source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39897 | You probably want to find $a$ and $b$ such that, for all $x$, $$(3-3b^2)\sin(bx)+3a\cos(2x)=6\cos(2x)$$ If this is the problem, then you can plug in whatever value of $x$; for instance, when $x=0$, we get $$(3-3b^2)\sin(b\cdot0)+3a\cos(2\cdot0)=6\cos(2\cdot0)$$ that becomes $$3a=6$$ so $a=2$. With this information we remain with $$(3-3b^2)\sin(bx)=0$$ for all $x$. If $b=0$, the equality surely holds. If $b^2\ne1$ and $b\ne0$, there surely are values of $x$ such that $\sin(bx)\ne0$, for instance $x=\frac{\pi}{2b}$. So we have the equality holding for all $x$ if and only if $b^2=1$ or $b=0$.
The following is multiple choice question (with options) to answer.
If b < 3 and 3x - 3b = 0, which of the following must be true? | [
"x > -3",
"x < 2",
"x = 3",
"x < 3"
] | D | 3x - 3b = 0
Therefore, 3x = 3b
As, b<3
3x < 3(3)
i.e x < 3
Answer : D |
AQUA-RAT | AQUA-RAT-39898 | classical-mechanics, scaling, dimensional-analysis
$Q \propto g^{1/2} (a-d)^{5/2}$
Now, this becomes invalid when $H$ and $D$ are of the same order, but we can get an idea of the time to drain $T$ by saying that as the flow rate does not depend of $H$, $Q T = V = 4 \pi (D/2)^2 H = \pi D^2 H$ so, dropping the $\pi$ we get
$T \propto g^{1/2} \, \frac{(a-d)^{5/2}}{D^2 \, H}$
and thus for everything but the size of sand grains $d$ constant
$\frac{T(d_1)}{T(d_2)} = \left( \frac{a - d_1}{a - d_2} \right)^{5/2}$
The following is multiple choice question (with options) to answer.
Bucket A is 1/4 full of sand, and Bucket B, which has one-half the capacity of Bucket A, is 3/8 full of sand. Jana pours the contents of Bucket B into Bucket A. Bucket A is now filled to what fraction of its capacity? | [
"3/8",
"7/16",
"1/2",
"5/8"
] | B | let capapcity of a = x
sand = x/4
capacity of b = x/2
sand = 3x/16
A total = x/4 + 3x/16 = 7x/16
ANSWER:B |
AQUA-RAT | AQUA-RAT-39899 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good. | [
"125/176",
"125/129",
"125/126",
"125/120"
] | C | Required probability
= 1 - 1/126
= 125/126
Answer:C |
AQUA-RAT | AQUA-RAT-39900 | algorithms, natural-language-processing, counting, mathematical-analysis
Goal is to implement this as a function in a programming language like Python or PHP and to use the (personalised) result for creating an output like "this article can be read in approximately 12-14 minutes".
It would be nice if such information is publicly available. There is a similar question, so I will add a bit over the answer.
There are too many factors independent of language but solely based on individual reading abilities that such estimate would not be possible.
I assume perfect text without spelling errors and unclear grammar.
Giving it a try:
People can read 20, 50, 200, 2000... pages per hour, it depends on reading skills - only prescreening test will give you answer to that: give 2 pages, measure time, or use eye tracker and do the same.
This depends on language used, but the reading speed is measured in native language, so the differences are in the next point as for language (or decrease the speed for non-native by... proficiency at that language).
People use Cohorts imagine that the dictionary forms trie, and the speed of recognizing given word depends on amount of words with the same prefix.
Cognitive Psychology: A Student's Handbook, fifth edition by Michael W. Eysenck,Mark T. Keane, page 347.
For example: "Av..." What are possible words?
Average, Avalanche... Avengers if someone seen the movie, comics or ads.
"Ave.." - word on it's own, Average fits, Avengers also.
If person reads about statistics, "average" is already good choice, about Roman Empire - Ave it is, about movies played - Avengers.
But reading continues to the next syllabe.
It depends on mood, recent readings, coherence of text, familiarity with topic, recent topics seen (not only read).
About topic: reading novel is faster than technical text. Technical text for someone proficient will not decrease reading speed.
So the task given is important, if someone is checked for understanding text, reading speed decreases.
The language itself is not a factor, you should not take mean reading speed but median of your target group.
Yes age is factor, but not that usable as it gives fraction of other factors.
With dyslexia the decrease also vary per person, $30%$ decrease is very optimistic case.
The following is multiple choice question (with options) to answer.
Varma can read a book in k minutes. What part of the book can he read in 8 minutes?(k>8) | [
"6/k",
"8/k",
"83k",
"2/k"
] | B | Part of the book he can read in 1 minute = 1/k
Part of the book he can read in 8 minutes =8/k.
Answer: B |
AQUA-RAT | AQUA-RAT-39901 | # 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 began business with Rs.27000 and was joined afterwards by B with Rs.36000. When did B join if the profits at the end of the year were divided in the ratio of 2:1? | [
"1.5",
"6.5",
"7.5",
"8"
] | C | 27*12 : 36*x = 2: 1
x = 4.5
12 -4.5 = 7.5
Answer: C |
AQUA-RAT | AQUA-RAT-39902 | ============================
Rewrite the equation as $(a+b)(a+c)=4bc$. Now \begin{eqnarray*} a+b= 2 \alpha \beta \\ a+c = 2 \gamma \delta \\ b= \alpha \gamma \\ c= \beta \delta \end{eqnarray*} will satisfy this provided $\alpha ( 2 \beta - \gamma)=a= \delta(2 \gamma - \beta)$. So choose \begin{eqnarray*} \alpha=2 \gamma - \beta \\ \delta= 2 \beta - \gamma \end{eqnarray*} and so we have faimily of solutions generated by \begin{eqnarray*} a= (2 \gamma - \beta)( 2 \beta - \gamma) \\ b= \gamma (2 \gamma - \beta)\\ c= \beta ( 2 \beta - \gamma).\\ \end{eqnarray*}
$$c=\frac{a^2+ab}{3b-a}$$
The problem then becomes: for which $a,b$ do we have $3b-a|a(a+b)$.
Let $d= gcd(a,b)$ then $a=da', b=db'$ with $gcd(a',b')=1$. Then
$$3b'-a'|da'(a'+b')$$
Now, let us observe that $$gcd(3b'-a', a')|3gcd(a',b')=3$$ and $$gcd(3b'-a', a'+b')|4gcd(a',b')=4 \,.$$
The following is multiple choice question (with options) to answer.
If a × b = 2a - 3b + ab, then 3 × 4 + 4 × 3 is equal to : | [
"17",
"23",
"24",
"25"
] | A | Explanation:
3 × 4 + 4 × 3 = (2 × 3 - 3 × 4 + 3 × 4) + ( 2 × 4 - 3 × 3 + 4 × 3)
= (6 + 8 - 9 + 12) = 17.
Answer: A |
AQUA-RAT | AQUA-RAT-39903 | 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 crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train? | [
"150m",
"278m",
"286m",
"121m"
] | A | Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
Answer: A |
AQUA-RAT | AQUA-RAT-39904 | # Probability a 9-digit number has the digits 2,4, and 6 next to each other.
The integers $1,2,3,....,9$ are arraned (at random) in a row, resulting in a $9$-digit integer (without replacement). What is the probability that:
The result is even? $\frac49$ or $\frac{4(8!)}{9!}$
The result is divisible by $5$? The number must end in $0$ or $5$. Edit: As Andre pointed out we have no $0$. So, $\frac19$. or $\frac{(8!)}{9!}$
The digits $2, 4,$ and $6$ are next to each other (in any order)? The above two I have confidence in, it is this last one I'm a little confused on.
$9\choose 3$ ways to position $2,4,6$ in the 9-digit number and $3!$ ways they can be arranged. This doesn't appear to be right as after you divide by $9!$ you get $.13$% which seems unreasonably low. So, I thought to multiple by $6!$ to account for the number of ways the other $6$ numbers can be arranged. This gives you:
$9\choose 3$$3!6!/9!$
which equals $1$, and obviously isn't right. Any suggestions as to where I went wrong would be great.
-
You're right! Silly me. – Vincent Jun 16 '14 at 22:47
For $2,4,6$ next to each other, imagine choosing the $3$ spots that will be reserved for our three favoured guests. This can be done in $\binom{9}{3}$ equally likely ways.
The three spots can be in a row in $7$ ways, for the leftmost of the spots can be in any of the positions $1$ to $7$. Thus the required probability is $\frac{7}{\binom{9}{3}}$. This simplifies nicely.
The following is multiple choice question (with options) to answer.
John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? | [
"15/16",
"11/16",
"11/12",
"1/2"
] | B | XXXX(1}{1}{1}
X can be a prime number or non prime number
Prime number:2,3,5,7
Non prime-4,6,8,9(1 is already taken)
Probability(X is a Prime number)=4/8=1/2
Probability(X is a non-prime number)=4/8=1/2
So,Atleast 2 prime numbers=P P NP NP=4!/2!2!*1/2*1/2*1/2*1/2=6/16
Three prime numbers and one Non prime=P P P NP=4!/3!*1//2^4=4/16
All prime=PPPP=1/16
Total=11/16.
Answer:B |
AQUA-RAT | AQUA-RAT-39905 | java
Title: Beginner question in Java I am just starting to study Java and was wondering why the following code
if(x>2) x*=2;
if(x>4)x=0;
System.out.println(x);
returns 0 for any value of x greater than 2. For example, when x=3, why would 6 not be returned? Let me rewrite your program in pseudocode:
If $x > 2$ then $x \gets 2x$.
If $x > 4$ then $x \gets 0$.
Print $x$.
Let us denote the value of $x$ at the beginning of line $i$ by $x_i$. If $x_1 > 2$ then the condition $x > 2$ in line 1 holds, and so $x_2 = 2x_1$. But then $x_2 = 2x_1 > 2 \cdot 2 = 4$, and so the condition $x > 4$ in line 2 holds. As a consequence, $x_3 = 0$, and so line 3 prints 0.
The following is multiple choice question (with options) to answer.
Which of the following could be the value of x, if |4x – 2| = 10? | [
"-3",
"-2",
"1",
"2"
] | B | |4x – 2| = 10
=> 4x - 2 = 10
or 4x - 2 = -10
=> 4x = 12 or 4x = -8
=> x = 3 or x = -2
Answer B |
AQUA-RAT | AQUA-RAT-39906 | # 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 having a length 6 m and breadth 3 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is: | [
"100 kg",
"120 kg",
"89 Kg",
"180 Kg"
] | D | Volume of water displaced = (6 x 3 x 0.01) m3
= 0.18 m3.
Mass of man = Volume of water displaced x Density of water
= (0.18 x 1000) kg
= 180 kg.
