source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38797 | c#, beginner
public static void AskForBookPage(int bookNumber_in)
{
Console.Write($"{bookNumber_in + 1}. book - How many pages does the book have? --> ");
}
}
static class BookList
{
public const int MAX_SIZE = 100;
public static string option;
public static string confirmRemove;
public static int bookNumber;
public static string bookTitle;
public static string bookAuthor;
public static int bookPage;
public static bool page_ok;
public static bool size_ok;
public static bool number_ok;
public static int howMany;
public static bool empty = BookList.IsListEmpty(books);
public static List<Book> books = new List<Book>();
public static bool IsListEmpty(List<Book> list)
{
if ((list != null) && (!list.Any()))
{
return false;
}
return true;
}
public static bool HowManyBooks(string size_in, ref int howMany_in, int max)
{
return int.TryParse(size_in, out howMany_in) && (howMany_in >= 0 && howMany_in <= max);
}
public static string GetBookTitle(string bookTitle_in)
{
do
{
bookTitle_in = Console.ReadLine();
} while (string.IsNullOrEmpty(bookTitle_in));
return bookTitle_in;
}
public static string GetBookAuthor(string bookAuthor_in)
{
do
{
bookAuthor_in = Console.ReadLine();
} while (string.IsNullOrEmpty(bookAuthor_in));
return bookAuthor_in;
}
public static int GetBookPage(int bookPage_in, bool? page_ok_in)
{
do
{
page_ok_in = int.TryParse(Console.ReadLine(), out bookPage_in);
} while (page_ok_in == false);
return bookPage_in;
}
The following is multiple choice question (with options) to answer.
The average number of printing error per page in a book of 512 pages is 4. If the total number of printing error in the first 302 pages is 1,208, the average number of printing errors per page in the remaining pages is | [
"0",
"4",
"840",
"90"
] | B | Remaining pages = 512 – 302 = 210
Let average printing error in remaining pages = x
Then, 1208+210×x /512=4
⇒ 210x = 840 ⇒ x = 4
Answer B |
AQUA-RAT | AQUA-RAT-38798 | 5. Why we can't solve it by 100+15/2=125
6. Why we can't solve it by 100+15/2=125
7. nice
8. The speed of the train going from Nagpur to Allahabad is 100 km/h while when coming back from Allahabad to Nagpur, its speed is 150 km/h and again going to the same way Nagpur to Allahabad at a speed of 120km/hr.Find the average speed during whole journey...
9. Average Speed = 2ab/(a + b)
Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. On similar lines, you can modify this formula for one-third distance.
10. Yes
11. A+B=100
B+A=150
(100+150) /2= 120
12. Dear Sir,
Can I get pdf file of aptitude part to learn it withous using internet.
13. So the formula of average speed with same distance can be broken as such:
Let the distance be n
So the average speed = total distance covered/total time taken
= 2n/[(n/100)+(n/150)]
= 2n/[(n/x)+(n/y)]
= 2n[(nx+ny)/xy]
= 2n[n(x+y)/xy]
= 2(x+y)/xy
14. Can you please point me to some article which would help understanding this formulae for finding the average speed . Or some different formula for calculating train speed average
15. Let the distance between them is 300 km,so total distance covered is 600 km...from Nagpur to Allahabad speed is 100 kmph so time taken is 3 hrs. And while retiring it's speed is 150kmph so time taken is 2 hrs...so total distance is 600 km nd total time taken is 5 hrs and by applying the basic formula we will get 120 kmph
16. Its is the formula of the finding average speed of two variant of speed for the same distance. where 'x' and 'y' are the speed of the train receptively
17. Why we should multiply 2 before it?
The following is multiple choice question (with options) to answer.
X and Y are two towns. Ganesh covers the distance from X to Y at an average speed of 60 Km/hr. However, he covers the distance from Y to X at an average speed of 36 Km/hr. His average speed during the whole journey in km/hr. is : | [
"45",
"43",
"40",
"38"
] | A | Solution: Average speed = 2XY / X+Y
= 2*60*36 / 60+36
= 45
Answer : A |
AQUA-RAT | AQUA-RAT-38799 | Nice observation! Something else you might notice, which turns out to imply your observation, is that all of the odd prime factors of your numbers are $$1 \bmod 4$$: $$\{ 5, 13, 17, 29, \dots \}$$. And a final thing you might notice is that all of your numbers are themselves congruent to $$2 \bmod 4$$, or equivalently are even but not divisible by $$4$$. This turns out to be an exact characterization:
Proposition: If $$n$$ is an even positive integer, the following are equivalent:
1. There exists an integer $$m$$ such that $$m^n \equiv -1 \bmod n$$.
2. There exists an integer $$x$$ such that $$x^2 \equiv -1 \bmod n$$.
3. $$n$$ is twice a product of primes congruent to $$1 \bmod 4$$.
4. There exist integers $$x, y$$ such that $$\gcd(x, y) = 1$$ and $$n = x^2 + y^2$$.
Proof. $$1 \Rightarrow 2$$: if $$n$$ is even then $$m^n = (m^{n/2})^2$$.
$$2 \Rightarrow 3$$: if $$x^2 \equiv -1 \bmod n$$ then $$x$$ is either even, in which case $$n$$ is odd, or odd (in which case $$x^2 + 1 \equiv 2 \bmod 4$$, so if $$n$$ is even then $$n \equiv 2 \bmod 4$$, meaning $$2$$ divides $$n$$ but $$4$$ doesn't.
The following is multiple choice question (with options) to answer.
If c and d are unique prime integers, which of the following must be odd? | [
"a−b",
"a−2b",
"2c−2d",
"2c/2d"
] | C | The answer is C. Let's assume that a is an even prime integer and b is an odd prime integer, their subtraction results in an odd integer and an odd integer multiplied by an even integer yields an even integer.
Answer C |
AQUA-RAT | AQUA-RAT-38800 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When the positive integer C is divided by 5 and 7, the remainder is 3 and 4, respectively. When the positive integer Dis divided by 5 and 7, the remainder is 3 and 4, respectively. Which of the following is a factor of C-D? | [
" 12",
" 24",
" 35",
" 16"
] | C | If I have a number n which when divided by 5 gives a remainder 3 and when divided by 7 gives a remainder 4, the number is of the form:
n = 5c + 3
n = 7d + 4
I will need to check for the smallest such number.
I put d= 1. n = 11. Is it of the form 5c + 3? No.
Put d= 2. n = 18. Is it of the form 5c + 3? Yes.
When 18 is divided by 5, it gives a remainder of 3. When it is divided by 7, it gives a remainder if 4.
Next such number will be 35 + 18 because 35 will be divisible by 5 as well as 7 and whatever is the remainder from 18, will still be the remainder
Next will be 35*2 + 18
and so on...
Difference between such numbers will be a multiple of 35 so
your answer is C |
AQUA-RAT | AQUA-RAT-38801 | # 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.
What is the difference between the compound interests on $5000 for 1 years at 4% per annum compounded yearly and half-yearly? | [
"$2.04",
"$3.06",
"$4.80",
"$8.30"
] | A | A.
$. 2.04
B.
$. 3.06
C.
$. 4.80
D.
$. 8.30
Answer: Option A
Explanation:
C.I. when interest
compounded yearly = $ 5000 x 1 + 4 x 1 + x 4
100 100
= $. 5000 x 26 x 51
25 50
= $. 5304.
C.I. when interest is
compounded half-yearly = $ 5000 x 1 + 2 3
100
= $ 5000 x 51 x 51 x 51
50 50 50
= $5306.04
Difference = $. (5306.04 - 5304) =$. 2.04
ANSWER : A $2.04 |
AQUA-RAT | AQUA-RAT-38802 | python, beginner, algorithm, strings, regex
Test 8: 1e is an invalid number.
--------------------------------------------------
Test 9: e3 is an invalid number.
--------------------------------------------------
Test 10: 6e-1 is a valid number.
--------------------------------------------------
Test 11: 99e2.5 is an invalid number.
--------------------------------------------------
Test 12: 53.5e93 is a valid number.
--------------------------------------------------
Test 13: --6 is an invalid number.
--------------------------------------------------
Test 14: -+3 is an invalid number.
--------------------------------------------------
Test 15: 95a54e53 is an invalid number.
The following is multiple choice question (with options) to answer.
Find the invalid no.from the following series 3, 7, 15, 27, 63, 127, 255 | [
"18",
"19",
"27",
"29"
] | C | Go on multiplying the number by 2 and adding 1 to it to get the next number.So, 27 is wrong.
C |
AQUA-RAT | AQUA-RAT-38803 | # 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.
A boat can travel with a speed of 16 km/hr in still water. If the rate of stream is 5 km/hr, then find the time taken by the boat to cover distance of 147 km downstream. | [
"4 hours",
"5 hours",
"6 hours",
"7 hours"
] | D | Explanation:
It is very important to check, if the boat speed given is in still water or with water or against water. Because if we neglect it we will not reach on right answer. I just mentioned here because mostly mistakes in this chapter are of this kind only.
Lets see the question now.
Speed downstream = (16 + 5) = 21 kmph
Time = distance/speed = 147/21 = 7 hours
Option D |
AQUA-RAT | AQUA-RAT-38804 | # Making Friends around a Circular Table
I have $n$ people seated around a circular table, initially in arbitrary order. At each step, I choose two people and switch their seats. What is the minimum number of steps required such that every person has sat either to the right or to the left of everyone else?
To be specific, we consider two different cases:
1. You can only switch people who are sitting next to each other.
2. You can switch any two people, no matter where they are on the table.
The small cases are relatively simple: if we denote the answer in case 1 and 2 for a given value of $n$ as $f(n)$ and $g(n)$ respectively, then we have $f(x)=g(x)=0$ for $x=1, 2, 3$, $f(4)=g(4)=1$. I’m not sure how I would generalize to larger values, though.
(I initially claimed that $f(5)=g(5)=2$, but corrected it based on @Ryan’s comment).
If you’re interested, this question came up in a conversation with my friends when we were trying to figure out the best way for a large party of people during dinner to all get to know each other.
Edit: The table below compares the current best known value for case 2, $g(n)$, to the theoretical lower bound $\lceil{\frac{1}{8}n(n-3)}\rceil$ for a range of values of $n$. Solutions up to $n=14$ are known to be optimal, in large part due to the work of Andrew Szymczak and PeterKošinár.
The moves corresponding to the current best value are found below. Each ordered pair $(i, j)$ indicates that we switch the people in seats $(i, j)$ with each other, with the seats being labeled from $1 \ldots n$ consecutively around the table.
The following is multiple choice question (with options) to answer.
Find the number of ways of arranging the host and 5 guests at a circular table so that the boy always sits in a particular seat? | [
"5!",
"8!",
"11!",
"12!"
] | A | Ans.(A)
Sol. Total number of persons = 6 Host can sit in a particular seat in one way. Now, remaining positions are defined relative to the host. Hence, the remaining can sit in 5 places => 5! Ways ... The number of required arrangements = 5! x 1= 5! = 5! ways |
AQUA-RAT | AQUA-RAT-38805 | # 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.
A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by: | [
"3",
"5",
"9",
"11"
] | D | Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
ANSWER : D |
AQUA-RAT | AQUA-RAT-38806 | mass, newtonian-gravity, statics
Title: How do you weigh a box on a scale whose limit is too low? As you will see I know nothing about physics and after being asked to solve a physics problem in a recent interview wanted to ask it of professionals and see what the response would be:
I have a set of domestic scales (actually just one scale) that weigh up to the maximum weight of 5kg and a large package that weighs more than 5kgs but less than 10kgs. How can I tell the exact weight using the inadequate scales.
