RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: How many integers from 1 to 100 (both inclusive) have odd number of factors?
A.7
B.8
C.9
D.10
Answer: D.10
Explanation: All perfect squares have odd # of factors. Squaring integers 1-10 produces 1, 4, 9, 16, ... 100 which all have odd # of factors. Answer is D
Question: Find the area of the quadrilateral of one of its diagonals is 20 cm and its off sets 9 cm and 6 cm?
A.287 cm2
B.150 cm2
C.1678 cm2
D.776 cm2
Answer: B.150 cm2
Explanation: 1/2 * 20(9 + 6)
= 150 cm2
Answer: B
Question: If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of
the boat in still water is
A.12 kmph
B.13 kmph
C.14 kmph
D.15 kmph
Answer: B.13 kmph
Explanation: Explanation:
Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 3 kmph.
Let speed in still water is x km/hr
Then, speed upstream = (x �3) km/hr.
x-3 = 10 or x = 13 kmph
Answer: B
Question: One soap can be made out of the scraps obtained while preparing 11 soaps. How many soaps can be prepared out of the scrap obtained out of 251 soaps. (Take all possibilities)
A.24
B.25
C.26
D.27
Answer: B.25
Explanation: From 242 soap scraps, we can make 242/11 = 22 soaps., which results in two more soaps. So, it now becomes 24 soaps and scraps of 2 soaps from 242 soaps. From the scrap of remaining 9 (=251-242) soaps, and together with the scraps of the two soaps already present, we can prepare one more soap. So total 25 soaps are possible.
ANSWER:B
Question: In a group of 8 engineers, 3 engineers have a PHD degree; the others have only an MS or BS degree. A team of 3 engineers is to be chosen which must have at least 1 engineer with a PHD, how many different teams can be chosen?
A.56
B.46
C.64
D.65
Answer: B.46
Explanation: The problem asks for a combination, since order doesn't matter. Now, selecting r items from a set of n gives the Combination Formula: nCr= n!/r!(n-r)!
n = 8
r = 3
So, total teams is 8C3 = 8!/(3!(8 - 3)!) = 56, and
n = 8 - 3 = 5
r = 3
for teams without a PHD is 5C3 = 5!/(3!(5 - 3)!) = 10,
so, teams with at least 1 PHD = 56 - 10 = 46
Answer: B
Question: A toothpaste manufacturer has set an annual production target to gain more profits. This year the target reached is 1/10 of last year's target. If the target production this year is increased 5% of the last year's target, how much percentage of the last year's target does he still need to reach to get the profit?
A.4/5
B.3/6
C.20/20
D.10/5
Answer: C.20/20
Explanation: Given 1/10 and 5% that is 1/10
Let the last year's target be 20
This year's target is 5% more of last year=22
The target reached this year=1/10*20=2
Remaining target is 22-2=20
Required 20/20
Answer (C)
Question: What is the units digit of 17^83 × 13^52 × 11^87?
A.4
B.5
C.6
D.7
Answer: B.5
Explanation: To find : the units digit of 17^83 × 13^82 × 11^87
Let's reduce the clutter and simplify the product
(7^83) (3^82) (1^87)
7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4th power
so 7^5 has the same last digit as 7^1, 7^9, 7^13
Thus , 7^83 has the same last digit as 7^3, 7^7, 7^11 i.e.3
3 has a cyclicity of 4 : exactly the same routine as above
Thus , 3^82 has the same last digit as 3^2, 3^6, 3^10 i.e.9
Any power of 1 will result in1as the last digit
So, product of our last digits = 3 x 9 x 1 = 27....last digit is 5
Correct Option : B
Question: The sum of two numbers is 50 and their difference is 10. Find their product.
A.600
B.114
C.315
D.325
Answer: A.600
Explanation: Let the numbers be x and y.
Then x + y = 50 and x - y = 10
x = 30; y = 20
xy = 30 * 20= 600
ANSWER:A
Question: Which of the following has most number of divisors?
A.99
B.101
C.176
D.182
Answer: C.176
Explanation: 99=1*3*3*11; 101=1*101
176=1*2*2*2*2*11; 182=1*2*7*13;
So, divisors of 99 are 1,3,9,11,33 and 99;
divisors of 101 are 1 and 101;
divisors of 176 are 1,2,4,8,16,22,44,88 and 176;
divisors of 182 are 1,2,7,13,14,26,91 and 182;
Hence, 176 as the most number of divisors.
