Class 8 Maths

Algebra Chapter 7 Graphs

⇒ Graphs are pictorial representations of mathematical data and relations.

⇒ To construct a graph some sort of graph paper is used. The graph paper is divided into small squares.

Read and Learn More WBBSE Solutions For Class 8 Maths

⇒ To locate the exact position of a point we need two number lines that are perpendicular to each other, they are called co-ordinate axes.

⇒ The horizontal number line XOX’ is called X-axis and the vertical number line YY’ is called Y axis.

⇒ The point O at which the two axes intersect is called the origin.

⇒ The region within angle XOY is called 1st quadrant.

⇒ The region within angle YOX’ is called 2nd quadrant.

⇒ The region within the angle X’OY’ and Y’OX are called 3rd and 4th quadrants respectively.

⇒ We get the distance from Y-axis is X coordinate and the distance from X-axis is Y.

⇒ The X coordinate is called the abscissa and the Y coordinate is called the ordinate.

⇒ The coordinate of origin O is (0, 0).

⇒ Signs of abscissa and ordinate in the quadrants is shown.

To find out the points (6, 5), (-4, 5), (- 10, -5), (7,-4)

Solution:

1. To plot the point (6,5)

1. I drew two axis X’OX and YOY’ on the graph paper. The intersection of two axes (0, 0).
2. From O (0, 0) we moved 6 units to the right side along X-axis, and then from the point we moved 5 units above along the line parallel to Y-axis and reached at the point P its position is (6, 5).

2. To plot the point (-4, 5) from O(0, 0) we moved 4 units to the left side along X-axis, and then from the point we moved 5 units above along the line parallel to Y-axis and reached at the point Q, its position is (-4, 5).

3. To plot the point (- 10, 5) from 0 (0, 0) we moved 10 units to the left side along X-axis, and then from the point we moved 5 units below along the line parallel to Y-axis and reached at the point R, its position is (-10, -5).

4. To plot the point (7,-4) from O(0, 0) we moved 7 units to the right side along X-axis, and then from the point we moved 4 units below along the line parallel to Y-axis and reached at the point S, its position is (-7,-4).

Plot the points (5, 0), (0, 4), (- 3, 0), and (0, -6),

1. point A (5, 0) is at 5 unit away from origin (0, 0) on the X-axis.

∴ The distance of point A from the Y-axis is 5 units and from X-axis is 0 unit.

∴ The coordinates of point A is (5, 0).

2. The point B (0, 4) is at 4 unit away from the origin (0, 0) on Y-axis.

The distance of point B from the X-axis is 4 unit and from Y-axis is 0 unit.

∴ The coordinate of point B is (0, 4).

3. The point C (-3, 0) is at 3 unit away from origin on X-unit towards left.

∴ The distance of point C from Y-axis is 3 unit and from X-axis is 0 unit.

∴ The coordinate of C is (-3, 0).

4. The distance of point D from X-axis is 6 unit and from Y-axis is 0 unit.

∴ The coordinate of point D is (0,-6).

Algebra Chapter 7 Graphs Examples:

Example 1. If

1. Plot points (3, 4), (-6, -1), and (8, 7) on the graph paper and see whether they are collinear.
2. Put three non-collinear points on the graph paper.
3. Put three more collinear points other than D the above three collinear points on the graph paper and write their coordinates.

Solution:

1. The points (3, 4), (-6, -1), and (8, 7) are plotted on the graph paper.

⇒ We join these three points by a straight line AB.

⇒ Hence these three points are collinear.

2. Three non-collinear points (-9, 4), (3, 8), and (12, 3) are plotted on the graph paper.

3. Three more collinear points (- 5, 7), (9, 4), and (11, 8) other than the above three collinear points are plotted and these three points are joint by a straight line CD.

Example 2. See the graph on the graph paper beside and find the answers to the questions below:

1 guava = length of the smallest squares along X-axis and 1 rupee = length of a side of one smallest square along Y-axis.

1. Write the relation between the number of guava and the price of guava from the graph.
2. Write the price of 4 guava.
3. Write the number of guavas in 30 from the graph.
4. Write the number of guavas in 9.

Solution:

1. As the number of guava increases then the price of guava also increases. So the price of guava is proportional to the number of guavas.
2. The price Price of 4 guavas is 12.
3. The number of guavas in 30 is 10.
4. The number of guavas in 9 is 3.

Example 3. Draw the graph on the graph paper of the following data and see whether it is a linear graph. Also, find out the price of 6 books and the number of exercise books for Rs. 100 from the graph.

Solution:

1. At first drawing X-axis and Y-axis we took a scale along the two axis.
2. Let’s we consider the length of a side of the smallest square along X-axis = 2 books and length of the side of the smallest square along Y-axis = 5 rupees.
3. From given data let’s plot (2, 50), (3, 75), (5, 125), (8, 200), and (10, 250) in the graph paper.
4. Joining the points, we got a straight line, segment PQ. So it is a linear graph.

From the graph, the price of 6 books is 150 and the number of books for 100 is 4.

Example 4. If Draw the graph of y = x + 2.

Solution: y = x + 2

Tabulation:

1 small division on X-axix and Y-axis = 1 unit.

⇒ First, we plot the point A (1, 3), B (3, 5) C (5, 7), then we find A, B, and C which are colinear (PQ).

Example 5. Draw the graph, y = 6.

Solution: y = 6 means for any real value of x, value of y is always 6.

Tabulation:

Taking 1 small division 1 unit on both the axes we plot P (5, 6), Q (0, 6), R (7, 6) and join P, Q, R which is a linear graph.

Example 6. Choose the correct answer:

1. Value of abscissa of each point on the y-axis is

1. 0
2. 1
3. -1
4. None of these

Solution: Correct answer is 1. 0

2. Value of the ordinate of each point on the X-axis is

1. 1
2. 0
3. -1
4. None of these

Solution: Correct answer is 2. 0

3. (2, 3) lies on

1. 2nd quadrant
2. 3rd quadrant
3. 1st quadrant
4. None of these

Solution: Correct answer is 3. 1st quadrant

Example 7. Write ‘True’ or ‘False’:

1. (-2, 0) lies on the 2nd quadrant.

Solution: The statement is true.

2. A lies on X-axis, its abscissa is always zero.

Solution: The statement is false.

3. A (5, 7) and B (7, 5) are same points.

Solution: The statement is false.

Example 8. Fill in the blanks:

1. (0, 2) lies on the _________ side of Y axis.

Answer: Positive.

2. The line joining (1, 1), (2, 2) passes through _________

Answer: Origin.

3. Position of a point was first thought by _________

Answer: Rene Descartes.

Algebra Chapter 5 HCF And LCM Of Algebraic Expression

To find the H.C.F. (Highest Common Factor) or G.C.D. (Greatest Common Divisor) the given expressions should first be resolved into elementary factors.

The H.C.F. or G.C.D. will be the product of the common elementary factors of the highest power that divide each of the given expressions.

Algebra Chapter 5 HCF And LCM Of Algebraic Expression Examples

Example 1. Find the H.C.F. of 4a²b², 12ab³ and 20a²b⁴.

Solution:

H.C.F. of 4a²b², 12ab³ and 20a²b⁴ is 2 x 2 x a x b x b or 4ab²

Alternate method:

G.C.D. of 4, 12 and 20 = 4

Lowest power of a is 1 and b is 2 in the expression 4a²b², 12ab^3, and 20a²b⁴.

∴ The required G.C.D. is 4a¹b² or 4ab².

Read and Learn More WBBSE Solutions For Class 8 Maths

Example 2. Find the G.C.D. of the following algebraic expression. 24x³yz², 18x²y³z³, 36x⁴y²z⁴

Solution:

The required G.C.D. = 2 x 3 x x x x x y x y x z x z = 6x²y²z²

Example 3. Find G.C.D. of the following expressions:

1. a² + 3a + 2, a² + 4a + 3, a² + 5a +6
2. 3 (a + b)³, 6 (a + b)², 9(a² -b²)
3. x³ – 3x² – 10x, x³ + 6x² + 8x, x⁴ – 5x³ – 14x²

Solution:

1. 1st expression,

a² + 3a + 2

= a² + (2 + 1) a + 2 = a² + 2a + a +2

= a (a + 2) + 1 (a + 2) = (a + 2)(a + 1)

a² + 3a + 2, a² + 4a + 3, a² + 5a +6 = a (a + 2) + 1 (a + 2) = (a + 2)(a + 1)

2nd expression,

a² + 4a + 3

= a² + (3 + 1) a +3 = a² + 3a + a + 3

= a (a + 3)+ 1 (a + 3) = (a + 3)(a + 1)

3 (a + b)³, 6 (a + b)², 9(a² -b²) = a (a + 3)+ 1 (a + 3) = (a + 3)(a + 1)

3rd expression,

a² + 5a + 6

= a² + (3 + 2) a + 6 = a² + 3a + 2a + 6

= a (a + 3) + 2 (a + 3) = (a + 3)(a + 2)

x³ – 3x² – 10x, x³ + 6x² + 8x, x⁴ – 5x³ – 14x² = a (a + 3) + 2 (a + 3) = (a + 3)(a + 2)

∴ The required GCD is 1.

2. 1st expression

3 (a + b)³ = 3 (a + b) (a + b)(a + b)

2nd expression

6 (a + b)² = 2 x 3 x (a + b)(a + b)

3rd expression

9 (a² – b²) = 3 x 3 x (a + b)(a – b)

∴ The required GCD = 3 x (a + b) = 3 (a + b)

3. 1st expression

x³ – 3x² – 10x = x (x² – 3x – 10)

= x {x² – (5 – 2) x 10} = x{x² – 5x + 2x – 10}

= x {x(x – 5) + 2(x – 5)} = x(x – 5)(x + 2)

2nd expression

x³ + 6x² + 8x = x(x² + 6x + 8)

= x {x² + (4 + 2)x + 8}

= x {x² + 4x + 2x + 8)

= x {x (x + 4) + 2(x + 4)}

= x (x + 4)(x + 2)

3rd expression

x⁴ – 5x³ – 14x² = x² (x² – 5x – 14)

= x² (x² – (7 – 2) x 14} = x² (x² – 7x + 2x – 14}

= x² {x(x – 7) + 2(x – 7)} = x x x x (x – 7)(x + 2)

∴ The required GCD is x x x x (x + 2) = x(x + 2)

To find the L.C.M. each expression is first to be resolved into factors and the product of the factors having the highest powers in those factors will be the L.C.M.

Example 4. Find L.C.M. of the following expressions:

1. 20 a²b³c⁴, 25a³b²c², 60 abc⁵
2. 4x⁵y²z, 8xy²z³, 12yz⁴
3. xy⁴ – 8xy, x²y⁴ + 8x²y, xy⁴ – 4xy²

Solution:

1. 1st expression,

20a²b³c⁴ = 2 x 2 x 5 x a x a x b x b x b x c x c x c x c

2nd expression,

25a³b²c² = 5 x 5 x a x a x a x b x b x c x c

3rd expression,

60 abc⁵ = 2 x 2 x 3 x 5 x a x b xc x c x c x c x c

The highest power of 2 is 2, highest power of 5 is 2, highest power of 3 is 1, the highest power of a is 3, the highest power of b is 3.

The highest power of C is 5.

∴ The required L.C.M.

= 2 x 2 x 5 x 5 x 3 x a x a x a x b x b x b x c x c x c x c x c = 300 a³b³c⁵

Alternate method:

The common factor of 20a²b³c⁴, 25a³b²c² and 60abc5 is 5 x a x b x c x c = 5abc²

∴ The required LCM = 5abc² x 2 x 2 x a x b x b x c x c x 5 x a x 3 x c = 300a³b³c⁵

2. 4x⁵y²z = 2 x 2 x x x x x x x x x x x y x y x z

8xy²z³ = 2 x 2 x 2 x x x y x y x z x z x z

12yz⁴ = 2 x 2 x 3 x y x z x z x z x z

LCM = 2 x 2 x 2 x 3 x x x x x x x x x x x y x y x z x z x z = 24 x⁵y²z⁴

3. 1st expression,

xy⁴ – 8xy = xy(y³ – 8) = xy(y³ – 2³)

= xy(y – 2)(y² + y.2 +2²) = xy(y – 2)(y² + 2y + 4)

2nd expression,

x²y⁴ + 8x²y = x²y (y³ + 8)

= x²y(y³ +2³) = x²y (y + 2) (y² + y.2 + 2²)

= x.x.y(y + 2)(y² – 2y + 4)

3rd expression,

xy⁴ – 4xy² = xy²(y² – 4)

= x x y x y(y + 2)(y – 2)

The required LCM = x x x x y x y x (y + 2)(y² – 2y + 4) (y – 2)(y² + 2y + 4)

= x²y²(y + 2)(y – 2)(y² – 2y + 4)(y² + 2y + 4)

Example 5. Find the H.C.F and L.C.M. of the following algebraic expressions:

1. 8(x² – 4), 12(x³ + 8), 36(x² – 3x-10)
2. a⁴ + a²b² + b⁴, a³b + b⁴, (a² – ab)³
3. x² + y² – z² + 2xy, z² + x² – y² + 2zx, y² + z² – x² + 2yz
4. 3a² – 15a + 18, 2a2 + 2a – 24, 4a² + 36a + 80

Solution:

1. 1st expression,

8(x² – 4) = 8(x² – 2²) = 2 × 2 × 2 × (x + 2)(x − 2)

2nd expression,

12(x³ + 8)= 12 (x³ + 2³) = 12(x + 2)(x² – x·2 + 2²) = 2 × 2 × 3 × (x + 2)(x² – 2x + 4)

3rd expression,

36(x² – 3x – 10) = 36 {x² – (5 – 2)x – 10}

= 36 {x² – 5x + 2x – 10} = 36{x² – 5x + 2x – 10}

= 36 {x(x – 5) + 2(x – 5)} = 2 x 2 x 3 x 3 x (x – 5)(x + 2)

H.C.F.= 2 x 2 x (x + 2) = 4(x + 2)

L.C.M. = 2 x 2 x 2 x 3 x 3(x + 2)(x – 2)(x – 5)(x² – 2x + 4) = 72(x + 2)(x – 2)(x – 5)(x² – 2x + 4)

2. 1st expression,

a⁴ + a²b² + b⁴ = (a²)² + 2a²b² + (b²)² – a²b²

= (a² + b²) – (ab)²

= (a² + b² + ab)(a² + b² – ab)

2nd expression,

a³b + b⁴ = b(a³ + b³) = b(a + b)(a² – ab + b²)

3rd expression,

(a² – ab)³ = {a(a – b)}³ = a x a x a x(a – b)(a – b)(a – b)

∴ The required HCF = 1 and

LCM= a x a x a x b x (a + b)(a – b)(a – b)(a – b)(a² – ab + b²)(a² + ab + b²)

= a³b(a + b)(a – b)³(a² – ab + b²)(a² + ab + b²)

3. 1st expression,

x² + y² – z² + 2xy

= (x² + 2xy + y²) – z²

= (x + y)² – z² = (x + y + z) (x + y − z)

2nd expression,

z² + x² – y² + 2zx

= (z² + 2zx + x²) – y² = (z + x)² – y²

= (z + x + y) (z + x − y).

