RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: #88
A necklace is made by stringing Q no individual beads together in the repeating pattern red bead, green bead, white bead, blue bead, and yellow bead. If the necklace design begins with a red bead and ends with a white bead, then Q could equal
A.16
B.28
C.41
D.54
Answer: B.28
Explanation: You can just write out the pattern and count: RGWBYRGWBYRGWBY... But to save time a good test taker will just look for a pattern. Min # is 3, because W is the third one. Then every 5 beads another white comes, so it must be 3+5+5+5.. and so on...
3+5 = 8
3+5+5 = 13
3+5+5+5=18
3+5+5+5+5=23
So you see it ends in either 8 or 3. Pick an answer that ends in either 8 or 3. Only one answer does, b.
Question: A swimmer jumps from a bridge over a canal and swims one kilometer stream up. After that first kilometer, he passes a floating cork. He continues swimming for half an hour, then turns around, and swims back to the bridge. The swimmer and the cork arrive at the bridge at the same time. The swimmer has been swimming with constant effort. How fast does the water in the canal flow?
A.3 km/h
B.2 km/h
C.1 km/h
D.4 km/h
Answer: C.1 km/h
Explanation: C
1 km/h
If you have written down a full paper of mathematical formulas, you have been thinking too complicatedly...
It is obvious that the cork does not move relatively to the water (i.e. it has the same speed as the water). Therefore, if the swimmer is swimming away from the cork for half an hour (upstream), it will take him another half hour to swim back to the cork again. Because the swimmer is swimming with constant effort, his speed is constant relatively to the speed of the water. You can look at it as if the water in the river does not move, the cork does not move, and the swimmer swims a certain time away from the cork and then back. Consequently, in that one hour time, the cork has floated from 1 kilometer up stream to the bridge.
Conclusion: The water in the canal flows at a speed of 1 km/h.
Question: Three years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now?
A.18 years
B.24 years
C.27 years
D.28 years
Answer: B.24 years
Explanation: Solution
Present age of (A+B)	= (18x2+3x2) years
= 42 years.
Present age of (A+B+C)	= (22x3) years
= 66 years.
∴ C's age	= (66 - 42) years
= 24 years.
Answer B
Question: If x+|x|+y=5 and x+|y|-y=6 what is x+y=?
A.1
B.-1
C.3
D.12
Answer: D.12
Explanation: If x<0 and y<0, then we'll have x-x+y=7 and x-y-y=6. From the first equation y=7, so we can discard this case since y is not less than 0.
If x>=0 and y<0, then we'll have x+x+y=7 and x-y-y=6. Solving gives x=4>0 and y=-1<0 --> x+y=3. Since in PS questions only one answer choice can be correct, then the answer is C (so, we can stop here and not even consider other two cases).
Answer: C.
Adding both eqn we get 2x +IxI+IyI = 13
Now considering X <0 and y>0
2x-x +y = 13
we get x+y =12
Hence answer should be D
Question: A man purchases 2 watches for $ 7650. He sells one at 25% profit and the other at a loss of 20%. He neither gains nor loses on the whole. Thus, the cost price of two watches are :
A.$ 3400 and $ 7650
B.$7650and $ 4250
C.$3400 and $4250
D.$7650 and $3400
Answer: C.$3400 and $4250
Explanation: x + y = 7650 --- (1)
1.25x + 0.8y = 7650 --- (2)
(2) - (1) --> 0.25x - 0.2y = 0
25x = 20y
x/y = 4/5 = (4*850)/(5*850) = 3400/4250
Answer: C
Question: n and p are integers greater than 1;
5n is the square of a number;
75 np is the cube of a number.
The smallest value for n + p is
A.14
B.18
C.20
D.30
Answer: A.14
Explanation: n>1
p>1
5∗n=x^2
75 np=y3
n+p=?
Well, a square number is usually the result of himself times himself so in the case of n the least value possible (being 5 a prime) is 5*5
n=5
as for p
75∗5∗p=y3
we know 375 is equal = 53∗31 therefore to obtain a cube we need 32 or 9 -> p=9p=9
p+n=14
ANSWER:A
Question: A truck covers a distance of 376 km at a certain speed in 8 hours. How much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?
A.6 hours
B.5 hours
C.7 hours
D.8 hours
Answer: A.6 hours
Explanation: Explanation :
Speed of the truck = Distance/time = 376/8 = 47 kmph
Now, speed of car = (speed of truck + 18) kmph = (47 + 18) = 65 kmph
Distance travelled by car = 376 + 14 = 390 km
Time taken by car = Distance/Speed = 390/65 = 6 hours.
Answer – A
Question: A man sells 45 lemons for Rs 40 and loses 20%. At how much price should he sell 24 lemons to the next customer to make a 20% profit?
A.37
B.299
C.72
D..32
Answer: D..32
Explanation: Explanation:
Let cost price of lemons be x.
Selling price of lemons becomes CP - loss.
=>SP=x-(20/100)x
=>40=x(80/100)
=>50
So 45 lemons cost Rs 50
Cost of 1 lemon = 50/45 Rs
Cost of 24 lemons = (50/45)*24 =80/3 Rs
Selling Price of 24 lemons =(80/3)+(20/100)(80/3)=Rs 32
ANSWER:D
Question: What least number should be added to 1021, so that the sum is completely divisible by 25?
A.4
B.3
C.2
D.0
Answer: A.4
Explanation: 1021 Ã· 25 = 40 with remainder = 21
21 + 4 = 25.
Hence 4 should be added to 1021 so that the sum will be divisible by 25
Answer: Option A
Question: A and B put in Rs.300 and Rs.500 respectively into a business. A reinvests into the business his share of the first year's profit of Rs.200 where as B does not. In what ratio should they divide the second year's profit?
A.39:40
B.39:49
C.15:16
D.14:18
Answer: C.15:16
Explanation: Explanation:
3: 5
A = 3/8*200 = 75
375: 400
39:40
Answer:C
Question: Calculate the average mark scored by Nancy if she obtained the following scores in an examination: 65, 78, 89, 51 and 69 marks (out of 100) in English Language, Physics, Mathematics, Chemistry and Biology?
A.40.4
B.60.4
C.50.4
D.70.4
Answer: D.70.4
Explanation: Explanation:
Average mark = 65+78+89+51+69/5=70.4
Answer: Option D
Question: If 183 is divisible by 11, find the value of the smallest natural no. N?
A.5
B.7
C.9
D.10
Answer: B.7
Explanation: In aptitude tests , we get questions on divisibility by 11.
A number is divisible by 11, when the difference between the sum of digits at even places and at odd places is 0 or multiple of 11
The given number is N183.
(Sum of digits at EVEN places) – (sum of digits at ODD places)=0
(8 + N)- (3+1)= 0
(8 + N) - 4 = 0
Here the value of N must be 7
B
Question: The salary of a person was reduced by 35%. By what percent should his reduced salary be raised so as to bring it at par with his original salary?
A.50%
B.32%
C.25%
D.54%
Answer: D.54%
Explanation: Let the original salary be $100
New salary = $65
Increase on 65 = 35
increase on 100 = 35/65 *100 = 54%(APPROXIMATELY)
Answer is D
Question: A hostel had provisions for 250 men for 36 days. If 50 men left the hostel, how long will the food last at the same rate?
A.30
B.45
C.50
D.60
Answer: B.45
Explanation: A hostel had provisions for 250 men for 36 days
If 50 men leaves the hostel, remaining men = 250 - 50 = 200
We need to find out how long the food will last for these 200 men.
Let the required number of days = x days
More men, Less days (Indirect Proportion)
(men) 250 : 200 :: x : 36
250×36=200x
5×36=4x
x=5×9=45
ANSWER B
Question: 288 Ã· ? Ã— 15 + 270 = 405
A.24
B.25
C.32
D.39
Answer: C.32
Explanation: Explanation:
288 Ã· ? Ã— 15 = 405 - 270 = 135
(288 Ã— 15)/? = 135
? = (288Ã— 15)/135 = 32
Answer: Option C
Question: Ramesh can finish a piece of work in 14 days. Rohan is twice as efficient as Ramesh.If they work together how many days will they need to finish the same amount of work?
A.8/3 days
B.9/3 days
C.10/3 days
D.14/3 days
Answer: D.14/3 days
Explanation: Ramesh can finish a piece of work in 14 days. Rohan is twice as efficient as Ramesh.
