divelab/Sys2Bench huggingface dataset:
Question: Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
Options:
A)5/7
B)5/4
C)5/1
D)5/6
E)5/3
Rationale: P(at least one graduate) = 1 - P(no graduates)
= 1 - ⁶C₃/¹⁰C₃
= 1 - (6 * 5 * 4)/(10 * 9 * 8)
= 5/6
Answer:D
Answer: D
Question: Photography changed drastically _________ introduced in the market.
Options:
A)when polariod
B)when polariod
C)when polariod
D)as polaroid
E)1312
Rationale: Explanation:
There is no adequation information is given about details of subsequent changes. So "when" is the right word. Option a. "As" inditates reason.
Answer:D
Answer: D
Question: If |e|=−e, which of the following must be true?
Options:
A)x≥0
B)e≤0
C)x2>x
D)x3<0
E)2x<x
Rationale: |e| =-e means absolute value of e is equal to negative of e. Since absolute value cannot be negative hence negative of e should result in a non negative number. It means e is a non positive number i.e. e<0. Answer is B
Answer: B
Question: What was the day of the week on, 19th July, 1776?
Options:
A)Friday
B)Wednesday
C)Monday
D)Saturday
E)SUNDAY
Rationale: 16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day.
100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days.
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan Feb Mar Apr May Jun Jul
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 5 days)
Total number of odd days = (0 + 5) = 5.
Required day was 'Friday'.
Answer:A
Answer: A
Question: If A lends Rs.3500 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is?
Options:
A)157.56
B)157.51
C)157.5
D)157.59
E)157.52
Rationale: (3500*1.5*3)/100 => 157.50.Answer: OC
Answer: C
Question: 1,000^2 + 1,001^2 +1,002^2 + 1,003^2 =
Options:
A)6,030,053
B)6,030,054
C)6,030,055
D)6,030,056
E)6,012,012
Rationale: Interesting problem.
I think key is to notice that all the given answer choices differs in last two digits. Therefore, our entire focus should be to figure out how the given terms contribute to last two digits of total.
1000^2 -> 00
1001^1 -> 01
.
.
1005^3 -> 14
Total -> *55
Answer E.
Answer: E
Question: If it takes a tub 2 minutes to drain 5/7 of its content, how much more time will it take for the tub to be empty?
Options:
A)48 seconds
B)1 minute, 12 seconds
C)1 minute, 50 seconds
D)2 minutes, 14 seconds
E)4 minutes, 12 seconds
Rationale: If 5/7 of tub's content is drained 2/7 th of tub still needs to be drained.
If it takes 2 minutes to drain 5/7 th of tub it takes 2*(7/5) minutes to drain the entire tub and 2*(7/5)*(2/7) min to drain 2/7 th of the tub which is 4/5 minutes or 48 minutes
So answer is A
Answer: A
Question: the sum of all two digit numbers divisible by 5 is
Options:
A)945
B)3487
C)28
D)279
E)1287
Rationale: Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.
Answer: A) 945
Answer: A
Question: A building contractor employs 20 male, 15 female and 5 child workers. To a male worker he pays Rs.25 per day, to a female worker Rs.20 per day and a child worker Rs.8 per day. The average wage per day paid by the contractor is?
Options:
A)11
B)21
C)88
D)77
E)61
Rationale: 20 15 5
25 20 8
500 + 300 + 40 = 840/40 = 21
Answer: B
Answer: B
Question: Two cars cover the same distance at the speed of 50 and 66 kmps respectively. Find the distance traveled by them if the slower car takes 1 hour more than the faster car.
Options:
A)212 km
B)214 km
C)188 km
D)216 km
E)204 km
Rationale: 50(x + 1) = 66x
X = 3.125
60 * 3.125 = 188 km
ANSWER:C
Answer: C
Question: Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?
Options:
A)2388
B)2088
C)3600
D)2028
E)1911
Rationale: Let the beds be numbered 1 to 7.Case 1 : Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5´5!ways = 600 ways.
Case 2 : Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6.
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways.
The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways
Answer: C
Answer: C
Question: Weights of two friends Ram and Shyam are in the ratio 1:5. If Ram's weight is increased by 10% and total weight of Ram and Shyam become 82.8 kg, with an increases of 15%. By what percent did the weight of Shyam has to be increased?
Options:
A)19%
B)10%
C)21%
D)16%
E)None
Rationale: Solution:
given ratio of Ram and Shayam's weight = 1:5
Hence, (x-15)/(15-10) = 1/5
Or, x = 16%.
