source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38197 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 10 hours, pipe B in 20 hours, and pipe C in 60 hours. If all the pipes are open, in how many hours will the tank be filled? | [
"6",
"7",
"8",
"9"
] | A | The part filled by A + B + C in 1 hour is 1/10 + 1/20 + 1/60 = 1/6
All the three pipes together will fill the tank in 6 hours.
The answer is A. |
AQUA-RAT | AQUA-RAT-38198 | human-biology, reproduction, anthropology
Note: here is a link about multiple pregnancy : https://www.healthline.com/health/pregnancy/chances-of-having-twins#assisted-reproduction. 49 - 19 = 30
So the 1st Child would be 30 when the last children would be born, as long as the parents stay healthy and the mother delivers the child in a natural way (No Surgical Procedure to deliver the child) and the Mother has not had Menopause (Menopause usually happens around 50 Years of age) the sibling age gap would be very possible, but lets take a look at how many children the mother had, if I am correct they had about 16 Children, many people in countries that have a considerably low life expectancy have many children, lets take a look at the world record for most children born from one mother...a person named Valentina Vassilyev, gave birth to 69 Children! Thats a lot of children, so giving birth to 16 children with a 30 age gap should be very possible and realistic as well.
The following is multiple choice question (with options) to answer.
The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is | [
"33",
"40",
"38",
"27"
] | B | Explanation:
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
Answer: B) 40 |
AQUA-RAT | AQUA-RAT-38199 | # 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 running downstream covers a distance of 20 km in 2 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water? | [
"6",
"7",
"8",
"9"
] | B | Explanation:
Rate downstream = (20/2) kmph = 10 kmph;
Rate upstream = (20/5) kmph = 4 kmph
Speed in still water = 1/2 (10 + 4) kmph = 7 kmph
ANSWER: B |
AQUA-RAT | AQUA-RAT-38200 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man sells a car to his friend at 12% loss. If the friend sells it for Rs.54000 and gains 20%, the original C.P.of the car was : | [
"Rs.21136.36",
"Rs.31136.36",
"Rs.51136.36",
"Rs.61136.36"
] | C | Explanation:
S.P = Rs.54,000. Gain earned = 20%
C.P = Rs.[100/120×54000]
=Rs. 45000
This is the price the first person sold to the second at at loss of 12%.
Now S.P = Rs.45000 and loss = 12%
C.P. Rs.[100/88×45000]= Rs.51136.36
Correct Option : C |
AQUA-RAT | AQUA-RAT-38201 | 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 an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 5500, the number of valid votes that the other candidate got, was: | [
"2800",
"1980",
"2900",
"2200"
] | B | B
Number of valid votes = 80% of 5500 = 4400.
Valid votes polled by other candidate = 45% of 4400
= (45/100 x 4400) = 1980. |
AQUA-RAT | AQUA-RAT-38202 | \limits_{C_{1}}=\frac{5}{6}.\frac{4^6}{4^4}+\frac{1}{5}.\frac{4^5}{4^2}-12=\frac{40}{3}+\frac{64}{5}-12$ For C2 we have x=2, dx=0 $\therefore \int\limits_{C_{2}}=\int\limits_{-4}^{4}(4y+3)dy=[2y^2+3y]_{-4}^{4}=32+12-32+12=24$ for C3 we have $\int\limits \limits_{C_{3}}=\int\limits_{4}^0(\frac{5y^5}{4^4}-\frac{y^4}{4^2}+3)dy=[\frac{5y^6}{6\times4^4}-\frac{y^5}{5\times4^2}+3y]_{4}^{0}$ $\int\limits \limits_{C_{3}}=-(\frac{5}{6}.\frac{4^6}{4^4}-\frac{1}{5}.\frac{4^5}{4^2}+12)=-\frac{40}{3}+\frac{64}{5}-12$ $\therefore \oint \limits_{C}=\frac{40}{3}+\frac{64}{5}-12+24-\frac{40}{3}+\frac{64}{5}-12=\frac{128}{5}$ Let $\phi(x,y)=x^2-2xy, \psi(x,y)=x^2y+3$ $\therefore \frac{\partial \psi}{\partial x}=2xy \space \space \space \space , \space
The following is multiple choice question (with options) to answer.
How many meters are there in 330 centimeters? | [
"550 meters",
"3.3 meters",
"100 meters",
"3.5 meters"
] | B | Solution:
1 meter = 100 centimeters
Just divide 330 by 100
330 ÷ 100 = 3.3, so there are 3.3 meters in 330 centimeters
Option B |
AQUA-RAT | AQUA-RAT-38203 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can? | [
"40",
"44",
"47",
"83"
] | B | Explanation:
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.
Answer: Option B |
AQUA-RAT | AQUA-RAT-38204 | solution: solution to get the median: the mean compared to sets a mean questions with solutions b Question! To new CBSE exam Pattern, MCQ questions for Class 10to help you are presented 6 ICSE.. A- a ; each being equal to the common difference we will help you sum must 12! Step 3: the given mean of a set of numbers deep understanding of maths.... Mean μ ' and the new standard deviation σ MCQ questions for Class 10to help.... To get the mean compared to sets a, a, b and C. a given data set you. Is 8 of all 5 numbers about Harmonic progression formula for nth term, sum of 5! Multiply all data values by a constant k and Calculate the mean been designed to test for understanding! Μ ' and the new standard deviation σ understanding of maths concepts b – =. Averages problem is a vey easy Question so the sequence is 2, 3 8... Mean is also called average of 56, 41, 59, 52, 42 and.... ' and the new standard deviation σ exercises on calculating the mean of given sets and word problems on mean. Set, you may need to determine the possible values of the 3 numbers in. Aggarwal Solutions Class 10 Chapter 9 - Benefits the workers earn a salary of Rs 6 to get the for. Set has a mean μ ' and the new standard deviation of the page 7 +11 ) /5 5.6! Using the average of the given classes will give the sum is 18 Divide by! Their sum must be 12, so you have x+y+z=12 ⇒ a = A- a ; each being equal the. To sets a, b are in A.P x in the middle solution: Question 21 properties... One of the 5 numbers using the average of 56, 41, 59, 52, and..., a, b and C. a given data set a = 3 if r1, r2 r3... B are in A.P a ) Calculate the mean example 2: Compute sum of set! So you have x+y+z=12 a complete solution 2 # the mean questions with solutions mean (... Lower part of the 20 people and Harmonic mean the minimum number 1! Between the first and second must be equal to the difference second and third, giving the x-y=y-z! =
The following is multiple choice question (with options) to answer.
The sum of the mean, the median, and the range of the set {1,2,3} equals which one of the following values? | [
"3",
"4",
"5",
"6"
] | D | Here Mean => 1+2+3/3 => 2
median => 2
and range => 3-1 => 2
hence sum => 2+2+2 => 6
Answer: D |
AQUA-RAT | AQUA-RAT-38205 | special-relativity, speed-of-light, reference-frames
Title: Relative speed when approaching the speed of light According to this chart of the Lorentz factor as a function of speed:
If a spacecraft neared (roughly) 0.85c, would it appear to be traveling at 1.7x the speed of light from the perspective of those on board - i.e. covering 0.85c distance, but in half the time amount of time? There is a spacecraft moving with velocity $v = 0.85c$ in a frame of reference where, helpfully, someone has put out 'mile' posts exactly 1 light-second apart (in that frame), along the direction of the spacecraft's travel.
In this frame of reference, the spacecraft just passes 85 mile posts in 100 seconds according to clocks at rest in this frame.
However, according to the spacecraft, the mile posts are less than 1 light-second apart (length contraction) and, in fact, measure
$$\sqrt{1 - (0.85)^2} = 0.527... $$
light-seconds apart.
So, even though the spacecraft's clock reads an elapsed time of just about 52.7 seconds (time dilation) when passing the "85" mile post, the distance covered, according to the spacecraft, is just
$$d = 85 \cdot 0.527$$
light-seconds and thus, as must be the case, the spacecraft observer calculates a relative speed of $0.85c$ too.
The following is multiple choice question (with options) to answer.
The speed of light is approximately 1,86 * 10^4 miles per second. This approximate speed is how many miles per hour? | [
" 1,11 * 10^7",
" 6,70 * 10^7",
" 1,11 * 10^8",
" 1,86 * 10^8"
] | B | The easiest way to answer this question is by POE, it does not require any calculation (answered in 30seconds). In one hour there are 3600 seconds, therefore speed in miles/hour will be.
(1.86*10^4)*3600 = (1.86*10^4)*(3.6*10^3) = some number*10^7...
The key is realizing that thesome numberhas to be bigger thatn 1.86 (because 1.86*3.6), and the only answer that fits that is B |
AQUA-RAT | AQUA-RAT-38206 | 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 280 m long is running at a speed of 54 km/hr. In what time will it pass a bridge 200 m long? | [
"32 sec",
"42 sec",
"45 sec",
"18 sec"
] | A | total length= 280+200=480
speed=54km/h=(54*5)/18= 15/1m/s
time=(480*1)/15=32 sec.
ANSWER:A |
AQUA-RAT | AQUA-RAT-38207 | a-5/a+2 = -1
a-5= -a-2
0= -2a+3
subtract 3 from both sides:
-3=-2a
Divide by -2
3/2=a
I know that the answer is (-2, 3/2), but I'm not sure where the -2 in the answer comes from. Thanks!
You have to be wary with inequalities because it changes direction if you multiply or divide by a negative so ideally we don't want to multiply across the inequality.
$\dfrac{a-5}{a+2} < -1$
$\dfrac{a-5}{a+2} + \dfrac{a+2}{a+2} < 0$
$\dfrac{2a-3}{a+2} < 0$
For the LHS to be negative the numerator and denominator must have different sign. Points to check are $\frac{3}{2}$ and $-2$ leading to the intervals
$a < -2$
$-2 < a < \frac{3}{2}$
$a > \frac{3}{2}$
The following is multiple choice question (with options) to answer.
If 4/(a - 5) = 5/(a + 5), then a = ? | [
"40",
"45",
"35",
"55"
] | B | Multiply all terms of the given equation by (a - 5)(a + 5), simplify and solve
(a - 5)(a + 5)[ 4/(a - 5) ] = (a - 5)(a + 5) [5/(a + 5) ]
4 (a + 5) = 5 (a - 5)
a =45
correct answer B |
AQUA-RAT | AQUA-RAT-38208 | 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.
