Percentages: 1. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

A. 57% B. 60% C. 65% D. 90%

Answer: Option A
Explanation: Total number of votes polled = (1136 + 7636 + 11628) = 20400. ∴Required percentage = [latex]\frac{11628}{20400}[/latex] x 100% = 57%
2. Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

A. Rs. 200 B. Rs. 250 C. Rs. 300 D. None of these

Answer: Option B
Explanation: ⇒z + [latex]\frac{120}{5}[/latex]z = 550 ⇒z + [latex]\frac{11}{5}[/latex]z = 550 ⇒z = [latex]\frac{550 x 5 }{11}[/latex]z = 250
3. Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

A. Rs. 15 B. Rs. 15.70 C. Rs. 19.70 D. Rs. 20

Answer: Option C
Explanation:
Let the amount taxable purchases be Rs. x. Then, 6% of x = [latex]\frac{30}{100}[/latex] ⇒x = [[latex]\frac{30}{100}[/latex] x [latex]\frac{100}{6}[/latex]] = 5 ∴Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70
4. Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.

A. Rs. 6876.10 B. Rs. 6999.20 C. Rs. 6654 D. Rs. 7000

Answer: Option A
Explanation: Rebate = 6% of Rs. 6650 = Rs. {[latex]\frac{6}{100}[/latex] x 6650} = Rs. 399. Sales tax = 10% of Rs. (6650 - 399) = Rs.{[latex]\frac{10}{100}[/latex] x 6251} = Rs. 625.10 ∴Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
5. The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

A. 4.37% B. 5% C. 6% D. 8.75%

Answer: Option A
Explanation: Increase in 10 years = (262500 - 175000) = 87500. Increase% = {[latex]\frac{87500 }{175000}[/latex] x 100}% = 50% ∴Required average = {[latex]\frac{50}{10}[/latex] }%= 5%
Profit & Loss:
1. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

Answer: Option D
Explanation: ⇒(C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls) ∴Gain% = {[latex]\frac{0.50}{3.50}[/latex] x 100}% = [latex]\frac{0.50}{3.50}[/latex] ⇒C.P. of 12 balls = S.P. of 17 balls = Rs.720. ⇒C.P. of 1 ball = Rs.[latex]\frac{720}{12}[/latex] = Rs.60
2. When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?

A. Rs. 21,000 B. Rs. 22,500 C. Rs. 25,300 D. Rs. 25,800

Answer: Option C
Explanation: 85 : 18700 = 115 : x ⇒x = [latex]\frac{18700 x 115}{85}[/latex] = 25300 Hence, S.P. = Rs. 25,300.
3. 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:

A. 14[latex]\frac{2}{7}[/latex]% gain B. 15% gain C. 14[latex]\frac{2}{7}[/latex]% loss D. 15% loss

Answer: Option A
Explanation: C.P. of 1 orange = Rs. [latex]\frac{350}{100}[/latex] =Rs. 3.50 S.P. of 1 orange = Rs.[latex]\frac{48}{12}[/latex] = Rs.4 ∴Gain% = {[latex]\frac{0.50}{3.50}[/latex] x 100}% = [latex]\frac{100}{7}[/latex]% = 14[latex]\frac{2}{7}[/latex]%
4. A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:

A. 5[latex]\frac{15}{17}[/latex]% loss B. 5[latex]\frac{15}{17}[/latex]% gain C. 6[latex]\frac{2}{3}[/latex]% gain D. None of these

Answer: Option B
Explanation: C.P. of [latex]{1}^{st}[/latex] transistor = Rs [[latex]\frac{100}{120}[/latex] x 840] = Rs. 700 C.P. of [latex]{2}^{nd}[/latex] transistor = Rs [[latex]\frac{100}{96}[/latex] x 840] = Rs. 1000 So, total C.P. = Rs. (700 + 1000) = Rs. 1700. Total S.P. = Rs. (840 + 960) = Rs. 1800. ∴Gain% = [[latex]\frac{100}{1700}[/latex] x 100]% = 5 [latex]\frac{15}{17}[/latex]%
5. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A. No profit, no loss B. 5% C. 8% D. 10%

Answer: Option B
Explanation: C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600. S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680. ∴Gain% = [[latex]\frac{100}{1700}[/latex] x 100]% = 5%
Time & Work:
1. A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:

A. 4 days B. 6 days C. 8 days D. 12 days

Answer: Option B
Explanation: Suppose A, B and C take x, [latex]\frac{x }{2}[/latex] and [latex]\frac{x }{3}[/latex] days respectively to finish the work. Then, [latex]\frac{1 }{x}[/latex] + [latex]\frac{2 }{x}[/latex] + [latex]\frac{3 }{x}[/latex] = [latex]\frac{1 }{2}[/latex] ⇒[latex]\frac{6 }{x}[/latex] = [latex]\frac{1 }{2}[/latex] ⇒ x = 12 So, B takes ([latex]\frac{12 }{2}[/latex]) = 6 days to finish the work.
2. A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in:

A. 8 days B. 10 days C. 12 days D. 15 days

Answer: Option C
Explanation: (A + B)'s 1 day's work = [latex]\frac{1 }{15}[/latex] + [latex]\frac{1 }{10}[/latex] = [latex]\frac{1}{6}[/latex] Work done by A and B in 2 days = 1 - [latex]\frac{1 }{3}[/latex] = [latex]\frac{2 }{3}[/latex] Now, [latex]\frac{1 }{15}[/latex] work is done by A in 1 day. ∴[latex]\frac{2 }{3}[/latex] work will be done by a in 15 x [latex]\frac{2}{3}[/latex] = 10 days Hence, the total time taken = (10 + 2) = 12 days.
3. A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

