divelab/Sys2Bench huggingface dataset:
Question: If integers x and b are distinct factors of 30, which of the following CANNOT be a factor of 30?
I. xb + b^2
II. (x + b)^2
III. x + b
Options:
A)I only
B)II only
C)III only
D)I and II only
E)I, II, and III
Rationale: Factors of 30 are: 1, 2, 3, 5, 6, 10, 15, and 30.
I. xb + b^2 --> if x=1 and b=2 then xb + b^2=6, which is a factor of 30. OK;
II. (x + b)^2 --> (x + b)^2 is a perfect square, no perfect square (but 1) is a factor of 30. DISCARD;
III. x + b --> if x=1 and b=2 then x + b=3, which is a factor of 30. OK.
Answer: B.
Answer: B
Question: If r and s are negative integers, which of the following statements must be true?
I. (-r)^s is positive.
II. (-r)^(-s) is positive.
III. r^(-s) is positive.
Options:
A)None
B)II only
C)I and II
D)I and III
E)I, II and III
Rationale: Let's r = -2 and s = -3
I. (-r)^s = (2)^(-3) = 1/(2^3) --> Positive
II. (-r)^(-s) is positive. = 2^3 --> Positive
III. r^(-s) is positive. = -2^3 --> Negative
Answer: C
Answer: C
Question: A boat can travel with a speed of 24 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 168 km downstream
Options:
A)3 hours
B)4 hours
C)5 hours
D)6 hours
E)7 hours
Rationale: Explanation:
Speed of the boat in still water = 24 km/hr
speed of the stream = 4 km/hr
Speed downstream = (22+5) = 28 km/hr
Distance travelled downstream = 168 km
Time taken = distance/speed = 168/28 = 6 hours
Answer: Option D
Answer: D
Question: A portion of the 85% solution of chemicals was replaced with an equal amount of 30% solution of chemicals. As a result, 40% solution of chemicals resulted. What part of the original solution was replaced?
Options:
A)5/11
B)6/11
C)7/11
D)9/11
E)13/6
Rationale: This is a weighted average question. Say x% of the solution was replaced --> equate the amount of chemicals: 0.85(1-x)+0.3*x=0.4 --> x=9/11.
Answer: D.
Answer: D
Question: Three friends Dhoni, Rahul and Aswin divide $1540 amongs them in such a way that if $10, $12 and $18 are removed from the sums that Dhoni, Rahul and Aswin received respectively, then the share of the sums that they got will be in the ratio of 15: 17: 18. How much did Aswin receive?
Options:
A)$560
B)$562
C)$564
D)$558
E)$570
Rationale: A+B+C = 1540
Given ratio 15:17:18
Let us say the shares of A,B,C deducting 10,12,18 be a,b,c
a+b+c = 1540-40= 1500 = 50K
c share = (1500X18)/50 = 540
C = Aswin share = 540+18= 558
Option D
Answer: D
Question: Every student at Tarcy School is in at least one of three clubs: horseback riding, embroidery, and country dancing, which are the only clubs in existence at the school. The ratio of the number of students in exactly two clubs to the number of students in exactly one club is 4:3, while the ratio of the number of students in exactly two clubs to the number of students in at least two clubs is 5:7. Which of the following could be the total number of students at Tarcy School?
Options:
A)63
B)69
C)74
D)82
E)86
Rationale: the total number of students at Tarcy School The answer is 86 or any multiple of 43.E
Answer: E
Question: Suresh started a business with Rs.20,000. Kiran joined him after 4 months with Rs.30,000. After 2 more months, Suresh withdrew Rs.5,000 of his capital and 2 more months later, Kiran brought in Rs.20,000 more. What should be the ratio in which they should share their profits at the end of the year?
Options:
A)32:21
B)12:17
C)17:12
D)21:32
E)15:19
Rationale: Here capital is not the same.
Suresh invested 20000 for initial 6 months and 15000 for the next 6 months. Hence his term of ratio
=(20000×6+15000×6)
Kiran invested Rs.30000 for 4 months and Rs.50000 for next 4 months. Hence his term of ratio
=(30000×4+50000×4)
Suresh : Kiran
=(20000×6+15000×6) :(30000×4+50000×4)
=(20×6+15×6):(30×4+50×4)
=(20×3+15×3):(30×2:50×2)
=105:160
=21:32
Answer is D.
