divelab/Sys2Bench huggingface dataset:
Question: Gary’s Gas Station serves an average of 16 cars per hour on Saturdays, 10 cars per hour on Sundays, and 9 cars per hour on all other days of the week. If the station is open from 6 a.m. to 10 p.m. every day, how many cars does Gary’s station serve over the course of a typical week?
Options:
A)1,136
B)1,200
C)1,240
D)1,280
E)1,320
Rationale: 6 a.m. to 10 p.m. = 16 hours
Number of cars serviced on weekdays = (16 * 9 * 5)
Number of cars serviced on Saturday = (16 * 16)
Number of cars serviced on Sunday = (16 * 10)
Number of cars served in a week = 16(45 + 16 + 10) = 16 * 71 = 1136
Answer: A
Answer: A
Question: Two pipes A and B can fill a cistern in 12 and 15 minutes respectively. Both are opened together but after 5 minutes A is turned off. After how much more time will the cistern be filled?
Options:
A)8 1/7
B)8 1/1
C)3 3/4
D)8 1/2
E)8 3/7
Rationale: 5/12 + (5 + x)/15 = 1
x = 3 3/4
Answer: C
Answer: C
Question: Along a yard 434 metres long, 32 trees are palnted at equal distances, one tree being at each end of the yard. What is the distance between two consecutive trees
Options:
A)14
B)16
C)10
D)11
E)12
Rationale: Explanation:
32 trees have 31 gaps between them,
Required distance (434/31) = 14
Option A
Answer: A
Question: A 70 cm long wire is to be cut into two pieces so that one piece will be 2/5th of the other, how many centimeters will the shorter piece be?
Options:
A)22
B)20
C)92
D)82
E)28
Rationale: 1: 2/5 = 5: 2
2/7 * 70 = 20.Answer:B
Answer: B
Question: Find the least number must be subtracted from 378461 so that remaining no.is divisible by 13?
Options:
A)4
B)5
C)7
D)6
E)8
Rationale: On dividing 378461 by 13 we get the remainder 5, so 5 should be subtracted
B
Answer: B
Question: Private Benjamin is a member of a squad of 10 soldiers, which must volunteer 4 of its members for latrine duty. If the members of the latrine patrol are chosen randomly, what is the probability T that private Benjamin will be chosen for latrine duty?
Options:
A)1/10
B)1/5
C)2/5
D)3/5
E)4/5
Rationale: Easier way : Desired probability = 1- 'excluded' probability
In this case, Excluded probability = probability of Benjamin not being a part of the 4 volunteers. We can choose 4 out of 9 remaining soldiers in 9C4 ways. total ways possible = 10C4.
Thus excluded probability = 9C4/10C4 = 3/5
Thus, the desired probability T= 1- 3/5 = 2/5. Thus C is the correct answer.
Answer: C
Question: Speed of a boat in standing water is 10kmph and speed of the stream is 2kmph. A man can rows to a place at a distance of 100km and comes back to the starting point. The total time taken by him is?
Options:
A)24hours
B)21hr 30min
C)20hr 50min
D)18hr 15min
E)22hrs
Rationale: Speed upstream = 8kmph
Speed downstream = 12kmph
Total time taken = 100/8 + 100/12 = 20 hours 50min
Answer is C
Answer: C
Question: A whale goes on a feeding frenzy that lasts for 9 hours. For the first hour he catches and eats x kilos of plankton. In every hour after the first, it consumes 3 kilos of plankton more than it consumed in the previous hour. If by the end of the frenzy the whale will have consumed a whopping accumulated total 270 kilos of plankton, how many kilos did he consume on the sixth hour?
Options:
A)33
B)47
C)50
D)53
E)62
Rationale: If you list the amount eaten each hour, you'll get an equally spaced list, increasing by 3 each hour. In any equally spaced list, the median equals the mean. Here, the mean is 270/9 = 30, so the median is also 30, and that is the amount eaten in the 5th hour. We need to add 3 to find the total eaten in the next hour, so the answer is 33.
option A
Answer: A
Question: If 8 men or 12 women can do a piece of work in 15 days, in how many days can the same work be done by 6 men and 11 women?
Options:
A)9 days
B)11 days
C)13 days
D)15 days
E)17 days
Rationale: 8men = 12women (i.e 2men = 3women)
12women 1 day work= 1/15
soln:
6men(9women)+ 11women=20women=?
