source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34097 | Including leading zeros you can see this as follows.
Let $d$ be a digit from $0,\ldots , 9$
• there is exactly 1 3-digit number $ddd$ $\Rightarrow 3$ occurrences
• there are $\binom{3}{2}\cdot 9$ 3-digit numbers, where $d$ occurs exactly twice $\Rightarrow 3\cdot 9 \cdot 2 = 54$ occurrences
• there are $\binom{3}{1}\cdot 9^2$ 3-digit numbers, where $d$ occurs exactly once $\Rightarrow 3\cdot 9^2 = 243$ occurrences
All together $300$ occurrences.
Here is another shorter argument:
If you write down all 3-digit codes with digits $0,\ldots , 9$, you write all together $3000$ digits. Each digit occurs equally often. So, each digit occurs $$\frac{3000}{10}= 300 \mbox{ times}$$
The following is multiple choice question (with options) to answer.
log3 N+log9 N what is 3 digit number N that will be whole number | [
"629",
"729",
"829",
"929"
] | B | No of values N can take is 1
9^3=729
ANSWER:B |
AQUA-RAT | AQUA-RAT-34098 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
What least number must be added to 3000 to obtain a number exactly divisible by 19 ? | [
"2",
"19",
"9",
"4"
] | A | On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 - 17) = 2.
ANSWER A 2 |
AQUA-RAT | AQUA-RAT-34099 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in upstream is 20 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"30 kmph",
"13 kmph",
"65 kmph",
"55 kmph"
] | A | Speed of the boat in still water
= (20+80)/2
= 50 kmph. Speed of the stream
= (80-20)/2
= 30 kmph.
Answer: A |
AQUA-RAT | AQUA-RAT-34100 | 1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2)
3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$
So I would answer problem 1 as a is not equal to either 0 or 1.
In the second problem, it is clear that $x \ne 3.$
Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$
$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$
Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.
Does this help
5. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by JeffM
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
The following is multiple choice question (with options) to answer.
If 3^a - 3^(a-1) = 162 then a(a-1) = | [
"12",
"16",
"20",
"30"
] | C | It seems like you were really close but you solved for a - 1 instead. I am curious - how did you reason through the question? Perhaps you figured the answer has to be in the form of a(a -1), e.g. (3)(4) = 12. Even then, had you figured a has to be a little bigger (plugging in 4 for a gives you 81 which is too low), So when you reasoned the answer to be 4 you may have plugged the 4 in the (a - 1) exponent place.
Again, a quick plugging in should get you (C) 20 as, the only other answer with consecutive integers as factors is far too big. |
AQUA-RAT | AQUA-RAT-34101 | 5. 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...
Here is an alternative "logical" way to consider this.
There are 4 possibilities in regard to the children being boy or girl.
The probability of having a boy and a girl is twice the probability of having 2 girls,
and is also twice the probability of having 2 boys.
(i) 1st and 2nd children are girls
(ii) 1st child is a girl and the 2nd is a boy
(iii) 1st child is a boy and the 2nd is a girl
(iv) 1st and 2nd children are boys
GG
GB
BG
BB
One is female.
This reduces to...
GG
GB
BG
3 cases of equal probability and in only 1 of these cases is the other child also a girl.
Hence the probability of the 2nd child also being a girl is 1/3.
Also note that in 2 of these 3 cases, the 2nd child is a boy,
therefore, if one of the children is a girl, the probability that the other is a boy is 2/3.
6. ## Solutions
This was my solution, but alas my friend disagrees:
I have 2 children, 1 is female. This gives four possible scenarios:
1. The female has a younger sister
2. The female has an older sister
3. The female has a younger brother
4. The female has an older brother
Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
---
My friend, with the PhD in Maths thinks the following:
The following is multiple choice question (with options) to answer.
A family has two children. find the probability that both the children are girls given that at least
one of them is a girl? | [
"1/3",
"1/5",
"1/7",
"1/9"
] | A | Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3
A) |
AQUA-RAT | AQUA-RAT-34102 | inside the circle can use the kite properties show! Triangle ABC is a diameter of the circle and the circle and touching the sides of the of! Is theta, this is also a cool trick to impress your less mathematically inclined or! Tannu ( 53.0k points ) circles ; class-10 ; 0 votes such that BC = 12 cm and =... My circle right there 've already labeled it, it 's also a cool trick to your... 5 cm distance over here also has this distance right here be and what is the basis trigonometry. And we can find the lengths of AB centre O has been inscribed the triangle if you know three! If we can find the circle of center O and radius r = 10 cm center is the. Which One angle is a diameter of the circle is called an inscribed angle is a of... 1, 2018 in Mathematics by Tannu ( 53.0k points ) circles I have question... From akshaya circle inscribed in a right triangle a student: a quadrilateral can beinscribed in a right triangle or right-angled triangle with length! Of 5 cm and AB = 8 cm ; Start date May 14, ;! That form the right angle if and only if AC is a chord through the center of incircle... Circle ) are opposite each other, they lie on the bottom is.. ' I$ is right ( 47.2k points ) circles I have a right is. 0 votes is πr², so these two sides of it hypotenuse that determined! If and only if its opposite angles aresupplementary at these points the radius of the radius of triangle!, find the circle ’ circle inscribed in a right triangle asking for: area of the circle tangent... That BC = 6 cm the Pythagorean theorem to show that ΔBOD is triangle.: a circle is 39.19 square centimeters, and so $\angle AC ' I$ is right is also..., diameter is 2 in., find the value of each variable,...
The following is multiple choice question (with options) to answer.
What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 5, 12 and 13? | [
"2.5",
"6.5",
"5",
"6.0"
] | B | Some of pyhtagron triplets we need to keep it in mind.
Like {( 2,3,5) , ( 5,12,13) ,( 7, 24,25), ( 11, 60,61).
So now we know the triangle is an right angle triangle. The circle circumscribes the triangle.
The circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 13 / 2 = 6.5
Ans. B |
AQUA-RAT | AQUA-RAT-34103 | # If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
My work:
We know the sample space $S:$"The set of number of 1 to 1000000" and $|S|=1000000$
Let $E$ the event such that $E:$"The set of number contains the digit 5 in $[1000000]$" We need calculate $|E|$.
I know in $[100]$ we have $5,15,25,35,45,50,51,52,53,54,55,56,57,58,59,65,75,85,95$ then we have 19 numbers contains the digit $5$ in the set $[100]$
Then in $[1000]-[500]$ we have 171 numbers have the digit 5. this implies [1000] have 271 number contains the digit 5.
. . .
Following the previous reasoning we have to $[10000]$ have 3439 number contains the digit 5.
Then, $[100000]$have 40951 number contains the digit 5.
Moreover, $[1000000]$ have 468559 number contain the digit 5.
In consequence the probability of we pick a digit contain the number 5 in the set $[1000000]$ is 0.468
Is correct this?
How else could obtain $|E|$?
Thanks
The following is multiple choice question (with options) to answer.
{-10, -6, -5, -4, -2.5, -1, 0, 2.5, 4, 6, 7, 10}
A number is to be selected at random from the set above. What is the probability that the number will be a solution to the equation (x-5)(x+4)(2X+5) = 0? | [
"a) 1/12",
"b) 1/6",
"c) 1/4",
"d) 1/3"
] | A | x = -1
Prob = 1/12
Answer - A |
AQUA-RAT | AQUA-RAT-34104 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
One drier dries certain quantity of material in 24 minutes. Another drier does the same work in 2 minutes how much time will it take to do the same job when both driers are put to work ? | [
"1.00 minutes",
"1.20 minutes",
"1.50 minutes",
"1.85 minutes"
] | D | By guess it is clear that the time taken will be less than 2 minutes and more than 1.5 mintes Therefore, answer 1.85 minutes will be correct. Answer-D |
AQUA-RAT | AQUA-RAT-34105 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
How many integers from 101 to 700, inclusive, remains the value unchanged when the digits were reversed? | [
" 50",
" 60",
" 70",
" 80"
] | B | question is asking for palindrome
first digit possibilities - 1 through 6 = 6
7 is not possible here because it would result in a number greater than 7 (i.e 707 , 717..)
second digit possibilities - 0 though 9 = 10
third digit is same as first digit
=>total possible number meeting the given conditions = 6*10 =60
Answer is B. |
AQUA-RAT | AQUA-RAT-34106 | 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 restaurant meal cost $55 and there was no tax. If the tip was more than 20 percent but less than 30 percent of the cost of the meal, then total amount paid must have been between: | [
"$65 and $72",
"$66 and $71",
"$64 and $71",
"$63 and $70"
] | A | let tip=t
meal cost=55
range of tip = from 20% of 55 to 30% of 55 = 11 to 16.5
hence range of amount paid= 55+T= 66 to 71.5 i.e. A |
AQUA-RAT | AQUA-RAT-34107 | 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 180 m long is running with a speed of 60 km/hr. In what time will it pass a man who is running at 6 km/hr in the direction opposite to that in which the train is going? | [
"7",
"10",
"8",
"2"
] | B | Speed of train relative to man = 60 + 6 = 66 km/hr.
= 66 * 5/18 = 55/3 m/sec.
Time taken to pass the men = 180 * 3/55
= 10 sec.
Answer B |
AQUA-RAT | AQUA-RAT-34108 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
List I: 3, 6, 8, 10
List II: x, 3, 6, 8, 10 | [
"7",
"8",
"9",
"10"
] | A | List I has even number of terms, thus its median is the average of two middle terms (when arranged in ascending/descending order), so median=(6+8)/2=7.
List II has odd number of terms, thus its median is the middle term (when arranged in ascending/descending order). As no other number in the list equal to 7, then x=7.
Answer: A. |
AQUA-RAT | AQUA-RAT-34109 | c#, algorithm, palindrome
In my tests this is 3 times faster.
I hate to break this to you, but there is a bug in your code, which will probably mean rewriting your algorithm. You algorithm assumes that the palindrome will be an odd number of characters. However, a palindrome can be an even number of characters and your code won't find it if it is.