Correct Option: C
Question: A man takes 5 hours 45 min. In walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is :
A.11 hrs 45 min
B.3 hrs 45 min
C.7 hrs 30 min
D.7 hrs 45 min
Answer: A.11 hrs 45 min
Explanation: Sol.
Let the distance be x km. Then,
(time taken to walk x km) + (Time taken to ride x km) = 23/4 hrs.
⇔ (Time taken to walk 2x km) + (Tie taken to ride 2x km) = 23/2 hrs.
But, time taken to ride 2x km = 15/4 hrs. ∴ Time taken to walk 2x km [23/2 - 15/4] hrs = 31/4 hrs = 7 hrs 45 min.
Answer A
Question: The no. of gift pack bought by Dexter is 1 more than the price, in rupees, of each gift pack. The amount of Rs.380 which Dexter had, fell short of the required amount. What is theamount by which he might have fallen short.
A.30
B.40
C.50
D.60
Answer: B.40
Explanation: Let the price of gift pack be 'aa'
Then number of packs bought =a+1=a+1
Hence total cost is a(a+1)a(a+1)
It is given that 380<a(a+1)380<a(a+1)
If a=19a=19, the total cost =19×20=380=19×20=380
Dexter would not have fallen short, If:
a=20a=20, a(a+1)=420a(a+1)=420
So he would have fallen short by Rs 40.
This is the minimum amount by which he may have fallen short.
B)
Question: Given that 268x74= 19332, find the value of 2.68x.74.
A.1.9332
B.1.0025
C.1.5693
D.1.0266
Answer: A.1.9332
Explanation: Solution
Sum of decimals places	=(2+2)
= 4.
Therefore,	= 2.68×.74
= 1.9332
Answer A
Question: After successive discounts of 20%, 10% and 5% a certain good is sold for Rs. 6700. Find the actual price of the good.
A.s. 6000
B.s. 9000
C.s. 10800
D.s. 9795
Answer: D.s. 9795
Explanation: Let actual price was 100.
After three successive discount this will become,
100 ==20% discount => 80 ==10% discount => 72 ==5% discount = 68.4
Now Compare,
68.4 = 6700
1 = 6700/68.4
100 = (6700*100)/68.4 = Rs. 9795.
Answer: Option D
Question: The sum and product of two numbers are 12 and 35 respectively. The sum of their reciprocals is
A.12/35
B.1/35
C.35/8
D.35/9
Answer: A.12/35
Explanation: X + Y = 12
XY = 35
1/X + 1/Y = (X + Y)/Xy = 12/35
Answer A.
Question: Given f(x) = 2x – 3, for what value of x does 2*[f(x)] – 11 = f(x – 2) ?
A.2
B.3
C.4
D.5
Answer: D.5
Explanation: 2(2x-3)-11 = 2(x-2) - 3
2x = 10
x = 5
The answer is D.
Question: The speed of a boat in still water in 20 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 24 minutes is:
A.9.6
B.10.2
C.9.2
D.10.4
Answer: C.9.2
Explanation: Speed downstream = (20 + 3) = 23 kmph
Time = 24 minutes =24/60 hour = 2/5 hour
Distance travelled = Time × speed =2/5×23= 9.20 km
Answer is C.
Question: A bag contains seven white and four red balls. Two balls are picked at random from the bag. What is the probability that they both are different color?
A.25/28
B.28/55
C.25/26
D.25/23
Answer: B.28/55
Explanation: Two balls can be picked from eleven balls in 11Câ‚‚ ways.
We select one white ball and one red ball from five white balls and four red balls. This can be done âµ ways.
The required probability = (7 * 4) = 28/55 = 28/55
Answer: B
Question: Suresh started a business, investing Rs.18000. After 3 months and 4 months respectively, Rohan and Sudhir joined him with capitals of 12000 and 9000. At the end of the year the total profit was Rs.3960. What is the difference between Rohan’s and Sudhir’s share in the profit?
A.s.345
B.s.350
C.s.352
D.s.360
Answer: D.s.360
Explanation: Suresh : Rohan : Sudhir
Ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8
= 6 : 3: 2
The difference between Rohan’s and Sudhir’s share = 1 share
:.i.e. = Rs. 3960 × 1/11 = Rs.360.
