RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Five years ago, the average age of A and B was 15 years. Average age of A, B and C today is 20 years. How old will C be after 14 years?
A.30
B.34
C.40
D.50
Answer: B.34
Explanation: Explanation :
(A + B), five years ago = (15 * 2) = 30 years.
(A+B), now = (30+5*2)years=40 years.
(A + B + C), now = (20 x 3) years =60 years.
C, now = (60 - 40) years = 20 years.
C, after 14 years = (20 + 14) years =34 years.
Answer : B
Question: A soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price. During the annual summer sale, everything in the store is an additional 20 percent off the original list price. If a replica jersey's list price is $80, approximately what W percent of the list price is the lowest possible sale price?
A.20
B.25
C.30
D.40
Answer: D.40
Explanation: Let the list price be 2x
for min sale price, the first discount given should be 50%, 2x becomes x here
now, during summer sale additional 20% off is given ie sale price becomes 0.8x
it is given lise price is $80
=> 2x = 80
=> x = 40
and 0.8x = 32
so lowest sale price is 32, which W is 40% of 80
hence, D is the answer
Question: The compound interest accrued on an amount of Rs.44000 at the end of two years is Rs.1193.60. What would be the simple interest accrued on the same amount at the same rate in the same period?
A.10522
B.10528
C.10560
D.10529
Answer: C.10560
Explanation: Let the rate of interest be R% p.a.
4400{[1 + R/100]2 - 1} = 11193.60
[1 + R/100]2 = (44000 + 11193.60)/44000
[1 + R/100]2 = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]2 = 784/625 = (28/25)2
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560
Answer: C
Question: By selling a pen for Rs.15, a man loses one sixteenth of what it costs him.The cost price of the pen is
A.Rs.16
B.Rs.18
C.Rs.20
D.Rs.21
Answer: A.Rs.16
Explanation: Solution
Let the C.P be Rs.x.Then,
x-15=x/16	‹=›x-x/16 = 15
15x/16=15
x = 16.
Therefore, C.P= Rs.16.
Answer A
Question: A and B invests Rs.10000 each, A investing for 6 months and B investing for all the 12 months in the year. If the total profit at the end of the year is Rs.4500, find their shares?
A.1500,6000
B.4500,3000
C.3500,3000
D.1500,3000
Answer: D.1500,3000
Explanation: The ratio of their profits A:B = 6:12 =1:2
Share of A in the total profit = 1/3 * 4500 = Rs.1500
Share of B in the total profit = 2/3 * 4500 = Rs.3000
Answer:D
Question: Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 1w widgets?
A.4
B.6
C.8
D.10
Answer: B.6
Explanation: Let Y produce w widgets in y Days
Hence, in 1 day Y will produce w/y widgets.
Also, X will produce w widgets in y+2 days (given, X takes two more days)
Hence, in 1 day X will produce w/y+2 widgets.
Hence together X and Y in 1 day will produce {w/y + w/y+2} widgets.
Together X and Y in 3 days will produce = 3 * [{w/y + w/y+2}] widgets.
It is given that in 3 days together they produce (5/4)w widgets.
Equating, 3 * [{w/y + w/y+2}] = (5/4)w
Take out w common and move 3 to denominator of RHS
w{1/y + 1/(y+2)} = (5/12)w
Canceling w from both sides
{1/y + 1/(y+2)} = 5/12
2y+2/y(y+2) = 5/12
24y+24=5y^2 + 10y
5y^2-14y-24=0
5y^2-20y+6y-24=0
5y(y-4)+6(y-4)=0
(5y+6)+(y-4)=0
y=-6/5 or y=4
Discarding y=-6/5 as no of days cannot be negative
y=4
hence it takes Y, 4 days to produce w widgets.
Therefore, it will take X (4+2)=6 days to produce w widgets.
Hence it will take X 1*6=6 days to produce 1w widgets.
Answer : B
Question: 1200 boys and 800 girls are examined in a test; 4% of the boys and 30% of the girls pass. The percentage of the total who failed is ?
