mathewhe/aqua_rat huggingface dataset:
Question: Of the 12,210 employees of the anvil factory, 2/7 are journeymen. If half of the journeymen were laid off, what percentage of the total remaining employees would be journeymen?
(A)  14.3%
(B)  13.67%
(C)  33%
(D)  28.6%
(E)  49.67%
Rationale: The exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers (also because otherwise the ratio would be incorrect). Since the question is about percentages, the actual numbers will be meaningless, as only the ratio of that number versus others will be meaningful. Nonetheless, for those who are curious, each 1/7 portion represents (14210/7) 2,030 employees. This in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers.
If half the journeymen were laid off, that would mean 1/7 of the total current workforce would be removed. This statistic is what leads many students to think that since half the journeymen are left, the remaining journeymen would represent half of what they used to be, which means 1/7 of the total workforce. If 1/7 of the workforce is journeymen, and 1/7 is roughly 14.3%, then answer choice A should be the right answer. In this case, though, it is merely the tempting trap answer choice.
What changed between the initial statement and the final tally? Well, you let go of 1/7 of the workforce, so the total number of workers went down. The remaining workers are still 1/7 of the initial workers, but the group has changed. The new workforce is smaller than the original group, specifically 6/7 of it because 1/7 was eliminated. The remaining workers now account for 1/7 out of 6/7 of the force, which if we multiply by 7 gives us 1 out of 6. This number as a percentage is answer choice B, 13.67%.
Using the absolute numbers we calculated before, there were 4,060 journeymen employees out of 14,210 total. If 2,030 of them are laid off, then there are 2,030 journeyman employees left, but now out of a total of (14,210-2,030) 12,180 employees. 2,030/12,180 is exactly 1/6, or 16.67%. The answer will work with either percentages or absolute numbers, but the percentage calculation will be significantly faster and applicable to any similar situation.
The underlying principle of percentages (and, on a related note, ratios) can be summed up in the brainteaser I like to ask my students: If you’re running a race and you overtake the 2nd place runner just before the end, what position do you end up in?
The correct answer is 2nd place.
Percentages, like ratios and other concepts of relative math, depend entirely on the context. Whether 100% more of something is better than 50% more of something else depends on the context much more than the percentages quoted. When it comes to percentages on the GMAT, the goal is to understand them enough to instinctively not fall into the traps laid out for you.
B
Answer: (B)  13.67%
Question: A confectioner decides to sell all of his pastry due to the coming holiday. His pastry goods are equally divided among a group of 25 regular customers. If only 49 customers come to the bakery, each one will receive 6 less pastry goods. How much pastry does the confectioner needs to sell?
(A) 300.
(B) 412.
(C) 432.
(D) 502.
(E) 522.
Rationale: Pastry is divided in 25 customers equally. So,total number of pastry must be a multiple of 25
Only option A satisfies the condition, and hence is the answer
Answer: (A) 300.
Question: A company has 50 employees, of which 22 studied accounting, 15 studied finance and 14 studied marketing. Eight employees have done just two courses and only one made ​​the three courses. How many employees have no training?
(A) A)0
(B) B)9
(C) C)10
(D) D)11
(E) E)26
Rationale: Total employees=50
studied accounting=22
studied finance=15
studied marketing=14
8 employees have been double counted and one employee has been counted thrice.
Therefore employees who have received training=22+15+14-8-(2*1)
=51-10
=41
Employees with no training
=50-41
=9
Ans=B
Answer: (B) B)9
Question: A trader marks his articles 20% more than the cost price. If he allows 20% discount, then find his gain or loss percent?
(A) 7
(B) 6
(C) 8
(D) 5
(E) 4
Rationale: Let CP of an article = RS. 100
MP= Rs. 120
Discount = 20%
SP = M[(100 - d%)/100] = 120(80/100) = Rs. 96
Clearly, the trader gets 4% loss.
Answer: E
Answer: (E) 4
Question: 529 is a perfect square. When x is added to the prime factor of 529, the result will give a perfect cube.then what is x?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: 529 is 23*23
here
A 23+1 = 24
B 23+2 = 25
C 23+3 = 26
D 23+4 = 27
E 23+5= 28
so, D is the only addition that gives a perfect cube. so x=4
then the answer is D
Answer: (D) 4
Question: In a three digit number, the hundred digit is 2 more than the tens digit and the units digit is 2 less than the tens digit. If the sum of the digits is 18, find the number?
