Profit and Loss Questions in English

(C) Total $C.P.$ of $500 \text{ m}$ wire $= 500 \times 0.50 = ₹ 250$.
He sold $50 \%$ of the wire,which is $250 \text{ m}$. The $C.P.$ of this $250 \text{ m}$ is $250 \times 0.50 = ₹ 125$.
Profit on this $50 \%$ is $5 \%$,so $S.P.$ $= 125 \times 1.05 = ₹ 131.25$.
To gain $10 \%$ on the whole transaction,the total $S.P.$ must be $110 \% \text{ of } 250 = ₹ 275$.
$S.P.$ required for the remaining $250 \text{ m}$ wire $= 275 - 131.25 = ₹ 143.75$.
Profit on the remaining wire $= 143.75 - 125 = ₹ 18.75$.
Required profit percentage $= (18.75 / 125) \times 100 = 15 \%$.

(A) माना कुल क्रय मूल्य $(C.P)$ = $ 300$ है।
वस्तु का $1/3$ भाग = $ 100$,जिसे $15 \%$ हानि पर बेचा गया है।
अतः,विक्रय मूल्य = $100 \times (1 - 0.15) = 85$ है।
शेष वस्तुओं का $C.P$ = $ 200$ है।
पूरे लेन-देन पर $10 \%$ लाभ प्राप्त करने के लिए,कुल विक्रय मूल्य = $300 \times 1.10 = 330$ होना चाहिए।
शेष वस्तुओं के लिए आवश्यक विक्रय मूल्य = $330 - 85 = 245$ है।
लाभ प्रतिशत = $((245 - 200) / 200) \times 100 = (45 / 200) \times 100 = 22.5 \%$.
अतः,शेष वस्तुओं को $22 \frac{1}{2} \%$ के लाभ पर बेचा जाना चाहिए।

(B) Cost Price $(CP)$ of each article = $Rs. 4000$.
Total $CP$ of two articles = $4000 + 4000 = Rs. 8000$.
Selling Price $(SP)$ of the first article with $12.5\%$ gain = $4000 \times (1 + 0.125) = 4000 \times 1.125 = Rs. 4500$.
Selling Price $(SP)$ of the second article with $20\%$ loss = $4000 \times (1 - 0.20) = 4000 \times 0.80 = Rs. 3200$.
Total $SP$ = $4500 + 3200 = Rs. 7700$.
Total Loss = Total $CP$ - Total $SP$ = $8000 - 7700 = Rs. 300$.
Overall Loss Percent = $(\text{Total Loss} / \text{Total } CP) \times 100 = (300 / 8000) \times 100 = 3.75\%$.

(C) Step $1$: Calculate the Cost Price $(C.P.)$ of the article.
Given that the Selling Price $(S.P.)$ is $Rs. 170$ and the loss is $15\%$.
$C.P. = \frac{S.P. \times 100}{100 - \text{Loss}\%} = \frac{170 \times 100}{100 - 15} = \frac{170 \times 100}{85} = Rs. 200$.
Step $2$: Calculate the required Selling Price $(S.P.)$ to gain $20\%$.
Required $S.P. = C.P. \times \frac{100 + \text{Gain}\%}{100} = 200 \times \frac{100 + 20}{100} = 200 \times \frac{120}{100} = Rs. 240$.

(A) Cost Price $(C.P.)$ of $80$ ball pens $= 140 \times \frac{100}{100 - 30} = 140 \times \frac{100}{70} = ₹ 200$.
To earn a profit of $30 \%$,the required Selling Price $(S.P.)$ for $80$ ball pens $= 200 \times \frac{130}{100} = ₹ 260$.
Since $₹ 260$ is the price for $80$ ball pens,the number of ball pens to be sold for $₹ 104$ is calculated as:
Number of pens $= \frac{80}{260} \times 104 = \frac{8}{26} \times 104 = 8 \times 4 = 32$ ball pens.

(A) Let the cost price $(C.P)$ of $1$ article be $x$.
Then,the $C.P$ of $50$ articles $= 50x$.
Given that the selling price $(S.P)$ of $40$ articles $= 50x$.
Therefore,the $S.P$ of $1$ article $= \frac{50x}{40} = 1.25x$.
Since $S.P > C.P$,there is a gain.
Gain $= S.P - C.P = 1.25x - x = 0.25x$.
Gain $\% = \left( \frac{\text{Gain}}{C.P} \right) \times 100 = \left( \frac{0.25x}{x} \right) \times 100 = 25 \%$.

