MU-NLPC/Calc-aqua_rat huggingface dataset:
At a florist shop on a certain day, all corsages sold for either $20 or $30. If 10 of the corsages that sold for $30 had instead sold for $20, then the store's revenue from corsages that day would have been reduced by 20 percent. What was the store's actual revenue from corsages that day?  Choose the correct option:  A) $200  B) $400  C) $500  D) $800  E) $1000

Solution: Let, No. of corsages @ $20 = x, no. of corsages @ $30 = y and revenue= r   so, 20x+30y=r.........(1)   Now,   Given the situation, 20(x+10) + 30(y-10)= r-.2r   => 20x+200+30y-300 = .8r   => 20x+30y = .8r+100............(2)   so, r = .8r+100   => r = 500   The answer is $500.
Answer: C
If a rod is sold for Rs.34.80, there is a loss of 25%. Find out the CP of the rod?	Choose the correct choice from the following options:
A) 45
B) 46
C) 46.4
D) 47.23
E) 47.34

Solution: SP=34.80Loss=25%46.4P=100(100−Loss%)×SP=100(100−25)×34.80=10075×34.80=4×34.803=4×11.60=46.40   46.4
Answer: C
In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 18 owned rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three?	Choose the correct choice:
A) 2
B) 4
C) 8
D) 12
E) 32

Solution: You are considering a case when cat dogs and rabbit are not exactly equal to 12 .   The solution shall be 60=30+40+18-12-2x   and hence x shall be 8.   Answer 8
Answer: C
Which is greater than 4?
Choices.
A) 5	B) -5	C) -1/2	D) -25	E) 1/2

Solution: 5 greater than 4.   5nswer: 5
Answer: A
A shopkeeper sold an article offering a discount of 5% and earned a profit of 23.5%. What would have been the percentage of profit earned if no discount was offered?  Choose the correct option:
A) 24  B) 28  C) 30  D) 32  E) 34

Solution: Let C.P. be Rs. 100.   Then, S.P. = Rs. 123.50   Let marked price be Rs. x. Then, 95/100 x = 123.50   x = 12350/95 = Rs. 130   Now, S.P. = Rs. 130, C.P. = Rs. 100   Profit % = 30%.   ANSWER:30
Answer: C
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km Select the correct option: A) 200 Km B) 222 Km C) 224 Km D) 248 Km E) None of these

Solution: Explanation:   Let time taken to travel the first half = x hr   Then time taken to travel the second half = (10 - x) hr   Distance covered in the the first half = 21x [because, distance = time*speed]   Distance covered in the the second half = 24(10 - x)   Distance covered in the the first half = Distance covered in the the second half   So,   21x = 24(10 - x)   => 45x = 240   => x = 16/3   Total Distance = 2*21(16/3) = 224 Km [multiplied by 2 as 21x was distance of half way]   Option 224 Km
Answer: C
Ravi bought 5 pencil and 10 eraser for Rs 4 . Sheetal bought 3 pencil and 9 eraser for Rs 3 then what would be the cost of pencil and eraser.
Answers: A) .15p, .30p B) .50p,.27p C) .40p,.20p D) .20p,.40p E) .50p,.40p

Solution: Let pencil be x and eraser be y so   5x+10y=4   3x+9y=3   By solving the above two equations we get x=.40p and y=.20p correct ans is .40p,.20p
Answer: C
A coin is tossed live times. What is the probability that there is at the least one tail?	Choose the correct choice from the following answers:
A) 31/32  B) 31/17  C) 31/19  D) 31/12  E) 31/13

Solution: Let P(T) be the probability of getting least one tail when the coin is tossed five times.   = There is not even a single tail.   i.e. all the outcomes are heads.   = 1/32 ; P(T) = 1 - 1/32   =31/32   31/32nswer:31/32
Answer: A
The average of a batsman for 40 innings is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, his average drops by 2 runs. Find his highest score.  Choose the correct answer:
A) 172
B) 173
C) 174
D) 175
E) 176

Solution: Total runs = 40 × 50 = 2000   Let his highest score be = x   Then his lowest score = x – 172   Now 200−x−(x−172) /38=48   ⇒ 2x = 2172 – 1824   ⇒ x = 174   Answer 174
Answer: C
The product of the digits of the four-digit number T is 36. No two digits of are identical. How many different numbers are possible values of T ? Choose the correct choice from the following options:	A) 6	B) 12	C) 24	D) 36	E) 48