Answer: D
Question: A trained covered x km at 40 kmph and another 2x km at 20 kmph. Find the average speed of the train in covering the entire 3x km?
Options:
A)43 kmph
B)65 kmph
C)24 kmph
D)14 kmph
E)61 kmph
Rationale: Total time taken = x/40 + 2x/20 hours
= 5x/40 = x/8 hours
Average speed = 3x/(x/8) = 24 kmph
Answer:C
Answer: C
Question: I chose a number and divide it by 2. Then I subtracted 100 from the result and got 4. What was the number I chose?
Options:
A)200
B)208
C)400
D)500
E)100
Rationale: Let x be the number I chose, then
x/2−100=4
x/2=104
x=208
Answer is B.
Answer: B
Question: Jack and Jill work at a hospital with 7 other workers. For an internal review, 2 of the 9 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?
Options:
A) 1/3
B) 1/4
C) 1/15
D) 3/8
E) 1/36
Rationale: 1/9C2=1/36.
Answer: E.
Answer: E
Question: A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is
Options:
A)83.33 ml
B)83.03 ml
C)93.33 ml
D)63.03 ml
E)82.33 ml
Rationale: Explanation:
Profit (%) = 9.09 % = \inline \frac{1}{11}
Since the ratio of water and milk is 1 : 11,
Therefore the ratio of water is to mixture = 1:12
Thus the quantity of water in mixture of 1 liter = \inline 1000\times \frac{1}{12} = 83.33 ml]
Answer: A) 83.33 ml
Answer: A
Question: If the LCM of two integers c, d (where d> c and c>1) is c*d, then which of the following can not be true?
Options:
A)Both c and d can be consecutive integers.
B)Both c and d can be Prime Numbers.
C)All Prime factors of c must be Prime factors of d.
D)c and d do not have a common factor.
E)c and d do not share any Prime factors.
Rationale: If LCM of two numbers C and D is the product CD then both numbers are Co- prime.
(Co-prime numbers are any two numbers which have an HCF of 1 i.e. when two numbers have no common prime factor apart from the number 1).
Numbers that can be Co-Prime are :
1) Two consecutive natural numbers.
2) Two consecutive odd numbers.
3) Two prime numbers.
4) One prime number and the other a composite number such that the composite number is not a multiple of the prime number.
With above four statements it can be concluded that Options A, B, D and E are all true.
Only Option C cannot be true.
Hence answer is C
Answer: C
Question: Vishal invested 10% more than Trishul. Trishul invested 10% less than Raghu. If the total sum of their investments is Rs. 6647, how much amount did Raghu invest ?
Options:
A)2300
B)2887
C)2000
D)1129
E)1192
Rationale: Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 6647
= x= 6647/2.89 = 2300
Therefore, amount invested by Raghu is Rs. 2300.
Answer: A
Answer: A
Question: At what time between 9 and 10 o’clock will the hands of a watch be together?
Options:
A)49 1/11 min. past 9
B)40 min. past 9
C)45 min. past 9
D)50 min. past 9
E)None
Rationale: Sol.
To be together between 9 and 10 o’clock, the minute hand has to gain 45 min. spaces. 55 min. spaces gined in 60 min.
45 min. spaces are gained in [60 / 55 * 45] min. or 49 1/11 min.
∴ The hands are together at 49 1/11 min. past 9.
Answer A
Answer: A
Question: A sum of Rs. 427 is to be divided among A, B and C such that 3 times A’s share, 4 tunes B’s share and 7 times C’s share are all equal. The share of C is :
Options:
A)84
B)140
C)140
D)240
E)None of these
Rationale: Explanation:
3A = 4B = 7C = k,Then A = k/3, B = k/4 and C= k/7.
A : B : C = k/3 : k/4 : k/7 = 28:21 :12.
Cs share = Rs. [427 x (12/61)] = Rs. 84
Answer: A
Answer: A
Question: The weights of one liter vegetable ghee packet of two brands ‘A’ and ‘B’ are 900 gm and 850 gm respectively. If they are mixed in the ratio of 3 : 2 by volumes to form a mixture of 4 liters, what is the weight (in kg) of the mixture?
