divelab/Sys2Bench huggingface dataset:
Question: In Mike's opinion, his weight is greater than 65kg but leas than 72 kg. His brother does not agree with Mike and he thinks that Mike's weight is greater than 60kg but less than 70kg. His mother's view is that his weight cannot be greater than 68 kg. If all of them are correct in their estimation, what is the average of different probable weights of Mike?
Options:
A)45.5 kg
B)50.5 kg
C)55.5 kg
D)66.5 kg
E)72.5 kg
Rationale: Let Mike's weight be X kg.
According to Mike, 65 < X < 72.
According to Mike's brother, 60 < X < 70.
According to Mike's mother, X < 68.
The values satisfying all the above conditions are 66 and 67.
Required average = (66 + 67) / 2 = 66.5 kg
D
Answer: D
Question: Tom reads at an average rate of 30 pages per hour, while Jan reads at an average rate of 37 pages per hour. If Tom starts reading a novel at 4:00, and Jan begins reading an identical copy of the same book at 4:56, at what time will they be reading the same page?
Options:
A)7:26
B)7:56
C)8:26
D)8:56
E)9:26
Rationale: Since Tom reads an average of 1 page every 2 minutes, Tom will read 28 pages in the first 56 minutes. Jan can catch Tom at a rate of 7 pages per hour, so it will take 4 hours to catch Tom.
The answer is D.
Answer: D
Question: There are 50 students in a class. If 10 % are absent on a particular day, find the number of students present in the class.
Options:
A)43
B)36
C)28
D)45
E)11
Rationale: Number of students absent on a particular day = 10 % of 50
i.e., 10/100 × 50 = 5
Therefore, the number of students present = 50 - 5 = 45 students.
Answer:D
Answer: D
Question: How many rectangles can be formed out of a chess board ?
Options:
A)204
B)1230
C)1740
D)1296
E)none of these
Rationale: A chess board consists 9 parallel lines x 9 parallel lines.
For a rectangle we need to select 2 parallel lines and the other two parallel lines that are perpendicular to the first set.
Hence, 9C2 x 9C2 = 1296 rectangles
ANSWER:D
Answer: D
Question: A, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.200 when the rate to profit rises from 5 to 7 percent. Find the Capital of B?
Options:
A)399
B)238
C)200
D)288
E)291
Rationale: A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 200
B capital = 15000*1/6 = 2500 .Answer: C
Answer: C
Question: How many quarters are equal to 4 dollars?
Options:
A)16
B)8
C)12
D)9
E)7
Rationale: 4*4 = 16 quarters
Answer : A
Answer: A
Question: Look at this series: 36, 34, 30, 28, 24, ... What number should come next?
Options:
A)20
B)24
C)22
D)26
E)30
Rationale: alternating no subtraction series first 2 is subtracted and then 4 and 2 so on
22 is the next one
ANSWER C
Answer: C
Question: Bradley owns b video game cartridges. If Bradley’s total is one-third the total owned by Andrew and seven times the total owned by Charlie, how many video game cartridges do the three of them own altogether, in terms of b?
Options:
A)(16/3)b
B)(17/4)b
C)(13/4)b
D)(19/12)b
E)(29/7)b
Rationale: Step 1: Categorize the problem
This problem is testing basic algebra and equations, with an extra layer of complexity by having variables in the answer choices.
Step 2: Think like the test maker
What is the key detail or key details in the problem? The key detail to immediately recognize is that the problem gives you Andrew and Charlie’s totals in relation to Bradley’s total rather than giving you Bradley’s total in relation to the other two; and then the problem asks you to solve the problem in terms of b. This means that you have to relate Andrew and Charlie’s values in relation to Bradley’s total. The test maker is attempting to force you to structure the problem in the opposite way that most people are used to structuring information. By gaining this insight, it makes it easier to solve the problem.
