the power of 3: 23/2 = 2√(23) = 2.828 2. Simplifying fractions with exponents: Fractions... read more

1. Simplifying fractional exponents:

The base b raised to the power of n/m is equal to:

bn/m = (mb)n = m(bn)

Example:

The base 2 raised to the power of 3/2 is equal to 1 divided by the base 2 raised to the power of 3:

23/2 = 2(23) = 2.828

2. Simplifying fractions with exponents:

Fractions with exponents:

(a / b)n = an / bn

Example:

(4/3)3 = 43 / 33 = 64 / 27 = 2.37

3. Negative fractional exponents:

The base b raised to the power of minus n/m is equal to 1 divided by the base b raised to the power of n/m:

b-n/m = 1 / bn/m = 1 / (mb)n

Example:

The base 2 raised to the power of minus 1/2 is equal to 1 divided by the base 2 raised to the power of 1/2:

2-1/2 = 1/21/2 = 1/2 = 0.7071

4. Fractions with negative exponents:

The base a/b raised to the power of minus n is equal to 1 divided by the base a/b raised to the power of n:

(a/b)-n = 1 / (a/b)n = 1 / (an/bn) = bn/an

Example:

The base 2 raised to the power of minus 3 is equal to 1 divided by the base 2 raised to the power of 3:

(2/3)-2 = 1 / (2/3)2 = 1 / (22/32) = 32/22 = 9/4 = 2.25

5. Multiplying fractional exponents:

Multiplying fractional exponents with same fractional exponent:

a n/mb n/m = (a b) n/m

Example:

23/2 ⋅ 33/2 = (2⋅3)3/2 = 63/2 =(63) = 216 = 14.7

Multiplying fractional exponents with same base:

a n/ma k/j = a (n/m)+(k/j)

Example:

23/2 ⋅ 24/3 = 2(3/2)+(4/3) = 7.127

Multiplying fractional exponents with different exponents and fractions:

a n/mb k/j

Example:

23/2 ⋅ 34/3 = (23) ⋅ 3(34) =2.828 ⋅ 4.327 = 12.237

6. Multiplying fractions with exponents:

Multiplying fractions with exponents with same fraction base:

(a / b) n ⋅ (a / b) m = (a / b)n+m

Example:

(4/3)3 ⋅ (4/3)2 = (4/3)3+2 = (4/3)5 = 45 / 35 = 4.214

Multiplying fractions with exponents with same exponent:

(a / b) n ⋅ (c / d) n = ((a / b)⋅(c / d)) n

Example:

(4/3)3 ⋅ (3/5)3 = ((4/3)⋅(3/5))3= (4/5)3 = 0.83 = 0.8⋅0.8⋅0.8 = 0.512

Multiplying fractions with exponents with different bases and exponents:

(a / b) n ⋅ (c / d) m

Example:

(4/3)3 ⋅ (1/2)2 = 2.37 / 0.25 = 9.481

7. Dividing fractional exponents:

Dividing fractional exponents with same fractional exponent:

a n/m / b n/m = (a / b) n/m

Example:

33/2 / 23/2 = (3/2)3/2 = 1.53/2 =√(1.53) = 3.375 = 1.837

Dividing fractional exponents with same base:

a n/m / a k/j = a (n/m)-(k/j)

Example:

23/2 / 24/3 = 2(3/2)-(4/3) = 2(1/6) =62 = 1.122

Dividing fractional exponents with different exponents and fractions:

a n/m / b k/j

Example:

23/2 / 34/3 = (23) / 3(34) =2.828 / 4.327 = 0.654

8. Dividing fractions with exponents:

Dividing fractions with exponents with same fraction base:

(a / b)n / (a / b)m = (a / b)n-m

Example:

(4/3)3 / (4/3)2 = (4/3)3-2 = (4/3)1 = 4/3 = 1.333

Dividing fractions with exponents with same exponent:

(a / b)n / (c / d)n = ((a / b)/(c / d))n = ((a⋅d / b⋅c))n

Example:

(4/3)3 / (3/5)3 = ((4/3)/(3/5))3= ((4⋅5)/(3⋅3))3 = (20/9)3 = 10.97

Dividing fractions with exponents with different bases and exponents:

(a / b) n / (c / d) m

Example:

(4/3)3 / (1/2)2 = 2.37 / 0.25 = 9.481

9. Adding fractional exponents:

Adding fractional exponents is done by raising each exponent first and then adding:

an/m + bk/j

Example:

