mathewhe/aqua_rat huggingface dataset:
Question: A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
(A) 65 hrs
(B) 99 hrs
(C) 35 hrs
(D) 45 hrs
(E) 32 hrs
Rationale: Suppose pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.
1/x + 2/x + 4/x = 1/5
7/x = 1/5 => x = 35 hrs.Answer: C
Answer: (C) 35 hrs
Question: How many ways could three bottle caps can be used to close five bottles of which two of the bottles will remain opened?
(A) 8
(B) 12
(C) 60
(D) 118
(E) 120
Rationale: ways in which 3 bottles out of 5 can be selected = 5C3
ways in which 3 caps can be used to close 3 bottles = 3!
Total ways of arrangement = 5C3*3!
=5*4/2*6
=10*6
=60
Ans = C
Answer: (C) 60
Question: If the simple interest on a sum of money for 4 years at 8% per annum is Rs. 560, what is the compound interest on the same sum at the rate and for the same time?
(A) Rs. 640.86
(B) Rs. 430.86
(C) Rs. 330.86
(D) Rs. 630.86
(E) Rs. 130.86
Rationale: Sum = (50 * 100) / (2 * 5) = Rs. 1,750.00
C.I. on Rs. 1,750.00 for 4 years at 8% = Rs. 2,380.86.
= Rs. 2,380.86 - 1,750.00 = Rs. 630.86
ANSWER:D
Answer: (D) Rs. 630.86
Question: A certain industrial loom weaves 0.128 meters of cloth every second. Approximately how many seconds will it take for the loom to weave 25 meters of cloth?
(A) 205
(B) 200
(C) 180
(D) 195
(E) 175
Rationale: Explanation:
Let the required number of seconds be x
More cloth, More time, (direct proportion)
Hence we can write as
(cloth) 0.128 : 25 :: 1 : x
⇒0.128x=25
⇒x=25/0.128
=25000/128
=3125/16≈195
Answer: Option D
Answer: (D) 195
Question: How many 5 letter combinations can be made from the letters of the word VERMONT if the first letter has to be a vowel and the last letter has to be a consonant, and each letter can be used only once?
(A) 21
(B) 42
(C) 120
(D) 600
(E) 720
Rationale: VERMONT consists of 2 vowels and 5 consonants:
The number of way to choose a vowel for the first letter = C12==2.
The number of way to choose a consonant for the last letter= C15==5.
The number of ways to choose middle three letters out of 5 remaining letters = C35==10.
The number of arrangements of these three letters = 3! = 6.
Total = 2*5*10*6 = 600.
Answer: D.
Answer: (D) 600
Question: A and B can do a piece of work in 6 2/3 days and 5 days respectively. They work together for 2 days and then A leaves. In how many days after that B will complete the work alone.
(A) 2 days
(B) 1 ½ days
(C) 3 days
(D) 3 ½ days
(E) 4 days
Rationale: Explanation:
3/20 * 2 + (2 + x)/5 = 1
x = 1 ½ days
Answer: Option B
Answer: (B) 1 ½ days
Question: 10 people are assigned a library of audio clips to be transcribed in 100 days. If 2 people have to take 25 days leave in between, how many more people should be added on the last day to complete the task in 100 days itself?
(A) 25
(B) 50
(C) 75
(D) 100
(E) 15
Rationale: B
For 25 days, instead of 10, only 8 people were transcribing the audio clips.
If 1 person transcribes N number of audio clips in 1 day, additional backlog, 25 X N X 2
10 : M :: 10 X N : 25 X 2 X N = 10 : 50
Thus, M = 10 X 50 / 10 = 50
Answer: (B) 50
Question: In a group of 120 people, 90 have an age of more 30 years, and the others have an age of less than 20 years. If a person is selected at random from this group, what is the probability the person's age is less than 20?
