Profit and Loss Questions in English

(A) Given: Cost Price $(CP)$ = $Rs. 27.50$,Selling Price $(SP)$ = $Rs. 28.60$.
Gain = $SP - CP = 28.60 - 27.50 = Rs. 1.10$.
Gain percent is calculated using the formula: $\text{Gain} \% = \left( \frac{\text{Gain}}{CP} \times 100 \right) \%$.
Substituting the values: $\text{Gain} \% = \left( \frac{1.10}{27.50} \times 100 \right) \% = \left( \frac{110}{2750} \times 100 \right) \% = \left( \frac{11000}{2750} \right) \% = 4 \%$.

(B) Given: Cost Price $(CP)$ = $Rs. 490$,Selling Price $(SP)$ = $Rs. 465.50$.
Since $CP > SP$,there is a loss.
$\text{Loss} = CP - SP = 490 - 465.50 = Rs. 24.50$.
$\text{Loss} \% = \left( \frac{\text{Loss}}{CP} \times 100 \right) \%$.
$\text{Loss} \% = \left( \frac{24.50}{490} \times 100 \right) \% = \left( \frac{2450}{490} \right) \% = 5\%$.

(B) The formula to calculate the Selling Price $(SP)$ when Cost Price $(CP)$ and Gain percentage are given is:
$SP = \left[ \frac{100 + \text{Gain}\%}{100} \right] \times CP$
Given:
$CP = Rs. 56.25$
$\text{Gain}\% = 20\%$
Substituting the values into the formula:
$SP = \left[ \frac{100 + 20}{100} \right] \times 56.25$
$SP = \left[ \frac{120}{100} \right] \times 56.25$
$SP = 1.2 \times 56.25$
$SP = 67.5$
Thus,the Selling Price is $Rs. 67.5$.

(C) The formula for Selling Price $(SP)$ when there is a loss is given by:
$SP = \left[\frac{100 - \text{loss}\%}{100}\right] \times CP$
Given $CP = Rs. 80.40$ and $\text{loss}\% = 5\%$.
Substituting the values into the formula:
$SP = \left[\frac{100 - 5}{100}\right] \times 80.40$
$SP = \left[\frac{95}{100}\right] \times 80.40$
$SP = 0.95 \times 80.40$
$SP = Rs. 76.38$

(A) The formula to calculate the Cost Price $(CP)$ when the Selling Price $(SP)$ and gain percentage are given is:
$CP = \frac{100 \times SP}{100 + \text{gain}\%}$
Given:
$SP = Rs. 40.60$
$\text{gain}\% = 16\%$
Substituting the values into the formula:
$CP = \frac{100 \times 40.60}{100 + 16}$
$CP = \frac{4060}{116}$
$CP = 35$
Therefore,the Cost Price is $Rs. 35$.

(A) The formula to calculate the Cost Price $(CP)$ when the Selling Price $(SP)$ and loss percentage are given is:
$CP = \frac{100 \times SP}{100 - \text{loss}\%}$
Given:
$SP = Rs. 51.70$
$\text{loss}\% = 12\%$
Substituting the values in the formula:
$CP = \frac{100 \times 51.70}{100 - 12}$
$CP = \frac{5170}{88}$
$CP = Rs. 58.75$

(C) Let the Cost Price $(CP)$ of the watch be $C$.
Given that the Selling Price $(SP_1)$ is $Rs. 1140$ with a loss of $5 \%$.
Using the formula $SP = CP \times (1 - \text{loss} \% / 100)$:
$1140 = C \times (1 - 5/100) = C \times (95/100)$.
$C = (1140 \times 100) / 95 = 1200$.
Now,to earn a $5 \%$ profit,the new Selling Price $(SP_2)$ is calculated as:
$SP_2 = CP \times (1 + \text{profit} \% / 100) = 1200 \times (1 + 5/100) = 1200 \times (105/100) = 1260$.
Therefore,the watch should be sold for $Rs. 1260$ to earn a $5 \%$ profit.

