## Quantitative Aptitude Notes - Profit and Loss

### Quantitative Aptitude Notes - Profit and Loss

Points to Remember:
Cost Price (CP): The price at which an article is brought, including all costs such as transportation, taxes, etc.
Selling Price (SP): The price at which an article is sold.
Profit or Gain: If selling price is greater than cost price, the seller makes profit or gain. Gain = SP – CP.
Loss: If the SP is less than the CP then seller makes loss. Loss = CP – SP
Basic Formulae:
i) Gain = SP - CP
ii) Loss = CP – SP
iii) Gain% = {[(SP – CP) x 100]/CP}
iv) Loss% = {[(CP – SP) x 100]/CP}
v) A person sells an item at x% loss (+ sign) or x% profit (- sign). Had he sold that item for Rs. X more his profit would have been y%. Then
CP = [X/(y ± x) × 100)
vi) If there is a loss of x% on selling an item at selling price (SP1) and also there is a profit of y% on selling price (SP2) then
S.P2 = [SP1(100 + y)/(100 – x)]
vii) When a person sells two similar items, one at a gain of say x%, and at a loss of x%, then the seller always incurs a loss given by:
Loss% = (Common Loss and Gain%/10)2 = (x/10)2
viii) If a trader professes to sell his goods at CP, but uses false weights, then,
Gain% = [Error/ (True value – error)] x 100%
ix) If the Cost price of M articles is equal to the selling price of N articles then the profit percent is given by
P = [(M – N)/N] x 100
Marked Price or List Price: Price that is indicated or marked on the article is called marked price or MP.
Discount: It is reduction given on the Marked Price or List Price of an article.
Discount% = discount/MP x 100%;
Selling Price = (100 – d)/100 x MP
x) If a trader gets x% profit and x% loss in selling two different articles, then in overall transaction, there is always loss which is given by Loss% = (x/2)2
xi) If an article is sold after allowing two successive discounts of d1% and d2% then selling price (S.P) is given by: S.P = [(100 – d1)/100] x [(100 – d2)/100] x MP
xii) Two successive discounts of d1 and d2 are equivalent to a single discount of
d = d1 + d2 – (d1 x d2/100)

MEMORY BASED SOLVED EXAMPLE PROBLEMS BASED ON VARIOUS TYPES
Type 1: Desired Price or Profit
1. If by selling an article for Rs 60, a person loses 1/7 of outlay (cost), what would he have gained or lost per cent by selling it for Rs77?
Solution:
C.P = S.P/(1 – 1/7) = 60 x 7/6 = 70
Profit% = 77 – 70/70 x 100 = 10%
2. If books bought at prices ranging from Rs. 200 to Rs. 350 are sold at prices ranging from Rs. 300 to Rs. 425, what is the greatest possible profit that might be made in selling eight books?
Solution:
Least Cost Price = Rs. (200 * 8) = Rs. 1600.
Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.
Required profit = Rs. (3400 - 1600) = Rs. 1800.
3. An article was purchased for Rs. 78,350/-. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price?
Solution:
Cost Price = Rs.78350
Marked Price = 78350 x 130/100 = Rs.101855
Selling Price = 101855 x 80/100 = Rs.81484
Profit = 81484 – 78350 = 3134
Therefore, Profit% = 3134/78350 x 100 = 4%
4. A man bought an article listed at Rs. 1500 with a discount of 20% offered on the list price. What additional discount must be offered to man to bring the net price to Rs. 1,104?
Solution:
SP of the article = MP – Discount = 1500 – 20% of 1500 = 1500 – 300 = Rs 1200
Therefore, additional discount = (1200 – 1104)/1200 x 100 = (96/1200) x 100 = 8%
5. A man sold one watch and one pen for Rs. 492 and Rs. 168 respectively, gaining 10% of the total C.P of the two articles. Had he sold the watch for Rs. 435 and the pen at its C.P, he would have lost 5% on the total C.P. Find the C.P of the watch.
Solution:
In the 1st case: Total S.P = 492 + 168 = 660
Profit earned = 10%  C.P of both articles = 660 * 100/ 110 = 570.
In the second case: Total loss = 5%
Total SP at 5% loss = 600 – 5% of 600 = 570
Therefore, S.P of watch + C.P of Pen = 570 => 435+ Cost of Pen =570
Therefore, Cost of Pen = 570 – 435 = 135.
Therefore, C.P of watch = Total C.P – C.P of Pen = 660 -135 = 465.