Here's some code that will find the longest palindrome regardless:
static string LargestPalindrome(string input)
{
string output = "";
int minimum = 2;
for(int i = 0; i < input.Length - minimum; i++)
{
for(int j = i + minimum; j < input.Length - minimum; j++)
{
string forstr = input.Substring(i, j - i);
string revstr = new string(forstr.Reverse().ToArray());
if(forstr == revstr && forstr.Length > minimum)
{
output = forstr;
minimum = forstr.Length;
}
}
}
return output;
}
EDIT
The above code has a bug. Here it is reworked:
static string LargestPalindrome(string input)
{
int longest = 0;
int limit = input.Length;
for (int i = 0; i < limit; i++)
{
for (int j = limit-1; j > i; j--)
{
string forStr = input.Substring(i, j - i);
string revStr = new string(forStr.Reverse().ToArray());
if (forStr == revStr && forStr.Length > longest)
{
return forStr;
}
}
}
return "";
}
The following is multiple choice question (with options) to answer.
Consider the word RMTMR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5 letter palindromes. | [
"15343",
"16423",
"17576",
"17809"
] | C | The first letter from the right can be chosen in 26 ways because there are 26 alphabets.
Having chosen this, the second letter can be chosen in 26 ways
The first two letters can chosen in 26 x 26 = 676 ways
Having chosen the first two letters, the third letter can be chosen in 26 ways.
All the three letters can be chosen in 676 x 26 =17576 ways.
It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.
C |
AQUA-RAT | AQUA-RAT-34110 | 1-1/3+1/5-1/7+...
I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.
Homework Helper
Okay I got (2), (thanks for the correction), by comparing (2) to $$\frac{\sqrt{n}}{n^2}=\frac{1}{n^\frac{3}{2}}$$ which converges (p-series w/ p>1), therefore (2) converges. Is this correct?
As for (1) I wrote out the terms and discovered it is an alternating series with the even terms all equal to 0 as follows:
1-1/3+1/5-1/7+...
I can't seem to write a series to satisfy the requirements however...any guidance would be appreciated.
Yes, that's right for (2). For (1) you have an alternating series with decreasing terms. How about the alternating series test?
malindenmoyer
Yea I figured it had to do with that.
If we define the alternating series to be: $$\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n$$
then I can't seem to determine a $$b_n$$ which defines the series for all n, especially since every odd $$n$$ changes sign and all the even $$n$$'s are 0...I must be missing something...
If we define the alternating series to be: $$\Sigma_{n=1}^{\infty} (-1)^{n-1}b_n$$
then I can't seem to determine a $$b_n$$ which defines the series for all n, especially since every odd $$n$$ changes sign and all the even $$n$$'s are 0...I must be missing something...
The following is multiple choice question (with options) to answer.
Look at this series: (1/9), (1/3), 1, ____ , 9, ... What number should fill the blank? | [
"3",
"4",
"1",
"6"
] | A | A
3
This is a multiplication series; each number is 3 times the previous number. |
AQUA-RAT | AQUA-RAT-34111 | To MINIMIZE the time required to fill the tank, we must MAXIMIZE the time for P, since P's input rate is greater than Q's input rate.
Since P must close for $$\frac{1}{4}$$ hour after each hour of work, P's time will be maximized if -- over a 5-hour period -- P works for 4 hours and takes 4 quarter-hour breaks, with Q working for $$\frac{1}{4}$$ hour during each break.
Thus, the input rate for each 5-hour period = (4 hours of work for P) + (1 hour of work for Q) - (5 hours of work for R) $$= (4*20) + (1*15) - (5*12) = 35$$ liters for every 5 hours of work.
Since the volume increases by 35 liters every 5 hours, the hourly rate $$≈ \frac{35}{5} ≈ 7$$ liters per hour.
(The hourly rate is an approximation because it increases when P works but decreases when Q works.)
Since the hourly rate is about 7 liters per hour, the approximate time to fill the 60-liter tank $$= \frac{60}{7} ≈ 8.5$$ hours.
10am + about 8.5 hours ≈ 6:30pm.
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Originally posted by GMATGuruNY on 18 Jul 2018, 03:43.
Last edited by GMATGuruNY on 18 Jul 2018, 12:09, edited 1 time in total.
##### General Discussion
Math Expert
Joined: 02 Aug 2009
Posts: 7211
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
13 Jul 2018, 08:09
2
2
EgmatQuantExpert wrote:
Question of the Week #7
The following is multiple choice question (with options) to answer.
A pipe takes a hours to fill the tank. But because of a leakage it took 4 times of its original time. Find the time taken by the leakage to empty the tank | [
"50 min",
"60 min",
"90 min",
"80 min"
] | D | pipe a can do a work 60 min.
lets leakage time is x;
then
1/60 -1/x=1/240
x=80 min
ANSWER:D |
AQUA-RAT | AQUA-RAT-34112 | computational-chemistry
( 5 0 | 4 0) = 0.000862792421254
( 5 2 | 1 0) = 0.000000000000000
( 5 3 | 1 0) = 0.049151047350638
( 5 4 | 1 0) = 0.041228155506758
( 5 1 | 2 0) = 0.000000000000000
( 5 1 | 3 0) = 0.005091750233326
( 5 1 | 4 0) = 0.004270986718219
( 5 0 | 2 1) = 0.000000000000000
( 5 0 | 3 1) = 0.001923027401919
( 5 0 | 4 1) = 0.001613045439387
( 5 2 | 1 1) = 0.000000000000000
( 5 3 | 1 1) = 0.184701361441018
( 5 4 | 1 1) = 0.154928467698285
( 5 1 | 2 1) = 0.000000000000000
( 5 1 | 3 1) = 0.038683534029509
( 5 1 | 4 1) = 0.032447950603007
( 5 2 | 2 0) = 0.009336058122476
( 5 2 | 3 0) = 0.000000000000000
( 5 2 | 4 0) = 0.000000000000000
( 5 3 | 2 0) = 0.000000000000000
( 5 3 | 3 0) = 0.011813922734513
( 5 3 | 4 0) = 0.002078445792233
( 5 4 | 2 0) = 0.000000000000000
( 5 4 | 3 0) = 0.002078445792233
( 5 4 | 4 0) = 0.011079469320272
( 5 0 | 2 2) = 0.054104219459936
( 5 0 | 3 2) = 0.000000000000000
The following is multiple choice question (with options) to answer.
The G.C.D. of 1.08, 0.36 and 0.5 is: | [
"0.02",
"0.9",
"0.18",
"0.108"
] | A | Given numbers are 1.08, 0.36 and 0.50. H.C.F. of 108, 36 and 50 is 18,
H.C.F. of given numbers = 0.02.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34113 | # 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.
A man buys an article for 10% less than its value and sells it for 10% more than its value. His gain or loss percent is: | [
"70%",
"80%",
"90%",
"20%"
] | D | Explanation:
Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 - 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%
Answer:D |
AQUA-RAT | AQUA-RAT-34114 | 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.
A can do a piece of work in 16 days which B can do in 12 days. B worked at it for 9 days. A can finish the remaining work in ? | [
"3 days",
"4 days",
"6 days",
"7 days"
] | B | B's 9 day's work = 9 x (1/12) = 3/4
Remaining work = (1 - 3/4) = 1/4
1/4 work is done by A in = 16 x (1/4) = 4 days.
answer : B |
AQUA-RAT | AQUA-RAT-34115 | +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}
The following is multiple choice question (with options) to answer.
Find the sub-triplicate ratio of 27:125 | [
"3:5",
"3:9",
"3:6",
"3:1"
] | A | The sub-triplicate ration of 27 : 125 = 3:5
Answer:A |
AQUA-RAT | AQUA-RAT-34116 | # 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 of certain article is 25 percent.had it been bought at 10 percent discount and sold for 20 percent more than the earliest cost,the profit would have been rs600.what is the cost price of the article. | [
"1800",
"2000",
"2200",
"2400"
] | B | Let CP be x
Gain% = 25%
Then SP = x*125/100
SP = 5x/4
If it had been bought at 10% discount
Then CP = x*90/100
CP = 9x/10
sold at 20% more that the earliest cost... [earliest cost means x]
Then SP = x*120/100 = 600+9x/10
6x/5 = 6000+9x/10
12x = 6000+9x
3x = 6000
x = 2000
ANSWER:B |
AQUA-RAT | AQUA-RAT-34117 | # 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.
A dishonest shopkeeper deceives by 15% at the time of purchase of article and also 15% at the time of sale.Find out the profit percentage
Profit% | [
"32.25%.",
"32",
"42.25%",
"42"
] | A | Solution
=15+15+(15*15 / 100)=30 + 225/100 =30+2.25=32.25%.
Answer A |
AQUA-RAT | AQUA-RAT-34118 | # 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.
A shop owner sells 56 apples and gains SP of 16 apples. Find the gain %? | [
"30%",
"40%",
"50%",
"56%"
] | B | Gain=16 apples
Total 56 apples
CP 56-16=40
(16/40)*100
=40%
Ans B |
AQUA-RAT | AQUA-RAT-34119 | ### Show Tags
05 Sep 2017, 11:01
2
In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hoursat 60 mph. What was his average speed for the whole trip?
(A) 50
(B) 53.5
(C) 55
(D) 56
(E) 57.5
The total distance traveled from City A to City B is 50*1 + 60*3 = 230
Since John drove for 4 hours to travel this distance,
the average speed of the whole trip is $$\frac{Distance}{Time} = \frac{230}{4}$$ = 57.5(Option E)
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Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink]
### Show Tags
17 Jan 2019, 03:38
How does this work using the harmonic mean?
$$\frac{1/50+3/60}{4}$$=$$\frac{7/100}{4}$$=57.1
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink] 17 Jan 2019, 03:38
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The following is multiple choice question (with options) to answer.
