divelab/Sys2Bench huggingface dataset:
Question: At a meeting, 3 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group?
Options:
A)2
B)7
C)C)120
D)D)144
E)E)720
Rationale: The number of arrangements of n distinct objects in a row is given by n!
The number of arrangements of n distinct objects in a circle is given by (n−1)!
"The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:
R=n!-n=(n−1)!
(n−1)!=(7−1)!=720
Answer: E
Answer: C
Question: Having received his weekly allowance, John spent 3/5 of his allowance at the arcade. The next day he spent one third of his remaining allowance at the toy store, and then spent his last $0.96 at the candy store. What is John’s weekly allowance?
Options:
A)$2.40
B)$3.00
C)$3.20
D)$3.60
E)$4.80
Rationale: Total allowance=X
Amount spent at the arcade=3/5X
Amount Remaining=2/5X
Amount Spent at the toy store=2/5*1/3X=2/15X
Amount Remaining=2/5X-2/15X=4/15X
Now, 4/15X=$0.96
Therefore, X=$3.60. Answer D
Answer: D
Question: If (c - a)/(c - b) = 3, then (5b - 5a)/(c - a) =
Options:
A)0.5
B)1
C)1.5
D)3.33
E)2.5
Rationale: Let's say c=7, b=5, a=1 so that our 1st expression holds true. Now, ibsert those numbers in the second expression and we'll get 3.33
Answer D (hopefully )))
Answer: D
Question: In the faculty of Reverse-Engineering, 220 second year students study numeric methods, 423 second year students study automatic control of airborne vehicles and 134 second year students study them both. How many students are there in the faculty if the second year students are approximately 80% of the total?
Options:
A)515.
B)545.
C)618.
D)644.
E)636.
Rationale: Answer is E :636
Solution : Total number of students studying both are 423 + 220 - 134 = 509 (Subtracting the 134 since they were included in the both the other numbers already).
So 80% of total is 509, so 100% is approx. 636.
Answer: E
Question: Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively The ratio of their speeds is
Options:
A)2 : 3
B)4 : 3
C)6 : 7
D)9 : 16
E)None
Rationale: Solution
Let us name the trains
as A and B.	= (A'S speed) : (B's speed)
=√b : √a
= √16 : √9
= 4 : 3
Answer B
Answer: B
Question: 10x + 2y = -6
2x + y = 3
In the system of equations above, what is the value of x?
Options:
A)3
B)2
C)1
D)-1
E)-2
Rationale: 10x + 2y = -6 can be written as 5x + y = -3
Let's subtract the second equation from this equation.
3x = -6
x = -2
The answer is E.
Answer: E
Question: What is the value of n if the sum of the consecutive odd intergers Q from 1 to n equals 169?
Options:
A)47
B)25
C)37
D)33
E)29
Rationale: Before you tackle this question you must first understand that the question is comprised of two key parts, 1st is finding out how manytermsis in that sequence and 2nd whatactual number valuethat term is. In an arithmetic progression, in this case consecutive odd integers 1, 3, 5, ...., there are two set of rules.
Rule #1 (Arithmetic Sequence): xn = a + d(n-1) Identifies what the actual # in the sequence would be. Each number in the sequence has a term such as 1(is the first term), 3(is the second term) and so on. So if I were to ask you to find out what the 10th term is of that sequence you would use that formula to find that value.
a=1 (first term)
d=2 (the common difference) remember in the sequence 1, 3, 5, 7 the common difference is always 2
*On a side note we use n-1 because we don't have d in the first term, therefore if we were solving for the first term we would get 0 as n-1 and 0 times d would give us 0, leaving only the first term. This works regardless what your first term is in any sequence.
But remember the question asksWhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169?which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is, in this case it would be the last number in that sequence. In order to find that we first need to knowhow many terms(how many of the n there is) in order to be able to plug n in this formula given we know what the sum is. For that to happen we need to use Rule #2.