6. Two tables are purchased for the total cost of Rs. 5000. 1st table is sold at 40% profit and 2nd at 40% loss. If S.P is same for both the tables, what is the cost price of the table that was sold at profit?
Solution:
140% of C.P of the 1st table = 60% of the C.P of 2nd table
Therefore, C.P of 1st table: C.P of 2nd table = 60: 140 = 3: 7
Therefore, C.P of 1st table = 3/10 * 5000 = Rs. 1500.
7. A vendor sells lemons at the rate of 5 for Rs. 14, gaining thereby 40%. For how much did he buy a dozen lemons?
Solution:
S.P of 1 lemon = 14 /5 => C.P of 1 lemon = 14 /5 * 100 / 140
C.P of 12 lemons = 14/ 5 * 100 * 140 *12 = Rs. 24
8. A loss of 19% gets converted into a profit of 17% when the S.P is increased by Rs. 162. The C.P of the article is:
Solution:
Difference in two S.P = 17% - (- 19%) = 36% of C.P
Actual difference in two S.P = 162 => C.P = 162/36*100 =Rs. 450
Type 2: Marked Price
1. The cost price of an article is Rs. 800. After allowing a discount of 10%, a gain of 12.5% was made. Then the marked price of the article is
Solution:
CP = Rs 800
Gain% = 12.5%
SP = 800 + 12.5% of 800 = 800 + 25/200 x 800 = 800 + 100 = Rs 900
MP = SP + Discount
MP = 900 + 10% of MP
MP – MP/10 = 900
9MP/10 = 900
MP = (900 x 10)/9 = Rs 1000
2. The marked price of an article is 50% above cost price. When marked price is increased by 20% and selling price is increased by 20%, the profit doubles. If original marked price is Rs.300, then original selling price is
Solution:
Original MP = 300; Therefore Original CP = 300/150 x 100 = 200
Let the original SP be x. Then Original profit = x – 200
If the SP is increased by 20% then new SP = 120/100 × x = 1.2x
Then new Profit = 1.2 x – 200
Given 1.2 x – 200 = 2(x – 200) =>2x – 1.2 x = 400 – 200 =>0.8 x = 200 => x = 200/0.8 = 250
3. Arun bought a computer with 15% discount on the labelled price. He sold the computer for Rs.2880 with 20% profit on the labelled price. At what price did he buy the computer?
Solution:
SP = 2880
Arun sold the computer with 20% profit on the Labelled Price
=> 2880 = 120 × Labelled Price/100 => Labelled price = 28800/12 = 2400
Arun bought a computer with 15% discount on the labelled price
=> CP = LP × 85/100 = 2400 × 85/100 = 2040
4. By selling an article at 80% of its marked price, a trader makes a loss of 10%. What will be the profit percentage if he sells it at 95% of its marked price?
Solution:
80% of marked price = 100% - 10% = 90%
Therefore, 95% of the marked price = 95/80*90 = 106. 9% (approx) of the C.P
Profit = (106.9 -100 )% = 6.9 % of the C.P
5. A cloth merchant has announced 25% rebate on marked price. If one needs to have a rebate of Rs. 40, then how many shirts each with a marked price of Rs. 32, should be purchased?
Solution:
Shirts Purchased = (40/25%) / 32 = 160 /32 = 5.
6. While selling a cooler, a shopkeeper gives a discount of 10% on the marked price. If he gives a discount of 12%, he earns Rs. 35 less as profit. The marked price of the cooler is:
Solution:
Let MP = Rs. 100  S.P in 1st case = 100-10=Rs. 90
SP in 2nd case = 100 – 12 = Rs. 88 => Diff. Between two SP= 90-88 = Rs. 2
If difference is Rs. 2, MP = 100 => If difference is Rs. 35, MP = 35 * 100/2 = Rs. 1750
Alternative Method:
12% -10% = 2% on MP = Rs. 35 => MP = 35 * 100/2 = Rs. 1750.
Type 3: Formula based method
1. If the cost price of 22 items is equal to the selling price of 20 items then the profit or loss incurred is
Solution:
SP of 20 items = CP of 22 items (It should be profit)
Profit % = [(22 – 20)/20] x 100% = 10% profit
2. A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5: 3. If he sells the blended variety at Rs 21 per kg, then his gain percent is
Solution:
Suppose he bought 5 kg and 3 kg of tea
Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150
Selling price = Rs. (8 x 21) = Rs. 168.