Solution: The main question here is how can you arrange 4 different digits T. We don't even need to find the number. from the question itself we know that there are 4 different digits and none of them is 0 as the product is non-zero.   Therefore no of ways to arrange 4 diff. digits is 4! which is 24   hence 24
Answer: C
Set A of 8 positive integers may have the same element and have 37. And set B of 8 positive integers must have different elements and have 37. When m and n are the greatest possible differences between 37 and other elements’ sums in set A and set B, respectively, m-n=? Choices:
A) 21	B) 22	C) 23	D) 25	E) 26

Solution: This is maximum-minimum. Hence, 37-(1+1+1+1+1+1+1)=30 and   37-(1+2+3+4+5+6+7)= 
  37-(1+2+3+4+5+6+7) = 9
9. So, 30-9=21. The correct answer is 22.
Answer: B
A man swims downstream 72 km and upstream 45 km taking 9 hours each time; what is the speed of the current?
Choose the correct choice from the following answers:
A) 1.6 B) 1.5 C) 1.9 D) 1.2 E) 1.7

Solution: 72 --- 9 DS = 8   ? ---- 1   45 ---- 9 US = 5   ? ---- 1 S = ?   S = (8 - 5)/2= 
(8 - 5)/2 = 3/2 = around 1.5
1.5   Answer: 1.5
Answer: B
A boats man can row in still water at speed of 7 km/hr. It takes 6 hours more to travel the same distance in upstream than in downstream if the speed of the river is 3 km/hr. what is the distance between the two destinations?	Choose the correct choice from the following options: A) 40 B) 87 C) 126 D) 18 E) 91

Solution: Explanation:   x = 7 km/hr ; y = 3 km/hr   Ds = 10 km/hr ; Us = 4 km/hr   Distance (d) is same. Therefore, if time taken for downstream is t hours, the time for upstream is (t + 6) hours.   10*t = 4*(t + 6)   6t = 24 ; t = 4 hours   d = 10*4= 
10*4 = 40
40 km   40NSWER: 40
Answer: A
If 7^r is a factor of the product of the integers from 1 to 100, inclusive, what is the largest value of r?
Choose the correct option:
A) 12
B) 13
C) 14
D) 15
E) 16

Solution: So the question just means that we have to find all the multiples r of 7 between 1 to 100   So there are 14 multiples of 7 ( 7 - 98)   But 49 and 98 contain two 7's as factors   so we have to add  14 + 2= 
 14 + 2 = 16
16   16
Answer: E
How many integers between 100 and 190, inclusive, cannot be evenly divided by 3 nor 5? Choose the correct choice from the following answers
A) 42  B) 44  C) 46  D) 48  E) 50

Solution: The total numbers between 100 and 190, inclusive, is 91.   3*34= 
3*34 = 102
102 and   3*63= 
  3*63 = 189
189 so the number of multiples of 3 is 30.   5*20= 
5*20 = 100
100 and   5*38= 
  5*38 = 190
190 so the number of multiples of 5 is 19.   However, the multiples of 15 have been counted twice.   15*7= 
15*7 = 105
105 and   15*12= 
  15*12 = 180
180 so the number of multiples of 15 is 6.   Thus the total number is  91-30-19+6= 
 91-30-19+6 = 48
48.   The answer is 48.
Answer: D
If selling price is doubled, the profit triples. Find the profit percent: Choose the correct choice from the following options
A) 227
B) 100
C) 278
D) 266
E) 181

Solution: Let C.P. be Rs. x and S.P. be Rx. y.   Then, 3(y - x) = (2y - x) => y = 2x   Profit = (y - x) = Rs. (2x - x) = Rs. x   Profit % = x/x * 100 = 100%.   Answer: 100
Answer: B
Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?  Choose the correct choice from the following
A) 256 B) 260 C) 316 D) 320 E) 350

Solution: there are total of 435 bags and 210 bags contain almonds.   From the diagram 20y=5x --> y=x/4. Now, Total=435={Almonds}+10x+y+x --> 435=210+10x+x/4+x --> x=20 --> # of bags that contain only one kind of item is the sum of yellow segments: 10x+x+5x=16x=320.   Answer: 320.
Answer: D
If the price of a certain bond on May 1st was 2/3 the price of the bond on June 1st and the price of the bond on July 1st was 25% greater than the price of the bond on May 1st. then the price of the bond on June 1st st was what percent of the average (arithmetic mean) price of the bond on May 1st and July 1st? Choose the most appropriate option:
A) 50%
B) 75%
C) 120%
D) 133 1/3%
E) 150%