(C) Given that the loss is $20 \%$.
Let the cost price $(C.P.)$ be $100$.
Since there is a loss of $20 \%$,the selling price $(S.P.)$ is $C.P. - 20 \% \text{ of } C.P. = 100 - 20 = 80$.
We need to find a fraction $x$ such that $S.P. \times x = C.P.$
$80 \times x = 100$
$x = \frac{100}{80} = \frac{5}{4}$.
Therefore,the selling price must be multiplied by $\frac{5}{4}$ to get the cost price.

(B) Let the cost price $(C.P.)$ of $100 \, g$ of rice be $₹ 100$.
Since the shopkeeper sells at a $10 \%$ profit,the selling price $(S.P.)$ for $100 \, g$ is $₹ 110$.
However,the shopkeeper uses a weight that is $30 \%$ less than the actual measure. Therefore,the actual amount of rice sold is $100 \, g - 30 \% \text{ of } 100 \, g = 70 \, g$.
The cost price of $70 \, g$ of rice is $₹ 70$.
The shopkeeper sells this $70 \, g$ of rice for $₹ 110$.
Profit = $S.P. - C.P. = 110 - 70 = ₹ 40$.
Gain percent = $\frac{\text{Profit}}{C.P.} \times 100 = \frac{40}{70} \times 100 = \frac{400}{7} = 57 \frac{1}{7} \%$.

(C) Cost Price $(C.P)$ of $4$ dozen eggs $= 4 \times 24 = ₹ 96$.
Cost Price $(C.P)$ of $2$ dozen eggs $= 2 \times 32 = ₹ 64$.
Total Cost Price $(C.P)$ $= 96 + 64 = ₹ 160$.
Total quantity of eggs $= 4 + 2 = 6$ dozen.
To gain $20\%$,the total Selling Price $(S.P)$ must be $120\%$ of the total $C.P$.
Total $S.P = 160 \times \frac{120}{100} = ₹ 192$.
Selling Price per dozen $= \frac{192}{6} = ₹ 32$ per dozen.

(D) Let the Cost Price $(C.P.)$ of the cloth be $x$ per meter.
Given that the shopkeeper loses $10 \%$ by selling it at $Rs. 9$ per meter.
So,$90 \%$ of $C.P. = 9$.
$0.90 \times C.P. = 9 \Rightarrow C.P. = \frac{9}{0.90} = 10$.
The cost price of the cloth is $Rs. 10$ per meter.
To earn a profit of $15 \%$,the Selling Price $(S.P.)$ should be $115 \%$ of $C.P.$
$S.P. = 1.15 \times 10 = 11.5$.
Therefore,the cloth should be sold at $Rs. 11.5$ per meter to earn a $15 \%$ profit.

(C) Let the number of apples Kamal ate be $x$.
He sold $40\%$ more than he ate,which means the number of apples sold is $x + 0.40x = 1.4x$.
Given that he sold $70$ apples,we have the equation: $1.4x = 70$.
Solving for $x$: $x = \frac{70}{1.4} = \frac{700}{14} = 50$.
Therefore,Kamal ate $50$ apples.

(B) Let the profit of $A$ be $A$ and the profit of $B$ be $B$.
Given that $A + B = 1650$.
According to the problem,$\frac{1}{3} A = \frac{2}{5} B$.
From this,we can express $A$ in terms of $B$: $A = \frac{2}{5} \times 3 \times B = \frac{6}{5} B$.
Substitute this into the total profit equation: $\frac{6}{5} B + B = 1650$.
$\frac{6B + 5B}{5} = 1650$.
$\frac{11B}{5} = 1650$.
$B = \frac{1650 \times 5}{11}$.
$B = 150 \times 5 = 750$.
Therefore,the profit of $B$ is $Rs. 750$.

(A) The ratio of profit is determined by the product of investment and time period.
Anil's investment $= Rs. 25,000$ for $12$ months.
Vishal's investment $= Rs. 30,000$ for $(12 - 3) = 9$ months.
Ratio of profit (Anil : Vishal) $= (25,000 \times 12) : (30,000 \times 9)$.
$= 300,000 : 270,000 = 30 : 27 = 10 : 9$.
Total ratio parts $= 10 + 9 = 19$.
Anil's share in profit $= \frac{10}{19} \times 19,000 = Rs. 10,000$.

