mathewhe/aqua_rat huggingface dataset:
Question: If 15 machine can finish a job in 36 days, then how many more machines would be needed to finish the job in one -fourth less time?
(A) a. 5
(B) b. 8
(C) c. 10
(D) d. 12
(E) e. 16
Rationale: You might think of this in a management context - we can use the principle of 'person-hours' to solve any problem where we have identical workers. So, using simpler numbers, suppose you know that 6 identical employees, working simultaneously, would finish a job in 5 hours. Then that job requires 6*5 = 30 total hours of person-work. If instead you wanted the job done in 3 hours, you'd assign 30/3 = 10 employees to do the job, because you want to get a total of 30 hours of work from the employees.
We can solve this problem identically. If 15 machines (identical ones, I assume) work simultaneously for 36 days, they will do a total of 15*36 machine-days of work. So the job requires 15*36 days of machine work in total. We instead want the job done in 1/4 less time, so in 27 days. So we'll need 15*36/27 = 20 machines, or 5 additional machines.
A
Answer: (A) a. 5
Question: If k is a multiple of 3 and k=(m2)nk=(m2)n, where m and n are prime numbers, which of the following must be a multiple of 9?
(A) m^2
(B) n^2
(C) mn
(D) mn^2
(E) (mn)^2
Rationale: must be a multiple of 9
m and n are prime...lets have 2 and 3
now either of then can be a 3...
a) m can be 2 and n = 3...out
b) n can be 2 and m = 3...out
c) m = 2 and n=3 ...out
d) n can be 2 and m = 3...out
e) m = 2/3 or n = 3/2 both satisfies .......must be a multiple of 9
ANSWER:E
Answer: (E) (mn)^2
Question: A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 200 kilometers?
(A) 4.5
(B) 5
(C) 6.5
(D) 7.5
(E) 8.5
Rationale: Each 40 kilometers, 1 gallon is needed. We need to know how many 40 kilometers are there in 180 kilometers?
200 ÷ 40 = 5 × 1 gallon = 5 gallons
correct answer B
Answer: (B) 5
Question: A series of number is given: 5, 6, 12, 5, 7, 13, 5,__14, 5, 9, 15.. What number fills the blank?
(A) 6
(B) 5
(C) 9
(D) 8
(E) 11
Rationale: Looking at the series we can see 1 is added to 6, then to 7, then to 8 and so on...
The answer is option D (8)
Answer: (D) 8
Question: A, B and C start a business each investing 2,000. After 6 months A withdrew 1000, B withdrew 1000 and C invests 1000 more. At the end of the year, a total profit of 6600 was recorded. Find the share of B.
(A) 20,000
(B) 21,200
(C) 1,800
(D) 20,500
(E) None of these
Rationale: Ratio of the capitals of A, B and C
= 2000 Ã— 6 + 1000 Ã— 6 : 2000 Ã— 6 + 1000 Ã— 6 : 2000 Ã— 6 + 3000 Ã— 6
= 18000 : 18000 : 30000 = 18 : 18 : 30
Bâ€™s share = (6600 Ã— 18â„66) = 1800
Answer C
Answer: (C) 1,800
Question: A worker is paid x rupees for painting every 5 metres of a wall and y rupees for painting every extra metre. During one week, he painted 5 metres on Monday, 6 metres on Tuesday, 7 metres on Wednesday, 8 metres on Thursday and 9 metres on Friday. What is his average daily earning rupees for the five day week?
(A) x + 2y
(B) (5x + 9y)/5
(C) 10x + (8/5)y
(D) None of these
(E) Cannot be determined
Rationale: Explanation :
On Monday, the payment = Rs x
On Tuesday, the payment = Rs (x + y)
On Wednesday, the payment = Rs (x + 2y)
On Thursday, the payment = Rs (x + 3y)
On Friday, the payment = Rs (x + 4y)
Average daily earning during the five-day week = (5x + 10y)/5
= x + 2y
Answer : A
Answer: (A) x + 2y
Question: If x is a positive, single-digit integer such that 2/3*x, 2x, x, and x + 2, and 3x – 2 form a non-ordered list of consecutive integers, which of the following could be the median of that list?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
Rationale: The big hints are thatx is a 1-digit integerAND (2/3)x, 2x, x, and x + 2, and 3x – 2 are ALLintegers
For (2/3)x to be an integer, x MUST BE A MULTIPLE OF 3. So, x can equal 3, 6, or 9.