Solution: The price on June 1st = 12 (assume);   The price on May 1st = 2/3*12= 
2/3*12 = 8
8;   The price on July 1st= 8*1.25= 
8*1.25 = 10
10.   The average price of the bond on May 1st and July 1st= (8+10)/2= 
(8+10)/2 = 9
9.   The price of the bond on June 1st (12) is 4/3 times (134%) the average price of the bond on May 1st and July 1st.   Answer: 133 1/3%.
Answer: D
I. x2 + 3x - 18 = 0,
II. y2 + y - 30 = 0 to solve both the equations to find the values of x and y? Options:
A) If x < y
B) If x > y
C) If x ≤ y
D) If x ≥ y
E) If x = y or the relationship between x and y cannot be established.

Solution: I. x2 + 6x - 3x - 18 = 0   =>(x + 6)(x - 3) = 0 => x = -6, 3   II. y2 + 6y - 5y - 30 = 0   =>(y + 6)(y - 5) = 0 => y = -6, 5   No relationship can be established between x and y.   ANSWIf x = y or the relationship between x and y cannot be established.R:If x = y or the relationship between x and y cannot be established.
Answer: E
In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color? Choose the correct choice from the following answers.
A) a)	8
B) b)	23
C) c)	29
D) d)	32
E) e)	53

Solution: Thats 29 in c)	29.
Answer: C
Alice and Heather are 20 miles apart and walk towards each other along the same route. Alice walks at constant rate that is 1 mile per hour faster than heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Alice, how far from the original destination has Heather walked when the two meet? Options
A) 4 miles  B) 8 miles  C) 10 miles  D) 6 miles  E) 12 mile

Solution: Original distance between S and H = 20 miles.   Speed of A = 5+1= 
5+1 = 6
6 mph, Speed of H= 5 mph.   Time traveled by H = t hours ---> time traveled by A = t+24/60 = t+2/5 hours.   Now, the total distances traveled by A and H = 20 miles ---> 6*(t+2/5)+5*t=20 ---> t= 8/5 hours. Thus H has traveled for 8/5 hours giving you a total distance for H = 5*8/5= 
5*8/5 = 8
8 miles.   B is thus the correct answer.   P.S.: based on the wording of the question, you should calculatehow far from theoriginal destination has Heather walkedwhen the two meet. 'Original destination' for H does not make any sense. Original destination for H was situated at a distance of 20 miles.
Answer: B
Let p be a prime number greater than 2, and let N = 32p. How many odd divisors does N have?	Answers:
A) 1 B) 2 C) 3 D) 4 E) 5

Solution: The answer is A, there is exactly one. Since is the largest (and only) even prime number, any p larger than 2 is odd. Then since 32 = 2*2*2*2, all of which are even, this means that p must be the only odd prime factor.
Answer: A
Given two fair dice, what is the probability Z that the sum of their numbers is 4 if exactly one die shows a 3?  Pick one
A) a) 2/11 B) b) 1/18 C) c) 3/11 D) d) 2/39 E) e) 1/5

Solution: If exactly one die shows 3, there are 10 such combinations (given that we have 2 fair dice).   1:3, 2:3, 4:3, 5:3, 6:3,   3:1, 3:2, 3:4, 3:5, 3:6   The sum of numbers would be 4 for two events: 1:3, 3:1.   Z=2/11 would be the answer if 'at least one of the die shows a 3'.a) 2/11
Answer: A
If x and y are integers such that |y + 3| ≤ 3 and 2y – 3x + 6 = 0, what is the least possible value W of the product xy? Answers:
A) -12
B) -3
C) 0
D) 2
E) None of the above

Solution: How to deal with inequalities involving absolute values? First example shows us the so callednumber case   In this case we have |y + 3| ≤ 3 which is generalized |something| ≤ some number. First we solve as if there were no absolute value brackets:   y + 3 ≤ 3   y ≤ 0   So y is 0 or negative   Second scenario - remove the absolute value brackets. Put a negative sign around the other side of the inequality,ANDflip the sign:   y + 3 >= -3   y >= -6   Therefore we have a possible range for y: -6=<y<=0   Ok, so far so good, we're half way through. What about x?   Here's the formula: 2y – 3x + 6 = 0, rewrite it as 2y + 6 = 3x.   You can say that 2y + 6 is a multiple of 3 (=3x). So all values which must be integer must also satisfy this constraint. I'm just saying that, so it's easier to evaluate all the possible numbers W(-6, -3, 0). If you plug in y = 0, x will be 2 and xy = 0 as the lowest possible value.   Hence, Answer 0hoice 0 is the one to go.
Answer: C
One-third of 1206 is what percent of 200?	Choose the correct answer.
A) 3 B) 201 C) 300 D) None of these E) cannot be determined

