Profit and Loss Questions in English

(C) Step $1$: Calculate the total cost price $(C.P.)$ of the mixture.
$C.P. = (20 \times 5) + (35 \times 7) = 100 + 245 = Rs. \, 345$.
Step $2$: Calculate the total selling price $(S.P.)$ of the mixture.
Total weight of the mixture $= 20 + 35 = 55 \, kg$.
$S.P. = 55 \times 6.5 = Rs. \, 357.5$.
Step $3$: Calculate the total profit.
Profit $= S.P. - C.P. = 357.5 - 345 = Rs. \, 12.5$.
Step $4$: Calculate the profit percentage.
Profit percentage $= (\frac{\text{Profit}}{C.P.}) \times 100 = (\frac{12.5}{345}) \times 100 = \frac{1250}{345} = \frac{250}{69} = 3 \frac{43}{69} \%$.

(C) Let the cost price $(CP)$ of the article be $₹ 100.$
Since the profit earned is $23.5 \%,$ the selling price $(SP)$ is $₹ 123.50.$
Let the marked price $(MP)$ of the article be $x.$
Given that a discount of $5 \%$ is offered on the marked price,the selling price is $95 \%$ of $x.$
So,$0.95x = 123.50.$
$x = \frac{123.50}{0.95} = 130.$
Thus,the marked price is $₹ 130.$
If no discount is offered,the selling price would be equal to the marked price,i.e.,$SP = ₹ 130.$
Profit percentage $= \frac{SP - CP}{CP} \times 100 = \frac{130 - 100}{100} \times 100 = 30 \%.$

(B) Labelled price of the cupboard $= Rs. 6500$.
Selling price $(S.P.)$ after $5\%$ discount $= 6500 \times (1 - 0.05) = 6500 \times 0.95 = Rs. 6175$.
Let the cost price $(C.P.)$ be $x$.
Given that the profit percentage is $15\%$,the relationship between $S.P.$ and $C.P.$ is $S.P. = C.P. \times (1 + \text{Profit}\%)$.
$6175 = x \times (1 + 0.15)$.
$6175 = 1.15x$.
$x = \frac{6175}{1.15} \approx 5369.56$.
Rounding to the nearest given option,the cost price is approximately $Rs. 5350$.

(B) Let the cost price ($C$.$P$.) of the article be $Rs. 100$.
Since the marked price ($M$.$P$.) is $10 \%$ above the cost price,$M$.$P$. $= 100 + 10 = Rs. 110$.
Given that the trader suffers a loss of $1 \%$,the selling price ($S$.$P$.) $= 100 - 1 = Rs. 99$.
Let the discount percentage be $x \%$.
The formula for $S$.$P$. in terms of discount is: $S.P. = M.P. \times (1 - \frac{x}{100})$.
Substituting the values: $99 = 110 \times (1 - \frac{x}{100})$.
Dividing both sides by $110$: $\frac{99}{110} = 1 - \frac{x}{100}$.
Simplifying the fraction: $0.9 = 1 - \frac{x}{100}$.
Rearranging the terms: $\frac{x}{100} = 1 - 0.9 = 0.1$.
Therefore,$x = 0.1 \times 100 = 10 \%$.
Thus,the trader allowed a discount of $10 \%$.

(B) Let the marked price of the radio be $x$.
Given,Cost Price $(C.P.)$ $= Rs. 1200$.
Discount $= 20 \%$ of $x$,so Selling Price $(S.P.)$ $= x - 0.20x = 0.8x$.
Profit required $= 25 \%$.
We know that $S.P. = C.P. \times (1 + \frac{\text{Profit } \%}{100})$.
$0.8x = 1200 \times (1 + \frac{25}{100})$.
$0.8x = 1200 \times 1.25$.
$0.8x = 1500$.
$x = \frac{1500}{0.8} = 1875$.
Therefore,the marked price should be $Rs. 1875$.

(C) The selling price $(S.P.)$ is given by $S.P. = C.P. \times (1 + \frac{\text{profit } \%}{100})$.
Given $S.P. = 300$ and profit $\% = 20$,we have $300 = C.P. \times (1 + 0.20)$.
Therefore,$C.P. = \frac{300}{1.2} = Rs. 250$.
The marked price is $Rs. 300$. During the sale,a discount of $10 \%$ is allowed.
New $S.P. = 300 \times (1 - \frac{10}{100}) = 300 \times 0.9 = Rs. 270$.
Gain $\% = \frac{S.P. - C.P.}{C.P.} \times 100 = \frac{270 - 250}{250} \times 100 = \frac{20}{250} \times 100 = 8 \%$.

