divelab/Sys2Bench huggingface dataset:
Question: Two persons start running simultaneously around a circular track of length 100 m from the same point at speeds of 5 km/hr and 13 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?
Options:
A)20 sec
B)37 sec
C)17 sec
D)23 sec
E)27 sec
Rationale: Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 100 / (5 + 13)5/18 = 100* 18 / 18 * 5 = 20 seconds.
Answer : A
Answer: A
Question: At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?
Options:
A)5 min. past 7
B)5(2/11) min. past 7
C)5(3/11) min. past 7
D)5(5/11) min. past 7
E)5(5/11) min. past 8
Rationale: When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min. spaces are gained in 5(5/11) min.
ANSWER:D
Answer: D
Question: A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest were 3% higher, then revenue earned would be 20% higher. What is the rate of interest?
Options:
A)24% p.a.
B)4% p.a.
C)12% p.a.
D)15% p.a.
E)18% p.a.
Rationale: Let the rate of interest be x%.
Now it will be (x + 3)%
% increase = (((x + 3) - x)/x) * 100 = 20
=> x = 15
ANSWER:D
Answer: D
Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?
Options:
A)5/21
B)5/22
C)5/12
D)5/29
E)5/21
Rationale: Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅
= (10 * 6)/252 = 5/21
Answer: A
Answer: A
Question: A company’s four cars running 8 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?
Options:
A)1200 lt
B)1555 lt
C)1664 lt
D)1728 lt
E)2160 lt
Rationale: Given that
4 cars running 8 hrs a day consume 1200 lts. of fuel in 10 days.
1 car consumption per hour per day = 1200 /4 *8 *10 = 3.75 litre
Now question say new car consume 20% less fuel than the company’s four cars = 80/100 of 3.75 = 3 ( 20 percent less than 3.75)
Hence we calculate total consumption for next 6 days, the company will need to run 5 new cars for 12 hrs = 3*12 *6*5 =1080
similarly = old 4 car consumption for next 6 days for 12 hrs = 3.75*6*12*4 = 1080
hence total is = 1080+1080 = 2160 lt
Ans is E .
Answer: E
Question: Sandy walks around a circular field at the rate of one round per hour while Molly runs around it at the rate of three rounds per hour. If they start in the same direction from the same point, how many minutes will it take until they meet again?
Options:
A)15
B)30
C)45
D)60
E)75
Rationale: Since Sandy and Molly move in the same direction along the circle, they will first meet each other when there is a difference of one round between the two.
The relative speed of Sandy and Molly is 3 - 1 = 2 rounds per hour.
The time required to complete one round at this speed is 1/2 hours = 30 minutes.
The answer is B.
Answer: B
Question: If it takes 10 days to dig a trench 100 m long, 50 m broad and 10 m deep, what length of trench, 25 m broad and 15 m deep can be dug in 30 days ?
Options:
A)400 m
B)200 m
C)100 m
D)88(8 / 9) m
E)89(8 / 9) m
Rationale: We can dig a trench of 100*50*10 in 10 days. The volume of this trench is 50 000m^3. This means that 50 000 m^3 of dirt is moved every 10 days. Thus, it is found that 3(50 000) of dirt, or 150 000 m^3 of dirt is moved in 30 days. Since we know that the dimensions of this hole are 25, 15, and x and the volume is 150 000, we can create the equation 15*25*x=150 000. Solving for x, we find that the missing dimension is 400m
ANSWER:A
Answer: A
Question: A rectangular field is 63 yards long and 21 yards wide. A fence is needed for the perimeter of the
field. Fencing is also needed to divide the field into three square sections. How many feet of fencing
are needed? (It is a good idea to make a drawing for this one.)
Options:
A)630 ft
B)530 ft
C)430 ft
D)330 ft
E)230 ft
Rationale: 63 + 21 + 63 + 21 = 168 yds. for
outside of field.