3rd expression,

y² + z² – x² + 2yz

= y² + 2yz + z² – x² = (y + z)² – x²

=(y + z + x) (y + z – x)

4. 1st expression,

3a² – 15a + 18

= 3(a² – 5a + 6) = 3{a² – (3 + 2)a + 6}

∴ HCF = (x + y + z)·

LCM = (x + y + z)(x + y = 2); (x – y + z) (y + z – x)

= 3(a² -3a – 2a + 6} = 3 {a(a – 3)- 2(a – 3)}

= 3(a – 3)(a – 2)

2nd expression,

2a² + 2a – 24

= 2 (a² + a – 12) = 2 {a² + (4 – 3)a – 12}

= 2 (a² + 4a – 3a – 12)

= 2 {a(a + 4) – 3(a + 4)}

= 2 (a + 4) (a-3)

3rd expression,

4a² + 36a + 80

= 4(a² + 9a + 20) = 4(a² + (5 + 4)a + 20)

= 4 {a² + 5a + 4a + 20)

4 {a(a + 5) + 4(a + 5)} = 2 × 2 × (a + 5) (a + 4)

HCF = 1

LCM = 2 x 2 x 3 x (a – 3)(a – 2)(a + 4)(a + 5)

= 12(a – 3)(a – 2)(a + 4)(a + 5)

Example 6. Choose the correct answer:

1. HCF of 3xy²z², 2yz²x², x²y²z is

1. xyz
2. x²y²z²
3. xyz²
4. None of these

Solution: The HCF of 3xy²z², 2yz²x^2, and x²y²z² is xyz.

∴ So the correct answer is 1. xyz

2. LCM of a²b²c², a³b²c³ and ab³c⁴ is

1. a³b³c³
2. a²b²c²
3. abc²
4. abc³

Solution: LCM is a³b³c⁴.

∴ So the correct answer is 1. a³b³c³

3. If 0 < x < 1, then HCF of x², x⁵, x is

1. x
2. x²
3. x⁵
4. None of these

Solution: Let x = \(\frac{1}{2}\)

x² = \(\frac{1}{4}\)

x⁵ = \(\frac{1}{32}\)

HCF of \(\frac{1}{32}\), \(\frac{1}{8}\) and \(\frac{1}{2}\) is \(\frac{\text { HCF of } 1,1,1}{\text { LCM of } 92,8 \text { and } \frac{1}{8}}=\frac{1}{32}=x^5\)

∴ So the correct answer is 3. x⁵

Example 7. Write ‘True’ or ‘False’:

1. If a, and b are prime numbers, then their LCM is ab.

Solution: The statement is true.

2. If a, and b are prime numbers then their HCF is 1.

Solution: The statement is true.

3. LCM of x^m and x^m+p is x^m.

Solution: \(x^{m+r}=x^m \cdot x^r\)

LCM of \(x^m and x^m \cdot x^r\) is

∴ So the statement is false.

Example 8. Fill in the blanks:

1. HCF of m² + 9m + 20 and m² + 13m + 36 is _______

Solution: m² + 9m + 20 = (m + 5)(x + 4)

m² + 13m + 36= (x + 9)(x + 4)

HCF is (x + 4).

HCF of m² + 9m + 20 and m² + 13m + 36 is (x + 4).

2. LCM of (a – b) and (b – a) is _______

Solution: (a – b) or (b – a).’

LCM of (a – b) and (b – a) is (a – b) or (b – a).’

3. HCF of ax², a²x³, a⁴x is _________

Solution: ax.

HCF of ax², a²x³, a⁴x is ax.

Arithmetic

• Chapter 1 Ratios And Proportion
• Chapter 2 Pie Chart
• Chapter 3 Rational Number
• Chapter 4 Rule of Three
• Chapter 5 Percentage
• Chapter 6 Mixture
• Chapter 7 Time and Work

Algebra

• Chapter 1 Simplify Revision Of Old Lessons
• Chapter 2 Multiplication and Division of Polynomials
• Chapter 3 Cubes
• Chapter 4 Factorisation of Algebraic Expression
• Chapter 5 HCF and LCM of Algebraic Expression
• Chapter 6 Simplification of Algebraic Expression
• Chapter 7 Graphs
• Chapter 8 Formation of an Equation and Its Solution

Geometry

• Chapter 1 Revision Of Quadrilateral
• Chapter 2 Complementary Angles, Supplementary Angles and Adjacent Angles
• Chapter 3 Concept of Vertically Opposite Angles
• Chapter 4 Properties of Parallel Lines and Their Transversal
• Chapter 5 Relation Between Two Sides of A Triangle and Their Opposite Angels
• Chapter 6 Verification of the Relation Between the Angles and Sides of a Traingle
• Chapter 7 Geometrical Proofs
• Chapter 8 Construction

Algebra Chapter 4 Factorization Of Algebraic Expression

⇒ Suppose area of a rectangle is (9x² – 16y²) sq. unit.

⇒ To find out the length of each side of a rectangle it is required to factorisation of the algebraic expression first.

⇒ Area = 9x² – 16y² = (3x)²– (4y)² = (3x + 4y) (3x – 4y)

⇒ So the length and breadth of the rectangle are (3x + 4y) unit and (3x – 4y) unit respectively.

Read and Learn More WBBSE Solutions For Class 8 Maths

Necessary identities which help to factorise.

1. a² – b² = (a + b) (a – b)
2. a³ + b³ = (a + b) (a² – ab + b²)
3. a³ – b³ = (a – b) (a² + ab + b²)
4. x² +(a + b)x + ab = (x + a) (x + b)

Algebra Chapter 4 Factorization Of Algebraic Expression Examples

Example 1. Comparing the following algebraic expression with the identity. x² + (a + b)x + ab = (x + a) (x + b), find the value of a and b and factorise

1. x² – 5x + 6
2. x² – 5x – 6
3. (x + y)² – 7(x + y) + 12

Solution:

1. x² – 5x + 6

Given

f(x) = x² – 5x + 6

= x² + {(-3)+(-2)}x + (-3) x (-2) = (x – 3)(x – 2)

Here a = -3, b = -2

2. x² – 5x – 6

Given

f(x) = x² – 5x – 6

= x² + {(-6) + 1}x + (-6) x 1

= (x – 6) (x + 1)

Here a = -6, b = 1

3. (x + y)² -7(x + y) + 12

Given

f(x) = (x + y)2 -7(x + y) + 12

= (x + y)² + {(-4)+(-3)}(x + y) + (-4) x (-3)

= (x + y = 4)(x + y − 3)

Here a = -4, b = -3

Example 2. Resolve into factors by middle-term break. x² + 3x – 40

Solution: The factors of 40 are (4, 10), (2, 20), (5, 8), (1, 40)

Again, 3 = 8 – 5.

x² + 3x – 40

= x² + (8 – 5)x – 40

= x² + 8x – 5x – 40

= x(x + 8) – 5 (x + 8)

= (x + 8)(x – 5)

x² + 3x – 40 = (x + 8)(x – 5)

Example 3. x² – 13x + 40

Solution: x² – (8 + 5)x + 40

= x²– 8x – 5x + 40

40 = 5 x 8
13 = 5 + 8

= x(x – 8) – 5(x – 8)

= (x – 8)(x – 5)

x² – 13x + 40 = (x – 8)(x – 5)

Example 4. (a + 1)(a + 3)(a – 4)(a – 6) + 24

Solution: (a + 1)(a + 3)(a – 4)(a – 6) + 24

= (a + 1)(a – 4)(a + 3)(a − 6) + 24

= (a² – 4a + a – 4)(a² – 6a + 3a – 18) + 24

= (a² – 3a -4)(a² – 3a – 18) + 24

= (x – 4)(x – 18)+ 24

= x² – 18x – 4x + 72 + 24

= x² – 22x + 96

= x² – (16 + 6)x + 96

= x – 16x – 6x + 96

96 = 6 x 16

22 = 6 + 16

= x(x – 16) – 6(x – 16)

= (x – 16)(x – 6)(a² – 3a – 16)(a² – 3a – 6)

(a + 1)(a + 3)(a – 4)(a – 6) + 24 = (x – 16)(x – 6)(a² – 3a – 16)(a² – 3a – 6)

Example 5. x² + 4pqx – (p² – q²)²

Solution:

Given

x² + 4pqx – (p + q)² (p − q)²

= x² + {(p + q)² – (p − q)² x – (p + q)²(p − q)²

= x2 + (p + q)² x – (p − q)² x – (p + q)²(p − q)²

= x{x + (p + q)²} – (p − q)² {x + (p + q)²}

= {x + (p + q)²} {x – (p − q)²}

x² + 4pqx – (p² – q²)² = {x + (p + q)²} {x – (p − q)²}

Example 6. To factorize ax² + bx + c, find two quantities such that their algebraic sum is b (co-efficient of x) and their product is ac. [The product of coefficient of x² and the term free from x] Resolve 6y² – y – 15

Solution: Here, a = 6, b = -1, c = -15

a x c = 6x – 15 = 90

90= 1 x 90 = 2 x 45 = 3 x 30 = 5 x 18 = 6 x 15 = 9 x 10

10 – 9 = 1

∴ 6y² – y – 15

= 6y² – (10 – 9)y – 15

= 6y² – 10y+ 9y – 15

= 2y(3y – 5) + 3(3y – 5)

= (3y – 5)(2y + 3)

10 × 9 = 90

10 – 9 = 1

Example 7. Resolve into factors: 6(x + y)² + 5 (x² – y²) – 6 (x − y)²

Solution:

Given

6 (x + y)² + 5(x + y)(x – y) – 6(x – y)²

[Let x + y = a;x – y = b]

[9 x 4 = 6 x 6 = 36; 9 – 4 = 5]

= 6a² + 5ab – 6b²

= 6a² + (9 – 4)ab – 6b²

= 6a² + 9ab – 4ab – 6b²

= 3a(2a+3b) – 2b (2a + 3b)

= (2a + 3b)(3a – 2b)

[Substituting the value of a and b]

= {2(x + y) + 3 (x – y)} {3(x + y) – 2(x – y)}

= (2x + 2y + 3x – 3y)(3x + 3y – 2x + 2y) = (5x – y) (x + 5y)

6(x + y)² + 5 (x² – y²) – 6 (x − y)² = (2x + 2y + 3x – 3y)(3x + 3y – 2x + 2y) = (5x – y) (x + 5y)

Example 8. Resolve into factors: x² – 2abx + (a² – b²)

Solution:

Given

x² -2abx + (a² – b²)

= x² – 2ax + (a + b)(a – b)

= x² – {(a + b) + (a – b)}x + (a + b) (a – b)

= x2 – (a + b)x – (a – b)x + (a + b)(a – b)

= x(x – a – b) – (a – b) (x – a – b) = (x – a – b) (x – a + b)

x² – 2abx + (a² – b²) = x(x – a – b) – (a – b) (x – a – b) = (x – a – b) (x – a + b)

Example 9. Resolve into factors: a² + 1 – \(\frac{6}{a^2}\)

Solution: \(a^2+\frac{3 a}{a}-\frac{2 a}{a}-\frac{6}{a^2}\)

= \(a\left(a+\frac{3}{a}\right)-\frac{2}{a}\left(a+\frac{3}{a}\right)=\left(a+\frac{3}{a}\right)\left(a-\frac{2}{a}\right)\)

Example 10. Resolve into factors: \(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7\)

Solution:

Given

= \(2\left\{\left(a-\frac{1}{a}\right)^2+2 \cdot a \cdot \frac{1}{a}\right\}-\left(a-\frac{1}{a}\right)-7=2\left(a-\frac{1}{a}\right)^2+4-\left(a-\frac{1}{a}\right)-7\)

= \(2\left(a-\frac{1}{a}\right)^2-\left(a-\frac{1}{a}\right)-3=2\left(a-\frac{1}{a}\right)^2-\{3-2\}\left(a-\frac{1}{a}\right)-3\)

= \(2\left(a-\frac{1}{a}\right)^2-3\left(a-\frac{1}{a}\right)+2\left(a-\frac{1}{a}\right)-3\)

= \(\left(a-\frac{1}{a}\right)\left\{2\left(a-\frac{1}{a}\right)-3\right\}+1\left\{2\left(a-\frac{1}{a}\right)-3\right\}\)

= \(\left\{2\left(a-\frac{1}{a}\right)-3\right\}\left(a-\frac{1}{a}+1\right)=\left\{2 a-\frac{2}{a}-3\right\}\left(a-\frac{1}{a}+1\right)\)

= \(\left(2 a-3-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)=\left(2 a-(4-1)-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)\)

= \(\left(2 a-4+1-\frac{2}{a}\right)\left(a-\frac{1}{a}+1\right)\)

= \(\left\{2(a-2)+\frac{1}{a}(a-2)\right\}\left(a-\frac{1}{a}+1\right)=(a-2)\left(2+\frac{1}{a}\right)\left(a-\frac{1}{a}+1\right)\)

Example 11. Resolve into factors: a (a + 1) x² – x – a (a− 1)

Solution:

Given

a (a + 1)x² – x – a(a – 1)

= a (a + 1)x² – {a² – (a² – 1)}x – a(a− 1)

= a (a + 1)x² – a² x + (a² – 1)x – a(a – 1)

= ax {(a + 1)x – a} + (a – 1) {(a + 1)x – a}

= {(a + 1)x – a} (ax + a – 1) = (ax + x – a)(ax + a – 1)

a (a + 1) x² – x – a (a− 1) = {(a + 1)x – a} (ax + a – 1) = (ax + x – a)(ax + a – 1)

Example 12. Factorising the expressions into factors by expressing them as the difference of two sqaures

1. x² + px + q

Solution: \(x^2+2 \cdot x \cdot \frac{p}{2}+\left(\frac{p}{2}\right)^2+q-\left(\frac{p}{2}\right)^2\)

= \(\left(x+\frac{p}{2}\right)^2-\left(\frac{p^2}{4}-q\right)^2=\left(x+\frac{p}{2}\right)^2-\left(\sqrt{\frac{p^2}{4}-q}\right)^2\)

= \(\left(x+\frac{p}{2}+\sqrt{\frac{p^2}{4}-q}\right)\left(x+\frac{p}{2}-\sqrt{\frac{p^2}{4}-q}\right)\)

2. x² + 5x + 6

= \(\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+\frac{5}{2}+\frac{1}{2}\right)\left(x+\frac{5}{2}-\frac{1}{2}\right) \)

= \(\left(x+\frac{6}{2}\right)\left(x+\frac{4}{2}\right)=(x+3)(x+2)\)

3. \(7 x^2-30 x+8\)

= \(7\left(x^2-\frac{30}{7} x+\frac{8}{7}\right)\)

= \(7\left\{x^2-2 \cdot x \cdot \frac{15}{7}+\left(\frac{15}{7}\right)^2-\left(\frac{15}{7}\right)^2+\frac{8}{7}\right\}\)

= \(7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{225}{49}-\frac{8}{9}\right)\right\}=7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{225-56}{49}\right)\right\}\)

= \(7\left\{\left(x-\frac{15}{7}\right)^2-\frac{169}{49}\right\}=7\left\{\left(x-\frac{15}{7}\right)^2-\left(\frac{13}{7}\right)^2\right\}\)

= \(7\left(x-\frac{15}{7}+\frac{13}{7}\right)\left(x-\frac{15}{7}-\frac{13}{7}\right)=7\left(x-\frac{2}{7}\right)\left(x-\frac{28}{7}\right)=(7 x-2)(x-4)\)

