mathewhe/aqua_rat huggingface dataset:
Question: Donna, Chris, and Austin have a total of $93 in their wallets. Donna has $7 more than Chris. Austin has 3 times what Donna has. How much Austin has in his wallet?
(A) $10
(B) $20
(C) $12
(D) $60
(E) $50
Rationale: Let the amount in Chris’s wallet be $x
Donna has $7 more than Chrish's wallet = $(x + 7)
Austin has 3 times than Donna's wallet = $3(x + 7)
According to the problem, Donna, chris, and Austin have a total of $93 in their wallets.
Therefore,
x + (x + 7) + 3(x + 7) = 93
x + x + 7 + 3x + 21 = 93
5x + 28 = 93
- 28 -28 (subtract 28 from both sides)
5x = 65
or, 5x/5 = 65/5 (Divide both sides by 5)
x = 13
Amount in Chris wallet = $x = $13
Amount in Donna's wallet = $(x + 7) = $(13 + 7) = $20
Amount in Austin 's wallet = $3(x + 7) = $3(13 + 7) = $3(20) = $ 60
correct answer D
Answer: (D) $60
Question: A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
(A) 31:46
(B) 32:44
(C) 32:45
(D) 32:41
(E) 32:42
Rationale: Explanation:
A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
Answer: C
Answer: (C) 32:45
Question: At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
(A) 7%
(B) 2%
(C) 4%
(D) 9%
(E) 5%
Rationale: 150 = (750*5*R)/100
R = 4%
Answer: C
Answer: (C) 4%
Question: A tank with a volume of 45 cubic feet has one inlet pipe and 2 outlet pipes. The inlet pipe fills water into the tank at the rate of 5 cubic inches/min and the 2 outlet pipes empty it out at the rates of 14 cubic inches/min and 6 cubic inches/min respectively. If all 3 pipes are opened when the tank is full, how many minutes does it take to empty the tank? (1 foot = 12 inches)
(A) 5184
(B) 5276
(C) 5368
(D) 5460
(E) 5552
Rationale: The tank is emptied at this rate: 14 + 6 - 5 = 15 cubic inches/min
The tank has a volume of 45*12*12*12 = 77760 cubic inches.
The time it takes to empty the tank is 77760/15 = 5184 minutes.
The answer is A.
Answer: (A) 5184
Question: How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?
(A) 3
(B) 2
(C) 1
(D) 0
(E) 4
Rationale: The left side of the equation is obviously not negative and can only be 00 or 11 and plugin the values in and around the boundary in |x−1|+|x−6| and none satisfies.
Answer is D.
Answer: (D) 0
Question: In a certain corporation, there are 300 male employees and 150 female employees. It is known that 50% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 450 employees is chosen at random, what is the probability this employee has an advanced degree or is female?
(A) 1/2
(B) 2/3
(C) 3/4
(D) 4/5
(E) 5/6
Rationale: P(female) = 150/450 = 1/3
P(male with advanced degree) = 0.5*300/450 = 150/450 = 1/3
The sum of the probabilities is 2/3
The answer is B.
Answer: (B) 2/3
Question: A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.100 more. What was the Sum?
(A) s.350
(B) s.700
(C) s.200
(D) s.600
(E) s.500
Rationale: At 5% more rate, the increase in S.I for 10 years = Rs.100(given)
So, at 5% more rate, the increase in SI for 1 year = 100/10 = Rs.10/-
i.e. Rs.10 is 5% of the invested sum
So, 1% of the invested sum = 10/5
Therefore, the invested sum = 10 × 100/5 = Rs.200 Answer: C
Answer: (C) s.200
Question: 3889 + 13.952 - ? = 3854.002
(A) 48.95
(B) 49.75
(C) 45.97
(D) 47.59
(E) 45.79
Rationale: Let 3889 + 13.952 - x = 3854.002.
Then x = (3889 + 13.952) - 3854.002
= 3902.952 - 3854.002
= 48.95.