Profit = 168 - 150 = 18
So, Profit % = (18/150) * 100 = 12%
3. A shopkeeper purchases 12 balloons for Rs.10 and sells them at 10 balloons for Rs.12.Thus, he earns a profit of
Solution:
Number of Balloons purchased and sold be = LCM (12,10) = 60
CP = 60/12 x 10 = 50; SP = 60/10 x 12 = 72
Profit = (72 – 50)/50 x 100 = 22/50 x 100 = 44%
4. Ravi buys some toffees at 2 for a rupee and sells them at 5 for a rupee. His loss per cent is:
Solution:
Let toffees bought = LCM of 2 and 5 = 10
Then, C.P = 5; S.P = 2; Loss= 5-2=3 => Loss% = 3/5*100 = 60%
Alternative Method:
CP: SP = 1/2: 1/5 = 5: 2 => Loss % = 3/5*100 = 60%
Type 4: Dishonest Dealer
1. A trader has a weighing balance that shows 1200 gm for a Kg. He further marks up his C.P by 10%. Then the net profit percentage is:
Solution:
The merchant gives 1000 gms charging the price of 1320 (=1200+10%) gm.
Therefore, Net Profit = 320 / 1000*100= 32%
2. A retailer professes to sell his goods at C.P. If using a false weight, he still gains 25%, find the weight he uses in place of 1 Kg.
Solution:
Let defraud = x grams per Kg. Then x/(1000-x) * 100 =25 => 4x = 1000-x
=> 5x = 1000 => x =200 => He uses 1000 -2300 = 800 grams for one Kg.
Trick: x / (1000-x) * 100 =25 => x / (1000-x) =25 / 100 = 1/4 => x /1000 = 1 /5 = 200g
Type 5: Profit & Loss
1. A machine is sold at a loss of 10%. Had it been sold at a profit of 15%, it would have fetched Rs. 50 more. The cost price of the machine is:
Solution:
Difference in two selling prices = 10% - (-15%) = 10% + 15% = 25% of cost price
Actual difference in two selling price = Rs. 50 (i.e. 2 times of 25)
Therefore, Cost Price = 2 × Rs. 100 = Rs. 200
2. A person sold a horse at a gain of 15%. Had he bought it for 25% less and sold it for Rs. 600 less, he would have made a profit of 32%. The cost price of the horse was:
Solution:
Let the cost price of the horse be Rs. 100. Then its SP = 100 + 15 =Rs. 115
In the second case: CP = 100- 25= Rs. 75
Profit = 75 * (32/100) =Rs. 24; SP = 75 + 24 = Rs. 99;
Difference between two selling prices = 115 – 99 = Rs. 16
But actual difference = Rs. 600 => CP = 600/16 * 100= Rs. 3750
Alternative Method:
115% of CP – Rs. 600 = 75% of 132% of CP
Therefore, 115% of CP - 600 = 99% of CP
Therefore, 16% of CP = 600 CP = 600/ 16 *100 = Rs. 3750
3. A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs. 56 on one item, his advertised price of the item, (in Rs.) is:
Solution:
Profit =56 => C.P (at 10% profit) =56 * 10 =Rs. 560
=> S.P = 560 * 11/10 => Advertised price = 560 * 11/ 10 *100/ 77 = Rs. 800
4. By selling 8 dozen pencils, a shopkeeper gains the S.P of 1 dozen pencils. What is the gain?
Solution:
Gain% = 1/ (8-1)*100 = 1/7 *100 = 14 2/7%
5. A merchant sells his goods at 10% profit but he mixes low quality of goods to one – fourth of good quality of goods. If the price of low quality of goods is four-fifth of that of good quality of goods, what is his profit percentage?
Solution:
Good quality goods : Low quality goods = 4 : 1
Total CP of goods whose SP is 110 = 100*4/5 + 100*1/5*4/5 = 80+16 = Rs. 96
Therefore, Profit = Rs. 14 on Rs. 96 = 14 / 96 *100% = 175/12% = 14 7/12%
6. A shopkeeper sold one- third of his articles at 16% profit and the remaining at 10% profit. Find his overall profit Per cent.
Solution:
Let total articles sold are 3 of Rs. 100(C.P) each.
Then, Total gain = 1 *16 + 2* 10 = Rs. 36 => Profit Percent = 36/3 = 12%
Alternative Method:
Overall Profit = 1/3 * 16% + 2/3 * 10% = 36 / 3 = 12%
7. The profit earned after selling an article for Rs. 1754 is same as loss incurred after selling the article for Rs. 1492. What is the C.P of the article?