21 + 21 = 42 yds. for dividing sections
168 + 42 = 210 yds. of fencing
210 x 3 = 630 ft. of fencing
correct answer A
Answer: A
Question: The equation (M + )/21 = (p – 7)/21 relates two temperature scales, where M is the number of degrees on one scale and P is the number of degrees on the other scale. Which of the following equations can be used to convert temperatures from the P scale to the M scale?
Options:
A)P
B)P-7
C)P-8
D)P-6
E)P-5
Rationale: M/21 =(P-7)/21
M = (P-7)*21/21
=(P-7)*1 = P-7
B
Answer: B
Question: Series E(n) is such that i*E(i) = j*E(j) for any pair of positive integers (i, j). If E(1) is a positive integer, which of the following is possible?
I. 2*E(100) = E(99) + E(98)
II. E(1) is the only integer in the series
III. The series does not contain negative numbers
Options:
A)I only
B)II only
C)IIII only
D)IIIII only
E)I, IIIII
Rationale: First thing I want to understand is this relation: i*E(i) = j*E(j) for any pair of positive integers. I will take examples to understand it.
When i = 1 and j = 2, E(1) = 2*E(2)
So E(2) = E(1)/2
When i = 1 and j = 3, E(1) = 3*E(3)
So E(3) = E(1)/3
I see it now. The series is: E(1), E(1)/2, E(1)/3, E(1)/4 and so on...
II and III are easily possible. We can see that without any calculations.
II. E(1) is the only integer in the series
If E(1) = 1, then series becomes 1, 1/2, 1/3, 1/4 ... all fractions except E(1)
III. The series does not contain negative numbers
Again, same series as above applies. In fact, since E(1) is a positive integer, this must be true.
I. 2*E(100) = E(99) + E(98)
2*E(1)/100 = E(1)/99 + E(1)/98 (cancel E(1) from both sides)
2/100 = 1/99 + 1/98
Not true hence this is not possible
Answer (D)
Answer: D
Question: If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office.
Options:
A)1⁄2 hr
B)1 hr
C)3⁄4 hr
D)2 hrs
E)None of these
Rationale: If new speed is a⁄b of original speed, then usual time × (b⁄a - 1) = change in time
∴ usual time × (4⁄3 - 1) = 1⁄3
⇒ usual time = 1⁄3 × 3 = 1 hr
Answer B
Answer: B
Question: Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?
Options:
A)250 m
B)260 m
C)240 m
D)270 m
E)280 m
Rationale: D
270 m
Relative speed = (72 - 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.
Answer: D
Question: An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 11 o'clock in the morning?
Options:
A)30
B)60
C)90
D)120
E)150
Rationale: The hour hand will move 3/12 of a circle.
(3/12)*360 = 90 degrees.
The answer is C.
Answer: C
Question: There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
Options:
A)120
B)380
C)400
D)800
E)250
Rationale: Explanation:
The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in ²⁰P₂ ways.
20P2 = 20 * 19 = 380.
Answer: B
Answer: B
Question: Two cars are traveling in the same direction along the same route. The red car travels at a constant speed of 20 miles per hour, and the black car is traveling at a constant speed of 50 miles per hour. If the red car is 20 miles ahead of the black car, how many hours will it take the black car to overtake the red car?
Options:
A)0.1
B)0.6
C)1
D)1.5
E)2
Rationale: Option B
20+20t=50t
t=0.6
Answer: B
Question: The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is?
Options:
A)288
B)208
C)200
D)480
E)261
Rationale: 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48 = 480
Answer: D
Answer: D
Question: Jane earned enough money by selling seashells at 25 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold ?
Options:
A)5
B)10
C)17
D)22
E)30
Rationale: Let's TEST Answer A: 5 seashells...
With 5 seashells, Jamie would have 5(25) = 125 cents. This would allow him to buy 2 books for 110 cents total, but he would then have 15 cents left over. We're told that he spent ALL of his money on books, so this cannot be the correct answer.