Answer is A
Answer: (A) 48.95
Question: Find the percentage gain if a shop owner sells 20 mtr of fabric and gains Cost Price of 5 meters?
(A) 15%
(B) 24%
(C) 35%
(D) 14%
(E) 25%
Rationale: Let cost of each metre be Rs. 100.
Therefore, cost price of 5 m cloth = 5*100 = 500
cost price of 20m cloth = 20*100 = 2000
Since the trader made the CP of 5m from the sale of 20m therefore
Profit on sale of 20m = cost price of 5 m cloth = 500
Selling price of 20m = 2000+500 = 2500
Profit % = 25.00%
E
Answer: (E) 25%
Question: Machine–A produces 50% of the total output and Machine-B produces 75% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 500 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective?
(A) 0.07
(B) 0.06
(C) 0.0048
(D) 0.04
(E) 0.05
Rationale: Suppose machine A produce 1000 units
total defective = 9
Machine B then produces=1500 units
total defective=3
probability=12/2500
=0.0048
Answer : C
Answer: (C) 0.0048
Question: The average weight of 47 students in a class is 52 kg. 5 of them whose average weight is 48 kg leave the class and other 5 students whose average weight is 54 kg join the class. What is the new average weight (in kg) of the class ?
(A) 51 1⁄47
(B) 52 30/47
(C) 52 1⁄30
(D) 43.42
(E) None of these
Rationale: Total weight of 47students
= 47 × 52 = 2444 kg
Total weight of 5 students who leave
= 5 × 48 = 240 kg
Total weight of 5 students who join
= 5 × 54 = 270 kg
Therefore, new total weight of 47 students
= 2444 – 240 + 270 = 2474
⇒ New average weight = 2474⁄47 = 52 30/47 kg
Answer B
Answer: (B) 52 30/47
Question: The number of rectangles that can be formed on a 8X8 chessboard is
(A) 2194
(B) 1284
(C) 1196
(D) 1296
(E) 1523
Rationale: 8 rows and 8 columns are separated by 9 vertical and 9 horizontal lines. Any 2 vertical line and any 2 horizontal line will make a rectangle, so total # of rectangles possible is C29∗C29=36∗36=1296C92∗C92=36∗36=1296.
Answer: D.
Answer: (D) 1296
Question: Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.6300. How much third person will get?
(A) 2400
(B) 2800
(C) 2699
(D) 2662
(E) 1821
Rationale: First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 6300 = 2800
Answer: B
Answer: (B) 2800
Question: There are 5 houses on a street: house A, B, C, D and E. The distance between any two adjacent houses is 100 feet. There are 2 children living in house A, 3 children living in house B, 4 children living in house C, 5 children living in house D and 6 children living in house E. If the school bus can only make one stop on that street, in front of which house should the bus stop so that the sum of walking distance among all children will be the least?
(A) house A
(B) house B
(C) house C
(D) house D
(E) house E
Rationale: If we move the stop from house A to house B, 2 children (who live in house A) will have to walk 100 feet more each, but 18 children (who live in house B & C & D & E) will walk 100 feet less each. Therefore, it will be 1800 – 200 =1600 feet less walk if we move the stop from house A to house B.
If we move the stop from house B to house C, 5 children (who live in house A & B) will walk 100 feet more each, but 15 children (who live in house C & D & E) will walk 100 feet less each. Therefore, it will save a total of 1500 – 500 =1000 feet if we move the stop from house B to house C.
If we move the stop from house C to house D, 9 children (who live in house A & B & C) will walk 100 feet more each, but 11 children (who live in house D & E) will walk 100 ft. less each. Therefore, it will be 1100 – 900 =200 feet less walk if we move the stop from house C to house D.
If we move the stop from house D to house E, 14 children (who live in house A & B & C & D) will walk 100 feet more each, but 6 children (who live in house E ) will walk 100 feet less each. Therefore, it will be 1400 – 600 = 800 feet more walk if we move the stop from house D to house E.
