51. If each edge of a cube is increased by 50%, find the percentage increase in its surface area.
Sol: Let the original length of each edge = a
Then, Original surface area = 6a^2
New surface area = 6 * (3a/2)^2 = 27a^2/2
Increase percent in surface area = (15/2a^2 * 1/(6a^2) * 100)% = 125%
52. Find the number of the bricks, each measuring 25 cm by 12.5 cm by 7.5 cm, required to build a wall 6 m long, 5 m high and 50cm thick, while the mortar occupies 5% of the volume of the wall.
Sol: Volume of the Wall = (600 * 500 * 50) cu. Cm.
Volume of the bricks = 95% of the volume of the wall.
= (95/100 * 600 * 500 * 50) cu. Cm.
Volume of 1 brick = (25 * 25/2 * 75/2) cu. Cm.
Therefore, Number of bricks = (95/100 * (600 * 500 * 50 * 2 * 10)/(25 * 25 * 75))=6080
53. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Sol: Area of the field = Total cost/Rate = (333.18/24.68) hectares =13.5 hectares.
= (13.5*10000) m^2 =135000m^2.
Let altitude = x meters and base = 3x meters.
Then, ½ *3x* x= 135000 or x^2 = 9000 or x= 300.
Therefore, base =900 m & altitude = 300m.
54. Find the area of a rhombus one side of which measures 20cm and one diagonal
24cm.
Sol: Let, other diagonal = 2x cm,
Since halves of diagonals and one side of rhombus form a right angled triangle
with side as hypotenuse, we have:
(20)^2 =(12)^2+x^2 or x=Ö(20)^2-(12)^2 =Ö256=16 cm.
Therefore, other diagonal = 32 cm.
55. A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are fill pipes?
Sol. Let the number of fill pipes be ‘n'. Therefore, there will be 8-n, waste pipes.
Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fill pipes will fill 1/8th
of the tank in an hour.
Hence, n fill pipes will fill n/8th of the tank in an hour.
Similarly, each of the waste pipes will drain the full tank in 6 hours. That is, each of the
waste pipes will drain 1/6th of the tank in an hour.
Therefore, (8-n) waste pipes will drain ((8-n)/6)th of the tank in an hour.
Between the fill pipes and the waste pipes, they drain the tank in 6 hours. That is, when all
8 of them are opened, 1/6th of the tank gets drained in an hour.
(Amount of water filled by fill pipes in 1 hour - Amount of water drained by waste pipes 1
hour)
= 1/6th capacity of the tank drained in 1 hour.
56. A pump can be used either to fill or to empty a tank. The capacity of the tank is 3600 m3. The emptying capacity of the pump is 10 m3/min higher than its filling capacity. What is the emptying capacity of the pump if the pump needs 12 more minutes to fill the tank than to empty it?
Sol. Let ‘f’ m3/min be the filling capacity of the pump. Therefore, the emptying capacity of the
pump will be = (f + 10 ) m3 / min.
The time taken to fill the tank will be = minutes
And the time taken to empty the tank will be = .
We know that it takes 12 more minutes to fill the tank than to empty it
i.e => 3600 f + 36000 - 3600 f = 12 (f2 + 10 f)
=> 36000 = 12 (f2 + 10 f) => 3000 = f2 + 10 f => f2 + 10 f - 3000 = 0.
Solving for positive value of ‘f’ we get, f = 50.
Therefore, the emptying capacity of the pump = 50 + 10 = 60 m3 / min
57. X alone can do a piece of work in 15 days and Y alone can do it in 10 days. X and Y undertook to do it for Rs. 720. With the help of Z they finished it in 5 days. How much is paid to Z?
Sol. In one day X can finish 1/15th of the work.
In one day Y can finish 1/10th of the work.
Let us say that in one day Z can finish 1/Zth of the work.
When all the three work together in one day they can finish 1/15 + 1/10 + 1/Z = 1/5th of the
work.
Therefore, 1/Z = 1/30.
Ratio of their efficiencies = 1/15: 1/10: 1/30 = 2: 3: 1.Therefore Z receives 1/6th of the total
money.
According to their efficiencies money is divided as 240: 360: 120.
Hence, the share of Z = Rs. 120.
58. Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut?
Ans:10 hours
Sol. Pipe A fills the tank normally in 2 hours. Therefore, it will fill ½ of the tank in an hour.
Let the leak take x hours to empty a full tank when pipe A is shut. Therefore, the leak will
empty of the tank in an hour.
The net amount of water that gets filled in the tank in an hour when pipe A is open and when
there is a leak = of the tank. — (1)
When there is a leak, the problem states that Pipe A takes two and a half hours to fill the tank. i.e.
hours. Therefore, in an hour, of the tank gets filled. – (2)
Equating (1) and (2), we get => => x = 10 hours.
The problem can also be mentally done as follows.
Pipe A takes 2 hours to fill the tank. Therefore, it fills half the tank in an hour or 50%
of the tank in an hour.
When there is a leak it takes 2 hours 30 minutes for the tank to fill. i.e hours to fill the
tank or or 40% of the tank gets filled.