Answer D: 22 seashells....
With 22 seashells, Jamie would have 22(25) = 550 cents. This would allow him to buy 10 books for 550 cents total, with no money left over. This is an exact MATCH for what we were told, so this MUST be the answer.
D
Answer: D
Question: Evaluate :	136.09 + 43.9
Options:
A)1000
B)3000
C)7000
D)5000
E)2000
Rationale: 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000
Answer is D
Answer: D
Question: A manufacturer of a certain product can expect that between 0.3 percent and 0.5 percent of the units manufactured will be defective. If the retail price is $1,500 per unit and the manufacturer offers a full refund for defective units, how much money can the manufacturer expect to need to cover the refunds on 20,000 units?
Options:
A) Between $90,000 and $150,000
B) Between $30,000 and $50,000
C) Between $60,000 and $100,000
D) Between $150,000 and $250,000
E) Between $300,000 and $500,000
Rationale: Number of defective units is between = .3 % of 20000 and .5% of 20000
= 60 and 100
Retail Price per unit = 1500 $
Expected price of refund is between = 1500 x 60 and 1500 x 100
=90,000 and 1,50,000 dollars
Answer A
Answer: A
Question: If the height of an equilateral triangle WBC is equal in length to the hypotenuse of an isosceles right triangle DEF, a side of ΔWBC is how many times as long as a leg of the right triangle DEF?
Options:
A)(√2)/(√3)
B)(√3)/(√2)
C)√2
D)(2√2)/(√3)
E)2/(√3)
Rationale: Side of the equilateral triangle WBC = 2m ----> height = m * sqtr(2)
Hipotenuse of the right triangle WBC = m * sqrt(3) ----> side = m * sqtr(3) / sqtr(2)
Question is 2m/m * sqtr(3) / sqtr(2) = 2*sqrt(2)/sqrt(3)=D
Answer: D
Question: If the length of a rectangle is increased by 10% and the area is unchanged , then its corresponding breadth must be decreased by?
Options:
A)9 1/19
B)9 1/15
C)9 1/16
D)9 1/11
E)9 1/10
Rationale: Suppose length = 100m and breadth = x m
Area = 100 m
New length = 110m and breadth = (x-y% of x)
Then, \inline 110\times (x-\frac{y}{100}x)=100\times x
\inline \Rightarrow \inline 110\times (1-\frac{y}{100})=100
\inline \Rightarrow \inline 1-\frac{y}{100}=\frac{100}{110}
\inline \Rightarrow \frac{y}{100}=1-\frac{100}{110}=\frac{1}{11}
\inline y=\frac{100}{11}=9\frac{1}{10}%
Answer:E
Answer: E
Question: Find the remainder when 6^50 is divided by 215
Options:
A)34
B)35
C)36
D)37
E)38
Rationale: (6^3)^16*6^2 /215
(215 + 1)^16 * 6^2/215
1^16 * 6^2/215
= 36
ANSWER:C
Answer: C
Question: Payal finished 2/3 of a book. She calculated that she finished 20 more pages than she has yet to read. How long is her book?
Options:
A)A)60
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=20
1/3*x=20
x=60
So the book is 270 pages long.
Answer is A.
Answer: A
Question: If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :
Options:
A)603
B)604
C)611
D)511
E)601
Rationale: If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.
ANSWER:E
Answer: E
Question: A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
Options:
A)44 km/hour
B)46 km/hour
C)48 km/hour
D)50 km/hour
E)None of these
Rationale: Explanation:
Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr
Average speed =
2xy /x+y
=2∗40∗60 /40+60=4800/100=48Km/hr
Option C
Answer: C
Question: In a 2000 m race, A beats B by 180 meters or 50 seconds. Find the speed of B?
Options:
A)3.6m/s
B)2m/s
C)5m/s
D)7m/s
E)3m/s
Rationale: Since A beats B by 180 m or 50 seconds,
it implies that B covers 180 m in 50 seconds.
