source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38497 | ### Show Tags
19 Aug 2015, 01:34
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Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
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Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
After replacing an old member by a new member, it was found that the average age of eight members of a club is the same as it was two years ago.What is the difference between the ages of the replaced and the new member? | [
"2 years",
"4 years",
"8 years",
"16 years"
] | D | Explanation :
Age decreased = (8 x 2) years
16 years
So, required difference =16 years.
Answer : D |
AQUA-RAT | AQUA-RAT-38498 | the probability of getting exactly two (3 Marks) heads. (i) If two coins are tossed simultaneously, there are three possible outcomes, two heads, two tails or one of each. A fair coin is flipped three times in row. X is the number of trials and P(x) is the probability of success. Show that, if the coin is thrown Il-times the event will support 2n. no heads iii. Get an answer for 'The probability that a coin turns up heads when it is tossed is 1/2. Two coins are tossed. Given : Two coins are tossed simultaneously If two coins are tossed then the possible outcomes are : HT, HH, TT, TH Total no. b Posted 2 years ago. Get an answer for 'A coin is tossed and a single 6-sided die is rolled. Find (a) p (both cards are diamonds). Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of outcomes. Plotting with ggplot2. The probability of one or the other of two mutually exclusive events happening is the sum of the separate probabilities of these events. number of steps. Therefore, the probability that three flips of a coin will produce exactly one head is 3/8 or 0. Use a tree diagram to find the probability of rolling 3 when a dice is rolled. When two coins are tossed simultaneously then the possible outcomes obtained are {HH, HT, TH, and TT}. PROBABILITY C1 X,Y ILLUSTRSTION SHEET (2) 13. Exercise For Practice: Three coins are tossed simultaneously. Anil Kumar 9,748 views. Two coins are tossed together. (vi) A coin is tossed and a die is thrown simultaneously : P is the event of getting head and a odd number. Use this information to solve these problems. Expected number of tosses till first head comes up to coin tosses. A pair of fair dice is rolled and the sum of the faces showing is recorded. Important Questions for Class 10 Maths Chapter 15 Probability Probability Class 10 Important Questions Very Short Answer (1 Mark) Question 1. What Is The Probability Of Getting A Three On One Of The Cubes? Collecting. (Note: We exclude the possibility. Exactly one head. Last time we talked about independence of a pair of outcomes, but we can easily go on and talk about independence of a longer sequence of
The following is multiple choice question (with options) to answer.
Nine coins are tossed simultaneously. In how many of the outcomes will the first coin turn up a head? | [
"2^8",
"2^10",
"3 * 2^8",
"3 * 2^9"
] | A | Fix the first coin as H. The remaining8 coins have 2^8 outcomes.
Ans:A |
AQUA-RAT | AQUA-RAT-38499 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many W ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order? | [
" 736",
" 768",
" 792",
" 840"
] | D | three letters can be arranged in 3! ways.
only one combination EIU is required.
7 letters can be arranged in 7! ways.
thus W=7!/ 3! * 1 = 840.
D |
AQUA-RAT | AQUA-RAT-38500 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 600 meter long train crosses a signal post in 30 seconds. How long will it take to cross a 3 kilometer long bridge, at the same speed? | [
"4 min",
"2 min",
"8 min",
"3 min"
] | D | S = 600/30 = 20 mps
S = 3600/20 = 180 sec = 4 min
Answer: D |
AQUA-RAT | AQUA-RAT-38501 | # 2008 AMC 12A Problems/Problem 17
## Problem
Let $a_1,a_2,\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$?
$\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004$
## Solution
All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.
• If $a_1=4n$, then $a_2=\frac{4n}{2}=2n.
• If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1.
• If $a_1=4n+2$, then $a_2=2n+1.
• If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.
The following is multiple choice question (with options) to answer.
A student writes a computer program that generates a sequence of numbers a1, a2, a3, ... such that a1=a2=1 and ak=a(k-1)*4+1 for 2<k<n. If n=15, find a5. | [
"85",
"87",
"89",
"93"
] | A | a1=1
a2=1
a3=1*4+1=5
a4=5*4+1=21
a5=21*4+1=84+1=85
85 |
AQUA-RAT | AQUA-RAT-38502 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
The original price of a car was $30,000. Because the car owner thought he could get more money for the car, he set a new price by increasing the original price of the car by 10%. After a week, the car had not sold, so the owner discounted the new price by 10%, and the car was finally sold. What price was the car sold for? | [
"$29,660",
"$29,680",
"$29,700",
"$29,720"
] | C | The car was sold for (0.9)(1.1)($30,000)=$29,700.
The answer is C. |
AQUA-RAT | AQUA-RAT-38503 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
The average marks in mathematics scored by the pupils of a school at the public examination were 39. If four of these pupils who actually scored 5, 12, 15 and 19 marks at the examination had not been sent up, the average marks for the school would have been 44. Find the number of pupils sent up for examination from the school? | [
"12",
"25",
"28",
"245"
] | B | 39x = 5 + 12 + 15 + 19 + (x – 4)44
x = 25
Answer: B |
AQUA-RAT | AQUA-RAT-38504 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
Six children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children AB must sit next to each other, and child C must be somewhere to the right of AB. How many possible configurations are there for the children? | [
"600",
"720",
"1440",
"4320"
] | A | A, B, C, D, E, F, G- seven children, of which AB must sit next to each other. Considering them as one X (A, B), we have X, C, D, E, F, G. These can be arranged in 6! ways. But A,B can arrange themselves in 2! ways. So a total of 6!*2! ways = 1440. Since in exactly half of them, C will be to the right of AB, and exactly half, C will be to the left of AB, therefore 1440/2 = 600.
A is the right answer. |
AQUA-RAT | AQUA-RAT-38505 | ## (i)AB, BC and AC are tangents to the circle at E, D and F.$$BD = 30 cm,DC = 7 cm,\angle BAC = 90$$From the theorem stated,$$BE = BD = 30 cm$$Also $$FC = DC = 7 cm$$Let $$AE = AF = x$$ …. (1)Then $$AB = BE + AE = (30 + x)$$$$AC = AF + FC = (7 + x)$$$$BC = BD + DC = 30 + 7 = 37 cm$$Consider right trianlge ABC, by Pythagoras theorem we have$$BC^2 = AB^2 + AC^2$$$$(37)^2 = (30 + x)^2 + (7 + x)^2$$$$1369 = 900 + 60x + x^2 + 49 + 14x + x^2$$$$2x^2 + 74x + 949 – 1369 = 0$$$$2x^2+ 74x – 420 = 0$$$$x^2 + 37x – 210 = 0$$$$x^2 + 42x – 5x – 210 = 0$$$$x (x + 42) – 5 (x + 42) = 0$$$$(x – 5) (x + 42) = 0$$$$(x – 5) = 0 or (x + 42) = 0$$$$x = 5 or x = – 42$$$$x = 5$$[Since x cannot be negative]$$\therefore AF = 5 cm$$[From (1)]Therefore$$AB =30 +x = 30 + 5 = 35 cm$$(ii)$$AC = 7 + x = 7 + 5 = 12 cm$$Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.Join point O, F; points O, D and points O, E.From the figure,$$\dfrac{1}{2} \times AC \times AB = \dfrac{1}{2} \times AB \times OE+ \dfrac{1}{2} \times
The following is multiple choice question (with options) to answer.
A, B, C, D and E are 5 consecutive points on a straight line. If BC = 3CD, DE = 7, AB = 5 and AC = 11, what is the length of AE? | [
"18",
"19",
"20",
"21"
] | C | AC = 11 and AB = 5, so BC = 6.
BC = 3CD so CD = 2.
The length of AE is AB+BC+CD+DE = 5+6+2+7 = 20
The answer is C. |
AQUA-RAT | AQUA-RAT-38506 | algorithms, optimization
but I still want all items to have 25% share.
Assuming amount can be decimal, how do I find the optimal way of rebalancing this portfolio back to 25% shares? How to determine how many of each item to sell so we won't have to buy it back again (sell everything and spend 25% on each item is not an option)?
What if I have extra $100 to spend while rebalancing?
Upd: what I mean by "optimal" is to keep the amount of transactions as less as possible, assuming each transaction costs something. Your problem specification is incomplete. What do you mean by optimal? Do you want to, say, minimize transactions costs? Do you want to make sure you do not cross the spread?
The original portfolio was worth \$800. The new portfolio is worth \$804. Assuming your actions do not change the market price, and assuming no transactions costs, that is a given.
You had \$200 allocated to each security. Now, you want \$201 allocated to each security:
| Name | Price | Amount | Notional |
|--------|-------|--------|----------|
| Item 1 | 220 | 1 | $220 |
| Item 2 | 130 | 2 | $260 |
| Item 3 | 17 | 4 | $68 |
| Item 4 | 32 | 8 | $256 |
For every security whose notional is above the desired amount, sell that many units:
Item 1: Sell 19/220 = 0.0863636363636364 units
Item 2: Sell 59/130 = 0.453846153846154 units
Item 4: Sell 55/32 = 1.71875 units
The following is multiple choice question (with options) to answer.
You hold some gold in a vault as an investment. Over the past year the price of gold increases by 52%. In order to keep your gold in the vault, you must pay 6% of the total value of the gold per year. What percentage has the value of your holdings changed by over the past year. | [
"42.88$",
"43%",
"45%",
"45.5%"
] | A | (100%+52%)*(100%-6%) = 152*0.94 = 142.88% an increase of 42.88%
Your gold holdings have increased in value by 42.88%.
The Answer is A |
AQUA-RAT | AQUA-RAT-38507 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
In how many ways can you seat 4 people on a bench if one of them, Rohit, does not want to sit on the middle seat or at either end? | [
"720",
"1720",
"2880",
"6"
] | D | Since Rohit does not want to sit on the middle seat or at either end (3 chairs), then he can choose 1 chairs to sit. The remaining 3 people can sit in 3! ways. Thus the # of arrangements is 1*3! = 6.
Answer: D. |
AQUA-RAT | AQUA-RAT-38508 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A sum of Rs. 395 was divided among A, B, and C in such a way that B gets 25% more than A and 20% more than C. What is the share of A? | [
"Rs.195",
"Rs.180",
"Rs. 98",
"Rs. 120"
] | D | Let each one’s share is A,B and C respectively, then
B =125A/100 = 120C/100
A =100B/125 =4/5B, C = 100B/120 = 5/6B
4/5B +B +5/6B =395, 79B/30 =395, B =395*30/79 =150
A = 4*150/5 = 120
ANSWER:D |
AQUA-RAT | AQUA-RAT-38509 | # Another probability question from my textbook
Suppose that we have a tennis tournament with 32 players. Players are matched in a completely random fashion, and we assume that each player always has probability 1/2 to win a match. What is the probability that two given players meet each other during the tournament.
-
I am trying to teach myself probability. Just to clarify, not exactly a student looking for homework solutions - but still a student – user669083 Jan 23 '12 at 22:29
Easy general answer for $n$ players in a knockout tournament (and here $n=32$):
There are $\dfrac{n(n-1)}{2}$ potential pairs for matches.
To have one winner, $n-1$ players must be knocked out, so there are $n-1$ actual matches.
So the probability that a particular pair actually have a match is $\dfrac{n-1}{{n(n-1)/2}} = \dfrac{2}{n}$.
-
Very elegant! – Rasmus Jan 24 '12 at 7:27
yes it seems to be most elegant one. – user669083 Jan 24 '12 at 16:15
Hint 1: Consider the number of other players a particular player meets with what probability: one other with probability $1/2$; two others with probability $1/4$; three others with probability $1/8$; etc.
Hint 2: What is the expected number of other players a particular player meets?
Hint 3: How does Hint 2 relate to the original question?