Solution: Answer   Let one-third of 1206 is N% of 200.   ∵ 1206/3 = (N x 200)/100   ∴ N = (402 x 100) / 200 = 201   Correct Option: 201
Answer: B
In a clock (Conventional clock with numbers 1 to 12 in order)is cut into 3 pieces such that the sum of each piece in AP with a common difference of 1. What is the count of numbers in each piece? and what are the numbers? Choose the correct choice from the following options.
A) 25	B) 26	C) 27	D) 28	E) 29

Solution: Sum of the 12 numbers=1+2+3+......+12=12*13/2= 
12*13/2 = 78
78   So we have to divide it into 3 slots such that the sum is 78 and sum of each slot is in AP, thus the 3 slots are 25,26,27. So the numbers can be:     3+4+5+6+7= 
  3+4+5+6+7 = 25
25   8+9+10= 
8+9+10 = 27
27   11+12+1+2= 
11+12+1+2 = 26
26   ANSWER:26
Answer: B
In how many different ways can the letters of the word ‘PREPTY’ be arranged?
Choose one:
A) 120  B) 230  C) 330  D) 340  E) 360

Solution: Number of ways = 61/21 6X5X4X3X2X1 - 360   360
Answer: E
Mr. loyd wants to fence his Square shaped land of 150Sqft each side.If a pole is laid every 30ft how many poles do he need?  Choose the correct choice:
A) 20  B) 30  C) 40  D) 50  E) 60

Solution: If each side is 120 feet..then total perimeter is 150*4= 
150*4 = 600
600   poles every 20 feet hence no of poles= 600/20= 
600/20 = 30
30   ANSWER:30
Answer: B
Find the number of ways in which 8 different beads can be arranged to form a necklace.  Choose the correct choice
A) 30 B) 20 C) 2520 D) 60 E) 80

Solution: Since in forming a necklace clockwise and anticlockwise arrangements are not different, therefore,8 beads can be arranged to form a necklace in(8-1)! /2ways   = 7*6*5*4*3*2/2= 
7*6*5*4*3*2/2 = 2_520
2520 ways   Answer is 2520
Answer: C
Double of Quarter of 1 percent written as a decimal is :  Choose the correct option.
A) 0.003	B) 0.0005	C) 0.25	D) 0.005	E) None of these

Solution: Explanation :   Solution: (2)*(1/4)*1% = 2*(1/4 * 1/100) = 1/400 = 2*0.0025 = 0.005.   Answer : 0.005
Answer: D
A cubical block of metal weighs 6 pounds. How much will another cube of the same metal weigh if its sides are twice as long?
Choose the correct choice:
A) 48
B) 12
C) 36
D) 60
E) None of these

Solution: Explanation:   If you double the sides of a cube, the ratio of the surface areas of the old and new cubes will be 1: 4. The ratio of the volumes of the old and new cubes will be 1: 8.   Weight is proportional to volume. So, If the first weighs 6 pounds, the second weighs 6x8 pounds =48.   48nswer: 48
Answer: A
What will come in place of the x in the following Number series? 11, 14, 19, 22, 27, 30, 35, 38, 43 , 46,51 ,x Options:
A) 23  B) 26  C) 54  D) 35  E) 45

Solution: (54)   The pattern is + 3, + 5, + 3, + 5, …………   So the missing term is = 51 + 3= 
51 + 3 = 54
54.
Answer: C
John has 10pairs of dark blue socks and 10pairs of black socks. He keeps them all in the same bag. If he picks out 3socks at random, then what is the probability thathe will get a matching pair?  Choose the correct choice from the following choices.
A) 0
B) 1
C) 4
D) 5
E) 6

Solution: If he draws any combination of 3 socks he will definitely have the matching pair of either colour.   1)
Answer: B
105 candies are distributed to children with the same number of candies for each child. What can’t be the range which includes the number of children?	Choose the correct option: A) 1~10. B) 10~20. C) 20~30. D) 30~40. E) 40~50.