Rule #2 (Summing an arithmetic series): 169 = n/2(2a+(n-1)d). Given the question gives us what the sum is (169 in this case) we would simply use this formula to solve for n. Once we solve for n (13 in this case) we can simply plug n into the first formula (rule 1) and find the value.
It feels very confusing and difficult at first, but once you identify the steps all you need to do is plug and play. We have the sum (169) of a sequence, the number of terms in that sequence is (unknown). Rule #2 tells us how many numbers there are in that sequence and Rule #1 gives us what that last term is.
Answer: B
Question: If 0 < X < Y, X is an odd number and Y is a prime number, which of the following can be the value of X + Y ?
Options:
A)11.
B)13.
C)17.
D)10
E)7.
Rationale: If x is odd ,implies x(odd)+y(odd/even prime)=(Even/Odd)
If Y is even, the 2 is the even prime , therefore y=2 & x=1 since 0<X<Y therefore X+Y =3. But 3 is not an option
which implies y is odd,
Now X(odd)+Y(odd)=Even
10 is the only option that is even.
Ans is D: 10
Answer: D
Question: If the complement of a certain angle is twice the measure of that certain angle, then what is the measure of that certain angle?
Options:
A)45°
B)30°
C)22.5°
D)18°
E)15°
Rationale: Thecomplementof angle A is the angle which, when added to angle A, gives 90 degrees. The two acute angles of a right triangle are complements, for example.
The original angle is x, so the complement is 2x, and together, these add up to 90 degrees.
x + 2x = 90
3x = 90
x = 30°
Answer = (B)
Answer: B
Question: Seven students are trying out for the school soccer team, on which there are three available positions: fullback, sweeper, and goalie. Each student can only try out for one position. The first two students are trying out for fullback. The next two students are trying out for sweeper. The remaining three students are trying out for goalie. However, the fourth student will only play if the second student is also on the team, and the third student will only play if the fifth student is on the team. How many possible combinations of students are there to fill the available positions?
Options:
A)3
B)5
C)7
D)10
E)12
Rationale: Based on the information in the prompt, we have seven players (A,B,C,D,E,F,G). We're asked for the total groups of 3 that can be formed with the following restrictions:
1) Only 1 player per position
2) A and B are trying out for fullback
C and D are trying out for sweeper
E, F and G are trying out for goalie
3) D will only play if B is also on the team
C will only play if E is also on the team
The big restrictions are in the two 'formal logic' rules....
-We can put E with ANYONE, but we can only put in C if E is ALSO there.
-We can put B with ANYONE, but we can only put in D if B is ALSO there.
By extension....
D can NEVER be with A (because then B would not be in the group)
C can NEVER be with F or G (because then E would not be in the group)
As such, there are only a few possibilities. We can have...
ACE
BCE
BDE
BDF
BDG
Final Answer:5 ANSWER:B
Answer: B
Question: The cricket team of 11members is 26yrs old & the wicket keeper is 3 yrs older. If the ages ofthese 2are excluded, the average age of theremaining players is 1 year less than the average age of the whole team. What is the average age of the team ?
Options:
A)20
B)21
C)23
D)26
E)29
Rationale: Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
C
Answer: C
Question: An equilateral triangle T2 is formed by joining the mid points of the sides of another equilateral triangle T1. A third equilateral triangle T3 is formed by joining the mid-points of T2 and this process is continued indefinitely. If each side of T1 is 45 cm, find the sum of the perimeters of all the triangles.
Options:
A)180 cm
B)220 cm
C)240 cm
D)270 cm
E)300 cm
Rationale: We have 45 for first triangle, when we join mid-points of first triangle we get the second equilateral triangle then the length of second one is 22.5 and continues.
So we have 45,22.5,11.25,...
We have ratio = 1/2, and it is GP type.
Sum of infinite triangle is a/1-r = 45/1-(1/2) = 90
Equilateral triangle perimeter is 3a = 3*90 = 270.
So option D.