So Rohan can finish a piece of work in 7 days.
days needed to finish the same amount of work = 1/(1/7+1/14)=14/3 days
ANSWER:D
Question: If a town of 15,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
A.a.16000
B.b.17000
C.c.18000
D.d.19000
Answer: A.a.16000
Explanation: 1% is quite small and hence the answer is A)
Question: In a renowned city, the average birth rate is 6 people every two seconds and the death rate is 3 people every two seconds. Estimate the size of the population net increase that occurs in one day.
A. 129,500
B. 129,600
C. 129,700
D. 129,800
Answer: B. 129,600
Explanation: This question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds.
For this particular question : Increase in the population every 2 seconds = 6-3 = 3 people.
Total 2 second interval in a day = 24*60*60/2 = 43,200
Population increase = 43,200 * 3 = 129,600. Hence B.
Question: How many seconds will a 600 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
A.11
B.36
C.99
D.88
Answer: B.36
Explanation: Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 600 * 3/50 = 36 sec.
Answer: B
Question: The value of (7x10^7)(7x10^8) is closest to which of the following?
A.10^11
B.10^12
C.10^17
D.10^14
Answer: C.10^17
Explanation: multiplying these we get,
49 x 10^7 x 10^8
= 4.9 x 10^16
which is closer to 10^17
Answer: C
Question: In a rectangular coordinate system, what is the area of a quadrilateral whose vertices have the coordinates (3,-1), (3, 8), (12, 2), (12,-7)?
A.69
B.75
C.81
D.87
Answer: C.81
Explanation: By graphing the points, we can see that this figure is a trapezoid. A trapezoid is any quadrilateral that has one set of parallel sides, and the formula for the area of a trapezoid is:
Area = (1/2) × (Base 1 + Base 2) × (Height), where the bases are the parallel sides.
We can now determine the area of the quadrilateral:
Area = 1/2 × (9 + 9) × 9 = 81.
The answer is C.
Question: If the product of two numbers is 51216 and their H.C.F. is 33, find their L.C.M.
A.600
B.680
C.582
D.560
Answer: C.582
Explanation: Explanation:
HCF * LCM = 51216, because we know
Product of two numbers = Product of HCF and LCM
LCM = 51216/88 = 582
Option C
Question: A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the six-digit integer is divisible by 3, and n is of the form 2k1,k24, where k represents a digit that occurs twice, how many values could n have?
A.1
B.2
C.3
D.4
Answer: D.4
Explanation: 2k1k24
Taking sum of the numericals = 2 + 1 + 2 + 4 = 9
We require the values of K such that they are also divisible by 3
201024
231324
261624
291924
Answer = 4 =D
Question: 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
A.360
B.384
C.432
D.470
Answer: B.384
Explanation: Solution:
Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
The various combinations of chairs that ensure that no two men are sitting together are listed.
(1, 3, 5,__), The fourth chair can be 5,6,10,11 or 12, hence 5 ways.
(1, 4, 8, __), The fourth chair can be 6,10,11 or 12 hence 4 ways.
(1, 5, 8, __), the fourth chair can be 10,11 or 12 hence 3 ways.
(1, 6, 8,__), the fourth chair can be 10,11 or 12 hence 3 ways.
(1,8,10,12) is also one of the combinations.
Hence, 16 such combinations exist.
In case of each these combinations we can make the four men inter arrange in 4! ways.
Hence, the required result =16*4!= 384.
Answer: Option B
Question: If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
A.25% increase
B.25% decrease
C.50% decrease
D.50% increase
Answer: D.50% increase
Explanation: Let original length =10
original breadth =10
Then, original area =10×10=100
Length is halved
⇒ New length =10/2=5
breadth is tripled.
⇒ New breadth =10×3=30
New area =5×30=150
Increase in area= new area - original area
=150−100=50
Percentage increase in area=increase area/original area×100
=50/100×100
=50%
ANSWER:D
Question: Ann's mother is 20 years older than her. In 2 years, however, Ann's mother will be three times older than Ann. How old is Ann now?
A.7 years
B.8 years
C.9 years
D.10 years
Answer: B.8 years
Explanation: At the present:
Ann's mother is 20 years older than Ann, thus
Ann's age is x
Ann's mother's age is y=x+20
In 2 years from now:
Ann's mother is three times older than Ann, thus
Ann's age is x+2
Ann's mother's age is y+2=3 (x+2)
To calculate Ann's age, we'll take both equations:
y=x+20
y+2=x+20+2=3 (x+2)
x+22=3x+6
2x=16
x=8
Ann is currently 8 years old.