D
Question: There are 200 rats in Rat-City. Out of the 200, 70 are street rats and the rest are domestic rats. 110 rats are gray, 30 out of the gray cats are NOT domestic ones. How many domestic rats are there which are not gray in Rat-City?
A.90
B.50
C.60
D.40
Answer: B.50
Explanation: SR=70,DR=130
GR=110,NGR=90
So SR,GR30 and DR,GR=80 NGR,SC=40 and NGR,DC=50
SR=stray rats
DR=domesticated rats
GR=Gray rats
NGR=Non Gray rats
Answer is B)
Question: The editor of an anthology of short stories has to select authors from a pool of 12 authors, of which 4 are female. If the anthology is to include the work of 8 authors of which at least 2 must be female, in how many ways can the editor select the authors?
A.168
B.201
C.294
D.462
Answer: D.462
Explanation: we have total 12 authors. we need to select 8. we have 8 males and 4 females.
we need to select in a such way, that at least 2 females are selected.
my first step:
12C8 = 495.
since this is a total number of combinations that we can have, regardless of the restrictions, we can eliminate right away A, B, C, and E.
why A, B, C - because the number of combinations when the restrictions are not respected is way smaller than the number of combinations in which the restrictions are respected.
nevertheless, I wanted to check further, so that to make sure I get the right answer:
second step, to get "at least 2 females", we need to deduct from 495 number of combinations when we have 8 males 0 females and when we have 7 males and 1 female.
8 males 7 females:
8C8 * 4C0 = 8
7 males and 1 female: 8C7*4C1=32.
Thus, the required number of combinations = Total - case with 0 females - case with 1 female = 495-1-32=462.
D is the correct answer.
Question: Line Q has the equation 3x + y = 7. Which of the following lines is perpendicular to Line Q?
A. y = 3x + 4
B. y = –3x – 6
C. y = (1/3)x – 1
D. y = (–1/3)x + 2
Answer: C. y = (1/3)x – 1
Explanation: I first rewrote the equation in the standard y=mx+b form. Therefore, Line Q as presented, 3x + y = 7, can be rewritten as follows: y=-3x+7.
Thought process next is what line would be perpendicular to Line Q? Any line with a reciprocal of the slope but in the opposite direction. The reciprocal of any fraction/integer is 1 over that number/integer. Therefore the reciprocal of -3 is -1/3 - need to drop the negative sign because the line would kinda run parallel and we want perpendicular. Scan the answers choices and notice C as the only one.
Question: 0.01 is what percent of 0.1?
A.1%
B.10%
C.100%
D.50%
Answer: B.10%
Explanation: Required percentage = 0.01*100/0.1 = 100/10 = 10%
Answer is B
Question: if 5 a = 3125, then the value of 5(a-3) is :
A.25
B.125
C.625
D.1625
Answer: A.25
Explanation: Solution
5 a = 3125 ⇔ 5 a = 5 a ⇔ a = 5.
∴ 5(a - 3) = 5(5 - 3) = 52 = 25.
Answer A
Question: A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?
A.378
B.277
C.299
D.646
Answer: D.646
Explanation: 840 = P [1 + (10*3)/100]
P = 646
Answer: D
Question: A train is 360 meter long is running at a speed of 56 km/hour. In what time will it pass a bridge of 140 meter length.
A.20.14 seconds
B.12.14 seconds
C.22.14 seconds
D.32.14 seconds
Answer: D.32.14 seconds
Explanation: Explanation:
Speed = 56 Km/hr = 56*(5/18) m/sec
= 140/9 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
=500/140/9=32.14 seconds
Answer: D
Question: If the sum of a number and its square is 132, then what is the number?
A.10
B.11
C.12
D.13
Answer: B.11
Explanation: Explanation:
Let the number be x. Then,
x2 + x = 132
=> x2 + x - 132 = 0
=> (x + 12)(x - 11) = 0
=> x = 11
Answer: Option B
Question: . How many shots of 1cm radius can be prepared from a sphere of 3cm radius?
A.20
B.72
C.27
D.83
Answer: C.27
Explanation: Explanation:
4/3 π * 3 * 3 * 3 = 4/3 π * 1 * 1 * 1 * x
x = 27
Answer: Option C
Question: If c<0, which of the following represents a positive number?