(C) Let the initial price of the article be $100$.
After the $1st$ rise of $30 \%$,the price becomes $100 \times 1.30 = 130$.
After the $1st$ discount of $10 \%$,the price becomes $130 \times (1 - 0.10) = 130 \times 0.9 = 117$.
After the $2nd$ discount of $10 \%$,the price becomes $117 \times (1 - 0.10) = 117 \times 0.9 = 105.3$.
The final price is $105.3$.
Percentage increase $= \frac{105.3 - 100}{100} \times 100 = 5.3 \%$.
Therefore,the price is increased by $5.3 \%$.

(C) Let the amount of the bill be $Rs$ $x$.
First discount $= 35 \%$.
Two successive discounts of $20 \%$ each.
The equivalent discount for two successive discounts of $a \%$ and $b \%$ is given by $\left(a + b - \frac{ab}{100}\right) \%$.
Equivalent discount $= 20 + 20 - \frac{20 \times 20}{100} = 40 - 4 = 36 \%$.
The difference between the two discount schemes is $36 \% - 35 \% = 1 \%$.
According to the problem,$1 \%$ of $x = 22$.
$\frac{1}{100} \times x = 22$.
$x = 22 \times 100 = 2200$.
Therefore,the amount of the bill is $Rs$ $2200$.

(A) Let the marked price be $Rs. 100$.
After a discount of $10 \%$,the price becomes $100 \times (1 - 0.10) = Rs. 90$.
After a further discount of $12 \%$,the price becomes $90 \times (1 - 0.12) = 90 \times 0.88 = Rs. 79.2$.
After a final discount of $15 \%$,the price becomes $79.2 \times (1 - 0.15) = 79.2 \times 0.85 = Rs. 67.32$.
The total single discount is $100 - 67.32 = 32.68 \%$.

(A) Listed price $= Rs. 1500$.
First discount $= 20 \%$.
Price after first discount $= 1500 \times (1 - 0.20) = 1500 \times 0.80 = Rs. 1200$.
Let the additional discount be $x \%$.
According to the problem,the final net price is $Rs. 1104$.
So,$1200 \times (1 - \frac{x}{100}) = 1104$.
$1 - \frac{x}{100} = \frac{1104}{1200}$.
$1 - \frac{x}{100} = 0.92$.
$\frac{x}{100} = 1 - 0.92 = 0.08$.
$x = 8 \%$.
Therefore,the additional discount required is $8 \%$.

(B) Let the marked price of each article be $Rs. x$.
Since there are two articles,the total marked price is $2x$.
The discount given is $15 \%$,so the selling price of the pair is $85 \%$ of the total marked price.
Given that the total selling price is $Rs. 37.40$,we have:
$2x \times \frac{85}{100} = 37.40$
$2x \times 0.85 = 37.40$
$1.70x = 37.40$
$x = \frac{37.40}{1.70}$
$x = 22$
Therefore,the marked price of each article is $Rs. 22$.

(B) Let the cost price $(C.P.)$ of each horse be $Rs. x$ and that of each cow be $Rs. y$.
According to the problem,the total cost for $4$ horses and $9$ cows is $Rs. 13400$:
$4x + 9y = 13400$ $...(1)$
Profit earned on $4$ horses at $10 \%$ and $9$ cows at $20 \%$ is $Rs. 1880$:
$\frac{10}{100} \times 4x + \frac{20}{100} \times 9y = 1880$
$0.4x + 1.8y = 1880$
Multiply by $10$ to simplify:
$4x + 18y = 18800$ $...(2)$
Subtract equation $(1)$ from equation $(2)$:
$(4x + 18y) - (4x + 9y) = 18800 - 13400$
$9y = 5400$
$y = 600$
Substitute $y = 600$ into equation $(1)$:
$4x + 9(600) = 13400$
$4x + 5400 = 13400$
$4x = 8000$
$x = 2000$
Therefore,the cost price of a horse is $Rs. 2000$.

