RS Aggarwal 2020 2021 Solutions for Class 8 Maths Chapter 9 Percentage are provided here with simple step-by-step explanations. These solutions for Percentage are extremely popular among class 8 students for Maths Percentage Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 8 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 8 Maths are prepared by experts and are 100% accurate.

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#### Question 7:

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#### Question 8:

#### Question 11:

$=\left(x×\frac{35}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{35x}{100}$

#### Question 12:

$=\left(x×\frac{35}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{35x}{100}$

#### Question 13:

$=\left(\mathrm{x}×\frac{85}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{85\mathrm{x}}{100}$

#### Question 14:

$=\left(\mathrm{x}×\frac{85}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{85\mathrm{x}}{100}$

$=Rs125$

#### Question 15:

$=Rs125$

$=\frac{1}{11}\mathrm{unit}$
$=9\frac{1}{11}%$

#### Question 16:

$=\frac{1}{11}\mathrm{unit}$
$=9\frac{1}{11}%$

$=\left(x×\frac{108}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{27x}{25}$

#### Question 17:

$=\left(x×\frac{108}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{27x}{25}$

$=\mathrm{Rs}\left(\mathrm{x}×\frac{80}{100}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}\frac{4\mathrm{x}}{5}$

#### Question 18:

$=\mathrm{Rs}\left(\mathrm{x}×\frac{80}{100}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}\frac{4\mathrm{x}}{5}$

$=28%$

#### Question 19:

$=28%$

#### Question 20:

#### Question 21:

#### Question 22:

$=91\frac{2}{3}%$

#### Question 23:

$=91\frac{2}{3}%$

Let the original salary be Rs 100
Then, after increment of 25% the salary becomes
=

To restore the original salary, let the new salary be decreased by x%.
Thus, we get

Therefore, the new salary must be reduced by 20% to restore the original salary.

#### Question 1:

Let the original salary be Rs 100
Then, after increment of 25% the salary becomes
=

To restore the original salary, let the new salary be decreased by x%.
Thus, we get

Therefore, the new salary must be reduced by 20% to restore the original salary.

(d) 60%

(d) 60%

(b) 0.008

(b) 0.008

(c) 120%

(c) 120%

(d) 180

#### Question 5:

(d) 180

(c) $133\frac{1}{3}%$

#### Question 6:

(c) $133\frac{1}{3}%$

(d) 2.5%

(d) 2.5%

(b) 600

(b) 600

(c) 15

(c) 15

(d) 560

#### Question 10:

(d) 560

(a) Rs 160

#### Question 11:

(a) Rs 160

(c) 175

$=\frac{56x}{100}$

#### Question 12:

(c) 175

$=\frac{56x}{100}$

(b) decrease by 1 %

#### Question 13:

(b) decrease by 1 %

(a) $18\frac{3}{4}%$

#### Question 14:

(a) $18\frac{3}{4}%$

(c) 1200

$=\left(x×\frac{35}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{35x}{100}$

#### Question 15:

(c) 1200

$=\left(x×\frac{35}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{35x}{100}$

(a) 50

(a) 50

(c) 120

(c) 120

(c) 0.25%

(c) 0.25%

(d) 300%

(d) 300%

(a) x

(a) x

(a) x

(a) x

#### Question 3:

#### Question 4:

$=28%$

#### Question 5:

$=28%$

$=\left(x×\frac{40}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{40x}{100}\phantom{\rule{0ex}{0ex}}$

#### Question 6:

$=\left(x×\frac{40}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{40x}{100}\phantom{\rule{0ex}{0ex}}$

(d) 10%

(d) 10%

(c) 120

(c) 120

(b) 15%

(b) 15%

(d) 120%

(d) 120%

(c) 80

(c) 80

(c) 240

#### Question 13:

(c) 240

#### Question 14: