RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 600 m long?
A.176 Sec
B.180 Sec
C.178 Sec
D.267 Sec
Answer: B.180 Sec
Explanation: L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1800 m
T = D/S
T = 1800/10
T = 180 Sec
Answer:B
Question: A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 25 percent of the pack would be outfielder cards?
A.10
B.5
C.6
D.7
Answer: A.10
Explanation: let the number of O card to be removed=X
then remaning cards=12-X
now this 12-X cards should count less than 25% of toatal Ocards
12-X<.25*12
X>12-3
X>9
X=10
Ans A
Question: The average of the marks of 10 students in a class is 80. If the marks of each student are doubled, find the new average?
A.80
B.120
C.160
D.270
Answer: C.160
Explanation: Sum of the marks for the 10 students
= 10 * 80= 800. The marks of each student are doubled, the sum also will be doubled.
The new sum = 800 * 2 = 1600. So, the new average
= 1600/10 = 160.
Answer:C
Question: What least number must be subtracted from 427398 so that remaining no.is divisible by 11
A.3
B.4
C.6
D.7
Answer: B.4
Explanation: Explanation:
On dividing 427398 by 11 we get the remainder 4, so 4 should be subtracted
Answer: Option B
Question: Water tax is increased by 20% but its consumption is decreased by 20%. Then, the increase or decrease in the expenditure of the money is :
A.No change
B.5% decrease
C.4% increase
D.4% decrease
Answer: D.4% decrease
Explanation: Explanation:
Let tax=Rs.100 and
Consumption=100 units
Original expenditure = Rs.(100×100)=Rs.10000(100×100)=Rs.10000
New expenditure = Rs.(120×80)=Rs.9600(120×80)=Rs.9600
Decrease in expenditure
= (400/10000×100)%=4%
Correct Option: D
Question: A shopkeeper buys mangoes at the rate of 5 a rupee and sells them at 3 a rupee. Find his net profit or loss percent?
A.66 2/3 %
B.33 7/3 %
C.33 1/3 %
D.33 5/8 %
Answer: A.66 2/3 %
Explanation: The total number of mangoes bought by the shopkeeper be 15.
If he buys 5 a rupee, his CP = 3
He selling at 3 a rupee, his SP = 5
Profit = SP - CP = 5 - 3 = 2
Profit Percent = 2/3 * 100 = 66 2/3 %
Answer: A
Question: For real number x, int(x) denotes integer part of x.int(x) is the largest integer less than or equal to x.int(1,2)=1,int(-2,4)=-3. Find the value of int(1/2)+int(1/2+ 100)+int(1/2+2/100)+....+int(1/2+99/100)
A.150
B.98
C.278
D.277
Answer: A.150
Explanation: int (1/2) = 0
int (1/2 + 100 ) = 100
into (1/2 + 2/100) = 0
......
int ( 1/2 + 50/100 ) = 1
int (1/2 + 51 /100) = 1
.......
int (1/2 + 99/100) = 1
So 100 + 1 + 1 + .....50 times = 150
Answer:A
Question: when fair coin is tossed 9 times find the probability of getting head at least once.
A.1/512
B.511/512
C.9/512
D.503/51
Answer: B.511/512
Explanation: 1-1/2^9= 1-1/512 = 511/512
ANSWER:B
Question: A certain sum amount to $1400 in 2 yrs and $2000 in five years on certain fixed simple interest. In how many yrs would the amount be $2040 had it been put at the rate of interest compounded annually?
A.5
B.4
C.3
D.2
Answer: C.3
Explanation: We can form equations as following:
Initial amount = P, Total amount after interest = T, Interest per year = I
T = P+I
Given T1 = P+2I = 1400T2 = P+ 5I = 2000
From this we get, P = 1000I = 200 $
Now we know that simple interest is<Compound interest ( CI = SI only after 1st year if CI is compounded annually)
consider simple interest - after 1st yer it is 200$ , after 2nd years it is 400$ , after 3rd year it is 600$.and after 4rd year it is 800$.
To get a 640 $ compound interest the duration have to be > 2 year and < 4 years.
Only option fitting : Option C.
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.57600?