Answer: $$\dfrac{1 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2^2} + 3 \times \dfrac{1}{2^3} + 4 \times \dfrac{1}{2^4} + 5 \times \dfrac{1}{2^4}}{31} = \dfrac{1}{16}$$
The following is multiple choice question (with options) to answer.
Let us say that a table tennis tournament was going on with knock out terms which means the one who loses the match is out of the tournament. 50 players took part in that tournament.
How many matches were played? | [
"99 matches.",
"88 matches.",
"49 matches.",
"66 matches."
] | C | Solution:
49 matches.
The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 49 matches were played.
Answer C |
AQUA-RAT | AQUA-RAT-38510 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two cars start at the same time from A and B which are 120 km apart. If the two cars travel in opposite direction they meet after one hour and if they travel in same direction ( from A towards B) then A meets B after 6 hours. What is the speed of the car starting from A? | [
"70 kmph",
"120 kmph",
"60 kmph",
"Data Inadequate"
] | A | Let, speed of car from A towards B = x km/hr
Let, speed of car from B towards A = y km/hr
Two cars start at same time which 120km apart = x + y = 120-----(i)
Travel in same direction, A meets B after 6hr = 6x-6y = 120-----(ii)
By Solving equation (i) and (ii), we get
12x = 840
x = 840/12
x = 70km/hr
Substituting x value in eqn (i) or (ii), we get
y = 50km/hr
So, speed of car starting from 70kmph.
ANSWER:A |
AQUA-RAT | AQUA-RAT-38511 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The profit obtained by selling an article for Rs. 57 is the same as the loss obtained by selling it for Rs. 43. What is the cost price of the article? | [
"Rs. 40",
"Rs. 50",
"Rs. 49",
"Rs. 59"
] | B | S.P 1- C.P = C.P – S.P 2
57 - C.P = C.P - 43
2 C.P = 57 + 43;
C.P = 100/2 = 50
ANSWER:B |
AQUA-RAT | AQUA-RAT-38512 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Ram professes to sell his goods at the cost price but he made use of 925 grms instead of a kg, what is the gain percent? | [
"2 9/13%",
"7 9/13%",
"3 9/13%",
"4 9/13%"
] | B | 925 --- 75
75 --- ? =>
=75/925 *100= 7 9/13%
Answer: B |
AQUA-RAT | AQUA-RAT-38513 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Two men Amar and Bhuvan have the ratio of their monthly incomes as 6 : 5. The ratio of their monthly expenditures is 3 : 2. If Bhuvan saves one-fourth of his income, find the ratio of their monthly savings?
A. 3 : 5 | [
"3:9",
"3:10",
"3:21",
"3:12"
] | B | Let the monthly incomes of Amar and Bhuvan be 6x and 5x respectively.
Let the monthly expenditure of Amar and Bhuvan be 3y and 2y respectively.
Savings of Bhuvan every month = 1/4(5x)
= (His income) - (His expenditure) = 5x - 2y.
=> 5x = 20x - 8y => y = 15x/8.
Ratio of savings of Amar and Bhuvan
= 6x - 3y : 1/4(5x) = 6x - 3(15x/8) : 5x/4 = 3x/8 : 5x/4
= 3 : 10.
Answer: Option B |
AQUA-RAT | AQUA-RAT-38514 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
The total of a company's assets in 1994 was 200% greater than the total in 1993, which in turn was 400% greater than the total in 1992. If the total of the company's assets in 1992 was N dollars, what were the company's assets in 1994? | [
"5N",
"7N",
"10N",
"15N"
] | D | In 1992, the company's assets were N.
In 1993, the company's assets were N + 4N = 5N.
In 1994, the company's assets were 5N + 10N = 15N.
The answer is D. |
AQUA-RAT | AQUA-RAT-38515 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Three men rent a pasture for Rs. 660. The first used it for 50 sheep for 4 months, the 2nd for 40 sheep for 3 months and the 3rd for 46 sheep for 5 months. How much should the first pay ? | [
"Rs. 220",
"Rs. 235",
"Rs. 240",
"Rs. 276"
] | D | The payment should be in ratio of 200:120:230 or 20:12:23
First person should pay 20*660/55 = 240 Rs.
ANSWER:D |
AQUA-RAT | AQUA-RAT-38516 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
In how many ways can 5 boys and 2 girls be seated on 8 chairs so that no two girls are together? | [
"5760",
"14400",
"480",
"5670"
] | D | Let us take opposite of the constraint.
2 girls sitting together: :
1 case is GGBGBBBB.
Total number of ways=3!*5!*5 with just shifting the rightmost girl.
Then the 2 leftmost girls can shift one position , and using the above reasoning, the total number of ways = 3!*5!*4 and so on till the rightmost girl has 1 position.
So total number of ways = 3!*5!(5+4+3+2+1)=120*90=10800
Similarly another case is:
GBGGBBBB.
Using the above reasoning, the total number of cases is: 3!*5!*(15) =10800
Let us take 3 girls sitting together
GGGBBBBB
There are 3! *5! Ways. The 3 leftmost girls can shift 6 positions. So there are a total of 3!*5!*6=4320 ways
So total is 2*10800 + 4320=25920
The total number of possibilities = 8! Ways =40,320
Answer is 40320-25920=5670
Hence D. |
AQUA-RAT | AQUA-RAT-38517 | ## Extra 01
How many arrangements can be made using the letters from the word COURAGE? What if the arrangements must contain a vowel in the beginning?
• $$4 \times 6!$$
## Extra Problem 02
How many arrangements are possible using the words
• EYE
• CARAVAN?
## 3(a)
There are (p+q) items, of which p items are homogeneous and q items are heterogeneous. How many arrangements are possible?
## 2(j)
There are 10 letters, of which some are homogeneous while others are heterogeneous. The letters can be arranged in 30240 ways. How many homogeneous letters are there?
Let, $$m = \text{number of homogeneous items}$$
• n(arrangements) = 30240 = $$\frac {10!}{m!}$$
• $$m! = \frac{10!}{30240}=120$$
• m = 5
## 2(k)
A library has 8 copies of one book, 3 copies of another two books each, 5 copies of another two books each and single copy of 10 books. In how many ways can they be arranged?
Total books = $$1 \times 8+3 \times 2+5 \times 2 + 8 \times 1 + 10$$ = 42
• n(arrangements) = $$\frac{42}{8!(3!)^2(5!)^2}$$
## 2(l)
A man has one white, two red, and three green flags; how many different signals can he produce, each containing five flags and one above another?
Flags: W = 2, R = 2, G = 3, Total = 7
Answer
## 2 (m)
A man has one white, two red, and three green flags. How many different signals can he make, if he uses five flags, one above another?
## 3(a)
How many different arragnements can be made using the letters of the word ENGINEERING? In how many of them do the three E’s stand together? In how many do the E’s stand first?
i
ii
iii
## 3(b)
In how many ways can the letters of the word CHITTAGONG be arranged, so that all vowels are together?
The following is multiple choice question (with options) to answer.
In how many ways can the letters of the word 'LEADER' be arranged? | [
"739",
"73",
"360",
"840"
] | C | The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways =6!/(1!)(2!)(1!)(1!)(1!)= 360.
Answer:C |
AQUA-RAT | AQUA-RAT-38518 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes? | [
"100 m",
"150 m",
"190 m",
"200 m"
] | A | Explanation:
Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr
Distance covered in 6 minutes = (1/60 × 6) km = 1/10 km = 100 m
∴ Distance between the thief and policeman = (200 – 100) m = 100 m.
Answer: A |
AQUA-RAT | AQUA-RAT-38519 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
If 21520 books need to be placed in boxes and each box can contain at most 9 books. How many books will be in the last unfilled box? | [
"9 books",
"8 books",
"1 book",
"4 books"
] | C | The number of books that can be placed in boxes of 9 books is the first number that is divisible by 9 that occurs before 21520.
In order to divide the sum in 9 parts, the amount must be divisible by 9
Divisibility rule of 9: The sum of the digits must be divisible by 9
Sum of digits of 21520 = 10 and 9 is divisible by 9.
Hence, we need to remove 1 to this number for it to be divisible by 9
Correct Option: C |
AQUA-RAT | AQUA-RAT-38520 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a train 120 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph? | [
"27 sec",
"23 sec",
"24 sec",
"25 sec"
] | A | D = 120 + 150 = 270
S = 36 * 5/18 = 10 mps
T = 270/10 = 27 sec
A |
AQUA-RAT | AQUA-RAT-38521 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 sec and a man standing on the platform in 20 sec. If the speed of the train is 54 km/hr. What is the length of the platform? | [
"167",
"240",
"881",
"278"
] | B | Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 20 = 300 m.
Let the length of the platform be x m . Then,
(x + 300)/36 = 15 => x = 240 m.
Answer: B |
AQUA-RAT | AQUA-RAT-38522 | +\frac1{271} +\frac1{273} +\frac1{275} -\frac1{66} +\frac1{277} +\frac1{279} +\frac1{281} +\frac1{283} -\frac1{68} +\frac1{285} +\frac1{287} +\frac1{289} +\frac1{291} -\frac1{70} +\frac1{293} +\frac1{295} +\frac1{297} +\frac1{299} +\frac1{301} -\frac1{72} +\frac1{303} +\frac1{305} +\frac1{307} +\frac1{309} -\frac1{74} +\frac1{311} +\frac1{313} +\frac1{315} +\frac1{317} -\frac1{76} +\frac1{319} +\frac1{321} +\frac1{323} +\frac1{325} -\frac1{78} +\frac1{327} +\frac1{329} +\frac1{331} +\frac1{333} +\frac1{335} -\frac1{80} +\frac1{337} +\frac1{339} +\frac1{341} +\frac1{343} -\frac1{82} +\frac1{345} +\frac1{347} +\frac1{349} +\frac1{351} -\frac1{84} +\frac1{353} +\frac1{355} +\frac1{357} +\frac1{359} -\frac1{86} +\frac1{361} +\frac1{363} +\frac1{365} +\frac1{367} -\frac1{88}
The following is multiple choice question (with options) to answer.
287 x 287 + 269 x 269 - 2 x 287 x 269 = ? | [
"466",
"646",
"666",
"324"
] | D | Given Exp. = a2 + b2 - 2ab, where a = 287 and b = 269
= (a - b)2 = (287 - 269)2
= (182)
= 324
Correct Option is D |
AQUA-RAT | AQUA-RAT-38523 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
If Re.100 amounts to Rs.300 over a period of 10 years. What is the rate of simple interest? | [
"20%",
"30%",
"15%",
"10%"
] | A | 200 = (100*10*R)/100
R = 20%
Answer:A |
AQUA-RAT | AQUA-RAT-38524 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When N is divided by 10 the remainder is 1 and when N is divided by 3 the remainder is 2. What is the remainder G when N is divided by 30? | [
"(a) 4",
"(b) 7",
"(c) 11",
"(d) 13"
] | C | N ends in the digit 1 because when N is divided by 10, the remainder is 1. Since N ends in 1, the remainder G when N is divided by 30 also ends in a 1. 11 is the only choice which ends in a 1.C |
AQUA-RAT | AQUA-RAT-38525 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
A recycling facility is staffed by 12 floor workers and one manager. All of the floor workers are paid equal wages, but the manager is paid n times as much as a floor worker. If the manager’s wages account for 1/11 of all wages paid at the facility, what is the value of n? | [
"1.1",
"2.1",
"2.2",
"2.3"
] | A | Say each floor worker is paid $x, then the manager is paid $xn.
