divelab/Sys2Bench huggingface dataset:
Question: Horse started to chase a dog as it relived stable two hours ago. And horse started to ran with average speed of 22kmph, horse crossed 10 meters road and two ponds with depth 3 meters and it crossed 2 small streets with 200 meters length. After travelling 6 hours, 2 hours after the sunset it got dog. Compute the speed of dog?
Options:
A)16.5 kmph
B)16.6 kmph
C)16.4 kmph
D)16.9 kmph
E)16.2 kmph
Rationale: In an average speed of 22 kmph horse got dog in 6 hours
So distance=speed*time
distance=22*6=132km
Dog takes 8 hours to travel 132 km. So the speed if dog is
speed=distance/time
speed=132/8=16.5 kmph
Answer: 16.5 kmph
Answer A
Answer: A
Question: A & B can make paintings in 6 days, B & C can can make those paintings in 10 days. If A, B & C together can finish the work in 4 days, then A & C together will do it in ________ days.
Options:
A)4 (2/7) days
B)1/ 8 days
C)2 (2/5) days
D)6 (3/ 8) days
E)1/ 9 days
Rationale: EXPLANATION:
Hint:
We are given that, A,B, & C together complete the work in 4 days.
We can write, (A+B+C) 's 1 day work =	1/4
Similarly, (A+B) 's 1 day work = 1/6 days & (B+C)'s 1 day work =	1/10
Since the work is divided in combination and we are asked to find out the combined work of (A + C), so we can find out,
(A + C)'s 1 day work = [2 (A+B+C)'s 1 day work] – [(A+ B) 's 1 day work + (B+C)'s 1 day work]
=[2*1/4]-[(1/6)+(1/10)]
=	1/2	–	16/60	=	1/2	–	4/15	=	7/30
Hence, A & C together can complete the work in 30 /7 days = 4(2/7)days.
ANSWER IS A
Answer: A
Question: a = 4^15 - 625^3 and a/x is an integer, where x is a positive integer greater than 1, such that it does NOT have a factor p such that 1 < p < x, then how many different values for x are possible?
Options:
A)None
B)One
C)Two
D)Three
E)Four
Rationale: This is a tricky worded question and I think the answer is should be D not C...
Here is my reason :
The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime
Number has no factor grater than 1 and Itself .
On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )
AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is
An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get 2 different values and answer is C....
Answer: C
Question: the average distance between the sun and a certain planet is approximatly 2.6 10^9 inches. which of the following is the closest to the average distence between sun and the planet in kelometers? (1 km is approx 3.9 x 10^4 inches)
Options:
A)7 x (10)^5
B)7 x (10)^7
C)6.7 x (10)^4
D)7 x (10)^8
E)7 x (10)^12
Rationale: No need to solve entire problem.
Distance in KM = 26*(10)^9/39*(10)^4 which is -260*(10)^8/39*(10)^4 = 6.7 x(10)^4
Answer : C
Answer: C
Question: If W is the set of all the integers between 59 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?
Options:
A)27
B)32
C)33
D)34
E)35
Rationale: Multiples of 2 from 59 to 99=Multiples of 2 from 1 to 99-Multiples of 2 from 1 to 58=[99/2]-[58/2]=49-29=20
Multiples of 3 from 59 to 99=Multiples of 3 from 1 to 99-Multiples of 3 from 1 to 58=[99/3]-[58/3]=33-169=14
Multiples of2 and 3 bothi.e.6 from 59 to 99=Multiples of 6 from 1 to 99-Multiples of 6 from 1 to 58=[99/6]-[58/6]=16-9=7
These 8 Numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3
i.e. Total Numbers in W = 20 + 14 - 7 = 27
Answer Option A
Answer: A
Question: Which of the following equations has a solution in common with x^2 + 5x + 4 = 0?
Options:
A)2x^2 + 5x +4 = 0
B)x^2 - 3x - 4 = 0
C)2x^2 - 14 1/2x + 7/2 = 0
D)x^2 + 4x - 21 = 0
E)none of the above
Rationale: Given : x^2 + 5x + 4 = 0 => (x+1) (x+4) = 0 => -1 and -4 are the roots..