A.81000
B.81028
C.27772
D.72900
Answer: D.72900
Explanation: 57600* 9/8 * 9/8 = 72900.Answer:D
Question: In what time will a train 100 meters long cross an electric pole, if its speed is 360 km/hr
A.1 second
B.4.5 seconds
C.3 seconds
D.2.5 seconds
Answer: A.1 second
Explanation: Explanation:
First convert speed into m/sec
Speed = 360*(5/18) = 100 m/sec
Time = Distance/speed
= 100/100 = 1 second
Answer: A
Question: The tax on a commodity is diminished by 25% and its consumption increased by 13%. The effect on revenue is?
A.18
B.16
C.10
D.15
Answer: D.15
Explanation: 100 * 100 = 10000
75 * 113 = 8475
-----------
10000-----------1525
100-----------? => 15% decrease
Answer:D
Question: Two employees A and B are paid a total of Rs. 550 per week by their employer. If A is paid 120 percent of the sum paid to B, how much is B paid per week?
A.Rs.220
B.Rs.250
C.Rs.260
D.Rs.270
Answer: B.Rs.250
Explanation: Let the amount paid to A per week = x
and the amount paid to B per week = y
Then x + y = 550
But x = 120% of y = 120y/100 = 12y/10
∴12y/10 + y = 550
⇒ y[12/10 + 1] = 550
⇒ 22y/10 = 550
⇒ 22y = 5500
⇒ y = 5500/22 = 500/2 = Rs.250
B
Question: Machine A takes 100 hours to complete a certain job and starts that job at 7AM. After two hour of working alone, machine A is joined by machine B and together they complete the job at 12PM. How long would it have taken machine B to complete the jobif it had worked alone for the entire job?
A.100.11
B.10.98
C.18.07
D.45.01
Answer: C.18.07
Explanation: Let us assume total job = 100 units
A finishes 100 units in 100 hrs(given)
Hence A (working rate)= 1 units /hr
Now Given that A works for 2 hr( so 2 units done) then A and B finish total work in 15 hours. Hence A and B finish 98 units in 15 hours. Of these 1x15=15 units were done by A. Hence B did 83units in 15 hours.
Hence B(working rate)= 83/15 units /hr
Hence B takes 100x15/83 = 18.07 hours to complete the job.
Answer C.
Question: A certain university will select 2 of 6 candidates eligible to fill a position in the mathematics department and 3 of 5 candidates eligible to fill 2 identical positions in the computer science department. If none of the candidates is eligible for a position in both departments, how many different sets of 3 candidates are there to fill the 3 positions?
A.42
B.70
C.150
D.165
Answer: C.150
Explanation: 6C2*5C3 = 15*10 = 150
the answer is (C)
Question: If 0 < L < Y, L is an odd number and Y is a prime number, which of the following can be the value of L + Y ?
A.11.
B.13.
C.17.
D.10.
Answer: D.10.
Explanation: If L is odd ,implies l(odd)+y(odd/even prime)=(Even/Odd)
If Y is even, the 2 is the even prime , therefore y=2l=1 since 0<L<Y therefore L+Y =3. But 3 is not an option
which implies y is odd,
Now L(odd)+Y(odd)=Even
10 is the only option that is even=D
Question: An amount of Rs.2,000 is to be distributed amongst P, Q, R and S such that “P” gets twice as that of “Q” and “S” gets four times as that of “R”. If “Q” and “R” are to receive equal amount, what is the difference between the amounts received by S and P?
A.2378
B.268
C.269
D.500
Answer: D.500
Explanation: Explanation:
We have, P = 2Q & S = 4R
Further Q = R & P + Q + R + S = 2,000
Thus we get, 2Q + Q + Q + 4Q = 2,000
8Q = 2,000 or Q = Rs. 250
Thus, R = Rs. 250, P = 500 & S = Rs. 1000
Hence, the required difference = (S – P) = (1000 – 500) = Rs. 500
ANSWER:D
Question: Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50%. What is his current speed Z?
A. 6
B. 8
C. 10
D. 12
Answer: B. 8
Explanation: Let the current speed be x miles per hour.
Time taken if speed is 50% faster (i.e. 3x/2 = 1.5x) = 96/1.5x
Time taken if speed is reduced by 4 miles/hr (i.e. (x-4)) = 96/(x-4)
As per question, 96/(x-4) - 96/1.5x = 16
Solving this Z we get x = 8.
