divelab/Sys2Bench huggingface dataset:
Question: If 3 men or 4 women can reap a field in 10 days how long will 5 men and 6 women take to reap it?
Options:
A)19/60
B)9/60
C)2/3
D)5/18
E)6/24
Rationale: Explanation:
3 men reap 1/10 field in 1 day
1 man reap 1/(3 x 10)
4 women reap 1/10 field in 1 day
1 woman reap 1/(10 x 4)
5 men and 6 women reap (5/(3 x 10)+ 6/(4 x 10)) =19/60 in 1 day
5 men and 6 women will reap the field in 19/60 days
Answer: Option A
Answer: A
Question: What is the greatest possible value of integer n if 100! is divisible by 15^n
Options:
A)20
B)21
C)22
D)23
E)24
Rationale: 15^n = 5^n * 3^n
Highest prime factor will be the limiting factor.
100/5 +100/25 =20+4 = 24
ANSWER:E
Answer: E
Question: Find the invalid no.from the following series 15, 16, 34, 105, 424, 2124, 12756
Options:
A)15
B)16
C)34
D)105
E)2124
Rationale: 2nd term = (1st term) * 1 + 1 = 15 *1 + 1 = 16.
3rd term = (2nd term) * 2 + 2 = 16 * 2 + 2 = 34
4th term = (3th term) *3 + 3 = 34 * 3 + 3 = 105
5th term = (4th term) * 4 + 4 = 105 * 4 + 4 = 424
6th term = (5th term) * 5 + 5 = 424 * 5 + 5 = 2125
6th term should 2125 instead of 2124.
E
Answer: E
Question: On Monday, a certain animal shelter housed 45 cats and dogs. By Friday, 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday?
Options:
A)10
B)11
C)12
D)13
E)14
Rationale: To maximize the number of pets adopted, we should maximize the number of dogs adopted since 1/4 > 1/5.
The greatest multiple of 4 less than 45 is 40 (such that the remaining number is divisible by 5), so let's assume there were 40 dogs and 5 cats.
Then the maximum number of adopted pets is 1/4*(40) + 1/5*(5) = 10 + 1 = 11.
The answer is B.
Answer: B
Question: A shopkeeper gave an additional 20 per cent concession on the reduced price after giving 30 per cent standard concession on an article. If Arun bought that article for 1,120, what was the original price?
Options:
A)3,000
B)4,000
C)2,400
D)2,000
E)None of these
Rationale: Original price = 1120 × 100⁄70 × 100⁄80 = 2000
Answer D
Answer: D
Question: The total age of A and B is 10 years more than the total age of B and C. C is how many years younger than A.?
Options:
A)16
B)12
C)10
D)20
E)10
Rationale: (A+B)-(B-C)=10
A-C=10
Answer is C
Answer: C
Question: In an election between two candidates, 10% of votes are were declares invalid. First candidate got 3600 votes which were 60% of the total valid votes . The total number of votes enrolled in that election was:
Options:
A)9000
B)98000
C)20000
D)11000
E)12000
Rationale: 100 % - 10 % = 90 %
54%-36%=18%
18% ------> 3600 ( 18 × 200 = 3600)
100% -------> 20000 votes (100 × 200 = 20000)
C
Answer: C
Question: Find a 8 digit number that if multiplied by the number 9 or any of its multiplications products (18, 27, 36, 45,..) will result in the multiplication factor repeated (n) number of times.
Options:
A)12345679
B)42345675
C)22347379
D)52345673
E)21345679
Rationale: A
The number is: 12345679
12345679 * 9 = 111111111
12345679 * 18 = 222222222
12345679 * 27 = 333333333
And so on...
Answer: A
Question: A and B invests Rs.3000 and Rs.7000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit?
Options:
A)9:6
B)9:8
C)9:14
D)9:9
E)9:5
Rationale: (3*6 + 6*6): (7*12)
54:84 => 9:14.Answer:C
Answer: C
Question: A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 20%, he is only able to save Rs. 200 per month. What is his monthly salary?
