divelab/Sys2Bench huggingface dataset:
Question: A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 80 minutes later?
Options:
A)2(10^5)
B)8(10^5)
C)(2^8)(10^5)
D)(10^8)(10^5)
E)(10^5)^8
Rationale: Every 10 minutes, the population is double the previous population.
In 80 minutes, the population doubles 8 times.
The population then is 2^8*10^5.
The answer is C.
Answer: C
Question: three men invested sum. and their ratio is 5:7:6. profit is 900.then b's share
Options:
A)2000
B)2100
C)2200
D)350
E)2400
Rationale: b's share =7/18*900=350
ANSWER:D
Answer: D
Question: ABCD is a parallelogram. BD = 2. The angles of triangle BCD are all equal. What is the perimeter of the parallelogram?
Options:
A)12
B)9√3
C)9
D)8
E)3√3
Rationale: So, BC = BD = CD = BA = AD = 2
The parallelogram has 4 equal sides
BC = CD = BA = AD = 2
So The perimeter of the parallelogram will be 2*4 =8
Hence answer will be (D) 2
Answer: D
Question: What is 35% of 4/13 of 585?
Options:
A)52
B)63
C)74
D)85
E)96
Rationale: This problem can be solved easily if we just use approximation: 35% is a little over 1/3, while 4/13 is a little less than 4/12, which is 1/3. Thus, the answer is about 1/3 of 1/3 of 585, or 1/9 of 585. Since the first 1/3 is a slight underestimate and the second 1/3 is a slight overestimate, the errors will partially cancel each other out. Our estimate will be relatively accurate.
The number 585 is between 540 and 630, so (1/9)*585 will be about 65.
Keeping track not only of your current estimate, but also of the degree to which you have overestimated or underestimated , can help you pinpoint the correct answer more confidently.
The closest answer is 63, so this is the answer to choose.
The answer is B.
Answer: B
Question: A travel agent offers a vacation plan which costs 'a' dollars for the first day, and a/8 for each additional day. How much does a vacation of b days cost, where b > 1.
Options:
A)(ab+7a)/4
B)(ab+8a)/8
C)(ab+a)/8
D)(ab+7a)/8
E)(ab+7a)/7
Rationale: Cost = a + (b-1)*a/8 = (8a + (b - 1)*a)/8 = (ab+7a)/8
Answer is D
Answer: D
Question: A man can row upstream at 25 kmph and downstream at 35 kmph, and then find the speed of the man in still water?
Options:
A)11
B)77
C)30
D)88
E)34
Rationale: US = 25
DS = 35
M = (35 + 25)/2 = 30
Answer:C
Answer: C
Question: a pole rests against a wall, and the tip of the pole is at a height of 8m. after sometime the pole slips back by 4m and falls on the ground.what is the length of the pole?
Options:
A)10
B)12
C)15
D)18
E)20
Rationale: let the distance from of the ladder from the pole be x meter. A/c to Pythagoras th., ladder= root_under( x^2 + 64).
In the second case, when the ladder moves 4 m , i.e distance from the pole = x+4, it touches the ground, means the length of ladder = x+4.
Now, equating the both lengths we get x=6, length = x+4 =10
ANSWER:A
Answer: A
Question: A shopkeeper forced to sell at cost price, uses a 800 grams weight for a kilogram. What is his gain percent?
Options:
A)10%
B)25%
C)11.11%
D)12 %
E)None of these
Rationale: Shopkeeper sells 800g instead of 1000g.
So, his gain = 1000 - 800 = 200g.
Thus, %Gain = 200 *100)/800 = 25%.
answer : OPTION B
Answer: B
Question: 60 is increased by 50%. Find the final number.
Options:
A)100
B)110
C)90
D)30
E)160
Rationale: Final number = Initial number + 50%(original number) = 60 + 50%(60) = 60 + 30 = 90.
Answer C
Answer: C
Question: (p/q-q/p)=21/10. Then find 4p/q+4q/p?
Options:
A)58/9
B)58/5
C)58/7
D)58/2
E)58/1
Rationale: Let p/q = a, then (a - 1/a) = 21/10
⇒a2−1=a.2110⇒a2−1=a.2110
⇒10a2−21a−10⇒10a2−21a−10 = 0
Roots of the equation = −b±b2−4ac−−−−−−−√2a−b±b2−4ac2a
a = 21±441+400−−−−−−−−√2021±441+40020
a = 21±292021±2920 = 5/2 or -2/5
For a = 5/2, 4p/q+4q/p = 58/5
For a = -2/5, 4p/q+4q/p = -58/5
Answer:B
Answer: B
Question: Ravi can do a piece of work in 25 days while Prakash can do it in 35 days. In how many days will they finish it together?
