mathewhe/aqua_rat huggingface dataset:
Question: Three friends were having a party and they ordered garlic bread sticks from a nearby restaurant. Upon arrival, they divided the number of sticks among them equally. After each one of them have had four breadsticks, the total number of breadsticks that remained were equal to the breadsticks that each one of them had upon equal division.
With the data given, can you calculate the original number of garlic bread sticks ?
(A) 13
(B) 26
(C) 22
(D) 19
(E) 18
Rationale: E
18
Let us suppose that there were (3 * G) garlic bread sticks originally.
On dividing equally, the number of bread sticks with each of the friend = G
Each one of them ate four. The remaining sticks with each one of them = (G - 4)
Now, the total number of breadsticks left = 3 * (G - 4)
This number is equal to the breadsticks each one of them had after division.
=> 3 * (G - 4) = G
(3 * G) - 12 = G
2 * G = 12
G = 6.
Therefore, the number of breadsticks originally = 3 * G = 3 * 6 = 18.
Answer: (E) 18
Question: In what time will a train 200 meters long completely pass an electric pole, if it travels at a speed of 80kmph.
(A) 7
(B) 9
(C) 5
(D) 10
(E) 3
Rationale: Sol.
Speed = [ 80x5/18] m/sec =22.2 m / sec.
Time taken = (200/22.2)sec = 9 sec.
Answer B
Answer: (B) 9
Question: A student multiplied a number by 3/5	instead of	5/3.What is the percentage error in the calculation?
(A) 34%
(B) 44%
(C) 54%
(D) 64%
(E) 74%
Rationale: Let the number be x.
Then, error =(5/3)x -	(3/5)	x =(16/15)x.
Error% =((16/15)x)	x (3/5x)x 100% = 64%.
Answer:D
Answer: (D) 64%
Question: Rose went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
(A) 19
(B) 19.7
(C) 19.9
(D) 20.1
(E) 20.4
Rationale: Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (25−30/100−x)6/100 = 30/100
⇒ 6(25 −0.3 −x) = 30
⇒ (25 − 0.3 − x) = 5
⇒ x = 25 − 0.3 − 5 = 19.7
B)
Answer: (B) 19.7
Question: If y ≠ 3 and 2h/y is a prime integer greater than 2, which of the following must be true?
I. h = y
II. y = 1
III. h and y are prime integers.
(A)  None
(B)  I only
(C)  II only
(D)  III only
(E)  I and II
Rationale: Note that we are asked which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
So we should try to make the listed statements not true for some h and y (which satisfy y≠3 and 2h/y=prime>2).
I. h = y --> not necessarily true: h=3 and y=2 (2h/y=3=prime>2);
II. y=1 --> not necessarily true: h=3 and y=2 (2h/y=3=prime>2);
III. h and y are prime integers --> not necessarily true: h=10 and y=4 (2h/y=5=prime>2).
Answer: A.
Answer: (A)  None
Question: In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg?
(A) 5/6
(B) 5/9
(C) 5/1
(D) 5/3
(E) 7/6
Rationale: Let us say the ratio of the quantities of cheaper and dearer varieties
= x : y
By the rule of allegation, x/y
= (87.5 - 7.50) / (7.50 - 6) =5/6
Answer: A
Answer: (A) 5/6
Question: If m > 1 and n = 2^(m−2), then 4^m =
(A) 16n^2
(B) 4n^2
(C) n^2
(D) n^2/4
(E) n^2/16
Rationale: n = 2^(m-2) = 2^m/4
2^m = 4n
4^m = (2^m)^2 = (4n)^2 = 16n^2
The answer is A.
Answer: (A) 16n^2
Question: If a light flashes every 6 seconds, how many times will it flash in 1/2 of an hour?
(A) 301 times
(B) 351 times
(C) 401 times
(D) 451 times
(E) 481 times
Rationale: In 1/2 of an hour there are 30*60 = 1800 seconds
The number of 6-second intervals = 1800/6 = 300
After the first flash, there will be 300 more flashes for a total of 301.
