Profit and Loss: CAT Previous Year Questions

Ans: 175

Price per chair for purchase = 100
The combined purchase cost for 13 chairs amounts to $13 \times 100 = 1300$.
A profit margin of 26% implies a total selling revenue of
$ 1.26 \times 1300 = 1638$.
This revenue is stated to be equivalent to the sale price of 12 chairs after a discount. Consequently, the reduced selling price per chair is calculated as $\text{SP}_{\text{disc}} = \frac{1638}{12} = 136.5$.
With a discount of 22% applied, the marked price (MP) can be determined as:
MP = $\frac{136.5}{0.78}$ = 175

Ans: C

Let the list price be denoted by $L$ and the initial discount percentage be $x$. The purchase price is $1650$. A profit of $20\%$ implies the sale price is:
$SP = 1650 \times 1.20 = 1980$
With discount $x$, we have:
$L(1-x) = 1980$
If the discount is doubled (i.e., $2x$), the sale price becomes $L(1-2x)$, and the profit is $110$. Therefore:
$L(1-2x) – 1650 = 110 \implies L(1-2x) = 1760$
Subtracting the two derived equations:
$L(1-x) – L(1-2x) = 1980 – 1760$
$Lx = 220 \implies L = \frac{220}{x}$
Substitute this expression for $L$ into the first equation:
$\frac{220}{x}(1-x) = 1980 \implies \frac{1-x}{x} = \frac{1980}{220} \implies \frac{1-x}{x} = 9$
$1-x = 9x \implies 1 = 10x \implies x = 0.1$
So, the initial discount percentage is $10\%$, and the list price is:
$L = \frac{220}{0.1} = 2200$
Now, let the new discount percentage be $y$ such that the profit percentage is equal to the discount percentage. The equation for this condition is:
$\frac{2200(1-y) – 1650}{1650} = y$
Rearranging and simplifying:
$2200(1-y) – 1650 = 1650y$
$2200 – 2200y – 1650 = 1650y$
$550 = 3850y$
$y = \frac{550}{3850} \approx 0.142857$
Thus, the required discount percentage is approximately $14\%$.

Ans: B

Explanation:Let the marked price per unit be $M$.
The cost price per unit is Rs. 240.

Monthly sales:
January: 120 units
February: 135 units
March: 150 units
April: 165 units

Discounts:
January: 20%
February: 10%
March: 5%
April: 0%

Selling price per unit in each month:
January selling price ($SP_J$) = $M \times (1 – 0.20) = 0.80M$
February selling price ($SP_F$) = $M \times (1 – 0.10) = 0.90M$
March selling price ($SP_M$) = $M \times (1 – 0.05) = 0.95M$
April selling price ($SP_A$) = $M \times (1 – 0.00) = M$

Profit per unit in each month:
Profit per unit in January ($P_J$) = $SP_J – CP = 0.80M – 240$
Profit per unit in February ($P_F$) = $SP_F – CP = 0.90M – 240$
Profit per unit in March ($P_M$) = $SP_M – CP = 0.95M – 240$
Profit per unit in April ($P_A$) = $SP_A – CP = M – 240$

Total profit from January to April = Rs. 138825

Total profit = (Sales in Jan * Profit per unit in Jan) + (Sales in Feb * Profit per unit in Feb) + (Sales in Mar * Profit per unit in Mar) + (Sales in Apr * Profit per unit in Apr)

$138825 = 120(0.80M – 240) + 135(0.90M – 240) + 150(0.95M – 240) + 165(M – 240)$

$138825 = (120 \times 0.80M – 120 \times 240) + (135 \times 0.90M – 135 \times 240) + (150 \times 0.95M – 150 \times 240) + (165M – 165 \times 240)$

$138825 = (96M – 28800) + (121.5M – 32400) + (142.5M – 36000) + (165M – 39600)$

Combine the terms with $M$:
$96M + 121.5M + 142.5M + 165M = (96 + 121.5 + 142.5 + 165)M = 525M$

Combine the constant terms:
$-28800 – 32400 – 36000 – 39600 = -(28800 + 32400 + 36000 + 39600) = -136800$

So, the equation becomes:
$138825 = 525M – 136800$

Add 136800 to both sides:
$138825 + 136800 = 525M$
$275625 = 525M$

Now, solve for $M$:
$M = \frac{275625}{525}$

Let’s perform the division:
$275625 \div 525 = 525$

So, the marked price per unit is Rs. 525.

Let’s check the options:
Option A: 520
Option B: 525
Option C: 510
Option D: 515

The calculated marked price matches Option B.

Correct_Option:B

Ans: B

Explanation:Let the original cost price be C and the original selling price be S.
Given that the selling price is fixed to ensure 40% profit.
So, S = C + 0.40C = 1.40C.

If the product had cost 40% less, the new cost price (C’) would be:
C’ = C – 0.40C = 0.60C.

If the product had been sold for 5 rupees less, the new selling price (S’) would be:
S’ = S – 5.

The resulting profit would have been 50%. This means the new selling price is 50% more than the new cost price.
So, S’ = C’ + 0.50C’ = 1.50C’.

Now we have a system of equations:
1) S = 1.40C
2) C’ = 0.60C
3) S’ = S – 5
4) S’ = 1.50C’

Substitute equation (2) into equation (4):
S’ = 1.50 * (0.60C)
S’ = 0.90C.

Now substitute equation (3) and the expression for S’ in terms of C into the equation S’ = 0.90C:
(S – 5) = 0.90C.

Now substitute the expression for S from equation (1) into this equation:
(1.40C – 5) = 0.90C.

Now, solve for C:
1.40C – 0.90C = 5
0.50C = 5
C = 5 / 0.50
C = 10.

The original cost price is 10 rupees.
The question asks for the original selling price.
Using equation (1):
S = 1.40C
S = 1.40 * 10
S = 14.

The original selling price is 14 rupees.

Let’s verify the conditions:
Original Cost Price (C) = 10
Original Selling Price (S) = 14
Profit = S – C = 14 – 10 = 4
Profit percentage = (4/10) * 100 = 40%. This matches the first condition.