A. 2x^2 + 4x + 3/2 = 0 => (x+1/2) (2x+3)
B. x^2 - 3x - 4 = 0 => (x-7) (2x-1/2)
C. 2x^2 - 14 1/2x + 7/2 = 0 => (x-7) (x-3)
D. x^2 + 4x - 21 = 0 => (x+7) (x-3)
None of the answer equations have exactly both the same roots as the question equation, but at least one of them does share one of the two roots.
So IMO "B" is correct answer... because it shares the one root (part of the solution), -1, from the (x+1) portion of its simplification.
Answer: B
Question: A is faster than B. A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then, A's speed is equal to?
Options:
A)5
B)3
C)4
D)7
E)9
Rationale: Let A's speed = x km/hr. Then, B's speed = (7 - x) km/ hr.
So, 24/x + 24/(7 - x) = 14
x2 - 98x + 168 = 0
(x - 3)(x - 4) = 0 => x = 3 or 4.
Since, A is faster than B, so A's speed = 4 km/hr and B's speed = 3 km/hr.
Answer: B
Answer: B
Question: A train running at the speed of 52 km/hr crosses a pole in 9 sec. What is the length of the train?
Options:
A)130 m
B)786 m
C)566 m
D)546 m
E)440 m
Rationale: Speed = 52 * 5/18 = 130/9 m/sec
Length of the train = speed * time = 130/9 * 9
= 130 m
Answer: A
Answer: A
Question: Approximately how many revolutions will be made by a car tire with a 14-inch diameter if the car travels 1 mile?
Options:
A)120
B)180
C)1,231
D)1,433
E)1,440
Rationale: Diameter of car tire = 14 inches
Radius of car tire = 7 inches = 7*2.54 cm = 17.78 cm
Circumference of the car tire = 2*pi * 17.78 = 2 *3.14 *17.78= 111.66 cm
Distance traveled by car = 1 mile = 1 *1.6 km = 1.6 km = 1600 m = 1.6 * 10^5 cm
Number of revolutions =Distance traveled / Circumference of tire = 1.6*10^5 / 111.66 = 1432.92
Since the answer choices are far apart we should use estimation in this question .
Ideally ,in the GMAT the conversion between units would be provided .
Answer D
Answer: D
Question: A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
Options:
A)29798
B)27908
C)78902
D)25000
E)27991
Rationale: 25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x
= 25000
Answer: D
Answer: D
Question: In X game of billiards, X can give Y 20 points in 60 and he can give Z 50 points in 60. How many points can Y give Z in X game of 100?
Options:
A)30
B)20
C)25
D)40
E)75
Rationale: X scores 60 while Y score 40 and Z scores 10.
The number of points that Z scores when Y scores 100 = (100 * 50)/40 = 25.
In X game of 100 points, Y gives (100 - 25) = 75 points to C.
E
Answer: E
Question: A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 5 : 6 : 7 : 8 : 9. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is
Options:
A)2
B)3
C)4
D)5
E)1
Rationale: Here we can assume that one subject is of 100 marks so total there are 5 subjects => 100*5 = 500.
Now according to the question he secured 60 % of those which is 60%of 500 = 300 marks in total.
The ratio between the marks is given as 5 : 6 : 7 : 8 : 9 , so now we can distribute 300 marks according to the ratio.
Total ratio = 35
For 6 : (300/35)*6 = 51.4
Similarly , we will get for others as 60 , 68.57 , 77.14
Hence , there are 5 subject where he secured more that 50 %.
The answer is 5.
D
Answer: D
Question: 20, 19, 17, ?, 10, 5
Options:
A)15
B)14
C)13
D)12
E)11
Rationale: The Pattern is - 1, - 2, -3, ...
Answer : B.
Answer: B
Question: .05% of seeds in a packet do not germinate. How many seeds must be in the packet for 6 seeds not to germinate?
Options:
A)10000
B)11000
C)12000
D)14000
E)6000
Rationale: Let the number of total seeds be x
Then, .05% of x=6
(5/100)*(1/100)*x=6
x=12000
Answer is C
Answer: C
Question: Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?
