**Simple Interest**
**1. In how many years will a sum of money doubles itself at 5% per annum on simple interest?**
**A.** 22 years
**B.** 20 years
**C.** 21 years
**D.** 29 years

** Answer**: Option B

**Explanation**:
P = [latex]\frac{(P*5*R)}{100}[/latex]
R = 20%

**2. A part of the certain sum of money is invested at 9% per annum and the rest at 12% per annum, if the interest earned in each case for the same period is equal, then the ratio of the sums invested is?**
**A.** 2:3
**B.** 3:4
**C.** 4:3
**D.** None

** Answer**: Option C

**Explanation**:
12:9
= 4:3

**3. Rs.800 amounts to Rs.920 in 3 years at simple interest. If the interest is increased by 3%, it would amount to how much?**
**A.** Rs.1056
**B.** Rs.1112
**C.** Rs.1182
**D.** Rs.992

** Answer**: Option D

**Explanation**:
[latex]\frac{(800*3*3)}{100}[/latex] = 72
920 + 72 = 992

**Simplification**
**4. 860 + [latex]\frac{4}{7}[/latex] of 21.21 - 41.4 = ?**
**A.** 1830.72
**B. **1829.28
**C.** 1831.72
**D.** 1831.28
**E.** None of these

** Answer**: Option A

**Explanation**:
1860 + [latex]\frac{4}{7}[/latex] of 21.21 - 41.4 = 1860 + 4(3.03) - 41.4
= 1860 + 12.12 - 41.4 = 1872.12 - 41.4
= 1830.72

**5. [latex]\frac{(22 * 495 + 891 * 44)}{(33 * 176 + 55 * 264)}[/latex]= ?**
**A.** 1
**B.** 2
**C.** [latex]\frac{69}{28}[/latex]
**D. ** [latex]\frac{66}{17}[/latex]
**E.** None of these

** Answer**: Option C

**Explanation**:
[latex]\frac{(22 * 495 + 891 * 44)}{(33 * 176 + 55 * 264)}[/latex] = [latex]\frac{(22 * 33 * 15 + 9 * 99 * 2 * 22)}{(33 * 22 * 8 + 5 * 11 * 22 * 12) }[/latex]
= [latex]\frac{(22 * 33 * 15 + 22 * 33 * 54)}{(33 * 22 * 8 + 33 * 22 * 20)}[/latex] = [latex]\frac{[22 * 33(15 + 54)]}{[33 * 22(8 + 20)]}[/latex] = [latex]\frac{69}{28}[/latex]

**6. 8.008 + 0.8 + 80.8 + 800.8 + 0.08 = ?**
**A.** 888.888
**B.** 889.688
**C. ** 890.488
**D.** 980.488
**E.** None of these

** Answer**: Option C

**Explanation**:
8.008 + 0.8 + 80.8 + 800.8 + 0.08
= 8 + 0.008 + 0.8 + 80 + 0.8 + 800 + 0.8 + 0.08
= 8 + 80 + 800 + 0.008 + 0.08 + 0.8 + 0.8 + 0.8
= 888 + 2.488 = 890.488

**Time and Distance**
**7. A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?**
**A.** 50 kmph
**B.** 52 kmph
**C.** 54 kmph
**D.** 56 kmph

** Answer**: Option C

**Explanation**:
S = [latex]\frac{240}{16}[/latex] * [latex]\frac{18}{5}[/latex] = 54 kmph

**8. The speed of a car is 90 km in the first hour and 60 km in the second hour. What is the average speed of the car?**
**A.** 72 kmph
**B.** 75 kmph
**C.** 30 kmph
**D.** 80 kmph

** Answer**: Option B

**Explanation**:
S = [latex]\frac{(90 + 60)}{2}[/latex] = 75 kmph

**9. A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. another man leaves the point give at 8 a.m. and reaches the point P at 12 noon. At what time do they meet?**
**A.** 8 a.m.
**B.** 8 p.m.
**C.** 9 a.m.
**D.** 9 p.m.

