RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: If x is the interest on y and y is the interest on z, the rate and time is the same on both the cases. What is the relation between x, y and z?
A.Y2 = xz
B.Y2 = xy
C.Y2 = Yz
D.Y3 = xz
Answer: A.Y2 = xz
Explanation: X = (Y*NR)/100 Y = (Z*NR)/100
X/Y = NR/100 Y/Z = NR/100
X/Y = Y/Z
Y2 = XZ.Answer:A
Question: H(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2). Which of the following can be the point H?
A.(1, 2).
B.(2, 0).
C.(0, -2).
D.(-2, 0).
Answer: D.(-2, 0).
Explanation: A quick glance at the options, H(x, y)and we can see that A and E do not satisfy the circle equation. Now we are left with B, C and D. Now B and C do not satisfy the line equation. So we are just left with D. And thats the answer=D
Question: Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1344?
A.1000
B.2217
C.1200
D.2777
Answer: C.1200
Explanation: 1344 = P [1 + (5*12/5)/100]
P= 1200
Answer: C
Question: If k is the greatest positive integer such that 4^k is a divisor of 16! then k =
A.3
B.4
C.5
D.6
Answer: C.5
Explanation: 16/4=4
16/16=1
4+1=5
k=5
Answer: C
Question: How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
A.577
B.567
C.360
D.376
Answer: D.376
Explanation: Explanation :
The smallest number in the series is 1000, a 4-digit number.The largest number in the series is 4000, the only 4-digit number to start with 4.The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.Hence, there are 3×5×5×5 or 375 numbers from 1000 to 3999.Including 4000, there will be 376
Answer : D
Question: If m^(-1) = -1/4 then m^(-2) is equal to
A.-9
B.-3
C.-1/9
D.1/16
Answer: D.1/16
Explanation: You made a mistake here.
You're right till
m^-2 = 1/m^2 and m^-1 = 1/m^1 = 1/m
given m^-1 = -(1/4) so, 1/m = -(1/4) solving this, m = -4
Now, m^-2 = 1/m^2 = 1/(-4)^2 = 1/16 ANS D
Question: A package contains 8 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
A.3/5
B.1/2
C.6/8
D.7/9
Answer: B.1/2
Explanation: Let number of balls = (8+8) = 16.
Number of white balls = 8.
P (drawing a white ball) =	8/16	= 1/2
B
Question: How many different 5 letter combinations can be created from the letters A,B,C,D and E?
A.625
B.125
C.3125
D.25
Answer: C.3125
Explanation: This is a case of number of arrangements possible with repetition.
Thus you have 5 options each for the 1st,2nd,3rd,4th and 5th letter giving you a total of 5*5*5*5*5 = 3125 number of possible combinations.
Answer : C
Question: Find the odd man out. 7,13,19,25,29,37,43
A.19
B.29
C.25
D.43
Answer: C.25
Explanation: Explanation :
All given numbers except 25 are prime numbers.
Answer : Option C
Question: When Greenville State University decided to move its fine arts collection to a new library, it had to package the collection in 20-inch by 20-inch by 15-inch boxes. If the university pays $0.60 for every box, and if the university needs 3.06 million cubic inches to package the collection, what is the minimum amount the university must spend on boxes?
A.$306
B.$275
C.$510
D.$1,250
Answer: A.$306
Explanation: Total no. Of boxes = 3060000/(20×20×15) = 510
Total cost = 510×$0.6=$306
Answer A
Question: Nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw. If the total weight of one screw and one nail is half that of 6 screws and one nail, then the total weight of 3 screws, and 3 nails is how many times that of 4.5 screws and 1.5 nails?
A.a. 4/3
B.b. 5/2
C.c. 2
D.d. 3/2
Answer: A.a. 4/3
Explanation: let the weight of nail be n and that of screw be s..
so s+w=1/2 * (6s+1n)...
or 1n = 4s..
lets see the weight of 3 s and 3n= 3s+3*4s=15s..
and weight of 4.5s and 1.5n= 4.5s+1.5*4.5s=11.25s..
ratio= 15s/11.25s=15/11.25=4/3
A
Question: Find the area of circle whose radius is 4m?