Options:
A)15 7/9 days
B)16 17/9 days
C)15 1/19 days
D)14 7/12 days
E)14 7/9 days
Rationale: 1/25 + 1/35 = 12/175
175/12 = 14 7/12 days
ANSWER:D
Answer: D
Question: In what time will a train 400 m long cross an electric pole, it its speed be 144 km/hr?
Options:
A)2.5 sec
B)12.9 sec
C)11.3 sec
D)21.9 sec
E)10.0 sec
Rationale: Speed = 144 * 5/18
= 40 m/sec
Time taken = 400/40
= 10.0 sec.
Answer:E
Answer: E
Question: Frank the Fencemaker needs to fence in a rectangular yard. He fences in the entire yard, except for one full side of the yard, which equals 40 feet. The yard has an area of 500 square feet. How many feet offence does Frank use?
Options:
A)14
B)47
C)70
D)180
E)240
Rationale: Area = Length x Breadth
500 = 40 x Breadth
So, Breadth = 15 Units
Fencing required is -
Breadth + Breadth + Length
15 + 15 + 40 => 70 feet
Answer must be (C) 70
Answer: C
Question: Kevin sold an item for Rs.5000/- and incurred a loss of 50%. At what price should he have sold the item to have gained a profit of 50%?
Options:
A)15000
B)14000
C)13000
D)12000
E)11000
Rationale: Explanation:
50% loss means SP = 50% of CP
:. CP = Rs.5000x100/50 = Rs.10000.
The SP for a gain of 50% = 10000+50% of 10000 = 10000+5000 =Rs15000.
Answer: Option A
Answer: A
Question: A train passes a station platform in 38 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)180 m
B)270 m
C)260 m
D)280 m
E)380 m
Rationale: Speed = 54 x 5/18 = 15 m/s
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
Then, (x + 300)/38 = 15
--> x + 300 = 570
x = 270m.
Answer : B.
Answer: B
Question: In a forest 120 deer were caught, tagged with electronic markers, then released. A week later, 50 deer were captured in the same forest. Of these 50 deer, it was found that 5 had been tagged with the electronic markers. If the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest, and if no deer had either left or entered the forest over the preceding week, what is the approximate number of deer in the forest?
Options:
A)1,200
B)750
C)1,250
D)1,500
E)2,500
Rationale: The percentage of tagged deer in the second sample = 5/50*100=10%.
So, 120 tagged deers comprise 10% of total # of deers --> total # of deers = 120*10 = 1,200
Answer: A
Answer: A
Question: When 1/10 percent of 3,000 is subtracted from 1/10 of 3,000, the difference is
Options:
A) 0
B) 50
C) 450
D) 397
E) 500
Rationale: We can break this problem into two parts:
1) What is 1/10 percent of 3,000?
2) What is 1/10 of 3,000?
To calculate 1/10 percent of 3,000 we must first remember to DIVIDE 1/10 BY 100.
So we have: (1/10)/(100)
To divide a number by 100 means to multiply it by 1/100, so we have:
1/10 x 1/100 = 1/1,000
Thus, 1/10 percent of 3,000 = 1/1,000 x 3,000 = 3.
Now let's concentrate on part 2. We need to calculate 1/10 of 3,000. To do this we simply multiply 1/10 by 3,000.
1/10 x 3,000 = 300
The answer to part 1 is 3, and the answer to part 2 is 300. Their difference is 300 – 3 = 397.
Answer D.
Answer: D
Question: A work as fast as B. If B can complete a work in 8days independently,the number of days in which A and B can together finish the work in?
Options:
A)2days
B)2 2/3days
C)4days
D)5days
E)6days
Rationale: Ratio of rates of working of A and B = 2:1
ratio of times taken = 1:2
A's 1day work = 1/4
B's 1day work = 1/8
A+B 1day work = 1/4 + 1/8 = 3/8 => 8/3 = 2 2/3
A and B can finish the work in 2 2/3days
Answer is B
Answer: B
Question: Three taps A,B and C can fill a tank in 12,15 and 20. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in?