New Cost Price (C’) = 0.60 * C = 0.60 * 10 = 6
New Selling Price (S’) = S – 5 = 14 – 5 = 9
Profit = S’ – C’ = 9 – 6 = 3
Profit percentage = (3/6) * 100 = 50%. This matches the second condition.

Correct_Option:B

Ans: 2160

Explanation:
Let the purchase price of Bina be $B_{cost}$.
Let the selling price of Bina be $B_{sell} = 4860$.
Bina incurs a 19% loss, so her selling price is 100% – 19% = 81% of her purchase price.
$B_{sell} = 0.81 \times B_{cost}$
$4860 = 0.81 \times B_{cost}$
$B_{cost} = \frac{4860}{0.81} = \frac{486000}{81} = 6000$.
So, Bina’s purchase price was Rs. 6000.

Shyam’s purchase price is Bina’s selling price, which is Rs. 4860.
Let Shyam’s selling price be $S_{sell}$.
Let Hari’s purchase price be $H_{cost}$, which is equal to Shyam’s selling price, $S_{sell}$.

We are given that if Bina had sold the product to Shyam at Hari’s purchase price, she would have obtained a 17% profit.
This means that Hari’s purchase price ($H_{cost}$) is 17% more than Bina’s purchase price ($B_{cost}$).
$H_{cost} = B_{cost} + 0.17 \times B_{cost}$
$H_{cost} = 1.17 \times B_{cost}$
$H_{cost} = 1.17 \times 6000$
$H_{cost} = 7020$.

Since Shyam sells the product to Hari, Shyam’s selling price is Hari’s purchase price.
$S_{sell} = H_{cost} = 7020$.

Shyam’s purchase price is Rs. 4860.
Shyam’s selling price is Rs. 7020.
The profit made by Shyam is $S_{sell} – S_{cost}$.
Shyam’s profit = $7020 – 4860 = 2160$.

Final_Answer:2160

Ans: A

Explanation:Let the cost price of the product be CP.
Gopi marks a price on the product in order to make 20% profit. So, the marked price (MP) is such that if sold at MP, Gopi would get a 20% profit. This implies that MP is the selling price if there were no discount.
Let the marked price be MP.
Ravi gets a 10% discount on the marked price.
Discount amount = 10% of MP = 0.10 * MP.
We are given that Ravi saves Rs 15, which means the discount amount is Rs 15.
So, 0.10 * MP = 15.
MP = 15 / 0.10 = 150.
The marked price is Rs 150.
Gopi marks the price to make a 20% profit. This means that if the product was sold at the marked price, the profit would be 20% of the cost price.
Let CP be the cost price.
MP = CP + 20% of CP = CP + 0.20 * CP = 1.20 * CP.
We found that MP = 150.
So, 1.20 * CP = 150.
CP = 150 / 1.20 = 150 / (12/10) = 150 * (10/12) = 1500 / 12 = 125.
The cost price of the product is Rs 125.
Ravi gets a 10% discount on the marked price.
Selling Price (SP) to Ravi = MP – Discount = 150 – 15 = 135.
The profit made by Gopi by selling the product to Ravi is the difference between the selling price and the cost price.
Profit = SP – CP = 135 – 125 = 10.
The profit made by Gopi is Rs 10.

Let’s recheck the initial assumption about the marked price. The problem states “Gopi marks a price on a product in order to make 20% profit.” This phrasing usually implies that the intended selling price is the marked price, and that intended selling price should yield a 20% profit over the cost price.

Let CP be the Cost Price.
Gopi marks a price MP.
The intention is to make 20% profit, so if sold at MP, the profit would be 20% of CP.
So, MP = CP + 0.20 * CP = 1.20 * CP.

Ravi gets a 10% discount on the marked price.
Discount = 10% of MP = 0.10 * MP.
This discount amount is Rs 15.
0.10 * MP = 15
MP = 15 / 0.10 = 150.

Now, we can find the Cost Price using MP = 1.20 * CP.
150 = 1.20 * CP
CP = 150 / 1.20 = 125.

The selling price to Ravi is the marked price minus the discount.
SP = MP – Discount = 150 – 15 = 135.

The profit made by Gopi is the selling price minus the cost price.
Profit = SP – CP = 135 – 125 = 10.

Let’s consider an alternative interpretation where the marked price is set to achieve a 20% profit on the selling price, but this is less common. The standard interpretation is profit on cost price.

The initial interpretation aligns with standard business terminology.

The profit made by Gopi is Rs 10.

Correct_Option:A

Ans: C

Explanation:Let the cost price of object A be $C_A$ and the cost price of object B be $C_B$.
Let the initial selling price of both objects be $S$.

Case 1: Initial selling
On object A, Gita makes a profit of 20%.
$S = C_A + 0.20 C_A = 1.20 C_A$
$C_A = S / 1.20 = S / (6/5) = 5S/6$

On object B, Gita makes a loss of 10%.
$S = C_B – 0.10 C_B = 0.90 C_B$
$C_B = S / 0.90 = S / (9/10) = 10S/9$

Case 2: Increased selling price
Let the new selling price be $S’$.
Objects A and B are still sold at an equal price $S’$.
A profit of 10% is made on object B.
$S’ = C_B + 0.10 C_B = 1.10 C_B$
Substitute the value of $C_B$ from Case 1:
$S’ = 1.10 \times (10S/9) = (11/10) \times (10S/9) = 11S/9$

Now, we need to find the profit made on object A with the new selling price $S’$.
Let the profit percentage on object A be $P_A$.
$S’ = C_A + (P_A/100) C_A = C_A (1 + P_A/100)$
Substitute the values of $S’$ and $C_A$:
$11S/9 = (5S/6) (1 + P_A/100)$

Divide both sides by $S$ (assuming $S \neq 0$):
$11/9 = (5/6) (1 + P_A/100)$

Now, solve for $(1 + P_A/100)$:
$1 + P_A/100 = (11/9) \times (6/5)$
$1 + P_A/100 = (11 \times 2) / (3 \times 5)$
$1 + P_A/100 = 22/15$

Now, solve for $P_A/100$:
$P_A/100 = 22/15 – 1$
$P_A/100 = 22/15 – 15/15$
$P_A/100 = 7/15$

Now, solve for $P_A$:
$P_A = (7/15) \times 100$
$P_A = 700 / 15$
$P_A = 140 / 3$
$P_A \approx 46.67\%$

We need to find the nearest option.
Option A: 42%
Option B: 45%
Option C: 47%
Option D: 49%

The nearest value to 46.67% is 47%.

