divelab/Sys2Bench huggingface dataset:
Question: Machine A can do a certain job in 8 hours. Machine B can do the same job in 10 hours. Machine C can do the same job in 12 hours. All three machines start the job at 9:00 a.m. Machine A breaks down at 11:00 a.m., and the other two machines finish the job. Approximately what time R will the job be finished?
Options:
A)Noon
B)12:30 p.m.
C)1:00 p.m.
D)1:30 p.m.
E)2:00 p.m.
Rationale: Ans: C (1:00pm)
From 9:00-11:00c(2 hrs)
A can do 1/8 Job/hr. so in 2 hours he can do 1/4 job.
B can do 1/10 job/hr so in 2 hrs. he can do 1/5 Job.
C can do 1/12 job/hr so in 2 hours he can do 1/6 job.
Total = 1/4+1/5+1/6 = 37/60 Job
Now balance work needs to be done by B and C.
Balance Work= 23/60 Job
Combined rate of B and C = 1/10+1/12 = 11/60 job/hr.
So they will do 22/60 work in 2 hrs (approx. to the balance work of 23/60)
Hence the Job will be completed 2hrs. after 11:00 i.e. @ 1:00pm=R
Ans: C (1:00pm)
Answer: C
Question: The perimeter of a triangle is 32 cm and the inradius of the triangle is 2.5 cm. What is the area of the triangle?
Options:
A)76 cm2
B)56 cm2
C)18 cm2
D)98 cm2
E)40 cm2
Rationale: Area of a triangle = r * s
Where r is the inradius and s is the semi perimeter of the triangle.
Area of triangle
= 2.5 * 32/2
= 40 cm2
Answer:E
Answer: E
Question: 6 friends went to a hotel and decided to pay the bill amount equally. But 5 of them could pay Rs. 100 each as a result 6th has to pay Rs. 100 extra than his share. Find the amount paid by him.
Options:
A)156
B)220
C)130
D)240
E)None
Rationale: Explanation:
Average amount paid by 5 persons = Rs. 100
Increase in average due to Rs. 120 paid extra by the 6th men = Rs. 100/5= Rs. 20
Therefore, Average expenditure of 6 friends = Rs. 100 + Rs. 20 = Rs. 120
Therefore, Amount paid by the 6th men = Rs. 120+ Rs. 100 = Rs. 220
Correct Option: B
Answer: B
Question: What is x if x + 2y = 12 and y = 3?
Options:
A)A)10
B)B)8
C)C)6
D)D)4
E)E)2
Rationale: x = 12 - 2y
x = 12 - 6.
x = 6
Answer : C
Answer: C
Question: The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4,then the middle number is?
Options:
A)11.5
B)12.6
C)15.3
D)9.3
E)10.6
Rationale: Middle number = 10.5*6 + 11.4*6 - 10.9*11 = 131.4-119.9 = 11.5
Answer is A
Answer: A
Question: If the mode of Set Q is equal to the median of Set Q, Set Q could be
Options:
A)[0, 1, 2, 2, 3, 4, 5, 6]
B)[2, 2, 5, 7, 9]
C)[x, x+1, x+2, x+2, x+3, x+3]
D)[10, 102, 105, 105]
E)[4, 7, 10, 11, 25/2, 13, 13, 17, 29, 51, 51, 51]
Rationale: Correct Answer: (C)
Set Q could be The mode is the most common number; the median is the middle number when the numbers are arranged from least to greatest. Any number plus an integer value is greater than the original number, so x + 1 is greater than x, x + 2 is greater than x + 1, etc., so set (C) is arranged from least to greatest. Its median is the average of (x+2) and (x+2), which must be (x+2). (x+2) is also the most common number, so the mode of set C is equal to the median of set C.
Answer: C
Question: If 5<x<8<y<13, then what is the greatest possible positive integer difference of x and y?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: Let x=5.1 and y=12.1
Greatest possible difference=12.1-5.1=7
Answer E
Answer: E
Question: How many different 3 letter combinations can be created from the letters A,B and C?