Answer: D |
AQUA-RAT | AQUA-RAT-39907 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Walking with 4/5 of my usual speed, I miss the bus by 5 minutes. What is my usual time? | [
"30 min",
"20 min",
"25 min",
"40 min"
] | B | Speed Ratio = 1:4/5 = 5:4Time Ratio = 4:51 -------- 5 4 --------- ? 20
ANSWER B |
AQUA-RAT | AQUA-RAT-39908 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price? | [
"1000",
"3329",
"2776",
"2988"
] | A | 90%
104%
--------
14% ---- 140
100% ---- ? => Rs.1000
Answer: A |
AQUA-RAT | AQUA-RAT-39909 | - 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 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
Twelve years ago, P was half of Q in age. If the ratio of their present ages is 3:4, what will be the total of their present ages | [
"A)35",
"B)34",
"C)42",
"D)25"
] | C | Explanation:
Let the present age of P and Q be 3x and 4x respectively.
Twelve years ago, P was half of Q in age
=> 2(3x – 12) = (4x – 12)
=> 6x – 24 = 4x – 12
=> 2x = 12
=> x = 6
7x = 7*6 = 42
Answer: Option C |
AQUA-RAT | AQUA-RAT-39910 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Status: QA & VA Forum Moderator
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average of five consecutive odd numbers A, B, C, D and E is 25. What percent of E is C ? | [
"86.2",
"87.5",
"56.8",
"88.9"
] | A | Explanation :
In such a case the middle number (C) is the average
∴ C = 25 and F = 29
Required percentage = 25/29 x 100 = 86.2
Answer : Option A |
AQUA-RAT | AQUA-RAT-39911 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio 3 : 5 expressed as a percent equals | [
"60%",
"40%",
"80%",
"125%"
] | A | Solution
3 : 5 = 3/ 5
= (3 / 5 x 100)%.
= 60 %.
Answer A |
AQUA-RAT | AQUA-RAT-39912 | However, if the phrase "randomly tidied up" instead meant "choose a random arrangement of books", then you get a different answer.
• There are $5$ ways where they are all on the same shelf.
• There are $5\cdot 4=20$ ways where there are two on shelf, and the other on a different shelf.
• Three are $\binom{5}{3}=10$ ways where they are all on different shelves.
Assuming each of these arrangements is equally likely, the desired probability is $\frac{5}{35}=\frac17$.
• where did you come up with $35$ – Educ Jan 17 '16 at 18:50
• $5+10+20{}{}{}{}{}$ – Jorge Fernández Hidalgo Jan 17 '16 at 18:52
• @dREaM Thanks again – Educ Jan 17 '16 at 18:53
The following is multiple choice question (with options) to answer.
10 books are placed at random in a shelf. The probability that a pair of books will always be together is - | [
"1/10",
"9/10",
"1/5",
"3/10"
] | C | Explanation:
10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!
Required probability = 1/5
ANSWER IS C |
AQUA-RAT | AQUA-RAT-39913 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains start at same time from two stations and proceed towards each other at the rate of 20 km/hr and 25 km/hr respectively. When they meet, it is found that one train has traveled 50 km more than the other. What is the distance between the two stations? | [
"540 km",
"767 km",
"276 km",
"450 km"
] | D | Explanation:
Let us assume that trains meet after 'x' hours
Distance = speed * Time
Distance traveled by two trains = 20x km and 25x km resp.
As one train travels 50 km more than the other,
25x – 20x =50
5x = 50
x = 10 hours
As the two trains are moving towards each other, relative speed = 20 + 25 = 45 km/hr
Therefore, total distance = 45*10 = 450 km.
ANSWER: D |
AQUA-RAT | AQUA-RAT-39914 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
If Josh, Doug, and Brad have a total of $77 between them, and Josh has three times as much money as Brad but same as much as Doug, how much money does Doug have? | [
"$8",
"$9",
"$27",
"$33"
] | D | Josh + Doug + Brad = 77 ; Joush = 3 brad, Josh = Doug
Josh +1/3 Josh+ Josh = 77 (substituted the given values)
Josh = 33. 33=Doug =33
Answer is D. |
AQUA-RAT | AQUA-RAT-39915 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is | [
"14 years",
"18 years",
"20 years",
"22 years"
] | D | Solution
Let the son's present age be x years .Then, man's present age = (x+24)years.
Then ‹=›(x+24)+2= 2(x+2)
‹=›x+26=2x+4
x= 22.
Answer D |
AQUA-RAT | AQUA-RAT-39916 | Shanonhaliwell April 8th, 2018 03:31 PM
Quote:
Originally Posted by romsek (Post 591335) outstanding, you seem to be getting the hang of things.
Thanks to you, I was able to do it.
All times are GMT -8. The time now is 12:30 AM.
The following is multiple choice question (with options) to answer.
8 : 4 seconds :: ? : 4 minutes | [
"10",
"8",
"20",
"25"
] | B | 8*4=4*x
x=8
ANSWER:B |
AQUA-RAT | AQUA-RAT-39917 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B? | [
"5000",
"3677",
"2977",
"2678"
] | A | Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Answer: A |
AQUA-RAT | AQUA-RAT-39918 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 14 cm, then its perimeter is | [
"42",
"48",
"52",
"56"
] | B | Side of the square = 12 cm. Area of the rectangle = [(12 x 12) – 4]
cm2 = 140 cm2
Now, area = 140 cm², length = 14 cm.
Breadth = A/L = 140/14 cm. = 10 cm.
Hence, Perimeter = 2 (l + b) = 2 (14 + 10) cm = 48 cm.
Answer :B |
AQUA-RAT | AQUA-RAT-39919 | Alternatively, the lcm of 54 and 72 can be found using the prime factorization of 54 and 72:
• The prime factorization of 54 is: 2 x 3 x 3 x 3
• The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(54,54) = 216
In any case, the easiest way to compute the lcm of two numbers like 54 and 72 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 54,72. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 54 and 72
What is the least common multiple of 54 and 72 used for? Answer: It is helpful for adding and subtracting fractions like 1/54 and 1/72. Just multiply the dividends and divisors by 4 and 3, respectively, such that the divisors have the value of 216, the lcm of 54 and 72.
$\frac{1}{54} + \frac{1}{72} = \frac{4}{216} + \frac{3}{216} = \frac{7}{216}$. $\hspace{30px}\frac{1}{54} – \frac{1}{72} = \frac{4}{216} – \frac{3}{216} = \frac{1}{216}$.
## Properties of LCM of 54 and 72
The most important properties of the lcm(54,72) are:
• Commutative property: lcm(54,72) = lcm(72,54)
• Associative property: lcm(54,72,n) = lcm(lcm(72,54),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The following is multiple choice question (with options) to answer.
The LCM and HCF of two numbers are 8 and 48 respectively. If one of them is 24, Find the other ? | [
"12",
"14",
"15",
"16"
] | D | HCF x LCM = Product of Numbers
8 x 48 = 24 x the other number
other number = (8 x 48) / 24
other number = 16
ANSWER : D |
AQUA-RAT | AQUA-RAT-39920 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
In March, Kurt ran an average of 1.5 miles an hour. If by June he had increased his pace by 20 seconds per mile, then which of the following expresses the number of hours it would take Kurt to complete one mile in June? | [
"2380/60^2",
"2410/60^2",
"2390/60^2",
"3586/60"
] | A | Kurt ran at an average of 1.5 miles/hour in march. So to run 1 mile he would take 1/1.5 hours = (60 * 60)/1.5 seconds = 2400 seconds.
If he increases his speed by 20 seconds, he will complete a mile in 2380 seconds. Converting in hours = 2380/(60 * 60) = 2380/(60^2)
ANSWER: A |
AQUA-RAT | AQUA-RAT-39921 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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Posts: 2898
Joined: 07 Sep 2017
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
### GMAT/MBA Expert
GMAT Instructor
Posts: 16162
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770
### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
8 people decided to split the restaurant bill evenly. If the bill was $214.15 dollars, how much money did they 1 cent is the smallest unit? | [
"$214.16",
"$214.17",
"$214.18",
"$214.19"
] | A | If the last three digits of a whole number are divisible by 8, then the entire number is divisible by 8
the last 3 digit 415 not divisible by a
Hence, we need to add 1 to this number for it to be divisible by 8
Correct Option: A |
AQUA-RAT | AQUA-RAT-39922 | the number of committees that exist where the man and women serve together is given by,
$$\begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$
so the total number of committees in this case amounts to,
$$1120 - 210 = 910$$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
The following is multiple choice question (with options) to answer.
In how many ways a committee, Consisting of 2 men and 6 women can be formed from 8 men and 10 women? | [
"2660",
"5880",
"11760",
"5040"
] | B | To choose Required number of ways to form a committee:
Select 2 men out of 8 men and 6 women out of 10 women.
That is 8C2 * 10C6.
8C2=8!/2!*6!=28
10C6=10!/6!*4!
=210
The Required number of ways=210*28=5880
B |
AQUA-RAT | AQUA-RAT-39923 | -
(+1) Well, clearly $\theta=\pi$ makes it the smallest, but that's nonsense. ;-) – Cameron Buie Apr 17 '13 at 15:37
Although I can see your point and thank you for that (+1) these seem like a long workaround. But thanks. – Randolf R-F Apr 17 '13 at 15:39
We minimize the square of the norm of $w$: $x^2+y^2$ with the constraint $x+2y=5$ $$f(y)=x^2+y^2=(5-2y)^2+y^2=5y^2-20y+25$$ so $$f'(y)=10y-20=0\iff y=2$$ hence $w=(1,2)$
The following is multiple choice question (with options) to answer.
what is the least number. which should be added to 0.0320 to make it a perfect square? | [
"0.0005",
"0.0004",
"0.0006",
"0.0002"
] | B | 0.0320+0.0004=0.0324
(0.18)^2
ANSWER:B |
AQUA-RAT | AQUA-RAT-39924 | Define a sequence of the number of water $$x_0=1$$ Now you take out 3/4 $$x_1=1/4$$ Now you take out another 3/4 but only out of the 1/4 so you leave only 1/4 of the water $$x_2=1/4*1/4$$ You continue that way, so in the n-th day you have $$x_n=(1/4)^n$$ If you want percentages so multiply by 100.
• Here is another explanation: you take 3/4 of the water, only 1/4 out of the 3/4 is old water, so you take out 3/16 old water. Therefore the amount of old water is $1/4-3/16=1/16$ – Yanko Jun 10 '17 at 15:57
The following is multiple choice question (with options) to answer.