Correct_Option:C

Ans: C

Explanation:Let the cost price of sunglasses for Minu be CP_Minu1 = Rs. 1000.
Minu sells to Kanu at 20% profit.
Profit made by Minu in the first transaction = 20% of 1000 = 0.20 * 1000 = Rs. 200.
Selling price to Kanu (SP_Kanu) = CP_Minu1 + Profit = 1000 + 200 = Rs. 1200.
So, Kanu buys the sunglasses for Rs. 1200.

Kanu sells it back to Minu at 20% loss.
The selling price for Kanu is the cost price for Minu in the second transaction (CP_Minu2).
Loss made by Kanu = 20% of 1200 = 0.20 * 1200 = Rs. 240.
Selling price for Kanu (which is the price Minu buys back at) = 1200 – 240 = Rs. 960.
So, Minu buys back the sunglasses for Rs. 960.

Minu sells the same pair of sunglasses to Tanu. Let the selling price to Tanu be SP_Tanu.
The profit made by Minu in the second transaction = SP_Tanu – CP_Minu2 = SP_Tanu – 960.

The total profit made by Minu from all her transactions is Rs. 500.
Total Profit = (Profit from first transaction) + (Profit from second transaction)
500 = 200 + (SP_Tanu – 960)
500 – 200 = SP_Tanu – 960
300 = SP_Tanu – 960
SP_Tanu = 300 + 960 = Rs. 1260.

Now we need to find the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu.
The cost price for Minu in the second transaction (when selling to Tanu) is the price she bought it back for, which is Rs. 960.
Profit made by Minu in the second transaction = SP_Tanu – CP_Minu2 = 1260 – 960 = Rs. 300.
Percentage Profit = (Profit / Cost Price) * 100
Percentage Profit (selling to Tanu) = (300 / 960) * 100
Percentage Profit = (30000 / 960)
Percentage Profit = (3000 / 96)
Percentage Profit = (1000 / 32)
Percentage Profit = (250 / 8)
Percentage Profit = (125 / 4)
Percentage Profit = 31.25%

Correct_Option:C

Ans: 407

Explanation:Let $w$ be the number of white shirts and $b$ be the number of blue shirts.
The cost of white shirts is $1000w$.
The cost of blue shirts is $1125b$.
The total cost is $C = 1000w + 1125b$.
The total number of shirts is $N = w + b$.

The average cost of all the shirts is $A = \frac{1000w + 1125b}{w+b}$.
The market price for each shirt is 25% higher than the average cost: $M = A \times 1.25 = \frac{5}{4}A$.
The selling price of each shirt is 10% less than the market price: $S = M \times 0.90 = \frac{9}{10}M$.
So, the selling price of each shirt is $S = \frac{9}{10} \times \frac{5}{4}A = \frac{9}{8}A$.

The total selling price is $T_S = S \times N = \frac{9}{8}A \times (w+b) = \frac{9}{8} \times \frac{1000w + 1125b}{w+b} \times (w+b) = \frac{9}{8}(1000w + 1125b)$.
The total profit is $P = T_S – C = \frac{9}{8}(1000w + 1125b) – (1000w + 1125b)$.
$P = (\frac{9}{8} – 1)(1000w + 1125b) = \frac{1}{8}(1000w + 1125b)$.
We are given that the total profit is Rs. 51000.
So, $\frac{1}{8}(1000w + 1125b) = 51000$.
$1000w + 1125b = 51000 \times 8 = 408000$.
Divide by 25: $40w + 45b = 16320$.
Divide by 5: $8w + 9b = 3264$.

We are looking for the maximum possible total number of shirts, $N = w + b$, given that $w > 0$ and $b > 0$ (since he bought both colors of shirts).
We can express $w$ in terms of $b$: $8w = 3264 – 9b$, so $w = \frac{3264 – 9b}{8} = 408 – \frac{9b}{8}$.
For $w$ to be an integer, $9b$ must be divisible by 8. Since 9 and 8 are coprime, $b$ must be divisible by 8.
Let $b = 8k$ for some positive integer $k$.
Then $w = 408 – \frac{9(8k)}{8} = 408 – 9k$.

Since $w > 0$, we have $408 – 9k > 0$, which means $9k < 408$, so $k < \frac{408}{9} \approx 45.33$.
Since $b > 0$, we have $8k > 0$, which means $k > 0$.
So, $k$ can be any integer from 1 to 45.

We want to maximize $N = w + b = (408 – 9k) + 8k = 408 – k$.
To maximize $N$, we need to minimize $k$.
The minimum possible value for $k$ is 1.
When $k=1$:
$b = 8 \times 1 = 8$.
$w = 408 – 9 \times 1 = 399$.
In this case, $N = w + b = 399 + 8 = 407$.
Both $w$ and $b$ are positive.

Let’s check the other extreme for $k$. The maximum value for $k$ is 45.
When $k=45$:
$b = 8 \times 45 = 360$.
$w = 408 – 9 \times 45 = 408 – 405 = 3$.
In this case, $N = w + b = 3 + 360 = 363$.

The total number of shirts is $N = 408 – k$. To maximize $N$, we need to minimize $k$. The smallest possible integer value for $k$ is 1.
This gives $N = 408 – 1 = 407$.

Final_Answer:407

Ans: C

A cloth dealer acquires fabric at a price of Rs. 100 per meter. For every meter of fabric bought, he is given an additional 5 cm of cloth at no extra charge.
This means the expense incurred for 105 cm of fabric is Rs. 100.

