divelab/Sys2Bench huggingface dataset:
Question: A car rental company charges for mileage as follows: 5 dollars per mile for the first n miles and 6 dollars per mile for each mile over n miles. How much will the mileage charge be, in dollars, for a journey of d miles, where d > n ?
Options:
A)8d-n
B)10d-n
C)6d-n
D)11d-n
E)5*(6d-n)
Rationale: Total Distance in miles = d
Millage Charge = 5n + (d-n) 6
5n --> For the first n Miles
(d-n)6--> Total no of miles after traveling the first n miles is d-n and the rate after n miles is 6 .
If we solve the equation , We will get 6d-n
Answer : C
Answer: C
Question: A scientist used a unique two-color code to identify each of the test subjects involved in a certain study. If the scientist found that choosing from among seven colors produced enough color codes to identify all but 5 of the test subjects, how many test subjects were in the study? (Assume that the order of the colors in the codes does not matter.)
Options:
A)25
B)26
C)27
D)28
E)29
Rationale: 7C2 = 21
The number of test subjects is 21 + 5 = 26
The answer is B.
Answer: B
Question: In a city where all streets run east-to-west, all avenues run north-to-south, and all intersections are right angles as shown below, Jenn needs to walk from the corner of 1st Street and 1st Avenue to the corner of 6th Street and 3rd Avenue. If her friend Amanda is sitting on a bench on 4th Street halfway between 1st and 2nd Avenues, and Jenn chooses her path randomly from any route that will allow her to walk exactly seven blocks to her destination, what is the probability that Jenn will walk down 4th St. past Amanda?
Options:
A)1/42
B)1/21
C)1/7
D)1/3
E)1/2
Rationale: All routes between (1,1) to (3,6) = 7!/(2!*5!) = 21
Routes which pass 4th st. between 1st and 2nd avenue = routs between (1,1) to (1,4) * routs between (2,4) to (3,6)
= 1 * (3!/(2!*1!)) = 3
the probability that Jenn will walk down 4th St. past Amanda= 3 / 21 = 1 / 7
C is correct.
Answer: C
Question: The figure above shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane, what should be the limit T on the height of vehicles that are allowed to use the tunnel?
Options:
A)5½ ft
B)7½ ft
C)8 ½ ft
D)9½ ft
E)10 ft
Rationale: Let's label the midpoint of the circle O. Since the base of the semi-circle is 20, we know that the diameter is 20 and, accordingly, the radius is 10.
We also know that the traffic lane is 12 feet long and there's an equal amount of space on either side, so the traffic lane extends 6 feet on either side of O. Let's call the leftmost point on the base of the traffic lane A. So, the distance OA is 6.
Now draw a line straight up from A to the top of the tunnel. Let's label the point at which the line intersects the circle B. The answer to the question will, therefore, be the height AB - .5 feet (we need to leave .5 feet of clearance).
Here's the key to solving the question: if we draw a line from O to B, that line is a radius of the circle and, therefore, has length 10. We now have right triangle OAB (the right angle is at point A), with leg OA=6 and hypotenuse OB=10. We can now solve for leg AB=8 (either by applying the pythagorean theorum or by applying the 3/4/5 special right triangle ratio).
Finally: AB=8, so the correct answer T is 8 - .5 = 7.5... choose (B)!
From a strategic guessing point of view, as soon as we realize that the height of the tunnel is 10 in the middle, we should quickly eliminate (D) and (E) as too big; worse case you have a 1/3 shot at picking up the points.B
Answer: B
Question: For all integers x>1. <x>=2x+(2x-1)+(2x-2)+......2+1. What is the value of <2>*<4>?
Options:
A) 60
B) 116
C) 210
D) 263
E) 360
Rationale: Since <x>=2x+(2x-1)+(2x-2)+......2+1=1+2+..+(2x-2)+(2x-1)+2x then<x> is basically the sum of all integers from 1 to 2x, inclusive.
Hence <2> is the sum of all integers from 1 to 2*2=4 and <4> is the sum of all integers from 1 to 2*4=8 -> <4>=36 and <2>=10 --> <3>*<2>=21*10=360
Answer: E
Answer: E
Question: A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?
