ML AGGARWAL CLASS 8 CHAPTER 7 Percentage Exercise 7.3

    Exercise 7.3

Question 1

Sol :

(i)

Marked price = 575, discount =12 %

Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$

$\begin{array}{l}12=\frac{\text { Discont }}{575} \times 100 \\ \text { Discount }=\frac{12 \times 575}{100}\end{array}$

Discount =₹69

Selling price = Marked price - Discount =575-69

=₹506

(ii)  Printed price = 12,750, discount $=8 \frac{1}{3} \%$

Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$

$\frac{25}{3}=\frac{D \text { discount }}{12,750} \times 100$

Discount $=\frac{25}{3} \times \frac{12,750}{1004}=1062.5$

Selling price = Marked price or printed price - discount

=12,750-1062.5

=11,687.5

Question 2

Sol :

(i) Marked price=780, 

Selling price =2721.5

Selling price = Marked price - Discount

Discount = Marked price - Selling price

=780-721.5

=58.5

Discount percentage $=\frac{\text { Discount }}{\text { Marked price }} \times 100 \%$

$=\frac{58.5}{780} \times 100 \%$

=7.5 %

(ii) Advertised price (or)

Marked price =₹28,500

selling price =₹24510$.

Selling price = Marked rice - Discount

$\begin{aligned} \text { Dis count } &=\text { Marked price - selling price } \\ &=28,500-24,510 \\ &=3990 \\ \text { Discount. Percentage } &=\frac{\text { Discount }}{\text { marked price }} \times 100 \end{aligned}$

$=\frac{3990}{28500} \times 100 \%$

=14 %

Question 3

Sol :

Marked price =₹ 30. (each)

Discount % $=\frac{\text { Discount }}{\text { Marked price }} \times 100$

Discount $=\frac{15 \times 30}{100}=4.5$

Discount on one note book =4.5

Discount on dozen note books $=12 \times 4.5=54$

Marked price on note books $=12 \times 30=360$

Selling price = Marked price - Discount

=360-54

=₹ 306

Question 4

Sol :

Selling price = Rs 728    

Discount = 9 % 

$\begin{aligned} s \cdot p &=\left(1-\frac{d}{100}\right) \text { of } M \cdot p \\ 7 28 &=\left(1-\frac{9}{100}\right) \times M \cdot \rho \\ M \cdot P &=\frac{728 \times 100}{91}=₹ 800\end{aligned}$

Question 5

Sol :

Marked price =₹ 800

Discount =20 %

(i) Selling price =?

$\begin{aligned} \text { Discount } & \%=\frac{\text { Discant }}{M \cdot \rho} \times 100 \\ &(O R) \\ S \cdot P &=\left[1-\frac{d}{100}\right] \text { of } M \cdot p \\=&\left[1-\frac{20}{100}\right] \times 800 \\ &=\frac{80}{100} \times 800 \\ S \cdot p &=₹640\end{aligned}$

(ii)  profit = 25 %

SP=\left[1+\frac{p}{100}\right]$ of P

$640=\left[1+\frac{25}{100}\right] \times(p)$

$CP=\frac{640 \times 100}{125}$

C . P= ₹512

Question 6

Sol :

Marked price = 2,250 

Discount =12 % profit =10 %

(i) $\begin{aligned} SP &=\left[1-\frac{d}{100}\right] \text { of } M \cdot p \\ &=\left[1-\frac{12}{100}\right] \times 2250 \\ &=\frac{88}{100} \times 2250 \end{aligned}$

=₹ 1980

(ii) $S.P=\left[1+\frac{P}{100}\right]$ of $C.P$

$1980=\left[1+\frac{10}{100}\right] \times C .P$

$\begin{aligned} C.P&=\frac{1980 \times 100}{110} \\ &=₹1800 \end{aligned}$

Question 7

Sol :

Cost price = Rs 650 , 

Discount = 20% ,  

Profit = 20 % 

(i) $\begin{aligned} S.P &=\left[1+\frac{p}{100}\right] \text { of } C \cdot p \\ &=\left[1+\frac{20}{100}\right] \times 650 \\ &=\frac{120}{109} \times 65 \\ S \cdot P &=780 \end{aligned}$

(ii) $S \cdot P=\left[1-\frac{d}{100}\right]$ of MP

$780=\left[1-\frac{20}{100}\right] \times MP$

$M.P=\frac{780 \times 100}{80}$

M.P=₹975

Question 8

Sol :