On account of the leak, (50 - 40)% = 10% of the water gets wasted every hour.
Therefore, the leak will take 10 hours to drain a full tank.
59. How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000?
Sol:7 does not occur in 1000. So we have to count the number of times it appears between 1 and
999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other
9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second
or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3-
digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with
the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first
or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54
+ 3 = 300
60. There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?
Sol: There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3, then the three objects can be
chosen in the following 23 ways - 3C0 ways of choosing none of the three, 3C1 ways of
choosing one out of the three, 3C2 ways of choosing two out of the three and 3C3 ways of
choosing all three.
In the given problem, there are 5 Rock songs. We can choose them in 25 ways. However, as
the problem states that the case where you do not choose a Rock song does not exist (at
least one rock song has to be selected), it can be done in 25 - 1 = 32 - 1 = 31 ways.
Similarly, the 6 Carnatic songs, choosing at least one, can be selected in 26 - 1 = 64 - 1 =
63 ways.
And the 3 Indi pop can be selected in 23 = 8 ways. Here the option of not selecting even
one Indi Pop is allowed.
Therefore, the total number of combinations = 31 * 63 * 8 = 15624
Friday, January 9, 2009
Aptitiude Questions Y5
41. A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times?
Ans.45 years.
Sol: P [1 + (R/100)]^15 = 2P è [1 + (R/100)]^15 = 2……….(i)
Let P [1 + (R/100)]^n = 8P è P [1 + (R/100)]^n = 8 = 2^3
= [{1 + (R/100)}^15]^3.
è [1 + (R/100)]^n = [1 + (R/100)]^45.
è n = 45.
Thus, the required time = 45 years.
42. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.
Ans: Sum = Rs. 5400,Rate=16 2/3 %.
Sol: S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.
Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.
Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.
è x * 7/6 * 7/6 = 7350.
è x = [7350 * 36/49] = 5400.
Therefore, Sum = Rs. 5400.
43. A, B and C start a business each investing Rs. 20000. After 5 months A withdrew Rs. 5000, B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.
Ans. A’s share = Rs. 20,500
B’s share = Rs. 21200
C’s share = Rs. 28200
Sol: Ratio of the capitals of A, B and C
= (20000*5+ 15000*7) : (20000*5+16000*7): (20000*5+26000*7)
=205000: 212000 : 282000 = 205:212:282
Therefore, A’s share = Rs. ( 69900*205/699) = Rs. 20,500
B’s share = Rs. (69900*212/699) = Rs. 21200,
C’s share = Rs. (69900*282/699) = Rs. 28200.
44. Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out an annual profit of Rs. 37100, find the share of each?
Sol: Ratio of their capitals= 36000*12:36000*9 = 4:3
Sanjiv’s share= Rs. ( 37100*4/7) = Rs. 21200.
Rajiv’s share = Rs. ( 37100*3/7) = Rs.15900.
45. If 20 men can build a wall 56m long in 6 days, what length of a similar wall can be built by 35 men in 3 days?
Ans. Length=49m.
Sol: Since the length is to be found out, we compare each item with the length as shown below.
More men, more length built (Direct).
Less days, less length built (Direct).
Men 20:35 :: 56: x
Similarly, days 6:3 :: 56: x.
Therefore, 20*6*x= 35*3*56 or x= 49.
Hence, the required length= 49m.
46.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.
Ans:26metric tonnes.
Sol: We shall compare each item with the quantity of coal.
Less engines, less coal consumed (direct)
More working hours, more coal consumed (direct)
If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
Less rate of consumption, less coal consumed (direct).
Therefore, number of engines 9:8 :: 24:x
Working hours 8:13 :: 24:x
Rate of consumption 1/3:1/4 :: 24:x.
9*8*1/3*x= 8*13*1/4*24 or x= 26.
Therefore, required consumption of coal 26 metric tonnes.
47. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
Ans.81
Sol: Remaining work = 1-4/7 =3/7.
Remaining period = (46-33) days =13 days.
Less work, less men (direct)
Less days, more men (indirect).
More hours per day, less men (indirect)
Therefore, work 4/7:3/7 ::117/x
Days 13:33 :: 117/x
Hrs/day 9:8:: 117/x
Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.
Therefore, additional men to be employed =(198-117) =81.
48. A garrison of 3300 men had provisions for 32 days, when given at the rate of 850gms per head. At the end of 7 days, reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825gms per head. What is the strength of the reinforcement?
Ans: 1700
Sol: The problem becomes:
3300 men taking 850gms per head have provisions for (32-7) or 25 days. How many
men taking 825gms each have provisions for 17 days?
Less ration per head, more men (indirect).
Less days, more men (indirect)
Ration 825:850::3300:x
Days 17:25::3300:x
Therefore, 825*17*x= 850*25*3300 or x= 5000.
Therefore, strength of reinforcement = 5000-3300 = 1700.
49. Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.