A certain car traveled twice as many miles from Town A to Town B as it did from Town B to Town C. From Town A to Town B, the car averaged 12 miles per gallon, and from Town B to Town C, the car averaged 16 miles per gallon. What is the average miles per gallon that the car achieved on its trip from Town A through Town B to Town C? | [
"13",
"13.5",
"14.4",
"14.5"
] | C | Ans is C
Given d_ab = 2*d_bc
let d_ab = d and d_bc = x so d=2x
for average miles per gallon = (d+x)/((d/12)+(x/16)) = 14.4 (formula avg speed = total distance/ total time) |
AQUA-RAT | AQUA-RAT-34120 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
The H.C.F. of two numbers is 12 and their L.C.M. is 600. If one of the number is 45, find the other? | [
"100",
"160",
"120",
"200"
] | B | Other number = 12*600/45= 160
Answer is B |
AQUA-RAT | AQUA-RAT-34121 | # whole numbers and division
Consider the whole number with one thousand digits that can be formed by writing the digits 2772 two hundred and fifty time in succession. Is it divisible by 9? Is it divisible by 11?
-
I answered yes to both because 2772 is divisible by 9 and 11. I am just trying to make sure this is correct! – SNS Feb 27 '12 at 23:19
If the sum of the digits of $n$ is divisible by 9, then (and only then) $n$ is divisible by 9. From this, your number is divisible by 9 (the sum of its digits is $250\cdot18$). I've forgotten the divisibility test for 11... – David Mitra Feb 27 '12 at 23:21
Can you show that your number is divisible by 2772? And can you finish up from there? – Gerry Myerson Feb 27 '12 at 23:26
2772/9=308 2772/11=252 I think that if the original number is divisible by 9 and 11 then if you continue adding the same numbers over and over it should always be divisible by 9 and 11. This is the part I want to double check on. – SNS Feb 27 '12 at 23:29
@DavidMitra If the alternating sum of the digits (add, subtract, add, subtract, etc.) is divisible by 11 then the number is divisible by 11. This is because $10^n\equiv (-1)^n\mod 10$. – Alex Becker Feb 27 '12 at 23:29
The answer is yes; but, in my opinion, you did not give enough information in your comment for a justification.
One way to show it is to use the divisibility tests for 9 and 11.
Let's call your number, obtained by writing "$2772$" two hundred and fifty times in succession, $y$.
A number $n$ is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $y$ is $250\cdot(2+7+7+2)=250(18)$, so $y$ is divisible by 9.
The following is multiple choice question (with options) to answer.
Which of the following numbers is divisible by 9? | [
"23274",
"25181",
"31001",
"50123"
] | A | 23274. This is the only option with last two digits divisible by 9
Answer:A |
AQUA-RAT | AQUA-RAT-34122 | More generally, let's denote the books as $A,B,C,\dots,Y,Z$ with corresponding probability of being chosen as $a,b,c,\dots,y,z$. Of course, $a+b+c+\cdots+y+z=1$.
Let's take a particular ordering, that I will label as $\beta$: $$\beta:=DJG\cdots CW.$$ The formula for the invariant probability of $\beta$ is the product $$\pi(\beta):={d\over d+j+g+\cdots+c+w}\,{j\over j+g+\cdots+c+w}\, {g\over g+\cdots+c+w} \cdots {c\over c+w}\, {w\over w}.$$
To check this, let's calculate the $\beta$th entry in the vector $\pi P$, that is, $\sum_\alpha \pi(\alpha) P(\alpha,\beta)$. The only states $\alpha$ from which it is possible to jump to state $\beta$ are those that look just like $\beta$ but with book $D$ moved. That is, $$\alpha\in\{ DJG\cdots CW, JDG\cdots CW, JGD\cdots CW, \dots, JG\cdots CWD\}.$$ For all such $\alpha$, the transition probability is simply $P(\alpha,\beta)=d$.
The following is multiple choice question (with options) to answer.
If C.P. of 25 books is equal to S.P of 30 books , then your gain or loss is ? | [
"23 1/3 gain %",
"30 10/23 loss %",
"16.7% loss %",
"30 1/3 loss %"
] | C | Say the cost price of 25 books is $25 so $1 for a book.
The cost of 30 books would be $30, and the selling price of 30 books would be 25$, hence the loss = ($30 - $25)/$30 = ~16.7%.
Answer: C. |
AQUA-RAT | AQUA-RAT-34123 | 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.
At what rate of interest is an amount doubled in two years, when compounded annually? | [
"41.4%",
"41.7%",
"41.4%",
"42.4%"
] | B | Let the rate of interest be r. The amount gets doubled in two years
=> P(1 + r/100)2 = 2p => 1 + r/100 = √2 => r/100 = √2 - 1
= 0.414 => r = 100(0.414)
= 41.4%
Answer:B |
AQUA-RAT | AQUA-RAT-34124 | # Faulty Combinatorial Reasoning?
I have 10 books, 4 of which are biographies while the remaining 6 are novels. Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies. How many different combinations of choosing 4 books in such a way are there?
The following line of reasoning is faulty, but I can't figure out why:
First we figure out how many ways there are of choosing 2 biographies from 4. Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8. This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices. Then we have:
1. BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).
2. REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order. This leaves us with (8*7)/2! choices.
3. Overall we have [(4*3)/2!]*[(8*7)/2!] = 168 total choices.
Where did I go wrong?
-
How could I adjust for the over-counting I did here? (Rather than constructing the answer of 115 by adding together the discrete cases of choosing 2 bios, 3 bios, and 4 bios)? – George Sep 4 '12 at 22:15
In your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_£$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.
To solve the problem:
The following is multiple choice question (with options) to answer.
John has on his shelf two books of poetry, four novels, and two reference works. Suppose from these ten books, we were to pick two books at random. What is the probability that we pick one novel and one reference work? | [
"1/2",
"2/5",
"3/10",
"2/7"
] | D | When we are picking two books, one novel and one reference work, we could either pick a novel first and then a reference book or pick a reference book and then a novel. Therefore the answer is 4/8*2/7 + 2/8*4/7 = 2/7.
Answer: D |
AQUA-RAT | AQUA-RAT-34125 | 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
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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 D of 2x and 3y? | [
"1",
"D=Cannot be determined",
"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-34126 | 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.
In what time will a train 55 m long cross an electric pole, it its speed be 36 km/hr? | [
"5.5",
"4.5",
"3.5",
"2.5"
] | A | Speed = 36 * 5/18 = 10 m/sec
Time taken = 55/10 = 5.5 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-34127 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
A car traveled 448 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city? | [
"14",
"18",
"21",
"22"
] | B | Let the speed in highway be h mpg and in city be c mpg.
h = c+6
h miles are covered in 1 gallon
462 miles will be covered in 462/h.
Similarly c miles are covered in 1 gallon
336 miles will be covered in 336/c.
Both should be same (as car's fuel capacity does not change with speed)
=> 336/c = 448/h
=> 336/c = 448/(c+6)
=> 336c+336*6=448c
=>c=336*6/112=18
Answer B. |
AQUA-RAT | AQUA-RAT-34128 | 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.
A is 50% as efficient as B.C does half of the work done by A and B together .If C along does the work in 40 days then A,B,C
together can do the work in? | [
"15 1/4 days",
"16 days",
"13 1/3 days",
"25 2/5 days"
] | C | A's 1 day's work:B's 1 day's work=150:100=3:2
let A's and B's 1 days work be 3x and 2x respectively
then C's 1st days work=(3x+2x)/2=5x/2
5 x/2=1/40
x=1/100
A's 1 day work=3/100
B's 1 day work=1/50
C's 1 day work=1/40
(A+B+C)'s 1 day work=(3/100+1/50+1/40)=15/200=3/40
so A,B,C together can do the work in 40/3=13 1/3 days
Answer(C) |
AQUA-RAT | AQUA-RAT-34129 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
On a sum of money, the simple interest for 2 years is Rs. 325, while the compound interest is Rs. 340, the rate of interest being the same in both the cases. The rate of interest is | [
"15%",
"14.25%",
"9.23%",
"10.5%"
] | C | Explanation:
The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
= (R × SI)/(2 × 100)
Difference between the compound interest and simple interest = 340 - 325 = 15
(R × SI)/(2 × 100) = 15
(R × 325)/(2 × 100) = 15
R = 9.23%
Answer: Option C |
AQUA-RAT | AQUA-RAT-34130 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Alex invested an amount of $500 for two years. Find the rate of compound interest that will fetch him an amount of $83.20 at the end of two years? | [
"8%",
"10%",
"12%",
"7%"
] | A | Let the rate of interest be R% p.a.
500[1 + R/100]^2 = 583.20
[1 + R/100]2 = (583.20)/500
= 11664/10000 = [108/100]2
[1 + R/100] = 108/100
R=8%
Answer is A |
AQUA-RAT | AQUA-RAT-34131 | ### 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, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.400 when the rate to profit rises from 5 to 7 percent. Find the Capital of C? | [
"3377",
"2899",
"5000",
"2778"
] | C | A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 400
C's capital = 30000*1/6 = 5000
Answer: C |
AQUA-RAT | AQUA-RAT-34132 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
Average of first seven multiples of 2 is | [
"8",
"9",
"13",
"15"
] | A | Explanation:
Average=2(1+2+3+4+5+6+7) / 7=56 / 7=8
Option A |
AQUA-RAT | AQUA-RAT-34133 | 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.