(A) 867
(B) 864
(C) 385
(D) 946
(E) 549
Rationale: B
864
Let the three digit numbers be 100a + 10b + c
a = b + 2
c = b - 2
a + b + c = 3b = 18 => b = 6
So a = 8 and b = 4
Hence the three digit number is: 864
Answer: (B) 864
Question: Two numbers are in the ratio of 7:6. If 18 be subtracted from each, they are in the ratio of 6:5. Find the numbers?
(A) 5,9
(B) 7,9
(C) 4,7
(D) 7,6
(E) 6,9
Rationale: (7x-18):(6x-18) = 6:5
x = 1 => 7,6
Answer: D
Answer: (D) 7,6
Question: A person bought apples at 10 for $1 and sold them at 4 for $1. Find his gain or loss percent?
(A) 150%
(B) -100%
(C) 100%
(D) -50%
(E) 50%
Rationale: Number of bananas bought = L.C.M of 10 and 4 = 20
C.P. = 1/10 * 20 = $2
S.P. = 1/4 * 20 = $5
Gain% = 3/2 * 100 = 150%
Answer is A
Answer: (A) 150%
Question: If n ≠ 0, which of the following must be greater than n?
I. 6n
II. n^2
III. 2 - n
(A) I only
(B) None
(C) I and II only
(D) II and III only
(E) II only
Rationale: n ≠ 0 => n = negative or n = positive
if n = negative:
2n < n (n = -1 then 2n = -2 => -2 < -1)
n^2 > n (n =-1 then n^2 = 1 => n^2 > n)
2 - n > n (n = -1 then 2--1 = 2+1=3 => 2-n > n)
but if n = positive:
n^2 > n (n = 2 then n^2 = 4 => n^2 > n) => II
2 - n < n(n =2 then 2-2 = 0)
=> only II
=>answer is E
Answer: (E) II only
Question: Line Q has the equation 5y – 3x = 75. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
(A) 48
(B) 52
(C) 56
(D) 60
(E) 64
Rationale: 5y - 3x = 75 and so y = 3x/5 + 15
When x = 0, then y = 15.
When y = 0, then x = -25
The slope is 3/5, so the slope of line S is -5/3.
Through the point (-25, 0), 0 = -5(-25)/3 + c
The y-intercept is c = -125/3 < -41.
Thus the perpendicular line S can have y-intercepts from -41 up to 14.
The number of possible lines is 41 + 14 + 1 = 56
The answer is C.
Answer: (C) 56
Question: In a certain match all the teams were to play with each other. If there are 12 teams, how many matches will have to be played?
(A) 40
(B) 45
(C) 66
(D) 90
(E) None of the above
Rationale: 12c2
=66
ANSWER:C
Answer: (C) 66
Question: A question paper has 2parts, A & B, each containing 10 questions. If a student has to choose 5 from part A &10 from part B, in how many ways can he choose the questions?
(A) 100
(B) 120
(C) 125
(D) 350
(E) 252
Rationale: There 5 questions in part A out of which 10 question can be chosen as = 10C5
Similarly, 10 questions can be chosen from 10 questions of Part B as = 10C10.
Hence, total number of ways,
=10C5*10C10
=[10!/(5!5!)]*[10!/(0!*10)]
= {10*9*8*7*6/(5*4*3*2*1)}* 10!/1*10!
= 252
E
Answer: (E) 252
Question: A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
(A) 100
(B) 120
(C) 130
(D) 150
(E) 160`
Rationale: Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train = (Speed x Time) = (50/3 * 9) m = 150 m.
Answer: Option D
Answer: (D) 150
Question: If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 7
Rationale: Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = -(- 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3.
B
Answer: (B) 3
Question: Find the missing figures :
?% of 30 = 2.125
(A) 8.5
(B) 6.5
(C) 8.07
(D) 7.08
(E) 4.05
Rationale: (i)	Let x% of 30 = 2.125. Then , (x/100)*30 = 2.125
X = (2.125 * 100/30) = 7.08
Answer is D.
