myhelper: RS Aggarwal solution class 8 chapter 8 Linear Equations Test Paper 8

Test Paper 8

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Q1 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 1:

Subtract 4a² + 5b² − 6c² + 8 from 2a² − 3b² − 4c² − 5.

Answer 1:

$2 a^{2} - 3 b^{2} - 4 c^{2} - 5 - (4 a^{2} + 5 b^{2} - 6 c^{2} + 8)$

$= 2 a^{2} - 3 b^{2} - 4 c^{2} - 5 - 4 a^{2} - 5 b^{2} + 6 c^{2} - 8$

$= 2 a^{2} - 4 a^{2} - 3 b^{2} - 5 b^{2} - 4 c^{2} + 6 c^{2} - 5 - 8$

$= - 2 a^{2} - 8 b^{2} + 2 c^{2} - 13$

Q2 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 2:

Find each of the following products:
(i) (4a + 5b) × (5a − 6b)
(ii) (6x² − x + 8) × (x² − 3)

Answer 2:

$(i) (4 a + 5 b) \times (5 a - 6 b) = 20 a^{2} - 24 a b + 25 a b - 30 b^{2}$

$= 20 a^{2} + a b - 30 b^{2}$

$(i i) (6 x^{2} - x + 8) \times (x^{2} - 3)$

$= 6 x^{4} - 18 x^{2} - x^{3} + 3 x + 8 x^{2} - 24$

$= 6 x^{4} - x^{3} - 18 x^{2} + 8 x^{2} + 3 x - 24$

$= 6 x^{4} - x^{3} - 10 x^{2} + 3 x - 24$

Q3 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 3:

Divide (5a³ − 4a² + 3a + 18) by (a² − 2a + 3).

Answer 3:

Therefore, $(5 a + 6)$ is the quotient.

Q4 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 4:

If $(x - \frac{1}{x}) = 4$ , find the value of
(i) $(x^{2} + \frac{1}{x^{2}}) .$
(ii) $(x^{4} + \frac{1}{x^{4}}) .$

Answer 4:

$(i) Given, (x - \frac{1}{x}) = 4 Squaring both the sides : {(x - \frac{1}{x})}^{2} = 4^{2}$

$\Rightarrow {(x)}^{2} - 2 \times (x) \times (\frac{1}{x}) + {(\frac{1}{x})}^{2} = 16$

$\Rightarrow x^{2} - 2 + \frac{1}{x^{2}} = 16$

$\Rightarrow x^{2} + \frac{1}{x^{2}} = 16 + 2$

$∴ (x^{2} + \frac{1}{x^{2}}) = 18$ ² + 2

⇒ 4² + 2

⇒ 16 + 2 ⇒ 18

^{$(i i) From the first part : (x^{2} + \frac{1}{x^{2}}) = 18$}

^{$Squaring both the sides : {(x^{2} + \frac{1}{x^{2}})}^{2} = 18^{2}$}

^{$\Rightarrow (x^{2})^{2} + 2 \times (x^{2}) \times (\frac{1}{x^{2}}) + {(\frac{1}{x^{2}})}^{2} = 324$}

^{$\Rightarrow x^{4} + 2 + \frac{1}{x^{4}} = 324$}

^{$\Rightarrow x^{4} + \frac{1}{x^{4}} = 324 - 2$}

^{$∴ x^{4} + \frac{1}{x^{4}} = 322$}

Q5 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 5:

Evaluate {(83)² − (17)²}.

Answer 5:

${(83)}^{2} - {(17)}^{2} = (83 + 17) (83 - 17) [according to the formula {(a)}^{2} - {(b)}^{2} = (a + b) (a - b)] = (100) (66) = 6600$

Q6 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 6:

Factorise:
(i) x³ − 3x² + x − 3
(ii) 63x²y² − 7
(iii) 1 − 6x + 9x²
(iv) 7x² − 19x − 6

Answer 6:

$(i) x^{3} - 3 x^{2} + x - 3 = x^{2} (x - 3) + 1 (x - 3) = (x^{2} + 1) (x - 3) (ii) 63 x^{2} y^{2} - 7 = 7 (9 x^{2} y^{2} - 1) = 7 ((3 x y)^{2} - (1)^{2}) [according to the formula a^{2} - b^{2} = (a + b) (a - b)] = 7 (3 x y + 1) (3 x y - 1) (iii) 1 - 6 x + 9 x^{2} = 9 x^{2} - 6 x + 1 = 9 x^{2} - 3 x - 3 x + 1 = 3 x (3 x - 1) - 1 (3 x - 1) = (3 x - 1) (3 x - 1) = {(3 x - 1)}^{2} (iv) 7 x^{2} - 19x - 6 = 7 x^{2} - 21 x + 2 x - 6 = 7 x (x - 3) + 2 (x - 3) = (7 x + 2) (x - 3)$

Q7 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 7:

Solve: $\frac{2 x + 7}{3 x + 5} = \frac{15}{17} .$

Answer 7:

$\frac{2 x + 7}{3 x + 5} = \frac{15}{17}$

$\Rightarrow 17 (2 x + 7) = 15 (3 x + 5)$

$\Rightarrow 34 x + 119 = 45 x + 75$

$\Rightarrow 119 - 75 = 45 x - 34 x$

$\Rightarrow 44 = 11 x$

$\Rightarrow x = \frac{44}{11} = 4$

$∴ x = 4$

Q8 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 8:

5 years ago a man was 7 times as old as his son. After 5 years he will be thrice as old as his son. Find their present ages.