Answer: D
Question: If a man buys 10 lollipops for $80 and sold them for $4 dollars determine his loss.
Options:
A)50
B)40
C)70
D)80
E)90
Rationale: Cost of 10 lollipops = $80
Sold each lollipop for $4
So he sold 10 lollipops for $(10 × 4) = $40
Loss = $(80 – 40) = $40
answer :B
Therefore, loss = $ 40.
Answer: B
Question: What should come in place of the question mark (?) in the following equation?
47(7.5) ÷ 47 3⁄2 × 4 7-3 = (√47)?
Options:
A)3
B)2 1⁄2
C)6
D)3.5
E)None of these
Rationale: (47)3 = [(47)1/2]6
Answer C
Answer: C
Question: A 250 m long train running at the speed of 120 km/hr crosses another train running in opposite direction at the speed of 80 km/hr in 9 sec. What is the length of the other train?
Options:
A)230
B)288
C)250
D)256
E)298
Rationale: Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 250)/9 = 500/9 => x = 250.
Answer: C
Answer: C
Question: a men walked 30m west 5m south 10m west 35m north to tollgate find the distance from the start gate?
Options:
A)40m
B)50m
C)60m
D)70m
E)80m
Rationale: The man is toward West = 30+10 = 40m
The man is towards North = 35m(N) - 5m(S) = 30m
Therefore, the distance from the start gate = sqrt(40^2+30^2)
= sqrt(2500) = 50m
ANSWER:B
Answer: B
Question: A man can row a boat at 20 kmph in still water. If the speed of the stream is 4 kmph, what is the time taken to row a distance of 60 km downstream?
Options:
A)30/18 hours
B)37/13 hours
C)80/13 hours
D)5/2 hours
E)30/43 hours
Rationale: Speed downstream = 20 + 4 = 24 kmph.
Time required to cover 60 km downstream
= d/s = 60/24
= 5/2 hours.
Answer: D
Answer: D
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)11.9
B)11.0
C)11.2
D)11.5
E)11.7
Rationale: 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
Answer:D
Answer: D
Question: X and Y invested in a business. They earned some profit which they divided in the ratio of 1 : 3. If X invested Rs.10,000. the amount invested by Y is
Options:
A)Rs.30,000
B)Rs.50,000
C)Rs.60,000
D)Rs.80,000
E)None
Rationale: Solution
Suppose Y invested Rs.y
Then,	10000 /y = 1 / 3
â€¹=â€º y=(10000Ã—3 / 1).
â€¹=â€º y=30000.
Answer A
Answer: A
Question: The average of first six prime numbers which are odd is?
Options:
A)7.55
B)9.33
C)6.56
D)8.23
E)3.5
Rationale: Explanation:
Sum of first 6 prime no. which are odd = 56
Average = 56/6 = 9.33
Answer:B
Answer: B
Question: In a certain city, 75 percent of the households have cable television, and 60 percent of the households have videocassette recorders. If there are 150,000 households in the city, then the number of households that have both cable television and videocassette recorders could be any number from:
Options:
A)52,500 to 90,000 inclusive
B)37,500 to 120,000 inclusive
C)60,000 to 90,000 inclusive
D)72,500 to 120,000 inclusive
E)90,000 to 120,000 inclusive
Rationale: The maximum percentage with both is 60% which is 90,000 households.
This means that every household with a recorder also has cable television.
The minimum percentage with both is 35% which is 52,500 households.
This means that 25% of households have only recorders but not cable television.
The other 35% of households with recorders have cable television.
The answer is A.
Answer: A
Question: A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
Options:
A)4800
B)4991
C)5004
D)5000
E)6000
Rationale: Explanation:
Let the sale in the sixth month =x
Then 6435+6927+6855+7230+6562+x/6=6500
x=39000−34009=4991
ANSWER IS B
Answer: B
Question: Two pipes can fill a tank in 10 minutes and 15 minutes. An outlet pipe can empty the tank in 45 minutes. If all the pipes are opened when the tank is empty, then how many minutes will it take to fill the tank?