The package is a long item akin to a steal girder but not in weight obviously.
I know that stack exchange has rules about this type of question so please treat it as light entertainment from someone in awe of your intellect. You put the package horizontally across three scales and add up the weight you see. If the center scale registers more than the limit, move the box, or put some pieces of paper under the scales which are registering a small weight to redistribute the weight.
A more physics-y question would be how to determine the weight when you have only one scale. This can be done by tilting the package at a small angle, by placing one end on a scale (perhaps propped up on something light, like your shoe, so that the weight momentum flow goes through the scale to the floor). To weigh this way requires knowing where the CM of the box is, and this can be determined by balancing it precisely on the edge of the scale (or on your finger, or on something else narrow).
Then the weight registered by the scale (minus the weight of the shoe) is the ratio of the length of the box to the distance from the end which is on the floor to the center of mass. If this is still too heavy, you can tilt the box up to 45 degrees, which will cut down the weight registered by the scale further by a factor of .707, which can be undone by multiplying by 1.414.
The following is multiple choice question (with options) to answer.
The ratio by weight, measured in pounds, of books to clothes to electronics in a suitcase initially stands at 7:5:3. Someone removes 5 pounds of clothing from the suitcase, thereby doubling the ratio of books to clothes. How many pounds do the electronics in the suitcase weigh? | [
"3",
"4",
"5",
"6"
] | D | The weights of the items in the suitcase are 7k, 5k, and 3k.
If removing 5 pounds of clothes doubles the ratio of books to clothes, then 5 pounds represents half the weight of the clothes.
2.5k = 5 pounds and then k = 2 pounds.
The electronics weigh 3(2) = 6 pounds.
The answer is D. |
AQUA-RAT | AQUA-RAT-38807 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is? | [
"22",
"88",
"48",
"99"
] | C | Relative speed = 60 + 90 = 150 km/hr.
= 150 * 5/18 = 125/3 m/sec.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 * 3/125 = 48 sec.
Answer: C |
AQUA-RAT | AQUA-RAT-38808 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
One half of a two digit number exceeds its one third by 5. What is the sum of the digits of the number? | [
"A)3",
"B)5",
"C)7",
"D)9"
] | A | x/2 – x/3 = 5 => x =30
3 + 0 = 3
ANSWER:A |
AQUA-RAT | AQUA-RAT-38809 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
If the price of sugar rises from Rs. 10 per kg to Rs. 13 per kg, a person, to have no increase in the expenditure on sugar, will have to reduce his consumption of sugar by | [
"15%",
"20%",
"23%",
"30%"
] | C | Sol.
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 10 = x × 13 = x = 77 kg.
∴ Reduction in consumption = 23%.
Answer C |
AQUA-RAT | AQUA-RAT-38810 | ### Show Tags
09 Oct 2015, 14:30
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
I found an almost identical question with the same question stem, but different statements.
Here it is: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Manager
Joined: 12 Sep 2015
Posts: 80
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
08 Apr 2016, 21:31
MrSobe17 wrote:
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
z can take any value in that case. Think of a case in which z = 12.
$$5^2 * 3 * 2^2 = 3 * x^y$$
Here x = 10
_________________
The following is multiple choice question (with options) to answer.
If x > y^5 > z^6, which of the following statements could be true? I.x=y II. x>y III.x>y>z | [
"I only",
"II only",
"II,III only",
"III only"
] | C | Answer : C |
AQUA-RAT | AQUA-RAT-38811 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Ram, who is half as efficient as Krish, will take 30 days to complete a task if he worked alone. If Ram and Krish worked together, how long will they take to complete the task? | [
"16 days",
"10 days",
"8 days",
"6 days"
] | B | Number of days taken by Ram to complete task = 30
Since Ram is half as efficient as Krish , Amount of work done by Krish in 1 day = Amount of work done by Ram in 2 days
If total work done by Ram in 30 days is 30w
Amount of work done by Ram in 1 day = w
Amount of work done by Krish in 1 day = 2w
Total amount of work done by Krish and Ram in a day = 3w
Total amount of time needed by Krish and Ram to complete task = 30w/3w = 10 days
Answer B |
AQUA-RAT | AQUA-RAT-38812 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
trisha bakes 6 cookies for 6 friends. 4 of the friend arrive early and eat 3 cookies each. How many cookies are left over if the other friends all have the same number of cookies? | [
"6",
"0",
"4",
"3"
] | B | trisha made 6*6 cookies = 36 cookies. 4 friends eat 3 cookies, leaving 24 remaining for 2 friends.
24 is divisible by 2. The 2 friends ate a cumulative number of cookies divisible by 2, because the number is the same.
24-24=0;
B is the correct answer |
AQUA-RAT | AQUA-RAT-38813 | 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 12 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in | [
"5days",
"6days",
"7.5days",
"8.5days"
] | B | Explanation:
B's 5 days work =
1/10∗5=1/2
Remaining work =1−1/2=1/2
A can finish work =12∗1/2= 6 days
Option B |
AQUA-RAT | AQUA-RAT-38814 | # Math Help - Finding the values of a and b
1. ## Finding the values of a and b
Hello everyone. This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
The keyword in this irksome question would be the word 'or'. So it denotes that that this involves quadratic formulas.
Can anyone give me a clue so that I may make a breakthrough in understanding this problem? Thank you so much!
2. The wording is a bit iffy, but they actually meant that the solutions for that equation is x = 3 AND x = -4.
3. Originally Posted by PythagorasNeophyte
This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
$x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$
We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:
$3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $-4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$
Now solve for a and b using simultaneous equation.
4. Originally Posted by Educated
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
The following is multiple choice question (with options) to answer.
From the given equation find the value of x: x² − 3x + 2 | [
"0",
"2",
"4",
"5"
] | B | (x − 1)(x − 2)
x = 1 or 2.
B |
AQUA-RAT | AQUA-RAT-38815 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Sam is training for the marathon. He drove 12 miles from his home to the Grey Hills Park and then ran 6 miles to Red Rock, retraced his path back for 4 miles, and then ran 3 miles to Rock Creek. If he is then n miles from home, what is the range of possible values for n? | [
"1 ≤ n ≤23",
"7 ≤ n ≤17",
"5 ≤ n ≤19",
"6 ≤ n ≤18"
] | B | ANSWER: C To find the maximum and minimum range for his distance from home, assume that he traveled either directly toward his home or directly away from his home. The range then is between 12+6-4+3=17 for the maximum, and 12-6+4-3=7 for the minimum, so B is the answer |
AQUA-RAT | AQUA-RAT-38816 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of five numbers is 18. The average of first two numbers is 14 and the average of last two numbers is 19. What is the middle number? | [
"24",
"27",
"29",
"32"
] | A | The total of seven numbers = 5X18 = 90
The total of first 2 and last 2 numbers is = 2 X 14+2 X 19 = 66
So, the middle number is (90 - 66 ) = 24
A |
AQUA-RAT | AQUA-RAT-38817 | Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers.
When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.
## Related Chapters
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The following is multiple choice question (with options) to answer.
About the number of pairs which have 16 as their HCF and 125 as their LCM, the conclusion can be | [
"only one such pair exists",
"no paid is exists",
"many such pairs exist",
"pair exists"
] | B | Explanation:
HCF is always a factor of LCM. ie., HCF always divides LCM perfectly.
Correct Option:B |
AQUA-RAT | AQUA-RAT-38818 | 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.
Ram, who is half as efficient as Krish, will take 24 days to complete a task if he worked alone. If Ram and Krish worked together, how long will they take to complete the task? | [
"16 days",
"12 days",
"8 days",
"6 days"
] | C | Answer
Ram takes 24 days to complete the task, if he works alone.
Krish is twice as efficient as Ram is. So, working alone, Krish will take half the time to complete the task.i.e., 12 days.
Ram will complete 1/24th of the task in a day.
Krish will complete 1/12th of the task in a day.
When they work together, they will complete 1/24+1/12=1/8th of the task in a day.
Therefore, when they work together they will complete the task in 8 days.
Choice C |
AQUA-RAT | AQUA-RAT-38819 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
Find the fraction which has the same ratio to 1/13 that 5/34 has to 7/48. | [
"110/1547",
"120/1547",
"140/1547",
"160/1547"
] | B | P : 1/13 = 5/34 : 7/48
As the product of the means is equal to the product of the extremes.
P*7/48 = 1/13 * 5/34
P*7/48 = 5/442
P = 240/3094 => P = 120/1547
ANSWER:B |
AQUA-RAT | AQUA-RAT-38820 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
The measurements obtained for the interior dimensions of a rectangular box are 100 cm by 100 cm by 150cm. If each of the three measurements has an error of at most 1 centimeter, which of the following is the closes maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements? | [
"50,000",
"40,000",
"45,000",
"50,000"
] | B | The options are well spread so we can approximate.
Changing the length by 1 cm results in change of the volume by 1*100*150 = 15,000 cubic centimeters;
Changing the width by 1 cm results in change of the volume by 100*1*150 = 15,000 cubic centimeters;
Changing the height by 1 cm results in change of the volume by 100*100*1 = 10,000 cubic centimeters.
So, approximate maximum possible difference is 15,000 + 15,000 + 10,000 = 40,000 cubic centimeters.
Answer: B. |
AQUA-RAT | AQUA-RAT-38821 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
At Company X senior sales representatives visit the home office once every 30 days, and junior sales representatives visit the home office once every 20 days. The number of visits that a junior sales representative makes in a 2-year period is approximately what percent greater than the number of visits that a senior representative makes in the same period? | [
"10%",
"25%",
"33%",
"50%"
] | D | When you say Senior is 1/30, what do you mean by it? It is the rate of visit. The time of visit is 30 days. The rate is 1/30. Similarly, the rate of visit for junior is 1/20.
The 2 yrs has no significance in this case. The rate stays the same for every time interval.
{1/20} - {1/30}/ {1/30} * 100 = 50%
ANSWER:D |
AQUA-RAT | AQUA-RAT-38822 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? | [
"12",
"14",
"36",
"38"
] | C | There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
ANSWER C |
AQUA-RAT | AQUA-RAT-38823 | electricity, batteries
A battery's capacity is the amount of electric charge it can deliver
at the rated voltage. The more electrode material contained in the
cell the greater its capacity. A small cell has less capacity than a
larger cell with the same chemistry, although they develop the same
open-circuit voltage.[30] Capacity is measured in units such as
amp-hour (A·h). The rated capacity of a battery is usually expressed
as the product of 20 hours multiplied by the current that a new
battery can consistently supply for 20 hours at 68 °F (20 °C), while
remaining above a specified terminal voltage per cell. For example, a
battery rated at 100 A·h can deliver 5 A over a 20-hour period at room
temperature. The fraction of the stored charge that a battery can
deliver depends on multiple factors, including battery chemistry, the
rate at which the charge is delivered (current), the required terminal
voltage, the storage period, ambient temperature and other
factors.[30]
The higher the discharge rate, the lower the capacity.[31] The
relationship between current, discharge time and capacity for a lead
acid battery is approximated (over a typical range of current values)
by Peukert's law:
${\displaystyle t={\frac {Q_{P}}{I^{k}}}}$
$Q_P$ is the capacity when discharged at a rate of 1 amp.
