RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: If a particular is never chosen, in how many ways can a cricketbe chosen out of 15 players?
A.345
B.350
C.364
D.370
Answer: C.364
Explanation: A particular players is never chosen, it means that 11 players are selected out of 14 players.
=> Required number of ways = 14C11
= 14!/11!x3! = 364
C
Question: A rectangle having length 150 cm and width 40 cm. If the length of the rectangle is increased by fifteen percent then how much percent the breadth should be decreased so as to maintain the same area.
A.25%
B.9.09%
C.40%
D.75%
Answer: B.9.09%
Explanation: Explanation :
Solution: (15/(150+15) * 100)% = 9.09%
Answer : B
Question: Johnny makes $7.35 per hour at his work. If he works 6 hours, how much money will he earn?
A.$30
B.$54
C.$48.32
D.$44.10
Answer: C.$48.32
Explanation: 4.75*6=28.50. Answer is C.
Question: A bag contains 3 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.
A.19/20
B.17/20
C.21/40
D.7/10
Answer: D.7/10
Explanation: Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
P(A) = (1/2) x (6/10) = 3/10.
P(B) = (1/2) x (4/10) = 1/5.
Hence, the required probability is P(A) + P(B) which is nothing but 7/10
ANSWER:D
Question: The mean daily profit made by a shopkeeper in a month of 30 days was Rs. 350. If the mean profit for the first fifteen days was Rs. 275, then the mean profit for the last 15 days would be
A.Rs. 200
B.Rs. 350
C.Rs. 275
D.Rs. 425
Answer: D.Rs. 425
Explanation: Average would be : 350 = (275 + x)/2
On solving, x = 425.
Answer: D
Question: A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A.5.6
B.8.9
C.7.2
D.2.5
Answer: C.7.2
Explanation: Explanation:
Speed = Distance/Time
Speed =(600/(5*60))m/sec
Converting m/sec to km/hr
=(2*18/5)km/hr=7.2 km/hr
Answer C
Question: In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons?
A.200
B.210
C.230
D.250
Answer: B.210
Explanation: 7p3 = 7 × 6 × 5 = 210
B)
Question: 11 speakers gathered to deliver speeches on two topics. Even though 11 speeches were initially planned to be delivered, due to lack of time only 2 speeches were finally decided to be delivered. Topics are to be spoken in a specific order and the speeches differ with respect to speaker or topic. How many such two speeches are possible?
A.2
B.121
C.50
D.100
Answer: B.121
Explanation: 1)there are 11 speeches in ONE topic and 11 in OTHER..
2)there is a specific order for topics, so
a) both the topics are to be spoken on- so ONE from each
b) the order does not matter since there is ONLY one specific order..
so 11C1∗11C1=11∗11=121
ANSWER:B
Question: 1!+2!+3!+4!+5!+..100! is divided by 24. Find the remainder?
A.2
B.5
C.9
D.10
Answer: C.9
Explanation: By applying rule 2, we divide the terms of the above expression individually, and add them to get the final remainder. But from 4! onwards all the terms leave a remainder 0 when divided by 24.
So the remainder = 1 + 2 + 6 + 0 + 0....... =
C
Question: The simple interest on Rs. 20 for 6 months at the rate of 5 paise per rupeeper month is
A.1.2
B.1.4
C.6
D.7
Answer: C.6
Explanation: Sol.
S.I. = Rs. [20 * 5/100 * 6] = Rs. 6
Answer C
Question: C and D started a business investing Rs. 85,000 and Rs. 15,000 respectively. In what ratio the profit earned after 2 years be divided between P and Q respectively?
A.10 : 3
B.12 : 3
C.17 : 3
D.20 : 15
Answer: C.17 : 3
Explanation: Sol.
C : D = 85000 : 15000 = 85 : 15 = 17 : 3.