4. \(12 a^2+13 a b-4 b^2\)

= \(12\left(a^2+\frac{13}{12} a b-\frac{4}{12} b^2\right)\)

= \(12\left\{a^2+2 \cdot a \cdot \frac{13 b}{24}+\left(\frac{13 b}{24}\right)^2-\left(\frac{13 b}{24}\right)^2-\frac{b^2}{3}\right\}\)

= \(12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{169 b^2}{576}+\frac{b^2}{3}\right)\right\}=12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{169 b^2+192 b^2}{576}\right)\right\}\)

= \(12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{361 b^2}{576}\right)\right\}=12\left\{\left(a+\frac{13 b}{24}\right)^2-\left(\frac{19 b}{24}\right)^2\right\}\)

= \(12\left(a+\frac{13 b}{24}+\frac{19 b}{24}\right)\left(a+\frac{13 b}{24}-\frac{19 b}{24}\right)=12\left(a+\frac{32 b}{24}\right)\left(a-\frac{6 b}{24}\right)\)

= \(12\left(a+\frac{4 b}{3}\right)\left(a-\frac{b}{4}\right)=3\left(a+\frac{4 b}{3}\right) 4\left(a-\frac{b}{4}\right)=(3 a+4 b)(4 a-b)\)

Example 13. Resolve into factors: ax² – (a² – 2)x – 2a

Solution:

Given

ax² – (a² – 2)x – 2a

= ax² – a²x + 2x – 2a = ax(x – a) + 2(x − a)

= (x – a)(ax + 2)

ax² – (a² – 2)x – 2a = (x – a)(ax + 2)

Example 14. Choose the correct answer:

1. (a – b) is equal to

1. \(\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)
2. \((\sqrt{a}-\sqrt{b})^2\)
3. \(\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)\left(a^{\frac{1}{3}}-b^{\frac{1}{3}}\right)\)
4. None of these

Solution: \(a-b=\left(a^{\frac{1}{2}}\right)^2-\left(b^{\frac{1}{2}}\right)^2=\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)

∴ So the correct answer is 1. \(\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)\)

2. Factorised form of (p² – 17p – 38) is

1. (p – 19)(p – 2)
2. (p – 19)(p + 2)
3. (p + 19)(p – 2)
4. (p + 19)(p + 2)

Solution: p² – 17p – 38

= p² – (19 – 2)p – 38 = p² – 19p + 2p – 38

= p(p – 19)+ 2(p – 19)= (p – 19)(p + 2)

∴ The correct answer is 2. (p – 19)(p + 2)

3. If (5x² – 4x – 9) = (x + 1) (5x + p), then the value of p is

1. 9
2. 5
3. -9
4. None Of these

Solution: 5x² – 4x – 9

= 5x² – 9x + 5x – 9 = x(5x – 9) + 1 (5x – 9)

= (5x – 9)(x + 1)

(x + 1)(5x + p) = (x + 1)(5x – 9)

5x + p = 5x – 9 ⇒ p = -9

∴ The correct answer is 3. -9

Example 15. Write ‘True’ or ‘False’:

1. The factors of (x² – xy – 30y²) is (x + 5y) (x − 6y).

Solution: x² – xy – 30y² = x² – 6xy + 5xy – 30y²

= x(x – 6y) + 5y(x – 6y) = (x – 6y)(x + 5y)

∴ The statement is true.

2. Difference of twp square of the expression x² – x -12 is \(\left(\dot{x}-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

Solution: \(x^2-x-12\)

= \(x^2-2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2-12\)

= \(\left(x-\frac{1}{2}\right)^2-\left(\frac{1}{4}+12\right)=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}\)

= \(\left(x-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)

∴ The statement is true.

3. Factorised form of (a + b)³ – c³ is (a + b+ c)(a² + b² + c² + 2ab + ac + bc)

Solution: (a + b)³ – c³

= (a + b – c) {(a + b)² + (a + b) c + c²}

= (a + b – c) (a² + 2ab + b² + ac + bc + c²)

∴ The statement is false.

Example 16. Fill in the blanks:

1. (x + a) (x + b) = x² + (a + b) x + ________

Solution: (x + a) (x + b) = x(x + b) + a(x + b)

= x² + bx + ax + ab = x2 + (a + b)x + ab.

(x + a) (x + b) = x² + (a + b) x + = x² + bx + ax + ab = x2 + (a + b)x + ab.

2. 67² – 37² = 30 x _________

Solution: 67² – 37² = 30 x ________

= (67 – 37)(67 +37) = 30 x 104

67² – 37² = 30 x = (67 – 37)(67 +37) = 30 x 104

3. (a + b + c)³ = a³ + b³ + c³ + ________

Solution: 3 (a + b)(b + c)(c + a).

(a + b + c)³ = a³ + b³ + c³ + 3 (a + b)(b + c)(c + a).

Algebra Chapter 3 Cubes

⇒ If length of each side of a cube is x unit then its volume is x³ cubic units.

⇒ If length of each side of a cube is (a + b) unit then its volume is (a + b)³

⇒ (a + b)³ = (a + b) (a + b)²

⇒ (a + b)³ = (a + b) (a² + 2ab + b²)

⇒ (a + b)³  = a(a² + 2ab + b²) + b(a² + 2ab + b²)

⇒ (a + b)³  = a³ + 2a²b + ab² + a²b + 2ab² + b³

∴ (a + b)³  = a³ + 3a²b + 3ab² + b³

⇒ (a – b)³ = (a – b) (a – b)²

⇒ (a – b)³ = (a – b) (a² – 2ab + b²)

⇒ (a – b)³ = a(a² – 2ab + b²) – b(a² – 2ab + b²)

⇒ (a – b)³= a³ – 2a²b + ab² – a²b + 2ab² – b³

⇒ (a – b)³= a³ – 3a²b + 3ab² – b³

Necessary Formulae:

Read and Learn More WBBSE Solutions For Class 8 Maths

1. (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab (a + b) = a³ + b³ – 3ab (a – b)
2. (a – b)³ = a³ – 3a²b + 3ab² – b³
3. a³+ b³ = (a + b)³ – 3ab (a + b) = (a + b) (a² – ab + b²)
4. a³ – b³ = (a – b)³ + 3ab (a – b) = (a – b) (a² + ab + b²)
5. a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c³ – ab- bc – ca) = \(\frac{1}{2}\)(a + b + c) {(a – b)² + (b − c)² + (c − a)²}

Algebra Chapter 3 Cubes Examples

Example 1. Find the value of

1. (105)³
2. (99)³

Solution:

1. (105)³ = (100+ 5)³

(a + b)³ = (100+ 5)³

Here a = 100, b= 5.

∴ (a + b)³ = a³ + 3a²b + 3ab² + b³

= (100)³ + 3 x (100)² x 5 + 3 x 100 x (5)² + (5)³

= 1000000 + 150000+ 7500 +125

= 1157625

∴(105)³ = 1157625

The value of (105)³ = 1157625

2. (99)³ = (100 – 1)³  = (a – b)³

Here a = 100, b= 1.

∴(a – b)³ = a³ – 3a²b + 3ab² – b³

= (100)³ -3 x (100)² x 1 + 3 x 100 x (1)² – ( 1)³

= 1000000 – 30000 + 300 – 1

= 970299

∴ (99)³ = 970299

The value of (99)³ = 970299

Example 2. Find the cube of

1. 3a + 5b
2. 9x – 7y
3. \(\frac{a}{2}\)–\(\frac{b}{3}\)
4. a +2b – 3c

Solution:

1. (3a + 5b)³

∴ (a + b)³ = a³ + 3a²b + 3ab² + b³

(3a + 5b)³ = (3a)³ + 3x (3a)² x 5b + 3 x 3a x (5b)² + (5b)³

(3a + 5b)³ = 27a³ + 135a²b + 150ab² + 125b³

2. (9x – 7y)³

∴(a – b)³ = a³ – 3a²b + 3ab² – b³

(9x – 7y)³ = (9x)³ – 3 x (9x)² x 7y + 3 x 9x × (7y)² – (7y)³

(9x – 7y)³ = 729x³ – 1701x²y + 1323xy³ – 343y³

3. \(\left(\frac{a}{2}-\frac{b}{3}\right)^3\)

= \(\left(\frac{a}{2}\right)^3-3 \times\left(\frac{a}{2}\right)^2 \times \frac{b}{3}+3 \times \frac{a}{2} \times\left(\frac{b}{3}\right)^2-\left(\frac{b}{3}\right)^3\)

= \(\frac{a^3}{8}-3 \times \frac{a^2}{4} \times \frac{b}{3}+3 \times \frac{a}{2} \times \frac{b^2}{9_3}-\frac{b^3}{27}=\frac{a^3}{8}-\frac{a^2 b}{4}+\frac{a b^2}{6}-\frac{b^3}{27}\)

4. (a + 2b – 3c)³

= {(a + 2b) – 3c)³

= (a + 2b)³ – 3 x (a + 2b)² x 3c + 3 x (a + 2b) x (3c)² – (3c)³

= a³ + 3 x a² × 2b + 3 x a x (2b)² + (2b)³ – 3 {a² + 2.a.2b + (2b)²} x 3c +(3a + 6b) x 9c³ – 27c³

= a³ + 6a²b + 12ab² + 8b³ – 9a²c – 36abc – 36b²c + 27ac² + 54bc² – 27c³

= a³ + 8b³ – 27c³ + 6a²b + 12ab² – 9a²c – 36abc – 36b²c + 27ac² + 54bc² – 27c³

Example 3. If a – 2b = 5, then find the value of a³ – 8b³ – 30ab

Solution: a³ – 8b³ – 30ab

= (a)³ – (2b)³ – 30ab = (a – 2b)³ + 3 x a x 2b (a – 2b) – 30ab

= (5)³ + 6ab x 5 – 30ab [a – 2b = 5]

= 125 + 30ab – 30ab = 125

Alternating method: a³ – 8b³ – 30ab

= (a)³ – (2b)³ – 3 x a x 2b. (5)

=a³ – (2b)³ – 3 x a x 2b x (a – 2b)

= (a – 2b)³ = (5)³ = 125

The value of a³ – 8b³ – 30ab = 125

Example 4. Simplify: 7.4 x 7.4 x 7.4 + 3 x 7.4 x 7.4 x 2.6 + 3 x 7.4 x 2.6 x 2.6 + 2.6 x 2.6 x 2.6

Solution: 7.4 x 7.4 x 7.4 + 3 x 74 x 74 x 2.6 + 3 x 74 x 26 x 2.6+ 2.6 x 2.6 x 2.6

= (7.4)³ + 3 x (7.4)² x 2.6 + 3 x 74 x (2.6)² + (2.6)² = (7.4 + 2.6)³ = (10)³ = 1000

7.4 x 7.4 x 7.4 + 3 x 7.4 x 7.4 x 2.6 + 3 x 7.4 x 2.6 x 2.6 + 2.6 x 2.6 x 2.6 = 1000

Example 5. If a³ + b³ + c³ = 3abc and a≠b≠c, then find the value of (a + b + c)

Solution: a³ + b³ + c³ = 3abc

⇒ a³ + b³ + c³ – 3abc = 0

⇒ (a + b + c)(a² + b² + c² – ab- bc – ca) = 0

either, a + b + c = 0 or, a² + b² + c² – ab – bc- ca = 0

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

If sum of two or more square are equal to zero then each square will be zero.

(a – b)² = 0
(a – b)² = 0
⇒ a – b = 0
⇒ a = b

(b – c)² = 0
(b – c)² = 0
⇒ b – c = 0
⇒ b = c

(c – a)² = 0
⇒ (c – a)² = 0

∴ a = b = c

But according to condition a ≠ b ≠ c

i.e. a² + b² + c² – ab- bc – ca ≠ 0

∴ a + b + c = 0

The value of (a + b + c) = 0

Example 6. If 2a + \(\frac{1}{3a}\) = 4, then find the value of 27a³ + \(\frac{1}{8 a^3}\)

Solution: \(2 a+\frac{1}{3 a}=4\Rightarrow \frac{3}{2}\left(2 a+\frac{1}{3 a}\right)=4 \times \frac{3}{2} \Rightarrow 3 a+\frac{1}{2 a}=6\)

= \((6)^3-\frac{9}{2} \times 6^3=216-27=189\)

Example 7. If m + n = 5 and mn = 6, then prove that (m² + n²) (m³ + n³) = 455

Solution: (m² + n²) (m³ + n³) = {(m + n)² – 2mn} {(m + n)³ – 3mn (m + n)}

= {(5)² – 2 x 6} {(5)³ – 3 x 6 x 5}= (25 – 12) (125 – 90)= 13 x 35 = 455

Example 8. Find the value of 64a³ – 144a² + 108a – 50 if a = \(\frac{3}{2}\)

Solution:

let f(a)= 64a³ – 144a² + 108a – 50

f(a) = (4a)³ – 3 x (4a)² x 3 + 3 x 4a x (3)² – (3)² + (3)³ – 50

f(a)= (4a – 3)³ + 27 – 50  [a = \(\frac{3}{2}\)]

f(\(\frac{3}{2}\)) = \(\left(4^2 \times \frac{3}{2}-3\right)^3-23\)

f(\(\frac{3}{2}\)) = (6-3)³ – 23 = (3)³ – 23 = 27 – 23 = 4

The value of 64a³ – 144a² + 108a – 50 = 4

Example 9. Simplify:

1. (4a² – 25b²)(4a² + 10ab + 25b²)(4a² – 10ab + 25b²)

Solution: (4a² – 25b²) (4a² + 10ab + 25b²) (4a² – 10ab + 25b²)

= {(2a)² – (5b)²} {(2a)² + 2a.5b + (5b)²} {(2a)² – 2a.5b + (5b)²}

= (2a +5b) (2a – 5b) {(2a)² + 2a.5b + (5b)²) {(2a)² – 2a.5b + (5b)²}

= (2a +5b) {(2a)² – 2a.5b + (5b)²} (2a – 5b) {(2a)² + 2a.5b + (5b)²}

= {(2a)³ + (5b)³} {(2a)³ – (5b)³}

= (8a³ + 125b³)(8a³ – 125b³) Here (a+b × a-b) = a²-b²

= (8a³)² -(125b³)² = 64a⁶ – 15625b⁶.

(4a² – 25b²)(4a² + 10ab + 25b²)(4a² – 10ab + 25b²) = (8a³)² -(125b³)² = 64a⁶ – 15625b⁶.