Furthermore, the dealer sets a marked price of Rs. 110 for every 100 cm of fabric and offers a 5% reduction on this price. Crucially, he shortchanges his patrons by supplying only 95 cm of fabric for every 100 cm a customer purchases.

Consequently, the realized price for 95 cm of fabric amounts to Rs. 110 multiplied by (19/20).
Based on this, the revenue generated from selling 105 cm of fabric is Rs. 115.5.

The resultant gain is 15.5%.
The correct choice is C.

Ans: 160

Here’s the explanation rephrased to be copyright-free while maintaining the original structure and logic:

Total syrup quantity – 110 kg
Total juice quantity – 120 kg
The purchase price of syrup per kilogram is 20% lower than the purchase price of juice per kilogram.
Assume the purchase price of juice per kg is 10x.
Then, the purchase price of syrup per kg is 8x.
Cost of 10 kg of syrup = 10 kg * 8x/kg = 80x.
Given, 10 kg of syrup is sold with a 10% gain. This means the selling price = 1.1 * 80x = 88x.
Cost of 20 kg of juice = 20 kg * 10x/kg = 200x.
Given, 20 kg of juice is sold with a 20% gain. This means the selling price = 1.2 * 200x = 240x.
Given, Amal sells the mixture, made by combining the remaining juice and syrup, at ₹ 308.32 per kg.
The selling price of the remaining mixture = 308.32 * 200 = ₹ 61664.
Total Selling Price = 61664 + 328x.
Total Purchase Price = 880x + 1200x = 2080x.
Overall gain = 64%.
$61664+328x=\frac{164}{100}\left(2080x\right)$
Solving this equation yields x = 20.
Purchase price for syrup per kg = 8x = 8 * 20 = ₹ 160.

Ans: 1000

Suppose the quantity of pens acquired is $n$.
The initial outlay for these pens amounts to $8n$.
The aggregate expenditure, encompassing the wages, is represented by $8n + W$, where $W$ denotes the wage sum.

In the initial scenario, the selling price (SP) is calculated as:
$SP_1 = 12 \times 100 + 11 \times (n – 100)$
With a profit of 300 in this instance, the equation becomes:
$(12 \times 100 + 11 \times (n – 100)) – (8n + W) = 300$
$1200 + 11n – 1100 – 8n – W = 300$
$3n – W = 200 \quad (Equation \ 1)$

For the second scenario, where a loss of 300 is incurred:
$SP_2 = 12 \times 100 + 9 \times (n – 100)$
The equation for the loss is:
$(8n + W) – (12 \times 100 + 9 \times (n – 100)) = 300$
$8n + W – (1200 + 9n – 900) = 300$
$8n + W – 1200 – 9n + 900 = 300$
$-n + W – 300 = 300$
$W – n = 600 \quad (Equation \ 2)$

Adding Equation 1 and Equation 2:
$(3n – W) + (W – n) = 200 + 600$
$2n = 800$
$n = 400$

Substituting the value of $n$ into Equation 2:
$W – 400 = 600$
$W = 1000$

Therefore, the number of pens is 400, and the total wages amount to 1000.

Ans: B

Explanation:
Let the amounts of tea bought be 2x, 3x, and 5x kg for the qualities priced at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively.

The cost of the first quality of tea is $2x \times 800 = 1600x$.
The cost of the second quality of tea is $3x \times 500 = 1500x$.
The cost of the third quality of tea is $5x \times 300 = 1500x$.

The total cost of the mixture is $1600x + 1500x + 1500x = 4600x$.
The total weight of the mixture is $2x + 3x + 5x = 10x$ kg.

The cost price per kg of the mixture is $\frac{4600x}{10x} = 460$ INR per kg.

The person wants to make an overall profit of 50%.
The total selling price required for a 50% profit is $4600x \times (1 + 0.50) = 4600x \times 1.50 = 6900x$.

One-sixth of the mixture is sold at ₹ 700 per kg.
The weight of the mixture sold is $\frac{1}{6} \times 10x = \frac{10x}{6} = \frac{5x}{3}$ kg.
The revenue from selling one-sixth of the mixture is $\frac{5x}{3} \times 700 = \frac{3500x}{3}$.

The remaining weight of the mixture is $10x – \frac{5x}{3} = \frac{30x – 5x}{3} = \frac{25x}{3}$ kg.

Let the price at which the remaining tea is sold be $P$ INR per kg.
The revenue from selling the remaining tea is $\frac{25x}{3} \times P$.

The total revenue from selling the entire mixture is the sum of the revenue from the two parts:
Total Revenue = Revenue from the first part + Revenue from the remaining part
$6900x = \frac{3500x}{3} + \frac{25x}{3} \times P$

Divide the entire equation by $x$ (assuming $x \neq 0$):
$6900 = \frac{3500}{3} + \frac{25P}{3}$

Multiply the entire equation by 3 to eliminate the denominators:
$6900 \times 3 = 3500 + 25P$
$20700 = 3500 + 25P$

Subtract 3500 from both sides:
$20700 – 3500 = 25P$
$17200 = 25P$

Now, solve for $P$:
$P = \frac{17200}{25}$

To simplify the division:
$P = \frac{17200 \times 4}{25 \times 4} = \frac{68800}{100} = 688$

So, the price at which she should sell the remaining tea is ₹ 688 per kg.

The final answer is $\boxed{688}$.
Correct_Option:B

Ans: B

Explanation:Let $F$ be the fixed monthly expense and $V$ be the variable expense per boarder.
The total monthly expense $E$ can be represented as $E = F + V \times n$, where $n$ is the number of boarders.
The hostel collects ₹ 1600 per month from each boarder. So, the total income $I$ is $I = 1600 \times n$.
The profit $P$ is the difference between income and expense: $P = I – E = 1600n – (F + Vn) = (1600 – V)n – F$.
The profit per boarder is $\frac{P}{n} = \frac{(1600 – V)n – F}{n} = (1600 – V) – \frac{F}{n}$.