Answer C
Question: Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 4 times his original investment. What is the value of n
A.55/2
B.65/2
C.75/2
D.85/2
Answer: C.75/2
Explanation: PA=p
t=n
r=8%
According to question
p+(pnr/100)=4p
p[1+n8/100]=4p
n8/100=3
n=300/8
n=75/2
ANSWER:C
Question: By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 5 p.c?
A.38008
B.38000
C.38029
D.380219
Answer: B.38000
Explanation: CP + CP/8 = 45000
CP = 40000
SP = 40000*(95/100) = 38000
Answer:B
Question: I bought two books; for Rs.480. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss?
A.188
B.280
C.166
D.199
Answer: B.280
Explanation: x*(85/100) = (480 - x)119/100
x = 280
Answer:B
Question: Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 20 miles apart. One and a half hours later, they are again 20 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour?
A.a) 80-x
B.b) 40-x
C.c) 26.67-x
D.d) 120-x
Answer: C.c) 26.67-x
Explanation: Let's say the two cars have speeds V1 and V2. The fact that they are moving in opposite direction means that their relative speed is (V1 + V2). In other words, any gap between them will be changing in size at a rate of (V1 + V2). It doesn't matter whether they are moving toward each other or away from each other. If they are approaching each other, the gap between them is decreasing at a rate of (V1 + V2). If they are moving away from each other, the gap between them is increasing at a rate of (V1 + V2). Either way, the number for the rate of change remains the same.
Here, the two cars approach a distance 20 mi, then move away from each other another distance of 20 miles. That's a total distance of 40 miles in 1.5 hr, which gives a rate of:
R = (40 mi)/(1.5) = 26.67 mph
That's the rate of change of the gap, so it must equal the sum of the speeds of the two cars.
One of the speeds is x, and let's call the other y. We want y.
x + y = 26.67
y = 26.67 - x
Answer =(C)
Question: Ann invested a certain sum of money in a bank that paid simple interest. The amount grew to $240 at the end of 2 years. She waited for another 3 years and got a final amount of $300. What was the principal amount that she invested at the beginning?
A.$150
B.$175
C.$200
D.$210
Answer: C.$200
Explanation: Amount at end of 2 years = 240$
Amount at end of 5 years = 300$
Difference in amount for time period of 3 years = 60$
Annual Interest earned = 60/3 = 20$
Principal amount invested at the beginning = 240 - 2*20 = 200$
Answer C
Question: A shopkeeper sells 400 metres of cloth for Rs. 18000 at a loss of Rs.5 per metre. Find his cost price for one metre of cloth?
A.Rs.55
B.Rs.57
C.Rs.54
D.Rs.50
Answer: D.Rs.50
Explanation: SP per metre = 18000/400
= Rs. 45 Loss per metre
= Rs. 5 CP per metre = 45 + 5
= Rs.50
Answer: D
Question: Find large number from below question The difference of two numbers is 1335. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder
A.1254
B.1376
C.1456
D.1599
Answer: D.1599
Explanation: Let the smaller number be x. Then larger number = (x + 1335).
x + 1335 = 6x + 15
5x = 1320
x = 264
Large number = 264+1335 = 1599
D
Question: ((7.3)*(10^2))-((7)*(10^2))=?
A.300
B.30
C.730
D.7300
Answer: B.30
Explanation: 10^2 is nothing but a 100.
7.3*(10^2)=7.3*100=730.
7*(10^2)=7*100=700
=730-700=30
The answer is option B
Question: Jadeja has x dollars more than Sharma has, and together they have a total of y dollars. Which of the following represents the number of dollars that Sharma has?
A.3(y – x)/2
B.5(y – x)/2
C.(y – x)/5
D.(y – x)/2
Answer: D.(y – x)/2
Explanation: To solve, we will set up two equations. Let's start by defining two variables.
A = number of dollars Sharma has
B= number of dollars Jadeja has
We are given that Jadeja has x dollars more than Sharma. We set up an equation:
B= x + A
We are next given that together they have a total of y dollars. We can set up our second equation:
A + B = y
Since we know that B = x + A, we can substitute x + A for B into the second equation A+ B = y.