1 women 1 day work =12*15= 1/180
so, 20 women work= 20/180=1/9
ans: 9 days
ANSWER:A
Answer: A
Question: The difference between the place values of 9 and 4 in the number 927435 is
Options:
A)899600
B)856200
C)875200
D)896200
E)None
Rationale: Sol.= (Place value of 9) – (place Value of 4)= (900000 - 400)= 899600 Answer A
Answer: A
Question: Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs. 73. What are the fares for cities B and C from A ?
Options:
A)Rs. 4, Rs. 23
B)Rs. 13, Rs. 17
C)Rs. 15, Rs. 14
D)Rs. 17, Rs. 13
E)Rs. 17, Rs. 15
Rationale: If fares for cities B and C from A are x and y respectively, then
2x+3y=77 or 6x+9y= 231
3x+2y =73 or 6x+4y= 146
subtracting
5y = 85
y= Rs 17.
Hence B. Rs. 13, Rs. 17
ANSWER:B
Answer: B
Question: Last year a worker saved 5% of her annual salary. This year, she made 20% more money than last year and she saved 6% of her salary. The amount she saved this year was what percent of the amount she saved last year?
Options:
A)136%
B)140%
C)144%
D)148%
E)152%
Rationale: Let x be the worker's salary last year.
Last year, she saved 0.05x.
This year, she saved 0.06(1.2x)=0.072x
0.072x/0.05x=144%
The answer is C.
Answer: C
Question: A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 liters per minute. When the tank is full in inlet is opened and due to the leak the tank is empties in 8 hours. The capacity of the tank is?
Options:
A)2777
B)5760
C)2766
D)8987
E)2681
Rationale: 1/x - 1/6 = -1/8
x = 24 hrs
24 * 60 * 4 = 5760
Answer: B
Answer: B
Question: An empty wooden vessel weighs 20% of its total weight when filled with paint. If the weight of a partially filled vessel is one half that of a completely filled vessel, what fraction of the vessel is filled.
Options:
A)3/5
B)5/9
C)1/24
D)9/19
E)2/5
Rationale: An empty wooden vessel weighs 20% of its total weight when filled with paint:
Vessel = 0.2(Vessel + Paint);
20V = V + P (so the weight of completely filled vessel is 10V)
P = 19V (so the weight of the paint when the vessels is completely filled is 19V).
The weight of a partially filled vessel is one half that of a completely filled vessel:
V + P' = 1/2*20V;
P' = 9V (so the weight of the paint when the vessels is partially filled is 9V).
What fraction of the vessel is filled?
So, we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled:
P'/P = 9V/19V = 9/19.
Answer: D.
Answer: D
Question: How many ways can a selection be done of 5 letters out of 5 A's, 4B's, 3C's, 2D's and 1 E.
Options:
A)60
B)75
C)71
D)121
E)221
Rationale: We have the following letters: {AAAAA}, {BBBB}, {CCC}, {DD}, {E}
There are 7 different cases of 5 letter selections possible:
(5) - all letters are alike - 1 way, all A's;
(4, 1) - 4 letters are alike and 1 different - C12∗C14=8C21∗C41=8, where C12C21 is # of ways to choose which letter provides us with 4 letters from 2 (A or B) and C14C41 is # of ways to choose 5th letter from 4 letters left;
(3, 2) - 3 letters are alike and other 2 are also alike - C13∗C13=9C31∗C31=9, where C13C31 is # of ways to choose which letter provides us with 3 letters from 3 (A, B or C) and C13C31 is # of ways to choose which letter provides us with 2 letters from 3 (for example if we choose A for 3 letters then we can choose from B, C or D for 2 letters);
(3, 1, 1) - 3 letters are alike and other 2 are different - C13∗C24=18C31∗C42=18, where C13C31 is # of ways to choose which letter provides us with 3 letters from 3 (A, B or C) and C24C42 is # of ways to choose which 2 letters provides us with one letter each;
(2, 2, 1) - 2 letters are alike, another 2 letters are also alike and 1 is different - C24∗C13=18C42∗C31=18, where C24C42 is # of ways to choose which 2 letters provides us with 2 letters from 4 (A, B, C or D) and C13C31 is # of ways to choose which provides us with 5th letter from 3 letters left;
(2, 1, 1, 1) - 2 letters are alike and other 3 are different - - C14∗C34=16C41∗C43=16, where C14C41 is # of ways to choose which letter provides us with 2 letters from 4 (A, B, C or D) and C34C43 is # of ways to choose which 3 letters out of 4 provides us with one letter each;
(1, 1, 1, 1, 1) - all letters are distinct - 1 way (A, B, C, D, E).