The answer is A.
Answer: (A) 301 times
Question: If the average (arithmetic mean) of a and b is 50 and the average of b and c is 70, what is the value of c − a?
(A) 40
(B) 50
(C) 90
(D) 140
(E) It cannot be determined from the information given.
Rationale: -(a + b = 100)
b + c=140
c-a=40
A.40
Answer: (A) 40
Question: The average weight of a group of 30 friends increases by 2 kg when the weight of their football coach was removed. If average weight of the group after removing the weight of the football coach is 28 kg, what is the weight of their football coach?
(A) 31 kg
(B) 60 kg
(C) 61 kg
(D) 62 kg
(E) 88 kg
Rationale: let c=coach's weight
original total weight=(30 friends)(30 kg avge)=900kg
(900-c)/(30-1)=28 kg avge
c=88 kg
Answer - E
Answer: (E) 88 kg
Question: If y is an integer and x = 3y + 2, which of the following CANNOT be a divisor of x?
(A) 4
(B) 6
(C) 8
(D) 9
(E) 10
Rationale: Just to add some more to Bunuel's explanation, if a number is a multiple of 3, it will be in a form:
n = 3*k, where k is an integer
And if a number is a multiple of 6, it will be in a form :
m = 6*l, where l is an integer
so m = 2 * 3 * l = (2*l) * 3 = 3 * p, where p is an integer.
In other words, the prime factorization of m must have 3 as a factor, if it's a multiple of 6, but the expression can't have that (as explained by Bunuel), so it can't be a multiple of 6 either.
Answer is B
Answer: (B) 6
Question: Of the 800 employees of Company X, 90 percent have been with the company for at least ten years. If y of theselong-termmembers were to retire and no other employee changes were to occur, what value of y would reduce the percent oflong-termemployees in the company to 60 percent ?
(A) 200
(B) 160
(C) 112
(D) 80
(E) 600
Rationale: The # oflong-termemployees is 90%*800=720.
After y of them retire new # oflong-termemployees would become 720-y.
Total # of employees would become 800-y.
We want 720-y to be 60% of 800-y --> 720-y=(800 -y)*60% --> y = 600.
Answer: E.
Answer: (E) 600
Question: If 1/a + 1/a + 1/a is an integer, which of the following must also be an integer?
(A) 12/a
(B) 3/a
(C) 6/a
(D) 9/a
(E) 15/a
Rationale: 1/a + 1/a + 1/a
=3/a
Answer B
Answer: (B) 3/a
Question: List I: { y, 2, 4, 7, 10, 11}
List II: {3, 3, 4, 6, 7, 10}
If the median W of List I is equal to the sum of the median of list II and the mode of list II, then y equals
(A) 5
(B) 7
(C) 8
(D) W=9
(E) 10
Rationale: mode of list ii =3
median W of list ii =4+6/2 =5
sum of mode + mean = 3+ 5=8
now to make 8 as the median we need to find a value of y such that if the no. of terms in list 1 are odd then y=8 else if even then 7+y /2=8
here its even so
7+y/2=8 from this y= 9 (D)
Answer: (D) W=9
Question: If y exceeds x by 26%, then x is less than y by?
(A) 16 2/8%
(B) 20 8/3%
(C) 20 40/63%
(D) 76 2/3%
(E) 20 2/3%
Rationale: X=100	y=126
126------26
100-------? => 20 40/63%
Answer:C
Answer: (C) 20 40/63%
Question: The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.5 the sixth number is?
(A) 11.4
(B) 11.3
(C) 11.8
(D) 12.1
(E) 11.1
Rationale: 1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.5 = 69
63 + 69 = 132 – 119.9 = 12.1
6th number = 12.1
Answer: D
Answer: (D) 12.1
Question: A 70 cm long wire is to be cut into two pieces so that one piece will be 2/3th of the other, how many centimeters will the shorter piece be?
