divelab/Sys2Bench huggingface dataset:
Question: How many odd integers from 1 to 100 (both inclusive) have odd number of factors?
Options:
A)13
B)14
C)5
D)6
E)7
Rationale: Integers having odd number of factors will be perfect squares. Odd numbers will have odd perfect squares. Thus, the possible values for the perfect squares are :
1,9,25,49,81 and the corresponding integers are 1,3,5,7,9 (more than 4). Thus C is the correct answer .
Answer: C
Question: At the faculty of Aerospace Engineering, 302 students study Random-processing methods, 232 students study Scramjet rocket engines and 112 students study them both. If every student in the faculty has to study one of the two subjects, how many students are there in the faculty of Aerospace Engineering?
Options:
A)444
B)432
C)346
D)422
E)448
Rationale: 302 + 232 - 112 (since 112 is counted twice )= 422
D is the answer
Answer: D
Question: 20 beavers, working together in a constant pace, can build a dam in 18 hours. How many hours will it take 12 beavers that work at the same pace, to build the same dam?
Options:
A)22.
B)24.
C)30.
D)36
E)38.
Rationale: Total Work = 20*18= 360 Beaver hours
12 Beaver * x =360 Beaver hours
x=360/12=30
ANSWER:C
Answer: C
Question: An Office used three types of envelopes: small,medium and large. They used 80 small envelopes. The no.of Medium envelopes used was 5 times, the no.of small envelopes plus 5/20 of the number of envelopes. The no.of.large envelopes was equals to the no.of small envelopes plus 5/20 of the no.of medium envelope. Altogether, How many envelope did they use?
Options:
A)950
B)960
C)972
D)970
E)980
Rationale: Let the total no. of envelopes be x
No. of small envelopes = 80
No. of medium envelopes =5*80+5/20*x = 400+5/20*x
No. of large envelopes = 80 +5/20(400+5/20*x) = 180+25/400*x
Sum of small, medium and large envelopes = x
80 + 400+5/20*x + 180+25/400*x = x
660 + 5/16*x = x
11/16*x = 660
x = 960
ANSWER:B
Answer: B
Question: Simplify: 1005-1000+10
Options:
A)900
B)926
C)1001
D)905
E)None of them
Rationale: 1005-1000+10
=1005-(1000/10)
=1005-100
=905.
Answer is D
Answer: D
Question: One ball will drop from a certain height. The height it will reach after rebounding from the floor is 50 percent of the previous height. The total travel is 250cm when it touches the floor on third time. What is the value of the original height?
Options:
A)80cm
B)90cm
C)100cm
D)120cm
E)130cm
Rationale: when Ball comes down, Then i have indicated the distance covered in green
when Ball goes up, Then i have indicated the distance covered in red
Distance travelled uptil the ball touches the floor 3rd time:h+0.5h+0.5 h+0.5*0.5 h+0.5*0.5 h
h + 2 * 0.5 * h + 2 * 0.25 * h = h (1+2*0.5+2* 0.25) = h (1+1+0.5) = 250
2.5 h = 250
h = 100.
C is the answer.
Answer: C
Question: A drum has a combination of 2 liquids X and Y in the proportion 7:5 when 9 L of combination are taken out and the Drum is filled with Y, the proportion of X and Y becomes 7:9. Then how many litres of liquids X was obtained by the Drum initially?
Options:
A)21 litres
B)22 litres
C)23 litres
D)24 litres
E)25 litres
Rationale: Explanation :
=> If the Drum has 7a and 5a litres of combinations X and Y correspondingly. Amount of Combination taken out is 9 litres
Some of X in Combination left = [ 7a - (7/12)*9] litres
= [ 7a - (21/4) ] litres
Some of Y in combination left = [ 5a - (5/12)*9 ] litres
= [ 5a - (15/4) ] litres
=> Since the combination is in the ratio 7:9,
{ [7a - (21/4)] / [ 5a - (15/4) ]+9 } = (7/9) => [ (28a - 21) / (20a + 21) ] = (7/9)
=>252a – 189 = 140a + 147
=> 112a = 336
=> a = 3
Some of liquid X, has in the Drum initially is (7*3) = 21 litres
Answer : A
Answer: A
Question: The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 10 minutes, then the area of the park (in sq. m) is:
Options:
A)153601
B)153600
C)153602
D)153603
E)240000
Rationale: Perimeter = Distance covered in 10 min. =(12000/60) x 10m = 2000 m.