Now let's plug each possible value of x into the terms and see which one yields a list ofconsecutive integers
When x =3, we get the following terms: (2/3)(3), 2(3),3,3+ 2, and 3(3) – 2.
When we evaluate these terms we get: 4, 6, 3, 5 and 7
Rearrange to get: 2, 4, 5, 6, 7. So, when x = 3, we getconsecutive integers.
In this case, the MEDIAN = 5
Since 5 is among the answer choices, we need not test the other 2 values of x (i.e., x = 6 and x = 9)
Answer:
C
Answer: (C) 5
Question: The average age of father and his two sons is 38 Years. Five years ago, the average age of the two sons was 16 Years. If the difference between the ages of the two sons is four years, what is the present age of the father?
(A) 45
(B) 46
(C) 72
(D) 48
(E) 49
Rationale: The total present age of father and two sons is 3x38 = 114 yrs
The total present age of sons is (16+5) X 2 = 42Years
so, present age of father is 114 â€“ 42 = 72 yrs
Answer C
Answer: (C) 72
Question: Sushil got thrice as many marks in English as in Science. His total marks in English, Science and Maths are 162. If the ratio of his marks in English and Maths is 3:5, find his marks in Science?
(A) 18
(B) 77
(C) 66
(D) 55
(E) 41
Rationale: S:E = 1:3
E:M = 3:5
------------
S:E:M = 3:9:15
3/27 * 162 = 18
Answer:A
Answer: (A) 18
Question: Three persons A, B and C divide a certain amount of money such that A's share is Rs.4 less than half of the total amount. B's share is Rs.8 more than half of what is left and finally C takes the which is Rs.14. Find the total amount they initially had with them?
(A) Rs.68
(B) Rs.85
(C) Rs.80
(D) Rs.56
(E) Rs.29
Rationale: Let the total amount be Rs.P
Let shares of A and B be Rs.X and Rs.Y respectively.
Given, C's share was Rs.14.
X + Y + 14 = P --- (1)
From the given data, X = P/2 - 4 --- (2)
Remaining amount = P - (P/2 - 4) = P/2 + 4
Y = 1/2(P/2 + 4) + 8
=> P/4 + 10 --- (3)
From (1), (2) and (3)
P/2 - 4 + P/4 + 10 + 14 = P
=> 3P/4 + 20 = P
=> P - 3P/4 = 20
=> P/4 = 20 => P = Rs.80
Answer: C
Answer: (C) Rs.80
Question: 40 is subtracted from 60% of a number, the result is 50. Find the number?
(A) 150
(B) 388
(C) 278
(D) 277
(E) 268
Rationale: (60/100) * X – 40 = 50
6X = 900
X = 150
Answer: A
Answer: (A) 150
Question: In what ratio must rice of Rs.32per kg be mixed with rice of Rs.42 per kg so that cost of mixture is Rs.20 per kg?
(A) 13:6
(B) 13:5
(C) 11:6
(D) 11:5
(E) 12:5
Rationale: (20-42)/(32-20)=22/12=11:6
ANSWER:C
Answer: (C) 11:6
Question: The average weight of a class is x pounds. When a new student weighing 60 pounds joins the class, the average decreases by 1 pound. In a few months the student’s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds. None of the other students’ weights changed. What is the value of x?
(A) 85
(B) 86
(C) 88
(D) 90
(E) 92
Rationale: When the student weighs 80 pounds the average weight is x - 1 pounds;
When the student weighs 110 pounds the average weight is x + 4 pounds.
So, the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of (x + 4) - (x - 1) =5 pounds, which means that there are 30/5 = 6 students (including the new one). So, initially there were 5 student.
Total weight = 5x + 80 = 6(x-1) --> x = 88 pounds.
Answer: C.
Answer: (C) 88
Question: A man's age is 125% of what it was 10 years ago, but 83 1/3%
of what it will be after 10 years. What is his present age?
(A) 70
(B) 60
(C) 50
(D) 40
(E) 35
Rationale: Age before 10 years =x
125x/100=x+10
125x=100x+1000
x=1000/25=40
Present age =x+10=40+10=50
answer :C
Answer: (C) 50
Question: If the speed of X inch per second is equivalent to the speed of Y meter per hour, what is Y in terms of X? (1 meter=39.3701 inch)
(A) 21.44X
(B) 33.35X
(C) 67.77X
(D) 91.44X
(E) 111.3X
Rationale: X inch per second -->
--> 3,600X per hour(as there are 3,600 seconds in one hour);
--> 3,600X/39.3701=91.44X meter per hour (as there are 39.3701 inch in one meter).