Total salary would be 12x+xn and we are told that it equals to 11xn:
11x+xn=11xn --> reduce by x: 11+n=11n -->10n=11
n = 11/10 = 1.1
Answer: A |
AQUA-RAT | AQUA-RAT-38526 | java, xml, swing
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/12 00:00:00</date>
<riddle>ipso lorem 7</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/19 00:00:00</date>
<riddle>ipso lorem 8</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/26 00:00:00</date>
<riddle>ipso lorem 9</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/02 00:00:00</date>
<riddle>ipso lorem 10</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/09 00:00:00</date>
<riddle>ipso lorem 11</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/16 00:00:00</date>
<riddle>ipso lorem 12</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/23 00:00:00</date>
<riddle>ipso lorem 13</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/30 00:00:00</date>
<riddle>ipso lorem 14</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
</weeks>
The following is multiple choice question (with options) to answer.
8th Dec 2007 was Saturday, what day of the week was it on 8th Dec, 2006? | [
"Sunday",
"Tuesday",
"Friday",
"Tuesday"
] | C | Given that 8th Dec 2007 was Saturday
Number of days from 8th Dec, 2006 to 7th Dec 2007 = 365 days
365 days = 1 odd day
Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday
Correct answer is Friday |
AQUA-RAT | AQUA-RAT-38527 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In a school with 5 classes, each class has 2 students less than the previous class. How many students are there in the largest class if the total number of students at school is 115? | [
"25",
"26",
"27",
"28"
] | C | Let x be the number of students in the largest class.
Then x + (x-2) + (x-4) + (x-6) + (x-8) = 115
5x - 20 = 115
5x = 135
x = 27
The answer is C. |
AQUA-RAT | AQUA-RAT-38528 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Posts: 9558
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Karishma
Veritas Prep GMAT Instructor
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can? | [
"44",
"58",
"74",
"12"
] | A | Explanation:
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.
Answer:A |
AQUA-RAT | AQUA-RAT-38529 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 4 litres from vessel P is mixed with 4 litres from vessel Q, the ratio of alcohol and water in the resulting mixture is? | [
"1:2",
"1:3",
"3:1",
"1:5"
] | C | Quantity of alcohol in vessel P = 62.5/100 * 4 = 5/2 litres
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/2 + 7/2 = 6 litres
As 8 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 6 : 2 = 3:1.
Answer:C |
AQUA-RAT | AQUA-RAT-38530 | 10000000x= 1532216.22162216....
Again, that each block in that infinitely repeating decimal has been replaced by the next one. The decimal parts of 1532216.22162216... and 153.22162216... are exactly the same.
Subracting 1000x= 153.22162216... from 10000000x= 1532216.22162216..., the decimal parts cancel and we have
(10000000- 1000)x= 1532216- 153 or
9999000x= 1532063.
Dividing both sides by 9999000 we get x= 1632064/9999000 which perhaps can be reduced but we have succeeded in writing the number as a fraction, proving that it is a rational number.
Now apply the same ideas to x= 0.3712371237.. where "3712" is repeating and
x= 0.99999999.... where "9" is repeating.
4. Jul 29, 2008
### powerless
Well, thank you, I try one of them now.
In 0.3712437124... it is the 37124 which is repeating, it has five digits so we multiply by $$10^{5}$$ ie. 100000 to get one of those on the other side: 37124.37124...
Now we subtract: 37124.37124... - 0.3712437124 = 37124
Now what should we devide it by? 100000?
In your examples I didn't quite understand how you got the 999 bit!
5. Jul 29, 2008
### HallsofIvy
Staff Emeritus
Subtract one equation from another. If x= .3712437124... then 100000x= 37124.371234.... Subtract one equation from another.
6. Jul 29, 2008
### powerless
That's what I did!
=> $$10^{5}x-x = 37124.37124... - 0.3712437124...= 37124$$
The following is multiple choice question (with options) to answer.
If the digits 35 in the decimal 0.00035 repeat indefinitely, what is the value of (10^5-10^3)(0.00035)? | [
"37",
"41",
"3.5e-05",
"35"
] | B | 99*0.35=34.65 approx. 35
Answer : B |
AQUA-RAT | AQUA-RAT-38531 | Then we come to the 3-1-1 and 2-2-1 combinations.
For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.
Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.
Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.
Could you please explain why we ignored 1-1-1 combination?
Hey jhabib,
You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.
The following is multiple choice question (with options) to answer.
Coins are to be put into 10 pockets so that each pocket contains at least one coin. At most 6 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets? | [
"7",
"13",
"20",
"22"
] | C | Since at most 6 of the pockets are to contain the same number of coins then minimize # of coins in each, so let each contain just 1 coin;
Next, we are told that no two of the remaining 4 pockets should contain an equal number of coins, so they should contain 2, 3, 4, and 5 coins each (also minimum possible);
Total: 1+1+1+1+1+1+2+3+4+5=20
Answer: C. |
AQUA-RAT | AQUA-RAT-38532 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row his boat with the stream at 6 km/h and against the stream in 4 km/h. The man's rate is? | [
"1",
"2",
"3",
"7"
] | A | DS = 6
US = 4
S = ?
S = (6 - 4)/2 = 1 kmph Answer:A |
AQUA-RAT | AQUA-RAT-38533 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If a = 2, what is the value of –(a^2 + a^3 + a^4 + a^5)? | [
"-14",
"-60",
"0",
"4"
] | B | if a = -1, then putting values in equation
= -[ (2)^2 + (2)^3 + (2^4) + (2^5) ]
= -[ 4 + 8 + 16 + 32 ]
= -60
Answer = B = -60 |
AQUA-RAT | AQUA-RAT-38534 | and number of copies; this app basically calc. a 7) or one from a certain suit (e. Aesther Diviner. Blackjack is a card comparison game between a player and a dealer with a deck of standard playing cards. Using the combinations formula 52 choose 5 shown here, we get:. Let us denote that as P(c1)=1/3. Probability first card (black) = 26/52 or 1/2; probability second card (black) = 25/51 (you have to change the denominator because you did not replace your original pulled card which forces the. Probability. Logarithmic normal distribution random number. Probability Tools. Calculate Probability (Odds) for a Blackjack or Natural 21 First capture by the WayBack Machine (web. This is illustrated in the following problem. I draw three cards out of the deck at random. There are 4 king cards in total. So, on a single draw the probability of selecting a club is 13/52 and the probability of selecting an ace is 4/52. Generally, a player is dealt two cards. Draw one card from a deck without replacement and then draw another card. The diagram below shows the possible ways in which the event sets can overlap, represented using Venn diagrams:. What is the probability that two cards drawn at random from a deck of playing cards will both be aces? Show Solution It might seem that you could use the formula for the probability of two independent events and simply multiply $\frac{4}{52}\cdot\frac{4}{52}=\frac{1}{169}$. Given a deck of 52 cards, there are 2,598,960 ways to select a subset of five cards. You have 2 decks of cards (each deck contains both red and black cards). A card is drawn at random from a well-shuffled deck of playing cards Find the probability that the card drawn is: (i) a card of spades or an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or a queen - Math - Probability. Since there is a total number of 52 cards in the deck, that is the total number of outcomes. This makes it very easy for bookies to calculate to returns for a bet like this. All kinds of probabilities come into play in card games, but they're all
The following is multiple choice question (with options) to answer.
Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen? | [
"55/119",
"55/221",
"55/223",
"55/227"
] | B | We have n(s) = \inline {\color{Black}52C_{2}} = = 1326.
Let A = event of getting both black cards
B = event of getting both queens
A∩B = event of getting queen of black cards
n(A) = \inline {\color{Black}26C_{2}} = = 325, n(B)= \inline {\color{Black}4C_{2}} = = 6 and n(A∩B) = \inline {\color{Black}2C_{2}} = 1
P(A) = n(A)/n(S) = 325/1326;
P(B) = n(B)/n(S) = 6/1326 and
P(A∩B) = n(A∩B)/n(S) = 1/1326
P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 =
B |
AQUA-RAT | AQUA-RAT-38535 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
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The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
When the price of an article was reduced by 15% its sale increased by 80%. What was the net effect on the sale? | [
"44% increase",
"44% decrease",
"53% increase",
"66% increase"
] | C | if n items are sold for $p each, revenue is $np. If we reduce the price by 15%, the new price is 0.85p. If we increase the number sold by 80%, the new number sold is 1.8n. So the new revenue is (0.85p)(1.8n) = 1.53np, which is 1.53 times the old revenue, so is 53% greater.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38536 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
Which number comes next in the series
5 12 24 36 ? | [
"52",
"55",
"60",
"120"
] | A | A
52
sum of consecutive prime numbers
(2+3 , 5+7 , 11+13 , 17+19 , 23+29) |
AQUA-RAT | AQUA-RAT-38537 | # How many ways can you put 8 red, 6 green and 7 blue balls in 4 indistinguishable bins?
1. Assume all balls with the same color are indistinguishable.
2. The order in which balls are put in a bin does not matter.
3. No bins are allowed to have the same distribution of balls!
For example, this configuration
{RRGGGB} {RRRRGBBBB} {RGB} {RGB}
is not ok, because the two last bins contains the same distribution of balls: one Red, one Green, one Blue.
Actualy I'm most intressted in the procedure for solving this problem where the number of balls, colors and bins are much larger numbers.
• Is this from some programming contest? – Aryabhata Jan 10 '15 at 18:43
• No, not what I know of. But I think the solution to this problem can be used to find all ways a number can be written as a product of k distinct integers, where k is the number of bins and balls corresponds to prime factors. – Penlect Jan 10 '15 at 19:04
• What exactly do you mean by indistinguishable bins? – Nicholas Pipitone Jan 10 '15 at 19:24
• I just mean that the bins are unlabeled. If two bins switched position, that should not be counted as a "new way". – Penlect Jan 10 '15 at 20:18
This problem is a straightforward application of the Polya Enumeration Theorem. Suppose we treat the case of $r$ red balls, $g$ green balls and $b$ blue balls and $n$ indistinguishable bins where no bins are left empty.
Recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is $$Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$$
The following is multiple choice question (with options) to answer.
How many ways are there to lay four balls, colored red, black, blue and green in a row? | [
"11",
"33",
"44",
"23"
] | B | Answer:B |
AQUA-RAT | AQUA-RAT-38538 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? | [
"25200",
"210",
"120",
"1000"
] | A | Required selection = select 3 consonants from 7 and 2 vowels frm 4
= 7C3 * 4C2
= 210
As we have to form 5 letter word the above combination of consonants and vowels can be done in 5! = 120 different ways
Required no. of ways = 210 * 120
= 25200
Ans - A |
AQUA-RAT | AQUA-RAT-38539 | Now we calculate $B$. 5 is prime, so $XYZ$ is divisible by 5 if and only if one of $X$, $Y$, or $Z$ is divisible by 5. We can use inclusion-exclusion: $B$ is the sum of the cases where (at least) $X$, $Y$, or $Z$ is divisible by 5, minus the cases where (at least) two are divisible by 5, plus the cases where all three are divisible by 5. That is, $$\begin{eqnarray}B&=&D_x + D_{y} + D_z \\ &&- D_{xy} - D_{xz} - D_{yz} \\ &&+ D_{xyz}\end{eqnarray}$$
Where $D_{xy}$ denotes the number of choices of $(X,Y,Z)$ where $5\mid X$ and $5\mid Y$, and similarly for the others.
By symmetry, $D_x = D_y = D_z$, and $D_{xy} = D_{xz} = D_{yz}$. Also, it is impossible to have two of $(X,Y,Z)$ divisible by 5 without the third also being divisible by 5, so $D_{xy} = D_{xyz}$. So the previous equation reduces to:
$$B = 3D_x - 2D_{xyz}$$
We calculate $D_x$: $X$ will be a multiple of 5 whenever $X\in\{0,5,10,15,20\}$, so we want $$\begin{eqnarray} &&\sum_{5\mid X} C(X) \\ &=& \sum_{X\in\{0,5,10,15,20\}} (21-X) \\ &=& 21\cdot5 - (0+5+10+15+20) \\ &=& 105 - 50 \\ &=& 55 \end{eqnarray}$$
The following is multiple choice question (with options) to answer.