Answer: (D) 7.08
Question: A can run 3 Kms in 3 min 18 sec and B can run same distance in 3 min 40 sec, then by how much distance A can beat B?
(A) 300 m
(B) 220 m
(C) 280 m
(D) 195 m
(E) 295 m
Rationale: Distance travelled=3000m
Time taken by A=3 m 18 s
Time taken by B=3 m 40 sec=220 sec
speed of B=3000m/220 s=150/11 m/s
Difference between time of A and B=22 sec
when time will be 3 m 18 s A will have travelled 3000m but B need 22 sec more to finish.
Distance travelled by B in 22 sec=(150/11)*22=300 m
Hence, A beat B by 300 m
ANSWER:A
Answer: (A) 300 m
Question: In how many ways can the letters of word CAREER be arranged such that all the vowels as well as all the Consonants always appear together?
(A) 3!*3!*2!
(B) 5!*4!
(C) 5!*5!
(D) 5!*4!*2!
(E) 6!*4!
Rationale: Since all the vowels and consonants have to be together there can only be there are 3! way to order the vowels and 3! to order the consonants. But they must be together either vowels come first or consonants so you multiply by 2.
So 3!3!2! (A)
Answer: (A) 3!*3!*2!
Question: A major airplane manufacturer announced the sale of 150 small airliners at $125 million each, 75 medium sized airliners at $175 million each, and 60 large airliners at $220 million each. What percentage of the company's income came from large and medium airliners?
(A) 29%
(B) 41%
(C) 58%
(D) 69%
(E) 70%
Rationale: Answer is (money from mid and large plane)/(total money from all the planes)x 100
=(175∗75+220∗60)/(125∗150+175∗75+220∗60)∗100(175∗75+220∗60)/(125∗150+175∗75+220∗60)∗100
= 58.4%
you could just multiply add divide to get answer.
However sometimes its much better to make it easy we should see if we have any common terms
125*150=(5*5*5)*(2*3*5*5)
175*75=(5*5*7)*(3*5*5)
220*60=(2*2*11*5)*(2*2*3*5)
3*5*5 is common cancel it
Ans becomes
(175+176)/(176+175+250)*100 =~=~ 350/600*100=~=~7/12*100=60% >>> 58% is closest.
ANSWER:C
Answer: (C) 58%
Question: The length of each side of an equilateral triangle having an area of 4√3 cm2 is?
(A) 1/8
(B) 1/9
(C) 1/5
(D) 1/23
(E) 1/2
Rationale: Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5
Answer: C
Answer: (C) 1/5
Question: A tank is filled by 3 pipes A,B,C in 7 hours. Pipe C is twice as fast as B and B is twice as fast as A. How much will pipe A alone take to fill the tank ?
(A) 25hr
(B) 35hr
(C) 40hr
(D) 49hr
(E) 50hr
Rationale: Suppose pipe A alone take x hours to fill the tank
Then pipe B and C will take x/2 and x/4 hours respectively to fill the tank.
1/x + 2/x + 4/x = 1/7
7/x = 1/7
x = 49 hours
Answer is D
Answer: (D) 49hr
Question: The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
(A) 1
(B) 3
(C) 4
(D) 2
(E) 6
Rationale: Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l - b = 8 - 6 = 2m.
Answer:D
Answer: (D) 2
Question: In a psychology school the grade of the students is determined by the following method: At the end of the first year the grade equals to twice the age of the student. From then on, the grade is determined by twice the age of the student plus half of his grade from the previous year. If Joey’s grade at the end of the first year is 36, what will be his grade at the end of the third year?
(A) 44
(B) 56
(C) 62
(D) 75
(E) 61
Rationale: Cool question!
There's a lot of information given here, so let's eat slow here.
Joey scores a 36 at the end of his first year. That means that he's currently 16 years old (since given per instruction, after the first year one's grade is determined by doubling his/her age).
Now, come next year, or his second year at this psychology school, he's technically 17 years old. To find out the second year's grade, we want to double his age PLUS half of his previous grade.