Sol: Slant Height, l = v(r^2 + h^2) =v(21^2 + 28^2) = v1225 = 35 cm
Volume = 1/3?r^2h = (1/3 * 22/7 * 21 * 21 * 28) cm^3 = 12936 cm^3
Curved surface area = ?rl = 22/7 * 21 *35 cm^3 = 2310 cm^2
Total Surface Area = (?rl + ?r^2) = (2310 + 22/7 * 21 * 21) cm^2 = 3696 cm^2
50. If the radius of the sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.
Sol: Let the original radius = R. Then, new radius = 150/100 R = 3R/2
Original Volume = 4/3?R^3, New volume = 4/3?(3 R/2)^3 = 9?R^3/2
Original surface area = 4?R^2 , New surface area = 4?(3R/2)^2 = 9?R^2
Increase % in surface area = (5?R^2/4?R^2 * 100)% = 125%
Ans.45 years.
Sol: P [1 + (R/100)]^15 = 2P è [1 + (R/100)]^15 = 2……….(i)
Let P [1 + (R/100)]^n = 8P è P [1 + (R/100)]^n = 8 = 2^3
= [{1 + (R/100)}^15]^3.
è [1 + (R/100)]^n = [1 + (R/100)]^45.
è n = 45.
Thus, the required time = 45 years.
42. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.
Ans: Sum = Rs. 5400,Rate=16 2/3 %.
Sol: S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.
Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.
Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.
è x * 7/6 * 7/6 = 7350.
è x = [7350 * 36/49] = 5400.
Therefore, Sum = Rs. 5400.
43. A, B and C start a business each investing Rs. 20000. After 5 months A withdrew Rs. 5000, B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.
Ans. A’s share = Rs. 20,500
B’s share = Rs. 21200
C’s share = Rs. 28200
Sol: Ratio of the capitals of A, B and C
= (20000*5+ 15000*7) : (20000*5+16000*7): (20000*5+26000*7)
=205000: 212000 : 282000 = 205:212:282
Therefore, A’s share = Rs. ( 69900*205/699) = Rs. 20,500
B’s share = Rs. (69900*212/699) = Rs. 21200,
C’s share = Rs. (69900*282/699) = Rs. 28200.
44. Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out an annual profit of Rs. 37100, find the share of each?
Sol: Ratio of their capitals= 36000*12:36000*9 = 4:3
Sanjiv’s share= Rs. ( 37100*4/7) = Rs. 21200.
Rajiv’s share = Rs. ( 37100*3/7) = Rs.15900.
45. If 20 men can build a wall 56m long in 6 days, what length of a similar wall can be built by 35 men in 3 days?
Ans. Length=49m.
Sol: Since the length is to be found out, we compare each item with the length as shown below.
More men, more length built (Direct).
Less days, less length built (Direct).
Men 20:35 :: 56: x
Similarly, days 6:3 :: 56: x.
Therefore, 20*6*x= 35*3*56 or x= 49.
Hence, the required length= 49m.
46.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.
Ans:26metric tonnes.
Sol: We shall compare each item with the quantity of coal.
Less engines, less coal consumed (direct)
More working hours, more coal consumed (direct)
If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
Less rate of consumption, less coal consumed (direct).
Therefore, number of engines 9:8 :: 24:x
Working hours 8:13 :: 24:x
Rate of consumption 1/3:1/4 :: 24:x.
9*8*1/3*x= 8*13*1/4*24 or x= 26.
Therefore, required consumption of coal 26 metric tonnes.
47. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?
Ans.81
Sol: Remaining work = 1-4/7 =3/7.
Remaining period = (46-33) days =13 days.
Less work, less men (direct)
Less days, more men (indirect).
More hours per day, less men (indirect)
Therefore, work 4/7:3/7 ::117/x
Days 13:33 :: 117/x
Hrs/day 9:8:: 117/x
Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.
Therefore, additional men to be employed =(198-117) =81.
48. A garrison of 3300 men had provisions for 32 days, when given at the rate of 850gms per head. At the end of 7 days, reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825gms per head. What is the strength of the reinforcement?
Ans: 1700
Sol: The problem becomes:
3300 men taking 850gms per head have provisions for (32-7) or 25 days. How many
men taking 825gms each have provisions for 17 days?
Less ration per head, more men (indirect).
Less days, more men (indirect)
Ration 825:850::3300:x
Days 17:25::3300:x
Therefore, 825*17*x= 850*25*3300 or x= 5000.
Therefore, strength of reinforcement = 5000-3300 = 1700.
49. Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.
Sol: Slant Height, l = v(r^2 + h^2) =v(21^2 + 28^2) = v1225 = 35 cm
Volume = 1/3?r^2h = (1/3 * 22/7 * 21 * 21 * 28) cm^3 = 12936 cm^3
Curved surface area = ?rl = 22/7 * 21 *35 cm^3 = 2310 cm^2
Total Surface Area = (?rl + ?r^2) = (2310 + 22/7 * 21 * 21) cm^2 = 3696 cm^2
50. If the radius of the sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.
Sol: Let the original radius = R. Then, new radius = 150/100 R = 3R/2
Original Volume = 4/3?R^3, New volume = 4/3?(3 R/2)^3 = 9?R^3/2
Original surface area = 4?R^2 , New surface area = 4?(3R/2)^2 = 9?R^2
Increase % in surface area = (5?R^2/4?R^2 * 100)% = 125%
Aptitude Questions Y5
31. What was the day of the week on 12th January, 1979?
Ans: Friday
Sol: Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.