Answer: D.
Answer: (D) 91.44X
Question: A, B and C started a business with capitals of Rs. 8000, Rs. 10000 and Rs. 12000 respectively. At the end of the year, the profit share of B is Rs. 3500. The difference between the profit shares of A and C is?
(A) 621
(B) 1400
(C) 236
(D) 600
(E) 211
Rationale: Explanation:
Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6
And also given that, profit share of B is Rs. 3500
=> 5 parts out of 15 parts is Rs. 3500
Now, required difference is 6 - 4 = 2 parts
Required difference = 2/5 (3500) = Rs. 1400
Answer:B
Answer: (B) 1400
Question: If x = a!, which of these following values of a is the least possible value (in this list) for which the last 9 digits of the integer x will all be zero?
(A) 26
(B) 32
(C) 37
(D) 44
(E) 49
Rationale: 2*5 will give one 0 at the end. We need 2^9*5^9 to get 9 zeroes at the end.
There are many 2's, so we need to find the first 9 appearances of 5.
These multiples of 5 are: 5, 10, 15, 20, 5*5, 30, 35, 40...which gives us 9.
To have at least 9 zeroes at the end, a => 40.
The answer is D.
Answer: (D) 44
Question: 36 men can complete a piece of work in 18 days. In how many days will 54 men complete the same work ?
(A) 24
(B) 77
(C) 88
(D) 29
(E) 12
Rationale: Explanation:
Less Men, means more Days {Indirect Proportion}
Let the number of days be x
then,
54 : 36 :: 18 : x
x = 12
Answer: E) 12 days
Answer: (E) 12
Question: The function f(Z) represents the number of ways that prime numbers can be uniquely summed to form a certain number Z such that Z = a + b + c + d… where those summed variables are each prime and a ≤ b ≤ c ≤ d ... For instance f(8) = 3 and the unique ways are 2 + 2 + 2 + 2 and 2 + 3 + 3 and 3 + 5. What is f(12)?
(A) 4
(B) 5
(C) 6
(D) 9
(E) 7
Rationale: Its better to start with 2 and check whether sum of two primes is primes is even.
1) 2(6 times)
2) 2(3 times) + 3(2 times)
3) 2(2 times) + 3 + 5
4) 2 + 3 + 7
5) 2 + 5 + 5
6) 3(4 times)
7) 5 + 7
Answer: E
Answer: (E) 7
Question: Machine A can finish a job in 4 hours, Machine В can finish the job in 12 hours, and Machine С can finish the job in 6 hours. How many hours will it take for A, B, and С together to finish the job?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: The combined rate is 1/4+1/12+1/6 = 1/2 of the job per hour.
The time to complete the job is 2/1 = 2 hours.
The answer is B.
Answer: (B) 2
Question: A certain protective additive increases from 20 days to 25 days the time between required maintenance checks on an industrial vehicle. By what percent is the time between maintenance checks increased by using the additive?
(A) 25%
(B) 33 1/3%
(C) 50%
(D) 66 2/3%
(E) 75%
Rationale: General formula for percent increase or decrease, (percent change):
percent=Change/Original∗100
So, the time between maintenance checks increased by 25−20/20∗100=25
Answer: A.
Answer: (A) 25%
Question: Rahul went to a shop and bought things worth Rs. 30, 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) Rs 15
(B) Rs 12.10
(C) Rs 24.70
(D) Rs 36.80
(E) None of these
Rationale: Explanation :
Let the cost of the tax items = x
Given that tax rate = 6%
Cost of tax items, x*(6/100) = 30/100
x = 5 rs
Now non taxable item = total cost - cost for taxable item - tax
= 30 - 5 - 0.30
=24.70
Answer : C
Answer: (C) Rs 24.70
Question: A 360 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?
(A) 229
(B) 108
(C) 278
(D) 216
(E) 112
Rationale: S = 360/6 * 18/5 = 216 kmph
Answer: D
Answer: (D) 216
Question: 7 points are marked on a straight line and another 5 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points?
(A) 91
(B) 105
(C) 175
(D) 21
(E) 14
Rationale: getting the answer 175 which is not in the choices, what is the source of the problem?
7C2* 5C1 + 5C2* 7C1 = 175
C
Answer: (C) 175
Question: It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill 1/32 of the dish?
(A) 25
(B) 26
(C) 27
(D) 28
(E) 29
Rationale: The bacteria doubles each day, so after 29 days, the dish was half full.