${\displaystyle I}$ is the current drawn from battery (A).
${\displaystyle t}$ is the amount of time (in hours) that a battery
can sustain.
${\displaystyle k}$ k is a constant around 1.3.
Using this law, you can see that for lower current, the discharge (drain) time is larger. Vice versa, for higher current, the discharge time is smaller.
In other words, $t_1(T_1) < t_2(T_2)$ or for higher temperature the discharge time is longer.
The following is multiple choice question (with options) to answer.
A mobile battery in 1 hour charges to 20 percent.How much time (in minute) will it require more to charge to 60 percent. | [
"145",
"150",
"175",
"180"
] | D | 1 hr =20 percent.Thus 15 min=5 percent .Now to charge 60 percent 180 min.Answer:D |
AQUA-RAT | AQUA-RAT-38824 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How long does a train 250 meters long running at the rate of 72km/hr take to cross a bridge 150 meters in length? | [
"10sec",
"20sec",
"25sec",
"30sec"
] | B | Distance = length of train + length of bridge = 250+150 = 400
speed = 72km/hr = 72*5/18 = 20m/s
Required time = 400/20 = 20 seconds
Answer is B |
AQUA-RAT | AQUA-RAT-38825 | # Average length of the longest segment
This post is related to a previous SE post If a 1 meter rope …. concerning average length of a smallest segment.
A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment. My answer is 11/18. Here is how I do it:
Here we have two independent random variables $X,Y$, both uniform on $[0,1]$. Let $A=\min (X,Y), B=\max (X,Y)$ and $C=\max (A, 1-B, B-A)$. First we want to find the probability density function $f_C(a)$ of $C$. Let $F_C(a)$ be the cumulative distribution function. Then $$F_C(a) = P(C\le a)=P(A\le a, 1-B\le a, B-A\le a).$$ By rewriting this probability as area in the unit square, I get $$F_C(a)=\left\{\begin{array}{ll} (3a-1)^2 & \frac{1}{3}\le a\le \frac{1}{2}\\ 1-3(1-a)^2 & \frac{1}{2}\le a\le 1\end{array}\right.$$ from which it follows that $$f_C(a)=\left\{\begin{array}{ll} 6(3a-1) & \frac{1}{3}\le a\le \frac{1}{2}\\ 6(1-a) & \frac{1}{2}\le a\le 1\end{array}\right.$$ Therefore the expected value of $C$ is $$\int_{1/3} ^{1/2}6a(3a-1) da+\int_{1/2} ^{1}6a(1-a) da= \frac{11}{18}.$$
My questions are:
(A) Is there a "clever" way to figure out this number 11/18?
The following is multiple choice question (with options) to answer.
The average length of 6 strings is 80 cm. If the average length of one third of the strings is 70 cm, what is the average of the other strings? | [
"75.",
"85.",
"90.",
"94."
] | B | Edit:
Given ( X1 + X2 ... + X6 ) / 6 = 80
( X1 + X2 ... + X6 ) = 480 --> Eq 1.
Now given avg length of one third strings is 70. That means out 6 / 3 = 2 strings.
let the avg length of two strings be ( X1 + X2 ) / 2 = 70 .
( X1 + X2 ) = 140. --> eq 2.
Now we are asked to find the average of the remaining i.e. ( X3 + X4 + X5 + X6 )
Substitute eq 2 in eq 1 then we get
140 + X3 + X4 + X5 + X6 = 480
=> X3 + X4 + X5 + X6 = 340
Now divide 340 by 4 we get 85.
=> ( X3 + X4 + X5 + X6) / 4 = 85 = avg length of remaining strings.
IMO correct option is B. |
AQUA-RAT | AQUA-RAT-38826 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
A sum of money is distributed among w,x,y,z in the proportion of 1:6:2:4. If X gets $1500 more than Y, what is the W's share? | [
"$1500",
"$150",
"$375",
"$200"
] | C | Let the shares of w,x,y,z are 1a,6a,2a,4a
6a-2a = 1500
4a = $1500, A=375
W's share = 1a = $375
Answer is C |
AQUA-RAT | AQUA-RAT-38827 | A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 Let's plug in some NICE VALUES that satisfy the given information... Last year the manufacturer sold twice as many units of Q as P. So, let's say the manufacturer sold 1 P and 2 Q's There were 3 units sold. So, average price =$54/3 = $18 Answer: E GMATPrepNow how do we deduce this statement? I get confused with this statement and it often appears on word problems. can you help me understand this and its various iterations on the exam? GMAT Club Legend Joined: 11 Sep 2015 Posts: 4320 Location: Canada Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 23 Oct 2019, 05:17 Top Contributor Shrey9 wrote: GMATPrepNow wrote: AbdurRakib wrote: A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is$20.00 and the revenue from the sale of each unit of Q is $17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year? A.$28.50
B. $27.00 C.$19.00
D. $18.50 E.$18.00
Let's plug in some NICE VALUES that satisfy the given information...
Last year the manufacturer sold twice as many units of Q as P.
So, let's say the manufacturer sold 1 P and 2 Q's
There were 3 units sold.
So, average price = $54/3 =$18
GMATPrepNow
how do we deduce this statement? I get confused with this statement and it often appears on word problems.
can you help me understand this and its various iterations on the exam?
Which statement are you referring to?
Cheers,
Brent
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Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
The following is multiple choice question (with options) to answer.
A manufacturer of a certain product can expect that between 0.3 percent and 0.5 percent of the units manufactured will be defective. If the retail price is $2,500 per unit and the manufacturer offers a full refund for defective units, how much money can the manufacturer expect to need to cover the refunds on 10,000 units? | [
" Between $15,000 and $25,000",
" Between $30,000 and $50,000",
" Between $75,000 and $125,000",
" Between $150,000 and $250,000"
] | C | Number of defective units is between = .3 % of 10000 and .5% of 10000
= 30 and 50
Retail Price per unit = 2500 $
Expected price of refund is between = 2500 x 30 and 2500 x 50
=75,000 and 125,000 dollars
Answer C |
AQUA-RAT | AQUA-RAT-38828 | 1) Show that the tangent line to the parabola y=Ax^2, A not equal to 0, at the point x=c will intersect the x-axis at the point (c/2, 0)
2) Determine where this line intersects the y-axis
Originally Posted by erimat89
Maybe I'm just confused but the question wasn't asking where it intersected the x-axis the x-axis intersection is given at the point (c/2, 0) it was only asking to determine where it crossed the y-axis in part 2 of the question. Why is this point (c/2, 0) given?
It asks you to show that the line intersects the x-axis at the point (c/2,0).
Since I got the equation of the line to be $y=2Acx-Ac^2$, at y=0, the line crosses the x-axis.
Thus $0=2Acx-Ac^2\implies 2Acx=Ac^2\implies x=\frac{Ac^2}{2Ac}\implies x=\frac{c}{2}$. So we see that it crosses the x-axis at $\left(\tfrac{1}{2}c,0\right)$
$\mathbb{Q.E.D.}$
Now, it intersects the y axis when x=0. Thus, we see that $y=2Ac(0)-Ac^2\implies y=-Ac^2$.
Thus, the y intercept is $\left(0,-Ac^2\right)$.
Does this make sense?
--Chris
5. ## Thanks
Makes alot of sense you clarified things for me. I've never been good with any sort of math problem that involves words for some reason I get thrown off.
The following is multiple choice question (with options) to answer.
On the x-y coordinate plane there is a parabola, y=x(4-x). Which of the following points is in the region that is enclosed by this parabola and the x-axis?
I. (1, 2) II. (2, 5) III. (3, 2) | [
"I only",
"II only",
"III only",
"I and III"
] | D | Since co-efficient of x^2 is negative, the parabola will be downward facing. So the parabola and the x axis will enclose some points where the y-values of the parabola are positive.
When x = 1, y = 3 lies on the parabola, so (1, 2) will be enclosed by the parabola and x axis.
When x = 2, y = 4 lies on the parabola, so (2, 5) will not be enclosed by the parabola and x axis.
When x = 3, y = 3 lies on the parabola, so (3, 2) will be enclosed by the parabola and x axis.
The answer is D. |
AQUA-RAT | AQUA-RAT-38829 | # Show that $\prod_{k=1}^n (n+k) = 2^n \prod_{k=1}^n (2k-1)$ for all $n\in Z^+$
Show that $$\prod_{k=1}^n (n+k) = 2^n \prod_{k=1}^n (2k-1)$$ for all $$n\in Z^+$$.
I think I'm supposed to use induction to prove this, but I can't seem to figure out how. Any help would be appreciated, thanks in advance!
• The proof basically amounts to factoring out $2$ from every even number in the original product, thus leaving you with a product of odd numbers. Every other term in the product will ultimately be even. It feels like it might be easier to demonstrate in cases specifying where $n$ is even in one case, and inducting on the even integers, and then inducting on odd $n$ in another case, since it might make the math a little cleaner and the logic/wording in the proof less cumbersome, but that's just a guess. – Eevee Trainer Nov 4 '18 at 17:18
• Welcome to math.SE. Please see How to Ask a Good Question for information about what we look for on this site. In particular, you should try to the extent possible to include the source and motivation of each problem, not just a statement of a problem that you are trying to solve. – Carl Mummert Nov 4 '18 at 19:20
Induction is not necessarily needed.
The following is multiple choice question (with options) to answer.
If n is the product of integers from 1 to 20 inclusive what is the greatest integer k for which 2^k is a factor of n?
. | [
"10",
"12",
"15",
"18"
] | D | 20! = 1*2*3*4*5...*19*20 (This is 20 factorial written as 20!)
n = 1*2*3*4*5*6*7.....*19*20
How many 2s are there in n?
One 2 from 2
Two 2s from 4
One two from 6
Three 2s from 8
and so on...
When you count them all, you get 18.
ANSWER:D |
AQUA-RAT | AQUA-RAT-38830 | the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand
The following is multiple choice question (with options) to answer.
The sum of the ages of 4 children is 52 years. The age of the youngest child is 4. If the intervals of years each born is same find the interval? | [
"4",
"8",
"10",
"6"
] | D | Explanation:
Let the interval is x .The youngest child's age is 4. Each of the other 3 children will then be 4+x, 4+2x, 4+3x
We know that the sum of their ages is 52 .
so, 4+(4+x)+(4+2x)+(4+3x) = 52
x= 6
Answer: D |
AQUA-RAT | AQUA-RAT-38831 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by: | [
"20 m",
"10 m",
"40 m",
"30 m"
] | A | To reach the winning point , A will have to cover a distance of 500 - 140= 360 metre
ratio of the speeds of two contestants A and B is 3 : 4
i.e., when A covers 3 metre, B covers 4 metre.
When A covers 360 metre, B covers 4/3×360 = 480 metre.
Remaining distance B have to cover = 500-480= 20 metre
A wins by 20 metre
Answer isA |
AQUA-RAT | AQUA-RAT-38832 | homework-and-exercises, kinematics
Title: Train problem question in kinematics engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last
compartment with velocity v. The middle point of the train passes past the same pole with a velocity of.
My thinking:
Q1 Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
Q2 Shouldn’t the middle part also cover it with u velocity.?
Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
You are right in saying that two ends of a same rigid body can't have different values of speed but you should note that your line is correct only if you are talking about the speed of engine and the last compartment at a given point of time.
In your question , the engine passes the pole with speed $u$ and note that since the train can't elongate or compress , the speed of the middle part as well as the last compartment at that instant is $u$ .