Well, his second year is then going to be 17(2) + 16 = 50.
Now let's tackle the main question: what's his grade going to be the third year?
Well, he will be at an age of 22 on his third year at this school. Again, to find out his third year's grade, you're going to double his age (18x2), plus half of his previous year's grade (as calculated for his second year, his grade is 50; therefore, 50/2)
Having said that, the final answer would be 18(2) + 50/2 = 36 + 25 = 61 --> (E)
Answer: (E) 61
Question: A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?
(A) 165 ways
(B) 176 ways
(C) 197 ways
(D) 160 ways
(E) 162 ways
Rationale: Total number of persons in the committee
= 5 + 6 = 11
Number of ways of selecting group of eight persons
= ¹¹C₈
= ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 ways.
Answer:A
Answer: (A) 165 ways
Question: A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
(A) 0
(B) 1/9
(C) 2/9
(D) 1/3
(E) 1
Rationale: Probability = (favorable cases)/(total number of cases)
Total number of ways in which a person can ride car = 3*3*3 = 27
(In first ride he has 3 options to sit, in second right again he has 3 seats available to sit and so on)
Number of favorable cases, i.e., when he rides on different cars;
He can choose seat car in 3 ways in his 1st ride.
He can choose seat car in 2 ways in his 2nd ride.
He can choose seat car in 1 ways in his 3rd ride.
So, 3*2*1 = 6 ways
Thus, probability of choosing different seats = 6/27 = 2/9
Answer: C
Answer: (C) 2/9
Question: In what ratio must tea at Rs.60 per Kg be mixed with tea at Rs.70 per Kg so that the mixture must be worth Rs.63 per Kg?
(A) 3 : 1
(B) 3 : 2
(C) 4 : 3
(D) 7 : 3
(E) None
Rationale: Required ratio = 700 : 300 = 7 : 3
Answer D
Answer: (D) 7 : 3
Question: Mr. Hernandez, who was a resident of State X for only 9 months last year, had a taxable income of $42,500 for the year. If the state tax rate were 4 percent of the year’s taxable income prorated for the proportion of the year during which the taxpayer was a resident, what would be the amount of Mr. Hernandez’s State X tax for last year?
(A)  $900
(B)  $1275
(C)  $1200
(D)  $1300
(E)  $1060
Rationale: Total tax for the year = 42,500 x 4% = 1700
As stated Annual tax is prorated as per the duration of stay.
Prorated Tax = 1700 (9/12) = 1275
Answer B
Answer: (B)  $1275
Question: A train running at the speed of 120 km/hr crosses a pole in 12 seconds. What is the length of the train?
(A) 200 m
(B) 250 m
(C) 300 m
(D) 350 m
(E) 400 m
Rationale: Speed=(120 * 5/18) m/sec
= (100/3) m/sec Length of the train
= (Speed x Time)
= (100/3 * 12) m = 400 m.
Answer:E
Answer: (E) 400 m
Question: A and B started a partnership business investing some amount in the ratio of 6 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
(A) 4 : 9 : 8
(B) 5 : 7 : 4
(C) 12 : 10 : 5
(D) 7 : 9: 4
(E) 5 : 4 : 8
Rationale: Let the initial investments of A and B be 6x and 5x.
A : B : C = (6x x 12) : (5x x 12) : (5x x 6) = 72 : 60 : 30
= 12 : 10 : 5.
ANSWER :C
Answer: (C) 12 : 10 : 5
Question: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Rationale: The factors of 10 are 1, 2, 5, and 10.
To find what k is, find how many times you can make 10 with the numbers in 1-40.
First, find the 10's as they are going to be most restrictive. Only 10, 20, 30 and 40 have 10 as a factor. 4 total.
Next find 5's. 5, 15, 25 and 35 each have a 5, with 25 having 2. 5 total.
You don't need to find the twos as you know there are more than 5's.
Add the two you get 9.
ANSWER:C
Answer: (C) 9
Question: 7 men, 5 women and 8 children were given as assignment of distributing 2000 books to students in a school over a period of 3 days. All of them distributed books on the 1st day. One of the 2nd day 2 women and 3 children remained absent and on the 3rd day 3 men and 5 children remained absent. If the ratio of the no. of books distributed in a day by a man, a woman & a child was 5 : 4 : 2 respectively, a total of approximately how many books were distributed on the second day?
