mathewhe/aqua_rat huggingface dataset:
Question: The guidelines at a certain college specify that for the introductory English class, the professor may choose one of four specified novels, and choose two plays from a list of 5 specified plays. Thus, the reading list for this introductory class has one novel and two plays. How many different reading lists could a professor create within these parameters?
(A) 15
(B) 40
(C) 90
(D) 150
(E) 360
Rationale: 4C1*5C2 = 4*10 = 40
The answer is B.
Answer: (B) 40
Question: Consider a right triangle AQC with length of sides being x,y and z where y represents the hypotenuse , Which of these represent the radius of the circle which is inscribed in triangle AQC
(A) [A] x+y+z/3
(B) [B] x-y+z/3
(C) [C] x-y+z/2
(D) [D] (x^2 + y^2 + z^2)/3
(E) [E] cannot be determined
Rationale: Radius of a circle inscribed in the right triangle AQC r = (x + z - y)/2 where y is the hypotenuse and x and y are other two sides.
Answer: C
Answer: (C) [C] x-y+z/2
Question: If a man buys 20 lollipops for $90 and sold them for $2 dollars determine his loss.
(A) $20
(B) $25
(C) $30
(D) $45
(E) $50
Rationale: Cost of 20 lollipops = $90
Sold each lollipop for $2
So he sold 20 lollipops for $(20 × 2) = $40
Loss = $(90 – 40) = $50
Therefore, loss = $ 50
correct answer E)$50
Answer: (E) $50
Question: A charitable association sold an average of 66 raffle tickets per member. Among the female members, the average was 70 raffle tickets. The male to female ratio of the association is 1:2. What was the average number E of tickets sold by the male members of the association
(A) 50
(B) 56
(C) 58
(D) 62
(E) 66
Rationale: Given that, Total average E sold is 66, Male/Female = 1/2 and Female average is 70. Average of Male members isX.
(70*F+X*M)/(M+F) = 66 -> Solving this equation after substituting 2M=F, X = 58. ANS C.
Answer: (C) 58
Question: 36 identical chairs must be arranged in rows with the same number of chairs in each row. Each row must contain at least three chairs and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible?
(A) 4
(B) 2
(C) 8
(D) 9
(E) 5
Rationale: Three conditions have to be satisfied.
1. The number of students per row has to be at least 3.
2. Number of row has to be at least 3.
3. Equal number of students has to be seated in a row.
The following arrangements satisfy all 3 conditions.
Arrangement 1: 3 students to a row; 12 rows.
Arrangement 2: 4 students to a row; 9 rows.
Arrangement 3: 6 students to a row; 6 rows.
Arrangement 4: 9 students to a row; 4 rows.
Arrangement 5: 12 students to a row; 3 rows.
You will observe that the number of students in a row is a factor of 36.
So, an alternative and faster approach is to list down factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, and 36.
And then start from 3 and quickly find out if the number of rows is at least 2.
Both the conditions are satisfied for the following factors : 3, 4, 6, 9, and 12. i.e., 5 arrangements.
Ans:E
Answer: (E) 5
Question: A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined
by B after 10 days. The number of days taken in completing the wotk were ?
(A) 1423kmph
(B) 1523kmph
(C) 1623kmph
(D) 1723kmph
(E) None of these
Rationale: Explanation:
Work done by A in l0 days = (1/25) *10 = 2/5
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100
9/100 work is done by them in 1 day
hence 3/5 work will be done by them in (3/5)*(100/9)
= 20/3days.
Total time taken = (10 + 20/3) = 16*(2/3) days
Answer: C
Answer: (C) 1623kmph
Question: The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?
(A) 20 years
(B) 21 years
(C) 23 years
(D) 25 years
(E) 26 years
Rationale: Explanation:
Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years. Answer: C
Answer: (C) 23 years
Question: A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
(A) 767 kmph
(B) 55 kmph
(C) 54 kmph
(D) 33 kmph
(E) 37 kmph
Rationale: D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
Answer: C
Answer: (C) 54 kmph
Question: Maxwell leaves his home and walks toward Brad's house at the same time that Brad leaves his home and runs toward Maxwell's house. If the distance between their homes is 80 kilometers, Maxwell's walking speed is 4 km/h, and Brad's running speed is 6 km/h, what is the distance traveled by Brad?