Options:
A)30 minutes
B)8 minutes
C)15 minutes
D)10 minutes
E)12 minutes
Rationale: Part of the filled by all the three pipes in one minute
= 1/10 + 1/15 - 1/45 = (9 + 6 - 2)/90 = 13/90
So, the tank becomes full in 8 minutes.
Answer: B
Answer: B
Question: An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq m?
Options:
A)Rs.4082.29
B)Rs.4082.27
C)Rs.4082.26
D)Rs.4082.40
E)Rs.4082.45
Rationale: Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 * 7)
= 315 (13 - 0.04)
= 4095 - 12.6
= Rs.4082.40
Answer: D
Answer: D
Question: if ratio of total amount and principal amount is 3:2.rate of interest 5%.calculate the time.
Options:
A)10
B)6
C)5
D)8
E)data not available
Rationale: s.i=p*r*t/100
(total amount-p)/p=5t/100
t.a/p-1=5t/100
1.5-1=5t/100
50=5t
t=10
answer A
Answer: A
Question: How many real roots does the equation x^2y+16xy+62y=0 have if y < 0?
Options:
A)0
B)1
C)2
D)3
E)Infinite
Rationale: x^2y+16xy+62y=0
=> y ( x^2 + 16x + 62) = 0
=> y (x+8)^2 = 0
if y<0 , then x=-8
So although there are 2 factors , they are the same x=-10 .
The equations has 2 distinct real root .
Answer C
Answer: C
Question: If the average (arithmetic mean) of the four numbers K, 2K + 3, 3K – 4 and 5K + 1 is 63, what is the value of K?
Options:
A) 22 10/11
B) 15 3/4
C) 22
D) 23
E) 25 3/10
Rationale: K + 2K +3 + 3K - 4 + 5K +1 = 11K
(11K -1)/4 = 63
11K = 63 * 4 = 252
K = 252 / 11 = 22 10/11
Answer A.
Answer: A
Question: A and B together can complete a work in 12 days. A alone can complete it in 20 days. If B does the work only for half a day daily, then in how many days A and B together will complete the work?
Options:
A)11 days
B)77 days
C)15 days
D)18 days
E)19 days
Rationale: B's 1 day work = (1/12 - 1/20) = 1/30
Now, (A + B)'s 1 day work = (1/20 + 1/60) = 1/15
So, A and B together will complete the work in 15 days.
Answer:C
Answer: C
Question: The distance from the x-axis to point P is half the distance from the y-axis to point P. If the coordinates of P are (x,-6), how many units is P from the y-axis?
Options:
A)3
B)6
C)9
D)10.5
E)12
Rationale: The x-axis is 6 units from the point P.
Thus the y-axis is 12 units from the point P.
The answer is E.
Answer: E
Question: In digging a pond 30m*10m*5m the volumes of soil extracted will be ?
Options:
A)2000 cu.m
B)2500 cu.m
C)15000 cu.m
D)1500 cu.m
E)3000 cu.m
Rationale: 30*10*5= 1500Answer : D
Answer: D
Question: When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 9 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster?
Options:
A)30
B)34
C)42
D)46
E)55
Rationale: Total # of students 26+28+18-(9+7+10)+9=55.
Answer: E.
Answer: E
Question: All of the stocks on the over-the-counter market are designated by either a 5-letter or a 6-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes?
Options:
A)2(26^5)
B)26(26^4)
C)27(26^4)
D)26(26^5)
E)27(26^5)
Rationale: Number of 4-letter codes: 26 * 26 * 26 * 26 * 26= 26^5
Number of 5-letter codes: 26 * 26 * 26 * 26 * 26 * 26 = 26^6
Total Number of codes: 26^5 + 26^6 = 26^5 + 26*(26^5) = 27*(26^5)
Therefore, the answer isE: 27*(26^5).