If 16% of 40% of a number is 3, then the number is | [
"200",
"225",
"46.875",
"320"
] | C | Explanation:
Let 16/100×40/100×a=3
a = 3×100×100/16×40=46.875
Correct Option: C |
AQUA-RAT | AQUA-RAT-39925 | Then $-x<0$. We thus have
$$a>b\Rightarrow -ax<-bx\Rightarrow ab-ax<ab-bx\Rightarrow a(b-x)<b(a-x)$$
$$\Rightarrow \frac{a}{b}<\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}>% \frac{a}{b}.$$
For the second question. If $x>0$, we have:
$$a<b\Rightarrow ax<bx\Rightarrow ab+ax<ab+bx\Rightarrow a(b+x)<b(a+x)$$
$$\Rightarrow \frac{a}{b}<\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}>% \frac{a}{b}.$$
Then $-x<0$. We thus have
$$a<b\Rightarrow -ax>-bx\Rightarrow ab-ax>ab-bx\Rightarrow a(b-x)>b(a-x)$$
$$\Rightarrow \frac{a}{b}>\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}<% \frac{a}{b}.$$
-
Did you mean in your first implication, to write $a > b \rightarrow ax > bx$? – amWhy Jun 18 '11 at 20:57
@amWhy: Thanks! corrected. – Américo Tavares Jun 18 '11 at 21:07
The following is multiple choice question (with options) to answer.
If a > 0 and b < 0, which of the following are true?
I. a/b < 0
II. b/a < 0
III. a + b < 0 | [
"I only",
"II only",
"III only",
"I and II only"
] | D | I. True, a is positive and b is negative so a/b must be negative
II. True, a is positive and b is negative so b/a must be negative
III. Not true, example: a = 5, b = -3, so a + b = 2
D |
AQUA-RAT | AQUA-RAT-39926 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years? | [
"2 : 3 : 4",
"3 : 2 : 4",
"3: 2 : 1",
"3: 4 : 7"
] | D | Sol.
A : B : C = (40000 * 36) : (80000 * 12 + 40000 * 24)
: (120000 * 24 + 40000 * 12) =144 : 192 : 336 = 3: 4 : 7
Answer D |
AQUA-RAT | AQUA-RAT-39927 | metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of the sector is. - 1133971 2:15 75.0k LIKES. Dec 15,2020 - If the perimeter of a sector of a circle of radius 5.2 cm. Let the angle subtended by the arc of the sector at the centre of the circle (radius assumed to be $r$) be $\theta$. What multiple of the radius is the area of the sector?a)5thb)3rdc)4thd)2ndCorrect answer is option 'B'. (a) Write down, in terms of r and θ, expressions for P and A. A sector of a circle is like a slice of pizza or pie. Units Metric units. In the Given Figure, Radius of Circle is 3.4 Cm and Perimeter of Sector P-abc is 12.8 Cm . Thus the perimeter of the sector is r+θr+r. Problem 6 : Find the length of arc, if the perimeter of sector is 45 cm and radius is 10 cm. Or P = d + L ( arc) => P = 2r + {(β*2 pi r)/360°} , where β is sector angle. = π* 25 * 42 /180+2* 25. thus the perimeter of the sector is L+2r units. Sector of a circle ↺ Theta is an angle that can be defined as the figure formed by two rays meeting at a common endpoint. The most common sector of a circle is a semi-circle which represents half of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. Perimeter of a Circle Sector A region bounded by two radii and an arc is called a sector of a circle. 15.8k SHARES. Click hereto get an answer to your question ️ The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Working with ratios (one question) Angles in quadrilateral (one question) We are learning about: The area and perimeter of a sector. If and the area of the
The following is multiple choice question (with options) to answer.
The area of sector of a circle whose radius is 12 metre and whose angle at the center is 42° is? | [
"52.7 m2",
"57.8 m2",
"52.8 metre sq",
"72.8 m2"
] | C | 42/360 * 22/7 * 12 * 12
= 52.8 m2
Answer: C |
AQUA-RAT | AQUA-RAT-39928 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 1st 3 of 4 numbers is 6 and of the last 3 are 5. If the sum of the first and the last number is 11. What is the last numbers? | [
"4",
"5",
"6",
"7"
] | A | A + B + C = 18
B + C + D = 15
A + D = 11
A – D = 3
A + D = 11
2D = 8
D = 4
ANSWER:A |
AQUA-RAT | AQUA-RAT-39929 | • Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
The following is multiple choice question (with options) to answer.
In a box of 12 pens, a total of 4 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective? | [
"1/6",
"2/33",
"14/33",
"9/16"
] | C | Method- 1
There are 9 fine pieces of pen and 4 defective in a lot of 12 pens
i.e. Probability of first pen NOT being defective = (8/12)
i.e. Probability of Second pen NOT being defective = (7/11) [11 pen remaining with 8 defective remaining considering that first was defective]
Probability of Both pen being NON-defective = (8/12)*(7/11) = 14/33
Answer: option C |
AQUA-RAT | AQUA-RAT-39930 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Insert the missing number
1, 8, 27, 64, 125, 216, (....) | [
"554",
"343",
"354",
"392"
] | B | Series is 1^3,2^3,3^3,4^3,5^3,6^3,7^3
7^3=343
ANSWER:B |
AQUA-RAT | AQUA-RAT-39931 | 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.
If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the
stream is | [
"5 km/hr",
"4 km/hr",
"2 km/hr",
"1 km/hr"
] | D | Explanation:
Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph
Answer: D |
AQUA-RAT | AQUA-RAT-39932 | python, python-3.x, datetime, reinventing-the-wheel
Furthermore, you can make is_leap_year a one-liner, since the if-else blocks and zero-comparisons are a bit verbose for my taste of "pythonicness":
def is_leap_year(year):
return not year % 4 and (year % 100 or not year % 400)
Also I find it a bit intricately, that you pass a tuple date to find_day and then access its members via indexes.
I suggest changing that to:
def find_day(month, day, year):
total = calc_month(month, year) + calc_day(day) + calc_year(year)
if total < 0:
return day_names[total % -7]
else:
return day_names[total % 7]
def main():
date = input("Enter a day like so <MM, DD, YYYY>: ").split()
month = int(date[0])
day = int(date[1])
year = int(date[2])
if month > 12 or month <= 0:
print("Invalid month")
elif day > month_days[month - 1] or day <= 0:
if not (day == 29 and month == 2):
print("Invalid day")
else:
print(find_day(month, day, year))
However, if you really want to handle date as such a tuple for later re-use, I recommend that you use tuple unpacking in find_day instead of index accessing for better readability:
def find_day(date):
month, day, year = date
total = calc_month(month, year) + calc_day(day) + calc_year(year)
if total < 0:
return day_names[total % -7]
else:
return day_names[total % 7]
The following is multiple choice question (with options) to answer.
Which of following is a leap year ? | [
"2000",
"2800",
"1800",
"2600"
] | B | The century year which is completely divisible by 400, is a leap year. Thus, the year 2800 is a leap year.
Answer : B |
AQUA-RAT | AQUA-RAT-39933 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
if Juan takes 12 seconds to run y yards, how many seconds will it take him to run x yards at the same rate? | [
"11x/y",
"11y/x",
"x/11y",
"12x/y"
] | D | Answer is D
The rate of running hards: 12 minutes per y hards.
Thus the time for running x hards = x hards * 12 minutes/y hards = 12x/y |
AQUA-RAT | AQUA-RAT-39934 | $$p \mid 3q$$ and $$3q \mid r$$.
pm = 3q, $$m \in \mathbb{Z}$$
3qn = r, $$n \in \mathbb{Z}$$
$$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$
$$m(1+n) \in \mathbb{Z}$$
So $$p | (3q + r)$$
### Poor style 2
Here's a more subtle example of poor style. It does have connector words, but the author has forgotten everything he learned in high school English class. Paragraphs, punctuation, and whitespace are critical to making technical arguments readable. Everyone makes occasional mistakes, and you'll make a few extras if your first language isn't English. But all of you can do a better job than this.
let p, q, and r be integers (p non-zero) suppose that $$p \mid 3q$$ and $$3q \mid r$$.
by definiton divides, $$p | 3q$$ means pm = 3q, m integer. $$3q | r$$ means 3qn = r for some integer n. now substitute these two equations into the expression 3q+r $$3q + r = 3q + 3qn = 3q(1 + n) = pm(1+n)$$ We know m(1+n) is an integer, since m and n are integers solet s=m(1+n) then 3q+r = ps therefore by the definiton of divides $$p | (3q + r)$$ QED
The following is multiple choice question (with options) to answer.
If QR = 3(Q+ 1) + R and Q and R are integers,Q could be any of the following values EXCEPT: | [
"2",
"3",
"12",
"4"
] | C | QR = 3(Q + 1) + R
QR - R = 3(Q + 1)
R = 3(Q + 1)/(Q-1)
R won't be an integer only if Q = 12 (from the options).
Answer: C |
AQUA-RAT | AQUA-RAT-39935 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Present ages of X and Y are in the ratio 5:6 respectively. Seven years hence this ratio will become 6:7 respectively. What is X's present age in years? | [
"35 years",
"17 years",
"77 years",
"88 years"
] | A | Let the present ages of X and Y be 5x and 6x years respectively.
Then, (5x + 7)/(6x + 7) = 6/7
7(5x + 7) = 6(6x + 7) => x = 7
X's present age = 5x = 35 years.
Answer:A |
AQUA-RAT | AQUA-RAT-39936 | (which of course we knew at the beginning)
1. 👍
2. 👎
3. There are 4 triangles
1. 👍
2. 👎
4. All four right triangles are 30-60-90 triangles. Therefore, the length of the shorter leg in each triangle is half the hypotenuse, and the length of the longer leg is sqrt3 times the length of the shorter leg. We apply these facts to each triangle, starting with triangle AOB and working clockwise.
From $\triangle AOB$, we find $AB = AO/2 = 4$ and $BO = AB\sqrt{3}=4\sqrt{3}$.
From $\triangle BOC$, we find $BC = BO/2 =2\sqrt{3}$ and $CO = BC\sqrt{3} =2\sqrt{3}\cdot\sqrt{3} = 6$.
From $\triangle COD$, we find $CD = CO/2 = 3$ and $DO = CD\sqrt{3} = 3\sqrt{3}$.
From $\triangle DOE$, we find $DE = DO/2 = 3\sqrt{3}/2$ and $EO =DE\sqrt{3} = (3\sqrt{3}/2)\cdot \sqrt{3} = (3\sqrt{3}\cdot \sqrt{3})/2 = \boxed{9/2}$
1. 👍
2. 👎
1. 👍
2. 👎
## Similar Questions
1. ### geometry
Find the value of x. Round the length to the nearest tenth. Diagram is not drawn to scale. A right triangle is drawn with right angle between a right vertical leg and lower horizontal leg. A dashed line is drawn horizontally from
2. ### Geometry
Find the value of x. Round to the nearest tenth of a unit. A right triangle. The hypotenuse is 580 yards. The opposite leg (not the base) is x. The exterior degree on the top of the triangle is 27. 263.3 yd 295.5 yd 516.8 yd
3. ### math
The following is multiple choice question (with options) to answer.
If 10 and 15 are the lengths of two sides of a triangular region, which of the following can be the length of the third side?
