Simple Interest
1. A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?
A. Rs. 7000
B. Rs. 8500
C. Rs. 8000
D. Rs. 7500
E. None of these
Answer: Option A
Explanation:
Let the sum be Rs. x.
(x * 18 * 2)/100 - (x * 12 * 2)/100 = 840 => 36x/100 - 24x/100 =840
=> 12x/100 = 840 => x = 7000.
2. Vijay lent out an amount Rs. 10000 into two parts, one at 8% p.a. and the remaining at 10% p.a. both on simple interest. At the end of the year he received Rs. 890 as total interest. What was the amount he lent out at 8% pa.a?
A. Rs. 6000
B. Rs. 5500
C. Rs. 4500
D. Rs. 5000
E. None of these
Answer: Option B
Explanation:
Let the amount lent out at 8% p.a. be Rs. A
=> (A * 8)/100 + [(10000 - A) * 10]/100 = 890
=> A = Rs. 5500.
Height and Distance
1. The actual height of a man is 6 feet but due to the position of Sun he casts a shadow of 4 feet. He is standing next to an electricity tower and notices that it casts a shadow of 36 feet. Find the height of the electricity tower.
A. 48 ft
B. 54 ft
C. 63 ft
D. 72 ft
Answer: Option B
Explanation:
Both the man and tower are near each other and are illuminated by the same sun from the same direction.
So the angle of elevation for the sun is the same for both.
So the ratio of object to shadow will be same for all objects. (Proportionality Rule)
[latex]\frac {Object height}{Shadow length}[/latex]=[latex]\frac {6}{4}[/latex]=[latex]\frac {H}{36}[/latex]
∴ H = 54 ft = Height of tower.
2. On the two sides of a road are two tall buildings exactly opposite to each other. The height of the taller building is 60 m. If the angle of elevation from the top of the smaller building to the top of the taller one is 30° and the angle of depression from the top of the taller building to the foot of the smaller one is 30°, then find the height of the smaller building.
Answer: Option C
Explanation:
Let AB be the taller building of height 60m and CD be the smaller one of height h m.
tan30= DB/AB = DB/60 = 1/[latex]\sqrt{3}[/latex]
DB = 34.64 m
tan30= AE/CE = AE/DB = 1/[latex]\sqrt{3}[/latex] = AE/34.64
AE = 60 – h = 20
h = 40 m
3. A man standing on the terrace of a building watches a car speeding towards him. If at that particular instant the car is 200 m away from the building makes an angle of depression of 60° with the man’s eye and after 8 seconds the angle of depression is 30°, what is the speed of the car?
A. 15 m/s
B. 16.67 m/s
C. 25 m/s
D. Cannot be determined
Answer: Option B
Explanation:
Let AB be the building and the man is standing at A.
When the car is 200m away from the building, the angle of depression is 60°
tan60 = BD/AB = [latex]\sqrt {3}[/latex]
200/AB = [latex]\sqrt {3}[/latex]
AB = 115.47 m
tan30° =BC/AB = x/115.47 = 1/[latex]\sqrt {3}[/latex]
x = 66.67
Now, the car travels distance CD in 8 seconds
CD = BD – BC = 200 – 66.67 = 133.33 m
Speed = 133.33/8 = 16.67 m/s
Volume and Surface Area
1. If the length, breadth and the height of a cuboid are in the ratio 6: 5: 4 and if the total surface area is 33300 cm[latex]^{2}[/latex], then length breadth and height in cms, are respectively?
A. 90, 85, 60
B. 85, 75, 60
C. 90, 75, 70
D. 90, 75,60
Answer: Option D
Explanation:
Length = 6x
Breadth = 5x
Height = 4x in cm
Therefore, 2(6x × 5x + 5x × 4x + 6x × 4x) = 33300
148x[latex]^{2}[/latex] = 33300
=> x[latex]^{2}[/latex] = 33300/148 = 225
=> x = 15.
Therefore, Length = 90cm,
Breadth = 75cm,
Height = 60cm
90, 75 , 60 cm
2. A wooden box of dimensions 8m x 7m x 6m is to carry rectangular boxes of dimensions 8cm x 7cm x 6cm. The maximum number of boxes that can be carried in 1 wooden box is?