(A) 16
(B) 18
(C) 40
(D) 24
(E) 30
Rationale: Time taken = Total Distance / Relative Speed
Total Distance = 80 kms
Relative Speed (Opposite side) (As they are moving towards each other speed would be added ) = 6+ 4 = 10 kms/hr
Time taken = 80 / 10 = 8 hrs
Distance traveled by Brad = Brad's speed * time taken = 6 * 8 = 40 kms... Answer - C
Answer: (C) 40
Question: The average salary of all the workers in a workshop is Rs. 9500. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is?
(A) A)23
(B) B)21
(C) C)52
(D) D)56
(E) E)12
Rationale: Let the total number of workers be x. Then,
9500x
= (12000 * 7) + 6000(x - 7)
=> 3500x = 42000
= x =12.
Answer:E
Answer: (E) E)12
Question: Car A runs at the speed of 42km/hr & reaches its destination in 6hr. Car B runs at the speed of 52 km/h & reaches its destination in 9h. What is the respective ratio of distances covered by Car A & Car B?
(A) 11 : 6
(B) 7 : 13
(C) 13 : 7
(D) 15 : 6
(E) 13 : 6
Rationale: Sol. Distance travelled by Car A = 42 × 6 = 252 km
Distance travelled by Car B = 52 × 9 = 468 km
Ratio = 252/468= 7 : 13
B
Answer: (B) 7 : 13
Question: From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
(A) 2/1223
(B) 1/122
(C) 1/221
(D) 3/1253
(E) 2/153
Rationale: Let s is the sample space. then
n(s ) = 52C2 = 1326
E = event of getting 2kings out of 4
n(E) = 4C2 = 6
P(E) = 6/1326 = 1/221
Answer is C
Answer: (C) 1/221
Question: IF
1= O
2= T
3= T
4=F
5= F
THEN 6=?
(A) S
(B) J
(C) K
(D) L
(E) M
Rationale: IF
1= O... First letter of ONE.
2= T.... First letter of TWO.
3= T
4=F
5= F
THEN 6=S ... First letter of SIX.
ANSWER:A
Answer: (A) S
Question: The sum of the mean, the median, and the range of the set {1, 2, 4,6,8,10,12} equals which one of the following values?
(A) 2.6
(B) 14.23
(C) 22.14
(D) 8.6
(E) 10
Rationale: set {1, 2, 4,6,8,10,12}
Mean = (1+2+4+6+8+10+12)/7 = 6.14
Meadian = Middle term = 6
Range = Highest - Lowest = 12-1 = 10
Mean+Median+Range = 6.14+6+10 = 22.14
Answer: Option C
Answer: (C) 22.14
Question: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
(A) 9:1
(B) 9:2
(C) 9:8
(D) 9:5
(E) 9:0
Rationale: Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.
Answer: Option A
Answer: (A) 9:1
Question: 5, 11, 17, 25, 33, 43, ?
(A) 49
(B) 51
(C) 52
(D) 53
(E) 54
Rationale: The sequence is +6, +6, +8, +8, +10, ....
Answer : D.
Answer: (D) 53
Question: If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (x - 2y) is odd
II. xy^2 is an even negative integer
III. (y^2 - x) is an odd negative integer
(A) I only
(B) II only
(C) I and II
(D) I
(E) II and III
Rationale: If x is an odd negative integer and y is an even integer, which of the following statements must be true?
I. (x - 2y) is odd
x - is always negative and odd
2y - can be 0, when y=0, or always even integer (positivie or negative)
odd integer +/- even integer = always odd
I statement is always true
II. xy^2 is an even negative integer
y can be 0 => xy^2 = 0 (non-negative even) => II statement is not true
III. (y^2 - x) is an odd negative integer
y can be 0 => -x will be positive => III statement is not true
basing on above, only I statement will always be true
answer is D
Answer: (D) I
Question: What is the remainder if 7^8 is divided by 100?
