Simple Interest.
1. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
A. Rs. 650
B. Rs. 690
C. Rs. 698
D. Rs. 700
Answer: Option (C)
Explanation:
S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.
2. A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
A. 3%
B. 4%
C. 5%
D. 6%
E. None of these
Answer: Option (D)
Explanation:
S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate = ([latex]\frac{100 \times 3000}{12500 \times 4}[/latex])% = 6%
Height and Distance.
1. Guddi was standing on a road near a mall. She was 1000m away from the mall and able to see the top of the mall from the road in such a way that top of the tree, which is in between her and the mall, was exactly in line of sight with the top of the mall. The tree height is 10m and it is 20m away from Guddi. How tall is the mall?
A. 423.5 m
B. 450 m
C. 470 m
D. 495 m
Answer: Option (C)
Explanation:
2. There is a tree between houses of A and B. If the tree leans on A’s House, the tree top rests on his window which is 12 m from the ground. If the tree leans on B’s House, the tree top rests on his window which is 9 m from the ground. If the height of the tree is 15 m, what is the distance between A’s and B’s house?
A. 21 m
B. 25 m
C. 16 m
D. 12 m
Answer: Option (A)
Explanation:
Volume and Surface Area
1. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is:
A. 12 kg
B. 60 kg
C. 72 kg
D. 96 kg
Answer: Option (B)
Explanation:
Volume of water displaced = (3 x 2 x 0.01) m3 = 0.06 m3.
Mass of man = Volume of water displaced x Density of water
= (0.06 x 1000) kg
= 60 kg.
2.A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
A. 49 m2
B. 50 m2
C. 53.5 m2
D. 55 m2
Answer: Option (A)
Explanation:
Area of the wet surface = [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2
= 49 m2.
Logarithm
1.Which of the following statements is not correct?
A. log10 10 = 1
B. log (2 + 3) = log (2 x 3)
C. log10 1 = 0
D. log (1 + 2 + 3) = log 1 + log 2 + log 3
Answer: Option B
Explanation:
(a) Since loga a = 1, so log10 10 = 1.
(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3
log (2 + 3) log (2 x 3)
(c) Since loga 1 = 0, so log10 1 = 0.
(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.
So, (b) is incorrect.
2. If log10 2 = 0.3010, the value of log10 80 is:
A. 1.6020
B. 1.9030
C. 3.9030
D. None of these
Answer: Option B
Explanation:
log10 80 = log10 (8 x 10)
= log10 8 + log10 10
= log10 (23 ) + 1
= 3 log10 2 + 1
= (3 x 0.3010) + 1
= 1.9030.
Races and Games
1. At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?
A. 22 points
B. 20 points
C. 12 points
D. 10 points
Answer: Option D
Explanation :
"A can give 15 points to B in 60".
In a game of 60, A starts with 0 points where B starts with 15 points.
"A can give 20 points to C in 60".
In a game of 60, A starts with 0 points where C starts with 20 points.
ie, in a game of 60-15-45, B can give C 20-15=5 points.
ie, in a game of 90, B can give C 5×2=10 points.
2. In a game of 90 points, A can give B 15 points and C 30 points. How many points can B give C in a game of 100 points?
A. 140
B. 20
C. 300
D. 50
Answer: Option B
Explanation :
While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points
i.e., when B scores 75 points, C scores 60 points
When B scores 100 points, C scores ([latex]\frac{60}{75}[/latex]) ×100 = 80 points
i.e., in a game of 100 points, B can give C (100-80)=20 points.
Simplification
1. (764 × ?) ÷ 250 = 382
A. 115
B. 145
C. 135
D. 125
E. None of these
Answer: D
Solution:
([latex]\frac{764 \times?}{250}[/latex]) = 382
? = ([latex]\frac{382 \times250}{764}[/latex]) = 125.
2. 9643 – 7750 +? = 4990
A. 3079
B. 3097
C. 3090
D. 4010
E. None of these
Answer: B
Solution:
? = 4990 – 9643 + 7750 = 3097.
Time and Distance
1. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Answer: Option B
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km =([latex]\frac{9}{54}[/latex]) x 60 min = 10 min.
2. A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 3.6
B. 7.2
C. 8.4
D. 10
Answer: Option B
Explanation:
Speed = ([latex]\frac{600}{5 \times60}[/latex]) 600 m/sec.
= 2 m/sec.
Converting m/sec to km/hr (see important formulas section)
= 2 x([latex]\frac{18}{5}[/latex]) km/hr
= 7.2 km/hr.