Options:
A)5hours
B)7hours
C)6hours
D)8hours
E)10hours
Rationale: A+B 1 hour work = 1/12 + 1/15 = 9/60 = 3/20
A+C 1 hour work = 1/12 + 1/20 = 8/60 = 2/15
Part filled in 2 hours = 3/20 + 2/15 = 17/60
Part filled in 6 hours = 3* 17/60 = 17/20
Remaining part = 1 - 17/20 = 3/20
Now it is the turn of A and B and 3/20 part is filled by A and B in 1hour
Total time taken to fill the tank = 6+1 = 7 hours
Answer is B
Answer: B
Question: Seven consecutive traffic signals each show either red or green. How many different arrangements of the seven signals are possible?
Options:
A)40
B)69
C)88
D)128
E)156
Rationale: Since the answer choices to the question are so small (relatively speaking), you can get to the correct answer without having to do any fancy calculations. With a drawing and a bit of 'brute force', you can answer this question relatively quickly.
We're told that 7 consecutive traffic signals each show either red or green. We're asked how many different arrangements of the 7 signals are possible.
Total Options: 128
Final Answer:
D
Answer: D
Question: In how many different number of ways 3 boys and 3 girls can sit on a bench such that girls always sit together.
Options:
A)A)720
B)B)730
C)C)740
D)D)154
E)E)144
Rationale: 144
Option 'E'
Answer: E
Question: If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is?
Options:
A)50 km
B)76 km
C)18 km
D)16 km
E)97 km
Rationale: Let the actual distance traveled be x km. Then,
x/10 = (x + 20)/14
4x - 200 =>x = 50 km.
Answer: A
Answer: A
Question: A straight line in the xy- plane has y-intercept of 20 . On this line the x-coordinate of the point is 150 and y-coordinate is 600 then what is the slope of the line?
Options:
A)3.9
B)4.6
C)5.2
D)6
E)8.1
Rationale: Eq of line = Y = MX +C
C= 20
X=150
Y=150M + 20,
Substitute Y by 600 as given in question.
600=150M+20 , M =3.9
correct option is A
Answer: A
Question: The population of a town increased from 1,75,000 to 2,27,500 in a decade. The average percent increase of population per year is
Options:
A)4.37%
B)5%
C)3%
D)8.75%
E)None
Rationale: Solution
Increase in 10 years	= (227500 - 175000)
= 52500.
Increase%	= (52500/175000Ã—100)%
= 30%.
Required average	= (30/10)%
= 3%.
Answer C
Answer: C
Question: Five years ago the average of the ages of A and B was 40 years and now the average of the ages of B and C is 51 years. What will be the age of the B ten years hence?
Options:
A)39
B)23
C)28
D)102
E)26
Rationale: Let the present ages of A, B and C be a, b and c respectively.
Given, [(a - 5) + (b - 5)] / 2 = 40 => a + b = 90 --- (1)
(b + c)/2 = 51 => b + c = 102 --- (2)
Answer: D
Answer: D
Question: If 170!/10^n is an integer, what is the largest possible value of n?
Options:
A)47
B)46
C)41
D)48
E)45
Rationale: The question actually asks the highest power of 10 which divides 300! ( For a number to be an integer - Without any remainder all the trailing zeroe's must be divided by the denominator)
10 = 2 x 5
170 factorial will have 41 as -
170/5 = 34
34/5 = 6
6/5 =1
So answer will be (C) 41
Answer: C
Question: By the how much is two-fifth of 300 greater than three -fifths of 125?
Options:
A)15
B)3
C)5
D)45
E)None of these
Rationale: Reqd no. = 2â„5 Ã— 300 - 3â„5 Ã— 125
= 120 - 75 = 45
Answer D
Answer: D
Question: Den Bosch and Eastbourne are 2 famous cities 300 kilometer apart. Mary starts from Den Bosch at 8:24 am . An hour later pele starts from Den Bosch. After travelling for 1 hour , Pele reaches Nottingham that Mary had passed 40 minutes earlier. Nottingham falls on the way from Den Bosch to Eastbourne. If pele and Mary just reaches Eastbourne at the same time, what are the speeds of the Mary and Pele Respectively?