(D) Let the selling price $(SP)$ be $Rs. 100$.
Given that the cost price $(CP)$ is $96 \%$ of the selling price.
Therefore,$CP = 96 \% \text{ of } 100 = Rs. 96$.
Profit is calculated as $Profit = SP - CP$.
$Profit = 100 - 96 = Rs. 4$.
Profit percentage is calculated using the formula: $\text{Profit } \% = (\frac{\text{Profit}}{CP}) \times 100$.
$\text{Profit } \% = (\frac{4}{96}) \times 100 = \frac{1}{24} \times 100 = 4.166... \% \approx 4.17 \%$.
Since $4.17 \%$ is not among the given options,the correct choice is $D$.

(C) True weight $= 1000 \, g$.
False weight $= 960 \, g$.
Profit is earned on the weight actually sold (the false weight).
Gain $= 1000 - 960 = 40 \, g$.
Gain $\% = \left( \frac{\text{Gain}}{\text{False weight}} \right) \times 100 \%$.
Gain $\% = \left( \frac{40}{960} \right) \times 100 \%$.
Gain $\% = \frac{1}{24} \times 100 \% = \frac{25}{6} \% \approx 4.166 \% \approx 4.17 \%$.

(D) When two items are sold at the same selling price,and there is a gain of $x\%$ on one and a loss of $x\%$ on the other,the overall transaction always results in a loss.
The formula for the net loss percentage is given by:
$\text{Loss } \% = \left( \frac{x}{10} \right)^2 = \frac{x^2}{100}$
Given that $x = 10$,we substitute the value into the formula:
$\text{Loss } \% = \frac{10^2}{100} = \frac{100}{100} = 1\%$
Therefore,the man incurs a loss of $1\%$.

(C) To find the equivalent single discount for two successive discounts of $a \%$ and $b \%$,we use the net discount formula:
$\text{Net Discount} = \left(a + b - \frac{a \times b}{100}\right) \%$
Here,$a = 40$ and $b = 20$.
Substituting these values into the formula:
$\text{Net Discount} = \left(40 + 20 - \frac{40 \times 20}{100}\right) \%$
$= (60 - \frac{800}{100}) \%$
$= (60 - 8) \% = 52 \%$
Therefore,two successive discounts of $40 \%$ and $20 \%$ are equivalent to a single discount of $52 \%$.

(D) Total Cost Price $(CP)$ for $5$ watches = $Rs. 9450$.
Total Selling Price $(SP)$ for $5$ watches = $Rs. 9700$.
Total Profit = $SP - CP = 9700 - 9450 = Rs. 250$.
Profit per watch = $\frac{\text{Total Profit}}{\text{Number of watches}} = \frac{250}{5} = Rs. 50$.
Since $Rs. 50$ is not among the given options,the correct answer is $D$ (None of these).

(A) Let the cost of one chair be $x$ and the cost of one table be $y$.
Given: $12x + 8y = 676$.
Divide the entire equation by $4$:
$3x + 2y = \frac{676}{4} = 169$.
We need to find the price of $21$ chairs and $14$ tables,which is $21x + 14y$.
Factor out $7$ from the expression:
$21x + 14y = 7(3x + 2y)$.
Substitute the value of $(3x + 2y) = 169$ into the expression:
$7 \times 169 = 1183$.
Therefore,the price of $21$ chairs and $14$ tables is $Rs. 1183$.

(D) Let the cost price $(CP)$ of the $TV$ be $Rs. 100.$
Since Aditya sold it at $12 \%$ more than the $CP,$ the selling price $(SP)$ for Sanjay is $100 + 12 = Rs. 112.$
Given that Sanjay paid $Rs. 17696,$ we have $SP = Rs. 17696.$
Using the unitary method:
If $SP$ is $112,$ then $CP$ is $100.$
If $SP$ is $17696,$ then $CP = (100 / 112) \times 17696.$
$CP = 158 \times 100 = Rs. 15800.$
Therefore,the original price of the $TV$ was $Rs. 15800.$

(B) Total Selling Price $(SP)$ $= Rs. 1220$.
Total Cost Price $(CP)$ of $13$ chairs $= 13 \times 115 = Rs. 1495$.
Since $CP > SP$,there is a loss on the transaction.
Loss $= CP - SP = 1495 - 1220 = Rs. 275$.

