1. A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

A. 2 hours B. 3 hours C. 4 hours D. 5 hours

Answer: Option C
Explanation: Speed downstream = (13 + 4) km/hr = 17 km/hr. Time taken to travel 68 km downstream = [latex]\frac {68}{17}[/latex]hrs = 4 hrs
2. A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:

A. 8.5 km/hr B. 9 km/hr C. 10 km/hr D. 12.5 km/hr

Answer: Option C
Explanation: Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr. Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.
3. What least number must be added to 1056, so that the sum is completely divisible by 23 ?

A. 2 B. 3 C. 18 D. 21

Answer: Option A
Explanation: 23) 1056 (45 92 --- 136 115 --- 21 --- Required number = (23 - 21) = 2.
4. How many of the following numbers are divisible by 132 ? 264, 396, 462, 792, 968, 2178, 5184, 6336

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

Answer: Option A
Explanation: 132 = 4 x 3 x 11 So, if the number divisible by all the three numbers 4, 3 and 11, then the number is divisible by 132 also. 264 11,3,4 (/) 396 11,3,4 (/) 462 11,3 (X) 792 11,3,4 (/) 968 11,4 (X) 2178 11,3 (X) 5184 3,4 (X) 6336 11,3,4 (/) Therefore the following numbers are divisible by 132: 264, 396, 792 and 6336. Required number of number = 4.
5. 331, 482, 551, 263, 383, 362, 284

A. 263 B. 383 C. 331 D. 551

Answer: Option B
Explanation: In each number except 383, the product of first and third digits is the middle one.

1. 835, 734, 642, 751, 853, 981, 532

A. 751 B. 853 C. 981 D. 532

Answer: Option A
Explanation: In each number except 751, the difference of third and first digit is the middle one.
2. A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

A. 0% B. 5% C. 7.5% D. 10%

Answer: Option A
Explanation: C.P. = Rs. 3000. S.P. = Rs. [latex]{3600 × 100}{100 + (10 × 2}[/latex] = Rs. 3000. 100 + (10 x 2) Gain = 0%.
3. If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is:

A. Rs. 20 B. Rs. 21.81 C. Rs. 22 D. Rs. 18.33

Answer: Option D
Explanation: S.I. on Rs. (110 - 10) for a certain time = Rs. 10. S.I. on Rs. 100 for double the time = Rs. 20. T.D. on Rs. 120 = Rs. (120 - 100) = Rs. 20. T.D. on Rs. 110 = Rs. [latex]{20}{120}[/latex] x 110 = Rs. 18.33
4. Rs. 20 is the true discount on Rs. 260 due after a certain time. What will be the true discount on the same sum due after half of the former time, the rate of interest being the same?

A. Rs. 10 B. Rs. 10.40 C. Rs. 15.20 D. Rs. 13

Answer: Option B
Explanation: S.I. on Rs. (260 - 20) for a given time = Rs. 20. S.I. on Rs. 240 for half the time = Rs. 10. T.D. on Rs. 250 = Rs. 10. T.D. on Rs. 260 = Rs. [latex]\frac {10 }{250}[/latex] X 260 = = Rs. 10.40 Sagar's present age = (5x + 6) = 16 years.
5. The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is:

A. Rs. 1800 B. Rs. 1450 C. Rs. 1360 D. Rs. 6800

Answer: Option C
Explanation: Sum = [latex]\frac {S.I. × T.D. }{(S.I.) - (T.D.)}[/latex] = [latex]\frac {85 × 80 }{(85 - 80) }[/latex] = Rs. 1360.

1. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276 B. 299 C. 322 D. 345

Answer: Option C
Explanation: Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322.
2. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4 B. 10 C. 15 D. 16

Answer: Option D
Explanation: L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together [latex]\frac {30}{2}[/latex] + 1 = 16 times.
3. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A.4 B. 5 C. 6 D. 8

Answer: Option A
Explanation: N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
4. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000 B. 9400 C. 9600 D. 9800

Answer: Option C
Explanation: Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600.
5. In order to obtain an income of Rs. 650 from 10% stock at Rs. 96, one must make an investment of:

A. Rs. 3100 B. Rs. 6240 C. Rs. 6500 D. Rs. 9600

Answer: Option B
Explanation: To obtain Rs. 10, investment = Rs. 96. To obtain Rs. 650, investment = Rs [latex]\frac {96}{10}[/latex] x 650 = Rs. 6240.