Options:
A)165
B)878
C)166
D)199
E)234
Rationale: Total number of persons in the committee = 5 + 6 = 11
Number of ways of selecting group of eight persons
= 11C8 = 11C3
= (11 * 10 * 9)/(3 * 2)
= 165 ways.
Answer: A
Answer: A
Question: Two cyclists are running on a circular track of 500m.They start from the same point A at the same time.If they meet for the first time at a point 150 m from A, which of the following could be the possible ratio of their speeds?
Options:
A)1:3
B)2:3
C)3:5
D)3:7
E)7:11
Rationale: well, the key aspect of this question is the distance mentioned, when they met for the first time, which is 150m away from the starting point. This distance between their first meeting can be less than 500 in only two circumstance.
1) if both are travelling in the same direction and one of them has been given a head start say by few meters.
2) or if both are travelling in the opposite direction.
now since they start their journey at the same time. hence option 1 is ruled out. hence they must be running in the opposite direction. hence answer should be 2:3.
Why?? here is a solution
As here time is constant, therefore, distance traveled will be directly proportional to the the respective velocities/speed of the two runners.(d1/d2=v1/v2)
now one of them has covered 150m, therefore other must have covered 350m, therefore ratio of their speeds/velocities will be 3:7
D
Answer: D
Question: Walking with 4/5 of my usual speed, I arrive at the bus stop 10 minutes later than normal. How many minutes does it take to walk to the bus stop at my usual speed?
Options:
A)32
B)36
C)40
D)44
E)48
Rationale: Let t = usual time = distance / usual speed
t + 10 = distance / (4* usual speed/5) = (5 * distance) / (4 * usual speed) = 5t/4
t = 40
The answer is C.
Answer: C
Question: AMOUNT OF BACTERIA PRESENT
Time Amount
1:00 P.M. 8.0 grams
4:00 P.M. x grams
7:00 P.M. 12.4 grams
Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?
Options:
A)12.0
B)12.1
C)12.2
D)10.0
E)10.4
Rationale: The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours.
Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:
Fractional increase from 1 PM to 4 PM = X / 8.0
Fractional increase from 4 PM to 7 PM = 12.4 / X
X \ 8 = 12.4 \ X
X^2 = 12.4 * 8
X = 10
answer : D
Answer: D
Question: A man can swim in still water at 6 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is?
Options:
A)3.9
B)4.2
C)5.3
D)2
E)5.2
Rationale: M = 5
S = x
DS = 6 + x
US = 6 - x
6 + x = (6 - x)2
6 + x = 12 -2x
3x = 6
x = 2
Answer:D
Answer: D
Question: The residents of Town X participated in a survey to determine the number of hours per week each resident spent watching television. The distribution of the results of the survey had a mean of 24 hours and a standard deviation of 12 hours. The number of hours that Pat, a resident of Town X, watched television last week was between 1 and 2 standard deviations below the mean. Which of the following could be the number of hours that Pat watched television last week?
Options:
A)30
B)20
C)18
D)12
E)6
Rationale: Given: mean=24 and SD=12
The number of hours that Pat watched television was between 1 and 2 standard deviations below the mean: 1 SD below the mean ismean-1*SD=12and 2 SD below the mean ismean-2*SD=0, so the number of hours that Pat watched television was between 0 and 12 hours.
Answer: E
Answer: E
Question: Let G represent a set of four distinct prime numbers. If the sum of the numbers in G is even and x is a member of G, then what is the least possible value that x can be?
Options:
A)1
B)2
C)3
D)5
E)7
Rationale: We are told that G is a set of four distinct prime numbers, and the sum of the numbers in G is even. Since the question involves odds, evens and the smallest possible prime, you should immediately think of 2, the smallest prime number and the only even prime number.
If 2 were a member of the set, then we would have one even number and three odd numbers. Since such sum would always be odd, 2 cannot be a member of the set. Thus, all the members of G must be odd to yield an even sum.
Since all the members of the set must be odd prime numbers, the answer is the smallest odd prime number, which is 3.
The correct answer is choice (C).