A.|c|/c
B.|c|c
C.-c|c|
D.c/ |c|
Answer: C.-c|c|
Explanation: Answer : C
Question: Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A.8%
B.6%
C.4%
D.7%
Answer: D.7%
Explanation: Explanation :
Let rate = R%
Then, Time, T = R years
P = Rs.1400
SI = Rs.686
SI= PRT/100
⇒686 = 1400 × R × R /100
⇒686=14 R^2
⇒49=R^2
⇒R=7
i.e.,Rate of Interest was 7%
Answer : Option D
Question: Find the largest number of five digits which is exactly divisible by 12 and 15 ?
A.A)12960
B.B)13050
C.C)10025
D.D)11080
Answer: A.A)12960
Explanation: The largest five digit numbers are
13050,12960,12080
13050 is not divisible by 12
12960 is divisible by both 12 and 15
ANSWER:A
Question: Jansi’s age is one-sixth of her mother’s age. Jansi’s mother’s age will be two times of Lenin’s age after 10 years. If Lenin’s 8-th birthday was celebrated 2 years before, then what is Jansi’s current age?
A.2 years
B.5 years
C.7 years
D.8 years
Answer: B.5 years
Explanation: B
5 years.
Lenin’s age after 10 years = (8 + 2 + 10) years = 20 years
Jansi’s mom’s age after 10 years = 40 years.
Jansi’s mom’s current age = 30 years.
Therefore, Jansi’s age = (1/6)x 30 years
= 5 years.
Question: If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 440, what is the least possible value of n?
A.10
B.11
C.12
D.13
Answer: B.11
Explanation: is it 11 ?
440 = 2 * 2 * 2 * 5 * 11
so the least value forNcan be 11.
B
Question: If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
A.7
B.9
C.4
D.13
Answer: C.4
Explanation: Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2
Solving these two equations, we get : x = 1/100 and y = 1/200
15 men + 20 boy)'s 1 day's work = (15/100 + 20/200) = 1/4
15 men and 20 boys can do the work in 4 days.
Answer: C
Question: Different breeds of dogs get older at different rates in “dog years.” Livonian wolfhounds age 6 times as fast as humans, whereas Khazarian terriers age 5 times as fast and Akkadian retrievers age 4 times as fast. If Dan bought a newborn Akkadian on January 1, 2010, a newborn Khazarian 1 year later, and a newborn Livonian 1 year after that, in what year will the sum of the dog-year ages of the Akkadian and the Khazarian first be exceeded by twice the age of the Livonian in dog years, rounding all ages down to the nearest integer?
A.2017
B.2014
C.2015
D.2016
Answer: A.2017
Explanation: chart starting at 2011: A=4
2012 = A=8 K=5 A+K =13
L=0.
Then I added 9 to a+k for each year and 12 to L for each year.
2013:AK=22 L=12
2014:AK=31 L=24
2015:AK=40 L=36
2016 :AK=49 L=48
2017: AK=59 L=60
A
Question: Presently a mixture in a tub contains substances A and B in the ratio of 7:5. If 9 litres of this solution is drawn out and B is added to the mixture in certain quantity the resulting mixture becomes 7:9. How much litres of liquid A was kept in the tub initially?
A.87
B.265
C.88
D.21
Answer: D.21
Explanation: Explanation:
Currently the mixture is in ratio of A/B: 7:5
=> This means fraction of A in the solution: 7/12
=> Fraction of B in the solvent: 5/12
=>New concentration of A= 7-(7/12)9
=>New concentration of B= 5-(5/12)9+x
=> (7-(7/12)9)/(5-(5/12)9+x) = 7/9
=>x=21
ANSWER: D
Question: A and B can do a work in 25 days and 10 days respectively. A starts the work and B joins him after 2 days. In how many days can they complete the remaining work?
A.4.28 days
B.4.38 days
C.5.28 days
D.6.28 days
Answer: A.4.28 days
Explanation: Work done by A in 2 days = 2/5
Remaining work = 3/5
Work done by both A and B in one day = 1/25 + 1/10 = 7/50
Remaining work = 3/5 * 50/7 = 4.28 days.
ANSWER:A
Question: A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio Q of milk and water in the resultant mixture?