When the last compartment reaches the pole, in that time interval the train has accelerated (first line of your question). So, at that instant, the speed of the engine , the middle part and the last compartment is $v$.
So your first question is just a mere confusion.
For the center of the train to pass the pole , the train has to travel for some time and since it is an accelerated motion , the speed with which it passes the pole is different than $u$.
Note: The value of speed with which the center passes the pole can be calculated using the three equation of motion involving constant acceleration :
$$ v = u + at$$
$$s = ut + \frac{1}{2} a t^2$$
$$v^2 = u^2 + 2as$$
Hope it helps .
The following is multiple choice question (with options) to answer.
In what time will a train 100 meters long cross an electric pole, if its speed is 72 km/hr | [
"5 seconds",
"2.8 seconds",
"3.5 seconds",
"2.5 seconds"
] | A | First convert speed into m/sec
Speed = 72*(5/18) = 20 m/sec
Time = Distance/speed
= 100/20 = 5 seconds
Answer: A |
AQUA-RAT | AQUA-RAT-38833 | newtonian-mechanics, buoyancy, weight
Title: What is the sensation by man carrying a water bucker with an object floating in it? A man is carrying a bucket with water. If a object is kept in it which floats in water, man will feel
1) heavier
2) lighter
3) none
I think when the body floats its weight is balanced by upthrust so no further be felt. Is it right or wrong? The man will have to carry the extra weight.
The object floats on the water, because the forces on it balance out. Gravity pulls the object down, the water pushes it up.
Now hold part of that thought:
The object pushes against the water.
The water pushes against the bucket.
The man is using force to hold the bucket.
Thus the bucket and its content are heavier when you add extra stuff to it. (Even if it floats).
Not using answer 1) because The man feels heavier is a bit weird way to say that. The bucket the man is carrying is heavier. The man himself did not change.
The following is multiple choice question (with options) to answer.
The weight of an iron bucket increases by 33.33% when filled with water to 50% of its capacity. Which of these may be 50% of the weight of the bucket when it is filled with water( assume the weight of bucket and its capacity in kg to be integers? | [
"7 kg",
"6 kg",
"5 kg",
"8 kg"
] | C | Say Iron = 6 kg EMPTY so filled with 50% water you have a 1/3*6 = 2 kg water weight. so adding the rest of the bucket with water will naturally increase it by another 2kg so you have 2+2+6 = 10 thus 50% of 10 = 5.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38834 | 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....
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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.
The average (arithmetic mean) of 16 students first quiz scores in a difficult English class is 61.5. When one student dropped the class, the average of the remaining scores increased to 64.0. What is the quiz score of the student who dropped the class? | [
"a) 10",
"b) 24",
"c) 40",
"d) 55"
] | B | Total Score of 16 students is 16*61.50 = 984
Total Score of 15 students is 15*64 = 960
So, the score of the person who left is 24 ( 984 - 960 )
Answer will be (B) |
AQUA-RAT | AQUA-RAT-38835 | d=18: (3,-3), (-3,-3), (-3,3), (3,3)
d=20: (4,2), (4,-2), (-4,-2), (2,-4), (-4,2), (-2,-4), (-2,4), (2,4)
d=25: (-3,-4), (-5,0), (5,0), (4,3), (-3,4), (-4,3), (0,-5), (4,-3), (-4,-3), (3,-4), (3,4), (0,5)
The following is multiple choice question (with options) to answer.
18! is equal to which of the following? | [
"6,402,373,705,728,125",
"6,402,373,705,728,216",
"6,402,373,705,728,624",
"6,402,373,705,728,000"
] | D | After 4!, the units digit of every factorial is 0.
5!=120
6!=720
etc...
The answer is D. |
AQUA-RAT | AQUA-RAT-38836 | # The division of a fraction - Whole or Part?
My example is 1/2 divide by 8.
The correct answer given : 1/16
However, in my mind this answer is 1 divide by 16, which to me is not what the question is asking.
When I construct the argument, it seems it should equal 1/8 as the question is asking "What is the division of 1/2 into 8 pieces?"
Not
"What is the total division of the whole when 1/2 is divided into 8 pieces?"
Why give the answer to the whole, when it is the fraction that is in question?
If I have a box, and I cut it into half, and then I ask "What is the division of this box into 8?" Why would I give a division of the whole box?
Edit
Thank you for the many answers, I believe it boils down to absolute quantity vs relative measurement, and the mass of the object.
So, 1/3 of half a tea spoon is an absurd but correct measurement.
but, say you have a house divided into 2 apartments, both having 8 rooms.
You would say : "An apartment has 8 rooms" not "The house has 16 rooms"
Edit
Absolute measurement, as a method of reaching an answer that is limited in complexity, is preferable to an infinite number of sets of fractions.
1/18 = Absolute Measurement
1/8 = Sets of fractions into infinity
This has really helped me appreciate set theory, thank you for your time.
The following is multiple choice question (with options) to answer.
A straight pipe 1 meter in length was marked off in halves and also in thirds. If the pipe was then cut into separate pieces at each of these markings, which of the following gives all the different lengths of the pieces, in fractions of a meter? | [
" 1/6 and 1/3 only",
" 1/4 and 1/3 only",
" 1/6, 1/4, and 1/3",
" 1/12, 1/6 and 1/4"
] | A | Generally fast way to solve such problem is writing the different marks in ascending/descending order with same denominator:
Here 4th : 0/4, 1/4, 2/4, 3/4, 4/4
and 3rd : 0/3, 1/3, 2/3, 3/3
Generally fast way to solve such problem is writing the different marks in ascending/descending order with same denominator:
Here 2th : 0/2, 1/2, 2/2
and 3rd : 0/3, 1/3, 2/3, 3/3
Now with understood common denominator 6 write the numbers : for 2th : 0,3,6 and for 3rd : 0,2,4,6
Now comine : 0,2,3,4,6
Now find the cut with denominator 6 (Substracrt adjacent terms : 1/3, 1/6, 1/6, 1/3 i.e. 1/3 and 1/6 after removing duplicates.
answer : A |
AQUA-RAT | AQUA-RAT-38837 | # 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 sold an article for 500 after giving 10% discount on the labelled price and made 20% profit on the cost price. What would have been the percentage profit,had he not given the discount’? | [
"33.33%",
"30%",
"23%",
"28%"
] | A | Cost price =500×100/120= 416.67
S.P. at no discount = 500×100/90 = 555.55
∴ % profit = 138.88×100/416.67 = 33.33 %
Answer A |
AQUA-RAT | AQUA-RAT-38838 | ### Question 3
Multiple choice!
The decimal $0.23$ equals:
(A) $\dfrac {23}{10}$
(B) $\dfrac {23}{100}$
(C) $\dfrac {23}{1000}$
(D) $\dfrac {23}{10000}$
### Question 4
The decimal $0.0409$ equals:
(A) $\dfrac {409}{100}$
(B) $\dfrac {409}{1000}$
(C) $\dfrac {409}{10000}$
(D) $\dfrac {409}{100000}$
## SIMPLE FRACTIONS AS DECIMALS
Some decimals give fractions that simplify further.
For example,
$0.5 = \dfrac{5}{10} = \dfrac{1}{2}$
and
$0.04 = \dfrac{4}{100} = \dfrac{1}{25}$.
Conversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.
For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and
$\dfrac{13}{20} = \dfrac{13 \times 5}{20 \times 5} = \dfrac{65}{100} = 0.65$.
What fractions (in simplest terms) do the following decimals represent?
$0.05$, $0.2$, $0.8$, $0.004$
### Question 6
Write each of the following fractions as a decimal.
$\dfrac {2} {5}$, $\dfrac {1} {25}$, $\dfrac {1} {20}$, $\dfrac {1} {200}$, $\dfrac {2} {2500}$
### Question 7
MULTIPLE CHOICE!
The decimal $0.050$ equals
The following is multiple choice question (with options) to answer.
Which of the following fractions is greater than 1/3 and less than 1/5?
1.1/3 2. 1/5 3. 1/2 4. 2/3 5. 4/5 6. 9/10 | [
"1/5",
"1/3",
"1/4",
"1/9"
] | C | 1/5 = 0.2, 1/3 = 0.33, 1/9= 0.11, 1/8= 0.125, 4/5= 0.8, 1/4= 0.25.
Clearly, 0.25 lies between 0.20 and 0.33.
Therefore, 1/4 Lies between 1/5 and 1/3.
Answer is C. |
AQUA-RAT | AQUA-RAT-38839 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When positive integer x is divided by positive integer y, the remainder is 9. If x/y = 96.25, what is the value of y? | [
"96",
"75",
"36",
"25"
] | C | Guys, one more simple funda.
5/2= 2.5
now .5 x2 =1 is the remainder
25/4 = 6.25
now .25x4=1 is the remainder
32/5=6.4
now.4x5 = 2 is the remainder
given x/y = 96.25 and remainder is 9
So .25 X y = 9
hence y= 36
Ans C |
AQUA-RAT | AQUA-RAT-38840 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
Which is the middle even number between 5 and 15? | [
"5",
"10",
"25",
"30"
] | B | 5+x = 15-x
2 x = 10
X = 5
Now 5+x = 10
ANSWER:B |
AQUA-RAT | AQUA-RAT-38841 | # 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.
Meera purchased two three items from a shop. Total price for three items is Rs.5200/- She have given Rs. 7000/- What is the balance amount meera got? | [
"650",
"400",
"1350",
"1800"
] | D | Total cost of Items : 5200/-
amount Paid : 7000/-
Balance receivable : 6000 - 7000= 1800/-
Answer is D |
AQUA-RAT | AQUA-RAT-38842 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
The speed of a car is 90 km in the first hour and 60 km in the second hour. What is the average speed of the car? | [
"25",
"75",
"27",
"25"
] | B | S = (90 + 60)/2
= 75 kmph
Answer: B |
AQUA-RAT | AQUA-RAT-38843 | # STRNO - Editorial
Setter: Kritagya Agarwal
Tester: Felipe Mota
Editorialist: Taranpreet Singh
Easy
Number theory
# PROBLEM:
Given two integers X and K, you need to determine whether there exists an integer A with exactly X factors and exactly K of them are prime numbers.
# QUICK EXPLANATION
A valid choice for A exists if the prime factorization of X has the number of terms at least K. So we just need to compute prime factorization of X.
# EXPLANATION
Let’s assume a valid A exists, with exactly K prime divisors.
The prime factorization of A would be like \prod_{i = 1}^K p_i^{a_i} where p_i are prime factors of A and a_i are the exponents. Then it is well known that the number of factors of A are \prod_{i = 1}^K (a_i+1). Hence we have X = \prod_{i = 1}^K (a_i+1) for a_i \geq 1, hence (a_i+1) \geq 2
So our problem is reduced to determining whether is it possible to write X as a product of K values greater than 1 or not.
For that, let us find the maximum number of values we can split X into such that each value is greater than 1. It is easy to prove that all values shall be prime (or we can further split that value). If the prime factorization of X is \prod r_i^{b_i}, then \sum b_i is the number of terms we can split X into.
For example, Consider X = 12 = 2^2*3 = 2*2*3. Hence we can decompose 12 into at most 3 values such that their product is X. However, if K < \sum b_i, we can merge the values till we get exactly K values. Suppose we have X = 12 and K = 2, so after writing 2, 2, 3, we can merge any two values, resulting in exactly two values.