(A) Given that the marked price $(MP)$ of the book is $Rs. 500$.
The discount percentage is $20 \%$.
Discount amount $= 20 \% \text{ of } 500 = \frac{20}{100} \times 500 = Rs. 100$.
Amount paid $= MP - \text{Discount} = 500 - 100 = Rs. 400$.
Therefore,Aditya paid $Rs. 400$ for the book.

(A) Given: Selling Price $(SP)$ = $Rs. 360$ and Profit percentage = $20\%$.
The formula for profit percentage is: $\text{Profit}\% = \left[\frac{SP - CP}{CP}\right] \times 100$.
Substituting the values: $20 = \left[\frac{360 - CP}{CP}\right] \times 100$.
Dividing both sides by $100$: $0.2 = \frac{360 - CP}{CP}$.
Multiplying by $CP$: $0.2 \times CP = 360 - CP$.
Adding $CP$ to both sides: $1.2 \times CP = 360$.
Solving for $CP$: $CP = \frac{360}{1.2} = 300$.
Therefore,the Cost Price $(CP)$ of the book is $Rs. 300$.

(A) Let the Cost Price be $CP$.
Given that the profit earned by selling at $Rs. 1630$ is equal to the loss incurred by selling at $Rs. 1320$.
Profit $= SP_1 - CP = 1630 - CP$.
Loss $= CP - SP_2 = CP - 1320$.
Since Profit $=$ Loss,we have:
$1630 - CP = CP - 1320$.
$2 \times CP = 1630 + 1320$.
$2 \times CP = 2950$.
$CP = \frac{2950}{2} = 1475$.
Therefore,the Cost Price is $Rs. 1475$.

(C) Given that the $CP$ of $50$ items is equal to the $SP$ of $40$ items.
Let the $CP$ of $1$ item be $x$.
Then,$CP$ of $50$ items $= 50x$.
$SP$ of $40$ items $= 50x$.
Therefore,$SP$ of $1$ item $= \frac{50x}{40} = 1.25x$.
Since $SP > CP$,there is a profit.
Profit $= SP - CP = 1.25x - x = 0.25x$.
Profit $\% = \left( \frac{\text{Profit}}{CP} \right) \times 100 = \left( \frac{0.25x}{x} \right) \times 100 = 25 \%$.

(A) Cost of $1$ banana $= Rs. 1.25$
Cost of $1$ apple $= Rs. 1.75$
Since $1$ dozen $= 12$ units,$2$ dozen bananas $= 2 \times 12 = 24$ bananas.
Cost of $2$ dozen bananas $= 24 \times 1.25 = Rs. 30$
$3$ dozen apples $= 3 \times 12 = 36$ apples.
Cost of $3$ dozen apples $= 36 \times 1.75 = Rs. 63$
Total cost $= Rs. 30 + Rs. 63 = Rs. 93$

(C) Let the marked price of the watch be $MP$.
Given that the discount is $24 \%$.
The selling price $(SP)$ is calculated as: $SP = MP \times (1 - \text{Discount} \%)$.
$779 = MP \times (1 - 0.24)$.
$779 = MP \times 0.76$.
$MP = \frac{779}{0.76}$.
$MP = \frac{77900}{76} = 1025$.
Therefore,the marked price of the watch is $Rs. 1025$.

(B) Let the cost of $1$ calculator be $Rs. C$ and the cost of $1$ pen be $Rs. P$.
According to the problem:
$3C + 4P = 2140$ $(i)$
$1C + 5P = 1355$ $(ii)$
To solve for $C$,multiply equation $(ii)$ by $4$:
$4C + 20P = 5420$ $(iii)$
Multiply equation $(i)$ by $5$:
$15C + 20P = 10700$ $(iv)$
Subtract equation $(iii)$ from $(iv)$:
$(15C - 4C) + (20P - 20P) = 10700 - 5420$
$11C = 5280$
$C = 5280 / 11 = 480$
Therefore,the cost of $1$ calculator is $Rs. 480$.