A.17 : 3
B.9 : 1
C.3 : 17
D.5 : 3
Answer: B.9 : 1
Explanation: he 20 litre mixture contains milk and water in the ratio of 3 : 2. Therefore, there will be 12 litres of milk in the mixture and 8 litres of water in the mixture.
Step 1.
When 10 litres of the mixture is removed, 6 litres of milk is removed and 4 litres of water is removed. Therefore, there will be 6 litres of milk and 4 litres of water left in the container. It is then replaced with pure milk of 10 litres. Now the container will have 16 litres of milk and 4 litres of water.
Step 2.
When 10 litres of the new mixture is removed, 8 litres of milk and 2 litres of water is removed. The container will have 8 litres of milk and 2 litres of water in it. Now 10 litres of pure milk is added. Therefore, the container will have 18 litres of milk and 2 litres of water in it at the end of the second step. Therefore, the ratio of milk and water is 18 : 2 or 9 : 1.
Shortcut.
We are essentially replacing water in the mixture with pure milk.
Let W_o be the amount of water in the mixture originally = 8 litres.
Let W_r be the amount of water in the mixture after the replacements have taken place.
Then,{W_r}/{W_o}= (1-R/M)^n
where R is the amount of the mixture replaced by milk in each of the steps, M is the total volume of the mixture and n is the number of times the cycle is repeated.
Hence, {W_r}/{W_o} =(1/2)^2  =1/4
Therefore Q,W_r ={W_o}/4= 8/4 = 2 litres.B
Question: The ratio of the ages of Anil and his son at present is 7:3. Six years hence, the ratio of the ages of the Anil's wife and the son will be 2:1. Find the ratio of the present ages of Anil and his wife?
A.2:2
B.2:9
C.2:4
D.Cannot be determined
Answer: D.Cannot be determined
Explanation: Let the present ages of Anil and his son be 7x and 3x years respectively.
Let the present age of the son's mother be m years.
(m + 6)/(3x + 6) = 2/1
m = 6x + 6
Ratio of the present ages of Anil and his son's mother = 7x/(6x + 6).
This cannot be found uniquely.
Answer:D
Question: Twenty six years ago,a father was three times as old as his son.Now the father is only twice as old as his son. Then the sum of the present ages of the son and the father is :
A.120
B.156
C.108
D.105
Answer: B.156
Explanation: Let the present ages of the father and son be 2x and x years respectively.
then,(2x-26)=3 (x-26)
x=52.
Required sum =(2x+x)=3x=156 years.
Answer is B
Question: 55% of respondents to a survey consisted of those who used both M and N and those who used neither M nor N. 47% of respondents did not use M. 40% used N. What percent of respondents used only N?
A.16%
B.21.5%
C.20%
D.18.5%
Answer: A.16%
Explanation: From the question, we can summarize the information as follows:
N Not N
M X
Not M Ans y 47
40 100
Then we can set a system of 3 equations 3 unknowns described below:
X + Y = 55
X + Ans = 40
Y + Ans = 47
Sum up all three equations to get 2X + 2Y + 2Ans = 142
Divide both sides by 2 to get X + Y + Ans = 71
Replace X + Y with 55 to get 55 + Ans = 71
Substract 55 from both sides to get Ans = 71 - 55 = 16
The answer must be A.
Question: A palindrome is a word or a number that reads the same forward and backward. For example, 2442 and 111 are palindromes. If 5-digit palindromes are formed using one or more of the digits 1, 2, 3, 4, 5, and 6, how many palindromes are possible?
A.36
B.96
C.132
D.216
Answer: D.216
Explanation: There are 6 choices for each of the first three digits.
The number of possible palindromes is 6^3=216.
The answer is D.
Question: A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
A.40 sec
B.32 sec
C.12 sec
D.85 sec
Answer: A.40 sec
Explanation: Answer: Option A
Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 2/25 = 40 sec
Question: A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m. ?
A.9.38 a.m
B.9.36 a.m
C.9.30 a.m
D.9.10 a.m
Answer: C.9.30 a.m
Explanation: Relative speed = 2 + 3 = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9.30 a.m.
Answer:C
Question: Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1120?
A.1000
B.289
C.209
D.280
Answer: A.1000
Explanation: 1120 = P [1 + (5*12/5)/100]
P= 1000
Answer: A
Question: One train is traveling 45 kmph and other is at 10 meters a second. Ratio of the speed of the two trains is?