Therefore, children will walk the least if the bus stop is at House D.
correct answer D
Answer: (D) house D
Question: 8 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
(A) 4125
(B) 5040
(C) 1620
(D) 2430
(E) 2560
Rationale: it is like a circular arrangement.
total ways for n people to arrange in a circle is = factorial (n-1)
in this case n= 8
hence ans = 7 factorial = 5040
hence B
Answer: (B) 5040
Question: Q and R are two-digit positive integers that have the same digits but in reverse order. If the positive difference between Q and R is less than 30, what is the greatest possible value of Q minus R?
(A) 23
(B) 24
(C) 25
(D) 26
(E) 27
Rationale: A two-digit integer "ab" can be expressed algebraically as 10a+b.
Q-R=(10a+b)-(10b+a)=9(a-b)<30.
The greatest multiple of 9 which is less than 30 is 27.
The answer is E.
Answer: (E) 27
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
(A) 99
(B) 26
(C) 88
(D) 26
(E) 18
Rationale: 30 * 33/2 = 5/2 * 2 * x => x
= 99
Answer: A
Answer: (A) 99
Question: John & Mary go for an interview for 2 vacancies. The probability for the selection of John is 1/3 & whereas the probability forthe selection of Mary is 1/5. What is the probabilitynone of them are selected?
(A) 2/11
(B) 3/7
(C) 8/15
(D) 4/9
(E) 5/7
Rationale: Let A = the event that John is selected and B = the event that Dani is selected.
Given that P(A) = 1/3 and P(B) = 1/5
We know that A¯ is the event that A does not occur and B¯ is the event that B does not occur
Probability that none of them are selected
=P(A¯∩B¯)(∵ Reference : Algebra of Events)
=P(A¯).P(B¯) (∵ Here A and B are Independent Events and refer theorem on independent events)
=[ 1 - P(A) ][ 1 - P(B)]
=(1−13)(1−15)
=23×45=8/15
C
Answer: (C) 8/15
Question: The HCF of two numbers is 50 and the other two factors of their LCM are 11 and 12. What is the largest number.
(A) 462
(B) 450
(C) 480
(D) 504
(E) 600
Rationale: Explanation:
HCF of the two numbers = 50
HCF will be always a factor of LCM
50 is factor of LCM
other two factors are 11 & 12
Then the numbers are (50* 11) and (50 x 12)
= 550 and 600
Answer: Option E
Answer: (E) 600
Question: A car rental company charges for mileage as follows: 5 dollars per mile for the first n miles and 6 dollars per mile for each mile over n miles. How much will the mileage charge be, in dollars, for a journey of d miles, where d > n ?
(A) 8d-n
(B) 10d-n
(C) 6d-n
(D) 11d-n
(E) 5*(6d-n)
Rationale: Total Distance in miles = d
Millage Charge = 5n + (d-n) 6
5n --> For the first n Miles
(d-n)6--> Total no of miles after traveling the first n miles is d-n and the rate after n miles is 6 .
If we solve the equation , We will get 6d-n
Answer : C
Answer: (C) 6d-n
Question: A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among seven colors produced enough color codes to identify all but 5 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)
(A) 25
(B) 26
(C) 27
(D) 28
(E) 29
Rationale: 7C2 = 21
The number of test subjects is 21 + 5 = 26
The answer is B.
Answer: (B) 26
Question: In a city where all streets run east-to-west, all avenues run north-to-south, and all intersections are right angles as shown below, Jenn needs to walk from the corner of 1st Street and 1st Avenue to the corner of 6th Street and 3rd Avenue. If her friend Amanda is sitting on a bench on 4th Street halfway between 1st and 2nd Avenues, and Jenn chooses her path randomly from any route that will allow her to walk exactly seven blocks to her destination, what is the probability that Jenn will walk down 4th St. past Amanda?