Notice that, after the substitution, we will only have variables of A, x, and y. Thus, we have:
A + x + A = y
2A + x = y
2A = y – x
A = (y – x)/2
Answer : D
Question: The number of people who purchased book A is twice the number of people who purchased book B. The number of people who purchased both books A and B is 500, which is twice the number of people who purchased only book B. What is the number of people Z who purchased only book A?
A.250
B.500
C.750
D.1000
Answer: D.1000
Explanation: This is best solved using overlapping sets or a Venn Diagram. We know that A = 2B, and that 500 people purchased both A and B. Further, those purchasing both was double those purchasing B only. This gives us 250 people purchasing B only. With the 500 that pruchased both, we have a total of 750 that purchased B and this is 1/2 of those that purchased A. So, 1500 purchased A. Less the 500 that purchased both, Z=1000 purchased A only. (This is much simpler to solve using the Venn diagram).
Correct answer is D. 1000
Question: What number has a 150:1 ratio to the number 2?
A.130
B.300
C.200
D.30
Answer: B.300
Explanation: 150:1 = x: 2
x = 150*2
x=300
ANSWER:B
Question: What is the greatest number of four digits which is divisible by 8, 16, 32 and 40 ?
A.9930
B.9940
C.9920
D.9950
Answer: C.9920
Explanation: Greatest number of four digits = 9999
LCM of 8, 16, 32 and 40 = 160
9999 ÷ 160 = 62, remainder = 79
Hence, greatest number of four digits which is divisible by 8, 16, 32 and 40
= 9999 - 79 = 9920
answer : C
Question: Find the expenditure on digging a well 14m deep and of 3m diameter at Rs.15 per cubic meter?
A.Rs.1185
B.Rs.1285
C.Rs.1385
D.Rs.1485
Answer: D.Rs.1485
Explanation: 22/7 * 14 * 3/2 * 3/2 = 99 m2
99 * 15 = 1485
ANSWER:D
Question: 5 men and 11 boys finish a job in 13 days, 11 men and 11 boys finish it in 10 days. 6 men and 6 boys shall finish it in how many days?
A.12
B.17
C.10
D.13
Answer: D.13
Explanation: 5 M + 11 B ----- 13 days
11 M + 11 B ------- 10 days
06 M + 06 B -------?
55 M + 121 B = 66 M +66 B
55 B = 11 M => 5 B = 1 M
25 B + 11 B = 36 B ---- 13 days
30 B + 6 B = 36 B -----? => 13 days
Answer: D
Question: If the price of a certain computer increased 30 percent from b dollars to 351 dollars, then 2b =
A.540
B.570
C.619
D.649
Answer: A.540
Explanation: Before Price increase Price = b
After 30% Price increase Price = b+(30/100)*b= 1.3b = 351 (Given)
i.e. b= 351/1.3 = $270
i.e. 2b = 2*270 = 540
Answer: option A
Question: Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 20 white marbles. How many red marbles could be in bag A?
A.1
B.3
C.4
D.6
Answer: C.4
Explanation: 6 is the answer.
Bag A-
R:W:B = 2:6:9
Let W in bag A be 6K
Bab B -
R:W = 1:4
Let W in bag B be 4k
W = 20 = 6K+4k
k=2
Total Red's in bag A will be 2K = 4
C
Question: On a scale that measures the intensity of a certain phenomenon, a reading of g+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of g. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?
A. 5
B. 50
C. 10^5
D. 5^10
Answer: C. 10^5
Explanation: To solve this problem we need to examine the information in the first sentence. We are told that “a reading of g + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of g.”
Let’s practice this idea with some real numbers. Let’s say g is 2. This means that g + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.
Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.
Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.
We have found a pattern, which can be applied to the problem presented in the stem:
3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.
4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.
5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.
We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.
Answer C.
