RS Aggarwal solution class 8 chapter 10 Profit and Loss Exercise 10B

Exercise 10B

Page-138

Q1 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 1:

The marked price of a water cooler is Rs 4650. The shopkeeper offers an off-season discount of 18% on it. Find its selling price.

Answer 1:

Marked price = $\mathrm{Rs} 4650$ and discount = 18%
Discount = 18% of marked price
              $=18\% of Rs 4650$
             $$\begin{gathered} =Rs \left(4650\times \frac{18}{100}\right) \\ =Rs 837 \end{gathered}$$
Selling price = marked price − discount
                     $$\begin{gathered} =\mathrm{Rs} (4650 - 837) \\ =\mathrm{Rs} 3813 \end{gathered}$$

Therefore, the selling price of the cooler is $\mathrm{Rs} 3813$.

Q2 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 2:

The price of a sweater was slashed from Rs 960 to Rs 816 by a shopkeeper in the winter season. Find the rate of discount given by him.

Answer 2:

Marked Price = Rs 960
Selling Price = Rs 816
Discount = MP − SP
                = Rs (960 − 816)
                  = Rs 144
$\mathrm{Rate} \mathrm{of} discount=144\times \frac{100}{960}=15\%$

Therefore, the discount on the sweater is 15%.

Q3 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 3:

Find the rate of discount being given on a shirt whose selling price is ₹ 1092 after deducting a discount of ₹ 208 on its marked price.

Answer 3:

SP of the shirt = ₹1,092
Discount = ₹208
MP = SP + Discount = ₹1,092 + ₹208 = ₹1,300
∴ Rate of discount = $\frac{\mathrm{Discount}}{\mathrm{MP}}\times 100\%=\frac{208}{1300}\times 100\%=16\%$

Q4 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 4:

After allowing a discount of 8% on a toy, it is sold for Rs 216.20. Find the marked price of the toy.

Answer 4:

Selling Price = Rs 216.20
Rate of discount = 8%
Marked Price = ?
SP = MP − discount
Let the MP be Rs $x$.

$$\begin{gathered} \mathrm{Now}, x-\frac{8}{100}\times x=216.20 \\ \Rightarrow \frac{92x}{100}=216.20 \\ \Rightarrow 92x=21620 \\ \Rightarrow x=\frac{21620}{92} \\ \Rightarrow x=235 \end{gathered}$$

∴ Marked price = $Rs 235$

Q5 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 5:

A tea set was bought for Rs 528 after getting a discount of 12% on its marked price. Find the marked price of the tea set.

Answer 5:

Cost price = Rs 528
Rate of discount = 12%
Marked price = ?
SP= MP − discount
Let the MP be Rs $x$.
$$\begin{gathered} \mathrm{Now}, \frac{x-12}{100\times x}=528 \\ \Rightarrow \frac{88x}{100}=528 \\ \Rightarrow 88x=52800 \\ \Rightarrow x=\frac{52800}{88} \\ \Rightarrow x=\mathrm{Rs} 600 \end{gathered}$$

Therefore, the marked price of tea set is Rs 600.

Q6 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 6:

A dealer marks his goods at 35% above the cost price and allows a discount of 20% on the marked price. Find the gain or loss per cent.

Answer 6:

Let Rs 100 be the CP.
Then, marked price = $Rs 135$
Discount = 20% of MP
                $$\begin{gathered} =\frac{20}{100}\times 135 \\ =27 \end{gathered}$$
Selling price = marked price − discount
                     = 135 − 27
                     = Rs 108
Now, gain = SP − CP
                  =108 − 100
                   =Rs 8

$\therefore \mathrm{Gain} \mathrm{percentage}=\frac{\mathrm{gain}}{\mathrm{CP}}\times 100$

                            = $$\begin{gathered} \frac{8}{100}\times 100 \\ = 8\% \end{gathered}$$

Q7 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 7:

A cellphone was marked at 40% above the cost price and a discount of 30% was given on its marked price. Find the gain or loss per cent made by the shopkeeper.

Answer 7:

Let Rs 100 be the CP.
Then, marked price = $Rs 140$
Discount = 30% of MP
                $$\begin{gathered} =\frac{30}{100}\times 140 \\ =42 \end{gathered}$$
Selling Price = marked price − discount
                      = 140 − 42
                      = Rs 98
Now, loss = CP − SP
                 = 100 − 98
                 = Rs 2
$$\begin{gathered} \therefore Loss percentage=\frac{Loss\times 100}{CP} \\ =\frac{2\times 100}{100} \\ =2\% \end{gathered}$$

Therefore, the shopkeeper had a loss of 2%.

Q8 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 8:

A dealer purchased a fan for Rs 1080. After allowing a discount of 25% on its marked price, he gains 25%. Find the marked price of the fan.

Answer 8:

Cost price of the fan = $Rs 1080$
Gain percentage = 25%
$$\begin{gathered} \therefore \mathrm{Selling} \mathrm{price} = \left\{\frac{(100 +\hspace{0.17em}\mathrm{gain} \%)}{100}\times \mathrm{CP}\right\} \\ =\left\{\frac{100+25}{100}\times 1080\right\} \\ =\frac{125}{100}\times 1080 \\ =\mathrm{Rs} 1350 \end{gathered}$$

Let the marked price be Rs $x$.
Discount = 25% of $Rs x$

             $=\frac{25x}{100}$

SP = MP − discount
⇒ 1350 = $X$ − $\frac{25X}{100}$

$$\begin{gathered} \Rightarrow 1350=\frac{100x-25x}{100} \\ \Rightarrow 135000=75x\Rightarrow x=\frac{13500}{75}\Rightarrow x=1800 \end{gathered}$$

Therefore, the marked price of the fan is $Rs 1800$.