In August,a cricket team that played 120 matches won 26% of the games it played.After a continuous winning streak,this team raised its average to 52%.How many matches did the team win to attain this average? | [
"40",
"52",
"65",
"80"
] | C | let the no of matches played more=x
so,
(120+x)*52/100=31.2+x
by solving we get x=65
ANSWER:C |
AQUA-RAT | AQUA-RAT-38209 | 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.
A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 72 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk? | [
"20",
"30",
"48",
"56"
] | D | Total distance = 72
Distance = Speed * Time
Walking speed = s1 = 8
Walking time = t1
Bike speed = s2 = 16
Time traveled in bike = t2
d1 + d2 = 72
s1t1 + s2t2 = 72
8*t1 + 16*t2 = 72
t1 + 2*t2 = 9 ----- (1)
Given: t1 + t2 = 8 ----- (2)
(1) - (2) --> t2 = 1 and t1 = 8 - 1 = 7
Walking distance = s1*t1 = 8*7 = 56
Answer: D |
AQUA-RAT | AQUA-RAT-38210 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
If it is possible to from a word with the first, fourth, seventh and eleventh letters in the word 'SPHERVLVODS' write the second letter of thet word. Otherwise, X is the answer.? | [
"D",
"E",
"S",
"P"
] | B | The first, fourth, seventh and eleventh letters of the word 'SPHERVLVODS'
The word formed is LESS
The second letter is E.
ANSWER B |
AQUA-RAT | AQUA-RAT-38211 | 2-4-8-6
2-4-8-6
2-4
The next digit in the pattern will be 8, which will belong to $$2^{11}$$.
In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base.
So $$652^{11}$$ will end in $$8,1896782^{11}$$ will end in 8, and so on…
A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits.
Units digit 3:
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on…
Units digit 4:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
The pattern here is 4, 6, 4, 6, 4, 6, and so on…
Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is:
$$1545^{23} = ……..5$$
$$1650^{19} = ……..0$$
$$161^{28} = ………1$$
Hope you get the point.
Units digit 7:
7^1 = 7
7^2 = 49
7^3 = 343
7^4 = ….1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit)
7^5 = ….7 (Now multiply 1 from above by 7 to get 7 as the units digit)
7^6 = ….9
The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on…
Units digit 8:
8^1 = 8
8^2 = 64
8^3 = …2
8^4 = …6
8^5 = …8
8^6 = …4
The following is multiple choice question (with options) to answer.
What is the units' digit of the following expression (11)^5*(1)^3*(21)^5? | [
"0",
"1",
"3",
"5"
] | B | We do not have to do any calculations or find units digit of remaining numbers...
all three terms - 11,1,21 - are ODD and since the PRODUCT contains 1, the units digit of the product will remain 1
B |
AQUA-RAT | AQUA-RAT-38212 | induction that the product of three consecutive integers is divisible by 6. Real numbers class 10. In mathematics, the least common multiple, also known as the lowest common multiple of two (or more) integers a and b, is the smallest positive integer that is divisible by both. Click 'show details' to verify your result. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Prove that n2-n is divisible by 2 for every positive integer n. Exercise: 2. For any positive integer n, prove that n3 – n is divisible by 6. The number is divisible by 6 means it must be divisible by 2 and 3. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. What are the two odd integers? 12. 111 is the smallest possible magic. Exercise 12. case (3) z is a multiple of three. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Some other very important questions from real numbers chapter 1 class 10. Prove that the product of any three consecutive positive integers is divisible by 6. Since n is a perfect square, n is congruent to 0 or 1 modulo 4. Solution for Determine whether the statement is true or false. 1, which is divisible by 9. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. : Therefore: n = 3p or 3p+1 or 3p+2, where p is some integer If n = 3p, then n is divisible by 3 If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3. Consider three consecutive integers, n, n + 1, and n+ 2. 7: given non-empty nite sets X and Y with jXj= jYj, a function X !Y is an injection if and only if it
The following is multiple choice question (with options) to answer.
The sum of three consecutive odd natural numbers, each divisible by 3 is 87. What is the largest among them? | [
"21",
"24",
"27",
"32"
] | D | Let the 3 consecutive odd numbers each divisible by 3 be X, X+3, X+6
Given, X + X+3 + X+6 =87
3X = 78
X = 26
The numbers are 26, 29, 32
Therefore the largest number is 32
Answer D. |
AQUA-RAT | AQUA-RAT-38213 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The current ofa stream runs at the rate of 2 km/hour. A motor boat goes 10 km upstream & back again to the starting point in 55 min. Find the speed of the motor boat in still water? | [
"20 km/hr",
"22 km/hr",
"28 km/hr",
"30 km/hr"
] | B | Let the speed of the boat in still water =x km/hr
Speed of the current = 2 km/hr
Then, speed downstream =(x+2) km/hr
speed upstream =(x−2) km/hr
Total time taken to travel 10 km upstream and back = 55 minutes =5560 hour = 1112 hour
⇒10x−2+10x+2=1112 120(x+2)+120(x−2)=11(x2−4) 240x=11x2−44 11x2−240x−44=0 11x2−242x+2x−44=0 11x(x−22)+2(x−22)=0 (x−22)(11x+2)=0 x=22 or −211
Since x cannot be negative, x = 22
i.e., speed of the boat in still water = 22 km/hr
B |
AQUA-RAT | AQUA-RAT-38214 | Probability of more than 1 goal at the end of a match
Probability of a match ending with
- 0 goals is 40%,
- 1 goal is 45% and
- more than 1 goal is 15%.
Now at half-time, the score is 1-0.
What is the probability of more than 1 goal at the end of the match?
A. Is it still 15%, or
B. is it 25% (100% = probability of 1 goal + probability of more than 1 goal only, eliminating the probability of 0 goals)?
Or is there anything I missed out?
Thanks for the help.
• There's something troubling with this question, and that is your implicit assumption that being 1-0 at half time maintains the original probabilities in some way. It's possible that being 1-0 at half time makes the teams behave differently. The leading team being more defensive now, making the probability of new goals very unlikely. Or maybe it drives the losing team into more aggressive tactics, increasing the probability that one of the teams will make a new goal.
I will have to make some simplifying assumptions in order to give a reasonable answer. I will have to assume that:
1. the score in each half follows the same distribution, and
2. the score in the second half is independent of the score in the first half.
In real life, I doubt this is entirely true. Maybe if one team scores in the first half, it?s more likely that there?s a goal in the second half because the team who?s behind is playing more aggressively. But I think I need to make this assumption to answer your question based on the information given.
Let?s define two random variables: $$X_{1}$$ = the number of goals in the first half and $$X_{2}$$ = the number of goals in the second half. Again, I?m assuming that the X's are independent and identically distributed.
The following is multiple choice question (with options) to answer.
Which of the following is the median number of goals scored per student? | [
"0",
"1",
"2",
"3"
] | B | Total number of students = 8+2+5= 15
so median must be the middle term out of the 15 terms i.e., 8th term when arranged in ascending order
arranging in ascending order we have {1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 3}
median number of goals scored per student = 1
ANSWER:B |
AQUA-RAT | AQUA-RAT-38215 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
A bag contains 25p coins, 50p coins and 1 rupee coins whose values are in the ratio of 8:4:2. The total values of coins are 840. Then find the total number of coins | [
"220",
"240",
"260",
"280"
] | D | Value is given in the ratio 8:4:2.
(8x/0.25) + (4x/0.5) + (2x/1) = 840.
X = 20. Total amount = 14*20 = 280
Answer:D |
AQUA-RAT | AQUA-RAT-38216 | Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1
Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.
Thus there are 60+90 = 150 ways we can place the marbles
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Joined: 30 May 2019
Posts: 82
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
### Show Tags
04 Oct 2019, 22:44
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
Solved it a little differently than others from what I read . So will share.
I first made sure that each of the pocket has at least 1 marble.
That is 5C3 * 3!
The following is multiple choice question (with options) to answer.
In his pocket, a boy has 3 red marbles, 4 blue marbles, and 5 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color? | [
"7",
"8",
"9",
"10"
] | D | The worst case scenario would be that he has taken 4 blue and 5 green, a total of 9 marbles, and still doesn't have 3 distinct colors. But the next draw will surely be red as only the red marbles are left in his pocket.
The answer is D. |
AQUA-RAT | AQUA-RAT-38217 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A man invests a certain sum of money at 6% per annum simple interest and another sum at 7% per annum simple interest. His income from interest after 2 years was Rs. 354. One-forth of the first sum is equal to one-fifth of the second sum. The total sum invested was : | [
"Rs.3100",
"Rs.2700",
"Rs.2200",
"Rs.1800"
] | B | Explanation :
Let the man invests Rs.x at 6% and Rs.y at 7%
Simple Interest on Rs.x at 6% for 2 years + Simple Interest on Rs.y at 7% for 2 years = Rs.354
x × 6 ×2 /100+y×7×2/100=354
x × 6 × 2+y×7×2=354×100
x × 6+y × 7=177×100
6x+7y=17700⋯(1)
One-forth of the first sum is equal to one-fifth of the second sum
=>x/4=y/5
=>x=4y/5⋯(2)
Solving (1) and (2),
6x+7y=17700
6(4y/5)+7y=17700
24y+35y=17700×5
59y=17700×
y=300×5=1500
x=4y/5
=4×1500/5
=4×300
=1200
total sum invested = x + y = 1500 + 1200 = 2700
Answer : Option B |
AQUA-RAT | AQUA-RAT-38218 | When I tested my answer and the book's, I found out that the book's answer gave the fraction that I was asked to break up.
Is there a reason why I must factorise the denominator into (1-x)^2 (1+x) and not (x-1)^2 (1+x)?
hi
you are correct and the book is also correct. You just play with the algebra here.
-1/4(x-1)= -1/4(-(1-x))=1/4(1-x)
It doesn't matter since both forms are correct.
#### HallsofIvy
MHF Helper
$$\displaystyle (1- x)^2= (-(x-1))^2= (-1)^2(x-1)^2= (x-1)^2$$
Last edited by a moderator:
#### dd86
Great! Thanks for the quick response, everyone!
The following is multiple choice question (with options) to answer.
If 1/(x+2)+1/(x-2)=1/(x+2), what is the value of x? | [
"-1",
"2",
"1",
"-2"
] | D | If we solve the question, we get x=-2.