Note:1 is not a prime number.
Answer: C
Question: How many numbers between 11 and 90 are divisible by 7?
Options:
A)10
B)7
C)11
D)19
E)15
Rationale: The required numbers are 14,21,28,...........,84.
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
ANSWER C
Answer: C
Question: The sum of all two digit numbers divisible by 7 is
Options:
A)910
B)510
C)728
D)628
E)810
Rationale: Required numbers are 14,21,28 . . . . . 98
This is an A.P. in which a=14,d=7 and l=98.
Let the number of terms in it be n.Then t=98
So a+(n-1)d=98.
14+(n-1)*7=98,
14+7n-7=98
7+7n=98
7n=98-7
n = 91/7
then n=13.
Required sum
=n/2(a+l)
=13/2(14+98)
=728
Answer is C.
Answer: C
Question: If a2 + b2 + c2 = 390 and ab + bc + ca = 5, then a + b + c is
Options:
A)19
B)20
C)25
D)30
E)64
Rationale: by formula ,
(a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ca),
since, a^2 + b^2 + c^2 = 390 and ab + bc + ca = 5,
(a + b + c)^2 = 390 + 2(5)
= 400
= 20^2
therefore: a + b + c = 20
ANSWER:B
Answer: B
Question: A & B started a partnership business. A's investment was thrice the investment of B and the period of his investment was two times the period of investments of B. If B received Rs 5000 as profit , what is their total profit?
Options:
A)28000
B)30000
C)32000
D)35000
E)None of these
Rationale: Explanation :
Suppose B's investment = x. Then A's investment= 3x
Suppose Bs period of investment = y, then A's period of investment = 2y
A : B = 3x * 2y : xy
= 6 : 1
Total profit * 1/7 = 5000
=> Total profit = 5000*7 = 35000. Answer : Option D
Answer: D
Question: If the circus were to sell all of its 220 tickets for this month's performance at its usual price, the revenue from sales would be 10% greater than that collected last month. If the circus raised the ticket price by 6% and sold only 200 tickets as a result, what percent less would last month's revenue be compared to this month's revenue?
Options:
A)2
B)5
C)300/53
D)110/53
E)9/4
Rationale: Let price be 10/ticket
then 220 tickets for this month's performance at its usual price=2200
which is 10%greater than last month(let X)---->x(1+10/100)=2200----->x=2000
now new revenue this month=200*10.6=2120
as per question asked 2000=2120(1-A/100)----->A=300/53%
Ans C
Answer: C
Question: The radius of a cylinder is 2 r units and height is 3 r units. Find the curved surface?
Options:
A)862
B)298
C)440
D)278
E)278
Rationale: r = 7
h = 3
2πr(h + r) = 2 * 22/7 * 7(10)
= 440
Answer: C
Answer: C
Question: A man buys a cycle for Rs.1400 and sells it at loss of 15%. What is the selling price of the cycle?
Options:
A)1192
B)1191
C)1118
D)1190
E)1112
Rationale: Explanation:
S.P = 85% of Rs.1400 ⇒ Rs.(8510085100 ×1400) = Rs.1190.
Answer: D
Answer: D
Question: We can rent a boat for x cents per minute for the first 60 minutes and then y cents for each additional minute. How much will it cost to rent a boat for z minutes, where z>60?
Options:
A)x + 60y
B)60x+(z-60)y
C)x + zy
D)60y+(z-60)x
E)60x + zy
Rationale: The first 60 minutes cost 60x.
The next (z-60) minutes cost (z-60)y.
The total cost is 60x+(z-60)y.
The answer is B.
Answer: B
Question: The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.4 the sixth number is?
Options:
A)9.5
B)10
C)10.5
D)11.5
E)12
Rationale: Explanation:
1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 119.9 = 11.5
6th number = 11.5
Option D
Answer: D
Question: 4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number?
Options:
A)311
B)312
C)313
D)314
E)315
Rationale: Let no. be x
4/15*5/7*x - 4/9*2/5*x =8
by further solving
20x/105- 8x/45=8
4x/315=8
x=630
we have to find x/2
=630/2=315
ANSWER:E
Answer: E
Question: A certain stock exchange designates each stock with a two-letter or three-letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes without using the letter A?