Solution:
Let Loss = gain =x => C.P = 1754 – x = 1492 + x => 2x = 1754 -1492 =262 => x =131
Therefore, C.P = S.P of 2nd + Loss = 1492 + 131 = 1623
Or, C.P = S.P of 1st - Profit = 1754 - 131 = 1623
Alternative Method:
Let Loss = gain = x
Then, difference in two S.P = 2x = 1754 – 1452 => 2x = 262 => x =131
Therefore, C.P = S.P of 2nd + Loss = 1492 + 131 = 1623
8. Raghavan purchased a scooter at 13/15 of its S.P and sold it at 12% more than its S.P. His gain is:
Solution:
Let marked price be Rs. X. Then C.P = (13/15)x; S.P = (112/100)x = (28/25)x
(SP/CP) * 100 = (28/25)x ÷ (13/15)x * 100 = 28/25 * 15/13 * 100 = 1680 /13 = 129 3/13%
Therefore, Profit = 129 3/13% - 100% = 29 3/13%
Type 6: Successive Selling
1. An article listed at Rs. 800 is sold at successive discounts of 25% and 15%. The buyer desires to sell it off at a profit of 20% after allowing a 10% discount. What would be his list price?
Solution:
SP = 800 x 75/100 x 85/100 = Rs 510
CP for the buyer = Rs 510
Profit = 20%
SP = 510 + 20% of 510 = 510 + 102 = Rs 612
SP = MP –Discount
612 = MP – 10% of MP
612 = MP – MP/10 = 9MP/10
MP = (612 x 10)/9 = Rs 680
2. Rajesh brought an old scooter and spent 1000 on its repair after one month he sold the scooter for 6500 and earned a profit of 10% on the cp at what cp did Rajesh purchase the scooter?
Solution:
Let the price at which Rajesh purchased the scooter be x
Total Cost Price = (1000 + x)
Total Selling Price = 6500
Profit = 10%
(1000+x)*110/100 = 6500
=> 1000+x = 6500*100/110 = 5909.09
x = 5909.09 - 1000 = 4909.09
Required Price = Rs.4909.09
3. A man sold an article at a loss of 10%. Had he bought it for 20% less and sold it for Rs. 55 more, he would have made a profit of 40%. Original cost price of the article is:
Solution:
Let CP of the article = Rs. 100. Then its SP = 100- 10 = 90.
In the second case, CP = 100- 20 =80. SP = 80 *140/100 = 112.
Difference between two SP = 112 – 90 = 22
But actual difference = 55 => CP = 55/ 22* 100= 250.
Alternative Method:
90% of CP + 55 = 140% of 80% of CP
90% of CP + 55 = 112% of CP => 22% of CP = 55 => CP = 55 * 100 /22 =250.
4. A sells a table costing Rs. 2000 to B and earns a profit of 6%. B sells it to C at a loss of 5%. At what price did B sell the table?
Solution:
Difference = 6 – 5 – 6*5 /100 = 0.7%
Therefore, Required Price = 2000 + 00.7 % of 2000 = 2000 + 14 = 2014
Alternative Method 1:
Required price = 2000 * (100+6) / 100 * (100-5) / 100 = 2000 * 106/100 * 95/ 100 =Rs. 2014
Alternative Method 2:
S.P for A = 2000 + 6% of 1200 = 2120
S.P for B = 2120 – 54% of 2120 = 2120 – 106 = Rs. 2014
Type 7: Same Selling Price for Two Articles
1. A man sold two articles for Rs. 600 each. On selling first, he gains 20% and on the other, he gains 30%. What is profit per cent in the transaction?
Solution:
Let S.P = LCM of 12 and 13 = Rs. 156 of each article => Total S.P = 2 *156 = Rs. 312
Total cost price = 156 * (100/120) + 156 *(100/130) = 130 + 120 = Rs. 250
Total Profit = 312 – 250 = Rs. 62
% Profit = 62/250 * 100 = 24.8 %
Alternate Method:
Profit % = [2 × (100 + x) (100 + y)]/[(100 + x) + (100 + y)] – 100 = = 2*120*130 / (120+130) – 100 = 124.8 – 100 = 24.8%
Type 8: Miscellaneous
1. Pankaj purchased an item for Rs.7500 and sold it at the gain of 24%. From that amount he purchased another item and sold it at the loss of 20%. What is his overall gain or loss?
Solution:
CP1 = Rs.7500; SP1 = 7500 x 124/100 = Rs.9300 = CP2
SP2 = 9300 x 80/100 = 7440. Here CP1 > SP2 (Hence, loss is incurred here)
Therefore Loss = 7500 – 7440 = Rs.60
2. Vijay purchased a Washing machine and a Television for Rs.15,400 and Rs.19,600respectively. He sold washing machine for a profit of 15 percent and the television for a loss of 20 percent. What is his overall loss/profit?