(A) 1/42
(B) 1/21
(C) 1/7
(D) 1/3
(E) 1/2
Rationale: All routes between (1,1) to (3,6) = 7!/(2!*5!) = 21
Routes which pass 4th st. between 1st and 2nd avenue = routs between (1,1) to (1,4) * routs between (2,4) to (3,6)
= 1 * (3!/(2!*1!)) = 3
the probability that Jenn will walk down 4th St. past Amanda= 3 / 21 = 1 / 7
C is correct.
Answer: (C) 1/7
Question: The figure above shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit T on the height of vehicles that are allowed to use the tunnel?
(A) 5½ ft
(B) 7½ ft
(C) 8 ½ ft
(D) 9½ ft
(E) 10 ft
Rationale: Let's label the midpoint of the circle O. Since the base of the semi-circle is 20, we know that the diameter is 20 and, accordingly, the radius is 10.
We also know that the traffic lane is 12 feet long and there's an equal amount of space on either side, so the traffic lane extends 6 feet on either side of O. Let's call the leftmost point on the base of the traffic lane A. So, the distance OA is 6.
Now draw a line straight up from A to the top of the tunnel. Let's label the point at which the line intersects the circle B. The answer to the question will, therefore, be the height AB - .5 feet (we need to leave .5 feet of clearance).
Here's the key to solving the question: if we draw a line from O to B, that line is a radius of the circle and, therefore, has length 10. We now have right triangle OAB (the right angle is at point A), with leg OA=6 and hypotenuse OB=10. We can now solve for leg AB=8 (either by applying the pythagorean theorum or by applying the 3/4/5 special right triangle ratio).
Finally: AB=8, so the correct answer T is 8 - .5 = 7.5... choose (B)!
From a strategic guessing point of view, as soon as we realize that the height of the tunnel is 10 in the middle, we should quickly eliminate (D) and (E) as too big; worse case you have a 1/3 shot at picking up the points.B
Answer: (B) 7½ ft
Question: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <2>*<4>?
(A)  60
(B)  116
(C)  210
(D)  263
(E)  360
Rationale: Since <x>=2x+(2x-1)+(2x-2)+......2+1=1+2+..+(2x-2)+(2x-1)+2x then<x> is basically the sum of all integers from 1 to 2x, inclusive.
Hence <2> is the sum of all integers from 1 to 2*2=4 and <4> is the sum of all integers from 1 to 2*4=8 -> <4>=36 and <2>=10 --> <3>*<2>=21*10=360
Answer: E
Answer: (E)  360
Question: A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?
(A) 165
(B) 878
(C) 166
(D) 199
(E) 234
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
Question: Two cyclists are running on a circular track of 500m.They start from the same point A at the same time.If they meet for the first time at a point 150 m from A, which of the following could be the possible ratio of their speeds?
(A) 1:3
(B) 2:3
(C) 3:5
(D) 3:7
(E) 7:11
Rationale: well, the key aspect of this question is the distance mentioned, when they met for the first time, which is 150m away from the starting point. This distance between their first meeting can be less than 500 in only two circumstance.
1) if both are travelling in the same direction and one of them has been given a head start say by few meters.
2) or if both are travelling in the opposite direction.
now since they start their journey at the same time. hence option 1 is ruled out. hence they must be running in the opposite direction. hence answer should be 2:3.
Why?? here is a solution
As here time is constant, therefore, distance traveled will be directly proportional to the the respective velocities/speed of the two runners.(d1/d2=v1/v2)
now one of them has covered 150m, therefore other must have covered 350m, therefore ratio of their speeds/velocities will be 3:7
D
Answer: (D) 3:7
Question: Walking with 4/5 of my usual speed, I arrive at the bus stop 10 minutes later than normal. How many minutes does it take to walk to the bus stop at my usual speed?
(A) 32
(B) 36
(C) 40
(D) 44
(E) 48
Rationale: Let t = usual time = distance / usual speed
t + 10 = distance / (4* usual speed/5) = (5 * distance) / (4 * usual speed) = 5t/4
t = 40
The answer is C.