Options:
A)15
B)10
C)81
D)71
E)12
Rationale: 4/12 + x/15 = 1
x = 10
Answer: B
Answer: B
Question: Lucy invested $10,000 in a new mutual fund account exactly three years ago. The value of the account increased by 10 percent during the first year, increased by 9 percent during the second year, and decreased by 10 percent during the third year. What is the value of the account today?
Options:
A)$10,350
B)$10,395
C)$10,791
D)$11,500
E)$12,705
Rationale: the amount at the end of first year will become the principle for 2nd year, applying this trick and calculating we get the amount at the end of third year to be 10791. So the answer is (C).
Answer: C
Question: The average age of 25 students in a class is 12years. If teacher's age is also included then average increases 1year then find the teacher's age?
Options:
A)40
B)35
C)29
D)38
E)42
Rationale: Total age of 25 students = 25*12 = 300
Total age of 26 persons = 26*13 = 338
Age of teacher = 338-300 = 38years
Answer is D
Answer: D
Question: A house wife saved $4 in buying an item on sale. If she spent $29 for the item, approximately how much percent she saved in the transaction?
Options:
A)8%
B)9%
C)10%
D)11%
E)12%
Rationale: Actual price = 29+4 = $33
Saving = 4/33* 100 = 12% approximately
Answer is E
Answer: E
Question: Uncle Bruce is baking chocolate chip cookies. He has 36 ounces of dough (with no chocolate) and 13 ounces of chocolate. How many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20% chocolate?
Options:
A) 4
B) 6
C) 7.2
D) 7.8
E) 9
Rationale: Answer is A.
x/x+36=1/5
x=9
13-9=4
Answer: A
Question: In a certain production lot, 40 percent of the toys are red and the remaining toys are green. Half of toys are small and half are large. If 10 percent of the toys are red and small, and 40 toys are green and large. How many of the toys are red and large?
Options:
A)40
B)60
C)70
D)80
E)90
Rationale: 40% toys are red.
50% toys are small - 50% are large
Red and small - 10%
Red and large must be 30% (since there are total 40% red toys)
But 50% toys are large. If 30% are red and large, the rest 20% large toys must be green. So large and green toys are 20%.
These 20% large and green toys are actually 40 so total number of toys must be 200.
Since red and large toys are 30%, they must be 60.
ANSWER:B
Answer: B
Question: Find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm.
Options:
A)81
B)91
C)99
D)108
E)111
Rationale: AB² = AC² - BC²
= 15² - 12²
= 225 - 144
= 81
ANSWER :OPTION A
Answer: A
Question: During a two-week period, the price of an ounce of silver increased by 8 percent by the end of the first week and then decreased by 12 percent of this new price by the end of the second week. If the price of silver was x dollars per ounce at the beginning of the two-week period, what was the price, in dollars per ounce, by the end of the period?
Options:
A)0.93x
B)0.95x
C)0.97x
D)0.99x
E)1.01x
Rationale: The price at the end is 0.88(1.08x)=0.9504x which is approximately 0.95x
The answer is B.
Answer: B
Question: Soja finished 2/3 of a book. She calculated that she finished 100 more pages than she has yet to read. How long is her book?
Options:
A)A)300
B)B)150
C)C)130
D)D)90
E)E)210
Rationale: Let x be the total number of pages in the book, then she finished 2/3*x pages.
Then she has x−2/3*x=1/3*x pages left.
2/3*x−1/3*x=100
1/3*x=100
x=300
So the book is 300 pages long.
Answer is A.
Answer: A
Question: A clock is started at noon. By 10 minutes past 2, the hour hand has turned through how many degrees?
Options:
A)55Â°
B)60Â°
C)65Â°
D)70Â°
E)75Â°
Rationale: The angle traced by the hour hand in 12 hrs is 360Â°
The angle traced by the hour hand in 1 hour is 30Â°
The angle traced by the hour hand in 10 minutes is 5Â°
The angle traced by the hour hand in 2 hours and 10 minutes is 2*30Â° + 5Â° = 65Â°
The answer is C.
Answer: C
Question: If g is the smallest positive integer such that 3,150 multiplied by g is the square of an integer, then g must be
Options:
A) 2
B) 5
C) 6
D) 7
E) 14
Rationale: Solution:
This problem is testing us on the rule that when we express a perfect square by its unique prime factors, every prime factor's exponent is an even number.