Solution: 105 = 3*5*7   There are factors in each range except for 40~50   The answer is 40~50..
Answer: E
A, B and C invested Rs.6300, Rs.4200 and Rs.10500 respectively, in a partnership business. Find the share of A in profit of Rs.12600 after a year?  Choose the correct choice from the following choices
A) 3630	B) 2881	C) 2887	D) 9977	E) 3780

Solution: 6300:4200:10500   3:2:5   3/10 * 12600= 
3/10 * 12600 = 3_780
3780.Answer: 3780
Answer: E
A box contains 5 green, 4 yellow and 3 white marbles, 3 marbles are drawn at random. What is the probability that they are not of the same colour?
Select the correct option.
A) 13⁄44  B) 41⁄44  C) 13⁄55  D) 52⁄55  E) None of these

Solution: Total no. of ways of drawing 3 marbles   = 12C3 = 12×11×10/1×2×3=220   Total no. of ways of drawing marbles, which are of same colour = 5C3 + 4C3 + 3C3 = 10 + 4 + 1= 
10 + 4 + 1 = 15
15   \Probability of same colour= 15⁄220 = 3⁄44   \ Probability of not same colour = 1 - 3⁄44 =   Answer 41⁄44
Answer: B
The product of all the prime numbers less than 12 is closest to which of the following powers of 10 ?
Pick:
A) 10^9
B) 10^3
C) 10^7
D) 10^6
E) 10^5

Solution: P = 2*3*5*7*11   ~ 10^ 3   Answer : 10^3
Answer: B
In a class of 20 students, 3 students did not borrow any books from the library, 9 students each borrowed 1 book, 4 students each borrowed 2 books, and the rest borrowed at least 3 books. If the average number of books per student was 2, what is the maximum number of books any single student could have borrowed? Choose one:	A) 13	B) 14	C) 15	D) 16	E) 17

Solution: The total number of books the students borrowed is 20*2= 
20*2 = 40
40.   The students who borrowed zero, one, or two books borrowed   9*1+4*2= 
  9*1+4*2 = 17
17 books.   The 4 students who borrowed at least three books borrowed   40-17= 
  40-17 = 23
23 books.   If 3 of these students borrowed exactly 3 books, then the maximum that one student could have borrowed is  23-9= 
 23-9 = 14
14 books.   The answer is 14.
Answer: B
9773 + x = 13200, then x is ?	Answers:
A) 3427  B) 3237  C) 3337  D) 2337  E) None of these

Solution: 3427nswer   x = 13200 - 9773= 
13200 - 9773 = 3_427
3427   Option: 3427
Answer: A
8 coins are tossed. Whats is the probability that one and only one turns up head ?
Choose the correct choice:
A) 1/256	B) 1/32	C) 1/64	D) 1/128	E) 1/189

Solution: Explanation :   Possible outcomes(Sample space) when 8 coins are tossed are: (H, H, H, H, H, H, H, H), (H, H, H, H, H, H, H, T), (H, H, H, H, H, H, T, H), (H, H, H, H, H, H, T, T), ....... (T, T, T, T, T, T, T, T) = 28   Required outcomes: (H, T, T, T, T, T, T, T), (T, H, T, T, T, T, T, T), (T, T, H, T, T, T, T, T), (T, T, T, H, T, T, T, T), (T, T, T, T, H, T, T, T), (T, T, T, T, T, H, T, T), (T, T, T, T, T, T, H, T), (T, T, T, T, T, T, T, H) = 8   So the required probability = 8/28 = 1/32   Answer : 1/32
Answer: B
The average age of A, B and C is 29 years. If the average age of A and C is 32 years, what is the age of B in years ?	Options:
A) 19
B) 35
C) 20
D) 32
E) 23

Solution: Age of B = Age of (A + B + C) â€“ Age of (A + C) = 29 Ã— 3 â€“ 32 Ã— 2 = 87 â€“ 64 = 23 years   23
Answer: E
The S.I. on a certain sum of money for 4 years at 12% per annum is half the C.I. on Rs. 6000 for 2 years at 15% per annum. The sum placed on S.I. is?
Choose the correct option:  A) 2019.625  B) 2017.625  C) 2013.625  D) 2015.625  E) 2011.625