Answer: E
Question: If the units digit of n^26 is 7, which of the following could be the value of n?
I. n = 41
II. n = 49
III. n = 47
Options:
A)Only I
B)Only II
C)Only III
D)I and II
E)II and III
Rationale: I tried in this way.
Given n^26 units digit is 7.
If at all we have to get 7 as units digits only 3 and 7 series has 7
3^! = 3 , 3^2 = 9 , 3^3 = 27 , 3^4 = 81
7^1 = 7, 7^2 = 49 , 7^3 = 3 (units ) , 7^4 = 1 (units)
We have n^26 - then divide 26 by 4 = reminder 2.
Then we have for only 7^2 - we have units digit as 9.
Hence 47^26 = 7 (units).
Option B is correct answer , and with 41^26 we always get 1 as units digit and 49^26- we get units as 9.
Answer: B
Question: A bag contains 3 red balls, 5 green balls and 6 white balls. A ball is drawn at
random from the box. What is the probability that the ball drawn is either red or green?
Options:
A)2/5
B)3/5
C)4/7
D)7/15
E)8/15
Rationale: Total number of balls = (3 + 5 + 6) = 14.
Therefore, n(S) = 14.
Let E1 = event of drawing a red ball.
and E2 = event of drawing a green ball.
Then, E1 n E2 = f.
P (E1 n E2) = P(E1) + P(E2) = (3/14 + 5/14) = 4/7.
ANSWER:C
Answer: C
Question: Narendra started a business, investing 30000. Six months later, Chandan joined him with 15000. If they make a profit of 10000 at the end of the year, what should be the share of Narendra?
Options:
A)Rs.7000
B)Rs.8000
C)Rs.9000
D)Rs.10000
E)Rs.11000
Rationale: The ratio of their investments = 30000 × 12 : 15000 × 6 = 4 : 1.
:. Narendra’s share on profit of Rs.10000 = 10000 × 4/5 = Rs.8000.
B)
Answer: B
Question: The fraction 101 27/100000 in decimal for is:
Options:
A)101.027
B)101.00027
C)101.0027
D)10100027
E)1010027
Rationale: 101 27/100000
= 101 + 27/ 100000
= 101 + .00027
= 101.00027
Answer is B.
Answer: B
Question: If 0.4 of a number is equal to 0.07 of another number, the ratio of the numbers i
Options:
A)2 : 3
B)7 : 40
C)3 : 20
D)20 : 3
E)None
Rationale: Sol.
0.4A = 0.08 B â‡” A / B = 0.07 / 0.40 = 7/40.
âˆ´ A : B = 7 : 40.
Answer B
Answer: B
Question: What is the largest 4 digit number exactly divisible by 48?
Options:
A)4676
B)4678
C)9984
D)9504
E)9936
Rationale: Largest 4 digit number = 9999
9999 ÷ 48 = 208, remainder = 15
Hence largest 4 digit number exactly divisible by 48
= 9999 - 15 = 9984
answer :C
Answer: C
Question: Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $189 million, which represented 9 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars?
Options:
A)17,010
B)2,100
C)1,890
D)1,701
E)210
Rationale: 9% ---- 189 millions
for 100% => ( 189 * 100% )/9% = 2100.
Option B.
Answer: B
Question: A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 70 km downstream?
Options:
A)30/83 hours
B)33/13 hours
C)31/13 hours
D)35/13 hours
E)30/14 hours
Rationale: Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 70 km downstream = d/s = 70/26
= 35/13 hours.
Answer:D
Answer: D
Question: In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. States B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State ?
Options:
A)7600
B)8000
C)8400
D)Data inadequate
E)None of these
Rationale: Solution
Let the number of candidates appeared from each State be x.
Then, 7 % of x -6% of x = 80 ⇔ 1% of x = 80 ⇔ x = 80 × 100 = 8000.