(B) Let the cost price of the article be $Rs. x$.
The selling price at $4 \%$ profit is $SP_1 = x \times (1 + \frac{4}{100}) = 1.04x$.
The selling price at $6 \%$ profit is $SP_2 = x \times (1 + \frac{6}{100}) = 1.06x$.
Given that the difference between the selling prices is $Rs. 3$:
$1.06x - 1.04x = 3$
$0.02x = 3$
$x = \frac{3}{0.02} = 150$.
The ratio of the two selling prices is:
$\frac{SP_1}{SP_2} = \frac{1.04x}{1.06x} = \frac{1.04}{1.06} = \frac{104}{106} = \frac{52}{53}$.
Thus,the required ratio is $52: 53$.

(D) Let the cost price of the fan be $Rs. x$.
The reduction in selling price is $Rs. 400 - Rs. 380 = Rs. 20$.
According to the problem,this reduction in selling price causes the loss to increase by $2 \%$ of the cost price.
Therefore,$2 \%$ of $x = 20$.
$\frac{2}{100} \times x = 20$.
$x = \frac{20 \times 100}{2}$.
$x = 1000$.
Thus,the cost price of the fan is $Rs. 1000$.

(D) Let the cost price of the bicycle for $A$ be $Rs. x$.
$A$ sells it to $B$ at a profit of $20 \%$,so the selling price for $A$ (which is the cost price for $B$) is $x \times (1 + \frac{20}{100}) = x \times \frac{120}{100}$.
$B$ sells it to $C$ at a profit of $25 \%$,so the selling price for $B$ (which is the cost price for $C$) is $(x \times \frac{120}{100}) \times (1 + \frac{25}{100}) = x \times \frac{120}{100} \times \frac{125}{100}$.
Given that $C$ pays $Rs. 225$,we have:
$x \times \frac{120}{100} \times \frac{125}{100} = 225$
$x \times \frac{6}{5} \times \frac{5}{4} = 225$
$x \times \frac{30}{20} = 225$
$x \times 1.5 = 225$
$x = \frac{225}{1.5} = 150$
Therefore,the cost price for $A$ is $Rs. 150$.

(D) Let the true weight be $1000 \text{ g}$.
When buying,the shopkeeper cheats by $10 \%$,meaning he receives $1100 \text{ g}$ while paying for $1000 \text{ g}$.
When selling,the shopkeeper cheats by $10 \%$,meaning he gives $900 \text{ g}$ while charging for $1000 \text{ g}$.
Let the cost price of $1 \text{ g}$ be $1$.
Cost Price $(CP)$ for $900 \text{ g}$ is $900$.
Selling Price $(SP)$ for $900 \text{ g}$ is $1100$ (since he sells $1100 \text{ g}$ worth of goods for the price of $1000 \text{ g}$).
Alternatively,using the formula for successive percentage gain: $\text{Total Gain } \% = \left( x + y + \frac{xy}{100} \right) \%$,where $x = 10$ and $y = 10$.
$\text{Total Gain } \% = 10 + 10 + \frac{10 \times 10}{100} = 20 + 1 = 21 \%$.
Thus,his total gain is $21 \%$.

(D) The dealer uses a scale of $90 \, cm$ instead of $100 \, cm$ (a metre scale).
This means for every $100 \, cm$ he should sell,he only gives $90 \, cm$.
His gain in length is $100 \, cm - 90 \, cm = 10 \, cm$.
Since he sells at cost price,his profit percentage is calculated on the actual amount given (the cost to him).
Profit $\% = \left( \frac{\text{Gain}}{\text{Actual quantity given}} \times 100 \right) \%$.
Profit $\% = \left( \frac{10}{90} \times 100 \right) \% = \frac{100}{9} \% = 11 \frac{1}{9} \%$.

(A) Let the cost price of milk be $Rs. 6.40$ per litre and the cost price of water be $Rs. 0$ per litre.
Profit percentage $= 37.5\%$.
Selling price of the mixture $= Rs. 8$ per litre.
We know that,$\text{Selling Price} = \text{Cost Price} \times (1 + \frac{\text{Profit}\%}{100})$.
$8 = \text{Cost Price} \times (1 + \frac{37.5}{100}) = \text{Cost Price} \times (1 + 0.375) = \text{Cost Price} \times 1.375$.
Mean cost price $= \frac{8}{1.375} = \frac{8000}{1375} = \frac{64}{11} \approx Rs. 5.82$ per litre.
Using the method of alligation:
Cost of Water $(Rs. 0)$ : Cost of Milk $(Rs. 6.40)$ = Mean Price $(Rs. 64/11)$.
Ratio of Water to Milk $= (6.40 - 64/11) : (64/11 - 0)$.
$= (70.4/11 - 64/11) : 64/11 = (6.4/11) : (64/11) = 6.4 : 64 = 1 : 10$.