(A) 35
(B) 20
(C) 28
(D) 36
(E) 30
Rationale: 1: 2/3 = 3: 2
2/5 * 70 =28
Answer:C
Answer: (C) 28
Question: If the area of a circle decreases by 30%, then the radius of a circle decreases by
(A) 20%
(B) 16%
(C) 36%
(D) 64%
(E) None of these
Rationale: If area of a circle decreased by x % then the radius of a circle decreases by
(100−10√100−x)%=(100−10√100−30)%
= (100−10√70)%
= 100 - 84 = 16%
Answer B
Answer: (B) 16%
Question: The ratio of two numbers is 2 : 5 and their H.C.F. is 5. Their L.C.M. is
(A) 50
(B) 22
(C) 56
(D) 27
(E) 67
Rationale: Explanation:
Let the numbers be 2x and 5x. Then, their H.C.F. = x. So, x = 5.
So, the numbers 10 and 25.
L.C.M. of 10 and 25 = 50.
Option A
Answer: (A) 50
Question: 60 is subtracted from 50% of a number, the result is 10. Find the number?
(A) 150
(B) 120
(C) 266
(D) 288
(E) 271
Rationale: (50/100) * X – 50 = 10
5X = 600
X = 120
Answer:B
Answer: (B) 120
Question: Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.130. How much was lent at 5%?
(A) 2000
(B) 2777
(C) 2688
(D) 1000
(E) 2871
Rationale: (x*5*1)/100 + [(2500 - x)*6*1]/100 = 130
X = 2000
Answer: A
Answer: (A) 2000
Question: Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
(A) 2 : 3 : 4
(B) 6 : 7 : 8
(C) 6 : 8 : 9
(D) 6 : 8 : 7
(E) None of these
Rationale: let 100% seats are there in all then mathematics increased by 40% then
5*140/100=7
physics is increased by 50% then
7*150/100=21/2
biology increased by 75% then
8*175/100=14
then ratios of increasing seats are 7:21/2:14
14:21:28
2:3:4
ANSWER:A
Answer: (A) 2 : 3 : 4
Question: 7M-20=2M,THEN M+7 IS EQUAL TO?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
Rationale: 7m-20=2m
so, 5m=20
so, m+7=11
ANSWER:C
Answer: (C) 11
Question: Let A = {a, b, c, d, e, f} and B = {b, d, f, g}.Find B-A
(A) {A}
(B) {G}
(C) {L}
(D) {B}
(E) {F}
Rationale: A = {a, b, c, d, e, f}
B = {b, d, f, g}
correct answer :B - A = {g)
B
Answer: (B) {G}
Question: Deepak deposited $9,000 to open a new savings account that earned 12 percent annual interest, compounded semi-annually. If there were no other transactions in the account, what the amount of money in Deepak account one year after the account was opened?
(A)  $10,200
(B)  $10,092
(C)  $10,105
(D)  $9,405
(E)  $10,260
Rationale: Approach #1:
12 percent annual interest compounded semi-annually --> 6% in 6 moths.
For the first 6 moths interest was 6% of $9,000, so $540;
For the next 6 moths interest was 6% of $9,000,plus6% earned on previous interest of $540, so $540+$12=$552;
Total interest for one year was $540+$552=$1092, hence balance after one year was $9,000+ $1,092=$10,092.
Answer: B.
Answer: (B)  $10,092
Question: Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 miles per hour. If both drivers maintained their speeds and both remained on the highway for at least 4 hours, how long after he passed the gas station did Paul catch up with Mary?
(A) 2
(B) 1.5
(C) 1.6
(D) 1.7
(E) 1.25
Rationale: d = rt
M: r = 50mph, t = t + 1/4hr
d = 50 (t + 1/4)
P: r = 60, t = t
d = 60t
since they went the same distance:
50t + 50/4 = 60t
10t = 50/4
t = 1.25 or 1 hr, 15min
E
Answer: (E) 1.25
Question: A certain bag contains 100 balls â€” 50 white, 20 green, 10 yellow, 17 red, and 3 purple. If a ball is to be chosen at random, what is the probability that the ball will be neither red nor purple?