Options:
A)3.84
B)1.75
C)3.52
D)2.72
E)None of these
Rationale: 1L A = 900 gm
1L B = 850 gm
We are mixing five parts (3 parts A plus 2 parts B, 5 parts total) to get 4L, so 5x = 4 ---> x = 4/5. Each "part" is 4/5 of a liter.
So if we have 3 parts A, we have 900*3*(4/5) = 2160
If we have 2 parts B, we have 850*2*(4/5) = 1360
2160 + 1360 = 3520
Solving for units gives us 3.52
So the answer is C
Answer: C
Question: What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 10%?
Options:
A)20%
B)35%
C)13 1/30%
D)12%
E)13%
Rationale: SP2 = 2/3 SP1
CP = 100
SP2 = 90
2/3 SP1 = 90
SP1 = 135
100 --- 35 => 35%
ANSWER:B
Answer: B
Question: While planning their outing, Abhishek understood that their boat could travel with a speed of 12kmph in still water.The group decided to travel 24 minutes downstream. Aarti had prior experience in that route and mentioned that the speed of stream was 4kmph.How many kms will the group travel downstream?
Options:
A)6.4
B)3.2
C)4
D)4.8
E)4.6
Rationale: downstream speed = speed of the boat + speed of the river
12+4 = 16
distance travelled = 16x24/60 = 6.4
ANSWER:A
Answer: A
Question: What will come in place of the x in the following Number series? 46080, 3840, 384, ?, 8, 2, 1
Options:
A)1
B)3
C)5
D)48
E)9
Rationale: 46080 /12 = 3840
3840 /10 = 384
384 /8 = 48
48 /6 = 8
8/4 = 2
2 /2 = 1
D
Answer: D
Question: What is the solution of the equations x - y = 0.08 and 1(x + y)-1 = 2 ?
Options:
A)x = 3.2, y = 2.3
B)x = 1, y = 0.1
C)x = 2, y = 1.1
D)x = 0.065, y = 0.735
E)None
Rationale: Answer
x - y = 0.8 ...(i)
and 1(x + y)-1=2
⇒ 1/ (x + y) = 2
⇒ 2(x + y) =1
⇒ x + y = 1/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 0.065
and y = 0.735
Correct Option: D
Answer: D
Question: John purchased a grinder and a mobile for Rs. 15000 &Rs. 10000 respectively. He sold the grinder at a loss of 4% and the mobile phone at a profit of 10%. Overall how much he make a profit.
Options:
A)s. 400
B)s. 120
C)s. 420
D)s. 450
E)s. 290
Rationale: Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 - 4/100) = 15000 - 600
m = 10000(1 + 10/100) = 10000 + 1000
Total SP - Total CP = r + m - (15000 + 10000) = -600 + 1000 = Rs. 400
As this is positive, an overall profit of Rs. 400 was made.
A
Answer: A
Question: A survey of n people in the town of Eros found that 50% of them preferred Brand A. Another survey of 110 people in the town of Angie found that 60% preferred Brand A. In total, 55% of all the people surveyed together preferred Brand A. What is the total number of people surveyed?
Options:
A)50
B)100
C)150
D)200
E)220
Rationale: It is simply a weighted average question. Since the given average of 50% and 60% is 55% (right in the middle), it means the number of people surveyed in Eros (n) is same as the number of people surveyed in Angie.
So n = 110
Total = 110 + 110 = 220
Answer (E)
Answer: E
Question: There are A girls in a class, and their average height is M inches. In the same class, there are Y boys with an average height of N inches. What is the average height of all the students in the class?
Options:
A)(X + Y)/(M + N)
B)(M + N)/(X + Y)
C)(XM + YN)/(M + N)
D)(AM + YN)/(A + Y)
E)MN/XY
Rationale: Total height = AM+YN
No of students = X+Y
Avg = (AM+YN)/(A+Y)
Answer is D
Answer: D
Question: Sum of the squares of three numbers is 179 and the sum of their products taken two at a time is 131. Find the sum?
Options:
A)20
B)22
C)26
D)28
E)29
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 179 + 2* 131
a + b + c = √441 = 21
B
Answer: B
Question: There are 6 ladies and 5 gents to dine at a round table. In how many ways can they seat themselves so that no two gents are seated together.
Options:
A)85400
B)86400
C)36400
D)65600
E)88600
Rationale: First arrange the ladies. This is a circular arrangment. 6 Gents can be arranged in 5! ways.