So a valid A exists when the number of prime factors of X with repetition is at least K.
The following is multiple choice question (with options) to answer.
What is the product of all the prime factors of 14? | [
"13",
"12",
"14",
"11"
] | C | factors : 2,7
2*7 =14
Answer : C |
AQUA-RAT | AQUA-RAT-38844 | software
9.0175 11.2135 2.3156 H 0
8.0176 9.4560 2.2492 C 0
6.3410 11.0331 3.1948 C 0
6.8195 11.7005 2.6977 H 0
6.1061 11.3837 4.0569 H 0
5.5420 10.7916 2.7197 H 0
6.8948 4.6534 2.5072 C 0
7.5354 3.9263 2.4577 H 0
6.3296 4.5991 1.7205 H 0
6.0338 4.4553 3.7359 C 0
5.2898 5.0768 3.7165 H 0
6.5567 4.6337 4.5332 H 0
5.5123 3.0333 3.7738 C 0
5.0259 2.8601 2.9524 H 0
6.2680 2.4268 3.8010 H 0
4.6053 2.7307 4.9451 C 0
3.8347 3.3179 4.9116 H 0
5.0816 2.9111 5.7713 H 0
4.1326 1.2872 4.9481 C 0
4.8942 0.7100 5.1149 H 0
3.7873 1.0713 4.0679 H 0
3.0575 0.9929 5.9842 C 0
2.3293 1.6247 5.8729 H 0
2.7039 0.1030 5.8300 H 0
3.5713 1.0798 7.3895 C 0
6.3200 5.1537 -1.1307 Cl 0
1 15 1
2 3 1
2 4 1
2 8 1
2 12 1
5 55 1
6 7 1
6 55 1
8 32 1
8 33 1
9 10 1
9 11 1
9 21 1
9 37 1
12 13 1
12 20 1
13 14 1
13 15 1
15 16 1
16 17 1
16 18 1
18 19 1
18 20 1
20 21 1
21 22 1
21 23 1
23 24 1
23 32 1
24 25 1
24 26 1
26 27 1
26 28 1
28 29 1
28 30 1
30 31 1
30 32 1
33 34 1
33 35 1
33 36 1
The following is multiple choice question (with options) to answer.
The G.C.D of 1.2, 0.36 and 0.9 is | [
"0.19",
"0.1",
"0.06",
"0.11"
] | C | Explanation:
Given numbers are 1.2 , 0.36 and 0.90
H.C.F of 120, 36 and 90 is 6 [ because G.C.D is nothing but H.C.F]
therefore H.C.F of given numbers = 0.06
Answer: C) |
AQUA-RAT | AQUA-RAT-38845 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
You collect pens. Suppose you start out with 25. Mike gives you another 22 pens. Since her father makes pens, Cindy decides to double your pens. Since you're nice, you give Sharon 19 pens. How many pens do you have at the end? | [
"39",
"40",
"41",
"75"
] | D | Solution
Start with 25 pens.
Mike gives you 22 pens: 25 + 22 = 47 pens.
Cindy doubles the number of pens you have: 47 × 2 = 94 pens.
Sharon takes 19 pens from you: 94 - 19 = 75 pens.
So you have 75 at the end.
Correct answer: D |
AQUA-RAT | AQUA-RAT-38846 | 7. wayki says:
Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).
I hate imperial so please allow me to convert it to metric.
0.07366 mm = 30.480m
Multiply all of this up by 1000 =
73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?
Those that come up with 8inches a mile are much closer to the truth.
8. wayki says:
Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.
9. Alex A says:
I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???
Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.
If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.
• mathscinotes says:
Thank you. Very nicely put.
mathscinotes
• Johnny Emerson Neeley says:
I think 18 miles is level at a 6' height😜😨, on a perfect sphere, 131,477,280 ft.
10. Steve says:
So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?
• mathscinotes says:
The following is multiple choice question (with options) to answer.
On a map, 1.5 inches represent 24 miles. How many miles approximately is the distance if you measured 47 centimeters assuming that 1-inch is 2.54 centimeters? | [
"174.2",
"212",
"288.1",
"296"
] | D | 1.5 inch = 2.54*1.5 cm.
So, 2.54*1.5 represents 24 miles.
So for 47 cm.:
47/(2.54*1.5) = x / 24 ---> x = 24*47/(4.81) = 296
Answer will be D. |
AQUA-RAT | AQUA-RAT-38847 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
The rear–most end of a 66 foot truck exits a 330 foot tunnel exactly 6 seconds after the front–most end of the truck entered the tunnel. If the truck traveled the entire tunnel at a uniform speed, what is the speed of the truck in miles per hour (1 mile = 5,280 feet)? | [
"225",
"90",
"45",
"37.5"
] | C | total length will be 330+66=396
time=6 secs
speed=396/6=66 feet/sec
conversion to miles per hour=66*3600/5280=45.
Answer C |
AQUA-RAT | AQUA-RAT-38848 | Example: Take $k=2002$, then $n=63$ and hence $2002$ lies in $(T_{63}-2002+1,63+2002-T_{63})=(15,49)$.
መለሰ
The following is multiple choice question (with options) to answer.
What will be the reminder when (1234567890123456789)^24 is divided by 6561 ? | [
"0",
"1",
"2",
"3"
] | A | The digital sum of number is
(1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9)=90=(9+0)=9
So it is divisible by 3 or 9
The given divisor 6561 = 9^4
So
9^24/9^4 Reminder is 0
ANSWER:A |
AQUA-RAT | AQUA-RAT-38849 | # Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?
A bag has:
$3$ red
$5$ black
$8$ green
marbles.
Total of 16 marbles.
You select a marble, and then another one right after. (without replacement).
What is the probability that $both$ are red?
Probability that first pick is red: $\frac{3}{16}$
Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)
Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$
How do I do this using combinations only?
• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59
Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$
Hope this helps you.
The following is multiple choice question (with options) to answer.
A bag contains 4 white marbles and 4 black marbles. If each of 4 girls and 4 boys randomly selects and keeps a marble, what is the probability that all of the girls select the same colored marble? | [
"1/35",
"1/9",
"1/10",
"1/20"
] | A | first, total ways to select for all boys and girls, i.e 8!/(4!*4!) = 8*7*6*5*4*3*2*1/4*3*2*1*4*3*2*1=70
then there are one two way girls can have all same colors, either white or black.
The number of ways in which 4 girls can select 4 white balls = 4C4 = 1
The number of ways in which 4 girls can select 4 black balls = 4C4 = 1
Therefore, total favorable outcomes/total outcomes = 2/70= 1/35
A |
AQUA-RAT | AQUA-RAT-38850 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A 2000 liter tank, half-full of water is being filled from a pipe with a flow rate of 1 kiloliter every 2 minutes. At the same time, the tank is losing water from two drains at a rate of 1 kiloliter every 4 minutes and every 6 minutes. How many minutes does it take to fill the tank completely? | [
"8",
"12",
"18",
"24"
] | B | In: we have: 1,000/2min = 500 litres per minute
Out: we have: 1,000/4 + 1,000/6
Then do: IN - OUT to figure out the net inflow per minute (you get 83.3). Then divide the total number of litres you need (1,000 by that net inflow to get the minutes) - 12 min. Answer B. |
AQUA-RAT | AQUA-RAT-38851 | 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.
Joe can clean the house in 4 hours. Cindy can clean the house in 3 hours. How many hours will it take for both of them working together to clean the house? | [
"1",
"2 5/7",
"2 1/5",
"1 5/7"
] | D | Work hrs=AB/(A+B)= 12/7 =1 5/7
Answer is D |
AQUA-RAT | AQUA-RAT-38852 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
The speed of a car is 145 km in the first hour and 60 km in the second hour. What is the average speed of the car? | [
"89 kmph",
"92 kmph",
"75 kmph",
"102.5 kmph"
] | D | S = (145 + 60)/2 = 102.5 kmph
D |
AQUA-RAT | AQUA-RAT-38853 | Similar Questions
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A). 11! / 2 B). 10! / 2 C). 12! / 2 D). Couldn't be determined
-- View Answer
7). 20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?
A). 18! B). 19! C). 20! D). Couldn't be determined
-- View Answer
8). A committee of 5 members is going to be formed from 3 trainees, 4 professors and 6 research associates. How many ways can they be selected, if in a committee, there are 2 trainees and 3 research associates?
A). 15 B). 45 C). 60 D). 9
-- View Answer
9). In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
The following is multiple choice question (with options) to answer.
In how many ways Chief Minister and Minister be elected from a team of 12 members? | [
"128",
"130",
"132",
"143"
] | C | To do this, if captain is elected first, then we have 12 ways of doing this.
For election of vice-captain, we have only 11 ways left, coz 1 way is already consumed. (Situations like this is called dependent situation. One selection depends upon other selection.)
So, the ans is 12*11 = 132 ways.
C |
AQUA-RAT | AQUA-RAT-38854 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are moving in the same direction at 144 kmph and 72 kmph. The faster train crosses a man in the slower train in 19 seconds. Find the length of the faster train? | [
"320",
"340",
"360",
"380"
] | D | Relative speed = (144 - 72) * 5/18 = 4 * 5
= 20 mps.
Distance covered in 19 sec
= 19 * 20
= 380 m.
The length of the faster train
= 380 m.
Answer:D |
AQUA-RAT | AQUA-RAT-38855 | graph-theory, co.combinatorics
9 meals are given by slopes $a = 0,1,\ldots,8$. The intersections with the 4 empty degenerate lines reduce the table size to 9-4=5.
An additional meal is given by the remaining degenerate lines $\{(b,x) | x \in F\}$ for every $b = 4,5,6,7,8$. Here the table size is 9. However (in any solution) we can break a table of size 9 into a table of size 5 and one of size 4.
If there are a few more people one can use the field of size 11.
The following is multiple choice question (with options) to answer.
Village A Village B Village C Village D Village E Village F ;
Village A
Village B
Village C
Village D
Village E
Village F
In the table above, what is the least number of table entries that are needed to show the mileage between each Village and each of the other five Villages? | [
"16",
"15",
"18",
"20"
] | B | Easy way to go about this problem is we have 6*6 = 36 enteries in table the least number of enteries would be (36 - 6) /2 since 6 enteries represent the distances between same points .
Alternatively this can be solved as combination problem.
Correct Answer B |
AQUA-RAT | AQUA-RAT-38856 | 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.
Myra drove at an average speed of 30 miles per hour for T hours and then at an average speed of 60 miles/hr for the next T hours. If she made no stops during the trip and reached her destination in 2T hours, what was her average speed in miles per hour for the entire trip? | [
"The average speed is 40 miles per hour.",
"The average speed is 45 miles per hour.",
"The average speed is 48 miles per hour.",
"The average speed is 50 miles per hour."
] | B | Here, time for which Myra traveled at the two speeds is same.
Average Speed = (a + b)/2 = (30 + 60)/2 = 45 miles per hour
Answer (B) |
AQUA-RAT | AQUA-RAT-38857 | We see that: $b^2-1 \text{ is even }\quad\Rightarrow\quad b^2\text{ is odd }\quad\Rightarrow\quad b\text{ is odd}$
Let: $b \:=\:2c+1$
Then [1] becomes: . $k \:=\:\frac{(2c+1)^2-1}{4} \:=\:c^2 + c \quad\Rightarrow\quad k \:=\:c(c+1)$
Hence, $k$ must be the product of two consecutive integers.