After 28 days, the dish was one quarter full.
After 27 days, the dish was one eighth full.
After 26 days, the dish was one sixteenth full.
After 25 days, the dish was 1/32 full.
The answer is A.
Answer: (A) 25
Question: Mr. Evans will states that each of his children will receive an equal share of his estate and that his grandchildren will split a portion of the estate that is equal to the share received by each of his children. If Mr. Evans has 3 children and 5 grandchildren, then approximately what percentage of Mr. Evans estate will each grandchild receive?
(A) 20%
(B) 17%
(C) 5.0%
(D) 3.3%
(E) 2.8%
Rationale: 25/5=5
Answer : C
Answer: (C) 5.0%
Question: X and Y started a business with capitals Rs. 20000 and Rs. 25000. After few months Z joined them with a capital of Rs. 30000. If the share of Z in the annual profit of Rs. 50000 is Rs. 14000, then after how many months from the beginning did Z join?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 5
Rationale: Investments of X, Y and Z respectively are Rs. 20000, Rs. 25000 and Rs. 30000
Let investment period of Z be x months.
Ratio of annual investments of X, Y and Z is (20000 * 12) : (25000 * 12) : (30000 * x)
= 240 : 300 : 30x = 8 : 10 : x
The share of Z in the annual profit of Rs. 50000 is Rs. 14000.
=> [x/ (18 + x)] 50000 = 14000 => [x/ (18 + x)] 25 = 7
=> 25x = 7x + (18 * 7) => x = 7 months.
Z joined the business after (12 - 7) months. i.e., 5 months.
Answer:E
Answer: (E) 5
Question: Farmer Charlie has a chicken farm. On a certain day, Charlie calculates in how many days he will run out of chicken-food. He notices that if he would sell 75 of his chickens, he could feed the remaining chickens twenty days longer with the chicken-food he has, and that if he would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier.
How many chickens does farmer Charlie have?
(A) 100
(B) 500
(C) 300
(D) 400
(E) 600
Rationale: C
300
Let n be the number of chickens and let d be the number of days in which Charlie will run out of chicken-food with this number of chickens. The total amount of chicken-food always equals the number of chickens multiplied by the number of days the food lasts. Therefore, the total amount of chicken-food equals n × d.
If Charlie would sell 75 of his chickens, he could feed the remaining chickens twenty days longer, so the total amount of chicken-food would then equal
( n - 75 ) × ( d + 20 ).
But the total amount of chicken-food does not change, so
( n - 75 ) × ( d + 20 ) = n × d
which can be rewritten to equation 1:
d = 4/15 × n - 20.
If Charlie would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier, so the total amount of chicken-food would then equal
( n + 100 ) × ( d - 15 ).
But the total amount of chicken-food does not change, so
( n + 100 ) × ( d - 15 ) = n × d
which can be rewritten to equation 2:
d = 3/20 × n + 15.
Combining equations 1 and 2 gives:
4/15 × n - 20 = 3/20 × n + 15.
Solving this equations gives n = 300, so farmer Charlie has 300 chickens.
Answer: (C) 300
Question: If the perimeter of a rectangular garden is 800 m, its length when its breadth is 100 m is?
(A) 300 m
(B) 777 m
(C) 200 m
(D) 167 m
(E) 128 m
Rationale: 2(l + 100) = 800 => l
= 300 m
Answer:A
Answer: (A) 300 m
Question: At an election 2 candidates are participated a candidate got 30% of votes and defeated by 5000 votes. 100 votes are invalid. Find the total polled votes?
(A) 11520
(B) 12600
(C) 15000
(D) 13560
(E) 12560
Rationale: winners votes = 100-30 = 70%
invalid votes = 100
polled votes = [(100*5000)/(2*70-100)]+100 = 12600
Answer is B
Answer: (B) 12600
Question: The sum of how many terms of the series 6 + 12 + 18 + 24 + ... is 1800 ?
(A) 19
(B) 23
(C) 13
(D) 26
(E) 24
Rationale: This is an A.P. in which a = 6, d = 6 and Sn = 1800
Then,	n/2*	[2a + (n - 1)d] = 1800
n/2*	[2 x 6 + (n - 1) x 6] = 1800
2
3n (n + 1) = 1800
n(n + 1) = 600
n^2 + n - 600 = 0
n^2 + 25n - 24n - 600 = 0
n(n + 25) - 24(n + 25) = 0
(n + 25)(n - 24) = 0
n = 24
Number of terms = 24
Answer is E.