A.50.28
B.51.54
C.51.77
D.52.88
Answer: A.50.28
Explanation: 22/7 * 4 * 4
= 50.28
Answer:A
Question: If a 4-digit integer is selected at random from the integers 6000 through 6999, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to three times the tens digit and two times the hundreds digit?
A. 1/999
B. 3/6000
C. 1/1000
D. 3/2000
Answer: C. 1/1000
Explanation: Thousand digit has to be 6 . So units digit needs to 6 as well
Both Hundreds and Units digit needs to be 3 times the tens digit
6/3 = 2
The tens digit is 2
Both Hundreds and Units digit needs to be 2 times the hundreds digit
6/2 = 3
The hundreds digit is 3
The number is 6326
Total numbers between 6000 and 6999 = (6999-6000)+1 = 1000
Probability = 1/1000
Answer C
Question: If 7 and 6 are factors of 60n, what is the minimum value of n?
A. 2
B. 3
C. 7
D. 14
Answer: C. 7
Explanation: 60n/7*6 should be integer => 2*2*3*5*n/7*2*3 = 2*5*n/7 must be an integer
for this to be true n must multiple of 7, thus min of n = 7
hence C
Question: A retailer sold 2 houses at each $10000. On 1st house he got 30% profit and on 2nd house he loss 10% then find his net profit or loss?
A.15%
B.17%
C.21%
D.19%
Answer: B.17%
Explanation: A = 30%
B = -10%
profit% = 30-10 + (-300/100) = 17%
Answer is B
Question: Bill drives from city A to city B. After two hours of driving he noticed that he covered 80 km and calculated that, if he continued driving at the same speed, he would end up been 15 minutes late. So he increased his speed by 10 km/hr and he arrived at city B 36 minutes earlier than he planned.
Find the distance between cities A and B.
A.132 km
B.250 km
C.390 km
D.440 km
Answer: B.250 km
Explanation: Let x be the distance between A and B. Since Bill covered 80 km in 2 hours, his speed was V=80/2=40 km/hr.
If he continued at the same speed he would be 15 minutes late, i.e. the planned time on the road is x/40−15/60x hr. The rest of the distance is (x−80)) km. V=40+10=50 km/hr.
So, he covered the distance between A and B in 2+x−80/50 hr, and it was 36 min less than planned. Therefore, the planned time was 2+x−80/50+36/60
When we equalize the expressions for the scheduled time, we get the equation:
x/40−15/60=2+x−80/50+36/60
x−10/40=100+x−80+30/50
x−10/4=x+50/5
5x−50=4x+200
x=250
So, the distance between cities A and B is 250 km.
correct answer is B)250km
Question: Sum of 3 consecutive even no.'s is 24 more than the 1st no. of the series. Find the middle no.?
A.11
B.12
C.14
D.16
Answer: A.11
Explanation: Let the numbers be x, x+2 and x+4 then
x + x+2 + x+4 = x+24
3x + 6 = x + 24
2x = 18
\inline \therefore x = 9
\inline \therefore Middle number = x + 2 = 9+2 = 11
A
Question: If a,b are positive integers and a/b=85.75 which of the following could be the value of b?
A.1. 25
B.4.24
C.3. 27
D.4. 27
Answer: B.4.24
Explanation: Given a/b = 85.75.
this is 85and 3/4 = 343/4
Since a and b are integers we know that whatever is the value of b, it must be divisible by 4 completely(to make a an integer)
Ans:B
Question: On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday two of his spiders, Wednesday would then have seven times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences?
A.2
B.4
C.6
D.8
Answer: B.4
Explanation: If Pugsley were to give Wednesday two of his spiders, Wednesday would then have seven times as many spiders as Pugsley does:
(w + 2) = 7(p - 2)
If Wednesday were to give Pugsley four of her spiders, Pugsley would now have four fewer spiders than Wednesday had before they traded:
p + 4 = w - 4
Solving gives p = 4 and w = 12.
Answer: B.
Question: A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct
A.2 p.m. on Monday
B.2 p.m. on Tuesday
C.2 p.m. on Wednesday
D.3 p.m. on Thursday
Answer: C.2 p.m. on Wednesday
Explanation: Sol.
Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.
∴ The watch gains [2 + 4 4/5] min. or 34/5 min. in 170 hrs.
Now, 34/5 min. are gained in 170 hrs.
∴ 2 min. are gained in [170 * 5/34 * 2] hrs = 50 hrs.
∴ Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e. it will be correct at 2 p.m. on Wednesday.
Answer C
Question: A can finish a piece of work in 15 days. B can do it in 30 days. They work together for 6 days and then A goes away. In how many days will B finish the work?
A.11 days
B.12 days
C.14 days
D.10 days
Answer: B.12 days
Explanation: 6/15 + (6 + x)/30 = 1 => x = 12 days
Answer: B
Question: Find the average of first 40 natural numbers.
A.15.5
B.20.5
C.40.5
D.25
Answer: B.20.5
Explanation: Explanation:
Sum of first n natural numbers = n(n+1)/2
Hence, sum of first 40 natural numbers = (40 x 41)/2 = 820
Therefore, required average of = 820/40 = 20.5
ANSWER: B
Question: A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
A.6
B.8
C.10
D.12
Answer: C.10
Explanation: If students are A, B, C, D, E and F; we can have 6C3 groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get 5C2 = 10 ways of doing it.
ANSWER:C
Question: A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 17 seconds. What is the length of the platform in meters?
A.240 meters
B.260 meters
C.420 meters
D.600 meters
Answer: B.260 meters
Explanation: Speed of the train in metres/sec = 72000/3600 = 20
Distance travelled by train to cross the platform = 30 * 20 = 600 = Length of train + Length of platform
Distance travelled by train to cross the man = 17 * 20 = 340 = Length of train
Length of platform = 600 - 340 = 260
Answer: B
Question: If the cost price of 121 chocolates is equal to the selling price of 77 chocolates, the gain percent is :
A.56.14%
B.57.44%
C.57.14%
D.56.34%
Answer: C.57.14%
Explanation: Explanation :
Solution: Let C.P. of each chocolate be Re. 1. then, C.P. of 77 chocolates = Rs. 77;
S.P. of 77 chocolates =Rs. 121.
.'. Gain% = 44*100/77 = 57.14%
Answer : C
Question: Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately?
A.7
B.9
C.6
D.2
Answer: C.6
Explanation: Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x2 - 2x - 24 = 0
(x - 6)(x + 4) = 0 => x = 6.
Answer: C
Question: From a group of 6boys&4girls a Committee of 4 persons is to be formed. In how many different ways can it be done, so that the committee has at least 2 boys?
A.120
B.140
C.185
D.190
Answer: C.185
Explanation: The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
(i) 2m. 2w in 6C2 x 4 C1=6X5 - 2X1 X 3X3 2X1 =n an (ii) 3m. 1w in 6C3 x 4C1 =6X5X4 3X2X1 X4 _n - a ''''
(iii) 4m in 6C4 -2x1 = 15 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men. = 90 + 18 + 15 = 185
C
Question: Donna, Chris, and Austin have a total of $93 in their wallets. Donna has $7 more than Chris. Austin has 3 times what Donna has. How much Austin has in his wallet?
A.$10
B.$20
C.$12
D.$60
Answer: D.$60
Explanation: Let the amount in Chris’s wallet be $x
Donna has $7 more than Chrish's wallet = $(x + 7)
Austin has 3 times than Donna's wallet = $3(x + 7)
According to the problem, Donna, chris, and Austin have a total of $93 in their wallets.