(C) Cost Price $(CP)$ of $6$ toffees $= Rs. 1$.
Cost Price $(CP)$ of $1$ toffee $= Rs. \frac{1}{6}$.
Let the number of toffees sold for $Rs. 1$ be $x$.
Selling Price $(SP)$ of $x$ toffees $= Rs. 1$.
Selling Price $(SP)$ of $1$ toffee $= Rs. \frac{1}{x}$.
Given,$\text{Gain} \% = 20 \%$.
We know that $\text{Gain} \% = \frac{SP - CP}{CP} \times 100$.
$20 = \frac{\frac{1}{x} - \frac{1}{6}}{\frac{1}{6}} \times 100$.
$\frac{20}{100} = \frac{\frac{6-x}{6x}}{\frac{1}{6}}$.
$\frac{1}{5} = \frac{6-x}{6x} \times 6$.
$\frac{1}{5} = \frac{6-x}{x}$.
$x = 5(6 - x)$.
$x = 30 - 5x$.
$6x = 30$.
$x = 5$.
Therefore,the vendor must sell $5$ toffees for a rupee to gain $20 \%$.

(B) When two items are sold at the same selling price,and there is a profit percentage $P\%$ on one and a loss percentage $L\%$ on the other,where $P = L$,there is always a net loss.
The formula for net loss percentage is given by: $\text{Net Loss } \% = \left( \frac{x^2}{100} \right) \%$,where $x$ is the common profit or loss percentage.
Given $x = 10$,
$\text{Net Loss } \% = \frac{10^2}{100} \% = \frac{100}{100} \% = 1\%$.
Therefore,the man incurs a $1\%$ loss.

(D) Let the cost price $(CP)$ be $Rs. 1$.
Given that the selling price $(SP)$ is $\frac{4}{3}$ of the cost price,so $SP = \frac{4}{3} \times 1 = Rs. \frac{4}{3}$.
Profit is calculated as $SP - CP = \frac{4}{3} - 1 = \frac{1}{3}$.
Profit percentage is calculated as $\left( \frac{\text{Profit}}{CP} \right) \times 100$.
Profit $\% = \left( \frac{1/3}{1} \right) \times 100 = \frac{100}{3} = 33.33\% \approx 33.3\%$.

(B) Step $1$: $A$ sells the house to $B$ at a $10\%$ loss.
Cost Price $(CP)$ for $A = Rs. 10,00,000$.
Selling Price $(SP)$ for $A = 10,00,000 - (10\% \text{ of } 10,00,000) = 10,00,000 - 1,00,000 = Rs. 9,00,000$.
So,$B$ buys the house for $Rs. 9,00,000$.
Step $2$: $B$ sells the house back to $A$ at a $10\%$ profit.
$CP$ for $B = Rs. 9,00,000$.
$SP$ for $B = 9,00,000 + (10\% \text{ of } 9,00,000) = 9,00,000 + 90,000 = Rs. 9,90,000$.
So,$A$ buys the house back for $Rs. 9,90,000$.
Step $3$: Calculate the net result for $A$.
Initially,$A$ had the house and $Rs. 0$ cash. After selling,$A$ had $Rs. 9,00,000$ cash.
After buying it back,$A$ has the house and $(9,00,000 - 9,90,000) = -Rs. 90,000$ cash.
This means $A$ spent $Rs. 90,000$ more than what he received,effectively losing $Rs. 90,000$.

(D) Let the initial cost price of the goods be $100$ units.
$1$. The shopkeeper marks up the price by $10 \%$,so the expected selling price is $100 + 10 = 110$ units.
$2$. He loses $20 \%$ of his goods due to theft. This means he only has $80 \%$ of the goods left to sell.
$3$. The cost price of the remaining goods is $80$ units.
$4$. Since he sells the remaining goods at the marked price,the total revenue generated is $80 \times (1 + 10/100) = 80 \times 1.1 = 88$ units.
$5$. The total cost price was $100$ units,and the final revenue is $88$ units.
$6$. Loss = $\text{Cost Price} - \text{Revenue} = 100 - 88 = 12$ units.
$7$. Loss percentage = $(12 / 100) \times 100 = 12 \%$.
Alternatively,using the net percentage change formula for successive changes: $x + y + (xy/100)$,where $x = +10$ (profit) and $y = -20$ (loss of goods):
$\Rightarrow 10 - 20 + \frac{10 \times (-20)}{100} = -10 - 2 = -12 \%$.
The negative sign indicates a loss of $12 \%$.

