divelab/Sys2Bench huggingface dataset:
Question: A cylindrical water tower with radius 28 m and height 10 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π m3. From noon, how many hours will it take to drain the entire tank?
Options:
A)840 2/7
B)820 6/7
C)821
D)821 3/7
E)840
Rationale: initial volume = (3/4)×∏×28²×10 = 5880∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr
Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“
So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain
What i did initially was formed an equation 5880∏ = 7∏×n (n is the number of hrs) so ended up with 840. E
Answer: E
Question: Which of the following numbers has four distinct prime factors?
Options:
A)204
B)231
C)308
D)320
E)390
Rationale: 204=2^2*3*17
231=3*7*11
308=2^2*77
320=2^6*5
390=2*3*5*13
390 has 4 distinct prime factors
Answer E
Answer: E
Question: A new apartment complex purchased 60 toilets and 20 shower heads. If the price of a toilet is three times the price of a shower head, what percent of the total cost was the cost of all the shower heads?
Options:
A)9%
B)10%
C)11%
D)13%
E)15%
Rationale: LET PRICE OF SHOWER HEADS=X
THEN ,, ,, TIOLETS=3X
TOTALCOST=20*X+60*3X=200x
% OF TOTAL COST WAS THE COST OF ALL THE SHOWER HEADS=(20X/200X)*100=10%
ANSWER:B
Answer: B
Question: For the symbol, m”n=n^2−m for all values of m and n. What is the value of 4”2?
Options:
A)5
B)3
C)2
D)1
E)0
Rationale: 4”2= 4 - 4=0
Answer: E
Answer: E
Question: If 7 spiders make 7 webs in 7 days, then how many days are needed for 1 spider to make 1 web?
Options:
A)1
B)7
C)3
D)14
E)15
Rationale: Explanation :
Let, 1 spider make 1 web in x days.
More spiders, Less days (Indirect proportion)
More webs, more days (Direct proportion)
⇒ 7 × 1 × 7 = 1 × 7 × x
⇒x=7. Answer : Option B
Answer: B
Question: In a party every person shakes hands with every other person. If there were a total of 190 handshakes in the party then what is the number of persons present in the party ?
Options:
A)15
B)16
C)17
D)18
E)20
Rationale: Explanation :
Let the number of persons be n
âˆ´ total handshakes = nC2 = 190
n(n-1)/2 = 190 âˆ´ n = 20
Answer : Option E
Answer: E
Question: A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water?
Options:
A)7
B)8
C)5
D)6
E)3
Rationale: 30 --- 3 DS = 10
? ---- 1
18 ---- 3 US = 6
? ---- 1 M = ?
M = (10 + 6)/2
= 8
Answer:B
Answer: B
Question: A trader marks his articles 20% more than the cost price. If he allows 20% discount, then find his gain or loss percent?
Options:
A)6%
B)8%
C)5%
D)2%
E)4%
Rationale: Let CP of an article = RS. 100
MP= Rs. 120
Discount = 20%
SP = M[(100 - d%)/100] = 120(80/100) = Rs. 96
Clearly, the trader gets 4% loss.
Answer: E
Answer: E
Question: The difference between the place values of two threes in the numerical 40378273 is
Options:
A)29997
B)299997
C)297
D)0
E)None of them
Rationale: Required difference = (300000-3)
=299997.
Answer is B
Answer: B
Question: Find the C.I. on Rs. 15,625 for 9 months at 16% per annum compounded quarterly?
Options:
A)Rs.1978
B)Rs.1979
C)Rs.1951
D)Rs.1968
E)Rs.1929
Rationale: P = Rs. 15625, n = 9 months = 3 quarters, R = 16% p.a. per quarter.
Amount = [15625 * (1 + 4/100)3]
= (15625 * 26/25 * 26/25 * 26/25)
= Rs. 17576 C.I.
= 17576 - 15625 = Rs. 1951.
Answer:C
Answer: C
Question: Kiran, an eight years old boy has 27 toys. He gave 19 toys to his brother Gourav, while Gourav playing all but 6 got destroyed. While Kiran playing all but 3 got destroyed. Finally how many toys left with both of them?