Therefore,
x + (x + 7) + 3(x + 7) = 93
x + x + 7 + 3x + 21 = 93
5x + 28 = 93
- 28 -28 (subtract 28 from both sides)
5x = 65
or, 5x/5 = 65/5 (Divide both sides by 5)
x = 13
Amount in Chris wallet = $x = $13
Amount in Donna's wallet = $(x + 7) = $(13 + 7) = $20
Amount in Austin 's wallet = $3(x + 7) = $3(13 + 7) = $3(20) = $ 60
correct answer D
Question: A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
A.31:46
B.32:44
C.32:45
D.32:41
Answer: C.32:45
Explanation: Explanation:
A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
Answer: C
Question: At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
A.7%
B.2%
C.4%
D.9%
Answer: C.4%
Explanation: 150 = (750*5*R)/100
R = 4%
Answer: C
Question: A tank with a volume of 45 cubic feet has one inlet pipe and 2 outlet pipes. The inlet pipe fills water into the tank at the rate of 5 cubic inches/min and the 2 outlet pipes empty it out at the rates of 14 cubic inches/min and 6 cubic inches/min respectively. If all 3 pipes are opened when the tank is full, how many minutes does it take to empty the tank? (1 foot = 12 inches)
A.5184
B.5276
C.5368
D.5460
Answer: A.5184
Explanation: The tank is emptied at this rate: 14 + 6 - 5 = 15 cubic inches/min
The tank has a volume of 45*12*12*12 = 77760 cubic inches.
The time it takes to empty the tank is 77760/15 = 5184 minutes.
The answer is A.
Question: How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?
A.3
B.2
C.1
D.0
Answer: D.0
Explanation: The left side of the equation is obviously not negative and can only be 00 or 11 and plugin the values in and around the boundary in |x−1|+|x−6| and none satisfies.
Answer is D.
Question: In a certain corporation, there are 300 male employees and 150 female employees. It is known that 50% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 450 employees is chosen at random, what is the probability this employee has an advanced degree or is female?
A.1/2
B.2/3
C.3/4
D.4/5
Answer: B.2/3
Explanation: P(female) = 150/450 = 1/3
P(male with advanced degree) = 0.5*300/450 = 150/450 = 1/3
The sum of the probabilities is 2/3
The answer is B.
Question: A sum was put at simple interest at a certain rate for 10 years . Had it been put at 5% higher rate , it would have fetched Rs.100 more. What was the Sum?
A.s.350
B.s.700
C.s.200
D.s.600
Answer: C.s.200
Explanation: At 5% more rate, the increase in S.I for 10 years = Rs.100(given)
So, at 5% more rate, the increase in SI for 1 year = 100/10 = Rs.10/-
i.e. Rs.10 is 5% of the invested sum
So, 1% of the invested sum = 10/5
Therefore, the invested sum = 10 × 100/5 = Rs.200 Answer: C
Question: 3889 + 13.952 - ? = 3854.002
A.48.95
B.49.75
C.45.97
D.47.59
Answer: A.48.95
Explanation: Let 3889 + 13.952 - x = 3854.002.
Then x = (3889 + 13.952) - 3854.002
= 3902.952 - 3854.002
= 48.95.
Answer is A
Question: Machine–A produces 50% of the total output and Machine-B produces 75% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 500 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective?
A.0.07
B.0.06
C.0.0048
D.0.04
Answer: C.0.0048
Explanation: Suppose machine A produce 1000 units
total defective = 9
Machine B then produces=1500 units
total defective=3
probability=12/2500
=0.0048
Answer : C
Question: The average weight of 47 students in a class is 52 kg. 5 of them whose average weight is 48 kg leave the class and other 5 students whose average weight is 54 kg join the class. What is the new average weight (in kg) of the class ?
A.51 1⁄47
B.52 30/47
C.52 1⁄30
D.43.42
Answer: B.52 30/47
Explanation: Total weight of 47students
= 47 × 52 = 2444 kg
Total weight of 5 students who leave
= 5 × 48 = 240 kg
Total weight of 5 students who join
= 5 × 54 = 270 kg
Therefore, new total weight of 47 students
= 2444 – 240 + 270 = 2474
⇒ New average weight = 2474⁄47 = 52 30/47 kg
Answer B
Question: The number of rectangles that can be formed on a 8X8 chessboard is
A.2194
B.1284
C.1196
D.1296
Answer: D.1296
Explanation: 8 rows and 8 columns are separated by 9 vertical and 9 horizontal lines. Any 2 vertical line and any 2 horizontal line will make a rectangle, so total # of rectangles possible is C29∗C29=36∗36=1296C92∗C92=36∗36=1296.
Answer: D.