A.A)63.3%
B.B)52.4%
C.C)62.8%
D.D)75.4%
Answer: C.C)62.8%
Explanation: Total number of students = 1200+800 = 2000
Number of students passed = (42% of 1200+30% of 800) = 504+240 = 744
number of failures = 1256*100/2000= 62.8%
Answer is C
Question: Q and R are two-digit positive integers that have the same digits but in reverse order. If the positive difference between Q and R is less than 70, what is the greatest possible value of Q minus R?
A.63
B.64
C.65
D.66
Answer: A.63
Explanation: A two-digit integer "ab" can be expressed algebraically as 10a+b.
Q-R=(10a+b)-(10b+a)=9(a-b)<70.
The greatest multiple of 9 which is less than 70 is 63.
The answer is A.
Question: A salt manufacturing company produced a total of 5000 tonnes of salt in January of a particular year. Starting from February its production increased by 100 tonnes every month over the previous months until the end of the year. Find its ave66rage daily production for that year?
A.180
B.185
C.190
D.195
Answer: B.185
Explanation: Total production of salt by the company in that year = 5000 + 5100 + 5200 + .... + 6100 = 66600.
Average monthly production of salt for that year = 66600/365 = 185
ANSWER:B
Question: If points A and B are (2+3) units apart, which of the following can have only one possible value?
I. The area of a square region with side AB
II. The circumference of a circle passing through points A and B
III. The area of a rectangular region with diagonal AB
A.I only
B.II only
C.III only
D.I and II only
Answer: A.I only
Explanation: I. The area of a square region with side AB --> side=AB=(2+3) or 5 --> the area of a square is 5^2=25, so only one possible value;
II. The circumference of a circle passing through points A and B --> circle is not defined by only two points, hence the circumference of a circle passing through points A and B can take infinitely many values. For example consider when AB=diameter and when points A and B are not the endpoints of the diameter;
III. The area of a rectangular region with diagonal AB --> consider the case when AB is the diagonal of a square (the area is 25/2) and the case when AB is the diagonal of a rectangle with the sides equal to 3 and 4 (the area 12).
Answer: A.
Question: Find the missing number in the given sequence : 1,6,11,16,?,26,31,?
A.20 & 35
B.19 & 34
C.18 & 33
D.21 & 36
Answer: D.21 & 36
Explanation: 1+5=6
6+5=11
11+5=16
16+5=21
21+5=26
26+5=31
31+5=36
ANSWER:D
Question: If a square mirror has a 20-inch diagonal, what is the approximate perimeter R of the mirror, in inches?
A. 40
B. 60
C. 80
D. 100
Answer: B. 60
Explanation: if you draw the square and diagonal inside the square. u can see square becomes part of two triangles opposite to each other.
And We know the property of the triangle, addition of two sides of triangle must be greater than its diagonal in order to complete the triangle. And each side must be less than 20 and perimeter R must be less than 80, so we can eliminate answer choice C, D and E.
so Side 1 + side 2 > 20, that means Side 1 or Side 2 must be > 10. so we can eliminate the answer choice A.
Now we are left with is B
Question: The number 50 can be written as the sum of squares of 3 integers. Which of the following could be the difference between the largest and smallest integers of the 3?
A.2
B.5
C.7
D.8
Answer: A.2
Explanation: Notice that the question asks which of the followingcouldbe the difference between the largest and smallest integers, not must be. The 3 integers could be: +/-3,+/-4and +/-5, so the difference could be 2,-2 or 1. Since only one of them is among the choices, then it must be the correct answer.
Answer: A
Question: At what rate percent per annum will a sum of money double in 7 years.
A.12.5%
B.13.5%
C.14.2%
D.14.5%
Answer: C.14.2%
Explanation: Let principal = P, Then, S.I.=P and Time=8 years
Rate = [(100 x P)/ (P x 8)]% = 14.2% per annum. Answer: C
Question: When the price of a product was increased by 15%, the number sold was decreased by 20%. What was the net effect?
A.8% gain
B.5% loss
C.8% loss
D.Cannot be determined
Answer: C.8% loss
Explanation: Net effect = + 15 – 20 – 15×20/100=−8%
–ve sign indicates loss.
Answer C
Question: If y exceeds x by 24%, then x is less than y by?