In a division sum, the quotient is 120, the divisor 456 and the remainder 333, find the dividend? | [
"55053",
"50553",
"5553",
"40532"
] | A | 120 * 456 + 333 = 55053
ANSWER A |
AQUA-RAT | AQUA-RAT-34134 | filters
a6 = 128 + (224 *w2 * w1 * w3) - (56 *w2 * w1*w1 * w3) + (448 *w2 * w1 * w4) - (112 *w2 * w1*w1 * w4) + (448* w1 * w3 * w4) + (224 *w2 * w3 * w4) - (640 *w1) - (320 *w2) - (320 *w3) - (640 *w4) + (64 *w2 * w1) + (112 *w2 * w1*w1) + (32 *w1*w1) - (224 *w2 * w1 * w3 * w4) - (168 *w2 * w1*w1 * w3 * w4) + (64 *w1 * w3) + (32 *w2 * w3) + (112 *w1*w1 * w3) + (128 *w1 * w4) + (64 *w2 * w4) + (64 *w3 * w4) + (224 *w1*w1 * w4) - (112 *w1*w1 * w3 * w4) + (32 *w4*w4) + (224 *w4*w4 * w1) - (56 *w4*w4 * w1*w1) - (168 *w4*w4 * w1 * w2 * w3) + (210 *w4*w4 * w1*w1 * w2 * w3) - (112 *w4*w4 * w1 * w2) - (84 *w4*w4 * w1*w1 * w2) + (112 *w4*w4 * w2) + (112 *w4*w4 * w3) - (112 *w4*w4 * w1 * w3) - (84 *w4*w4 * w1*w1 * w3) - (56 *w4*w4 * w2 * w3)
The following is multiple choice question (with options) to answer.
select the no. which is different from series
1,4,9,16,23,25,36 | [
"25",
"36",
"23",
"9"
] | C | answer is 23 only rest every no. is square of any no.
answer C |
AQUA-RAT | AQUA-RAT-34135 | javascript, datetime
// a simple helper function
function nextDay(date) {
return new Date(date.getFullYear(), date.getMonth(), date.getDate() + 1);
}
// as long we're in the same year, keep adding 1 day,
// and store the ones that match the weekday we're looking for
while(current.getFullYear() === year) {
if(current.getDay() === weekday) {
dates.push(current);
}
current = nextDay(current);
}
return dates;
}
No need to build ranges, map, and filter them. Just a while loop and push.
Incidentally, if you want to learn more about why programming calendar and time things are just hideously complex, check out this video on youtube
1) Happy Guy Fawkes day everyone
The following is multiple choice question (with options) to answer.
Today is Tuesday. The day after 59 days will be | [
"Monday",
"Tuesday",
"Saturday",
"Friday"
] | D | Explanation :
59 days = 8 weeks 3 days = 3 odd days
Hence if today is Tuesday, After 59 days, it will be = (Tuesday + 3 odd days)
= Friday. Answer : Option D |
AQUA-RAT | AQUA-RAT-34136 | 4,639 views
In a tournament with $7$ teams, each team plays one match with every other team. For each match, the team earns two points if it wins, one point if it ties, and no points if it loses. At the end of all matches, the teams are ordered in the descending order of their total points (the order among the teams with the same total are determined by a whimsical tournament referee). The first three teams in this ordering are then chosen to play in the next round. What is the minimum total number of points a team must earn in order to be guaranteed a place in the next round?
1. $13$
2. $12$
3. $11$
4. $10$
5. $9$
I think possible with 9 only.
no 9 will not guarantee!!!
Let the $7$ Teams be $A,B,C,D,E,F,G$ and so each team plays total $6$ matches.
Suppose, Team $A$ wins over $E,F,G$ and draws with $B,C,D$ hence total points scored by Team A $= 9$ points
Now, Team $B$ wins over $E,F,G$ and draws with $A,C,D$ hence total points scored by Team B $= 9$ points
Similarly, happens for next two teams $C$ and $D$ .
Hence, Finalized scores are $\Rightarrow$
A = 9
B = 9
C = 9
D = 9
E = ? (Less than or equal to 4)
F = ? ("...")
G = ? ("...")
Given that the order among the teams with the same total are determined by a whimsical tournament referee.
So, He/She can order the top $3$ teams like $ABC,ABD,BCD,ACD,\ldots$
But, Question says " team must earn in order to be guaranteed a place in the next round "
Hence, Not to depend on that whimsical referee, the minimum total number of points a team must earn in order to be guaranteed a place in the next round = $9+1 = 10$ points
Correct Answer: $D$
by
So only 2 teams have 10 point right ??
why u considered A has won with 3 teams and tie with 3 teams?
The following is multiple choice question (with options) to answer.
A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was | [
"665",
"287",
"297",
"255"
] | A | Explanation:
Let the total score be x.
(x + 92 - 85) / 8 = 84.
So, x + 7 = 672 => x = 665.
Answer: A |
AQUA-RAT | AQUA-RAT-34137 | 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's speed with the current is 18 km / hr and the speed of the current is 2 km / hr. The man's speed against the current is | [
"8.5 km / hr",
"9 km / hr",
"14 km / hr",
"12.5 km / hr"
] | C | Sol.
Man's rate in still in water = (18 - 2) km / hr = 16 km / hr.
Man's rate against the current = (16 - 2) km / hr = 14 km / hr.
Answer C |
AQUA-RAT | AQUA-RAT-34138 | 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.
A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm. | [
"223",
"160",
"297",
"276"
] | B | Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am.
B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km
Answer:B |
AQUA-RAT | AQUA-RAT-34139 | 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 the county of Veenapaniville, there are a total of 50 high schools, of three kinds: 25 public schools, 16 parochial schools, and 9 private independent schools. These 50 schools are divided between three districts: A, B, and C. District A has 18 high schools total. District B has 17 high schools total, and only three of those are private independent schools. If District C has an equal number of each of the three kinds of schools, how many private independent schools are there in District A? | [
"1",
"3",
"4",
"5"
] | A | Total Private Schools = 9
Dist A: High Schools = 18 ==> Private Schools = ?
Dist B: High Schools = 17 ==> Private Schools = 3
Dist C: High Schools = 15 ==> Private Schools = 5
Therefore, 9 - 3 - 5 ==> 1
Answer A) |
AQUA-RAT | AQUA-RAT-34140 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4383
Location: India
GPA: 3.5
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
Thanks and Regards
Abhishek....
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 49 hrs and a machine of type S does the same job in 14 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used? | [
"49/9",
"44/9",
"46/9",
"41/9"
] | A | Rate of machine R =1/49
Rate of machine S =1/14
since same no of machines used for R and S to do the same work in 2 hrs
So collective rate needed to finish the work in 2 hrs= 1/2
Let the no of machine be x
So, x/49 +x/14 =1/2
9x=49
x=49/9
So no of machine R is 49/9
Answer A |
AQUA-RAT | AQUA-RAT-34141 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
The simple interest on Rs.10000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for two years at the same rate of interest. | [
"3087.82",
"3087.89",
"3087.85",
"3087.36"
] | D | R = 100 I / PT
=> R = (100 * 7200)/ (10000 * 5) = 14.4%
CI = P{ [1 + R /100]n - 1}
= 10000 { [ 1 + 14.4 / 100]^2 - 1} = Rs.3087.36
Answer:D |
AQUA-RAT | AQUA-RAT-34142 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
Two numbers are 30% and 37% are less than a third number .How much percent is the second number less than the first? | [
"10",
"15",
"20",
"30"
] | A | I II III
70 63 100
70 -------- 7
100 ------ ? => 10%
ANSWER A |
AQUA-RAT | AQUA-RAT-34143 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If 4-X < (2-5X)/3, which of the following is correct? | [
"X < -5.",
"X > -5.",
"X > 5.",
"-5 < X < 0."
] | A | 4-X < (2-5X)/3
12-3X < 2- 5X
10 < -2X
-10/2>X
-5>X
A is the answer |
AQUA-RAT | AQUA-RAT-34144 | 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 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long? | [
"40",
"74",
"38",
"76"
] | A | Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 140
= 500 m
Required time = 500 * 2/25
= 40 sec
Answer: A |
AQUA-RAT | AQUA-RAT-34145 | javascript, mathematics
Now for my next approach
Everything you have seen above remains the same, but now I don't calculate the offsets. Instead I calculate the total number of elements.
Since I have my forwards variable set to my absolute difference... It is actually the length of my line segment from 2 to 5 on my number scale.
Then I can just subtract this segment from number scale to get the reverse distance
1-2-3-4-5-6-7-8
| |
|~~3~~|
|~seg~|
1-2-3-4-5-6-7-8
|~~~~~~8~~~~~~|
|~ num-scale ~|
seg - num-scale
|~|~ 5 ~|~~~~~|
Everything else remains the same...
Hence here is the code (Yes, I also made this code to be series extensible):
function findClosestTwo(from, to, min, max){
let smaller = (from > to) ? to : from,
bigger = (from < to) ? to : from;
let totalNumberOfElements = max - min + 1;
let forwards = bigger - smaller,
backwards = totalNumberOfElements - forwards;
let result = (forwards < backwards) ? forwards : -backwards;
result = (from > to) ? -result : result;
The following is multiple choice question (with options) to answer.
0---:-|---:--|-:--:-|--:---|-:----1
On the number line above, the segment from 0 to 1 has been divided into fifths, as indicated by the large tick marks, and also into sevenths, as indicated by the small tick marks. What is the LEAST possible distance X between any two of the tick marks? | [
"1/70",
"1/35",
"2/35",
"1/12"
] | B | divisions are 0 1/7 2/7 3/7 4/7 5/7 6/7 1
1/5 2/5 3/5 4/5 5/5
expressing the same in terms of the lcm of 5,7 ,i.e 35
0 7/35 14/35 21/35 28/35 35/35
5/35 10/35 15/35 20/35 25/35 30/35
by comparing the divisions with each other we can see 1/35 is the shortest possible distance between any two selected divisions.
X=1/35
Answer is B. |
AQUA-RAT | AQUA-RAT-34146 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A sum of money at simple interest amounts to Rs. 825 in 3 years and to Rs. 846 in 4 years. The sum is? | [
"s. 738",
"s. 638",
"s. 650",
"s. 762"
] | D | S.I. for 1 year = (846 - 825) = Rs. 29
S.I. for 3 years = 21 * 3 = Rs. 63
Principal = (825 - 63) = Rs. 762.
ANSWER:D |
AQUA-RAT | AQUA-RAT-34147 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
A box contains either blue or red flags. The total number of flags in the box is an even number. A group of children are asked to pick up two flags each. If all the flags are used up in the process such that 60% of the children have blue flags, and 60% have red flags, what percentage of children have flags of both the colors? | [
"5%",
"10%",
"15%",
"20%"
] | D | Solution: let the total number of flags be 100(even number)
let the total number of 'blue' flags alone be 'a'
let the total number of 'red' flags alone be 'b'
let the total number of 'both' flags be 'c'
We have given,
total number of blue flags = 60% = 60 = a+c
total number of red flags=55%=55=b+c
total number of flags = a+b+c=100 (since all the flags have been utilized)
So, substituting for c in the third equation, we have,
60-c+c+60-c=100
c=20
Option D. |
AQUA-RAT | AQUA-RAT-34148 | control, motion-planning, algorithm
The issue I have to treat is, how to optimally bypass the blue obstacle even I don't know how deep it is. Driving to the left and to the right only to capture better data points (to generate a 3D model) is not possible.