Answer: Option D
Answer: D
Question: The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.80 per metre.
Options:
A)2399
B)6336
C)2999
D)5808
E)2888
Rationale: Explanation:
Area = (13.86 x 10000) sq.m = 138600 sq.m
Circumference =
Cost of fencing = Rs. (1320 x 4.80) = Rs. 6336.
Answer: B) 6336
Answer: B
Question: How long will a boy take to run round a square field of side 50 meters, if he runs at the rate of 10km/hr?
Options:
A)52sec
B)45sec
C)72sec
D)25sec
E)39sec
Rationale: Speed = 10 km/hr = 10*5/18 = 25/9 m/sec
Distance = 50*4 = 200m
Time taken = 200*9/25 = 72sec
Answer is C
Answer: C
Question: A firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009?
Options:
A)700
B)1000
C)1300
D)1600
E)2000
Rationale: Profit=Revenue-Cost
2007:
-1000=R1-C1
2008:
0=(1+2X%)R1-(1+X%)C1
2009:
P3=(1+2 X%)^2*R1-(1+X%)^2*C1
we've already known that R3=1.44R1, so (1+2 X%)^2=1.44, X=10
we put X=10 back to the equation of 2008:
0=1.2R1-1.1C1, and we have 1.2R1=1.1C1, which we could combine with the equation of 2007.
Then we have C1=12000, R1-11000
Finally, take C1 and R1 into the equation of 2009 and we got P3=15840-14520=1320 ;Answer: C
Answer: C
Question: The length of the bridge, which a train 140 meters long and travelling at 45 km/hr can cross in 30 seconds, is?
Options:
A)235
B)288
C)245
D)776
E)232
Rationale: Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (140 + X)/30 = 25/2 ==> 2(140 + X) = 750 ==> X
= 235 m.
Answer:A
Answer: A
Question: A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose?
Options:
A)120
B)335
C)784
D)217
E)210
Rationale: Let x be the number he chose, then
2*x*138=102
2x=240
x=120
correct answer A
Answer: A
Question: The average of five numbers is 27. If one number is excluded, the average becomes 21. The excluded number is?
Options:
A)48
B)49
C)50
D)51
E)52
Rationale: Excluded number = (27 * 5) - (21 * 4)
= 135 - 84
= 51.
Answer: D
Answer: D
Question: Julie decided to save a certain amount of her monthly salary each month and her salary was unchanged from month to month. If Julie 's savings by the end of the year from these monthly savings were four times the amount she spent per month, what should be the fraction of her salary that she spent each month?
Options:
A)5/3
B)3/9
C)3/2
D)3/5
E)3/4
Rationale: Let Julie 's monthly savings = S
Julie 's monthly pay = P
Julie 's monthly expenditure = P-S
Julie 's savings by the end of the year from these monthly savings were Four times the amount she spent per month
12S = 4*(P-S)
=>3S = P-S
=> P = 4S
Julie's monthly expenditure = P-S = 4S-S = 3S
fraction of her salary that Julie spent each month = 3S / 4S=P-S/P = 3/4. Answer is E
Answer: E
Question: A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 539, what was his profit?
Options:
A)s. 64
B)s. 99
C)s.72
D)s.75
E)s.90
Rationale: C.P. = Rs. (100/122.5)x539
= Rs. (1000/1225)x539
= Rs. 440
Profit = Rs. (539 - 440) = Rs. 99.
Answer:B
Answer: B
Question: A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 24 cm. What is the length of the wire?
Options:
A)6 cm
B)4 cm
C)8 cm
D)3 cm
E)9 cm
Rationale: Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(24)^2 * h = (4/3)π (12)^3 => h
= 4 cm
Answer:B
Answer: B
Question: If you graph the lines 3x + 2y = –2 and
5x – y = 14 on a standard xy-graph, at
which point will the lines intersect?
Options:
A)(1, –7)
B)(2, –4)
C)(–5, 4)
D)(–9, 3)
E)(11, 2)
Rationale: Solve the two equations, 3x + 2y = –2 and 5x – y = 14, as a system of equations. To do
this, multiply the second equation by 2 and then add it to the first equation:
3x+2y=-2+10x-2y=28=>13x=26
Now solve for x:
x = 2
Replace x with 2 in either equation and solve for y:
3(2) + 2y = –2
6 + 2y = –2
2y = –8
y = –4
At the point of intersection, x = 2 and y = –4, so this point is (2, –4).
correct answer B) (2, –4).