Options:
A)16,250
B)16,500
C)16,750
D)17,000
E)17,250
Rationale: The number of two-letter codes is 25*25=625.
The number of three-letter codes is 25*25*25=15,625.
The total number of codes is 625+15,625=16,250.
The answer is A.
Answer: A
Question: The ages of two person differ by 20 years. If 8 years ago, the elder one be 5 times as old as the younger one, their present ages(in years) are respectively
Options:
A)33, 13
B)25, 5
C)29, 9
D)50, 30
E)20,10
Rationale: Let their ages be x and (x + 20) years.
Then, 5(x - 8) = (x + 20 - 8)
=> 4x = 52
=>x =13
Their present ages are 33 years and 13 year.
ANSWER:A
Answer: A
Question: If 6 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is?
Options:
A)20:15:15
B)20:15:18
C)20:15:12
D)20:15:14
E)20:15:11
Rationale: 6A = 8B = 10C
A:B:C = 1/6:1/8:1/10
= 20:15:12
Answer: C
Answer: C
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 72 kilometers, Maxwell's walking speed is 6 km/h, and Brad's running speed is 12 km/h. What is the distance traveled by Maxwell when they meet in the middle?
Options:
A)14
B)24
C)16
D)18
E)20
Rationale: Consider Max starts from Point A and Brad Starts from Point B and move towards each other.
Assume they shall meet at point O after Time 'T'. The question asks us to find OA.
From the question stem we can make out :- Distance OA = 50km - Distance OB
=> 6xT = 72 - 12xT (i.e distance =Speed x Time)
=> 18T = 72
Hence T = 4
OA = 6 x 4 = 24 Km
Answer : B
Answer: B
Question: How many prime numbers between 1 and 100 are factors of 67830?
Options:
A)4
B)5
C)6
D)3
E)2
Rationale: factor of 67830 = 2*3*5*7*17*19--- 6 prime numbers
C
Answer: C
Question: What is the sum of 20 consecutive integers from -9 inclusive, in a increasing order?
Options:
A)-9
B)9
C)-10
D)10
E)20
Rationale: From -9 to -1 --> 9 nos.
zero --> 1 number
from +1 to +9 --> 9 nos.
When we add up nos. from -9 to +9 sum will be zero. Total 19 nos will be added.
20th number will be 10.
Sum of these 20 nos. = 10.
D is the answer.
Answer: D
Question: The area of a triangle is with base 4m and height 5m?
Options:
A)15
B)19
C)56
D)10
E)36
Rationale: Explanation:
Area=1/2*b*h
1/2 * 4 * 5 = 10 m^2
Answer : D
Answer: D
Question: Car A runs at the speed of 65km/hr & reaches its destination in 8hr. Car B runs at the speed of 70 km/h & reaches its destination in 4h. What is the respective ratio of distances covered by Car A & Car B?
Options:
A)3 : 7
B)4 : 9
C)13 : 7
D)5 : 7
E)6 : 11
Rationale: Sol. Distance travelled by Car A = 65 × 8 = 520 km
Distance travelled by Car B = 70 × 4 = 280 km
Ratio = 520/280 = 13 : 7
C
Answer: C
Question: There are 8 periods in each working day of a school. In how many ways can one organize 7 subjects such that each subject is allowed at least one period?
Options:
A)112456
B)145678
C)141120
D)11456
E)None of these
Rationale: 7 subjects can be arranged in 8 periods in 8P7 ways.
Remaining 1 period can be arranged in 7P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (8P7 x 7P1)/2! = 141120
ANSWER:C
Answer: C
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.1600?
Options:
A)3000
B)4200
C)2025
D)1205
E)3251
Rationale: 1800* 9/8 * 9/8
= 2025
Answer: C
Answer: C
Question: (1000)7÷(10)17=?
Options:
A)10
B)100
C)1000
D)10000
E)None of these
Rationale: Explanation:
=(103)7/(10)17
=(10)21/(10)17=10(4)=10000
Option D
Answer: D
Question: If u and v are both negative and uv < v^2, which of the following must be true?