Total: 1+8+9+18+18+16+1=71.
Answer: C.
Answer: C
Question: There are 100 lights which are functional and each is controlled by a separate On/Off switch. Two children A and B start playing with the switches. A starts by pressing every third switch till he reaches the end. B, thereafter, presses every fifth switch till he too reaches the end. If all switches were in Off position at the beggining, How many lights are switched On by the end of this operation?
Options:
A)13
B)41
C)47
D)48
E)53
Rationale: Number of switches = 100
Number of switches turned on by A: 3, 6, ... 99 = 33
Number of switches turned on by B: 5, 10, .... 100 = 20
Few switches are turned on by A and later turned off by B: LCM(3,5) = 15x = 15, 30,....90 = 6.
Subtract the above 6 switches from both A and B as they are turned off.
Number of switches that are turned on = (33 - 6) + (20 - 6) = 41
Answer: B
Answer: B
Question: ((-1.9)(0.6) – (2.6)(1.2))/7.0 = ?
Options:
A)-0.61
B)1.0
C)1.07
D)1.71
E)2.71
Rationale: Dove straight into calculation but quickly realized that the sum of two negatives is a negative so there is only one option.
-0.61
Answer A
Answer: A
Question: Trapezoid JKLM in the x-y plane has coordinates J = (–2, –5), K = (–2, 1), L = (6, 7), and M = (6, –5). What is its perimeter?
Options:
A) 34
B) 35
C) 38
D) 40
E)(F) 42
Rationale: JK = 6
LM = 11
KL =using distance formula 10
JM = using distance formula 8
Sum of all is 35
B
Answer: B
Question: According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 28, by how much does the temperature in degrees Celsius (C) increase?
Options:
A)9
B)15.55
C)47
D)48 3/5
E)59
Rationale: You can plug in values.
C = 5/9*(F-32)
F=32 --> C=0;
F=32+28=60 --> C=5/9*28=15.55.
Increase = 15.55 degrees.
Answer: B.
Answer: B
Question: Look at this series: 90, 87, 92, 89, 94, 91, ... What number should come next?
Options:
A)94
B)96
C)90
D)98
E)92
Rationale: Explanation:
In this simple alternating subtraction and addition series; 3 is subtracted, then 5 is added, and so on.
Answer: Option B
Answer: B
Question: Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made?
Options:
A)Rs. 16.58
B)Rs. 16.51
C)Rs. 16.59
D)Rs. 16.50
E)Rs. 16.52
Rationale: Explanation:
Let the quantities of A and B mixed be 3x kg and 7x kg.
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
Answer:D
Answer: D
Question: The manager of a theater noted that for every 10 admissions tickets sold, the theater sells 3 bags of popcorn at 2.00 $ each, 4 sodas at 1.50 $ each and 2 candy bars at 1.00$ each. To the nearest cent, what is the average (arithmetic mean) amount of these snacks sales per ticket sold?
Options:
A)1.40$
B)1.58$
C)1.60$
D)1.64$
E)1.70$
Rationale: For every 10 tickets amount of snacks sold is 3*2+4*1.5+2*1=$14, hence amount of the snacks sales per ticket is $14/10=~$1.4.
Answer: A.
Answer: A
Question: A shop sells bicycles and tricycles. In total there are 9 cycles (cycles include both bicycles and tricycles) and 19 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels.
Options:
A)b=8,t=1
B)b=5,t=3
C)b=6,t=2
D)b=2,t=5
E)b=3,t=4
Rationale: Let b be the number of bicycles and let t be the number of tricycles.
Set up the equations
b+t=9...........(1)
2b+3t=19............(2)
Rearrange equation (1) and substitute into equation (2)
2b+27-3b=19
-b=-8
b=8
Calculate the number of tricycles t
t=9−b
=9−8
=1
There are 1 tricycles and 8 bicycles.
Answer is A.
Answer: A
Question: In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a four or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
Options:
A)1/3
B)4/9
C)5/9
D)2/3
E)3/4
Rationale: We're looking for the number of outcomes that include AT LEAST a 1 or a 4.