(C) Total Cost Price $(CP)$ = (Cost of oranges) + (Transportation cost)
$CP = (200 \times 10) + 500 = 2000 + 500 = Rs. 2500$
Total Selling Price $(SP)$ = (Number of oranges) $\times$ (Selling price per orange)
Since $1$ dozen = $12$ oranges,total oranges = $200 \times 12 = 2400$.
$SP = 2400 \times 1 = Rs. 2400$
Since $CP > SP$,there is a loss.
Loss = $CP - SP = 2500 - 2400 = Rs. 100$
Loss $\%$ = $\frac{\text{Loss}}{CP} \times 100 = \frac{100}{2500} \times 100 = 4 \%$

(B) Cost Price $(CP)$ of $11$ mangoes $= Rs. 10$.
Cost Price $(CP)$ of $1$ mango $= Rs. \frac{10}{11}$.
Selling Price $(SP)$ of $10$ mangoes $= Rs. 11$.
Selling Price $(SP)$ of $1$ mango $= Rs. \frac{11}{10}$.
Since $SP > CP$,there is a gain.
Gain $= SP - CP = \frac{11}{10} - \frac{10}{11} = \frac{121 - 100}{110} = Rs. \frac{21}{110}$.
Gain $\% = \left( \frac{\text{Gain}}{CP} \times 100 \right) \% = \left( \frac{21/110}{10/11} \times 100 \right) \% = \left( \frac{21}{110} \times \frac{11}{10} \times 100 \right) \% = \left( \frac{21}{100} \times 100 \right) \% = 21\%$.

(B) Let the cost price $(CP)$ of $1$ article be $1$ unit.
Then,the $CP$ of $20$ articles $= 20$ units.
According to the problem,the selling price $(SP)$ of $x$ articles is equal to the $CP$ of $20$ articles,so $SP$ of $x$ articles $= 20$ units.
Therefore,the $SP$ of $1$ article $= \frac{20}{x}$ units.
Profit percentage is given as $25 \%$.
Profit $\% = \frac{SP - CP}{CP} \times 100$
$25 = \frac{(\frac{20}{x} - 1)}{1} \times 100$
$0.25 = \frac{20}{x} - 1$
$1.25 = \frac{20}{x}$
$x = \frac{20}{1.25} = \frac{2000}{125} = 16$.
Thus,the value of $x$ is $16$.

(B) Let the cost price be $CP$ and the initial selling price be $SP$. Let the initial profit be $P$.
We know that $P = SP - CP$ $....(i)$
According to the problem,when the selling price is doubled $(2SP)$,the profit becomes $3P$.
So,$2SP - CP = 3P$ $....(ii)$
From equation $(i)$,we have $SP = P + CP$.
Substitute this into equation $(ii)$:
$2(P + CP) - CP = 3P$
$2P + 2CP - CP = 3P$
$CP = P$
Since $CP = P$,the profit percentage is calculated as:
Profit $\%$ $= (P / CP) \times 100 = (CP / CP) \times 100 = 100 \%$.

(D) Cost Price $(CP)$ of $6$ articles $= Rs. 5$.
Therefore,$CP$ of $1$ article $= Rs. \frac{5}{6}$.
Selling Price $(SP)$ of $5$ articles $= Rs. 6$.
Therefore,$SP$ of $1$ article $= Rs. \frac{6}{5}$.
Gain $= SP - CP = \frac{6}{5} - \frac{5}{6} = \frac{36 - 25}{30} = \frac{11}{30}$.
Gain percent $= \left( \frac{\text{Gain}}{CP} \right) \times 100 = \left( \frac{11/30}{5/6} \right) \times 100 = \left( \frac{11}{30} \times \frac{6}{5} \right) \times 100 = \left( \frac{11}{25} \right) \times 100 = 44\%.$

(C) Let the cost price $(CP)$ of $1$ table be $x$.
Then,the $CP$ of $12$ tables $= 12x$.
According to the problem,the selling price $(SP)$ of $16$ tables $= 12x$.
Therefore,the $SP$ of $1$ table $= \frac{12x}{16} = 0.75x$.
Since $SP < CP$,there is a loss.
Loss $= CP - SP = x - 0.75x = 0.25x$.
Loss percent $= (\frac{\text{Loss}}{CP}) \times 100 = (\frac{0.25x}{x}) \times 100 = 25\%$.