Options:
A)u < v < u^2 < v^2
B)u < v < v^2 < u^2
C)v < u < u^2 < v^2
D)u^2 < v^2 < v < u
E)v^2 < u^2 < v< u
Rationale: Because uv < v^2 and both are negative, I thought u < v. So I crossed off answers c), d), and e).
And because u< v , u^2 < v^2.
ans C
Answer: C
Question: The greatest ratio out of 2:3, 5:8, 3:1, 8:9 and 8:5 is?
Options:
A)2:3
B)3:1
C)5:8
D)8:5
E)8:9
Rationale: 2/3=0.666
5/8=0.625
3/1=3
8/9=0.888
8/5=1.6
ANSWER:B
Answer: B
Question: A fruit shake costs R 2,00 more than a chocolate milk shake. If 3 fruit shakes and 5 chocolate milk shakes cost R 78,00, determine the individual prices.
Options:
A)x=6,y=9
B)x=4,y=4
C)x=7,y=5
D)x=5,y=11
E)x=9,y=11
Rationale: Let the price of a chocolate milkshake be x and let the price of a fruitshake be y.
Set up a system of equations
y=x+2.........(1)
3y+5x=78............(2)
Substitute equation (1) into equation (2)
3(x+2)+5x=78
Rearrange and solve for x
3x+6+5x=78
8x=72
x=9
Substitute the value of x back into equation (1) and solve for y
y=x+2
=9+2
=11
One chocolate milkshake costs R 9,00 and one fruitshake costs R 11,00.
Answer is E.
Answer: E
Question: How many two-digit numbers yield a remainder of 1 when divided by both 4 and 17?
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: Easier to start with numbers that are of the form 17p+1 ---> 18,35,52,69,86.
Out of these, there is only one (69) is also of the form 4q+1. Thus 1 is the answer. B is the correct answer.
Answer: B
Question: In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag- the least chance of drawing a red ball ?
Options:
A)1/32
B)2/32
C)3/32
D)4/32
E)5/32
Rationale: Least Chance.
P= 1/2 * 5/16 + 1/2 * 0 = 5/32
ANSWER:E
Answer: E
Question: A man sells two articles for Rs.3600 each and he gains 35% on the first and loses 35% on the next. Find his total gain or loss?
Options:
A)9% loss
B)400
C)12.25%loss
D)324
E)8% LOSS
Rationale: (35*35)/100 = 12.25%loss
ANSWER:C
Answer: C
Question: Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
Options:
A)Rs. 8600
B)Rs. 8620
C)Rs. 8820
D)Rs. 8000
E)None of these
Rationale: Explanation:
Amount
= Rs.[8000x(1+5/100)2]
=Rs.(8000x21/20x21/20)
=Rs.8820.
ANSWER IS C
Answer: C
Question: There are two box,one containing 39 red balls and the other containing 26 green balls. You are allowed to move the balls between the boxes so that when you choose a box at random and a ball at random from the chosen box, the probability of getting a red ball is maximized. This maximum probability is
Options:
A).60
B).50
C).70
D).80
E).90
Rationale: Here we are allowed to transfer the balls between the two boxes, so we keep only one red ball in first box and move all the remaining balls to the second box, so that we get maximum probability of drawing red ball from first box
So first box--- only one red ball
second box--- 38 red and 26 green
Probability of choosing any box is 1/ 2.
probability of choosing the red ball from first box is 1
probability of choosing the red ball from second box is 38/64
So probability of taking one red ball = [(1/2)×(1)]+[(1/2)*(38/64)]≃0.8
ANSWER:D
Answer: D
Question: If a car went the first third of the distance at 80 kmh, the second third at 30 kmh, and the last third at 48 kmh, what was the average speed of the car for the entire trip?