A.16 2/8%
B.16 8/3%
C.19 11/31%
D.76 2/3%
Answer: C.19 11/31%
Explanation: X=100	y=124
124------24
100-------? => 19 11/31%
Answer:C
Question: A certain basketball team that has played 2/3 of its games has a record of 13 wins and 7 losses. What is the greatest number of the remaining games that the team can lose and still win at least 3/4 of all of its games?
A. 0
B. 1
C. 2
D. 4
Answer: A. 0
Explanation: 13 wins, 7 losses - total 20 games played.
The team has played 2/3rd of all games so total number of games = 30
3/4th of 30 is 22.5 so the team must win 23 games and can afford to lose at most 7 total games. It has already lost 7 games so it can lose another 0 at most.
Answer (A)
Question: The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
A.2010
B.2011
C.2012
D.2013
Answer: B.2011
Explanation: Suppose commodity X will cost 40 paise more than Y after z years.
Then, (4.20 + 0.40z) - (6.30 + 0.15z) = 0.40
0.25z = 0.40 + 2.10
z = 2.50/0.25
z = 250/25
z =10
X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.
Answer = B
Question: The H.C.F of two numbers is 13 and their L.C.M is 2200. If one of the numbers is 286, then the other is?
A.85
B.90
C.95
D.100
Answer: D.100
Explanation: Other number = (13 * 2200)/286
= 100.
Answer: D
Question: A shopkeeper sells a badminton racket, whose marked price is Rs. 30, at a discount of 15% and gives a shuttle cock costing Rs. 1.50 free with each racket. Even then he makes a profit of 20%. His cost price per racket is :
A.23.5
B.22.5
C.21.0
D.20.0
Answer: D.20.0
Explanation: Sol.
Marked price = Rs. 30. S.P. = Rs.[(85/100 * 30) - 1.50] = Rs. (25.50 - 1.50) = Rs. 24.
Let C.P. be Rs. x Then, 120% of x = 24 ⇒ x (24 * 100 / 120) = Rs. 20.
Answer D
Question: The value of (34.31*0.473*1.5)/(0.0673*23.25*7.57) is close to
A.2
B.1.15
C.2.01
D.2.06
Answer: D.2.06
Explanation: (34.31*0.473*1.5)/(0.0673*23.25*7.57) = 24.343/11.845
= 2.06
ANSWER:D
Question: A contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days B will the work get over?
A.60
B.70
C.B=75
D.B=80
Answer: C.B=75
Explanation: We can also use the concept of man-days here
100 days -->10men so the job includes 100*10=1000 man-days
After 20 days
1/4 of Job is completed so 1/4 X 1000 man-days=250 man-days Job is done
Now the Balance Job=1000-250=750 man-days worth of Job
Since 2 men are fired so B/L men=8
Therefore Total no. of days of Job=750 man-day/8 days = 375/4=94 days (approx.)
Now since this is total and Ques. is asking for additional no. of days,
So 94-20=74 days
The nearest approx. to answer is 75
Ans: C (75 days)
Question: In an encryption code 'MATTER' is written as 'B"EEQW' when encoded and 'POTATO' is written as 'IUE"EU' when encoded. How is ''TOMATO" written in that encryption code?
A.BEQWU"
B.UEBUEB
C.BE"QBU
D.EUB"EU
Answer: D.EUB"EU
Explanation: T = E
O = U
M = B
A = "
T = E
O = U
Correct Answer: D
Question: Find the area of a rhombus one side of which measures 20 cm and one diagonal is 26 cm.
A.600cm2
B.500cm2
C.516cm2
D.416cm2
Answer: D.416cm2
Explanation: Explanation: Let other diagonal = 2x cm.
Since diagonals of a rhombus bisect each other at right angles, we have:
(20)2 = (12)2 + (x)2 =>x =√(20)2 – (12)2= √256= 16 cm. _I
So, other diagonal = 32 cm.
Area of rhombus = (1/2) x (Product of diagonals) =(1/2× 26 x 32) cm2 = 416 cm2
Answer: Option D
Question: Tom opened a shop investing Rs.30000. Jose joined him 2 months later, investing Rs.45000. They earned a profit of Rs.45000 after completion of one year. What will be Jose's share of profit?