(D) Step $1$: Calculate the total cost price $(C.P.)$ of the mixture.
Total $C.P. = (30 \times 11.50) + (20 \times 14.25) = 345 + 285 = Rs. \, 630$.
Step $2$: Calculate the cost price per $kg$ of the mixture.
Total weight $= 30 \, kg + 20 \, kg = 50 \, kg$.
$C.P.$ per $kg = 630 / 50 = Rs. \, 12.60$.
Step $3$: Calculate the selling price $(S.P.)$ per $kg$ for a $30 \%$ profit.
$S.P. = C.P. \times (1 + \text{Profit} \% / 100) = 12.60 \times 1.30 = Rs. \, 16.38$.
Rounding to the nearest given option,the price is approximately $Rs. \, 16.30$ per $kg$.

(A) Cost Price $(C.P.)$ of $1$ apple = $Rs. \frac{34}{8} = Rs. 4.25$.
Selling Price $(S.P.)$ of $1$ apple = $Rs. \frac{57}{12} = Rs. 4.75$.
Profit per apple = $S.P. - C.P. = 4.75 - 4.25 = Rs. 0.50$.
Let $x$ be the number of apples sold to earn a total profit of $Rs. 45$.
Total Profit = (Profit per apple) $\times$ $x$.
$45 = 0.50 \times x$.
$x = \frac{45}{0.50} = 90$.
Therefore,$90$ apples must be sold to earn a net profit of $Rs. 45$.

(C) Let the selling price of $1$ notebook be $SP$ and the cost price be $CP$.
Given that the profit on selling $12$ notebooks is equal to the selling price of $4$ notebooks.
Profit $= SP_{12} - CP_{12} = SP_4$.
Rearranging the terms: $SP_{12} - SP_4 = CP_{12}$.
This implies $SP_8 = CP_{12}$.
Profit percentage is calculated on the cost price: $\text{Profit } \% = \frac{\text{Profit}}{CP} \times 100$.
Since Profit $= SP_4$ and $CP_{12} = SP_8$,we have:
$\text{Profit } \% = \frac{SP_4}{CP_{12}} \times 100 = \frac{SP_4}{SP_8} \times 100 = \frac{4}{8} \times 100 = 50 \%$.

(B) Let the cost price $(C.P.)$ of $1$ article be $1$ unit.
Then,the $C.P.$ of $20$ articles $= 20$ units.
According to the problem,the selling price $(S.P.)$ of $x$ articles $= 20$ units.
Therefore,the $S.P.$ of $1$ article $= \frac{20}{x}$ units.
Profit percentage is given as $25 \%$.
Profit $= S.P. - C.P. = \frac{20}{x} - 1 = \frac{20-x}{x}$.
Profit percentage $= \left( \frac{\text{Profit}}{C.P.} \right) \times 100 = 25 \%$.
$\frac{(20-x)/x}{1} \times 100 = 25$.
$\frac{20-x}{x} = \frac{25}{100} = \frac{1}{4}$.
$4(20-x) = x$.
$80 - 4x = x$.
$5x = 80$.
$x = 16$.

(A) Let $x$ be the cost price $(C.P.)$ of the article in $Rs.$
According to the problem,the percentage profit at $Rs. 1920$ equals the percentage loss at $Rs. 1280$.
Profit percentage = $\frac{1920 - x}{x} \times 100$
Loss percentage = $\frac{x - 1280}{x} \times 100$
Equating the two: $\frac{1920 - x}{x} = \frac{x - 1280}{x}$
Since the denominators are the same,$1920 - x = x - 1280$.
$2x = 1920 + 1280 = 3200$
$x = 1600$
The cost price of the article is $Rs. 1600$.
To earn a $25 \%$ profit,the selling price $(S.P.)$ should be:
$S.P. = C.P. \times (1 + \frac{25}{100}) = 1600 \times 1.25 = Rs. 2000$.