(A) 0.25
(B) 0.55
(C) 0.65
(D) 0.75
(E) 0.85
Rationale: Number of people whose age is less than 20 is given by
120 - 90 = 30
Probability P that a person selected at random from the group is less than 20 is gieven by
30 / 120 = 0.25
correct answer A
Answer: (A) 0.25
Question: A car travels first half distance between two places with a speed of 40 kmph and the rest half distance with a speed of 60 kmph. The average speed of the car is
(A) 48 kmph
(B) 37 kmph
(C) 44 kmph
(D) 45 kmph
(E) None of these
Rationale: we know that speed = distance traveled/time taken
let the total distance traveled by the car is 2x km.
then time taken by it to cover first half is x/60 hour.
and for second half is x/40 hour.
then average speed= total distance traveled/ total time
taken.
=2x/[(x/60) + (x/40)]
=2x/[5x/120] (on simplification)
=48 kmph
ANSWER:A
Answer: (A) 48 kmph
Question: Ron begins reading a book at 2: 30 p.m. and reads at a steady pace of 30 pages per hour. Michelle begins reading a copy of the same book at 4: 00 p.m. If Michelle started 5 pages behind the page that Ron started on and reads at an average pace of 50 pages per hour, at what time would Ron and Michelle be reading the same page?
(A) 5: 00 p.m.
(B) 6: 30 p.m.
(C) 7: 45 p.m.
(D) 8: 00 p.m.
(E) 8: 30 p.m.
Rationale: M gains 50-30=20 pages per hour on R
at 4pm R has read 45 pages and M is 45+5=50 pages behind him
time=50/20=2.5 hours
they are on the same page at 6:30pm
B
Answer: (B) 6: 30 p.m.
Question: In a class there are 20 boys and 21 girls. In how many ways can a boy and a girl be selected?
(A) 508
(B) 420
(C) 800
(D) 570
(E) 503
Rationale: We can select one boy from 20 boys in 20 ways.
We select one girl from 21 girls in 21 ways
We select a boy and girl in 20 * 21 ways i.e., = 420 ways.
Answer:B
Answer: (B) 420
Question: In a class of 36 students 26 play football and play 20 long tennis, if 17 play above, many play neither?
(A) 7
(B) 8
(C) 10
(D) 12
(E) 14
Rationale: 26 + 20 - 17 = 29
36 - 29 = 7 play neither
Answer is A
Answer: (A) 7
Question: if (p/q)-(q/p)=21/10
then what ia the value of(4p/q)+(4q/p)
(A) 58/5
(B) 59/5
(C) 121/105
(D) 101/105
(E) 1/2
Rationale: p/q =a then q/p = 1/a
a+1/a=21/10
by solving this we get a=5/2
then 4*5/2+4*2/5 = 58/5
ANSWER:A
Answer: (A) 58/5
Question: The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table?
(A) 2299
(B) 2699
(C) 2670
(D) 6725
(E) 2601
Rationale: CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.
Answer:D
Answer: (D) 6725
Question: Average weight of 10 people increased by 3 kg when one person of 45 kg is replaced by a new
man. Then weight of the new man is
(A) 50
(B) 55
(C) 60
(D) 65
(E) 75
Rationale: Explanation:
Total weight increased is 3 * 10 = 30.
So weight of new person is 45+30 = 75
Answer: Option E
Answer: (E) 75
Question: The average age of 32 students in a group is 16 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years?
(A) cannot be determined
(B) None of these
(C) 48
(D) 49
(E) 50
Rationale: Age of the teacher = (33 × 17 – 32 × 16) years
= 49 years.
Answer D
Answer: (D) 49
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 ?
(A) 11
(B) 13
(C) 44
(D) 55
(E) 61
Rationale: Explanation:
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. Then, 2x + 3y = 77 ...(i) and 3x + 2y = 73 ...(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17. Putting y = 17 in (i), we get: x = 13.
Answer: B
Answer: (B) 13
Question: A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
(A) 6%
(B) 7%
(C) 8%
(D) 4%
(E) 1%
Rationale: SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100-L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CP-SP = (6250/3)- 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100
=0.04 * 100
= 4%
Answer: D
Answer: (D) 4%
Question: Two trains 100 m and 60 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
(A) 10.9 sec
(B) 5.76 sec
(C) 10.6 sec
(D) 10.8 sec
(E) 20.8 sec
Rationale: Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 100 + 60 = 160 m.