X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers? | [
"4",
"6",
"8",
"9"
] | C | PRIME # is the # that has only 2 factors: One is 1 and another is the # itself.
infer that
FOR any given # 1 and the # itself are the definite factors.
Knowing above concepts:
product of X,Y, and Z = XYZ
divisible by 1, xyz, x, y, z, xy,yz,xz ==> total 8
ANSWER "C" |
AQUA-RAT | AQUA-RAT-38540 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Jane and Ashley take 10 days and 40 days respectively to complete a project when they work on it alone. They thought if they worked on the project together, they would take fewer days to complete it. During the period that they were working together, Jane took an eight day leave from work. This led to Jane's working for four extra days on her own to complete the project. How long did it take to finish the project ? | [
"10.2 Days",
"15.2 Days",
"16.2 Days",
"18 Days"
] | B | Let us assume that the work is laying 40 bricks.
Jane = 4 bricks per day
Ashley = 1 brick per day
Together = 5 bricks per day
Let's say first 8 days Ashley works alone,
No of bricks = 8
Last 4 days Jane works alone,
No. of bricks = 16
Remaining bricks = 40 - 24= 16
So together, they would take 16/5 = 3.2
Total no. of days = 8 + 4 + 3.2 = 15.2
Answer is B |
AQUA-RAT | AQUA-RAT-38541 | Just to check each case: \begin{align*} f(S_T)=\begin{cases} 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + 0 + 0 = 3 & \text{if }S_T\leq 30, \\ 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + (30 - S_T) + 0 = 33- S_T & \text{if }30
I know @KeSchn already answered but hope this helps since this is how I usually do these. Of course, you can do this multiple ways but this gets to a correct answer relatively quickly.
Edit: Gordon's answer is definitely the way to go if you're comfortable with indicator functions. It does everything the graphical methods do without requiring any visualizing etc
The following is multiple choice question (with options) to answer.
Evaluate 35 / .07 | [
"400",
"500",
"505",
"None of these"
] | B | Explanation:
35/.07 = 3500/7 = 500
Option B |
AQUA-RAT | AQUA-RAT-38542 | (ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED.
The following is multiple choice question (with options) to answer.
The number N is 7,6H4, where H represents the ten's digit. If N is divisible by 9, what is the value of H? | [
"1",
"3",
"5",
"7"
] | A | Integer is divisible by 9 - Sum of digits is divisible by 9
Answer: A |
AQUA-RAT | AQUA-RAT-38543 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Rs.585 is divided amongst A, B, C so that 4 times A's share, 6 times B's share and 3 times C's share are all equal. Find C's share? | [
"260",
"150",
"817",
"716"
] | A | A+B+C = 585
4A = 6B = 3C = x
A:B:C = 1/4:1/6:1/3
= 3:2:4
4/9 * 585
= Rs.260
Answer: A |
AQUA-RAT | AQUA-RAT-38544 | Our goal is to find possible values for $x$, then use the equation above to find $n$. The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 121\cdot 7,$ and $77\cdot 11$. The differences between these factors are $846$, $114$, and $66$ - those are all possible values for $2x$. Thus the possibilities for $x$ are $423$, $57$, and $33$.
Now plug in these values into the equation $n = \frac{-1 + x}{2}$, so $n$ can equal $211$, $28$, or $16$, hence the answer is $\boxed{\textbf{(A)}\ 255}$.
~Edits by BakedPotato66
## Solution 2
As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$. Clearly $m > n$, so let $m = n + a$, where $a$ is some positive integer. We have:
$(n+a)^2 = n(n+1) + 212$. Expanding, grouping like terms and factoring, we get: $n = \frac{(212 - a^2)}{(2a - 1)}$.
We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ ($14^2 = 196 < 212 < 15^2 = 225$). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$. These gives $n$ values of $211, 28,$ and $16$, respectively. $211 + 28 + 16 = 255$. Hence, the answer is $\boxed{\textbf{(A)}\ 255}$.
## Solution 3
The following is multiple choice question (with options) to answer.
a = 2^15 - 625^3 and a/x is an integer, where x is a positive integer greater than 1, such that it does NOT have a factor p such that 1 < p < x, then how many different values for x are possible? | [
"None",
"One",
"Two",
"Three"
] | B | This is a tricky worded question and I think the answer is should be D not C...
Here is my reason :
The stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . The stem wants to say that X is a PRIME NUMBER . because any prime
Number has no factor grater than 1 and Itself .
On the other hand the stem says that X COULD get how many different number NOT MUST get different number ( this is very important issue )
AS our friends say, if we simplify Numerator more we can obtain : 5^12 ( 5^3-1) = 5^12 (124) = 5^12 (31*2*2) divided by x and we are told that this fraction is
An integer . so, X COULD Be ( not Must be) 5 , 31 ,or 2 !!! so , X could get 1 different values and answer is B.... |
AQUA-RAT | AQUA-RAT-38545 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 16 seconds. What is the length of the platform in meters? | [
"240 meters",
"360 meters",
"280 meters",
"600 meters"
] | C | Speed of the train in metres/sec = 72000/3600 = 20
Distance travelled by train to cross the platform = 30 * 20 = 600 = Length of train + Length of platform
Distance travelled by train to cross the man = 16 * 20 = 320 = Length of train
Length of platform = 600 - 320 = 280
Answer: C |
AQUA-RAT | AQUA-RAT-38546 | # Runner's High (Speed)
I find the following mind-boggling.
Suppose that runner $$R_1$$ runs distance $$[0,d_1]$$ with average speed $$v_1$$. Runner $$R_2$$ runs $$[0,d_2]$$ with $$d_2>d_1$$ and with average speed $$v_2 > v_1$$. I would have thought that by some application of the intermediate value theorem we can find a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$. This is not necessarily so!
Question. What is the smallest value of $$C\in\mathbb{R}$$ with $$C>1$$ and the following property?
Whenever $$d_2>d_1$$, and $$R_2$$ runs $$[0,d_2]$$ with average speed $$Cv_1$$, then there is a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$.
The following is multiple choice question (with options) to answer.
IN Common Wealth Games an athlete runs 400 meters in 24 seconds. Hisspeed is ? | [
"25 km/hr",
"27 km/hr",
"60.60 km/hr",
"32 km/hr"
] | C | Explanation :
(400/1000)/(24/3600)=60.60 km/hr
Answer : C |
AQUA-RAT | AQUA-RAT-38547 | The base case is
$$n=1:\;\;\;\;\;1^3+3^3=28\stackrel ?=(1+1)^2(2+4+1)=28\ldots\ldots good$$
Assume for $\,n\,$ and show for $\,n+1\,$:
$$1^3+3^3+\ldots+(2n+1)^3+(2n+3)^3\stackrel{\text{Ind. hypothesis}}=(n+1)^2(2n^2+4n+1)+(2n+3)^3=\ldots$$
-
The following is multiple choice question (with options) to answer.
If 2^2n + 2^2n + 2^2n + 2^2n = 4^28, then n = | [
"3",
"6",
"12",
"27"
] | D | 2^2n + 2^2n + 2^2n + 2^2n = 4^28
=> 4 x 2^2n = 4^28 = 2^56
=> 2^2 x 2^2n = 2^56
=> 2^(2n+2) = 2^56
=> 2n+2 = 56=> n =27
So. Answer will be D. |
AQUA-RAT | AQUA-RAT-38548 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
It takes Jim N minutes to prepare one omelet. It takes Cole N/2 minutes to prepare one omelet. It takes Mitch N/2 minutes to prepare 2 omelets. If Jim, Cole and Mitch together prepare a total of 30 omelets in 40 minutes, then how many minutes would it take Mitch alone to prepare 30 omelets? | [
"70",
"120",
"140",
"160"
] | A | Cole is twice as fast as Jim and Mitch is 4 times as fast as Jim (it takes Mitch N/2 minutes to prepare 2 omelets, thus N/4 minutes to prepare one omelet).
Say the rate of Jim is x omelet/minute, the rate of Cole 2x omelet/minute, and the rate of Mitch 4x omelet/minute.
Then, from (time)(rate) = (job done):
(40 minutes)(x + 2x + 4x) = (30 omelets);
(40 minutes)(7x) = (30 omelets);
Since it takes 40 minutes to prepare 30 omelets at the rate of 7x omelet/minute, then it would take 40*7/4 = 70 minutes to prepare the same 30 omelets at the rate of 7x*4/7 = 4x omelet/minute (the rate of Mitch).
Or:
(40 minutes)(7x) = (30 omelets);
(40*7/4 minutes)(7x*4/7) = (30 omelets);
(70 minutes)(4x) = (30 omelets);
Answer: A. |
AQUA-RAT | AQUA-RAT-38549 | You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.
• \begin{align}a+b+c&=(m^2-n^2)+(2mn)+(m^2+n^2)\\ &=(m^2+m^2)+2mn+(n^2-n^2) \\ &=2m^2+2mn \\ &=2m(m+n).\end{align} And now $abc=2mn(m^4-n^4)$ so I must show that $m+n\mid n(m^4-n^4)$ which means $n\mid m$. But $m$ and $n$ are arbitrary. Am I doing something wrong? – user477343 Jun 20 '18 at 11:30
• $x^4-y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x ^2+y^2)$ – Stefan Jun 20 '18 at 11:38
• $a+b+c=2m^2+2mn=2m(m+n)$, while $abc={\bf 2m}n(m^2+n^2)(m-n){\bf (m+n)}$. – Berci Jun 20 '18 at 11:40
• @Stefan thank you very much. I was able to put to-and-to together :) – user477343 Jun 20 '18 at 11:42
• Congratulations, Michael! You have a tick! $$\color{green}{\checkmark} \ \ (+1)$$ – user477343 Jun 20 '18 at 11:43
The following is multiple choice question (with options) to answer.
If a, b and c are even integers, which of the following could be the value of a^2 + b^2 + c^2? | [
"a)82",
"b)84",
"c)85",
"d)87"
] | B | a, b and c are even integers which means that, supposing that x, y and z are integers, then a=2x, b=2y and c=2z
Therefore we have: a^2+b^2+c^2=2^2*x^2+2^2*y+2^2*z=4x^2+4y^2+4z^2=4(x^2+y^2+z^2).
Which means that the total a^2+b^2+c^2 must be divisible by 4.
The only choice which is divisible by 4 is: B)
84=8^2+4^2+2^2
Answer: B. |
AQUA-RAT | AQUA-RAT-38550 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains 8 red and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? | [
"4/7",
"2/3",
"5/8",
"1/4"
] | A | Total number of balls = 8+6 = 14
number of white balls = 8
Probability = 8/14 = 4/7
Answer is A |
AQUA-RAT | AQUA-RAT-38551 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A farmer planned to plough a field by doing 120 hectares a day. After two days of work he increased his daily productivity by 25% and he finished the job two days ahead of schedule. What is the area of the field? | [
"1420",
"1430",
"1450",
"1440"
] | D | First of all we will find the new daily productivity of the farmer in hectares per day: 25% of 120 hectares is 25100⋅120=30 hectares, therefore 120+30=150 hectares is the new daily productivity. Lets x be the planned number of days allotted for the job. Then the farm is 120⋅x hectares. On the other hand, we get the same area if we add 120⋅2 hectares to 150(x−4) hectares. Then we get the equation
120x=120⋅2+150(x−4)
x=12
So, the job was initially supposed to take 12 days, but actually the field was ploughed in 12 - 2 =10 days. The field's area is 120⋅12=1440 hectares.