Answer: B
Question: A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is :-
Options:
A)a gain of Rs 27
B)a loss of Rs 37
C)neither gain nor a loss
D)a gain or a loss depending upon the order in which the wins and losses occur
E)None of these
Rationale: Explanation :
As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply initial amount by 1.5 thrice and by 0.5 thrice (in any order).
The overall resultant will remain same.
So final amount with the person will be (in all cases):
=64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs 27
Hence the final result is:
64−27=37.
Answer : B
Answer: B
Question: T divided by S equals 405.24 what is a possible remainder?
Options:
A)54
B)2
C)9
D)40
E)4
Rationale: With the decimal being 24/100, the simplest form is 6/25. If the divisor is a multiple of 25, the remainder must be a multiple of 6. The only number that is so is 54. Answer A.
Answer: A
Question: A rectangular lawn of dimensions 110 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m?
Options:
A)s.3988
B)s.4800
C)s.3228
D)s.3922
E)s.3928
Rationale: Area = (l + b – d) d
(110 + 60 – 10)10 => 1600 m2
1600 * 3
= Rs.4800
Answer:B
Answer: B
Question: You hold some gold in a vault as an investment. Over the past year the price of gold increases by 42%. In order to keep your gold in the vault, you must pay 4% of the total value of the gold per year. What percentage has the value of your holdings changed by over the past year.
Options:
A)38%
B)46%
C)47.68%
D)44.32%
E)36.32%
Rationale: (100%+42%)*(100%-4%) = 1.42*0.96 = 136.32% an increase of 36.32%
Your gold holdings have increased in value by 36.32%.
The Answer is E
Answer: E
Question: Of the 85 house in a development, 50 have a two-car garage, 40 have an in-the-ground swimming pool, and 35 have both a two-car garage and an in-the-ground swimming pool. How many houses in the development have neither a two-car garage nor an in-the-ground swimming pool?
Options:
A)10
B)15
C)20
D)25
E)30
Rationale: Neither Car nor Garage
= Total - Garage - (Swim - Common)
= 85 - 50 - (40-35) = 85-55 = 30
Answer E
Answer: E
Question: After his first semester in college, Thomas is applying for a scholarship that has a minimum Grade Point Average (GPA) requirement of 3.5. The point values of pertinent college grades are given in the table below. If Thomas took 7 courses, each with an equal weight for GPA calculations, and received two grades of A-, one grade of B+, and one grade of B, what is the lowest grade that Thomas could receive for his fifth class to qualify for the scholarship?
Point Values of Select Grades
Grade: A | A- | B+ | B | B- | C+ | C | C-
Value: 4 | 3.7 | 3.3 | 3 | 2.7 | 2.3 | 2 | 1.7
Options:
A) A
B) B+
C) B
D) B-
E) C+
Rationale: Grade: A | A- | B+ | B | B- | C+ | C | C-
Value: 4 |3.7|3.3|3| 2.7 | 2.3 | 2 | 1.7
7.4 + 3.3 + 3 + x = 3.5 * 5
x = 17.5 - 13.7 = 3.8
Grade A is required to qualify
Answer = B-
D
Answer: D
Question: The average runs scored by a batsman in 35 matches is 36. In the next 14 matches the batsman scored an average of 15 runs. Find his average in all the 30 matches?
Options:
A)31
B)35
C)29
D)25
E)30
Rationale: Total score of the batsman in 35 matches = 1260.Total score of the batsman in the next 14 matches = 210.Total score of the batsman in the 30 matches = 1470.Average score of the batsman = 1470/49 = 30.Answer: E
Answer: E
Question: A man buys a cycle for Rs. 1600 and sells it at a loss of 10%. What is the selling price of the cycle?
Options:
A)1410
B)1420
C)1430
D)1440
E)1540
Rationale: S.P. = 90% of Rs. 1600 = 90/100 x1600= Rs. 1440
answer : D
Answer: D
Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Options:
A)299
B)277
C)240
D)88
E)112
Rationale: Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Answer:C
Answer: C
Question: A boy has fourteen trousers and 13 shirts. In how many different ways can he select a trouser and a shirt?
Options:
A)182 ways
B)287 ways
C)266 ways
D)108 ways
E)121 ways
Rationale: Explanation:
The boy can select one trouser in 14 ways.
The boy can select one shirt in 13 ways.
The number of ways in which he can select one trouser and one shirt is 14 * 13 = 182 ways.