Now there are 6 spaces (space in between two ladies) and we have to seat 5 gents there.
First select 5 places out of 6 available places. can be done in 6C5 ways. Then arrange 5 gents there in 5! ways.
Total: 5!*6C5*5! = 86400
B is the answer.
Answer: B
Question: In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
Options:
A)3
B)9
C)3^5
D)5^3
E)2^4
Rationale: The first letter can be posted in any of the 3 post boxes. Therefore, we have 3 possibilities.
Similarly, the second, the third, the fourth and the fifth letter can each be posted in any of the 3 post boxes.
Each of the 5 letters has 3 possibilities because we can post any number of letters in all of the boxes.
Therefore, the total number of ways the 5 letters can be posted in 3 boxes is
3 * 3 * 3 * 3 * 3 = 3^5
Ans:C
Answer: C
Question: In a class, there are 15 boys who are over 160 cm tall. If these constitute three-fourths of the boys and the total number of boys is two-thirds of the total number of students in the class, what is the number of girls in the class ?
Options:
A)10
B)12
C)18
D)24
E)26
Rationale: Explanation:
Let the number of boys be x. Then, (3/4)x = 15 or x = 15 x(4/3) = 20.
If total number of students is y, then (2/3) y = 20 or y = 20 x (3/2) = 30.
Therefore Number of girls in the class = (30 - 20) = 10.
Answer: A
Answer: A
Question: The average earning of amechanic for the first 4days of a week is Rs.60 and for the last four days is Rs.22. If heearns Rs.20 on the fourth day, his average earning forthe whole week is?
Options:
A)s. 10
B)s. 44
C)s. 30
D)s. 40
E)s. 50
Rationale: Total earning for the week = Sum of earning of first four days + Sum of earning of last four days - Earning of 4th day
= 4 x 60 + 4 x 22 -20
= Rs. 308
âˆ´ Average earning = 308/ 7
= Rs. 44
B
Answer: B
Question: What sum of money will produce Rs.70 as simple interest in 4 years at 3 1/2 percent?
Options:
A)Rs.525
B)Rs.500
C)Rs.550
D)Rs.555
E)Rs.545
Rationale: 70 = (P*4*7/2)/100
P = 500
Answer:B
Answer: B
Question: Roberto has three children: two girls and a boy. All were born on the same date in different years. The sum of the ages of the two girls today is smaller than the age of the boy today, but a year from now the sum of the ages of the girls will equal the age of the boy. Two years from today, the difference between the age of the boy and the combined ages of the girls will be
Options:
A)1
B)2
C)3
D)–2
E)–1
Rationale: Approach I (Plugin's)
Girl I ............... Girl II ................ Boy
1 ....................... 1 ........................ 3 (Assume the current ages)
1 + 1 < 3 .......... (Satisfies the given condition)
1 Year later there ages are
2 ....................... 2 ......................... 4
2 + 2 = 4 ............ (Satisfies the given condition)
After 2 years there ages are
3 ....................... 3 ............................ 5
Difference = 5 - (3+3) = 5 - 6 = -1
Answer = E
Answer: E
Question: A man can row 9 kmph in still water. When the river is running at 3.1 kmph, it takes him 1 hour to row to a place and black. What is the total distance traveled by the man?
Options:
A)2.21
B)2.48
C)9.24
D)7.29
E)7.94
Rationale: M = 9
S = 3.1
DS = 12.1
US = 5.9
x/12.1 + x/5.9 = 1
x = 3.97
D = 3.97 * 2 = 7.94
Answer: E
Answer: E
Question: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 80 percent of books that were loaned out are returned and there are 69 books in the special collection at that time, how many books of the special collection were loaned out during that month?
Options:
A)20
B)30
C)35
D)40
E)55
Rationale: there are 6 books less(75-69) which represents 20% of the loaned books (100-80)
So total loaned out books = 30
Answer B
Answer: B
Question: Let y = 2m + x^2 and m = 5x + 5. If m^2 =100, then which of the following can be a value of 2y + 3m?
Options:
A)83
B)63
C)93
D)73
E)53
Rationale: m^2 = 100 => m=10
10= 5x+5 => x=1
y= 2*10+1^2 = 21
2y+3m = 2*21+3*10 = 93
Answer : C
Answer: C
Question: In a simultaneous throw of 2 dice what is the probability of getting a total of 3?