Question: 17 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would 21 women take to do the same. If 3 women do as much work as 2 men?
A.34
B.87
C.30
D.99
Answer: A.34
Explanation: 3W = 2M
17M ------ 21 * 8 hours
21 W ------ x * 6 hours
14 M ------ x * 6
17 * 21 * 8 = 14 * x * 6
x = 34
Answer: A
Question: If n is a positive integer and the greatest common divisor of n and 16 is 4, and the greatest common divisor of n and 15 is 3, which of the following is the greatest common divisor of n and 96?
A.20
B.22
C.24
D.12
Answer: D.12
Explanation: The greatest common divisor of n and 16 is 4 --> n is a multiple of 4 but not a multiple of 8.
The greatest common divisor of n and 15 is 3 --> n is a multiple of 3 but not a multiple of 5.
96=2*3*4*4 is divisible by 12. therefore the greatest common divisor is 12
D
Question: A is thrice efficient as B and C is twice as efficient as B. what is the ratio of number of days taken by A,B and C, when they work individually?
A.2:6:3
B.2:6:9
C.2:6:6
D.2:6:1
Answer: A.2:6:3
Explanation: Answer: A) 2:6:3
Question: If a, b, c are consecutive positive integers and a<b<c, which of the following must be true?
I. c-a=2
II. abc is an even integer
III. a-b is positive integer
A.I
B.II
C.I and II
D.II and III
Answer: C.I and II
Explanation: If a, b, c are consequitive positive integers and a<b<c, which of the following must be true?
1. c-a=2
2. abc is an even integer
3. (a+b+c) is an integer
1.
a=a
b=a+1
c=a+2
c-a=a+2-a=2.
Correct
2. abc
a can be either even or odd
if a=even; abc=even;
if a=odd,b=even; abc=even
Rule:
multiple of any number of integers if atleast once multiplied by an even number will result in even.
even*even=even
even*odd=even
Correct
3. a-b = 1-2 = -1 => negative
Answer : C
Question: The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 4900 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units?
A.140 sq.units
B.280 sq.units
C.167sq.units
D.178sq.units
Answer: B.280 sq.units
Explanation: Given that the area of the square = 4900 sq.units
=> Side of square = √4900 = 70 units
The radius of the circle = side of the square = 70 units Length of the rectangle = 2/5 * 70 = 28 units
Given that breadth = 10 units
Area of the rectangle = lb = 28 * 10 = 280 sq.units
Answer:B
Question: In a Question paper there are 4 subjects. Each subject has has 5 question sets with each subject having one single question set that is the toughest of all. What is the total number of ways in which a candidate will not get all the four toughest question sets?
A.19
B.85
C.120
D.624
Answer: D.624
Explanation: A candidate can get a question set in 5*5*5*5=5^4 number of ways (each aubject has 5 question sets and we have total of 4 subjects). Now, out of these cases there will be only one case when the candidate got all the four toughest question sets. Therefore the total number of ways in which a candidate will NOT get all the four toughest question set is is 5^4-1=624.
Answer: D.
Question: A card is randomly drawn from a deck of 52 cards. What is the probability getting an Ace or King or Queen?
A.7/13
B.9/13
C.3/13
D.4/13
Answer: C.3/13
Explanation: Total number of cards = 52
Total number of Ace cards = 4
P(Ace) = 4/52=1/13
P(King) = 4/52=1/13
P(Queen) = 4/52=1/13
P(Ace or King or Queen) = P (Ace) + P (King)+ P(Queen)
=1/13+1/13+1/13=3/13
Answer C
Question: What percent of 5.2 kg is 16 gms ?
A.30
B.66
C.58
D.29
Answer: A.30
Explanation: Explanation:
Required percentage = (16/5200 * 100)% = 3/10% = 0.30%
Answer: A) .30%
Question: On rainy mornings, Mo drinks exactly N cups of hot chocolate (assume that N is an integer). On mornings that are not rainy, Mo drinks exactly 4 cups of tea. Last week Mo drank a total of 26 cups of tea and hot chocolate together. If during that week Mo drank 14 more tea cups than hot chocolate cups, then how many rainy days were there last week?