Question: Three persons invested Rs.9000 in a joint business. The second person invested Rs.1000 more than the first and the third Rs.1000 more than second. After two years, they gained Rs.6300. How much third person will get?
A.2400
B.2800
C.2699
D.2662
Answer: B.2800
Explanation: First persons investment = x
Second persons investment = x + 1000
Third persons investments = x + 2000
x + x + 1000 + x + 2000 = 9000
3x = 6000
x = 2000
Ratio = 2000 : 3000 : 4000
2:3:4
4/9 * 6300 = 2800
Answer: B
Question: There are 5 houses on a street: house A, B, C, D and E. The distance between any two adjacent houses is 100 feet. There are 2 children living in house A, 3 children living in house B, 4 children living in house C, 5 children living in house D and 6 children living in house E. If the school bus can only make one stop on that street, in front of which house should the bus stop so that the sum of walking distance among all children will be the least?
A.house A
B.house B
C.house C
D.house D
Answer: D.house D
Explanation: If we move the stop from house A to house B, 2 children (who live in house A) will have to walk 100 feet more each, but 18 children (who live in house B & C & D & E) will walk 100 feet less each. Therefore, it will be 1800 – 200 =1600 feet less walk if we move the stop from house A to house B.
If we move the stop from house B to house C, 5 children (who live in house A & B) will walk 100 feet more each, but 15 children (who live in house C & D & E) will walk 100 feet less each. Therefore, it will save a total of 1500 – 500 =1000 feet if we move the stop from house B to house C.
If we move the stop from house C to house D, 9 children (who live in house A & B & C) will walk 100 feet more each, but 11 children (who live in house D & E) will walk 100 ft. less each. Therefore, it will be 1100 – 900 =200 feet less walk if we move the stop from house C to house D.
If we move the stop from house D to house E, 14 children (who live in house A & B & C & D) will walk 100 feet more each, but 6 children (who live in house E ) will walk 100 feet less each. Therefore, it will be 1400 – 600 = 800 feet more walk if we move the stop from house D to house E.
Therefore, children will walk the least if the bus stop is at House D.
correct answer D
Question: 8 baseball players each stand at different corners of a baseball diamond. The sides of the diamond are all of equal length. Two arrangements of baseball players are considered different only when the relative positions of the players differ. How many different ways can the baseball players arrange themselves around the diamond?
A.4125
B.5040
C.1620
D.2430
Answer: B.5040
Explanation: it is like a circular arrangement.
total ways for n people to arrange in a circle is = factorial (n-1)
in this case n= 8
hence ans = 7 factorial = 5040
hence B
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
A.99
B.26
C.88
D.26
Answer: A.99
Explanation: 30 * 33/2 = 5/2 * 2 * x => x
= 99
Answer: A
Question: John & Mary go for an interview for 2 vacancies. The probability for the selection of John is 1/3 & whereas the probability forthe selection of Mary is 1/5. What is the probabilitynone of them are selected?
A.2/11
B.3/7
C.8/15
D.4/9
Answer: C.8/15
Explanation: Let A = the event that John is selected and B = the event that Dani is selected.
Given that P(A) = 1/3 and P(B) = 1/5
We know that A¯ is the event that A does not occur and B¯ is the event that B does not occur
Probability that none of them are selected
=P(A¯∩B¯)(∵ Reference : Algebra of Events)
=P(A¯).P(B¯) (∵ Here A and B are Independent Events and refer theorem on independent events)
=[ 1 - P(A) ][ 1 - P(B)]
=(1−13)(1−15)
=23×45=8/15
C
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 ?
A.8d-n
B.10d-n
C.6d-n
D.11d-n
Answer: C.6d-n
Explanation: 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
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.)
A.25
B.26
C.27
D.28
Answer: B.26
Explanation: 7C2 = 21
The number of test subjects is 21 + 5 = 26
The answer is 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?
A.1/42
B.1/21
C.1/7
D.1/3
Answer: C.1/7
Explanation: 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.
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?
A.5½ ft
B.7½ ft
C.8 ½ ft
D.9½ ft
Answer: B.7½ ft
Explanation: 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
Question: A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons?
A.165
B.878
C.166
D.199
Answer: A.165
Explanation: 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