Profit and Loss Questions with Solution (Advanced)

Solution:

CP = 80 × 40

Profit from the n objects = n% × 40 × n.

Profit from the remaining objects = (100 – n)% × 40 × (80 – n). We need to find the minimum possible value of n% × 40 × n + (100 – n)% × 40  × (80 – n).

Or, we need to find the minimum possible value of n² + (100 – n) (80 – n). Minimum of n² + n² – 180n + 8000

Minimum of n² – 90n + 4000

Minimum of n² – 90n + 2025 – 2025 + 4000

We add and subtract 2025 to this expression in order to crate an expression that  can be expressed as a perfect square.

This approach is termed as the "Completion of Squares" approach. We keep

revisiting this in multiple chapters.

Minimum of n² – 90n + 2025 + 1975 = (n – 45)² + 1975

This reaches minimum when n = 45.

When n = 45, the minimum profit made

45% × 40 × 45 + 55% × 40 × 35

18 × 45 + 22 × 35 = 810 + 770 = 1580

Solution:

It is given that a merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him.

Hence, the cost price of 105 cm clothes is 100 rupees.

It is also known that he marked the price of 100 cm clothes as 110 rupees,  and gave a 5% discount, and he cheated his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers.

Hence, the selling price of 95 cm clothes is 110*(19/20) rupees. Therefore, the selling price of 105 cm clothes is 115.5 rupees. Hence, the profit is 15.5%

Solution:

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A  is 5/6 p

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of  B is 10/9 p

Now, she sold them at a price such that a 10% profit is made on B Selling price = s = 11/10 * 10/9 p → 11/9 p

Profit % on A = [(11/9 - 5/6) / (5/6)] * 100 = 46.666% → 47%

Solution:

The cost price of the sunglass for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the

selling price of the sunglass is 1200 rupees, which Kanu purchased. Hence,  the profit made by Meenu is (1200-1000) = 200 rupees.

Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.

The cost price of the same sunglass for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.

Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.

Hence, the profit made by Minu in the second transaction is

(300/960) * 100% = 31.25%

Solution:

The cost price of the sunglass for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the

selling price of the sunglass is 1200 rupees, which Kanu purchased. Hence,  the profit made by Meenu is (1200-1000) = 200 rupees.

Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.

The cost price of the same sunglass for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.

Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.

Hence, the profit made by Minu in the second transaction is

(300/960) * 100% = 31.25%

Solution:

Let the number of white shirts be m, and the number of blue shirts be n.  Hence, the total cost of the shirts = (1000m+1125n), and the number of  shirts is (m+n)

The average price of the shirts is (1000m + 1125n)/(m+n)

It is given that he set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10%.

Hence, the average selling price of the shirts = [(1000m + 1125n)/(m+n)] *  5/4 * 9/10 = 9/8 * (1000m + 1125n)/(m+n)

The average profit of the shirts = 1/8 * (1000m + 1125n)/(m+n) →  1/8  *(1000m + 1125n) = 51000

1000m + 1125n = 51000 × 8 = 408000

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero.  Therefore, m has to be maximum.

m = (408000 – 1125n)/1000

The maximum value of m such that m, and both are integers is m = 399, and  n = 8 (by inspection)

Hence, the maximum number of shirts = m + n = 399 + 8 = 407

Solution:

profit and loss formula

Let SP of the car = x.

Discount of 10%. Profit = 8%.

∴ ((90/100)x - 135000/135000)×100 = 8

⇒ (90/100)x - 135000 = 8×1350 ⇒ (90/100)x ⇒ x = 162000

Solution:

CP = 100. Profit = 10%. SP = 110. The dealer could pay only 72 paisa in the rupee. ∴Net amount paid to merchant by the dealer would be = 0.72 × 110 =  Rs 79.2. Now as the merchant has sold Rs. 100 item to the dealer at Rs.  79.2,

hence the net loss to the merchant would be = 100 – 79.2 = 20.8%

Solution:

CP of 2 pens = 740. Let CP of 1^st pen is x and CP of 2^nd pen is y. Since there is no profit and loss in the whole transaction, so 12% of x = 8% of y ⇒x: y = 2: 3

Hence the cost of first pen = (2/3) × 740 = Rs296 and that of the second pen =  (3/5) × 740 = Rs444

Solution:

SP of first TV = Rs. 4800, the CP of first TV = 4800 × 100/120 = Rs 4000,  so the profit earned is Rs. 800.

Now he must get a loss of Rs. 800 by selling second TV as he neither gaining nor losing on the whole.

Now Rs. 800 must be 16% of the CP of the second TV i.e. 800 × 100/16 = Rs.  5000,

but the question is asking the SP of the second TV, which will be 5000 – 800 = Rs.  4200.

Solution:

As profit and loss are calculated of the same figure i.e. the cost price. Hence you can straightaway find 12% is what percent of 8% i.e. 100 × 12/8  =150%.

Solution:

If L represents the loss, then 14 L represents the profit and the sum of the two is difference of the two selling prices.

Hence 15L = 1050 – 375 = Rs 675 ⇒ L = 675/15 = Rs45

Since the second SP = Rs 375, so the CP = 375 + 45 = 420.  The SP to have 20% profit is 420 × 1.2 = Rs. 504

Solution:

Let CP of the article = 100 ∴ Old SP = 108. New CP = 90.

As the profit is 15%, so the new SP = 90 × 115/100 = 103.5.  The difference in the two selling prices = 108 – 103.5 = Rs 4.5  If difference in SP is 4.5 then CP = 100,

If difference in SP is 18 then CP = (100/4.5) × 18 = Rs 400

Solution:

Uses 833.33 gms instead of 1000 gms

∴ % profit = (1000-833.33/833.33) × 100 = 20%

Shopkeeper also states that he makes a gain of 15%.

There will be a mutual impact of the two as well.

So the formula that can be applied is x + y + xy/100.

Applying that you get the answer as 20 + 15 + (20)(15)/100 = 38%

Solution:

Let CP = 100 ∴ SP = 120. 10% of 120 = 120 × (10/100) = Rs 12 ∴ Real CP = 120 – 12 = 108.

Here the difference between the CP and real CP can be taken as overheads =  108 – 100 = Rs 8.

Now apply unitary method to find the answer i.e. If overheads is 8 then SP =  120

∴ If overheads is 33200, then SP = (120/8) × 33220 = Rs 498000

Solution:

Amount paid in 1^st service = Rs.100

Amount paid in 2^nd service = Rs.90

Amount paid in 3^rd service = Rs.81

Amount paid in 4^th service = Rs. 72.9

Amount paid in 5^th service = Rs.60

Total amount paid= Rs. 403.9

Discount = 500 – 403.9= 96.1

Discount % = 96.1/500= 19.42%