(C) To find the equivalent single discount for two successive discounts of $a \%$ and $b \%$,we use the formula:
$\text{Net Discount} = \left( a + b - \frac{a \times b}{100} \right) \%$
Given $a = 4$ and $b = 4$:
$\text{Net Discount} = \left( 4 + 4 - \frac{4 \times 4}{100} \right) \%$
$= \left( 8 - \frac{16}{100} \right) \%$
$= (8 - 0.16) \% = 7.84 \%$
Therefore,two successive discounts of $4 \%$ are equivalent to a single discount of $7.84 \%$.

(B) Let the cost price $(CP)$ for $A$ be $Rs. x$.
$A$ sells the bicycle to $B$ at a profit of $20 \%$,so the selling price for $A$ (which is the cost price for $B$) is:
$CP_B = x + 0.20x = 1.20x$.
$B$ sells the bicycle to $C$ at a profit of $25 \%$,so the selling price for $B$ (which is the cost price for $C$) is:
$CP_C = 1.20x + 0.25(1.20x) = 1.20x + 0.30x = 1.50x$.
Given that $C$ pays $Rs. 1500$,we have:
$1.50x = 1500$.
Solving for $x$:
$x = \frac{1500}{1.50} = 1000$.
Therefore,$A$ paid $Rs. 1000$ for the bicycle.

(D) The total cost price $(CP)$ is the sum of the purchase price,repair cost,and cartage.
$CP = 70 + 17 + 0.50 = 87.50$ $Rs$.
Selling price $(SP)$ = $100$ $Rs$.
Profit = $SP - CP = 100 - 87.50 = 12.50$ $Rs$.
Profit percentage = $\left( \frac{\text{Profit}}{CP} \right) \times 100$.
Profit percentage = $\left( \frac{12.50}{87.50} \right) \times 100 = \frac{1}{7} \times 100 \approx 14.2857 \%$.
Rounding to one decimal place,the approximate profit percentage is $14.3 \%$.

(D) Let the $CP$ be $Rs. 100$.
Since the shopkeeper marks his goods $20 \%$ above $CP$,the $MP = 100 + 20 = Rs. 120$.
He allows a $30 \%$ discount on the $MP$.
$SP = MP \times (1 - \text{Discount } \%) = 120 \times (1 - 0.30) = 120 \times 0.70 = Rs. 84$.
Since $SP < CP$,there is a loss.
$\text{Loss} = CP - SP = 100 - 84 = 16$.
$\text{Loss } \% = \frac{\text{Loss}}{CP} \times 100 = \frac{16}{100} \times 100 = 16 \%$.

(D) To find the equivalent single discount for two successive discounts of $d_1$ and $d_2$,we use the formula:
$D_{eq} = d_1 + d_2 - \frac{d_1 \times d_2}{100}$
Given $d_1 = 40$ and $d_2 = 30$.
Substituting these values into the formula:
$D_{eq} = 40 + 30 - \frac{40 \times 30}{100}$
$D_{eq} = 70 - \frac{1200}{100}$
$D_{eq} = 70 - 12 = 58 \%$
Therefore,the single equivalent discount is $58 \%$.

(D) Given that the $CP$ of $13$ bats is $Rs. 390$.
First,calculate the $CP$ of $1$ bat: $CP = 390 / 13 = Rs. 30$.
However,the question asks for the total selling price of the $13$ bats at a $10 \%$ loss.
$\text{Loss} \% = 10 \%$.
$\text{Loss} = 10 \% \text{ of } 390 = (10 / 100) \times 390 = Rs. 39$.
$\text{Selling Price (SP)} = CP - \text{Loss} = 390 - 39 = Rs. 351$.
Alternatively,$SP = CP \times (100 - \text{Loss} \%) / 100 = 390 \times (90 / 100) = 39 \times 9 = Rs. 351$.