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 2000 or x = 200.
Length = 600 m and Breadth = 400 m.
Area = (600 x 400) m2 = 240000 m2.
Answer: E
Answer: E
Question: The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
Options:
A)76 kg
B)80 kg
C)85 kg
D)90 kg
E)None
Rationale: Solution
Total weight increased	= (8 x 2.5)kg
= 20 kg.
Weight of new person	= (65 + 20)kg
= 85 kg.
Answer C
Answer: C
Question: A soccer store typically sells replica jerseys at a discount of 30 percent to 70 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 percent of the list price is the lowest possible sale price?
Options:
A)20
B)25
C)30
D)40
E)50
Rationale: Let the list price be 2x
for min sale price, the first discount given should be 70%, 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 is 40% of 80
hence, D is the answer
Answer: D
Question: 21 ball numbered 1 to 21. A ballis drawn and then another ball is drawn without replacement.
Options:
A)2/31
B)4/35
C)9/42
D)8/41
E)9/67
Rationale: The probability that first toy shows the even number,
=1021=1021
Since, the toy is not replaced there are now 9 even numbered toys and total 20 toys left.
Hence, probability that second toy shows the even number,
=920=920
Required probability,
=(1021)×(920)=(1021)×(920)
=9/42
C
Answer: C
Question: In a local school district, the high school and middle school each received r dollars toward funding for the student arts program. The high school enrolled 400 students and the middle school enrolled 200 students. Later, the middle school transferred s dollars to the high school so that they would have received the same funding per student. Which of the following is equivalent to s?
Options:
A)r/2
B)r/3
C)r/4
D)r/5
E)r/8
Rationale: Total per head = 2r/600
After s transfer both schools have total for head.
So at High School it will be:
r+s = (400)* (2r/600) = 8r/6=4r/3
i.e. s = 4r/3 - r/1 =4r-3r/3=r/3
Answer : B
Answer: B
Question: What is the perimeter of a square with area p^2/16 ?
Options:
A)3p/4
B)3p^2/4
C)p
D)3p^2
E)4p/3
Rationale: Area of square, (side)^2 = (p/4)^2
Therefore side of the square = p/4
Perimeter of square = 4*side = 4* (p/4) = p
Answer is C.
Answer: C
Question: A spirit and water solution is sold in a market. The cost per liter of the solution is directly proportional to the part (fraction) of spirit (by volume) the solution has. A solution of 1 liter of spirit and 1 liter of water costs 50 cents. How many cents does a solution of 1 liter of spirit and 3 liters of water cost?
Options:
A)13
B)33
C)56
D)50
E)52
Rationale: C. 50 cents
Yes, ensure that you understand the relation thoroughly!
Cost per liter = K * Fraction of Spirit
50 cents is the cost of 2 liters of solution (1 part water, 1 part spirit). So cost per liter is 25 cents. Fraction of spirit is 1/2.
25 = K * (1/2)
K = 50
Cost per liter = 50 * (1/4) (1 part spirit, 3 parts water)
Cost for 4 liters = 50 * (1/4) * 4 = 50 cents
D. 50 cents
Answer: D
Question: The difference between the simple interest received from two different sources on Rs. 1000 for 3 years is Rs. 13.50. The difference between their rates of interest is?
Options:
A)A)2.4%
B)B)2.6%
C)C)0.3%
D)D)0.25%
E)E)1.3%
Rationale: (1000 * R1 * 3)/100 - (1000 * R2 * 3)/100 = 13.50 4000(R1 - R2) = 1350
R1 - R2 = 0.25%
Answer:D
Answer: D
Question: Find the compound interest on $1200 for 6 years at 20% p.a. if CI is component yearly?
Options:
A)$120
B)$150
C)$2383
D)$250
E)$300
Rationale: A = P(1 + R/100)^T
= 1200(1 + 20/100)^6
=$3583
CI = $2383
Answer is C
Answer: C
Question: A bank offers 5% C.I. calculated on half-yearly basis . A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is?