I. 5
II. 15
III. 20 | [
"I only",
"II only",
"I and II only",
"II and III only"
] | D | The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.
Let x be the length of the third side.
15-10 < x < 15+10
5 < x < 25
The answer is D. |
AQUA-RAT | AQUA-RAT-39937 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy? | [
"343",
"147",
"189",
"063"
] | C | •P(Buy) = 3/10, P(No Buy) = 7/10
•2 Yes, 1 No = 3/10*3/10*7/10 = 63/1000
•3 Possibilities = YYN + YNY + NYY = 3(63/1000) = 189/1000 = .189%
Answer is C. |
AQUA-RAT | AQUA-RAT-39938 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
Jane bought some pairs of pants on sale for 15% off the original price and received an additional 20% discount for buying three pairs. If one of the pairs of pants cost $10 after both discounts, what was the original price of that pair? | [
"$32.26",
"$20.25",
"$20",
"$14.71"
] | D | Let f be the original price. Then the first discounted price, g, is given by
g = f - 15%f
and the second discounted price, h, is given by
h = g - 20%g
With h = $10, solve for f
$10 = (f - 15%f) - 20%(f - 15%f)
f = $14.71
Correct answer is D |
AQUA-RAT | AQUA-RAT-39939 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Ram sold two bicycles, each for Rs.990. If he made 10% profit on the first and 10% loss on the second, what is the total cost of both bicycles? | [
"Rs.2000",
"Rs.2028",
"Rs.2020",
"Rs.2720"
] | A | (10*10)/100 = 1%loss
100 --- 99
? --- 1980 => Rs.2000
Answer:A |
AQUA-RAT | AQUA-RAT-39940 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In an engineering class that contained 75 students, the final exam consisted of 2 questions. Three fifths of the students answered the first question correctly. If four fifths of the remainder answered the second question correctly, how many students answered both questions incorrectly? | [
"4",
"6",
"10",
"12"
] | B | 1st Question : Three fifths of the students answered the first question correctly - so 3/5∗75=45
2nd Question : four fifths of the remainder answered the second question correctly so 4/5∗(75−45)=4/5∗30=24
total students who answered at least one Q correct =45+24=69
so students who answered BOTH incorrectly = 75−69=6
ANSWER:B |
AQUA-RAT | AQUA-RAT-39941 | Of course not! It is actually due to the fact that $102 = 6 \cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \cdot 17$, then $391 = 17 \cdot (17 + 6) = 17 \cdot 23$, and $493 = 17 \cdot (23 + 6) = 17 \cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.
This sequence of multiples of 17 starts from $17 \cdot 5$, and if we continue it further we see that it contains several more numbers from our exceptional set as well:
$\begin{array}{rcl}85 &=& 17 \cdot 5 \\ 187 &=& 17 \cdot 11 \\ \mathbf{289} &=& 17 \cdot 17 \\ \mathbf{391} &=& 17 \cdot 23 \\ \mathbf{493} &=& 17 \cdot 29 \\ 595 &=& 17 \cdot 35 \\ \mathbf{697} &=& 17 \cdot 41 \\ \mathbf{799} &=& 17 \cdot 47 \\ \mathbf{901} &=& 17 \cdot 53 \end{array}$
What about if we start with $17 \cdot 7$ and keep adding $102$?
The following is multiple choice question (with options) to answer.
What is the total number of prime numbers less than 70? | [
"17",
"18",
"19",
"20"
] | C | Solution
Prime numbers less than 70 are =12.
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,and 67
Their number is 19.
Answer C |
AQUA-RAT | AQUA-RAT-39942 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
What amount does Kiran get if he invests Rs.5600 at 9% p.a. compound interest for two years, compounding done annually? | [
"6653",
"6500",
"6400",
"5678"
] | A | A= P{1 + R/100}n
=> 5600{1 + 9/100}2
= Rs.6653
Answer: A |
AQUA-RAT | AQUA-RAT-39943 | newtonian-mechanics, drag, relative-motion
Title: Relative motion-Acceleration My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.
Say a boat travels at 10 m/s in a water channel.
the water speed relative to ground is 0.
so the boat travels at 10 m/s relative to the ground.
now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
As after a while the boat speed relative to ground has become 0,
then from the ground-based observer's point of view, the boat has undergone a deceleration.
My question is;
Is this deceleration always necessarily equal to minus the water acceleration?
In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?
PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?
My question is; Is this deceleration always necessarily equal to minus the water acceleration?
The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:
$$f=kv\qquad \text{ for low speed}\\
f=kv^2\qquad \text{ for high speed}$$
where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.
The following is multiple choice question (with options) to answer.
A boat can move upstream at 20 kmph and downstream at 35 kmph, then the speed of the current is? | [
"5 kmph",
"7.5 kmph",
"8 kmph",
"9 kmph"
] | B | US = 20
DS = 35
M = (35 - 20)/2 = 7.5
Answer:B |
AQUA-RAT | AQUA-RAT-39944 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
How many times in a day, are the hands of a clock in straight line but opposite in direction? | [
"20",
"22",
"24",
"48"
] | B | The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o'clcok only).
So, in a day, the hands point in the opposite directions 22 times. Answer: Option B |
AQUA-RAT | AQUA-RAT-39945 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
Four car rental agencies A, B, C and D rented a plot for parking their cars during the night. A parked 15 cars for 12 days, B parked 12 cars for 20 days, C parked 18 cars for 18 days and D parked 16 cars for 15 days. If A paid Rs. 1125 as rent for parking his cars, what is the total rent paid by all the four agencies? | [
"Rs.6158",
"Rs.6120",
"Rs.6119",
"Rs.6150"
] | D | The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1125
45k = 1125 => k = 25
246k = 246(25) = Rs. 6150
Thus the total rent paid by all the four agencies is Rs.6150.
Answer:D |
AQUA-RAT | AQUA-RAT-39946 | general-relativity, time, equivalence-principle
the one near the earth will have a greater acceleration than the one on the mountain
Yes; as shown in the indicated answer, the two participants which remain rigid to each other in a flat region move at ("slightly") different constant proper accelerations, too;
namely "the top" moving with proper acceleration $ k \, e^{(\frac{-k}{c^2} L )} $,
compared to "the bottom" moving with proper acceleration $k$,
where $\frac{2 L}{c}$ is the constant ping duration of "the bottom" to "the top" and back.
The following is multiple choice question (with options) to answer.
Two persons start from two places 160 km apart at the same time with velocities 50 km/hr and 30 km/hr. move in opposite direction. Where they meet ? | [
"90 km from A",
"50 km from B",
"100 km from A",
"70 km from B"
] | C | Explanation :
Solution: d = 160, v1 = 50, v2 = 30
relative speed vr = v1 + v2 = 50 + 30 = 80.
the distance covered by them will be indirectly proportional to their distance
of the persons of meeting from A = 160*50/(50+30) km
= 160*5/8 = 100 km.
Answer : C |
AQUA-RAT | AQUA-RAT-39947 | heat-transfer, energy-storage
Title: How do you use an insulation jacket's R-value for calculating heat transfer from a hot water tank? I know how to use thermal resistance circuits to calculate heat transfer in Cartesian and cylindrical coordinates, but I don't know how to deal with real-world insulation values, and I was not able to find much about this online.
For example, I have this hot water tank that has an insulation jacket with 2" R-12.5 insulation. For simplicity, let's assume the insulation is the only thing that separates the hot water from cold outside air (no heat transfer from bottom of tank, top of tank, etc. Just through the insulation jacket). Assume water tank thickness aside from the 2" insulation jacket is negligible.
So it goes hot water -> insulation -> cold air outside. I can look up a free convection coefficient for the tank's hot water and can look up a convection coefficient for air as well. That allows me to use the formula
$$
R=1/(h*2*pi*r*heightofinsulationjacket)
$$
twice (once for water and once for air).
So the problem is how to deal with the insulation jacket. How do I interpret 2" R-12.5 insulation? How do I factor in the thickness of the insulation? The surface area is taken care of in the heat transfer equation.
$$
Q''=(Tf-Ti)/R
$$
or
$$
Q=(Tf-Ti)/R*A
$$
Updated: How do I use an R-value in cylindrical coordinates?
e.g.
$$
R/A=1/(hW*2*pi*r1)+Rins/(whatAreaHere)+1/(hA*2*pi*r2)
$$
I am missing the area for my R-value insulation The insulation R-value already accounts for thickness, so you can ignore it. You just have to account for the effect of surface area and temperature difference.
The SI units of R are $\frac{K \cdot m^2}{W}$
Heat transfer increases as the surface area increases, so Q is calculated as:
$$Q_{insulation}=\frac{A (T_f-T_i)}{R}$$ and
The following is multiple choice question (with options) to answer.
A rectangular tank needs to be coated with insulation. The tank has dimensions of 3 feet, 5 feet, and 2 feet. Each square foot of insulation costs $20. How much will it cost to cover the surface of the tank with insulation? | [
"$1100",
"$1240",
"$1360",
"$1480"
] | B | The total surface area is 2(2*3 + 3*5 + 2*5) = 62 square feet
The total cost is 62*$20 = $1240
The answer is B. |
AQUA-RAT | AQUA-RAT-39948 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If A^2 + B^2 = 10 and AB = 10, what is the value of the expression (A - B)^2 + (A + B)^2? | [
"10",
"20",
"30",
"60"
] | B | (a-b)^2 = a^2 + b^2 - 2ab = 10-20= -10
(a+b)^2 = a^2 + b^2 + 2ab = 10+20=30
so (a+b)^2 + (a-b)^2 = 30 - 10 =20
B |
AQUA-RAT | AQUA-RAT-39949 | Hint: if $$8$$ is the largest single digit factor than doesn't it make sense that the largest such three digit number would be in the $$8$$ hundreds.
Now, that was the hard way. Were there any handy mathematical observations you might have used to make this easier? Would knowing that $$24 = 2^3*3$$ is the unique prime factorization of $$24$$ have helped you find the ways to find three term factorizations? Would knowing there are $$k!$$ ways to arrange $$k$$ objects have helped? What if some of the objects were indistinguishable?
$$xyz= 2^3 .3$$
powers of $$2$$ can be divided among $$x,y,z$$ in
$$^{3+(3-1)} \ C _{3-1}= 10$$ ways
and power of $$3$$ can be divided into $$^{1+(3-1)} \ C _{3-1}= 3$$ ways
thus giving total $$10\times 3 =30$$ways
but here we also counted arrangements like $$2 \times 1\times 12$$ and $$1\times 1\times 24$$ which violates 3 digit number policy so, by subtracting these i.e, $$\left(3! + \dfrac{3!}{2!}\right)=9$$ ways
we get
total number of $$3$$ digit numbers having product $$4 != (30-9)=21$$
and out of all such $$3$$ digit numbers largest is $$8 3 1$$
The following is multiple choice question (with options) to answer.
