RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Insert the missing number. 2, 7, 10, 22, 18, 37, 26
A.42
B.52
C.46
D.62
Answer: B.52
Explanation: Explanation :
There are two series here
2, 10, 18, 26, ... (Increase by 8)
7, 22, 37, ... (Increase by 15)
Hence, next term is 37+15 = 52
Answer : Option B
Question: Working at a constant rate, P can finish a job in 4 hours. Q, also working at a constant rate, can finish the same job in 15 hours. If they work together for 3 hours, how many more minutes will it take P to finish the job, working alone at his constant rate?
A.5
B.8
C.10
D.12
Answer: D.12
Explanation: Each hour they complete 1/4 + 1/15 = 19/60 of the job.
In 3 hours, they complete 3(19/60) = 19/20 of the job.
The time for P to finish is (1/20) / (1/4) = (1/5) hour = 12 minutes
The answer is D.
Question: 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length if the wire in meters will be:
A.76 m
B.80 m
C.84 m
D.88 m
Answer: C.84 m
Explanation: Explanation:
Let the length of the wire be h
Radius=1/2mm=120cm
πr2h=66
22/7∗1/20∗1/20∗h=66
=>h=66∗20∗20∗7/22=8400cm=84m
Option C
Question: What is the sum of the first 23 natural numbers (starting from 1)?
A.256
B.266
C.276
D.286
Answer: C.276
Explanation: The sum of n natural numbers = n*(n+1) / 2 = 23*24/2 = 276
The answer is C.
Question: If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by
A.10%
B.20%
C.25%
D.30%
Answer: B.20%
Explanation: Solution
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 6 = x × 7.50
x = 80 kg.
Reduction in consumption = 20%.
Answer B
Question: Of 40 applicants for a job, 32 had at least 5 years of prior work experience, 24 had advanced degrees, and 16 had at least 5 years of prior work experience and advanced degrees. How many of the applicants had neither 5 years of prior work experience nor advanced degrees?
A.0
B.2
C.4
D.8
Answer: A.0
Explanation: soln:
The 16 with both experience and degrees are the first group;
we can subtract the 16 from the 24 with advanced degrees to determine that 8 have advanced degrees but not 5 years of experience,
we can subtract the 16 from the 32 with experience to determine that there are 16 applicants with 5 years of experience but no advanced degrees
We can then subtract these three groups from the total pool of 40 applicants to determine how many lack both
experience and advanced degrees; 40-16-16-8=0,
so there were no such applicants among the 40
ANSWER:A
Question: What is the ratio whose term differ by 30 and the measure of which is 2/7?
A.A)12:42
B.B)16:56
C.C)71:85
D.D)32:39
Answer: A.A)12:42
Explanation: Let the ratio be x:(x+30)
Then, x/(x+30) = 2/7
x=12
required ratio = 12:42
Answer is A
Question: There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 530. The number of hares in the zoo is?
A.90
B.99
C.77
D.65
Answer: D.65
Explanation: 200*2 = 400
530
-----
130
1----2
?----130 = 65
Answer:D
Question: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
A.Rs. 20
B.Rs. 60.
C.Rs. 40
D.Rs. 30
Answer: B.Rs. 60.
Explanation: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
C.P. of 1 ball = Rs. 720/12 = Rs. 60.
Answer:B
Question: What sum of money put at C.I amounts in 2 years to Rs.8840 and in 3 years to Rs.9261?
A.8054
B.8877
C.2877
D.2678
Answer: A.8054
Explanation: 8840 ---- 421
100 ---- ? => 4.76%
x *104.76/100 * 104.76/100 = 8840
x*1.0975=8840
x=8840/1.0975 => 8054.54
Answer: A
Question: The population of a bacteria colony doubles every day. If it was started 3 days ago with 3 bacteria and each bacteria lives for 12 days, how large is the colony today?
A.512
B.1024
C.768
D.2048
Answer: B.1024
Explanation: 9 days ago- 2
8 days ago- 4
7 days ago- 8
6 days ago- 16
5 days ago- 32
4 days ago- 64
3 days ago- 128
2 days ago- 256
Yesterday- 512
Today- 1024
Answer:B
Question: Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveler sets out from city X in the same direction as the first traveler and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveler be ahead of the first?
A.2 days
B.6 days
C.From the 2nd day after the 2nd traveler starts
D.From the 3rd day after the 2nd traveler starts
Answer: B.6 days
Explanation: Explanatory Answer
Travel Pattern of Traveller 2
Day 1: 1 km; Day 2: (1 + 2) = 3 km; Day 3: (1 + 2 + 2) = 5km and so on
i.e. on each of these days he covers the following distance Day 1: 1km, Day 2: 3km, Day 3: 5km and so on.