Step 3: Solve the problem
b = Bradley’s total
3b = Andrew’s total
(1/7)b = Charlie’s total
Add each total
b + 3b+ (1/7)b = 4b + (1/7)b = (29/7)b
Therefore,choose E.
Answer: E
Question: A man can row his boat with the stream at 6 km/h and against the stream in 4 km/h. The man's rate is?
Options:
A)5 kmph
B)1kmph
C)7 kmph
D)9 kmph
E)11 kmph
Rationale: Explanation:
DS = 6
US = 4
S = ?
S = (6 - 4)/2 = 1 kmph
Answer: B
Answer: B
Question: What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12% p.a.?
Options:
A)10123.298
B)10123.22
C)10123.2
D)10123.21
E)10123.54
Rationale: Amount = [25000 * (1 + 12/100)3]
= 25000 * 28/25 * 28/25 * 28/25 = Rs. 35123.20
C.I. = (35123.20 - 25000) = Rs. 10123.20
Answer: C
Answer: C
Question: A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 11.30 a.m. ?
Options:
A)7
B)5
C)9
D)17
E)12
Rationale: D
17
Relative speed = 2 + 3 = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 17 times before 11.30 a.m.
Answer: D
Question: A bicyclist travels uphill from town A to town B for 2 hours at an average speed of 4 miles per hour and returns along the same road at an average speed of 8 miles per hour. What is the bicyclist’s average speed for the round trip, in miles per hour?
Options:
A)(a) 24/5
B)(b) 5
C)(c) 26/5
D)(d) 27/5
E)(e) 16/3
Rationale: from the uphill we can find out the distance = 4*2 = 8
average speed = total distance/total time
= 16/ (2+ (8/8)) = 16/3
Answer is E.
Answer: E
Question: A jar contains 10 marbles consisting of an equal number of red, green, and blue marbles. Three marbles are removed from the jar and discarded. What is the probability that only two colors will remain in the jar after the Three marbles have been removed?
Options:
A)1/40
B)1/165
C)1/81
D)1/30
E)1/20
Rationale: Jar contains {3-red, 3-green, 3-blue} marbles. We are asked to find the probability that after removing 3 marbles only 2 colours remain in the jar, i.e., the probability of removing EITHER red, OR green, OR blue.
Sample space = 10C3 = 120.
Probability of choosing 3 red = 3C3/10C3 = 1/120.
Probability of choosing 3 green = 3C3/10C3 = 1/120.
Probability of choosing 3 blue = 3C3/10C3 = 1/120.
Required probability = 1+1+1/120 = 3/120 = 1/40. Ans (A).
Answer: A
Question: A,B,C together can earn $600 per day. while A and C together earn $400 and B and C together can earn $300. The daily earnings of C?
Options:
A)$100
B)$200
C)$250
D)$300
E)$500
Rationale: B's daily earnings = 600-400 = $200
A's daily earnings = 600-300 = $300
C's daily earnings = 600-200-300 = $100
Answer is A
Answer: A
Question: A profit of Rs. 800 is divided between X and Y in the ratio of 1/2:1/3 . What is the difference between their profit shares ?
Options:
A)s. 260
B)s. 160
C)s. 360
D)s. 50
E)s. 90
Rationale: A profit of Rs. 800 is divided between X and Y in the ratio of 1/2 : 1/3 or 3:2.
so profits are 480 and 320.
difference in profit share = 480-320=160
ANSWER:B
Answer: B
Question: If each side of a square is increased by 25%, find the percentage change in its area.
Options:
A)46.25%
B)56.25%
C)56%
D)50.25%
E)36.25%
Rationale: Let each side of the square be a. Then, area = a2.
New side =(125a/100) =(5a/4). New area = (5a/4) 2=(25a2)/16.
Increase in area = ((25 a2)/16)-a2=(9a2)/16.
Increase% = [((9a2)/16)*(1/a2)*100] % = 56.25%.