** Answer**: Option C

**Logarithm**
**10. If log 27 = 1.431, then the value of log 9 is:**
** A. ** 0.934
** B.** 0.945
**C.** 0.954
** D.** 0.958

**Answer**: Option C

**Explanation**:
log 27 = 1.431
log (3[latex]^ {3}[/latex]) = 1.431
3 log 3 = 1.431
log 3 = 0.477
log 9 = log(3[latex]^ {2}[/latex] ) = 2 log 3 = (2 x 0.477) = 0.954.

**11. If [latex]log_ {10}[/latex] 2 = 0.3010, the value of [latex]log_ {10}[/latex] 80 is:**
**A.** 1.6020
**B.** 1.9030
**C.** 3.9030
**D.** None of these

**Answer**: Option B

**Explanation**:
[latex]log_ {10}[/latex] 80 = [latex]log_ {10}[/latex] (8 x 10)
= [latex]log_ {10}[/latex] 8 + [latex]log_ {10}[/latex] 10
= [latex]log_ {10}[/latex] (2[latex]^{3}[/latex] ) + 1
= 3 [latex]log_ {10}[/latex] 2 + 1
= (3 x 0.3010) + 1
= 1.9030.

**12. If [latex]log_ {10}[/latex] 5 + [latex]log_ {10}[/latex] (5x + 1) = [latex]log_ {10}[/latex] (x + 5) + 1, then x is equal to:**
**Answer**: Option B

**Explanation**:
[latex]log_ {10}[/latex] 5 + [latex]log_ {10}[/latex] (5x + 1) = [latex]log_ {10}[/latex] (x + 5) + 1
[latex]log_ {10}[/latex] 5 + [latex]log_ {10}[/latex] (5x + 1) = [latex]log_ {10}[/latex] (x + 5) + [latex]log_ {10}[/latex] 10
[latex]log_ {10}[/latex] [5 (5x + 1)] = [latex]log_ {10}[/latex] [10(x + 5)]
5(5x + 1) = 10(x + 5)
5x + 1 = 2x + 10
3x = 9
x = 3.

**Permutation and Combination**
**13. Arrange the letters of the word "DARKER" so that the three vowels do not appear together. In how many ways can this be done?**
**A.** 240

**B.** 360

**C.** 500

**D.** 720

**Answer**: Option A

**Explanation**:
'DARKER' has 6 letters.
We can arrange 'n' things in n! ways.
While arranging letters/things/numbers, if there are two same things or things get repeated twice, then we need to divide by 2!
If there are 3 same things or things get repeated thrice, then divide by 3! And so on..

Thus, we can arrange 6 letters in 6! ways.
But R gets repeated. There are 2 Rs. So divide by 2!
∴ Total ways =[latex]\frac {6!}{2!}[/latex] = 360
Vowels not together = Total ways - Vowels together
Consider the 2 vowels (A and E) as one group.
We have 4 letters and 1 group = 5
We can arrange them in 5! Ways.
But again here R comes twice. So we will have[latex]\frac {5!}{2!}[/latex]
Also, the 2 vowels can be arranged in 2! Ways.
So, Number of ways with vowels together = 2! x[latex]\frac {5!}{2!}[/latex] = 120
∴ Number of ways with vowels not together = 360 - 120 = 240

**14. Priyanka wants to attend Grammy Awards at London and National awards at Delhi before heading to the UNICEF Summit in Berlin. 8 flights come from London to Delhi through different routes and 6 flights fly from Delhi to Berlin through 6 different routes. Find the number of ways in which Priyanka can reach from London to Berlin through Delhi. **
**Answer**: Option C

**Explanation**:
This is very easy to solve
Say out of 8, Priyanka chooses one way to go from London to Delhi.
Now, From Delhi to Berlin she has 6 routes i.e. 6 options
Similarly, for the 2nd route between London and Delhi, Priyanka will again have 6 options from Delhi to Berlin.
This is true for all 8 routes between London and Delhi.
So, answer = 8 x 6 = 48 possible ways to travel from London to Berlin via Delhi.