Options:
A)81
B)3
C)9
D)27
E)243
Rationale: This is a case of number of arrangements possible with repetition.
Thus you have 3 options each for the 1st,2nd and 3rd letter giving you a total of 3*3*3 = 27 number of possible combinations.
Answer : D
Answer: D
Question: Today is Monday. A person wants to meet a lawyer and as that lawyer is busy he asks him to come three days after the before day of the day after tomorrow? On which day the lawyer asks the person to come?
Options:
A)SUNDAY
B)MONDAY
C)TUESDAY
D)WEDNESDAY
E)FRIDAY
Rationale: Today is Monday
after three days--->Thursday
now, (thursday)after before day--->Wednesday
then, (Wednesday)day after tomorrow--->Friday
ANSWER:E
Answer: E
Question: A farmer had d animals, all of which the farmer bought for the same price. The farmer decided to sell the animals. 16 of the animals were sold at a profit of 10% each while the rest were sold at a loss of 20% each. Overall, the farmer made neither a profit nor a loss. Which of the following is equal to d?
Options:
A)20
B)24
C)28
D)32
E)36
Rationale: Let P be the price the farmer paid for each animal.
The total profit on the 16 animals is 16*0.1*P.
The total loss on the other (d-16) animals is (d-16)*0.2*P.
(d-16)*0.2*P = 16*0.1*P
0.2*d = 1.6 + 3.2
d = 24
The answer is B.
Answer: B
Question: How many ways can six girls be selected from a group of ten girls?
Options:
A)⁶C₄
B)¹⁰C₄
C)4C₄
D)¹⁰P₄
E)¹⁰P₅
Rationale: Six members can be selected from ten members in
¹⁰C₆ = ¹⁰C₄ ways
B
Answer: B
Question: What is the sum of the multiples of 4 between 34 and 135 inclusive?
Options:
A)2000
B)2100
C)2150
D)2200
E)2500
Rationale: the fastest way in an AP is to find the average and multiply with total integers..
between 34 and 135, the smallest multiple of 4 is 36 and largest =132..
average=(36+132)/2=84..
total numbers = (132-36)/4 + 1= 24+1=25..
sum=25*84=2100
ans B
Answer: B
Question: A total of $20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 12 percent, respectively. If the total interest on the two certificates was $1,440 at the end of one year, what fractional part of the $20.000 was invested at the higher rate?
Options:
A) 3/8
B) 2/5
C) 1/5
D) 3/5
E) 3/4
Rationale: x * 6/100 * 1 + (20000 - x) * 12/100 * 1 = 1440
6x - 12x = 144000 - 240000
=> -6x = -96000
=> x = 16000
So 4000/20000 = 1/5
Answer - C
Answer: C
Question: How many seconds will a train 70 meters long take to cross a bridge 80 meters long if the speed of the train is 36 kmph?
Options:
A)22 sec
B)27 sec
C)25 sec
D)15 sec
E)11 sec
Rationale: Explanation:
D = 70 + 80 = 150
S = 36 * 5/18 = 10 mps
T = 150/10 = 15 sec
Answer: Option D
Answer: D
Question: What are the last two digits of (301*402*503*604*645*547*448*349)^2
Options:
A) 96
B) 76
C) 56
D) 36
E) 00
Rationale: ((301*402*503*604*645)*(547*448*349))^2
If you observe above digits, last digit are: 1,2,3,4,5,7,8,9; 6 is missing; so I have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01,02,03,04,45 and final three as 47*48*49.
Solving for only last two digits and multiplying them we get:( (06*04*45)(56*49))^2 = (00*44)^2=00^2=00
Hence answer is E
Answer: E
Question: What is the difference between the place values of two sevens in the numeral 54207479 ?