A.2
B.3
C.4
D.5
Answer: A.2
Explanation: T= the number of cups of tea
C= the number of cups of hot chocolate
T+C = 26 T-C=14 -> T= 20. C=6.
Mo drinks 4 cups of tea a day then number of days that are not rainy = 20/4 = 5
So number of rainy days = 7-5 = 2
A is the answer.
Question: Carl can wash all the windows of his house in 7 hours. His wife Maggie can wash all the windows in 4 hours. How many hours will it take for both of them working together to wash all the windows?
A.2
B.2 1/4
C.11 6/2
D.4 1/2
Answer: C.11 6/2
Explanation: Work hrs=AB/(A+B)= 28/11 =11 6/2
Answer is C
Question: If Sharon's weekly salary increased by 16 percent, she would earn $406 per week. If instead, her weekly salary were to decreased by 10 percent, how much would she earn per week?
A.Rs.415
B.Rs.325
C.Rs.352
D.Rs.315
Answer: D.Rs.315
Explanation: Soln:-
(406/116)90 =315
In this case long division does not take much time.
(406/116)=3.5
3.5*90=Rs.315
ANSWER:D
Question: Three numbers are in the ratio 3:5:7. The largest number value is 70. Find difference between Smallest & largest number is?
A.40
B.45
C.35
D.30
Answer: A.40
Explanation: == 3:5:7
Total parts = 15
= The largest number value is 70
= The largest number is = 7
= Then 7 parts -----> 70 ( 7 * 10 = 70 )
= smallest number = 3 & Largest number = 7
= Difference between smallest number & largest number is = 7 - 3 = 4
= Then 4 parts -----> 40 (4 * 10 = 40)
A
Question: 106 Ã— 106 + 94 Ã— 94 = ?
A.20072
B.20062
C.10072
D.20172
Answer: A.20072
Explanation: Explanation :
(a + b)2 + (a âˆ’ b)2 = 2(a2 + b2)
(Reference : Basic Algebraic Formulas)
1062 + 942 = (100 + 6)2 + (100 âˆ’ 6)2 = 2(1002 + 62) = 2(10000 + 36) = 20072. Answer : Option A
Question: A train 110 m long is running with a speed of 40 km/h. In how many seconds will the train pass a man who is running at 4 km/h in the direction opposite to that in which the train is going?
A.6
B.7
C.8
D.9
Answer: D.9
Explanation: The speed of the train relative to the man = 40 + 4 = 44 km/h.
44000 m/h * 1h/3600 s = (440/36) m/s
(110 m) / (440/36 m/s) = (110 * 36) / 440 = 36/4 = 9 seconds
The answer is D.
Question: A certain bag contains 100 balls â€” 50 white, 30 green, 8 yellow, 9 red, and 3 purple. If a ball is to be chosen at random, what is the probability that the ball will be neither red nor purple?
A.0.9
B.0.75
C.0.6
D.0.88
Answer: D.0.88
Explanation: According to the stem the ball can be white, green or yellow, so the probability is (white + green + yellow)/(total) = (50 + 30 + 8)/100 = 88/100 = 0.88.
Answer is D
Question: A car takes 6 hours to cover a distance of 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
A.40 Kmph
B.52 Kmph
C.60 Kmph
D.76 Kmph
Answer: C.60 Kmph
Explanation: Time = 6
Distance = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
C)
Question: The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
A.32%
B.34%
C.42%
D.44%
Answer: D.44%
Explanation: Explanation:
Let original length = x metres and original breadth = y metres.
Original area =xy m2New Length =120/100x=6/5x
New Breadth =120/100y=6/5y
=>New Area =6/5x∗6/5y
=>New Area =36/25xy
Area Difference=36/25xy−xy
=11/25xy
Increase%=Differnce/Actual∗100
=11xy/25∗1/xy∗100=44%
ANSWER IS D
Question: What is the greatest prime factor of 4^17 - 2^30?