Required time = 160 * 9/250 = 5.76
= 5.76 sec.
Answer: B
Answer: (B) 5.76 sec
Question: From a group of 25 astronauts that includes 13 people with previous experience in space flight, a 3-person crew is to be selected so that exactly 1 person in the crew has previous experience in space flight. How many different crews of this type are possible?
(A) 785
(B) 858
(C) 864
(D) 964
(E) 722
Rationale: number of astronauts with previous experience = 13
Remaining =12
we need to pick exactly one from the first group
and 2 from the second group
(13C1)*(12C2)
=13*66= 858
Answer : B
Answer: (B) 858
Question: A high school has 400 students 1/2 attend the airthmetic club, 5/8 attend the biology club and 3/4 attend the chemistry club. 1/4 attend all 3 clubs. If every student attends at least one club how many students attend exactly 2 clubs.
(A) 100
(B) 150
(C) 140
(D) 120
(E) 110
Rationale: A-Club has 200 members (1/2 of 400)
B-Club has 250 members (5/8 of 400)
C-Club has 300 members (3/4 of 400)
We can create an equation to solve this:
200+250+300 = n + x + 2y
where n is the number of students, x is the number of students in two clubs, and y is the number of students in three clubs.
The question provides y for us (100).
750 = 400 + x + 200
x = 150
B
Answer: (B) 150
Question: Tough and Tricky questions: Combinations.
Five contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?
(A) 6
(B) 7
(C) 12
(D) 15
(E) 24
Rationale: Four contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded?
We have 2 slots to be filled using 5 contestants:5 options for slot1*3 option for slot2
= 5* 3
= 15
Ans. D) 15
Answer: (D) 15
Question: If -2 < h < 0, which of the following has the greatest value?
(A) a) 1-h
(B) b) 1+h
(C) c) 1+h^2
(D) d) 1-1/h
(E) e) 1-1/(h^2)
Rationale: easiest way of solving this would be to substitute a value.
lets say h= - 1/2
a) 3/2
b) 1/2
c) 5/4
d) 3
e) - 3
the value you chose is wrong . h cannot be 0.9 as -1<h<0. if you chose -0.9 answer will still be D.
Answer is E.
Answer: (E) e) 1-1/(h^2)
Question: A local business made a $125,000 profit on its first $1 million in sales and a $80,000 profit on its next $2 million in sales. By approximately what percent did the ratio of profit to sales decrease from the first $1 million in sales to the next $2 million in sales?
(A) 100%
(B) 150%
(C) 200%
(D) 250%
(E) 300%
Rationale: Solution:
This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.
We first set up the ratios of profits to sales. The first ratio will be for the first 1 million in sales, and the second ratio will be for the next 2 million in sales. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.
First 1 Million
profit/sales = 125/1000 = 1/8
Next 2 Million
profit/sales = 8/200 = 2/50
We can simplify each ratio by multiplying each by the LCM of the two denominators, which is 400. Keep in mind that we are able to do this only because our answer choices are expressed in percents.
First 1 Million
royalties/sales = (1/8) x 400 = 50
Next 2 Million
royalties/sales = (2/50) x 400 = 16
We can plug 5 and 6 into our percent change formula:
(new – old)/old x 100
[(16 – 50)/16] x 100
=-34/16 x 100 so approximately a 200% decrease.
Answer C.
Answer: (C) 200%
Question: A 6 litre sol is 20% alcohol .How many litres of pure alcohol must be added to produce a sol that is 50% alcohol ?
(A) A.0.6
(B) B.1
(C) C.2.1
(D) D.3.6
(E) E.5.4
Rationale: 20% of 6 = 1.2
50% of 6 = 3
shortage is 1.8
so we need to have 1.8/50% to get 50% alcohol content.
= 3.6 D
Answer: (D) D.3.6
Question: A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?