A. 1200000
B. 1000000
C. 9800000
D. 7500000
Answer: Option B
Explanation:
Number of Boxes
= [latex]\frac {volume of wooden box in cm ^{3}}{volume of 1 small box}[/latex]
=[latex]\frac {800 \times 700 \times 600}{8 \times 7 \times 6}[/latex]
= 1000000
Logarithm
1. 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.
2. If [latex]log_{10}[/latex] 2 = 0.3010, then [latex]log_{2}[/latex] 10 is equal to:
A. [latex]\frac {699}{301}[/latex]
B. [latex]\frac {1000}{301}[/latex]
C. 0.3010
D. 0.6990
Answer: Option B
Explanation:
[latex]log_{2}[/latex] 10 = [latex]\frac {1}{log_{10} 2}[/latex] = [latex]\frac {1}{0.3010}[/latex] = [latex]\frac {10000}{3010}[/latex] = [latex]\frac {1000}{301}[/latex].
Races and Games
1. A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by:
A. 100 m
B. 75 m
C. 25 m
D. 50m
Answer: Option A
Explanation:
When B runs 25 m, A runs 45/2 m
When B runs 1000 m, A runs (45/2*1/25*1000)m = 900 m.
B beats A by 100 m.
2. In a race of 1000 meters, A can beat B by 100 meters, in a race of 800 meters, B can beat C by 100 meters. By how many meters will A beat C in a race of 600 meters?
A. 125.5 meters
B. 126.5 meters
C. 127.5 meters
D. 128.5 meters
Answer: Option C
Explanation:
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 - 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
Simplification
1. Find the value of (25 × (10 + 5) – 15) ÷ 62
A. 20
B. 10
C. 60
D. 0
E. None of these
Answer: Option B
Explanation:
Follow BODMAS rule to solve this question, as per the order is given below,
Step-1- Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, the BODMAS rule must be followed,
Step-2- Any mathematical 'Of' or 'Exponent' must be solved next,
Step-3-Next, the parts of the equation that contains 'Division' and 'Multiplication' are calculated,
Step-4-Last but not least, the parts of the equation that contains 'Addition' and 'Subtraction' should be calculated.
The given expression is,
(25 × (10 + 5) – 15) ÷ 6[latex]^ {2}[/latex]
= (25 × 15 – 15) ÷ 6[latex]^ {2}[/latex]
= (375 - 15) ÷ 6[latex]^ {2}[/latex]
= 360 ÷ 6[latex]^ {2}[/latex]
= 360 ÷ 36
= 10
2. What approximate value will come in place of question mark in the following question?
23.003 × 22.998 + 100.010 =?
A. 630
B. 720
C. 840
D. 520
E. 900
Answer: Option A
Explanation:
Solve the given expression taking the nearest approximate value (round off) of the numbers given in the expression –
? = 23.003 × 22.998 + 100.010
⇒ ? ≈ 23 × 23 + 100 = 629
⇒ ? ≈ 630
Time and Distance
1. A child goes to school on bike at a speed of 18 km/h. While coming back, he take school bus which runs faster by 27 km/h. What is his average speed for the entire journey?
A. 25.71 km/hr
B. 27 km/hr
C. 27.39 km/hr
D. 31.5km/hr
Answer: Option A
Explanation:
Speed = S = [latex]\frac {D}{T}[/latex]
Average Speed = [latex]\frac {Total distance traveled}{Total time taken}[/latex]
Let the distance between start and end be D
Total distance travelled = D+D = 2D
The bus runs 27 kmph faster, so its speed is 45 kmph.
Total Time = t1+t2 =[latex]\frac {d1}{s1}[/latex] + [latex]\frac {d2}{s2}[/latex] = [latex]\frac {D}{18}[/latex]+[latex]\frac {D}{45}[/latex]
∴ Average speed =[latex]\frac {2D}{D/18 + D/45}[/latex] = 25.71 km/hr
2. In how much time will a train of length 100 m, moving at 36 kmph cross an electric pole?
A. 15 sec
B. 22 sec
C. 18 sec
D. 10 sec
Answer: Option D
Explanation:
Convert kmph to mps. 36 kmph = 36 * 5/18 = 10 mps.