The following is multiple choice question (with options) to answer.
You are riding a horse. In front of you, there is a fire engine. A helicopter is following you. To your left a sports car is driving. To your right there is a depth.How can you arrange that you will all stop simultaneously, without crashing and without mutual communication? | [
"helicopter man stop it",
"you stop it",
"man go to stop it.",
"no idea"
] | C | You ask the man of the merry-go-round to stop it.
Answer C |
AQUA-RAT | AQUA-RAT-34149 | At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
5. Hmmm....I'm not a maths expert but let's see what we get if we reverse the ratios. Perhaps there was a communication error.
$p:a=3:5$
$\frac {1}{10}$of a=wooden
$\frac{1}{10}\times\frac{5}{8}=\frac{5}{80}$
= number of wooden.
number of brick = 1-number of wooden
$=\frac{75}{80} = \frac{15}{16}$
Edit: Are you sure you wrote the ratio the correct way around? The way you wrote it suggests that the person above has the correct answer.
I don't know how it got 15/16. I actually can't figure out why that is.
At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
The following is multiple choice question (with options) to answer.
How much 90% of 40 is greater than 80% of 30? | [
"12",
"14",
"16",
"17"
] | A | (90/100) * 40 – (80/100) * 30
36 - 24 = 12
Answer: A |
AQUA-RAT | AQUA-RAT-34150 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman scored 18 runs in his 18th innings and that makes his average 18. Find his average upto the 17th innings? | [
"17",
"18",
"19",
"20"
] | B | total innings=18
avg=18
so total run at 18th innings is 18*18=324
total at 17th inning=324-18=306
now 306/17=18
ANSWER:B |
AQUA-RAT | AQUA-RAT-34151 | ## Circular Combination
1. Arrange 8 dancers in circular fashion.
2. Use 8 pearls in a band to make a necklace.
3. Arrange 8 science and 7 arts students circularly so no two arts students are together.
• 7!
• $$\frac{(8-1)!}{2}$$
• $$^8P_7 \times \space 7!$$
# Combination
## Concept
A, B, C
In how many ways can we select 2?
• AB, AC, BC
• When making teams $$AB \equiv BA$$
## Formulae And Notation
Formulae (Don’t Memorize!)
• $$^nC_r = {n! \over r!(n-r)!}$$
• $$^nC_r \times ^rP_r = ^nP_r = ^nP_r \times r!$$ (Permutation-combination relationship)
• $$^{n+1}C_r = ^nC_r + ^nC_{r-1}$$
• $$^nC_r=?$$ (from above, expanding twice up to n-2)
• $$^nC_r = ^nC_{n-r}$$
Notation
• $$^nC_r \equiv$$ $$n \choose r$$
• Select 5 people from 6 $$\rightarrow ^6C_5 =$$ $$6 \choose 5$$
## Expaniosn of $$^nC_r$$
The following is multiple choice question (with options) to answer.
There are ten different models that are to appear in a fashion show. Two are from Europe, two are from South America, and two are from North America. If all the models from the same continent are to stand next to each other, how many ways can the fashion show organizer arrange the models? | [
"72",
"48",
"64",
"24"
] | C | Since we have 3 continental pairs (EU, SA, NA), these 3 pairs have 3*2*1 = 8 Combinations. Within each pair, you have however 2 different ways to put them together for each of the pair (2*2*2 = 8). So we have 8*8 = 64.
Please correct me if im wrong.
Answer C. |
AQUA-RAT | AQUA-RAT-34152 | dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
_________________
-----------------------
tusharvk
Manager
Joined: 27 Oct 2008
Posts: 180
Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
The following is multiple choice question (with options) to answer.
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue? | [
"33/99",
"33/91",
"36/91",
"33/97"
] | B | Explanation:
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. When four marbles are picked at random, then the probability that none is blue is
= 33/91
Answer: B |
AQUA-RAT | AQUA-RAT-34153 | 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.
Five women can do a work in ten days. 7 men can complete the same work in 5 days. What is the ratio between the capacity of a man and a woman? | [
"10:7",
"11:10",
"2:3",
"3:2"
] | A | Explanation:
(50 × 10) women can complete the work in 1 day.
∴ 1 woman's 1 day's work = 1/50
(7 × 5) men can complete the work in 1 day.
∴ 1 man's 1 day's work = 1/35
so, required ratio =1/50 : 1/35= 10:7
Answer: A |
AQUA-RAT | AQUA-RAT-34154 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
Last year the price per share of Stock N increased by a percent and the earnings per share of Stock N increased by b percent, where a is greater than b. By what percent did the ratio of price per share to earnings per share increase, in terms of a and b? | [
"100*(a-b) %",
"100*(a-b)/(100+b) %",
"(100+b) %",
"(a-b)/(100+b) %"
] | B | Let P = the old price per share; E = the old earning per share. Thus P/E is the price to earning ratio before the increases
After the increase the new price is: P*(1+a/100) and the new earning is: E*(1 +b/100)
The new P/E is: (1+a/100)P/(1+b/100)E
The Percent of P/E increase = (new P/E - P/E)/(P/E). Subsititute new P/E to the equation we have:
[(1+a/100)/(1+b/100)*P/E - P/E]/(P/E)*100%. Simplifly the expression and you should get the answer to be:
100*(a-b)/(100+b) %
Answer : B |
AQUA-RAT | AQUA-RAT-34155 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 100 m race,A covers the distance in 36 seconds and B in 45 second.In this race A beats B by : | [
"20 m",
"25 m",
"22.5 m",
"9 m"
] | A | Solution
Distance covered by B in 9 sec. = (100/45x9)m =20m.
∴ A beats B by 20 metres.
Answer A |
AQUA-RAT | AQUA-RAT-34156 | algorithms
Title: How to calculate least common multiple (LCM) for two numbers with constants (shifts)? What's the most effective algorithm to calculate LCM for 2 numbers where each of them has their own "shift"?
Example:
We want to find LCM for 25 + 5 as c and 30 + 10 as c
Then for the first case, the sequence will look like this:
25*1+5, 25*2+5, 25*3+5, 25*4+5...
And the second one:
30*1+10, 30*2+10, 30*3+10, 30*4+10...
In this case, the result should be 130 - 5th member of the first sequence and 4th of the second one. Given integers $a,b \geq 1$ and $c,d \geq 0$, you are looking for the set of solutions of $ax + c = by + d$ over integers $x,y \geq 0$. We can assume without loss of generality that $d = 0$ (replace $c,d$ with $c-d,0$ or $0,d-c$) and that $(a,b,c) = 1$ (otherwise, divide everything by the GCD).
I claim that if there is any solution then necessarily $(a,b) = 1$. Indeed, suppose that $(a,b) = g > 1$. If $ax + c = by$ then $g \mid ax,by$ implies $g \mid c$, and so $(a,b,c) = g > 1$, contrary to assumption. Therefore $(a,b) = 1$.
If $ax + c = by$ then $ax + c \equiv 0 \pmod{b}$ and so $x \equiv -c/a \pmod{b}$ (note that we can divide by $a$ since $(a,b)=1$). Similarly, $y \equiv c/b \pmod{a}$. These equations essentially give us both the minimal solution and all solutions.
The following is multiple choice question (with options) to answer.
What is the least common multiple of 15, 16, and 24? | [
"60",
"120",
"240",
"360"
] | C | Let us first write the numbers in the form of prime factors:
15 = 3*5
16 = 2^4
24 = 2 * 17^1
The LCM would be the largest powers of the prime numbers from all these three numbers.
Hence LCM = 240
Option C |
AQUA-RAT | AQUA-RAT-34157 | 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.
Angela, Bernie, and Colleen can complete a job, all working together, in 2 hours. Angela and Bernie, working together at their respective rates, can complete the same job in 5 hours. How long would it take Colleen, working alone, to complete the entire job? | [
"8 hours",
"10/3 hours",
"12 hours",
"16 hours"
] | B | A+B+C complete the job in 2 hours.
A+B complete the job in 5 hours.
A+B and C complete the job in 2 hours -->1/(A+B) + 1/C = 1/2-->1/5+1/C=1/2 ----> C=10/3 hours. ANS B |
AQUA-RAT | AQUA-RAT-34158 | base-conversion
Title: What does it mean when asking to Convert 16-bit binary numbers to two 8-bit signed integers (two’s complement) in decimal The question is
Convert the following 16-bit binary numbers to two 8-bit signed
integers (two’s complement) in decimal.
i. 0b 0011 0110 1101 0100
I first flipped it and added 1 to it which equaled
0b 1100 1001 0011 1100 which is 51516 in decimal
Is this right? I don't really understand what the question is asking for when it says "two" 8-bit signed integers (two’s complement) in decimal. That question is not the clearest, but I don't think that's what it asked you to do. "Two’s complement" means two things: a system of mapping between bit-strings and integers in a useful way, and an operation (which you performed) that corresponds to negation under that mapping. This question is not asking you to negate the number, it's asking you to "convert" it.
What sort of conversion we're supposed to be doing is not that clear, but a prime candidate for "converting" a 16 bit number into two 8 bit numbers is to split it down the middle.
So I think the question wants this:
Take 0011 0110, interpret it as a signed 8-bit number, using the two's complement system. So eg 1111 1111 would mean -1, instead of -127 (in the sign and magnitude system) or -0 (in the ones' complement system) or something else. 0011 0110 would mean 54. 1101 0100 is more interesting since it's negative.
The following is multiple choice question (with options) to answer.
In a certain code language, '3a, 2b, 7c' means 'Truth is Eternal';
'7c, 9a, 8b, 3a' means 'Enmity is not Eternal' and
'9a, 4d, 2b, 8b' means 'Truth does not perish'.