Hence speed of B = 180/50 = 3.6 m/s.
Answer: A
Answer: A
Question: A starts a business with a capital of Rs. 8500. B joins in the business with Rs.8500 after some time. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1?
Options:
A)4 months
B)6 months
C)7 months
D)8 months
E)None of these
Rationale: Explanation :
Let B joins for x months. Then
A:B = 8500*12 : x* 8500 = 3 : 1
=> 85*12 : 85x= 3 : 1
=> 85*12/ 85x = 3/1 = 3
=> 85*4 /85x = 1
=> 4/x = 1
=> x = 4. Answer : Option A
Answer: A
Question: How many times does the minute hand to go around the clock in one day?
Options:
A)1355
B)1416
C)1432
D)1400
E)1300
Rationale: The minute hand will go around once. Thus, the second hand will pass the minute hand 59 times in an hour. This means they will meet 59Ã—24=1416 times in a day.
Answer B
Answer: B
Question: A person travels equal distances with speeds of 9 km/hr, 10km/hr and 11 km/hr and takes a total time of 47 minutes. The total distance is?
Options:
A)6 km
B)3 km
C)8 km
D)9 km
E)2 km
Rationale: Let the total distance be 3x km.
Then, x/9 + x/10 + x/11 = 47/60
29x/96 = 47/60 => x = 2.59
Total distance = 3 *2.59 = 7.78 km.
Answer:C
Answer: C
Question: How many cases do you need if you have to pack 240 pairs of shoes into cases that each hold 24 shoes?
Options:
A)35
B)15
C)20
D)25
E)30
Rationale: C
20
240 pairs of shoes = 480 shoes.
480÷ 24 = 20.
Answer: C
Question: The average weight of 19 students is 15kg. by the admission of a new student the average weight is reduced to 14.9 kg. the weight of the new student is ?
Options:
A)10.6 kg
B)10.8 kg
C)11 kg
D)13 kg
E)None
Rationale: Answer
Weight of new student = Total weight of all 20 students - Total weight of initial 19 students
= (20 x 14.9 - 19 x 15) kg
= 13 kg.
Correct Option: D
Answer: D
Question: In a group of ducks and cows, the total number of legs are 28 more than twice the no. of heads. Find the total no.of buffaloes.
Options:
A)10
B)12
C)13
D)15
E)14
Rationale: Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 28
=> 2x = 28 => x = 14
E
Answer: E
Question: A type R machine can complete a job in 5 hours and a type B machine can complete the job in 7 hours. How many hours will it take 2 type R machines and 3 type B machines working together and independently to complete the job?
Options:
A)1/5
B)29/35
C)5/6
D)35/29
E)35/12
Rationale: now D should be the answer.
R need 5 hours to complete
and B needs 7 hours to compete
so 2R + 3B will complete 2/5 + 3/7 or 29/35 portion of the job in 1 hour
so the whole job will take 35/29 hours....=D
Answer: D
Question: 0.0008?=0.02
Options:
A)40
B)4
C)0.4
D)None of these
E)0.04
Rationale: Explanation :
Required Answer = 0.0008/0.02 = 0.08/2 = 0.04. Answer : Option E
Answer: E
Question: 3 pumps working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?
Options:
A)34
B)77
C)36
D)49
E)38
Rationale: Explanation:
Let the required length be x meters
More men, More length built (Direct proportion)
Less days, Less length built (Direct Proportion)
\inline \fn_jvn \left.\begin{matrix} Men\: \: \: 20:35\\ Days\: 6:3 \end{matrix}\right\}::56:x
\inline \fn_jvn \therefore (20 x 6 x X)=(35 x 3 x 56) \inline \fn_jvn \Leftrightarrow \inline \fn_jvn x=\frac{35\times 3\times 56}{120} =49
Hence, the required length is 49 m.