Options:
A)Rs.181
B)Rs.121
C)Rs.126
D)Rs.128
E)Rs.129
Rationale: Amount = [1600 * (1 + 5/(2 * 100)2 + 1600 * (1 + 5/(2 * 100)]
= [1600 * 41/40(41/40 + 1)
= [(1600 * 41 * 81)/(40 * 40)] = Rs. 3321.
C.I. = 3321 - 3200 = Rs. 121.
Answer: B
Answer: B
Question: What is the value of the following expression?
(1/ (2^2 – 1)) + (1/ (4^2 – 1)) + (1/ (6^2 – 1)) + .... + (1/ (24^2 – 1))
Options:
A)a. 9/19
B)b. 10/19
C)c. 12/25
D)d. 11/21
E)e. 11/19
Rationale: it can be written into
1/3+1/15+ 1/35.....1/575
i.e
1/3+1/2*(1/3-1/5)+1/2*(1/5-1/7).... til last term - 1/2*(1/25)
all terms from 1/5 .. to second last term get cancelled
left with
1/3+1/6-1/50=1/2-1/50=24/50=12/25 C
Answer: C
Question: If y ≠ 3 and 2m/y is a prime integer greater than 2, which of the following must be true?
I. m = y
II. y = 1
III. m and y are prime integers.
Options:
A) None
B) I only
C) II only
D) III only
E) I and II
Rationale: Note that we are asked which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.
So we should try to make the listed statements not true for some m and y (which satisfy y≠3 and 2m/y=prime>2).
I. m = y --> not necessarily true: m=3 and y=2 (2m/y=3=prime>2);
II. y=1 --> not necessarily true: m=3 and y=2 (2m/y=3=prime>2);
III. m and y are prime integers --> not necessarily true: m=10 and y=4 (2m/y=5=prime>2).
Answer: A.
Answer: A
Question: The sum of two numbers is 75 and their difference is 20. Find the difference of their squares.
Options:
A)3000
B)2700
C)1500
D)1254
E)3421
Rationale: If the sum of two numbers is x and their difference is y, then the difference of their squares is xy.
Here x= 75 and y = 20
75 × 20 = 1500
Answer: C
Question: The value for G=(0.889 × 55)/9.97 to the nearest tenth is
Options:
A)0.5
B)4.63
C)4.9
D)7.7
E)49.1
Rationale: if we read the Q, we can easily home on to the answer,,
the numerator is clearly between 40 and 50..
denomiator is close to 10..
so G=(0.889 × 55)/9.97 is between 4 and 5..
It may tempt us to solve it since ther are two values between 4 and 5..
but the catch is innearest tenth
4.63 can be called nearest to hundreth and no tenth, so can be eliminated..
4.9 is our answer...
C
Answer: C
Question: Two quarts containing 6⁄7 water and 1/7 formula are mixed with three quarts containing 7⁄10 water and 2⁄10 formula. Approximately what percent of the combined five-quart mixture is water?
Options:
A)74%
B)76%
C)73%
D)72%
E)60%
Rationale: 6/7 of 2 quarts = 12/7 quarts of water
7/10 of 3 quarts = 21/10 quarts of water
12/7 = 1.714
21/10 = 2.100
Total water = 1.714 + 2.100= 3.814
Total quarts = 5
3.814/5 = 76.28% approx. 76%
Answer : B
Answer: B
Question: The price of a book is increased from $300 to $360. What is the % of increase in its price?
Options:
A)10%
B)20%
C)35%
D)40%
E)50%
Rationale: Explanation: Change in the price = Rs 360 – Rs 300
= Rs 60
Percentage of increase = Change in the price Initial Price
* 100.
Percentage increase in price =( 60 300
) *100 = 20%
B
Answer: B
Question: Linda and Angela contract to paint a neighbor's house. Even though Linda
spends 75% more time painting the house than Angela, each receives a payment
of m dollars when the work is completed. If Angela decides to pay Linda n
dollars so that they would have received the same compensation per hour
worked, what is n in terms of m?