B.
Question: A 180 meter long train crosses a man standing on the platform in 6 sec. What is the speed of the train?
A.48 kmph
B.52 kmph
C.108 kmph
D.100 kmph
Answer: C.108 kmph
Explanation: S = 180/6 * 18/5 = 108 kmph
answer :C
Question: For some integer y, y^2 - 5 is divisible by all of the following EXCEPT
A.29
B.30
C.31
D.38
Answer: B.30
Explanation: Hint: y^2-5 (y is an integer) is never multiple of 3 (try to prove this), hence 30 is out.
Answer: B
Question: A searchlight on top of the watchtower makes 2 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?
A.5/6
B.5/12
C.7/12
D.11/12
Answer: A.5/6
Explanation: The searchlight completes one revolution every 30 seconds.
The probability that the man's area will be lit up is 5/30=1/6.
The probability that he will stay in the dark is 1-1/6=5/6
The answer is A.
Question: The height of the wall is 4 times its width and length of the wall is 3 times its height .if the volume of the wall be 10368 cu.m.its width is
A.4m
B.5m
C.6m
D.7m
Answer: C.6m
Explanation: Explanation:
Let width = x
Then, height=4x and length=12x
12x Ã— 4x Ã— x = 10368
x = 6
Answer: C
Question: A military garrison in a wild frontier has 500 men who have provisions to last 26 weeks. At the end of 6 weeks they are joined by another 300 men. How long will the provisions last?
A.13 1/3
B.12 1/3
C.11 3/3
D.12 1/2
Answer: D.12 1/2
Explanation: Total provisions=500*26*7=91000
Provisions used for in 6 weeks=6*7*500=21000
remaining provisions=70000
remaining provisions need to be split between 800(500+300) people.
Number of provisions per person= 70000/800
we need to find the answer in terms of weeks=(70000/800)/7=25/2
Answer:D
Question: A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
A.1000
B.3783
C.1837
D.3799
Answer: A.1000
Explanation: 90%
104%
--------
14% ---- 140
100% ---- ? =>
Rs.1000
Answer: A
Question: A family has two children. find the probability that both the children are girls given that at least
one of them is a girl?
A.1/2
B.1by3
C.3/5
D.3/7
Answer: B.1by3
Explanation: Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3
B)
Question: Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 25 hours respectively. The ratio of their speeds is
A.4:1
B.4:2
C.3:5
D.5:3
Answer: D.5:3
Explanation: Ratio of their speeds = Speed of first train : Speed of second train
= √25−−√9
= 5:3
Answer is D.
Question: A sum was put at simple interest at a certain rate for 6 years Had it been put at 4% higher rate, it would have fetched 144 more. Find the sum.
A.500
B.600
C.700
D.800
Answer: B.600
Explanation: Difference in S.I. = P×T/100(R1−R2)
⇒ 144 = P×6x4/100(∵ R1 - R2 = 2)
⇒ P = 144×100/6×4=600
Answer B
Question: A bus travel 5/7 of its speed covers 42 km in 1 hr 40 min48 sec. What is theactual speed of the bus ?
A.25 km/hr
B.35 km/hr
C.42 km/hr
D.46 km/hr
Answer: B.35 km/hr
Explanation: Time = 1 hr 40 min 48 sec = 1hr +4060hr+483600hr=1+23+175=12675hrdistance = 42 kmspeed=distancetime=42(12675)=42×75126⇒57 of the actual speed = 42×75126⇒actual speed = 42×75126×75=42×1518=7×153=7×5=35 km/hr
B
Question: Avg cost of 5 pens and 4 pencils isRs. 36.The avg cost of 5 pens and 9 pencils is Rs.56.Find the total cost of 25 pens and 25 pencils?
A.80
B.90
C.100
D.110
Answer: C.100
Explanation: Avg cost of 5 pens and 4 pencils is Rs. 36
(5 pen + 4 pencils) 9=36
therefore 1 pen cost is 36/9=4, so 5 pen cost is 5x4=20 and 4 pencils cost is 4x4=16
Again, avg cost of 5 pens and 9 pencils is Rs.56
(5 pen + 9 pencils) 14=56
therefore 1 pen cost is 56/14=4, so 5 pen cost is 5x4=20 and 9 pencils cost is 9x4=36
So 25 pen cost is (25/a)=100 and 25 pencils cost is (25x4)=100
ANSWER:C
Question: The average weight of a group of boys is 30 kg. After a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ?