Answer: D
Answer: D
Question: A started a bussiness with a sum of Rs. 14000. After 4 months, B entered with an amount of Rs. 12000 and A added another 2,000 in the business. After 6 months, C entered with Rs. 20000 and 'B' withdrew Rs. 2000 from the business. What will be C's share of profit out of a total profit of Rs. 94000 at the end of 2 years.
Options:
A)Rs. 376000
B)Rs. 204000
C)Rs. 36000
D)Rs. 18000
E)Rs. 260000
Rationale: Investments--
A=14000x4+16000x20=376000
B=12000x2+10000x18=204000
C=20000x18=360000
So C's Share = 94000xC/(A+B+C)=36000
ANSWER:C
Answer: C
Question: A box contains nine slips that are each labeled with one number: 1, 3, 5, 8, 13, 21, 34 and 55. Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?
Options:
A)7/72
B)1/6
C)5/28
D)15/36
E)21/36
Rationale: Probability = No: of desired outcomes / Total No: of outcomes.
You are picking two slips out of 8 slips. So
Total No: of outcomes = 8C2 = 28
Desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box. How many such outcomes are there?
If you look at the numbers closely, you will see that the following pair of numbers will give you the desired outcome.
(3,5) (5,8) (8,13) (13,21) (21,34) . There are 7 such pairs. If the two numbers which I pick is from any of these 5 pairs, then I get my desired outcome.
So No: of desired outcomes = 5
Probability = 5/28
Answer :C
Answer: C
Question: Given that there are 11 cricket players per team, how many ways can you select 10 cricket players from 2 teams if no more than 2 players can be selected from each team?
Options:
A)3025
B)1500
C)2600
D)3500
E)4200
Rationale: Now from each group we have to select 5 persons only: 11C2 * 11C2= 55 * 55 = 3025
Total ways=313444 ways.
Answer A
Answer: A
Question: The game 'Chunk-a-Luck' is played at carnivals in some parts of Europe. Its rules are as follows:
If you pick a number from 1 to 6 and the operator rolls three dice.
If the number you picked comes up on all three dice, the operator pays you Rs. 3.
If it comes up on two dice, you are paid Rs. 2.
And it comes up on just one dice, you are paid Rs. 1.
Only if the number you picked does not come up at all, you pay the operator Rs. 1.
The probability that you will win money playing in this game is:
Options:
A)0.52
B)0.753
C)0.42
D)0.68
E)None of the above
Rationale: Explanation :
If one picks up a number out of six numbers then the only case in which he/she will lose money if none of the three dice shows the picked number on the top surface.
Required probability of losing the game:
=>(5/6) x (5/6) x (5/6).
=>125/216.
Hence, the required probability of winning the game is :-
=>1-(125/216).
=>91/216.
=>0.42.
Answer : C
Answer: C
Question: Jack, Jill, and Sandy each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/6, 1/7, and 1/8 respectively, what is the probability that Jill and Sandy will make a basket but Jack will miss?
Options:
A)3/168
B)5/168
C)7/168
D)5/336
E)7/336
Rationale: The probability that Jill and Sandy will make a basket but Jack will miss is 5/6*1/7*1/8=5/336.
The answer is D.
Answer: D
Question: If A = 625 * 7776 * 16 / 1296, How many digits are in A?
Options:
A)5
B)2
C)3
D)4
E)0
Rationale: A = 625 * 7776 * 16 / 1296
A = 5^4 * 6^5 * 2^4 / 6^4
A = 10^4*6 = 60000
Option: 5
Answer: A
Question: 34.95+240.016+23.98=?
Options:
A)298.0946
B)298.111
C)298.946
D)299.09
E)None of them
Rationale: 34.95
240.016
23.98
----------
298.946
Answer is C
Answer: C
Question: a,b,c,d,e,f,g sitting in a row what is the probability that a,b & c sitting together.