Options:
A)120 km/hr
B)158 km/hr
C)98 km/hr
D)102 km/hr
E)100 km/hr
Rationale: Let Pele x km in 1 hour. so Maradona takes (2h-40min)=1 h 20 min to cover x km. Let speed of Maradona and Pele be M and P respectively than
x = \inline M\times \frac{4}{3} and x= P x 1
\inline \Rightarrow \inline \frac{M}{P}=\frac{3}{4}
Again \inline \frac{300}{M}=\frac{300}{P}=1
\inline \Rightarrow \: \: \frac{300}{3k}-\frac{300}{4k}=1
\inline \Rightarrow k= 25
\inline \Rightarrow M= 3k = 75 km/h
and P= 4k = 100 km/h
E
Answer: E
Question: An 80-liter solution of cool-drink is made from 10% jasmine water. If 5 liters of jasmine and 15 liters of water were added to the solution, what percent of the solution is jasmine?
Options:
A)11%
B)13%
C)15%
D)17%
E)19%
Rationale: The percent of jasmine in the resulting solution is: (amount of jasmine) / (total volume)
(0.1(80) + 5) / 100 = 13/100 = 13%
The answer is B.
Answer: B
Question: If -11 and -8 are negative integers, then -11*-8 + 5 is
Options:
A)-14
B)14
C)33
D)83
E)93
Rationale: Answer : E
Answer: E
Question: Set X consists of all two-digit primes and set Y consists of all positive multiples of 6 less than 100. If the two sets are combined into one, what will be the range of the new set?
Options:
A) 84
B) 89
C) 90
D) 92
E) 91
Rationale: Set X = {11, 13, 17,............., 83, 89, 97}
Set Y = {6, 12, 18,..............., 84, 90, 96}
Combining two sets, say Set Z
Set Z = {6, 11, 12, 13, 17, 18,..................., 83, 84, 89, 90, 96,97}
Range = Max Value - Min Value
Range (Z) = 97 - 6 = 91
OA E is the answer.
Answer: E
Question: If you have 8 New Year greeting cards and you want to send them to 6 of your friends, in how many ways can this be done?
Options:
A)20160
B)22100
C)23000
D)24500
E)None
Rationale: Solution
We have to find number of permutations of 6 objects out of 8 objects. This number is 8P6=8(8-1)(8-2)(8-3)(8-4)(8-5)=8*7*6*5*4*3=20160
Answer A
Answer: A
Question: The diameter of a cylindrical tin is 6cm and height is 5 cm. Find the volume of the cylinder?
Options:
A)33
B)45
C)66
D)77
E)21
Rationale: r = 3 h = 5
π * 3 * 3 * 5 = 45 π cc
Answer:B
Answer: B
Question: How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.
Options:
A)2154
B)2454.
C)2254
D)2354
E)1245
Rationale: There are 8 distinct letters: M-A-T-H-E-I-C-S. 3 letters M, A, and T are represented twice (double letter).
Selected 4 letters can have following 3 patterns:
1. abcd - all 4 letters are different:
8P4=16808 (choosing 4 distinct letters out of 8, when order matters) or 8C4∗4!=16808 (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters);
2. aabb - from 4 letters 2 are the same and other 2 are also the same:
3C2∗4!/2!2!=18 choosing which two double letter will provide two letters (out of 3 double letter - MAT), multiplying by 4!/2!2! to get different arrangements (for example MMAA can be arranged in 4!/2!2! # of ways);
3. aabc - from 4 letters 2 are the same and other 2 are different:
3C1∗7C2∗4!/2!=756 choosing which letter will proved with 2 letters (out of 3 double letter - MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by 4!/2!to get different arrangements (for example MMIC can be arranged in 4!/2!# of ways).
1680+18+756=2454
Answer: B
Answer: B
Question: If the weight of 13 meters long rod is 13.4 kg. What is the weight of 6 meters long rod ?
Options:
A)7.2 kg.
B)10.8 kg.
C)12.4 kg.
D)6.18 kg
E)None
Rationale: Answer
∵Weight of 13 m long rod = 13.4 kg
∴ Weight of 1 m long rod = 13.4 / 13 kg
∴ Weight of 6 m long rod = 13.4 x 6 / 13
= 6.18 kg
Option: D
Answer: D
Question: A salt manufacturing company produced a total of 4000 tonnes of salt in January of a particular year. Starting from February its production increased by 100 tonnes every month over the previous months until the end of the year. Find its ave66rage daily production for that year?