(A) 340
(B) 450
(C) 680
(D) 650
(E) 700
Rationale: Sol. Let the books distributed by man, a woman and a child be 5x, 4x and 2x respectively.
∴ No. of books distributed in 1st day
= 7×5x+5×4x+8×2x=71x
No. of books distributed in 1Ind day
=7×5x+3×4x+5×2x=57x
And no. of books distributed in IIIrd day
=4×5x+5×4x+3×2x=46x
71x + 57x + 46x = 2000, x = 2000/174
57x=2000/174×57=650
D
Answer: (D) 650
Question: A train 55 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
(A) 5
(B) 6
(C) 7
(D) 3
(E) 5
Rationale: Speed of train relative to man = (60 + 6) km/hr = 66 km/hr
[66 * 5/18] m/sec = [55/3] m/sec.
Time taken to pass the man = [55 * 3/55] sec = 3 sec
Answer: D
Answer: (D) 3
Question: What is the greatest possible length which can be used to measure exactly the lengths 10 m 50 cm, 14 m 55 cm and 50 cm?
(A) 20 cm
(B) 24 cm
(C) 30 cm
(D) 10 cm
(E) 5 cm
Rationale: Required length = HCF of 1050 cm, 1455 cm, 50cm = 5 cm
Answer is E
Answer: (E) 5 cm
Question: Which of the following is a prime number
(A) 18
(B) 20
(C) 19
(D) 16
(E) 24
Rationale: Clearly, 19 is a prime number.
Option C
Answer: (C) 19
Question: A right circular cylinder has a height of 23 and a radius of 5. A rectangular solid with a height of 15 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. What is the volume of the liquid?
(A) 500(π – 3)
(B) 575(π – 1.5)
(C) 500(π – 2)
(D) 500(π – 1.5)
(E) 500(π – 1)
Rationale: [quote=Bunuel]A right circular cylinder has a height of 20 and a radius of 5. A rectangular solid with a height of 15 and a square base, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. Liquid is then poured into the cylinder such that it reaches the rim. What is the volume of the liquid?
The square base has sides of sqrt(50) due to the 45-45-90 triangle
23*25*pi - 15*sqrt(50)^2
=575(π – 1.5)
B. 500(π – 1.5)
Answer: (B) 575(π – 1.5)
Question: Teas worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 155 per Kg , the price of the third variety per Kg will be?
(A) Rs. 147.50
(B) Rs. 785.50
(C) Rs. 177.50
(D) Rs. 258.50
(E) None of these
Rationale: Explanation :
Since first and second varieties are mixed in equal proportions.
So, their average price =Rs. (126+135)/2.
=>Rs. 130.50.
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1.
We have to find x.
By the rule of alligation, we have:
Cost of 1 kg Cost of 1 kg
of 1st kind of 2nd kind
(Rs. 130.50) (Rs. x)
\ /
Mean Price
(Rs. 155)
/ \
xâˆ’155 22.50
=>xâˆ’(155/22.50)=1.
=>xâˆ’155=22.50.
=>x= 177.50 Rs.
Answer : C
Answer: (C) Rs. 177.50
Question: A person borrows Rs.5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 7% p.a for 2 years. Find his gain in the transaction per year.
(A) 167.5
(B) 150
(C) 225
(D) 112.5
(E) 212.5
Rationale: Explanation :
The person borrows Rs. 5000 for 2 years at 4% p.a. simple interest
Simple interest that he needs to pay = PRT/100=5000×4×2/100=400
He also lends it at 7% p.a for 2 years
Simple interest that he gets = PRT/100=5000×7×2/100=700
His overall gain in 2 years = Rs.700 - Rs.400 = Rs.300
His overall gain in 1 year = 300/2 = Rs.150
Answer : Option B
Answer: (B) 150
Question: $9,000 is deposited in a savings account that pays 14% annual interest compounded semiannually. To the nearest dollar, how much is in the account at the end of the year?