The distance to be covered is equal to the length of the train.
Required time t = d/s = 100/10 = 10 sec.
Chain Rule
1. A trip was organized for 75 travelers for 35 days and the food supply was arranged accordingly. However, some travelers showed up with friends and there were 15 additional people from day one of the trip. Now, for how many days will the food supply last for all these people?
A. 11.25 days
B. 29.1 days
C. 30 days
D. 32 days
Answer: Option B
Explanation:
Total people after guests come = 75+15 = 90
75 people x 35 days x 1 = 90 people x ? days x 1
∴ ? = 29.1 days = Supplies will last for these many days.
2. 3 men can build a stage in 18 days while 6 boys can also do it in 18 days. If 4 men and 4 boys work on the assignment together, how many days will they take to finish it?
A. 4.5 days
B. 6 days
C. 9 days
D. 27 days
Answer: Option C
Explanation:
Men = M; Days = D; Time/Hours = T; Work = W
M1D1T1W2 = M2D2T2W1
Note that - W2 is on left side and W1 is on right side
To complete work in 18 days we need either 3 men or 6 boys.
∴ 1 man = 2 boys
Take work done = 1
∴ 3 men x 18 days x 1 = (4 men + 4 boys) x ? days x 1
Remember - Convert either all men to boys or all boys to men.
∴ 6 boys x 18 days x 1 = (8 boys + 4 boys) x ? days x 1
∴ ? = 9 days = they will need these many days.
3. A tank has water for 72 trees and can last for 54 days for them. If each tree is given 10% less water than 90 trees can get the water for how many days?
A. 24 days
B. 36 days
C. 42 days
D. 48 days
Answer: Option D
Explanation:
Let each tree take T amount of water every day.
So 72 trees take 72T water in one day.
With 10% reduction, each tree will consume[latex]\frac {90T}{100}[/latex]amount each day
So 90 Trees take 90[latex]\frac {90T}{100}[/latex]amount in one day
Total water quantity is constant
∴ 72T x 54 = 90 x [latex]\frac {90T}{100}[/latex] x D
∴ D = 48 days = Number of days 90 trees can use the water.
Permutation and Combination
1. A question paper had 10 questions. Each question could only be answered as true(T) of False(F). Each candidate answered all the questions, Yet no two candidates wrote the answers in an identical sequence. How many different sequences of answers are possible?
A. 20
B. 40
C. 512
D. 1024
Answer: Option D
Explanation:
Each question can be answered in 2 ways.
10 Questions can be answered = 2[latex]^ {10}[/latex]= 1024 ways.
2. In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?
A. 120
B. 240
C. 360
D. 480
Answer: Option D
Explanation:
As there are six players, So total ways in which they can be arranged = 6!ways =720.
A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.
Therefore, Number of ways when they don’t remain together = 720 -240 =480.
Surds and Indices
1. Simplify : ([latex]y^ {4b-a }[/latex]. [latex]y^ {4c-b}[/latex] . [latex]y^ {4a-c}[/latex]) / [latex](y^ {a}. y^ {b}. y^ {c})^{3}[/latex]
A. 0
B. 1
C. y
D. y[latex]^ {2}[/latex]
E. None of these
Answer: Option B
Explanation:
([latex]y^ {4b-a }[/latex]. [latex]y^ {4c-b}[/latex] . [latex]y^ {4a-c}[/latex]) / [latex](y^ {a}. y^ {b}. y^ {c})^{3}[/latex]
=([latex]y^ {4b-a }[/latex]. [latex]y^ {4c-b}[/latex] . [latex]y^ {4a-c}[/latex]) / [latex](y^ {a + b + c} )^{3}[/latex]
=([latex]y^ {4b-a+4c-b+4a-c}[/latex] / [latex]y ^{3(a + b + c)}[/latex]
=[latex]y ^{3(a + b + c)}[/latex] / [latex]y ^{3(a + b + c)}[/latex]
= 1.