A.4
B.5
C.6
D.7
Answer: A.4
Explanation: Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 35 kg joins the group, total weight of boys = 30x + 35
So 30x + 35 + 31(x + 1) = > x = 4.
Answer: A
Question: Three coins are tossed. Find the probability of at least 1 head and 1 tail ?
A.1/2
B.1/4
C.3/4
D.7/8
Answer: C.3/4
Explanation: n(s) = 2^3 = 8 ....(3 coins tossed)
let E is the event of getting at least 1 head and 1 tail
n(E) = TTH , HTT, THT, HHT, HTH, THH = 6
P(E) = n(E) / n(s)
= 6/8
= 3/4
Ans- C
Question: Calculate the value of 5.11 x 10^8 ?
A.511000000
B.5110000
C.511000
D.51100000
Answer: A.511000000
Explanation: 5.11 x 10^6 = 5.11 x 100000000 = 511000000.
Answer = A
Question: In how many ways a committee, Consisting of 3 men and 5 women can be formed from 8 men and 10 women?
A.266
B.86400
C.11760
D.14112
Answer: D.14112
Explanation: To choose Required number of ways to form a committee:
Select 3 men out of 8 men and 5 women out of 10 women.
That is 8C3 * 10C6.
8C3=8!/3!*5!
=56
10C5=10!/5!*5!
=252
The Required number of ways=252*56=14112
D
Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron
weighs 8g/ cm^3 , then the weight of the pipe is
A.4.996 kg
B.5.696 kg
C.3.796 kg
D.3.696 kg
Answer: D.3.696 kg
Explanation: External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron = [ 22/7 x (4 ^2 x 3 ^2) 21] cm^3 = 462 cm^3
Weight of iron = (462 x 8)gm = 3696 gm = 3.696 kg
ANSWER:D
Question: Mario and Nina each have a bag of marbles, each of which contains 10 blue marbles, 10 red marbles, an 10 white marbles. If Mario and Nina each select one marble from their respective bags, what is the probability that either Mario or Nina select a red marble?
A.3/4
B.2/9
C.1/4
D.1/8
Answer: B.2/9
Explanation: Mario and Nina each select one marble from their respective bags.
Probability that either Mario or Nina select a red marble = Probability that Mario selects a red marble + Probability that Nina selects a red marble
Probability that either Mario or Nina select a red marble = (10/30)*(10/30) + (10/30)*(10/30) = 2*(1/9)
Probability that either Mario or Nina select a red marble = 2/9
Answer would be B.
Question: John tossed a fair coin 5 times. What is the probability that the coin landed heads up exactly twice?
A.0.156
B.0.256
C.0.35
D.0.456
Answer: A.0.156
Explanation: The probability is exactly same as the cion landed tail up once. That is 5*(1/2)^5 = 5/32 =0.156.
The answer is, therefore, (A).
Question: The bankers discount and the true discount of a sum at 20% per annum simple interest for the same time are Rs.200 and Rs.150 respectively. What is the sum and the time?
A.Sum = Rs.600 and Time = 1.6 years
B.Sum = Rs.500 and Time = 1.6 years
C.Sum = Rs.400 and Time = 1.6 years
D.Sum = Rs.600 and Time = 3 years
Answer: A.Sum = Rs.600 and Time = 1.6 years
Explanation: Explanation :
BD = Rs.200
TD = Rs.150
R = 20%
F = BD ×TD/(BD – TD)=200×150/(200–150)=200×150/50=Rs. 600
BD = Simple Interest on the face value of the bill for unexpired time =FTR/100
⇒200 = (600×T×20)/100
⇒200 = 6 × T × 20 ⇒20 = 12 × T
⇒ T = 20/12 = 1.6 years Answer : Option A
Question: P and Q had the same number of candies. P gave Q 16 candies so that now Q has five times as many candies as P has. How many candies does Q have now?
A.25
B.30
C.35
D.40
Answer: D.40
Explanation: Let P and Q be the number of candies they had at the start.