Options:
A)5000
B)2889
C)2779
D)6257
E)5362
Rationale: Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80*(20/100) = Rs. 96
Present Savings = 100 – 96 =Rs. 4
100 ------ 4
? --------- 200 => 5000
Answer: A
Answer: A
Question: A boat covers a certain distance downstream in 1 hour, while it comes back in 11â„2 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
Options:
A)31 kmph
B)16 kmph
C)19 kmph
D)15 kmph
E)14 kmph
Rationale: Let the speed of the water in still water =x
Given that speed of the stream = 3 kmph
Speed downstream =(x+3) kmph
Speed upstream =(xâˆ’3) kmph
He travels a certain distance downstream in 1 hour and come back in 11â„2 hour.
i.e., distance travelled downstream in 1 hour = distance travelled upstream in 11â„2 hour
Since distance = speed Ã— time, we have
(x+3)Ã—1=(x-3)3/2
2(x+3)=3(x-3)
2x+6=3x-9
x=6+9=15 kmph
Answer: D
Answer: D
Question: A and B invests Rs.4000 and Rs.6000 in a business. After 4 months, A withdraws three-fourth of his capital and 4 months later, B withdraws half of his capital. In what ratio should they share the profits at the end of the year?
Options:
A)2:5
B)3:5
C)4:5
D)3:7
E)5:7
Rationale: A : B
(4000*4)+(1000*8) : (6000*8)+(3000*4)
24000 : 60000
2 : 5
Answer: A
Answer: A
Question: 0.5, 0.55, 0.65, 0.8, ?
Options:
A)0.7
B)0.9
C)0.95
D)1
E)2
Rationale: The pattern is + 0.05, + 0.10, + 0.15, .....
Answer : D.
Answer: D
Question: If 15 women or 10 men can complete a project in 55 days, in how many days will 5 women and 4 men working together complete the same project ?
Options:
A)75
B)8
C)9
D)85
E)None of these
Rationale: 15 W = 10 M
Now, 5W + 4M = 5W + 4×15/10W = 5W + 6W = = 11 W
If 15 women can complete the project in 55 days,
11 women can complete the same project in
55×15/11=75days
Answer A
Answer: A
Question: If n is a positive integer and n^2 is divisible by 450, then what is the largest positive integer that must divide n?
Options:
A)10
B)20
C)30
D)40
E)50
Rationale: 450 = 2 * 3^2 * 5^2
If 450 divides n^2, then n must be divisible by 2 * 3 * 5 = 30
The answer is C.
Answer: C
Question: Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when fill?
Options:
A)6
B)8
C)7
D)4
E)1
Rationale: 1/10 + 1/15 = 1/6 * 4 = 2/3
1 - 2/3 = 1/3
1/10 + 1/15 - 1/x = 1/3
x = 8
Answer:B
Answer: B
Question: A certain club has 14 members, including Harry. One of the 14 members is to be chosen at random to be the president, one of the remaining 13 members is to be chosen at random to be the secretary, and one of the remaining 12 members is to be chosen at random to be the treasurer. What is the probability that Harry will either be the secretary or the member chosen to be the treasurer?
Options:
A)5/7
B)4/7
C)3/7
D)1/7
E)2/7
Rationale: Each member out of 14, including Harry, has equal chances to be selected for any of the positions (the sequence of the selection is given just to confuse us). The probability that Harry will be selected to be the secretary is 1/14 and the probability that Harry will be selected to be the treasurer is also 1/14. So, the probability that Harry will be selected to be either the secretary or the the treasurer is 1/14+1/14=2/14.
Answer: D.
Answer: D
Question: Afreena walks 8 km towards East and then walks 13 km back, then she turns left and walks 4 km; then walks 5 km after turning left; she turns left again and walks 3 km. How far is she from the starting point ?
Options:
A)3 km
B)2 km
C)1 km
D)6 km
E)7 km
Rationale: Explanation:
From given details we know that she walks in square shaped path and will be in 1 km away from starting point.