Option: D |
AQUA-RAT | AQUA-RAT-38219 | 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 after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late . Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late. What is the speed Q of the train. | [
"a) 45",
"b) 33",
"c) 48",
"d) 55"
] | C | Let y be the balance distance to be covered and x be the former speed.
A train after traveling for 50km meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 minutes late
so, y/(3x/4) - y/x = 35/60
4y/3x - y/x = 7/12
y/x(4/3-1)=7/12
y/x*1/3=7/12
y/x=7/4
4y-7x=0 ........ 1
Had the accident occurred 24 km farther, it would have reached the destination only 25 minutes late
so, (y-24)/(3x/4) - (y-24)/x = 25/60
4(y-24)/3x - (y-24)/x = 5/12
(y-24)/x (4/3-1) = 5/12
(y-24)/x *1/3 = 5/12
(y-24)*12 = 3x*5
(y-24)*4 = 5x
4y-5x = 96 ....... 2
eq2 - eq1
2x=96
x=48=Q
Ans = C |
AQUA-RAT | AQUA-RAT-38220 | Thus,
200 x 0.3 = 60 received student loans
200 x 0.4 = 80 received scholarships
We are trying to determine what percent of those surveyed said that they had received neither student loans nor scholarships.
Let’s fill all this information into a table. Note that each row sums to create a row total, and each column sums to create a column total. These totals also sum to give us the grand total, designated by 200 at the bottom right of the table.
Statement One Alone:
25 percent of those surveyed said that they had received scholarships but no loans.
Using statement one we can determine the number of students who received scholarships but no loans.
200 x 0.25 = 50 students who received scholarships but no loans.
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
We are given that 50 percent of those surveyed who said they had received loans also said that they had received scholarships. From the given information we know that 60 students received loans; thus, we can determine the number of these 60 students who also received scholarships.
60 x 0.5 = 30 students who received loans who also received scholarships
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement two is sufficient to answer the question.
_________________
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Re: In a survey of 200 college graduates, 30 percent said they [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
If a survey shows that 40 citizens out of a sample of 200 support a particular Senate bill, what percent of the sample does not support the bill? | [
"56%",
"64%",
"80%",
"82%"
] | C | Those who support = 40.
% of those who don't support it is (200-40)/ 200 = 80%
Answer C |
AQUA-RAT | AQUA-RAT-38221 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
Evaluate: 2344 - 12*3*2 =? | [
"2337",
"1227",
"3327",
"5457"
] | A | According to order of operations, 12?3?2 (division and multiplication) is done first from left to right
12**2 = 4* 2 = 8
Hence
2344 - 12*3*2 = 2344 - 8 = 2337
correct answer A |
AQUA-RAT | AQUA-RAT-38222 | 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.
Total of the ages of A, B ahd C at present is 90 years. Ten years ago, the ratio of their ages was 1: 2: 3. What is the age of B at present | [
"25",
"30",
"45",
"60"
] | B | Explanation:
Let their ages 10 years ago is x, 2x and 3x years.
10 + 2x + 10 + 3x + 10 = 90 hence x= 10
B’s present age = (2x + 10) =30 years
Option B |
AQUA-RAT | AQUA-RAT-38223 | # Calculate the number of ways to paint $5$ buildings with $4$ colours such that all $4$ colours must be used
A developer has recently completed a condominium project in a valley. There are blocks of buildings $A$, $B$, $C$, $D$ and $E$ as shown in the diagram below.
The developer has colours available to paint the buildings. Each block can only be painted using a single colour. Find the number of ways to paint all the blocks if all $4$ colours must be used.
My attempt: We have $5$ ways to choose $4$ buildings with $4$ different colours. Among the $4$ buildings, we have $4!$ ways to paint them using $4$ different colours.
So my answer is $120$. However, the answer given is $240$. What is my mistake?
• Um, what do paint the fifth building? Don't you have 4 chooses for that? I'd get that you method out to give 4*120 =480. Which is also wrong as we double counted. We'd choose a building to be a duplicate color. there are 5 choices for that. We'd paint the other four. 24 four that. We'd pick one of the four buildings to duplicate the fifth one. There are four choices of that. Which ever building we choose to duplicate is a double counting as we could have choosen that building to be a duplicate. so divide by two. – fleablood May 12 '17 at 15:24
You're on the right track. After choosing the four buildings with different colours, what about the fifth? It will be a repeated colour, and there are four colours to choose from. However, we will have overcounted by a factor of $2$, so the final answer will be $$120\cdot\frac42 = 240$$ To explain the overcounting, suppose that you first choose $A,B,C,D$ as the set of four buildings. Then $E$ is the same as one of them, say $A$. But then we will also later consider $E,B,C,D$ as the set of four, with $A$ the same as $E$.
The following is multiple choice question (with options) to answer.
A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time W it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week? | [
" 0.8(80 – y)",
" 0.8 + 0.0025y",
" 80/y – 1.25",
" 80/1.25y"
] | B | This may be a case of bad numberpicking but it worked for me.
80 houses. Let's say that y=40 houses are painted at the rate x=10 houses per week <=> 1/10 week per house. 40*1/10 = 4 houses per week will be painted at this rate.
80-y = 80-40 = 40 houses are to be painted at the faster rate. X*1,25=12,5 houses per week <=> 1/12,5 weeks per house * 40 houses = 40/12,5 = 80/25 = 320/100 = 3,2 weeks.
Which means finishing all houses at normal rate W=> 2*4 = 8 weeks. Faster rate = 4+3,2 = 7,2 weeks.
7,2/8 = 9/10 = 0,9. Insert y=40 in equations and it is clear that only (B) gives us 0,9. |
AQUA-RAT | AQUA-RAT-38224 | • so $x=12$ or $x=-14$
Let $$\mathbb S$$ be a set of all even integers, i.e:
$$\mathbb S = \{x | x = 2k, k\in \mathbb Z \}$$
Suppose we have two consecutive elements $$x_1$$ and $$x_2$$ in the set $$\mathbb S$$, i.e:
$$x_1 = 2k$$ ...(1)
$$x_2=2k+2$$ ...(2)
...such that their product is 168, i.e:
$$(2k)(2k+2)=168$$
Solving for k:
$$4k^2 + 4k -168 = 0$$
$$k^2 + k - 42 = 0$$
$$(k-6)(k+7) = 0$$
$$k= 6$$ or $$k = -7$$
Substituting the above into (1) and (2) gives us two solutions:
$$x_1 =12$$ and $$x_2 = 14$$, or,
$$x_1 =-14$$ or $$x_2 = -12$$
The following is multiple choice question (with options) to answer.
Set M contains number that satisfy the condition that, if integer x is in the set then x + 2 will also be in the set M. If -4 is one value in the set, which of the following values must also be present in the set M?
I) -6
II) -2
III) 2 | [
"I only",
"II only",
"I and II only",
"II and III only"
] | D | Since -4 is in the set, then the set must also include -2, 0, 2, 4, etc...
Depending on whether -4 is the starting element in the set or not, we can have -6...or not.
The answer is D. |
AQUA-RAT | AQUA-RAT-38225 | F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
F6 = { 0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 1/1 }
F7 = { 0/1, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1/1 }
F8 = { 0/1, 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8, 1/1 }
Centered
F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
The following is multiple choice question (with options) to answer.
What is the HCF of 2/3, 4/12 and 6/5 | [
"7/45",
"2/45",
"4/15",
"1/30"
] | D | Explanation:
HCF of Fractions = HCF of Numerators/LCM of Denominators
= (HCF of 2, 4, 6)/(LCM of 3, 12, 5) = 2/60 = 1/30
Answer: Option D |
AQUA-RAT | AQUA-RAT-38226 | # If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
My work:
We know the sample space $S:$"The set of number of 1 to 1000000" and $|S|=1000000$
Let $E$ the event such that $E:$"The set of number contains the digit 5 in $[1000000]$" We need calculate $|E|$.
I know in $[100]$ we have $5,15,25,35,45,50,51,52,53,54,55,56,57,58,59,65,75,85,95$ then we have 19 numbers contains the digit $5$ in the set $[100]$
Then in $[1000]-[500]$ we have 171 numbers have the digit 5. this implies [1000] have 271 number contains the digit 5.
. . .
Following the previous reasoning we have to $[10000]$ have 3439 number contains the digit 5.
Then, $[100000]$have 40951 number contains the digit 5.
Moreover, $[1000000]$ have 468559 number contain the digit 5.
In consequence the probability of we pick a digit contain the number 5 in the set $[1000000]$ is 0.468
Is correct this?
How else could obtain $|E|$?
Thanks
The following is multiple choice question (with options) to answer.
From (1, 2, 3, 4, 5, 6), one number is picked out and replaced and one number is picked out again. If the sum of the 2 numbers is 3, what is the probability that the 2 numbers included the number 1? | [
"1/5",
"2/5",
"1",
"3/5"
] | C | We know that the sum of two numbers picked was 3. So, there could be the following cases:
(1, 2);
(2, 1);
Total of 2 cases out of which 2 have 1, thus the probability is 2/2=1.
Ans C |
AQUA-RAT | AQUA-RAT-38227 | {BB, BG, GB, GG}
From these possible combinations, we can eliminate the GG combination since we know that one child is a boy. The three remaining possible combinations are:
{BB, BG, GB}
In these combinations there are four boys, of whom we have chosen one. Let's identify them from left to right as B1, B2, B3 and B4. So we have:
{B1B2, B3G, GB4}
Of these four boys, only B3 and B4 have a sister, so our chance of randomly picking one of these boys is 2 in 4, and the probability is 1/2 - as you have indicated.
So, we put all our two-child families into that room, and half the boys will be from two-boy families (two of them supplied by each such family!), and the other half will be from one-boy families. Everything is as we expect. But here, we counted boys, not families.
But now let's look at a different way of selecting the "boy" in the problem. Suppose we randomly choose the two-child _family_ first. Once the family has been selected, we determine that at least one child is a boy. (For example, from all the mothers with two children, we select one and ask her whether she has at least one son.) In this case, an unambiguous statement of the question could be:
From the set of all families with two children, a family is selected at random and is found to have a boy. What is the probability that the other child of the family is a girl?
Note that here we have a pool of families (all of whom are two-child families) and we're pulling one family out of the pool. Once we've selected the family, we determine that there is, in fact, at least one boy.