Options:
A)9
B)1
C)8
D)76
E)2
Rationale: While Gourav playing 'All but 6 got destroyed' means 'All except 6 got destroyed'. i.e., 13 toys got destroyed. It means now Gourav has 6 toys. While Kiran is playing 'All but 3 got destroyed' means 'except 3 all are not working i.e; 3 toys are good working so total toys at both of them is 6 + 3 = 9.
Answer:A
Answer: A
Question: What is the units digit of (5!*5! + 6!*5!)/3?
Options:
A)4
B)3
C)2
D)1
E)0
Rationale: (5!*5! + 6!*5!)/3
=5!(5! + 6!)/3
=120 ( 120 + 720)/3
=(120 * 840)/3
= 120 * 280
Units digit of the above product will be equal to 0
Answer E
Answer: E
Question: A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
Options:
A)3 days.
B)2 days.
C)4 days.
D)1 days.
E)6 days.
Rationale: B's 10 day's work =(1/15 x 10)=2/3.
Remaining work =(1 -2/3)=1/3.
Now,	1/18	work is done by A in 1 day.
1/3	work is done by A in	(18 x 1/3)= 6 days.
Answer: Option E
Answer: E
Question: (x —1)(x + 3) > 0
(x +5)(x—4) < 0
Which of the following values of x satisfy both inequalities shown?
I. 3
II. -4
III. 8
IV. 5
Options:
A)I and II only
B)I and III only
C)II and III only
D)II and IV only
E)I, II, III, and IV
Rationale: (x - 1)(x + 3) > 0 --> x<-3 or x>1.
(x + 5)(x - 4) < 0 --> -5<x<4.
Among choices only -4 and 3 satisfy both inequalities.
Answer: A.
Answer: A
Question: If x < 0 and 0 < y < 1, which of the following has the greatest value W?
Options:
A)x^2
B)(xy)^2
C)(x/y)^2
D)x^2/y
E)x^2*y
Rationale: Given, x < 0 and 0 < y < 1
Let, x = -2 and y = 1/2
A. x^2 = (-2)^2 =4
B. (xy)^2 = (-2*1/2)^2 =1
C. (x/y)^2 = {-2/(1/2)}^2 = (-4)^2 =16
D. x^2/y = (-2)^2 / (1/2) = 4*2 =8
E. x^2*y = (-2)^2*(1/2) =2
Answer: option C
Answer: C
Question: Find the invalid no.from the following series 13, 18, 25, 30, 37, 40
Options:
A)23
B)25
C)40
D)46
E)48
Rationale: The differences between two successive terms from the beginning are 7, 5, 7, 5, 7, 5. So, 40 is wrong.
C
Answer: C
Question: A class average mark in an exam is 70. The average of students who scored below 60 is 50. The average of students who scored 60 or more is 75. If the total number of students in this class is 20, how many students scored below 60?
Options:
A)5
B)10
C)15
D)17
E)20
Rationale: Let n the number of students who scored below 60 and N the number of students who scored 60 or more. Xi the grades below 60 and Yi the grades 60 or above.
[sum(Xi) + sum(Yi)] / 20 = 70 : class average
sum(Xi) / n = 50 : average for less that 60
sum(Yi) / N = 75 : average for 60 or more
50n + 75N = 1400 : combine the above equations
n + N = 20 : total number of students
n = 4 and N = 16 : solve the above system
Answer: E
Question: In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and chuck are not seated next to each other?
Options:
A)72
B)48
C)56
D)76
E)96
Rationale: When the constraint on an arrangement says,Two people should not be seated together,we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit!
Let's see why....
These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.
Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
Answer :A
Answer: A
Question: The true discount on Rs. 1760 due after a certain time at 12% per annum is Rs. 160. The time after which it is due is :
Options:
A)6 months
B)8 months
C)9 months
D)10 months
E)None
Rationale: Solution
P.W = Rs. (1760 - 160) = Rs. 1600.
∴ S.I on Rs.1600 ar 12% is Rs. 160.
∴ Time = 100 x 160/1600x12 ) = 5/6years = ( 5/6x12 ) months = 10 months.
Answer D
Answer: D
Question: A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Options:
A)532 m
B)537 m
C)500 m
D)846 m
E)536 m
Rationale: Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (800 + x)/60 = 65/3
x = 500 m.