(C) The actual ratio is $\frac{1}{5}: \frac{1}{4}: \frac{1}{8}$. To simplify,multiply by the $LCM$ of $5, 4, 8$,which is $40$. The ratio becomes $8: 10: 5$. The sum of the ratio parts is $8 + 10 + 5 = 23$.
Actual shares:
$A = \frac{8}{23} \times 391 = 136$
$B = \frac{10}{23} \times 391 = 170$
$C = \frac{5}{23} \times 391 = 85$
The cookies were actually divided in the ratio $5: 4: 8$. The sum of the ratio parts is $5 + 4 + 8 = 17$.
New shares:
$A = \frac{5}{17} \times 391 = 115$
$B = \frac{4}{17} \times 391 = 92$
$C = \frac{8}{17} \times 391 = 184$
Comparing the gains:
$A$ gains: $115 - 136 = -21$ (Loss)
$B$ gains: $92 - 170 = -78$ (Loss)
$C$ gains: $184 - 85 = 99$ (Gain)
Therefore,$C$ gains the most,which is $99$ cookies.

(B) Let the total quantity of sugar be $x \text{ kg}$.
After selling $5 \%$ of the sugar,the remaining quantity is $(100 \% - 5 \%) = 95 \%$ of the total quantity.
Given that the remaining quantity is $5 \text{ kg}$.
Therefore,$95 \% \text{ of } x = 5 \text{ kg}$.
$\frac{95}{100} \times x = 5$.
$x = \frac{5 \times 100}{95}$.
$x = \frac{500}{95} = \frac{100}{19} \text{ kg}$.
Converting this to a mixed fraction,we get $5 \frac{5}{19} \text{ kg}$.

(A) Let the cost price of the first item be $x$ and the cost price of the second item be $(520 - x)$.
Since there is no overall profit or loss,the gain on the first item equals the loss on the second item.
Gain on the first item = $16\%$ of $x = 0.16x$.
Loss on the second item = $10\%$ of $(520 - x) = 0.10(520 - x)$.
Equating them: $0.16x = 0.10(520 - x)$.
$0.16x = 52 - 0.10x$.
$0.26x = 52$.
$x = 52 / 0.26 = 200$.
So,the cost price of the first item is $Rs. 200$ and the cost price of the second item is $520 - 200 = Rs. 320$.
The second item is sold at a loss of $10\%$.
Selling price of the second item = $320 - (10\% \text{ of } 320) = 320 - 32 = Rs. 288$.

(A) Let the marked price $(M.P)$ be $Rs. 5000$.
Mr. $X$ bought the article at a discounted price,which becomes his cost price $(C.P)$.
He sold the article at $Rs. 5000$,which is the selling price $(S.P)$.
The profit percentage is $11 \frac{1}{9} \% = \frac{100}{9} \% = \frac{1}{9}$.
We know that $Profit = S.P - C.P$,and $Profit \% = \frac{S.P - C.P}{C.P} \times 100$.
Given $Profit \% = \frac{1}{9}$,we have $\frac{S.P}{C.P} - 1 = \frac{1}{9}$,so $\frac{S.P}{C.P} = 1 + \frac{1}{9} = \frac{10}{9}$.
Since $S.P = 5000$,then $C.P = 5000 \times \frac{9}{10} = 4500$.
The discount is the difference between $M.P$ and $C.P$: $Discount = 5000 - 4500 = 500$.
Discount percentage $= \frac{Discount}{M.P} \times 100 = \frac{500}{5000} \times 100 = 10 \%$.
Thus,the discount percentage allowed was $10 \%$.

(C) Let the cost price of the table be $T$ and the cost price of the chair be $C$.
Given that the total cost price is $T + C = 500$ ... $(I)$.
The table is sold at a loss of $10\%$ and the chair at a gain of $10\%$,resulting in a total gain of $Rs. 10$.
This can be expressed as: $0.10C - 0.10T = 10$.
Dividing the entire equation by $0.10$,we get $C - T = 100$ ... $(II)$.
Adding equation $(I)$ and $(II)$:
$(T + C) + (C - T) = 500 + 100$
$2C = 600$
$C = 300$.
Thus,the cost price of the chair is $Rs. 300$.

(B) Let the marked price $(MP)$ of the book be $100$.
Given that the commission is $10 \%$,the selling price $(SP_1)$ for the publisher is $100 - 10 = 90$.
The publisher gains $20 \%$,so the cost price $(CP)$ is calculated as:
$CP = \frac{SP_1}{1 + \text{Gain} \%} = \frac{90}{1.20} = 75$.
Now,if the commission is increased to $15 \%$,the new selling price $(SP_2)$ becomes $100 - 15 = 85$.
The new gain is $SP_2 - CP = 85 - 75 = 10$.
The new gain percent is $\frac{\text{Gain}}{CP} \times 100 = \frac{10}{75} \times 100 = \frac{2}{15} \times 100 = \frac{40}{3} = 13 \frac{1}{3} \%$.