The distances covered each day by the first traveller is in an AP with the first term being 1 and the common difference being 2
After 3 days, the first traveler would have traveled 9 kms. He is ahead of the second traveler at the end of 3 days by 9 kms.
Travel Pattern of Traveller 2
The second traveler travels as follows: On his first day he travels 12 kms, his second day he travels 13 kms and so on.
The distances covered each day by the second traveller is in an AP with the first term being 12 and the common difference being 1
Step 2
Compute total distance covered by each of the travellers
Let the number of days travelled by the second traveller be n.
Therefore, the first traveller would have travelled for (n + 3) days. Traveler 1 has covered a total distance of 1+ 3 + 5 + 7 + ..... + (n + 3) km.
The total distance covered by traveler 1 is the summation of the AP.
Sum of an AP = (n/2)(2a1 + (n-1)d)
Total distance travelled by traveller 1 in (n + 3) days = n+3/2[2(1)+(n+3−1)(2)]
The above term simplifies as (n + 3)2
The total distance covered by the second traveler = 12 + 13 + 14 + ... + n terms
Total distance travelled by traveller 2 in n days = n2[2(12)+(n−1)(1)] = n(n+23)2
On the days that the second traveller is ahead of the first one, the total distance covered by the second traveller till that day will be more than the total distance covered by the first traveller.
Note: All values calculated are with respect to the day from which the second traveller started travelling
Therefore, let us solve the inequality n(n+23)/2 > (n + 3)2
n2 + 23n > 2(n2 + 6n + 9)
Or n2 - 11n + 18 < 0
The quadratic factorizes as (n - 2)(n - 9) < 0
Possibility 1: (n - 2) > 0 AND (n - 9) < 0
i.e., n > 2 AND n < 9
Therefore, 2 < n < 9
Possibility 2: (n - 2) < 0 AND (n - 9) > 0
i.e., n < 2 AND n > 9
Infeasible solution
Therefore, starting the 3rd day and up to the 8th day the second traveller was ahead of the first traveller
The correct answer is Choice (2). 6 days
Answer B
Question: After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
A.2 years
B.4 years
C.8 years
D.15 years
Answer: D.15 years
Explanation: Solution
Age decreased	= (5 x 3) years
15 years
So, required difference =15years.
Answer D
Question: At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. The 3 restaurants are the family buffet, the dining room, and the snack bar. Exactly 19 employees are trained to work in the family buffet, 16 are trained to work in the dining room, and 12 are trained to work in the snack bar. If 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants?
A.2
B.3
C.4
D.5
Answer: A.2
Explanation: 39 = 19 + 16 + 12 - 4 - 2x
2x = 19 + 16 + 12 - 4 - 39
= 43- 39
=4
x = 2
A
Question: If log1087.5 = 5.9421, then the number of digits in (875)10 is?
A.30
B.70
C.27
D.26
Answer: B.70
Explanation: X = (875)10 = (87.5 x 10)10
Therefore, log10X = 10(log1087.5 + 1)
= 10(5.9421 + 1)
= 10(6.9421) = 69.421
X = antilog(69.421)
Therefore, number of digits in X = 70.
ANSWER:B
Question: The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?
A.1:96
B.1:98
C.1:95
D.1:45
Answer: A.1:96
Explanation: Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 2
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x =1:96.
Answer: A
Question: In how many ways can the letters of the word REGULATION be arranged such that all the vowels always appear together?
A.10!*5!
B.10!*6!
C.10!
D.6!*5!
Answer: D.6!*5!
Explanation: The 10 letters can be grouped into 5 consonants and one set of 5 vowels.
The number of ways to arrange 6 units is 6!
Then, for each arrangement, the 5 vowels can be arranged in 5! ways.
The total number of arrangements is 6!*5!
The answer is D.
Question: Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the24 cannot be adjacent digits in the integer, how many 5-digit integers are possible?
A. 48
B. 72
C. 76
D. 78
Answer: B. 72
Explanation: Number of arrangements using 5 distinct digits = 5!
Number of arrangements in which 3 and 4 are adjacent - consider 3 and 2 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 3 and 2 can be arranged as 32 or 23.
Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72
Answer : B
Question: A girl has exactly 6 quarters, 8 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?