Options:
A)A)(1/2)m
B)B)(1/3)m
C)C)(9/33)m
D)D)(1/5)m
E)E)(1/6)m
Rationale: let angela work for 12 hrs and get 1650 rs. (m =1650)
According to question, linda works 75% more time and gets the same amount. so let linda work 21 hrs and get 1650 rs.
so they do a total of 21 +12 =33 hrs of work and get 3300 rs together . The question basically asks to equalise their payment. so a total of 3300 rs is received by them and they do 10 hrs work together. average rate will be 3300/33 = 100 rs per hr. so linda should get 21*100 =2100 rs. But she got only 1650. so angela should give 450 rs to linda. answer = 450/1650 =9/33
answer C
Answer: C
Question: The speed of a car is 65 km in the first hour and 45 km in the second hour. What is the average speed of the car?
Options:
A)72 kmph
B)55 kmph
C)50 kmph
D)80 kmph
E)82 kmph
Rationale: S = (65 + 45)/2
= 55 kmph
ANSWER:B
Answer: B
Question: 4, 16, 36, 49, 64 ?, F ind the missing number(?).
Options:
A)88
B)74
C)110
D)99
E)100
Rationale: The series consists of squares of consecutive EVEN numbers.
Answer : E.
Answer: E
Question: A train is running at a uniform speed of 80 kmph. It passes a railway platform in 15 seconds. If the length of the platform is 130 m,then the length of the train is
Options:
A)160 m
B)203.33 m
C)140.5 m
D)150.5 m
E)None of these
Rationale: Let the length of train be X m.
We know that Speed = Distance / Time.
Thus, (X + 130)/15 = 80 x 5/18
or, (X + 130)/15 = 400/18
or, (X + 130) = 1000/3
or X= 1000/3 - 130
X = 203.33
ANSWER:B
Answer: B
Question: Sophika can buy watches at a price of P dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of X by selling Y watches, then in terms of P and X and Y, what is the percent of the markup from her buy price to her sell price?
Options:
A)(YP)/100X
B)100X/(YP)
C)(YP)+100X
D)(YP)-100X
E)(YP)*100X
Rationale: Algebraic Solution:If she makes a total profit of X for Y watches, then that must be a profit of X/Y for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (X/Y)/P *100 = 100X/(YP)
Answer : B
Answer: B
Question: Fence X is twice as long as fence Y, and fence Y is 2 feet shorter than fence Z. If 2 feet were added to each fence, which of the following must be true?
I) X is twice as long as Y.
II) Y is 2 feet shorter than Z.
III) X is longer than Z.
Options:
A)I only
B)II only
C)II and III
D)I and II
E)III only
Rationale: If X= 6
Y= 3
Z= 5
Then, after adding 3 feet to each:
X= 8
Y=5
Z=7
And the answer would be C, which is wrong.
Answer: C
Question: A man is 24 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:
Options:
A)14 Years
B)18 Years
C)20 Years
D)22 Years
E)17 Years
Rationale: Let the son's present age be x years. Then, man's present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
ANSWER D
Answer: D
Question: If a > 1, which of the following is equal to (2a + 6)/(a^2 + 4a + 3)?
Options:
A)a
B)a + 3
C)2/(a - 1)
D)2a/(a - 3)
E)2/(a + 1)
Rationale: Here's the algebraic approach:
(2A + 6)/(A^2 + 4A + 3) can be rewritten as...
2(A + 3)/(A+3)(A+1)
We can simplify the fraction, which leaves us with...
2/(A+1)
Ans : E
Answer: E
Question: A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is ?
Options:
A)37.9
B)37.53
C)37.5
D)37.1
E)37.4
Rationale: Explanation:
Average speed = (2xy) /(x + y) km/hr
= (2 * 50 * 30) / (50 + 30) km/hr.
37.5 km/hr.
Answer: C
Answer: C
Question: A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. IN what ratio will the profit be shared at the end of 3 years?
Options:
A)3:4:9
B)3:4:7
C)3:4:2
D)3:4:4
E)3:4:3
Rationale: A:B:C = (40000 * 36) : (80000 * 12 + 40000 * 24) : (120000 * 24 + 40000 * 12)
= 144:192:336 = 3:4:7
Answer: B
Answer: B
Question: Two different primes may be said to “rhyme” around an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it?