ANSWER: B
Question: The radius of a semicircular plot is 25. What is the approximate perimeter of the plot?
A.114
B.150
C.129
D.25
Answer: C.129
Explanation: perimeter of a circle =2pi*r
perimeter of a semicircle = pi*r + 2r
aprox perimiter = 3.14*25 + 2*25 = 128.50 approximately 129
answer C
Question: The LCM and HCF of two numbers are 10000 and 1000 respectively. Find the larger of the two numbers if their sum is 11000.
A.10000
B.562
C.352
D.576
Answer: A.10000
Explanation: There are 2 approaches in solving this.
Methode 1.
HCF * LCM = The Actual Number.
10000* 1000= 10000000
So the answer which we are looking for has to be a factor of 10000000.
So among the options shortlist the answers by eliminating those numbers which is not divisible by 10000000. and then take the highest number as the answer as the question asks abt the highest number.
answer is A
Question: A box contains 5 pairs of shoes (10 shoes in total). If two shoes are selected at random, what it is the probability that they are matching shoes?
A.1/190
B.1/20
C.1/9
D.1/10
Answer: C.1/9
Explanation: The problem with your solution is that we don't choose 1 shoe from 20, but rather choose the needed one AFTER we just took one and need the second to be the pair of it. So, the probability would simply be: 1/1*1/9(as after taking one at random there are 9 shoes left and only one is the pair of the first one)=1/9
Answer: C.
Question: The average weight of 19 students is 15kg. by the admission of a new student the average weight is reduced to 14.6 kg. the weight of the new student is ?
A.10.6 kg
B.10.8 kg
C.7 kg
D.14.9 kg
Answer: C.7 kg
Explanation: Answer
Weight of new student = Total weight of all 20 students - Total weight of initial 19 students
= (20 x 14.6 - 19 x 15) kg
= 7 kg.
Correct Option: C
Question: A man engaged a servant on the condition that he would pay him Rs.900 and auniform after 1year service. He served only for 9months and received uniform and Rs.650, find the price of the uniform?
A.Rs.80
B.Rs.100
C.Rs.120
D.Rs.145
Answer: B.Rs.100
Explanation: 9/12 = 3/4 * 900 = 675
650
-------------
25
1/4 -------- 25
1 --------- ? => Rs.100
B
Question: A lent Rs. 5000 to B for 2 years and Rs 3000 to C for 4 years on simple interest at the same rate of interest and received Rs 2200 in all from both of them as interest. The rate of interest per annum is
A.9%
B.10%
C.11%
D.12%
Answer: B.10%
Explanation: Explanation:
Let R% be the rate of simple interest then,
from question we can conclude that
(5000∗R∗2/100)+(3000∗R∗4/100)=2200
<=>100R+120R=2200
<=>R=10%
Option B
Question: A child was looking for his father. He went 90 m in the east before turning to his right. he went 20 m before turning to his right afain to lok for his father at his uncles place 30 m from this point. His father was not there. From there he went 100m north before meeting hiss father in a street. How far did the son meet his father from the starting point.
A.137
B.100
C.2977
D.126
Answer: B.100
Explanation: From the diagram, AB = 90 - 30 = 60 and BD = 100 - 20 = 80
AD=AB2+BD2−−−−−−−−−−√=602+802−−−−−−−−√=100AD=AB2+BD2=602+802=100
Answer:B
Question: Enrollment in City College in 1980 was 85/3 percent of enrollment in 1990. What was the percent increase in the college’s enrollment from 1980 to 1990?
A.100%
B.16 2/3 %
C.200%
D.253%
Answer: D.253%
Explanation: Assume 100 enrollments present in 1990
85/3 % of 1990 = enrollments on 1980
enrollments on 1980 = 28.33
% change = (enrollment in 1990 - enrollment in 1980)*100/(enrollment in 1980)
=(100 - 28.33) *100 / 28.333
= 7166.67 / 28.333
= 253 %
Ans - D
Question: The 15 homes in a new development are each to be sold for one of three different prices so that the developer receives an average (arithmetic mean) of $200,000 per home. If 4 of the homes are to be sold for $170,000 each and 5 are to be sold for $200,000 each, what will be the selling price of each of the remaining 12 homes?
