1. Which of the following has fractions in ascending order?

A. [latex]\frac{2}{3}[/latex], [latex]\frac{3}{5}[/latex], [latex]\frac{1 }{3}[/latex], [latex]\frac{4 }{7}[/latex], [latex]\frac{5}{6}[/latex] B. [latex]\frac{1}{3}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{3 }{5}[/latex], [latex]\frac{5 }{6}[/latex], [latex]\frac{4}{7}[/latex] C. [latex]\frac{1}{3}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{5}{6}[/latex], [latex]\frac{4 }{7}[/latex], [latex]\frac{3}{5}[/latex] D. [latex]\frac{1}{3}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{4 }{7}[/latex], [latex]\frac{3 }{5}[/latex], [latex]\frac{5}{6}[/latex]

Answer: Option D
Explanation: Convert fractions into decimal form. These calculations are to be solved quickly in mind otherwise will require a lot of time. Option 1: 0.4, 0.6, 0.33, 0.5, 0.8 --- (wrong) Option 2: 0.3, 0.4, 0.6, 0.8, 0.5 --- (wrong) Option 3: 0.3, 0.4, 0.8 , 0.5, 0.6 --- (wrong) Option 4: 0.33, 0.4, 0.5, 0.6, 0.8 --- (correct) 0.33 < 0.4 < 0.5 < 0.6 < 0.8 ---- (Ascending order)
2. Which of the following has fractions in descending order?

A. [latex]\frac{5}{6}[/latex], [latex]\frac{4}{7}[/latex], [latex]\frac{2 }{5}[/latex], [latex]\frac{3 }{5}[/latex], [latex]\frac{1}{3}[/latex] B. [latex]\frac{5}{6}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{4 }{7}[/latex], [latex]\frac{2 }{5}[/latex], [latex]\frac{1}{3}[/latex] C. [latex]\frac{4}{7}[/latex], [latex]\frac{1}{3}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{5 }{6}[/latex], [latex]\frac{3}{5}[/latex] D. [latex]\frac{1}{3}[/latex], [latex]\frac{2}{5}[/latex], [latex]\frac{4 }{7}[/latex], [latex]\frac{3 }{5}[/latex], [latex]\frac{5}{6}[/latex]

Answer: Option C
Explanation: Growth rate of rat population in 3 months = 20 *(3/12) = 5% Increase in first 3 months = 3200 x 1.05 = 3360 Also, net decrease in 3 months = 160 Rat population = 3360 - 160 = 3200 In the same way, after every 3 months, rat population remains the same Hence, even after 3 x 8 months i.e., 2 years, the population is maintained
3. Convert 0.737373… into vulgar fraction?

A. [latex]\frac{73}{99}[/latex] B. [latex]\frac{73}{100}[/latex] C. [latex]\frac{73}{90}[/latex] D. [latex]\frac{73}{900}[/latex]

Answer: Option C
Explanation: In a decimal fraction, if there are n numbers of repeated numbers after a decimal point, then just write one repeated number in the numerator and in denominator take n number of nines equal to repeated numbers you observe after the decimal point. 0.737373… is written as [latex]0. \bar{73}[/latex] Numerator = 73 ---- (one repeated number) Denominator = 99 ---- (73 is the number which is repeated) Vulgar fraction = [latex]\frac{73}{99}[/latex]
4. Find the number of shares that can be bought for Rs.8200 if the market value is Rs.20 each with brokerage being 2.5%.

A. 450 B. 500 C. 400 D. 410

Answer: Option B
Explanation: Cost of each share = (20 + 2.5% of 20) = Rs.20.5 Therefore, number of shares = [latex]\frac{8200}{20.5}[/latex] = 400
5. Find the market value of the stock if 6% yields 10%.

A. 60 B. 70 C. 80 D. 100

Answer: Option C
Explanation: Let the investment be Rs.100 for an income of Rs.10 Therefore, for an income of Rs.6, the investment = [latex]\frac{600}{10}[/latex] = Rs.60

1. What is the investment made if one invests in 15% stock at 50 and earns Rs.2000?

A. 8000 B. 7000 C. 5000 D. 6000

Answer: Option C
Explanation: To earn Rs.15, investment = Rs.50. Hence, to earn Rs.1500, investment = [latex]\frac{(1500*50)}{15}[/latex] = Rs.5000
2. A company pays a 12.5% dividend to its investors. If an investor buys Rs.50 shares and gets 25% on the investment, at what price did the investor buy the shares?