Let $P_n$ be the profit per boarder when there are $n$ boarders.
We are given:
When $n = 50$, $P_{50} = 200$.
So, $200 = (1600 – V) – \frac{F}{50}$ (Equation 1)

When $n = 75$, $P_{75} = 250$.
So, $250 = (1600 – V) – \frac{F}{75}$ (Equation 2)

Let $C = 1600 – V$. Then the profit per boarder equation becomes $P_n = C – \frac{F}{n}$.

From Equation 1: $200 = C – \frac{F}{50}$
From Equation 2: $250 = C – \frac{F}{75}$

Subtract Equation 1 from Equation 2:
$250 – 200 = (C – \frac{F}{75}) – (C – \frac{F}{50})$
$50 = -\frac{F}{75} + \frac{F}{50}$
$50 = F (\frac{1}{50} – \frac{1}{75})$
$50 = F (\frac{3 – 2}{150})$
$50 = F \times \frac{1}{150}$
$F = 50 \times 150 = 7500$.

Now, substitute the value of $F$ back into Equation 1 to find $C$:
$200 = C – \frac{7500}{50}$
$200 = C – 150$
$C = 200 + 150 = 350$.

We know $C = 1600 – V$, so $350 = 1600 – V$, which gives $V = 1600 – 350 = 1250$.

Now we need to find the total profit when the number of boarders is 80.
The profit per boarder when $n=80$ is $P_{80} = C – \frac{F}{80} = 350 – \frac{7500}{80}$.
$P_{80} = 350 – \frac{750}{8} = 350 – 93.75 = 256.25$.

The total profit when $n=80$ is $P_{total} = P_{80} \times 80$.
$P_{total} = 256.25 \times 80 = 256.25 \times 8 \times 10 = 2050 \times 10 = 20500$.

Alternatively, we can find the total profit directly using the formula $P = (1600 – V)n – F$.
We found $F = 7500$ and $V = 1250$.
So, $1600 – V = 1600 – 1250 = 350$.
When $n=80$, the total profit $P = 350 \times 80 – 7500$.
$P = 28000 – 7500 = 20500$.

Correct_Option:B

Ans: 20000

Explanation:Let the purchase price of the desktop computer be $D$ rupees and the purchase price of the laptop computer be $L$ rupees.
We are given that the total purchase price is Rs 50000.
So, $D + L = 50000$ (Equation 1)

The desktop was sold at a 20% profit.
Selling price of desktop = $D + 0.20D = 1.20D$

The laptop was sold at a 10% loss.
Selling price of laptop = $L – 0.10L = 0.90L$

The overall profit was 2%.
Total selling price = $1.20D + 0.90L$
Overall profit = 2% of Rs 50000 = $0.02 \times 50000 = 1000$

The total selling price is also equal to the total purchase price plus the overall profit.
Total selling price = $50000 + 1000 = 51000$

So, $1.20D + 0.90L = 51000$ (Equation 2)

Now we have a system of two linear equations:
1) $D + L = 50000$
2) $1.20D + 0.90L = 51000$

From Equation 1, we can express $L$ in terms of $D$:
$L = 50000 – D$

Substitute this expression for $L$ into Equation 2:
$1.20D + 0.90(50000 – D) = 51000$
$1.20D + 45000 – 0.90D = 51000$
$0.30D = 51000 – 45000$
$0.30D = 6000$
$D = \frac{6000}{0.30}$
$D = \frac{6000}{\frac{3}{10}}$
$D = 6000 \times \frac{10}{3}$
$D = 2000 \times 10$
$D = 20000$

The purchase price of the desktop is Rs 20000.

We can also find the purchase price of the laptop:
$L = 50000 – D = 50000 – 20000 = 30000$

Let’s verify the selling prices:
Selling price of desktop = $1.20 \times 20000 = 24000$
Selling price of laptop = $0.90 \times 30000 = 27000$
Total selling price = $24000 + 27000 = 51000$
Total purchase price = $20000 + 30000 = 50000$
Overall profit = $51000 – 50000 = 1000$
Overall profit percentage = $\frac{1000}{50000} \times 100 = 2\%$
The conditions are satisfied.

Final_Answer:20000

Ans: A

Explanation:
Let the labeled price of each toy be L.
Anil buys 12 toys, so the total cost price (CP) of 12 toys is 12 * CP_per_toy.
He sells 8 toys at a 20% discount on the labeled price.
The selling price (SP1) of these 8 toys is L * (1 – 20/100) = 0.8L.
The revenue from these 8 toys is 8 * 0.8L = 6.4L.

He sells the remaining 4 toys at an additional 25% discount on the discounted price.
The discounted price for the remaining 4 toys is the SP1, which is 0.8L.
The additional 25% discount is on this 0.8L.
So, the selling price (SP2) of these 4 toys is 0.8L * (1 – 25/100) = 0.8L * 0.75 = 0.6L.
The revenue from these 4 toys is 4 * 0.6L = 2.4L.

The total revenue Anil gets is the sum of revenue from both sets of toys:
Total Revenue = 6.4L + 2.4L = 8.8L.
We are given that the total revenue is Rs 2112.
So, 8.8L = 2112.
L = 2112 / 8.8 = 21120 / 88 = 240.
The labeled price of each toy is Rs 240.

Anil makes a 10% profit on the total sale.
Total Revenue = CP * (1 + Profit Percentage)
2112 = CP * (1 + 10/100)
2112 = CP * 1.1
CP = 2112 / 1.1 = 21120 / 11 = 1920.
The total cost price of 12 toys is Rs 1920.
The cost price per toy is 1920 / 12 = 160.

Now, we need to find the percentage of profit if there were no discounts.
If there were no discounts, the selling price of each toy would be the labeled price, which is L = Rs 240.
The total revenue without discount would be 12 * 240 = 2880.
The cost price of 12 toys is Rs 1920.
The profit without discount = Total Revenue without discount – Total CP
Profit = 2880 – 1920 = 960.

The percentage of profit without discounts = (Profit / Total CP) * 100
Percentage Profit = (960 / 1920) * 100 = (1/2) * 100 = 50%.

Correct_Option:A

Ans: C

Explanation:Let the cost price (CP) of 1 kg of sugar be Rs. x.
The total cost price of 35 kg of sugar is 35x.