Answer is D. |
AQUA-RAT | AQUA-RAT-38552 | temperature, agriculture
Title: Minimum temperature in Growing Degree Day (GDD) - from daytime or from the whole day? I want to calculate growing degree days.
I have hourly data of temperature between the hours of 00:00 and 23:59.
I want to calculate the growing degree days and I saw that I need the minimum,maximum and mean temperatures. However, I don't know if I should filter my data to contain temperature data only for the daylight time or to use the whole day temperature.
What is the right data to use for the GDD calcualtion? I have no actual familiarity myself with implementing growing degree days itself.
That said, in reading through some references like (1), (2), and (3) I see no indication that it should be filtered.
The terminology also sounds to stem from the longer used heating degree day and cooling degree day, which I know is done from just the max and min over the 24 hour day (as this NOAA page suggests).
And in addition, the lowest temperature of the day usually occurs right near or just after dawn, so filtering it to "daytime" temperatures being when the sun is up would basically include roughly the coldest (and hottest) temperature of the day anyways.
So while I've never interacted with this specific metric, everything suggests to me that you would just use the values over a standard 24 hour day.
The following is multiple choice question (with options) to answer.
The average temperature for Monday, Tuesday, Wednesday and Thursday was 48 degrees and for Tuesday, Wednesday, Thursday and Friday was 46 degrees. If the temperature on Monday was 42 degrees. Find the temperature on Friday? | [
"28",
"27",
"22",
"34"
] | D | M + Tu + W + Th = 4 * 48 = 192
Tu + W + Th + F = 4 * 46 = 184
M = 42
Tu + W + Th = 192 -42 = 150
F = 184 – 150 = 34
Answer:D |
AQUA-RAT | AQUA-RAT-38553 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 7 employees to 2 different offices? | [
"215",
"310",
"343",
"243"
] | C | Each of three employee can be assigned to either of offices, meaning that each has 2 choices --> 7*7*7=7^3=343.
Answer: C. |
AQUA-RAT | AQUA-RAT-38554 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres? | [
"8",
"15",
"16",
"18"
] | D | Capacity of the tank = (12 x 13.5) liters = 162 liters.
Capacity of each bucket = 9 liters.
Numbers of buckets needed = (162/9) = 18
ANSWER:D |
AQUA-RAT | AQUA-RAT-38555 | $⌊100/2⌋+⌊100/4⌋+⌊100/8⌋+⌊100/16⌋+⌊100/32⌋+⌊100/64⌋=50+25+12+6+3+1=97 .$
Hence, 2 divides $S$ to an odd power. So we need to divide $S$ by $k!$ which 2 divides to an odd power to get a perfect square quotient. This reduces the possibilities for $k$ to 50 or 51. Since
$S={2}^{99}·{3}^{98}·{4}^{97}\dots {99}^{2}·100=\left(2·4\dots 50\right)\left({2}^{49}·{3}^{49}·{4}^{48}\dots 99\right){}^{2}=50!·{2}^{50}\left(\dots \right){}^{2} ,$
The following is multiple choice question (with options) to answer.
Which of the following number will completely divide (36^9)-1? | [
"5",
"6",
"7",
"9"
] | C | x^n - 1 will be divisible by x+1 only when n is even.
36^9 - 1 = (6^2)^9 - 1 = 6^18 - 1 which is divisible by 6+1 i.e. 7
Answer is C |
AQUA-RAT | AQUA-RAT-38556 | Can you solve it from here?
-Dan
Is it:
x - 3/2 = + square root 1 / 4 and x - 3/2 = - square root 1/4
then
x = 3/2 + square root 1 /4 and x = 3/2 - square root 1/4
so x = 1.75 or x = 1.25
4. Originally Posted by princess_anna57
Is it:
x - 3/2 = + square root 1 / 4 and x - 3/2 = - square root 1/4
then
x = 3/2 + square root 1 /4 and x = 3/2 - square root 1/4
so x = 1.75 or x = 1.25
no.. what is $\sqrt{\frac{1}{4}}$?
5. Originally Posted by princess_anna57
Is it:
x - 3/2 = + square root 1 / 4 and x - 3/2 = - square root 1/4
then
x = 3/2 + square root 1 /4 and x = 3/2 - square root 1/4
so x = 1.75 or x = 1.25
$x=\frac{3}{2}+\frac{1}{2}=2$
$x=\frac{3}{2}-\frac{1}{2}= 1$
The following is multiple choice question (with options) to answer.
If x is a positive number and 1/3 the square root of x is equal to 3x, then x = | [
"1/3",
"1/9",
"1/81",
"1"
] | C | 1/3 of sqrt(x) = 3x, which means that sqrt(x) = 9x
or x = 81x^2 -> divide by x
1=81x
x=1/81
C. |
AQUA-RAT | AQUA-RAT-38557 | c++, beginner, object-oriented, quiz
__ __ ____ ____
\ \/ /___ __ __ / __ \____ ______________ ____/ / /
\ / __ \/ / / / / /_/ / __ `/ ___/ ___/ _ \/ __ / /
/ / /_/ / /_/ / / ____/ /_/ (__ |__ ) __/ /_/ /_/
/_/\____/\__,_/ /_/ \__,_/____/____/\___/\__,_(_)
)" << "\n";
std::cout << "\n";
std::cin.get();
std::cin.ignore();
return 0;
}
else
{
std::cout << "You failed... Sorry, better luck next time.\n";
std::cout << "\n";
}
std::cin.get();
std::cin.ignore();
return 0;
}
The following is multiple choice question (with options) to answer.
In a software company 4/10 of people know C++, 2/4 of them know java, 2/8 of them know neither,find the total possibility to know C++? | [
"1/4",
"1/3",
"1/2",
"2/3"
] | A | Assume 100 persons in a company
person knows c++=4/10=100*4/100=20 persons
person knows java=2/4=100*2/4=50 persons
neither=2/8=100*2/8=25 persons
according to our assumption 100 person=20+50+25=95 person covered(5 remaining)
so,total possibility=20 person(c++)+5(remaining)=25 persons
which is 25/100=1/4
ANSWER:A |
AQUA-RAT | AQUA-RAT-38558 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Percentage of profit earned by selling a book for $1380 is equal to the percentage loss incurred by selling the same book for $1020. What price should the book be sold to make 20% profit? | [
"$1260",
"$1440",
"$1500",
"$1680"
] | B | Let C.P. be Rs. x.
Then, (1380 - x) x 100 =(x - 1020) x 100
1380 - x = x - 1020
2x = 2400
x = 1200
Required S.P. = 120% of 1200 = 1.20 x 1200 = $1440
Answer : B |
AQUA-RAT | AQUA-RAT-38559 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
If (2^x)(3^y) = 216, where x and y are positive integers, then (2^x-1)(3^y-2) equals: | [
"12",
"24",
"48",
"96"
] | A | So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=3, y=3. now quickly 2^2(3^1)=12. Than answer should be number 1.
A |
AQUA-RAT | AQUA-RAT-38560 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
John walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 8 minutes earlier. How far is the office from his house? | [
"2 1/5 Km",
"3 1/5 Km",
"3 2/5 Km",
"3 3/5 Km"
] | B | Formula = S1*S2/S2-S1 * T1+T2/60
= 4*6/2 * 16/6
= 24/2 * 16/60
= 8 * 2/5
= 16/5 = 3 1/5 Km
B |
AQUA-RAT | AQUA-RAT-38561 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
Village A’s population is 300 greater than Village B's population. If Village B’s population were reduced by 600 people, then Village A’s population would be 4 times as large as Village B's population. What is Village B's current population? | [
"900",
"1000",
"1100",
"1200"
] | A | A = B+300.
A=4(B-600).
4(B-600)=B+300.
3B=2700.
B=900.
The answer is A. |
AQUA-RAT | AQUA-RAT-38562 | c#, beginner, io
Example run:
Welcome to Buzzway Subs!
May I take your order?
Catering Menu:
Sandwich Platter: $39.99
Cookie Platter: $19.99
Purchase how many of Sandwich Platter for $39.99 each?
1
Purchase how many of Cookie Platter for $19.99 each?
2
Subtotal: 79.97
Tax: 4.80
Total: 84.77
Your total due is $84.77.
Pay how much?
85
Your change is $0.23.
--- Receipt ---
1 Sandwich Platter $39.99 ea. $39.99
2 Cookie Platter $19.99 ea. $39.98
Subtotal: $79.97
Tax: $4.80
Total: $84.77
Payment: $85.00
Your change is $0.23.
Thank you for shopping at Buzzway! First, keep track of your instances of a class:
new BuzzwaySubs();
BuzzwaySubs.processCustomer();
That should be:
BuzzwaySubs restaurant = new BuzzwaySubs(); // or `var restaurant`
restaurant.processCustomer();
As it is, the first line is utterly useless. Additionally, this only works because your methods are all static.
Keeping track of your instances is important because what happens when you have two restaurants? You need to know which restaurant is controlled by which class instance so you can manage them appropriately.
Second, declare your variables in the tightest scope possible:
int itemQty = 0;
foreach (var item in order)
{
itemQty = item.Value;
decimal costOfItems = itemQty * cateringMenu[item.Key];
subTotal += costOfItems;
}
That variable should be declared in the foreach loop. You have this probably in a great many places.
foreach (var item in order)
{
int itemQty = item.Value;
decimal costOfItems = itemQty * cateringMenu[item.Key];
subTotal += costOfItems;
}
This is important for many reasons, including keeping your variables from leaking information to other sections of the program, releasing memory when you aren't using it, and more.
Third, your naming does not follow standard C# naming practices:
public static void printCateringMenu()
The following is multiple choice question (with options) to answer.
A business executive and his client are charging their dinner tab on the executive's expense account.The company will only allow them to spend a total of 50$ for the meal.Assuming that they will pay 7% in sales tax for the meal and leave a 15% tip,what is the most their food can cost? | [
"39.55$",
"40.63$",
"41.63$",
"42.15$"
] | B | Let x be the amount the most their food can cost.
x + 0.07x + 0. 15x = 50
Solving for x, x = 40.98$, this is the maximum amount that can be used for food.
Hence 40.63$ is the maximum value less than 40.98$
Hence Option B |
AQUA-RAT | AQUA-RAT-38563 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
The total of the ages of Mahesh, Suresh, Ramesh is 102 years. Ten years ago, the ratio of their ages was 2:3:4. What is the present age of Mahesh? | [
"32years",
"42years",
"45years",
"35years"
] | B | Let ages of Mahesh, Suresh, Ramesh 10years ago be 2x,3x,4x
2x+10 + 3x+10 +4x+10 = 102
x = 8
Ramesh present age = 4*8+10 = 42 years
Answer is B |
AQUA-RAT | AQUA-RAT-38564 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
How many digits will be there to the right of the decimal point in the product of 98 and .08216 ? | [
"5",
"6",
"9",
"7"
] | A | Product of 98 and .08216 is 8.05168.
Therefore number of digits to right of decimal point is 5
Answer is A. |
AQUA-RAT | AQUA-RAT-38565 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
The king of a country and 4 other dignitaries are scheduled to sit in a row on the 5 chairs represented above. If the king must sit in the center chair, how many different seating arrangements are possible for the 5 people? | [
" 4",
" 5",
" 24",
" 20"
] | C | IMO C
Given that one chair is taken, i think the remaining 4 dignitaries can be arranged 4!=24. |
AQUA-RAT | AQUA-RAT-38566 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
X, Y and Z, each working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn $2350, what will be Z's share of the earnings? | [
"$525",
"$550.50",
"$1080.02",
"$522.3"
] | D | The dollars earned will be in the same ratio as amount of work done
1 day work of Z is 1/12 (or 2/24)
1 day work of the combined workforce is (1/6 + 1/8 + 1/12) = 9/24
Z's contribution is 2/9 of the combined effort
Translating effort to $ = 2/9 * 2350 = $522.3
hence:D |
AQUA-RAT | AQUA-RAT-38567 | # Seating couples around 2 tables
Here's my question and possible answer.