The largest 4 digit number exactly divisible by 88 is: | [
"4252",
"6287",
"7631",
"9944"
] | D | D
Divide largest four digit number 9999 by 88. You get 113.625.
Obviously 113 would be exactly divisible so we want to know what that number is.
We get this by multiplying 113 with 88 = 9944 |
AQUA-RAT | AQUA-RAT-39950 | 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 of runs of a cricket player of 10 innings was 30. How many runs must he make in his next innings so as to increase his average of runs by 4? | [
"87",
"74",
"10",
"76"
] | B | Average after 11 innings = 34
Required number of runs
= (34 * 11) - (30 * 10) = 374- 300
= 74.
Answer: B |
AQUA-RAT | AQUA-RAT-39951 | The symmetry between $A$ and $B$ tells you that they each have the same chance of winning the tournament. If you can figure out the probability that the tournament is drawn, you can compute the chance that $A$ wins by taking half of the chance the tournament is not drawn. There are fewer cases to consider than you had to work with. This explains the book approach.
In your calculation, the second and third lines from the end are not correct. When $A$ does not win four games, there is no need for him to win the last game. For the next to last line, the chance that $A$ wins two and there are five draws should be $(\frac 13)^7{7 \choose 2}=(\frac 13)^7\frac {7!}{5!2!}$ because $A$ could have won any two. It looks like you have that problem in all the lines that do not result in $4$ wins for $A$.
• Thank you for the answer! This really helped me realize the fault in my method (along with JMoravitz's comment). This reminds me, for the bazillionth time, "check if you are really including all the cases. Try damn hard to find a counterexample, or prove that there is no case missed"! – FreezingFire Jul 15 '16 at 16:52
Approach similarly to yours except allow the teams to continue playing games even after a winner is inevitable.
We have the following possibilities:
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Combinations.
Six contestants representing four different countries advance to the finals of a fencing championship. Assuming all competitors have an equal chance of winning, how many possibilities are there with respect to how a first-place and second-place medal can be awarded? | [
"6",
"7",
"12",
"18"
] | D | Number of ways First-place medal can be awarded to four contestants = 6
Number of ways Second-place medal can be awarded to contestants after awarding First-place medal =3
Therefore number of possibilities = 6 *3 =18
Answer:D |
AQUA-RAT | AQUA-RAT-39952 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A can build a wall in the same time in which B and C together can do it. If A and B together could do it in 25 days and C alone in 35 days, in what time could B alone do it? | [
"275 days",
"178 days",
"185 days",
"175 days"
] | D | Explanation:
No explanation is available for this question!
ANSWER: D |
AQUA-RAT | AQUA-RAT-39953 | "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$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When positive integer n is divided by 3, the remainder is 2. When n is divided by 5, the remainder is 5. How many values less than 100 can n take? | [
" 0",
" 2",
" 3",
" 4"
] | A | a quick approac to this Q is..
the equation we can form is.. 3x+2=7y+5..
3x-3=7y... 3(x-1)=7y...
so (x-1) has to be a multiple of 7 as y then will take values of multiple of 3..
here we can see x can be 1,8,15,22,29 so 5 values till 100 is reached
as (29-1)*3=84 and next multiple of 7 will be 84+21>100..
ans 0.. A |
AQUA-RAT | AQUA-RAT-39954 | Question
# From a two digit number N, we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:
A
N cannot end in 5
B
N can end in any digit other than 5
C
N does not exist
D
there are exactly 7 values for N
E
there are exactly 10 values for N
Solution
## The correct option is D there are exactly 7 values for NLet the two digit number be 10a+bN = 10a+bN' = reversed number = 10b+aN-N' = 9(a-b)N-N' is positive perfect cube.$$\therefore$$ a>bFor 9(a-b) be perfect cube, a-b = 3$$\therefore$$ b $$\epsilon$$ [0, 6] $$\longrightarrow$$ 7 values$$\therefore$$ a $$\epsilon$$ [3, 9] $$\longrightarrow$$ 7 values$$\therefore$$ Total 7 values are possible for N.Mathematics
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The following is multiple choice question (with options) to answer.
the largest four-digit number which is a perfect cube, is | [
"8000",
"9261",
"9999",
"None of these"
] | B | Solution
Clearly, 9261 is a perfect cube satisfying the given property.
Answer B |
AQUA-RAT | AQUA-RAT-39955 | homework-and-exercises, reference-frames, angular-velocity
Title: Reference Frame and Angular Speed Related? I am given the following problem:
If an airplane propeller rotates at 2000 rev/min while the airplane flies at a speed of 480 km/h relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius 1.5m, as seen by (a) the pilot and (b) an observer on the ground? The plane’s velocity is parallel to the propeller’s axis of rotation.
The following is multiple choice question (with options) to answer.
The end of a blade on an airplane propeller is 20 feet from the center. If the propeller spins at the rate of 1,320 revolutions per second, how many miles will the tip of the blade travel in one minute? (1 mile = 5,280 feet) | [
"200π",
"240π",
"600π",
"480π"
] | C | Distance traveled in 1 revolution = 2πr= 2π20/5280
Revolutions in one second= 1320
Revolutions in 60 seconds (one minute)= 1320*60
Total distance traveled= total revolutions *distance traveled in one revolution
1320*60 *2π20/5280= 600π
C is the answer |
AQUA-RAT | AQUA-RAT-39956 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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 which has 390 m long, is running 45 kmph. In what time will it cross a person moving at 9 kmph in same direction ? | [
"26 sec",
"39 sec",
"36 sec",
"29 sec."
] | B | Time taken to cross a moving person = length of train/ relative speed
Time taken = 390/((45-9) (5/18)
= 390/ 36*(5/18)
= 390/10
= 39 sec
ANSWER:B |
AQUA-RAT | AQUA-RAT-39957 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In an engineering class that contained 25 students, the final exam consisted of 2 questions. Three fifths of the students answered the first question correctly. If four fifths of the remainder answered the second question correctly, how many students answered both questions incorrectly? | [
"2",
"6",
"10",
"12"
] | A | 1st Question : Three fifths of the students answered the first question correctly - so 3/5∗25=15
2nd Question : four fifths of the remainder answered the second question correctly so 4/5∗(25−15)=4/5∗10=8
total students who answered at least one Q correct =15+8=23
so students who answered BOTH incorrectly = 25−23=2
ANSWER:A |
AQUA-RAT | AQUA-RAT-39958 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
Presently the ratio between the ages of Dan and James is 6:5. after 4 years Dan will be 28. What is the present age of James? | [
"20",
"19",
"21",
"18"
] | A | Let the present ages Dan and James be 6x years and 5x years respectively
6x+4 = 28
6x = 24
x = 4
Kim's age = 5x = 20 years
Answer is A |
AQUA-RAT | AQUA-RAT-39959 | => 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.
Of the 3,600 employees of Company X, 1/4 are clerical. If the clerical staff were to be reduced by 1/4, what percent of the total number of the remaining employees would then be clerical? | [
" 25%",
" 22.2%",
" 20%",
" 12.5%"
] | C | Welcome, just post the question and the choices
Let's see, the way I did it was
1/4 are clerical out of 3600 so 900 are clerical
900 reduced by 1/3 is 900*1/3
so it reduced 300 people , so there is 600 clerical people left
but since 600 people left, it also reduced from the total of 3600 so there are 3000 people total
since 600 clerical left / 300 people total
you get (A) 20%
Answer : C |
AQUA-RAT | AQUA-RAT-39960 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
The banker's gain on a bill due due 1 year hence at 12% per annum is Rs.8.4. The true discount is | [
"Rs.70",
"Rs.36",
"Rs.54",
"Rs.50"
] | A | Solution
T.D = [B.G x 100 / R x T]
= Rs.(8.4 x 100 / 12 x 1)
= Rs.70.
Answer A |
AQUA-RAT | AQUA-RAT-39961 | => 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.
Working individually, Allison can sew dresses for 9 hours and Al can sew dresses in 12 hours. If Allison and Al work together but independently at the task for 3 hours, at which point Al leaves, how many remaining hours will it take Allison to complete the task alone? | [
"45/12",
"23/12",
"40/33",
"34/12"
] | A | In first 3 hrs Al will finish 3/12 = 1/4 of work and Allison will finish 3/9=1/3 work so total 1/4 + 1/3 = 7/12 work is finished and 1-7/12 = 5/12 work remaining. Now Allison will take (5/12)*9 = 45/12 hrs to finish it.
So answer is A. |
AQUA-RAT | AQUA-RAT-39962 | 5. Originally Posted by Nath
"I have 2 children. One is female. What is the probability the other is also female?"
This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.
I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.
I'm interested to get some more opinions...
Here is an alternative "logical" way to consider this.
There are 4 possibilities in regard to the children being boy or girl.
The probability of having a boy and a girl is twice the probability of having 2 girls,
and is also twice the probability of having 2 boys.
(i) 1st and 2nd children are girls
(ii) 1st child is a girl and the 2nd is a boy
(iii) 1st child is a boy and the 2nd is a girl
(iv) 1st and 2nd children are boys
GG
GB
BG
BB
One is female.
This reduces to...
GG
GB
BG
3 cases of equal probability and in only 1 of these cases is the other child also a girl.
Hence the probability of the 2nd child also being a girl is 1/3.
Also note that in 2 of these 3 cases, the 2nd child is a boy,
therefore, if one of the children is a girl, the probability that the other is a boy is 2/3.
6. ## Solutions
This was my solution, but alas my friend disagrees:
I have 2 children, 1 is female. This gives four possible scenarios:
1. The female has a younger sister
2. The female has an older sister
3. The female has a younger brother
4. The female has an older brother
Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
---
My friend, with the PhD in Maths thinks the following:
The following is multiple choice question (with options) to answer.
What is the probability for a family with four children to have two boys and two girls (assuming the probability of having a boy or a girl is equal)? | [
"1/2",
"1/4",
"1/8",
"3/8"
] | D | One possible case is: girl - girl - boy - boy
The probability of this is 1/2*1/2*1/2*1/2 = 1/16
There are 4C2 = 6 such cases so we should multiply by 6.
P(two boys and two girls) = 6/16 = 3/8
The answer is D. |
AQUA-RAT | AQUA-RAT-39963 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Out of four numbers ,the average of first three is 16 and that of the last three is 15. If the last number is 20,the first number is : | [
"22",
"23",
"77",
"99"
] | B | Explanation:
Let the numbers be a,b,c,d
Given, a + b + c = 48, b + c + d = 45
Now, d = 20
thus, b + c + 20 = 45 ⇒ b + c = 25
Putting the value of b + c in a + b + c = 48
a + 25 = 48 ⇒ a = 23
Answer: B |
AQUA-RAT | AQUA-RAT-39964 | # System of three equations
$$3x+2y-z=4$$ $$4y-2z+6x=8$$ $$x-2y=5$$
I found that the first two equations were the same. Then, looking at the second and third equations, I thought the answer was No Solution, because I couldn't find a way to solve it, but instead the answer was Infinitely Many Solutions. Can someone explain this to me?