Options:
A)699990
B)99990
C)99980
D)699930
E)None of these
Rationale: Explanation:
Required Difference
= 7000 - 70 = 6930
Answer is E
Answer: E
Question: Find a two digit number which is doubled and after that add +2 in it. and then reverse the number we would get the actual number.
Options:
A)23
B)24
C)25
D)26
E)27
Rationale: Let the two digit number be 10x+y
Adding 2 after doubling the number is reverse of the number.
So 2(10x+y)+2 = 10y+x
19x -8y +2=0
Above equation satisfies for the value, x=2, y=5
So the number is 10*2+5= 25
ANSWER:C
Answer: C
Question: Ten cards numbered 1 to 10 are placed in a box and then one card is drawn randomly. If it is known that the number on the drawn card is more than 4, what is the probability that it is an even number?
Options:
A)1/2
B)1/5
C)1/7
D)1/9
E)1/10
Rationale: There are 6 numbers higher than 4 and 3 of them are even.
The probability is 3/6=1/2.
The answer is A.
Answer: A
Question: Running at the same constant rate, 6 identical machines can produce a total of 300 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
Options:
A)648
B)1800
C)2000
D)10800
E)10900
Rationale: Let the required number of bottles be x.
More machines, More bottles (Direct Proportion)
More minutes, More bottles (Direct Proportion)
Machines	6	:	10	:: 300 : x
Time (in minutes)	1	:	4
6 x 1 x x = 10 x 4 x 300
x =	(10 x 4 x 300)/(6)
x = 2000.
ANSWER:C
Answer: C
Question: If P represents the product of the first 16 positive integers, then P is NOT a multiple of
Options:
A)99
B)84
C)72
D)65
E)51
Rationale: Answer is E.
Since prime factor of 51 is 3x17. 17 is a prime number and not part of the first 15 positive integers.
A) 9 x 11
B) 12 x 7
C) 12 x 6
D) 13 x 5
E) 3 x 17
Answer: E
Question: A mixture of 50 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 15% water?
Options:
A)15
B)20
C)25
D)30
E)10
Rationale: Quantity of milk in the mixture = 90/100 (50) = 45 litres.
After adding water, milk would form 60% of the mixture.
Hence, if quantity of mixture after adding x liters of water, (60) / 100 x = 45 => x = 75
Hence 75 - 50 = 25 litres of water must be added.
Answer: C
Answer: C
Question: What is the value of 3x^2 − 1.8x + 0.3 for x = 0.3?
Options:
A)−0.3
B)0
C)0.03
D)1.08
E)2.46
Rationale: 3x^2-1.8x+0.3 for x=0.6
=3(0.3*0.3)-6*0.3*(0.3)+0.3
=-3*0.3*0.3+0.3
=-0.27+0.3
= 0.03
Correct Option : C
Answer: C
Question: How many positive three-digit integers are divisible by both 4 and 5?
Options:
A)40
B)42
C)45
D)50
E)60
Rationale: A number to be divisible by both 4 and 5 should be divisible by the least common multiple of 4 and 5 so by 20.
Multiples of 20 between 100 and 999, inclusive is
(last-first)/multiple+1=(980-100)/20+1=44+1=45
Answer: C.
Answer: C
Question: A can finish a piece of work in 5 days. B can do it in 10 days. They work together for two days and then A goes away. In how many days will B finish the work?
Options:
A)4
B)5
C)6
D)8
E)2
Rationale: 2/5 + (2 + x)/10 = 1 => x = 4 days
Answer: A
Answer: A
Question: If four machines working at the same rate can do 3/4 of a job in 30 minutes, how many minutes would it take two machines working at the same rate to do 3/5 of the job?
Options:
A)48
B)60
C)75
D)80
E)100
Rationale: using the std formula
m1d1h1/w1=m2d2h2/w2
substituting the values we have
4*1/2*4/3=2*5/3*x (converted 30 min into hours =1/2)
8/3=10/3*x
x=4/5 hour
so 48 minutes
ANSWER:A
Answer: A
Question: Pipe A can fill a tank in 32 minutes. Pipe B can fill the same tank 4 times faster than pipe A. If both the pipes are connected to the tank so that they fill the tank simultaneously, how long will it take for the empty tank to overflow?