Options:
A)1/4
B)1/18
C)2/11
D)3/13
E)1/5
Rationale: n(S) = 6*6 = 36
E = {(1,2),(2,1)}
Probability = 2/36 = 1/18
Answer is B
Answer: B
Question: Working alone at its constant rate, machine A produces x boxes in 10 minutes and working alone at its constant rate, machine B produces x boxes in 5 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce 3x boxes?
Options:
A)3 minutes
B)4 minutes
C)10 minutes
D)6 minutes
E)12 minutes
Rationale: Rate = Work / Time
Given Rate of Machine A = X / 10 min
Machine B Produces 2x boxes in 5 min hence , Machine B produces 4x boxes in 10 min .
Rate of Machine B = 2x / 10
we need tofind the combined time that machines A and B, working simultaneouslytakeat their respective constant rates
let's first find the combined Rate of Machine A and B
Rate of Machine A = X / 10 min + Rate of Machine B = 2x / 10 = 3X/10
Now combine Time = combine work needs to be done / Combine Rate = 3x/3x * 10 = 10 Min
Ans: C
Answer: C
Question: If y exceeds x by 20%, then x is less than y by?
Options:
A)16 2/9%
B)16 3/3%
C)16 2/3%
D)12 2/3%
E)16 1/3%
Rationale: X=100	y=120
120------20
100-------? => 16 2/3%
Answer:C
Answer: C
Question: The respective ages of father and his son are 41 and 16 years. In how many years will the father be twice as old as his son ?
Options:
A)19 years
B)9 years
C)10 years
D)15 years
E)12 years
Rationale: Suppose x years later the father will be twice as old as his son.
x + 41 = 2 (x + 16)
x = 41 - 32 = 9 years
answer : B
Answer: B
Question: If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours day, bow much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?
Options:
A)24 metric tonnes
B)36 metric tonnes
C)14 metric tonnes
D)43 metric tonnes
E)26 metric tonnes
Rationale: Let 3 engines of former type consume 1 unit in 1 hour.
Then, 4 engines of latter type consume 1 unit in 1 hour.
Therefore 1 engine of former type consumes(1/3) unit in 1 hour.
1 engine of latter type consumes(1/4) unit in 1 hour.
Let the required consumption of coal be x units.
Less engines, Less coal consumed (direct proportion)
More working hours, More coal consumed (direct proportion)
Less rate of consumption, Less coal consumed(direct prportion)
Number of engines 9: 8
Working hours	8 : 13 } :: 24 : x
Rate of consumption (1/3):(1/4)
[ 9 x 8 x (1/3) x x) = (8 x 13 x (1/4) x 24 )
24x = 624
x = 26.
Hence, the required consumption of coal = 26 metric tonnes.
Answer is E.
Answer: E
Question: The probability of having a girl is identical to the probability of having a boy. In a family with six children, what is the probability that all the children are of the same gender?
Options:
A)1/32
B)1/9
C)1/16
D)1/25
E)1/2
Rationale: There are two cases. All boys and all girls
case 1. all boys.
1/2*1/2*1/2*1/2*1/2*1/2= 1/64
case 2 .all girls.
1/2*1/2*1/2*1/2*1/2*1/2=1/64
We will add the cases to get 1/32
Answer: A
Question: Vijay bought 160 shirts at the rate of Rs. 225 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%?
Options:
A)Rs. 258
B)Rs. 270
C)Rs. 277.5
D)Rs. 285
E)None of these
Rationale: Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 20)/100] = Rs. 285.
ANSWER:D
Answer: D
Question: 3897 x 999 = ?
Options:
A)3883203
B)3893103
C)3639403
D)3791203
E)None of them
Rationale: =3897 x 999
=3897 x (1000 - 1)
=3897 x 1000 - 3897 x 1
=3897000 - 3897
=3893103
Answer is B
Answer: B
Question: A can complete a task in 12 days. B can complete the task in 18 days. If A and B work on this same task in alternate days starting with A, in how many days do they finish the entire task?
Options:
A)10.8 days
B)14.33 days
C)11 days
D)8.4 days
E)9.6 days
Rationale: DETAILED SOLUTION
A can complete a task in 12 days and B can complete the task in 18 days. So,
In one day, A can do 1/12, in one day B can do 1/18.