• "I couldn't find a way to solve it": there are zillion equations that have solutions though we don't know how to address them ! – Yves Daoust Sep 20 '16 at 12:49
• Usually I can solve such problems though – thunderbolt Sep 21 '16 at 2:59
## 4 Answers
As one of the equations is a duplicate, you can drop it and solve.
$$\begin{cases}3x+2y-z=4\\ x-2y=5\end{cases}$$
Let us move $z$ to the RHS and consider it as known to obtain a system of two equations in two unknowns,
$$\begin{cases}3x+2y=z+4\\ x-2y=5.\end{cases}$$
Then by elimination
$$\begin{cases}4x=z+9\\ 8y=z-11.\end{cases}$$
We can assign $z$ any value, yet get values of $x,y,z$ that satisfy all equations. For instance $(3,-1,3)$ or $(1,-2,-5)$. There are infinitely many others. In the solution space, $\mathbb R^3$, the solutions describe a straight line.
This is the same as$$\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\8\\5\end{bmatrix}$$
The following is multiple choice question (with options) to answer.
6x – 5y + 3z = 26
4x + 8y – 11z = 7
5x – 6y + 2z = 12
Given the equations above, x + y + z = ? | [
" 11",
" 12",
" 13",
" 14"
] | D | (6x – 5y + 3z) - (5x – 6y + 2z ) = 26-12
or, x+y+z = 14
Option D is the ans |
AQUA-RAT | AQUA-RAT-39965 | This has been offset by payments, whose value at the time of the $$(k)$$-th payment is
$$\displaystyle P[(1+j)^{k-1} + (1+j)^{k-2} + \cdots + 1] = P\left[\frac{(1+j)^k - 1}{(1 + j) - 1}\right] = P\left[\frac{(1+j)^k - 1}{j}\right].$$
This means that the loan balance, immediately after your $$(k)$$-th payment is
$$\displaystyle L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right].$$
During the period between the payments $$(k)$$ and $$(k+1)$$, the interest on this loan balance is
$$\displaystyle \left\{L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right]\right\} \times j$$
$$\displaystyle = \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}.$$
Therefore, the principal reduction for payment $$(k+1)$$ is
$$\displaystyle P - \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}$$
$$\displaystyle = P - L(1 + j)^k j + P\left[(1+j)^k - 1\right]$$
Using equation (1) above, this equals
$$\displaystyle = P - P \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right](1 + j)^k j + P\left[(1+j)^k - 1\right]$$
The following is multiple choice question (with options) to answer.
Consider a lady took a loan from a bank at the rate of 12% p.a. simple interest. After 3 years she had to pay Rs. 9000
interest only for the period. The principal amount borrowed by her was | [
"Rs. 25000",
"Rs. 10000",
"Rs. 15000",
"Rs. 20000"
] | A | Explanation :
Principal = Rs. (100×9000 / 12×3)
=> Rs. 25,000.
Answer : A |
AQUA-RAT | AQUA-RAT-39966 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In a certain city, 60 percent of the registered voters are Democrats and the rest are Republicans. In a mayoral race, if 75 percent of the registered voters who are Democrats and 20 percent of the registered voters who are Republicans are expected to vote for Candidate A, what percent of the registered voters are expected to vote for Candidate A ? | [
"50%",
"53%",
"54%",
"55%"
] | B | Registered voters = 100
D = 60
R = 40
75% of D ( 60) = 45
20% of R ( 40 ) = 8
Total voter% of registered voters = 53 / 100
53%
Answer: B |
AQUA-RAT | AQUA-RAT-39967 | The distance from $A$ to $B$ is $B-A$. If we start at $A$, and then walk half that distance, we should end up precisely at the midpoint $M$ between them. After all, that's what "midpoint" means. Remembering that "walking along the number line" means adding the distance walked to the starting point, we get $$M = A + \frac12(B-A) = A + \frac{B-A}{2}\\ = \frac{2A+B-A}{2} = \frac{A+B}{2} = \frac A2 + \frac B2$$
• Ah awesome this is the explanation which I needed. I also thought "the distance between $A, B$ is $B-A$, so …" but then got stuck what to make of it.
– Max
Jan 24 '18 at 10:25
• If it's not out of scope for this question, could you do me a favor and expand your answer with an example where the ratio would be 1/3A + 2/3B (so that $M$ would be closer to $B$ than to $A$?
– Max
Jan 24 '18 at 10:34
• @Max Well, then you're walking a bit more than half the distance from $A$. Specifically, $\frac23(B-A)$, which is longer than $\frac12(B-A)$. If you do the arithmetic the same way, you get $\frac13A + \frac23B$. Jan 24 '18 at 10:40
The following is multiple choice question (with options) to answer.
On a scale of a map 0.6 cm represents 6.6km. If the distance between two points on the map is 80.5 cm , what is the the actual distance between these points? | [
"885.5 km",
"860 km",
"892.5 km",
"825 km"
] | A | Explanation :
Let the required actual distance be x km
More scale distance, More actual distance(direct proportion)
Hence we can write as
scale distance.6:80.5}::6.6:x
⇒.6x = 80.5 × 6.6
⇒ .1x = 80.5 × 1.1
⇒ x = 80.5 × 11 = 885.5. Answer : Option A |
AQUA-RAT | AQUA-RAT-39968 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
Calculate how much time could be saved if the train would run at its usual speed, given that it ran at 6/8 of its own speed and got to its destination in 11 hours? | [
"2.5 hrs",
"2.9 hrs",
"4.5 hrs",
"3.5 hrs"
] | A | New speed = 6/8 of usual speed
New time = 6/8 of usual time
6/8 of usual time = 10 hrs
Usual time = 10*6/8 = 7.5 hrs
Time saved = 10-7.5 = 2.5 hrs
Answer is A |
AQUA-RAT | AQUA-RAT-39969 | Is this correct? Or can I not just solve this in $\mathbb{Z}_8$?
Yep. All correct!
Note that it is slightly easier in $\mathbb Z_{10}$.
What are the possibilities for the last digit of any square?
#### Deveno
##### Well-known member
MHB Math Scholar
As ILS points out, the last digit (in base 10, our usual base system) of a perfect square must be either:
0,1,4,5,6 or 9.
3 is not on this list.
#### mathmari
##### Well-known member
MHB Site Helper
Yep. All correct!
Note that it is slightly easier in $\mathbb Z_{10}$.
What are the possibilities for the last digit of any square?
As ILS points out, the last digit (in base 10, our usual base system) of a perfect square must be either:
0,1,4,5,6 or 9.
3 is not on this list.
The following is multiple choice question (with options) to answer.
How many two-digit numbers satisfy this property.: The last digit (unit's digit) of the square of the two-digit number is 3 ? | [
"4",
"0",
"3",
"2"
] | B | A number ending with 3 can never be a perfect square.
Answer: B |
AQUA-RAT | AQUA-RAT-39970 | javascript, datetime
// a simple helper function
function nextDay(date) {
return new Date(date.getFullYear(), date.getMonth(), date.getDate() + 1);
}
// as long we're in the same year, keep adding 1 day,
// and store the ones that match the weekday we're looking for
while(current.getFullYear() === year) {
if(current.getDay() === weekday) {
dates.push(current);
}
current = nextDay(current);
}
return dates;
}
No need to build ranges, map, and filter them. Just a while loop and push.
Incidentally, if you want to learn more about why programming calendar and time things are just hideously complex, check out this video on youtube
1) Happy Guy Fawkes day everyone
The following is multiple choice question (with options) to answer.
Today is Tuesday. After 75 days, it will be: | [
"Saturday",
"Sunday",
"Friday",
"Thursday"
] | B | Each day of the week is repeated after 7 days.
So, after 77 days, it will be Tuesday.
After 75 days, it will be Sunday.
Answer :B |
AQUA-RAT | AQUA-RAT-39971 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
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The following is multiple choice question (with options) to answer.
Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in? | [
"7 min",
"13 min",
"23 min",
"9 min"
] | D | 1/10 + 1/15 - 1/x = 1/18
x = 9
ANSWER:D |
AQUA-RAT | AQUA-RAT-39972 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
What is the rate percent when the simple interest on Rs.2000 amount to Rs.500 in 2 Years? | [
"5%",
"6%",
"12.5%",
"95%"
] | C | Interest for 2yrs = 500
Interest for 1yr = 250
Interest rate = 250/2000 x 100 = 12.5%
Answer : C |
AQUA-RAT | AQUA-RAT-39973 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
Workers at a campaign office have 2000 fliers to send out. If they send out 1/10 of them in the morning and 1/4 of the remaining ones out during the afternoon, how many are left for the next day? | [
"300",
"1350",
"1100",
"1200"
] | B | (1/10)*2000 = 200
Remaining = 2000-200 = 1800
(1/4) of remaining = (1/4)*1800 = 450
Remaining now = 1800-450 = 1350
Answer: Option B |
AQUA-RAT | AQUA-RAT-39974 | Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
-
M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
The following is multiple choice question (with options) to answer.
The number 70 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers? | [
"17",
"16",
"15",
"14"
] | D | I think brute force with some common sense should be used to solve this problem.
Write down all perfect squares less than 70: 1, 4, 9, 16, 25, 36, 49, 64.
Now, 70 should be the sum of 3 of those 8 numbers. Also to simplify a little bit trial and error, we can notice that as 70 is an odd numbers then either all three numbers must be odd (odd+odd+odd=odd) OR two must be even and one odd (even+even+odd=odd).
We can find that 60 equals to 9+25+36=3^2+5^2+6^2=70 --> 3+5+6=14.
Answer: D. |
AQUA-RAT | AQUA-RAT-39975 | 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.
In what ratio must sugar at Rs.7.10 be mixed with sugar at Rs.9.20 so that the mixture may be worth Rs.8 per kg? | [
"6:3",
"7:3",
"1:7",
"4:3"
] | D | CP of 1 kg sugar of 1st kind =7.1
CP of 1 kg sugar of 2nd kind=9.2
Mean Price=8
CP of 1 kg sugar of 1st kind - Mean Price=8-7.1=0.9
CP of 1 kg sugar of 2nd kind - Mean Price=9.2-8=1.2
Required ratio = 1.2 : 0.9 = 12 : 9 = 4 : 3
Answer:D |
AQUA-RAT | AQUA-RAT-39976 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
5,9,20,32,52,__ | [
"79",
"85",
"81",
"82"
] | B | (9+5) +6 = 20
(16+9) +7 = 32
(28+16)+8 = 52
(48+28)+9 = 85
ANSWER:B |
AQUA-RAT | AQUA-RAT-39977 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A can complete a project in 20 days while B can complete same project in 30 days. If A and B start working together and A leaves the work 10 days before completion of project, then in how many days the project will be completed? | [
"15",
"16",
"17",
"18"
] | D | A's 1day work=1/20;
B's 1day work=1/30;
(A+B) 1day work=(1/20+1/30)=1/12;
It is given that A leaves the work 10 days before completion of the project..
