.1. Four friends (A, B, C and D) started a business in partnership by investing capitals in the proportion 3 : 5 : 4 : 6. During the period of one-year, these capitals were utilized in the proportion of 6 : 4 : 5 : 3 respectively. If, at the end of the year, a profit of Rs. 15,550 was made, what will be the share of D?

A. Rs. 3250.5 B. Rs. 3750 C. Rs. 4062.59 D. Rs. 3682.89

Answer: Option D
Explanation: A = (Investment * capital utilized) = (3 * 6) = 18 B = (5 * 4) = 20 C = (4 * 5) = 20 D = (6 * 3) = 18 Thus, out of Rs. 76 (18 +20 + 20 + 18), D's share is Rs. 18. Hence, out of Rs. 15,500 the share of D would be given as (18/76) * 15550 = Rs. 3682.89
2. A boy has to cover a total distance of 300 km. in 6 hours. He travels at a rate of 60 kmph. for the first 90 minutes and next 100 kms. at the rate of 50 kmph. At what average speed must he travel now in order to complete the journey in 6 hours?

A. 25 kmph. B. 32 kmph C. 44 kmph D. 58 kmph

Answer: Option C
Explanation: The boy travels 60 kmph for 90 min, this means he travels 90 km. Next 100 km at the rate of 50 kmph, this means he travels 100 km. Total distance traveled till now = 100 + 90 = 190 km Time spent = 1.5 + 2 = 3.5 hr Required speed = Remaining distance/ Time = [latex]\frac{(300 – 190)}{(6 – 3.5) }[/latex] = [latex]\frac{110}{2.5 }[/latex] = 44 kmph
3. A certain sum earns simple interest of Rs. 800 in 2 years at a certain rate of interest. If the same sum earns compound interest of Rs. 845 in the same period of 2 years, What must be the rate of interest?

A. 5% p.a. B. 7.5% p.a. C. 10% p.a. D. 12.5% p.a.

Answer: Option C
Explanation: Given: 800 = [latex]\frac{(P * R * 2)}{100 }[/latex] S.I. For 1 year = Rs. 400 Thus, (840 – 800) = S.I. on Rs. 400 for 1 year 40 = [latex]\frac{(400 * R * 1)}{100 }[/latex] R = 10% p.a.
4. An amount of Rs.20,000 is to be distributed amongst P, Q, R and S such that “P” gets twice as that of “Q” and “S” gets four times as that of “R”. If “Q” and “R” are to receive equal amount, what is the difference between the amounts received by S and P?

A. Rs. 5000 B. Rs. 4570 C. Rs. 2750 D. Rs. 2950

Answer: Option A
Explanation: We have, P = 2Q & S = 4R Further Q = R & P + Q + R + S = 20,000 Thus we get, 2Q + Q + Q + 4Q = 20,000 8Q = 20,000 or Q = Rs. 2500 Thus, R = Rs. 2500, P = 5000 & S = Rs. 10000 Hence, the required difference = (S – P) = (10000 – 5000) = Rs. 5000
5. By giving Rs. 50 to M, A would have the amount equal to what M had earlier. If the sum of the amounts with A and M is Rs. 650. What is the ratio of the amount with A to that with M earlier?

A. 7 : 4 B. 5 : 3 C. 2: 1 D. 7: 6

Answer: Option C
Explanation: Let the amounts with A and M be Rs. “X” and Rs. “Y” respectively. Thus, we have, X + Y = 650 and X – 50 = Y X – Y = 50. Hence, X = 350 & Y = 300 Thus the required ratio is 350 : 300 = 7 : 6

1. The length and breadth of a rectangular floor are 16.25 metre and 12.75 metre respectively. Find how many minimum number of square tiles would be required to cover it completely?

A. 375 B. 2570 C. 2800 D. 3315

Answer: Option D
Explanation: Since we require minimum number of square tiles, the size of the tile is given as the H.C.F. of two sides of the room. The H.C.F. Of 1625 cm & 1275 cm. is 25 cms. Hence, we get, Required Number = [latex]\frac{(1625 * 1275)}{(25 * 25) }[/latex] = 3315
2. P, Q, R subscribe Rs. 50,000 for a business. P subscribes Rs. 4000 more than Q and Q Rs. 5000 more than R. Out of a total profit of Rs. 25,000, what will be P's share?