2. (x + 3) (x² – 3x + 9)- (2x – 5) (4x² + 10x + 25) – (x – 4) (x² + 4x + 16)

Solution: (x + 3)(x² – 3x+9)- (2x-5) (4x²+10x + 25) – (x – 4) (x² + 4x + 16)

= (x + 3) (x² – x.3 + 3²) (2x – 5) {(2x)² + 2x.5+ (5)²} – (x − 4) {x² + x.4 + (4)²}

= (x³ + 3³) – {(2x)³ – (5)³} – {x³ – 4³}

= x³ + 27 – 8x³ + 125 – x³ + 64

= 216 – 8x³

(x + 3) (x² – 3x + 9)- (2x – 5) (4x² + 10x + 25) – (x – 4) (x² + 4x + 16) = 216 – 8x³

Example 10. If \(\frac{a}{b}+{b}{a}\) = 1, then find the value of (a³ – b³)

Solution: \(\frac{a}{b}+{b}{a}\) = -1

⇒ \(\frac{a^2+b^2}{a b}=-1\)

⇒ a² + b² = -ab

⇒  a² + ab + b² = 0

a³ – b³ = (a – b) (a² + ab + b²) = (a – b) (0) = 0

The value of (a³ – b³) = 0

Example 11. Resolve into factors:

1. a³ – 9b³ – 3ab(a – b)

Solution: a³ – 9b³ – 3ab (a – b) = a³ – b³ – 3ab(a – b) – 8b³

= (a – b)³ – (2b)³

= (a – b-2b) {(a – b)² + (a – b).2b + (2b)²}

= (a – 3b) (a² – 2ab + b² + 2ab – 2b² + 4b²)

= (a – 3b) (a² + 3b²)

a³ – 9b³ – 3ab(a – b) = (a – 3b) (a² + 3b²)

2. a¹² – b²⁼¹²

Solution: \(\left(a^6\right)^2-\left(b^6\right)^2=\left(a^6+b^6\right)\left(a^6-b^6\right)\)

= \(\left\{\left(a^2\right)^3+\left(b^2\right)^3\right\}\left\{\left(a^3\right)^2-\left(b^3\right)^2\right\}\)

= \(\left(a^2+b^2\right)\left\{\left(a^2\right)^2-a^2 b^2+\left(b^2\right)^2\right\}\left(a^3+b^3\right)\left(a^3-b^3\right)\)

= \(\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)(a+b)\left(a^2-a b+b^2\right)(a-b)\left(a^2+a b+b^2\right)\)

= \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

3. \(\frac{a^3}{64}-\frac{64}{a^3}\)

Solution: = \(\left(\frac{a}{4}\right)^3-\left(\frac{4}{a}\right)^3=\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}\right)^2+\frac{a}{4} \cdot \frac{4}{a}+\left(\frac{4}{a}\right)^2\right\}\)

= \(\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}\right)^2+2 \cdot \frac{a}{4} \cdot \frac{b}{4}+\left(\frac{4}{a}\right)^2-1\right\}=\left(\frac{a}{4}-\frac{4}{a}\right)\left\{\left(\frac{a}{4}+\frac{4}{a}\right)^2-(1)^2\right\}\)

= \(\left(\frac{a}{4}-\frac{4}{a}\right)\left(\frac{a}{4}+\frac{4}{a}+1\right)\left(\frac{a}{4}+\frac{4}{a}-1\right)\)

4. \(x^3-3 x^2 y+3 x y^2-28 y^3\)

Solution: \(\left(x^3-3 x^2 y+3 x y^2-y^3\right)-27 y^3\)

∴(a – b)³ = a³ – 3a²b + 3ab² – b³

= \((x-y)^3-(3 y)^3=(x-y-3 y)\left\{(x-y)^2+(x-y) \cdot 3 y+(3 y)^2\right\}\)

= \((x-4 y)\left\{x^2-2 x y+y^2+3 x y-3 y^2+9 y^2\right\}\)

= \((x-4 y)\left(x^2+x y+7 y^2\right)\)

Example 12. Simplify: \((a-c)\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right\}\)

Solution: \(\{(a+b)-(b+c)\}\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right.\)

= \((a+b)^3-(b+c)^3=a^3+3 a^2 b+3 a b^2+b^3-b^3-3 b^2 c-3 b c^2-c^3\)

= \(a^3+3 a^2 b+3 a b^2-3 b^2 c-3 b c^2-c^3\)

\((a-c)\left\{(a+b)^2+(a+b)(b+c)+(b+c)^2\right\}\) = \(a^3+3 a^2 b+3 a b^2-3 b^2 c-3 b c^2-c^3\)

Example 13. Simplify: If \(x^2+\frac{1}{16 x^2}=4 \frac{1}{2}\), then \(8 x^3-\frac{1}{8 x^3}=\)?

Solution: \(x^2+\frac{1}{16 x^2}=4 \frac{1}{2} \Rightarrow(x)^2+\left(\frac{1}{4 x}\right)^2=\frac{9}{2} \Rightarrow\left(x-\frac{1}{4 x}\right)^2+2 \cdot x \cdot \frac{1}{4 x}=\frac{9}{2}\)

⇒ \(\left(x-\frac{1}{4 x}\right)^2=\frac{9}{2}-\frac{1}{2}=4 \Rightarrow x-\frac{1}{4 x}=\sqrt{4}=2\)

⇒ \(2\left(x-\frac{1}{4 x}\right)=2 \times 2 \Rightarrow 2 x-\frac{1}{2 x}=4\)

= (4)³ + 3.4 = 64 + 12 = 76

Example 14. If \(a+\frac{1}{a}=\sqrt{3}, \text { then } a^3+\frac{1}{a^3}=\text { ? }\)

Solution: \(a+\frac{1}{a}=\sqrt{3}\)

Example 15. If a + b + c = 0, then \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) =?

Solution: \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}=\frac{a^3+b^3+c^3}{a b c}=\frac{\left(a^3+b^3+c^3-3 a b c\right)+3 a b c}{a b c}\)

= \(\frac{(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)+3 a b c}{a b c}\)

= \(\frac{(0)\left(a^2+b^2+c^2-a b-b c-c a\right)+3 a b c}{a b c}=\frac{0+3 a b c}{a b c}=3\)

Example 16. Choose the correct answer:

1. If a + b + c = 0, then a³ + b³ + c³ =?

1. 3abc
2. abc
3. c
4. None of these

Solution: a³ + b³ + c³

= (a + b)³ – 3ab (a + b) + c³

= (-c)³ – 3ab (-c) + c³ [ a + b + c = 0]

= c³ + 3abc + c³ = 3abc

∴ So the correct answer is 1. 3abc

2. If a – b = 1 and a³ – b³ = 61, then the value of ab is

1. 10
2. 20
3. 30
4. None of these

Solution: a³ – b³ = 61

⇒ (a – b)³ + 3ab (a – b) = 61

⇒ (1)³ + 3ab (1) = 61

⇒ 3ab = 61 – 1 = 60

⇒ ab = \(\frac{60}{3}\) = 20

∴ So the correct answer is 2. 20

3. If \(x^2+\frac{1}{16 x^2}=3 \frac{1}{2}\) then tha value of \(8 x^3+\frac{1}{8 x^3}\) is

1. 66
2. 56
3. 76
4. 52

Solution: \(x^2+\frac{1}{16 x^2}=\frac{7}{2}\)

⇒ \((x)^2+\left(\frac{1}{4 x}\right)^2=\frac{7}{2} \Rightarrow\left(x+\frac{1}{4 x}\right)^2-2 \cdot x \cdot \frac{1}{24 x}=\frac{7}{2} \)

⇒ \(\left(x+\frac{1}{4 x}\right)^2=\frac{7}{2}+\frac{1}{2}=4 \Rightarrow x+\frac{1}{4 x}=\sqrt{4}=2\)

⇒ \(2\left(x+\frac{1}{4 x}\right)=2 \times 2 \Rightarrow 2 x+\frac{1}{2 x}=4\)

= \(\left(2 x+\frac{1}{2 x}\right)^3-3 \cdot 2 x \cdot \frac{1}{2 x}\left(2 x+\frac{1}{2 x}\right)\)

= \((4)^3-3 \cdot 4=64-12=52\)

∴ The correct answer is 4. 52

Example 17. Write ‘True’ or ‘False’:

1. 1729 is a Hardy Ramanujam number.

Solution: The statement is true.

2. If \(3 x+\frac{3}{x}=2\), then the value od \(x^3+\frac{1}{x^3}=\frac{27}{89}\)

Solution: \(3 x+\frac{3}{x}=2 \Rightarrow 3\left(x+\frac{1}{x}\right)=2 \Rightarrow x+\frac{1}{x}=\frac{2}{3}\)

= \(\frac{8}{27}-2=\frac{8-54}{27}=-\frac{46}{27}\)

∴ Statement is false

3. \(p^3 q^3+1=(p q-1)\left(p^2 q^2+p q+1\right)\)

Solution: \(p^3 q^3+1=(p q)^3+(1)^3=(p q+1)\left\{(p q)^2-p q \cdot 1+(1)^2\right\}\)

= \((p q+1)\left(p^2 q^2-p q+1\right)\)

∴ The statement is false.

Example 18. Fill in the blanks:

1. If a + \(\frac{9}{a}\) = 3, then a³ + 27 = _________

Solution: a + \(\frac{9}{a}\) = 3

⇒ a² + 9 = 3a

⇒ a² – 3a + 9 = 0

⇒ a³ + 27 = (a)³ + (3)³

⇒ a³ + 27 = (a + 3) (a² – a·3 + 3²)

⇒ a³ + 27 = (a + 3) (a² – 3a + 9)= (a + 3) (0) = 0

2. If t² – 4t + 1 = 0, then t³ + \(\frac{1}{t^3}\) = ______

Solution: \(t^2-4 t+1=0 \Rightarrow t^2+1=4 t\)

⇒ \(\frac{t^2+1}{t}=4 \Rightarrow t+\frac{1}{t}=4\)

⇒ \(t^3+\frac{1}{t^3}=\left(t+\frac{1}{t}\right)^3-3 \cdot t \cdot \frac{1}{t}\left(t+\frac{1}{t}\right)=(4)^3-3 \cdot 4=52\)

3. (a + b)³ = a³ + b³ + _________

Solution: 3ab (a + b)

(a + b)³ = a³ + b³  = 3ab (a + b)

Algebra Chapter 1 Revision Of Old Lessons

Necessary Formulae:

1. (a + b)² = a² + 2ab + b² = (a – b)² + 4ab
2. (a – b)² =a² – 2ab + b² = (a + b)² – 4ab
3. (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
4. a² + b² = (a + b)² – 2ab = (a – b)² + 2ab
5. a² – b² = (a + b) (a – b)
6. 2 (a²+ b²) = (a + b)² + (a – b)²
7. 4ab = (a + b)² – (a – b)²
8. ab = \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)

Algebra Chapter 1 Revision Of Old Lessons Examples

Example 1. Simplify: – 3 – [-5 – 6- {3 – 5 – (2 – \(\overline{3-a}\))}]

Solution:

Given That:

f(x) = – 3 – [- 5 – 6 – {3 – 5 – (2 – \(\overline{3-a}\)}]

= – 3 – [- 5 – 6 – {3 – 5 – (2 – 3 + a)}]

= – 3 – [- 5 – 6 – {3 – 5 – (1 + a)}]

Read and Learn More WBBSE Solutions For Class 8 Maths

= – 3- [- 5 – 6 – {3 – 5 + 1 – a}]

= – 3 – [- 5 – 6 – {-1 – a}]

= – 3 – [- 5 – 6 + 1 + a]

= – 3 – [10 + a]

= – 3 + 10 – a

f(x) = – 3 – [- 5 – 6 – {3 – 5 – (2 – \(\overline{3-a}\)}] = 7 – a

Example 2. Find the summation of 3x² – 7xy – 5y², x² + 2xy + 7y², 9x² – 10xy – 15y²

Solution:

Given That:

let us consider equations as f(x),f(y),f(z).

f(x) = (3x² – 7xy – 5y²)

f(y) = (x² + 2xy + 7y²)

f(z) = (9x² – 10xy – 15y²)

Here summation means adding i.e., adding of equation 1 and equation 2 and Equation 3.

f(x + y + z) = (3x² – 7xy – 5y²) + (x² + 2xy + 7y²) + (9x² – 10xy – 15y²)

f(x + y + z) = 3x² -7xy – 5y² + x² + 2xy + 7y² + 9x² – 10xy – 15y²

f(x + y + z) = 3x² + x² + 9x² – 7xy + 2xy – 10xy – 5y² + 7y² – 15y²

f(x + y + z) = 13x² – 15xy – 13y²

Example 3. Subtract: (2x – 3y + 4z) from (-3x + 8y – 8z)

Solution:

Given That

f(x) = (- 3x + 8y – 8z)_______ eq (1)
f(y) = (2x – 3y + 4z)  _______ eq (2)

Now substraction of Equation 2 from equation 1

f(x – y) = (- 3x + 8y – 8z) – (2x – 3y + 4z)

f(x – y)  = – 3x + 8y – 8z – 2x + 3y – 4z

f(x – y)  = – 3x – 2x + 8y + 3y – 8z – 4z

f(x – y)  = 5x + 11y – 12z

Example 4. Simplify: (x − y) (x² + xy + y²) + (y − z) (y² + yz + z²) + (z − x) (z² + zx + x²)

Solution:

Given That :

f(x) = (x – y) (x² + xy + y²) + (y − z) (y² + yz + z²) + (z – x) (z² + zx + x²)

= x (x² + xy + y²) − y (x² + xy + y²) + y (y² + yz + z²) − z (y² + yz + z²) + z (z² + zx + x²) − x (z² + zx + x²)

= x³ + x²y + xy² – x²y – xy² – y³ + y³ + y²z + yz² -y²z – yz² – z³ + z³ + xz² + x²z – xz² – ²z – x³

f(x)  = 0

(x − y) (x² + xy + y²) + (y − z) (y² + yz + z²) + (z − x) (z² + zx + x²) = 0

Example 5. Divide: (15 a³b²c – 10 ab²c – 25 a²bc²) by (- 5 abc)

Solution: \(\frac{15 a^3 b^2 c-10 a b^2 c-25 a^2 b c^2}{-5 a b c}\)

= \(-\frac{15 a^3 b^2 c}{5 a b c}+\frac{10 a b^2 c}{5 a b c}+\frac{25 a^2 b c^2}{5 a b c}\)

= \(-3 a^{3-1} b^{2-1} c^{1-1}+2 a^{1-1} b^{2-1} c^{1-1}+5 a^{2-1} b^{1-1} c^{2-1}\)

= \(-3 a^2 b^1 c^0+2 a^0 b^1 c^0+5 a^1 b^0 c^1\)

= \(-3 a^2 b+2 b+5 a c\)  [a⁰ =1, b⁰ =1, c⁰ =1]

(15 a³b²c – 10 ab²c – 25 a²bc²) by (- 5 abc) = \(-3 a^2 b+2 b+5 a c\)  [a⁰ =1, b⁰ =1, c⁰ =1]

Example 6. Find the square of a + 2b – c + 3d

Solution: (a + 2b – c + 3d)²

= {(a + 2b) – (c-3d)}²

= (a + 2b)² – 2 (a + 2b) (c – 3d) + (c – 3d)²

= a² + 2.a.2b + (2b)² – 2 (ac – 3ad + 2bc – 6bd) + c² – 2.c.3d + (3d)²

= a² + 4ab + 4b² -2ac + 6ad – 4bc + 12bd + c² – 6cd + 9d²

= a² + 4b² + c² + 9d² + 4ab – 2ac + 6ad – 4bc + 12bd – 6cd

The square of a + 2b – c + 3d = a² + 4b² + c² + 9d² + 4ab – 2ac + 6ad – 4bc + 12bd – 6cd

Example 7. Express as perfect square and find the value. 25a² – 30 (b – 2c) + 9 (b – 2c)² where a = 1, b = -2, c = 3

Solution: 25a² – 30 (b – 2c) + 9 (b – 2c)²

= (5a)² – 2.5a.3 (b – 2c) + {3 (b- 2c)}²

= {5a – 3 (b – 2c)}²

= (5a – 3b + 6c)²

= (5.1 – 3(-2) + 6.3)²

= (5 + 6 + 18)² = (29)² = 841

25a² – 30 (b – 2c) + 9 (b – 2c)²  = 841

Example 8. If a + b = 5 and ab= 6, then find the value of (a² – b²)

Solution: a + b = 5, ab = 6

(a – b)² = (a + b)² – 4ab

= (5)² – 4 x 6 = 25 – 24 = 1

= a – b ± √1 = ±1

a² – b² = (a + b) (a – b) = 5 x (± 1) = ± 5

The value of (a² – b²) = ± 5

Example 9. If x + y = 12 and x – y = 4, then find the value of 3xy (x² + y²).

