MU-NLPC/Calc-aqua_rat huggingface dataset:
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? Choose the correct choice from the following answers:  A) 15%  B) 17%  C) 21%  D) 19%  E) 25%

Solution: A = 30%   17% = -10%   profit% = 30-10 + (-300/100)= 
30-10 + (-300/100) = 17
17%   Answer is 17%
Answer: B
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.  Answers:  A) 132 km  B) 250 km  C) 390 km  D) 440 km  E) 578 km

Solution: Let x be the distance between A and B. Since Bill covered 80 km in 2 hours, his speed was V=80/2= 
80/2 = 40
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= 
40+10 = 50
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 250 km)250km
Answer: B
Sum of 3 consecutive even no.'s is 24 more than the 1st no. of the series. Find the middle no.?	Choose the correct choice from the following answers:
A) 11	B) 12	C) 14	D) 16	E) 18

Solution: 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= 
9+2 = 11
11   11
Answer: A
If a,b are positive integers and a/b=85.75 which of the following could be the value of b? Choose the most appropriate option
A) 1. 25
B) 4.24
C) 3. 27
D) 4. 27
E) 5. 30

Solution: 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:4.24
Answer: B
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?
Pick one:
A) 2  B) 4  C) 6  D) 8  E) 10

Solution: 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: 4.
Answer: B
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	Options:
A) 2 p.m. on Monday
B) 2 p.m. on Tuesday
C) 2 p.m. on Wednesday
D) 3 p.m. on Thursday
E) None

Solution: 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= 
 170 * 5/34 * 2 = 50
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 2 p.m. on Wednesday
Answer: C
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?
Choose the correct choice:
A) 11 days	B) 12 days	C) 14 days	D) 10 days	E) 15 days

Solution: 6/15 + (6 + x)/30 = 1 => x = 12 days   Answer: 12 days
Answer: B
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?	Select the correct option
A) 6	B) 8	C) 10	D) 12	E) 15

Solution: 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:10
Answer: C
In a graduate physics course, 70 percent of the students are male and 40 percent of the students are married. If two-sevenths of the male students are married, what fraction of the female students is single?
Choose the correct choice from the following options.
A) 2/7  B) 5/3  C) 1/2  D) 2/3  E) 1/3

Solution: let assume there are 100 students of which 70 are male and 30 are females   if 40 are married then 60 will be single.   now its given that two-sevenths of the male students are married that means 2/7 of 70 = 20 males are married   if 40 is the total number of students who are married and out of that 20 are males then the remaining 20 will be females who are married.   total females = 30   married females = 20   then single females = 30-20= 
30-20 = 10
10   we need to find the fraction of female students who are single i.e single female students / total female student= 10/30 = 1/3 [1/3]
Answer: E
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? Answers:
A) 240 meters  B) 260 meters  C) 420 meters  D) 600 meters  E) Cannot be determined

Solution: Speed of the train in metres/sec = 72000/3600= 
72000/3600 = 20
20   Distance travelled by train to cross the platform= 30 * 20= 
30 * 20 = 600
600= Length of train + Length of platform   Distance travelled by train to cross the man = 17 * 20= 
17 * 20 = 340
340= Length of train   Length of platform = 600 - 340= 
600 - 340 = 260
260   Answer: 260 meters
Answer: B
If the cost price of 121 chocolates is equal to the selling price of 77 chocolates, the gain percent is :
Choose the correct choice from the following choices
A) 56.14%
B) 57.44%
C) 57.14%
D) 56.34%
E) 57.87%

Solution: 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 : 57.14%
Answer: C
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?	Options:
A) 7 B) 9 C) 6 D) 2 E) 3

Solution: 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: 6
Answer: C
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? Choices.
A) 120
B) 140
C) 185
D) 190
E) 210

Solution: 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   185
Answer: C
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?  Select the correct option:
A) $10  B) $20  C) $12  D) $60  E) $50

Solution: 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 ($60ivide both sides by 5)   x = 13   Amount in Chris wallet = $x = $13   Amount in $60onna's wallet = $(x + 7) = $(13 + 7)= 
(13 + 7) = 20
$20   Amount in Austin 's wallet= $3(x + 7) = $3(13 + 7) = $3(20) = $ 60   correct answer $60
Answer: D
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? Choose the correct choice from the following answers.
A) 31:46 B) 32:44 C) 32:45 D) 32:41 E) 32:42

Solution: Explanation:   A : B   (8000*4)+(4000*8) : (9000*6)+(6000*6)   64000 : 90000   32 : 45   Answer: 32:45
Answer: C
At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years? Choose the correct choice from the following answers:
A) 7% B) 2% C) 4% D) 9% E) 5%

Solution: 150 = (750*5*R)/100   R = 4%   Answer: 4%
Answer: C
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)  Choose the correct choice from the following.
A) 5184	B) 5276	C) 5368	D) 5460	E) 5552

Solution: The tank is emptied at this rate: 14 + 6 - 5= 
14 + 6 - 5 = 15
15 cubic inches/min   The tank has a volume of  45*12*12*12= 
 45*12*12*12 = 77_760
77760 cubic inches.   The time it takes to empty the tank is 77760/15= 
77760/15 = 5_184
5184 minutes.   The answer is 5184.
Answer: A
How many solutions are possible for the inequality | x - 1 | + | x - 6 | < 2?  Choose the correct choice from the following
A) 3 B) 2 C) 1 D) 0 E) 4