A.$200,000
B.$210,000
C.$215,000
D.$220,000
Answer: B.$210,000
Explanation: IMO the answer has to be D.
5 houses are being sold for 200,000. 4 houses are being sold for $30,000 less, resulting in a loss of $120,000. To make the average selling price intact i.e. $200,000, the remaining 12 houses must be sold at such a profit that it compensates for the loss of 120,000. Hence 12x=120,000.
x=10,000.
The 12 houses are sold at $10,000 profit or at $210,000. (ANSWER B)
Question: Abbey's income is 30 percent more than Abigail's income, and Abigail's income is 35 percent less than Jenny's income. What percent of Jenny's income is Abbey's income?
A.84.5%
B.81.5%
C.82.5%
D.83.5%
Answer: A.84.5%
Explanation: Jenny's income = 100 (assume);
Abigail's income = 65 (40 percent less than Jenny's income);
Abbey's income = 84.5 (30 percent more than Abigail's income).
Thus, Abbey's income (84.5) is 84.5% of Juan's income (100).
Answer: A
Question: Thirty percent of a number is 21 less than four fifth of that number. What is the number ?
A.42
B.40
C.230
D.130
Answer: A.42
Explanation: Let the number be x.
Then, 4*x/5 â€“30% of x) = 21
4x/5 â€“30x/100 = 21
x = 42.
Answer A.
Question: In a company there are 5 partners A,B,C,D,E. They invested money in the ratio of 1:2:3:4:5. If after one year if there is a profit of Rs.15,000. Then share of E will be ?
A.3000
B.4000
C.5000
D.6000
Answer: C.5000
Explanation: 5 partners A,B,C,D,E are invested money in the ratio of 1:2:3:4:5
Total profit = 15000
share of E = Profit*ratio(E)/(ratio(A)+ratio(B)+ratio(C)+ratio(D)+ratio(E))
= 15000*5/(1+2+3+4+5)=5000
ANSWER:C
Question: What is 15% of 2/3 of 0.4?
A.0.09
B.0.9
C.9
D.90
Answer: A.0.09
Explanation: The best way to solve these questions is to convert every term into fraction
(15/100)*(2/3)*(4/10) = 120/3000 = 0.04
Option A
Question: what is the remainder when 30^72^87 is divided by 11.
A.5
B.7
C.1
D.3
Answer: A.5
Explanation: let us take N=72^87
eq is..30^N/11=8^N/11
here cyclicity of 8 is 10
then N/10=72^87/10=2^87/10
here cyclicity of 2 is
then 87%4=3(3rd cycle of 2)....i.e2^3%10=8(8th cycle of8)
therefore final ans is 8^8%11=5
ANSWER:A
Question: Two friends want to buy a snack. They combine their money and find they have $3.45, consisting of quarters, dimes, and nickels. If they have 31 coins and the number of quarters is half the number of nickels, how many quarters do they have?
A.5
B.6
C.7
D.8
Answer: C.7
Explanation: Let x be the number of quarters. Then 2x is the number of nickels.
(1) 25x + 10x + 10D = 345. Then 7x + 2D = 69
(2) 3x + D = 31. Then 6x + 2D = 62
Let's subtract equation (2) from equation (1).
x = 7
The answer is C.
Question: The average age of a husband and a wife is 28 years when they were married eight years ago but now the average age of the husband, wife and child is 26 years(the child was born during the interval). What is the present age of the child?
A.2 years
B.6 years
C.9 years
D.4 years
Answer: B.6 years
Explanation: 36 * 2 = 72
26 * 3 = 78
-----------
6 years
Answer:B
Question: Rodrick mixes a martini that has a volume of 'n' ounces having 36% Vermouth and 60% Gin by volume. He wants to change it so that the martini is 25% Vermouth by volume. How many ounces of Gin must he add?
A.n/6
B.n/3
C.4n/9
D.5n/6
Answer: C.4n/9
Explanation: total V G
1 ounce 0.36 0.6
n ounce 0.36n 0.6n -------------initial expression
lets say g ounces of gin is added to this mixture
n+g 0.36n 0.6n+g --------------final expression
given that after adding g ounces of gin , V should become 25% of the total volume.