The man wants to make a 20% profit.
Marked Price (MP) per kg for a 20% profit = CP * (1 + Profit%)
MP per kg = x * (1 + 20/100) = x * 1.20 = 1.2x

He sells 5 kg at this marked price.
Revenue from the first 5 kg = 5 * 1.2x = 6x

He sells 15 kg at a 10% discount.
Discounted price per kg = MP * (1 – Discount%)
Discounted price per kg = 1.2x * (1 – 10/100) = 1.2x * 0.90 = 1.08x
Revenue from the next 15 kg = 15 * 1.08x = 16.2x

3 kg of sugar is wasted.

Remaining sugar = 35 kg – 5 kg – 15 kg – 3 kg = 12 kg

He sells the remaining 12 kg by raising the marked price by p percent.
The initial marked price was 1.2x.
The new marked price per kg = 1.2x * (1 + p/100)

Revenue from the remaining 12 kg = 12 * [1.2x * (1 + p/100)] = 14.4x * (1 + p/100)

The overall profit is 15%.
Total Selling Price (SP) = CP * (1 + Overall Profit%)
Total SP = 35x * (1 + 15/100) = 35x * 1.15 = 40.25x

Total SP is the sum of revenues from all sales:
Total SP = Revenue from first 5 kg + Revenue from next 15 kg + Revenue from remaining 12 kg
40.25x = 6x + 16.2x + 14.4x * (1 + p/100)
40.25x = 22.2x + 14.4x * (1 + p/100)

Divide by x (since x cannot be 0):
40.25 = 22.2 + 14.4 * (1 + p/100)
40.25 – 22.2 = 14.4 * (1 + p/100)
18.05 = 14.4 * (1 + p/100)

Divide both sides by 14.4:
18.05 / 14.4 = 1 + p/100
1.25347… = 1 + p/100

p/100 = 1.25347… – 1
p/100 = 0.25347…
p = 0.25347… * 100
p = 25.347…

The value of p is nearest to 25.

Correct_Option: C

Ans: C

Explanation:Let the cost price of the pen be P and the cost price of the book be B.

Case 1: Karim sells the pen at 5% loss and the book at 15% gain.
Loss on pen = 0.05P
Gain on book = 0.15B
Net gain = Gain on book – Loss on pen
7 = 0.15B – 0.05P (Equation 1)

Case 2: He sells the pen at 5% gain and the book at 10% gain.
Gain on pen = 0.05P
Gain on book = 0.10B
Net gain = Gain on pen + Gain on book
13 = 0.05P + 0.10B (Equation 2)

Now we have a system of two linear equations with two variables:
1) 0.15B – 0.05P = 7
2) 0.10B + 0.05P = 13

We can add Equation 1 and Equation 2 to eliminate P:
(0.15B – 0.05P) + (0.10B + 0.05P) = 7 + 13
0.15B + 0.10B = 20
0.25B = 20

To find B, divide both sides by 0.25:
B = 20 / 0.25
B = 20 / (1/4)
B = 20 * 4
B = 80

So, the cost price of the book is Rs. 80.

We can also find the cost price of the pen by substituting B = 80 into either equation. Let’s use Equation 2:
13 = 0.10(80) + 0.05P
13 = 8 + 0.05P
13 – 8 = 0.05P
5 = 0.05P
P = 5 / 0.05
P = 5 / (5/100)
P = 5 * (100/5)
P = 100

Cost price of pen = Rs. 100.
Cost price of book = Rs. 80.

Let’s verify with the given conditions:
Case 1: Pen at 5% loss (0.05 * 100 = 5 loss), Book at 15% gain (0.15 * 80 = 12 gain). Net gain = 12 – 5 = 7. Correct.
Case 2: Pen at 5% gain (0.05 * 100 = 5 gain), Book at 10% gain (0.10 * 80 = 8 gain). Net gain = 5 + 8 = 13. Correct.

The cost price of the book is Rs. 80.

Correct_Option:C

Ans: A

Explanation:Let the cost price of each table be $p$.

Amal buys a table at cost price $p$ and sells it to Bimal at a profit of 20%.
The selling price of the table to Amal = $p * (1 + 20/100) = p * (1 + 0.20) = 1.2p$.
Amal sells his table to Bimal at a profit of 30%.
The amount paid by Bimal, $x$, is the selling price of Amal’s table.
$x$ = Selling price of Amal’s table = $1.2p * (1 + 30/100) = 1.2p * (1 + 0.30) = 1.2p * 1.30 = 1.56p$.

Asim buys a table at cost price $p$ and sells it to Barun at a loss of 20%.
The selling price of the table to Asim = $p * (1 – 20/100) = p * (1 – 0.20) = 0.8p$.
Asim sells his table to Barun at a loss of 30%.
The amount paid by Barun, $y$, is the selling price of Asim’s table.
$y$ = Selling price of Asim’s table = $0.8p * (1 – 30/100) = 0.8p * (1 – 0.30) = 0.8p * 0.70 = 0.56p$.

We need to find the value of $(x – y) / p$.
Substitute the values of $x$ and $y$:
$(x – y) / p = (1.56p – 0.56p) / p$
$(x – y) / p = (1.00p) / p$
$(x – y) / p = 1$.

Correct_Option:A

Ans: C

Explanation:Let CP be the purchase price of one bicycle.
Mukesh purchased 10 bicycles, so the total purchase price is 10 * CP.

He sold six bicycles at a profit of 25%.
Profit per bicycle = 25% of CP = 0.25 * CP
Selling price of one bicycle (at profit) = CP + 0.25 * CP = 1.25 * CP
Total selling price of six bicycles = 6 * (1.25 * CP) = 7.5 * CP

He sold the remaining four bicycles at a loss of 25%.
Loss per bicycle = 25% of CP = 0.25 * CP
Selling price of one bicycle (at loss) = CP – 0.25 * CP = 0.75 * CP
Total selling price of four bicycles = 4 * (0.75 * CP) = 3 * CP

The total selling price of all 10 bicycles = (7.5 * CP) + (3 * CP) = 10.5 * CP

The total profit made is the total selling price minus the total purchase price.
Total Profit = Total Selling Price – Total Purchase Price
Rs. 2000 = 10.5 * CP – 10 * CP
Rs. 2000 = 0.5 * CP

To find CP, we can rearrange the equation:
CP = Rs. 2000 / 0.5
CP = Rs. 4000

Therefore, the purchase price of a bicycle was Rs. 4000.