**15. Out of the 35 players attending the opening ceremony of an event, there are 12 cricketers, 10 wrestlers, 5 badminton players, and 8 hockey players. The management wants a cricketer or a wrestler to lead the group in the parade. In how many ways can this be done?**
**A.** [latex]\frac {1}{2}[/latex]
**B.** 13
**C.** 22
**D.** 35

**Answer**: Option D

**Explanation**:
The leader has to be from cricketers or wrestlers i.e. from 12 cricketers and 10 wrestlers.
So, there are only 12 cricketers + 10 wrestlers = 22 ways to select a leader
Going Further:
If it is said that - The organizer of the group wants any one person to be the leader of the group, then
The number of ways simply would be 35.
Because there are 35 people and anyone can be selected as a leader.

**Area**
**16. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?**
**A.** 40
** B.** 50
**C.** 120
** D.** Data inadequate
**E.** None of these

**Answer**: Option E

**Explanation**:
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = [latex]\frac {5300}{26.50}[/latex] m = 200 m.
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

**17. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?**
**Answer**: Option D

**Explanation**:
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

**18. A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:**
**A.** Rs. 456
**B.** Rs. 458
**C.** Rs. 558
**D.** Rs. 568

**Answer**: Option C

**Explanation**:
Area to be plastered = [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m[latex]^ {2}[/latex]
= (444 + 300) m[latex]^ {2}[/latex]
= 744 m[latex]^ {2}[/latex].
Cost of plastering = Rs.(744 x [latex]\frac {75}{100}[/latex] = Rs. 558.

**Problems on HCF and LCM. **
**19. The least square number which divides 8, 12 and 18 is?**
**A.** 100
**B.** 121
**C.** 64
**D.** 144

**Answer**: Option D

**Explanation**
LCM = 72
72 * 2 = 144

**20. The least number which when divided by 8, 12 and 16, leave in each remainder 3, is?**
**Answer**: Option D

**Explanation**
LCM = 48 + 3 = 51

**21. The least number which when diminished by 7 is divisible by 21, 28, 36 and 45 is?**
**A.** 1267
**B.** 1265
**C.** 1261
**D.** 68

**Answer**: Option A

**Explanation**
LCM = 1260
1260 + 7 = 1267

**Problem On Ages **
**22. Rajan got married 8 years ago. His present age is 6/5 times his age at the time of his marriage. Rajan's sister was 10 years younger to him at the time of his marriage. The age of Rajan's sister is:**
**A.** 32 years
**B.** 36 years
**C.** 38 years
**D.** 40 years

**Answer**: Option C

**Expalnation**
Let Rajan's present age be x years.
Then, his age at the time of marriage = (x - 8) years.
x = 6/5 (x - 8)
5x = 6x - 48 => x = 48
Rajan's sister's age at the time of his marriage = (x - 8) - 10 = 30 years.
Rajan's sister's present age = (30 + 8) = 38 years.

**23. A father said to his son, "I was as old as you are at present at the time of your birth." If the father's age is 38 years now, the son's age five years back was:**
**A.** 19 years
**B.** 14 years
**C.** 33 years
**D.** 38 years

**Answer**: Option B

**Expalnation**
Let the son's present age be x years.
Then, (38 - x) = x
2x = 38 => x = 19
Son's age 5 years back = (19 - 5) = 14 years.

**24. The age of Arvind's father is 4 times his age if 5 years ago father’s age was 7 times the age of his son at the time. What is Arvind's father’s present age?**
**A.** 35 years
**B.** 40 years
**C.** 70 years
**D.** 84 years

**Answer**: Option B

**Expalnation**
Arvind’s age be x years Father’s age = 4x years.
Therefore (4x -5) =7(x -5) or 3x = 30
= x =10
Arvind’s father’s age is 40 years

**Allegation or Mixture**
**25. Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made?**
**A.** Rs. 13.50
**B.** Rs. 14.50
**C.** Rs. 15.50
**D.** Rs. 16.50
**E.** None of these

** Answer**: Option D

**Explanation**:
Let the quantities of A and B mixed be 3x kg and 7x kg.
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = [latex]\frac {132x}{10x}[/latex](5) = Rs. 66
Profit made in selling 5 kg of the mixture = [latex]\frac {25}{100}[/latex] (cost of 5 kg of the mixture) = [latex]\frac {25}{100}[/latex] * 66
= Rs. 16.50