ANSWER IS C
Answer: C
Question: A coed soccer team has W women and M men on the team. If 4 women and 2 men are added to the team roster, and if one person on the team is selected at random to serve as team captain, then the probability that the team captain will be a woman can be represented as
Options:
A)(W + 4)/(W + M + 6)
B)(W + 4)/(W + M + 2)
C)(W + 4)/(M + 2)
D)(W + 4)/(W + M)
E)W/M
Rationale: W women and M men..now 4 women and 2 men are added then total women = W+4 and total men = M+2.. if we club both as total one team (W + M + 6)
We need to find the probability that the team captain will be a woman => W+4 / (W + M + 6)..
IMO option A is correct answer...
Answer: A
Question: A man is 30 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is
Options:
A)14 years
B)28 years
C)20 years
D)22 years
E)None
Rationale: Solution
Let the son's present age be x years .Then, man's present age	= (x+30)years.
Then	â€¹=â€º(x+30)+2= 2(x+2)
â€¹=â€ºx+32=2x+4
x= 28.
Answer B
Answer: B
Question: Chris mixed 3 pounds of raisins with 3 pounds of nuts. If a pound of nuts costs 3 times as much as a pound of raisins, then the total cost of the raisins was what fraction of the total cost of the mixture?
Options:
A) 1/7
B) 1/5
C) 1/4
D) 1/3
E) 3/7
Rationale: 1 lbs of raisin = $1
3 lbs of raisin = $3
1 lbs of nuts = $3
3 lbs of nuts = $9
Total value of mixture = 9+3 = 12
Fraction of the value of raisin = 3/12 = 1/4
Ans:C
Answer: C
Question: If 2^(2w) = 8^(w−2), what is the value of w?
Options:
A)3
B)6
C)9
D)12
E)15
Rationale: 2^(2w) = 8^(w−2)
2^(2w) = 2^(3*(w−2))
2^(2w) = 2^(3w-6)
Let's equate the exponents as the bases are equal.
2w = 3w-6
w = 6
The answer is B.
Answer: B
Question: 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 Tuesday
B)2 p.m. on Wednesday
C)3 p.m. on Thursday
D)1 p.m. on Friday
E)None
Rationale: Solution
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 x4/5)min
= 34 / 5 min.in 170 hrs.
Now,34/5 min are gained in 170 hrs.
2 min are gained in(170x 5/34 x2)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 B
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 40 miles apart. One and a half hours later, they are again 40 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?
Options:
A)a) 80-x
B)b) 53.33-x
C)c) 80-2x
D)d) 120-x
E)e) 40-x/2
Rationale: 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 40 mi, then move away from each other another distance of 40 miles. That's a total distance of 80 miles in 1.5 hr, which gives a rate of:
R = (80 mi)/(1.5) = 53.33 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 = 53.33
y = 53.33 - x
Answer =(B)
Answer: B
Question: The angle between two hands at 3.45 is
Options:
A)110 degree
B)115 degree
C)112 1/2 degree
D)117 degree
E)157 1/2 degree
Rationale: theta degree=11/2m-30h
=11/2(45)-30(3)
=247.5-90
=157.5 or 157 1/2 degree
ANSWER:E
Answer: E
Question: A jeep takes 4 hours to cover a distance of 620 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
Options:
A)148 Kmph
B)152 Kmph
C)106 Kmph
D)103 Kmph
E)165 Kmph
Rationale: Time = 4
Distance = 620
3/2 of 4 hours = 4 * 3/2 = 6 Hours
Required speed = 620/6 = 103 Kmph
D)
Answer: D
Question: Find at what time between 8 and 9 o'clock will the hands of a clock be in the same straight line but not together.
Options:
A)10x11/10
B)10x10/11
C)11x10/11
D)12x10/11
E)None
Rationale: Solution
At 8 o'clock, the hour is at 8 and the minute hand is at 12,i.e. the two hands are 20 min. Spaces apart.
So, the minute hand gain	= (30 - 20)minute
55 minutes will be gained in 60 min.
10 minutes spaces will be gained in (60/55 x10) min.	= 10x10/11 min.
The hands will be in the same straight line but not together at 10x10/11 min.past 8.