Answer: A
Answer: A
Question: Excluding stoppages, the speed of a train is 45 kmph and including stoppages it is 34 kmph. Of how many minutes does the train stop per hour?
Options:
A)E982
B)27
C)12
D)14.6
E)28
Rationale: Explanation:
T = 11/45 * 60 = 14.6
Answer: Option D
Answer: D
Question: In a certain pet shop, the ratio of dogs to cats to bunnies in stock is 3 : 7 : 13. If the shop carries 352 dogs and bunnies total in stock, how many dogs are there?
Options:
A)42
B)66
C)98
D)112
E)154
Rationale: Let us assume the number of dogs, cats and bunnies to be 3x, 7x and 13x
Total dogs and bunnies = 16x.
And we are given that 16x = 352.
Hence x = 22.
Dogs = 3x = 3*22 = 66 (Option B)
Answer: B
Question: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $0.30 per glass on the first day, what was the price per glass on the second day?
Options:
A)$0.15
B)$0.20
C)$0.30
D)$0.40
E)$0.45
Rationale: On the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade;
On the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade;
So, the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3. Naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3.
We are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day.
Say the price of the glass of the orangeade on the second day was $x then 2*0.3=3*x --> x=$0.2.
Answer: B.
Answer: B
Question: What is x if x + 3y = 33 and y = 10?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Substitute y by 10 in x + 3y = 33
x + 3(10) = 33
x + 30 = 33
If we substitute x by 3 in x + 30 = 33, we have 3 + 30 = 33. Hence
x = 3 correct answer C
Answer: C
Question: A cylindrical water tower with radius 10 m and height 16 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π m3. From noon, how many hours will drain the entire tank?
Options:
A)170 2/7
B)170 6/7
C)171
D)171 3/7
E)172
Rationale: initial volume = (3/4)×∏×10²×16 = 1200∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr
Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“
So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain
What i did initially was formed an equation 1200∏ = 7∏×n (n is the number of hrs) so ended up with 171.428. This wrong
Look at this way
after 171 hrs the amount of water drain will be 171×7∏ = 1197∏ m³
Left over water in the tank after 171 hrs = 3∏ m³
From above we know that it take 1 more hour to drain that 3∏ m³.
So ans is 172hrs
E
Answer: E
Question: A train running at the speed of 60 km/hr crosses a pole in 6 sec. What is the length of the train?
Options:
A)118
B)150
C)277
D)258
E)100
Rationale: Speed = 60 * 5/18 = 50/3 m/sec
Length of the train = speed * time = 50/3 * 6 = 100 m
Answer: E
Answer: E
Question: If f(x) = 2x^2 + y, and f(2) = 30, what is the value of f(10)?
Options:
A)104
B)60
C)52
D)50
E)222
Rationale: f(x) = 2x^2 + y
f(2) = 30
=> 2*(2)^2 + y = 30
=> 8 + y = 30
=> y = 22
f(10)= 2*(10)^2 + 22
= 222
Answer E
Answer: E
Question: A, B and C work on a task. To complete the task alone, B takes twice the time that A would take to complete the task alone and 1/2rd the time that C would take to complete the task alone. If B actually worked for half the number of days that A worked and 3/2 times the number of days that C worked, what proportion of the total work was completed by B?
Options:
A)1/3
B)2/9
C)9/49
D)7/19
E)4/21
Rationale: The first thing to notice is that A is faster than B and B is faster than C.
Since work is proportional to time, in 1 day lets say if A does 2 works, B does 1 work and C does 2/3rd of a work.
If A works for 2 days, B works for 1 day and C works for only 2/3 of the day.
Therefore total work done = (2*2) + (1*1) + (1/2*1/2) = 21/4
Proportion of work done by B = (1*1) / (21/4) = 4/21
Hence Answer E.
Answer: E
Question: If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 10, respectively, what is the probability that b/a will not be an integer?
Options:
A)1/2
B)1/3
C)1/4
D)1/6
E)1/8
Rationale: a is from the following set {1,2,3,4}
b is from the following set {10,12,14}
The total number of ways of choosing 2 integers, one from each set is 4*3 = 12 ways.
The possibilities where b/a is not an integer is for the following outcomes:
{b,a} = {10,3},{10,4},{14,3},{14,4} = 4 ways
Thus the probability is 4/12 = 1/3
The answer is B.
Answer: B
Question: Company P had 15 percent more employees in December than it had in January. If Company P had 480 employees in December, how many employees did it have in January?