(A) 7
(B) 1
(C) 9
(D) 10
(E) 11
Rationale: 7^8 can be written as (7^4)^2
If we divide 7^4 by 100 the reminder is 1
So, (7^4)^2 by 100, the reminder is 1^2 = 1
Answer : B
Answer: (B) 1
Question: If it takes a machine 2⁄3 minute to produce one item, how many items will it produce in 1 hours?
(A) 1⁄3
(B) 4⁄3
(C) 90
(D) 120
(E) 180
Rationale: 1 item takes 2/3 min so it takes 60 min to produce x
2x/3=60 the x=90
Answer:C
Answer: (C) 90
Question: Two identical trains AA and BB running in opposite direction at same speed tale 2 min to cross each other completely. The number of bogies of AA are increased from 12 to 16. How much more time would they now require to cross each other?
(A) 40 Sec
(B) 50 sec
(C) 60 sec
(D) 20 sec
(E) 25 sec
Rationale: Total initial bogies is 12+12=2412+12=24
Additional bogies =16âˆ’12=4
24 bogies is 12+12=24
additional bogies = 16 - 12 = 4
24 bogies take 2 min
2*60/24 *4=20 sec
ANSWER D
Answer: (D) 20 sec
Question: In how many Q ways can a four-letter password be chosen, using the letters A, B, C, D, E, and/or F, such that at least one letter is repeated within the password?
(A) 720
(B) 864
(C) 900
(D) 936
(E) 1,296
Rationale: Total number of four letter passwords = 6*6*6*6= 1296 ------(1)
Total number of passwords in which no letter repeats = 6C4*4!= 15*24=360------(2)
therefore required value Q= (1)-(2)= 1296-360=936.D
Answer: (D) 936
Question: When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of W?
(A)  7
(B)  8
(C)  9
(D)  10
(E)  11
Rationale: A simple and elegant solution.
As addition of 39, shifts mean by 2, and addition of 15, shifts mean by 1 to the other side, we have the mean lying between 3915, and in a ratio of 2:1
39-15 = 24
24 divide by 3 is 8.
Meaning mean of the n terms is 15+8 = 39-16 = 23
Now, from first statement, When a person aged 39 is added to a group of n people, the average age increases by 2.
W*23 +39 = 25*(W+1)
W = 7
Ans. (A)
Answer: (A)  7
Question: An item is being sold for $10 each. However, if a customer will “buy at least 3” they have a promo discount of 12%. Also, if a customer will “buy at least 10” items they will deduct an additional 8% to their “buy at least 3” promo price. If Sam buys 10 pcs of that item how much should he pay?
(A) $92.00
(B) $88.00
(C) $87.04
(D) $80.96
(E) $80.00
Rationale: Without any discount Sam should pay 10*10=$100. Now, the overall discount would be slightly less than 20%, thus he must pay slightly more than $80. Only answer choice D fits.
Answer: D.
Answer: (D) $80.96
Question: The labeled price of a table is Rs. 6,000. The shopkeeper sold it by giving 5% discount on the labeled price and earned a profit of 15%. What approximately is the cost price of the table?
(A) Rs. 4956
(B) Rs. 6195
(C) Rs. 6425
(D) Rs. 7200
(E) None of these
Rationale: Explanation:
Labeled price = Rs. 6,000
By giving 5% discount on labeled price, the selling price is
= 95 / 100 * 6,000 = Rs. 5700
By earning 15% profit on the selling of price Rs. 5700 the cost price is
= 100 / 115 * 5700 = Rs. 4956
Therefore, approximate cost is 4956
ANSWER A
Answer: (A) Rs. 4956
Question: A certain amount earns simple interest of 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
(A) 35
(B) 350
(C) 245
(D) Cannot be determined
(E) None of these
Rationale: Let p and r be the principal amount and rate of interest respectively.
Then, p×r×7/100=1750
or, pr = 25000
Now, SI = p×(r+2)×7/100=1750
We have to find the value of
p×(r+2)×7/100−p×r×7/100= M - 1750
M = SI when the rate of interest is 2% more. When we solve this equation, we find that we have two variables and one equation. Therefore, can’t be determined the correct answer.