Since we're told that one child (we don't know which) is a boy, we can eliminate the GG combination. Thus, our remaining possible combinations are:
{BB, BG, GB}
Each of these combinations is still equally likely because we picked one of the four families.
Now we want to count the combinations in which the "other" child is a girl. There are two such combinations: BG and GB.
The following is multiple choice question (with options) to answer.
Each family in a locality has atmost two adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls and more girls than families. Then the minimum possible number of families in the locality is | [
"4",
"5",
"2",
"3"
] | D | Explanation :
As per the question, we need to satisfy three
conditions namely:
1. Adults (A) > Boys (B)
2. Boys (B) > Girls (G)
3. Girls (G) > Families (F)
Clearly, if the number of families is 2, maximum number of adults can only be 4. Now, for the second condition to be satisfied, every family should have atleast two boys and one girl each.
This will result in non-compliance with the first condition because adults will be equal to boys. If we consider the same conditions for 3 families, then all three conditions will be satisfied.
Answer : D |
AQUA-RAT | AQUA-RAT-38228 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A runs twice as fast as B and gives B a start of 50m.How long should the racecourse be so that A and B might reach in the same time? | [
"75 m.",
"80 m.",
"150 m.",
"100 m."
] | D | ratio of speeds of a and b is 2:1
b is 50m away from a but we know that a covers 1 meter(2-1) more in every second than b
the time taken for a to cover 50m is 50/1=50m
so the total time taken by a and b to reach =2*50=100m
ANSWER:D |
AQUA-RAT | AQUA-RAT-38229 | In other words, the measured rectified average voltage will be multiplied by about 1.11 to get the indicated value.
To answer the first part of the question, you need to find $V_{avg}$ for the sawtooth waveform, which you have already correctly done and found it to be 5V.
What the meter will then do is multiply that by 1.11 and it will indicate 5.55V (not 6.5V -- your given answer is not correct, which was probably the source of your problem).
To determine the percentage of error is simply to compare that value with the true RMS value, which again, you have correctly calculated as 5.77V. $$err = \frac{5.55 - 5.77}{5.77} = -0.0381 = -3.81\%$$
So the percentage of error is -3.81%, with the negative sign meaning that the indicated value is lower than the actual. Here again, the given answer was simply not correct.
If this is from a textbook, you might want to see if there are published errata that corrects that. If not, you might send the author a note -- although it's too late for you, it could save countless hours of frustration for future students.
The following is multiple choice question (with options) to answer.
An inspector rejects 0.08% of the meters as defective, How many meters he examine to reject 2 meteres | [
"1300",
"1400",
"2400",
"2500"
] | D | It means that 0.08% of x = 2
(8100×100 ×x)=2
x=2×100×1008
x=2500
ANSWER :D |
AQUA-RAT | AQUA-RAT-38230 | a) How long does it take to stop (time)?
b) How far does it travel before it stops?
Can you work that out?
9. Oct 25, 2016
Yes.
a)
v1=v0+a0(t1-t0)
t1=1.93 s
b)
v12=v02+2a0(x1-x0)
-182.25=-14x1
x1=13.02 m
10. Oct 26, 2016
### PeroK
So, you need to start decelerating $13.02m$ before the point you want to stop. And what is $100m - 13.02m$?
11. Oct 26, 2016
So 100m - 13.02m = 86.98 m would be the position in which the driver must switch from constant velocity to decelerating in order to stop the car at 100m correct?
12. Oct 26, 2016
### PeroK
Yes, exactly. And now you have a much simpler solution to the original problem.
13. Oct 26, 2016
The following is multiple choice question (with options) to answer.
At 3:00 pm, a car has driven 10 miles east. It will continue to drive east at 0.8 minutes per mile and then turn around and drive at 0.8 minutes per mile back to its original starting point. How far can it drive before turning around in order to arrive back to its original starting point by 3:40 pm? | [
"a) 10",
"b) 11",
"c) 12",
"d) 20"
] | D | 0.8 minutes --> 1 mile
1 minute --> 1/0.8 = 10/8 = 1.25 miles/minute
Distance covered in 40 minutes = 1.25 * 40 = 12.5 * 4 = 50 miles
Distance covered in the current direction = Distance covered from the opposite direction (since car returns back to starting point)
Let x be the miles driven before turning
10 + x = 50 - x
2x = 40
x = 20
Answer: D |
AQUA-RAT | AQUA-RAT-38231 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
The prices of tea and coffee per kg were the same in june. In july the price of coffee shot up by 20% and that of tea dropped by 20%. if in july , a mixture containing equal quantities of tea and coffee costs 60/kg. how much did a kg of coffee cost in june ? | [
"50",
"60",
"80",
"100"
] | B | Let the price of tea and coffee be x per kg in June.
Price of tea in July = 1.2x
Price of coffee in July = 0.8x .
In July the price of 1/2 kg (600gm) of tea and 1/2 kg (600gm) of coffee (equal quantities) = 60
1.2x(1/2) + 0.8x(1/2) = 60
=> x =60
Thus proved...option B. |
AQUA-RAT | AQUA-RAT-38232 | # 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 women buys an article and sells it at a profit of 20%. If she had bought it at 20% less and sold it for $75 less, she could have gained 25%. What is the cost price? | [
"392",
"424",
"375",
"273"
] | C | C
375
CP1 = 100 SP1 = 120
CP2 = 80 SP2 = 80 * (125/100) = 100
20 ----- 100
75 ----- ? => 375 |
AQUA-RAT | AQUA-RAT-38233 | electric-circuits, electric-current, electrical-resistance
Title: Resistors in parallel - finding total current
I am given this circuit, where $R_1 = 4.00 \Omega$, $R_2=8.00 \Omega$, and $R_3 =5.00\Omega$. I was given by the problem statement that $I_2$, the current through $R_2$ is $I_2=4\text{A}$. I found the current through $R_1$ to be $I_1 = 8 \text{A}$. I used the relation $\frac{I_1}{I_2} = \frac{R_2}{R_1}$ to solve for $I_1$.
The next part of the problem asks us to find the current $I_3$ through $R_3$. Here is what I tried to do:
I considered the part of $R_1$ and $R_2$ as one combined parallel component to $R_3$. Since $R_1$ and $R_2$ are parallel (WRT each other), I did $\frac{1}{R_{eq}} = \frac{1}{R_1}+\frac{1}{R_2}$. Then I summed their currents, $I_{eq} = I_1 + I_2$.
I am not sure if that last step where I summed their currents is legal.
Is the right idea to treat the $R_1$ and $R_2$ parts as one component parallel to $R_3$? I wanted to use the relation $\frac{I_{eq}}{I_3}=\frac{R_3}{R_{eq}}$, where $I_{eq}$ and $R_{eq}$ denote the whole $R_1$ and $R_2$ component. Yes, what you did is correct. You have rediscovered Kirchoff's current law.
The following is multiple choice question (with options) to answer.
If two resistors, A(R1) and B(R2) stand in parallel with each other in electrical wire, the total resistor appears as R1R2/(R1+R2). If three resistors, A(R1), B(R2), and C(2R2) stand in parallel in electrical wire, what is the ratio W of the resistors’ sum of A and C to the resistors’ sum of A and B? | [
"2(R1+R2):(R1+2R2)",
"(R1+R2):(R1+2R2)",
"(2R1+R2):(R1+2R2)",
"2(R1+R2):(2R1+R2)"
] | A | two resistors A (r1) and B (r2).
total or sum of two resistors appear as
r1r2/r1+r2. It is looks like inversion of sum of rates.
1/r1+1/r2= r1+r2/r1r2.
same way sum of A(r1) and c(2r2)=1/r1+1/2r2
=2r2+r1/r12r2.
inversion rate = r12r2/2r2+r1.
ratio W of sum of a and c/sum of a and b=2r2r1/2r2+r1*r1+r2/r1r2
=2(r1+r2)/2r2+r1.A |
AQUA-RAT | AQUA-RAT-38234 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 50 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne? | [
"1 hour",
"1 hour 20 minutes",
"2 hours 30 minutes",
"1 hour 40 minutes"
] | B | v1 and V2 are speeds.
v1.t /50 = v2
v2.t/40 = v1
v1/v2 = 5/4
which train A would 50. 4/5 mins to cover the same distance
40 + 40 = 80 mins (ANS B) |
AQUA-RAT | AQUA-RAT-38235 | 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 of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long? | [
"5 mins",
"3 mins",
"7 mins",
"9 mins"
] | B | D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec = 3 mins
Answer: B |
AQUA-RAT | AQUA-RAT-38236 | 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 area of a circular field is 17.56 hectares. Find the cost of fencing it at the rate of Rs. 4 per metre approximately | [
"4457",
"4567",
"5943",
"4547"
] | C | Explanation:
Area = (17.56 x 10000) m2= 175600 m2.
ΠR2 = 175600 ⇔ (R)2 = (175600 x (7/22)) ⇔ R = 236.37 m.
Circumference = 2ΠR = (2 x (22/7) x 236.37) m =1485.78 m.
Cost of fencing = Rs. (1485.78 x 4) = Rs. 5943.
Answer: Option C |
AQUA-RAT | AQUA-RAT-38237 | Since we are not interested on the numerical value of this equation, but that the result of the equation to be an integer, not a fraction, the n outside of the parenthesis is irrelevant.
So the equation could be written as, (3n ± 1)/2, and the result of this equation is always an integer for any odd or even value of n.
-
We know that every prime $> 3$ is of the form $6n+1$ or $6n-1$. Now find two numbers $a$ and $b$ such that $a+b = p$ and $a-b =1$, where $p$ is either $6n+1$ or $6n-1$. Suppose $p = 6n+1$ Solving $a+b=6n+1$ and $a-b = 1$, we get $a= 3n+1$ and $b=3n$ Thus $ab$ is a multiple of $6$. Therefore $4ab$ is a multiple of $24$. But $4ab = (a+b)^2 – (a-b)^2$. Thus we see that $p^2-1$ is a multiple of $24$.
-
Note that any prime $p$ which is $>3$ we have $p-1$ or $p+1$ should be divisible by $3.$ $$(2m+1)^2=4m(m+1)+1\equiv1\mod(8)$$ Hence $p^2-1=(p-1)(p+1)$ is divisible by $3\times 8=24.$
-
The great clue, I believe, is the perennially useful fact that p^2-1 is (p+1)(p-1). Since p>3, p must be odd and cannot be divisible by three; since there is one number divisible by three in any set of three consecutive numbers either p+1 or p-1 must be divisible by three; and since every other even number is divisible by four, one of these factors must be divisible by two and the other by four. Simple arithmetic follows: 4x3x2=24.