Solution: x + y = 12, x – y = 4

3xy (x² + y²)

= \(\frac{3}{8}\).4xy.2 (x² + y²)

= \(\frac{3}{8}\){(x + y² – (x − y)²} {(x + y)² + (x − y)²}

= \(\frac{3}{8}\) x ((12)²– (4)²) ((12)² + (4)²)

= \(\frac{3}{8}\) x (144 – 16) (144 + 16)

= \(\frac{3}{8}\) x128 x 160 = 7680

The value of 3xy (x² + y²) = 7680

Example 10. If \(\left(\frac{a^2}{b^2}+t a+\frac{b^2}{25}\right)\) is a perfect square, then what will be the value of t where
a ≠ 0.

Solution: \(\frac{a^2}{b^2}+t a+\frac{b^2}{25}\)

= \(\left(\frac{a}{b}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{t b}{2}+\left(\frac{t b}{2}\right)^2-\left(\frac{t b}{2}\right)^2+\frac{b^2}{25}\)

= \(\left(\frac{a}{b}+\frac{t b}{2}\right)^2-\frac{t^2 b^2}{4}+\frac{b^2}{25}\)

As, given expression is a perfect square

so \(-\frac{t^2 b^2}{4}+\frac{b^2}{25}=0 \Rightarrow-\frac{t^2 b^2}{4}=-\frac{b^2}{25} \Rightarrow \frac{t^2 b^2}{4}=\frac{b^2}{25}\)

⇒ \(t^2=\frac{b^2}{25} \times \frac{4}{b^2} \Rightarrow t^2=\frac{4}{25} \Rightarrow t= \pm \sqrt{\frac{4}{25}} \Rightarrow t= \pm \frac{2}{5}\)

Example 11. Resolve into Factors:

1. a² – b² + 2bc – c²
2. 64ax² – 49a (x – 2y)²
3. 2a²b² + 2b²c² + 2c²a² – a⁴ – b⁴ – c⁴
4. x⁴ + x² + 1
5. x² – 2 (a² + b²) x + (a² – b²)²

Solution:

1. \(a^2-b^2+2 b c-c^2\)

= \(a^2-\left(b^2-2 b c+c^2\right)\)

= \(a^2-(b-c)^2\)

= \((a+b-c)(a-b+c)\)

2. \(64 a x^2-49 a(x-2 y)^2\)

= \(a\left[64 x^2-49(x-2 y)^2\right]\)

= \(a\left[8 x^2-7(x-2 y)^2\right]\)

= \(a\left[(8 x)^2-(7 x-14 y)^2\right]\)

= \(a(8 x+7 x-14 y)(8 x-7 x+14 y)\)

= \(a(15 x-14 y)(x+14 y)\)

3. \(2 a^2 b^2+2 b^2 c^2+2 c^2 a^2-a^4-b^4-c^4\)

= \(4 a^2 b^2-2 a^2 b^2+2 b^2 c^2+2 c^2 a^2-a^4-b^4-c^4\)

= \(4 a^2 b^2-\left(a^4+b^4+c^4+2 a^2 b^2-2 b^2 c^2-2 c^2 a^2\right)\)

= \((2 a b)^2-\left(a^2+b^2-c^2\right)^2\)

= \(\left(2 a b+a^2+b^2-c^2\right)\left(2 a b-a^2-b^2+c^2\right)\)

= \(\left\{\left(a^2+2 a b+b^2\right)-c^2\right\}\left\{c^2-\left(a^2-2 a b+b^2\right)\right\}\)

= \(\left\{(a+b)^2-c^2\right\}\left\{c^2-(a-b)^2\right\}\)

= (a + b + c)(a + b – c)(c + a – b)(c – a + b)

4. \(x^4+x^2+1\)

= \(\left(x^2\right)^2+2 x^2 \cdot 1+(1)^2-x^2\)

= \(\left(x^2+1\right)^2-x^2\)

= \(\left(x^2+1+x\right)\left(x^2+1-x\right)\)

5. \(x^2-2\left(a^2+b^2\right) x+\left(a^2-b^2\right)^2\)

= \(x^2-2\left(a^2+b^2\right) x+(a+b)^2(a-b)^2\)

= \(x^2-\left\{(a+b)^2+(a-b)^2\right\} x+(a+b)^2(a-b)^2\)

= \(x^2-(a+b)^2 x-(a-b)^2 x+(a+b)^2(a-b)^2\)

= \(x\left\{x-(a+b)^2\right\}-(a-b)^2\left\{x-(a+b)^2\right\}\)

= \(\left\{x-(a+b)^2\right\}\left\{x-(a-b)^2\right\}\)

Example 12. Solve:

1. 4x – 6 {8 – (x – 5)+ 7x} – 50 = 72
2. \(\frac{1}{4}\)(x + 4) + \(\frac{1}{5}\)(x + 5) = \(\frac{1}{6}\)(x + 6) + \(\frac{1}{7}\)( x + 7)
3. \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3 a-3 b}{a+b}=0\)
4. 2 (x – 3) – (5 – 3x) = 3(x + 1)-5 (2 + x)

Solution:

1. 4x – 6 {8- (x – 5) + 7x} – 50 = 72

⇒ 4x – 6 {8x + 5 + 7x} = 72 + 50

⇒ 4x – 6 {6x + 13} = 122

⇒ 4x – 36x – 78 = 122

⇒ – 32x = 122 + 78

⇒ -32x = 200

⇒ x = –\(\frac{200}{32}\)

⇒ x = –\(\frac{25}{4}\)

2. \(\frac{1}{4}\)(x + 4) + \(\frac{1}{5}\)(x + 5) = \(\frac{1}{6}\)(x + 6) + \(\frac{1}{7}\)( x + 7)

⇒ \(\frac{x}{4}+1+\frac{x}{5}+1=\frac{x}{6}+1+\frac{x}{7}+1\)

⇒ \(\frac{x}{4}+\frac{x}{5}-\frac{x}{6}-\frac{x}{7}=1+1-1-1\)

⇒ \(\frac{105 x+84 x-70 x-60 x}{420}=0\)

⇒ \(\frac{59 x}{420}=0 \Rightarrow x=0\)

3. \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3 a-3 b}{a+b}=0\)

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3(a+b)}{a+b}=0\)

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x}{a+b}-3=0\)

⇒ \(\left(\frac{x-a}{b}-1\right)+\left(\frac{x-b}{a}-1\right)+\left(\frac{x}{a+b}-1\right)=0\)

⇒ \(\frac{x-a-b}{b}+\frac{x-b-a}{a}+\frac{x-a-b}{a+b}=0\)

⇒ (x – a – b)\(\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{a+b}\right)\)

⇒ x – a – b = 0  [\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}\right)\) ≠ 0]

⇒ x = a + b

4. 2 (x-3) – (5 – 3x) = 3(x + 1)- 5(2 + x)

2x – 6 – 5 + 3x = 3x + 3 – 10 – 5x

→ 2x + 3x – 3x + 5x = 3 – 10 + 6 + 5

⇒ 7x = 4

⇒ x = \(\frac{4}{7}\)

Example 13. Ten’s digit of a number of two digits is greater by 7 than the unit’s digit. Sum of the digits is \(\frac{1}{9}\) of the number. Find the number.

Solution: Let the digit at unit’s place be x. Then digit at ten’s place is (x + 7).

Number = 10 (x + 7) + x = 10x + 70 + x = 11x + 70

According to question,

x + (x + 7) = \(\frac{1}{9}\)(11x + 70)

⇒ 9(2x+7)= 11x + 70

⇒ 18x + 63 = 11x + 70

⇒ 18x – 11x = 70 – 63 7x = 7

⇒ x = \(\frac{7}{7}\) = 1

∴ The required number is (11 x 1 + 70) = 81

Example 14. Choose the correct answer:

1. If x + \(\frac{1}{x}\) = 4, then the value of \(x^2+\frac{1}{x^2}\)

1. 14
2. 16
3. 2
4. 4

Solution: x + \(\frac{1}{x}\) = 4

∴ The correct answer is 1. 14

2. If 4x² + tx + 9 is a perfect square, then the value of t is

1. ± 12
2. ± 6
3. ± 8
4. ± 36

Solution: 4x² + tx +9

= (2x)² + 2·2x.3 + (3)² ∓ 2·2x.3 + tx = (2x ± 3)² ∓ 12x + tx

∴ ∓ 12x + tx = 0

⇒ tx = ±12x

⇒ t = ± 12

∴ So the correct answer is 1. ± 12

3. The value of (2:31)² – (1.69)² is

1. 2.48
2. 4.24
3. 2.24
4. 4.48

Solution: (2:31)² – (1.69)² = (2.31 + 1.69) (2.31 – 1.69) = 4 x 0.62 = 2.48

∴ The correct answer is (a)

Example 15. Write ‘True’ or ‘False’:

1. If a + b = 0, then a² – b² ≠ 0

Solution: a² – b² = (a + b) (a – b) = (0) (a – b) = 0

∴ The statement is false.

2. x = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

Solution: x = x.1 = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

∴ The statement is true.

3. If \(\frac{a}{b}-{b}{a}\) = 4, then the value of \(\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)\) is 18

Solution: \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=\left(\frac{a}{b}\right)^2+\left(\frac{b}{a}\right)^2=\left(\frac{a}{b}-\frac{b}{a}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{b}{a}=(4)^2+2=18\)

∴ The statement is true

Example 16. Fill in the blanks:

1. (a – 2) (a + 2) (a² + 4) = ________

Solution: (a− 2) (a + 2) (a² + 4) = (a² – 22) (a² + 4)

= (a² – 4) (a² + 4) = (a²)² – (4)² = a⁴ – 16

(a – 2) (a + 2) (a² + 4) = a⁴ – 16

2. The value of \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\) is ________

Solution: \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\)

= \(\frac{(3.19)^2-(1.81)^2}{3.19-1.81}=\frac{(3.19+1.81)(3.19-1.81)}{(3.19-1.81)}\)

= 5.00 = 5

The value of \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\) is 5.

3. If (x + 3) (x − p) = x² – 9 then the value of p is _________

Solution: (x + 3) (x − p) ⇒ x² – 9

⇒ (x + 3) (x − p) = x² – 3²

⇒ (x + 3) (x − p)(x + 3) (x – 3) ⇒ x – p = x -3

⇒ – p = -3

⇒ p = 3

The value of p is 3.

Algebra Chapter 2 Polynomials

The Algebraic expression having one or more terms is called Polynomial Expression.

Multiplication of Algebra

Example 1. Let the length and breadth of a rectangle are (3x + 2y) unit and (x – 4y) unit respectively.

Solution: Area is (3x + 2y) x (x – 4y) sq. unit

= {3x (x – 4y) + 2y (x – 4y)} sq. unit

Read and Learn More WBBSE Solutions For Class 8 Maths

= (3x² – 12xy + 2xy – 8y²) sq. unit = (3x² – 10xy – 8y²) sq. unit

Example 2. Multiplication: (a² + ab + b²) by (a² – ab + b²)

1. Multiplicand: a² + ab + b²
2. Multiplier: a² – ab + b²

Solution: Multiplication:

The product = a⁴ + a²b² + b⁴

Alternating Method:

(a² + ab + b²) (a² – ab + b²)

= a² (a² – ab + b²) + ab (a² – ab + b²) + b² (a² – ab + b²)

= a⁴  – a³b + a²b² + a³b – a²b² + ab³ + a²b² – ab³ + b⁴ = a⁴+ a²b² + b⁴

Example 3. Find the product by successive multiplication. (x² – 1), (x² + 2x² – 5), (5 – x³ + x⁴)

Solution:

∴ The required product = x⁹ + x⁸ – 3x⁷ – 6x⁶ + 125x⁵ + 15x⁴ – 10x³ – 35x² + 25

Example 4. Find the continued product of (a + b + c), (a – b + c), (a + b – c) and (b + c – a)

Solution:

∴ The required product = 2a²b² + 2b²c² + 2c²a² – a⁴ – b⁴ – c⁴

Example 5. Find the product by successive multiplication: \(\left(\frac{a}{b}+\frac{b}{c}\right),\left(\frac{b}{c}+\frac{c}{a}\right),\left(\frac{c}{a}+\frac{a}{b}\right)\)

Solution: \(\left(\frac{a}{b}+\frac{b}{c}\right)\left(\frac{b}{c}+\frac{c}{a}\right)\left(\frac{c}{a}+\frac{a}{b}\right)=\left\{\frac{a}{b}\left(\frac{b}{c}+\frac{c}{a}\right)+\frac{b}{c}\left(\frac{b}{c}+\frac{c}{a}\right)\right\}\left(\frac{c}{a}+\frac{a}{b}\right)\)

= \(\left(\frac{a}{c}+\frac{c}{b}+\frac{b^2}{c^2}+\frac{b}{a}\right)\left(\frac{c}{a}+\frac{a}{b}\right)\)

= \(\frac{a}{c}\left(\frac{c}{a}+\frac{a}{b}\right)+\frac{c}{b}\left(\frac{c}{a}+\frac{a}{b}\right)+\frac{b^2}{c^2}\left(\frac{c}{a}+\frac{a}{b}\right)+\frac{b}{a}\left(\frac{c}{a}+\frac{a}{b}\right)\)

= \( 1+\frac{a^2}{b c}+\frac{c^2}{a b}+\frac{a c}{b^2}+\frac{b^2}{a c}+\frac{a b}{c^2}+\frac{b c}{a^2}+1\)

= \(2+\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}+\frac{a b}{c^2}+\frac{b c}{a^2}+\frac{c a}{b^2}\)

Example 6. Simplify: (a – b) (a² + ab + b²) + (b − c) (b² + bc + c²) + (c − a) (c² + ac + a²)

Solution:

Given

(a – b) (a² + ab + b²) + (b − c) (b² + bc + c²) + (c− a) (c² + ca + a²)

= a (a² + ab + b²) – b (a² + ab + b²) + b (b² + bc + c²) – c (b² + bc + c²) + c (c² + ca + a²) – a (c² + ca + a²)

= a³ + a²b + ab² – a²b – ab² – b³ + b³ + b²c + bc² – b²c – bc² – c³ + c3 + c²a + a²c – c²a – a²c – a³ = 0

Example 7. Simplify: \(\left(a^{-1}+b^{-1}\right)\left(a^{-1}-b^{-1}\right)\left(a^{-2}+b^{-2}\right)\)

Solution:

Given that

= \(\{\left(a^{-1}\left(a^{-1}-b^{-1}\right)+b^{-1}\left(a^{-1}-b^{-1}\right)\right\}\left(a^{-2}+b^{-2}\right)\)

= \(\left(a^{-2}-a^{-1} b^{-1}+ a^{-1} b^{-1}-b^{-2}\right)\left(a^{-2}+b^{-2}\right)\)

= \(\left(a^{-2}-b^{-2}\right)\left(a^{-2}+b^{-2}\right)\)

= \(a^{-2}\left(a^{-2}+b^{-2}\right)-b^{-2}\left(a^{-2}+b^{-2}\right)\)

= \(a^{-4}+q^{-2} b^{-2}-a^{-2} \cdot b^{-2}-b^{-4}\)

= \(a^{-4}-b^{-4}\)

\(\left(a^{-1}+b^{-1}\right)\left(a^{-1}-b^{-1}\right)\left(a^{-2}+b^{-2}\right)\) = \(a^{-4}-b^{-4}\)

Example 8. Simplify: (a + b + c) (a² + b² + c² – ab – bc – ca)

Solution:

Given

(a + b + c) (a² + b² + c² – ab- bc – ca)

= a (a² + b² + c² – ab- bc – ca)

+ b (a² + b² + c² – ab- bc – ca)

+ c (a² + b² + c² – ab- bc – ca)

= a³ + ab² + ac² – a²b – abc – a²c

+ a²b + b³ + bc² – ab² – b²c – abc

+ b²c + c³ – abc – bc² – ac²

(a + b + c) (a² + b² + c² – ab- bc – ca) = a³ + b³ + c³ – 3abc

Algebra Chapter 2 Division

Method of division are of two types

1. Exact division
2. Inexact division

In case of Exact division, the remainder zero.