(A) Rs 300
(B) Rs 250
(C) Rs 400
(D) Rs 500
(E) None of these
Rationale: Explanation :
Let x be the total no. of people the college will ask for donation.
The people already solicited 0.6x.
Amount raised from the people solicited 600 x 0.6x = 360x.
Now, 360 constitutes 75% of the amount.
Hence, the remaining 25% = 120x.
Hence, the average donation from remaining people
= 120x / 0.4x = 300.
Answer : A
Answer: (A) Rs 300
Question: For any two numbers X and Y , define X# Y=X+Y -XY . If X,Y>0.5, then X#X < Y#Y if
i. 1>X>Y
ii. 1>Y>X
iii. Y>X>1
(A) i only
(B) option ii only
(C) i and ii only
(D) ii and iii only
(E) i and ii only
Rationale: X#X = 2X-X^2 and Y#Y=2Y-Y^2
X#X <Y#X
=> 2X - X^2 < 2Y - Y^2
=> 2(X-Y) < (X-Y)(X+Y)
=> X+Y>2 and X-Y>0 or X+Y<2 and X-Y<0
=> X+Y>2 and X>Y or X+Y<2 and X<Y
Using (i), if 1>X>Y, then the first case applies but X+Y>2 is not satisfied. False.
Using (ii), if 1>Y>X, then the second case applies and X+Y<2 is satisfied. True.
Using (iii), if Y>X>1, then the second case applies but X+Y<2 is not satisfied. False.
Therefore only (ii) is true, and choice (B) is correct.
Answer: (B) option ii only
Question: Find the distance covered by a man walking for 24min at a speed of 10km/hr?
(A) 1km
(B) 3km
(C) 4km
(D) 5km
(E) 6km
Rationale: Distance = 10*24/60 = 4 km
Answer is C
Answer: (C) 4km
Question: There are 5 locks and 5 keys and each of the 5 keys matches each of the 5 locks. What is the minimum and the maximum trial numbers of attempts F needed to confirm that each of the 5 keys matches each of the 5 locks?
(A) 5,15
(B) 4,15
(C) 5,10
(D) 4,10
(E) 5,20
Rationale: Assume you have locks 1-5 and keys A-E.
Minimum: assume you are lucky to find the correct key/combo on the first try. So, 1 -> A, 2 -> B, 3 ->C, and 4 -> D, then 5 must match with E. Therefore, you only need to try 4 combos at a minimum.
Maximum: assume that it takes as many guesses as possible. So, with the first key you try A, B, C, and D with no success, therefore E must be the match (so 4 attempts). For key 2 you no longer have E available so you try A, B, and C, with no success, therefore D must be the match (3 attempts). And so on for key 3 (2 attempts) and key 4 (1 attempt). Key 5 matches with the remaining lock for a total of F= 4 + 3 + 2 + 1 = 10 attempts.D
Answer: (D) 4,10
Question: Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum K in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is
(A) 90%
(B) 80%
(C) 70%
(D) 60%
(E) 40%
Rationale: The problem asks for the approximate chance that no more than 1/3 of the original investment is lost. We can apply the “1 – x” technique: what’s the chance that more than 1/3 of the original investment is lost? There are two outcomes we have to separately measure:
(a) All 3 investments become worthless.
(b) 2 of the 3 investments become worthless, while 1 doesn’t.
Outcome (a): The probability is (0.2)(0.2)(0.2) = 0.008, or a little less than 1%.
Outcome (b): Call the investments X, Y, and Z. The probability that X retains value, while Y and Z become worthless, is (0.8)(0.2)(0.2) = 0.032. Now, we have to do the same thing for the specific scenarios in which Y retains value (while X and Z don’t) and in which Z retains value (while X and Y don’t). Each of those scenarios results in the same math: 0.032. Thus, we can simply multiply 0.032 by 3 to get 0.096, or a little less than 10%.
The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.
The correct answer is A.