Answer D
Answer: (D) Cannot be determined
Question: A and B walk around a circular track. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 10 a.m.?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 10
Rationale: Relative speed = Speed of A + Speed of B (? they walk in opposite directions)
= 2 + 3 = 5 rounds per hour
=> They cross each other 5 times in 1 hour
Time duration from 8 am to 10 am = 2 hour
Hence they cross each other 10 times before 10 am
Answer is E
Answer: (E) 10
Question: When the scores of the top four students are deleted, the average score of the remaining 66 students falls by 6 marks. Assume that it is possible for two or more students to have the same net score. What is the minimum score possible for the fourth ranking student if no student got a net score of more than 352?
(A) 280
(B) 320
(C) 300
(D) 340
(E) 350
Rationale: Explanation :
Sum of score of top four students = 1356
Maximum possible score possible for top three students = 352*3 = 1056
Hence the minimum possible score for the 4th student = 300
Answer : C
Answer: (C) 300
Question: A and B can do a work in 60 days; B and C can do it in 120 days; A and C can do it in 80 days. In what time A alone can do the work?
(A) 100
(B) 90
(C) 80
(D) 70
(E) 60
Rationale: (A+B)’s one day’s work = 1/60
(B+C)’s one day’s work = 1/120
(A+C)’s one day’s work = 1/80
Adding we get, 2(A+B+C)’ s one day’s work= 1/60 + 1/120 + 1/80
= 28/720
= 14/365
(A+B+C)’ s one day’s work = 14*1/365*2 = 14/720
So, A’s one day’s work = 14/720 - 1/120 = 1/90
A alone can do the work in 90 days.
ANSWER:B
Answer: (B) 90
Question: In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 200 m. Then, A wins by:
(A) 60 m
(B) 100 m
(C) 43m
(D) 20m
(E) 23m
Rationale: To reach the winning post A will have to cover a distance of (500 - 200)m, i.e., 300 m.
While A covers 3 m, B covers 4 m.
While A covers 300 m, B covers 4 x 300/3	m	= 400 m.
Thus, when A reaches the winning post, B covers 400 m and therefore remains 100 m behind.
A wins by 100 m.
Answer:	B
Answer: (B) 100 m
Question: IN Common Wealth Games an athlete runs 280 meters in 24 seconds. Hisspeed is ?
(A) 25 km/hr
(B) 42.42 km/hr
(C) 30 km/hr
(D) 32 km/hr
(E) None of these
Rationale: Explanation :
(280/1000)/(24/3600)=42.42 km/hr
Answer : B
Answer: (B) 42.42 km/hr
Question: In a throw of dice what is the probability of geÆ«ng number greater than 1
(A) 1/2
(B) 1/3
(C) 1/5
(D) 5/6
(E) None of these
Rationale: Explanation:
Number greater than 1 is 2, 3, 4, 5 & 6, so only 5 number
Total cases of dice = [1,2,3,4,5,6]
So probability = 5/6
Answer: E
Answer: (E) None of these
Question: 4^5^6^7^8^9^10 find unit digit
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: guys first find out unit digit of4^5ie is 4 then
4^6 unit digit its 6
then 6^7^8^9^10=6 bcz 6 unit digit alway 6
ANSWER:E
Answer: (E) 6
Question: Pipe A can fill a tank in 16 minutes and pipe B cam empty it in 24 minutes. If both the pipes are opened together after how many minutes should pipe B be closed, so that the tank is filled in 30 minutes?
(A) 11
(B) 21
(C) 33
(D) 51
(E) 61
Rationale: Explanation:
Let the pipe B be closed after x minutes.
30/16 - x/24 = 1 => x/24 = 30/16 - 1 = 14/16
=> x = 14/16 * 24 = 21.
Answer:B
Answer: (B) 21
Question: Simple interest on a sum at 9% per annum for 2 years is Rs.200. The C.I. on the same sum for the same period is?