Answer:C
Answer: C
Question: Amithab's average expenditure for the January to June is Rs. 4200 and he spends Rs.1200 in January and Rs.1500 in july.The average expenditure for the months of febraury to July is
Options:
A)Rs. 4250
B)Rs. 4228
C)Rs. 4128
D)Rs. 4988
E)Rs. 4192
Rationale: Amithab's total expenditure for Jan - June = 4200 x 6 = 25200
Expenditure for Feb - June = 25200-1200 = 24000
Expenditure for the months of Feb - July = 24000 + 1500 =25500
The average expenditure = {25500}/{6}=4250
Answer: A
Answer: A
Question: A taxi driver states that his cab number is divisible by the no 2,3,4,5,6 with a remainder of 1 and when the number is divided by 11 it does not give any remainder
Options:
A)101
B)11
C)121
D)111
E)131
Rationale: on finding the LCM of 2,3,4,5 & 6 we get the value as 60 ,on adding 1 to 60 satisfy the first condition
Now 61 not divisible by 11
Then multiply 60 by 2,
we get 120: 120+1, 121 is exactly divisible by 11
ANSWER:C
Answer: C
Question: Company S produces two kinds of radios: basic and deluxe. Of the radios produced by Company S last month, 2/3 were basic and the rest were deluxe. If it takes 7/5 as many hours to produce a deluxe radio as it does to produce a basic radio, then the number of hours it took to produce the deluxe radio last month was what fraction of the total number of hours it took to produce all the radios?
Options:
A)7/17
B)2/3
C)8/17
D)6/17
E)3/4
Rationale: No of basic radios was 2/3 of total and # of deluxe radios was 1/3 of total, let's assume total=12, then Basic=8 and Deluxe=4.
Now, if time needed to produce one Deluxe rdio is 1 unit than time needed to produce one Basic radio would be 7/5 units. Total time for Basic would be 8*1=8 and total time for Deluxe would be 4*7/5=28/5 --> total time for both of them would be 8+28/5 -->68/5 deluxe/total=28/5 / 68/5 = 28/68 =7/17
Answer: A.
Answer: A
Question: A train 100 meters long completely crosses a 300 meters long bridge in 12 seconds. What is the speed of the train is?
Options:
A)32 kmph
B)76 kmph
C)120 kmph
D)43 kmph
E)40 kmph
Rationale: S = (100 + 300)/12
= 400/12 * 18/5
= 120
Answer:C
Answer: C
Question: John has on his shelf four books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work?
Options:
A)1/2
B)2/5
C)3/10
D)7/20
E)probability= 8/45
Rationale: For the denominator, we are going to pick two books from among ten total: a combination of two from ten. Again, we will use the formula:
nC2 = n(n-1)/2
which, for profound mathematical reasons we need not address here, is also the formula for the sum of the first (n – 1) positive integers. Here
10C2=10(9)/2=5*9=45
That’s the total number of pairs of books we could pick from the ten on the shelf. That’s our denominator.
Now, the numerator. We want one novel and one reference work. Well, there are four novels and two reference works, so by the FCP, the number of ways we can pick this is 4 x 2 = 8. That’s the total possible number of pairs involving exactly one of these four novels and exactly one of these two reference works. That’s our numerator.
probability = 8/45
Answer = E
Answer: E
Question: One student to pass the subject as probability 1/4. another man to get driving licence and student passed subject probability 1/6. find the probability man get driving licence.
Options:
A)1/3
B)2/3
C)4/3
D)5/3
E)1/4
Rationale: Let the probability of the man get driving licence is x.
so, 1/4 * x = 1/6
x = 2/3.
ANSWER:B
Answer: B
Question: In how many ways can be 5 boys and 5 girls sit around circular table so that no two boys sit next to each other?
Options:
A)(5!)^2
B)(6!)^2
C)5!4!
D)11!
E)(5!)^2*6!
Rationale: first fix one boy and place other 4 in alt seats so total ways is 4!
now place each girl between a pair of boys... total ways of seating arrangement of girls 5!
total is 5!*4!
ANS C
Answer: C
Question: R is the set of all positive integers x such that x^2 is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer x in R?