(D) Let the total cost price $(C.P.)$ of the oranges be $₹ 1200$.
The overall profit required is $10\%$,so the total selling price $(S.P.)$ must be $1200 + (10\% \text{ of } 1200) = 1200 + 120 = ₹ 1320$.
$A$ sold $1/3$ of the oranges at a $20\%$ loss. The cost price of $1/3$ of the oranges is $1200 / 3 = ₹ 400$.
The selling price of this $1/3$ portion is $400 - (20\% \text{ of } 400) = 400 - 80 = ₹ 320$.
The remaining oranges have a cost price of $1200 - 400 = ₹ 800$.
The required selling price for the remaining oranges is $1320 - 320 = ₹ 1000$.
Profit on the remaining oranges $= 1000 - 800 = ₹ 200$.
Profit percentage $= (200 / 800) \times 100 = 25\%$.

(A) Let Bob's present age be $B$ and Abby's present age be $A$.
According to the first condition: $B = A - 8$,which implies $A = B + 8$.
Four years hence,Bob's age will be $B + 4$ and Abby's age will be $A + 4$.
The ratio of their ages will be $4:5$,so: $\frac{B + 4}{A + 4} = \frac{4}{5}$.
Substitute $A = B + 8$ into the equation: $\frac{B + 4}{(B + 8) + 4} = \frac{4}{5}$.
$\frac{B + 4}{B + 12} = \frac{4}{5}$.
Cross-multiply: $5(B + 4) = 4(B + 12)$.
$5B + 20 = 4B + 48$.
$5B - 4B = 48 - 20$.
$B = 28$.

(C) For equation $I$: $2x^2 + 19x + 45 = 0$
$2x^2 + 10x + 9x + 45 = 0$
$2x(x + 5) + 9(x + 5) = 0$
$(2x + 9)(x + 5) = 0$
$x = -4.5$ or $x = -5$
For equation $II$: $2y^2 + 11y + 12 = 0$
$2y^2 + 8y + 3y + 12 = 0$
$2y(y + 4) + 3(y + 4) = 0$
$(2y + 3)(y + 4) = 0$
$y = -1.5$ or $y = -4$
Comparing the values:
$x = -4.5, -5$
$y = -1.5, -4$
Since both values of $x$ are less than both values of $y$ (i.e.,$-5 < -4$ and $-4.5 < -1.5$),we conclude that $x < y$.

(C) For equation $I$: $3x^2 - 13x + 12 = 0$
$3x^2 - 9x - 4x + 12 = 0$
$3x(x - 3) - 4(x - 3) = 0$
$(3x - 4)(x - 3) = 0$
$x = 4/3 \approx 1.33$ or $x = 3$
For equation $II$: $2y^2 - 15y + 28 = 0$
$2y^2 - 8y - 7y + 28 = 0$
$2y(y - 4) - 7(y - 4) = 0$
$(2y - 7)(y - 4) = 0$
$y = 7/2 = 3.5$ or $y = 4$
Comparing the values:
$x = 1.33, 3$
$y = 3.5, 4$
Since $1.33 < 3.5$,$1.33 < 4$,$3 < 3.5$,and $3 < 4$,we conclude that $x < y$.

(D) Step $1$: Solve equation $I$: $x^{2} = 16$. Taking the square root on both sides,we get $x = 4$ or $x = -4$.
Step $2$: Solve equation $II$: $2y^{2} - 17y + 36 = 0$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,where $a = 2, b = -17, c = 36$.
$y = \frac{17 \pm \sqrt{(-17)^{2} - 4(2)(36)}}{2(2)}$
$y = \frac{17 \pm \sqrt{289 - 288}}{4}$
$y = \frac{17 \pm 1}{4}$
So,$y = \frac{18}{4} = 4.5$ or $y = \frac{16}{4} = 4$.
Step $3$: Compare the values of $x$ and $y$:
For $x = 4$,$y$ can be $4$ or $4.5$. Here $x \leq y$.
For $x = -4$,$y$ can be $4$ or $4.5$. Here $x < y$.
Combining these,we get $x \leq y$.

(B) For equation $I$: $6x^2 + 19x + 15 = 0$
$6x^2 + 9x + 10x + 15 = 0$
$3x(2x + 3) + 5(2x + 3) = 0$
$(3x + 5)(2x + 3) = 0$
$x = -5/3 \approx -1.67$ or $x = -3/2 = -1.5$
For equation $II$: $3y^2 + 11y + 10 = 0$
$3y^2 + 6y + 5y + 10 = 0$
$3y(y + 2) + 5(y + 2) = 0$
$(3y + 5)(y + 2) = 0$
$y = -5/3 \approx -1.67$ or $y = -2$
Comparing the values:
If $x = -1.5$ and $y = -1.67$,then $x > y$.
If $x = -1.5$ and $y = -2$,then $x > y$.
If $x = -1.67$ and $y = -1.67$,then $x = y$.
If $x = -1.67$ and $y = -2$,then $x > y$.
In all cases,$x \geq y$.