A.1
B.2
C.3
D.4
Answer: C.3
Explanation: When the probability of NOT drawing a quarter is 50% or less, then the probability of drawing a quarter is at least 50%.
P(not drawing a quarter with 2 coins) = 18/24 * 17/23 = 51/92 > 50%
P(not drawing a quarter with 3 coins) = 51/92 * 16/22 = 102/253 < 50%
The answer is C.
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
A.20 years
B.30 years
C.60 years
D.40 years
Answer: D.40 years
Explanation: D
40 years
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
Question: Two trains of equal lengths are running at speeds of 30 kmph and 60 kmph. The two trains crossed each other in 30 seconds when travelling in opposite direction. In what time will they cross each other when travelling in the same direction?
A.90
B.82
C.16
D.67
Answer: A.90
Explanation: Let the length of each train be x m.
(x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5)
= 90 sec.
Answer: A
Question: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 300, 300, and 360 as the side’s lengths, what is the triangle’s area?
A.34,200
B.36,200
C.43,200
D.42,200
Answer: C.43,200
Explanation: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2,
-> P=(300+300+360)/2=480, area=√(480(480-300)(480-300)(480-360))=43,200. Therefore, the answer is C.
Question: My brother is 3 years elder to me.My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born.If my sister was 4 years of age when my brother was born,then,what was the age of my father and mother respectively when my brother was born ?
A.32 yrs,23 yrs
B.32 yrs,29 yrs
C.35 yrs,29 yrs
D.35 yrs,33 yrs
Answer: A.32 yrs,23 yrs
Explanation: Solution
Clealy,my brother was born 3 years before I was born and 4 years after my sister was born.
So,father's age when brother was born =(28+4) years =32 years,
mother's age when brother was born =(26-3) years =23 years . Answer A
Question: Club X has more than 50 but fewer than 80 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?
A.1
B.2
C.3
D.4
Answer: C.3
Explanation: The number of members is 4j+3 = 5k+3.
Let's start at 4(12)+3 = 51 and keep adding 4 until we find a number in the form 5k+3.
51, 55, 59, 63 = 5(12) + 3
The club has 63 members.
63/6 = 6(10) + 3
The answer is C.
Question: What is the smallest positive integer x, such that 7000x is a perfect cube?
A.4
B.6
C.8
D.12
Answer: D.12
Explanation: Take out the factors of 7000x that will come 10^3*7. for perfect cube you need every no. raise to the power 3. for 7000x to be a perfect cube, need two 7 that means 49.
E is the answer.
Question: A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train?
A.79.5
B.39.5
C.79.9
D.79.2
Answer: D.79.2
Explanation: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/20 = y
y = 22
Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.
Answer:D
Question: In a college 1/5th of the girls and 1/8th of the boys took part in a social camp. What of the total number of student in the college took part in the camp ?
A.13/40
B.13/80
C.2/13
D.Data inadequate
Answer: C.2/13
Explanation: Answer
Out of the 5 girls, 1 took part in the camp. Out of the 8 boys, 1 took part in the camp.
∴ 2/13 of total number of students took part in the camp.
Correct Option: C
Question: A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then?
A.144
B.255
C.148
D.265
Answer: A.144
Explanation: If each of 66 tractors ploughed 120120 hectares a day and they finished the work in 44 days, then the whole field is: 120⋅6⋅4=720⋅4=2880120⋅6⋅4=720⋅4=2880 hectares. Let's suppose that each of the four tractors ploughed xx hectares a day. Therefore in 5 days they ploughed
5⋅4⋅x=20⋅x5⋅4⋅x=20⋅x hectares, which equals the area of the whole field, 2880 hectares.
So, we get 20x=288020x=2880
x=288020=144x=288020=144. Hence, each of the four tractors would plough 144 hectares a day.
correct answer is A)144
Question: Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
A.7
B.6
C.8
D.9
Answer: B.6
Explanation: There are five letters in the given word.
Consider 5 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6.
Answer:B
Question: Water consists of hydrogen and oxygen, and the approximate ratio, by mass, of hydrogen to oxygen is 2:16. Approximately how many grams of oxygen are there in 153 grams of water?
A. 16
B. 72
C. 112
D. 136
Answer: D. 136
Explanation: Since the ratio, by mass, of hydrogen to oxygen is 2:16, then oxygen is 16/(2+16)=8/9 of mass of water.
Therefore there are 153*8/9=136 grams of oxygen in 153 grams of water.
Answer: D.
Question: In a 240 meters race A beats B by 56 m or 7 seconds. A's time over the course is:
A.23 seconds
B.12 seconds
C.10 seconds
D.18 seconds
Answer: A.23 seconds
Explanation: B runs 56 m in 7 sec.