Answer: (E) 24
Question: In the number 11,0AB, A and B represent the tens and units digits, respectively. If 11,0AB is divisible by 55, what is the greatest possible value of B × A?
(A) 0
(B) 5
(C) 10
(D) 15
(E) 25
Rationale: You should notice that 55*2=110 so 11,000 is divisible by 55: 55*200=11,000 (or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55) --> B*A=0*0=0.
Next number divisible by 55 is 11,000+55=11,055: B*A=5*5=25 (next number won't have 110 as the first 3 digits so we have only two options 0 and 25).
Answer: E
Answer: (E) 25
Question: If the average (arithmetic mean) of 5 positive temperatures is y degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be:
(A) 6y
(B) 4y
(C) 5y/3
(D) 3y/2
(E) 3y/5
Rationale: The sum of three greatest should be more than sum of two lowest.
The total sum is; 5y
A. 6y; 6y is more than 5y. Not possible.
B. 4y; 5y-4y=y(Possible)
C. 5y/3; 10y/3; 10y/3 > 5y/3. Not possible
D. 3y/2; 7y/2; 7y/2 > 3y/2. Not possible
E. 3y/5; 22y/5; 22y/5 > 3y/5. Not possible.
Ans:B
Answer: (B) 4y
Question: A trade analyst announced 10% reduction in the unit price of bike. As a result, the sales volume went up by 5%. What was the net effect on the sales revenue?
(A) no change
(B) decreases by 5.5%
(C) increases by 5.5%
(D) increases by 5%
(E) None of these
Rationale: Explanation :
Reduction of price = (100 -10)% = 90% = 0.9
Increase of sale = (100+5)% = 105% = 1.05
Total effect = 0.9 X 1.05 =94.5%, Decreases by 5.5%
Answer : B
Answer: (B) decreases by 5.5%
Question: E is the sister of F and G. H is F's daughter. I is F's son. Which of the following statements is definitely false?
(A) E is F's sister
(B) G is E's uncle
(C) F is G's brother
(D) G is I's uncle
(E) I is G's nephew
Rationale: Explanation:
E,F and G are children of same parents. So. G cannot be E's uncle.
Answer: B
Answer: (B) G is E's uncle
Question: Three partner’s A,B,C start a business. Twice A’s capital is equal to thrice B’s capital and B’s capital is four times C’s capital. Out of a total profit of Rs.16500 at the end of the year, B’s share is:
(A) Rs.4000
(B) Rs.6000
(C) Rs 7500
(D) Rs 7500
(E) Rs.6600
Rationale: Let C = x. Then B = 4x and 2A = 3 * 4x = 12x. or A = 6x.
Therefore, A : B : C = 6x : 4x : x = 6 : 4 : 1.
Therefore, B’s capital = Rs. 16500 * 4/11
= Rs.6000.
ANSWER:B
Answer: (B) Rs.6000
Question: If 3<x<6<y<7, then what is the greatest possible positive integer difference of x and y?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: 3 < x < 6 < y < 7;
3 < x
y < 7
3 + y < x + 7
y - x < 4.
Positive integer difference is 3(for example y = 6.5 and x = 3.5)
Answer:A.
Answer: (A) 3
Question: N, which is a set of natural numbers is partitioned into subsets S1 = {1}, S2 = {2,3}, S3 = {4,5,6}, S4 = {7,8,9,10} and so on. The sum of the elements of the subset S50 is
(A) (a) 61250
(B) (b) 65525
(C) (c) 42455
(D) (d) 62525
(E) (e) 55425
Rationale: The question is based heavily on pattern-matching skills, and is far more 'layered' than what you would likely see on the Official GMAT on this subject. You'll likely find it easiest to define the patterns one-at-a-time, so that you can see how they impact the question.
To start, we're given information on the sub-sets of numbers:
-The 1st set has 1 number
-The 2nd set has 2 numbers
-The 3rd set has 3 numbers
Etc.
So the 50th set will have 50 numbers in it.
Since the numbers in the subsets are consecutive, we can figure out ALL of the numbers that show up BEFORE we get to the 50th set by usingbunching...
The 1st set has 1 number and the 49th set has 49 numbers....1 + 49 = 50 numbers
The 2nd set has 2 numbers and the 48th set has 48 numbers...2 + 48 = 50 numbers
Etc.
This pattern will continue all the way down to...