(D) For equation $I$: $2x^2 - 11x + 15 = 0$
$2x^2 - 6x - 5x + 15 = 0$
$2x(x - 3) - 5(x - 3) = 0$
$(2x - 5)(x - 3) = 0$
So,$x = 2.5$ or $x = 3$.
For equation $II$: $2y^2 - 11y + 14 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{11 \pm \sqrt{121 - 4(2)(14)}}{2(2)}$
$y = \frac{11 \pm \sqrt{121 - 112}}{4}$
$y = \frac{11 \pm \sqrt{9}}{4}$
$y = \frac{11 \pm 3}{4}$
So,$y = \frac{14}{4} = 3.5$ or $y = \frac{8}{4} = 2$.
Comparing the values:
If $x = 2.5$,then $x < y$ (since $y = 3.5$) and $x > y$ (since $y = 2$).
Since the relationship varies,the relationship between $x$ and $y$ cannot be determined.

(D) Let the marked price be $100 ₹$.
First discount of $10 \%$ on $100 ₹$ is $10 ₹$,so the price becomes $100 - 10 = 90 ₹$.
Second discount of $20 \%$ is applied on the remaining $90 ₹$.
Second discount $= 20 \% \text{ of } 90 = \frac{20}{100} \times 90 = 18 ₹$.
Final selling price $= 90 - 18 = 72 ₹$.
Total discount $= 100 - 72 = 28 ₹$.
Therefore,the total discount percentage is $\frac{28}{100} \times 100 = 28 \%$.

(C) Marked Price $(M.P.)$ = $Rs. 720$.
First discount = $10 \%$.
Price after the first discount = $720 - (10 \% \text{ of } 720) = 720 - 72 = Rs. 648$.
Let the second discount rate be $x \%$.
Selling Price $(S.P.)$ = $Rs. 550.80$.
Price after second discount = $648 - (x \% \text{ of } 648) = 550.80$.
$648 \times (1 - \frac{x}{100}) = 550.80$.
$1 - \frac{x}{100} = \frac{550.80}{648} = 0.85$.
$\frac{x}{100} = 1 - 0.85 = 0.15$.
$x = 15 \%$.
Therefore,the second discount rate is $15 \%$.

(C) Let the marked price be $M.P$.
Given successive discounts are $20 \%$ and $15 \%$.
The selling price $(S.P.)$ is calculated as:
$S.P. = M.P. \times (1 - \frac{20}{100}) \times (1 - \frac{15}{100})$
$3060 = M.P. \times (0.80) \times (0.85)$
$3060 = M.P. \times 0.68$
$M.P. = \frac{3060}{0.68}$
$M.P. = 4500$
Therefore,the marked price is $Rs. 4,500$.

(B) Given, discount percentage $= 20 \%$.
Selling Price $(S.P.) = Rs. 300$.
Let the Marked Price be $M.P$.
Since $S.P. = M.P. \times (1 - \text{Discount } \%)$, we have:
$300 = M.P. \times (1 - 0.20)$
$300 = M.P. \times 0.80$
$M.P. = \frac{300}{0.80} = Rs. 375$.
Now, if the shopkeeper sells the toy at $Rs. 405$, the new selling price is $Rs. 405$.
Since the cost price of the toy is the marked price (assuming no other costs), the gain is:
$\text{Gain} = 405 - 375 = Rs. 30$.
$\text{Gain } \% = \left( \frac{\text{Gain}}{C.P.} \right) \times 100 = \left( \frac{30}{375} \right) \times 100 = 8 \%$.

(C) Let the selling price $(S.P.)$ of each horse be $₹ S$.
For the first horse,profit is $10 \%$,so $C.P._1 = S / 1.10 = 10S / 11$.
For the second horse,loss is $10 \%$,so $C.P._2 = S / 0.90 = 10S / 9$.
Total $C.P. = 10S/11 + 10S/9 = (90S + 110S) / 99 = 200S / 99$.
Total $S.P. = 2S = 198S / 99$.
Since Total $C.P. >$ Total $S.P.$,there is a loss.
Loss $= 200S/99 - 198S/99 = 2S/99$.
Loss percentage $= (Loss / Total C.P.) \times 100 = (2S/99) / (200S/99) \times 100 = (2/200) \times 100 = 1 \%$.
Alternatively,for equal $S.P.$ with $x \%$ profit and $x \%$ loss,the net result is always a loss of $(x/10)^2 \% = (10/10)^2 \% = 1 \% \text{ loss}$.