Answer B
Answer: B
Question: Starting with 0, a mathematician labels every non-negative integer as one of five types: alpha, beta, gamma, delta, or epsilon, in that repeating order as the integers increase. For instance, the integer 8 is labeled delta. What is the label on an integer that is the sum of an alpha raised to the seventh power and another alpha raised to the seventh power?
Options:
A)alpha.
B)beta.
C)gamma.
D)delta.
E)epsilon.
Rationale: Let the two alphas be 5k and 5j.
(5k)^7 + (5j)^7 is a multiple of 5 so it is an alpha.
The answer is A.
Answer: A
Question: if log 6 = 0.778151, the number of digits in 6128 is
Options:
A)22
B)39
C)88
D)100
E)98
Rationale: Explanation:
log(6128) = 128log(6) = 128 × 0.778151 ≈ 99.60
ie, its characteristic = 99
Hence, number of digits in 2128 = 99+1 = 100
Answer: Option D
Answer: D
Question: Martha has the unique talent of being able to guess other people’s height and weight. For every four people that Martha meets, she consistently guesses the people’s correct height three times, and for every five people that she meets, she consistently guesses the people’s correct weight four times. If Martha meets three people and her success rate remains constant, what is the probability that Martha correctly guesses a person’s weight and height at least once?
Options:
A)8/27
B)117/125
C)49/81
D)98/125
E)125/144
Rationale: Probability that she guesses the height correctly P(h)=3/4
Probability that she guesses the weight correctlyP(w)=4/5
Probability that she guesses both weight and height correctly(P(h)*P(w))=3/4 * 4/5= 12/20
Now the Q asks about the probability of this happening atleast once.
We calculate it by finding the probability of not being able to guess in any of the three occasions.
Probability of not being able to guess any no. of times =1-12/20=8/20=2/5
For all three occasions P(A)=2/5 * 2/5 * 2/5=8/125
Probability of the event happening atleast once=1- P(A)=1-8/125=117/125
Ans should be B
Answer: B
Question: A train passes a platform in 16 seconds. The same train passes a man standing on the platform in 10 seconds. If the speed of the train is 54 km/hr, The length of the platform is
Options:
A)40
B)50
C)90
D)30
E)20
Rationale: Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s
Length of the train = speed × time taken to cross the man = 15×10 = 150 m
Let the length of the platform = L
Time taken to cross the platform = (150+L)/15
=> (150+L)/15 = 16
=> 150+L = 15×16 = 240
=> L = 240-150 = 90 meter
Answer is C
Answer: C
Question: The average of first five multiples of 4 is
Options:
A)3
B)6
C)9
D)12
E)15
Rationale: Solution
Average	= 4(1+2+3+4+5) / 5
= 60 / 5.
= 12
Answer D
Answer: D
Question: Subtracting 7% of A from A is equivalent to multiplying A by how much ?
Options:
A)0.93
B)9.4
C)0.094
D)94
E)None
Rationale: Answer
Let A - 7% of A = AB.
⇒ (93 x A) / 100 = AB
∴ B = 0.93
Correct Option: A
Answer: A
Question: In a certain game, you pick a card from a standard deck of 52 cards. If the card is a spade, you win. If the card is not a spade, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a spade, and the point is to measure how many draws did it take before the person picked a spade and won. What is the probability that there will be at least three draws involved in a win, i.e. someone picking her first spade on the third draw or later?
Options:
A) 1/2
B) 11/16
C) 9/16
D) 13/16
E) 15/16
Rationale: Probability of picking a spade on any draw = 1/4
Probability of NOT picking a spade on the first draw AND on the second draw = [1-(1/4)] X [1-(1/4)] = 3/4 X 3/4 = 9/16
Answer : C
Answer: C
Question: What is the number that should come in the place of question mark?
1050	420	168	67.2	?
Options:
A)29.88
B)10.752
C)4.3008
D)26.88
E)27.88
Rationale: 1050/2.5 = 420
420/2.5 = 168
168/2.5 = 67.2
so 67.2/2.5 = 26.88
ANSWER:D
Answer: D