A.20000
B.10000
C.25000
D.34000
Answer: C.25000
Explanation: SoL = ~s- So Anju’s share = [5/9] x 45000 = 25000
C
Question: A man is 20 years older than his son. In two years, his age will be twice the age of his son. The
present age of his son is
A.18 years
B.21 years
C.22 years
D.24 years
Answer: A.18 years
Explanation: Explanation:
Let the son's present age be x years. Then, man's present age = (x + 20) years
=> (x + 20) + 2 = 2(x + 2)
=> x + 22 = 2x + 4
So, x = 18
Answer: Option A
Question: In a set of numbers from 100 to 1000 inclusive, how many integers are even and do not contain the digit 5?
A.180
B.196
C.360
D.288
Answer: C.360
Explanation: We have to find the total number of 3-digit even numbers not having 5 as a digit.
Units digits will be among 0,2,4,6,8
Tenth digits will be among 0,1,2,3,4,6,7,8,9
Hundredth digits will be among 1,2,3,4,6,7,8,9
So total numbers = 5*9*8 =360
Answer : C
Question: A developer has land that has x feet of lake frontage. The land is to be subdivided into lots, each of which is to have either 80 feet or 100 feet of lake frontage. If 1/9 of the lots are to have 80 feet of frontage each and the remaining 64 lots are to have 100 feet of frontage each, what is the value of x ?
A.400
B.3,200
C.3,700
D.7,040
Answer: D.7,040
Explanation: Total number of plots,be T
Plots with 80ft is (1/9)th of total plots.
Plots with 100ft frontage = (8/9) X total number of plots
i.e 8/9 X T = 64 T = 72
Total lake frontage = 64X100 + 8X80 = 7040
Answer : D
Question: Rice weighing 25/2 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each container? (Note that 1 pound = 16 ounces)
A.40
B.50
C.60
D.70
Answer: B.50
Explanation: 25/2 ÷ 4 = 25/8 pounds in each container
25/8 pounds * 16 ounces / pound = 50 ounces in each container
The answer is B.
Question: A certain sum is invested at simple interest at 18% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 480. Find the sum?
A.7000
B.7029
C.4000
D.2800
Answer: C.4000
Explanation: Let the sum be Rs. x.
(x * 18 * 2)/100 - (x * 12 * 2)/100 = 480 => 36x/100 - 24x/100 =480
=> 12x/100 = 480 => x = 4000.Answer:C
Question: The tax on a commodity is diminished by 20% but its consumption is increased by 5%. Find the decrease percent in the revenue derived from it?
A.12%
B.14%
C.16%
D.20%
Answer: C.16%
Explanation: Explanation:
100 * 100 = 10000
80 * 105 = 8400
10000------- 1600
100 ------- ? = 16%
C)
Question: A room contains 180 people, 15 percent of whom are women. A group of people, 30 percent of whom are women, leaves the room. Of the people remaining in the room, 10 percent are women. How many people left the room?
A.10
B.20
C.45
D.60
Answer: C.45
Explanation: Say group of x people left the room.
# of women in the room was 0.15*180=27;
# of women in the group that left 0.3x;
# of women in the room after the group left it is 0.1(180-x);
So, 27-0.3x= 0.1(180-x) --> x=45.
Answer: C.
Question: A bus started its journey from Mumbai and reached Pune in 65 min with its average speed of 60km/hr. If the average speed of the bus is increased by 5 km/hr, how much time will it take to cover the same distance?
A.2 Hrs
B.3 Hrs
C.1 Hrs
D.1 1/2 Hrs
Answer: C.1 Hrs
Explanation: Sol. Distance between Ramgarh and Devgarh =(60*65)/60=60
Average speed of the bus is increased by 5 km/hr then the
speed of the bus = 60 km/hr
Required time = 60/60 = 1Hr
C
Question: Shawn and Raquel want to have gardens that are the same size. Shawn plants a garden 25 feet by 24 feet. Raquel wants one side of his garden to be 30 feet. How long should the other side be?
A.22
B.29
C.20
D.15
Answer: C.20
Explanation: S1 S2 = R1 R2
So 25*24 = 30*R2=> R2 = 20
Answer:C
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?