A.2
B.3
C.5
D.7
Answer: C.5
Explanation: 4^17 - 2^30
=(2^2)^17 - 2^30
= 2^34 - 2^30
= 2^30 (2^4 -1)
= 2^30 * (16-1)
= 2^30 * 15
= 2^30 * 3 * 5
The greatest prime factor is 5.
The answer is C.
Question: Pipe A can fill a tank in 10 hrs and pipe B can fill it in 8 hrs. If both the pipes are opened in the empty tank.There is an outlet pipe in 3/4 th of the tank. In how many hours will it be fill 3/4th of that tank?
A.3 hr
B.3.10hr
C.3.15 hr
D.3.20 hr
Answer: D.3.20 hr
Explanation: Part filled A in 1 hr= (1/10)
Part filled B in 1 hr= (1/8)
Part filled by (A+B) together in 1 hr=(1/10)+(1/8)=18/80
So, The tank will be full in 80/18 hrs.
Time taken to fill Exact 3/4th of the tank = (80/18) *(3/4) = 3.20 hrs
Answer : D
Question: If a farmer wants to plough a farm field on time, he must plough 160 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially?
A.600
B.490
C.448
D.435
Answer: C.448
Explanation: Let x be the number of days in the initial plan. Therefore, the whole field is 160â‹…x hectares. The farmer had to work for x+2 days, and he ploughed 85(x+2) hectares, leaving 40 hectares unploughed. Then we have the equation:
160x=85(x+2)+40
75x=210
x=2.8
So the farmer planned to have the work done in 6 days, and the area of the farm field is 160(2.8)=448 hectares
correct answer C
Question: A man has Rs. 176 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A.33
B.60
C.75
D.90
Answer: A.33
Explanation: Let number of notes of each denomination be x.
Then x + 5x + 10x = 176
16x = 176
x = 11.
Hence, total number of notes = 3x = 33.
Answer: Option A
Question: The number 0.8 is how much greater than 1/2 ?
A.½
B.3/10
C.1/50
D.1/500
Answer: B.3/10
Explanation: let x be the difference
then
.8-1/2=x
8/10-1/2=x
x= 3/10
ans B
Question: Find the value of x in each of the following equation:	[(17.28/x) / (3.6*0.2)] = 2
A.12
B.14
C.16
D.18
Answer: A.12
Explanation: (17.28/x) = 2*3.6*0.2  x = (17.28/1.44) = (1728/14) = 12.
Answer is A.
Question: Three men sit at a table. One dollar is placed in front of each of them. When a buzzer sounds, the first man to touch the money in front of him keeps the money that is in front of him, while the other two receive nothing. If none of the three men touches the money in front of them, then the amount of money doubles for the next round. After three rounds of this activity, what is the maximum amount of money that any one man could receive?
A.$4
B.$5
C.$10
D.$15
Answer: A.$4
Explanation: The Optimized scenario
After 1st Round - $1. Let nobody touches the amount will double in next round.
After 2nd Round - $2. Let Man A touches it and get$2 or nobody touches
After 3rd Round - $2 Let the same Man A touches it and get another$2.( $4 in case nobody touches in 2nd round).
So, after 3rd round, the maximum amount a man receives is $4.
Answer (A)
Question: Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 3?
A.2/5
B.2/7
C.33/83
D.51/250
Answer: D.51/250
Explanation: SOLUTION
The number of integers from 101 to 350, inclusive is 250, out of which 100 (from 200 to 299) will have a hundreds digit of 2. Thus the probability is 51/250.
Answer: D
Question: What is x if x + 2y = 20 and y = 5?
A.A)10
B.B)8
C.C)6
D.D)4
Answer: A.A)10
Explanation: x = 20 - 2y
x = 20 - 10.
x = 10
Answer : A