How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:
1) each couple must sit at the same table, and,
2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).
My solution:
Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)
$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)
$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$
$=2 * 8!$
Can someone verify this solution or provide the correct one?
First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$
Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4}$ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.
Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.
Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.
Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$
And again, if you do not differentiate between the tables, then divide this by $2$
The following is multiple choice question (with options) to answer.
In how many ways can a group of 16 friends be seated round 2 tables if one of the tables can seat 10 and the other can seat 6 people? | [
"15C5 * 9! *4!",
"15C3 * 8! *3!",
"15C4 * 9! *3!",
"16C4 * 9! *5!"
] | D | 10 people can be selected from 16 people in 16C10 ways.
Remaining 6 people can be selected from 6 people in 6C6 ways.
Now, arranging 10 people on a round table = (10-1) ! = 9!
Arranging 6 people on a round table = (6-1) ! = 5!
Hence, total number of ways = 16C10 * 9! * 6C6 * 5!
= 16C4 * 9! * 5!
D |
AQUA-RAT | AQUA-RAT-38568 | # For any two sets, $A - B = B - A$ implies $A = B$
Is the following statement True or False:
For any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$.
If it is true, prove it, otherwise provide a counterexample.
I am unable to come up with a counter example. I think the statement is true but how do I prove it?
• Suppose $x \in A - B$. Then $x \in B-A$. In particular $x \in B$, contradiction. So $A - B = \emptyset$ i.e. $A \subset B$. Same argument shows $B - A = \emptyset$ i.e. $B \subset A$. – hunter Jun 23 '17 at 6:55
If $A-B=B-A$ then for any $x\in A-B=B-A$ we $x\in A;x\in B; x\not \in A; x\not \in B$. That's a contradiction so $A-B=B-A$ is empty.
Thus there are no elements in $A$ that are not in $B$. In other words $A$ is a subset of $B$. Likewise there are no elements of $B$ that are in $A$. So $B$ is a subset of $A$.
So $A=B$.
If $A \setminus B = B \setminus A$, then
$A=A \setminus B \cup (A\cap B)= B \setminus A \cup (B \cap A) = B$.
Let’s use some Boolean algebra, in order to show a different point of view.
The following is multiple choice question (with options) to answer.
Which of the following is(are) true for a@b=b@a?
I. a@b=ab+ba
II. a@b=(a+b)(a-b)
III. a@b=(a/b)+(b/a) | [
"only Ⅰ",
"only Ⅱ",
"only Ⅲ",
"only Ⅰ& Ⅲ"
] | D | If I. a@b=ab+ba, a@b=ab+ba=ba+ab=b@a (O).
If II. a@b=(a+b)(a-b), a@b=(a+b)(a-b)≠(b+a)/(b-a)=b@a (X).
If III.a@b=(a/b)+(b/a), a@b=(a/b)+(b/a)=(b/a)+(a/b)=b@a (O).
Thus, Ⅰ& Ⅲ are the answers. Therefore, D is the answer. |
AQUA-RAT | AQUA-RAT-38569 | # Homework Help: Probability of choosing stale donuts out of 24
Tags:
1. Sep 2, 2016
### TheSodesa
1. The problem statement, all variables and given/known data
There are 6 stale donuts in a set of 24. What is the probability of:
a) there being no stale donuts in a sample of 10?
b) there being 3 stale donuts in a sample of 10?
c) What is the chance of a stale doughnut being found?
2. Relevant equations
$N \, permutations = N!$
3. The attempt at a solution
Let $X$ denote the number of stale donuts in a set of 10.
a) I used the idea of permutations like so:
$$P(X = 0) = \frac{\frac{18!}{8!}}{\frac{24!}{15!}} \approx 0.335$$
This was incorrect, according to the testing software
b) Here I followed the same idea:
$$P(X=3) = \frac{\frac{18!}{11!} \times \frac{6!}{3!}}{\frac{24!}{15!}} \approx 0.041$$
c) This is just the complement of part $a$:
$$P(X \, is \, at \, least \, 1) = 1 - 0.061 = 0.665$$
Parts $a$ and $b$ (and $c$ as a consequence of $a$ being wrong) are apparently wrong and I'm not sure whats wrong with my reasoning.
Last edited: Sep 2, 2016
2. Sep 2, 2016
### TheSodesa
I was completely wrong. I was supposed to use combinations instead:
a) $$P(X = 0) = \frac{18 \choose 10}{24 \choose 10} =39/1748 \approx 0.22$$
b) $$P(X = 3) = \frac{{18 \choose 7} \times {6 \choose 3}}{{24 \choose 10}} =1560/4807 \approx 0.325$$
The following is multiple choice question (with options) to answer.
The first doughnut is priced at $1 and then if you purchase additional doughnuts as dozens then the price is $6/dozen. What is the total number of doughnuts purchased if you paid $25? | [
"45",
"49",
"53",
"57"
] | B | $25 = 4 * $6 + $1
The number of doughnuts is 4*12 + 1 = 49
The answer is B. |
AQUA-RAT | AQUA-RAT-38570 | # Two points of a square $K$ determine a diagonal of another square that is contained in $K$
Let $$K:=[0,1]^2$$ be a square on $$\mathbb{R}^{2}$$. We select 2 random points $$A$$, $$B$$ $$\in [0,1]^{2}$$ in this square. What is the probability that the square whose diagonal is the line segment $$AB$$, is contained in $$K$$?
I found that if we fix coordinates of $$A=(x,y)\in [0,1]^{2}$$, then the probability equals $$\int\limits_{0}^{1}\int\limits_{0}^{1}[1-(x-y)^{2}-(1-x-y)^{2}\times\textbf{1}(x+y<1)]dxdy \, ,$$ where: $$\textbf{1}(x+y<1):=1$$ if $$x+y<1$$ and $$\textbf{1}(x+y<1):=0$$ otherwise.
Some attempts(or some elements of stream of consciousness)
I tackled other problems from geometric probability, but this problem cannot be solved by standard methods(i.e. by finding dependency between given information in question, then making, at least, rough plot on cartesian coordinate system and integrate the area under the graph of detected dependencies within specific constraints). I have completely, even intuitively, idea how to come towards such solution.
The following is multiple choice question (with options) to answer.
Square A has an area of 65 square centimeters. Square B has a perimeter of 16 centimeters. If square B is placed within square A and a random point is chosen within square A, what is the probability the point is not within square B? | [
"0.25",
"0.5",
"0.75",
"0.1"
] | C | I guess it's mean that square B is placed within square Aentirely.
Since, the perimeter of B is 16, then its side is 16/4=4 and the area is 4^2=16;
Empty space between the squares is 65-16=48 square centimeters, so if a random point is in this area then it won't be within square B: P=favorable/total=48/64 = 0.75.
Answer: C |
AQUA-RAT | AQUA-RAT-38571 | # Is it possible to solve for $a, b \in \mathbb{N}$?
I need to solve the following equation so that both $a$ and $b$ are natural numbers.
$$ab - 2a = 2b$$
I must also prove that the solutions found are the only ones possible.
Is it possible to do so, and if yes how can I do it? Can this be solved as some type of diophantine equation?
• $ab=2(a+b)$ means $a\mid 2b$ and $b\mid 2a$. No Diophantine equations needed. – Dietrich Burde Feb 23 '15 at 16:34
Hint:
$$ab-2a=2b\iff (a-2)(b-2)=4$$
Hint $\ b\neq 2\,\Rightarrow\ a = \dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$
$(0,0)$ is a solution.
It can't be that $b = 1$, neither can it be that $b = 2$. Assume $b > 2$.
Considering the equation modulo $b - 2$, we get that $b - 2$ divides $4$. We conclude that $b - 2$ is either $1$, or $2$, or $4$. So $b = 3$ or $b = 4$ or $b = 6$. It happens that all of these give solutions.
Therefore the solutions are $(0,0)$, and the pairs for which $b \in \{3,4,6\}$.
The following is multiple choice question (with options) to answer.
32 = a + 2b
|a| > 2
If ‘a’ and ‘b’ are both integers, then what is the smallest possible values of ‘a’ that can be used to solve the above equation. | [
"2",
"3",
"4",
"5"
] | C | Let us understand the meaning of |a| > 2
Mod is very easy concept if you solve mod question by considering as a distance. when a mod is written as |x-(a)| = b, this means the distance from point 'a' (both side left and right of 'a' on number line) is b. |x-(a)| < b means the distance is between the two extreme distance(left and right side of 'a' on number line, considering the max distance is 'b' from 'a' - as per this scenario.....hence the value of 'a' must be between these two extremes. |x-(a)| > b means the distance is greater than the distance of 'b'..i.e the value of a could be anywhere more than 'b' (i.e. current case).
Now come to the question. First its given |a| > 2 ==> a > 2 i.e. a can have any value bigger than 2 (i.e. 3, 4, 5…).
Now, lets move to equation a + 2b = 32 ==> b = (32 – a)/2 ==> b = 16 – (a/2). According to question, b is an integer, hence to make ‘b’ integer a must be divisible by 2. Minimum possible value of ‘a’ that is divisible by 2 is 4 (a is greater than 2 so the next number that is divisible by 2 is 4). So the answer is C (value 4). |
AQUA-RAT | AQUA-RAT-38572 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B enter into partnership with capital as 7:9. At the end of 8 months, A withdraws. If they receive the profits in the ratio of 8:9 find how long B's capital was used? | [
"4",
"5",
"34",
"7"
] | D | 7 * 8 : 9 * x = 8:9 => x= 7
Answer:D |
AQUA-RAT | AQUA-RAT-38573 | Going back to $9- x^2= (3- x)(3+ x)= -(x- 3)(x+ 3)\ge 0$, another and perhaps simpler way to see that is this: $a- b> 0$ if and only if $a> b$ and $a- b< 0$ if and only if a< b. If x< -3 then it is also less than 3 so both of x- 3 and x+ 3= x-(3) are negative. Their product (the product of two negative numbers) is positive and so -(x- 3)(x+ 3) is [b]negative[/tex]. If -3< x< 3, then x- 3 is still negative but x+ 3= x-(-3) is now positive. Their product (the product of a positive and negative number) is negative so -(x- 3)(x+ 3) is positive. Finally, if x> 3, it is also greater than -3 so both x- 3 and x+ 3 are positive. The product of two positive numbers is positive so -(x- 3)(x+ 3) is negative. Again, we have that $x^2- 9\ge 0$ if and only if $-3\le x\le 3$.
but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...
The following is multiple choice question (with options) to answer.
What is the greatest positive integer x such that 3^x is a factor of 9^9? | [
"5",
"9",
"10",
"18"
] | D | What is the greatest positive integer x such that 3^x is a factor of 9^9?
9^9 = (3^2)^9 = 3^18
D. 18 |
AQUA-RAT | AQUA-RAT-38574 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
each man in a gang of 100 investors has investments ineither equities or securities or both. Exactly 2525 of the investors in equities haveinvestments in securities, exactly 4040 of the investorsin securities have investments inequities. How many have investmentsin equities? | [
"70",
"80",
"90",
"100"
] | B | The investors can be categorized into three groups:
(1) Those who have investments in equities only.
(2) Those who have investments in securities only.
(3) Those who have investments in both equities and securities.