A. 18 days B. 24 days C. 30 days D. 36 days

Answer: Option A
Explanation:
2(A + B + C)'s 1 day's work = [latex]\frac{1 }{30}[/latex] + [latex]\frac{1 }{24}[/latex] + [latex]\frac{1 }{20}[/latex] = [latex]\frac{15 }{120}[/latex] Therefore, (A + B + C)'s 1 day's work = [latex]\frac{1 }{2 x8 }[/latex] = [latex]\frac{1 }{16 }[/latex] Work done by A, B, C in 10 days = [latex]\frac{10 }{16}[/latex] = [latex]\frac{5}{8}[/latex] Remaining work = 1 - [latex]\frac{5 }{8}[/latex] = [latex]\frac{3 }{8}[/latex] A's 1 day's work = [latex]\frac{1 }{16}[/latex] - [latex]\frac{1 }{24}[/latex] = [latex]\frac{3 }{8}[/latex] Now, [latex]\frac{1 }{48}[/latex] work is done by A in 1 day. So, [latex]\frac{3 }{8}[/latex] work will be done by A in 48 [latex]\frac{3}{8}[/latex]
4. A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in:

A. 4 days B. 6 days C. 8 days D. 18 days

Answer: Option A
Explanation:
Ratio of rates of working of A and B = 2 : 1. So, the ratio of times taken = 1: 2. B's 1 day's work = [latex]\frac{1 }{12}[/latex] ∴ A's 1 day's work = [latex]\frac{1 }{6}[/latex] ; (2 times of B's work) (A + B)'s 1 day's work = [latex]\frac{1 }{6}[/latex] + [latex]\frac{1 }{12}[/latex] = [latex]\frac{3 }{12}[/latex] = [latex]\frac{1 }{4}[/latex] So, A and B together can finish the work in 4 days.
5. Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

A. 3 : 4 B. 4 : 3 C. 5 : 3 D. Data inadequate

Answer: Option B
Explanation: (20 x 16) women can complete the work in 1 day. ∴1 woman's 1 day's work = [latex]\frac{1 }{320}[/latex] (16 x 15) men can complete the work in 1 day. ∴ 1 man's 1 day's work = [latex]\frac{1 }{240}[/latex] So, required ratio = [latex]\frac{1 }{240}[/latex] : [latex]\frac{1 }{320}[/latex] [latex]\frac{1 }{3}[/latex] : [latex]\frac{1 }{4}[/latex] = 4:3 (cross multiplied)
Time & Distance:
1. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

A. 5 kmph B. 6 kmph C. 6.25 kmph D. 7.5 kmph

Answer: Option A
Explanation: Let Abhay's speed be x km/hr. Then, [latex]\frac{30 }{x}[/latex] - [latex]\frac{30 }{2x}[/latex] = 3 ⇒6x = 30 ⇒x = 5 km/hr.
2. Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

A. 8 kmph B. 11 kmph C. 12 kmph D. 14 kmph

Answer: Option C
Explanation: Let the distance travelled by x km. Then, [latex]\frac{x }{10}[/latex] - [latex]\frac{x }{15}[/latex] = 2 ⇒3x - 2x = 60 ⇒x = 60 km. Time taken to travel 60 km at 10 km/hr = [latex]\frac{60 }{10}[/latex] = 6 hrs. So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M. ∴ Required speed = [latex]\frac{60 }{5}[/latex]kmph = 12 kmph.
3. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

A. 2 : 3 B. 3 : 2 C. 3 : 4 D. 4 : 3

Answer: Option C
Explanation: Let the speed of the train be x km/hr and that of the car be y km/hr. Then, [latex]\frac{120}{x}[/latex] + [latex]\frac{480}{y}[/latex] = 8 ⇒ [latex]\frac{1}{x}[/latex] + [latex]\frac{4}{y}[/latex] = [latex]\frac{1}{15}[/latex]....(i) And, [latex]\frac{20}{x}[/latex] + [latex]\frac{400}{y}[/latex] ⇒ [latex]\frac{1}{x}[/latex] + [latex]\frac{2}{y}[/latex] = [latex]\frac{1}{24}[/latex] ....(ii) Solving (i) and (ii), we get: x = 60 and y = 80. ∴Ratio of speeds = 60 : 80 = 3 : 4.
4. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:

A. 14 km B. 15 km C. 16 km D. 17 km

Answer: Option C
Explanation: Let the distance travelled on foot be x km. [latex]\frac{x}{4}[/latex] + [latex]\frac{(61-x)}{9}[/latex] = 9 Then, distance travelled on bicycle = (61 -x) km. ⇒9x + 4(61 -x) = 9 x 36 ⇒5x = 80 ⇒x = 16 km.
5. man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is

A. 35 B. 36 [latex]\frac{2}{3}[/latex] C. 37 [latex]\frac{1}{2}[/latex] D. 40

Answer: Option D
Explanation: Let distance = x km and usual rate = y kmph. Then, [latex]\frac{x}{y}[/latex] - [latex]\frac{x}{y+ 3}[/latex] = [latex]\frac{40}{60}[/latex] ⇒ 2y(y + 3) = 9x ....(i) And, [latex]\frac{x}{y-2}[/latex] - [latex]\frac{x}{y}[/latex] = [latex]\frac{40}{60}[/latex] ⇒ y(y - 2) = 3x ....(ii) y -2 y 60 On dividing (i) by (ii), we get: x = 40.