A. Rs. 8400 B. Rs. 10,500 C. Rs. 13,600 D. Rs. 14,700

Answer: Option B
Explanation: Let amount subscribed for R be x. Therefore, Q = x + 5000 and P = x + 5000 + 4000 = x + 9000 So, x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000 P : Q : R = 21000 : 17000 : 12000 = 21 : 17 : 12 P’s share = 25000 X [latex]\frac{21}{50 }[/latex] = Rs. 10,500
3. Pipe P can fill a tank in 4 hours, pipe Q in 8 hours and pipe R in 24 hours. If all the pipes are open, in how many hours will the tank be filled?

A. 2 hours B. 2.4 hours C. 3 hours D. 3.5 hours

Answer: Option B
Explanation: Part filled by (P + Q + R) in 1 hour = ([latex]\frac{1}{4 }[/latex] + [latex]\frac{1}{8 }[/latex] + [latex]\frac{1}{24 }[/latex]) = [latex]\frac{10}{24 }[/latex] All the three pipes together will fill the tank = 24/10 = 2.4 hours
4. A car is running at a speed of 90 kmph. What distance will it cover in 15 second?

A. 100 m B. 255 m C. 375 m D. Cannot be determined

Answer: Option C
Explanation: Speed = 108 kmph = (90 x ([latex]\frac{5}{18 }[/latex])) m/sec = 25 m/sec Distance covered in 15 second = (25 x 15) m = 375 m.
5. How long does a train 90 meters long running at the speed of 71 km/hr take to cross a bridge 132 meters in length?

A. 9.8 sec B. 11.26 sec C. 12.42 sec D. 14.3 sec

Answer: Option B
Explanation: Speed = (71 x [latex]\frac{5}{18 }[/latex]) m/sec = 19.72 m/sec Total distance covered = (90 + 132) m = 222 m. Required time = ([latex]\frac{222}{19.72 }[/latex]) sec = 11.26 sec

1. A boat running downstream covers a distance of 20 km in 2 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water?

A. 4 km / hr B. 7 km / hr C. 8 km / hr D. Data Inadequate

Answer: Option B
Explanation: Rate downstream = ([latex]\frac{20}{2 }[/latex]) kmph = 10 kmph; Rate upstream = ([latex]\frac{20}{5 }[/latex]) kmph = 4 kmph Speed in still water = [latex]\frac{1}{2}[/latex] (10 + 4) kmph = 7 kmph
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. If 500 gm of salt solution has 30% salt in it, how much salt must be added to make the concentration of salt 50% in the solution?

A. 200 gm B. 100 gm C. 90 gm D. 50 gm

Answer: Option A
Explanation: Amount of water in solution = 100 % - 30 % = 70 % of total Amount of water = 70 % x 500 = 350 gm Amount of water remains same for both the solutions. Therefore, let amount of new solution be x gm. x = 350 x (100/ 50) x = 700 Salt added = (700 – 500) = 200 gm Alternate solution: Quantity of sugar to be added : = [Solution (Required % value – Present % value)] / (100 – Required % value) =[latex]\frac{[500 (50 – 30)] }{100 - 50}[/latex] = [latex]\frac{(500 × 20)}{50}[/latex] = 200 gm
4. The L.C.M. Of 2, 2.7 and 0.09 is:

A. 5.4 B. 0.54 C. 0.054 D. 54

Answer: Option D
Explanation: Given numbers are 2.00, 2.70 and 0.09. L.C.M. of 200, 270 and 9 is 5400. L.C.M. Of given numbers = 54.00 = 54
5. The average of five consecutive odd numbers is 51. What is the difference between the highest and lowest number?

3 B. 7 C. 8 D. 11

Answer: Option C
Explanation: Let the numbers be x, x + 2, x + 4, x + 6 and x + 8. Then, [latex]\frac{[x + (x + 2) + (x + 4) + (x + 6) + (x + 8)]}{5}[/latex] = 51 5x + 20 = 255 x = 47 So, required difference = (47 + 8) – 47 = 8