Answer B
Answer: B
Question: On Sunday morning, Pugsley and Wednesday are trading pet spiders. If Pugsley were to give Wednesday four of his spiders, Wednesday would then have four times as many spiders as Pugsley does. But, if Wednesday were to give Pugsley five of her spiders, Pugsley would now have five fewer spiders than Wednesday had before they traded. How many pet spiders does Pugsley have before the trading game commences?
Options:
A)6
B)7
C)8
D)10
E)12
Rationale: If Pugsley were to give Wednesday four of his spiders, Wednesday would then have five times as many spiders as Pugsley does:
(w + 4) = 4(p - 4)
If Wednesday were to give Pugsley five of her spiders, Pugsley would now have five fewer spiders than Wednesday had before they traded:
p + 5 = w - 5
Solving gives p = 10 and w = 20.
Answer: D.
Answer: D
Question: To apply for the position of photographer at a local magazine, a photographer needs to include two or three photos in an envelope accompanying the application. If the photographer has pre-selected six photos representative of her work, how many choices does she have to provide the photos for the magazine?
Options:
A)32
B)35
C)36
D)40
E)42
Rationale: 6C2 + 6C3 = 15 + 20 = 35
The answer is B.
Answer: B
Question: If x is an integer and y = 3x + 2, which of the following CANNOT be a divisor of y?
Options:
A)4
B)5
C)6
D)7
E)8
Rationale: Just to add some more to Bunuel's explanation, if a number is a multiple of 3, it will be in a form:
n = 3*k, where k is an integer
And if a number is a multiple of 6, it will be in a form :
m = 6*l, where l is an integer
so m = 2 * 3 * l = (2*l) * 3 = 3 * p, where p is an integer.
In other words, the prime factorization of m must have 3 as a factor, if it's a multiple of 6, but the expression can't have that (as explained by Bunuel), so it can't be a multiple of 6 either
ANSWER:C
Answer: C
Question: A book store bought copies of a new book by a popular author, in anticipation of robust sales. The store bought 400 copies from their supplier, each copy at wholesale price W. The store sold the first 150 copies in the first week at 80% more than W, and then over the next month, sold a 100 more at 20% more than W. Finally, to clear shelf space, the store sold the remaining copies to a bargain retailer at 40% less than W. What was the bookstore’s net percent profit or loss on the entire lot of 400 books?
Options:
A)30% loss
B)10% loss
C)10% profit
D)20% profit
E)60% profit
Rationale: C.P. = 400W
S.P. = 1.8W*150 + 1.2W*100 + 0.6W*150 = 480W
Since, S.P. > C.P., there is net profit.
Profit = S.P. - C.P. = 480W - 400W = 80W
Profit % = (Profit/C.P.)*100 = (80W/400W)*100 = 20%
So, the correct answer is D.
Answer: D
Question: A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
Options:
A)1/4
B)1/3
C)2/3
D)6/7
E)8/9
Rationale: Total no of ways in which you can select a committee of 2 people out of 7 people is 7C2 ways= 21 ways
Requirement is at-least 1 preacher= 4C1*3C1+4C2 or 7C2-3C2=18 ways
Probability= 18/21 = 6/7
ANSWER:D
Answer: D
Question: If 73! has 16 zeroes at the end, how many zeroes will 80! have at the end?
Options:
A)16
B)17
C)18
D)19
E)20
Rationale: Prime Factoring: you need a 2 and a 5 to make a 10 (a "zero"), and there are TONS of 2's so let's skip these and focus on the (rarer) 5s:
80! = 1*2*3*4*5*6...*78*79*80
Since there are 80 consecutive numbers, there are 16 multiples of 5 in there, but if we're prime factoring, we need to remember that SOME multiples of 5 actually contain more than just one 5. Which? 25 comes to mind -- it's got two of them! So all the multiples of 25 actually contain two 5's (ie: 50 and 75)
So, to recap, we have 16 of them, plus 3 more (the additional 5's in 25, 50, and 75), so that makes 19, and since we have more than enough 2's, we know our number will have exactly 19 zeros at the end.