This problem illustrates the power of diversification in financial investments. All else being equal, it’s less risky to hold a third of your money in three uncorrelated (independent) but otherwise equivalent investments than to put all your eggs in one of the baskets. That said, be wary of historical correlations! Housing price changes in different US cities were not so correlated—and then they became highly correlated during the recent housing crisis (they all fell together), fatally undermining spreadsheet models that assumed that these price changes were independent.
Answer: (A) 90%
Question: A man is walking at the rate of 6 km/hr crosses a bridge in 15 minutes. The length of the bridge is
(A) 1000 meters
(B) 1050 meters
(C) 1200 meters
(D) 1500 meters
(E) None of these
Rationale: Explanation:
We need to get the answer in meters. So we will first of change distance from km/hour to meter/sec by multiplying it with 5/18 and also change 15 minutes to seconds by multiplying it with 60.
Speed=6âˆ—5/18=5/3m/sec
Time=15âˆ—60seconds=900seconds
Distance=Timeâˆ—Speed/Distance=5/3âˆ—900=1500meter
Option D
Answer: (D) 1500 meters
Question: if a three-digit number is selected at random from the integers 100 to 999, inclusive, what is the probability that the first digit and the last digit of the integer will both be exactly two less than the middle digit?
(A) 1:900
(B) 7:900
(C) 1:1,000
(D) 1:100
(E) 7:100
Rationale: There are 900 numbers between 100 and 999 (inclusive).
Numbers of the form ABA are: 131, 242,353,464,575,686,797 = total 7
Thus the probability asked = 7/900
B is the correct answer.
Answer: (B) 7:900
Question: Anil purchased 30 kg of wheat at the rate of Rs. 22 per kg and 20 kg of wheat at the rate of Rs. 24 per kg. He mixed the two. Approximately at what rate per kg should he sell the mixture to make 25% profit?
(A) Rs. 25
(B) Rs. 28.5
(C) Rs. 30
(D) Rs. 32.5
(E) None of these
Rationale: Explanation:
Total quantity of mixture = 30 + 20 = 50 kg.
By earning a profit of 25% on the total cost, the selling price of the mixture is
125/100 * (660 + 480 = 1140) = Rs. 1425
Selling price per kg = Total selling price / Total of mixture
= 1425 / 50
= Rs. 28.5
ANSWER B
Answer: (B) Rs. 28.5
Question: A pipe can empty 2/3rd of a cistern in 12mins. In 5 mins, what part of the cistern will be empty?
(A) 2/3
(B) 5/18
(C) 4/9
(D) 3/8
(E) 4/7
Rationale: 2/3 ---- 12
? ----- 5 ==> 5/18
B
Answer: (B) 5/18
Question: The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second?
(A) 69%
(B) 79%
(C) 60%
(D) 64%
(E) 24%
Rationale: r = 3 πr2 = 9
r = 5 πr2 = 25
25 π – 16 π
100 ---- ? => 64%
Answer: D
Answer: (D) 64%
Question: If a < b < c , which of the following must be true?
(I) a < b^2
(II) b − a > c
(III) a^2 < b^2 < c^2
(A) II only
(B) I only
(C) None
(D) III only
(E) II and III
Rationale: (I) a < b^2. Not necessarily true fractions: try a=1/3 and b=1/2.
(II) b − a < c. Not necessarily true: try a=-1/2, b=-1/3, and c=-1/4.
(III) a^2 < b^2 < c^2. Not necessarily true: try a=-1, b=0, and c=1.
None of the options must be true.
Answer: C.
Answer: (C) None
Question: According to a survey, at least 50% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?
(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%
Rationale: To minimize the percentage who like all three, let's maximize the percentage who like 2 out of 3.
80% like cherries and 20% don't like cherries.
Let's combine this with the people who like bananas.
20% don't like cherries but do like bananas.
55% like cherries and bananas.
25% like cherries but not bananas.
Now let's distribute the people who like apples.
20% like bananas and apples, but not cherries.
25% like cherries and apples, but not bananas.
5% like all three.
50% like cherries and bananas, but not apples.
The answer is A.
Answer: (A) 5%
Question: A pen factory stores its toys in either a big box or a small box. The big box can store 27 pens while the small box can store 18 pens. If the factory is to store 112 newly manufactured pens, what is the least number of pens that will be left unboxed?