#getting a one in 1st die = {(1,1), (1,2), ...(1,6) }= 6
#getting a one in 2nd die {(2,1), ...(5,1),(6,1) } = 4 [exclude 1,1 and 4,1] since it will be counted twice
#getting a three in 1st die {(3,1), (3,2), ...(3,6) }= 6
#getting a three in 1st die {(2,4), (3,4), ...(6,4) } = 4 [exclude 1,4 and 4,4] since it will be counted twice
Total possibities = 20
Probability of rolling at least a 1 or a 4 on two dice: 20/36 = 5/9
C
Answer: C
Question: A man whose speed is 4.5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph, find his average speed for the total journey?
Options:
A)5
B)4
C)8
D)9
E)2
Rationale: M = 45
S = 1.5
DS = 6
US = 3
AS = (2 * 6 * 3) /9 = 4
Answer: B
Answer: B
Question: In how many different number of ways 4 boys and 3 girls can sit on a bench such that girls always sit together.
Options:
A)720
B)740
C)760
D)790
E)800
Rationale: 720
Option 'A'
Answer: A
Question: A man buys a field of agricultural land for Rs. 3,60,000. He sells one-third at a loss of 20% and two-fifths at a gain of 25%. At what price must he sell the remaining field so as to make an overall profit of 10%?
Options:
A)Rs.1,00,000
B)Rs. 1,15,000
C)Rs. 1,20,000
D)Rs. 1,25,000
E)None of these
Rationale: Solution: First Method
CP = 3600000.
To gain 10% on whole land, SP = 360000 + 10% of 360000 = Rs. 396000.
1/3 of the land sold on 20% loss.
SP of 1/3 land = (360000/3) - 20% of (360000/3) = Rs. 96000.
SP of 2/5 of the land = (360000 * 2)/5 + 25% of (360000*2)/5 = Rs. 180000.
Thus, SP of the remaining land = 396000 - 96000 - 180000 = Rs. 120000.
Answer: Option C
Answer: C
Question: Two pens and three pencils cost $86. four Pens and a pencil cost $112. find the cost of a pen and that of a pencil.
Options:
A)25, 12
B)12, 30
C)30, 45
D)50, 48
E)15, 25
Rationale: Let the cost of a pen and a pencil be $ X and $ Y respectively.
Then, 2x + 3y = 86 ….(i) and 4x + y =112.
Solving (i) and (ii), we get: x = 25 and y = 12.
Cost of a pen =$ 25 and the cost of a pencil =$ 12.
Answer A.
Answer: A
Question: Which interchange of signs will make the following equation correct ?
35 + 7 × 5 ÷ 5 – 6 = 24
Options:
A)24
B)23
C)22
D)26
E)28
Rationale: A
24
35 ÷ 7 × 5 +5 – 6 = 24
5 × 5 + 5 – 6 = 24
25 – 1 = 24
24 = 24
Answer: A
Question: Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
Options:
A)42, 33
B)44, 36
C)44, 38
D)45, 40
E)44, 48
Rationale: Let the marks secured by them be x and (x + 9)
Then sum of their marks = x + (x + 9) = 2x + 9
Given that (x + 9) was 56% of the sum of their marks
=>(x+9) = 56/100(2x+9)
=>(x+9) = 14/25(2x+9)
=> 25x + 225 = 28x + 126
=> 3x = 99
=> x = 33
Then (x + 9) = 33 + 9 = 42
Hence their marks are 33 and 42
Option 'A'
Answer: A
Question: At what price must an Book costing $47.50 be marked in order that after deducting 15% from the list price. It may be sold at a profit of 25% on the cost price?
Options:
A)72.85
B)69.85
C)62.85
D)82.85
E)60.85
Rationale: C
$62.50
CP = 47.50
SP = 47.50*(125/100) = 59.375
MP*(85/100) = 59.375
MP = 69.85
B
Answer: B
Question: The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 85 and Rs. 80 respectively. The sum is:
Options:
A)1360
B)1450
C)1600
D)1800
E)None
Rationale: Sol.
Sum = S.I. * T.D. / (S.I)-(T.D.) = 85 * 80 / (85 - 80) = Rs.1360.
Answer A
Answer: A
Question: Louie takes out a three-month loan of $2000. The lender charges him 10% interest per month compounded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?
Options:
A)A)333
B)B)383
C)C)402
D)D)433
E)E)804
Rationale: Here's the calculation for that case, assume monthly payment is X.