Let’s start by prime factorizing 3,150.
3,150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2
3,150 = 2^1 x 3^2 x 5^2 x 7^1
(Notice that the exponents of both 2 and 7 are not even numbers. This tells us that 3,150 itself is not a perfect square.)
We also are given that 3,150 multiplied by g is the square of an integer. We can write this as:
2^1 x 3^2 x 5^2 x 7^1 x g = square of an integer
According to our rule, we need all unique prime factors' exponents to be even numbers. Thus, we need one more 2 and one more 7. Therefore, g = 7 x 2 = 14
Answer is E.
Answer: E
Question: A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is?
Options:
A)11
B)66
C)88
D)10
E)27
Rationale: Let the rate be R% p.a. Then,
(5000 * R * 2)/100 + (3000 * R * 4)/100 = 2200
100 R + 120 R = 2200
R = 10%
Answer:D
Answer: D
Question: A shopkeeper sold 30 articles at the cost price of 35 articles. Then find the profit% or lost%
Options:
A)12%
B)15%
C)16.67%
D)18%
E)22%
Rationale: here 30 articles selling price = 35 articles cost price
so the difference = 35-30 = 5
% of profit = 5*100/30 = 16.67%
correct option is C
Answer: C
Question: How many kg of pure salt must be added to 100kg of 10% solution of salt and water to increase it to a 30% solution?
Options:
A)A)6.7
B)B)1.3
C)C)9.6
D)D)12.5
E)E)28.6
Rationale: Amount salt in 100kg solution = 10*100/100 = 10 kg
Let x kg of pure salt be added
then(10+x)/(100+x)=30/100
100+10x=300+3x
7x = 200
x = 28.6
Answer is E
Answer: E
Question: A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?
Options:
A)8 days
B)10 days
C)15 days
D)11 3/7 days
E)16 days
Rationale: Explanation:
(x – 5)/15 + x/20 = 1
x = 11 3/7 days
Answer: D
Answer: D
Question: If n = (33)^43 + (53)^33 what is the units digit of n?
Options:
A)0
B)2
C)4
D)6
E)8
Rationale: First of all, the units digit of (33)^53 is the same as that of 3^53 and the units digit of (53)^33 is the same as that of 3^33. So, we need to find the units digit of 3^53 + 3^33.
Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}:
3^1=3 (the units digit is 3)
3^2=9 (the units digit is 9)
3^3=27 (the units digit is 7)
3^4=81 (the units digit is 1)
3^5=243 (the units digit is 3 again!)
...
Thus:
The units digit of 3^53 is the same as the units digit of 3^3, so 7 (53 divided by the cyclicity of 4 gives the remainder of 3).
The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).
Therefore the units digit of (33)^53 + (53)^33 is 7 + 7 = 4.
Answer: C.
Answer: C
Question: The annual interest rate earned by an investment increased by 10 percent from last year to this year. If the annual interest rate earned by the investment this year was 1 percent, what was the annual interest rate last year?
Options:
A)1%
B)1.1%
C)9.1%
D)10%
E)10.8%
Rationale: let i = interest rate
i(this year) = i(last year) + 0.10i(last year) = 1.1i(last year)
1 = 1.1 x i(last year)
i(last year) = 1/1.1 = 10/11 = 0.90909...
or i(last year) = 0.91%
ANSWER:C
Answer: C
Question: The average employee age of a certain company which has 80 employees is 38. One employee aged 60 got retired and replaced by a fresher of age 22. What is the new average age of employees?
Options:
A)35.5
B)36.2
C)36.8
D)37.5
E)38.2
Rationale: Sum of ages of employees before replacement= 80x38=3040
Sum of ages of employees after replacement=(80x38)+22-60=3002
The new average age=3002/80=37.5
Answer: D
Answer: D
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?
Options:
A)Rs:1000
B)Rs:1067
C)Rs:1278
D)Rs:1028
E)Rs:1027
Rationale: 90%
104%
--------
14% ---- 140
100% ---- ? =>
Rs:1000
Answer: A
Answer: A
Question: A committee of 4 people is to be selected from 8 people. How many different committees are possible?