Thus,B alone does the remaining job in 10 days.
So,In 10 days ,B can do 1/3 w ..
Thus,(A+B) have worked (1-1/3)=2/3 w..
(A+B)can do 1/12 work in 1 day...
They did 2/3 w in 8 days.
Total days=(8+10) = 18
ANSWER:D |
AQUA-RAT | AQUA-RAT-39978 | 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.
There are 7 players in a bowling team with an average weight of 85 Kg. If two new players join the team, one weighs 110 Kg and the second weighs 60 Kg, what will be the new average weight?
. | [
"75 Kg",
"80 Kg",
"85 Kg.",
"90 Kg"
] | C | New average weight= 110+60/2= 85 kgs
Since avg weight of 85 is added to avg weight of 85, the average will remain the same, i.e 85
C is the answer |
AQUA-RAT | AQUA-RAT-39979 | polygon Students will the! 12 ) pentagon whose apothem is also used sometimes to find the areas of each side.... Is derived, see Derivation of regular polygons can be broken down into a combination triangles!, it is easy to show it with an adjustable slider that increased the of. Angles have the Same Degree ( i.e a two-dimensional figure any side and it is perpendicular to that side that... This can give you a good estimate of the vertices of this polygon equal... The coordinates, or 8.66 multiplied by 60 divided by 2 length is 6 cm i do n't to. A segment that joins the polygon ’ s talk about the last of polygons = [ n/2 × L √! That joins the polygon outer circle and touches its circumference calculated using the following formula n. Into small sections of regular polygons into small sections of regular polygons be! In geometry, area of a regular polygon are equal is n't so clear to what... Up their areas we talk about geometry, we have, n =.. Measurement of area is square meters ( m2 ) is equal to the midpoint of any side and is. N'T subtracted ( grey ) is the total space or region bound by the sides need divide. The last digit is not always correct center of outer circle and touches circumference... Polygon is within a circle is given by only with the bounding polygonal chains of simple polygons other! Equal ) enclose the greatest area given a regular polygon 's sides all have the Same (. Following the cross product of the shapes of congruent isosceles triangles the length the. The greatest area given a regular hexagon the previous lessons region of polygon! Inner circle which inscribed in a regular polygon in Java shaded region of a shape for which 've. Have the Same Degree ( i.e does not intersect itself different angles ( Here =! Angle '' rectangle and BDC is a triangle is a Plane shape with straight lines & examples formed in polygon. ’ s talk about the last for area of triangles formed in the polygon into small sections regular. Of straight lines any polygon with perimeter of 44 cm and the side lengths an! Applying the Pythagoras theorem sides of a regular polygon has all angles equal and all
The following is multiple choice question (with options) to answer.
A HEXAGON IS FORMED ABCDEF WHICH PERIMETER 36cm. A POLYGON IS FORMED BY JOINING THE POINTS AE &BD . FIND THE AREA OF THE POLYGON ABDE? | [
"34√3 cm2",
"35√3 cm2",
"36√3 cm2",
"37√3 cm2"
] | C | A regular hexagon consist of 6 equilateral triangle. Each of side 6 cm.
So the area of 1 equilateral triangle is = ¼(√3 a2) = 9√3 cm2
The figure formed by joining polygon ABDE consist of 2 equilateral triangle and 4 half equilateral triangle
Which employs area of 4 equilateral triangle.
So, The area of polygon ABDE = 4*(area of equilateral traingle)
= 4* 9√3 = 36√3 cm2
ANSWER:C |
AQUA-RAT | AQUA-RAT-39980 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
The average marks of 10 students in a class is 100. But a student mark is wrongly noted as 90 instead of 10 then find the correct average marks? | [
"A)78",
"B)82",
"C)92",
"D)91"
] | C | correct avg marks = 100+(10-90)/10
avg = 100-8 = 92
Answer is C |
AQUA-RAT | AQUA-RAT-39981 | 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 man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is | [
"9",
"10",
"11",
"12"
] | B | Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.
Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.
Answer:B |
AQUA-RAT | AQUA-RAT-39982 | # Maximum number of guaranteed coins to get in a “30 coins in 3 boxes” puzzle
There are a total of 30 gold coins in three wooden boxes (but you do not know how many in each individual box). However, you know that one box has exactly 4 coins more than another box.
For each box, you can ask for a number of coins from that box, of your choice. If there are at least that many coins in that box, then you get as many coins as you asked for. Otherwise, you get nothing from that box. You must place all your demands simultaneously in the beginning.
What is the maximum number of coins that you can guarantee yourself to get?
Added for clarification: you don't know which box has 4 coins more.
Let $x, y, y+4$ be the coin contents $=> x+2y=26$ $=>$ solutions $(0,13,17),...,(24,1,5)$.
What next?
• You forgot 26, 0, 4 – QuIcKmAtHs Jan 13 '18 at 7:41
• Do we know which box has 4 more coins than other? (or one unknown box has 4 more coins than some other?) – Atbey Jan 13 '18 at 7:49
• @Atbey No, we do not know which box has 4 coins more. – ami_ba Jan 13 '18 at 8:10
• Fun puzzle. If one looks at it statistically, the best choice is actually $(8,8,8)$ as this has the highest expected value, which is 14.77. – Jens Jan 13 '18 at 14:19
• Since you're using quote block, may I know the source of this puzzle? (just curious, because sometimes users are misusing quote block and code block to highlight/emphasize, which is certainly not the correct purpose) – Andrew T. Jan 13 '18 at 17:56
If you ask for $13,13,13$, then you get nothing if the boxes are $10,8,12$.
If you ask for $12,12,12$, you will always get exactly $12$ coins, since in all cases, exactly one of the boxes has at least $12$ coins.
Claim $12$ is best possible.
Certainly it's best possible with all requests equal.
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Word Problems.
In a rare coin collection, one in six coins is gold, and all coins are either gold or silver. If 10 silver coins were to be subsequently traded for an additional 10 gold coins, the ratio of gold coins to silver coins would be 1 to 6. Based on this information, how many gold coins would there be in this collection after the proposed trade? | [
"50",
"60",
"180",
"200"
] | C | The correct answer is C. |
AQUA-RAT | AQUA-RAT-39983 | • Lingering in the background, is the question, whether there is such a "global constant" $C>1$ as asked for in the question. I think there must be, but I haven't been able to prove it... – Dominic van der Zypen Dec 5 '18 at 14:48
• Is the runner allowed to run backward? – Robert Israel Dec 5 '18 at 15:01
• I don't find this so bind-boggling: if $d_1$ is bigger then $d_2/2$, then there is a subset $S$ of $[0,d_2]$ included in every subinterval of length $d_1$. If the second runner is very slow on $S$, for example it needs more time than the total time for the first runner (meaning that it must be very fast outside $S$ to be faster, on average, then the first runner) we have the required situation. – Ricky Dec 6 '18 at 21:21
The constant is $$2$$. Let $$n=\lfloor d_2/d_1 \rfloor \geq 1$$, and let $$t_k$$ be the time which the long distance runner takes to arrive at the distance $$kd_1$$ from the origin, $$1\leq k\leq n$$.
Proving by contradiction, suppose that on every interval $$[(j-1)d_1,jd_1], j=1,...,k$$ the average speed of the long distance runner is less than $$v_2/2$$. Then $$t_n>2nd_1/v_2$$. On the other hand the total time of the long distance runner is $$d_2/v_2\geq t_n$$. Therefore $$2nd_1 < t_n v_2 \leq d_2 \leq (n+1)d_1,$$ which implies $$n<1$$, a contradiction.
The following is multiple choice question (with options) to answer.
An athlete decides to run the same distance in 1/4th less time that she usually took. by how much percent will she have to increase her average speed? | [
"40",
"44.4",
"33.3",
"22.2"
] | C | simple logic (1-1/4)=3/4 then reverse it so 4/3=1.333
1.33-1=33.33
ANSWER:C |
AQUA-RAT | AQUA-RAT-39984 | When morning comes and all $N$ pirates wake up, they see the greatly diminished pile and a happy fat monkey.
The question is: how many bananas could there have been in the original pile for this to occur?
• Is this for all sets of possible bananas? – awesomepi May 22 '14 at 15:58
• ...The number of possible bananas is limited depending on N, N can be any natural number (except 1). – kaine May 22 '14 at 16:27
• The way I remember it, concludes with the pirates waking up, finding a greatly diminished pile of bananas, each taking their share of $x$ bananas, and feeding the final banana to the monkey. That would give you a starting pint, but I'm not sure it's correct. – SQB May 23 '14 at 7:01
• @SQB I did not remember it ending that way but that could be my bad memory. Mathematically though, I am pretty sure that would be the same as N+1 pirates so Ross's answer still applies. – kaine May 23 '14 at 12:52
• @kaine it is not the same, since N+1 pirates would always divide by (N+1) here it is always a division by N, only it happens N+1 times with the waking up part. – Falco Aug 6 '14 at 11:22
The basic idea is to work backwards.
The last pirate must have found $N+1$ bananas, because he had to find enough bananas remaining for at there to be least 1 banana in the pile left for each pirate. He took 1, fed 1 to the monkey, and left $N-1$. That $N+1$ means that the next to last pirate found $\frac N{N-1}\cdot (N+1)+1$ and so on.
A cute trick is to recognize that there could have been
$-(N-1)$ bananas. Each pirate gives the monkey $1$, takes $-1$, and leaves the pile the same size as before. Since we need to divide it by $N$ for each pirate, the next solution is higher by $N^N$, so the minimal positive solution is $N^N-N+1$ bananas.
The following is multiple choice question (with options) to answer.
Suppose 3 monkeys take 3 minutes to eat 3 bananas. How many monkeys would it take to eat 6 bananas in 6 minutes? | [
"9",
"10",
"11",
"3"
] | D | One monkey takes 3 min to eat 1 banana , so in 6 mins 1 monkey will eat 2 bananas , so for 6 bananas in 6 min we need 6/2 =3 monkeys
ANSWER:D |
AQUA-RAT | AQUA-RAT-39985 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A car started running at a speed of 30 km/hr and the speed of the car was increased by 2 km/hr at the end of every hour. Find the total distance covered by the car in the first 10 hours of the journey. | [
"278 km",
"390 km",
"267 km",
"286 km"
] | B | The total distance covered by the car in the first 10 hours
= 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48
= sum of 10 terms in AP whose first term is 30 and last term is 48
= 10/2 [30 + 48] = 390 km.