Then: . $x \;=\;\frac{1 + \sqrt{1 + 4k}}{2} \;=\;\frac{1 + \sqrt{1+4c(c+1)}}{2} \;=\;\frac{1 + (2c+1)}{2} \;=\;c+1$, an integer.
6. Originally Posted by Soroban
Hello, starvin-marvin!
We have: . $4k + 1 \:=\:b^2$, for some integer $b$
. . Then: . $k \:=\:\frac{b^2-1}{4}\;\;{\color{blue}[1]}$
We see that: $b^2-1 \text{ is even }\quad\Rightarrow\quad b^2\text{ is odd }\quad\Rightarrow\quad b\text{ is odd}$
Let: $b \:=\:2c+1$
Then [1] becomes: . $k \:=\:\frac{(2c+1)^2-1}{4} \:=\:c^2 + c \quad\Rightarrow\quad k \:=\:c(c+1)$
Hence, $k$ must be the product of two consecutive integers.
The following is multiple choice question (with options) to answer.
If a,b and c are consecutive positive integers and a>b>c. What can be the value of (a^2-b^2)(b^2-c^2)? | [
"21",
"79",
"143",
"231"
] | C | Since a,b and c are consecutive positive integers and a>b>c, we can write them as (x + 1), x and (x - 1) respectively
(a^2-b^2)(b^2-c^2) can be simplified into (2x + 1)(2x - 1) = 4x^2 - 1
using the options to find the product,
only C. 143 satisfactorily gives us x as an integer
ANSWER:C |
AQUA-RAT | AQUA-RAT-38858 | Maybe the examiner didn't think through your method.
From 10,000 to 99,999 the numbers divisible by 5 end in 0 or 5.
The first (most significant) digit can be any of nine from 1 to 9.
The 2nd digit can be any of 10.
The 3rd digit can be any of 10.
The 4th digit can be any of 10.
The 5th digit can be either of 2.
That's 9(10)(10)(10)2=18,000
The following is multiple choice question (with options) to answer.
The number N is 5,1H4, where H represents the ten's digit. If N is divisible by 9, what is the value of H? | [
"1",
"3",
"8",
"7"
] | C | Integer is divisible by 9 - Sum of digits is divisible by 9
Answer: C |
AQUA-RAT | AQUA-RAT-38859 | homework-and-exercises, newtonian-mechanics, rotational-kinematics
Now your doubt is that why (2) holds it might happen that $\alpha$ can be positive or negative and thus $\omega_i$ can be greater than or less than $\omega_f$.
In the above problem it is given that wheel covers 90 revolutions in 15 sec. So on averge it angular velocity is $\frac{90}{15}=6$ revolutions/sec.
But final velcity is $10$ revolutions/sec.
This means that at some point of time the velocity is less than $6$ revolutions/sec only then we get an average of 6 revolution/sec.
But we are given that angular acceleration is constant. So velocity can either decrease or increase in that time interval. It can't be the case that at some interval velocity will decrease and then increase as for that angular acceleration has to change sign so it won't remains constant giving us contradiction.
As at some interval angular velocity has to take value less than $6$ revolutions/sec.
This suggests that over the whole interval velocity will increase continuously thus initial angular velocity will be less that final angular velocity.
So, the take home message is that in the equation $(2)$ the information of angular acceleration is inherent, though it might not be so much evident from the form of equation but we can see this from the derivation.
Hope that helps!
The following is multiple choice question (with options) to answer.
A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. When the smaller wheel has made 21 revolutions, then the number of revolutions mad by the larger wheel is: | [
"4",
"9",
"12",
"21"
] | B | More cogs less revolution,..
so
6:14::x:21
(21*6)/14=9
ANSWER:B |
AQUA-RAT | AQUA-RAT-38860 | Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
Hence answer will be (E) $693 The wage difference is not with respect to time , correct approach will be with respect to work done.. ( Errorr is highlighted in red ) Further explanation as provide to you via PM Wages/Salary can be of 2 types ( Something You will certainly admit ) - 1. Hourly Wage payment System ( Most White collar Jobs ) 2. Unit of Work Done ( Most Blue collar Jobs ) Here it is given the part of job completed by A and B , both have diff efficiency.. Suppose total Work is 18 units Efficiency of A = 11 and Efficiency of B = 7 Now, tell me , whom will you give more money A or B ( According to the amount of work completed ) Certainly A !!! Continuing with the same example - Time Taken by A = 18/11 = 1.63 Hours Time Taken by B = 18/7 = 2.57 Hours Now, tell me , whom will you give more money A or B ??? B is inefficient , straightay, he takes more time to complete less amount of work than A... Hence Hourly wage approach is not suited here, hope this helps.. _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 110 Location: United States GPA: 3.37 Re: Elana was working to code protocols for computer processing. She did [#permalink] ### Show Tags 02 Nov 2016, 11:44 Elena did 11/18 which means that Andrew did 7/18. They both earn the same hourly rate. The difference between the work they did is 4/18. We know that 4/18 equals$154 because that is the difference in pay. Divide $154 by 4 so we can see how much 1/11th of the job is worth.$154/4=\$38.5
The following is multiple choice question (with options) to answer.
A certain bus driver is paid a regular rate of $20 per hour for any number of hours that does not exceed 40 hours per week. For any overtime hours worked in excess of 40 hours per week, the bus driver is paid a rate that is 75% higher than his regular rate. If last week the bus driver earned $1000 in total compensation, how many total hours did he work that week? | [
"46",
"40",
"44",
"48"
] | A | For 40 hrs = 40*20=800
Excess = 1000-800=200
For extra hours =.75(20)=15+20=35
Number of extra hrs =200/35=5.71=6 approx.
Total hrs =40+6=46
Answer A 46 |
AQUA-RAT | AQUA-RAT-38861 | Hint: Start your proof like this:
Let $k$ be any arbitrary odd integer. Then by the definition of an odd integer, we have $k=2a+1$ for some integer $a$. Thus...
Then consider $-k=-(2a+1)$ and perform some algebraic manipulations. Your final step should look something like:
...hence, since $-k=2b+1$ where $b$ is an integer, it follows by definition that $-k$ is also an odd integer, as desired.
-
One could prove this inductively:
Assume that the $n^\text{th}$ odd positive integer, $2n-1$, has an odd negation. Then $-2n+1$ is odd, so $$\underbrace{-2n+1}_{\text{odd}}-\underbrace{2}_{\text{even}}=\underbrace{-2n-1}_{\text{odd}}.$$ Thus the statement holds for the $(n+1)^\text{th}$ positive integer, $2n+1$.
-
I like the simplicity of this approach. Pedantic: you still also need to prove the base case. – Lie Ryan Jun 23 '13 at 17:12
Here's my proof of the equivalent statement for even integers. I'll let you figure out what to do for odds.
Let $n$ be an even integer. By definition of "even", $\exists m \in \mathbb{Z}: n=2m$. Then $-n = -2m = 2(-m)$. Because $\mathbb{Z}$ is closed under additive inverse, $-m$ is an integer. By the definition of "even", $2(-m) = -n$ is also even, Q.E.D.
-
Let our odd number be $o$. Since $-1\equiv 1 \pmod{2}$, $-o\equiv o \pmod{2}$. By the definition of odd numbers, $o\equiv 1 \pmod{2}$. Combining this with the earlier statement, one gets
The following is multiple choice question (with options) to answer.
If a and b are odd integers, which of the following is an even integer? | [
"2ab+a",
"2ab-b",
"2ab+a+b",
"2ab+2a+b"
] | C | 2ab+a+b=even+odd+odd=even
The answer is C. |
AQUA-RAT | AQUA-RAT-38862 | design-patterns, bash, plugin
validateNumeric() {
BOOKLENGTH=$1
reg='^[0-9]+$'
if ! [[ $BOOKLENGTH =~ $reg ]] ; then
echo "Error: Argument not a number, try again:" >&2;
read BOOKLENGTH
validateNumeric $BOOKLENGTH
else
validateEven $BOOKLENGTH
fi
}
validateEven() {
BOOKLENGTH=$1
echo "Testing if ${BOOKLENGTH} is even now"
if [ $((BOOKLENGTH%2)) -eq 0 ] ; then
echo "Ok ....... Proceeding"
echo "Setting book length = $BOOKLENGTH"
return $BOOKLENGTH
else
echo "Error: Not an even number, try again:" >&2;
read BOOKLENGTH
validateEven $BOOKLENGTH
fi
}
setupProject() {
echo "Setting up $PROJECTNAME now ..."
mkdir -p "$1" && cd "$1" && touch README.md license.txt .gitignore && mkdir "trash" "cover" "templates" "images" "manuscript" || return $?
echo "# $1" >> README.md
cd "templates" && touch template.html head.html template.css template.js && cd ".."
}
createPages() {
PAGES=$1
cd "manuscript"
p=0
while [ "$p" -lt "$PAGES" ]; do
p=$((p+1))
mkdir -p "page-$p"
cd "page-$p"
touch "body.html"
touch "style.css"
echo "body{background:rgba(200, 235, 255, 0.99); margin:0 0; overflow:hidden;}" >> style.css
cd ".."
done
echo "Done!" && cd ".." #Head back to root
}
The following is multiple choice question (with options) to answer.
How many digits are required to number a book containing 260 pages? | [
"784",
"672",
"492",
"372"
] | B | 9 pages from 1 to 9 will require 9 digits.
90 pages from 10 to 99 will require 90*2=180 digits.
260-(90+9)=161 pages will require 161*3=483 digits.
The total number of digits is 9+180+483=672.
The answer is B. |
AQUA-RAT | AQUA-RAT-38863 | Note: Accept $$\frac{{15 – 14}}{{\sqrt {\frac{{3.03}}{{20}}} }}$$.
$$= – 2.569 \ldots$$ (A1)
Note: Accept $$2.569 \ldots$$
$$p{\text{ – value}} = 0.009392 \ldots \times 2 = 0.0188$$ A1
Note: Accept any answer that rounds to 0.019.
Note: Award (M1)(A1)A0 for any answer that rounds to 0.0094.
insufficient evidence to reject $${{\text{H}}_0}$$ (or equivalent, eg accept $${{\text{H}}_0}$$ or reject $${{\text{H}}_1}$$) R1
Note:FT on their p-value.
[4 marks]
c.
Examiners report
In (a), most candidates estimated the mean correctly although many candidates failed to obtain a correct unbiased estimate for the variance. The most common error was to divide $$\sum {{x^2}}$$ by $$20$$ instead of $$19$$. For some candidates, this was not a costly error since we followed through their variance into (b) and (c).
a.
In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.
b.
In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.
c.
Question
The following is multiple choice question (with options) to answer.
In a certificate by mistake a candidate gave his height as 25% more than actual height. In the interview panel, he clarified that his height was 5feet 5 nches. Find the % correction made by the candidate from his stated height to his actual height? | [
"10",
"20",
"30",
"43"
] | B | His height was = 5 feet 5 inch = 5 + 60 = 65 inch. Required % correction =65*(1.25-1)* 100 = 20
B |
AQUA-RAT | AQUA-RAT-38864 | 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 speeds past a pole in 15 seconds and a platform 100 m long in 40 seconds. Its length is? | [
"188 m",
"876 m",
"251 m",
"100 m"
] | D | Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 40 = x / 15
x = 100 m.
Answer: D |
AQUA-RAT | AQUA-RAT-38865 | In a bag there are 10 Black balls, 8 White balls and 5 Red balls.