(A) 0.9
(B) 0.75
(C) 0.6
(D) 0.8
(E) 0.5
Rationale: According to the stem the ball can be white, green or yellow, so the probability is (white + green + yellow)/(total) = (50 + 20 + 10)/100 = 80/100 = 0.8.
Answer is D
Answer: (D) 0.8
Question: A shopkeeper sold an article offering a discount of 5% and earned a profit of 38.7%. What would have been the percentage of profit earned if no discount had been offered?
(A) 28.5
(B) 46
(C) 30
(D) Data inadequate
(E) None of these
Rationale: Giving no discount to customer implies selling the product on printed price. Suppose the cost price of the article is 100.
Then printed price = 100Ã—(100+38.7)/(100âˆ’5) =146
Hence, required % Profit = 146 â€“ 100 = 46%
Answer B
Answer: (B) 46
Question: If line L passes through point (m, n) and (– m, – n), where m and n are not 0, which of the following must be true?
I.	The slope of L is need not be positive
II.	The slope of L is negative
III.	L exactly passes through 2 quadrants
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
Rationale: Let (m,n) = (2,3)
(-m,-n) = (-2,-3)
Slope = (3+3)/(2+2) = 6/4 = 3/2
So II may not be true
Let (m,n) = (2,-3) then (-m,-n) = (-2,3)
So slope = (3 +3)/(-2-2) = -3/2
so I may be true
So such a line would be -> (y - 3) = 3/2(x - 2)
=> 2y - 6 = 3x - 6
=> 2y - 3x = 0, hence no x or y intercept, so it passes through origin.
III is true.
Answer - E
Answer: (E) I and III only
Question: P alone can complete a piece of work in 12 days. Work done by Q alone in one day is equal to one-third of the work done by P alone in one day. In how many days can the work be completed if P and Q work together?
(A) 6 (1/4) days
(B) 9 days
(C) 7 (3/4) days
(D) 8 days
(E) 9 (3/4) days
Rationale: Work done by P alone in one day = 1/12th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/12 of the total) = 1/36 of the total.
Work done by P and Q, working together in one day = 1/12 + 1/36 = 4/36=1/9 of the total
They would take 9 days to complete the work working together.
ANSWER:B
Answer: (B) 9 days
Question: A restaurant meal cost $18.5 and there was no tax. If the tip was more than 30 percent but less than 10 percent of the cost of the meal, then the total amount paid must have been between
(A) $24 and $20
(B) $25 and $23
(C) $38 and $40
(D) $37 and $39
(E) $36 and $37
Rationale: The total amount for the meal was between 18.5*1.3=24.05 and 18.5*1.1=~20.35. Only option which covers all possible values of the meal is A.
Answer: A.
Answer: (A) $24 and $20
Question: A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
(A) Rs.1000
(B) Rs.1140
(C) Rs.860
(D) Rs.760
(E) Rs.500
Rationale: Explanation:
90%
104%
--------
14% ---- 140
100% ---- ? => Rs.1000
ANSWER IS A
Answer: (A) Rs.1000
Question: A man travels 250km by train at 50km/hr, 200km by ship at 20km/hr, 300km by aeroplane at 100km/hr and 100km by car at 50km/hr. What is the average speed for the entire distance?
(A) 35.3km/hr
(B) 42.5km/hr
(C) 48.9km/hr
(D) 52.6km/hr
(E) 27.6km/hr
Rationale: Total distance traveled = 250+200+300+100 = 850km
Total time taken = 250/50 + 200/20 + 300/100 + 100/50 = 20 hrs
Average speed =850/20 = 42.5km/hr approximately ;
Answer is B
Answer: (B) 42.5km/hr
Question: A train covers a certain distance at a speed of 150kmph in 4 hours. To cover the same distance in 2 hours, it must travel at a speed of
(A) 700km/hr
(B) 600km/hr
(C) 500km/hr
(D) 400km/hr
(E) 300km/hr
Rationale: Explanation:
Distance = 150×4 = 600km
Required speed = (600/2) = 300km/hr
Answer: Option E
Answer: (E) 300km/hr
Question: Rounded to three decimal places, 1.001^4 =
(A) 1.004
(B) 1.006
(C) 1.008
(D) 1.012
(E) 1.016
Rationale: As compared to 1, 0.001 is a very small quantity. Thus, we can write (1+0.001)^4 is nearly equal to (1+4*0.001) = 1.004.As the question asks for approximation to three decimal places, the further terms will anyways not come into picture.