2. If 3[latex]^ {4m+1}[/latex] = 3[latex]^ {7m-5}[/latex], solve for m.
A. 2
B. 3
C. 1/3
D. 1/2
E. None of these
Answer: Option A
Explanation:
3[latex]^ {4m+1}[/latex] = 3[latex]^ {7m-5}[/latex] equating powers of 3 on both sides, 4m+1 = 7m - 5 => 3m = 6 => m = 2.
3. Identify the greatest numbers: 4[latex]^ {50}[/latex], 2[latex]^ {100}[/latex], 16[latex]^ {25}[/latex]
A. 4[latex]^ {50}[/latex]
B. 2[latex]^ {100}[/latex]
C. 16[latex]^ {25}[/latex]
D. All are equal
E. 2[latex]^ {100}[/latex], 16[latex]^ {25}[/latex]
Answer: Option D
Explanation:
4[latex]^ {50}[/latex] = [latex](2 ^{(2)50})[/latex] = 2[latex]^ {100}[/latex] => 16[latex]^ {25}[/latex] = [latex](2 ^{(4)25})[/latex] = 2[latex]^ {100}[/latex]
Hence, 4[latex]^ {50}[/latex], 2[latex]^ {100}[/latex] and 16[latex]^ {25}[/latex]are all equal.
Pipes and Cistern
1. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is?
A. 6 hrs
B. 10 hrs
C. 15 hrs
D. 30 hrs
Answer: Option C
Explanation:
Suppose, first pipe alone takes x hours to fill the tank. Then, second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank.
1/x + 1/(x - 5) = 1/(x - 9)
(2x - 5)(x - 9) = x(x - 5)
x2 - 18x + 45 = 0
(x- 15)(x - 3) = 0 => x = 15
2. Three taps A, B, and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full is?
A. 6 hrs
B. 6 2/3 hrs
C. 7 hrs
D. 7 1/2 hrs
Answer: Option C
Explanation:
(A + B)'s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)'s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 - 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
3. One top can fill a cistern in 2 hours and another can empty the cistern in 3 hours. How long will they take to fill the cistern if both the taps are opened?
A. 5 Hours
B. 6 Hours
C. 7 Hours
D. 8 Hours
Answer: Option B
Explanation:
Net filling in 1 hour = (1/2 – 1/3) = 1/6
Time is taken to fill the cistern = 6 hours
Boats and Streams
1. A boat covers a distance of 30 km in 2 ½ hours running downstream while returning it covers the same distance in 3 ¾ hours. What is the speed of the boat in still water?
A. 8km/hr
B. 12km/hr
C. 14 km/hr
D. 15 km/hr
E. None of these
Answer: Option E
Explanation:
Speed downstream = (30 x 2/5)km/hr = 12km/hr
Speed upstream = (30 x 4/15) km/hr = 8 km/hr
Speed of boat in still water = [latex]\frac {1}{2}[/latex] (12 + 8)km/hr = 10 km/hr
2. The speed of a motorboat is that of the current of water as 36:5. The boat goes along with the current in 5 hours 10 min, it will come back in?
A. 5hrs 50 min
B. 6 hours
C. 6 hours 50 min
D. 12 hours 10 min
Answer: Option C
Explanation:
Let the speed of motorboat be 36x km/hr and that of the current of water be 5x km/hr
Speed downstream = (36x +5x)km/hr = 41x km/hr
Speed upstream = (36 -5x)km/hr = 31x km/hr
Distance covered downstream = (41x × 31/6)km
Distance upstream = [1271x/6 × 1/31x]hrs
= 41/6 hrs = 6hrs 50 min
3. A boat goes 6 km in an hour in still water. It takes thrice as much time in covering the same distance against the current. The speed of the current is?
A. 2 km/hr
B. 3km/hr
C. 4km/hr
D. 5 km/hr
Answer: Option C
Explanation:
Speed in still water = 6 km/hr
Speed against the current = 6/3 km/hr = 2 km/hr
Let the speed of the current be x km/hr
6-x = 2 => x =4 km/hr
Numbers
1. 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
2. The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction?
A. 2/5
B. 7/9
C. 3/11
D. 8/5
Answer: Option C
Explanation:
The denominator is P, the numerator will be (P - 8).