I. 9
II. 15
III. 27
Options:
A)I only
B)II only
C)I and II only
D)I and III only
E)I, II, and III
Rationale: Ans: C
R->{x} where x^2 is a multiple of both 27 and 375 means 3^3 and (5^3)*3
means x must contain 3^2 and 5^2
so with these conditions we know that 9=3^2 and 15=3*5 both have required factors for the divisibility of lowest int for x which is 9*25
but 27 is not a divisor because it can't divide 9*25 fully.
so Ans : C
Answer: C
Question: A waiter's salary consists of his salary and tips. During one week his tips were 5/4 of his salary. What fraction of his income came from tips ?
Options:
A)4/9
B)5/4
C)5/8
D)5/9
E)6/9
Rationale: Income = salary (s) +tips = s + s*5/4 = s*9/4
Tips = s*5/4
fraction of his income came from tips = (s*5/4)/(s*9/4)= 5/9
ANSWER:D
Answer: D
Question: What will come in place of (a) in the following expression?
a - 796.21 + 498.05 = 215.50 – 300.01
Options:
A)431.65
B)488.6
C)388.67
D)431.2
E)488.61
Rationale: Explanation:
a - 796.21 + 498.05 = 215.50 - 425.01
a = 413.65
ANSWER: A
Answer: A
Question: Mangala completes a piece of work in 30 days, Raju completes the same work in 45 days. If both of them work together, then the number of days required to complete the work is
Options:
A)18 days
B)6 days
C)8 days
D)10 days
E)11 days
Rationale: If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days.
That is, the required No. of days = 30 × 45/75 = 18 days.
Answer : A
Answer: A
Question: In what time will a train 120 m long cross an electric pole, it its speed be 144 km/hr?
Options:
A)2.5
B)3
C)4
D)5
E)6
Rationale: Speed = 144 * 5/18 = 40 m/sec
Time taken = 120/40 = 3 sec.
Answer: Option B
Answer: B
Question: 4 liters of a 20 percent solution of alcohol in water are mixed with 5 liters of a 10 percent alcohol in water solution. What is the percentage of alcohol in the new solution?
Options:
A)14.4%
B)14.8%
C)15.2%
D)15.6%
E)16.0%
Rationale: The total amount of alcohol is 0.2(4) + 0.1(5) = 1.3 liters.
The percentage is 1.3 / 9 = 13 / 90 which is about 14.4%
The answer is A.
Answer: A
Question: Amit and Ian paint a wall in alternating shifts. First Amit paints alone, then Ian paints alone, then Amit paints alone, etc. During each of his shifts, Amit paints 1/2 of the remaining unpainted area of the wall, while Ian paints 1/3 of the remaining unpainted area of the wall during each of his shifts. If Amit goes first, what fraction of the wall's area will remain unpainted after Amit has completed his 3th shift?
Options:
A)1/17
B)1/54
C)1/81
D)1/162
E)1/216
Rationale: Fraction of area unpainted after FIRST shift of Amit = 1-(1/2) = 1/2
Fraction of area unpainted after FIRST shift of Ian = (1/2)-(1/3)(1/2) = (2/3)(1/2)
Fraction of area unpainted after SECOND shift of Amit = (1/2)(2/3)(1/2)
Fraction of area unpainted after SECOND shift of Ian = (2/3)(1/2)(2/3)(1/2)
Fraction of area unpainted after THIRD shift of Amit = (1/2)(2/3)(1/2)(2/3)(1/2)=1/18
Answer: option A
Answer: A
Question: Company P had 15 percent more employees in December than it had in January. If Company P had 490 employees in December, how many employees did it have in January?
Options:
A) 391
B) 426
C) 410
D) 423
E) 445
Rationale: D = number of employees in December
J = number of employees in January
J x 1.15 = D
J x 1.15 = 490
J = 490/1.15
J = 49,000/115 = 426
Thus B is the correct answer.
Answer: B
Question: Today is Tuesday. A person wants to meet a lawyer and as that lawyer is busy he asks him to come three days after the before day of the day after tomorrow? On which day the lawyer asks the person to come?
Options:
A)Saturday
B)MONDAY
C)TUESDAY
D)WEDNESDAY
E)FRIDAY
Rationale: Today is Tuesday
after three days--->Friday
now, (Friday)after before day--->Thursday
then, (Thursday)day after tomorrow--->Saturday
ANSWER:A
Answer: A
Question: Two vessels A and B contain milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio 9:4
Options:
A)2:7
B)5:2
C)3:5
D)5:7
E)6:7
Rationale: Soln:-
Required Ratio = (5/7−9/13)/(9/13−8/13)
= 2/7
Answer: A
Answer: A
Question: There are 6 magazines lying on a table; 3 are fashion magazines and the other 3 are sports magzines. If 2 magazines are to be selected at random from the 8 magazines, what is the probability that at least one of the fashion magazine will be selected?