Correct_Option:C

Ans: C

The sale price of the blended tea is Rs.40 per kilogram.
Let ‘a’ represent the cost per kilogram of tea variety A, and ‘b’ represent the cost per kilogram of tea variety B.
When the two varieties are combined in a 3:2 proportion, a profit of 10% is achieved.
Let ‘x’ denote the cost price of one kilogram of this mixture.
The problem states that 1.1 times the cost price equals the selling price, so 1.1x = 40.
This gives us x = 40/1.1.
The cost price per kilogram of the mixture when A and B are in a 3:2 ratio is calculated as:
(3a + 2b) / 5
Therefore, (3a + 2b) / 5 = 40 / 1.1
Multiplying both sides by 5 and by 1.1, we get:
1.1(3a + 2b) = 200
3.3a + 2.2b = 200 ——–(1)

When the two varieties are combined in a 2:3 proportion, a profit of 5% is achieved.
The cost price per kilogram of the mixture when A and B are in a 2:3 ratio is calculated as:
(2a + 3b) / 5
Therefore, (2a + 3b) / 5 = 40 / 1.05
Multiplying both sides by 5 and by 1.05, we get:
1.05(2a + 3b) = 200
2.1a + 3.15b = 200 ——(2)

Now, we equate equations (1) and (2) since both represent the total value of 200 for different proportions:
3.3a + 2.2b = 2.1a + 3.15b
Rearranging the terms to group ‘a’ and ‘b’:
3.3a – 2.1a = 3.15b – 2.2b
1.2a = 0.95b
To find the ratio of ‘a’ to ‘b’:
a / b = 0.95 / 1.2
Simplifying the fraction by multiplying the numerator and denominator by 100:
a / b = 95 / 120
Further simplifying by dividing both by 5:
a / b = 19 / 24
Thus, the ratio of the cost prices of tea A to tea B is 19:24.

Ans: A

Let the base price of peanuts be represented by 100 units.
Consequently, the base price of walnuts is 300 units.

The shopkeeper’s acquisition cost for peanuts is 110 units per kg.
The shopkeeper’s acquisition cost for walnuts is 360 units per kg.

The total expenditure for the shopkeeper on acquiring the nuts was:
(8 kg peanuts * 110 units/kg) + (16 kg walnuts * 360 units/kg) = 880 units + 5760 units = 6640 units.

Due to transit losses, 5 kg of walnuts and 3 kg of peanuts were lost.
The remaining quantity of nuts with the shopkeeper is:
(8 kg peanuts – 3 kg lost) + (16 kg walnuts – 5 kg lost) = 5 kg peanuts + 11 kg walnuts = 16 kg total nuts.

The total revenue generated from selling the remaining nuts was Rs. 2656 (calculated as 16 kg * Rs. 166/kg).

A profit of 25% was achieved.
This means the selling price is 125% of the cost price.
Therefore, the cost price = Selling Price / 1.25 = Rs. 2656 / 1.25 = Rs. 2124.80.

The total cost incurred by the shopkeeper was Rs. 2124.80.
We established this cost as 6640 units.
So, 6640 units = Rs. 2124.80.

Solving for one unit:
1 unit = Rs. 2124.80 / 6640 = Rs. 0.32.

The original selling price of walnuts was Rs. 300 per kg (in base units).
Therefore, the actual price of walnuts per kg is:
300 units/kg * Rs. 0.32/unit = Rs. 96 per kg.

Thus, option A is the correct choice.

Ans: C

Let the cost of paint B be represented by ‘x’.
Consequently, the cost of paint A is ‘x + 8’.
Given that the quantity of paint B in the mixture is not more than that of paint A, paint B can constitute at most half of the total mixture.
The vendor markets 10 liters of paint for Rs. 264, achieving a 10% gain.
This implies the original expense for 10 liters of the paint blend was Rs. 240.
Thus, the expense for 1 liter of the blend is Rs. 24.
Our objective is to determine the maximum feasible expense of paint B.
As the expense of paint B rises, so does the expense of paint A. If the base expense of the blend approaches the expense of paint B, then the proportion of paint B in the blend ought to be larger than that of paint A.
The greatest achievable expense for paint B occurs when the quantities of paint A and paint B in the blend are identical.
Therefore, the equation becomes: ((x) + (x + 8)) / 2 = 24
Simplifying this yields: 2x + 8 = 48
2x = 40
x = Rs. 20
Hence, option C represents the correct solution.

Ans: D

Explanation:Let the cost price of the item be C and the retail price be R.
The seller gives a discount of 15% on the retail price, so the selling price (S1) is R * (1 – 0.15) = 0.85R.
She still makes a profit of 2% on this selling price. The profit is calculated on the cost price. So, S1 = C * (1 + 0.02) = 1.02C.
Therefore, 0.85R = 1.02C.
From this, we can find the ratio of R to C: R/C = 1.02 / 0.85 = 102 / 85 = 6/5.
This means R = (6/5)C = 1.2C.

Now, we want to find which option ensures a profit of 20%. A profit of 20% means the selling price (S2) should be C * (1 + 0.20) = 1.20C.

Let’s check each option:

Option A: Give a discount of 5% on retail price.
The selling price S2 = R * (1 – 0.05) = 0.95R.
Substitute R = 1.2C: S2 = 0.95 * (1.2C) = 1.14C.
This is not a 20% profit (which requires 1.20C).

Option B: Give a discount of 2% on retail price.
The selling price S2 = R * (1 – 0.02) = 0.98R.
Substitute R = 1.2C: S2 = 0.98 * (1.2C) = 1.176C.
This is not a 20% profit.