The fraction will be (P - 8)/P.
Adding 3 to the numerator and subtracting 3 from the denominator, (P - 8 + 3)/(P - 3) = 3/4.
(P - 5)/(P - 3) = 3/4
P = 20 - 9 => P = 11.
The fraction is 3/11.
Partnership
1. Three friends M, N and O stay in a business for 2 months, 5 months and 8 months respectively. They share the profit in the ration 5 : 8 : 9 respectively. Find the ratio of their investment.
A. 7 : 13 : 17
B. 10 : 40 : 72
C. 100 : 64 : 45
D. 110 : 74 : 64
Answer: Option C
Explanation:
Since they spend 2 months, 5 months and 8 months in the business respectively, their time period in business is in the ratio of 2: 5: 8.
The ratio of investment for M, N, and O is found by =[latex]\frac {5}{2}[/latex]: [latex]\frac {8}{5}[/latex]: [latex]\frac {9}{8}[/latex]
Making denominators common by multiplying by 40
∴ Ratio of Time periods for M, N, and O = [latex]\frac {40 x\times 5}{2}[/latex] : [latex]\frac {40 \times 8}{5}[/latex]: [latex]\frac {40 \times 9}{8}[/latex] = 100 : 64 : 45
2. Armaan and Gubbi started a cafe with Rs. 40000 and Rs. 80000, respectively. Gubbi got married to someone in other town and left after 7 months. But Jayaram immediately replaced her with an investment of Rs. 144000. At the end of the year, the business performed well and registered a profit of Rs. 50600. What is Gubbi’s share in this profit?
A. Rs. 13800
B. Rs. 16100
C. Rs. 16500
D. Rs. 16866.67
Answer: Option B
Explanation:
The ratio of Investment x Time = Ratio of Profit
∴ (A's investment x Time) : (B's investment x Time) = Profit of A : Profit of B
Total value of investment of Arman, Gubbi and Jayram after 12 months is =
Rs. 40000 x 12 months : Rs. 80000 x 7 months : Rs. 144000 x (12-7)months = 480000 : 560000 : 720000
∴ Profit ratio = 240000 : 280000 : 360000 = 6 : 7 : 9
∴ Share of Gubbi = [latex]\frac {7}{6+7+9}[/latex] x 50600 = Rs. 16100.
Ratio and Proportion
1. 1600 men have provisions for 28 days in the temple. If after 4 days, 400 men leave the temple, how long will the food last now?
A. 28 days
B. 30 days
C. 32 days
D. 35 days
Answer: Option C
Explanation:
1600 ---- 28 days
1600 ---- 24
1200----?
1600*24 = 1200*x
x = 32 days
2. There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 19 days more?
A. 60 men
B. 40 men
C. 25 men
D. 20 men
Answer: Option B
Explanation:
760 ---- 22
760 ---- 20
x ----- 19
x*19 = 760*20 = 800
x = 800 - 760 = 40
Problems on H.C.F and L.C.M.
1. Find the greatest 4-digit number exactly divisible by 3, 4 and 5?
A. 9985
B. 9960
C. 9957
D. 9975
Answer: Option B
Explanation:
The greatest 4-digit number is 9999. LCM of 3, 4 and 5 is 60. Dividing 9999 with 60, we get remainder 39. Thus, the required number is 9999 - 39 = 9960.
2. Find the greatest number which, while dividing 19, 83 and 67, gives a remainder of 3 in each case?
Answer: Option A
Explanation:
Subtract the remainder 3 from each of the given numbers: (19-3)=16, (67-3)=64 and (83-3)=80. Now, find the HCF of the results 16, 64 and 80, we get 16. Thus, the greatest number is 16.
Banker’s Discount
1. The banker's discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time is Rs. 60. The sum due is:
A. Rs. 360
B. Rs. 432
C. Rs. 540
D. Rs. 1080
Answer: Option A
Explanation:
Sum =[latex]\frac {B.D. \times T.D.}{B.D. - T.D.}[/latex] = Rs. [latex]\frac {72 \times 60}{72 - 60}[/latex] = Rs. [latex]\frac {72 \times 60}{12}[/latex] = Rs. 360.