Which of the following means 'enmity' in that language ? | [
"3",
"9",
"8",
"6"
] | C | Explanation:
Justification:
In the second and third statements, the common code is '9a' and the common word is 'not'.
So, '9a' means 'not'.
In the first and second statements, the common codes are '7c' and '3a' and the common words are 'is' and 'Eternal'.
So, in the second statement, '8b' means 'enmity'.
Answer: C) 8b |
AQUA-RAT | AQUA-RAT-34159 | 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.
Two pipes P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first is turned off. How much longer will the cistern take to fill? | [
"11 1/8",
"11 1/4",
"11 1/1",
"11 1/2"
] | B | Explanation:
3/12 + x/15 = 1
x= 11 1/4
Answer:B |
AQUA-RAT | AQUA-RAT-34160 | Positive counterpart geometry cubed root is always positive 2 = n × n × 2... Or cube number is..... a special value that when cubed gives the original number sure its asking that. Power of 3 is an integer that results from cubing another integer when cubed gives the number... The problem is cubed root can be used to find the length of a side a! Volume to ( Weight ) Mass Converter for Recipes, Weight ( Mass ) to volume to Converter for,... The answer for the square root, cubic root, cubic root, the cubed root be! Another integer of 1/27 multiplying that number by itself 3 times ( ^3 ) is...: n √ a = b b n = 3 is because cubing negative. Questions like: What is the cube root of -8 is written as \ ( \sqrt 3... Whole number, it is a cubed ( power of 3 one be. Because 3 x 3 is equal to 27 is said to be a perfect because... 3 = n × n × n × n 2 = n × 2. Is always positive to find the length of a cubed when the volume is known OUR cube of... Cubing a negative number results in an answer different to that of cubing it 's positive.... 1 to 1000 be 49 a negative number results in an answer different to that of cubing is to. Number multiplied by its square: whole number, it is a cubed ( power of 3 ).. Answer for the square root, N-th root Section to be a perfect square, a cube. It 's positive counterpart click on the above image, then choose link... The process of cubing is similar to squaring, only that the number multiplied by its square: known... Cubed root, where n = a. Estimating a root of 27/64 | Mathway the same as!, assuming real numbers to volume to ( Weight ) Mass Converter for Recipes Weight. A whole number, it is asking for that number by itself 3 (...
The following is multiple choice question (with options) to answer.
What positive number, when squared, is equal to the cube of the positive square root of 17? | [
"64",
"32",
"8",
"4"
] | C | Let the positive number be x
x^2 = ((17)^(1/2))^3
=>x^2 = 4^3 = 64
=> x = 8
Answer C |
AQUA-RAT | AQUA-RAT-34161 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Find the odd man out. 30, -5, -45, -90, -145, -195, -255 | [
"-5",
"-145",
"-255",
"-195"
] | B | Explanation :
30
30 - 35 = -5
-5 - 40 = -45
-45 - 45 = -90
-90 - 50 = -140
-140 - 55 = -195
-195 - 60 = -255
Hence, -145 is wrong. -140 should have come in place of -145
Answer : Option B |
AQUA-RAT | AQUA-RAT-34162 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Rita told Mani, "The girl I met yesterday at the beach was the youngest daughter of the brother-in-law of my friend's mother." How is the girl related to Rita's friend ? | [
"Cousin",
"Daughter",
"Niece",
"Friend"
] | A | Explanation:
Daughter of brother-in-law — Niece;
Mother's niece — Cousin.
So, the girl is the cousin of Rita's friend.
Answer: A |
AQUA-RAT | AQUA-RAT-34163 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
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.
Kudos [?]: 24 [2], given: 0
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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
1
This post was
BOOKMARKED
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.
The average age of the district level hockey team of eleven is 25 years.The average age gets increased by 1 year when the coach age is also included.what is the age of the coach ? | [
"36 yrs.",
"37 yrs.",
"34 yrs.",
"33 yrs."
] | B | Explanation:
Total age of players in team = 25 x 11 = 275
When coach is included, total members = 12, Avg age increases by one becomes 26
Total age when coach is included = 26x12 = 312
Age of coach = 312-275 = 37 yrs.
Answer: B |
AQUA-RAT | AQUA-RAT-34164 | # Calculate the number of ways to paint $5$ buildings with $4$ colours such that all $4$ colours must be used
A developer has recently completed a condominium project in a valley. There are blocks of buildings $A$, $B$, $C$, $D$ and $E$ as shown in the diagram below.
The developer has colours available to paint the buildings. Each block can only be painted using a single colour. Find the number of ways to paint all the blocks if all $4$ colours must be used.
My attempt: We have $5$ ways to choose $4$ buildings with $4$ different colours. Among the $4$ buildings, we have $4!$ ways to paint them using $4$ different colours.
So my answer is $120$. However, the answer given is $240$. What is my mistake?
• Um, what do paint the fifth building? Don't you have 4 chooses for that? I'd get that you method out to give 4*120 =480. Which is also wrong as we double counted. We'd choose a building to be a duplicate color. there are 5 choices for that. We'd paint the other four. 24 four that. We'd pick one of the four buildings to duplicate the fifth one. There are four choices of that. Which ever building we choose to duplicate is a double counting as we could have choosen that building to be a duplicate. so divide by two. – fleablood May 12 '17 at 15:24
You're on the right track. After choosing the four buildings with different colours, what about the fifth? It will be a repeated colour, and there are four colours to choose from. However, we will have overcounted by a factor of $2$, so the final answer will be $$120\cdot\frac42 = 240$$ To explain the overcounting, suppose that you first choose $A,B,C,D$ as the set of four buildings. Then $E$ is the same as one of them, say $A$. But then we will also later consider $E,B,C,D$ as the set of four, with $A$ the same as $E$.
The following is multiple choice question (with options) to answer.
A building of 6 floors is being painted by 2 painters, and they take 3 days to paint the first floor. If 3 men are then added to the work and all of them continue painting at the same rate, how many days take to them paint all the building, since the first day? | [
"5 days",
"8 days",
"9 days",
"10 days"
] | C | We have that: 2 painters ----- 3 days, (2+3) P ----- X d, that is: (X d/ 2P) = (3 d/5P), then: X d = (3 d/5P)2P = (6/5)d. Is to say, the 5 painters paint the same quantity in 6/5 days; as it is only 1/6 of the building, all the building would be: B = 3 d + 5(6/5) d = (3 + 6) d = 9 days. Answer C. |
AQUA-RAT | AQUA-RAT-34165 | 3 BIOGRAPHIES books: $\dfrac{4\cdot 3 \cdot 2}{32!} \cdot 6 =24$
4 BIOGRAPHIES books: 1
then the result is 115.
-
Suppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$. Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$. So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$. That made a contribution of $1$ to your $168$.
But among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your "two more" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$.
Both of these ways of choosing end us up with $A$, $B$, $C$, and $N$. So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next. Still another contribution of $1$ to your $168$.
So your product counts the set $\{A, B, C, N\}$ three times. This it does for every combination of three biographies and one novel. It also overcounts the set $\{A, B, C, D\}$.
One could adjust for the overcount. In some problems that is a useful strategy. Here it takes some care.
But a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.
-
The following is multiple choice question (with options) to answer.
Ravid has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d? | [
" 5/6*d",
" 7/3*d",
" 10/3*d",
" 7/2*d"
] | C | Ravid has d books;
Jeff has d/3 books;
Paula has 2d books;
Total = d+d/3+2d=10d/3.
Answer: C. |
AQUA-RAT | AQUA-RAT-34166 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
How many three-digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6? | [
"None",
"One",
"Two",
"Three"
] | A | The possible number N can be written as follow:
N = Multiple of LCM(6,10) + 1st such number
N = 30x + 1
Possible values = 1, 31, 61, 91,121,151,181,211,241,271.....991
Answer : None:A |
AQUA-RAT | AQUA-RAT-34167 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is: | [
"Rs. 650",
"Rs. 690",
"Rs. 698",
"Rs. 700"
] | C | S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.
Answer: Option C |
AQUA-RAT | AQUA-RAT-34168 | We can get rid of the double workaround by rewriting the equation to the following, which is not shorter, but saves one operator and operates in int32 the whole time:
Bottles_1 = n => 32 - Math.clz32(n - 2)
« Back to problem overview
The following is multiple choice question (with options) to answer.
Solution for 2.01+.3+.34 | [
"2.91",
"2.65",
"2.938",
"2.986"
] | B | 2.01+.3+.34=0
0=0-2.01-0.3-0.34
0=-2.65
answer :B |
AQUA-RAT | AQUA-RAT-34169 | ### 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 ratio of investments of two partners A and B is 7:5 and the ratio of their profits is 7:10. If A invested the money for 5 months, find for how much time did B invest the money ? | [
"11 months",
"9 months",
"7 months",
"10 months"
] | D | Explanation:
7x5: 5xk = 7:10
k = 10 months
ANSWER IS D |
AQUA-RAT | AQUA-RAT-34170 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
90 students represent x percent of the boys at Jones Elementary School. If the boys at Jones Elementary make up 60% of the total school population of x students, what is x? | [
"122",
"150",
"225",
"250"
] | A | 90=x/100*60/100*x=>x^2 = 9*10000/6 = > x = 122
A |
AQUA-RAT | AQUA-RAT-34171 | # 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.
A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit? | [
"Rs. 18.20",
"Rs. 70",
"Rs. 72",
"Rs. 88.25"
] | C | Explanation:
C.P. = Rs.(100/122.5 x 392)= Rs.(1000*1225 x 392)= Rs. 320
Profit = Rs. (392 - 320) = Rs. 72.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-34172 | What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$
The following is multiple choice question (with options) to answer.
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high ? | [
"13 m",
"14 m",
"15 m",
"16 m"
] | A | Explanation:
the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is=13mts.