(A) 6
(B) 2
(C) 4
(D) 3
(E) 1
Rationale: Total pens to be stored 112
The big box can store 27
small box can store 18
storing 112 pens in big box 112/27 =4 r 4
least number of pens that will be left unboxed=4
Answer: C
Answer: (C) 4
Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
(A) 278
(B) 240
(C) 88
(D) 21
(E) 192
Rationale: Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train = (15 x 20)m = 300 m. Let the length of the platform be x meters. Then, (x + 300)/36 = 15 ==> x + 300 = 540 ==> x = 240 m Answer:B
Answer: (B) 240
Question: If the price of petrol increases by 44, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?
(A) 25%
(B) 20%
(C) 16.67%
(D) 33.33%
(E) None of these
Rationale: Explanation :
let us assume before increase the petrol will be Rs. 100.
After increase it will be Rs (100+44) i.e 144.
Now, his consumption should be reduced to :-
=(144−100)/144 ∗ 100.
Hence, the consumption should be reduced to 30.6%.
Answer : E
Answer: (E) None of these
Question: The vertices of a rectangle in the standard (x,y) coordinate place are (0,0), (0,6), (12,0) and (12,6). If a line through (1,1) partitions the interior of this rectangle into 2 regions that have equal areas, what is the slope of this line?
(A) 1/2
(B) 2/3
(C) 1/4
(D) 2/5
(E) 5/12
Rationale: We should know that a rectangle can be divided into two equal areas by a straight line only when the straight line passes through the center of the rectangle.
The center of the rectangle is the point (6,3).
The slope of line passing through points (1,1) and (6,3) = (3-1)/(6-1) = 2/5
The answer is D.
Answer: (D) 2/5
Question: A envelop weight 8.5 gm,if 800 of these envelop are sent with an advertisement mail. How much wieght?
(A) 6.6kg
(B) 6.8kg
(C) 6.7kg
(D) 6.9kg
(E) 7.8kg
Rationale: 800*8.5
6800.0gm
6.8kg
ANSWER:B
Answer: (B) 6.8kg
Question: If XY denotes X is raised to the power Y, Find the last two digits of 19413843+ 19614181
(A) 12
(B) 22
(C) 42
(D) 82
(E) 92
Rationale: 1941^3843
unit digit= 1^3=1
10th digit=4*3=12=2 means last two digit =21
and 1961^4181
unit digit=1
10th digit=6*1=6 mean last two digit= 61
hence,
21+61= 82
ANSWER:D
Answer: (D) 82
Question: Ravi can do a piece of work in 20 days while Prakash can do it in 40 days. In how many days will they finish it together?
(A) 13 1/3 days
(B) 15 1/3 days
(C) 17 1/3 days
(D) 17 7/3 days
(E) 18 1/3 days
Rationale: 1/20 + 1/40 =3/40
40/3
= 13 1/13 days
Answer: A
Answer: (A) 13 1/3 days
Question: In a camp, there is a meal for 180 men or 200 children. If 150 children have taken the meal, how many men will be catered to with the remaining meal ?
(A) 38 men
(B) 32 men
(C) 34 men
(D) 45 men
(E) None of these
Rationale: Explanation:
There is meal for 200 children. 150 children have taken the meal.
Remaining meal is to be catered to 50 children.
Now, 200 children = 180 men
So, 50 Children = 180/200 * 50 = 45 men
Answer: D
Answer: (D) 45 men
Question: A set of numbers has the property that for any number t in the set, t + 2 is in the set. If -1 is in the set, which of the following must also be in the set?
I. 1
II. -3
III. 5
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
Rationale: The question is which of the following must be in the set, not could be in the set.
If -1 is in the set so must be -1+2=1, as 1 is in the set so must be 1+2=3, as 3 is in the set so must be 3+2=5 and so on. So basically knowing that -1 is in the set we can say that ALL odd numbers more than -1 are also in the set.
Answer: C.
Answer: (C) I and III only