The following is multiple choice question (with options) to answer.
If n is a natural number, then (6n2 + 6n) is always divisible by: | [
"Both 6 and 12",
"6 only",
"12 only",
"None of these"
] | A | Explanation :
6n2 + 6n = 6n(n + 1)
Hence 6n2 + 6n is always divisible by 6 and 12 (∵ remember that n(n + 1) is always even) Answer : Option A |
AQUA-RAT | AQUA-RAT-38238 | 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 150 m long is running at a speed of 62 kmph. How long does it take to pass a man who is running at 8 kmph in the same direction as the train? | [
"5 sec",
"10 sec",
"12 sec",
"18 sec"
] | B | Answer : B.
Speed of the train relative to man = (62 - 8) kmph
= (54* 5/18) m/sec = 15m/sec
Time taken by the train to cross the man
= Time taken by It to cover 150 m at 15 m / sec = 150 *1/ 15 sec = 10sec |
AQUA-RAT | AQUA-RAT-38239 | javascript, datetime, converting
if(minutes < 10) minutes = '0' + minutes;
return hours + ':' + minutes + ' ' + suffix;
}
totalMinutes rolls over after rounding, so passing in 23.999 is equivalent to passing in 0, and 27 becomes 3, etc.
hours also roll over but start at 12 instead of zero, then 1, 2, 3 and so on.
minutes should be straight-forward, and
suffix basically looks at whether the totalMinutes are greater than or equal to noon-in-minutes.
Snippet with some test cases below (format borrowed from rolfl).
function original(info) {
var suffix='AM';
var hrs = parseInt(Number(info));
var min = Math.round((Number(info)-hrs) * 60);
if (hrs >12) { hrs = hrs - 12; suffix='PM'; }
if (min < 10) {min='0'+min;}
return hrs + ':' + min + ' ' + suffix;
}
function reviewed(fractionalHours) {
var totalMinutes = Math.round(fractionalHours * 60) % (24 * 60),
hours = Math.floor(totalMinutes / 60) % 12 || 12,
minutes = totalMinutes % 60,
suffix = totalMinutes >= 12 * 60 ? 'PM' : 'AM';
if(minutes < 10) minutes = '0' + minutes;
return hours + ':' + minutes + ' ' + suffix;
}
var testCases = [
["12:00 AM", 0], // midnight
["12:30 AM", 0.5],
["1:00 AM", 0.99999999],
["11:30 AM", 11.5],
["12:00 PM", 11.9999999], // noon
["12:00 PM", 12],
["11:30 PM", 23.5],
["12:00 AM", 23.9999999],
["12:00 AM", 24],
["1:00 AM", 25]
];
The following is multiple choice question (with options) to answer.
There are how many hours between x minutes past 12 noon and 8:10 p.m. of the same day, where x < 60? | [
" (490-x)/60",
" (480-x)/60",
" (470-x)/60",
" 60 (60 – x + 7)"
] | A | Let's just say that X is equal to 10 so we will have that the number of hours will be 8, this is our target value. Now plug 10 in the answer choices to see which gives 8.
A is correct here. |
AQUA-RAT | AQUA-RAT-38240 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
P is three times as fast as Q and working together, they can complete a work in 12 days. In how many days can Q alone complete the work? | [
"16",
"88",
"29",
"14"
] | A | P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 12 days, which means Q can do the work in 48 days.
Hence, P can do the work in 16 days. Answer: A |
AQUA-RAT | AQUA-RAT-38241 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 60 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day? | [
" 100",
" 120",
" 140",
" 150"
] | B | Case 1:
let rate = R, Time = T, and Distance =D
so D=RT
Case 2:
(D+60)=(R+5)(T+1)
Case 3:
(D+X)=(R+10)(T+2)
X=120
Ans B |
AQUA-RAT | AQUA-RAT-38242 | We're asked for the price at which the TOTAL PROFIT from the sale of these units = $42,000. Since there are 600 units, then each unit has to bring in$42,000/600 = $70 in profit. The COST of each unit is$90, so to hit that total profit, we just need to increase that $90 by$70 on each unit.
$90 +$70 = $160 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ SVP Joined: 03 Jun 2019 Posts: 1739 Location: India Re: A television manufacturer produces 600 units of a certain model each [#permalink] ### Show Tags 16 Sep 2019, 09:23 RSOHAL wrote: A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of$90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least $42,000? A$110
B$120 C$140
D$160 E$180
A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of $90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least$42,000?
Let the selling price per unit be $x 600 (x - 90) >= 42,000 x -90 >= 70 x>=$160
IMO D
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."
Please provide kudos if you like my post. Kudos encourage active discussions.
My GMAT Resources: -
The following is multiple choice question (with options) to answer.
A BISCUIT manufacturer produces 300 units of a certain model each month at a cost to the manufacturer of £30 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus production costs) on the sales of these units will be at least £30,000? | [
"110",
"330",
"220",
"440"
] | B | 300(x-30)≥30,000
x-30≥300
x≥330
Answer: Option B |
AQUA-RAT | AQUA-RAT-38243 | ### Show Tags
07 Nov 2012, 05:44
breakit wrote:
Bunuel wrote:
breakit wrote:
Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)
Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622
Hope it helps.
Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)
[color=#ff0000] $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$,
$$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> cross-multiply (multiply both parts by 10*9): $$(10-w)(10-w-1)<9$$ --> $$(10-w)(9-w)<9$$.
Hope it's clear.
_________________
Intern
Joined: 09 Sep 2012
Posts: 23
Schools: LBS '14, IMD '16
Re: If 2 different representatives are to be selected at random [#permalink]
### Show Tags
13 Mar 2013, 16:44
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.
This is extreme value problem
for p > 1/2 , p1 * p2 > 1/2 ie p1 or p2 > 1/4
The following is multiple choice question (with options) to answer.
210 college students were asked in a survey if they preferred Windows or Mac brand computers. 60 Students claimed that they preferred Mac to Windows brand computers. One third as many of the students who preferred Mac to Windows, equally preferred both brands. 90 of the students had no preference. How many of the students in the survey preferred Windows to Mac brand computers? | [
"25",
"40",
"50",
"60"
] | B | 120 = 60(Mac) + x(Window) + 20(both) => x=40
ANSWER:B |
AQUA-RAT | AQUA-RAT-38244 | electromagnetism
Title: What happens if I crush a neodymium magnet into powder? There are a couple questions here. Will it retain its magnetism? Can I add the powder to the iron powder currently in my "magnetic" silly putty to make it truly magnetic? And finally, how best do I crush it? The magnet that came with the putty has chipped so I'm willing to sacrifice it in the name of recreational science. When you crush the magnet it just becomes smaller pieces of magnet, so I believe the individual granules should still retain their magnetism. However, the directions of the poles of the magnet granules when put would be just cancel each other out... (because they are in a mess in your powder form with some granules having poles pointing one direction, another some other direction) so your silly putty would not become truly magnetic.
(I believe...)
The following is multiple choice question (with options) to answer.
Magnabulk Corp sells boxes holding d magnets each. The boxes are shipped in crates, each holding b boxes. What is the price charged per magnet, if Magnabulk charges m dollars for each crate? | [
"100bd/m",
"100m/(bd)",
"bd/(100m)",
"m/(bd)"
] | D | crate has = bd magnates
C crate cost = m dollars = 100 m cents
each magnate cost for each crate = m / bd
Ans; D |
AQUA-RAT | AQUA-RAT-38245 | Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 52344
### Show Tags
31 Mar 2014, 00:26
1
karimtajdin wrote:
walker wrote:
jimjohn wrote:
1. z is prime and is 3. So, x=5 SUFF.
Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
Notice that we are told that x, y, and z are integers greater than 1, hence x=25 and y=1 is not possible.
Check here for a complete solution: if-x-y-and-z-are-integers-greater-than-1-and-57122.html#p1346892
Similar question to practice: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Hope it helps.
_________________
Intern
Joined: 14 Mar 2014
Posts: 2
### Show Tags
31 Mar 2014, 05:02
Bunuel wrote:
we are told that x, y, and z are integers greater than 1
Oh! Can't believe I missed that! It makes sense now . Thanks!
Manager
Joined: 10 Mar 2013
Posts: 195
GMAT 1: 620 Q44 V31
GMAT 2: 690 Q47 V37
GMAT 3: 610 Q47 V28
GMAT 4: 700 Q50 V34
GMAT 5: 700 Q49 V36
GMAT 6: 690 Q48 V35
GMAT 7: 750 Q49 V42
GMAT 8: 730 Q50 V39
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
01 Sep 2014, 16:11
Can anyone explain whether my approach is valid?
5^2*z = 3*x^y
(x^y)/z = (5^2)/3 = (5^2a)/(3a)
x^y = 5^2a
z = 3a
(1) z is prime, so a = 1 and x^y = 25 => x = 5
S
The following is multiple choice question (with options) to answer.
The overall age of X and Y is 12 year greater than the overall age of Y and Z. Z is how many decades younger that X? | [
"11",
"15",
"12",
"17"
] | C | C
12
(X + Y) – (Y + Z) = 12
X – Z = 12 |
AQUA-RAT | AQUA-RAT-38246 | 4. As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!
$(2V/pi)=4r^3$ He divided each side by 2 to get...
$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?
You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.
What would you do now?
5. Originally Posted by SuperTyphoon
As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!
$(2V/pi)=4r^3$ He divided each side by 2 to get...
$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?
You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.
What would you do now?
I didn't divide through by two! I actually multiplied through by two and then skipped some steps. May I should have included the extra steps...
$\frac{2V}{\pi}=4r^3\implies {\color{red}2}\cdot\frac{2V}{\pi}={\color{red}2}\c dot 4r^3\implies \frac{4V}{\pi}=8r^3\implies \frac{4V}{\pi}=(2r)^3\implies 2r=\sqrt[3]{\frac{4V}{\pi}}$
Does this clarify things?
--Chris
6. Ahhhh yessss. That makes a big difference! Thanks for clearing that up. My calc teacher always discourages skipping steps for the same reason that happened here.
Thanks again!
The following is multiple choice question (with options) to answer.