78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days
12 days of January have 5 odd days.
Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.
Therefore, the desired day was Friday.
32. Find the day of the week on 16th july, 1776.
Ans: Tuesday
Sol: 16th july, 1776 means = 1775 years + period from 1st january to 16th july
Now, 1600 years have 0 odd days.
100 years have 5 odd days.
75 years = 18 leap years + 57 ordinary years
= (36 + 57) odd days = 93 odd days
= 13 weeks + 2 odd days = 2 odd days
Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.
Now, days from 1st Jan to 16th july; 1776
Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days
= (28 weeks + 2 days) odd days
Therefore, total number of odd days = 2
Therefore, the day of the week was Tuesday
33 .Find the angle between the minute hand and hour hand of a click when the time is
7.20?
Ans: 100deg
Sol: Angle traced by the hour hand in 12 hours = 360 degrees.
Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.
Angle traced by minute hand in 60 min = 360 deg.
Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.
Therefore, required angle = (220 - 120) = 100deg.
34.The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?
Ans: the clock gains 10 10/43 minutes
Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60
minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand.
55 minutes are gained in 60 min.
60 min. are gained in [(60/55) * 60] min == 65 5/11 min.
But they are together after 65 min.
Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.
Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.
Therefore, the clock gains 10 10/43 minutes in 24 hours.
35.A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?
Ans. 48 min. past 12.
Sol: Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145/6 hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
36. At what time between 2 and 3 o’ clock will the hands 0a a clock together?
Ans: 10 10/11 min. past 2.
Sol: At 2 o’ clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10
min space apart.
To be together, the minute hand must gain 10 minutes over the other hand.
Now, 55 minutes are gained by it in 60 min.
Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.
Therefore, the hands will coincide at 10 10/11 min. past 2.
37. A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.
Ans: Rs. 4460
Sol: Let the Sum be Rs. P. Then
P [1 + (R/100)]^3 = 6690………..(i)
P [1 + (R/100)]^6 = 10035………..(ii)
On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.
P * (3/2) = 6690 or P = 4460.
Hence, the sum is Rs. 4460.
38. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, if both are numerically equal.
Ans: Rate = 8% and Time = 8 years
Sol: Let sum = X. Then S.I. = 16x/25
Let rate = R% and Time = R years.
Therefore, x * R * R/100 = 16x/25 or R^2 = 1600/25, R = 40/5 = 8
Therefore, Rate = 8% and Time = 8 years.
39. Find
i. S.I. on RS 68000 at 16 2/3% per annum for 9 months.
ii. S.I. on RS 6250 at 14% per annum for 146 days.
iii. S.I. on RS 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995.
Ans: i. RS 8500.
ii. RS 350.
iii. RS 108.
Sol:
i. P = 68000, R = 50/3% p.a. and T = 9/12 year = ¾ years
Therefore, S.I. = (P * Q * R/100)
= RS (68000 * 50/3 * ¾ * 1/100) = RS 8500.
ii. P = RS 6265, R = 14% p.a. and T = (146/365) year = 2/5 years.
Therefore, S.I. = RS (6265 * 14 * 2/5 *1/100) = RS 350.
iii. Time = (24 + 31 + 18) days = 73 days = 1/5 year
P = RS 3000 and R = 18% p.a.
Therefore, S.I. = RS (3000 * 18 * 1/5 * 1/100) = RS 108
40. A sum at simple interest at 13 ½% per annum amounts to RS 2502.50 after 4 years. Find the sum.
Ans: sum = RS 1625
Sol: Let sum be x. Then,
S.I. = (x * 27/2 * 4 * 1/100) = 27x/50
Therefore, amount = (x + 27x/50) = 77x/50
Therefore, 77x/50 = 2502.50 or x = 2502.50 * 50 / 77 = 1625
Hence, sum = RS 1625
Ans: Friday
Sol: Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.
78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days
12 days of January have 5 odd days.
Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.
Therefore, the desired day was Friday.
32. Find the day of the week on 16th july, 1776.
Ans: Tuesday
Sol: 16th july, 1776 means = 1775 years + period from 1st january to 16th july
Now, 1600 years have 0 odd days.
100 years have 5 odd days.
75 years = 18 leap years + 57 ordinary years
= (36 + 57) odd days = 93 odd days
= 13 weeks + 2 odd days = 2 odd days
Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.
Now, days from 1st Jan to 16th july; 1776
Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days
= (28 weeks + 2 days) odd days
Therefore, total number of odd days = 2
Therefore, the day of the week was Tuesday
33 .Find the angle between the minute hand and hour hand of a click when the time is
7.20?
Ans: 100deg
Sol: Angle traced by the hour hand in 12 hours = 360 degrees.
Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.
Angle traced by minute hand in 60 min = 360 deg.
Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.
Therefore, required angle = (220 - 120) = 100deg.
34.The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?
Ans: the clock gains 10 10/43 minutes
Sol: In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60
minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand.
55 minutes are gained in 60 min.
60 min. are gained in [(60/55) * 60] min == 65 5/11 min.
But they are together after 65 min.
Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.
Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.
Therefore, the clock gains 10 10/43 minutes in 24 hours.
35.A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?
Ans. 48 min. past 12.
Sol: Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145/6 hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
36. At what time between 2 and 3 o’ clock will the hands 0a a clock together?
Ans: 10 10/11 min. past 2.
Sol: At 2 o’ clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10
min space apart.
To be together, the minute hand must gain 10 minutes over the other hand.
Now, 55 minutes are gained by it in 60 min.
Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.
Therefore, the hands will coincide at 10 10/11 min. past 2.
37. A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.
Ans: Rs. 4460
Sol: Let the Sum be Rs. P. Then
P [1 + (R/100)]^3 = 6690………..(i)
P [1 + (R/100)]^6 = 10035………..(ii)
On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.
P * (3/2) = 6690 or P = 4460.
Hence, the sum is Rs. 4460.
38. Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, if both are numerically equal.
Ans: Rate = 8% and Time = 8 years
Sol: Let sum = X. Then S.I. = 16x/25
Let rate = R% and Time = R years.
Therefore, x * R * R/100 = 16x/25 or R^2 = 1600/25, R = 40/5 = 8
Therefore, Rate = 8% and Time = 8 years.
39. Find
i. S.I. on RS 68000 at 16 2/3% per annum for 9 months.
ii. S.I. on RS 6250 at 14% per annum for 146 days.
iii. S.I. on RS 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995.
Ans: i. RS 8500.
ii. RS 350.
iii. RS 108.
Sol:
i. P = 68000, R = 50/3% p.a. and T = 9/12 year = ¾ years
Therefore, S.I. = (P * Q * R/100)
= RS (68000 * 50/3 * ¾ * 1/100) = RS 8500.
ii. P = RS 6265, R = 14% p.a. and T = (146/365) year = 2/5 years.
Therefore, S.I. = RS (6265 * 14 * 2/5 *1/100) = RS 350.
iii. Time = (24 + 31 + 18) days = 73 days = 1/5 year
P = RS 3000 and R = 18% p.a.
Therefore, S.I. = RS (3000 * 18 * 1/5 * 1/100) = RS 108
40. A sum at simple interest at 13 ½% per annum amounts to RS 2502.50 after 4 years. Find the sum.
Ans: sum = RS 1625
Sol: Let sum be x. Then,
S.I. = (x * 27/2 * 4 * 1/100) = 27x/50
Therefore, amount = (x + 27x/50) = 77x/50
Therefore, 77x/50 = 2502.50 or x = 2502.50 * 50 / 77 = 1625
Hence, sum = RS 1625
Aptitude Questions Y4
21. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.
Ans:6km.
Sol: Let the required distance be x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.
Hence ,the required distance is 6 km
22. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?
Ans:50 min
Sol: New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) – usual time = 10min
Therefore Usual time = 50min
23.A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.
Ans. length of the train=160m
length of the platform=140 m.
Sol: Let the length of the train be x meters and length of the platform be y meters.
Speed of the train relative to man=(54-6) kmph =48 kmph.
=(48*5/18) m/sec =40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Therefore, length of the train=(Relative speed *Time)
=(40/3 * 12) m =160 m.
Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.
Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.
Therefore, Length of the platform=140 m.
24. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.
Ans: length of train=120m
Speed of train=54kmph
Sol: Let the length of the train be x meters
Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.
Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120
Therefore Length of the train = 120m
Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph
25. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?
Ans. Rs. 50.
Sol: Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.
26. A grosser purchased 80 kg of rice at Rs.13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?
Ans: Rs.17.40 per kg.
Sol: C.P of 200 kg of mix. = Rs (80*13.50+120*16) = Rs.3000.
S.P = 116% of Rs 3000= Rs (116*3000/100) = Rs.3480.
Rate of S.P of the mixture = Rs.3480/200.per kg. = Rs.17.40 per kg.
27. Two persons A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?
Ans:24hours.
Sol: (A+B)’s one hour’s work =1/8, A’s one hour’s work =1/12
Therefore, B’s one hour’s work = (1/8-1/12) =1/24.
Hence, B alone can dig the trench in 24 hours.
28. A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together?
Also, find the number of days taken by each to finish it working alone?
Ans:60 days
Sol: (A+B)’s one day’s work=1/12; (B+C)’s one day’s work=1/15 and (A+C)’s one day’s
work=1/20.
Adding, we get: 2(A+B+C)’s one day’s work = (1/12+1/15+1/20)=1/5.
Therefore, (A+B+C)’s one day’s work=1/10.
Thus, A, B and C together can finish the work in 10 days.
Now, A’s one day’s work
= [(A+B+C)’s one day’s work] – [(B+C)’s one day’s work]
= 1/10-1/15)
= 1/30.