2. The certain worth of a certain sum due sometime hence is Rs. 1600 and the true discount is Rs. 160. The banker's gain is:
A. Rs. 20
B. Rs. 24
C. Rs. 16
D. Rs. 12
Answer: Option C
Explanation:
B.G. =[latex]\frac {(T.D.)^{2}}{P.W.}[/latex] = Rs. [latex]\frac {160 \times 160}{1600}[/latex] = Rs. 16.
3. The present worth of a certain bill due sometime hence is Rs. 800 and the true discount is Rs. 36. The banker's discount is:
A. Rs. 37
B. Rs. 37.62
C. Rs. 34.38
D. Rs. 38.98
Answer: Option B
Explanation:
B.G. =[latex]\frac {(T.D.)^{2}}{P.W.}[/latex] = Rs. [latex]\frac {36 \times 36}{800}[/latex] = Rs. 1.62
B.D. = (T.D. + B.G.) = Rs. (36 + 1.62) = Rs. 37.62
Compound Interest
1. Rohit says that, an amount became 4 times in 6 years. His friend wanted to know, how many years will it take for the amount to become 64 times, provided the rate of interest remains same?
A. 12 years
B. 16 years
C. 18 years
D. 24 years
Answer: Option C
Explanation:
Initial Amount = P.
P becomes 4 times in 6 years ∴ P → 4P in 6 years
∴ 4P → 4 x (4P) = 16P in next 6 years.
Continuing this, 16P → 4 x (16P) = 64P in next 6 years.
So total time to get to 64 times i.e. 64P = 6+6+6 = 18 years.
2. A sum of Rs.4800 is invested at compound interest for three years, the rate of interest is 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years.
A. Rs.2520
B. Rs.3120
C. Rs.3320
D. Rs.2760
E. None of these
Answer: Option B
Explanation:
Let A be the amount received at the end of the three years.
A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4800 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.7920
So the interest = 7920 - 4800 = Rs.3120
Area
1. The perimeter of a semi circle is 144 cm then the radius is?
A. 25 cm
B. 28 cm
C. 30 cm
D. 35 cm
Answer: Option B
Explanation:
36/7 r = 144 => r = 28
2. The Length of a plot is four times is breadth. A playground measuring 1200 square meters occupies a third of the total area of the plot. What is the length of the plot in meters?
A. 20
B. 30
C. 60
D. None of these
Answer: Option D
Explanation:
Area of the plot = (3x1200)Sq.m = 3600 Sq.m
Breadth = x meters and length = 4x meters
4x × x = 3600 or x[latex]^ {2}[/latex] = 900 0r x = 30
Length of plot = 4x = (4×30) m = 120 m
Time and Work
1. A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in.
A. 420 days
B. 120 days
C. 105 days
D. 60 days
Answer: Option D
Explanation:
A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days
2. Anita, Indu, and Geeta can do a piece of work in 18 days, 27 days and 36 days respectively. They start working together. After working for 4 days. Anita goes away and Indu leaves 7 days before the work is finished. Only Geeta remains at work from beginning to end. In how many days was the whole work is done?
A. 16 days
B. 17 days
C. 18 days
D. 19 days
Answer: Option A
Explanation:
4/18 + (x -7)/27 + x/36 = 1
x = 16 days
Allegation or Mixture
1. A body massage oil mix contains 60 liters of Oil A with some liters of Oil B. The rate of Oil A is Rs. 32 per liter while the rate of Oil B is 9 Rs/liter less than Oil A. A shopkeeper sells this oil mix at Rs. 28/liter. How much Oil B should be mixed in the mixture to ensure that there is no profit, no loss in the whole process?
A. 24 litres
B. 36 litres
C. 48 litres
D. 50 litres
Answer: Option C
Explanation:
Let Oil B quantity be B liters.
Final mixture has '60+B' litres
There is no profit no loss
So, (Oil 1 quantity * rate) + (Oil 2 quantity * rate) = Mix quantity * rate
Oil B is 9 Rs/- cheap. So, the rate of Oil B per litre = 23 Rs/litre
∴ 60 x 32 + B x 23 = (60+B) x 28
∴ B = 48 litres = Oil B quantity
2. Three vessels contain a milk mixture 30 liter each. When put in a big vessel, they result in a mixture of milk and water in the ratio 2:1. If the ratio is to be reversed to make it 1: 2, how much more water should be added to the mix?