Let xx, yy, and zz denote the number of people in the respective categories. Since the total number of investors is 110, we have
x+y+z=110x+y+z=110 ------------- (1)
Also,
The number of people with investments in equities is x+zx+z and
The number of people with investments in securities is y+zy+z.
Since exactly 25%25% of the investors in equities have investments in securities, we have the equation
25100×(x+z)=z25100×(x+z)=z
25100×x=25100×z25100×x=25100×z
x=3zx=3z ------------------- (2)
Since exactly 40%40% of the investors in securities have investments in equities, we have the equation
40100×(y+z)=z40100×(y+z)=z
(y+z)=5z2(y+z)=5z2
y=3z2y=3z2
Substituting equations (2) and (3) into equation (1) yields
3z+3z2+z=1103z+3z2+z=110
11z2=11011z2=110
z=110×211=20z=110×211=20
Hence, the number of people with investments in equities is:
x+z=3z+z=3×20+20=60+20x+z=3z+z=3×20+20=60+20= 80
B |
AQUA-RAT | AQUA-RAT-38575 | solubility, alloy, teaching-lab
What other such situations are there is chemistry; when would one actually need to do such simultaneous calculations? You may also give the weight of a sample made of an alloy $\ce{Zn-Mg}$. Then you dip it into some diluted $\ce{HCl}$. Both metals will react simultaneously according to $$\ce{Zn + 2 HCl -> H_2 + ZnCl_2}$$ and $$\ce{Mg + 2 HCl -> H_2 + MgCl_2}$$ This will produce an important amount of $\ce{H_2}$ gas. You can measure its volume. So you have two measured data, the original mass and the volume of gas. It is enough for calculating the proportion of $\ce{Zn}$ and $\ce{Mg}$ in the original sample.
Example : Take a mixture of $0.1$ mol $\ce{Zn}$ + $0.3$ mol $\ce{Mg}$. Of course the molar masses of $\ce{Zn}$ and $\ce{Mg}$ are respectively $65.39$ g/mol and $24.32$ g/mol. The total mass is : $6.54$ g + $7.29$ g = $13.83$ g. Suppose the temperature and the pressure of the gas is such that $1$ mole gas occupies $24$ L. So $0.1$ mol $\ce{Zn}$ will produce $0.1$ mol $\ce{H_2}$ which occupies $2.4$ L gas. Then $0.3$ mol $\ce{Mg}$ will produce $0.3$ mol $\ce{H_2}$, or $7.2$ L gas. The total volume of $\ce{H_2}$ is $2.4$ L + $7.2$ L = $9.6$ L.
The following is multiple choice question (with options) to answer.
An alloy is to contain copper and zinc in the ratio 5:3. The zinc required to be melted with 40kg of copper is? | [
"15kg",
"20kg",
"8kg",
"24kg"
] | D | Let the required quantity of copper be x kg
5:3::40:x
5x = 40*3
x = 24kg
Answer is D |
AQUA-RAT | AQUA-RAT-38576 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 9 hours, pipe B in 18 hours and pipe C in 54 hours. If all the pipes are open,in how many hours will the tank be filled ? | [
"5.1",
"5.4",
"5",
"5.2"
] | B | 1/9+1/18+1/54=10/54=1/5.4. so 5.4 hrs
ANSWER:B |
AQUA-RAT | AQUA-RAT-38577 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
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Joined: 13 Mar 2017
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A student scored an average of 80 marks in 3 subjects: Physics, Chemistry and Mathematics. If the average marks in Physics and Mathematics is 90 and that in Physics and Chemistry is 70, what are the marks in Physics? | [
"87",
"37",
"80",
"36"
] | C | Given M + P + C = 80 * 3 = 240 --- (1)
M + P = 90 * 2 = 180 --- (2)
P + C = 70 * 2 = 140 --- (3)
Where M, P and C are marks obtained by the student in Mathematics, Physics and Chemistry.
P = (2) + (3) - (1) = 180 + 140 - 240 = 80
Answer: C |
AQUA-RAT | AQUA-RAT-38578 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
12 welders work at a constant rate they complete an order in 3 days. If after the first day, 9 welders start to work on the other project, how many more days the remaining welders will need to complete the rest of the order? | [
"5",
"2",
"8",
"4"
] | C | 1. We need to find out the time taken by 3 workers after day 1.
2. total no.of wokers * total time taken = time taken by 1 worker
3. Time taken by 1 worker = 12*3 = 36 days
4. But on day 1 twelve workers had already worked finishing 1/3 of the job. So 3 workers have to finish only 2/3 of the job.
5. Total time taken by 3 workers can be got from formula used at (2). i.e., 3* total time taken = 36. Total time taken by 3 workers to finish the complete job is 36/ 3 = 12 days.
6. Time taken by 6 workers to finish 2/3 of the job is 2/3 * 12 =8 days.
The answer is choice C |
AQUA-RAT | AQUA-RAT-38579 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be: | [
"175.5",
"170",
"169.5",
"180"
] | A | Since first and second varieties are mixed in equal proportions.
average price = 126+135/2 = 130.50
o, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
x - 153 = 22.50
x = 175.50
ANSWER A |
AQUA-RAT | AQUA-RAT-38580 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
A 12% stock yielding 8% is quoted at? | [
"165",
"170",
"175",
"150"
] | D | Assume that face value = Rs.100 as it is not given
To earn Rs.8, money invested = Rs.100
To earn Rs.12, money invested = 100×12/8 = Rs.150
ie, market value of the stock = Rs.150
Answer is D. |
AQUA-RAT | AQUA-RAT-38581 | Try to solve it as a diophantine equation
${19x + 96y = D\\ \gcd ( 19, 96 ) = 1}$
${ \text {One possible solution of }\\ 19x + 96y = 1\\ \text{is} \ (\bar x, \bar y) = (-5, 1)\\ \text{thus all the solution of}\\ 19x + 96y = D\\ \text{are } (x,y) = (-5D - 96k, D + 19k) \text{ where } k \in \mathbb{Z}}$
${ \text{in our case } D = 5xy \text{ so }\\ \{ \begin{array}{1} x = -25xy -96k\\y = 5xy + 19k \end{array}\\ \text{namely}\\\{ \begin{array}{1} x(1+25y) \equiv 0 \pmod{96}\\y(1-5x) \equiv 0\pmod{19} \end{array}\\ \text{Then you can solve the system of congruences.} }$
• Ooh, and how would you do that? – Quote Dave May 3 '19 at 19:45
Write it as $y=\frac {19x}{5x-96}$ and $x=\frac{96y}{5y-19}$ We must have $x \ge 20, y \ge 4$. $x$ and $y$ move in opposite directions. Plugging in $x=20$ gives $y=95$, so $4 \le y \le 95$. One can make a spreadsheet to check all these. I find six solutions. This is more cases than factoring, but copy down makes it easy.
Hint $\$ Apply the AC-method to reduce to the monic case, which follows easily by $\rm\,\color{#0a0}{\text{completing a product,}}$ in a similar way to completing a square, i.e.
The following is multiple choice question (with options) to answer.
3x + 2y = 19 , and x + 3y = 1. Find the value of 4x + 5y | [
"10",
"18",
"11",
"20"
] | D | Add these two equations
4x + 5y = 20
Divide by 2 ( to Get 4x + 5y)
Answer will be D. 20 |
AQUA-RAT | AQUA-RAT-38582 | a circle and the coordinates of its center? This definition can be used to find an equation of a circle in the coordinate plane. Re: Finding a point on a circle, given only anothe point, the radius, and the arc len The co-ordinates of your blue point cannot be determined with the data given. Then graph the circle. But the lengths of the legs aren't just plain old x and y anymore. SOLUTION: Rearrange and complete the square in both x and y:. To find this point, calculate the average of the x-coordinates of the endpoints, and also the average of the y. P is the point where the angle intersects the circle and is. Get angle in radians given a point on a circle. Given a circle on the coordinate plane, Sal finds its standard equation, which is an equation in the form (x-a)²+(y-b)²=r². Coordinates of a point on a circle. T must be the same point, so the radius from the center of the circle to the point of tangency is perpendicular to the tangent line, as desired. 7 Circles in the Coordinate Plane(continued) Name _____ Date _____ Work with a partner. Construct the perpendicular bisectors of two sides. x2 + y2 + 6x + 4y - 3 = 0 (x - 3)2 + (y-2) = 42, so the center is (3,2). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The center point of the circle is the center of the diameter, which is the midpoint between (−2,4) ( - 2. find the ratio in which the line segment joining A(2,-2)and B(-3,-5)is divided by the y axis. We can still draw a right triangle, so Pythagoras can still help us out. Destination point given distance and bearing from start point You have starting points: xcoordinate, a list of x-coordinates (longitudes) ycoordinate, a list of y-coordinates (latitudes) num_samples, the number of samples in the plane towards the destination point bearings: heading, a list of headings (degrees)
The following is multiple choice question (with options) to answer.
In the coordinate plane, the points F (-1,1), G (1,4), and H (4,1) lie on a circle with center P. What are the coordinates of point P ? | [
"(0,0)",
"(1,1)",
"(1.5,1.5)",
"(1,-2)"
] | D | I thought I need to find the 4th point that would lie on the circle
Hence my answer was D (1,-1)
But the the center of the circle will lie on (1.5,1.5)
D |
AQUA-RAT | AQUA-RAT-38583 | Case 1 : A pair of socks are present, hence exactly 2 draws for the socks to match. Case 2 : 2 pair of socks are present in the drawer. How?
To start with, instead of looking for a matching pair, let's find the probability that both socks are red.
The Sock Drawer: Probability and Statistics Problem A drawer contains red socks and black socks. Two drawn at random. When two socks are drawn at random, the probability that both are red is 1/2. That should be what you do first with an easy assignment as this. It would seem you can either make a pair or have a mismatched pair, and that both of those events would have equal chances, making for a 50 percent probability. (Deatsville, AL, USA). In a drawer $r$ red, $b$ blue, and $g$ green socks.
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The following is multiple choice question (with options) to answer.
George's drawer has 10 loose black socks, 13 loose blue socks, and 8 loose white socks. If George takes socks out of the drawer at random, how many would he need to take out to be sure that he had at least two matching pairs of socks in two different colors? | [
"7",
"9",
"11",
"16"
] | D | The worst case scenario is choosing 13 blue socks. Then he would need 3 more socks to be sure of a matching pair of black socks or white socks.
The answer is D. |
AQUA-RAT | AQUA-RAT-38584 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If 50 apprentices can finish a job in 4 hours, and 30 journeymen can finish the same job in 8 hours, how much of the job should be completed by 10 apprentices and 15 journeymen in one hour? | [
"1/9",
"29/180",
"9/80",
"1/5"
] | C | 50 apprentices can finish the job in 4 hours,thus:
10 apprentices can finish the job in 4*5 = 20 hours;
In 1 hour 10 apprentices can finish 1/20 of the job.
30 journeymen can finish the same job in 4,5 hours,thus:
15 journeymen can finish the job in 8*2 = 16 hours;
In 1 hour 15 journeymen can finish 1/16 of the job.
Therefore, in 1 hour 10 apprentices and 15 journeymen can finish 1/20+1/16=9/80 of the job.
Answer: C |
AQUA-RAT | AQUA-RAT-38585 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
A group of students was interviewed for that if it was asked whether or not they speak French and / or English. Among those who speak French, 10 speak English well, while 40 of them do not speak English. If 75% of students do not speak French, how many students were surveyed? | [
"210",
"225",
"200",
"250"
] | C | Number of students who speak French are 40 + 10 = 50
Of total students, the percentage of students who do not speak French was 75% --> percentage of who do is 25%
50-------25%
x ------- 100%
x = 50*100/25 = 200 = number of all students
Answer is C |
AQUA-RAT | AQUA-RAT-38586 | 3. Originally Posted by Nath
"I have 2 children. One is female. What is the probability the other is also female?"