A
Answer: (A) 1.004
Question: If 40% of a certain number is 160, then what is 90% of that number?
(A) 270
(B) 380
(C) 260
(D) 360
(E) 290
Rationale: Explanation:
40% = 40*4 = 160
90% = 90*4 = 360
Answer: Option D
Answer: (D) 360
Question: An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 12 hours to fill the tank. If the tank is full, how much time will the leak take to empty it?
(A) 4 hours
(B) 12hours
(C) 8hours
(D) 5hours
(E) 15hours
Rationale: Work done by the leak in 1 hour = 1/3 - 1/12 = 1/4
The leak will empty the tank in 4hours
Answer is A
Answer: (A) 4 hours
Question: How many seconds does Sandy take to cover a distance of 600 meters, if Sandy runs at a speed of 18 km/hr?
(A) 100
(B) 120
(C) 140
(D) 160
(E) 180
Rationale: 18 km/hr = 18000 m / 3600 s = 5 m/s
time = 600 / 5 = 120 seconds
The answer is B.
Answer: (B) 120
Question: Four cups of milk are to be poured into a 3-cup bottle and a 6-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 6-cup bottle?
(A)  2/3
(B)  7/9
(C)  24/9
(D)  8/3
(E)  3
Rationale: Let x be the # of cups going into the 6 cup bottle.
so.... x/6 = ((4-x)/3)
24 - 6x = 3x
24 = 9x
x = 24/9.
Answer : C
Answer: (C)  24/9
Question: An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 60 cm, find the sum of the perimeters of all the triangles.
(A) 180 cm
(B) 220 cm
(C) 240 cm
(D) 270 cm
(E) 360 cm
Rationale: We have 60 for first triangle, when we join mid-points of first triangle we get the second equilateral triangle then the length of second one is 30 and continues.
So we have 60,30,15,...
We have ratio = 1/2, and it is GP type.
Sum of infinite triangle is a/1-r = 60/1-(1/2) = 120
Equilateral triangle perimeter is 3a = 3*120 = 360.
So option E.
Answer: (E) 360 cm
Question: A single discount equivalent to the discount series of 20%, 10% and 5% is?
(A) 31.7
(B) 31.1
(C) 31.6
(D) 31.4
(E) 31.8
Rationale: 100*(80/100)*(90/100)*(95/100)
= 68.4
100 - 68.4
= 31.6
Answer: C
Answer: (C) 31.6
Question: Let y = 2m + x^2 and m = 5x + 4. If m^2 =81, then which of the following can be a value of 2y + 3m?
(A) 65
(B) 130
(C) 141
(D) 142
(E) 160
Rationale: m^2 = 81 => m=9
9= 5x+4 => x=1
y= 2*9+1^2 = 19
2y+3m = 2*19+3*9 = 65
Answer : A
Answer: (A) 65
Question: A sells a bicycle to B and makes a profit of 60%. B sells the same bicycle to C at a profit of 25%. If the final S.P. of the bicycle was Rs. 225, find out the cost price of the bicycle for A.
(A) 237
(B) 126
(C) 971
(D) 611
(E) 112.5
Rationale: Explanation:
Let CP be 100
A sells at 60% profit so SP=160
B sells at 25% profit = 160 x (1+25/100) = 200
CP SP
100--- 200
x --- 225
CP = 225 x 100/200 = 112.5
ANSWER: E
Answer: (E) 112.5
Question: The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 146, what is the difference between the ages of Patrick and Monica?