Options:
A)16/3 minutes
B)32/7 minutes
C)192/7 minutes
D)224/7 minutes
E)28 minutes
Rationale: Pipe A can fill a tank in 32 minutes --> the rate of A = 1/32 tank/minute;
Pipe B can fill the same tank 4 times fasterthan pipe A --> the rate of B = 1/32 + 4/32 = 5/32 tank/minute.
Their combined rate = 1/32 + 5/32 = 3/16 tank/minute;
Therefore, A and B can fill the tank in 16/3 minutes.
Answer: A.
Answer: A
Question: An emploer pays Rs. 20 for each day a worker works, and forfeits Rs. 3 for each day he is idle. At the end of 60 days, a worker gets Rs. 280 . For how many days did the worker remain idle?
Options:
A)30 days
B)20 days
C)10 days
D)50 days
E)40 days
Rationale: Suppose the worker remained idle for x days .Then, he worked for (60-x)days
20(60-x)-3x=280
1200-23x=280
23x=920
x=40
so, the worker remained idle for 40 days
ANSWER: E
Answer: E
Question: Each person who attended a company meeting was either a stockholder in the company, an employee of the company, or both. If 57 percent of those who attended the meeting were stockholders and 51 percent were employees, what percent were stockholders who were not employees?
Options:
A)47
B)49
C)51
D)53
E)55
Rationale: 57+51=108 so 8% were both stockholders and employees.
The percent who were stockholders but not employees is 57-8=49%
The answer is B.
Answer: B
Question: A Bus is running at 9/10 of its own speed reached a place in 22 hours. How much time could be saved if the bus would run at its own speed ?
Options:
A)8.2 hrs
B)6.2 hrs
C)2.2 hrs
D)1.2 hrs
E)3.2 hrs
Rationale: Explanation:
Let the original Speed be "s" kmph
And the usual time be "t" hrs
Given that if the bus is running at 9s/10 kmph the time is 22 hrs
=> [9s/10] x 22 = t x s
=> 99/5 = t
=> t = 19.8 hrs
Answer: C
Answer: C
Question: If -1 ≤ a ≤ 2 and 1 ≤ b ≤ 3, then least possible value of (2a – 3b) is:
Options:
A)-11
B)-3
C)-4
D)-5
E)-6
Rationale: -1 ≤ a; so a = -1
b ≤ 3; so b = 3
Now, 2a – 3b = -2 - 9 = -11
Therefore, the least value is -11
ANSWER:A
Answer: A
Question: The average age of a group of 5 members is 20 years. Two years later, a new member joins the group. The average age of the group becomes 23 years. What is the age of the new member ?
Options:
A)20 years
B)21 years
C)28 years
D)23 years
E)24 years
Rationale: The average age of a group of 5 members is 20 years --> the sum of the ages is 5*20=100;
Two years laterthe sum of the ages of these 5 members would be 100+5*2=110;
Now, say the age of the new member is x years, so the sum of the ages of a new 6 member group is 110+x. Since given that the average age of this group of 6 members is 23 years then: 23*6=110+x --> x=28.
Answer: C.
Answer: C
Question: The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?
Options:
A)10 cm
B)40 cm
C)20 cm
D)30 cm
E)50 cm
Rationale: Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is \fn_jvn {\color{Black} 2x^{2}}
after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=\fn_jvn {\color{Black} 2x^{2}+5x-25}
since new area is 75 units greater than original area thus
\fn_jvn {\color{Black}2x^{2}+75= 2x^{2}+5x-25}
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.
Answer: B
Answer: B
Question: The cost price of 20 books is the same as the selling price of ’X’ books. If the profit is 50%, then the value of X is ?