(B) Let the initial cost price $(C.P.)$ be $x$.
Since the profit is $320 \%$ of the cost, the profit $= 3.2x$.
Therefore, the initial selling price $(S.P.)$ $= C.P. + \text{Profit} = x + 3.2x = 4.2x$.
If the cost increases by $25 \%$, the new $C.P. = x + 0.25x = 1.25x$.
The selling price remains constant, so the new $S.P. = 4.2x$.
The new profit $= \text{New } S.P. - \text{New } C.P. = 4.2x - 1.25x = 2.95x$.
The required percentage of the profit with respect to the selling price is given by $\left( \frac{\text{New Profit}}{\text{New } S.P.} \times 100 \right) \%$.
Required percentage $= \left( \frac{2.95x}{4.2x} \times 100 \right) \% = \frac{295}{4.2} \% \approx 70.23 \%$.
Rounding to the nearest integer, the percentage is approximately $70 \%$.

(B) Let the cost price be $C.P.$ and the initial selling price be $S.P.$
Initial profit $= S.P. - C.P.$
According to the problem,if the selling price is doubled $(2 S.P.)$,the profit becomes three times the original profit.
So,$2 S.P. - C.P. = 3(S.P. - C.P.)$
$2 S.P. - C.P. = 3 S.P. - 3 C.P.$
Rearranging the terms,we get $3 C.P. - C.P. = 3 S.P. - 2 S.P.$
$2 C.P. = S.P.$
Now,the profit percent is calculated as $\frac{S.P. - C.P.}{C.P.} \times 100$.
Substituting $S.P. = 2 C.P.$,we get $\frac{2 C.P. - C.P.}{C.P.} \times 100 = \frac{C.P.}{C.P.} \times 100 = 100 \%$.

(D) Given that the ratio of the selling price $(S.P.)$ to the cost price $(C.P.)$ is $\frac{S.P.}{C.P.} = \frac{7}{5}$.
Profit is calculated as $Profit = S.P. - C.P.$
Therefore,the ratio of profit to the cost price is $\frac{Profit}{C.P.} = \frac{S.P. - C.P.}{C.P.}$.
Substituting the given values: $\frac{7 - 5}{5} = \frac{2}{5}$.
Thus,the ratio between the profit and the cost price is $2:5$.

(C) Given,Selling Price $(S.P.)$ $= Rs. 18700$ and Loss percentage $= 15 \%$.
Since the loss is $15 \%$,the $S.P.$ is $85 \%$ of the Cost Price $(C.P.)$.
$0.85 \times C.P. = 18700$
$C.P. = \frac{18700}{0.85} = Rs. 22000$.
To gain $15 \%$,the new $S.P.$ should be $115 \%$ of the $C.P.$
New $S.P. = 1.15 \times 22000 = Rs. 25300$.
Therefore,the plot must be sold at $Rs. 25300$ to gain $15 \%$.

(B) Selling Price $(S.P.)$ of one piece $= Rs. 25$.
Total pieces $= 2000$.
If $5\%$ fail,the number of accepted pieces $= 2000 \times (1 - 0.05) = 1900$.
Revenue from $1900$ pieces $= 1900 \times 25 = Rs. 47500$.
Since the manufacturer expects a $25\%$ profit on the Cost Price $(C.P.)$,we have $S.P. = C.P. \times (1 + 0.25)$.
$47500 = C.P. \times 1.25 \implies C.P. = 47500 / 1.25 = Rs. 38000$.
Now,if $50\%$ of components are rejected,the number of accepted pieces $= 2000 \times 0.50 = 1000$.
Actual Revenue $= 1000 \times 25 = Rs. 25000$.
Loss $= C.P. - \text{Actual Revenue} = 38000 - 25000 = Rs. 13000$.

(A) Step $1$: Calculate the Cost Price $(CP)$ of the mangoes.
Given that the Selling Price $(SP_1)$ is $Rs. 9/kg$ and the loss is $20\%$.
$SP = CP \times (1 - \text{Loss}\% / 100)$
$9 = CP \times (1 - 20/100)$
$9 = CP \times 0.8$
$CP = 9 / 0.8 = 11.25 \text{ per } kg$.
Step $2$: Calculate the required Selling Price $(SP_2)$ for a $5\%$ profit.
$SP_2 = CP \times (1 + \text{Profit}\% / 100)$
$SP_2 = 11.25 \times (1 + 5/100)$
$SP_2 = 11.25 \times 1.05 = 11.8125$.
Rounding to two decimal places,the required price is $Rs. 11.81/kg$.