(A) 27
(B) 48
(C) 45
(D) 72
(E) 18
Rationale: Ages of P and Mi=3x+5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=47x=146
x=3
Difference if ages of Pa and Mo=25x-9x=16x=16*3=48
Answer B
Answer: (B) 48
Question: At a certain conference, 150 students spoke one or more of the following languages: Spanish, English, and French. If 51 students spoke Spanish, 90 spoke English, 65 spoke French, and 32 spoke only 2 languages, how many students spoke three different languages?
(A) 7
(B) 9
(C) 12
(D) 15
(E) 18
Rationale: 150=51+90+65-32-2x
x=12
C. 12
Answer: (C) 12
Question: From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12?
(A) 36
(B) 60
(C) 84
(D) 72
(E) 86
Rationale: DETAILED SOLUTION
Any multiple of 12 should be a multiple of 4 and 3. First, let us look at the constraint for a number being a multiple of 3. Sum of the digits should be a multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit and still retain a multiple of 3, we should drop either 3 or 6.
So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
When the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 36 different numbers.
When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 24 different numbers.
Overall, there are 60 different 5-digit numbers possible.
Correct Answer: B
Answer: (B) 60
Question: David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?
(A) 5000
(B) 5007
(C) 5006
(D) 5003
(E) 5009
Rationale: Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Answer:A
Answer: (A) 5000
Question: A rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze?
(A) 1218
(B) 1212
(C) 1210
(D) 277
(E) 291
Rationale: π (232 – 122) = 1210
Answer: C
Answer: (C) 1210
Question: In ΔPQS above, if PQ =5 and PS = 6, then
(A)  9/4
(B)  12/5
(C)  16/5
(D)  30/7
(E) 50/3
Rationale: there are two ways to calculate area of PQS. area remains same, so both are equal.
5*6/2 = PR*7/2
PR = 30/7
D
Answer: (D)  30/7
Question: P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 12 days. How many days more will P take to finish the work?
(A) 12
(B) 15
(C) 18
(D) 20
(E) 22
Rationale: Let work done by P in 1 day = p,
Work done by Q in 1 day = q,
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 12 days = 12 × (1/16) = 12/16 = 3/4
Remaining work = 1 = 3/4 = 1/4
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (1/4) / (1/48) = 12
Option A
Answer: (A) 12
Question: Some of the 50% solution of acid was removed and this was replaced with an equal amount of 10% solution of acid. As a result, a 20% solution of acid was obtained. What fraction of the original solution was replaced?
(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/5
(E) 4/5
Rationale: Let x be the fraction of the original solution that was replaced.
0.5*(1-x)+0.1(x)=0.2
0.4x=0.3
x=3/4
The answer is C.
Answer: (C) 3/4
Question: Set C consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set C?
(A) 5
(B) 6
(C) 7
(D) 1
(E) 766
Rationale: Whenever a question asks you to deal with the MEDIAN of a group of numbers, you MUST put the numbers in order from least to greatest (or at the very least,group offthe numbers so that you can restrict the potential value of the median).
Here, we have 8 values (one of the values is the variable N). We're told that the values are UNIQUE, so N CANNOT be any of the other 7 values listed. We're asked which of the following answers COULD be the median, so if we determine that one of the answer choices is a POSSIBLE median, then we can stop working....
Since the group includes 8 values, the median will be the AVERAGE of the 4th and 5th values...
Putting the 7 numbers in order, we have:
-27, -3, -2, 2, 3, 15, 17
This means that the2will either be the 4th term or the 5th term (depending on the value of N).
IF....
N is REALLY BIG, then the 4th and 5th terms will be 2 and 3, so the median will = 2.5
N is REALLY SMALL, then the 4th and 5th terms will be -2 and 2, so the median will = 0
Neither of those answers is among the choices though, so N must be something relativelycloseto 2.....
IF....
N = 0, then the 4th and 5th terms will be 0 and 2, so the median will = 1. That answer IS among the choices, so we're done.
D
Answer: (D) 1