P = Q
5(P-16) = Q+16
5(Q-16) = Q+16
4Q = 96
Q = 24
Now person Q has 24+16 candies which is 40 candies.
The answer is D.
Question: D and E are two points respectively on sides AB and AC of triangle ABC such that DE is parallel to BC. If the ratio of area of triangle ADE to that of the trapezium DECB is 144:25 and DE=13 cm, then find the length of BC.
A.5
B.13
C.14
D.11
Answer: A.5
Explanation: ABC and ADE are similar triangles.
So (side of ABC/side of ADE)^2=25/169
side of ABC/side of ADE = 5/13
So the length of BC = 5
Answer - A
Question: 0.99999999/1.0001−0.99999991/1.0003
A.10^(-8
B.3*10^(-8)
C.3*10^(-4)
D.2*10^(-4)
Answer: D.2*10^(-4)
Explanation: look at the options closely
A. 0.00000001
B. 0.00000003
C. 0.0003
D. 0.0002
E. 0.0001
Only one option is an even number, the rest odd (an esoteric sort of even/odd number)
both fractions in the question are odds
odd minus odd is always even.
So answer must be even.
Only D is even.
Question: The banker's gain of a certain sum due 2 years hence at 10% per annum is Rs. 24. What is the present worth?
A.Rs. 600
B.Rs. 500
C.Rs. 400
D.Rs. 300
Answer: A.Rs. 600
Explanation: Explanation :
T = 2 years
R = 10%
TD = (BG ×100)/TR = (24 × 100)/(2×10) = 12 × 10 = Rs. 120
TD = (PW × TR)/100
⇒ 120 = (PW × 2 × 10)/100
⇒ 1200 = PW × 2
PW =1200/2 = Rs. 600 Answer : Option A
Question: In a 120 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?
A.49
B.47
C.25
D.43
Answer: C.25
Explanation: Solution simple
Out of 120 20% are male i.e 24 and 25% are female i.e 30 , so total homeowner is 54.
Now min number homeowner is 24 and max is 54 so question ask us to find least and 25 has least value among all option.
So ans is 25.
Answer : C
Question: Two pipes A and B can separately fill a tank in 12 and 15 minutes respectively. A third pipe C can drain off 45 liters of water per minute. If all the pipes are opened, the tank can be filled in 15 minutes. What is the capacity of the tank?
A.33
B.540
C.287
D.278
Answer: B.540
Explanation: 1/12 + 1/15 - 1/x = 1/15
x = 12
12 * 45 = 540
Answer:B
Question: John and Amanda stand at opposite ends of a straight road and start running towards each other at the same moment. Their rates are randomly selected in advance so that John runs at a constant rate of 2, 3, 4, or 5 miles per hour and Amanda runs at a constant rate of 4, 5, 6, or 7 miles per hour. What is the probability that John has traveled farther than Amanda by the time they meet?
A.1/16
B.5/16
C.3/8
D.1/2
Answer: A.1/16
Explanation: John will run farther if he runs at 5 mph and Amanda runs at 4 mph.
P(John runs farther) = 1/4 * 1/4 = 1/16
The answer is A.
Question: Oranges are bought at 7 for Rs.6. At what rate per hundred must they be sold to gain 33%?
A.57
B.68
C.95
D.114
Answer: D.114
Explanation: 7 oranges bought for rs 6
1 orange cost =6/7
therefore for 100 oranges =100*(6/7)
gain =33%
600/7+(33/100)*600/7=rate per hundred he want to sell
ans :114
ANSWER:D
Question: The traffic lights at 3 different roads change after every 12 sec, 72 sec, 108 sec respectively. If they all change simultaneously at 9:15 AM then at what time will they again change simultaneously.
A.9:18:36
B.9:42:15
C.10:15:24
D.9:54:12
Answer: A.9:18:36
Explanation: LCM of 12,72,108 =216 = 3mnts 36sec
The lines will change every 3m 36sec
The light will change again after 9:15 AM is = 9:15+0:03:36 = 9:18:36
Answer is A
Question: If the ratio of a to b is 11 to 3 and the ratio of b to c is 1 to 5, what is the ratio of a to c?