Three balls are chosen at random and one is found to be Black.
Find the probability that remaining two are white.
. $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$
I see it as a Conditional Probability problem . . .
Given that at least one ball is Black,
. . find the probability that we have one Black and two White balls.
Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$
There are ${23\choose3} = 1771$ possible ways to choose 3 balls.
To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways.
. . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$
The opposite of "at least 1 Black" is "NO Blacks".
There are: . ${13\choose3} = 286$ ways to choose no Blacks.
So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls.
. . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$
The following is multiple choice question (with options) to answer.
A bag contains 12white and 18black balls. 2balls are drawn in succession. What is the probability that first is white and second is black? | [
"31/145",
"36/145",
"32/145",
"36/147"
] | B | The probability that first ball is white:
=12C130C1=12C130C1
=1230=1230
=25=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C129C1=18C129C1
=1829=1829
Required probability,
=(25)×(1829)=(25)×(1829)
=36/145
B |
AQUA-RAT | AQUA-RAT-38866 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 9 hours and Mary worked 2 hours less than John. If Mary gave John y dollars of her payment so that they would have received the same hourly wage, what was the dollar amount, in terms of y, that John was paid in advance? | [
" 4y",
" 5y",
" 6y",
" 17y"
] | D | Let $x be the advance that both receive = 2x
Amount earned per hour by John and Mary = x/9 and x/8
Mary gives $y to John to make the wages earned equal
Hence John wage per hr = (x+y)10 which is now equal to Mary's wage (x-y)/8
Solve (x+y)9 = (x-y)/8
8x + 8y = 9x -9y
x = 17y
Ans. D |
AQUA-RAT | AQUA-RAT-38867 | 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 72 km and upstream 45 km taking 9 hours each time; what is the speed of the current ? | [
"1 kmph",
"3.2 kmph",
"1.5 kmph",
"2 kmph"
] | C | Explanation:
72 --- 9
? ---- 1
=> Down Stream = 8
45 ---- 9
? ---- 1
=> Up Stream = 5
Speed od current S = ?
S = (8 - 5)/2 = 1.5 kmph.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-38868 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Sachin is younger than Rahul by 7 years. If the ratio of their ages is 6:9, find the age of Sachin | [
"24.5",
"14",
"24.21",
"27"
] | B | Explanation:
If Rahul age is x, then Sachin age is x - 7,
so,
9x - 63 = 6x
3x = 63
x = 21
So Sachin age is 21 - 7 = 14
Answer: B) 14 |
AQUA-RAT | AQUA-RAT-38869 | ### 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.
A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is | [
"7hours 75 mins",
"7 hours",
"8 hours",
"5 hours"
] | A | Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
Answer is A. |
AQUA-RAT | AQUA-RAT-38870 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two goods trains each 850 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one? | [
"22 sec",
"81.6 sec",
"48 sec",
"18.3 sec"
] | B | Relative speed = 45 + 30 = 75 km/hr.
75 * 5/18 = 125/6 m/sec.
Distance covered = 850 + 850 = 1700 m.
Required time = 1700 * 6/125 = 81.6 sec.
Answer:B |
AQUA-RAT | AQUA-RAT-38871 | # 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.
Sujit incurred a loss of 45% on selling an article for Rs.3850/-. What was the cost price of the article? | [
"Rs.5725/-",
"Rs.5080/-",
"Rs.6250/-",
"Rs.7000/-"
] | D | Explanation:
45% loss means SP of Rs.3850/- is 55% of CP
:. CP = 3850x100/55 = Rs.7000/-
Answer: Option D |
AQUA-RAT | AQUA-RAT-38872 | Within a class of$M=30$children, knowing that the year counts$A=365$days... The probability that at least$n=2$children have their birthday the same day is: $$P(365,30,2)\simeq 70,6\%$$ The probability that at least$n=3$children have their birthday the same day is: $$P(365,30,3)\simeq 2,85\%$$ The probability that at least$n=4$children have their birthday the same day is: $$P(365,30,4)\simeq 0,0532\%$$ Nicolas Just like to point out that Trazom's answer is incorrect for the general case - the sets being counted in the outer sum overlap. I don't have enough reputation to comment. I wrote a blog post about the general case here : https://swarbrickjones.wordpress.com/2016/05/08/the-birthday-problem-ii-three-people-or-more/ Anyone looking for generalized birthday problem i.e. How many people are required such that M of them share same birthday with certain probability. This link explain various method for calculating probability of generalized birthday problem. http://mathworld.wolfram.com/BirthdayProblem.html Also this paper talk more about various kind of coincidences we face ib life, interesting read. https://www.stat.berkeley.edu/~aldous/157/Papers/diaconis_mosteller.pdf • "Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." – Shraddheya Shendre Sep 10 '17 at 5:35 I am looking at this question and the complicated answers and it's confusing me. Supposing I want to solve in a group of 100 people. what is the probability that at least 3 people share a birthday. So I start from very basic - if there are 3 people, the probability of them sharing a birthday is $$\frac{1}{365}
The following is multiple choice question (with options) to answer.
The sum of the ages of 4 children born at the intervals of 3 years each is 36 years. what is the age of the youngest child ? | [
"22",
"18",
"4.5",
"99"
] | C | Let x = the youngest child. Each of the other four children will then be x+3, x+6, x+9
We know that the sum of their ages is 36
so, x+(x+3)+(x+6)+(x+9)= 36
therefore The youngest child is 4.5 years old
Answer: C |
AQUA-RAT | AQUA-RAT-38873 | 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.
At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.5000? | [
"1655",
"2888",
"2776",
"2997"
] | A | A = 5000(11/10)^3
= 6655
= 5000
----------
1655
Answer: A |
AQUA-RAT | AQUA-RAT-38874 | Let P=[$(\sqrt{n}+0.7)^2$]
Second Hint
given $n \geq 1$, put n=1 gives P=2
n=2 gives P=4
n=3 gives P=5
n=4 gives P=7
n-5 gives P=8
n=6 gives P=9
n=7 gives P=11
Final Step
here missing number are
1,3,6,10,… which is following a certain pattern
1, 1+2, 3+3, 6+4, 10+5, 15+6, 21+7, 28+8, 36+9, 45+10, 55+11, 66+12.
so, $n_{12}$=78.
The following is multiple choice question (with options) to answer.
Can you find the missing number in the sequence given below?
10 18 26 11 19 ? 12 20 28 13 21 29 | [
"25",
"23",
"20",
"27"
] | D | Let's break the given series as below:
10 18 26
11 19 ?
12 20 28
13 21 29
Now read the number from left hand side from top to bottom as :
So the number that will replace '?' is 27
Answer : D |
AQUA-RAT | AQUA-RAT-38875 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
2,5,12,31,81,__ | [
"210",
"211",
"213",
"214"
] | C | 12= 5*2+2
31= 12*2+5+2
81= 31*2+12+5+2
Similarly 81*2+31+12+5+2 = 162+50=212
ANSWER:C |
AQUA-RAT | AQUA-RAT-38876 | ### Explanation of the last step
This is a property all rectangles in the coordinate plane have.
For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\left(\frac{2+8}2,\frac{5+3}2\right)=(5,4)$, therefore $\frac{a+c}2=5$, and $a+c=10$.
$[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label("(2,5)",X,W*1.5); label("(8,3)",Y,E*1.5); label("(a,b)",(4,7),N); label("(c,d)",(6,1),S); [/asy]$
### An alternate last step
We can easily compute $a$ and $c$ using our picture.
The following is multiple choice question (with options) to answer.
A rectangle PRSU, is divided into two smaller rectangles PQTU, and QRST by the line TQ. PQ = 10 cm. QR = 5 cm and RS = 10 cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A, C), (A, D), (A, E), (F, C), (F, D), (F, E), (B, C), (B, D), (B, E) are 10 3 cm apart.
Which of the following statements is necessarily true? | [
"The closest pair of points among the six given points cannot be (F, C)",
"Distance between A and B is greater than that between F and C.",
"The closest pair of points among the six given points is (C, D), (D, E), or (C, E).",
"none of the above"
] | D | Explanation :
We have not been given the distances between any two points.
Answer : D |
AQUA-RAT | AQUA-RAT-38877 | ## 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 |x + 2| = 10, what is the sum of all the possible values of x? | [
"-16",
"-13",
"-4",
"-2"
] | C | There will be two cases
x+2 = 10 or x+2 = -10
=> x = 8 or x= -12
sum of both the values will be -12 + 8 = -4
Answer: C |
AQUA-RAT | AQUA-RAT-38878 | Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1
Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.
Thus there are 60+90 = 150 ways we can place the marbles
Manager
Joined: 30 May 2019
Posts: 82
Location: United States
Concentration: Technology, Strategy
GPA: 3.6
Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
### Show Tags
04 Oct 2019, 22:44
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
Solved it a little differently than others from what I read . So will share.
I first made sure that each of the pocket has at least 1 marble.
That is 5C3 * 3!
The following is multiple choice question (with options) to answer.
How many Q ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl? | [
"6C4",
"6P4",
"4^6",
"6^4"
] | C | Each marble has 4 options, so there are total of Q= 4*4*4*4*4*4=4^6 ways.
Answer: C.
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is (n+r -1)C(r-1). |
AQUA-RAT | AQUA-RAT-38879 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
Ratio between Rahul and Deepak is 4:3, After 6 Years Rahul age will be 26 years. What is Deepak present age | [
"14",
"15",
"20",
"22"
] | B | Explanation:
Present age is 4x and 3x,
=> 4x + 6 = 26 => x = 5
So Deepak age is = 3(5) = 15
Answer: Option B |
AQUA-RAT | AQUA-RAT-38880 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock is set right at 1 p.m. If it gains one minute in an hour, then what is the true time when the clock indicates 6 p.m. in the same day? | [
"55 5⁄61 minutes past 5",
"5 minutes past 6",
"5 minutes to 6",
"59 1⁄64 minutes past 5"
] | A | Time interval indicated by incorrect clock
= 6 p.m – 1 p.m = 5hrs.
Time gained by incorrect clock in one hour
= + 1 min. = +1⁄60 hr.
Using the formula, True time interval /Time interval in incorrect clock
= 1 /1+hour gained in 1 hour by incorrect clock
⇒ Truetimeinterval/5=1/1+1/60
⇒ True time interval = 5×60/61=4 56/61
∴ True time = 1 p.m. + 4 56/61 hrs.
= 5 p.m. + 56⁄61 hrs. = 5 p.m. + 56⁄61 × 60 min.
= 555⁄61 minutes past 5.
Answer A |
AQUA-RAT | AQUA-RAT-38881 | 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.
what is the smallest integer that is multiple of 5,7,8 | [
"A)70",
"B)35",
"C)200",
"D)280"
] | D | It is the lcm of 5, 7 and 8 which is 280.
The answer is D. |
AQUA-RAT | AQUA-RAT-38882 | 0.00698 acre1 acre = 43560 square feet304 sq ft * 1 acre/43560 sq ft = 0.00698 acre
### Number of sq ft per acre?
1 sq mi=5280ft x 5280ft = 27878400 sq ft/sq mi/ 27,878,400 sq ft/sq mi divided by 640acres/sq mi = 43,560 sq ft /acre 1 acre = 43,560 sq ft
### How many square feet are in 1.8 acre?
1 acre = 43,560 sq ft → 1.8 acre = 1.8 × 43,560 sq ft = 78,408 sq ft
### What does 0.2 acres equals sq feet?
1 acre = 43,560 sq ft .2 acre * 43560 sq ft = 8712 sq ft
### What percentage of acres in 160 feet x 180 feet?
160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.