Options:
A)140
B)149
C)156
D)195
E)100
Rationale: Total production of salt by the company in that year = 4000 + 4100 + 4200 + .... + 5100 = 54600.
Average monthly production of salt for that year = 54600/365 â‰ˆ 149
ANSWER:B
Answer: B
Question: Arun's Ice Cream Shop sells ice cream at x cents a scoop. For an additional y cents, a customer can add 4 toppings to his or her sundae. How much would a sundae with 4 scoops and 4 toppings cost, in terms of x and y?
Options:
A)3x+y
B)5x+y
C)6x+y
D)4x+y
E)7x+y
Rationale: Answer A 4x+y
Answer: A
Question: A certain university will select 1 of 5 candidates eligible to fill a position in the mathematics department and 2 of 8 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
Options:
A)138
B)140
C)142
D)145
E)148
Rationale: 1C5*2C8 = 5*28 = 140
the answer is (B)
Answer: B
Question: If cost of sugar increases by 15%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed?
Options:
A)13%
B)50%
C)20%
D)80%
E)30%
Rationale: 100
115
-----
115 ----- 15
100 ------ ? => 13%
Answer: A
Answer: A
Question: A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 66 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is less than m + d ?
Options:
A) 16%
B) 32%
C) 83%
D) 84%
E) 92%
Rationale: This is easiest to solve with a bell-curve histogram. m here is equal to µ in the Gaussian normal distribution and thus m = 50% of the total population.
So, if 66% is one st.Dev, then on either side of m we have 66/2 = 33%. So, 33% are to the right and left of m (= 50%). In other words, our value m + d = 50 + 33 = 83% goingfrom the mean m, to the right of the distributionin the bell shaped histogram.. This means that 83% of the values are below m + d.
Like I said, doing it on a bell-curve histogram is much easier to fullygethow this works, or you could apply GMAT percentile jargon/theory to it
C
Answer: C
Question: In an examination, questions were asked in five sections. Out of the total students, 5% candidates cleared the cut-off in all the sections and 5% cleared none. Of the rest, 25% cleared only one section and 20% cleared four sections. If 24.5% of the entire candidates cleared two sections and 300 candidates cleared three sections. Find out how many candidates appeared at the examination?
Options:
A)1000
B)1200
C)1500
D)2000
E)1800
Rationale: Solution: Passed in none = 5%
Passed in all = 5%
Passed in four = 20% 0f 90% = 18%
Passed in three = 24.5%
Passed in three =(100-5-5-22.5-24.5-18) = 25%.
But given 300 students passed in three.
Hence, 25% = 300.
So, 100% = 1200.
1200 students must have appeared.
Answer: Option B
Answer: B
Question: A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
Options:
A)3
B)6
C)7
D)4
E)5
Rationale: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.
Answer:D
Answer: D
Question: In a certain candy store, 22% of the customers are caught sampling the candy and are charged a small fine, but 15% of the customers who sample the candy are not caught. What is the total percent of all customers who sample candy?
Options:
A)22%
B)23%
C)24%
D)26%
E)34%
Rationale: Since 15% of the customers who sample the candyare notcaught, then 88% of the customers who sample the candyarecaught:
{% of customers who sample candy}*0.85 = 0.22;
{% of customers who sample candy} = 0.259
Answer: D.
Answer: D
Question: The ratio of the ages of Anil and his son at present is 7:3. Six years hence, the ratio of the ages of the Anil's wife and the son will be 2:1. Find the ratio of the present ages of Anil and his wife?
Options:
A)5:4
B)6:5
C)4:3
D)Cannot be determined
E)None of these
Rationale: Let the present ages of Anil and his son be 7x and 3x years respectively.
Let the present age of the son's mother be m years.
(m + 6)/(3x + 6) = 2/1
m = 6x + 6
Ratio of the present ages of Anil and his son's mother = 7x/(6x + 6).
This cannot be found uniquely.
Answer: D
Answer: D
Question: A water tank is one-third full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
Options:
A)5 min. to empty
B)6 min. to full
C)9 min. to empty
D)9 min. to full
E)8 min. to full
Rationale: Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 1/3
Part emptied by (A + B) in 1 minute = (1/6 - 1/10) = 1/15
1/15 : 1/3 :: 1 : x
x = (1/3 * 1 * 15) = 5 min.
So, the tank will be emptied in 5 min.