Options:
A)25
B)18
C)16
D)15
E)6.7
Rationale: let S.P. of X books = Rs 100 = C.P. of 20 books
Therefore, C.P. of one book = Rs 10
Profit = 50 %
Therefore, S.P of one book = 15
S.P of X articles = 15 X
15 X = 100 => X = 6.7
ANSWER:E
Answer: E
Question: For any even integer P, 300 multiplied by P is square of an integer. What is the least value of P?
Options:
A)3
B)4
C)10
D)12
E)14
Rationale: p∗3∗10^2=s^2
So, s=√p∗3∗10^2
So, if p = 3 we get a perfect square number !!
Hence answer will be (A) 3
Answer: A
Question: 3^13 + 3^13 + 3^13 =
Options:
A)3^11
B)3^13
C)3^14
D)9^15
E)9^30
Rationale: Since we have 3^13 added together 3 times, we can factor out a 3 and we get 3(3^13) which is equal to 3^14.
The answer is C.
Answer: C
Question: If the annual increase in the population of a town is 10% and the present number of people is 12000, what will the population be in 2 years?
Options:
A)12100
B)15240
C)12456
D)11452
E)14520
Rationale: The required population is = 12000 (1 + 10/100)^2
= 12000 * 11/10 * 11/10
= 14520
Answer is E
Answer: E
Question: Bil and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Bil and Jill will both be chosen?
Options:
A)A)1/3
B)B)1/4
C)C)1/15
D)D)3/8
E)E)2/3
Rationale: Total number of people = 6
Probability of selecting Bil first and then Jill is 1/6∗1/5=1/30
Probability of selecting Jill first and then Bil is 1/6∗1/5=1/30
Therefore probability of selecting Bil and Jill for the review is 1/30+1/30=1/15
Answer is C
Answer: C
Question: 10 friends went to a hotel and decided to pay the bill amount equally. But 9 of them could pay Rs. 50 each as a result 10th has to pay Rs. 180 extra than his share. Find the amount paid by him.
Options:
A)130
B)140
C)110
D)120
E)None
Rationale: Explanation:
Average amount paid by 9 persons = Rs. 50
Increase in average due to Rs. 50 paid extra by the 10th men = Rs. 180/9= Rs. 20
Therefore, Average expenditure of 10 friends = Rs. 50 + Rs. 20 = Rs. 70
Therefore, Amount paid by the 10th men = Rs. 70 + Rs. 50 = Rs. 120
Correct Option: D
Answer: D
Question: Four people each roll a die once. Find the probability that at least two people will roll the same number?
Options:
A)3/18
B)13/18
C)14/18
D)15/18
E)16/18
Rationale: atleast 2 means
1-p(no one has same number)
p(no person has same number)=6*5*4*3/6^4=10/36
reqd probability=1-10/36
=26/36=13/18
ANSWER:B
Answer: B
Question: The average of first 20 natural numbers is?
Options:
A)5.6
B)10.5
C)10.1
D)5.8
E)5.2
Rationale: Sum of 20 natural no.
= 420/2 = 210
Average = 210/20
= 10.5
Answer:B
Answer: B
Question: What approximate value should come in place of question mark (?) in the following equation
287.532 + 1894.029 – 857.48 = 743.095 + ?
Options:
A)870
B)790
C)580
D)770
E)890
Rationale: ? = 285 + 1895 – 855 – 745 or, ? = 580
Answer C
Answer: C
Question: A train 260 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
Options:
A)40 sec
B)50 sec
C)32 sec
D)49 sec
E)60 sec
Rationale: Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 260 + 140 = 400 m
Required time = 400 * 2/25 = 32 sec
Answer:C
Answer: C
Question: Last year, for every 100 million vehicles that traveled on a certain highway, 40 vehicles were involved in accidents. If 2 billion vehicles traveled on the highway last year, how many of those vehicles were involved in accidents? (1 billion = 1,000,000,000)
Options:
A)800
B)900
C)100
D)1000
E)950
Rationale: To solve we will set up a proportion. We know that “100 million vehicles is to 40 accidents as 2 billion vehicles is to x accidents”. To express everything in terms of “millions”, we can use 2,000 million rather than 2 billion. Creating a proportion we have:
100/40 = 2,000/x
Cross multiplying gives us:
100x = 2,000 * 40
x = 20 * 40 = 800
Answer : A
Answer: A
Question: A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?