Options:
A)36 kmh
B)40 kmh
C)42 kmh
D)45 kmh
E)50 kmh
Rationale: Assume D/3 = 240 (this number is convenient because it is divisible by 80, 30 and 48)
So:
240 = 80*T1 = 3 hrs
240 = 30*T2 = 8 hrs
240 = 48*T3 = 5 hrs
T = T1 + T2 + T3 = 16 hrs
D = RT
(240*3) = R*16
R = 45
ANSWER: D
Answer: D
Question: Calculate the speed of a stream if a boat going upstream at
4km/hr and downstream at 8km/hr
Options:
A)1 kmph
B)2 kmph
C)3 kmph
D)6 kmph
E)8 kmph
Rationale: Sol.
Speed of stream = 1/2(8 -4) kmph = 2 kmph.
Answer B
Answer: B
Question: A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction R of the cookies that were placed in the other tins were placed in the green tin?
Options:
A)15/2
B)9/4
C)5/9
D)7/5
E)9/7
Rationale: Blue tin or red tin: 2/3 (N)
Red tin: (1/3)N
Blue tin: (1/4)N
What the last statment meant, is it wants this fraction: (# of cookies in green tin)/(# of cookies in red and green tin)
# of cookies in green tin = 2N/3 - N/4 = 8N- 3N/12 = 5N/12
# of cookies in red and green tin = N/3 + 5N/12 = 9N/12
Fraction R= 5N/12 * 12/9N = 5/9 (C)
Answer: C
Question: Which of the following is closest to the difference between sum Q of all proper fractions (fractions less than 1) in the form 1/x , where x is a positive digit, and the product of all proper fractions in the form y/(y+1), where y is a positive digit?
Options:
A)2.82
B)2.72
C)1.82
D)1.72
E)0.82
Rationale: Sum Q of all proper fractions (fractions less than 1) in the form 1/x, where x is a positive digit:
1/1 + 1/2 + 1/3 +.....+ 1/9
This is a harmonic progression. Harmonic progression is inverse of arithmetic progression.
Approximate sum of a harmonic progression with even number of terms = number of terms * (average of middle 2 terms)
Approximate sum of a harmonic progression with odd number of terms = number of terms * (middle term)
The actual sum will be slightly more than the approximation.
Here we have 9 terms (odd).
Sum = 9 * 1/5
= 9/5
Product of all proper fractions in the form y/(y+1), where y is a positive digit:
1/2 * 2/3 * 3/4 *.....*9/10
We will be left with 1/10.
Required = 9/5 - 1/10
= 1.8 - 0.1
= 1.7
Closest is 1.72
Answer (D).
Answer: D
Question: A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat?
Options:
A)400 m
B)500 m
C)250 m
D)100 m
E)150m
Rationale: Explanation:
A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
ANSWER D
Answer: D
Question: If x > 3000, then the value of (3x)/(2 x^1.21 - 1111^1.3) is closest to?
Options:
A)1/6
B)1/3
C)10/21
D)4/10
E)3/2
Rationale: assume x = 3002
(3x)/(2 x^1.21 - 1111^1.3) = 9006 / (2 * (3002)^1.21 - 1111^1.3)
=9006 / 23154
= = 4 /10
D
Answer: D
Question: Q. How many 9-digit even numbers less than 3,000,000 can be formed using the following digits: 1, 2, 2, 3, 5, 5, 6, 6, 7?
Options:
A)180
B)140
C)210
D)270
E)300
Rationale: the number is less than 3000000 so we will have the digit on the left most side to be either 1 or 2
i.e your number can be 1 _ _ _ _ _ _ or 2 _ _ _ _ _ _
case 1 > 1 _ _ _ _ _ _ here the units digit can be either 2 or 6
when the units digit is 2 i.e 1 _ _ _ _ _ 2
number of ways this can be done would be 5! / 2! (as 5 is repeated twice) = 60
when the units digit is 6 i.e. 1 _ _ _ _ _ 6, number of ways would be 5!/(2! * 2!) {both 2 and 5 repeat twice} = 30
case 2 > 2 _ _ _ _ _ _ (here units digit may be 2 or 6)
number of ways this can be done would be 5! / (2!) for 2 =60
and 5! / 2! for 6 ... = 60
adding all of these ... gives the answer 140 ...
the explanation looks difficult but this technique helps if you are familiar with the concept of permutations when the numbers/alphabets repeat
B
Answer: B