(A) $11304.12
(B) $12304.54
(C) $11604.14
(D) $10304.10
(E) $9304.41
Rationale: this is the case of semi-annual compoundingso, multiply time period by 2 and divide rate by 2
So,
new time=2periods and new rate=14/2=7%
Now,
CA=9000(1+7/100)^2=$10304.10
Hence answer is D
Answer: (D) $10304.10
Question: A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?
(A) 223
(B) 186
(C) 287
(D) 167
(E) 171
Rationale: Explanation:
When at least 2 women are included.
The committee may consist of 3 women, 2 men : It can be done in\inline \fn_jvn {\color{Black} 4C_{3}\times 6C_{2}} ways
or, 4 women, 1 man : It can be done in \inline \fn_jvn {\color{Black} 4C_{4}\times 6C_{1}} ways
or, 2 women, 3 men : It can be done in \inline \fn_jvn {\color{Black} 4C_{2}\times 6C_{3}} ways.
Total number of ways of forming the committees
= \inline \fn_jvn {\color{Black} 4C_{2}\times 6C_{3}+4C_{3}\times 6C_{2}+4C_{4}\times 6C_{1}}
= 6 x 20 + 4 x 15 + 1x 6
= 120 + 60 + 6 =186
Answer: B) 186
Answer: (B) 186
Question: A and B can do a piece of work in 9 days. With the help of C they finish the work in 6 days. C alone can do that piece of work in?
(A) 33
(B) 878
(C) 18
(D) 88
(E) 11
Rationale: C = 1/6 â€“ 1/9 = 1/18 => 18 days
Answer: C
Answer: (C) 18
Question: Pipe A can fill a pool in 12 hours at a constant rate. If Pipe A does the job for 8 hours and Pipe B does the rest of the job, which works at 1/3 constant rate of Pipe A. How long will it take for Pipe B alone to do the rest of the job?
(A) 12hrs
(B) 6hrs
(C) 8hrs
(D) 10hrs
(E) 24hrs
Rationale: Pipe A will do 8/12 in 8 hrs, so A does 2/3 of the work ..
Therefore, B will do the remaining 1/3 rd work alone..
As the speed of B is 1/3 rate of A, B will do the 1/3rd work in same time that A takes to complete full job...
ans 12
A
Answer: (A) 12hrs
Question: If (x^2 + 1)y=6, which of the following is not a possible value for y?
(A) 3/2
(B) 2
(C) 1/3
(D) 2/13
(E) 13/2
Rationale: Its simple
equation is (x^2 +1)y = 6 ===> x^2 = (6/y) - 1 . As , in LHS, it is x^2 the value has to be positive or zero in RHS.
Hence, (6/y) - 1>= 0 ===> y =< 6. Now only option E has greater value of 'y' than 6.
Thus, correct ans is E
Answer: (E) 13/2
Question: If logx (base 27) +logx (base 9) +logx (base3) =11 then find the value of x?
(A) 3^4
(B) 3^3
(C) 3^5
(D) 3^6
(E) 3^9
Rationale: logx(base 3)=Y
Now logx(base27)=logx(3^3 base)=1/3 logx(base3)=1/3 Y
Also logx(base 9)=1/2 logx(base3)=1/2 Y
Now as given,
1/3 Y + 1/2 Y+ Y = 11
So 11/6 Y = 11
Or Y =6
So logx(base3)=6
X=3^6
Answer D
Answer: (D) 3^6
Question: What is the greatest possible length which can be used to measure exactly the lengths 10 m, 14 m 50 cm and 20 m 40cm cm?
(A) 20 cm
(B) 10 cm
(C) 25cm
(D) 50 cm
(E) 28 cm
Rationale: Required length = HCF of 1000 cm, 1450 cm, 2040 cm = 10cm
Answer is B
Answer: (B) 10 cm
Question: If a, x, y, and z are positive integers and (x)×(y)×(z)=a^2, which of the following could be the values of x, y, and z?
(A) 3, 16, 25
(B) 9, 25, 24
(C) 2, 36, 32
(D) 2, 9, 16
(E) 8, 16, 36
Rationale: As it contains every no. 2 times
2
36=6*6
32=4*4*2
C is the answer
Answer: (C) 2, 36, 32