Here Dividend = Divisor x Quotient.

In case of Inexact division, after completion of division, we have quotient and remainder.

Here Dividend = Divisor x Quotient + Remainder.

Example 1. Divide (x² – 4x – 5) by (x – 5)

1. Divisor = (x- 5)
2. Dividend = x² – 4x – 5

Solution:

∴ The quotient is (x + 1) and remainder is 0.

Example 2. Divide (5x³y² – 7x²y³ + 3xy⁴) by xy⁴

Solution:

∴ The quotient is 5x² – 7xy + 3y² and remainder is 0.

Example 3. Divide the first expression by the second:

1. (a³ + b³ + c³ – 3abc), (a + b + c)
2. (x⁴ – 2x³ – 15x² + 16x + 35), (x² + x – 7)

Solution:

∴ Quotient = a² + b² + c² – ab- bc – ca

Remainder = 0

2.

∴ Quotient = x² –  3x – 5

Remainder = 0

Example 4. Divide (a – b) by \(a^{\frac{1}{3}}-b^{\frac{1}{3}}\)

Solution:

∴ Quotient = \(a^{\frac{2}{3}}+a^{\frac{1}{3}} b^{\frac{1}{3}}+b^{\frac{2}{3}}\)

Remainder = 0

Example 5. Find the quotient and the remainder of the following:

1. (x³ + x² + 25) by (x + 3),
2. (81x⁴ + 4) by (3x – 1)
3. (9a² – 4a³ – 5 + a4 -19a) by (7 + a² – a)

Solution:

∴ The required quotient = x² – 2x + 6 and remainder = 7

2.

∴ Quotient = 27x³ + 9x² + 3x + 1

Remainder = 5

3. Dividend = 9a² – 4a³ – 5+ a⁴ – 19a

= a⁴ – 4a³ + 9a² – 19a – 5 [In descending power of a]

Divisor = 7 + a² – a

= a² – a + 7 (In descending power of a)

∴ The Quotient = a² – 3a-  7

Remainder = 44-5a

Example 6. Divide (6x² + 11xy + 6y²) by (3x – 2y)

Solution:

∴ Quotient = 2x + 5y

Remainder = 16y²

Example 7. In a division problem the divisor is (x² – 3x + 5), the quotient is (2x – 7) and remainder.

Solution: Dividend = Divisor x Quotient + Remainder

= [(x² – 3x + 5) × (2x-7)] + (x + 3)

= [ 2x³ – 7x² – 6x² + 21x + 10x – 35] +[ x + 3 ]

= 2x³ – 13x² + 32x – 32

Example 8. Choose the correct answer:

1. (x – 3)(x + 2) =

1. x² – 5x – 6
2. x² – x – 6
3. x² – x + 6
4. x² -5x + 6

Solution: (x-3) (x+2)

= x(x+2)- 3(x+2)

= x² + 2x – 3x – 6 = x² – x – 6

∴ The correct answer is 2. x² – x – 6

2. (8x⁸ – 8x² – 36x) ÷ 4x =

1. 2x⁸ – 2x² – 9
2. 2x⁸ – 8x – 36
3. 2x⁷ – 2x – 9
4. 2x⁷ – 2x – 18

Solution: \(\frac{8 x^8-8 x^2-36 x}{4 x}=2 x^7-2 x-9\)

∴ The correct answer is 3. 2x⁷ – 2x – 9

3. (x² – 14x + 45 ÷ (x – 5) =

1. x – 9
2. x + 5
3. 9 – x
4. x + 9

Solution:

∴ So the correct answer is 1. x – 9

Example 9. Write ‘True’ or ‘False’:

1. (x – a) (x – b) (x – c)…. (x − z) = 1

Solution: (x – a) (x – b) (x – c)…… (x – z)

= (x − a) (x – b) (x − c)….. (x – x) (x − y) (x − z)

= (x – a) (x – b) (x – c)…. (0) (x − y) (x – z) = 0

∴ The statement is false.

2. \(\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{a}{b}-\frac{c}{d}\right)\left(\frac{a^2}{b^2}+\frac{c^2}{d^2}\right)\)=\(\frac{a^4}{b^4}-\frac{c^4}{d^4}\)

Solution:

Given that

= \(\left(\frac{a^2}{b^2}\right)^2-\left(\frac{c^2}{d^2}\right)^2=\frac{a^4}{b^4}-\frac{c^4}{d^4}\)

∴ So the statement is true.

3. (x² + 11x + 27) ÷ (x + 6) = x + 5 + \(\frac{(-3)}{x+6}\)

Solution:

∴ \(\left(x^2+11 x+27\right) \div(x+6)=(x+5)+\left(-\frac{3}{x+6}\right)\)

∴ So the statement is true.

Example 10. Fill in the blanks:

1. \((-2 a)^2 \times(-3 a)^3 \times\left(-\frac{1}{3} a\right)^3=\)________

Solution: \((-2 a)^2 \times(-3 a)^3 \times\left(-\frac{1}{3} a\right)^3\)

= \(4 a^2 \times-27 a^3 \times-\frac{1}{27} a^3=4 a^2\)

2. (x² + xy + y²) (x − y) = ________

Solution: (x² + xy + y²) (x − y)

= ( (x² + xy + y²) × (x))
– ((y) × (x² + xy + y²))

= x³ + x²y + xy²  – x²y – xy²  – y³

= x³ – y³

(x² + xy + y²) (x − y) = = x³ – y³

3. (5x² – 2x – 3) ÷ (x – 1) = ________

Solution:

Quotient is (5x+3).

(5x² – 2x – 3) ÷ (x – 1) = (5x+3).

Arithmetic Chapter 7 Time And Work

1. If the number of workers is fixed, then the amount of work done is directly proportional to the time spent on doing the work.
2. If the allotted time for doing the work is constant, then the number of workers required varies directly with the amount of work to be done.
3. If the amount of work to be done is fixed, then the time taken to do a work varies inversely to the number of workers doing it.

Arithmetic Chapter 7 Time And Work Examples

Example 1. 4 tractors can plough 28 bighas in a day. Calculate how many bighas can be ploughed by 8 tractors in a day.

Solution: 4 tractors in a day can plough 28 bighas.

1 tractors in a day can plough \(\frac{28}{4}\) bighas.

Read and Learn More WBBSE Solutions For Class 8 Maths

8 tractors in a day can plough \(\frac{28 \times 8}{4}\) bighas = 56 bighas

∴ 56 bighas can be ploughed by 8 tractors in a day.

Example 2. In a factory 120 parts of machine are made in 5 days. Calculate how many parts of the machine will be made in 9 days.

Solution: Let the number of parts of machine will be made in 9 days is x [x > 0].

In the mathematical language, the problem is:

Time (in days)
5
9

Number of parts
120
x

The number of parts of machine and time taken by a fixed number of labourers are directly proportional.

∴ 5: 9 : : 120: x

⇒ \(\frac{5}{9}\) = \(\frac{120}{x}\)

⇒ 5x = 9 × 120

⇒ x = \(\frac{9 \times 120}{5}\)

⇒ x = 216

∴ 216 parts of machine will be made in 9 days.

Example 3. If 45 men can do a piece of work in 5 days, then how many days will be needed to complete this work?

Solution: 45 men can do a piece of work in 5 days.

1 men can do this work in (5 x 45) days

15 men can do this work in \(\frac{5 \times 45}{15}\) = 15 days

∴ 15 days will be needed to complete this work.

Example 4. Mahim can do a piece of work in 10 days. Calculate the rest of the total work after 3 days.

Solution: Mahim completes the work in 10 days

In 1 day he can do the work \(\frac{1}{10}\) part

In 3 days he can do the work or part or (\(\frac{1 \times 3}{10}\)) part or \(\frac{3}{10}\)

∴ After 3 days the rest of the total work is (1-\(\frac{3}{10}\)) part or \(\frac{7}{10}\) part

∴ 3 days will be needed to complete this work.

Example 5. Aritra, Soumyadip, and Subhadip can do a piece of work separately in x days, y days, and z days respectively. If they do the work together, in how many days they together will complete the work?

Solution: Aritra alone does in 1 day the \(\frac{1}{x}\) part

Soumyadip alone does in 1 day the \(\frac{1}{y}\) part

Subhadip alone does in 1 day the \(\frac{1}{z}\) part

They together do in 1 day (\(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\)) part = \(\left(\frac{y z+x z+x y}{x y z}\right)\) part

∴ They do \(\left(\frac{y z+x z+x y}{x y z}\right)\) part of the work in 1 day.

∴ Total work i.e. 1 part is done in \(\left(1 \div \frac{x y+y z+z x}{x y z}\right)\) days = \(\frac{x y z}{x y+y z+z x}\)

∴ \(\frac{x y z}{x y+y z+z x}\) days they together will complete the work.

Example 6. B can do a piece of work in 12 days. If the efficiency of A is 60% more than the efficiency of B, then what time can A do it alone?

Solution: B does in 1 day \(\frac{1}{12}\) part

As the efficiency of A is 60% more than the efficiency of B

∴ A does in 1 day (\(\frac{1}{12}\)+\(\frac{1}{12}\)x\(\frac{60}{1200}\)) part = \(\left(\frac{1}{12}+\frac{1}{20}\right)=\left(\frac{5+3}{60}\right)\) part

= \(\frac{8}{60}\) part = \(\frac{2}{15}\) part

∴ A do \(\frac{2}{15}\) part of total work in 1 day

∴ Total work i.e. 1 part is done in (1 + \(\frac{2}{15}\)) days = 1 x \(\frac{15}{2}\) day = 7\(\frac{1}{2}\) day

∴ A can do it alone in 7\(\frac{1}{12}\) days.

Example 7. A does \(\frac{7}{10}\) part of a piece of work in 14 days and then with the help of B, A finish the work in 2 days. How long would B take to do the work alone?

Solution: A does the work in (14 + 2) days or 16 days.

A in 14 days do \(\frac{7}{10}\) part of the total work

A in 1 day do (\(\frac{7}{10 \times 14}\)) part of the total work

A in 16 days do (\(\left(\frac{7 \times 16}{10 \times 14}\right)\)) part or \(\frac{4}{5}\) part of the total work

∴ The remaining part is (1-\(\frac{4}{5}\)) part = \(\frac{1}{5}\) part

B do \(\frac{1}{5}\) part of the total work in 2 day.

B do 1 part in (2÷\(\frac{1}{5}\)) days = (2x\(\frac{5}{1}\)) days = 10 days

∴ B alone take 10 days to finish the works.

Example 8. The empty tank is filled with the first pipe in 45 minutes and the full tank is made empty with another pipe in 1 hour. If two pipes are opened together then find the time to fill the empty tank.

Solution: 1 hour 60 minutes

It takes 1 minute to fill up the \(\frac{1}{45}\) part of the empty tank by the first pipe

It takes 1 minute to make empty the \(\frac{1}{60}\) part of the tank by the second pipe

If the two pipes are open, it takes 1 minute to fill the (\(\frac{1}{45}\)–\(\frac{1}{60}\)) part = \(\frac{4-3}{180}\) part = \(\frac{1}{180}\) part

The \(\frac{1}{180}\) part is filled in in 1 minute 1 part is filled in (1÷\(\frac{1}{180}\)) minutes = (1 x 180) minutes = 3 hours

∴ If two pipe are open the full tank will be filled in 3 hours.

Example 9. A, B and C can do a piece of work in 10 days, 12 days, and 15 days respectively. If they do the work together then how many days they will complete the work? [Using proportion method]

Solution: A does in 1 day \(\frac{1}{10}\) part

B does in 1 day \(\frac{1}{12}\) part

C does in 1 day \(\frac{1}{15}\) part

Together they do in 1 day (\(\frac{1}{10}\)+\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{6+5+4}{45}\) part = \(\frac{15}{60}\) part = \(\frac{1}{4}\) part

In the mathematical language, the problem is:

Amount of work (in part)
\(\frac{1}{4}\)
1

Time taken (in days)
1
? (x) (say)

The amount of work and time taken are directly proportional if the number of man is fixed.

∴ \(\frac{1}{4}\): 1 : : 1: x

⇒ \(\frac{\frac{1}{4}}{1}=\frac{1}{x}\)

⇒ \(\frac{x}{4}\) = 1

⇒ x = 4

∴ The required time to complete the work is 4 days.

Example 10. If 50 men working 8 hours a day can do a piece of work in 12 days, in how many days can 60 men working 10 hours a day do twice as much work?

Solution: 50 men working 8 hours a day do the work in 12 days

1 man working 8 hours a day do it in (12 x 50) days

1 man working 1 hour a day do it in (12 x 50 x 8) days

60 men working 1 hour a day do it in \(\frac{12 \times 50 \times 8}{60}\) days

60 men working 10 hours a day do it in \(\frac{12 \times 50 \times 8}{60 \times 10}\) = 8 days

∴ 60 men working 10 hours a day do twice as much work in (8 x 2) days or 16 days.

Example 11. A and B together can do a piece of work in 3 days, B and C can do a piece of work in 4 days and A and C can do a piece of work in 5 days. In how many days can each of them do it separately?

Solution: In 1 day (A + B) can do \(\frac{1}{3}\) part of the work

In 1 day (B + C) can do \(\frac{1}{4}\) part of the work

In 1 day (A + C) can do \(\frac{1}{5}\) part of the work

In 1 day twice the work of (A + B + C) together = \(\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right) \text { part }=\left(\frac{20+15+12}{60}\right)\) part = \(\frac{47}{60}\) part.