(C) The total cost price of the machine includes the purchase price,repair costs,and transport costs.
Total Cost Price $= 80000 + 5000 + 1000 = Rs. 86000$.
To find the selling price at a $25\%$ profit,we use the formula: $\text{Selling Price} = \text{Cost Price} \times (1 + \frac{\text{Profit}\%}{100})$.
Selling Price $= 86000 \times (1 + \frac{25}{100}) = 86000 \times 1.25$.
Selling Price $= 86000 \times \frac{125}{100} = 860 \times 125 = Rs. 107500$.

(A) Cost Price $(C.P.)$ of $100$ oranges = $Rs. 350$.
$C.P.$ of $1$ orange = $\frac{350}{100} = Rs. 3.5$.
Selling Price $(S.P.)$ of $1$ dozen $(12)$ oranges = $Rs. 48$.
$S.P.$ of $1$ orange = $\frac{48}{12} = Rs. 4$.
Since $S.P. > C.P.$,there is a profit.
Profit = $S.P. - C.P. = 4 - 3.5 = Rs. 0.5$.
Profit percentage = $\left( \frac{\text{Profit}}{C.P.} \times 100 \right) \% = \left( \frac{0.5}{3.5} \times 100 \right) \% = \left( \frac{1}{7} \times 100 \right) \% = 14 \frac{2}{7} \% \text{ gain}$.

(C) Cost price $(CP)$ of $70$ $kg$ of potatoes $= Rs. 420$.
Cost price per $kg = 420 / 70 = Rs. 6$ per $kg$.
Selling price $(SP)$ per $kg = Rs. 6.5$ per $kg$.
Gain $= SP - CP = 6.5 - 6 = Rs. 0.5$ per $kg$.
Gain percent $= (\text{Gain} / CP) \times 100 = (0.5 / 6) \times 100 = (1 / 12) \times 100 = 100 / 12 = 25 / 3 = 8 \frac{1}{3} \%$.

(C) Given that the ratio of the cost price $(CP)$ to the selling price $(SP)$ is $4:5$.
Let $CP = 4x$ and $SP = 5x$.
Profit $= SP - CP = 5x - 4x = x$.
Profit percent $= (\text{Profit} / CP) \times 100$.
Profit percent $= (x / 4x) \times 100 = (1/4) \times 100 = 25\%$.

(B) Let $x$ be the cost price $(C.P.)$ of the article.
According to the problem,the profit at $Rs. 832$ is equal to the loss at $Rs. 448$.
Profit $= S.P. - C.P. = 832 - x$
Loss $= C.P. - S.P. = x - 448$
Equating the two: $832 - x = x - 448$
$2x = 832 + 448$
$2x = 1280$
$x = 640$
So,the cost price is $Rs. 640$.
To make a $50\%$ profit,the new selling price $(S.P.)$ should be:
$S.P. = C.P. \times (1 + \text{Profit } \% / 100)$
$S.P. = 640 \times (1 + 50 / 100) = 640 \times 1.5 = Rs. 960$.

(A) Let the cost price $(C.P.)$ of $1$ article be $x$.
Then,the $C.P.$ of $40$ articles $= 40x$.
According to the problem,the selling price $(S.P.)$ of $50$ articles $= 40x$.
Therefore,the $S.P.$ of $1$ article $= \frac{40x}{50} = 0.8x$.
Since $S.P. < C.P.$,there is a loss.
Loss $= C.P. - S.P. = x - 0.8x = 0.2x$.
Loss percentage $= \left( \frac{\text{Loss}}{C.P.} \right) \times 100 = \left( \frac{0.2x}{x} \right) \times 100 = 20 \%$.
Thus,there is a $20 \%$ loss.

(C) The retailer pays for $5$ dozen boxes but receives $5 + 1 = 6$ dozen boxes in total.
The free quantity is $1$ dozen.
The discount percentage is calculated as the ratio of free items to the total items received.
Discount percentage $= \left( \frac{\text{Free items}}{\text{Total items}} \right) \times 100$
Discount percentage $= \left( \frac{1}{5 + 1} \right) \times 100 = \frac{1}{6} \times 100 = 16 \frac{2}{3} \%$.