Answer: B |
AQUA-RAT | AQUA-RAT-39986 | meters (m), centimeters (cm) & millimeters (mm). This packet covers all you need to know about sectors of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. A worksheet where you need to find the perimeter of sectors given the radius and angle of the arc. The perimeter of a rectangle is 40 cm. The portion of the circle's circumference bounded by the radii, the arc , is part of the sector. The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. Find the area of the sector. Information that I knew just from looking at the diagram that could prove useful. Find A(P-abc). If the angle is θ, then this is θ/2π the fraction of the full angle for a circle. Sector-A sector is a portion of a circle which is enclosed between its two radii and the arc adjoining them. (2) Given that r = 2√2 and that P = A, (b) show that θ = 2 1 The formula for the area of a sector is (angle / 360) x π x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. Videos, worksheets, 5-a-day and much more Formula: s=r x θ l=s+(2*r) Where, r = Radius of Circle θ = Central Angle s = Sector of Circle Arc Length l = Sector of Circle Perimeter Related Calculator: Perimeter of sectors. 15.8k VIEWS. Answer is: units. Simplifying expressions. A FULL LESSON on calculating the area and perimeter of a sector. Practice Question: Calculate perimeter of sector of circle for the following problems: N.B. Perimeter is the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of
The following is multiple choice question (with options) to answer.
The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter? | [
"91.5 cm",
"93.5 cm",
"94.5 cm",
"92.5 cm"
] | A | Explanation:
Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42 = 91.5 cm
ANSWER IS A |
AQUA-RAT | AQUA-RAT-39987 | 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 running at the speed of 60 km/hr crosses a pole in 12 seconds. What is the length of the train? | [
"140",
"130",
"150",
"200"
] | D | Speed=(60x5/8)m/sec = (50/3) m/sec.
Length of the train = (Speed x Time).
Length of the train = (50/3 x 12)m = 200 m.
Answer is D. |
AQUA-RAT | AQUA-RAT-39988 | + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 14+3x \\ 8. & 7+2x &-2 a_1 + 2 b& 3 a_1 -b & 13+3x \end{array}$$
The following is multiple choice question (with options) to answer.
Simplify: (8-5)+(10-2+3)-5 | [
"8",
"9",
"-8",
"-9"
] | B | Solution:
(8-5)+(10-2+3)-5
=3+8+3-5
=14-5
=9
Answer: (B) |
AQUA-RAT | AQUA-RAT-39989 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
You are a cashier at a grocery store. Your customer, David, buys a pound of bacon
$1.39, one dozen eggs for $1.29, and a pound of coffee for $2.79. What is the sum
total of his purchases? | [
"$5.47",
"$6.47",
"$7.47",
"$8.47"
] | A | Add $1.39+ $1.29+$2.79
we have the sum $5.47
correct answer A |
AQUA-RAT | AQUA-RAT-39990 | the value of a. Physicists tend to blend conceptual physics with mathematical symbolism in a. E Maths Formula List / A Maths Formula Sheet Attached below are the Formula Lists for E Maths and A Maths (O Level) Do be familiar with all the formulas for Elementary Maths and Additional Maths inside, so that you know where to find it when needed! Wishing everyone reading this all the best for…. Must-have printable for all geometry and math classes! Use this page of diagrams and formulas for finding the perimeter, area, and volume of geometric shapes. Perimeter and Area - Summary 10-A A circle’s perimeter is called its circumference. 609 km 1yard= 0. Since the Renaissance, every century has seen the solution of more mathematical problems than the century before, yet many mathematical problems, both major and minor, still remain unsolved. Hence the year 3 figure is £500 1. Fraction-Decimal Conversion. 2015 Grade 5 PSSA MATHEMATICS GRADE 5 REFERENCE. 1300 Math Formulas = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = fp_k= =VVQVNMTTQN= = çéóêáÖÜí=«=OMMQ=^KpîáêáåK=^ää=oáÖÜíë. Maths All Formulas Needed for IGCSE Posted by Rahul at 08:28. Just scroll down or click on what you want and I'll scroll down for you! Circumference =. Select chapter to view Important Formulas chapter wise. E Maths Formula List / A Maths Formula Sheet Attached below are the Formula Lists for E Maths and A Maths (O Level) Do be familiar with all the formulas for Elementary Maths and Additional Maths inside, so that you know where to find it when needed! Wishing everyone reading this all the best for…. Write the formula for the total surface area and the. 86 (as shown in the mortgage calculator). Basic Electrical Formulas. com provides Maths Formulas, Mathematics Formulas, Maths Coaching Classes. Lines (Linear Functions) Consider the line that goes through points A(x1,y1) and B(x2,y2). GDP deflator: A price index used to adjust nominal GDP to arrive at real GDP. Continuity and Differentiability. Here's a breakdown of
The following is multiple choice question (with options) to answer.
The total marks obtained by a student in Mathematics and Physics is 32 and his score in Chemistry is 20 marks more than that in Physics. Find the average marks scored in Mathamatics and Chemistry together. | [
"40",
"26",
"27",
"28"
] | B | Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 32 and C - P = 20 M + C / 2 = [(M + P) + (C - P)] / 2 = (32 + 20) / 2 = 26.Answer:B |
AQUA-RAT | AQUA-RAT-39991 | 80&4352803871250866057941284067804134&18,17,27,27,29,30\\ 81&11395788481298729007027429355459050&18,17,27,29,29,30\\ 82&29834561572649264127826347038980028&18,17,29,29,29,30\\ 83&78107896236653244353987095675034922&18,17,29,29,29,32\\ 84&204489127137313012297066889013936544&18,17,29,29,31,32\\ 85&535359485175296566394760822973220276&18,17,29,31,31,32\\ 86&1401589328388601697072447570729328594&18,17,31,31,31,32\\ 87&3669408499990537951967408951471847094&18,17,31,31,31,34\\ \end{array}
The following is multiple choice question (with options) to answer.
A person is 80 years old in 490 and only 70 years old in 500 in which year is he born? | [
"400",
"550",
"570",
"440"
] | C | 490(80),500(70),510(60),520(50),530(40),540(30),550(20),560(10),570(0)
ANSWER:C |
AQUA-RAT | AQUA-RAT-39992 | By brute force, here are the possible sequences.
(8,0,...),(7,1,0,...),(6,2,0,...),(6,1,1,0,...),(5,3,0,...),(5,2,1,0,...)(5,1,1,1,0,...),(4,4,0,...),(4,3,1,0...),(4,2,2,0,...),(4,2,1,1,0,...),(4,1,1,1,1,0,...),(3,3,2,0,...),(3,3,1,1,0,...),(3,2,2,1,0,...),(3,2,1,1,1,0,...),(3,1,1,1,1,1,0,...),(2,2,2,2,0,...),(2,2,2,1,1,0,...),(2,2,1,1,1,1,0,...),(2,1,1,1,1,1,1,0,...),(1,1,1,1,1,1,1,1,0,...)
As you can see, there are 22 of them. I'm not sure how to generalize this result at the moment, perhaps someone who is more familiar with the problem will come around with a counting method. Perhaps a recurrence relation.
Looking for a pattern at the sequence of numbers for this problem with 1,2,3,... number of objects: I get the amounts: 1,2,3,5,7,11,15,22.
By google searching I find http://mathworld.wolfram.com/PartitionFunctionP.html which has a great deal of history about the problem and methods of calculating the sequence.
The following is multiple choice question (with options) to answer.
Complete the sequence:
6 7 9 8 6 8 ? | [
"9",
"2",
"2",
"8"
] | A | A
9
Number of alphabets in the days of a week. Monday contains 6 letters; Tuesday contains 7 letters and so on. |
AQUA-RAT | AQUA-RAT-39993 | units, geometry, dimensional-analysis
Title: Physical representation of volume to surface area I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.
My question is: what is the physical representation of length in this ratio? For the case of a sphere the ratio you found is:
$$ \frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3} $$
We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus.
One approach is then to ask "what is a function divided by its derivative". This is really similar to the area to perimeter ratio of a circle.
$$ \frac{A}{P} = \frac{ \pi R^2}{ 2 \pi R} = \frac{R}{2} $$
Of course you see the "2" because of the value of the exponent, which comes from there being two dimensions, just like the sphere. So now we have explained part of the answer, which is that the linear dimension is divided by the number of dimensions. This is still unsatisfactory because we have no clear sense of how we should define this particular "characteristic length".
One attempt at resolution of this problem would be to test the idea for a square-cube system.
$$ \frac{V}{S} = \frac{ R^3}{ 6 R^2} = \frac{R}{6} $$
$$ \frac{A}{P} = \frac{ R^2}{4 R} = \frac{R}{4} $$
The following is multiple choice question (with options) to answer.
The areas of the two spheres are in the ratio 1: 4. The ratio of their volume is? | [
"1:7",
"1:3",
"1:8",
"1:9"
] | C | 4πr12 : 4πr22 = 1 : 4
r1 : r2 = 1 : 2
4/3 πr13 : 4/3 πr23
r13 : r23 = 1:8
Answer:C |
AQUA-RAT | AQUA-RAT-39994 | 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
_________________
Test confidently with gmatprepnow.com
Intern
Joined: 14 Sep 2019
Posts: 16
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 0.1 percent and 0.3 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 20,000 units? | [
" Between $15,000 and $25,000",
" Between $50,000 and $150,000",
" Between $60,000 and $100,000",
" Between $150,000 and $250,000"
] | B | Number of defective units is between = .1 % of 20000 and .3% of 20000
= 20 and 60
Retail Price per unit = 2500 $
Expected price of refund is between = 2500 x 20 and 2500 x 60
=50,000 and 1,50,000 dollars
Answer B |
AQUA-RAT | AQUA-RAT-39995 | Sort by:
Solution for Q-3:
Using binomial theorem, one can note that,
$P(x)=(x-3)^4-10x+15$
We have, using Remainder-Factor Theorem and the fact that $a,b,c,d$ are roots of $P(x)$ that,
$P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}$
Using the last result, we have the sum as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)$
Using Vieta's formulas, we have,
$\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120$
Hence, the required sum evaluates as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}$
- 5 years, 11 months ago
Nicely done.
Staff - 5 years, 11 months ago
Your hint did most of the work. So, the credit actually goes to you. :)
- 5 years, 11 months ago
$Q-1$
The following is multiple choice question (with options) to answer.
What is the value of (P + Q)/(P - Q) if P/Q is 10? | [
"4/3",
"2/3",
"2/6",
"11/9"
] | D | (P + Q)/(P - Q) = [(P/Q) + 1]/[(P/Q) - 1]
= (10 + 1)/(10 - 1)
= 11/9
= 11/9
ANSWER:D |
AQUA-RAT | AQUA-RAT-39996 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
A bookseller has two display windows. She plans to display 5 new fiction books in the left window, and 6 new non-fiction books in the right window. Assuming she can put the six fiction books in any order, and separately, the three non-fiction books in any order, how many total configurations will there be for the two display windows? | [
"86400",
"65777",
"89000",
"34560"
] | A | Two displays books could be arranged n! # of total arrangements are combinations multiplied together,
5!*6! =120*720=86400
Answer is A |
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