∴ In 1 day (A + B + C) can do (\(\frac{47}{60 \times 2}\)) or \(\frac{47}{120}\) part of the work

∴ In 1 day A can do (\(\frac{47}{120}\)–\(\frac{1}{4}\)) part = \(\frac{47-30}{120}\) part = \(\frac{17}{120}\) part

∴ A can do the whole work in (1÷\(\frac{17}{120}\)) days = \(\frac{1 \times 120}{17}\) days = 7\(\frac{1}{17}\) days

Similarly, in 1 day B can do the work (\(\frac{47}{120}\)–\(\frac{1}{5}\)) part = \(\frac{47-24}{120}\) part = \(\frac{23}{120}\) part

∴ B can do the whole work in \(\frac{120}{23}\) days or 5\(\frac{5}{23}\) days

In 1 day C can do (\(\frac{47}{120}\)–\(\frac{1}{3}\)) part = \(\frac{47-40}{120}\) part = \(\frac{7}{120}\) part

∴ C can do the whole work in \(\frac{120}{7}\) days or 17\(\frac{1}{7}\) days.

Example 12. X, Y and Z individually can complete a work in 10, 12 and 15 days respectively. They started doing the work together. After 3 days X left away. Calculate in how many days Y and Z will complete the remaining work.

Solution: X does in 1 day \(\frac{1}{10}\) part

Y does in 1 day \(\frac{1}{12}\) part

Z does in 1 day \(\frac{1}{15}\) Part

Together they do in 1 day (\(\frac{1}{110}\)+\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{6+5+4}{60}\) part = \(\frac{1}{4}\) part

Together they do in 3 days \(\frac{3}{4}\) part

Remaining part of the work is (1-\(\frac{3}{4}\)) part = \(\frac{1}{4}\) part

After 3 days X left away

Y and Z together do in 1 day is (\(\frac{1}{12}\)+\(\frac{1}{15}\)) part = \(\frac{5+4}{60}\) part =\(\frac{9}{60}\) part = \(\frac{3}{20}\) part

In the mathematical language the problem is:

Amount of work (in part)
\(\frac{3}{20}\)
\(\frac{1}{4}\)

Time (in days)
1
? (x) [say]

If the number of men are fixed, the amount of work and time are in direct proportion

∴ \(\frac{3}{20}\): \(\frac{1}{4}\) : : 1: x

⇒ \(\frac{\frac{3}{20}}{\frac{1}{4}}=\frac{1}{x}\)

⇒ x x \(\frac{3}{20}\) = \(\frac{1}{4}\)

⇒ x = \(\frac{20}{3}\) x \(\frac{1}{4}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)

∴ Y and Z will complete the remaining work in 1\(\frac{2}{3}\) days.

Example 13. The inlet pipe of a roof tank can fill it in 30 minutes. If outlet pipe is opened to supply water to the house, the tank is completely emptied in 45 minutes. When the tank is half filled, if both the pipes are opened together then what time it take to fill the tank?

Solution: The inlet pipe of a roof tank can fill it in 1 minute \(\frac{1}{30}\) part of the tank.

The outlet pipe can empty the tank in 1 minute \(\frac{1}{45}\) part of the tank.

When both the pipes are opened together, then the tank filled in 1 minute is

(\(\frac{1}{30}\)–\(\frac{1}{45}\)) part = \(\frac{6-4}{180}\) part = \(\frac{2}{180}\) part = \(\frac{1}{90}\) part

Now, according to the problem, the tank is half filled.

So the part which is to be filled is (1-\(\frac{1}{2}\)) part = \(\frac{1}{2}\) part of the tank.

In the mathematical language, the problem is:

Part filled
\(\frac{1}{90}\)
\(\frac{1}{2}\)

Time taken (in minutes)
1
? (x) [say]

∴ Part filled and time is direct proportion

\(\frac{1}{90}\): \(\frac{1}{2}\) : : 1: x

⇒ \(\frac{\frac{1}{90}}{\frac{1}{2}}=\frac{1}{x}\)

⇒ \(\frac{x}{90}\) = \(\frac{1}{2}\)

⇒ x = \(\frac{90}{2}\) = 45

∴ Required time is 45 minutes.

Example 14. A tailor can make 12 shirts in 3 days. How long will he take to make 8 shirts?

Solution: In the mathematical language, the problem is:

No. of shirts made
12
8

Time taken (days)
3
?

So, we get in this problem is 12: 8: 3: Required time.

∴ Required time = \(\frac{8 \times 3}{12}\) = 2 days.

Example 15. Out of 100 trees, 20 wood-cutters cut 20 trees in 4 hours. After that 4 wood- cutters leave the job, then when will the whole job get finished?

Solution: 4 wood-cutters leave the job.

So number of wood-cutters left is (20-4) = 16.

Let the time taken by them be x hours.

According to the problem in mathematical language is:

No. of trees
20
(100 – 20) = 80

Number of Wood-cutters
20
(20-4) = 16

Time taken (hrs.)
4
x

So, we get that 80: 16 : : x: 4

or, \(\frac{80 \times 4}{16}\) = 20 hours

So that the total cutting is done after 4+20= 24 hours.

Example 16. Choose the correct answer:

1. In a factory 216 parts of machine are made in 3 days. The number of parts of the machine will be made in 7 days is

1. 72
2. 432
3. 648
4. 504

Solution: Let the number of parts of machine will be made in 7 days is x [x > 0]

In the mathematical language, the problem is:

Time (in days)
3
7

Number of parts
216
x

The number of parts time and time are directly proportional.

∴ 3: 7 : : 216: x ⇒ \(\frac{3}{7}\) = \(\frac{216}{x}\)

⇒ x = \(\frac{7 \times 216}{3}\) = 504

504 parts will be made in 7 days.

∴ The correct answer is 4. 504

The number of parts of the machine will be made in 7 days is 504.

2. If 3 tractors can plough 18 bighas in a day, then 7 tractors can plough in a day is

1. 42 bighas
2. \(\frac{54}{7}\) bighas
3. \(\frac{7}{6}\) bighas
4. None of these

Solution: 3 tractors in a day can plough 18 bighas

1 tractors in a day can plough \(\frac{18}{3}\) bighas

7 tractors in a day can plough \(\frac{18 \times 7}{3}\) bighas

∴ The correct answer is 1. 42 bighas.

7 tractors can plough in a day is 42 bighas.

Example 17. Write ‘True’ or ‘False’:

1. If 24 men need 12 days to cut a pond, then 36 men will be needed to did that pond in 8 days.

Solution: In 12 days to cut a pond required 24 men

In 1 day to cut that pond required (24 × 12) men

In 8 days to cut that pond required (\(\frac{24 \times 12}{8}\)) men = 36 men

∴ So the statement is true.

2. If two pipes can fill a cistern in 10 hours and 15 hours respectively, then in 5 hours they together fill it.

Solution: In 1 hour the 1st pipe fill \(\frac{1}{10}\) part and second pipe fill \(\frac{1}{15}\) part.

When both pipes are opened together it takes 1 hour to fill up the (\(\frac{1}{10}\)+\(\frac{1}{15}\)) part = \(\frac{6+4}{60}\) part = \(\frac{10}{60}\) part = \(\frac{1}{6}\) part

Thus \(\frac{1}{6}\) part is filled in 1 hour

1 part is filled in (1÷\(\frac{1}{6}\)) hours (1 x 6) hours = 6 hours

∴ So the statement is false.

Example 18. Fill in the blanks:

1. If x men can do a piece of work in y days, then z men can do it in _______ days.

Solution: x men can do the work in y days

1 man can do it in (x x y) days

z men can do it in \(\frac{xy}{z}\) days.

2. Amal and Kamal completes a work separately in x days and y days respectively. If they do the work together then they will completes the work in ______ days.

Solution: In 1 day Amal does \(\frac{1}{x}\) part

In 1 day Kamal does \(\frac{1}{y}\) part

Together they do in 1 day (\(\frac{1}{x}\)+\(\frac{1}{y}\)) part = \(\frac{y+x}{xy}\) part

∴ They do \(\frac{x+y}{xy}\) part of the total work in 1 day

Total work i.e. 1 part is done in (1÷\(\frac{x+y}{xy}\)) days

or 1 x \(\frac{xy}{x+y}\) days = \(\frac{xy}{x+y}\) days.

They do the work together then they will completes the work in 1 x \(\frac{xy}{x+y}\) days = \(\frac{xy}{x+y}\) days

Arithmetic Chapter 6 Mixture

If two or more articles of different values and in different quantities or volume are mixed to produce an article of a different value and quantity. This is known as mixture.

Arithmetic Chapter 6 Mixture Examples

Example 1. If the ratio of measurements of urea and potash in 25 kg of mixed fertilizer is 3:2, then find the amount of potash.

Solution: The ratio of measurements of urea and potash = 3:2

The proportional part of the quantity of potash = \(\frac{2}{3+2}=\frac{2}{5}\)

The amount of potash is 25 kg of mixed fertilizer is \(\left(25 \times \frac{2}{5}\right)\) kg = 10 kg

Read and Learn More WBBSE Solutions For Class 8 Maths

Example 2. If 2 kg of Darjeeling tea costing ₹300 per kg mixed with 3 kg of Assam tea kg of mixture tea. costing ₹200 per kg, then find the price of 1 kg of mixture tea.

Solution: The price of 2 kg of Darjeeling tea at the rate of  ₹300 per kg is ₹(2 x 300) or ₹600.

The price of 3 kg Assam tea is ₹(3 x 200) or ₹600.

The price of (2+3) kg or 5 kg of mixtured tea is ₹(600+ 600) or ₹1200.

∴ The price of 1 kg of mixtured tea is ₹\(\frac{1200}{5}\) or ₹240.

Example 3. In 48 litres of mixture the ratio of measurements of water and glycerine is 2 1. What value of glycerine should be added to mixture so that the ratio of measurements of water and glycerine becomes 3: 2?

Solution: The quantity of water is \(\left(48 \times \frac{2}{2+1}\right)\) litres = \(\left(48 \times \frac{2}{3}\right)\) litres = 32 litres.

The quantity of glycerine is (48 – 32) litres = 16 litres.

Let x litres of glycerine be added to the mixture.

According to questions,

⇒ \(\frac{32}{16+x}=\frac{3}{2}\)

⇒ 48 + 3x = 64

⇒ 3x = 64 – 48 = 16

⇒ x = \(\frac{16}{3}\) = 5\(\frac{1}{3}\)

∴ 5\(\frac{1}{3}\) litre of glycerine should be added.

Example 4. The ratio of measurements of two types of tea in a mixture is 2: 3. If 2 kg of first type of tea is mixed with 55 kg of blended tea then, find the ratio of measurements of two types of tea.

Solution: The quantity of 1st type of tea = \(\left(55 \times \frac{2}{2+3}\right) \mathrm{kg}=\left(55 \times \frac{2}{5}\right) \mathrm{kg}=22 \mathrm{~kg}\)

The quantity of 2nd type of tea = (55-22) kg = 33 kg.

If 2 kg of 1st types of tea is mixed, then the ratio of the measurement of two types of tea is (22 + 2): 33 = 24: 33 = 8: 11

8: 11 ratio of measurements of two types of tea.

Example 5. The ratio of measurements of copper, zinc and nickel in German silver is 4: 3: 2 respectively. Find out what weight of zinc should be added to 54 kg of German silver, so that the ratio of measurements become 6:5: 3.

Solution: In 54 kg of German silver the quantity of zinc = (54 x \(\frac{3}{4+3+2}\))kg

= \(\left(54 \times \frac{3}{9}\right)\) kg = 18 kg.

Let x kg of zinc be added.

∴ The quantity of zinc in (54 + x) kg of German silver is (54 + x) x \(\frac{5}{6+5+3}\) kg = \(\frac{5}{14}\)(54 + x) kg.

According to question, x + 18 = \(\frac{5}{14}\)(54 + x)

⇒ 14 (x + 18) = 5(54 + x)

⇒ 14x + 252 = 270+ 5x

⇒ 14x – 5x = 270 – 252

⇒ 9x = 18

⇒ x =\(\frac{18}{9}\) = 2

∴ 2 kg of zinc be added.

Example 6. In a vessel of beverage the ratio of measurement of syrup and water is 4:1. Find out what part of the drink should removed and replaced by water so that the volume of syrup and water becomes equal.

Solution: Let x unit of beverage is there in a vessel. y unit of it is removed and replaced by the same volume of water.

The quantity of syrup in x units of beverage is \(\left(\frac{4}{4+1} \times x\right)\) units = \(\frac{4x}{5}\) units and the quantity of water = \(\left(\frac{1}{4+1} \times x\right)\) units = \(\frac{x}{5}\) units

Again the quantity of syrup in y units = \(\frac{4y}{5}\) units and water = \(\frac{y}{5}\) units

In new beverage the quantity of syrup is \(\left(\frac{4 x}{5}-\frac{4 y}{5}\right)\) units and water is \(\left(\frac{x}{5}-\frac{y}{5}+y\right)\) units.

According to question,

⇒ \(\frac{4 x-x}{5}=\frac{4 y-y+5 y}{5} \Rightarrow \frac{3 x}{5}=\frac{8 y}{5} \Rightarrow 3 x=8 y \Rightarrow y=\frac{3 x}{8}\)

∴ \(\frac{3}{8}\) part of the beverage will be removed.

Example 7. In what ratio tea of price ₹35 per kg and ₹28 per kg be mixed, so that there will be neither loss or gain on selling mixed tea at the rate of 30 per kg?

Solution: Let x kg tea of price ₹35 per kg and y kg tea price ₹28 per kg be mixed.

Total price of mixture tea is ₹(35x + 28y).

According to the question:

35x + 28y = 30 (x + y)

⇒ 35x – 30x = 30y – 28y ⇒ 5x = 2y

⇒ \(\frac{x}{y}\) = \(\frac{2}{5}\)

⇒ x: y = 2:5

∴ The required ratio is 2: 5.

Example 8. \(\frac{1}{2}\) and \(\frac{1}{5}\) parts two similar vessels contain fruit juice. The remaining empty parts of these vessels are filled with water and poured the juice mixed with water of the two vessels into another big vessels. Find out the ratio of measurements of fruit and water in the new vessel.

Solution: In the 1st vessel contains \(\frac{1}{2}\) parts fruit juice and (1-\(\frac{1}{2}\)) parts or \(\frac{1}{2}\) parts water.

In the second vessel contains \(\frac{1}{5}\) parts fruit juice and (1-\(\frac{1}{5}\)) or \(\frac{4}{5}\) parts water.

The ratio of fruir juice and water in new vessel is (\(\frac{1}{2}\)+\(\frac{1}{5}\)): (\(\frac{1}{2}\)+\(\frac{4}{5})\) = \(\frac{7}{10}\): \(\frac{13}{10}\) = 7: 13

Example 9. Two different types of stainless steel contain chromium and steel in the ratio of measurement of 3 : 8 and 5: 17 respectively. Find out in what proportion these two types of steel should be mixed, so that the ratio of measurement of chromium and steel becomes 8: 25.

Solution: Let x unit of 1st type of stainless steel is mixed with y unit of 2nd type of stainless steel.

The quantity of chromium in x unit of 1st type stainless steel is (x+\(\frac{3}{3+8}\)) unit or \(\frac{3x}{11}\) unit and quantity of steel is (x x \(\frac{8}{3+8}\)) unit or \(\frac{8x}{11}\) unit.

The quantity of chromium in y unit of 2nd type stainless steel is (y x \(\frac{5}{5+17}\)) unit or \(\frac{5y}{22}\)

In the new mixture the quantity of chromium is (\(\frac{3x}{11}\)+\(\frac{5y}{22}\)) unit and steel is (\(\frac{8x}{11}\)+\(\frac{17y}{22}\)) unit.