=> B runs 240 m in 7/56*240 = 30 seconds
Since A beats B by 7 seconds, A runs 240 m in (30-7) = 23 seconds
Hence, A's time over the course = 23 seconds
ANSWER:A
Question: a and b are positive integers less than or equal to 6. If a and b are assembled into the six-digit number ababab, which of the following must be a factor of ababab?
A.3
B.4
C.5
D.2
Answer: D.2
Explanation: 242424 when divided by 2 gives the result 121212
CONCEPT: Rule of Divisibility of 2 isIf sum of the digits of the Number is divisible by 2 then the number will be divisible by 2
SUm of the digits of No. which is divisible by 2 hence the Number 242424 will be divisible by 2
D
Question: A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec fast at 2 p.m. on the following Monday. When was it correct?
A.2 p.m. on Tuesday
B.7 p.m. on Tuesday
C.8 p.m. on Tuesday
D.5 p.m. on Tuesday
Answer: A.2 p.m. on Tuesday
Explanation: 2 minutes slow at monday noon.
4 min 48 sec fast at 2 pm next monday.
It ran faster than normal time by 6 min 48 secs = 408 sec in 7 days 2 hours = 170 hours.
It was correct when it advanced 120 seconds than normal time.
The clock gains 408 seconds in 170 hours.
The clock gains 1 second in 170/408 hours
The clock gains 120 seconds in 170408×120170408×120 hours = 50 hours = 2 days 2 hours.
So the clock shows correct time on Wednesday 2 pm..
Answer:A
Question: By selling 12 toffees for a rupee, a man loses 20%. How many for a rupee should be sell to get a gain of 20%?
A.2
B.8
C.9
D.7
Answer: B.8
Explanation: Explanation:
Let S.P. of 12 toffees be Rs. x.
Then, 80:1 = 120:x or x = 120/180 = 3/2
For Rs. 3/2, toffees sold = 12.
For Re. 1, toffees sold = 12 * 2/3 = 8
Answer:B
Question: Four extra-large sandwiches of exactly the same size were ordered for m students, where m > 6. Three of the sandwiches were evenly divided among the students. Since 4 students did not want any of the fourth sandwich, it was evenly divided among the remaining students. If Carol ate one piece from each of the four sandwiches, the amount of sandwich that she ate would be what fraction of a whole extra-large sandwich?
A.(4m-15)/[m(m-5)]
B.(4m-15)/[m(m-8)]
C.(4m-18)/[m(m-5)]
D.(4m-20)/[m(m-5)]
Answer: A.(4m-15)/[m(m-5)]
Explanation: Three of the sandwiches were evenly divided among the students
i.e. Amount of Sandwiches with each students = 3/m
Amount of 4th Sandwich with remaining (m-6) students = 1/(m-6)
The amount of Sandwich that carol ate = 3/m + 1/(m-6) = (3m-18+m)/[m(m-5)] = (4m-15)/[m(m-5)]
Answer: A
Question: In one alloy there is 12% chromium while in another alloy it is 8%. 15 kg of the first alloy was melted together with 40 kg of the second one to form a third alloy. Find the percentage of chromium in the new alloy.
A.8.8%
B.9%
C.9.2%
D.9.1%
Answer: D.9.1%
Explanation: The amount of chromium in the new 15+40=55 kg alloy is 0.12*15+0.08*40=5 kg, so the percentage is 5/55*100=9.1%.
Answer: D.
Question: The cost price of an article is 64% of the marked price. Calculate the gain percent after allowing a discount of 12%?
A.37.5%
B.37.6%
C.38.5%
D.17.5%
Answer: A.37.5%
Explanation: Let marked price = Rs. 100.
Then, C.P. = RS. 64, S.P. = Rs. 88
Gain % = 24/64 * 100 = 37.5%.
Answer: A
Question: Running at the same constant rate, 5 identical machines can produce a total of 270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes?
A. 648
B. 1,800
C. 2,160
D. 10,800
Answer: C. 2,160
Explanation: 5 machines produce 270 bottles per minute;
1 machine produces 270/5=54 bottles per minute;
10 machines produce 54*10=540 bottles per minute;
In 4 minutes 10 machines produce 540*4=2,160 bottles.
Answer: C.
Question: An empty wooden vessel weighs 14% 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.