Q9 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 9:

A dealer bought a refrigerator for Rs 11515. After allowing a discount of 16% on its marked price, he gains 20%. Find the marked price of the refrigerator.

Answer 9:

Cost price of the refrigerator = $Rs 11515$
Gain percentage = 20%.
$$\begin{gathered} \therefore \mathrm{Selling} \mathrm{price\; =}\left\{\frac{(100 +\hspace{0.17em}\mathrm{gain} \%)}{100}\times C.P\right\} \\ =\left\{\frac{100+20}{100}\times 11515\right\} \\ =\frac{120}{100}\times 11515 \\ =\mathrm{Rs} 13818 \end{gathered}$$

Let the marked price be Rs $x$.
Discount = 16% of $Rs x$
                $=\frac{16x}{100}$
S.P = MP − Discount
⇒ 13818 = x − $\frac{16x}{100}$

$$\begin{gathered} \Rightarrow 13818=\frac{100x-16x}{100} \\ \Rightarrow 1381800=84x\Rightarrow x=\frac{1381800}{84}\Rightarrow x=16450 \end{gathered}$$

​Therefore, the marked price of the refrigerator is $Rs 16450$.

Q10 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 10:

A jeweller allows a discount of 16% to his customers and still gains 20%. Find the marked price of a ring which costs the jeweller Rs 1190.

Answer 10:

The cost price of the ring is $Rs 1190$.
Gain percentage = 20%.
$$\begin{gathered} \therefore \mathrm{Selling} \mathrm{price\; =}\left\{\frac{(100 +\hspace{0.17em}gain \%)}{100}\times C.P\right\} \\ =\left\{\frac{100+20}{100}\times 1190\right\} \\ =\frac{120}{100}\times 1190 \\ =\mathrm{Rs} 1428 \end{gathered}$$

Let the marked price be $x$.
Discount = 16% of $Rs x$
               $=\frac{16x}{100}$
SP = MP − Discount

$$\begin{gathered} \Rightarrow 1428=x-\frac{16x}{100} \\ \Rightarrow 1428=\frac{100x-16x}{100} \\ \Rightarrow 142800=84x \\ \Rightarrow \frac{142800}{84}=x \\ \Rightarrow x=1700 \end{gathered}$$

​Therefore, the marked price of the ring is $\mathrm{Rs} 1700$.

Q11 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 11:

After allowing a discount of 10% on the marked price, a trader still makes a gain of 17%. By what per cent is the marked price above the cost price?

Answer 11:

Let $Rs 100$ be the cost price.
Gain required = 17%
∴ Selling price = $Rs 117$
Let the marked price be $Rs x$.
Then, discount = 10% of x
                        $$\begin{gathered} =\frac{10}{100}\times x \\ =\frac{x}{10} \end{gathered}$$
Selling Price = MP − discount
$$\begin{gathered} \Rightarrow 117=x-\frac{x}{10} \\ \Rightarrow 117=\frac{9x}{10} \end{gathered}$$

$$\begin{gathered} \Rightarrow 9x=1170 \\ \Rightarrow x=\frac{1170}{9} \\ \Rightarrow x=130 \end{gathered}$$

∴ Marked price = $Rs 130$

Hence, the marked price is 30% above the cost price.

Q12 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 12:

How much per cent above the cost price should a shopkeeper mark his goods so that after allowing a discount of 10% on the marked price, he gains 8%?

Answer 12:

Let $Rs 100$ be the cost price.
Gain required = 8%
Therefore, the selling price is $Rs 108$.
Let $Rs x$ be the marked price.
Then, discount = 10% of x
                      $$\begin{gathered} =\frac{10}{100}\times x \\ =\frac{x}{10} \end{gathered}$$
Selling Price = MP − discount
$$\begin{gathered} \Rightarrow 117=x-\frac{x}{10} \\ \Rightarrow 117=\frac{9x}{10} \end{gathered}$$

$$\begin{gathered} \Rightarrow 9x=1080 \\ \Rightarrow x=\frac{1080}{9} \\ \Rightarrow x=120 \end{gathered}$$

∴ Marked price = $Rs 120$

Hence, the marked price is 20% above the cost price.

Q13 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 13:

The marked price of a TV is Rs 18500. A dealer allows two successive discounts of 20% and 5%. For how much is the TV available?

Answer 13:

Marked price of the TV = Rs 18500
First discount = 20%

$$\begin{gathered} \mathrm{Now}, 20\% of 18500 \\ =\frac{20}{100}\times 18500 \\ =\mathrm{Rs} 3700 \end{gathered}$$

Price after the first discount = Rs (18500 − 3700)= Rs 14800
Second discount = 5% of 14800
                                $$\begin{gathered} =\frac{5}{100}\times 14800 \\ =740 \end{gathered}$$
Price after the second discount = (14800 − 740)
                                                   = Rs 14060
The TV is available for $\mathrm{Rs} 14060.$

Q14 | Ex-10B | Class 8 | RS AGGARWAL | chapter 10 | Profit and Loss | myhelper

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Question 14:

Find the single discount which is equivalent to two successive discounts of 20% and 5%.

Answer 14:

​Let the marked price of the article be Rs 100.
First discount = 20%
Price after the first discount = (100 − 20) = Rs 80
Second discount = 5% of 80
                            $$\begin{gathered} =\frac{5}{100}\times 80 \\ =\mathrm{Rs} 4 \end{gathered}$$
Price after the second discount = (80 − 4) = Rs 76
Net selling price = Rs 76
∴ Single discount equivalent to the given successive discounts = (100 − 76)% = 24%