Important Formulae
1. Percentage Increase = (Actual Increase / Original Quantity) X 100Percentage Decrease = (Actual Decrease / Original Quantity) X 100
2. If the percentage increaseis p%, then the new value is (p/100 + 1) times the old value.
If the new value is k times the old value, then the percentage increase is (k - 1) X 100
3. If there are successive increases of p%, q% and r% in three stages, the effective percentage increase is
{([100+p]/100)([100+q]/100)([100+r]/100)-1} X 100
4. If the value of an item goes up/down by p%, the percentage reduction/increment to be now made to bring it back to the original level is 100p/(100±p)%
If A is p% more/less than B, then B is 100p/(100±p)% less/more than A.
If the price of an item goes up/down by p%, then the quantity consumed should be reduced/increased by 100x/(100±p)% so that the total expenditure remains the same.
5. Given the cost price (C.P.) and profit(+)/loss(-) percentage p%, the selling price will be given by
S.P. = C.P. X (100 ± p)/100
Given the selling price (S.P.) and profit(+)/loss(-) percentage p%, the cost price will be given by
C.P. = S.P. X 100/(100 ± p)
6. When two articles are Sold at the same price such that their Profit of p% on one article and a Loss of p% on the other (i.e., common profit or loss percentage), then, the net result of the transaction is Loss. This percentage loss is given by
Loss Percentage = (Common profit or loss)²/100 = p²/100
7. Selling Price = Marked Price - Discount
Discount percent = [(Marked Price - Selling Price)/Marked Price] X 100
8. If the successive discounts are p%, q% and r%, on a product whose selling price is S.P., then the effective price after all the discounts is given by
Discounted price = S.P X (100-p)(100-q)(100-r)/100X100X100
9. Profits among the partners are shared based on the investments (amount and time period) of partners.
10. Stocks: % of dividend = [Dividend Amount / Par Value] X 100
Examples
Example 1. Water in a cylindrical vessel is poured into another cylindrical vessel whose radius is 10% more than that of the first vessel. By what percent is the height of the water level in the second vessel more or less than that in the first vessel?(A) 9 ¹/₁₁ % less (B) 21% less (C) 17.4% less (D) 17.4% more
Solution:
Let the radius and the height of the water levels in the two vessels be r, h and R, H respectively.
Given, R = 11r/100
𝝅R² = 𝝅r² ⇒ H/h = (r/R)² = 100/121
So, H is 21/121 less than h i.e. 17.4% less than h. Answer: C
Example 2. A salesperson used to get a commission of 9% of his total sales for a month. With a change in the terms of employment, he now gets a fixed amount of Rs.3000 per month plus a commission of only 5% on the excess of his sales over Rs.10,000. If the difference in the amounts that he received last month and this month is Rs.400, though the total sales were the same, what were the total sales this month?
(A) Rs.52,500 (B) Rs.67,500 (C) Rs.57,500 (D) Cannot be determined
Solution:
For a sales of Rs.10000, the person would get Rs.900 and Rs.3000 as per old scheme and new scheme respectively. So, the difference between the amounts from the two schemes is Rs.2100. For the sales below Rs.10000, the difference would be even more. So, let the sales in the two months be Rs.s in each month.
Case I:
New scheme income - old scheme income = 400
⇒ 3000 + 5/100 (s - 10000) - 9/100 (s) = 400
⇒ 2500 - 0.4s = 400 ⇒ s = 2100/0.04 = 52500
Case II:
Old scheme income - new scheme income = 400
⇒ 9s/100 - [3000 + 5/100 (s - 10000)] = 400
⇒ 0.045 - 2400 = 400,
⇒ s = 72500
Since we have two possible answers satisfying required conditions, we cannot determine the answer uniquely. Answer: D
Example 3. All the water in a big tank P is emptied into two smaller empty tanks Q and R. The volume of water in R is 33 ¹/₃ % of that in P. If 300 liters of water, which had gone into Q had instead gone into R, Q would have 50% more water than R. What was the volume of water in P?