Option C: Increase the retail price by 2%.
The new retail price R’ = R * (1 + 0.02) = 1.02R.
If she sells at this new retail price, the selling price S2 = R’ = 1.02R.
Substitute R = 1.2C: S2 = 1.02 * (1.2C) = 1.224C.
This gives a profit of 22.4%, not 20%.

Option D: Sell at retail price.
The selling price S2 = R.
Substitute R = 1.2C: S2 = 1.2C.
This results in a profit of (1.2C – C) / C * 100% = 0.2C / C * 100% = 20%.

Therefore, selling at the retail price ensures a profit of 20%.

Correct_Option:D

Ans: B

Explanation:Let the price of good quality mangoes be $P$ per kg.
Then, the price of medium quality mangoes is $\frac{P}{2}$ per kg.

The shopkeeper buys 80 kg good mangoes, so the cost of good mangoes is $80 \times P$.
The shopkeeper buys 40 kg medium quality mangoes, so the cost of medium quality mangoes is $40 \times \frac{P}{2} = 20 \times P$.

The total cost price (CP) for the shopkeeper is the sum of the cost of good and medium quality mangoes:
Total CP = $80P + 20P = 100P$.

The shopkeeper sells all the mangoes (80 kg + 40 kg = 120 kg) at a common price.
This common selling price (SP) is 10% less than the price at which he bought the good ones.
Price of good ones = $P$.
10% less than $P$ = $P – 0.10P = 0.90P$.

So, the selling price per kg is $0.90P$.

The total selling price (SP) for 120 kg of mangoes is:
Total SP = $120 \times 0.90P = 108P$.

The overall profit is calculated as:
Profit = Total SP – Total CP
Profit = $108P – 100P = 8P$.

The overall profit percentage is:
Profit Percentage = $\frac{\text{Profit}}{\text{Total CP}} \times 100$
Profit Percentage = $\frac{8P}{100P} \times 100$
Profit Percentage = $8\%$.

Correct_Option:B

Ans: D

Let the initial retail value be $R$ and the wholesale cost be $W$.
It is stated that,
$0.6 R = 1.2 W$
$R = 2W$
The total wholesale cost for 60 units is $60W$.
However, only 50 units are left.
Therefore,
$R (1 – \text{discount rate}) \times 50 = 72W$
$1 – \text{discount rate} = \frac{72W}{50R}$
Substituting $R = 2W$:
$1 – \text{discount rate} = \frac{72W}{50(2W)}$
$1 – \text{discount rate} = \frac{72W}{100W}$
$1 – \text{discount rate} = 0.72$
$\text{discount rate} = 1 – 0.72$
$\text{discount rate} = 0.28$
Thus, the discount rate is 28%.

Ans: B

Explanation:Let the manufacturing cost of the table be $C$.

The manufacturer sells the table to a wholesale dealer at a profit of 10%.
So, the selling price of the manufacturer (which is the cost price for the wholesale dealer) is $C + 0.10C = 1.10C$.

The wholesale dealer sells the table to a retailer at a profit of 30%.
So, the selling price of the wholesale dealer (which is the cost price for the retailer) is $1.10C + 0.30(1.10C) = 1.10C(1 + 0.30) = 1.10C(1.30) = 1.43C$.

Finally, the retailer sells it to a customer at a profit of 50%.
So, the selling price of the retailer (which is the price paid by the customer) is $1.43C + 0.50(1.43C) = 1.43C(1 + 0.50) = 1.43C(1.50)$.

We are given that the customer pays Rs 4290 for the table.
Therefore, $1.43C(1.50) = 4290$.

To find the manufacturing cost $C$, we can solve the equation:
$1.43 \times 1.50 \times C = 4290$
$2.145 \times C = 4290$
$C = \frac{4290}{2.145}$

To simplify the division, we can multiply the numerator and denominator by 1000:
$C = \frac{4290000}{2145}$

Let’s perform the division:
$4290000 \div 2145$

We can see that $2145 \times 2 = 4290$.
So, $4290 \div 2145 = 2$.
Therefore, $4290000 \div 2145 = 2000$.

So, the manufacturing cost of the table is Rs 2000.

Let’s verify:
Manufacturing cost = 2000
Manufacturer’s selling price = $2000 \times 1.10 = 2200$
Wholesale dealer’s selling price = $2200 \times 1.30 = 2860$
Retailer’s selling price = $2860 \times 1.50 = 4290$

This matches the price paid by the customer.

Correct_Option:B

Ans: A

Explanation:Let the number of dozens of each type of candy Mayank bought be $x$.
The cost of the first type of candy is Rs 15 per dozen. So, the cost of $x$ dozens is $15x$.
The cost of the second type of candy is Rs 12 per dozen. So, the cost of $x$ dozens is $12x$.
The total cost price (CP) of $2x$ dozens of candies is $15x + 12x = 27x$.

Mayank sells all the candies for Rs 16.50 per dozen.
The total number of dozens sold is $x + x = 2x$.
The total selling price (SP) of $2x$ dozens of candies is $16.50 \times 2x = 33x$.

The profit made is the difference between the selling price and the cost price.
Profit = SP – CP
Profit = $33x – 27x = 6x$.

We are given that the profit is Rs 150.
So, $6x = 150$.
To find the value of $x$, we divide both sides by 6:
$x = \frac{150}{6}$
$x = 25$.

This means Mayank bought 25 dozens of the first type of candy and 25 dozens of the second type of candy.
The total number of dozens of candies he bought altogether is $x + x = 2x$.
Total dozens = $2 \times 25 = 50$.

Thus, Mayank bought 50 dozens of candies altogether.

Let’s verify:
Cost price of 50 dozens = (Cost of 25 dozens at Rs 15/dozen) + (Cost of 25 dozens at Rs 12/dozen)
CP = $(25 \times 15) + (25 \times 12) = 375 + 300 = 675$.

Selling price of 50 dozens at Rs 16.50/dozen:
SP = $50 \times 16.50 = 825$.

Profit = SP – CP = $825 – 675 = 150$.
This matches the given profit.

Correct_Option:A