ANSWER B 56.25%
Answer: B
Question: What day of the week was 7th January 1601
Options:
A)Sunday
B)Tuesday
C)Monday
D)Thursday
E)Saturday
Rationale: Explanation:
7th Jan 1601 = (1600 years + 7th Jan 1601)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
7th Jan 1601 = 7 odd day
Total number of odd days = (0 + 7) = 7 = 0
0 odd days = Sunday
Hence 7th January 1601 is Sunday.
Answer: Option A
Answer: A
Question: In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?
Options:
A)2438
B)8000
C)2379
D)2988
E)2791
Rationale: Explanation:
Let the number of candidates appeared from each state be x.
In state A, 6% candidates got selected from the total appeared candidates
In state B, 7% candidates got selected from the total appeared candidates
But in State B, 80 more candidates got selected than State A
From these, it is clear that 1% of the total appeared candidates in State B = 80
=> total appeared candidates in State B = 80 x 100 = 8000
=> total appeared candidates in State A = total appeared candidates in State B = 8000
Answer: B) 8000
Answer: B
Question: A bag contains 20 red jellybeans and 20 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
Options:
A)9/100
B)2/19
C)1/8
D)3/20
E)3/10
Rationale: Method-1
10 red jellybeans and 10 blue jellybeans
Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120
Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3
Probability = 120 / 1140 = 2/19
Answer: option B
Method-2
Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans]
Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining]
Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining]
Required Probability = (10/20)*(9/19)*(8/18) = 3/20
Answer: option D
Answer: D
Question: If a, b, and c are distinct positive prime numbers, then (a^3)(b)(c) has how many different positive factors greater than 1?
Options:
A)a. 3
B)b. 9
C)c. 12
D)d. 15
E)e. 27
Rationale: Let Number is (a^3)(b)(c) = (1^3)(2)(3) = 6
We can write 6 as product of two numbers in following ways
1*6
2*3
2 cases = 2*2 i.e. 4 factors (including 1)
Factors greater than 1 = 3
Answer: Option A
Answer: A
Question: A trader sold an item at a loss of 18%. if he had increased the price by $25 he would have made a gain of 5%. what is the cost price of the item?
Options:
A)$414.78
B)$424.78
C)$534.78
D)$434.78
E)$435.78
Rationale: Let C.P. be $x
Then 105% of x - 82% of x = 100
23% of x = 100
23/100 *X = 100
23x = 10000
x = 434.78
Answer is D
Answer: D
Question: There are 12 pieces of radioactive metal W that look identical. 11 of the pieces give the same radiation count when measured, the 12th piece is a counterfeit and gives a different radiation level, which may be more or less than the other 11. We are given a radiation scale, which can take 2 sets of samples and compare their added up radiation levels to tell us if the sums are the same or if different, which set has the higher level of radiation. What is the minimum number of comparisons we need on this scale to identify the counterfeit sample and to also determine whether it has more or less radiation than the other samples ?
Options:
A)2
B)3
C)4
D)5
E)6
Rationale: First of all if you are down to just 3 pieces and you know that if the offending piece is less or more active, then it takes exactly 1 measurement to find out the offending piece. So you know you have to reduce the problem to three.
Now when you are down to either A or B after measurement 1, you need the next measurement to (a) reduce the problem set to 3 and (b) to know whether anser is more or less. Now you cannot compare a group of 4 to 4, as in the best case it will only reduce the problem to 4 elements which is not good enough.