Solution: 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 0.
Answer: D
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?
Choose the correct choice from the following answers.
A) 1/2	B) 2/3	C) 3/4	D) 4/5	E) 5/6

Solution: 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 2/3.
Answer: B
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?
Choose the correct option
A) s.350
B) s.700
C) s.200
D) s.600
E) s.500

Solution: 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: s.200
Answer: C
Find the percentage gain if a shop owner sells 20 mtr of fabric and gains Cost Price of 5 meters?  Answers:
A) 15%
B) 24%
C) 35%
D) 14%
E) 25%

Solution: Let cost of each metre be Rs. 100.   Therefore, cost price of 5 m cloth = 5*100= 
5*100 = 500
500   cost price of 20m cloth= 20*100= 
20*100 = 2_000
2000   Since the trader made the CP of 5m from the sale of 20m therefore   Profit on sale of 20m= cost price of 5 m cloth = 500   Selling price of 20m = 2000+500= 
2000+500 = 2_500
2500   Profit %= 25.00%   25%
Answer: E
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? Choose the correct option: A) 0.07 B) 0.06 C) 0.0048 D) 0.04 E) 0.05

Solution: Suppose machine A produce 1000 units   total defective = 9   Machine B then produces=1500 units   total defective=3   probability=12/2500= 
12/2500 = 3/625 = around 0.0048
0.0048   Answer : 0.0048
Answer: C
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 ?
Choose the correct choice:  A) 51 1⁄47  B) 52 30/47  C) 52 1⁄30  D) 43.42  E) None of these

Solution: 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 52 30/47
Answer: B
The number of rectangles that can be formed on a 8X8 chessboard is
Choose the correct choice from the following answers:
A) 2194
B) 1284
C) 1196
D) 1296
E) 1523

Solution: 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: 1296.
Answer: D
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? Answers:
A) 2400
B) 2800
C) 2699
D) 2662
E) 1821

Solution: 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= 
4/9 * 6300 = 2_800
2800   Answer: 2800
Answer: B
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? Select the correct option.
A) house A  B) house B  C) house C  D) house D  E) house E

Solution: 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 house D
Answer: D
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?  Choose the correct choice from the following answers
A) 4125 B) 5040 C) 1620 D) 2430 E) 2560

Solution: 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 5040
Answer: B
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 30, what is the greatest possible value of Q minus R? Pick
A) 23
B) 24
C) 25
D) 26
E) 27

Solution: A two-digit integer "ab" can be expressed algebraically as 10a+b.   Q-R=(10a+b)-(10b+a)=9(a-b)<30.   The greatest multiple of 9 which is less than 30 is 27.   The answer is 27.
Answer: E
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? Choose the correct option:
A) 99 B) 26 C) 88 D) 26 E) 18

Solution: 30 * 33/2 = 5/2 * 2 * x => x   = 99   99nswer: 99
Answer: A
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? Choices:
A) 2/11
B) 3/7
C) 8/15
D) 4/9
E) 5/7

Solution: 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   8/15
Answer: C
The HCF of two numbers is 50 and the other two factors of their LCM are 11 and 12. What is the largest number.
Choose the correct choice from the following options
A) 462  B) 450  C) 480  D) 504  E) 600

Solution: 600xplanation:   HCF of the two numbers = 50   HCF will be always a factor of LCM   50 is factor of LCM   other two factors are 11 & 12   Then the numbers are (50* 11) and (50 x 12)   = 550 and 600   Answer: Option 600
Answer: E
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 ?
Select:
A) 8d-n
B) 10d-n
C) 6d-n
D) 11d-n
E) 5*(6d-n)

Solution: 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 : 6d-n
Answer: C
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.) Pick: A) 25 B) 26 C) 27 D) 28 E) 29

Solution: 7C2 = 21   The number of test subjects is  21 + 5= 
 21 + 5 = 26
26   The answer is 26.
Answer: B
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?  Choose the correct option.
A) 1/42	B) 1/21	C) 1/7	D) 1/3	E) 1/2

Solution: 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   1/7 is correct.
Answer: C
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?
Choose the correct answer
A) 5½ ft
B) 7½ ft
C) 8 ½ ft
D) 9½ ft
E) 10 ft

Solution: 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 (7½ ft)!   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.7½ ft
Answer: B
A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons? Choose the correct choice from the following answers:
A) 165 B) 878 C) 166 D) 199 E) 234

Solution: Total number of persons in the committee = 5 + 6= 
5 + 6 = 11
11   Number of ways of selecting group of eight persons= 11C8 = 11C3   = (11 * 10 * 9)/(3 * 2)= 
(11 * 10 * 9)/(3 * 2) = 165
165 ways.   165nswer: 165
Answer: A
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?  Choose the correct option: A) 1:3 B) 2:3 C) 3:5 D) 3:7 E) 7:11

Solution: 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   3:7
Answer: D
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?	Choose the correct choice from the following: A) 32 B) 36 C) 40 D) 44 E) 48

Solution: 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 40.
Answer: C