According to the questions,

⇒ \(\frac{6 x+5 y}{16 x+17 y}=\frac{8}{25}\)

⇒ 150x + 125y = 128x + 136y

⇒ 150x – 128x = 136y – 125y

⇒ 22x = 11y ⇒ \(\frac{x}{y}\)= \(\frac{11}{2}\)

⇒ x: y = 1: 2

∴ The two types of stainless steel are mixed up in the ratio of measurements of 1: 2.

Example 10. In what ratio is water to be mixed with milk at ₹30 per litre to make the price of the mixture ₹25 per litre?

Solution: Let x litre of water is mixed with y litre of milk.

The price of y litre of milk is ₹30y.

The price of (x + y) litre of mixture is ₹25 (x + y)

According to question,

25(x+y)= 30 y

⇒ 25x + 25y = 30y

⇒ 25x = 30y – 25y

⇒ 25x = 5y

⇒ \(\frac{x}{y}\)= \(\frac{5}{25}\)

⇒ x: y = 1:5

∴ The ratio of measurement of water and milk is 1: 5.

Example 11. Choose the Correct Answer:

1. If the ratio of measurement of milk and water in 30 litres is 4: 1, the measurement of milk is

1. 5 litres
2. 6 litres
3. 24 litres
4. 25 litres

Solution: The proportional part of the quantity of milk = \(\frac{4}{4+1}\) = \(\frac{4}{5}\)

The quantity of milk in 30 litres of mixture = (\(\frac{4}{5}\) x 30) litres = 24 litres

∴ So the correct answer is 3. 24 litres

The measurement of milk is 24 litres

2. 2/5 parts of a vessel contain syrup. If the vessel filled with water, then the ratio of syrup and water becomes

1. 2: 5
2. 3: 5
3. 2: 3
4. 4: 5

Solution: 2/5 parts of a vessel contain syrup. If the vessel filled with water, then the amount of water is (1- \(\frac{2}{5}\)) part or 3/5 part of that vessel.

The ratio of syrup and water = \(\frac{2}{5}\): \(\frac{3}{5}\) = 2:3

∴ The correct answer is 3. 2: 3

The ratio of syrup and water becomes 2: 3

3. In a alloy iron contains 20% copper contains 30% and the rest is nickel. The ratio of measurements of iron, copper and nickel is

1. 5: 2: 3
2. 2: 3: 5
3. 3: 2: 5
4. 5: 3: 2

Solution: The measurements of nickel is {100 (20+30)} % = 50%

∴ The ratio of measurements of iron, copper and nickel is 20%: 30%: 50% = \(\frac{20}{100}\): \(\frac{30}{100}\): \(\frac{50}{100}\) = 2: 3: 5

∴ The correct answer is 2. 2: 3: 5

The ratio of measurements of iron, copper and nickel is 2: 3: 5

Example 12. Write ‘True’ or ‘False’:

1. In 60 litres of a mixture of syrup and water contains 30% water. If 30 litres of syrup be added to the mixture, then the new mixture contains 20% water.

Solution: The quantity of water in 60 litres of a mixture is (60x\(\frac{30}{100}\)) litres or 18 litres.

So the quantity of syrup (60 – 18) litres = 42 litres.

If 30 litres of syrup be added to the mixture the total volume is (60+30) litres or 90 litres.

∴ The quantity of water = (\(\frac{18}{90}\)x100)% = 20%

∴ So the statement is true.

2. In 50 c.c. of concentrated sulphuric acid contains 2% water. The ratio of measurements of acid and water is 25: 1.

Solution: The quantity of water in 50 c.c. of concentrated sulphuric acid is

(50x\(\frac{2}{100}\)) = 1 c.c.

The quantity of acid is (50 – 1) c.c. = 49 c.c.

The ratio of acid and water is 49: 1.

∴ So the statement is false.

Example 13. Fill in the blanks:

1. If the ratio of measurements of sulphur, charcoal and potassium nitrate in 240 gm of gun powder is 1 : 3: 4, then the quantity of sulphur is ______ gm.

Solution: The quantity of sulphur = (240 x \(\frac{1}{1+3+4}\)) gm = (240 x \(\frac{1}{8}\)) gm = 30 gm

2. In air nitrogen contains 77.16%, oxygen contains 20.60%, carbon dioxide contains 0.04%, water vapour contains 1.40% and the rest is inert gas. The ratio of measurements of carbon dioxide and inert gas is ________

Solution: The quantity of inert gas is

{100 (77.16 + 20.60 + 0.04 + 1·40)}% = (100 – 99.20)% = 0.80%

The ratio of measurements of carbon dioxide and inert gas is

0.04%: 0.80% = 4: 80 = 1: 20.

The ratio of measurements of carbon dioxide and inert gas is 1: 20.

Arithmetic Chapter 5 Percentage

The term is per cent or per centum, per means divided by and cent means 100. So, percentage implies calculation per every hundred.

The words per cent is briefly expressed by the symbols %.

Read and Learn More WBBSE Solutions For Class 8 Maths

Express 12% as a fraction.

Express into percentage

Express 1% of 0.1 in decimal.

⇒ \(1 \% \text { of } 0 \cdot 1=0 \cdot 1 \times \frac{1}{100}=\frac{1}{10} \times \frac{1}{100}=\frac{1}{1000}=0.001\)

Equivalent discount: On a particular principal the equivalent discount is equal to more than one successive discounts on that principal.

The discount equivalent to successive discounts of q% and b% is (a + b – \(\frac{ab}{100}\))%

Arithmetic Chapter 5 Percentage Examples

Example 1. Calculate:

1. 45% of 900
2. 115% of 25.5 metre
3. 205% of 150 gm

Solution:

45% of ₹900

=

= ₹405

45% of ₹900 = ₹405

115% of 25.5 metre

=

= \(\frac{1173}{40}\) metre = 29.325 metre

115% of 25.5 metre = 29.325 metre

205% of 150 gm

=

= \(\frac{615}{2}\) gm = 307.5 gm

205% of 150 gm = 307.5 gm

Example 2. Express the following in percentage.

1. 2 kg 250 gm out of 0.72 quintal
2. ₹1.75 out of ₹3.50

Solution:

1. 2 kg 250 gm = 2250 gm

0.72 quintal = 0.72 x 100 x 1000 gm = 72000 gm.

2250 gm out of 72000 gm is

= \(\frac{25}{8}\)% = 3.125%

2 kg 250 gm out of 0.72 quintal = 3.125%

2. ₹1·75 out of ₹3.50 is

= 50%

₹1·75 out of ₹3.50 is 50%

Example 3. The length of each side of a square is increased by 20%. Find the increase of percentage area of the square.

Solution: Let the length of each side of a square is x units.

∴ Area is x² square units.

If the lengths of each side of a square is increased by 20% then the length is

Area of the square = \(\left(\frac{6 x}{5}\right)^2\) square units = \(\frac{36 x^2}{25}\) square units

Area increased = \(\left(\frac{36 x^2}{25}-x^2\right)\) square units = \(\frac{36 x^2-25 x^2}{25}\) square units = \(\frac{11 x^2}{25}\) square units.

∴ Area increased by

= 44%

The increase of percentage area of the square = 44%

Example 4. The ratio of measurements of copper and zinc in a brass is 4: 1. Find the percentage of copper and zinc.

Solution: The ratio of measurement of copper and zinc is 4: 1.

The percentage of copper is \(\left(\frac{4}{4+1} \times 100\right)=\left(\frac{4}{5} \times 100\right)=80\)

The percentage of zinc is \(\left(\frac{1}{4+1} \times 100\right) \text { or, } \frac{100}{5} \text { or, } 20\)

∴ Copper has 80% and zinc has 20%.

The percentage of copper and zinc 80% and 20%.

Example 5. If A’s income be 20% more than B’s income, how much per cent is B’s income less than A’s income?

Solution: If B’s income is ₹100 then A’s income is ₹(100+20) ⇒ ₹120.

If A’s income is ₹120, then B’s income less is ₹20

If A’s income is ₹1 then B’s income less is ₹\(\frac{20}{120}\)

If A’s income is ₹100 then B’s income less is ₹\(\frac{20 \times 100}{120}=₹ \frac{50}{3}=₹ 16 \frac{2}{3}\)

∴ B’s income is 16\(\frac{2}{3}\)% less than A’s income.

Example 6. If 40% of (2a – b) = 30% of (a + 3b), then find a : b.

Solution: 40% of (2a – b) 30% of (a + 3b)

⇒ (2a-b) x \(\frac{40}{100}\) = (a+3b) x \(\frac{30}{100}\)

⇒ 40 (2a – b) = 30 (a + 3b)

⇒ 4 (2a – b) = 3 (a + 3b)

⇒ 8a – 4b = 3a + 9b

⇒ 8a – 3a = 9a + 4b

⇒ 5a = 13b

⇒ \(\frac{a}{b}\) = \(\frac{13}{5}\)

⇒ a: b = 13: 5

Example 7. The price of sugar has increased by 20%. Find the percentage of decreased is consumption sugar if the monthly expenses of sugar remain same.

Solution: Let previously 100 units of sugar was used whose price was ₹100.

Now the cost of sugar increased by 20% i.e. now ₹120 costs for 100 units of sugar.

Now ₹120 costs 100 units of sugar

₹1 costs \(\frac{100}{120}\) units of sugar

₹100 costs \(\frac{100 \times 10}{120}\) units of sugar = \(\frac{250}{3}\) units of sugar

In order to keep the monthly expense for sugar same, it will reduce the use of sugar per month by (100-\(\frac{250}{3}\))units

= \(\frac{300 – 250 }{3}\) units = 16\(\frac{2}{3}\)

∴ It will reduce the use of sugar by 16\(\frac{2}{3}\)%

Example 8. In a certain school 85% of the candidates passed in Bengali, 70% passed in Mathematics and 65% passed in both the subjects in the annual examination. If the number of students are 120, then find the number of students failed in both the subjects.

Solution: 65% passed in both the subjects.

∴ Passed only in Bengali = (85 – 65)% = 20%

Passed only in Mathematics = (70 – 65)% = 5%

∴ Passed either in both the subjects or in any one subject only is (65 + 20 + 5) % = 90%

∴ Failed in both the subjects is (100 – 90)% = 10%

∴ The number of students failed in both the subjects is \(\left(120 \times \frac{10}{100}\right)\) or 12

Example 9. If money is devalued by 4%, find the percentage increase in money that will have to be spend to buy a particular article.

Solution: Let the marked price of particular article be ₹100

So the present value of a coin of previous valuation of ₹100 is ₹(100-4) or ₹96

To purchase a particular article of ₹100 is required more by ₹(100 – 96) or ₹4.

So,

If the present value of a coin is ₹96, then ₹4 more is required

If the present value of a coin is ₹1, then ₹\(\frac{4}{96}\) more is required

If the present value of a coin is ₹100, then ₹\(\frac{4 \times 100}{96}\) more is required

=₹\(\frac{25}{6}\) more is required

= ₹4\(\frac{1}{6}\) more is required

∴ Required increase is 4\(\frac{1}{6}\)%.

Example 10. 89% of pure milk is water. If 90% of a sample milk is water, then find the quantity of water mixed in 22 litres of such milk.

Solution: Let x litres of water mixed in 22 litres of such milk.

∴ The quantity of pure milk is (22 – x) litres.

The quantity of water in (22 – x) litres of pure milk is (22 – x) x \(\frac{89}{100}\) litre.

The quantity of water in 22 litres of sample milk is \(\left(\frac{22 \times 90}{100}\right)\) litre.

According to questions,

⇒ 11x + 1958 = 1980

⇒ 11x = 1980 – 1958

⇒ 11x = 22 ⇒ x = \(\frac{22}{11}\) = 2

∴ The 2 litres water mixed in 22 litres of such milk.

Example 11. When water freezes into ice, it increases in volume by 10%. Find out in percentage how much it will decreased in volume if the ice melts into water.

Solution: When 100 unit of water freezes into ice its volume increases to (100 + 10) unit or 110 units.

In the mathematical language, the problem is:

Volume of ice (unit)
110
100

Volume of water by melting of ice (unit)
100
? (x) (say)

Volume of ice and volume of water by melting of ice are directly proportional.

∴ 110: 100 : : 100: x

⇒ \(\frac{110}{100}=\frac{100}{x} \Rightarrow x=\frac{100 \times 100}{110} \Rightarrow x=\frac{1000}{11}\)

If 100 unit of ice melted to water, then the volume decrease to \(\frac{100}{11}\) unit

∴ Volume decreases = (100-\(\frac{1000}{11}\))% = (\(\frac{1100-1000}{11}\))% = \(\frac{100}{11}\)% = 9\(\frac{1}{11}\)%

Example 12. Due to use of high-yielding seed Gopalbabu has got 40% production high in paddy cultivation. But for this the cost of cultivation has increased by 30%. Previously a yield of ₹2500 was produced by investing 1000. Find out whether his income will be increased or decreased after using high-yielding seeds.

Solution: Previously the cost of cultivation was ₹1000.

At present cost of cultivation is ₹(1000+1000x\(\frac{30}{100}\)) = ₹(1000+300) = ₹1300

Previously yield ₹2500 was produced.

Due to use of high-yielding seed at present yield of ₹(25000+2500x\(\frac{40}{100}\)) =₹(2500 + 1000) = ₹3500

Previously his income was ₹(2500-1000) = ₹1500

At present his income is ₹(3500-1300) = ₹2200

His income increases by ₹(2200-1500) or ₹700.

Example 13. Choose the correct answer:

1. 15% of ₹80 is

1. ₹10
2. ₹12
3. ₹20
4. ₹24

Solution: 15% of ₹80

= ₹12

∴ So the correct answer is 2. 12

15% of ₹80 is 12.

2. The ratio of hydrogen and oxygen in water is 2: 1. The percentage of hydrogen in water is

1. 33\(\frac{1}{3}\)
2. 50
3. 25
4. 66\(\frac{2}{3}\)

Solution: The percentage of hydrogen is (\(\frac{2}{2+1}\)x100) = \(\frac{200}{3}\) = 66\(\frac{2}{3}\)

∴ The correct answer is 4. 66\(\frac{2}{3}\)

The percentage of hydrogen in water is 66\(\frac{2}{3}\)

3. Aloke has got 36 marks out of 50 in Bengali. He has got a percentage

1. 18
2. 36
3. 54
4. 72

Solution: Out of 50 marks Aloke has got 36 marks.

Out of 1 mark he has got \(\frac{36}{50}\) marks

Out of 100 marks he has got \(\frac{36 \times 100}{50}\) marks = 72 marks

He has got 72% marks.

∴ So the correct answer is 4. 72

He has got a percentage 72.

Example 14. Write ‘True’ or ‘False’:

1. 200% of 480 gm is 960 gm.

Solution: 200% of 480 gm

= 960gm

∴ So the statement is true.

2. 60% of 3.16 = 1.9

Solution: 60% of 3.16

= \(\frac{60}{100} \times \frac{316-31}{90}\)

=

∴ So the statement is true.

Example 15. Fill in the blanks:

1. If 30% of A = 50% of B. Then A = _______ % of B.

Solution: A x \(\frac{30}{100}\) = 50% of B

A = of B = \(\frac{500}{3}\)% of B

2. The expression of 125% in ratio is ________

Solution: 125% = \(\frac{125}{100}\) = \(\frac{5}{4}\) = 5: 4