(D) Let the cost price $(C.P.)$ of one ball be $x$.
Then,the $C.P.$ of $17$ balls $= 17x$.
The loss is equal to the $C.P.$ of $5$ balls,which is $5x$.
We know that $Loss = C.P. - S.P.$
Given that $S.P.$ of $17$ balls $= 720$.
So,$17x - 720 = 5x$.
$17x - 5x = 720$.
$12x = 720$.
$x = \frac{720}{12} = 60$.
Therefore,the cost price of a ball is $Rs. 60$.

(D) Cost Price $(C.P.)$ of $1$ article $= Rs. \frac{5}{6}$.
Selling Price $(S.P.)$ of $1$ article $= Rs. \frac{6}{5}$.
Gain $= S.P. - C.P. = \frac{6}{5} - \frac{5}{6} = \frac{36 - 25}{30} = Rs. \frac{11}{30}$.
Gain $\% = \left( \frac{\text{Gain}}{C.P.} \times 100 \right) = \left( \frac{11/30}{5/6} \times 100 \right)$.
$= \left( \frac{11}{30} \times \frac{6}{5} \times 100 \right) = \left( \frac{11}{25} \times 100 \right) = 44\%.$

(C) Cost Price $(C.P.)$ of $1$ fruit $= \frac{24}{16} = Rs. 1.50$.
Selling Price $(S.P.)$ of $1$ fruit $= \frac{18}{8} = Rs. 2.25$.
Profit $= S.P. - C.P. = 2.25 - 1.50 = Rs. 0.75$.
Profit $\% = \left( \frac{\text{Profit}}{C.P.} \right) \times 100$.
Profit $\% = \left( \frac{0.75}{1.50} \right) \times 100 = \frac{1}{2} \times 100 = 50\%$.

(C) Let the weights of the three varieties be $2x$,$4x$,and $3x$ kg respectively.
Total weight of the mixture $= 2x + 4x + 3x = 9x$ kg.
Total Cost Price $(CP) = (50 \times 2x) + (20 \times 4x) + (30 \times 3x) = 100x + 80x + 90x = 270x$.
Total Selling Price $(SP) = 33 \times 9x = 297x$.
Profit $= SP - CP = 297x - 270x = 27x$.
Profit $\% = (\text{Profit} / CP) \times 100 = (27x / 270x) \times 100 = 10\%$.

(B) Total quantity of rice $= 26 \text{ kg} + 30 \text{ kg} = 56 \text{ kg}$.
Cost Price $(C.P.)$ of $26 \text{ kg}$ rice $= 26 \times 20 = Rs. 520$.
Cost Price $(C.P.)$ of $30 \text{ kg}$ rice $= 30 \times 36 = Rs. 1080$.
Total $C.P. = 520 + 1080 = Rs. 1600$.
Selling Price $(S.P.)$ of $56 \text{ kg}$ rice at $Rs. 30$ per $\text{kg} = 56 \times 30 = Rs. 1680$.
Profit $= S.P. - C.P. = 1680 - 1600 = Rs. 80$.
Profit $\% = \left( \frac{\text{Profit}}{C.P.} \right) \times 100 = \left( \frac{80}{1600} \right) \times 100 = 5\%$.

(C) Cost Price $(C.P.)$ of $1$ toffee $= Rs. \frac{1}{6}$.
To gain $20 \%$,the Selling Price $(S.P.)$ of $1$ toffee should be $120 \%$ of the $C.P.$
$S.P. = \frac{120}{100} \times \frac{1}{6} = \frac{1.2}{6} = Rs. \frac{1}{5}$.
This means the vendor must sell $5$ toffees for $Rs. 1$ to gain $20 \%$.

(C) The shopkeeper uses $800 \text{ g}$ instead of $1000 \text{ g}$ $(1 \text{ kg})$.
Here,the cost price $(CP)$ is the cost of $800 \text{ g}$ and the selling price $(SP)$ is the cost of $1000 \text{ g}$ (since he sells at cost price).
Let the cost of $1 \text{ g}$ be $1$.
Then,$CP = 800$ and $SP = 1000$.
Profit $= SP - CP = 1000 - 800 = 200$.
Profit $\% = \left( \frac{\text{Profit}}{CP} \right) \times 100 = \left( \frac{200}{800} \right) \times 100 = \frac{1}{4} \times 100 = 25 \%$.