Answer 8:

$Let the present age of the son be x years and that of the father be f years . 5 years back, the father was 7 times as old as his son . ∴ (f - 5) = 7 (x - 5) f = 7 x - 35 + 5 f = 7 x - 30 . . . (1) After 5 years, ages of the father and son will be (f + 5) and (x + 5), respectively . After 5 years, the father will be three times older than his son . ∴ (f + 5) = 3 (x + 5) 7 x - 30 + 5 = 3 x + 15 [inserting the value of f from equation (1)] 7 x - 25 = 3 x + 15 7 x - 3 x = 25 + 15 4 x = 40 x = \frac{40}{4} = 10 Therefore, the present age of the son is 10 years . Father' s present age = (7 x - 30) = 7 (10) - 30 = 70 - 30 = 40 years$

Q9 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 9:

Mark (✓) against the correct answer:
ab − a − b + 1 = ?
(a) (1 − a)(1 − b)
(b) (1 − a)(b − 1)
(c) (a − 1)(b − 1)
(d) (a − 1)(1 − b)

Answer 9:

Q10 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 10:

Mark (✓) against the correct answer:
3 + 23x − 8x² = ?
(a) (1 − 8x)(3 + x)
(b) (1 + 8x)(3 − x)
(c) (1 − 8x)(3 − x)
(d) none of these

Answer 10:

(b) $(1 + 8 x) (3 - x)$

$3 + 23 x - 8 x^{2} = 3 + 24 x - x - 8 x^{2} = 3 (1 + 8 x) - x (1 + 8 x) = (1 + 8 x) (3 - x)$

Q11 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 11:

Mark (✓) against the correct answer:
7x² − 19x − 6 = ?
(a) (x − 3)(7x + 2)
(b) (x + 3)(7x − 2)
(c) (x − 3)(7x − 2)
(d) (7x − 3)(x + 2)

Answer 11:

(a) $(x - 3) (7 x + 2)$

$7 x^{2} - 19 x - 6 = 7 x^{2} - 21 x + 2 x - 6 = 7 x (x - 3) + 2 (x - 3) = (x - 3) (7 x + 2)$

Q12 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 12:

Mark (✓) against the correct answer:
12x² + 60x + 75 = ?
(a) (2x + 5)(6x + 5)
(b) (3x + 5)²
(c) 3(2x + 5)²
(d) none of these

Answer 12:

(c) $3 {(2 x + 5)}^{2}$

$12 x^{2} + 60 x + 75 = 3 (4 x^{2} + 20 x + 25) = 3 ({(2 x)}^{2} + 2 \times 2 x \times 5 + 5^{2}) = 3 {(2 x + 5)}^{2}$

Q13 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 13:

Mark (✓) against the correct answer:
10p² + 11p + 3 = ?
(a) (2p + 3)(5p + 1)
(b) (5p + 3)(2p + 1)
(c) (5p − 3)(2p − 1)
(d) none of these

Answer 13:

(b) $(5 p + 3) (2 p + 1)$

$10 p^{2} + 11 p + 3 = 10 p^{2} + 5 p + 6 p + 3 = 5 p (2 p + 1) + 3 (2 p + 1) = (5 p + 3) (2 p + 1)$

Q14 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 14:

Mark (✓) against the correct answer:
8x³ − 2x = ?
(a) (4x − 1)(2x − 1)x
(b) (2x² + 1)(2x − 1)
(c) 2x(2x − 1)(2x + 1)
(d) none of these

Answer 14:

Q15 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 15:

Mark (✓) against the correct answer:
$\frac{x + 5}{2} + \frac{x - 5}{3} = \frac{25}{6} gives$
(a) x = 3
(b) x = 4
(c) x = 5
(d) x = 2

Answer 15:

(b) $x = 4$

$\frac{x + 5}{2} + \frac{x - 5}{3} = \frac{25}{6} \Rightarrow \frac{3 (x + 5) + 2 (x - 5)}{6} = \frac{25}{6} (because L . C . M . of 2 & 3 is 6) \Rightarrow \frac{3 x + 15 + 2 x - 10}{6} = \frac{25}{6} \Rightarrow \frac{5 x + 5}{6} = \frac{25}{6} \Rightarrow 5 x + 5 = 25 (cancelling 6 from the denominator on both the sides) \Rightarrow 5 x = 25 - 5 \Rightarrow 5 x = 20 \Rightarrow x = \frac{20}{5} \Rightarrow x = 4$

Q16 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 16:

Fill in the blanks.
(i) x²− 18x + 81 = (......)
(ii) 4 − 36x² = (......)(......)(......)
(iii) x² − 14x + 13 = (......)(......)
(iv) 9z² − x² − 4y² + 4xy = (......)(......)
(v) abc − ab − c + 1 = (......)(......)