If you have to choose a set of 3 to compare, you cannot pick any 3 on the same side from the same set (A or B) because if you do this, a quick check will show you that in every choice there is a case where you can only get down to 4 elements. Eg. If you weighed {1,2,3} v/s {5,9,10} and they were equal you're problem would only reduce to {4,6,7,8}
The easiest way to solve this then is to compare 3 to 3, and make sure each side has elements from both AB such that whatever the measurement outcome in the worst case the problem reduces to 3 elements only. Which is why the sets {1,5,9} and {2,6,7} OR {A,B,C}{A,B,B}. The extra element from C is just taken to make the problem symmetric so to say, we have 8 elements and we make it 9, to compose 3 sets of 3 each.=B
Answer: B
Question: Of 70 players on a football team, 28 are throwers. The rest of the team is divided so one third are left- handed and the rest are right handed. Assuming that all throwers are right handed, how many right- handed players are there total?
Options:
A)56
B)59
C)63
D)71
E)92
Rationale: Total = 70
Thrower = 28
Rest = 70 - 28 = 42
Left Handed = 42/3 = 14
Right handed = 28
If all Thrower are right handed then total right handed is 28 + 28 = 56
so A. 56 is the right answer
Answer: A
Question: Company X sells a selection of products at various price points. Listed below are unit sales made for one particular day. How many unit sales on that day were greater than the mean sale price but less than the median sale price?
$50, $50, $97, $97, $97, $120, $125, $155, $155, $199, $199
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: Taking the prices of products in ascending order(already arranged)
$50, $50, $97, $97, $97, $120, $125, $155, $199, $199, $199
we see that median value is the 6th value as there in total 11 values given
Arithmetic mean = total/number of entries = 1388/11 =126.1818
We are asked to find how many unit sales on that day were greater than the mean sale price but less than the median sale price
as we can clearly see that there is one value between $120 and $126.1818, answer is 1 unit
Correct Answer - B
Answer: B
Question: A car takes 6 hours to cover a distance of 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
Options:
A)50 Kmph
B)60 Kmph
C)70 Kmph
D)80 Kmph
E)90 Kmph
Rationale: Time = 6
Distance = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
B
Answer: B
Question: In an examination, 30% of total students failed in Hindi, 35% failed in English and 35% in both. The percentage of these who passed in both the subjects is :
Options:
A)10%
B)20%
C)30%
D)70%
E)50%
Rationale: pass percentage = 100 - (30+35-35) = 100-30 = 70
ANSWER:D
Answer: D
Question: Find the invalid no.from the following series 15 46 63 71 75 77 78
Options:
A)15
B)46
C)63
D)71
E)75
Rationale: In this question, if you start from the beginning, firstly the difference is 31, then the difference is 17, then the difference is 8. So there is no logic. If you start backwards,you can see that firstly, 1 is added, then 2 is added, then 4 is added, then 8,16 and 32 should have been added. So you have got the right clue. So 15 + 32 the first number should be 47 and afterwards, the series becomes + 16, + 8, + 4and so on.
46
B
Answer: B
Question: At the end of a business conference the ten people present all shake hands with each other once. How many handshakes will there be altogether ?
Options:
A)20
B)45
C)55
D)90
E)85
Rationale: Explanation:
Clearly, total number of handshakes = (9+ 8 + 7 + 6 + 5 + 4 + 3 + 2+1) = 45.
Answer: B
Answer: B
Question: The probability of having a girl is identical to the probability of having a boy. In a family with five children, what is the probability that all the children are of the same gender?
Options:
A)1/16
B)1/4
C)1/8
D)1/12
E)1/25
Rationale: There are two cases. All boys and all girls
case 1. all boys.
1/2*1/2*1/2*1/2*1/2= 1/32
case 2 .all girls.
1/2*1/2*1/2*1/2*1/2=1/32
We will add the cases to get 1/16
Answer: A
Question: In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?
Options:
A)5
B)10
C)15
D)20
E)25
Rationale: Solution
Let us write the fraction for each group of students
Group A: younger than 10: 50% = 50/100 = 1/2
Group B: 10 years old: 1/20
Group C: older that 10 but younger than 12: 1/10
Group D: 12 years or older: 70 students
The fraction for group A, B and C together is given by
1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator
= 13 / 20 , add numerators
The fraction for group D is given by
20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students
If X is the total number of students, then
7 / 20 of X = 70
or (7 / 20) × X = 70
Solve for X
X = 70 × (20 / 7) = 200
Students who are 10 years old have a fraction of 1 / 20 of the total X and their number is equal to
(1 / 20) of 200 = 10 students
ANSWER IS B
Answer: B
Question: How many digits are in (8×10^10)(10×10^5)?