Options:
A)4/5
B)8/9
C)4/6
D)1/2
E)3/6
Rationale: 1- (3c2/6c2) = 1-1/5 = 4/5
Answer : A
Answer: A
Question: Two employees X and Y are paid a total of Rs. 660 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
Options:
A)s.150
B)s.200
C)s.250
D)s.350
E)s.300
Rationale: Let the amount paid to X per week = x
and the amount paid to Y per week = y
Then x + y = 660
But x = 120% of y = 120y/100 = 12y/10
âˆ´12y/10 + y = 660
â‡’ y[12/10 + 1] = 660
â‡’ 22y/10 = 660
â‡’ 22y = 6600
â‡’ y = 6600/22 = 600/2 = Rs.300
E
Answer: E
Question: A certain sum of money at simple interest amounted Rs.850 in 10 years at 3% per annum, find the sum?
Options:
A)651.85
B)652.85
C)653.85
D)654.85
E)655.85
Rationale: 850 = P [1 + (10*3)/100]
P = 653.85
Answer: C
Answer: C
Question: A person can swim in still water at 4 km/h. If the speed of water 2 km/h, how many hours will the man take to swim back against the current for 14km?
Options:
A)3
B)6
C)8
D)9
E)7
Rationale: M = 4
S = 2
US = 4 - 2 = 2
D = 14
T = 14/2 = 7
Answer:E
Answer: E
Question: 800 students took the test on English and Maths. 35% students failed in english and 45% failed in maths. 40% of those who passed in maths also passed in english, then how many students failed in both ?
Options:
A)162
B)138
C)16
D)38
E)12
Rationale: Passed in english = 65%
Passed in maths = 55%
Passed in both = 40% of 55% = 2/5 * (55%) = 22%
Passed in (English + Maths - Both + Neither) = 100%
65 + 55 - 22 + Neither = 100
Neither = 100 - 98 = 2% = 0.02 * 800 = 16
Answer: C
Answer: C
Question: If the selling price of 50 articles is equal to the cost price of 25 articles, then the loss or gain percent is:
Options:
A)45%
B)23%
C)20%
D)60%
E)50%
Rationale: C.P. of each article be Re. 1.
Then, C.P. of 50 articles = Rs. 50;
S.P. of 50 articles = Rs. 25.
Loss % = 25/50 * 100 = 50%
ANSWER E
Answer: E
Question: 3 women and a few men participated in a chess tournament. Each player played two matches with each of the other players. If the number of matches that men played among themselves is 78 more than those they played with the women, how many more men than women participated in the tournament?
Options:
A)20
B)15
C)11
D)10
E)19
Rationale: Let x be the number of men.
Number of matches men play among themselves is 2*(xC2)
Number of matches men play with women is 2*(3x)
2*(xC2) - 2*(3x) =78
x=13
Difference between men and women is 13-3=10.
ANSWER D
Answer: D
Question: A train 110 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: Speed of train relative to man = (60 + 6) km/hr = 66 km/hr
[66 * 5/18] m/sec = [55/3] m/sec.
Time taken to pass the man = [110 * 3/55] sec = 6 sec
Answer: Option B
Answer: B
Question: In the infinite sequence 50, 60, 70, 80, 90, . . . , where each term is 10 greater than the previous term, the 100th term is
Options:
A)1330
B)1807
C)1910
D)1040
E)2060
Rationale: T(n) = a + (n-1) d
Here d =10, a=50, n=100
Solving we get T(n) = 1040.
Answer: D
Answer: D
Question: If C.P. of 26 books is equal to S.P of 30 books , then your gain or loss is ?
Options:
A)23 1/3 gain %
B)30 10/23 loss %
C)13 1/3 loss %
D)30 1/3 loss %
E)40 1/3 loss %
Rationale: Say the cost price of 26 books is $26 so $1 for a book.
The cost of 30 books would be $30, and the selling price of 30 books would be 26$, hence the loss = ($30 - $26)/$30 = ~13.3%.
Answer: C.