ANSWER:D
Answer: D
Question: Mary's income is 60 percent more than Tim's income, and Tim's income is 40 percent less than Juan's income. What percent of Juan's income is Mary's income?
Options:
A)124%
B)120%
C)96%
D)80%
E)64%
Rationale: Juan's income = 100 (assume);
Tim's income = 60 (40 percent less than Juan's income);
Mary's income = 96 (60 percent more than Tim's income).
Thus, Mary's income (96) is 96% of Juan's income (100).
Answer: C
Answer: C
Question: Y and Z start a business with Rs.3000 and Rs.6000 respectively. How should they share their profits at the end of one year?
Options:
A)4:1
B)1:3
C)1:4
D)1:5
E)1:2
Rationale: They should share the profits in the ratio of their investments.
The ratio of the investments made by Y and Z =
3000 : 6000 => 1:2 .
Answer:E
Answer: E
Question: If 100!/x is not an integer, which of the following could be the value of x?
Options:
A)5^24
B)7^16
C)11^9
D)13^6
E)17^6
Rationale: he answer will be a number which doesn't completely divide 100!
No of 5's in 100! = 24
No of 7's in 100! = 16
No of 11's in 100! = 9
No of 13's in 100! = 7
No of 17's in 100! = 5
There are 5 seventeens in 100! , however we are dividing by 17^6 , so this can not divide 100!
Hence our answer will be E
Answer: E
Question: P is two years older than Q who is twice as old as R. The total of the ages of P, Q and R is 25. How old is Q?
Options:
A)7.2 years
B)8.2 years
C)9.8 years
D)9.2 years
E)9.4 years
Rationale: Let age of R =x. Then,
Age of Q =2x
Age of P =2+2x
Total age of P,Q and R =25
⇒(2+2x)+2x+x=25
⇒5x=23
⇒x=23/5=4.6 years
B's age =2x=2×4.6=9.2 years
ANSWER:D
Answer: D
Question: A man can row with a speed of 20 kmph in still water. If the stream flows at 7 kmph, then the speed in downstream is?
Options:
A)27
B)23
C)20
D)27
E)12
Rationale: M = 20
S = 7
DS = 20 + 7 = 27
Answer: D
Answer: D
Question: A theater box office sold an average (arithmetic mean) of 65 tickets per staff member to a particular movie. Among the daytime staff, the average number sold per member was 75, and among the evening staff, the average number sold was 60. If there are no other employees, what was the ratio of the number of daytime staff members to the number of evening staff members?
Options:
A)1:2
B)1:4
C)5:11
D)1:3
E)13:15
Rationale: Deviation from the mean for the daytime staff = 75 - 65 = 10.
Deviation from the mean for the evening staff = 65 - 60 = 5.
Thus, the ratio of the number of daytime staff members to the number of evening staff members is 5:10 = 1:2.
The answer is A.
Answer: A
Question: Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Four years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
Options:
A)A)24
B)B)89
C)C)67
D)D)32
E)E)45
Rationale: Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then,	(5x + 4/4x + 4)=11/9
9(5x + 4) = 11(4x + 4)
45x + 36 = 44x + 44
45x - 44x = 44 - 36
x = 8.
Anand's present age = 4x = 32 years.
Answer:D
Answer: D
Question: If a man can cover 18 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?
Options:
A)288
B)162
C)878
D)168
E)243
Rationale: 18 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 18 * 18/5 * 15/4 km = 243 km.
Answer: E
Answer: E
Question: How many meaningful English words can be formed with the letters ESRO using each letter only once in each word?
Options:
A)1
B)3
C)4
D)8
E)9
Rationale: Explanation:
Meaningful words are ROSE, SORE and EROS.
Note the meaning of the words SORE and EROS with the help of dictionary.
Answer: B) Three
Answer: B
Question: If S is the product of all positive integers less than 31, than what is the greatest integer k for which S/18^k is an integer?
Options:
A)3
B)6
C)7
D)14
E)26
Rationale: 30! S must be divisible by (3*3*2)^k.