Answer: (C) 40
Question: AMOUNT OF BACTERIA PRESENT
Time Amount
1:00 P.M. 8.0 grams
4:00 P.M. x grams
7:00 P.M. 12.4 grams
Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?
(A) 12.0
(B) 12.1
(C) 12.2
(D) 10.0
(E) 10.4
Rationale: The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.
Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:
Fractional increase from 1 PM to 4 PM = X / 8.0
Fractional increase from 4 PM to 7 PM = 12.4 / X
X \ 8 = 12.4 \ X
X^2 = 12.4 * 8
X = 10
answer : D
Answer: (D) 10.0
Question: A man can swim in still water at 6 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?
(A) 3.9
(B) 4.2
(C) 5.3
(D) 2
(E) 5.2
Rationale: M = 5
S = x
DS = 6 + x
US = 6 - x
6 + x = (6 - x)2
6 + x = 12 -2x
3x = 6
x = 2
Answer:D
Answer: (D) 2
Question: The residents of Town X participated in a survey to determine the number of hours per week each resident spent watching television. The distribution of the results of the survey had a mean of 24 hours and a standard deviation of 12 hours. The number of hours that Pat, a resident of Town X, watched television last week was between 1 and 2 standard deviations below the mean. Which of the following could be the number of hours that Pat watched television last week?
(A) 30
(B) 20
(C) 18
(D) 12
(E) 6
Rationale: Given: mean=24 and SD=12
The number of hours that Pat watched television was between 1 and 2 standard deviations below the mean: 1 SD below the mean ismean-1*SD=12and 2 SD below the mean ismean-2*SD=0, so the number of hours that Pat watched television was between 0 and 12 hours.
Answer: E
Answer: (E) 6
Question: Let G represent a set of four distinct prime numbers. If the sum of the numbers in G is even and x is a member of G, then what is the least possible value that x can be?
(A) 1
(B) 2
(C) 3
(D) 5
(E) 7
Rationale: We are told that G is a set of four distinct prime numbers, and the sum of the numbers in G is even. Since the question involves odds, evens and the smallest possible prime, you should immediately think of 2, the smallest prime number and the only even prime number.
If 2 were a member of the set, then we would have one even number and three odd numbers. Since such sum would always be odd, 2 cannot be a member of the set. Thus, all the members of G must be odd to yield an even sum.
Since all the members of the set must be odd prime numbers, the answer is the smallest odd prime number, which is 3.
The correct answer is choice (C).
Note:1 is not a prime number.
Answer: (C) 3
Question: How many numbers between 11 and 90 are divisible by 7?
(A) 10
(B) 7
(C) 11
(D) 19
(E) 15
Rationale: The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
ANSWER C
Answer: (C) 11
Question: The sum of all two digit numbers divisible by 7 is
(A) 910
(B) 510
(C) 728
(D) 628
(E) 810
Rationale: Required numbers are 14,21,28 . . . . . 98
This is an A.P. in which a=14,d=7 and l=98.
Let the number of terms in it be n.Then t=98
So a+(n-1)d=98.
14+(n-1)*7=98,
14+7n-7=98
7+7n=98
7n=98-7
n = 91/7
then n=13.
Required sum
=n/2(a+l)
=13/2(14+98)
=728
Answer is C.
Answer: (C) 728
Question: If a2 + b2 + c2 = 390 and ab + bc + ca = 5, then a + b + c is
(A) 19
(B) 20
(C) 25
(D) 30
(E) 64
Rationale: by formula ,
(a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ca),
since, a^2 + b^2 + c^2 = 390 and ab + bc + ca = 5,
(a + b + c)^2 = 390 + 2(5)
= 400
= 20^2
therefore: a + b + c = 20
ANSWER:B
Answer: (B) 20
Question: A & B started a partnership business. A's investment was thrice the investment of B and the period of his investment was two times the period of investments of B. If B received Rs 5000 as profit , what is their total profit?