(A) Given that the Selling Price $(SP)$ is $Rs. 924$ and the Profit percentage is $10\%$.
We know the formula: $SP = CP \times (1 + \frac{\text{Profit}\%}{100})$.
Substituting the values: $924 = CP \times (1 + \frac{10}{100})$.
$924 = CP \times (1 + 0.1) = CP \times 1.1$.
$CP = \frac{924}{1.1} = \frac{9240}{11}$.
$CP = 840$.
Therefore,the cost price is $Rs. 840$.

(B) Let the Selling Price $(SP)$ be the price before sales tax.
Given that the cost including sales tax is $Rs. 616$ and the sales tax rate is $10 \%$.
$SP + 10 \% \text{ of } SP = 616$
$1.10 \times SP = 616$
$SP = \frac{616}{1.10} = Rs. 560$
Now,the shopkeeper makes a profit of $12 \%$ on the Cost Price $(CP)$.
$SP = CP \times (1 + \frac{\text{profit } \%}{100})$
$560 = CP \times (1 + \frac{12}{100})$
$560 = CP \times 1.12$
$CP = \frac{560}{1.12} = Rs. 500$

(C) Let the total value of the consignment be $Rs. x$.
The value of the first part sold is $\frac{2}{3}x$,and it is sold at a profit of $5 \%$.
Profit on the first part $= \frac{2}{3}x \times \frac{5}{100} = \frac{10x}{300} = \frac{x}{30}$.
The value of the remaining part is $x - \frac{2}{3}x = \frac{1}{3}x$,and it is sold at a loss of $2 \%$.
Loss on the second part $= \frac{1}{3}x \times \frac{2}{100} = \frac{2x}{300} = \frac{x}{150}$.
Given that the total profit is $Rs. 400$,we have:
$\frac{x}{30} - \frac{x}{150} = 400$
To solve for $x$,find a common denominator:
$\frac{5x - x}{150} = 400$
$\frac{4x}{150} = 400$
$4x = 400 \times 150$
$4x = 60000$
$x = 15000$
Therefore,the total value of the consignment was $Rs. 15000$.

(D) Let the $CP$ of each article be $Rs. 100$.
Then the $CP$ of $16$ articles $= 16 \times 100 = Rs. 1600$.
Since the trader gains $35 \%$,the total $SP$ for $16$ articles is $1600 \times 1.35 = Rs. 2160$.
Since the customer gets $1$ article free for every $15$ bought,the customer receives $16$ articles for the price of $15$.
Thus,the $SP$ of $15$ articles is $Rs. 2160$,which means the $SP$ of $1$ article is $2160 / 15 = Rs. 144$.
Given a $4 \%$ discount on the marked price $(MP)$,we have $SP = MP \times (1 - 0.04) = 0.96 \times MP$.
$144 = 0.96 \times MP \implies MP = 144 / 0.96 = Rs. 150$.
The $MP$ is $Rs. 150$ and the $CP$ is $Rs. 100$.
Therefore,the $MP$ is $(150 - 100) / 100 \times 100 = 50 \%$ above the $CP$.

(C) Let the retail price be $Rs. 100$.
Initial concession $= 36 \%$ of $100 = Rs. 36$.
Initial selling price $(SP_1) = 100 - 36 = Rs. 64$.
Given profit $= 8.8 \%$.
Cost price $(CP) = \frac{100 \times SP_1}{100 + \text{gain } \%} = \frac{100 \times 64}{100 + 8.8} = \frac{6400}{108.8} = \frac{64000}{1088} = Rs. \frac{1000}{17}$.
The concession is reduced by $24 \%$. This means the new concession is $36 - 24 = 12 \%$.
New selling price $(SP_2) = 100 - 12 = Rs. 88$.
New profit $= SP_2 - CP = 88 - \frac{1000}{17} = \frac{1496 - 1000}{17} = Rs. \frac{496}{17}$.
New profit percentage $= \left( \frac{\text{Profit}}{CP} \right) \times 100 = \left( \frac{496/17}{1000/17} \right) \times 100 = \frac{496}{1000} \times 100 = 49.6 \%$.