Answer 16:

(i) ${(x - 9)}^{2}$

$x^{2} - 18 x + 81 = x^{2} - 2 \times x \times 9 + 9^{2} = {(x - 9)}^{2}$

(ii) $(4) (1 - 3 x) (1 + 3 x)$

$4 - 36 x^{2} = 4 (1 - 9 x^{2}) = 4 (1^{2} - {(3 x)}^{2}) = (4) (1 - 3 x) (1 + 3 x)$

(iii) $(x - 13) (x - 1)$

$x^{2} - 14 x + 13 = x^{2} - 13 x - x + 13 = x (x - 13) - 1 (x - 13) = (x - 13) (x - 1)$

(iv) $(3 z + x - 2 y) (3 z - x + 2 y)$

$9 z^{2} - x^{2} - 4 y^{2} + 4 x y = 9 z^{2} - (x^{2} - 4 x y + 4 y^{2}) = 9 z^{2} - (x^{2} - 2 \times x \times 2 y + {(2 y)}^{2}) = 9 z^{2} - {(x - 2 y)}^{2} = {(3 z)}^{2} - {(x - 2 y)}^{2} = (3 z + (x - 2 y)) (3 z - (x - 2 y)) = (3 z + x - 2 y) (3 z - x + 2 y)$

(v) $(c - 1) (a b - 1)$

$a b c - a b - c + 1 = a b (c - 1) - 1 (c - 1) = (c - 1) (a b - 1)$

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Q17 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Question 17:

Write 'T' for true and 'F' for false for each of the following:
(i) (5 − 3x²) is a binomial.
(ii) −8 is a monomial.
(iii) (5a − 9b) − (−6a + 2b) = (−a − 7b).
(iv) When x = 2 and y = 1, the value of $\frac{- 8}{7} x^{3} y^{4} is \frac{- 64}{7} .$
(v) $\frac{x}{4} + \frac{x}{6} - \frac{x}{2} = \frac{3}{4} \Rightarrow x = - 9 .$
(vi) $2 x - 5 = 0 \Rightarrow x = \frac{2}{5} .$

Answer 17:

(i) T
Binomial expression is an expression that shows the sum or the difference of two unlike terms. The above expression has two unlike terms, i.e. 5 and 3x².

(ii) T
Any expression that contains only one term is called a monomial. It can either be a constant or a variable.

(iii) F

$L . H . S . = (5 a - 9 b) - (- 6 a + 2 b) = 5 a - 9 b + 6 a - 2 b = (- a - 7 b) = 5 a + 6 a - 9 b - 2 b = (- a - 7 b) = (11 a - 11 b) This is not equal to (- a - 7 b) . Thus, the L . H . S . is not equal to the R . H . S .$

(iv) T

$\frac{- 8}{7} x^{3} y^{4} Given : x = 2 y = 1 Given expression = \frac{- 8}{7} {(2)}^{3} {(1)}^{4} = \frac{- 8}{7} \times 8 \times 1 = \frac{- 64}{7}$

(v) T

$\frac{x}{4} + \frac{x}{6} - \frac{x}{2} = \frac{3}{4} \Rightarrow \frac{3 x + 2 x - 6 x}{12} = \frac{3}{4} \Rightarrow \frac{- x}{12} = \frac{3}{4} \Rightarrow - 4 x = 36 \Rightarrow x = - \frac{36}{4} \Rightarrow x = - 9$

(vi) F

$2 x - 5 = 0 \Rightarrow 2 x = 5 \Rightarrow x = \frac{5}{2}$

RS Aggarwal solution class 8 chapter 8 Linear Equations Test Paper 8

Test Paper 8

Page-115

Q1 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 1:

Answer 1:

Q2 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 2:

Answer 2:

Q3 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 3:

Answer 3:

Q4 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 4:

Answer 4:

Q5 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 5:

Answer 5:

Q6 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Answer 6:

Q7 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

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Answer 7:

Q8 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 8:

Answer 8:

Q9 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 9:

Answer 9:

Q10 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 10:

Answer 10:

Q11 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 11:

Answer 11:

Q12 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 12:

Answer 12:

Q13 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 13:

Answer 13:

Q14 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 14:

Answer 14:

Q15 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 15:

Answer 15:

Q16 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 16:

Answer 16:

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Q17 | Test Paper 8 | Class 8 | RS AGGARWAL | chapter 8 | Linear Equations Solution

Question 17:

Answer 17:

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