Therefore, A alone can finish the work in 30 days.
Similarly, B’s 1 day’s work = (1/10 -1/20) = 1/20.
Therefore, B alone can finish the work in 20 days.
And, C’s 1 day’s work= (1/10-1/12) = 1/60.
Therefore, C alone can finish the work in 60 days.
29. A is twice as good a workman as B and together they finish a piece of work in 18
days.In how many days will A alone finish the work?
Ans:27 days.
Sol: (A’s 1 day’s work): (B’s 1 day’s work) = 2:1.
(A + B)’s 1 day’s work = 1/18.
Divide 1/18 in the ratio 2:1.
Therefore A’s 1 day’s work = (1/18 * 2/3) = 1/27.
Hence, A alone can finish the work in 27 days.
30. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?
Ans: 12 ½ days.
Sol: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work =y.
Then, 2x+3y=1/10 and 3x+2y=1/8.
Solving, we get: x=7/200 and y=1/100.
Therefore (2 men +1 boy)’s 1 day’s work = (2*7/200 + 1*1/100) = 16/200 = 2/25.
So, 2 men and 1 boy together can finish the work in 25/2 =12 ½ days.
Ans:6km.
Sol: Let the required distance be x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.
Hence ,the required distance is 6 km
22. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?
Ans:50 min
Sol: New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) – usual time = 10min
Therefore Usual time = 50min
23.A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.
Ans. length of the train=160m
length of the platform=140 m.
Sol: Let the length of the train be x meters and length of the platform be y meters.
Speed of the train relative to man=(54-6) kmph =48 kmph.
=(48*5/18) m/sec =40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Therefore, length of the train=(Relative speed *Time)
=(40/3 * 12) m =160 m.
Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.
Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.
Therefore, Length of the platform=140 m.
24. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.
Ans: length of train=120m
Speed of train=54kmph
Sol: Let the length of the train be x meters
Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.
Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120
Therefore Length of the train = 120m
Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph
25. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?
Ans. Rs. 50.
Sol: Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.
26. A grosser purchased 80 kg of rice at Rs.13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?
Ans: Rs.17.40 per kg.
Sol: C.P of 200 kg of mix. = Rs (80*13.50+120*16) = Rs.3000.
S.P = 116% of Rs 3000= Rs (116*3000/100) = Rs.3480.
Rate of S.P of the mixture = Rs.3480/200.per kg. = Rs.17.40 per kg.
27. Two persons A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?
Ans:24hours.
Sol: (A+B)’s one hour’s work =1/8, A’s one hour’s work =1/12
Therefore, B’s one hour’s work = (1/8-1/12) =1/24.
Hence, B alone can dig the trench in 24 hours.
28. A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together?
Also, find the number of days taken by each to finish it working alone?
Ans:60 days
Sol: (A+B)’s one day’s work=1/12; (B+C)’s one day’s work=1/15 and (A+C)’s one day’s
work=1/20.
Adding, we get: 2(A+B+C)’s one day’s work = (1/12+1/15+1/20)=1/5.
Therefore, (A+B+C)’s one day’s work=1/10.
Thus, A, B and C together can finish the work in 10 days.
Now, A’s one day’s work
= [(A+B+C)’s one day’s work] – [(B+C)’s one day’s work]
= 1/10-1/15)
= 1/30.
Therefore, A alone can finish the work in 30 days.
Similarly, B’s 1 day’s work = (1/10 -1/20) = 1/20.
Therefore, B alone can finish the work in 20 days.
And, C’s 1 day’s work= (1/10-1/12) = 1/60.
Therefore, C alone can finish the work in 60 days.
29. A is twice as good a workman as B and together they finish a piece of work in 18
days.In how many days will A alone finish the work?
Ans:27 days.
Sol: (A’s 1 day’s work): (B’s 1 day’s work) = 2:1.
(A + B)’s 1 day’s work = 1/18.
Divide 1/18 in the ratio 2:1.
Therefore A’s 1 day’s work = (1/18 * 2/3) = 1/27.
Hence, A alone can finish the work in 27 days.
30. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?
Ans: 12 ½ days.
Sol: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work =y.
Then, 2x+3y=1/10 and 3x+2y=1/8.
Solving, we get: x=7/200 and y=1/100.
Therefore (2 men +1 boy)’s 1 day’s work = (2*7/200 + 1*1/100) = 16/200 = 2/25.
So, 2 men and 1 boy together can finish the work in 25/2 =12 ½ days.
Aptitiude Questions Y2
11. Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by tain was???
Ans.800km
Sol: Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.
12. In a college ,1/5 th of the girls and 1/8 th of the boys took part in a social camp.What of the total number of students in the college took part in the camp?
Ans: 2/13
Sol: Out of 5 girls 1 took part in the camp
out of 8 boys 1 took part in the camp
so, out of 13 students 2 took part in the camp.
So, 2/13of the total strength took part in the camp.
13. On sports day,if 30 children were made to stand in a column,16 columns could be formed. If 24 children were made to stand in a column , how many columns could be formed?