=>Volume of V/total volume = 25/100
=> 0.36n /n+g = 1/4
=> 1.44n = n+g
=> g = 4n/9
Answer is C.
Note that after we add pure Gin, the volume of Vermouth will remain the same. Based on this set the equation:
0.36n=0.25(n+g) --> g=4n/9
Answer: C.
Question: The difference between the compounded interest and simple interest on a certain sum at 10% per annum for 2 years is Rs.631. Find the sum.
A.Rs. 52255
B.Rs. 50000
C.Rs. 63200
D.Rs. 63100
Answer: D.Rs. 63100
Explanation: Let the sum be Rs. X .Then, C.I.. = x( 1 + 10/100 )2 - x = 21x/100 S.I. = [ (x*10*2)/100] = x/5 (C.I.) – (S.I.) = [ (21x)/100 – (x/5) ] = x/100 => x/100 = 631 => x= 63100 Hence, the sum is Rs. 63100
Answer D
Question: A waiter's salary consists of his salary and tips. During one week his tips were 3/4 of his salary. What fraction of his income came from tips ?
A.3/7
B.5/4
C.5/8
D.5/9
Answer: A.3/7
Explanation: Income = salary (s) +tips = s + s*3/4 = s*7/4
Tips = s*3/4
fraction of his income came from tips = (s*3/4)/(s*7/4)= 3/7
ANSWER:A
Question: A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 5^12 and 5^14, respectively, what is the value of A?
A.-24(5^12)
B.-23(5^12)
C.-24(5^6)
D.23(5^12)
Answer: A.-24(5^12)
Explanation: Ans : A
Assume that the numbers appear as shown below on the number line
A-----B-----C-----D-----E-----F-----G-----H
(5^12) (5^13)
As the values for G and H are given , we can calculate the difference between any two terms of the series .
Common Difference ,d = (5^13) - (5^12)
= (5^12) *[ 5-1 ]
= (5^12)*(4)
Also F + d = G as the terms are in equidistant and in increasing order.
So F + (5^12)*(4) = (5^12).
That is , F = (5^12) - (5^12)*(4)
= (5^12)[ 1-4]
= (5^12) (-3)
Similarly , E = F - d
= (5^12)[-3-4]
= (5^12)*(-7)
You can see a -4 getting added to the non-exponent part of the values . That is , according to the pattern , D SHOULD BE (5^12)*(-7-4)= (5^12)*(-11)
Following this pattern , A = (5^12)*(-24)
Question: 10 people meet and shake hands. The maximum number of handshakes possible if there is to be no “cycle” of handshakes is (A cycle of handshakes is a sequence of k people a1, a2, ……, ak (k > 2) such that the pairs {a1, a2}, {a2, a3}, ……, {ak-1, ak}, {ak, a1} shake hands).
A.7
B.6
C.9
D.8
Answer: C.9
Explanation: The question is not written in proper manner but we can work in 3 methods
1) If these are random handshakes, then total methods = 10*9/2*1 = 45
2) If these are cyclic handshakes, then total methods = 10
3) If these persons are not cyclic but they are arranged in a straight bench, then number of methods = 9
ANSWER:C
Question: When n liters of fuel was added to a tank that was already 1/3 full, the tank was filled to 5/9 of its capacity. In terms of n, what is the capacity of the tank, in liters?
A.10/9 n
B.4/3 n
C.9/2 n
D.9/4 n
Answer: C.9/2 n
Explanation: C is the answer.
To solve this problem, draw diagram or use algebra. I'm more comfortable with algebra.
Given
- tank was already 1/3 full.
- When n ltr added, it became 5/9
So 1/3 + n = 5/9
n = 5/9 - 3/9
n = 2/9
capacity of the tank is full fraction . i.e. 1 i.e. 9/9
so the question is 9/9 is how much times n
= 9/9 = 1 = xn
so x = 9/2
and total = 9/2n
Question: The price of a DVD Player is marked at $ 12,000. If successive discounts of 15%, 10% and 5% be allowed, then at what price does a customer buy it?
A.$ 8721
B.$ 6383
C.$ 8739
D.$ 8766
Answer: A.$ 8721
Explanation: A
$ 8721
Actual price = 95% of 90% of 85% of $ 12000
= 95/100 * 90/100 * 85/100 * 12000 = $ 8721.