A. 20L
B. 40L
C. 90L
D. 100L
Answer: Option C
Explanation:
3 vessels of 30 litres each = 3 X 30 = 90 ml milk mixture
Amount of milk in mixture = [latex]\frac {2}{2+1}[/latex] x 90 = 60L
Amount of water = 90-60 = 30L
To make milk to water ratio 1:2, we simply need to make water double of milk.
By direct observation, we can say that we should have 60L x 2 = 120L water to make the required ratio
We already have 30L, we need (120-30) = 90L more water.
Decimal Fraction
1. How many digits will be there to the right of the decimal point in the product of 89.635 and .02218?
Answer: Option C
Explanation:
Sum of decimal places = 3 + 5 = 8
The last digit in the product(digit at the extreme right) is zero (Since 5 x 8 = 40)
Hence, there will be 7 significant digits to the right of the decimal point.
2. 24.39 + 562.093 + 35.96 = ?
A. 622.441
B. 622.243
C. 622.233
D. 622.443
Answer: Option D
Explanation:
24.39 + 562.093 + 35.96 = 622.443
3. What decimal of an hour is a second?
A. 0.00027
B. 0.00025
C. 0.00026
D. 0.00024
Answer: Option A
Explanation:
1 hour = 60 minutes = 3600 seconds
Hence, required decimal = 1/3600 = 0.00027
Probability
1. Ramesh throws two dices simultaneously. What is the probability that the sum of numbers on top faces in a throw is 9?
A. [latex]\frac {1}{12}[/latex]
B. [latex]\frac {1}{9}[/latex]
C. [latex]\frac {1}{3}[/latex]
D. [latex]\frac {2}{9}[/latex]
Answer: Option B
2. India is playing 3 sets of badminton matches against Japan at Singapore. The probability of India winning the three sets is 1/7, 1/5, 3/4. What is the probability of winning at least 1 set of the game?
A. [latex]\frac {7}{24}[/latex]
B. [latex]\frac {23}{24}[/latex]
C. [latex]\frac {23}{35}[/latex]
D. [latex]\frac {29}{35}[/latex]
Answer: Option D
3. If I roll two dices simultaneously, what is the probability of getting a sum less than 13?
A. [latex]\frac {1}{2}[/latex]
B. 0.75
C. 1
D. 1.25
Answer: Option C
Average
1. Find the average price of the goods he has bought.
A. Rs. 5.5
B. Rs. 6.05
C. Rs. 8.25
D. Rs. 9
Answer: Option B
Explanation:
Average =[latex]\frac {Sum of Observations}{Number of Observations}[/latex]
Average = [latex]\frac {Total cost of pencils + Total cost of pens + Total cost of sharpeners}{Total pencils + Total pens + Total sharpeners}[/latex]
∴ Average price = [latex]\frac {(35 \times 4)+(30 \times 11)+(25 \times 3)}{35+30+25}[/latex] = Rs. 6.05
2. If I do not consider the earnings from selling the six seater dining table, my average falls by Rs 2000/-. What is the cost of six seater dining table if average earnings from selling 11 four-seater dining tables and one six-seater dining table are Rs 18000?
A. Rs. 29000
B. Rs. 30000
C. Rs. 36000
D. Rs. 40000
Answer: Option D
Explanation:
Without six seater dining table, there are 11 four-seater dining tables.
Average = 18000 - 2000 = 16000.
Total cost of 11 small dining tables = 11 x 16000 = Rs. 176000
With big dining table, total cost of 12 tables = 12 x 18000 = Rs. 216000
Cost of six seater dining table = 216000 - 176000 = Rs. 40000
3. In an exam, the average marks for 80 students of Class V are 35. The average of marks in section A of the class is 55 while the average of marks in section B is 30. Find the number of students in Class V B.
