1. The cost price of a particular amount of articles is same as the number of articles. Selling price of the articles comes out to be 20. Over the whole transaction 25% is gained. What is X?

A. 15 B. 18 C. 16 D. 20

Answer: Option C
Explanation: According to the question let n be the number of articles. => Cost price of n articles = n => Selling price of n articles = 20 => Profit % = 25% => Profit % = [latex]\frac{(SP - CP)}{CP }[/latex] X 100 Putting the values in the formula [latex]\frac{(20-n)}{n }[/latex] x 100 = 25 => [latex]\frac{5n}{4 }[/latex] = 20 => n = 16
2. Akshay buys a motorcycle for Rs.25000. If he decides to sell the motorcycle for a profit of 7%, find the selling price.

A. 27000 B. 26750 C. 26000 D. 25750

Answer: Option B
Explanation: Profit = [latex]\frac{%profit}{100 }[/latex] * cost price P =[latex]\frac{7}{100 }[/latex] * 25000 = 1750. Selling price = cost price + profit = 25000 + 1750 = Rs.26750
3. If the money was put in for the same duration of time by A,B and C, three business partners and four times A’s capital is equal to 6 times B’s capital is equal to 10 times C’s capital. Determine that out of a total profit of Rs 4650 what is C’s share?

A. Rs 2250 B. Rs 1550 C. Rs 450 D. Rs 900

Answer: Option D
Explanation: It is given that business partnership is independent of time in this case as all the investors have invested their money for same amount of time. Let A’s capital be x, B’s capital is y and C’s capital be z. =>x+y+z = 4650 => 4x = 6y = 10z =>C’s share is z so let us solve the question w.r.t. z =>x = [latex]\frac{5z}{2 }[/latex] =>y = [latex]\frac{5z}{3 }[/latex] =>([latex]\frac{5z}{2 }[/latex])+([latex]\frac{5z}{3 }[/latex])+(z) = 4650 =>31z = 4650*6 =>z = 900
4. Shweta buys a product at a 25% discount rate. At what % over the cost price should she sell the product to make an overall 25% profit over the listed price?

A. 22 B. 33.333 C. 66.667 D. 40

Answer: Option C
Explanation: Hence at 25% discount rate the Cost Price will become: 3x/4 Now to make an overall profit of 25% over the listed price: SP = CP+(25/100)CP => SP = X + [latex]\frac{X}{4 }[/latex] =>SP = 5 [latex]\frac{X}{4 }[/latex] =>Selling Price over Cost price: (5 [latex]\frac{X}{4 }[/latex])-(3 [latex]\frac{X}{4 }[/latex]) =>x/2 =>%Selling price over Cost price: ( [latex]\frac{X}{2 }[/latex])/(3 [latex]\frac{X}{4 }[/latex])*100 =>66.67%
5. A man sells 45 lemons for Rs 40 and loses 20%. At how much price should he sell 24 lemons to the next customer to make a 20% profit?

A. 32 B. 20 C. 24 D. 16

Answer: Option A
Explanation: Let cost price of lemons be x. The selling price of lemons becomes CP - loss. =>SP = x- ([latex]\frac{20}{100 }[/latex])x =>40 = x([latex]\frac{80}{100 }[/latex]) =>50 So 45 lemons cost Rs 50 Cost of 1 lemon = [latex]\frac{50}{45 }[/latex] Rs Cost of 24 lemons = ([latex]\frac{50}{45 }[/latex])*24 = [latex]\frac{80}{3 }[/latex] Rs Selling Price of 24 lemons =([latex]\frac{80}{3 }[/latex])+([latex]\frac{20}{100 }[/latex])([latex]\frac{80}{3 }[/latex])=Rs 32

1. What is the CP of Rs 100 stock at 4 discount, with [latex]\frac{1}{5 }[/latex]% brokerage?

A. 99.6 B. 96.2 C. 97.5 D. 98.25

Answer: Option B
Explanation: Use the formula, CP = 100 – discount + brokerage% CP = 100 - 4 + [latex]\frac{1}{5 }[/latex] 96.2 Thus the CP is Rs 96.2.
2. An employer pays Rs. 30 for each day a worker works and forfeits Rs. 5 for each day he is idle. At the end of 60 days, a worker gets Rs. 500. For how many days did the worker remain idle?

A. 35 B. 48 C. 52 D. 58

Answer: Option C
Explanation: Suppose the worker remained idle for m days. Then, he worked for (60 - m) days. 30 (60 - m) – 5m = 500 1800 – 25m = 500 25m = 1300 m = 52 So, the worker remained idle for 52 days.
3. P and Q together can complete a piece of work in 4 days. If P alone can complete the same work in 20 days, in how many days can Q alone complete that work?