The 24th set has 24 numbers and the 26th set has 26 numbers...24 + 26 = 50 numbers
Thus, we have 24 'pairs' of 50 numbers and one set (the 25th set with 25 numbers) that doesn't have a 'pair-match'...
So 24(50) + 25 = 1225 consecutive numbers exist in the 49 sets BEFORE the 50th set - that's the numbers 1 to 1225, inclusive...
Thus, the 50th set starts with 1226 and ends with 1275. We can use 'bunching' again to figure out the sum of this group (although there's no 'middle term' this time)...
25(1226+1275) =
25(2501) =
62525
D
Answer: (D) (d) 62525
Question: 19! is equal to which of the following?
(A) 121,645,100,408,832,916
(B) 121,645,100,408,832,225
(C) 121,645,100,408,832,424
(D) 121,645,100,408,832,378
(E) 121,645,100,408,832,000
Rationale: After 4!, the units digit of every factorial is 0.
5!=120
6!=720
etc...
The answer is E.
Answer: (E) 121,645,100,408,832,000
Question: A train overtakes 2 girls who are walking inthe opposite direction in which the train is going at the rate of 3 km/hour & 6km/hour and passes them completely in 36sec & 30 sec respectively. Find the length of the train is?
(A) 90 m
(B) 100 m
(C) 150 m
(D) 180 m
(E) 190 m
Rationale: Let the length of the train e x meter, and let the speed of train be y km/h, then
\inline x=\left ( y+3 \right )\frac{5}{18}\times 36 ........(1)
and \inline x=\left ( y+6\right )\frac{5}{18}\times 30 ........(2)
From eq (1) and (2), we get
\inline (y+3)\times 36 = ( y+6)\times 30
y = 12 km/h
\inline \therefore \inline x=(y+3)\times \frac{5}{18}\times 36
x= 150 m
C
Answer: (C) 150 m
Question: Ram and Shakil run a race of 2000m. First Ram gives Shakil a start of 200m and beats him by 1 minute.Next Ram gives shakil a start of 6 min and is beaten by 1000 meters . Find the time in minutes in which Ram and Shakil can run the race seperately.
(A) 12,18
(B) 10,18
(C) 10,12
(D) 8,10
(E) 8,9
Rationale: let speed of ram and shakil be x and y resp.
case1: ram gives a start of 200m and beats shakil by 1 minute.
2000/x - 1800/y= -60
2000/x= 1800/y -60---------------------(1)
case2: ram gives shakil a start of 6 min and is beaten by 1000 meters
distance travelled by shakil in 6 minutes= (6*60*y) meters = 360y meters
1000/x = (2000- 360y)/y------------------(2)
solve for x, y..
answer is x=8minutes, y= 10minutes
ANSWER:D
Answer: (D) 8,10
Question: A company plans to award prizes to its top 3 salespeople, with the largest prize going to the top salesperson, the next-largest prize to the next salesperson, and a smaller prize to the third-ranking salesperson. If the company has 6 salespeople on staff, how many different arrangements of winners are possible?
(A) 120
(B) 1,440
(C) 1,320
(D) 220
(E) 6
Rationale: [quote=Bunuel]A company plans to award prizes to its top 3 salespeople, with the largest prize going to the top salesperson, the next-largest prize to the next salesperson, and a smaller prize to the third-ranking salesperson. If the company has 6 salespeople on staff, how many different arrangements of winners are possible?
6*5*4
=120
Answer : A
Answer: (A) 120
Question: If you roll one fair six-sided die, what is the probability that the number is even or greater than 3?
(A) 1/2
(B) 1/3
(C) 2/3
(D) 3/4
(E) 5/6
Rationale: The numbers which satisfy the conditions are 2, 4, 5, and 6.
The probability is 4/6=2/3
The answer is C.
Answer: (C) 2/3
Question: There are 30 doors marked with numbers 1 to 30. There are 30 individuals marked 1 to 30.
An operation on a door is defined as changing the status of the door from open to closed or vice versa. All the doors are closed to start with. One at a time, one randomly chosen individual goes and operates the doors. The individual however operates only those doors which are a multiple of the number he/she is carrying. For example, the individual marked with number 5 operates the doors marked with 5, 10, 15, 20, 25, and 30. If every individual in the group gets one turn, then how many doors are open at the end?
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5
Rationale: If a door is closed at the start, it requires an odd number of people to operate to be open at the end. Only the perfect squares have an odd number of factors.
The doors which are open at the end are: 1, 4, 9, 16, 25 for a total of 5 doors.
The answer is E.
Answer: (E) 5