Options:
A)12
B)15
C)17
D)18
E)20
Rationale: Let's test each option:
A. 12 --> 2*12=24 --> 24=5+19=7+17=11+13: 6 rhyming primes (start from the least prime and see whether we can get the sum of 24 by adding another prime more than 12 to it);
B. 15 --> 2*15=30 --> 30=7+23=11+19=13+17: 6 rhyming primes;
C. 17 --> 2*15=30 --> 34=7+23=11+19=13+17: 6 rhyming primes;
D. 18 --> 2*18=36 --> 36=5+31=7+29=13+23=17+19: 8 rhyming primes;
E. 20 --> 2*20=40 --> 40=3+37=11+29=17+23: 6 rhyming primes.
Answer: D.
Answer: D
Question: A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?
Options:
A)20%
B)30%
C)40%
D)50%
E)55%
Rationale: Let the side of the square(ABCD) be x metres.
Then, AB + BC = 2x metres.
AC = 2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.
Saving % = (0.59x)/2x	x 100	%	= 30% (approx.)
B)
Answer: B
Question: A and B together can do a work in 6 days. If A alone can do it in 15 days. In how many days can B alone do it?
Options:
A)10
B)99
C)77
D)65
E)55
Rationale: 1/6 – 1/15 = 1/10 => 10
Answer: A
Answer: A
Question: One night 25 percent of the female officers on a police force were on duty. If 500 officers were on duty that night and half of these were female officers, how many female officers were on the police force?
Options:
A)90
B)180
C)270
D)500
E)1,000
Rationale: Let total number of female officers in the police force = F
Total number of officers on duty on that night = 500
Number of female officers on duty on that night = 500/2 = 250
(25/100)*F = 250
=>F = 1000
Answer E
Answer: E
Question: In a group of Peacocks and cows, the total number of legs are 36 more than twice the no. of heads. Find the total no.of buffaloes.
Options:
A)10
B)19
C)12
D)16
E)18
Rationale: Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 36
=> 2x = 36 => x = 18
E
Answer: E
Question: Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions?
Options:
A)10
B)8
C)77
D)26
E)22
Rationale: RS = 60 -40 = 20 * 5/18 = 100/18
T = 50
D = 50 * 100/18 = 2500/9
RS = 60 + 40 = 100 * 5/18
T = 2500/9 * 18/500 = 10 sec.Answer: A
Answer: A
Question: Which of these must the factor W of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80
Options:
A)W=1,2 only
B)W=2,3 only
C)W=1,2,3 only
D)W=1,2,3,4 only
E)All of them
Rationale: Let the four variables be a< b< c< d.
Assume a worst case scenario where the a equal to a prime number (hence odd).
therefore
a = divisible by 1
b = a + 1 (divisible by 2)
c = a + 3 (divisible by 3)
d = a + 4 (divisible by 4)
therefore
each answer choice must be divisible by 2 x 3 x 4 = 24
only 80 is not divisible. Therefore answer = D
Answer: D
Question: There are 2 available positions and 50 candidates, one half of whom are democrats and another half are republicans. If it was decided that the positions would be filled at random, then what is the probability Q that the both positions will be taken by members of just one party?
Options:
A)1/25
B)12/49
C)1/4
D)24/49
E)1/2
Rationale: Q probability of one party having both spots:
(1/2) * (24/49) = 12/49
(1/2) or (25/50) because it does not matter which party or which person gets the first spot. (24/49) because after one person from a particular party is chosen, there are 24 members of the same party left out of 49 total candidates.
Since this result can happen for both parties, (12/49) + (12/49) = (24/49)
Answer: D
Answer: D
Question: If there are 4 peanuts in a box and Mary puts 12 more peanuts inside, how many peanuts are in the box?
Options:
A)8
B)16
C)10
D)11
E)12
Rationale: 12+4=16
correct answer is B)16
Answer: B
Question: A qualified worker digs a well in 4 hours. He invites 2 apprentices, both capable of working 3/4 as fast and 2 trainees both working 1/5 as fast as he. If the five-person team digs the same well, how much time does the team need to finish the job?