(C) Let the initial cost price $(C.P.)$ be $₹ 100$.
The initial selling price $(S.P.)$ is fixed at $33 \%$ above the cost price,so $S.P. = 100 + 33 = ₹ 133$.
If the cost of production increases by $12 \%$,the new cost price $(C.P.')$ becomes $100 + 12 = ₹ 112$.
The manufacturer raises the selling price by $10 \%$,so the new selling price $(S.P.')$ becomes $133 + (10 \% \text{ of } 133) = 133 + 13.3 = ₹ 146.3$.
The new profit is $S.P.' - C.P.' = 146.3 - 112 = ₹ 34.3$.
The new profit percentage is $\frac{\text{Profit}}{C.P.'} \times 100 = \frac{34.3}{112} \times 100 = \frac{3430}{112} = 30.625 \%$.
Converting $0.625$ to a fraction: $0.625 = \frac{625}{1000} = \frac{5}{8}$.
Thus,the profit percentage is $30 \frac{5}{8} \%$.

(D) Marked Price $(M.P.)$ $= ₹ 12,600$.
First discount $= 5 \%$.
Price after first discount $= 12600 \times (1 - 0.05) = 12600 \times 0.95 = ₹ 11,970$.
Second discount (for cash payment) $= 2 \%$.
Final Selling Price $(S.P.)$ $= 11970 \times (1 - 0.02) = 11970 \times 0.98$.
$S.P. = ₹ 11,730.60$.
Therefore,the cash payment to be made is $₹ 11,730.60$.

(D) Let the number of oranges purchased in each lot be the $LCM$ of $3$ and $5$,which is $15$.
Cost Price $(CP)$ of the first lot ($15$ oranges): $(40 / 3) \times 15 = ₹ 200$.
Cost Price $(CP)$ of the second lot ($15$ oranges): $(60 / 5) \times 15 = ₹ 180$.
Total Cost Price $(CP)$ for $30$ oranges: $200 + 180 = ₹ 380$.
Selling Price $(SP)$ of $30$ oranges: $(50 / 3) \times 30 = ₹ 500$.
Profit = $SP - CP = 500 - 380 = ₹ 120$.
Profit percentage = $(\text{Profit} / CP) \times 100 = (120 / 380) \times 100 = (12 / 38) \times 100 \approx 31.57 \%$.
Rounding to the nearest integer,the profit is $32 \%$.

(C) Let the cost of production $(C.P.)$ be $₹ 100$.
Since the dealer fixed the price $40 \%$ above the cost of production,the marked price $(M.P.)$ is $100 + 40 = ₹ 140$.
He allows a discount of $20 \%$ on the marked price,so the selling price $(S.P.)$ is $140 \times (1 - 0.20) = 140 \times 0.80 = ₹ 112$.
The profit is calculated as $S.P. - C.P. = 112 - 100 = ₹ 12$.
If the profit is $₹ 12$,the $C.P.$ is $₹ 100$.
If the profit is $₹ 48$,the $C.P.$ is $\frac{100}{12} \times 48 = ₹ 400$.

(B) Let the Cost Price $(C.P.)$ of rice be $x$ per $kg$.
Given that there is a $10 \%$ loss when sold at $Rs. 54$ per $kg$.
So,$90 \%$ of $C.P. = 54$.
$0.90 \times C.P. = 54$.
$C.P. = \frac{54}{0.90} = 60$.
Now,to earn a profit of $20 \%$,the Selling Price $(S.P.)$ should be $120 \%$ of $C.P$.
$S.P. = 1.20 \times 60 = 72$.
Therefore,the price of rice per $kg$ to earn a $20 \%$ profit is $Rs. 72$.

(A) Let the cost price $(CP)$ be $x$.
Initial selling price $(SP_1)$ $= x + 0.05x = 1.05x = \frac{21x}{20}$.
New cost price $(CP_2)$ $= x - 0.05x = 0.95x = \frac{19x}{20}$.
New selling price $(SP_2)$ $= CP_2 + 10\% \text{ of } CP_2 = 1.10 \times \frac{19x}{20} = \frac{1.10 \times 19x}{20} = \frac{20.9x}{20} = \frac{209x}{200}$.
According to the problem,the difference between the two selling prices is $Rs. 2$:
$SP_1 - SP_2 = 2$
$\frac{21x}{20} - \frac{209x}{200} = 2$
Multiply by $200$ to clear the denominator:
$210x - 209x = 400$
$x = 400$.
Therefore,the cost price of the article is $Rs. 400$.