Options:
A)28
B)40
C)56
D)420
E)336
Rationale: Because the order does not matter in selecting 4 people for the committee, this is a combination question.
The number of ways to select 4 people from 8 = 8C4 = 8!/4!*4! = 420. Hence D.
Answer: D
Question: A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
Options:
A)6
B)9
C)5
D)1
E)2
Rationale: C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5
For Rs. 6/5, toffees sold = 6. For re. 1.
Toffees sold = 6 * 5/6 = 5
AnswerC
Answer: C
Question: A, B, and C working alone can finish constructing a house in 6, 9, and 18 hours respectively. A starts alone and works for 1 hour. He is then joined by B and they work together for 1 hour. Finally C joins them and they all work together to finish the work. If they start working at 9 am, at what time will they finish?
Options:
A)11:45 am
B)11:55 am
C)12:10 pm
D)12:25 pm
E)12:40 pm
Rationale: A completes 1/6 of the job each hour.
B completes 1/9 of the work each hour.
C completes 1/18 of the work each hour.
A completes 1/6 of the work in the first hour.
Working together, A and B complete another 5/18 for a total of 8/18.
Working together, A, B, and C can complete 6/18 of the work each hour.
To complete 10/18, the time required is (10/18) / (6/18) = 10/6 hours.
The work will be completed at 12:00 plus 4*60/6 minutes which is 12:40 pm.
The answer is E.
Answer: E
Question: If k^3 = m^3, which of the following must be true?
Options:
A)k = m
B)k = -m
C)k = |m|
D)k = -|m|
E)|k| = |m|
Rationale: Solution:
We are given that k^3 = m^3, and we can start by simplifying the equation by taking the square root of both sides.
3√k^3 = 3√m^3
When we take the square root of a variable squared, the result is the absolute value of that variable. Thus:
3√k^3 = 3√m^3 is |k| = |m|
Note that answer choices A through D could all be true, but each of them would be true only under specific circumstances. Answer choice E is the only one that is universally true.
Answer: E
Answer: E
Question: A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
Options:
A)676
B)833
C)1023
D)839
E)1035
Rationale: Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
Answer:C
Answer: C
Question: 7122,6521,8332,4743,9911,387?
find the value of the question tag.
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: 7+1+2+2=12
6+5+2+1=14
so
9+9+1+1=20
3+8+7+x=22
so x=4
ANSWER:D
Answer: D
Question: The average of 25 result is 18.The average of 1st 12 of them is 14 & that of last 12 is 17.Find the 13th result.
Options:
A)68
B)80
C)78
D)60
E)87
Rationale: Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12)
=450-(168+204)
=450-372
=78.
Answer is C.
Answer: C
Question: Evaluate: 13 + sqrt(- 4 + 5Ã—3Ã·3) =?
Options:
A)12
B)13
C)14
D)15
E)16
Rationale: According to order of operations, inner brackets first where 5x3Ã·3 is first calculated since it has a multiplication and a division.
5Ã—3Ã·3 = 15Ã·3 = 5
Hence
13 + sqrt(- 4 + 5Ã—3Ã·3)= 13 + sqrt(- 4 + 5)
= 13 + sqrt(1) = 13 + 1 = 14
correct answer C)14
Answer: C
Question: If a tire rotates at 400 revolutions per minute when the car is traveling 120km/h, what is the circumference of the tire?
Options:
A)2
B)1
C)4
D)3
E)5
Rationale: 400 rev / minute = 400 * 60 rev / 60 minutes
= 24,000 rev / hour
24,000 * C = 120,000 m : C is the circumference
C = 5 meters
correct answer E
Answer: E
Question: How many 5-digit positive integers exist the sum of whose digits are odd?
Options:
A)36000
B)38000
C)45000
D)90000
E)None
Rationale: Solution:
There are 9�104 = 90000, 5-digit positive integers.
Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.
Hence, there are 90000 /2= 45000, 5-digit positive integers whose sum add up to an odd number.