Solution:
CP of WM = 15400, P = 15%; CP of Tel = 19600, L = 20%
SP of WM = 115/100 x 15400 = 17710; SP of Tel = 80/100 x 19600 = 15680
Total CP = 15400 + 19600 = 35000
Total SP = 17710 + 15680 = 33390
Since CP is greater than SP loss is occurred. Loss = 35000 – 33390 = 1610
3. If the selling price of an article is doubled, then its loss per cent is converted into equal profit per cent. The loss per cent on the article is
Solution:
Let CP = C, SP = S, Loss% = x %
=> x = (C – S)/C x 100 ------- (1)
When SP is doubled, loss% becomes profit%.
x = (2S – C)/C x 100 ------ (2)
From eqns (1) and (2)
2S – C = C – S
3S = 2C => S = 2/3 C
Substituting b = 2/3 a in eqn (1)
x = (C – 2/3 C)/C x 100= (C/3)/C x 100 = 100/3 = 33 1/3%
4. The Cost price of an item increases by 10% first then 10% next and by 10% yet again and finally the last 10% increases is equivalent to an increase of Rs.121. What was the first 10% equivalent to
Solution:
If CP is initially A then the value after increases are 1.1A, 1.21A and 1.331A
Since 0.121A = Rs.121 then A = Rs.1000. Therefore 0.1A is equivalent to Rs.100
5. A man purchased 30 kg of rice at the rate of Rs. 17.5. Per kg and another 30 kg rice at a certain rate. He mix the two and sold the entire quantity at the rate of Rs. 18.60/kg and made 20%overall profit. At what Price per kg did he purchased the lot of another 30 kg rice?
Solution:
Let the price per kg of the other 30 kg rice be x
Cost price is 30*17.5+30*x = 30(17.5+x)
Selling Price = 60*18.60 = Rs.1116
Profit is 20%
Cost Price * 120/100 = Selling Price
30(17.5+x) * 120/100 = 1116
17.5+x = 31 => x = 13.5
6. The profit of selling 10 shampoos is equal to the selling price of 3 deodorants and the loss of selling 10 deodorants is equal to the selling price of 4 shampoos. Profit % and the cost of shampoos is half the cost of a deodorant. Find the ratio of selling price of a shampoo to the selling price of a deodorant?
Solution:
Let cost price of a shampoo be x and cost price of a deodorant be 2x
Let selling price of a shampoo be s and selling price of a deodorant be d
Profit of selling 10 shampoos = Selling price of 3 deodorants
10(s-x) = 3d ------(1)
Loss of selling 10 deodorants = Selling price of 4 shampoos
10(2x-d) = 4s ------(2)
From (1)
10s - 10x = 3d
10x = 10s - 3d --------(A)
From (2)
20x - 10d = 4s => 10x - 5d = 2s
10x = 5d + 2s --------(b)
From (A) and (B)
10s - 3d = 5d + 2s
8s = 8d => s/d = 1 => Required ratio = 1 : 1
7. An article is sold at a profit of 20%. If both the cost price and selling price are Rs. 100 less, the profit would be 25%. The cost price of the article is:
Solution:
Let CP = Rs. X => SP = 1.2x => New CP = x – 100 and New SP= 1.2x – 100;
Now (5/4) (x-100) = 1.2x – 100 => 5x -500 = 4.8x – 400 => 0.2x = 100 => x = Rs. 500
TRICK: CP = New Profit%/Increase in profit % * Decrease in Amount = 25/5*100 = 500
Note: Above formula can be used when CP and SP are reduced by same Amount.
8. An article is sold for Rs. 500 and hence lost something. Had the article been sold for Rs. 700 the merchant would have gained three times the former loss. Find the C.P of the article:
Solution:
Let Loss in 1st case = Rs x; Then, Profit in 2nd case =Rs. 3x;
Therefore, 500 = C.P – x ----(1) and 700 = C. P + 3x ----(2)
Solving equations (1) and (2), we get: 4x =200 => x =50
Therefore, C.P = S.P of 1st +Loss= 500+50=550
Or, C.P = S.P of 2nd - Profit= 700 – 3*50=550
Alternative Method:
Let loss in the 1st case = Rs. X; Then, Profit in 2nd case =Rs. 3x;
Difference between two S.P = x +3x = 700 – 500 => 4x = 200 => x= 50
Therefore, C.P = S.P of 1st +Loss= 500+50=550.
More on Profit and Loss:

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