As I will find a lot of 2's among 30!'s factors I will focus on the 3's (that also occurs twice instead of only once as a factor in 18).
Every multiple of 3 will provide me with atleast one factor of 3. There aren't that many so I list them quickly.
3 6 9*
12 15 18*
21 24 27**
30
The stars are marked as they are multiples of 9* or 27** and thus provides two or three 3's respectively.
3=3 (occurs 10 times) +10
3*3=9 (occurs 3 times) +3(adding just one additional three each as they were not counted for in the multiples of 3)
3*3*3=27 (occurs 1 time) +1 (again, adding just one as they were previously not counted)
Total: 14.
For every k we add two 3's. So we must divide 14 by 2 to get the highest integer K before we receive a non-integer result.
Answer: 7 -> Answer choice: C.
Answer: C
Question: What is the ratio between perimeters of two squares one having 3 times the diagonal then the other?
Options:
A)3: 8
B)3: 6
C)3: 7
D)3: 1
E)3: 3
Rationale: d = 3d d = d
a√2 = 3d a√2 = d
a = 3d/√2 a = d/√2 => 3: 1
Answer:D
Answer: D
Question: Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
Options:
A)22
B)72
C)37
D)16
E)88
Rationale: Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes,they will together (30/2)+1=16 times
Answer: D) 16
Answer: D
Question: A train of length 150 metres takes 40.5 seconds to cross a tunnel of length 300 metres. What is the speed of the train in km/hr ?
Options:
A)13.33
B)26.67
C)40
D)66.67
E)50
Rationale: Explanation:
Speed = 150+300/40.5m/sec
=450/40.5=18/5km/hr = 40 km/hr.
ANSWER IS C
Answer: C
Question: The ratio of the number of boys and girls in a college is 7:8.If the percentage increase in the number of boys and girls be 20% and 10% respectively.What will be the new ratio?
Options:
A)22:21
B)19:22
C)23:22
D)22:19
E)21:22
Rationale: Let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number=(120% of 7x) and (110% of 8x)
==[(120/100)*7x] and [(110/100)*8x]
=(42x/5) and (44x/5)
The required ratio =(42x/5):(44x/5)
=21:22
Answer is option E
Answer: E
Question: Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 3. The numbers are
Options:
A)12, 24, 30
B)12, 24, 38
C)12, 24, 362
D)3, 6, 9
E)12, 24, 321
Rationale: Explanation:
Let the required numbers be x, 2x, 3x. Then, their H.C.F =x. so, x= 3
\inline \fn_jvn \therefore The numbers are 3, 6, 9
Answer: D) 3, 6, 9
Answer: D
Question: A miniature roulette wheel is divided into 10 equal sectors, each bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly determines the winning sector by settling in that sector. If the wheel is spun two times, approximately what is the probability that the product of the two winning sectors’ integers will be even?
Options:
A)88%
B)75%
C)67%
D)63%
E)50%
Rationale: The only way to have an odd product is if both integers are odd.
P(odd product) = 1/2*1/2 = 1/4
P(even product) = 1 - 1/4 = 3/4 = 75%
The answer is B.
Answer: B
Question: What is the lowest positive integer that is divisible by each of the integers 1 through 8, inclusive?
Options:
A)420
B)560
C)840
D)960
E)1080
Rationale: The integers are 1 , 2 , 3 , 4 , 5 , 6 , 7 and 8
Lowest positive integer that is is divisible by each of the integers 1 through 8
= 2^3 * 3 * 5 * 7
=8*3 *5 *7
=24 * 35
= 840
Answer C
Answer: C
Question: How many different subsets of the set {0, 1, 2, 3, 4, 5,6,} do not contain 0?
Options:
A)A.66
B)B.57
C)C.31
D)D.32
E)E.58
Rationale: Number of subset
Since we have 6 digits other than 0, we can take any numbers from the set of 6 to make a subset. Also it is a matter of selection and not arrangement.So we will consider combinations.
6c1+6c2+6c3+6c4+6c5+6c6=57
And one set is the NULL set having no elements in it so
57+1=58.
ANSWER E.
Answer: E