(A) Let the Cost Price $(CP)$ be $₹ 100$.
Since the profit is $25 \%$,the Marked Price $(MP)$ is $100 + 25 = ₹ 125$.
After allowing a discount,the new profit is $12\frac{1}{2} \% = 12.5 \%$.
Therefore,the new Selling Price $(SP)$ is $100 + 12.5 = ₹ 112.5$.
The discount amount is $MP - SP = 125 - 112.5 = ₹ 12.5$.
The discount percentage is calculated on the Marked Price:
Discount $\% = (\text{Discount} / MP) \times 100$
Discount $\% = (12.5 / 125) \times 100 = 0.1 \times 100 = 10 \%$.

(B) Let the cost price $(C.P.)$ be $₹ 100$.
Since the shopkeeper gains $17 \%$,the selling price $(S.P.)$ is $₹ 117$.
Let the marked price be $M.P.$. Given that a discount of $10 \%$ is allowed on the $M.P.$,we have:
$S.P. = M.P. \times (1 - \frac{10}{100}) = M.P. \times 0.9$
$117 = M.P. \times 0.9$
$M.P. = \frac{117}{0.9} = ₹ 130$.
If the article is sold at the marked price with no discount,the new selling price is $₹ 130$.
The profit is $S.P. - C.P. = 130 - 100 = ₹ 30$.
Profit percentage $= (\frac{\text{Profit}}{C.P.}) \times 100 = (\frac{30}{100}) \times 100 = 30 \%$.

(B) List price of one book $= ₹ 100$.
Total list price of three books $= 3 \times 100 = ₹ 300$.
Selling price of three books $= ₹ 274.50$.
Total discount $= 300 - 274.50 = ₹ 25.50$.
Discount rate in $\% = \left( \frac{\text{Total Discount}}{\text{Total List Price}} \right) \times 100$.
Discount rate in $\% = \left( \frac{25.50}{300} \right) \times 100 = \frac{25.50}{3} = 8.5 \%$.
Therefore,the rate of discount is $8.5 \%$.

(D) Let the cost price $(CP)$ of the article be $₹ 100$.
Since the printed price (marked price) is $40 \%$ higher than the cost price,the marked price $(MP)$ is $100 + 40 = ₹ 140$.
To gain a $12 \%$ profit,the selling price $(SP)$ must be $CP + 12 \% \text{ of } CP = 100 + 12 = ₹ 112$.
The discount is the difference between the marked price and the selling price: $Discount = MP - SP = 140 - 112 = ₹ 28$.
The discount percentage is calculated on the marked price: $\text{Discount } \% = (\frac{Discount}{MP}) \times 100$.
$\text{Discount } \% = (\frac{28}{140}) \times 100 = 0.2 \times 100 = 20 \%$.

(A) Let the cost price $(C.P.)$ of $1$ book be $x$.
Then,the $C.P.$ of $100$ books $= 100x$.
Given that the $C.P.$ of $100$ books is equal to the selling price $(S.P.)$ of $60$ books.
So,$S.P.$ of $60$ books $= 100x$.
$S.P.$ of $1$ book $= \frac{100x}{60} = \frac{5x}{3}$.
Since $S.P. > C.P.$,there is a gain.
Gain $= S.P. - C.P. = \frac{5x}{3} - x = \frac{2x}{3}$.
Gain percentage $= \frac{\text{Gain}}{C.P.} \times 100 = \frac{2x/3}{x} \times 100 = \frac{2}{3} \times 100 = 66 \frac{2}{3} \%$.

(D) Let the Cost Price $(CP)$ be $₹ 100$.
Since the gain is $20 \%$,the Selling Price $(SP_1)$ is $₹ 120$.
Given that the discount $(D_1)$ is $10 \%$,we have $SP_1 = MP \times (1 - 0.10)$,where $MP$ is the Marked Price.
$120 = MP \times 0.90 \implies MP = \frac{120}{0.90} = ₹ \frac{400}{3}$.
Now,if the discount $(D_2)$ is increased to $20 \%$,the new Selling Price $(SP_2)$ is:
$SP_2 = MP \times (1 - 0.20) = \frac{400}{3} \times 0.80 = \frac{400}{3} \times \frac{4}{5} = \frac{320}{3} = ₹ 106.66...$
Gain = $SP_2 - CP = 106.66 - 100 = ₹ 6.66...$
Gain $\% = \frac{6.66...}{100} \times 100 = 6.66... \% = 6 \frac{2}{3} \%$.