After 1st month: (2000)(1.1)-X = 2200-X
After 2nd month: (2200-X)(1.1)-X = 2420-2.21X
After 3rd month: (2420-2.21X)(1.1)-X = 2662-3.31X
Now, the amount after the last payment in 3rd month must bring the total to 0. Hence:
2662-3.31X = 0
X = 2662/3.31 = 804
Answer E
Answer: E
Question: If r=1 and (r-b)/c = 1 which of the following is NOT a possible value of b?
Options:
A)-2
B)-1
C)0
D)1
E)2
Rationale: Equation is:
(r-b)/c = 1 where r = 1
so plug in 1 for r and look at it again
(1-b)/c = 1
now C can be anything we want, so if b = -2 we can plug it in and make it work
(1-(-2))/c = 1
3/c = 1
c = 3 and it works
HOWEVER, if we have a 0 in the numerator there isn't a number on earth that makes 0/C = 1
(1-1)/c = 1
0/c = 1
there aren't any matches for C to make this true
B cannot be 1.
Answer is D
Answer: D
Question: A bag consists of 1 rupee, 50 paise and 25p coins and their total value in rupees is Rs 1120 .These three types of coins are in the ratio 3:20:4. How may 25 paisa coins are there?
Options:
A)30 coins
B)330 coins
C)320 coins
D)32 coins
E)10 coins
Rationale: The ratio is 100p: 50 p : 25 p =4 : 2: 1
The coins are in the ratio = 3 :20: 4
The ratio of their values in rupees=> 12 :40 :4
The value of 25 p coins in total Rs 1120 = 1120 x 4/56= 80 rupees
Number of 25 p coins are = 80x4 = 320 coins
Answer:C
Answer: C
Question: Anintegratedteam consists of 2 members from the engineering team and 2 members from the design team. If the engineering team has 8 members and the design team has 6 members, how manyintegratedteam formations are possible?
Options:
A) 36
B) 54
C) 176
D) 220
E) 360
Rationale: 2 out of 8 times 2 out of 6 => 8*7/2*1 * 6*5/2*1 = 360
E
Answer: E
Question: In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
Options:
A)30%
B)70%
C)100%
D)250%
E)300%
Rationale: cp = 100
profit = 320 ans SP = 420
new CP = 125% of RS 100 = Rs 125
New SP = 420
profit = (420-125) = 295
req % (295/420*100)% = 1475/21% = 70%
ANSWER B
Answer: B
Question: Tn = 2Tn-1 + 4 and Qn = 4Qn-1 + 8 for all n > 1. If T5 = Q4 and T7 = 316, what is the first value of n for which Qn is an integer?
Options:
A) 1
B) 2
C) 3
D) 4
E) 5
Rationale: Assuming I have understood the symbols used correctly answer is C- 3
T7 = 2S6 +4 = 316
T6 = 312/2 = 156 = 2T5 + 4
T5 = 152/2 = 76 = Q4
Q4 = 4Q3 + 8 = 76
Q3 = 68/4 = 17
Q3 = 4Q2 + 8 = 17
Q2 = 9/4 = 2.25
Q2 = 4Q1 + 8
Q1 will not be integer
Answer: C
Question: A train 55 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)7 sec
B)5 sec
C)4 sec
D)6 sec
E)3 sec
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
= [55 * 3/55] sec
= 3 sec
Answer:E
Answer: E
Question: A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that exactly three bulbs are good?
Options:
A)20/69
B)20/63
C)20/62
D)20/29
E)20/61
Rationale: Required probability
= (⁵C₃ . ⁴C₁)/⁹C₄
= (10 * 4)/126
= 20/63
Answer: B
Answer: B
Question: In how many ways the letters of the word RAM be arranged?
Options:
A)8
B)12
C)10
D)6
E)120
Rationale: Total ways of arranging the letters = 3! = 6 ways.
D)
Answer: D
Question: Country X imported approximately $1.44 billion of goods in 1996. If Country X imported $388 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year, by how much would Country Xs 1997 imports exceed those of 1996?
Options:
A) $24 million
B) $120 million
C) $144 million
D) $240 million
E) $888 million
Rationale: Convert units to millions as answer is in millions
1996 Imports = $1.44 bill = $1440 mill
i.e. 1440/12 = $120 mill/month
1997 Imports = $388 mill/2 month
i.e. $194 mill/month
Difference/month = 194-120 = 74
Difference/year = $74 mill * 12 = $988 mill
ANSWER: E
Answer: E
Question: A train 360 m long runs with a speed of 45 km/hr. What time will it take to pass a platform of 140 m long?