In two days, they can complete 1/12 + 1/18 = which is 5/36.
In 4 days, they can finish 10/36, and so on.
We can have 7 such sets of 2days each.
So, at the end of 14 days, they would have done 35/36 of task.
On the 15th day A would begin work with 1/36of the task to finish. He can finish 1/12 in a day. So, he would take one-third of a day. So, they can finish the whole task in 14.33 days.
CORRECT ANSWER B
Answer: B
Question: How many figures are required for numbering the pages of a book containing 1000 pages?
Options:
A)6489
B)3892
C)2893
D)2883
E)2983
Rationale: 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 999 = 900 * 3 =2700
1000 = 4
-----------
2893
ANSWER:C
Answer: C
Question: The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is:
Options:
A)4
B)6
C)8
D)10
E)12
Rationale: Sol.
P.W. = Rs. (1760 - 160) = Rs. 1600
∴ S.I. on Rs.1600 at 12% is Rs. 160.
∴ Time = [100 * 160 / 1600 * 12] = 5/6 years = [5/6 * 12] months = 10 months.
Answer D
Answer: D
Question: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 8, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
Options:
A)I only
B)II only
C)III only
D)I and III only
E)I, II, and III
Rationale: C
for 7 we will have mean as 7.16(approx) and median as 7 but for 3 we will have mean and median = 6.5 and for 12 we will have median and mean =8
Answer: C
Question: Tough and Tricky questions: Exponents.
If 5^(x+1)*4^(y-1) = 25^x*64^y, then x + y =
Options:
A)3/2
B)-1/2
C)-1
D)1/2
E)2/3
Rationale: Here is my solution.
5^(x+1)*4^(y-1) = 25^x*64^y
Here RHS 25^x*64^y= 5^(2x)*4^(3y)
Equating powers on both sides-->
x+1=2x , thus x=1 and
2y-1=3y giving y=-1/2
So, x+y=1/2
Option: D
Answer: D
Question: A cave is going to explode. A family is trapped inside that cave. The only way to get outside is a narrow tunnel. But the tunnel is weak and thus only two people can go at one time. Also, everybody is afraid of going alone.
The father takes one minute to go through till the end, the mother takes two, the son takes four and the daughter takes five. The cave and tunnel will explode in twelve minutes. Can they make it to the other end in that time frame?
PS: Keep in mind that when two people are moving, the total time taken will be as per the slower one.
Options:
A)10 minutes
B)12 minutes
C)18 minutes
D)19 minutes
E)16 minutes
Rationale: B
12 minutes
Explanation:
Yes, surely they can all make it out of the cave.
First, father and mother will go. Time taken = 2 minutes.
Father comes back. Time taken = 2 + 1 = 3 minutes.
Both children go. Time taken = 3 + 5 = 8 minutes.
Mother comes back. Time taken = 8 + 2 = 10 minutes.
Mother and Father go. Time taken = 10 + 2 = 12 minutes.
Thus in 12 minutes, all of them will be out of the cave through the tunnel.
Answer: B
Question: 11 + 12 + 13 + ... 51 + 52 + 53 = ?
Options:
A)1361
B)1376
C)1363
D)1364
E)1365
Rationale: Sum = 11 + 12 + 13 + ... 51 + 52 + 53
Sum of n consecutive positive integers starting from 1 is given as n(n+1)/2
Sum of first 53 positive integers = 53*54/2
Sum of first 10 positive integers = 11*10/2
Sum = 11 + 12 + 13 + ... 51 + 52 + 53 = 53*54/2 - 11*10/2 = 1376
Answer : B
Answer: B
Question: Z is a set of positive numbers. The median of Z is greater than the mean of Z. Which of the following Q must be true?
I. At least 50% of the numbers in Z are smaller than the median.
II. Less than 50% of the numbers in Z are greater than the median.
III. The median of Z is greater than the average of the largest and smallest numbers in Z.
Options:
A)I only
B)II only
C)III only
D)I and III only
E)Q=None of the above
Rationale: Condition 3 can be invalidated by: (1, 3, 4, 5)--median is equal to average of 1 and 5 and average is greater than 3.
Then, don't conditions 1 and 2 have to occur simultaneously? I reasoned this based on the word... maybe that's incorrect.
Even if not, we can rule them both out. So, E.
Answer: E