**26. In a can, there is a mixture of milk and water in the ratio 4: 5. If it is filled with an additional 8 liters of milk the can would be full and ratio of milk and water would become 6: 5. Find the capacity of the can?**
**A.** 40
**B.** 44
**C.** 48
**D.** 52
**E.** None of these

** Answer**: Option B

**Explanation**:
Let the capacity of the can be T liters.
Quantity of milk in the mixture before adding milk = [latex]\frac {4}{9}[/latex] (T - 8)
After adding milk, quantity of milk in the mixture = [latex]\frac {6}{11}[/latex] T.
[latex]\frac {6T}{11}[/latex] - 8 = [latex]\frac {4}{9}[/latex](T - 8)
10T = 792 - 352 => T = 44.

**27. All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?**
**A.** 648
**B.** 888
**C.** 928
**D. **1184
**E.** None of these

** Answer**: Option D

**Explanation**:
B has 62.5% or ([latex]\frac {5}{8}[/latex]) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.

**Numbers**
**28. How many natural numbers are there between 23 and 100 which are exactly divisible by 6?**
** A.** 8
**B.** 11
**C.** 12
** D.** 13
**E.** None of these

** Answer**: Option D

**Explanation**:
Required numbers are 24, 30, 36, 42, ..., 96
This is an A.P. in which a = 24, d = 6 and l = 96
Let the number of terms in it be n.
Then t[latex]_ {n}[/latex] = 96 => a + (n - 1)d = 96
24 + (n - 1) x 6 = 96
(n - 1) x 6 = 72
(n - 1) = 12
n = 13
The required number of numbers = 13.

**29. The sum of three consecutive even numbers is 42. Find the middle number of the three?**
** Answer**: Option A

**Explanation**:
Three consecutive even numbers (2P - 2), 2P, (2P + 2).
(2P - 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.

**30. The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?**
** Answer**: Option B

**Explanation**:
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28

**Compound Interest**
**31. Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?**
**A.** Rs.81000
**B.** Rs.80000
**C.** Rs.75000
**D.** Rs.64000

**Answer**: Option A

**Explanation**:
64000* [latex]\frac {(9}{8}[/latex]* [latex]\frac {(9}{8}[/latex] = 81000

**32. Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest?**
**A.** 18% p.a.
**B.** 22% p.a.
**C.** 20% p.a.
**D.** 24% p.a.
**E.** None of these.

**Answer**: Option C

**Explanation**:
Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year.
Rate of interest =[latex]\frac {(100 * 240)}{(100 * 1)}[/latex] = 20% p.a

**Probability**
**33. Three unbiased coins are tossed. What is the probability of getting at most two heads?**
** A.** [latex]\frac {3}{4}[/latex]
**B.** [latex]\frac {1}{4}[/latex]
**C.** [latex]\frac {3}{8}[/latex]
**D.** [latex]\frac {7}{8}[/latex]

**Answer**: Option D

**Explanation**:
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =[latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {7}{8}[/latex].

**34. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?**
**A.** [latex]\frac {1}{2}[/latex]
**B.** [latex]\frac {3}{4}[/latex]
**C.** [latex]\frac {3}{8}[/latex]
**D.** [latex]\frac {5}{16}[/latex]

**Answer**: Option B

**Explanation**:
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4),
(3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(E) = 27.
P(E) = [latex]\frac {n(E)}{n(S)}[/latex]= [latex]\frac {27}{36}[/latex]= [latex]\frac {3}{4}[/latex].

**35. Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart is:**
**A.** [latex]\frac {3}{20}[/latex]
**B.** [latex]\frac {29}{34}[/latex]
**C.** [latex]\frac {47}{100}[/latex]
**D.** [latex]\frac {13}{102}[/latex]

**Answer**: Option D

**Explanation**:
Let S be the sample space.
Then, n(S) = [latex]^{52}C_{2}[/latex] = [latex]\frac {(52 \times 51)}{(2 \times 1)}[/latex] = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13
= ([latex]^{13}C_{1}[/latex] x [latex]^{13}C_{1}[/latex])
= (13 x 13)
= 169.
P(E) = [latex]\frac {n(E)}{n(S)}[/latex] = [latex]\frac {169}{1326}[/latex] = [latex]\frac {13}{102}[/latex] .