(A) 201.1
(B) 201.5
(C) 200.4
(D) 109.1
(E) 81.1
Rationale: SI = 100 + 100
CI = 100 + 100 + 1.1 = 201.1
Answer: A
Answer: (A) 201.1
Question: A walks at 30 kmph and 30 hours after his start, B cycles after him at 55 kmph. How far from the start does B catch up with A?
(A) 1200km
(B) 1980km
(C) 2000km
(D) 2500km
(E) 3600km
Rationale: Suppose after x km from the start B catches up with A. Then, the difference in the time taken by A to cover x km and that taken by B to cover x km is 30 hours.
x/30 - x/55 = 30
x = 1980 km
Answer is B
Answer: (B) 1980km
Question: A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:
(A) 8.5 km/hr
(B) 10 km/hr
(C) 12.5 km/hr
(D) 9 km/hr
(E) 8 km/hr
Rationale: Man's speed with the current = 15 km/hr
=> speed of the man + speed of the current = 15 km/hr
speed of the current is 2.5 km/hr
Hence, speed of the man = 15 - 2.5 = 12.5 km/hr
man's speed against the current = speed of the man - speed of the current
= 12.5 - 2.5 = 10 km/hr
ANSWER:B
Answer: (B) 10 km/hr
Question: Jack and Jill are marathon runners. Jack can finish a marathon (42 km) in 6 hours and Jill can run a marathon in 4.2 hours.What is the ratio of their average running speed? (Jack: Jill)
(A) 7/10
(B) 15/14
(C) 4/5
(D) 5/4
(E) Cannot be determined
Rationale: Average Speed of Jack =Distance/ Time
= 42 /6= 7
Average Speed of Jill = 42/(4.2) = 10
Ratio of Average speed of Jack to Jill = 7/10
Answer A
Answer: (A) 7/10
Question: A man can row downstream at 22 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
(A) 13, 3
(B) 12, 6
(C) 15, 3
(D) 14, 4
(E) 16, 6
Rationale: Explanation:
Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 22 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 32 => x = 16, y = 6.
Answer: Option E
Answer: (E) 16, 6
Question: If x=y=2z and xyz=256 then what is the value of x?
(A) 12
(B) 8
(C) 16
(D) 6
(E) 2
Rationale: Given that x=y=z.
so,x=y,
x=2z,
y=2z.
given xyz=256
=>(2z)*(2z)*z=256
=>4z^3=256
=>z=4
then x=2z
=2(4)
=8
ANSWER:B
Answer: (B) 8
Question: Harkamal purchased 8kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 50 per kg. How much amount did he pay to the shopkeeper?
(A) A)1000
(B) B)1010
(C) C)1065
(D) D)1075
(E) E)1080
Rationale: Cost of 8 kg grapes = 70 × 8 = 560.
Cost of 9 kg of mangoes = 50 × 9 = 450.
Total cost he has to pay = 560 + 450 = 1010.
B)
Answer: (B) B)1010
Question: There are 1000 students in a school and among them 20% of them attends chess class. 10% of the students who are in the chess class are also enrolled for swimming. No other students are interested in swimming so how many will attend the swimming class if all enrolled attends?
(A) 1
(B) 10
(C) 100
(D) 50
(E) 20
Rationale: 20% of 1000 gives 200. so 200 attends chess and 10% of 200 gives 20.
so 20 enrolled for swimming
Answer: E
Answer: (E) 20
Question: Find the odd man out. 1,4,9,16,20
(A) 8
(B) 10
(C) 25
(D) 6
(E) 14
Rationale: Explanation :
1^2=1
2^2=4
3^2=9
4^2=16
5^2=25
Hence, it is clear that 25 should have come instead of 20
Answer : Option C
Answer: (C) 25
Question: Havid has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?
(A)  5/6*d
(B)  7/3*d
(C)  10/3*d
(D)  7/2*d
(E)  9/2*d
Rationale: Havid has d books;
Jeff has d/3 books;
Paula has 2d books;
Total = d+d/3+2d=10d/3.
Answer: C.
