myhelper: RS Aggarwal solution class 8 chapter 9 Percentage Test Paper 9

Test Paper 9

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Q1 | Test Paper 9 | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 1:

Express:
(i) 24% as a fraction;
(ii) 105% as a decimal;
(iii) 4 : 5 as a percentage;
(iv) 56% as a ratio.

Answer 1:

$(i) 24 % = \frac{24}{100} = \frac{6}{25}$

$(i i) 105 % = \frac{105}{100} = 1.05 (i i i) 4 : 5 = \frac{4}{5} = (\frac{4}{5} \times 100) % = 80 % (i v) 56 % = \frac{56}{100} = \frac{14}{25} = 14 : 25$

Q2 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 2:

If 34% of a number is 85, find the number.

Answer 2:

$Let the required number be x . Then, we have : (34 % of x) = 85 \Rightarrow (x \times \frac{34}{100}) = 85 \Rightarrow \frac{34 x}{100} = 85 \Rightarrow x = (85 \times \frac{100}{34}) \Rightarrow x = 250 Hence, the required number is 250 .$

Q3 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 3:

The value of a machine depreciates every year by 10%. If the present value of the machine is Rs 54000, what was its value last year?

Answer 3:

$Let the value of the machine last year be Rs x . Then, its present value = 90 % of Rs x$
$= R s (x \times \frac{90}{100}) = R s \frac{90 x}{100}$
$Now, \frac{90 x}{100} = 54000 \Rightarrow x = (54000 \times \frac{100}{90}) \Rightarrow x = Rs 60000$
$Hence, the value of the machine last year was Rs 60, 000 .$

Q4 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 4:

An alloy contains 30% copper, 42% nickel and rest zinc. Find the mass of zinc in 1 kg of alloy.

Answer 4:

$Percentage of copper = 30 % Percentage of nickel = 42 % Percentage of zinc = \{100 - (30 + 42)\} %$
      $= 28 %$

$∴ Mass of zinc in 1 kg of the alloy = (\frac{28}{100} \times 1) kg = 0.28 kg = 280 g$

Q5 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 5:

In a class, 60% of the total number of students are boys and there are 14 girls. How many students are there in the class?

Answer 5:

$Let the total number of students be x . Then, we have : Percentage of boys = 60 % Percentage of girls = 40 % ∴ Number of girls = 40 % of x$
$= (x \times \frac{40}{100}) = \frac{40 x}{100}$
$Now, \frac{40 x}{100} = 14 \Rightarrow x = (14 \times \frac{100}{40}) \Rightarrow x = 35$

$∴ Total number of students = 35$

Q6 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 6:

Which is largest in $8 \frac{1}{3} %, \frac{4}{25}$ and 0.15?

Answer 6:

$We have : 8 \frac{1}{3} % = \frac{25}{3} % = (\frac{25}{3} \times \frac{1}{100}) = \frac{1}{12} = 0.083$
$Also, \frac{4}{25} = 0.16 The third number is 0.15 . Clearly, 0.16 is the largest . i . e ., \frac{4}{25} is the largest .$

Q7 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 7:

Mark (✓) against the correct answer:
What per cent of $\frac{2}{9} is \frac{1}{45} ?$
(a) 2.5%
(b) 5%
(c) 7.5%
(d) 10%

Answer 7:

(d) 10%

$Required percentage = (\frac{1}{45} \times \frac{9}{2} \times 100) % = 10 %$

Q8 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 8:

Mark (✓) against the correct answer:
A number decreased by 30% gives 84. The number is
(a) 90
(b) 110
(c) 120
(d) 135

Answer 8:

(c) 120

$Let the required number be x$
$x - (30 % o f x) = 84 \Rightarrow \{x - (x \times \frac{30}{100})\} = 84 \Rightarrow (x - \frac{30 x}{100}) = 84 \Rightarrow \frac{70 x}{100} = 84 \Rightarrow x = (84 \times \frac{100}{70}) \Rightarrow x = 120$

Q9 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 9:

Mark (✓) against the correct answer:
(?)% of 320 is 48?
(a) 25%
(b) 15%
(c) 14%
(d) 9%

Answer 9:

(b) 15%

$Let the required number be x . Then, we have : (x % o f 320) = 48 \Rightarrow (320 \times \frac{x}{100}) = 48 \Rightarrow \frac{320 x}{100} = 48 \Rightarrow x = (48 \times \frac{100}{320}) \Rightarrow x = 15 %$