Options:
A) 391
B) 417
C) 410
D) 423
E) 445
Rationale: D = number of employees in December
J = number of employees in January
J x 1.15 = D
J x 1.15 = 480
J = 480/1.15
J = 48,000/115 = 417
Thus B is the correct answer.
Answer: B
Question: A jogger running at 9 km/hr along side a railway track is 360 m ahead of the engine of a 90 m long train running at 63 km/hr in the same direction. In how much time will the train pass the jogger?
Options:
A)76 sec
B)67 sec
C)30 sec
D)36 sec
E)23 sec
Rationale: Speed of train relative to jogger = 63 - 9 = 54 km/hr.
= 54 * 5/18 = 15 m/sec.
Distance to be covered = 360 + 90 = 450 m.
Time taken = 450/15 =30 sec.
Answer:C
Answer: C
Question: If a: b = 19: 3, b: c = 17: 13, find a: b: c?
Options:
A)323: 51: 39
B)333: 61: 49
C)343: 71: 59
D)353: 81: 69
E)363: 91: 79
Rationale: a: b = 19: 3
b: c = 17: 13
a: b: c = 323: 51: 39
ANSWER:A
Answer: A
Question: The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 90 metres?
Options:
A)16 seconds
B)76 seconds
C)26 seconds
D)12 seconds
E)18 seconds
Rationale: Speed of the boat downstream = 15 + 3
= 18 kmph
= 18 * 5/18 = 5 m/s
Hence time taken to cover 90 m
= 90/5
= 18 seconds.
Answer: E
Answer: E
Question: If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?
Options:
A)360
B)720
C)920
D)670
E)None
Rationale: Solution
We have to find number of permutations of 4 objects out of 6 objects. This number is 6P4=6(6-1)(6-2)(6-3)=6*5*4*3=360
Answer A
Answer: A
Question: What is the diffference between the place value and face value of 9 in the numeral 9184?
Options:
A)8891
B)9991
C)8881
D)8091
E)8991
Rationale: place value of 9 = 9 * 1000 = 9000
face value of 9 = 9
9000 - 9 = 8991
E
Answer: E
Question: Train A leaves the station traveling at 30 miles per hour. Two hours later train В leaves the same station traveling in the same direction at 36 miles per hour. How many miles from the station was train A overtaken by train B?
Options:
A)320
B)360
C)400
D)440
E)480
Rationale: After two hours, Train A is ahead by 60 miles.
Train B can catch up at a rate of 6 miles per hour.
The time to catch up is 60/6 = 10 hours.
In 10 hours, train A travels another 30*10 = 300 miles for a total of 360 miles.
The answer is B.
Answer: B
Question: A vessel of capacity 75 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?
Options:
A)23.89
B)72.9
C)58.08
D)78.3
E)79.3
Rationale: Explanation:
Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T
For the given problem, T = 75, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(75 - 9)/75]^2 (75) = 58.08 litres.
Answer: Option C
Answer: C
Question: Find the area of a cuboid of length 10cm , breadth 8 cm . and height 6cm.
Options:
A)400 cm cube
B)410 cm cube
C)420 cm cube
D)430 cm cube
E)480 cm cube
Rationale: Area of a cuboid = lxbxh =10cm x 8cm x 6cm = 480 cm cube
answer :E
Answer: E
Question: if cost of an apple is equal to cost of 4 oranges and he has 100$ with him. Cost of an orange is 1$. He want to buy same number of orange and apples. So how many apple and orange he would be able to buy in 100$?
Options:
A)10 apple an 10 oranges
B)20 apple and 20 oranges
C)27 apples and 27 oranges
D)30 apple and 30 oranges
E)34 apples and 34 oranges
Rationale: Since cost of an orange is 1$ and apple is 4 times costlier than orange so the cost of apple is 4$. Now if we add cost of an apple and orange it comes to be 5$ and hence if we divide 100 by 5 we get answer 20. So he will be able to buy 20 apple and 20 oranges. So option B is correct.
Answer: B
Question: Two men A and B start from place X walking at 4 ½ kmph and 5 ¾ kmph respectively. How many km apart they are at the end of 3 ½ hours if they are walking in the same direction?
Options:
A)7 3/8 km
B)5 7/8 km
C)17 7/8 km
D)7 7/8 km
E)2 7/8 km
Rationale: RS = 6 ¾ - 4 ½ = 2 ¼
T = 3 ½ h.
D = 9/4 * 7/2 = 63/8 = 7 7/8 km
ANSWER:D
Answer: D