In the formula V = (4r)^2, if r is halved, then V is multiplied by | [
"64",
"8",
"1/4",
"1/8"
] | C | Say r=2 => V1 = 16
when r=1; V2 = 4
V2 = 1/4*V1.
Answer : C |
AQUA-RAT | AQUA-RAT-38247 | Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$
-
Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring: \begin{align} \left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2 &=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\ &=8+2\sqrt{16-12}\\[6pt] &=12 \end{align} Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
-
Write as $\sqrt{4+2\sqrt{3}} = a+b\sqrt{3}$. Now square both sides, equate real and radical part. This gives two equations in $a$ and $b$. Now eliminate $a$, solve for $b$. Goes perfect. Same for the other term.
-
The following is multiple choice question (with options) to answer.
Find √? /11= 4 ? | [
"76",
"1936",
"1304",
"1296"
] | B | Answer
Let √N/11= 4
Then √N = 11 x 4 = 44
∴ N = 44 x 44= 1936.
Correct Option: B |
AQUA-RAT | AQUA-RAT-38248 | \begin{align*} =&0.2016+0.1296+0.756+0.336+\\ & 0.864+0.504+0.224+0.324+ \\ &0.144+0.0084+0.0216+0.0036+\\ &0.0096+0.0056+0.002 \end{align*}
$=0.6976$
Edit: Thank you Vaibhav for providing an alternative method.
Alternate Method:
probability that the plane is hit when all the four shots are fired,
$P=1- \text{probability of not hitting the target}$
$=1-(0.6\times 0.7 \times 0.8 \times 0.9)$
$=1-0.3024$
$=\textbf{0.6976}$
Edit 2: For another alternative method, check comments by Soma Sekhar.
## (7) Comment(s)
Pranjal Mishra
()
i guess this answer is wrong because as soon as plane is hit next shot will not be taken. So if the plane is hit in the 1st shot 2nd shot is not taken
in 1st shot p = .4
in 2nd shot p =(1 - .4) * (.3) = .18
in 3rd shot p = (1 - .18) *(.2) = .164
in 4th shot p =(1 - .164) * (.1) = .0836
so the final P = .4 + .18 + .164 + .0836 = .8276
Soma Sekhar
()
Alternate way
0.4+(0.6*0.3)+(0.6*0.7*0.7)+(0.6*0.7*0.8*0.1)
=0.6976
Jack
()
0.4 + 0.6x0.3 + 0.6x0.7x"0.2" + ....
Vaibhav
()
Alternative way
1-probability of not hitting the target
=1-(0.6* 0.7 * 0.8* 0.9)
=0.6976
Deepak
()
Sneha
()
good one...
The following is multiple choice question (with options) to answer.
An gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the 1st, 2nd, third and 4th shots are 1.4, 1.3, 1.2 & 1.1 respectively. What is the probability that the plane is hit when all the four shots are fired? | [
"0.6976",
"0.7013",
"0.789",
"0.8356"
] | A | Required probability:
=(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)=(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)
=0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002=0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002
=0.6976=0.6976
Edit: Thank you Vaibhav for providing an alternative method.
Alternate Method:
probability that the plane is hit when all the four shots are fired,
P=1−probability of not hitting the targetP=1−probability of not hitting the target
=1−(0.6×0.7×0.8×0.9)=1−(0.6×0.7×0.8×0.9)
=1−0.3024=1−0.3024
=0.6976
A |
AQUA-RAT | AQUA-RAT-38249 | Another way: If Alice is to win, it must be after an even number of tosses. So divide the sequence of tosses into pairs. Let D be the event two heads, and let M (mixed) be the event head then tail or tail then head. Each of D and M has probability $\frac{4}{9}$.
The ways Alice wins are D, MD, MMD, MMMD, and so on. So the probability Alice wins is $$\frac{4}{9}+\left(\frac{4}{9}\right)^2+\left(\frac{4}{9}\right)^3+\left(\frac{4}{9}\right)^4+\cdots.$$ This infinite geometric series has sum $\frac{4}{5}$.
Remark: We can use either analysis to find the probability Alice wins if the probability of head is $h$, where $0\lt h\lt 1$.
There are 5 possible states for this game: \begin{align} (C_A, C_B)= \begin{cases} (2,2)\\ (1,3)\\ (3,1)\\ (4,0)\\ (0,4)\\ \end{cases} \end{align}
This writes a Markov Chain with
\begin{align} P((C_A, C_B) \mapsto (C_A, C_B)+(i,j))= \begin{cases} 2/3 \qquad &\text{if } (i,j)=(1,-1) \;\&\&\; C_B \geq 1\\ 1/3 \qquad &\text{if } (i,j)=(-1,1) \;\&\&\; C_A \geq 1\\ 1 \qquad &\text{if } i=j \;\&\&\; (C_A =0 || C_B=0) \\ 0 \qquad &\text{Otherwise} \end{cases} \end{align}
The following is multiple choice question (with options) to answer.
Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 1/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose? | [
"3/140",
"1/280",
"27/280",
"3/35"
] | C | P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 1/5*1/8*5/7 + 1/5*7/8*2/7 + 4/5*1/8*2/7 = 27/280.
Answer: C |
AQUA-RAT | AQUA-RAT-38250 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
From a group of 3 MEN and 3 WOMEN, 4 ADULTS are to be randomly selected. What is the probability that equal numbers of men and women will be selected? | [
"1/5",
"2/5",
"3/5",
"4/5"
] | C | Using the first example, here is the probability of THAT EXACT sequence occurring:
MMWW = (3/6)(2/5)(3/4)(2/3) = 36/360 = 1/10
Each of the other 5 options will yield the exact SAME probability....
eg
MWMW = (3/6)(3/5)(2/4)(2/3) = 36/360 = 1/10
So we have 6 different options that each produce a 1/10 chance of occurring.
6(1/10) = 6/10 = 3/5
Final Answer:
C |
AQUA-RAT | AQUA-RAT-38251 | 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.
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get ? | [
"57%",
"60%",
"65%",
"90%"
] | A | Solution
Total number of votes polled= (1136 + 7636 + 11628) = 20400.
∴ Required percentage= ( 11628/20400x 100)%= 57%
Answer A |
AQUA-RAT | AQUA-RAT-38252 | $$(2)-(1')\Rightarrow f\, =\, -2e\, +\, 765 \: \: \: ---(3)\\ (3)\rightarrow (1)\Rightarrow 22e = 21(-2e+765)+319 \\ e=256 \\ f=253$$
The following is multiple choice question (with options) to answer.
(160)^2 - (161)^2 = | [
"1",
"100",
"229",
"323"
] | D | Using the formula:
(a+1)^2 - a^2 = 2a+1
So, Answer = 161*2 + 1
= 322+1 = 323 = Answer = D |
AQUA-RAT | AQUA-RAT-38253 | ### Show Tags
30 Oct 2018, 10:20
Bunuel wrote:
When a company places a wholesale order on coffee cups with the company logo, they pay $692 for 80 cups. When Phil wants to buy just one such cup, it costs him$12.50. How much above the wholesale price per cup is he paying?
A. $0.55 B.$3.85
C. $7.35 D.$10.65
E. $14.25 The wholesale price is 692/80 =$8.65 per cup.
Thus, Phil pays 12.5 - 8.65 = \$3.85 above the wholesale price if he buys just one cup.
_________________
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Re: When a company places a wholesale order on coffee cups with the compan &nbs [#permalink] 30 Oct 2018, 10:20
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The following is multiple choice question (with options) to answer.
At the wholesale store you can buy an 8-pack of hot dogs for $1.55, a 20-pack for $3.05, and a 400-pack for $22.95. What is the greatest number of hot dogs you can buy at this store with $200? | [
"1,108",
"3,300",
"2,108",
"2,124"
] | B | We have $200 and we have to maximize the number of hot dogs that we can buy with this amount.
Let's try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money, which in this case is 400 for $22.95.
For the sake of calculation, let's take $23. 23x8 gives 184 ,i.e. a total of 400x8 = 3200 hot dogs . We are left with ~$16. Similarly, let's use $3 for calculation. We can buy 5 20-pack hot dogs (3x5), a total of 20x5 = 100 hot dogs. So we have 3300 hot dogs.
2108 looks far-fetched (since we are not likely to be left with > $1.55). Hence, (b) 3300 (ANSWER B) |
AQUA-RAT | AQUA-RAT-38254 | 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.
A bike covers a certain distance at the speed of 62km/h in 8 hrs. If the bike was to cover the same distance in approximately 6 hrs, at what approximate speed should the bike travel? | [
"85 km/h",
"82.66 km/h",
"87.67 km/h",
"90.25 km/h"
] | B | Ans.(B)
Sol. Total distance = 62 × 8 = 496 km
Now speed = 496/6 = 82.66 km/h |
AQUA-RAT | AQUA-RAT-38255 | # 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.
Cindy paddles her kayak upstream at c kilometers per hour, and then returns downstream the same distance at n kilometers per hour. How many kilometers upstream did she travel if she spent a total of p hours for the round trip? | [
"cnp",
"cn/p",
"(c + n)/p",
"cnp/(c + n)"
] | D | let t1 and t2 be the time taken to row upstream and downstream respectively
now,
t1=distance/speed=d/c
similarly,
t2=d/n (as same distance has to be rowed)
also,
t1+t2=p
therefore,
p=(d/c)+(d/n)
=d(c+n)/cn
d=pcn/(c+n)=D |
AQUA-RAT | AQUA-RAT-38256 | The (10/3t)/(7/3t) = 10/7 then the work ratios is 10 to 7
Since Time Ratio is the inverse of work, the the answer is 7 to 10
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
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THANKS = KUDOS. Kudos if my post helped you!
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Originally posted by TudorM on 03 Feb 2015, 22:53.
Last edited by TudorM on 03 Feb 2015, 23:29, edited 2 times in total.
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Joined: 17 Dec 2013
Posts: 58
GMAT Date: 01-08-2015
Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
### Show Tags
04 Feb 2015, 04:14
1
x=0,5t
y=t
z=0,75t
t=4hours so we get:
x=2
y=4
z=3
x+z=1/2+3/4=5/6 -> they need 6/5 hours
y+z=1/4+1/3=7/12 -> they need 12/7 hours
ratio is 6/5 divided by 12/7, or multiplied by 7/12 -> we get 7/10
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Joined: 02 Aug 2009
Posts: 6961
Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Three workers have a productivity ratio of 2 to 4 to 8. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? | [
"20 to 3",
"6 to 1",
"5 to 1",
"1 to 6"
] | A | The fastest worker who does 8 units of job worked for all 10 hours, so he did 8*10=80 units of job;
The slowest worker who does 2 unit of job worked for only 4+2=6 hours (first 4 hours and last 2 hours), so he did 2*6=12 units of job;
The ratio thus is 80 to 12, or 20 to 3.