Chain Rule
1. If a quarter kg of potato costs 60 paise, how many paise will 200 gm cost?
A. 48 paise
B. 54 paise
C. 56 paise
D. 72 paise
Answer: Option A
Explanation:
Let the required weight be x kg.
Less weight, Less cost (Direct Proportion)
250 : 200 :: 60 : x 250 x x = (200 x 60)
x= ([latex]\frac{200 \times 60}{250}[/latex])
x = 48.
2. A wheel that has 6 cogs meshes with a larger wheel or 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions made by the larger wheel is:
Answer: Option B
Explanation:
Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Fewer revolutions (Indirect Proportion)
14 : 6 :: 21 : x
14 x x = 6 x 21
x = ([latex]\frac{6\times 21}{14}[/latex])
x = 9.
Permutation and Combination
1. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
A. 360
B. 480
C. 720
D. 5040
E. None of these
Answer: Option C
Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
2. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Answer: Option D
Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Surds and Indices
1. (17)3.5 x (17)? = 178
A. 2.29
B. 2.75
C. 4.25
D. 4.5
Answer: Option D
Explanation:
Let (17)3.5 x (17)x = 178.
Then, (17)3.5 + x = 178.
3.5 + x = 8
x = (8 - 3.5)
x = 4.5
2. If 3(x - y) = 27 and 3(x + y) = 243, then x is equal to:
Answer: Option C
Explanation:
3x - y = 27 = 33 x - y = 3 ....(i)
3x + y = 243 = 35 x + y = 5 ....(ii)
On solving (i) and (ii), we get x = 4.
Pipes and Cistern
1. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank at the same time 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 hours
B. 10 hours
C. 15 hours
D. 30 hours
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.
([latex]\frac{1}{x}[/latex])+([latex]\frac{1}{(x - 5)}[/latex]) = ([latex]\frac{1}{(x - 9)}[/latex])
([latex]\frac{x - 5 + x}{x(x - 5)}[/latex]) = ([latex]\frac{1}{(x - 9)}[/latex])
(2x - 5)(x - 9) = x(x - 5)
x2 - 18x + 45 = 0
(x - 15)(x - 3) = 0
x = 15. [neglecting x = 3]
2. A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
A. 4 ([latex]\frac{1}{3}[/latex]) hours
B. 7 hours
C. 8 hours
D. 14 hours
Answer: Option D
Explanation:
Work was done by the leak in 1 hour =([latex]\frac{1}{2}[/latex]) - ([latex]\frac{3}{7}[/latex]) =([latex]\frac{1}{14}[/latex]).
The leak will empty the tank in 14 hrs.
Boats and Streams
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. In one hour, a boat goes 11 km/hr along the stream and 5 km/hr against the stream. The speed of the boat in still water (in km/hr) is:
A. 3 km/hr
B. 5 km/hr
C. 8 km/hr
D. 9 km/hr
Answer: Option C
Explanation:
Speed in still water =([latex]\frac{1}{2}[/latex])(11 + 5) kmph
= 8 kmph.
Numbers
1. Which one of the following is not a prime number?
Answer: Option D
Explanation:
91 is divisible by 7. So, it is not a prime number.
2. What is the unit digit in {(6374)1793 x (625)317 x (341491)}?
Answer: Option A
Explanation:
Unit digit in (6374)1793 = Unit digit in (4)1793
= Unit digit in [(42)896 x 4]
= Unit digit in (6 x 4) = 4
Unit digit in (625)317 = Unit digit in (5)317 = 5
Unit digit in (341)491 = Unit digit in (1)491 = 1
Required digit = Unit digit in (4 x 5 x 1) = 0.
Partnership
1. A, B, and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
A. Rs. 1900
B. Rs. 2660
C. Rs. 2800
D. Rs. 2840
Answer: Option B
Explanation:
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs. 7030 x ([latex]\frac{14}{37}[/latex]) = Rs. 2660.
2. A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
A. 3 : 5 : 2
B. 3 : 5 : 5
C. 6 : 10 : 5
D. Data inadequate
Answer: Option C
Explanation:
Let the initial investments of A and B be 3x and 5x.
A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.
Ratio and Proportion
1. A sum of money is to be distributed among A, B, C, D in the proportion of 5: 2: 4 : 3. If C gets Rs. 1000 more than D, what is B's share?
A. Rs. 500
B. Rs. 1500
C. Rs. 2000
D. None of these
Answer: Option C
Explanation:
Let the shares of A, B, C, and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.
Then, 4x - 3x = 1000
x = 1000.
B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
2. Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40: 57. What is Sumit's salary?
A. Rs. 17,000
B. Rs. 20,000
C. Rs. 25,500
D. Rs. 38,000
Answer: Option D
Explanation:
Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.