Answer: C
Question: Beth received 3/12 of the votes cast in a certian election. What franction of the other votes cast would she have needed in order to have received 1/2 of the votes cast?
Options:
A)1/5
B)2/7
C)3/9
D)7/20
E)1/2
Rationale: Let the total votes be 12. Beth received 3 votes, other votes 12-3=9. In order to have received 1/2 of the votes she needed 6 votes so 3 more, which is 3/9 of the other votes.
Answer: C.
Answer: C
Question: Four barrels of water have an average capacity of 8 gallons. What is the maximum possible capacity of the smallest barrel Y if the median is 10 gallons?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Suppose 4 barrels are B1, B2, B3, B4 in ascending capacity.
(B1+B2+B3+B4)/4 = 8
B1+B2+B3+B4=32
median is 10 ....... as the no. of terms are even i.e 4, median = (B2+B3)/2
==> B2+B3 = 20
now we have B1+B4 = 32-20 =12
Only the following possible cases can be there :
1,9,11,11
1,10,10,11
2,10,10,10
Max value of smallest barrel Y= 2
Answer: B
Question: A circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside. Find the area of path.
Options:
A)1694 m2
B)1700 m2
C)1598 m2
D)1500 m2
E)None of thes
Rationale: Radius of a circular grass lawn (without path) = 35 m
∴ Area = πr2 = π(35)2
Radius of a circular grass lawn (with path)
= 35 + 7 = 42 m
∴ Area = πr2 = π(42)2
∴ Area of path = π(42)2 – π(35)2
= π(422 – 352) = π( 42 + 35)(42 – 35)
Answer A
Answer: A
Question: X starts a business with Rs.45000. Y joins in the business after 3 months with Rs.30000. What will be the ratio in which they should share the profit at the end of the year?
Options:
A)1:2
B)2:1
C)1:3
D)3:1
E)4:2
Rationale: Explanation:
Ratio in which they should share the profit
= Ratio of the investments multiplied by the time period
=45000Ã—12:30000Ã—9=45Ã—12:30Ã—9=3Ã—12:2Ã—9=2:1
ANSWER IS B
Answer: B
Question: In how many ways can three boys can be seated on five chairs?
Options:
A)30
B)40
C)60
D)80
E)90
Rationale: There are three boys.
The first boy can sit in any of the five chairs (5 ways).
5
Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).
5 4
Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).
5 4 3
Hence, total number of ways in which 3 boys can be seated on 5 chairs
=5×4×3=60
ANSWER C 60
Answer: C
Question: The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 600 resolutions?
Options:
A)844.8 m
B)704 m
C)774 m
D)714 m
E)744 m
Rationale: In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 600 resolutions.
= 600 * 2 * 22/7 * 22.4 = 84480 cm
= 844.8 m
Answer:A
Answer: A
Question: We have a 630 ml of mixture of milk and water in the ratio 7:2. How much water must be added to
make the ratio 7:3?
Options:
A)70 ml
B)60 ml
C)80 ml
D)50 ml
E)None of these
Rationale: Explanation :
concentration of water in mixture1 = 2/9 (Since the ratio of milk and water = 7:2) ---item(1)
concentration of water in pure water= 1 ---item(2)
Now the above mentioned items are mixed to form a mixture2 where milk and water ratio = 7: 3
=>concentration of water in mixture2 = 3/10
By the rule of alligation, we have
concentration of water
in mixture1 : 2/9 concentration of water
in pure water : 1
Mean concentration
3/10
1 - 3/10 = 7/10 3/10 - 2/9 = 7/90
=> Quantity of mixture1 : Quantity of water = 7/10 : 7/90 = 1/10 : 1/90 = 1 : 1/9
Given that Quantity of mixture1 = 630 ml
=> 630 : Quantity of water = 1 : 1/9
⇒Quantity of water = 630×19=70 ml
Answer : Option A
Answer: A
Question: When the integer y is divided by 12, the remainder is 3. Which of the following, when divided by 12, will have a remainder of 6 ?
I. 2y
II. 6y
III. 4y + 6
Options:
A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
Rationale: is it E.
y = 12*a + 3
2y = 12*2*a + 6
6y = 12*6*a + 18 ( 18 = 12+6)
4y + 6 = 12*4*a + 18 (18 = 12 + 6)
Answer: E