(B) The marked price $(M.P.)$ of the article is $Rs. 100$.
First,calculate the effective cost price after successive discounts of $10\%$ and $20\%$:
$Cost Price = 100 \times (1 - 0.10) \times (1 - 0.20) = 100 \times 0.90 \times 0.80 = Rs. 72$.
The dealer spends $10\%$ of this cost price on transportation:
$Transportation Cost = 10\% \text{ of } 72 = 0.10 \times 72 = Rs. 7.20$.
The total cost price $(C.P.)$ is the sum of the purchase price and transportation cost:
$Total C.P. = 72 + 7.20 = Rs. 79.20$.
To earn a profit of $15\%$,the selling price $(S.P.)$ should be:
$S.P. = Total C.P. \times (1 + 0.15) = 79.20 \times 1.15 = Rs. 91.08$.

(B) Given,Cost Price $(CP) = ₹ 600$.
Desired profit percentage $= 20 \%$.
Selling Price $(SP)$ required to earn $20 \%$ profit $= CP + (20 \% \text{ of } CP) = 600 + (0.20 \times 600) = 600 + 120 = ₹ 720$.
Let the Marked Price $(MP)$ be $x$.
The shopkeeper allows a discount of $10 \%$ on the $MP$,so the Selling Price is $90 \%$ of $MP$.
$0.90 \times x = 720$.
$x = \frac{720}{0.90} = \frac{7200}{9} = ₹ 800$.
Therefore,the marked price should be $₹ 800$.

(D) Let the initial cost price $(CP)$ of the bicycle be $₹ x$.
The initial selling price $(SP_1)$ at a profit of $10 \%$ is $SP_1 = x \times (1 + 0.10) = 1.1x$.
If the dealer had bought the bicycle at $10 \%$ less price,the new cost price $(CP_2)$ would be $CP_2 = x \times (1 - 0.10) = 0.9x$.
If he sold it at $Rs. 60$ more than the original selling price,the new selling price $(SP_2)$ would be $SP_2 = 1.1x + 60$.
Given that the gain in this scenario is $25 \%$,we have $SP_2 = CP_2 \times (1 + 0.25) = 0.9x \times 1.25$.
Equating the two expressions for $SP_2$:
$1.1x + 60 = 0.9x \times 1.25$
$1.1x + 60 = 1.125x$
$1.125x - 1.1x = 60$
$0.025x = 60$
$x = \frac{60}{0.025} = \frac{60000}{25} = 2400$.
Therefore,the cost price of the bicycle was $₹ 2400$.

(D) Let the marked price $(M.P.)$ be $x$.
For the first case,successive discounts of $40\%$ and $30\%$:
Effective discount $= 40 + 30 - \frac{40 \times 30}{100} = 70 - 12 = 58\%$.
Selling price $(S.P._1)$ $= (100 - 58)\% \text{ of } x = 0.42x$.
For the second case,successive discounts of $45\%$ and $20\%$:
Effective discount $= 45 + 20 - \frac{45 \times 20}{100} = 65 - 9 = 56\%$.
Selling price $(S.P._2)$ $= (100 - 56)\% \text{ of } x = 0.44x$.
The difference between the selling prices is given as $Rs. 12$:
$0.44x - 0.42x = 12$
$0.02x = 12$
$x = \frac{12}{0.02} = 600$.
Thus,the marked price of the article is $Rs. 600$.

(D) Let the Marked Price $(MP)$ be $100$.
After a discount of $10 \%$,the price becomes $100 \times (1 - 0.10) = 90$.
After a further discount of $20 \%$,the price becomes $90 \times (1 - 0.20) = 90 \times 0.8 = 72$.
After a final discount of $25 \%$,the price becomes $72 \times (1 - 0.25) = 72 \times 0.75 = 54$.
The final selling price is $54$.
The equivalent single discount is $(100 - 54) \% = 46 \%$.

(A) Let the cost price of $1$ book be $x$.
Then,the cost price of $100$ books $= 100x$.
According to the problem,the selling price of $60$ books $= 100x$.
Therefore,the selling price of $1$ book $= \frac{100x}{60} = \frac{5x}{3}$.
Since the selling price is greater than the cost price,there is a profit.
Profit $= \text{Selling Price} - \text{Cost Price} = \frac{5x}{3} - x = \frac{2x}{3}$.
Profit percentage $= \left( \frac{\text{Profit}}{\text{Cost Price}} \times 100 \right) \%$.
Profit percentage $= \left( \frac{2x/3}{x} \times 100 \right) \% = \frac{2}{3} \times 100 = \frac{200}{3} = 66 \frac{2}{3} \%$.

(D) Marked Price $(M.P.) = ₹ 975$
Selling Price $(S.P.) = ₹ 897$
Discount $= M.P. - S.P. = 975 - 897 = ₹ 78$
Discount percentage $= (\frac{\text{Discount}}{M.P.}) \times 100$
$= (\frac{78}{975}) \times 100$
$= 0.08 \times 100 = 8\%$