Answer: A. |
AQUA-RAT | AQUA-RAT-38257 | Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink] 21 Apr 2017, 03:36
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If a, b, c, and d are positive, is a/b > c/d? 1 04 May 2017, 19:50
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8 If a, b, c, and d are positive numbers and a/b < c/d , which of the fo 7 27 Mar 2017, 22:14
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
There are 3 numbers A, B and C. If A:B = 3/4, B:C = 4/5, C:D = 5/6, then A:D will be? | [
"1 : 2",
"3 : 5",
"5 : 7",
"6 : 11"
] | A | Sol. A : B = 3 : 4, B : C = 4 : 5, C : D = 5 : 6
∴A∶B∶C∶D= 3 : 4 : 5 : 6.
Thus, A : D = 3 : 6 or, 1 : 2
A |
AQUA-RAT | AQUA-RAT-38258 | • One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37
The following is multiple choice question (with options) to answer.
If (y - 1)(z - 3) + 5(y - 1) = 0, then which of the following must be true? | [
" y = -4 and z = -3",
" y = 4 or z = -3",
" y = -4 or z = 3",
" y = 1 and z = 3"
] | D | factor out:
(y-1)(z-3+5)=0
(y-1) (z+2) = 0
So by checking options we get ,
Ans - D |
AQUA-RAT | AQUA-RAT-38259 | 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.
Ravi's brother is 3 years senior to him. His father was 28 years of age when his sister was born while his mother was 26 years of age when he was born. If his sister was 4 years of age when his brother was born, what were the ages of Ravi's father and mother respectively when his brother was born ? | [
"32 years, 23 years",
"32 years, 29 years",
"35 years, 29 years",
"35 years, 33 years"
] | A | Explanation:
When Ravi's brother was born, let Ravi's father's age = x years and mother's age = y years.
Then, sister's age = (x - 28) years. So, x - 28 = 4 or x = 32.
Ravi's age = (y - 26) years. Age of Ravi's brother = (y - 26 + 3) years = (y - 23) years.
Now, when Ravi's brother was born, his age = 0 i.e. y - 23 = 0 or y = 23.
Answer: Option A |
AQUA-RAT | AQUA-RAT-38260 | Each of your methods gives you $1.496 \times 10^9 ~\rm{m}$ - accurate to 4 significant digits. That's better than you should expect.
The accuracy of your result is only ever as good as the accuracy of the inputs. Some of the data you have is given to 5 significant digits - that will give you the "more accurate" results.
I recommend that you learn about error propagation. There are millions of resources online for this - a relatively basic introduction can be found here
• Sorry for the late reply! I have been really busy with the project and school. I actually totally forgot about error propagation and simply kept the numbers intact without thinking!
– user156610
May 30, 2017 at 9:01
$a = 149598023$
$e = 0.0167086$
$p = a(1-e^2)$
$p = a(0.99972082268604) = 149,556,258.62576513369892$
$b = \frac{p}{ \sqrt{1-e^2}}$
$= \frac{p} {0.99986040159916324190990908617052}$
$= 149,577,139.35522085944399314340085$
$b = \sqrt{p*a}$ = 149,577,139.35522085944399314340085
$pa = rmax * rmin$
$22,373,320,617,691,160.86368910923516 = rmax * rmin$
$rmax = \frac{p}{(1-e)} = \frac{p}{0.9832914} =152,097,596.5270978$
$rmin = \frac{p}{(1+e)} = \frac{p}{1.0167086} =147,098,449.4729022$
The following is multiple choice question (with options) to answer.
In expressing a length of 81.472 km as nearly as possible with the three significant digits, find the percentage error | [
"0.35%",
"0.34%",
"0.034%",
"0.035%"
] | C | Explanation:
Error = (81.5 - 81.472) = 0.028
Required percentage =
0.028/ 81.472×100=0.034
Option C |
AQUA-RAT | AQUA-RAT-38261 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"10 kmph",
"13 kmph",
"65 kmph",
"55 kmph"
] | A | Speed of the boat in still water
= (60+80)/2
= 70 kmph. Speed of the stream
= (80-60)/2
= 10 kmph.
Answer: A |
AQUA-RAT | AQUA-RAT-38262 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
_________________
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
### Show Tags
07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
The following is multiple choice question (with options) to answer.
There are 400 wolfs at a large zoo. The number of wolfs is 4 times the number of all the other animals combined. How many more wolfs are there than non wolfs animals at the zoo? | [
"450",
"420",
"400",
"390"
] | D | Answer is D.
Total wolfs= 400 = 4 * Other Animals (X)
or X = 10.
So, Difference in wolfs and X = 400-10 = 390. |
AQUA-RAT | AQUA-RAT-38263 | 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 the following is not a leap year? | [
"1200",
"800",
"700",
"2000"
] | C | Explanation :
1. Every year divisible by 4 is a leap year, if it is not a century.
2. Every 4th century is a leap year, but no other century is a leap year.
800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).
Hence 800,1200 and 2000 are leap years
700 is not a 4th century, but it is a century. Hence it is not a leap year
Answer : Option C |
AQUA-RAT | AQUA-RAT-38264 | = 4 52. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Two fair dice are rolled and the sum of the points is noted. For example: 1 roll: 5/6 (83. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. and only one way to roll a 12 (6-6, or boxcars). EXPERIMENTAL PROBABILITIES Simulate rolling two dice 120 times. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Two dice are tossed. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Sample space S = {H,T} and n(s) = 2. That intuition is wrong. When you roll a pair of dice there are 36 possible outcomes. To find the probability we use the mutually exclusive probability formula P(A) + P(B). A sum less than or equal to 4. Isn’t that kind of cool?. Two dice are tossed. So the probability of getting a sum of 4 is 3/36 or 1/12. 10 5 13 ! Find the probability distribution. So 1/36 is part of the. Sum of Two Dice. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Rolling two dice. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. The fundamental counting principle tells us there are 6*6=36 ways to roll two dice, all of them equally likely if the dice are fair. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Let B be the event - The sum
The following is multiple choice question (with options) to answer.
In a simultaneous throw of 2 dice what is the probability of getting a total of 11? | [
"1/4",
"1/18",
"2/11",
"3/13"
] | B | n(S) = 6*6 = 36
E = {(5,6),(6,5)}
Probability = 2/36 = 1/18
Answer is B |
AQUA-RAT | AQUA-RAT-38265 | 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.
Find the odd man out. 1, 3, 9, 12, 19, 29 | [
"12",
"9",
"1",
"3"
] | A | 12 is an even number. All other given numbers are odd
Answer : Option A |
AQUA-RAT | AQUA-RAT-38266 | 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 certain number of men can do a work in 65 days working 6 hours a day. If the number of men are decreased by one-fourth, then for how many hours per day should they work in order to complete the work in 40 days? | [
"11",
"12",
"14",
"13"
] | D | D
13
Let the number of men initially be x. we have M1 D1 H1= M2 D2 H2
So, x * 65 * 6 = (3x)/4 * 40 * h2
=> h2 = (65 * 6 * 4)/(3 * 40) = 13. |
AQUA-RAT | AQUA-RAT-38267 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
In a certain college, 20% of the boys and 40% of the girls attended the annual college outing. If 35% of all the students are boys, what percent of all the employees went to the outing? | [
"33%",
"34%",
"35%",
"36%"
] | A | assume total students =100 in that 35 students are boys ,remaining 65 are girls........
20% of boys(35) =7
40% of girls(65) =26
totally (26+7)=33 mens went to outing...... 33 is 33% in 100
so, the answer is 33%
ANSWER:A |
AQUA-RAT | AQUA-RAT-38268 | algorithms, scheduling, algorithm-design
Title: Schedule two trains whose tracks overlap so they don't crash I've encountered scheduling problems in my algorithms class before like the type we use vertex cover to solve.
Recently I was asked this question and did not even know what algorithmic technique to use to answer it!
There are two trains, which run between stations and share a portion of track. For instance train 1 runs between stations A B C D E F G H and train 2 runs between stations I J K B C D E F L M so the trains share the track between B C D. At each end of the track the train turns around and goes the other way on its track. The trains go from station to station in 1 unit time. I need to prevent the trains from crashing head on on the shared track, I have the power to stop either train at any station and restart it again when I please.
How can I solve this problem without using a ton of if else clauses, what CS principles should I be using here?
edit
Some background, the first solution I wrote was just a simple check to see if the other train was on the shared track before sending the other train out. That seemed ad hoc to me and I'm sure it did to the company I interviewed for as well.
I figure there must be an algorithm or at least a school of thought for this type of problem, after all how do they build extensible systems for managing shared runway space at an airport, or managing traffic flow at stop lights.
My intuition is that this is a very introductory problem to a school of thought in computational problem solving I have not yet encountered.
I might be wrong, but could anybody clear the air for me? If you calculate the cycle, as there are only two trains and distance between every station is one, you could simply run them in order and appoint wait in some station to avoid crash.
If you have to implement some unfortunate events (first one stops for some reason, reschedule it to run on itd own time).
With two trains it will work without problems.
Otherwise use semaphores (it fits even literally ;)
There is no explicitly given synchronisation problem, so nasty trick works.
The following is multiple choice question (with options) to answer.
A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in: | [
"7 4/5",
"9 3/5",
"4 1/6",
"7 7/12"
] | B | A + B + C's 1 day's work = 1/4
A's 1 day's work = 1/16
B's 1 day's work = 1/12
Therefore C's 1 day's work = 1/4 - ( A's 1 day's work + B's 1 day's work )
=> C's 1 day's work = 1/4 - ( 1/16 + 1/12 )
=> C's 1 day's work = (12-3-4)/48 = 5/48
So, C alone can do the work in 48/5 = 9 3/5 days.