Options:
A)237
B)729
C)277
D)166
E)181
Rationale: Choose 5 starters from a team of 12 players. Order is not important.
12C_{5}}= 729
Answer: B
Answer: B
Question: For which of the following functions is f(d) = f(−d) for all values of d ?
Options:
A)f(x) = x^3 + 3
B)f(x) = −x
C)f(x) = 2x + 3
D)f(d) = −d^2 + 2
E)f(x) = 5x − 4
Rationale: f(d) = f(−d)
The above is a property of an even function .
f(d) = f^(n)
f(d) = f(−d) will hold when n is an even integer .
The only case when d is even is option D
Answer D
Answer: D
Question: Find Nikhil's present age, if
I. Five years ago, Nikhil's age was double that of his daughter's age at that time.
II. Present ages of Nikhil and his daughter are in the ratio of 11:6 respectively
III. Five years hence, the respective ratio of Nikhil's age and his daughter age will become 12:7
Options:
A)Any two of the three
B)8
C)7
D)6
E)4
Rationale: Explanation:
1) Present ages of Nikhil and his daughter will be 11x and 6x years respectively.
2) 5 years ago, Nikhil's age = 2 * His daughter's age
3) 5 years hence, Nikhil's age / daughter's age = 12/7
Clearly, any two of the above will give Nikhil's present age
Therefore, correct answer is (d).
ANSWER: A
Answer: A
Question: 4 male and 6 female can cultivate a field in 8 days, 7 male and 3 female can cultivate a field in 10 days. then how many days will take to cultivate a field if 10 female works?
Options:
A)120/19
B)130/19
C)140/19
D)120/18
E)120/17
Rationale: 4 male+6female in 8 days=> 32 male+48 female in 1day
7male+ 3 female in 10 days =>70 male+30 female in 1 day
.:32 male+48 female=70 male+30 female
18 female=38 male
1 female= 38/18 male=19/9 male
now,4 male+6 female=4 male+6*(19/9)=50/3 male=150/9 male
we have to find for 10 female workers= 10*(19/9)=190/9 male
more men ,fewer days wud be req to finish work
male days
150/9 8
190/9 x
so, x/8=(150/9)/(190/9)
x=120/19
ANSWER:A
Answer: A
Question: if x:y = 2:3, then find the value of
(7x+3y):(2x+y)
Options:
A)23:7
B)15:5
C)16:5
D)17:5
E)None of these
Rationale: Explanation:
let x = 2k and y = 3k, so
=7(2k)+3(3k)/ 2(2k)+1(3k)
=23k/7k
=23:7
Option A
Answer: A
Question: There are 50 boys in a class. Their average weight is 45 kg. When one boy leaves the class, the average reduces by 100 g. Find the weight of the boy who left the class.
Options:
A)40.9 kg
B)42.9 kg
C)49.9 kg
D)39.9 kg
E)None of these
Rationale: Here one boy is excluded and final average of the group decreases.
∴ change in average is (–)ve = – 0.1 kg.
Using the formula
Sum of the quantities excluded
=(Changein no.ofquantities×Origina lAverage)+(ChangeinAverage×Final no.ofquantities)
⇒ weight of the boy who left = (1 × 45) – (– 0.1 × 49)
= 49.9 kg
Answer C
Answer: C
Question: After a storm deposits 120 billion gallons of water into the city reservoir, the reservoir is 50% full. If the original contents of the reservoir totaled 280 billion gallons, the reservoir was approximately what percentage full before the storm?