Answer: (C)  10/3*d
Question: Mike needs 30% to pass. If he scored 212 marks and falls short by 25 marks, what was the maximum marks he could have got?
(A) 343
(B) 677
(C) 790
(D) 867
(E) 932
Rationale: If Mike had scored 25 marks more, he could have scored 30%
Therefore, Mike required 212 + 25 = 237 marks
Let the maximum marks be m.
Then 30 % of m = 237
(30/100) × m = 237
m = (237 × 100)/30
m = 23700/30
m = 790
Answer:C
Answer: (C) 790
Question: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 24 white marbles. How many red marbles could be in bag A?
(A) 1
(B) 3
(C) 4
(D) 6
(E) 8
Rationale: 6 is the answer.
Bag A-
R:W:B = 2:6:9
Let W in bag A be 6K
Bab B -
R:W = 1:4
Let W in bag B be 4P
W = 24 = 6K+4P
=> K =2, P=3
Total Red's in bag A will be 2K = 4
C
Answer: (C) 4
Question: The timing of a college is from 12 p.m to 4.20 p.m. Five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students. Find the duration of each lecture.
(A) 52 minutes
(B) 45 minutes
(C) 30 minutes
(D) 48 minutes
(E) 58 minutes
Rationale: Explanation:
Total time a student spends in college = 4 hours 20 minutes = 260 minutes
As there are 5 lectures, the number of breaks between lectures is 4. Total time of the break = 20 minutes
Hence, the duration of each lecture is = (260 – 20)/5
= 48 minutes
ANSWER D
Answer: (D) 48 minutes
Question: On dividing 12401 by a certain number, we get 76 as quotient and 13 as remainder. What is the divisor?
(A) 163
(B) 785
(C) 852
(D) 456
(E) 852
Rationale: Divisor *quotient+remainder=dividend
divisor=(dividend)-(remainder)/quotient
(12401-13)/76=163
Answer(A)
Answer: (A) 163
Question: Three people each took 5 tests. If the ranges of their scores in the 5 practice tests were 15, 25 and 30, what is the minimum possible range in scores of the three test-takers?
(A) 25
(B) 30
(C) 35
(D) 15
(E) 10
Rationale: I simply looked at the 3 different possible scores for each individual test: 15,30,25
We have to find the minimum range:
30-15=15
30-25=5
25-15=10
The find the minimum range, you have to make the set of the 5 scores as small as possible.
Which means that 4 of the 5 scores of each individual person is zero.
5*5 = 25.
Answer:A
Answer: (A) 25
Question: There are 30 people in a party. If everyone is to shake hands with one another, how many hand shakes are possible?
(A) 435
(B) 367
(C) 361
(D) 297
(E) 261
Rationale: Explanation:
Total number of persons = n = 30
Shakehands involve only 2 persons = r = 2
Number of shakehands = nCr = 30C2
30C2 = (30 * 29) /(2 * 1) = 435
nCr = (n!) / r! (n – r)!
= 30! / 2! 28!
= 435
ANSWER:A
Answer: (A) 435
Question: A pump can fill a tank with a water in 2 hours. Because of a leak, it took 2 and 4/5 hours to fill the tank. The leak can drain all the water of the full tank in how many hours?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Rationale: The rate of the pump + leak = 5/14
1/2 - leak's rate = 5/14
leak's rate = 1/2 - 5/14 = 1/7
The leak will empty the tank in 7 hours.
The answer is D.
Answer: (D) 7
Question: P alone can complete a piece of work in 6 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) 491/2)
(B) 4(1/2)
(C) 4(1/8)
(D) 7(1/2)
(E) 4(5/2)
Rationale: Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.
Answer:D
Answer: (D) 7(1/2)
Question: At present George is 10 years older than Jason. After 5 years George will be twice the age of Jason. Find their present age.
(A) 2 and 12
(B) 6 and 16
(C) 4 and 14
(D) 10 and 20
(E) 5 and 15
Rationale: Let the present age be x and x+10
After 5 years ages will be x+5 and x+15
2 (x + 5) = x + 15
x = 5
Hence the present ages are 5 and 15
Answer : E
Answer: (E) 5 and 15