ANSWER IS A |
AQUA-RAT | AQUA-RAT-34173 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Mr. Smitherly leaves Cedar Rapids at 8:00 a.m. and drives north on the highway at an average speed of 50 km per hour. Mr. Dinkle leaves Cedar Rapids at 9:00 a.m. and drives north on the same highway at an average speed of 60 km per hour. Mr. Dinkle will | [
"overtake Mr. Smitherly at 10:30 a.m.",
"overtake Mr. Smitherly at 11:30 a.m.",
"overtake Mr. Smitherly at 2:00 p.m.",
"be 30 miles behind at 9:35 a.m."
] | C | After one hour, Mr. Smitherly is ahead by 50 km.
Mr. Dinkle gains on Mr. Smitherly 10 km each hour.
Five hours after 9:00 a.m., Mr. Dinkle will catch Mr. Smitherly.
The answer is C. |
AQUA-RAT | AQUA-RAT-34174 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
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 $$0.3*1$$.
The following is multiple choice question (with options) to answer.
A 6-liter solution is 40% alcohol. How many liters of pure alcohol must be added to produce a solution that is 50% alcohol? | [
"1.2",
"1.5",
"1.8",
"2.1"
] | A | Let x be the amount of pure alcohol required.
0.4(6) + x = 0.5(x+6)
0.5x = 3 - 2.4
x = 1.2 liters
The answer is A. |
AQUA-RAT | AQUA-RAT-34175 | # 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 list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is? | [
"9%",
"4%",
"3%",
"6%"
] | B | 65*(90/100)*((100-x)/100) = 56.16
x = 4%
Answer: B |
AQUA-RAT | AQUA-RAT-34176 | $\implies \text{Var} \left(X\right) = \frac{1}{n} \cdot \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{1}{n} \cdot \frac{n \left(n + 1\right)}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{\left(n + 1\right) \left(2 n + 1\right)}{6} - {\left(\frac{n + 1}{2}\right)}^{2}$
$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \left[\frac{2 n + 1}{3} - \frac{n + 1}{2}\right]$
$\implies \text{Var} \left(X\right) = \frac{n + 1}{2} \cdot \frac{n - 1}{6}$
$\implies \text{Var} \left(X\right) = \frac{{n}^{2} - 1}{12}$
So, Standard deviation of $\left\{1 , 2 , 3 , \ldots . , n\right\}$ is ${\left[\text{Var} \left(X\right)\right]}^{\frac{1}{2}} = {\left[\frac{{n}^{2} - 1}{12}\right]}^{\frac{1}{2}}$
In particular, your case the standard deviation of $\left\{1 , 2 , 3 , 4 , 5\right\}$
$= {\left[\frac{{5}^{2} - 1}{12}\right]}^{\frac{1}{2}} = \sqrt{2}$.
The following is multiple choice question (with options) to answer.
For the set { 4, 4, 5, 5, x}, which of the following values of x will most increase the standard deviation? | [
"1",
"(-3.5)^2",
"3",
"4"
] | B | Standard Deviation Step 1, as pointed out by others, is to find out the mean = 4.5
Step 2, For each number: subtract the Mean and square the result =
(1-4.5)^2=(-3.5)^2
(2-4.5)^2=(-2.5)^2
(3-4.5)^2 =(-1.5)^2
(4-4.5)^2=(.5)^2
(5-4.5)^2=(.5)^2
Clearly (1-4.5)^2=(-3.5)^2 will give you the greatest value among all the other options.
Hence B |
AQUA-RAT | AQUA-RAT-34177 | # Find Smallest Positive Integer
#### anemone
##### MHB POTW Director
Staff member
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
#### Wilmer
##### In Memoriam
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Unfortunately, got no exotic formula for you
#### anemone
##### MHB POTW Director
Staff member
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Hi Wilmer,
Thanks for participating and thanks for the compliment to this problem as well.
Yes, 3987 is the answer to this problem but...
Unfortunately, got no exotic formula for you
30 minutes in the corner, please...
#### Wilmer
##### In Memoriam
30 minutes in the corner, please...
No fair! You only asked: "Find the smallest positive integer [FONT=MathJax_Math-italic]n" [/FONT]
#### anemone
The following is multiple choice question (with options) to answer.
N and M are each 3-digit integers. Each of the numbers 1, 2, 3,4,5 and 6 is a digit of either N or M. What is the smallest possible positive difference between N and M? | [
"59",
"49",
"58",
"113"
] | B | You have 6 digits: 1,2, 3, 4, 5, 6
Each digit needs to be used to make two 3 digit numbers. This means that we will use each of the digits only once and in only one of the numbers. The numbers need to be as close to each other as possible. The numbers cannot be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other.
The first digit (hundreds digit) of both numbers should be consecutive integers
Now let's think about the next digit (the tens digit). To minimize the difference between the numbers, the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . So let's not use 1 and 6 in the hundreds places and reserve them for the tens places Now what are the options?
Try and make a pair with (2** and 3**). Make the 2** number as large as possible and make the 3** number as small as possible.
265 and 314 (difference is 49) or
Try and make a pair with (4** and 5**). Make the 4** number as large as possible and make the 5** number as small as possible. We get 463 and 512 (difference is 49)
B |
AQUA-RAT | AQUA-RAT-34178 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
_________________
Please press +1 Kudos if this post helps.
Intern
Joined: 16 May 2017
Posts: 14
GPA: 3.8
WE: Medicine and Health (Health Care)
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A? | [
"648",
"888",
"928",
"1184"
] | D | B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
ANSWER:D |
AQUA-RAT | AQUA-RAT-34179 | = push-ups / seconds in air is close 340.1... { 450nm } * { f } = { 450nm } * { f } = { 450nm *... Page, or Hz for short bin numbers ( upper levels ) the... Is hit on one end with a speaker bolted to it is n't a concept 's! Calculate the frequency and vice versa vibrational frequency can easily be changed when dial! End with a period of a particular value occurs enrolling in a that... Will show a Sine wave that can be measured in seconds two years of college and save thousands your... Waves that pass a fixed point in unit time divided by a count of all values write our frequencies units. N'T measure something like push-ups in 30 seconds that repeats has a master degree! { 45Hz } =20,250nm/s < =20.25um/s } and the time it takes to do just one.! The 42 seconds per lap through some practice problems the velocity of life... The calculation can be applied to many situations Mean = 2 + 10 + 12 + 8 + 14. Diameter, is the frequency in hertz even if it is an essential feature engines... 1/2 inch in diameter, is the height from highest to lowest points and divide that by 2 your blocker. Means they are related like this: if you are just asked for ,., Types & uses, Mass and Weight: Differences and Calculations, 83,000... Event to occur or education level 83,000 lessons in all major subjects, { courseNav.course.mDynamicIntFields.lessonCount... At B2-B10 and use the first two years of college and save thousands off degree. Something repeats attend yet from a frequency associated with it measurement we use the fact that period is also inverse. Like this: Essentially, anything that repeats has a master 's degree physics. For many people it probably has something to do just one lap how to find frequency measurement! Many situations to connect it to a given wavelength college you want to yet. In how to find frequency proportion of the system frequency is a measure of cycles second! The 1 second of period or from any point to the trough ) an for... To 60 seconds we can see RPM represents a frequency and period can be to! One lap Mean
The following is multiple choice question (with options) to answer.
What is the average of four tenths and five thousandths | [
"0.2022",
"0.3325",
"0.2025",
"0.2012"
] | C | Explanation:
Four tenths = 0.4
Five thousandths = 0.005
The average is (0.4 + 0.005)/2 = 0.2025
Answer: C |
AQUA-RAT | AQUA-RAT-34180 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Walking with 4/5 of my usual speed, I miss the bus by 5 minutes. What is my usual time? | [
"16 min",
"26 min",
"34 min",
"20 min"
] | D | Speed Ratio = 1:4/5 = 5:4Time Ratio
= 4:51 -------- 5 4 --------- ? è 20
Answer: D |
AQUA-RAT | AQUA-RAT-34181 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
A train sets off at 2 p.m. at the speed of 70 kmph. Another train starts at 3:30 p.m. in the same direction at the rate of 85 kmph. At what time the trains will meet? | [
"10.37 p.m",
"10.20 p.m",
"10.30 p.m",
"10.38 p.m"
] | C | D = 70 * 1 ½ = 105 km
RS = 85 – 70 = 15
T = 105/15 = 7 h
3.30 + 7 h = 10.30 p.m.
Answer: C |
AQUA-RAT | AQUA-RAT-34182 | time, sun
d represents the day of the year for 2000. For example d=7
would be January 7th, 2000.
If you're willing to lose some precision in exchange for
convenience, you can compute the day of the current year (instead
of counting all the way back to 2000), since the sun's declination
repeats yearly (roughly speaking).
To calculate the day of the year, remember that December 31st
(noon) of the previous year is day 0. This means day 1 is January
1st, and day 31 is January 31st. Day 31 is also "February 0", so if
you need a date in February, just add. February 19th, for example,
would be 31+19 or 50. Since February has 29 days this year, February
29th would be day 31+29 or day 60, which is also "March 0". However,
at our level of precision, it doesn't matter whether you could the
leap day or not: the results will be approximately the same.
The formula above is accurate to about 0.1 degrees for this
century. Since the sun's declination can change by as much as 0.4
degrees in a day, this amount of precision should suffice.
Technically, the formula above computes the sun's declination at
Greenwich noon for a given day, which is the time when most of the
world is observing the same day. Again, the inaccuracies from using
Greenwich noon (instead of the actual, as yet unknown, time) are
small enough to ignore for our purposes.
The following is multiple choice question (with options) to answer.
On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004? | [
"Sunday",
"Friday",
"Saturday",
"Monday"
] | A | Explanation :
Given that 8th Feb, 2005 was Tuesday
Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366
(Since Feb 2004 has 29 days as it is a leap year)
366 days = 2 odd days
Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunday
Answer : Option A |
AQUA-RAT | AQUA-RAT-34183 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is $48. Then the sum is | [
"$600",
"$1000",
"$1200",
"$1500"
] | C | 20% of (Year 1 Interest) = 20% of (20% of Principal) = 48
Principal = $1200
Answer (C) |
AQUA-RAT | AQUA-RAT-34184 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
What is the height of the triangle?