(A) 4500 liters (B) 1800 liters (C) 2000 liters (D) 1200 liters
Solution:
From the first condition the water in R is 1/3 of the water in P (we take the volume in P as 3 parts)
From the other condition, we take the volume in Q and R as 3 parts and 2 parts respectively. (i.e. Volume in P is 5 parts)
Let the volume in P be 15x (L.C.M of 3 parts and 5 parts)
A B C
15x 10x 5x ------ I
15x 9x 6x ------ II
From the second condition
10x - 300 = 9x, i.e. x = 300
The volume of water in P is 4500 liters. Answer: A
Example 4: A man sold his computer at a loss of 5%. Had he sold it for Rs.2250 more, he would have made a profit of 10%. What is the cost price of his computer?
(A) Rs.16,000 (B) Rs.24,000 (C) Rs.15,000 (D) Rs.20,000
Solution:
The difference between 5% loss and 10% profit is 15% of the cost price which is Rs.2250.
∴ The cost price is 2250(100/15) = 15,000 Answer: C
Example 5. A sports dealer bought 2000 cricket bats for Rs.350 each. 10% of them were found to be defective and unfit to be sold. At what price should he sell the remaining bats to get an overall profit of 8%?
(A) Rs.400 (B) Rs.380 (C) Rs.420 (D) Rs.450
Solution:
Number of Non-defective bats = 2000 - 10%(2000) = 2000 - 200 = 1800
Total cost of all bats = 2000 X 350 = 7,00,000
Total amount received on selling bats at 8% profit = 7,00,000 + 8% (7,00,000) = 7,00,000 + 56,000 = 7,56,000
This total amount should be derived from selling only the non-defective 1800 bats.
Selling price of each bat = 756000/1800 = Rs. 420 Answer: C
Example 6. By selling 160 calculators, an electronics dealer makes a profit equal to the selling price of 30 calculators. Find his profit percentage.
(A) 13 ²/₁₃ % (B) 15 ¹⁵/₁₉ % (C) 18 ³/₄ % (D) 23 ¹/₁₃ %
Solution:
Let the selling price of each calculator be c.
Total sale = 160c, Profit = 30c, Cost price = 130c
Profit percentage = 30/130 = 3/13 = 300/13 % = 23 ¹/₁₃ % Answer: D
Example 7. A trader cheats both his supplier and customer by using faulty weights. When he buys from the supplier, he takes 10% more than the indicated weight. When he sells to his customer, he gives 10% less than the indicated weight. If he sells at the cost price of the indicated weight, what is his profit percent?
(A) 10 % (B) 22 ²/₉ % (C) 20 % (D) 11 ¹/₉ %
Solution:
Let the supplier sell his goods at Rs.100 per kg.
While buying, the trader takes 1100 gm from the supplier for Rs.100 and while selling, he gives only 900 gm and charges Rs.100.
So, by selling 1100 gm at this rate, the amount he would get = 1100/900 (100) = Rs. 122 ²/₉
∴ He makes a profit of Rs. 22 ²/₉ for every Rs.100 invested. i.e. a profit of 22 ²/₉ % . Answer: B
Example 8. Sarath, Praveen and Vinay start a business with Rs.30,000, Rs.40,000 and Rs.50,000 respectively. Sarath stays for the entire year. Praveen leaves the business after two months but rejoins after another 4 months but only with 3/4 of his initial capital. Vinay leaves after 3 months and rejoins after another 5 months but with only 4/5 of his capital. If the year end profit is Rs.27,900, how much more than Praveen did Vinay get?
(A) Rs.9300 (B) Rs.12,400 (C) Rs.1500 (D) Rs.3100
Solution:
Sarath invested Rs.30,000 for 12 months.
Praveen invested Rs.40,000 for 2 months and Rs.30,000 for 6 months.
Vinay invested Rs.50,000 for 3 months and Rs.40,000 for 4 months.
∴ The annual profit will be divided in the ratio
(30)(12):[(40)(2) + (30)(6)] : [(50)(3) + (40)(4)]
=360 : 260 : 310 = 36 : 26 : 31
Praveen's share = 26/90 (27900) = 7800
Vinay's share = 31/93 (27900) = 9300.
∴ Vinay gets Rs.1500 more than Praveen. Answer: C
Example 9. A wholesale vegetable vendor sold carrots, marked at Rs.1000. Four successive discounts of 10% each, instead of the promised 40%. By what amount did the vendor defraud the customer?