This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.
I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.
I'm interested to get some more opinions...
When you know that the speaker has two children you have the probability that they are BB is 1/4, GG is 1/4 and BG is 1/2.
Now you are told that one is a girl, which means that BB is not possible, but the relative probabilities of GG and BG are unchanged, so the probability of GG is 1/3.
Alternatively the result follows from Bayes' theorem:
P(GG|one or more girls)=P(one or more girls|GG)P(GG)/P(one or more girls)=1.(1/4)/((1/4)+(1/2))=1/3
CB
4. I agree with undefined that the wording is ambiguous.
If someone told me they had two children and one is a girl, I would conclude that they mean one is a girl and one is a boy.
But there are at least two other interpretations. (1) The child I am looking at (or thinking of) is a girl; (2) at least one child is a girl. The canonical interpretation of the problem is that (2) is intended, but it's very difficult (for me, at least) to think of a realistic situation in which you know that at least one child is a girl and that's all you know-- unless the speaker is being intentionally obscure.
5. Originally Posted by Nath
"I have 2 children. One is female. What is the probability the other is also female?"
The following is multiple choice question (with options) to answer.
Townville has 100 residents, and 40 of them are females. If half of Townville’s male residents are smokers, and 1/4 of the female residents are smokers, which of the following represents the total number of Townville’s residents who are NOT smokers? | [
"54",
"58",
"62",
"60"
] | D | Number of people =100
Females = 40
men = 100-40=60
Half of the men are smoker60/2=30, that means the remaining men 60-30=30 are non smokers.
1/4 females are smoker. i.e 1/4*40 = 10.
40-10=30 females are non smokers
So, total number of non smokers in the town are 30+30 =60
Answer : D |
AQUA-RAT | AQUA-RAT-38587 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A cricket player whose bowling average was 21.5 runs per wicket, takes 5 wicket for 52 runs in a match. Due to this his average decreases by 0.5. What will be the number of wickets taken by him till the last match? | [
"64",
"72",
"111",
"721"
] | C | Average = Total Runs / Total wickets
Total runs after last match = 21.5w + 52
Total wickets after last match = w + 5
(21.5w + 52) / (w + 5) = 21.5 - 0.5 = 21
w = 106
so total wickets aftr last match = w+5 = 111
ANSWER:C |
AQUA-RAT | AQUA-RAT-38588 | in a cantilever beam the deflection is $$\delta_{max} = \frac {PL^3}{3EI}$$
In this case assuming free sliding between the planks the load P is going to be supported equally between the 3 planks.
So the deflection will be $$\delta_{max} = \frac {(P/3)L^3}{3EI_{\text{single board}}}$$
Because $I$ is proportional to the cube of the board's height (in this case, its thickness), the single board's inertia will be $(1/3)^3=1/27$ that of the bonded boards. Therefore the unbonded deflection will be greater than the bonded boards by a factor of
$$\begin{gather} \dfrac{\left(\frac{1}{3}\right)}{\left(\frac{1}{27}\right)} = \frac{27}{3} = 9 \\ \therefore \delta_{unbonded}= 9\delta_{bonded} \end{gather}$$
• @Wasabi, thanks for the edit. I am liable to make spelling and arithmetic errors because I use my cell phone to write my answers. – kamran Sep 5 '18 at 19:06
• The three beams are subject to a point load at the end, but also act on each other unless the point load is perfectly distributed among perfectly equal beams (even if we ignore friction there will be a vertical component) – mart Aug 7 '20 at 5:23
Thanks to @kamran for his answer.
I simulated the problem in ANSYS student v19 to verify his approach. In the pictures below, the upper beam is solid, the middle one is split into two segments and the lower one is split into 3 segments. Each segment is allowed to slide in respect to its neighbors. It is clear that the deflection of the 3 segments beam is 9 times the full one.
In the case the segments are bonded together (i.e cannot slip in respect to each other) - We get the same results for all the case. All the beam act like a full solid body in bending:
Although I agree with @kamran, I have another way of thinking about it
The following is multiple choice question (with options) to answer.
Three pieces of timber 54 m, 63 m and 81 m long have to be divided into planks of the same length, What is the greatest possible length of each plank ? | [
"9 m",
"4 m",
"4 m",
"6 m"
] | A | Answer
Greatest possible length of each plank = H.C.F of 54, 63, 81 = 9 m
Correct Option: A |
AQUA-RAT | AQUA-RAT-38589 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
Pipe A fills a tank in 30 minutes. Pipe B can fill the same tank 5 times as fast as pipe A. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow? | [
"4 mins",
"6 mins",
"5 mins",
"3 mins"
] | C | Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by A = 3 liters.
Capacity of tank filled in 1 minute by B = 15 liters.
Therefore, capacity of the tank filled by both A and B in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38590 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Each month a retailer sells 100 identical items. On each item he makes a profit of $60 that constitutes 10% of the item's price to the retailer. If the retailer contemplates giving a 5% discount on the items he sells, what is the least number of items he will have to sell each month to justify the policy of the discount? | [
"191",
"213",
"221",
"223"
] | D | For this question, we'll need the following formula:
Sell Price = Cost + Profit
We're told that the profit on 1 item is $20 and that this represents 10% of the cost:
Sell Price = Cost + $60
Sell Price = $600 + $60
Thus, the Sell Price is $660 for each item. Selling all 100 items gives the retailer...
100($60) = $2,000 of profit
If the retailer offers a 5% discount on the sell price, then the equation changes...
5%(660) = $33 discount
$627 = $600 + $27
Now, the retailer makes a profit of just $27 per item sold.
To earn $2,000 in profit, the retailer must sell....
$27(X) = $2,000
X = 2,000/27
X = 222.222222 items
You'll notice that this is not among the answer choices.... 221 and 223 are.
Selling 221 items would get us 9(221) = $1989 which is NOT enough money. To get back to AT LEAST $2,000, we need to sell 223 items.
Final Answer:
D |
AQUA-RAT | AQUA-RAT-38591 | greatest common divisor
cehsu
104 views
Greatest Common Divisor
In mathematics, the greatest common divisor (gcd) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers. For example, the gcd of 8 and 12 is 4.
The greatest common divisor is also known as the greatest common factor (gcf), highest common factor (hcf), greatest common measure (gcm), or highest common divisor.
Check for the gcd
Write a getGCD function that takes two numbers as parameters and returns the gcd.
Sample Input and Output
Input: 8 12 Output: 4
Optimization
Check out Euclid's algorithm and see how much things speed up. How can you account for this in terms of time complexity?
See:
https://en.wikipedia.org/wiki/Euclidean_algorithm#Worked_example
https://en.wikipedia.org/wiki/Greatest_common_divisor#Complexity_of_Euclidean_method
Write the getGCD function
1
2
3
4
5
6
7
8
function getGCD (numOne, numTwo) {
}
module.exports = getGCD;
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Open Source Your Knowledge: become a Contributor and help others learn.
The following is multiple choice question (with options) to answer.
If x and y are positive integers, and 1 is the greatest common divisor of x and y, what is the greatest common divisor S of 2x and 3y? | [
"1",
"S=Cannot be determined",
"S=2",
"5"
] | B | My explanation: from question stem we know that nothing is common between X and Y , X and Y are two prime numbers eg: X=2, Y=3 and their GCD(2,3) =1 and so 2X and 3Y will have a GCD(2X,3Y) = 1 . what if either X or Y was 1, eg: X=1,Y=4 then GCD(1,4) =1 , but GCD(2,12) = 2.
and hence answer B |
AQUA-RAT | AQUA-RAT-38592 | time
Title: Is there a difference between 0.18 and 0.180 seconds? Q. Is there a difference between 0.18 and 0.180 seconds ?
In our databases we are collecting timestamps and there are 2 time instants which are the same for the hour, minute and seconds but the milliseconds field is being returned as:
0.18
0.177 When scientists report a measurement of 0.18, it usually means they are certain the actual value they've measured is between 0.175 and 0.185.
When scientists report a measurement of 0.180, it usually means they are certain the actual value they've measured is between 0.1795 and 0.1805.
When scientists report a measurement of 0.177, it usually means they are certain the actual value they've measured is between 0.1765 and 0.1775.
How you should interpret those numbers in your database is be a question for the person who put those numbers there, but without other available information, that's the interpretation I would default to.
The following is multiple choice question (with options) to answer.
0.0023*0.13/0.236=my age
0.022*0.12/0.220=my wife age
then what is difference b/w ages? | [
"0.01073",
"0.02073",
"0.03076",
"0.04078"
] | A | My age = 0.0023*0.13/0.236 = 0.000299/0.236 = 0.001266949
My wife age = 0.022*0.12/0.220 = 0.00264/0.220 = 0.012
difference b/w ages = 0.012 - 0.001266949 = 0.01073
ANSWER:A |
AQUA-RAT | AQUA-RAT-38593 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Ashok has Rs.1000, He invests a part of it at 3% per annum and remainder at 8% per annum, He earns Rs.100 in 2 years. Find the sum invested at different rates of interest? | [
"Rs.5000, Rs.500",
"Rs.600, Rs.400",
"Rs.700, Rs.300",
"Rs.300, Rs.700"
] | B | (600÷100)×3×2=34
(400÷100)×8×2=64
Total=100
ANSWER:B |
AQUA-RAT | AQUA-RAT-38594 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
what is the sum of all the 4 digit numbers that can be formed using all of the digits 2,3,5 and 7? | [
"113323",
"113322",
"114322",
"113321"
] | B | Actually there is the direct formula for this kind of problems.
1. Sum of all the numbers which can be formed by using the n digits without repetition is: (n-1)!*(sum of the digits)*(111…..n times).
2. Sum of all the numbers which can be formed by using the n digits (repetition being allowed) is: n^{n-1}*(sum of the digits)*(111…..n times).
so as per question:(4-1)!*(2+3+5+7)*1111=113322
ANSWER:B |
AQUA-RAT | AQUA-RAT-38595 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec. What is the length of the fast train? | [
"27 7/8",
"27 7/6",
"27 7/4",
"27 7/9"
] | D | Relative speed = (40 - 20) = 20 km/hr.
= 20 * 5/ 18 = 50/9 m/sec.
Length of faster train = 50/9 * 5 = 250/9 = 27 7/9 m.
Answer: D |
AQUA-RAT | AQUA-RAT-38596 | P(30 \leq X \leq 40) = P(29.5 \leq X \leq 40.5)
using the general conversion formula
Z=\frac{X-\mu_{X}}{\sigma_{X}}
where \sigma_{x_i}= \sqrt(350/12) we can write
P \left( \frac{29.5-35}{\sqrt(\frac{350}{12})} \leq \frac{X-35}{\sqrt(\frac{350}{12})} \leq \frac{40.5-35}{\sqrt(\frac{350}{12})} \right)
P \left( -1.01840 <Z < 1.01840 \right)
where
\phi(Z)=pnorm(1.01840)
and
1-2\phi(Z) = 0.6915
# R crosscheck
a1=(29.5-35)/(sqrt(350/12))
a2=(40.5-35)/(sqrt(350/12))
a3 = 1 -2*pnorm(a1)
The following is multiple choice question (with options) to answer.
If 25% of x is 30 less than 20% of 1000, then x is? | [
"188",
"216",
"156",
"680"
] | D | 25% of x = x/4 ; 20% of 1000 = 20/100 * 1000 = 200
Given that, x/4 = 200 - 30 => x/4 = 170 => x = 680.
Answer: D |
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