(A) Let the two investments be $3x$ and $5x$ respectively.
Total investment $= 3x + 5x = 8x$.
Gain on the first investment $= 15 \%$ of $3x = 0.15 \times 3x = 0.45x$.
Loss on the second investment $= 10 \%$ of $5x = 0.10 \times 5x = 0.50x$.
Net result $= 0.45x - 0.50x = -0.05x$.
The negative sign indicates a loss.
Loss percentage $= \frac{\text{Total Loss}}{\text{Total Investment}} \times 100 = \frac{0.05x}{8x} \times 100 = \frac{5}{8} = 0.625 \%$.

(D) Let the Cost Price $(CP)$ of the article be $Rs. x$.
When sold for $Rs. 900$,the profit is $(900 - x)$.
When sold for $Rs. 450$,the loss is $(x - 450)$.
According to the problem,the profit is double the loss:
$900 - x = 2(x - 450)$
$900 - x = 2x - 900$
$3x = 1800$
$x = 600$
So,the $CP$ is $Rs. 600$.
To earn a $25\%$ profit,the Selling Price $(SP)$ should be:
$SP = CP \times (1 + \frac{25}{100})$
$SP = 600 \times 1.25 = Rs. 750$.

(D) Let the Cost Price $(CP)$ of the article be $C$.
Given that the Selling Price $(SP_1)$ is $Rs. 35$ and the profit percentage is $40\%$.
The formula for $CP$ is $CP = \frac{SP \times 100}{100 + \text{Profit}\%}$.
So,$C = \frac{35 \times 100}{100 + 40} = \frac{3500}{140} = Rs. 25$.
Now,we want to earn a profit of $60\%$.
New Selling Price $(SP_2)$ $= CP \times \frac{100 + \text{Profit}\%}{100}$.
$SP_2 = 25 \times \frac{100 + 60}{100} = 25 \times \frac{160}{100} = 25 \times 1.6 = Rs. 40$.
Therefore,the article should be sold at $Rs. 40$ to earn a $60\%$ profit.

(A) Let the original price of sugar be $x$ per $kg$.
The new price of sugar after a $20\%$ increase is $1.2x$ per $kg$.
According to the problem,the difference in quantity purchased for $Rs.\, 135$ is $1.5\, kg$:
$\frac{135}{x} - \frac{135}{1.2x} = 1.5$
Multiply by $1.2x$ to solve for $x$:
$135(1.2) - 135 = 1.5(1.2x)$
$162 - 135 = 1.8x$
$27 = 1.8x$
$x = \frac{27}{1.8} = 15$
The original price was $Rs.\, 15$ per $kg$.
The increased price is $1.2 \times 15 = Rs.\, 18$ per $kg$.

(B) Total cost price $(CP)$ of the mixture $= (26 \times 20) + (30 \times 36) = 520 + 1080 = Rs. 1600$.
Total quantity of the mixture $= 26 + 30 = 56 \text{ kg}$.
Total selling price $(SP)$ of the mixture $= 56 \times 30 = Rs. 1680$.
Profit $= SP - CP = 1680 - 1600 = Rs. 80$.
Profit percent $= (\text{Profit} / CP) \times 100 = (80 / 1600) \times 100 = 5\%$.

(C) Let the selling price $(SP)$ of the $TV$ in Chandigarh be $Rs. x$.
The dealer bought the $TV$ in Delhi at a $20\%$ discount on the Chandigarh price,so the purchase price is $x - 0.20x = 0.8x$.
Total cost price $(CP)$ including transportation is $0.8x + 600$.
The profit percentage is given as $14 \frac{2}{7}\% = \frac{100}{7}\%$.
Using the formula: $\text{Profit}\% = \frac{SP - CP}{CP} \times 100$.
$\frac{100}{7} = \frac{x - (0.8x + 600)}{0.8x + 600} \times 100$.
Dividing both sides by $100$: $\frac{1}{7} = \frac{0.2x - 600}{0.8x + 600}$.
Cross-multiplying: $0.8x + 600 = 7(0.2x - 600)$.
$0.8x + 600 = 1.4x - 4200$.
$4200 + 600 = 1.4x - 0.8x$.
$4800 = 0.6x$.
$x = \frac{4800}{0.6} = 8000$.
Therefore,the value of $x$ is $Rs. 8000$.