Ans. 20
Sol: Total number of children=30*16=480
Number of columns of 24 children each =480/24=20.
14. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr.In how much time will they cross each other if they are running in the same direction.
Ans: 252sec
Sol: Relative speed=45-40=5km/hr=25/18 mt/sec
Total distance covered =sum of lengths of trains =350mts.
So, time taken =350*18/25=252sec.
15. 5/9 part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is:
Ans: (250/9)%
Sol: Let the population =x Males=(5/9)x
Married males = 30% of (5/9)x = x/6
Married females = x/6
Total females = (x-(5/9)x)=4x/9
Unmarried females = (4x/9 – x/6) = 5x/18
Required percentage = (5x/18 * 1/x * 100) = (250/9)%
16. From height of 8 mts a ball fell down and each time it bounces half the distnace back. What will be the distance travelled
Ans.: 24
Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24
17. First day of 1999 is sunday what day is the last day
Ans.: Monday
18. Increase area of a square by 69% by what percent should the side be incresed
Ans.: 13
Sol:Area of square=x2
Then area of increase=100+69=169
square root of 169 i.e 13 .
19. Ten years ago, chandrawathi’s mother was four times older than her daughter. After 10years, the mother will be twice older than daughter. The present age of Chandrawathi is:
Ans.20 years
Sol: Let Chandrawathi’s age 10 years ago be x years.
Her mother’s age 10 years ago = 4x
(4x+10)+10=2(x+10+10)
x=10
Present age of Chandrawathi = (x+10) = 20years
20. Finding the wrong term in the given series
7, 28, 63, 124, 215, 342, 511
Ans:28
Sol: Clearly, the correct sequence is
2^3 – 1, 3^3 – 1, 4^3 – 1, 5^3 – 1, ……….
Therefore, 28 is wrong and should be replaced by (3^3 – 1) i.e, 26.
Ans.800km
Sol: Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.
12. In a college ,1/5 th of the girls and 1/8 th of the boys took part in a social camp.What of the total number of students in the college took part in the camp?
Ans: 2/13
Sol: Out of 5 girls 1 took part in the camp
out of 8 boys 1 took part in the camp
so, out of 13 students 2 took part in the camp.
So, 2/13of the total strength took part in the camp.
13. On sports day,if 30 children were made to stand in a column,16 columns could be formed. If 24 children were made to stand in a column , how many columns could be formed?
Ans. 20
Sol: Total number of children=30*16=480
Number of columns of 24 children each =480/24=20.
14. Two trains 200mts and 150mts are running on the parallel rails at this rate of 40km/hr and 45km/hr.In how much time will they cross each other if they are running in the same direction.
Ans: 252sec
Sol: Relative speed=45-40=5km/hr=25/18 mt/sec
Total distance covered =sum of lengths of trains =350mts.
So, time taken =350*18/25=252sec.
15. 5/9 part of the population in a village are males. If 30% of the males are married, the percentage of unmarried females in the total population is:
Ans: (250/9)%
Sol: Let the population =x Males=(5/9)x
Married males = 30% of (5/9)x = x/6
Married females = x/6
Total females = (x-(5/9)x)=4x/9
Unmarried females = (4x/9 – x/6) = 5x/18
Required percentage = (5x/18 * 1/x * 100) = (250/9)%
16. From height of 8 mts a ball fell down and each time it bounces half the distnace back. What will be the distance travelled
Ans.: 24
Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24
17. First day of 1999 is sunday what day is the last day
Ans.: Monday
18. Increase area of a square by 69% by what percent should the side be incresed
Ans.: 13
Sol:Area of square=x2
Then area of increase=100+69=169
square root of 169 i.e 13 .
19. Ten years ago, chandrawathi’s mother was four times older than her daughter. After 10years, the mother will be twice older than daughter. The present age of Chandrawathi is:
Ans.20 years
Sol: Let Chandrawathi’s age 10 years ago be x years.
Her mother’s age 10 years ago = 4x
(4x+10)+10=2(x+10+10)
x=10
Present age of Chandrawathi = (x+10) = 20years
20. Finding the wrong term in the given series
7, 28, 63, 124, 215, 342, 511
Ans:28
Sol: Clearly, the correct sequence is
2^3 – 1, 3^3 – 1, 4^3 – 1, 5^3 – 1, ……….
Therefore, 28 is wrong and should be replaced by (3^3 – 1) i.e, 26.
Aptitude Questions Y1
1.A three digit number consists of 9,5 and one more number . When these digits are reversed and then subtracted from the original number the answer yielded will be consisting of the same digits arranged yet in a different order. What is the other digit?
Sol. Let the digit unknown be n.
The given number is then 900+50+n=950+n.
When reversed the new number is 100n+50+9=59+100n.
Subtracting these two numbers we get 891-99n.
The digit can be arranged in 3 ways or 6 ways.
We have already investigated 2 of these ways.
We can now try one of the remaining 4 ways. One of these is n 95
100n+90+5=891-99n
or 199n =796
so, n=4
the unknown digit is 4.