Cost price = 1200  ,  

Profit = 80 %  ,  

Discount = 15%  

(i) $\begin{aligned} \text { Marked price } &=1200+(80 \%+(p)\\ &=1200+\left(\frac{80}{100} \times 1200\right) \\ &=1200+960 \\ &=2160\end{aligned}$

$\left[1-\frac{d}{100}\right] \times MP$

$\left[1-\frac{15}{100}\right] \times 2160$

$\frac{85}{100} \times 2160$

=1836

(iii)

 $\begin{aligned} \text { profil percentage } &=\left(\frac{S \cdot P-C .P}{c .p} \times 100\right) \% \\ &=\left(\frac{1836-1200}{1200} \times 10\right) \% \\ &=\frac{636}{1200} \times 100 \\ &=53 \% \end{aligned}$

Question 9

Sol :

Cost price of an article = ₹ 1600

(i) since the cost price is 20 %  below the marked price 

C.P=MP-20% of r

$1600=M \cdot P-\frac{20}{100} \times M \cdot P$

$\begin{aligned} 1600 &=\left[1-\frac{20}{100}\right] \times \mathrm{M} \cdot P \\ M \cdot P &=\frac{1600 \times 100}{80} \\ M \cdot P &= 2000 \end{aligned}$

(ii) Discount =16%

$\begin{aligned} S \cdot P=&\left[1-\frac{d}{100}\right] \times \mathrm{M} \cdot \mathrm{p} \\=&\left[1-\frac{16}{100}\right] \times 2000 . \\ &=\frac{8 y}{100} \times 2000 \\ S \cdot p &=1680 \end{aligned}$

(iii) 

$\begin{aligned} \text { profit percentage } &=\left[\frac{s .p}{C{.p}} \times 100\right] \% \\ &=\left[\frac{1680-1100}{1600} \times 100\right] \% \\ &=\frac{80}{1600} \times 100 \% \\ &=5 \% \end{aligned}$

Question 10

Sol :

(i)

Discount =20 % ,Profit =20%   ,Selling Price = 360

$SP=\left[1-\frac{d}{100}\right]$ of MP

$360=\left[1-\frac{20}{100}\right] \times M \cdot 1$

$MP=\frac{360 \times 100}{80}$

MP=450

(ii) $SP=\left[1+\frac{p}{100}\right]$ of CP

$360=\left[1+\frac{20}{1 \cdot 0}\right] \times(\cdot p)$

$C \cdot p=\frac{360 \times 100}{120}$

CP=₹ 300

Question 11

Sol :

Marked price of a refrigerator =28,600

The selling price of a refrigerator is

$=\left[1-\frac{10}{100}\right]\left[1-\frac{5}{100}\right]$ of $\mathrm{M} \cdot \mathrm{p}$

$=\frac{90}{100} \times \frac{95}{10^{\circ}} \times 28600$

= 24,453

Question 12

Sol :

Let the market price be 'x'

first dealer 

$\begin{aligned} \text { S. } P &=\left[1-\frac{15}{100}\right]\left[1-\frac{5}{100}\right] \text { of } \mathrm{M} \cdot \mathrm{P} \\ &=\frac{85}{100} \times \frac{95}{100} \times x \\ &=\frac{17 \times 19 \times x}{20 \times 20}=0.8075 \mathrm{x} \end{aligned}$

Second dealer ;

$\begin{aligned} s \cdot p &=\left[1-\frac{20}{100}\right] \text { of } m \cdot p \\ &=\frac{80}{100} \times x \\ &=\frac{4 x}{5}=0.8 x \end{aligned}$

As the second dealer offer price is less compared to first order

So the second dealer is best offer

Question 13

Sol :

Let the marked price of an article be $x^{\prime}$.

and a single discount of $d \%$ be equivalent to

two given successive discounts of $30 \%$ and $10 \%$, then

$\left(1-\frac{d}{100}\right)$ of $\bar{₹} x=\left(1-\frac{36}{100}\right)\left(1-\frac{10}{100}\right) \overline{₹} x$

$1-\frac{d}{100}=\frac{78}{100} \times \frac{90}{100}$

$1-\frac{d}{100}=\frac{63}{100}$

$\frac{d}{100}=\frac{37}{100}$

d=37

hence a discount of 37 % is equivalent to two given successive discount