Then, ([latex]\frac{2x + 4000}{3x + 4000}[/latex])= ([latex]\frac{40}{57}[/latex])
57(2x + 4000) = 40(3x + 4000)
6x = 68,000
3x = 34,000
Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.
Problems on H.C.F and L.C.M.
1. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Answer: Option A
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
2. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then the sum of the digits in N is:
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
Banker’s Discount
1. The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is:
A. 3 months
B. 4 months
C. 6 months
D. 8 months
Answer: Option B
Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.
Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
Time = ([latex]\frac{100 \times 80}{1600 \times 15}[/latex]) year = ([latex]\frac{1}{3}[/latex])year = 4 months.
2. The banker's gain on a sum due 3 years hence at 12% per annum is Rs. 270. The banker's discount is:
A. Rs. 960
B. Rs. 840
C. Rs. 1020
D. Rs. 760
Answer: Option C
Explanation:
T.D. = ([latex]\frac{B.G.\times 100}{R\times T}[/latex]) = Rs. ([latex]\frac{270\times 100}{12\times 3}[/latex]) = Rs. 750.
B.D. = Rs.(750 + 270) = Rs. 1020.
1. There is a 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
A. Rs. 2160
B. Rs. 3120
C. Rs. 3972
D. Rs. 6240
E. None of these
Answer: Option C
Explanation:
Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = ([latex]\frac{100 \times 60}{100 \times 6}[/latex]) = 10% p.a.
Now, P = Rs. 12000. T = 3 years and R = 10% p.a.
C.I.
= Rs. [latex]1200 \times ((1 + \frac{10}{100})^{3} -1)[/latex]
= Rs. ([latex]\frac{12000 \times 331}{1000}[/latex])
= 3972.
2. What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?
A. Rs. 9000.30
B. Rs. 9720
C. Rs. 10123.20
D. Rs. 10483.20
E. None of these
Answer: Option C
Explanation:
Amount
= Rs. [latex]1200 \times (1 + \frac{12}{100})^{3} [/latex]
= Rs. 1200 x ([latex]\frac{28}{25}[/latex]) x ([latex]\frac{28}{25}[/latex]) x ([latex]\frac{28}{25}[/latex])
= Rs. 35123.20
C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20
Area
1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
A. 4.04%
B. 2.02%
C. 4%
D. 2%
Answer: Option A
Explanation:
Percentage error in calculated area =(2+2+([latex]\frac{2 \times 2}{100}[/latex]))%
= 4.04%
2. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage
decrease in the area?
A. 30%
B. 28%
C. 32%
D. 26%
Answer: B
Explanation:
percentage change in area
=(−20−10+([latex]\frac{20\times 10}{100}[/latex]))% = −28%
i.e., the area is decreased by 28%
Time and Work
1. A can do a certain work at the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
A. 15 days
B. 20 days
C. 25 days
D. 30 days
Answer: Option C
Explanation:
(A + B)'s 1 day's work = ([latex]\frac{1}{10}[/latex])
C's 1 day's work = ([latex]\frac{1}{50}[/latex])
(A + B + C)'s 1 day's work = ([latex]\frac{1}{10}[/latex] + [latex]\frac{1}{50}[/latex]) = ([latex]\frac{6}{50}[/latex]) = ([latex]\frac{3}{25}[/latex]) . .... (i)
A's 1 day's work = (B + C)'s 1 day's work .... (ii)
From (i) and (ii), we get: 2 x (A's 1 day's work) = ([latex]\frac{3}{25}[/latex])
A's 1 day's work = ([latex]\frac{3}{50}[/latex]).
B's 1 day's work ([latex]\frac{1}{10}[/latex] - [latex]\frac{3}{50}[/latex]) = ([latex]\frac{2}{50}[/latex]) = ([latex]\frac{1}{25}[/latex]) .
So, B alone could do the work in 25 days.
2. A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished?
A. 11:30 A.M.
B. 12 noon
C. 12:30 P.M.
D. 1:00 P.M.
Answer: Option D
Explanation:
(P + Q + R)it's 1 hour's work = ([latex]\frac{1}{8}[/latex] + [latex]\frac{1}{10}[/latex] + [latex]\frac{1}{12}[/latex]) = ([latex]\frac{37}{120}[/latex])
Work was done by P, Q, and R in 2 hours = (([latex]\frac{37}{120}[/latex]) x 2 ) = ([latex]\frac{37}{60}[/latex]).
Remaining work = ( 1 - ([latex]\frac{37}{60}[/latex])) = ([latex]\frac{23}{60}[/latex]).