Solution: C.I. = [6000 * (1 + 15/100)2 - 6000]= (6000 * 23/20 * 23/20 - 6000) = Rs. 1935.Sum = (967.5 * 100)/(4 * 12)= Rs.2015.625Answer:2015.625
Answer: D
$300 is divided amongst A,B and C so that A may get 2/3 as much as B and C together, B may get 6/9 as much as A and C together, then the share of A is  Select the correct option:
A) $100 B) $150 C) $125 D) $120 E) $250

Solution: A:(B+C) = 2:3   A's share = 300*2/5= 
300*2/5 = 120
$120   Answer is $120
Answer: D
An equal number of desks and bookcases are to be placed along a library wall that is 15 meters long. Each desk is 2 meters long, and each bookshelf is 1.5 meters long. If the maximum possible number of desks and bookcases are to be placed along the wall, then the space along the wall that is left over will be how many meters W long?	Pick one.
A) 0.5
B) 1
C) 1.5
D) 2
E) 3

Solution: Let x be the number of desks and bookcases that are placed along the library wall.   2x + 1.5x < 15   3.5x < 15   Since x is a non negative integer, the largest number x can be is 4.   When x is 4, the desks and bookcases take up 3.5 * 4= 
3.5 * 4 = 14
14m, leaving 1m of empty space.   Thus, I believe the answer is 1) 1
Answer: B
Some persons can do a piece of work in 16 days. Two times the number of these people will do half of that work in?
Choose the correct choice from the following answers:  A) 3  B) 4  C) 5  D) 6  E) 8

Solution: 16/ (2 * 2)= 
16/ (2 * 2) = 4
4 days   Answer:4
Answer: B
The arithmetic mean and standard deviation of a certain normal distribution are 10.5 and 1, respectively. What value is exactly 2 standard deviations less than the mean?
Choose the correct choice from the following answers
A) 6	B) 8.5	C) 11.5	D) 12	E) 12.5

Solution: Mean = 10.5   Two standard deviations is 1 + 1= 
1 + 1 = 2
2   There could be two calues for this.   Mean + two standard deviations= 12.5   Mean - Two standard deviations = 8.5   Answer choice has 8.5 and so 8.5 is the answer.
Answer: B
How many digits are in the number 50^8 × 8^3 × 11^2 x 10^2? Choose the correct choice from the following choices
A) 22 B) 20 C) 21 D) 18 E) 17

Solution: Whenever you are asked to find the number of digits, try to bring the number in multiples of 10. This way, we can wasily calculate the umber of 0's through the powers of 10   50^8 × 8^3 × 11^2 x 10^2 = (5^2*2)^8*2^9*11^2x 10^2 = 5^16*2^17*11^2 x10^2 = 2*11^2*10^16x10^2 = 242*10^18   Hence we would have 16 trailing 0's and the three digits from 242   Total digits = 3 + 18= 
3 + 18 = 21
21   21orrect Option: 21
Answer: C
McDonald's ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were "No Prize" signs. Beneath the other seven spots were names of prizes, two of which were identical.
For example, one card might have two pictures of a hamburger, one picture of a Coke, one of French fires, one of a milk shake, one of 5 Dollar, one of 1000 Dollar and three "No Prize" signs. For this card the customer could win a hamburger.
To win on any card, the customers had to uncover the two matching spots (which showed the potential prize for that card) before uncovering a "No Prize"; any card with a "No Prize" uncovered was automatically void. Assuming that the two matches and the three "No Prize" signs were arranged randomly on the cards, what is the probability of a customer winning?
Choices.
A) 0.1  B) 0.15  C) 0.12  D) 0.18  E) 0.13

Solution: Explanation :   As per the question there are 10 cover spots out of which   (i) Three spots are there with no prize (identical)   (ii) Two spots of the same sign (Prize)   (iii) Five other spots which are distinct   Let three spots of no prize are (x,x,x), two spots of same sign are b(P,P) and five other spots are (A,B,C,D,E).   Total number of cases without restriction:   =>10!/(3!x2!).   Total number of favourable cases happen in the following ways which shows sequence of can covering are as follows:-   CASE I: Second uncovering is P   =>P P _ _ _ _ _ _ _ _=8C3×5!×1.   CASE II:   =>_ _ P _ _ _ _ _ _ =7C3×5!×2.   CASE III:   =>6C3×5!×3.   CASE IV:   =>5C3×5!×4.   CASE V:   =>4C3×5!×5.   CASE VI:   =>3C3×5!×6.   Hence, the total number of favourable cases are :-   =>case (I+II+III+IV+V+VI)/(10!3!X2!).   =>0.10.   0.1nswer : 0.1
Answer: A