Options:
A)17
B)25
C)26
D)27
E)28
Rationale: The question simplfies to (8×10^10)(10^6)
=> 8*10^16
=> Will contain 16 zeros +1 digit 8
=>17
Ans A
Answer: A
Question: Set A is {1, 3, 5, 7, 9}
Set B is {16, 18, 20, 22, 24}
For the sets of numbers above, which of the following is true?
I. The mean of Set B is greater than the mean of Set A.
II. The median of Set B is greater than the median of Set A.
III. The standard deviation of Set B is greater than the standard deviation of Set A.
Options:
A)I only
B)I and II only
C)I and III only
D)II and III only
E)I, II, and III
Rationale: The mean and median of Set A is 5.
The mean and median of Set B is 20.
Thus statements I and II are both true.
The standard deviation is related to the distance from the mean.
Since the elements of both sets are spaced identically around the mean (+/- 2, +/- 4), the standard deviations of both sets are equal.
The answer is B.
Answer: B
Question: In the first half of the 20th century, the population of a particular country increased by 200 percent. In the second half of the century, the population increased by 300 percent. What was the percent increase for the 20th century as a whole?
Options:
A)500%
B)600%
C)800%
D)1100%
E)1200%
Rationale: Say initially population was 100.
What is 200% of 100? It is 200/100 * 100 = 200.
An increase of 200% means the new population became 100 + 200 = 300
What is 300% of 300? It is 300/100 * 300 = 900
An increase of 300% means the new population now is 300 + 900 = 1200
So from 100, the population increased to 1200 i.e. an increase of 1100.
1100 is what percent of 100?
1100 = x/100 * 100
i.e. it is 1100%
ANSWER:D
Answer: D
Question: A vendor bought Headphones at 5 for Rs 100.How many for Rs 100 must he sell to gain 50%?
Options:
A)8
B)3
C)5
D)6
E)None
Rationale: C.P. of 9 Headphones= Rs100
S.P. of 9 Headphones = 150% of Rs 100 = Rs150
For Rs.150 , Headphones sold = 9.
For Rs. 100,Pens sold=(9/150)*100=6
Answer: Option D
Answer: D
Question: (500 + 200+100) Ã— 4 Ã— (3 + 4) = ?
Options:
A)52000
B)32400
C)22400
D)22800
E)24200
Rationale: (500 + 200+100) Ã— 4 Ã— (3 + 4) = ?
or, ? = 800 Ã— 4 Ã— 7 = 22400
Answer C
Answer: C
Question: P is four times as old as Q, and Q is six years older than R. If R is z years old, what is P’s age in terms of z?
Options:
A)4z + 6
B)4z − 24
C)4z + 24
D)(z + 6)/4
E)(z − 6)/4
Rationale: P = 4Q
Q = R+6 = z+6
P = 4(z+6) = 4z + 24
The answer is C.
Answer: C
Question: By the how much is two-fifth of 400 greater than three -fifths of 150?
Options:
A)15
B)70
C)5
D)30
E)None of these
Rationale: Reqd no. = 2â„5 Ã— 400 - 3â„5 Ã— 150
= 160 - 90 = 70
Answer B
Answer: B
Question: Two cars start from the opposite places of a main road, 150 km apart. First car runs for 30 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?
Options:
A)65
B)38
C)60
D)28
E)21
Rationale: Answer: C) 60 km
Answer: C
Question: The ratio, by volume, of soap to alcohol to water in a certain solution is 2:50:100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved. If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?