(Q + R)it's 1 hour's work = ([latex]\frac{1}{10}[/latex] + [latex]\frac{1}{12}[/latex]) = ([latex]\frac{11}{60}[/latex]).
Now, ([latex]\frac{11}{60}[/latex]) work is done by Q and R in 1 hour.
So, ([latex]\frac{23}{60}[/latex]) work will be done by Q and R in ([latex]\frac{60}{11}[/latex] x [latex]\frac{23}{60}[/latex]) = [latex]\frac{23}{11}[/latex] hours = 2 hours.
So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.
Allegation or Mixture
1. How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?
A. 36 kg
B. 42 kg
C. 54 kg
D. 63 kg
Answer: Option D
Explanation:
S.P. of 1 kg of mixture = Rs. 9.24, Gain 10%.
C.P. of 1 kg of mixture = Rs. ([latex]\frac{100}{120}[/latex] x 9.24) = Rs. 8.40
By the rule of allylation, we have:
The ratio of quantities of 1st and 2nd kind = 14: 6 = 7 : 3.
Let x kg of sugar of 1st be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27
x = ([latex]\frac{7\times 27}{3}[/latex]) = 63 kg.
2. A container contains 40 liters of milk. From this container, 4 liters of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26.34 litres
B. 27.36 litres
C. 28 litres
D. 29.16 litres
Answer: Option D
Explanation:
Amount of milk left after 3 operations = [latex]40 \times (1 - \frac{4}{40})^{3} [/latex]
(40 x [latex]\frac{9}{10}[/latex] x [latex]\frac{9}{10}[/latex] x [latex]\frac{9}{10}[/latex]) = 29.16 litres.
Decimal Fraction
1. 3889 + 12.952 - ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
Answer: Option D
Explanation:
Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
= 3901.952 - 3854.002
= 47.95.
2. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y?
A. 2010
B. 2011
C. 2012
D. 2013
Answer: Option B
Explanation:
Suppose commodity X will cost 40 paise more than Y after z years.
Then, (4.20 + 0.40z) - (6.30 + 0.15z) = 0.40
0.25z = 0.40 + 2.10
z = ([latex]\frac{2.50}{0.25}[/latex]) = ([latex]\frac{250}{25}[/latex]) = 10
X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.
Probability
1. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A. [latex]\frac{1}{2}[/latex]
B. [latex]\frac{2}{5}[/latex]
C. [latex]\frac{8}{15}[/latex]
D. [latex]\frac{9}{20}[/latex]
Answer: Option D
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = [latex]\frac{n(E)}{n(S)}[/latex] = [latex]\frac{9}{20}[/latex].
2. 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].
Average
1. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. Rs. 4991
B. Rs. 5991
C. Rs. 6001
D. Rs. 6991
Answer: Option A
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991.
2. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
Answer: Option D
Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
Stocks and Share
1. 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.
2. A 6% stock yields 8%. The market value of the stock is:
A. Rs. 48
B. Rs. 75
C. Rs. 96
D. Rs. 133.33
Answer: Option B
Explanation:
For an income of Rs. 8, investment = Rs. 100.
For an income of Rs. 6, investment = Rs.([latex]\frac{100}{8}[/latex] x 6) = Rs. 75.
Market value of Rs. 100 stock = Rs. 75.
Square Root and Cube Root
1. The cube root of .000216 is:
Answer: Option B
Explanation:
[latex]{(.000216)^{\frac{1}{3}}}[/latex] = [latex]\frac{216}{10^{6}}^{\frac{1}{3}} [/latex]
= ([latex]\frac{6}{102} \times \frac{6}{102} \times \frac{6}{102})^{\frac{1}{3}}[/latex]
=[latex]\frac{6}{10^{2}}[/latex]
=[latex]\frac{6}{100}[/latex]
= 0.06
2. The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444
Answer: Option A
Explanation:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444.
Problems on Ages
1. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
A. 4 years
B. 8 years
C. 10 years
D. None of these
Answer: Option A
Explanation:
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
2. Present ages of Sameer and Anand are in the ratio of 5: 4 respectively. Three years hence, the ratio of their ages will become 11: 9 respectively. What is Anand's present age in years?
A. 24
B. 27
C. 40
D. Cannot be determined
E. None of these
Answer: Option A
Explanation:
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then,[latex]\frac{5x + 3}{4x + 3}[/latex] = [latex]\frac{11}{9}[/latex]
9(5x + 3) = 11(4x + 3)
45x + 27 = 44x + 33
45x - 44x = 33 - 27
x = 6.
Anand's present age = 4x = 24 years.