Answer: Option C
Answer: C
Question: In a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder. The dividend is:
Options:
A)86
B)90
C)92
D)94
E)95
Rationale: Divisor = (6 * 3) + 2 = 20
5 * Quotient = 20
Quotient = 4.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 4) + 6 = 86.
A
Answer: A
Question: Machines A and B produce bottles at their respective constant rates. Machine A produces k bottles in 3 hours and machine B produces k bottles in 5 hours. How many hours does it take machines A and B, working simultaneously, to produce 16k bottles?
Options:
A)24
B)26
C)28
D)30
E)32
Rationale: A's rate = k/3
B's rate = k/5
k/3 + k/5 = 8k/15
16k / (8k/15) = 30 hours
The answer is D.
Answer: D
Question: In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Options:
A)7/8
B)4/5
C)3/2
D)2/3
E)1/3
Rationale: Total number of balls = (8 + 7 + 6) = 21.
Let E	= event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7.
P(E) =n(E)/n(S)=7/21=1/3.
Ans: E
Answer: E
Question: A rectangular floor is covered by a rug except fo a strip p meters along each of the four edges. If the floor is m meters by g meteres, What is the area of the rug in square meters?
Options:
A)mn - p(m+n)
B)mn – 2p(m+n)
C)mn – p^2
D)(m - p)(n - p)
E)(m - 2p)(g - 2p)
Rationale: The answer must be E.
Plug in M 100
N 50
g 10.
Answer: E
Question: The credit card and a global payment processing companies have been suffering losses for some time now. A well known company recently announced its quarterly results. According to the results, the revenue fell to $48.0 billion from $69.0 billion, a year ago. By what percent did the revenue fall?
Options:
A)29.43
B)30.43
C)31.43
D)32.43
E)33.43
Rationale: $69-$48=21$
(21/69)*100=30.434%
ANSWER:B
Answer: B
Question: The closest approximation of A(69.28 × 0.004)/0.03 is
Options:
A)0.092
B)0.92
C)9.2
D)92
E)920
Rationale: A(69.28 × 0.004)/0.03
1. 0.004 = 4 × 10^(-3)
2. 0.03 = 3 × 10^(-2)
3. (a × b)/c = a × (b/c)
4. 0.004/0.03 = 4 × 10^(-3)/ (3 × 10^(-2)) = 4 × 10^( -3 - (-2))/3 = 4 × 10^(-1) / 3 = (4/3) × 10^(-1) = 1.333 × 10^(-1)
Therefore,
(69.28 × 0.004)/0.03 = 69.28 × (0.004/0.03) = 69.28 × 1.33 × 10^(-1) = 69.28 × 1.33/10 = 6.928*1.33
Now,
7 × 2 = 14
7 × 1 = 7
or better:
6.9 × 1 = 6.9
6.9 × 2 = 13.8
hence,
6.9 <6.928 × 1.33 < 13.8
9.2 is the only answer that satisfies this condition.C
Answer: C
Question: 97 students take part in a test and the median score is 85. Which of the following descriptions must be correct?
I.At least 49 students’ score are less than 85.
II. At least 49 students’ score are equal to or more than 85
III. At most 49 students’ score are equal to or more than 85.
Options:
A)I
B)II
C)III
D)III
E)IIII
Rationale: If we order the scores increasingly, then on the 49th place we have the score of 85. There must be 48 scores (those on the right of the 85 score) greater or equal to 85. So, I is necessarily correct.
The scores on the left of the 49th place, all have to be equal or less than 85. Since we can have the extreme case of all the scores 85, neither II, nor III is correct. The question asks for descriptions thatmustbe correct.
Answer B.
Answer: B
Question: A certain hall contains two cuckoo clocks. If the first clock chirps 20 times per hour and the second clock chirps 9 times per hour, and both clocks chirp for the first time at 2:03 pm, at what time will the first clock have chirped three times as many times as the second clock can chirp per hour?
Options:
A)2:42 pm
B)2:50 pm
C)3:07 pm
D)3:21 pm
E)3:30 pm
Rationale: The question is asking when the first clock will have chirped 27 times (3*9 chirps).
The first clock chirps once every 3 minutes.
If the first chirp is at 2:03, then the 27th chirp is at 81 minutes after 2:00.
The answer is D.
Answer: D