A.6/13
B.5/9
C.1/24
D.4/9
Answer: A.6/13
Explanation: An empty wooden vessel weighs 14% of its total weight when filled with paint:
Vessel = 0.14(Vessel + Paint);
14V = V + P (so the weight of completely filled vessel is 14V)
P = 13V (so the weight of the paint when the vessels is completely filled is 13V).
The weight of a partially filled vessel is one half that of a completely filled vessel:
V + P' = 1/2*14V;
P' = 6V (so the weight of the paint when the vessels is partially filled is 6V).
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 = 6V/13V = 6/13.
Answer: A.
Question: 2 numbers are there in the ratio 3:4. If 3 is to be added to both of the numbers the ratio will become 7:9 then find initial numbers?
A.18,24
B.15,12
C.3,2
D.21,22
Answer: A.18,24
Explanation: Let 2 numbers are 3x,4x
after adding 3 on both numbers 3x+3, 4x+3
3x+3:4x+3 = 7:9
9(3x+3) = 7(4x+3)
x = 6
Intial numbers = 3x,4x = 18,24
Answer is A
Question: Two trains of equal lengths take 10 sec and 14 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction?
A.10
B.12
C.13
D.11.7
Answer: D.11.7
Explanation: Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/14 = 8.6 m/sec.
Relative speed = 12 + 8.6 = 20.6 m/sec.
Required time = (120 + 120)/20.6 = 11.7 sec.
Answer: Option D
Question: In a zoo, the ratio of the number of cheetahs to the number of pandas is 1 : 3 and was the same five years ago. If the increase in the number of cheetahs in the zoo since then is 6, then what is the increase in the number of pandas?
A. 18
B. 3
C. 5
D. 10
Answer: A. 18
Explanation: One short cut to solve the problem is
c:p = 1:3
c increased to 7
=> 1:3 = 7:x
=> x = 21
=>p increased by 18
A is the answer
Question: A watch which gains uniformly is 2 minutes low at noon on Monday and is 4 min. 48 sec. fast at 2 p.m. on the following Monday. When was it correct?
A.2 p.m. on Tuesday
B.2 p.m. on Wednesday
C.3 p.m. on Thursday
D.1 p.m. on Friday
Answer: B.2 p.m. on Wednesday
Explanation: Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.
∴ The watch gains (2 + 44⁄5) min.
or 34⁄5 min. in 170 hrs.
Now, 34⁄5 min. are gained in 170 hrs.
∴ 2 min. are gained in (170 × 5⁄34 × 2)hrs. = 50 hrs
∴ Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e. it will be correct at 2 p.m. on Wednesday.
Answer B
Question: Seven machines at a certain factory operate at the same constant rate. If six of these machines, operating simultaneously, take 35 hours to fill a certain production order, how many fewer hours does it take all seven machines, operating simultaneously, to fill the same production order?
A.4
B.5
C.7
D.10
Answer: B.5
Explanation: The total work is 6*35 = 210 machine hours
The time required for seven machines is 210/7 = 30 hours, thus 5 fewer hours.
The answer is B.
Question: A student got twice as many sums wrong as he got right. If he attempted 75 sums in all, how many did he solve correctly ?
A.25
B.16
C.18
D.24
Answer: A.25
Explanation: Explanation:
Suppose the boy got x sums right and 2x sums wrong.
Then, x + 2x = 75 3x = 75 x = 25.
Answer:A
Question: A does a work in 10 days and B does the same work in 30 days. In how many days they together will do the same work ?
A.5.5 days
B.6.5 days
C.7.5 days
D.8.5 days
Answer: C.7.5 days
Explanation: Explanation:
Firstly we will find 1 day work of both A and B, then by adding we can get collective days for them,
So,
A's 1 day work = 1/10
B's 1 day work = 1/30
(A+B)'s 1 day work =
(1/10+1/30)=2/15
So together they can complete work in 7.5 days.
Answer: C
Question: Winners will be announced the following day at 9 AM Pacific/1 PM Eastern Time.
If A, X, Y, and Z are unique nonzero digits in the equation:
XY
*YX
____
AYZ
And Y > X, the 2-digit number XY = ?
A.13
B.21
C.23
D.24
Answer: A.13
Explanation: The reasoning is as follows:-
B) can't be answer because Y<X
D) can't be answer because it XY x YX will give 4 digit number. Thus wrong
E) can't be answer because it XY x YX will give 4 digit number. Thus wrong
A) can't be answer because it XY x YX will give 3 digit number but the middle digit will not be equal to Y. Thus wrong
C) is the correct answer because Y>xwill give 3 digit numbermiddle digit is Y
A