A.4/15
B.11/15
C.2/5
D.4/5
Answer: B.11/15
Explanation: a:b = 11:3 -- 1
b:c = 1:5
=>b:c = 3 :15 --2
From 1 and 2 , we get
a:c = 11:15
Answer B
Question: The face value of a share is Rs 100. It is sold at a discount of Rs 5. How many such shares can be bought by investing Rs 38,000?
A.360
B.380
C.420
D.400
Answer: D.400
Explanation: Face value of a share=Rs.100
Discount=5%
so,Share is bought at Rs.95
Investing amount=Rs.38,000,so 380 shares are bought without discount
With discount 400 shares can be bought
ANSWER:D
Question: If a traveller was offered 5 destinations in asia and 11 destinations in europe and asked to chose utmost 3 destinations in asia and overall a 11 destinations. How many ways can he travel
A.2256 ways
B.2356 ways
C.2456 ways
D.2556 ways
Answer: A.2256 ways
Explanation: He can choose atmost 3 out of 5 destinations in Asia and 11 destinations overall.
Also it is mentioned that there are 11 destinations in Europe:
No. of ways this can be done is :-
5c3*11c8+5c2*11c9+5c1*11c10+5c0*11c11= 2256 ways
ANSWER:A
Question: Each term of a certain sequence is 4 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?
A.1
B.13
C.15
D.19
Answer: C.15
Explanation: We know that it is a decreasing sequence i.e. the terms keep decreasing till 0 and then negative terms start.
What does this imply?If the sum of the first n terms of the sequence is n
Since number of terms will definitely be positive, we are looking for a positive sum.
19 + 16 + 13 + 10 + 7 + 4 + 1 -2 -5 -7 -10 .... and so on
Note that the first 7 terms are positive and all others negative. Every negative term has greater absolute value than the corresponding positive terms i.e. -2 absolute value is greater than 1 absolute value, -5 absolute value is greater than 4 absolute value, and so on...
Since we have 7 positive terms, we must have less than 7 negative terms to get the sum as positive. If we have 6 negative terms, we will have a total of 13 terms. Of the given options, only 13 is possible and hence it must be the answer.
Answer (C)
Question: You have a bag of 9 letters: 3 Xs, 3 Ys and 3 Zs. You are given a box divided into 3 rows and 3 columns for a total of 9 areas. How many Q different ways can you place one letter into each area such that there are no rows or columns with 2 or more of the same letter?
A.5
B.6
C.9
D.12
Answer: D.12
Explanation: Consider one particular arrangement of the first row: XYZ, then we can construct only two boxes with so that no rows or columns have 2 or more of the same letter:
XYZ
YZX
ZXY
And:
XYZ
ZXY
YZX
Now, the first row itself can be arranged in 3!=6 ways (since there are three distinct letter), hence the total number of boxes possible Q is 2*6=12.
Answer: D.
Question: John and Peter are among the eight players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter?
A.2/5
B.2/7
C.2/9
D.5/14
Answer: D.5/14
Explanation: The total possible ways of selecting a 5-member team is 8C5 = 56
The possible ways which include John and Peter is 6C3 = 20
The probability of choosing both John and Peter is 20/56 = 5/14
The answer is D.
Question: Randy can rebuild an automobile engine in b hours. Alvin can rebuild the same engine in a hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete?
A.ab/(a+b)
B.(a-b)/(a+b)
C.b/(b+a)
D.a/(a+b)
Answer: C.b/(b+a)
Explanation: Randy can do the job in a hrs, Alvin can do the job in b hrs.
When they work together, you need to find the fraction of work that Randy doesn't need to do i.e. the fraction that will be done by Alvin.
Ratio of Randy's speed:Alvin's speed = b:a (since time taken by them is in the ratio a:b)
So Randy does a/(a+b) of the work and Alvin does b/(a+b) of the work.
Answer (C).
Question: A train ride from Two P to Town Q costs $6.35 more than does a bus ride from Town P to Town Q. Together, the cost of one train ride and one bus ride is $9.85. What is the cost of a bus ride from Town P to Town Q?
A.$1.75
B.$2.50
C.$4.10
D.$4.70
Answer: A.$1.75
Explanation: Let x be the cost of a bus ride.
x + (x+635) = 985
2x = 350
x = $1.75
The answer is A.