A. 8 B. 7 C. 4 D. 5

Answer: Option D
Explanation: (P + Q)'s 1 day's work = [latex]\frac{1}{4 }[/latex], P's 1 day's work = [latex]\frac{1}{20 }[/latex] Q's 1 day's work = ([latex]\frac{1}{4 }[/latex] – [latex]\frac{1}{20 }[/latex]) = ([latex]\frac{4}{20 }[/latex]) = ([latex]\frac{1}{5 }[/latex]) Hence, Q alone can complete the work in 5 days.
4. If 10 bulls can plough 20 identical fields in 3 days working 10 hours a day, then in how many days can 30 bulls plough 32 same identical fields working 8 hours a day?

A. 2 B. 4 C. 8 D. 10

Answer: Option B
Explanation: M1*D1*W2 = M2*D2*W1 10*3*10*32 = 30*d*8*20 d = 2 days
5. A pack of people can complete a certain amount of work in 12 days. Two times the same number of persons will complete half of the work in?

A. 12 days B. 3 days C. 4 days D. 6 days

Answer: Option D
Explanation: More no of people: Less days (Inverse Relationship) More work: More days (Direct Relationship) The ratio is given: Persons: 1:: 2 Work 1:: [latex]\frac{1}{2 }[/latex] Time 12:: x The solution of this can be explained by solving this ratio by the sort of relationships they possess with time. 1*[latex]\frac{1}2 }[/latex]*12 = 2*1*x => x = 3 days

1. 12 men and 18 boys working 7.5 minutes an hour complete a particular piece of work in 60 hours. If a boy’s work is half as efficient as a man’s work, how many boys will be needed to help 21 men to achieve twice the work in 50 hours working 9 minutes in an hour.

A. 42 B. 46 C. 48 D. 38

Answer: Option A
Explanation: 1 man is equivalent to 2 boys in work capacity. 12 men + 18 boys = 12 x 2 + 18 = 42 boys. Let the required number of boys be x. So a Total number of people doing work = 21 men x 2 (according to work capacity) + x boys. Taking respective ratios as required: Given: => Hours 50:60 => Minutes per hour 9: [latex]\frac{15}2 }[/latex] => Work 1:2 => People: 42: 42 + x => 50 x 9 x 1 x (42+x) = 60 x [latex]\frac{15}{2 }[/latex] x 2 x 42 => 42 + x = 84 x = 42
2. 5/8th of a job is completed in 10 days. If a person works at the same pace, how many days will he take to complete the job?

A. 4 B. 5 C. 6 D. 7

Answer: Option C
Explanation: Solution: It is given that 5/8th of the work is completed in 10 days. => Remaining work = [latex]\frac{3}{8 }[/latex]th of total Applying the unitary method: Total work will be completed in 10 * [latex]\frac{8}{5 }[/latex] days => It takes 16 days to complete total work => Hence, remaining work days = 16 - 10 = 6 days
3. If the ratio of present ages of Jeet and Jay is 5:7 and after 6 years the ratio will be 3:4, what is the present age of Jay?

A. 42 B. 30 C. 36 D. None of these

Answer: Option A
Explanation: As the present age of Jeet and Jay are in the ratio 5:7, let their ages be 5x and 7x respectively. Therefore, their ages after 6 years will be (5x+6) and (7x+6) respectively. Now, it is given that [latex]\frac{(5x+6)}{(7x+6) }[/latex] = [latex]\frac{3}{4 }[/latex] 4*(5x+6) = 3*(7x+6) x = 6 Hence, the present age of Jay is 7x = 7*6 = 42 years
4. Find the value of x when y = 5, if x varies directly as 4y-1 and x = 14 when y = 2.

A. 38 B. 35 C. 10 D. 28

Answer: Option A
Explanation: Let z = 4y-1 When x = 14, y = 2, z = (4*2) – 1 = 7 Now, x varies directly as z = 4y-1 When y = 5, z = (4*5) – 1 = 19 x 14 7 y - 19 Therefore, x = [latex]\frac{(14*19)}{7 }[/latex]= 38
5. In what ratio must one add water to milk so as to gain 16.666% on selling this mixture at the cost price?

1:6 B. 1:3 C. 6:1 D. 3:1

Answer: Option A
Explanation: To start off this question let us assume that cost price of 1 litre milk is Rs 1 No need to make a mixture and sell this mixture at 1 Rs per liter such that the total gain on the mixture is 16.667%. Therefore, CP of 1 liter of the mixture becomes (quantity of milk)/ (quantity of mixture containing 1 L milk)*(the price of 1-liter milk). =>(100/(100+50/3))*1 =>CP of 1-litre milk of mixture: Rs 5/6 As the price of any amount of water is zero, and as 1-liter milk costs Rs 1.5/6litre of the mixture will comprise entirely of cost of milk which means,1 liter of the mixture will contain a 5/6th amount of milk. =>Water is added in the ratio of (1-5/6)=1/6