### How many ft are in a acre?
There are 43559.66 sq ft in an acre.
### What is a fifth of an acre?
an acre is 43,560 sq ft 1/5th of that is 8,712 sq ft
### How many acres is 11026 sq ft?
43,560 sq ft = 1 acre11,026 sq ft = 0.2531 acre (rounded)
### Figure how many acres in 60000 sq ft?
(60,000 sq ft) / (43,560 sq ft / acre) = 1.3774 acre (rounded)
### What is the convertion of 435 acre to square feet?
1 acre = 43,560 sq ft → 435 acre = 435 x 43,560 sq ft = 18,948,600 sq ft
### How many sq ft for one acre 14 guntas in India?
The following is multiple choice question (with options) to answer.
25% of the total cost of a plot of area 280 sq.ft. is Rs.132370. What is the rate per sq.ft of the plot? | [
"Rs.2091",
"Rs.1891",
"Rs.1981",
"Rs.1991"
] | B | Explanation:
25% = ¼
:. Cost of total area = Rs.132370x4
Rate per sq.ft = 132370x4/280 = Rs.1891/-
Answer: Option B |
AQUA-RAT | AQUA-RAT-38883 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
Irin, Ingrid and Nell bake chocolate chip cookies in the ratio of 9.18: 5.17: 2.05. If altogether they baked a batch of 148 cookies, what percent of the cookies did Ingrid bake? | [
"0.125%",
"1.25%",
"31.7%",
"125%"
] | C | 9.18x+5.17x+2.05x = 16.4x = 148 cookies
x= 148/16.4 = 9 (approx)
So, Ingrid baked 9*5.17 cookies or 47 cookies (approx)
% share = 47/148 = 31.7 approx
Hence, answer is C. |
AQUA-RAT | AQUA-RAT-38884 | 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.
X and Y started a business investing Rs. 50,000 and Rs 40,000 respectively. In what ratio the profit earned after 2 years be divided between X and Y respectively? | [
"3:2",
"9:2",
"5 : 4",
"1:4"
] | C | X:Y = 50000 : 40000 = 5 : 4
ANSWER:C |
AQUA-RAT | AQUA-RAT-38885 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A man whose bowling average is 12.4, takes 8 wickets for 26 runs and there by decreases his average by 0.4. the number of wickets taken by him before his last match is? | [
"173",
"174",
"175",
"176"
] | C | 12.4*x+26=(8+x)12
solve equation x=175
ANSWER:C |
AQUA-RAT | AQUA-RAT-38886 | # 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.
An article is bought for Rs.675 and sold for Rs.900, find the gain percent? | [
"6 %",
"8 %",
"5 5/11%",
"11 %"
] | C | Explanation:
Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.
Selling Price (S.P.) = Rs. 5800.
Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300.
Gain % =(300/5500 x 100)%= 5 5/11%
Answer:C |
AQUA-RAT | AQUA-RAT-38887 | WAIT!
S=11!
Because we have 11! ways of arranging the 11 items.
Sorry about that.
5. Hello, sweeetcaroline!
Edit: Plato is absolutely right . . . *blush*
6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order.
What is the probability that no two neighboring beads are the same color?
There are: . ${11\choose6,4,1} \:=\:2310$ possible orders.
Place the 6 Red beads in a row, leaving a space between them.
. . $\begin{array}{ccccccccccccc} R & \_ & R & \_ & R & \_ & R & \_& R & \_ & R\end{array}$
Place the Blue bead is any of the 5 spaces: . $5$ choices.
Drop the 4 White beads in the remaining 4 spaces: . $1$ way.
Hence, there are: . $5\cdot1 \,=\,5$ ways.
Therefore, the probability is: . $\frac{5}{2310} \;\;=\;\;\frac{1}{462}$
6. Originally Posted by sweeetcaroline
the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
Can someone give me a line of these beads in which no two neighboring beads are the same color other than the five listed below?
$RWRWRWRWRBR$
$RWRWRWRBRWR$
$RWRWRWBWRWR$
$RWRBRWRWRWR$
$RBRWRWRWRWR$
The following is multiple choice question (with options) to answer.
People standing in a straight line are wearing alternating colored shirts. The pattern of shirts begins with red, followed by green, blue, and yellow. This pattern continues (red, green, blue, and yellow) until it ends with a green shirt. If there are more than 4 people in line, which of the following cannot be the number of people in the line? | [
"18",
"22",
"42",
"41"
] | D | The correct answer should not take the form: 4x + 2. Hence, it is D |
AQUA-RAT | AQUA-RAT-38888 | substitute our 3rd conclusion:
$1 = x \oplus y \oplus w = 0 \oplus y \oplus 0 = 0 \oplus y = y \Longrightarrow y = 1$
-
The following is multiple choice question (with options) to answer.
If x/|y| = 1 which of the following must be true? | [
"x = -y",
"x = y",
"x = y^2",
"x^2 = y^2"
] | D | x/|y| = 1
x= 1 |y|
Squaring both sides
x^2= (1 |y|) ^2= y^2
D is the answer |
AQUA-RAT | AQUA-RAT-38889 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Two sacks together contained 40 kilograms of sugar. If after 1 kilogram of sugar was taken from the first sack and poured into the second the weight of the sugar in the first sack became 60% the weight of the sugar in the second, what was the original difference in the weights of the sacks? | [
"4",
"6",
"8",
"10"
] | C | Say the weight of the second sack after change is x kilograms, then the weight of the first sack after change would be 0.6x. Since the total weight of sugar in both sacks remained the same then x+0.6x=40, which gives x=25.
Now, if the weight of the second sack after change is 25 kilograms then initially it was 25−1=24 kilograms and the initial weight of the first sack was 40−24=16 kilograms, so the difference was 24−16=8 kilograms.
Answer: C |
AQUA-RAT | AQUA-RAT-38890 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
At a certain supplier, a machine of type A costs $20,000 and a machine of type B costs $60,000. Each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time. If the finance charges are equal to 40 percent of the remainder of the cost, how much less would 2 machines of type A cost than 1 machine of type B under this arrangement? | [
"$10,000",
"$26,400",
"$12,000",
"$12,800"
] | B | 1 machine of type B will cost: 20% down payment of 60,000 = 12,000plusremaining sum (50,000-12,000=48,000) with 40% of finance charges 48,000*1.4=67,200 --> 12,000+67,200=79,200;
2 machine of type A will cost: 20% down payment of 2*20,000 = 8,000plusremaining sum (40,000-8,000=32,000) with 40% of finance charges 32,000*1.4=44,800 --> 8,000+44,800=52,800;
Difference = 79,200 - 52,800 = 26,400.
Answer: B. |
AQUA-RAT | AQUA-RAT-38891 | +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.
If |5x-30| = 100, then find the sum of the values of x? | [
"1",
"-2",
"12",
"-3"
] | C | |5x-30| = 100
5x-30 = 100 or 5x-30 = -100
5x = 130 or 5x = -70
x = 26 or x = -14
sum = 26-14 = 12
Answer is C |
AQUA-RAT | AQUA-RAT-38892 | 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 walking towards each other at 5am at speed of 4kmph and 8kmph. They were initially 36km apart. At what time do they meet? | [
"8am",
"6am",
"7am",
"10am"
] | A | Time of meeting = distance / relative speed = 36/8 + 4 = 36/12 = 3 hrs after 5am = 8am
Answer is A |
AQUA-RAT | AQUA-RAT-38893 | 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.
a number divided by 44 leaves remainder 17 what is the remainder when same number divided by 8 | [
"5",
"7",
"9",
"10"
] | A | add 44+17=61
now 61 divided by 8 so we get 5 as reaminder
ANSWER:A |
AQUA-RAT | AQUA-RAT-38894 | $\text{ }$
$\displaystyle (9) \ \ \ \ \ E(X \wedge u)=\int_{-\infty}^u x \ f_X(x) \ dx+u \ S_X(u)$
$\displaystyle (10) \ \ \ \ \ E(X \wedge u)=\biggl(\sum \limits_{x < u} x \ P(X=x)\biggr)+u \ S_X(u)$
$\text{ }$
Interestingly, we have the following relation.
$\text{ }$
$\displaystyle (X-d)_+ + (X \wedge d)=X \ \ \ \ \ \ \text{and} \ \ \ \ \ \ E[(X-d)_+] + E(X \wedge d)=E(X)$
$\text{ }$
The above statement indicates that purchasing a policy with a deductible $d$ and another policy with a policy maximum $d$ is equivalent to buying full coverage.
Another way to interpret $X \wedge d$ is that it is the amount of loss that is eliminated by having a deductible in the insurance policy. If the insurance policy pays the loss in full, then the insurance payment is $X$ and the expected amount the insurer is expected to pay is $E(X)$. By having a deductible provision in the policy, the insurer is now only liable for the amount $(X-d)_+$ and the amount the insurer is expected to pay per loss is $E[(X-d)_+]$. Consequently $E(X \wedge d)$ is the expected amount of the loss that is eliminated by the deductible provision in the policy. The following summarizes this observation.
$\text{ }$
$\displaystyle (X \wedge d)=X-(X-d)_+ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ E(X \wedge d)=E(X)-E[(X-d)_+]$
The following is multiple choice question (with options) to answer.
Health insurance Plan Q requires the insured to pay $1000 or 50% of total cost, whichever is lower. Plan B requires the insured to pay the initial $300, but then pays 80% of the cost over $300. Which of the following is a cost level for which both insurance plans pay out the same amount? | [
"$600",
"$1000",
"$3800",
"$5300"
] | B | 0.5 * 600 = 300 where Q = B |
AQUA-RAT | AQUA-RAT-38895 | 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 300 meter long train crosses a plaƞorm in 39 seconds while it crosses a signal pole in 18 seconds.
What is the length of the plaƞorm. | [
"310 meter",
"335 meter",
"345 meter",
"350 meter"
] | D | Explanation:
Speed = Distance/Ɵme = 300/18 = 50/3 m/sec
Let the length of the plaƞorm be x meters
then
Distance=Speed
∗
Timex+300=503
∗
39=>3(x+300)=1950=>x=350 meters
Answer: D |
AQUA-RAT | AQUA-RAT-38896 | # Proof for divisibility by $7$
One very classic story about divisibility is something like this.
A number is divisible by $2^n$ if the last $n$-digit of the number is divisible by $2^n$. A number is divisible by 3 (resp., by 9) if the sum of its digit is divisible by 3 (resp., by 9). A number $\overline{a_1a_2\ldots a_n}$ is divisible by 7 if $\overline{a_1a_2\ldots a_{n-1}} - 2\times a_n$ is divisible by 7 too.
The first two statements are very well known and quite easy to prove. However I could not find the way on proving the third statement.
PS: $\overline{a_1a_2\ldots a_n}$ means the digits of the number itself, not to be confused with multiplication of number.
The following is multiple choice question (with options) to answer.
It is being given that (2^32 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number? | [
"16^2+1",
"12",
"(2^96+1)",
"24"
] | C | Explanation:
Let 2^32 = x. Then, (2^32 + 1) = (x + 1).
Let (x + 1) be completely divisible by the natural number N. Then,
(296 + 1) = [(2^32)3 + 1] = (x^3 + 1) = (x + 1)(x^2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
C |
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