Options:
A)3/7
B)5/7
C)6/7
D)2/7
E)1/7
Rationale: total number of arrangement is= 7!= 5040
Favorable event i.e a,b and c can be arranged together in 3! and the remaining can be arranged in5! since abc can be in any of the five positions.
so 5!*3!/7! = 2/5
ANSWER:E
Answer: E
Question: If A is a positive integer, which of the following must be divisible by 6?
Options:
A)A(A^2−2)
B)A^2*(A−5)
C)A(A+2)^3
D)A^2*(A−5)
E)(A−1)*A*(A+1)
Rationale: option E) (A−1)*A*(A+1) which represents the product of three consecutive integers
the product of 3 consecutive integers is divisible by both 2 and 3 and so also divisible by 6.
Answer is E
Answer: E
Question: The circulation for magazine P in 1971 was 4 times the average (arithmetic mean) yearly circulation for magazine P for the years 1972-1980. What is the ratio of the circulation in 1971 to the total circulation during 1971-1980 for magazine P ?
Options:
A)4/7
B)5/7
C)3/7
D)2/7
E)1/7
Rationale: There are 9 years from 1972-1980, inclusive. Let's say the average circulation every year between 1972-1980 inclusive is x.
So the total circulation is 9x from 1972-1980, inclusive.
In 1971, the circulation is 4x.
So total circulation for 1971-1980 is 4x + 9x = 13x.
Ratio of circulation in 1971 to total circulation during 1971-1980 is 4x to 14x or 4/14=2/7
Answer : D
Answer: D
Question: A telephone company needs to create a set of 3-digit area codes. The company is entitled to use only digits 6, 4 and 3, which can be repeated. If the product of the digits in the area code must be even, how many different codes can be created?
Options:
A)20
B)22
C)26
D)24
E)30
Rationale: Total # of codes possible is 3*3*3 = 27. Oit of those 27 codes only the product of 333 and will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even. Therefore the number of codes where the product of the digits is even = (total) - (restriction) = 27 - 1 = 26.
Answer: C
Answer: C
Question: Which is greater than 16
Options:
A)1.6
B)66
C)6
D)-6
E)6.1
Rationale: 66 greater than 16.
Answer: B
Answer: B
Question: A clothing store originally bought 100 jackets at a total cost of j dollars. If each jacket was sold for 30 percent more than its original cost, what was the individual price, in terms of j, of each jacket sold?
Options:
A)j/77
B)7j/500
C)140j
D)j/100+40
E)140j/100
Rationale: lets take j=1000, as decided by you..
now 30% increase makes it 1300..
this is the price for 100 jackets, so selling price for each jacket=1300/100=$13..
so ans is correct as A..
Answer: A
Question: John spent a total of $135 on baseball tickets. The price of a ticket is either $12 for the field seats or $5 for the balcony. What is the minimum amount W of dollars he could have spent for the $12 tickets?
Options:
A)$48
B)$60
C)$84
D)$96
E)$120
Rationale: 60 ..... 12 *5 W= 60. (min). Thus answer .B
Answer: B
Question: A certain stock exchange designates each stock with a one-, two-, 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?
Options:
A)2951
B)8125
C)15600
D)16302
E)18278
Rationale: Remember that Numbers can be repeated.
Number of 1 Digit Codes = 26
Number of 2 Digit Codes = 26 * 26 = 676
Number of 2 Digit Codes = 26 * 26 * 26 = 17576
Total = 26 + 676 + 17576 = 18278
ANS:E
Answer: E
Question: Out of three given numbers, the first one is twice the second and three times the third. If the average of these numbers is 88, then the difference between first and third is.
Options:
A)92
B)39
C)87
D)96
E)None
Rationale: Sum of three number is=88*3=264
Let three numbers are a, b,c and a is the highest and c is the lowest
Then,
2b=a so b=a/2 and 3c=a so c=a/3
we can write,
a+b+c=264
a+a/2+a/3=264
11a/6=264
a=144
So, c=144/3=48
So there difference is=144-48=96
Answer D
Answer: D