ANSWER:A
Answer: A
Question: Last year, a certain public transportation system sold an average (arithmetic mean) of 40,000 tickets per day on weekdays (Monday through Friday) and an average of 18,000 tickets per day on Saturday and Sunday. Which of the following is closest to the total number of tickets sold last year?
Options:
A)1 million
B)1.25 million
C)10 million
D)12.2 million
E)125 million
Rationale: If you type that into the calculator, it results in 12.3 mil
[D], 12.2 mil, is the correct answer... see 2 posts above for quick approx method
Answer: D
Question: If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same sum at the rate and for the same time?
Options:
A)Rs.51.25
B)Rs.51.27
C)Rs.51.79
D)Rs.51.8
E)Rs.51.93
Rationale: Sum = (50 * 100) / (2 * 5) = Rs. 500
Amount = [500 * (1 + 5/100)2] = Rs. 551.25
C.I. = (551.25 - 500)
= Rs.51.25.
Answer: A
Answer: A
Question: If A is to be chosen at random from the set {14, 7, 25, 6} and B is to be chosen at random from the set {2, 5, 8}, what is the probability that AB will be even?
Options:
A)2/3
B)8/3
C)5/6
D)1/3
E)4/3
Rationale: Method-1
AB will be even when
1) A is even and B is odd, Probability of A even is (2/4) and Probability of B odd is (1/3), so Probability of Case(1) = (2/4)*(1/3) = (2/12)
2) A is Odd and B is Even, Probability of A Odd is (2/4) and Probability of B Even is (2/3), so Probability of Case(2) = (2/4)*(2/3) = (4/12)
3) A is even and B is Even, Probability of A even is (2/4) and Probability of B Even is (2/3), so Probability of Case(1) = (2/4)*(2/3) = (4/12)
Total Favorable Probability = (2/12)+(4/12)+(4/12) = (32/12) = 8/3
Answer: option B
Answer: B
Question: A pack of 52 playing cards, 2cards are drawn together at random. Find the probability of both the cards being king?
Options:
A)1/113
B)2/213
C)1/221
D)3/341
E)3/215
Rationale: Two cards can be drawn from a pack of 52 playing cards in 52C2 ways. i,e., 52 x 51 / 2 = 1326 ways. The event that two kings appear in a single drawn of cards is 4C2 ways, i.e 6 ways.
∴ The probability that the two cards drawn from a pack of 52 cards are kings = 6/1326 = 1/221
C
Answer: C
Question: It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If two beavers start to build the same dam at 7:00 AM, and one beaver per hour is added beginning at 4:00 PM, at what time will the dam be complete?
Options:
A)7:30 PM
B)8:00 PM
C)9:00 PM
D)10:00 PM
E)11:00 PM
Rationale: The rate of 1 beaver is 1/60.
The amount of work done by 2 beavers in 9 hours is 2/60 *9 = 18/60
In the next hour, 3 beavers do 3/60 for a total of 21/60.
In the next hour, 4 beavers do 4/60 for a total of 25/60.
In the next hour, 5 beavers do 5/60 for a total of 30/60.
In the next hour, 6 beavers do 6/60 for a total of 36/60.
In the next hour, 7 beavers do 7/60 for a total of 43/60.
In the next hour, 8 beavers do 8/60 for a total of 51/60.
In the next hour, 9 beavers do 9/60 for a total of 60/60.
The dam will be completed at 11:00 pm.
The answer is E.
Answer: E
Question: A man sells an article at 15% loss. Had he sold it for Rs. 210/- more he could have gained 20%. What is the cost price of the article?
Options:
A)Rs.420/-
B)Rs.400/-
C)Rs.600/-
D)Rs.680/-
E)None of these
Rationale: Explanation:
Short cut method:
Loss =====> 15 %
Profit ====> 20 %
~
----------------
35 % ---------> Rs. 210/-
100 % --------> Rs. 600/-
Answer: Option C
Answer: C
Question: The banker's discount on $1600 at 15% per annum is the same as true discount on $1680 for the same time and at the same rate. The time is:
Options:
A)2 months
B)1 months
C)4 months
D)8 months
E)9 months
Rationale: S.I. on $1600 = T.D. on $1680.
THEREFORE $1600 is the P.W. of $1680, i.e., $80 is on $1600 at 15%.
Time =(100 x 80/ 1600 x 15) year	=	1/3	year = 4 months.
ANSWER:C
Answer: C