Options:
A)38 sec
B)35 sec
C)44 sec
D)40 sec
E)50 sec
Rationale: Explanation:
Speed = 45 km/hr = 45×(10/36) m/s
= 150/12 = 50/4 = 25/2 m/s
Total distance = length of the train + length of the platform
= 360 + 140 = 500 meter
Time taken to cross the platform = 500/(25/2) = 500×2/25 = 40 seconds
Answer: Option D
Answer: D
Question: In a student's election between two candidates, 75% of the voters cast thierthier votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.
Options:
A)16800
B)1600
C)1680
D)1700
E)17500
Rationale: Let the number of votes enrolled be x. Then ,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x] =9261.
X = [(9261*100*100*100)/(75*98*75)] =16800.
ANSWER A
Answer: A
Question: A certain manufacturer of cake, muffin, and bread mixes has 80 buyers, of whom 50 purchases cake mix, 40 purchase muffin mix, and 20 purchase both cake mix and muffin mix. If a buyer is to be selected at random from the 100 buyers, what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix?
Options:
A)a. 1/10
B)b. 3/8
C)c. 1/2
D)d. 7/10
E)e. 9/10
Rationale: C+M+B-CM-MB-CB-2CMB = 80
C - cake buyers, M - muffin and B - Bread buyers. CM,MB,CB and CMB are intersecting regions.
the question asks for people who have bought only bread mixes = B-CB-MB-2CMB has to be found out.
50 + 40 + B-CB-MB-20-2CMB = 80
B-CB-MB-2CMB = 30
hence the probability = 30/80 = 3/8.
B
Answer: B
Question: A man cycling along the road noticed that every 10 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?
Options:
A)5 minutes
B)6 minutes
C)8 minutes
D)9 minutes
E)48/7 minutes
Rationale: Let's say the distance between the buses is d. We want to determine Interval=\frac{d}{b}, where b is the speed of bus.
Let the speed of cyclist be c.
Every 10 minutes a bus overtakes cyclist: \frac{d}{b-c}=10, d=10b-10c;
Every 4 minutes cyclist meets an oncoming bus: \frac{d}{b+c}=4, d=4b+4c;
d=10b-10c=4b+4c, --> b=7c/3, --> d=12b-12b*3/7=48/7b.
Interval=\frac{d}{b}=\frac{48/7b}{b}=48/7
Answer: E (48/7 minutes).
Answer: E
Question: The H.C.F of two numbers is 8. Which of the following can never be their L.C.M?
Options:
A)55
B)44
C)66
D)sixty (60)
E)12
Rationale: H.C.F of two numbers divides their L.C.M exactly. Clearly, 8 is not a factor of 60.
Answer:D
Answer: D
Question: A lockbox has a 4 digit number passcode. How many different passwords are possible if repetition of numbers is allowed?
Options:
A)40
B)10000
C)1000
D)100
E)10
Rationale: 10 x 10 x 10 x 10 = 10,000
Answer: B
Answer: B
Question: A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.
Options:
A)70.11 km/hr
B)71.11 km/hr
C)72.11 km/hr
D)73.11 km/hr
E)None of these
Rationale: Explanation:
We know Time = Distance/speed
So total time taken =
(160/64+160/80)=9/2hours
Time taken for 320 Km = 320∗2/9=71.11km/hr
Option B
Answer: B
Question: A man saves Rs 300 at the end of each year and lends the money at 5% compound interest. How
much will it become at the end of 3 years.
Options:
A)Rs 662
B)Rs 333.01
C)Rs 663.02
D)Rs 993.03
E)None of these
Rationale: Explanation:
=[300(21/20×21/20×21/20)+300(21/20×21/20)+300(21/20)]=993.03
Answer: D
Answer: D
Question: The prime factorization of integer N is A x A x B x C, where A, B and C are all distinct prime integers. How many factors does N have?
Options:
A)11
B)12
C)14
D)16
E)18
Rationale: if N=ap+br+cr
then prime factor of N will be: (p+1)(q+1)(r+!)
here
p=2,q=1,r=1
(2+1)(1+1)(1+1)
3*2*2
ANSWER:B
Answer: B