(A) 28000
(B) 30000
(C) 32000
(D) 35000
(E) None of these
Rationale: Explanation :
Suppose B's investment = x. Then A's investment= 3x
Suppose Bs period of investment = y, then A's period of investment = 2y
A : B = 3x * 2y : xy
= 6 : 1
Total profit * 1/7 = 5000
=> Total profit = 5000*7 = 35000. Answer : Option D
Answer: (D) 35000
Question: If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 6% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
(A) 2
(B) 5
(C) 300/53
(D) 110/53
(E) 9/4
Rationale: Let price be 10/ticket
then 220 tickets for this month's performance at its usual price=2200
which is 10%greater than last month(let X)---->x(1+10/100)=2200----->x=2000
now new revenue this month=200*10.6=2120
as per question asked 2000=2120(1-A/100)----->A=300/53%
Ans C
Answer: (C) 300/53
Question: The radius of a cylinder is 2 r units and height is 3 r units. Find the curved surface?
(A) 862
(B) 298
(C) 440
(D) 278
(E) 278
Rationale: r = 7
h = 3
2πr(h + r) = 2 * 22/7 * 7(10)
= 440
Answer: C
Answer: (C) 440
Question: A man buys a cycle for Rs.1400 and sells it at loss of 15%. What is the selling price of the cycle?
(A) 1192
(B) 1191
(C) 1118
(D) 1190
(E) 1112
Rationale: Explanation:
S.P = 85% of Rs.1400 ⇒ Rs.(8510085100 ×1400) = Rs.1190.
Answer: D
Answer: (D) 1190
Question: We can rent a boat for x cents per minute for the first 60 minutes and then y cents for each additional minute. How much will it cost to rent a boat for z minutes, where z>60?
(A) x + 60y
(B) 60x+(z-60)y
(C) x + zy
(D) 60y+(z-60)x
(E) 60x + zy
Rationale: The first 60 minutes cost 60x.
The next (z-60) minutes cost (z-60)y.
The total cost is 60x+(z-60)y.
The answer is B.
Answer: (B) 60x+(z-60)y
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 11.4 the sixth number is?
(A) 9.5
(B) 10
(C) 10.5
(D) 11.5
(E) 12
Rationale: Explanation:
1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 119.9 = 11.5
6th number = 11.5
Option D
Answer: (D) 11.5
Question: 4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number?
(A) 311
(B) 312
(C) 313
(D) 314
(E) 315
Rationale: Let no. be x
4/15*5/7*x - 4/9*2/5*x =8
by further solving
20x/105- 8x/45=8
4x/315=8
x=630
we have to find x/2
=630/2=315
ANSWER:E
Answer: (E) 315
Question: A certain stock exchange designates each stock with a two-letter or three-letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes without using the letter A?
(A) 16,250
(B) 16,500
(C) 16,750
(D) 17,000
(E) 17,250
Rationale: The number of two-letter codes is 25*25=625.
The number of three-letter codes is 25*25*25=15,625.
The total number of codes is 625+15,625=16,250.
The answer is A.
Answer: (A) 16,250
Question: The ages of two person differ by 20 years. If 8 years ago, the elder one be 5 times as old as the younger one, their present ages(in years) are respectively
(A) 33, 13
(B) 25, 5
(C) 29, 9
(D) 50, 30
(E) 20,10
Rationale: Let their ages be x and (x + 20) years.
Then, 5(x - 8) = (x + 20 - 8)
=> 4x = 52
=>x =13
Their present ages are 33 years and 13 year.
ANSWER:A
Answer: (A) 33, 13
Question: If 6 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is?
(A) 20:15:15
(B) 20:15:18
(C) 20:15:12
(D) 20:15:14
(E) 20:15:11
Rationale: 6A = 8B = 10C
A:B:C = 1/6:1/8:1/10
= 20:15:12
Answer: C
Answer: (C) 20:15:12