A.11.4
B.11.3
C.11.8
D.11.5
Answer: D.11.5
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
Answer: D
Question: There are ten pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is at least one pair is
A.195/323
B.99/323
C.198/323
D.185/323
Answer: B.99/323
Explanation: Explanation :
We will solve this question in a moment. First let us understand one concept which is bit confusing.
Suppose there are 4 objects and I ask you to pick up 2 from them. In how many ways can you do that?
The answer is simple: 4C2 = 6
Can I do like this:
Out of 4 given objects first I will pick one object(4C1). Then again I will pick one more object in the next draw from the remaining 3 objects(3C1).
This way I have selected two objects from fours objects and number of ways to do this is: 4C1* 3C1 = 12
Do you see in both the way I am selecting two objects from four objects but in first way of selection the result is 6 while in the other way it is 12. Why so ??
This is because in the second way when we are selecting two objects one by one, their order of selection is being counted in the result. Didn't understand ?
OK, let's understand this way: Suppose we mark all 4 objects from 1 to 4.
One way is: In the first draw we select object marked 1 and in the second draw we pick object marked 2.
The other way is: the first selected object is marked 2 and second object is marked 1.
We are selecting the same two objects but the number of way is 2 because their order of selection (1,2) and (2,1) made it happen so.
But our ultimate goal was to pick up two object and the order does not matter to us.
Similarly if you select 3 objects from these 4 objects you can have following orders for objects marked 1,2,3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) = 6 ways= 3! ways. While it should be just 1 way and not 6 ways.
So we can conclude that we will have to divide the result by the factorial of the number of objects we are selecting.
Like in the first example when we are selecting 2 objects we will have to divide the result by 2! and in the second example when we are selecting 3 objects we will have to divide the result by 3!.
Now coming back to our given problem:
At least on pair of socks must be selected means either one pair or two pair.
So required probability will be: 1 - (no pair of socks are selected)
The way we can select no pair of socks is:
Select one sock from 20 sock in the first pick. In the second draw exclude the pair of the first selected sock and pick from the remaining 18 socks and so on.
= 20C1 * 18C1 * 16C1 * 14C1/4! (4! because of the above explained concept)
So required probability: 1 - ((20C1 * 18C1 * 16C1 * 14C1/4!)/20C4) = 99/323
Answer : B
Question: A list of numbers has six positive integers. Three of those integers are known: 4, 5 and 24 and three of those are unknown: x, y and z. The three unknowns are known to be distinct. It is also known that the mean of the list is 10 and the median lies between 7 and 8 (exclusive). Which of the following CANNOT be the value of any one of the unknowns?
A. 13
B. 12
C. 11
D. 10
Answer: C. 11
Explanation: VERITAS PREPOFFICIAL SOLUTION:
The question gives us concrete information about mean – it is 10 – but not about median – it is between 7 and 8 (exclusive). What can we say about median from this? That it cannot be 7 or 8 but anything in between. But we know that the list has all integers. When we have even number of integers, we know that the median is the average of the middle two numbers – when all are placed in increasing order. So can the average of the two middle numbers be, say, 7.1? Which two positive integers can average to give 7.1? None! Note that if the average of two integers is a decimal, the decimal must be (some number).5 such as 7.5 or 9.5 or 22.5 etc. This happens in case one number is odd and the other is even. In all other cases, the average would be an integer.
Since the median is given to be between 7 and 8, the median of the list of the six positive integers must be 7.5 only.
Now we know that the mean = 10 and median = 7.5
Method 1: Algebra/Logic
Let’s try to solve the question algebraically/logically first.
There are 6 elements in the list. The average of the list is 10 which means the sum of all 6 elements = 6*10 = 60
4 + 5 + 24 + x + y + z = 60
x + y + z = 27
Median of the list = 7.5
So sum of third and fourth elements must be 7.5 * 2 = 15
There are two cases possible:
Case 1: Two of the three integers x, y and z could be the third and the fourth numbers. In that case, since already 4 and 5 are less than 7.5, one of the unknown number would be less than 7.5 (the third number) and the other two would be greater than 7.5.