Options:
A)45%
B)48%
C)54%
D)58%
E)65%
Rationale: When the storm deposited 115 billion gallons, volume of water in the reservoir = 280 + 120 = 400 billion gallons
If this is only 80% of the capacity of the reservoir, the total capacity of the reservoir = 400/0.5 = 800 billion gallons
Therefore percentage of reservoir that was full before the storm = (280/800)*100 = 35%
Option C
Answer: C
Question: A envelop weight 8.5 gm,if 880 of these envelop are sent with an advertisement mail. How much wieght?
Options:
A)6.6kg
B)6.8kg
C)7.48kg
D)6.9kg
E)7.8kg
Rationale: 880*8.5
7480.0gm
7.48kg
ANSWER:C
Answer: C
Question: The product X of two prime numbers is between 15 and 90. If one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25, then X =
Options:
A)18
B)29
C)37
D)44
E)85
Rationale: Option BC can be ruled out as they themselves are prime numbers
18 = 2*9 = 3*6 >> Ignore
44 = 2 * 22 = 4 * 11 >> Ignore
85 = 5*17 >> Answer
Answer = E
Answer: E
Question: How many integers from 0 to 50 inclusive have a remainder of 3 when divided by 6?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: The numbers should be of the form 6c+3.
The minimum is 3 when c=0.
The maximum is 45 when c=7.
There are 8 such numbers.
The answer is D.
Answer: D
Question: If w is divisible by 15 and 9, w must be a multiple of which of the following?
Options:
A)12
B)18
C)25
D)36
E)45
Rationale: If w is divisible by 3*5 and 3*3, then w is divisible by 3*3*5 = 45.
The answer is E.
Answer: E
Question: A department of nine people - four men and five women - needs to send a team of five to a conference. If they want to make sure that there are no more than three members of the team from any one gender, how many distinct groups are possible to send?
Options:
A)100
B)120
C)150
D)180
E)200
Rationale: They can make a team of 3 men and 2 women.
The number of ways to do this is 4C3*5C2=4*10 = 40
They can make a team of 2 men and 3 women.
The number of ways to do this is 4C2*5C3=6*10 = 60
The total number of distinct groups is 100.
The answer is A.
Answer: A
Question: If m, n and r are positive integers greater than 1, and p and q are factors of r, which of the following must be the factor of r^(mn)?
I. m+n
II. n^m
III. m^2 * n^2
Options:
A)I only
B)II only
C)III only
D)I and II
E)II and III
Rationale: given that m and n are factors of r.
so we can picture it this way r=m*n*n (n-some another factor of r)
so, r^mn= (m*n*n)^mn
I. m+n .since the question is MUST BE TRUE, we eleminate this option
II. (m*n*n)^mn / n^m= integer YES!
III.(m*n*n)^mn/ m^2 * n^2 YEs, since we are said that integer m>1 and integer n>1
Answer: E
Question: A number when successively divided by 5, 3, 2 gives remainder 0, 2, 1 respectively in that order. What will be the remainder when the same number is divided successively by 2, 3, 5 in that order
Options:
A)22
B)87
C)26
D)25
E)08
Rationale: use this simple technique.[ (1 x 3) + 2] = 5
[(5 x 5) + 0] = 25
Answer:D
Answer: D
Question: How long does a train 110 m long running at the speed of 72 km/hr takes to cross a bridge 132 m length?
Options:
A)13.9 sec
B)12.1 sec
C)17.9 sec
D)61.9 sec
E)47.98 sec
Rationale: Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec.
Answer:B
Answer: B
Question: The cost price of 13 articles is equal to the selling price of 11 articles. Find the profit percent?
Options:
A)18 2/19%
B)18 2/11%
C)18 2/41%
D)18 4/11%
E)48 2/11%
Rationale: 13 CP = 11 SP
11 --- 2 CP
100 --- ? =>18 2/11%
Answer: B
Answer: B