33/2 + 25/2 = √(33) + √(25) = √(27) + √(32) = 5.196 + 5.657 = 10.853

Adding same bases b and exponents n/m:

bn/m + bn/m = 2bn/m

Example:

42/3 + 42/3 = 2⋅42/3 = 2 ⋅3√(42) = 5.04

10. Subtracting fractional exponents:

Subtracting fractional exponents is done by raising each exponent first and then subtracting:

an/m - bk/j

Example:

33/2 - 25/2 = √(33) - √(25) = √(27) - √(32) = 5.196 - 5.657 = -0.488

Subtracting same bases b and exponents n/m:

3bn/m - bn/m = 2bn/m

Example:

3⋅42/3 - 42/3 = 2⋅42/3 = 2 ⋅3√(42) = 5.04

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Lesson Posted on 23/05/2017 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

Understanding Quadrilaterals Class-8

PANKAJ PATHAK

More than 6 years experience in the field of teaching.

Dear Student, Try to solve these question, all are very easy and as per your Syllabus, I hope you all can solve easily. State whether True or False. (a) All rectangles are squares. (b) All rhombuses are parallelograms. (c) All square are rhombuses... read more

Dear Student, Try to solve these question, all are very easy and as per your Syllabus,

I hope you all can solve easily.

  1. State whether True or False.

           (a)  All rectangles are squares.                

           (b)  All rhombuses are parallelograms.

           (c)  All square are rhombuses and also rectangles.

           (d)  All squares are not parallelograms.

           (e)  All kites are rhombuses.

           (f)  All rhombuses are kites.

           (g)  All parallelograms are trapeziums.

           (h)  All squares are trapeziums.

  1. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1: 2: 3: 4. Find the measure of each angle of the quadrilateral
  2. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1: 2: 3: 4. Find the measure of each angle of the quadrilateral.
  3. The interior angle of a regular polygon is 108°. Find the number of sides of the polygon.
  4. The exterior angle of a regular polygon is one-fifth of its interior angle. How many sides have the polygon?
  5. The measures of two adjacent angles of a parallelogram are in the ratio 4: 5. Find the measure of each of the angles of the parallelogram.
  6. If an exterior angle of a regular polygon is 45°, then find the number of its sides.
  7. If an interior angle of a regular polygon is 162°, then find the number of its sides.
  8. Find the measure of an interior angle of a regular polygon having 15 sides.
  9. An angle of a parallelogram measures 70°. Find the measure of the remaining three angles.
  10. One angle of a quadrilateral is 111° and the remaining three angles are equal. Find three angles.
  11. What is the ratio of the interior angles of a pentagon and a decagon?
  12. The perimeter of a parallelogram is 150 cm. One of its side is greater than the other by 25 cm. Find length of all sides of the parallelogram.
  13. Lengths of adjacent sides of a parallelogram is 3 cm and 4 cm. Find its perimeter.
  14. In a parallelogram, the ratio of the adjacent sides is 4 : 5 and its perimeter is 72 cm then, find the sides of the parallelogram.
  15. Two opposite angles of a parallelogram are (5x – 8)° and (2x + 82)°. Find the measures of each angle of the parallelogram.

 Thanks.

 

 

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Lesson Posted on 26/04/2017 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

What are rational numbers and irrational numbers?

Rajiv Vadera

I am MBA Finance. I believe in conceptual learning and not just theoretical

What are rational numbers?
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Lesson Posted on 19/04/2017 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

Relating natural numbers, whole numbers and integers

Rajiv Vadera

I am MBA Finance. I believe in conceptual learning and not just theoretical

Understanding relationship between natural numbers, whole numbers and integers
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Answered on 29/08/2016 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

How much can be charged for class 8 icse maths tutions per month

Hanamant Kullur

Stock Market Trainer

For Everyday Daily Classes One can charge 4500-6000 rs Monthly as a Home Tutor . For Weekend Saturday and Sunday Classes 4000 rs Nearly could be the Nominal Charge.
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Answered on 29/08/2016 Tuition/Class VI-VIII Tuition Tuition/Class VI-VIII Tuition/Mathematics

How much can be charged for class 8 icse maths tutions per month if it is online ?

Hanamant Kullur

Stock Market Trainer

It would be Nearly 300-400 rs per Hour Maximum One can Charge
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