(D) Let the total quantity of goods be $100$ units and the cost price $(C.P.)$ per unit be $Rs. 1$.
Total $C.P. = 100 \times 1 = Rs. 100$.
Since he makes a $10 \%$ profit on his goods,the total selling price $(S.P.)$ for the entire stock would have been $Rs. 110$.
He lost $20 \%$ of his goods due to theft,meaning he only has $80$ units left to sell.
Since the profit margin is $10 \%$,the $S.P.$ per unit is $1.10$.
Total $S.P. = 80 \times 1.10 = Rs. 88$.
Loss $= C.P. - S.P. = 100 - 88 = Rs. 12$.
Loss percentage $= (\text{Loss} / C.P.) \times 100 = (12 / 100) \times 100 = 12 \%$.

(D) Let the original value of the article be $V = 100$.
The cost price $(CP)$ is $10 \%$ less than its value: $CP = 100 - 10 = 90$.
The selling price $(SP)$ is $10 \%$ more than its value: $SP = 100 + 10 = 110$.
Profit = $SP - CP = 110 - 90 = 20$.
Profit percentage = $\frac{\text{Profit}}{CP} \times 100 = \frac{20}{90} \times 100 = \frac{200}{9} \% = 22 \frac{2}{9} \%$.
Since $22 \frac{2}{9} \% > 20 \%$,the man makes more than $20 \%$ profit.

(C) Let the cost price of the article be $C.P.$
The gain when selling at $Rs. 350$ is $(350 - C.P.)$.
The gain when selling at $Rs. 340$ is $(340 - C.P.)$.
According to the problem,the gain at $Rs. 350$ is $5 \%$ more than the gain at $Rs. 340$ of the cost price.
So,$0.05 \times C.P. = 350 - 340$.
$0.05 \times C.P. = 10$.
$C.P. = \frac{10}{0.05} = \frac{1000}{5} = 200$.
Therefore,the cost price of the article is $Rs. 200$.

(B) When two items are sold at the same selling price,one at a profit of $x \%$ and the other at a loss of $x \%$,there is always a net loss in the transaction.
The formula for the net loss percentage is given by: $\text{Loss } \% = \left( \frac{x}{10} \right)^2$.
Here,$x = 12$.
Therefore,$\text{Loss } \% = \left( \frac{12}{10} \right)^2 = (1.2)^2 = 1.44 \%$.
Converting $1.44$ into a fraction: $1.44 = \frac{144}{100} = \frac{36}{25} = 1 \frac{11}{25} \%$.
Thus,the shopkeeper incurs a loss of $1 \frac{11}{25} \%$.

(C) Let the total cost price $(C.P.)$ of the cloth be $100$ units.
$1$. Half of the cloth ($50$ units) is sold at $20 \%$ profit:
$S.P._1 = 50 \times 1.20 = 60$ units.
$2$. Half of the remaining cloth ($25$ units) is sold at $20 \%$ loss:
$S.P._2 = 25 \times 0.80 = 20$ units.
$3$. The rest of the cloth ($25$ units) is sold at cost price:
$S.P._3 = 25 \times 1.00 = 25$ units.
Total Selling Price $(S.P.)$ $= 60 + 20 + 25 = 105$ units.
Since the total $S.P.$ $(105)$ is greater than the total $C.P.$ $(100)$,there is a profit.
Profit $\% = \frac{S.P. - C.P.}{C.P.} \times 100 = \frac{105 - 100}{100} \times 100 = 5 \%$ gain.

(D) The total discount is calculated in two parts.
First part: $4\%$ of $Rs. 200000 = \frac{4}{100} \times 200000 = Rs. 8000$.
Second part: $2.5\%$ of $Rs. 72000 = \frac{2.5}{100} \times 72000 = Rs. 1800$.
Total discount $= 8000 + 1800 = Rs. 9800$.
The actual price charged by the company $= \text{Marked Price} - \text{Total Discount}$.
Actual price $= 272000 - 9800 = Rs. 262200$.

(C) Let the labelled price be $x$.
First discount $= 20 \%$.
Price after first discount $= x \times (1 - 0.20) = 0.80x$.
Second discount $= 12 \%$ on the discounted price.
Final selling price $= 0.80x \times (1 - 0.12) = 0.80x \times 0.88 = 0.704x$.
Given that the final selling price is $Rs \, 704$.
So,$0.704x = 704$.
$x = \frac{704}{0.704} = 1000$.
Therefore,the labelled price is $Rs \, 1000$.