Answer: Option C
Explanation:
Total students i.e. Section A + Section B = 80
Section A = 80 - Section B = 80 - B
∴ 80 x 35 = B x 30 + (80 - B) x 55
2800 = 30B + 4400 - 55B
1600 = 25 B
B = 64
∴ B = 64 = Number of students in Class V B.
Stocks and Share
1. A man sells Rs.5000, 12 % stock at 156 and uinvests the proceeds parity in 8 % stock at 90 and 9 % stock at 108. He hereby increases his income by Rs. 70. How much of the proceeds were invested in eac
A. 4000
B. 4200
C. 4002
D. 4020
Answer: Option B
2. The cash realized on selling a 14% stock is Rs.106.25, brokerage being 1/4% is
A. 123
B. 106
C. 100
D. 156
Answer: Option B
3. A invested some money in 10% stock at 96. If B wants to invest in an equally good 12% stock, he must purchase a stock worth of :
A. Rs.80
B. Rs.115.20
C. Rs.120
D. Rs.125.40
Answer: Option B
Square Root and Cube Root
1. Find the smallest number by which 1780 must be added to make the sum a perfect square.
Answer: Option C
Explanation:
Step 1: Find the square root of 1780
[latex]\sqrt {1780}[/latex] = 42.190
Take the square of the next number after 42 i.e. 43.
Step 2: Find the least number
Subtract given number from 43[latex]^{2}[/latex]
43[latex]^{2}[/latex] – 1780 = Number to be added
1849 – 1780 = 69
Hence, the number to be added is 69.
2. If [latex]\sqrt {5}[/latex] = 2.236, then find the value of [latex]\sqrt {320}[/latex] –[latex]\frac {1}{5}[/latex] [latex]\sqrt {125 }[/latex] – [latex]\sqrt {180}[/latex]
A. 4.236
B. 2.236
C. 3.346
D. 1.566
Answer: Option B
Explanation:
= [latex]\sqrt {320}[/latex] – (1/5)[latex]\sqrt {125}[/latex] – [latex]\sqrt {180}[/latex]
= [latex]\sqrt {64 \times 5 }[/latex] – (1/5) [latex]\sqrt {25 \times 5}[/latex] – [latex]\sqrt { 36 \times 5}[/latex]
= 8[latex]\sqrt {5}[/latex] – (1/5)5[latex]\sqrt {5}[/latex] – 6[latex]\sqrt {5}[/latex]
= 8[latex]\sqrt {5}[/latex] – (1 + 6)[latex]\sqrt {5}[/latex]
= [latex]\sqrt {5}[/latex]
Value of[latex]\sqrt {5}[/latex] = 2.236
3. Find the least number by which 1470 must be divided to get a number which is a perfect square.
Answer: Option D
Explanation:
This numerical can be easily solved, if we solve it by considering the options.
Divide 1470 by each option and find out the number which is a perfect square.
[latex]\frac {1470}{6}[/latex] = 245, [latex]\frac {1470}{9}[/latex] = 163.33, [latex]\frac {1470}{12}[/latex]=122.5, [latex]\frac {1470}{30}[/latex] = 49
Hence, it is clear that the correct option is D = 30
Problems on Ages
1. A person was asked to state his age in years. His reply was, "Take my age three years hence, multiply it by 3 and then subtract three times my age three years ago and you will know how old I am." What was the age of the person?
A. 18 years
B. 20 years
C. 24 years
D. 32 years
Answer: Option A
Explanation:
Let the present age of the person be x years.
Then, 3(x + 3) - 3(x - 3) = x
3x + 9 - 3x + 9 = x => x = 18
2. The sum of the ages of a son and father is 56 years after four years the age of the father will be three times that of the son. Their ages respectively are:
A. 12years, 44 years
B. 16years, 42years
C. 16years, 48years
D. 18years, 36 years
Answer: Option A
Explanation:
Present ages of son and father are x years (56 -x)years
(56- x +4) = 3(x + 4) or 4x =48 or x = 12
Ages are 12 years, 44 years
3. In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, the present age of B is
A. 19 Years
B. 29 Years
C. 39 Years
D. 49 Years
Answer: Option C
Explanation:
Present ages of B and A be x years and (x +9) years
(x +9 +10) = 2 (x -10) or x =39