2.A farmer built a fence around his 17 cows,in a square shaped region.He used 27 fence poles on each side of the square. How many poles did he need altogether???
Ans.104 poles
Sol. Here 25 poles Must be there on each side .And around four corners 4 poles will be
present. 4*25+4=100+4=104 poles.
3.On the first test of the semester, kiran scored a 60. On the last test of the semester, kiran scored
75%
By what percent did kiran's score improve?
Ans: 25%
Sol. In first test kiran got 60
In last test he got 75.
% increase in test ( 60(x+100))/100=75
0.6X+60=75
0.6X=15
X=15/0.6=25%
4.A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group. Find the probability for the selected person to be an employee.
Ans:29/40
Sol: Assume men=100,women=100 then employed men & women r (100-10)+(100-45)=145
So probability for the selected person to be an employee=145/200=29/40
5. Randy's chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in
1999, by what percent did the number of cars sold decrease?
Ans: 4%
Sol. Let percentage of decrease is x , then
16400(100-x)/100=15744
16400-15744=164x
x=656/164=4%
6. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?
A) 260%
B) 150%
C) 100%
D) 50%
E) None of these
Ans: 260%
Sol. Let x be original cost of the radio.
The solding price = (100+20)x=120x
If , it is sold for thrice the price ,then 3*120x=360x
So, gain percent is (360-100)=260%.
7. Find the perimeter of the shape below.
1cm
Ans: 24cm
Sol: 2+4+5+1+7+5=24
8.If the Arithmetic mean is 34 and geometric mean is 16 then what is greates number in that series of numbers?
Ans. 64
Sol. Let two numbers be x, y;
Arthmetic mean=34=>( x+y)/2=34
x+y=68
geometric mean=16=>(xy)pow 1/2=16
xy=16*16=256
By trail and error 16*16=64*4
And 64+4/2=34
So the greatest number int hat series is 64.
9. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?
Ans. 250
Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.
10. The boys and girls in a college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is:??
Ans.78%
Sol: Suppose boys = 3x and girls = 2x
Not adults = (80*3x/100) + (75*2x/100) = 39x/10
Required percentage = (39x/10)*(1/5x)*100 = 78%
Sol. Let the digit unknown be n.
The given number is then 900+50+n=950+n.
When reversed the new number is 100n+50+9=59+100n.
Subtracting these two numbers we get 891-99n.
The digit can be arranged in 3 ways or 6 ways.
We have already investigated 2 of these ways.
We can now try one of the remaining 4 ways. One of these is n 95
100n+90+5=891-99n
or 199n =796
so, n=4
the unknown digit is 4.
2.A farmer built a fence around his 17 cows,in a square shaped region.He used 27 fence poles on each side of the square. How many poles did he need altogether???
Ans.104 poles
Sol. Here 25 poles Must be there on each side .And around four corners 4 poles will be
present. 4*25+4=100+4=104 poles.
3.On the first test of the semester, kiran scored a 60. On the last test of the semester, kiran scored
75%
By what percent did kiran's score improve?
Ans: 25%
Sol. In first test kiran got 60
In last test he got 75.
% increase in test ( 60(x+100))/100=75
0.6X+60=75
0.6X=15
X=15/0.6=25%
4.A group consists of equal number of men and women. Of them 10% of men and 45% of women are unemployed. If a person is randomly selected from the group. Find the probability for the selected person to be an employee.
Ans:29/40
Sol: Assume men=100,women=100 then employed men & women r (100-10)+(100-45)=145
So probability for the selected person to be an employee=145/200=29/40
5. Randy's chain of used car dealership sold 16,400 cars in 1998. If the chain sold 15,744 cars in
1999, by what percent did the number of cars sold decrease?
Ans: 4%
Sol. Let percentage of decrease is x , then
16400(100-x)/100=15744
16400-15744=164x
x=656/164=4%
6. A radio when sold at a certain price gives a gain of 20%. What will be the gain percent, if sold for thrice the price?
A) 260%
B) 150%
C) 100%
D) 50%
E) None of these
Ans: 260%
Sol. Let x be original cost of the radio.
The solding price = (100+20)x=120x
If , it is sold for thrice the price ,then 3*120x=360x
So, gain percent is (360-100)=260%.
7. Find the perimeter of the shape below.
1cm
Ans: 24cm
Sol: 2+4+5+1+7+5=24
8.If the Arithmetic mean is 34 and geometric mean is 16 then what is greates number in that series of numbers?
Ans. 64
Sol. Let two numbers be x, y;
Arthmetic mean=34=>( x+y)/2=34
x+y=68
geometric mean=16=>(xy)pow 1/2=16
xy=16*16=256
By trail and error 16*16=64*4
And 64+4/2=34
So the greatest number int hat series is 64.
9. The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?
Ans. 250
Sol. Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.
10. The boys and girls in a college are in the ratio 3:2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is:??
Ans.78%
Sol: Suppose boys = 3x and girls = 2x
Not adults = (80*3x/100) + (75*2x/100) = 39x/10
Required percentage = (39x/10)*(1/5x)*100 = 78%
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