(A) Rs.56.10 (B) Rs.66.10 (C) Rs.58.10 (D) Rs.65.10
Solution:
Four successive discounts of 10%, 10%, 10% and 10% mean the final offer is
(0.9)(0.9)(0.9)(0.9) M (where M is the marked price) = 0.6561M
The promised offer was 0.6M.
The excess amount charged was (0.0561)(1000) = 56.1 Answer: A
Example 10. I bought a refrigerator three years back for Rs.9000. The current price of a similar refrigerator is Rs.16,000. If I sell my refrigerator, in the condition it is in, I would get Rs.10,500. Instead, if I get it painted for Rs.3000, I can sell it for the price of a new one. What discount can I offer, if I expect to make the same percentage profit as I would have made, if I sell without painting?
(A) 8% (B) 9.6% (C) 12.5% (D) 10%
Solution:
The two options are
(1) Sell it in the condition it is in.
The profit is 1/6 = 16 ²/₃ %
(2) Paint & sell at 1/6 profit.
i.e. sell at (9000 + 3000) 7/6 = 14,000
The refrigerator can be marked at Rs.16,000
i.e. a discount of 2000/16,000 = 1/8 = 12.5% can be offered. Answer: C
Example 11. A spectacle frame was marked up by 40% and after that a discount of 20% is offered on it. If the frame cost Rs.500, what is the profit earned by selling it?
(A) Rs.24 (B) Rs.60 (C) Rs.36 (D) Rs.48
Solution:
The cost price is Rs.500
The marked price is Rs.700
The selling price is = 700 (1 - 20/100) = Rs.560
∴ The profit is Rs.60 Answer: B
Example 12. The population of a colony of insects increases by 20% everyday. If one Monday the population is 3000. On which day of the week is it 5184?
(A) Tuesday (B) Wednesday (C) Thursday (D) Friday
Solution:
Let the initial population be P = 3000
After n days it is P (1.2)ⁿ = 5184
3000 (1.2)ⁿ = 5184
(1.2)ⁿ = 1.728 ⇒ n = 3
On Thursday the population would be 5148. Answer: C
Example 13. One year 60% of the students of a school are boys. The next year the number of girls increases by 20% and the total number of students increases by 14%. By what percent does the number of boys increase?
(A) 8% (B) 11% (C) 10% (D) 9%
Solution:
The data is tabulated below:
One Year The Next Year
Boys 60 66
Girls 40 48
Total 100 (say) 114
∴ The number of boys increases by 6/60 = 10% Answer: C
Example 14. A person buys some potatoes at 5 for a rupee and an equal number at 25np each. He sells them at a rate of 9 for Rs.2 but incurs a loss of Rs.5 in the transaction. How many potatoes did the person purchase?
(A) 1,500 (B) 900 (C) 1,800 (D) None of these
Solution:
Let the person buys 2x potatoes in all.
x for 1/5th of rupee each and x for 1/4th of rupee each.
Total cost price = x/5 + x/4 = 9x/20
He sold 2x potatoes at 9 for rs.2
⇒ Selling price = 2/9 (2x),
As 9x/20 > 4x/9, it is a loss of 9x/20 - 4x/9.
Given, 9x/20 - 4x/9 = 5 ⇒ 81x - 80x = 180 X 5
⇒ x = 900
But potatoes purchased = 2x = 2 X 900 = 1800 Answer: C
Example 15. A person bought Rs.7200 worth of 4% stock quoting at Rs.90. When the stock is quoting at Rs.105 he sells it. He invests this money in 3% stock quoting at Rs.70 and sells it when the stock was quoting Rs.80. What is the percentage of profit he makes from the overall transaction?
(A) 33 ¹/₃ % (B) 37 ¹/₂ % (C) 25% (D) 50%
Solution:
Rs.7200 of stock at Rs.90 per share.
⇒ he gets 7200/90 = 80 shares
These 80 shares are sold at Rs.105 and he gets each Rs.105 x 80 = Rs.8400
He reinvests at Rs.70. So, the number of shares = 8400/70 = 120.
This stock is sold at Rs.80. So, he gets 80 x 120 i.e. = Rs.9600
∴ Percentage of profit = (9600-7200)/7200 (100%) = 2400/7200 (100%) = 33 ¹/₃ % Answer: A
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