Q10 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 10:

Mark (✓) against the correct answer:
What per cent of 45 is 54?
(a) $83 \frac{1}{3} %$
(b) 104%
(c) 108%
(d) 120%

Answer 10:

(d) 120%
$Required percentage = (\frac{54}{45} \times 100) % = 120 %$

Q11 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 11:

Mark (✓) against the correct answer:
A number exceeds 25% of itself by 60. The number is
(a) 75
(b) 45
(c) 80
(d) 65

Answer 11:

(c) 80

$Let the required number be x . Then, we have : (25 % o f x) + 60 = x \Rightarrow (x \times \frac{25}{100}) + 60 = x \Rightarrow \frac{25 x}{100} + 60 = x \Rightarrow (\frac{25 x}{100} - x) = - 60 \Rightarrow \frac{- 75 x}{100} = - 60 \Rightarrow x = (60 \times \frac{100}{75}) \Rightarrow x = 80$

Q12 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 12:

Mark (✓) against the correct answer:
5% of which number is 12?
(a) 120
(b) 180
(c) 240
(d) 320

Answer 12:

(c) 240

$Let the required number be x . Then, we have : (5 % of x) = 12 \Rightarrow (x \times \frac{5}{100}) = 12 \Rightarrow \frac{5 x}{100} = 12 \Rightarrow x = (12 \times \frac{100}{5}) \Rightarrow x = 240$

Q13 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 13:

Fill in the blanks.
(i) $7 \frac{1}{2} %$ of Rs 1200 = .........
(ii) 240 mL is ......... % of 3 L.
(iii) If x% of 35 is 42, then x = .........
(iv) $\frac{12}{5}$ = .........%.
(v) 120 = (.........)% of 80.

Answer 13:

$(i) 7 \frac{1}{2} % of Rs 1200 = (\frac{15}{2} % of Rs 1200)$
$= Rs (\frac{15}{2} \times \frac{1}{100} \times 1200) = Rs 90$
$Hence, 7 \frac{1}{2} % of Rs 1200 = Rs 90$

$(ii) Required percentage = (\frac{240}{3 \times 1000} \times 100) % = 8 % Hence, 240 ml is 8 % of 3 L .$

$(iii) (x % of 35) = 42 \Rightarrow (35 \times \frac{x}{100}) = 42 \Rightarrow \frac{35 x}{100} = 42 \Rightarrow x = (42 \times \frac{100}{35}) \Rightarrow x = 120 % ∴ If x % of 35 is 42, then x = 120 % .$

$(iv) (\frac{12}{5} \times 100) % = 240 % Hence, \frac{12}{5} = 240 %$

$(v) Let the required number be x . Then, we have : 120 = x % o f 80 \Rightarrow (80 \times \frac{x}{100}) = 120 \Rightarrow \frac{80 x}{100} = 120 \Rightarrow x = (120 \times \frac{100}{80}) \Rightarrow x = 150 % ∴ 120 = 150 % o f 80$

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Q14 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Question 14:

Write 'T' for true and 'F' for false for each of the following:
(i) 6% of 8 is 48.
(ii) 6 : 5 = 30%.
(iii) $\frac{3}{5} = 60 % .$
(iv) 6 hours = 25% of a day.

Answer 14:

$(i) 6 % of 8 = (8 \times \frac{6}{100}) = 0.48 Hence, it is false .$

$(ii) 6 : 5 = \frac{6}{5} = (\frac{6}{5} \times 100) % = 120 % Hence, it is false .$

$(iii) \frac{3}{5} = (\frac{3}{5} \times 100) % = 60 % Hence, it is true .$

$(iv) 6 h o u r s = (\frac{6}{24} \times 100) % = 25 %$
Hence, it is true.

RS Aggarwal solution class 8 chapter 9 Percentage Test Paper 9

Test Paper 9

Page-125

Q1 | Test Paper 9 | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Answer 1:

Q2 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Answer 2:

Q3 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q4 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q5 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q6 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Answer 6:

Q7 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q8 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q9 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Answer 9:

Q10 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q11 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q12 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Q13 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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Answer 13:

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Q14 | Ex-9A | Class 8 | RS AGGARWAL | chapter 9 | Percentage

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