Options:
A)1:24
B)1:34
C)1:44
D)1:36
E)2:14
Rationale: and answer is correct as you have calculated : 100/72 hrs that is 1.64 hrs - 1 hr and 36 minutes, so 1:36, hence, D.
Answer: D
Question: The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 90 kmph. Find the speed of the boat in still water and the speed of the stream?
Options:
A)10 kmph.
B)15 kmph.
C)18 kmph.
D)11 kmph.
E)12 kmph.
Rationale: Speed of the boat in still water = (60+90)/2 = 75 kmph. Speed of the stream = (90-60)/2
= 15 kmph.
Answer:B
Answer: B
Question: A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 4?
Options:
A)3/8
B)1/16
C)1/8
D)1/2
E)1/4
Rationale: Cases for sum to be greater than 4 are as mentioned below
Case 1: (1, 2, 2)
Case 2: (2, 1, 2)
Case 3: (2, 2, 1)
Case 4: (2, 2, 2)
Total Outcomes = 2 x 2 x 2 = 8
Probability = Favorable Outcomes / Total Outcomes
I.e. Probability = 4/8 = 1/2
Answer: Option D
Answer: D
Question: Vijay lent out an amount Rs. 10000 into two parts, one at 8% p.a. and the remaining at 10% p.a. both on simple interest. At the end of the year he received Rs. 850 as total interest. What was the amount he lent out at 8% pa.a?
Options:
A)Rs. 15000
B)Rs. 6000
C)Rs. 25000
D)Rs. 10000
E)Rs. 18000
Rationale: Let the amount lent out at 8% p.a. be Rs. A
=> (A * 8)/100 + [(10000 - A) * 10]/100 = 850
=> A = Rs. 15000.
ANSWER:A
Answer: A
Question: What is the next number of the following sequence
3^2,27,3^4,243
Options:
A)3^6
B)3^5
C)3^4
D)3^2
E)3^1
Rationale: 3^2,27,3^4,243,x
3^2,3^3,3^4,3^5,3^6
x=3^6
ANSWER:A
Answer: A
Question: Out of 15 consecutive numbers, 2 are chosen at random. The probability that they are both odds or both primes is?
Options:
A)1/8
B)1/7
C)1/9
D)1/2
E)1/1
Rationale: A leap year has 52 weeks and two days
Total number of cases = 7
Number of favourable cases = 1
i.e., {Saturday, Sunday}
Required Probability = 1/7
Answer: B
Answer: B
Question: The number of sequences in which 10 players can throw a ball, so that the youngest player may not be the last is ?
Options:
A)4300028
B)3265920
C)4320201
D)3786220
E)3101242
Rationale: x Not younger
The last ball can be thrown by any of the remaining 10 players. The first 6 players can throw the ball in 9P9 ways.
The required number of ways = 9(9!)
= 3265920
Answer:B
Answer: B
Question: A and B are two partially filled buckets of water. If 5 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 3 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?
Options:
A)11
B)13
C)17
D)49/5
E)23
Rationale: Let A contains A, B contains B liters
so, (A-5)/(B+5)= 1/3.......(1)
again, (B-3)/(A+3)= 1/2..........(2)
from (1)(2) we find A= 49/5
ans: (D)
Answer: D
Question: Two bullet trains of equal lengths take 10 seconds and 12 seconds respectively to cross a telegraph post. If the length of each bullet train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?
Options:
A)11 sec.
B)15 sec.
C)12 sec.
D)17 sec.
E)19 sec.
Rationale: Speed of the first bullet train = 120/10 m/sec = 12 m/sec.
Speed of the second bullet train = 120/12 m/sec = 10 m/sec.
Relative speed = (12 + 10) = 22 m/sec.
Required time = (120 + 120)/22 sec = 11 sec.
A
Answer: A
Question: A batsman scored 120 runs which included 10 boundaries and 6 sixes. What % of his total score did he make by running between the wickets
Options:
A)40%
B)50%
C)36.67%
D)70%
E)75%
Rationale: Number of runs made by running = 120 - (10 x 4 + 6 x 6)
= 120 - (76)
= 44
Now, we need to calculate 60 is what percent of 120.
=> 44/120 * 100 = 36.67 %
C
Answer: C