A.5:4
B.2:8
C.5:1
D.5:2
Answer: A.5:4
Explanation: 45 * 5/18 = 10
25:20 => 5:4
Answer:A
Question: The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the fathers age at that time. The present age of father and son
A.34,11
B.35,10
C.36,9
D.40,5
Answer: C.36,9
Explanation: Explanation:
Let sons age = x years. Then fathers age = (45 - x)years.
(x—5)(45—x—5) = 4(45- x - 5) hence (x—5) = 4 so x = 9
Their ages are 36 years and 9 years.
Answer: Option C
Question: A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 18 seconds. What is the length of the train in meters?
A.180 meters
B.1111 meters
C.1777 meters
D.1822 meters
Answer: A.180 meters
Explanation: Train overtakes a bike means that we are talking about total length of the train. (train's head is close to bike when it started and its tail crosses the bike when it overtakes the bike)
Relative speed= 100-64= 36Km/h = 36000 m/h
time= 18 seconds
Distance= speed*time
36000* 18/3600= 180 meters.
A is the answer.
Question: A, B and C can do a work in 7 , 14 and 21 days respectively. They completed the work and got Rs.242. What is the share of C?
A.33
B.44
C.55
D.11
Answer: B.44
Explanation: The ratio of their working rates = 1/7 : 1/14 : 1/21 = 6 : 3 : 2.
Since, they work together, the share of C
= 2/11 * 242 = Rs.44
Answer: B
Question: How many times the keys of a writer have to be pressed in order to write first 50 counting no's?
A.90
B.91
C.92
D.93
Answer: D.93
Explanation: 1 to 9 = 9 * 1 = 9
10 to 50 = 41 * 2 = 84
-----------
93
D
Question: An electric motor makes 3,000 revolutions per minutes. How many degrees does it rotate in one second?
A.19,000 degrees / second
B.17,000 degrees / second
C.16,000 degrees / second
D.18,000 degrees / second
Answer: D.18,000 degrees / second
Explanation: 3000 revolutions / minute
= 3000*360 degrees / 60 seconds
= 18,000 degrees / second.
Answer is D.
Question: Two pipes A and B can fill a tank in 10 hrs and 40 hrs respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?
A.8 hours
B.6 hours
C.4 hours
D.2 hours
Answer: A.8 hours
Explanation: Explanation :
Pipe A can fill 1⁄10 of the tank in 1 hr
Pipe B can fill 1⁄40 of the tank in 1 hr
Pipe A and B together can fill 1⁄10 + 1⁄40 = 1⁄8 of the tank in 1 hr
i.e., Pipe A and B together can fill the tank in 8 hours .Answer : Option A
Question: If 6th March, 2021 is Thursday, what was the day of the week on 6th March, 2020?
A.Sunday
B.Saturday
C.Tuesday
D.Wednesday
Answer: D.Wednesday
Explanation: Solution
The year 2020 is a leap year. So, it has 2 odd days.
But, Feb 2020 not included because we are calculating from March 2020 to March 2021. So it has 1 odd day only.
The day on 6th March, 2021 will be 1 day beyond the day on 6th March, 2020. Given that, 6th March, 2021 is Thursday.
6th March, 2020 is Wednesday (1 day before to 6th March, 2021)
Answer D
Question: A train 360 m long is running at a speed of 72 km/hr. In what time will it pass a bridge 140 m long?
A.40 sec
B.29 sec
C.25 sec
D.27 sec
Answer: C.25 sec
Explanation: Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 360 + 140 = 500 m
Required time = 500 * 1/20
= 25 sec
Answer: C
Question: The length of a rectangle is reduced by 22%. By what % would the width have to be increased to maintainthe original area?
A.15%
B.20%
C.28%
D.35%
Answer: C.28%
Explanation: Sol. Required change = (22*100)/(100-22)=28%
C
Question: The profit earned by selling an article for Rs. 832 is equal to the loss incurred when the same article is sold for Rs. 448. What should be the sale price for making 50% profit?
A.260.
B.960.
C.968.
D.9690.
Answer: B.960.
Explanation: Let C.P. = Rs. x.
Then, 832 - x = x - 448
2x = 1280 => x = 640
Required S.P. = 150% of Rs. 640 = 150/100 * 640 = Rs. 960.
Answer:B