I. The area of the triangle is 20 times its base.
II. The perimeter of the triangle is equal to the perimeter of a square of side 10 cm. | [
"I alone sufficient while II alone not sufficient to answer",
"II alone sufficient while I alone not sufficient to answer",
"Either I or II alone sufficient to answer",
"Both I and II are not sufficient to answer"
] | A | EXPLANATION
I. A = 20 x B ⇒1-div-1by2 x B x H = 20 x B⇒H = 40.
I alone gives the answer.
II gives, perimeter of the triangle = 40 cm.
This does not give the height of the triangle.
answer is (A). |
AQUA-RAT | AQUA-RAT-34185 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Dacid obtained 51, 65, 82, 67 and 85 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks? | [
"29",
"38",
"39",
"70"
] | D | Average = (51 + 65 + 82 + 67 + 85)/5
= 70
Answer: D |
AQUA-RAT | AQUA-RAT-34186 | # 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.
The population of a town is 7800. It decreases annually at the rate of 10% p.a. What was its population 2 years ago? | [
"9600",
"8000",
"8500",
"9500"
] | A | Formula :
( After =100 denominator
Ago = 100 numerator)
7800 × 100/90 × 100/90 = 9629
A) |
AQUA-RAT | AQUA-RAT-34187 | 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 swims downstream 120 km and upstream 50 km taking 5 hours each time; what is the speed of the current? | [
"3 kmph",
"7 kmph",
"13 kmph",
"6.5 kmph"
] | B | Explanation:
120 --- 5 DS = 24
? ---- 1
50 ---- 5 US = 10
? ---- 1 S = ?
S = (24 - 10)/2 = 7
Answer: Option B |
AQUA-RAT | AQUA-RAT-34188 | $$\{{}_{(10)}T_L, {}_9T_L, \cdots {}_2T_L\}.$$
I have glossed over a point that may be critical:
for large $$L$$, for $$n,m \in \{1,2,\cdots, 10\} ~: m \leq n,$$ it is unclear how feasible it will be to compute how many (non-distinct) solutions to $$(f_1 \times \cdots \times f_n) = L$$ will have exactly $$m$$ identical factors.
• what if it were very large number , would i examine each of them separately ? Jan 26, 2021 at 12:24
• @Bulbasaur When you are dealing with a very large number, assuming that you want an exact enumeration, and assuming that no one discovers a more elegant approach, then yes, you would take the same approach. Note that it should be fairly straightforward to write a computer program (e.g. in Java or C) to [1] identify all prime numbers less than or equal to [M], [2] compute the prime factorization of each separate $a \in \{1,2,\cdots, M\}$, and [3] then apply your combinatorics analysis to each separate value of $a$. For any $M \leq 100,000$, (for example), the computer pgm should run ok. Jan 26, 2021 at 12:29
• thanks for your effort , i appreciate you.$+1$ for this elegant but long explanation.However , i want to wait for seeing whether or not there is any brillant shortcut.Thanks for your time and effort again.. Jan 27, 2021 at 17:12
We can start with listing all the factors of $$180$$ so we have
$$x,y,z|180: x\land y\land z\in\big\{ 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30,36,45,60,90, 180 \big\}$$
The following is multiple choice question (with options) to answer.
Find the number of different prime factors of 1250 | [
"4",
"2",
"3",
"5"
] | B | Explanation:
L.C.M of 1250 = 2 x 5 x 5 x 5 x 5
2, 5
Number of different prime factors is 2.
Answer: Option B |
AQUA-RAT | AQUA-RAT-34189 | velocity, dimensional-analysis
Title: Why is meters/second the same as meters per second? In quantities such as speed where the derived (SI) unit is m/s, why do we pronounce it and interpret it as meters per second? My guess is that 1 m is associated with 1 second. Similarly, 5 m/s is pronounced and interpreted as 5 meters per second, because 5 meters are associated with 1 second. I am not sure whether this view is naive. It is an instantaneous value. The speed, in m/s (or any other unit) is:
$$\frac{ds}{dt} $$
where $s$ is distance/displacement. If speed is constant, it really doesn't matter: an object moving at 5 m/s will cover 5 m in every second.
It will be very different if the speed varies. Imagine a dropping stone, subject to a 10m/s/s acceleration. At 1 s, the instantaneous speed is 10 m/s, but that speed occurs for exactly zero time.
The following is multiple choice question (with options) to answer.
A speed of 24 metres per second is the same as : | [
"60.4 km/hr",
"86.4 km/hr",
"65 km/hr",
"68.5 km/hr"
] | B | Sol.
24 m/sec = [24 * 18/5] km/hr = 86.4 km/hr.
Answer B |
AQUA-RAT | AQUA-RAT-34190 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
The area of a rectangle is 15 square centimeters and the perimeter is 16 square centimeters. What are the dimensions of the rectangle? | [
"3 and 5",
"3 and 6",
"3 and 7",
"3 and 8"
] | A | Let x and y be the length and width of the rectangle. Using the formulas for the area and the perimeter, we can write two equations.
15 = x y and 16 = 2 x + 2 y
Solve the second equation for x
x = 8 - y
Substitute x in the equation 15 = x y by 8 - y to rewrite the equation as
15 = (8 - y) y
Solve for y to find
y = 3 and y = 5
Use x = 8 - y to find x
when y = 3 , x = 5 and when y = 5 , x = 3.
The dimensions of the rectangle are 3 and 5.
As an exercise, check that the perimeter of this rectangle is 16 and its area is 15.
Answer A |
AQUA-RAT | AQUA-RAT-34191 | 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 goods train runs at the speed of 72 km/hr and crosses a 250 m long platform in 26 sec. What is the length of the goods train? | [
"230",
"240",
"250",
"270"
] | D | Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 250)/26 = 20
x = 270 m.
Answer: Option D |
AQUA-RAT | AQUA-RAT-34192 | # Three Times Larger: Idiom or Error?
Having just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does “three times larger” mean? How about “300% more”? We’ll focus on one discussion that involved several of us, and referred back to other answers we’ve given.
## Percent increase vs. factor
The question, from 2006, started with the idea of a percent increase:
Percent Increase and "Increase by a Factor of ..."
A math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase.
But can't I also say there was an increase by a factor of 2? Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent? This is what is confusing me!
I'd never been confused about saying "increased by a factor of" and "increased by percent of" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know! Thank you for any help.
Unfortunately, Joseph didn’t directly quote from the page he had in mind, and we have never said exactly what he said; so we couldn’t be sure which page it was. Everything he said, however, was correct.
Doctor Rick was the first to answer:
Hi, Joseph.
Yes, this can be very confusing, because some statements about increases are ambiguous.
When we say "increased by a factor of 2," the word "factor" makes it clear that we mean "multiplied by 2."
When we say "increased by 10%," there is only one reasonable interpretation: the amount of the increase is 10% of the original amount. If we meant multiplication by 10%, that would be a decrease -- not an increase! Even when we say "increased by 100%," there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase.
The following is multiple choice question (with options) to answer.
A number increased by 20% gives 1080. The number is? | [
"800",
"700",
"500",
"900"
] | D | Formula = TOTAL=100% ,INCREASE = "+" DECREASE= "-"
A number means = 100 %
That same number increased by 20 % = 120 %
120 % -------> 1080 (120 × 9 = 1080)
100 % -------> 900 (100 × 9 = 900)
Option 'D' |
AQUA-RAT | AQUA-RAT-34193 | # 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 goes away with a SANTRO car at a speed of 40 kmph. The theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start? | [
"2 Hrs",
"3 Hrs",
"1 Hr",
"4 Hrs"
] | A | |-----------20--------------------|
50 40
D = 20
RS = 50 – 40 = 10
T = 20/10 = 2 hours
ANSWER A |
AQUA-RAT | AQUA-RAT-34194 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
John has taken four (4) tests that have an average of 82. In order to bring his course grade up to a ‘B’, he will need to have a final average of 86. What will he need to average on his final two tests to achieve this grade? | [
"87",
"90",
"92",
"94"
] | D | Traditional Method:
Total scored till now 82*4=328
Total score to avg 86 in 6 tests = 86*6=516
Total to be scored on 2 tests = 516-328=188
Avg on 2 tests = 188/2 = 94
Answer: D |
AQUA-RAT | AQUA-RAT-34195 | When is this difference positive and when is it negative? Let us see:
$\begin{array}{l|c|c|c} x&d_2(x)&d_2(x-6)&d_2(x)-d_2(x-6)\\\hline 2&1&0&1\\3&2&0&2\\4&3&0&3\\5&4&0&4\\6&5&0&5\\7&6&0&6\\8&5&1&4\\9&4&2&2\\10&3&3&0\\11&2&4&-2\\12&1&5&-4\\13&0&6&-6\\14&0&5&-5\\15&0&4&-4\\16&0&3&-3\\17&0&2&-2\end{array}$
Indeed, the difference is always positive until you get to $x=10$, when it goes to 0 and then negative for larger $x$. You can thus see that
$d_3(10)-d_3(9)=d_2(9)-d_2(3)=2>0$
$d_3(11)-d_3(10)=d_2(10)-d_2(4)=0$
$d_3(12)-d_3(11)=d_2(11)-d_2(5)=-2<0$
And therefore, it’s reasonable to conclude that sums of 10 and 11 have the most number of possibilities, since the changes in $d_3(x)$ become negative afterwards.
The following is multiple choice question (with options) to answer.
If 6<x<10<y<17, then what is the greatest possible positive integer difference of x and y? | [
"3",
"4",
"10",
"6"
] | C | Let x=6.1 and y=16.1
Greatest possible difference=16.1-6.1=10
Answer C |
AQUA-RAT | AQUA-RAT-34196 | 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.
Joshua is fifth from the bottom in a queue of 100 boys.
What is the position of Joshua from the top of the queue? | [
"51",
"52",
"53",
"54"
] | A | Solution:
51
Explanation:
The position of boys higher than Joshua in the queue is 100 - 50 = 50.
Therefore, the position of Joshua from the top is 50 + 1 = 51.
Answer A |
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