Options:
A)50
B)200
C)400
D)625
E)800
Rationale: soap:alcohol
Initial ratio soap:alcohol: water --> 2:50:100
Initial soap:alcohol = 2/50 =2:50
after doubled soap:alcohol =2* 2/50 = 4:50
Initial soap:water = 2/100=2:100
after halved soap:water: 1/2 * 2/100 = 1/100 = 1:100= 4:400
After soap: alcohol:water --> 4:50:400
given alcohol 100 cumeter.
ratio is 8:100:800 (4:50:400)
For 100 cu meter of Alcohol ---800 cu m water is required
ANSWER:E
Answer: E
Question: If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?
Options:
A)11
B)13
C)14
D)18
E)10
Rationale: Required difference = [3 ½ % of Rs.8400] – [3 1/3 % of Rs.8400]
= [(7/20-(10/3)]% of Rs.8400 =1/6 % of Rs.8400
= Rs. [(1/6)8(1/100)*8400] = Rs. 14.
Answer is C.
Answer: C
Question: P is four times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work?
Options:
A)35 days
B)15 days
C)16 days
D)25 days
E)18 days
Rationale: P = 4Q
P + Q = 4Q + Q = 5Q
These 5Q people can do the work in 12 days, which means Q can do the work in 60 days.
Hence, P can do the work in 15 days.
ANSWER:B
Answer: B
Question: Find the average of the series : 150 , 160, 328, 147 and 120?
Options:
A)174
B)178
C)181
D)180
E)186
Rationale: Average = (150 + 160 + 328 + 147 + 120)/5
= 905/5
= 181
Answer: C
Answer: C
Question: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.
Options:
A)42
B)48
C)72
D)67
E)54
Rationale: Sol: Let the third number be x.
Then second number = 3x.
First number=3x/2.
Therefore x+3x+(3x/2)=(44*3) or x=24
So largest number= 2nd number=3x=72
ANSWER :C
Answer: C
Question: The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?
Options:
A)11 year
B)12 year
C)23 year
D)25 year
E)27 year
Rationale: Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
Answer: C
Answer: C
Question: The average of 9 observations was 7, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation?
Options:
A)27
B)12
C)15
D)17
E)18
Rationale: Explanation:
1 to 9 = 9 * 7 = 63
1 to 5 = 5 * 10 = 50
5 to 9 = 5 * 8 = 40
5th = 50 + 40 = 90 – 63 = 27
Option A
Answer: A
Question: In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?
Options:
A)21
B)22
C)23
D)24
E)25
Rationale: n(A)=(60/100 ∗96)=288/5
n(B)=(30/100 ∗96)=144/
n(A∩B)=(15/100 ∗96)=72/5
People who have either or both lunchn(A∪B)=288/5 +144/5 −72/5 =360/5 =72
So People who do no have either lunch were = 96 -72 = 24 ANSWER : D
Answer: D
Question: Which one of the following numbers is divisible by 99?
Options:
A)3572404
B)135792
C)913464
D)114345
E)114375
Rationale: Explanation:
The sum of all the digits of 114345 is 18.
The sum of all digits at even places and odd places are 9 and 9 respectively.
The difference between the sums is 0.
So, the number is divisible by both 9 and 11 and hence by 99.
Answer:D
Answer: D
Question: In a circular track of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. when will all the three meet for the first time after the start?
Options:
A)90
B)100
C)110
D)120
E)130
Rationale: A takes (100m) / (10m/s) = 10s to finish one complte round.
B takes (100m) / (8m/s) = 12.5s to finish one complte round.
C takes (100m) / (15m/s) = (20/3)s to finish one complte round.
For them to meet, the time will be LCM of above three times ie LCM(10,12.5,20/3)
= LCM(10/1,25/2,20/3)
= LCM(10,25,20)/GCD(1,2,3)
=100/1
=100
ANSWER:B
Answer: B