The sum of the third and fourth elements of the list is 15 so
15 + z = 27
z = 12
So, two numbers whose sum is 15 such that one is less than 7.5 and the other greater than 7.5 could be
5 and 10
6 and 9
7 and 8
x, y and z could take values 5, 6, 7, 8, 9, 10 and 12.
Case 2: The known 5 could be the third number in which case one of the unknown numbers is less than 5 and two of the unknown numbers would be more than 7.5.
If the third number is 5, the fourth number has to be 10 to get a median of 7.5. Hence, 10 must be one of the unknown numbers.
The sum of the other two unknown numbers would be 27 – 10 = 17.
One of them must be less than 5 and the other greater than 10. So possible options are
4 and 13
3 and 14
2 and 15
1 and 16
x, y and z could take various values but none of them could be 11
Answer (C)
Question: If the sum of five numbers is between 100,000 and 101,000, then the average (arithmetic mean) of the five numbers could be which of the following?
A.20,010
B.25,010
C.30,010
D.35,010
Answer: A.20,010
Explanation: 100000/5=20000
101000/5=20200
Only A fits in between these two limits.
Answer: A
Question: What is the 4 digit number in which the 1st digit is 1/3 of the second, the 3rd is the sum of the 1st and 2nd, & the last is three times the second?
A.1234
B.1567
C.NO =1349
D.1456
Answer: C.NO =1349
Explanation: First digit is 1/3 second digit => The numbers can be 1 & 3, 2& 6, 3 & 9.
First + second = third => we can eliminate 3 & 9 since 3 + 9 = 12.
Last is 3 times the second => we can eliminate option 2 & 6 since 3 * 6 = 18.
Hence the number is 1349
C
Question: A woman is 4 times as old as her daughter. In 3 years she will be 3 times as old as her daughter. How old is the woman now?
A.3 years
B.6 years
C.12 years
D.24 years
Answer: D.24 years
Explanation: Let Woman's present age = w
and daughter's present age = d
w= 4d --- equation 1
w+3 = 3(d+3)
=>w = 3d +6 --- equation 2
From 1 and 2 , we get
d=6
w=24
Answer D
Question: If P is a prime number greater than 3, find the remainder when P^2 + 15 is divided by 12.
A.6
B.4
C.0
D.8
Answer: B.4
Explanation: Every prime number greater than 3 can be written 6N+1 or 6N-1.
If P = 6N+1, then P^2 + 15 = 36N^2 + 12N + 1 + 15 = 36N^2 + 12N + 12 + 4
If P = 6N-1, then P^2 + 15 = 36N^2 - 12N + 1 + 15 = 36N^2 - 12N + 12 + 4
When divided by 12, it must leave a remainder of 4.
The answer is B.
Question: A boat can travel with a speed of 16 km/hr in still water. If the rate of stream is 5 km/hr, then find the time taken by the boat to cover distance of 84 km downstream.
A.4 hours
B.5 hours
C.6 hours
D.7 hours
Answer: A.4 hours
Explanation: Explanation:
It is very important to check, if the boat speed given is in still water or with water or against water. Because if we neglect it we will not reach on right answer. I just mentioned here because mostly mistakes in this chapter are of this kind only.
Lets see the question now.
Speed downstream = (16 + 5) = 21 kmph
Time = distance/speed = 84/21 = 4 hours
ANSWER IS A
Question: The U.S. Defense Department has decided that the Pentagon is an obsolete building and that it must be replaced with an upgraded version: the Hexagon. The Secretary of Defense wants a building that is exactly 70 feet high and 200 feet on a side, and that has a hexagonal bull's-eye cutout in the center (somewhat like the current one) that is 60 feet on a side. What will be the volume of the new building in cubic feet?
A.3,937,500 cubic feet
B.6,619,898 cubic feet
C.11550√3 cubic feet
D.15,750√3 cubic feet
Answer: B.6,619,898 cubic feet
Explanation: Volume of the Hexagon with side 200 and height 70 = Area * height
= 6*(sqrt(3)/4)(200^2)(70)
Volume of the center bull's eye that is similar in shape of a hexagon but side 60
= 6 * (sqrt(3)/4)(60^2)*70
Volume of the building = 6*(sqrt(3)/4)(200^2)(70) - 6 * (sqrt(3)/4)(60^2)*70=6,619,898
Answer is B