A. 6.25 B. 25 C. 50 D. 12.5

Answer: Option A
Explanation: Dividend on 1 share = [latex]\frac{(12.5 * 50)}{100}[/latex] = Rs.6.25 Rs.25 is income on an investment of Rs.100 Rs.6.25 is income on an investment of Rs. [latex]\frac{(6.25 * 100)}{25}[/latex] = Rs.25
3. Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?

A. [latex]\frac{8}{35}[/latex] B. [latex]\frac{34}{35}[/latex] C. [latex]\frac{27}{35}[/latex] D. None of these

Answer: Option A
Explanation: P(A) = 2/5 P(B) = 4/7 E = {A and B both get selected} P(E) = P(A)*P(B) = [latex]\frac{2}{5}[/latex] * [latex]\frac{4}{7}[/latex] = [latex]\frac{8}{35}[/latex]
4. From a pack of 52 cards, one card is drawn at random. Find the probability that the drawn card is a club or a jack?

A. [latex]\frac{17}{52}[/latex] B. [latex]\frac{8}{13}[/latex] C. [latex]\frac{4}{13}[/latex] D. [latex]\frac{1}{13}[/latex]

Answer: Option C
Explanation: n(S) = 52 n(E) = 16 P(E) = [latex]\frac{ n(E)}{ n(S)}[/latex] = [latex]\frac{ 16}{ 52}[/latex] = [latex]\frac{ 4}{ 13}[/latex]
5. Tickets numbered 1 to 50 are mixed and one ticket is drawn at random. Find the probability that the ticket was drawn has a number which is a multiple of 4 or 7?

A. [latex]\frac{ 9}{25}[/latex] B. [latex]\frac{ 9}{ 50}[/latex] C. [latex]\frac{ 18}{ 25}[/latex] D. None of these

Answer: Option A
Explanation: S = {1, 2, 3, … , 49, 50} E = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 7, 14, 21, 35, 42, 49} n(S) = 50 n(E) = 18 P(E) = [latex]\frac{ n(E)}{ n(S)}[/latex] = [latex]\frac{18}{50}[/latex] = [latex]\frac{ 9}{25}[/latex]

1. 4 men and 5 boys can do a piece of work in 20 days while 5 men and 4 boys can do the same work in 16 days. In how many days can 4 men and 3 boys do the same work?

A. 10 days B. 15 days C. 20 days D. 25 days

Answer: Option C
Explanation: Assume 1 man's 1 day work = x & 1 boy's 1 day work = y From the given data, we can generate the equations as : 4x + 5y = 1/20 ---(1) & 5x + 4y = 1/16 ---(2) By solving the simultaneous equations (1) & (2), X = [latex]\frac{1}{80}[/latex] & y = 0 Therefore, (4 men + 3 boys ) 1-day work = 4 X [latex]\frac{1}{80}[/latex] + 3 X 0 = [latex]\frac{1}{20}[/latex] Thus, 4 men and 3 boys can finish the work in 20 days
2. If in a race of 80m, A covers the distance in 20 seconds and B in 25 seconds, then A beats B by:

A. 20m B. 16m C. 11m D. 10m

Answer: Option B
Explanation: The difference in the timing of A and B is 5 seconds. Hence, A beats B by 5 seconds. The distance covered by B in 5 seconds = [latex]\frac{(80 * 5)}{25}[/latex] = 16m Hence, A beats B by 16m.
3. If X can run 48m and Y 42m, then in a race of 1km, X beats Y by:

A. 140m B. 125m C. 100m D. 110m

Answer: Option B
Explanation: When X runs 48m, Y runs 42m. Hence, when X runs 1000m, Y runs [latex]\frac{(1000 * 42)}{48}[/latex] = 875m X beats Y by 125m.
4. In a race of 800m, if the ratio of the speeds of two participants P and R is 3:4, and P has a start of 120m, then R beats P by:

A. 100m B. 80m C. 90m D. 110m

Answer: Option B
Explanation: As P has a start of 120m, P has to cover 680m while R has to cover 800m. Now, when P covers 3m, R covers 4m. Hence, when R covers 800m, P covers [latex]\frac{(800 * 3)}{4}[/latex] = 600m Therefore, R beats P by 80m.
5. If in a race of 120m X can beat Y by 20m and Z by 35m, then Y can beat Z by

12m B. 10m C. 15m D. 18m

Answer: Option D
Explanation: In a race of 120m, as X beats Y by 20m, when X runs 120m, Y runs 100m. Similarly, as X beats Z by 35m, when X covers 120m, Z covers 85m. Hence, when Y runs 100m, Z runs 85m. When Y runs 120m, Z runs [latex]\frac{(120 * 85)}{100}[/latex] = 102m Hence, Y beats Z by 18m.