So correct answer is B |
AQUA-RAT | AQUA-RAT-38269 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If |4x+6| = 46, what is the sum of all the possible values of x? | [
"2",
"-2",
"4",
"-5"
] | B | There will be two cases
4x+6= 46 or 4x+6 = -46
=> x =10 or x= -12
sum of both the values will be -12 +10 = -2
Answer is B |
AQUA-RAT | AQUA-RAT-38270 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a cistern in 20 and 30 minutes respectively, and a third pipe C can empty it in 60 minutes. How long will it take to fill the cistern if all the three are opened at the same time? | [
"17 1/4 min",
"28 1/7 min",
"15 min",
"17 1/8 min"
] | C | 1/20 + 1/30 - 1/60 = 4/60 = 1/15
15/1 = 15
Answer:C |
AQUA-RAT | AQUA-RAT-38271 | We're asked for the price at which the TOTAL PROFIT from the sale of these units = $42,000. Since there are 600 units, then each unit has to bring in$42,000/600 = $70 in profit. The COST of each unit is$90, so to hit that total profit, we just need to increase that $90 by$70 on each unit.
$90 +$70 = $160 Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: Rich.C@empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ SVP Joined: 03 Jun 2019 Posts: 1739 Location: India Re: A television manufacturer produces 600 units of a certain model each [#permalink] ### Show Tags 16 Sep 2019, 09:23 RSOHAL wrote: A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of$90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least $42,000? A$110
B$120 C$140
D$160 E$180
A television manufacturer produces 600 units of a certain model each month at a cost to the manufacturer of $90 per unit and all of the produced units are sold each month. What is the minimum selling price per unit that will ensure that the monthly profit (revenue from sales minus the manufacturer's cost to produce) on the sales of these units will be at least$42,000?
Let the selling price per unit be $x 600 (x - 90) >= 42,000 x -90 >= 70 x>=$160
IMO D
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The following is multiple choice question (with options) to answer.
Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx2, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
How many units should Mr. David produce daily ? | [
"130",
"100",
"70",
"150"
] | B | Explanation :
Cost price function f(x) = cx2 + bx + 240.
When x = 20, f(20) = 400c + 20b + 240.
=> f(40) = 1600c + 40b + 240
66.67% is 2/3.
=> f(40) – f(20) = 2/3 * f(20).
Substituting f(40) and f(20), we get, 1200c+ 20b =(2/3) * (400c + 20b + 240)
Solving, we get 20b + 2800c = 480.
Now, f(60) = 3600c + 60b + 240.
f(60) – f(40) = ½ * f(40)
Substituting, and solving, we get, 360 + 60b + 2400c = 240 + 60b + 3600c
c = 1/10.
Substituting we get, b = 10.
So, the cost function f(x) = 0.1x2 + 10x + 240.
Selling price = 30 * x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x2 + 10x + 240) = -x2/10 + 20x – 240.
For maximum profit, differentiating this should give 0.
dp/dt = 0, dp/dt = -x/5 + 20
20 – x/5 = 0. x= 100.
Also double differentiating d2p / dt2, we get a negative number, so profit is maximum.
So, profit is maximum when 100 units are produced daily.
Answer : B |
AQUA-RAT | AQUA-RAT-38272 | homework-and-exercises, work, potential-energy
Title: Problem regarding finding work done This problem is given in my physics book. How much work will be done if anyone wants to stack up 12 bricks? Given that each brick is $10 cm$ high and each brick's mass is $2 kg$. ($g=9.8$)
Now in finding my answer I used the mass and height of all the bricks. But in my book they used the mass of all the bricks. But in using the height they used the height of only 11 bricks. So my result is $141.12 J$ and the book's result is $129.36 J$. Why is that? We are supposed to stack 12 bricks on one another. One brick is unmoved. The centre of mass of 11 bricks is 60 cm or 0.6m above initial level after stacking.
Now, the work done against the force of gravity is:
$W = mg \Delta h = 2(11)(9.8)(0.6) = 129.36J $
Another method is to calculate work done individually
$W = mg(\Delta h_1+\Delta h_2+\Delta h_3+\Delta h_4+\Delta h_5+\Delta h_6+\Delta h_7+\Delta h_8+\Delta h_9+\Delta h_{10}+\Delta h_{11})$
$\implies W = 2(9.8)(0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9+1+1.1)$
$\implies W = 129.36J$
The following is multiple choice question (with options) to answer.
In a building there are 5 rooms.Each having a equal area .The length of the room is 4m and breadht is 5 m.The height of the rooms are 2m. If 17 bricks are needed to make a square meter then how many bricks are needed to make the floor of a particular room ? | [
"320",
"340",
"300",
"310"
] | B | Area of the floor = 4m x 5m = 20 m2
17 bricks are needed to make 1 m2 area.
So to make 20 m2 no of bricks needed = 17 x 20 = 340 Answer : B |
AQUA-RAT | AQUA-RAT-38273 | -
You say the answer is 1/2 and seem to argue that the answer is 2/3. – Aryabhata Aug 18 '10 at 20:10
He says the answer is 2/3 for the cards example, which is correct, not for the boys/girls paradox, which should be 1/2 and he so states in the opening paragraph. – Abel Aug 18 '10 at 20:24
Ok, I clarified. – Nate Eldredge Aug 18 '10 at 20:32
Let A be the event "Family has at least one girl". Let B be the event "Boy opens door"
Pr(A and B)=1/4 *1/2 + 1/4*1/2 = 1/4
AND
Pr(B)=1*1/4 + 1/4*1/2 +1/4*1/2 = 1/2
So Pr(A|B) = 1/4 / ( 1/2) = 1/2.
-
We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).
If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on both the children, i.e. it is a function of both of them considered together.
However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.
Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.
Mathematically,
The following is multiple choice question (with options) to answer.
A boy & a girl appear for an interview against 2 vacant posts in an office. The probability of the boy's selection is 1/5 and that of the girl's selection is 1/3. What is the probability that only 1 of them is selected? | [
"1/3",
"2/5",
"2/7",
"3/8"
] | B | Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)
= = 2/5
B |
AQUA-RAT | AQUA-RAT-38274 | # Making Friends around a Circular Table
I have $n$ people seated around a circular table, initially in arbitrary order. At each step, I choose two people and switch their seats. What is the minimum number of steps required such that every person has sat either to the right or to the left of everyone else?
To be specific, we consider two different cases:
1. You can only switch people who are sitting next to each other.
2. You can switch any two people, no matter where they are on the table.
The small cases are relatively simple: if we denote the answer in case 1 and 2 for a given value of $n$ as $f(n)$ and $g(n)$ respectively, then we have $f(x)=g(x)=0$ for $x=1, 2, 3$, $f(4)=g(4)=1$. I’m not sure how I would generalize to larger values, though.
(I initially claimed that $f(5)=g(5)=2$, but corrected it based on @Ryan’s comment).
If you’re interested, this question came up in a conversation with my friends when we were trying to figure out the best way for a large party of people during dinner to all get to know each other.
Edit: The table below compares the current best known value for case 2, $g(n)$, to the theoretical lower bound $\lceil{\frac{1}{8}n(n-3)}\rceil$ for a range of values of $n$. Solutions up to $n=14$ are known to be optimal, in large part due to the work of Andrew Szymczak and PeterKošinár.
The moves corresponding to the current best value are found below. Each ordered pair $(i, j)$ indicates that we switch the people in seats $(i, j)$ with each other, with the seats being labeled from $1 \ldots n$ consecutively around the table.
The following is multiple choice question (with options) to answer.
Find the no.of ways of arranging the boy and 9 guests at a circular table so that the boy always sits in a particular seat? | [
"3!",
"8!",
"7!",
"9!"
] | D | Ans.(D)
Sol. Total number of persons = 10 Host can sit in a particular seat in one way. Now, remaining positions are defined relative to the host. Hence, the remaining can sit in 9 places in 9P9 = 9! Ways ... The number of required arrangements = 9! x 1= 9! = 9! ways |
AQUA-RAT | AQUA-RAT-38275 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 50 miles apart. One and a half hours later, they are again 50 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour? | [
"a) 66.67-x",
"b) 40-x",
"c) 80-2x",
"d) 120-x"
] | A | Let's say the two cars have speeds V1 and V2. The fact that they are moving in opposite direction means that their relative speed is (V1 + V2). In other words, any gap between them will be changing in size at a rate of (V1 + V2). It doesn't matter whether they are moving toward each other or away from each other. If they are approaching each other, the gap between them is decreasing at a rate of (V1 + V2). If they are moving away from each other, the gap between them is increasing at a rate of (V1 + V2). Either way, the number for the rate of change remains the same.
Here, the two cars approach a distance 50 mi, then move away from each other another distance of 50 miles. That's a total distance of 100 miles in 1.5 hr, which gives a rate of:
R = (100 mi)/(1.5) = 66.67 mph
That's the rate of change of the gap, so it must equal the sum of the speeds of the two cars.
One of the speeds is x, and let's call the other y. We want y.
x + y = 66.67
y = 66.67 - x
Answer =(A) |
AQUA-RAT | AQUA-RAT-38276 | \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{r,s} \ . \hspace{1cm} (4)$$ Note that we have also changed the order of the product operations over $i$ and $j$ here. Now the average over the $s_j$ can be performed easily: for each $j$, $$\left\langle \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{s_j} = \frac1 2 \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right) \ ,$$ which is independent of $j$, so that $$\Big\langle P(Ms=0) \Big\rangle_M = 2^{-n} \left\langle \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right)^n \right\rangle_{r} \ . \hspace{2cm} (5)$$ To evaluate the right hand side of $(5)$ we expand the $n^{\rm th}$ power, $$\Big\langle P(Ms=0) \Big\rangle_M = 2^{-n} \sum_{k=0}^n \begin{pmatrix} {n} \\ {k} \end{pmatrix} 2^{-km} \prod_{i=1}^m \left\langle \Big( \omega^{r_i} + \omega^{-r_i} \Big)^k \right\rangle_{r} \ , \hspace{2cm} (6)$$ and likewise the $k^{\rm th}$ power
The following is multiple choice question (with options) to answer.
A salt manufacturing company produced a total of 5000 tonnes of salt in January of a particular year. Starting from February its production increased by 150 tonnes every month over the previous months until the end of the year. Find its average monthly production for that year? | [
"2989",
"2765",
"5800",
"2989"
] | C | Total production of salt by the company in that year
= 5000 + 5150 + 5300 + .... + 6650 = 70050.
Average monthly production of salt for that year
= 70050/12
= 5837.
Answer:C |
AQUA-RAT | AQUA-RAT-38277 | kinematics, speed, distance
Title: How to calculate the speeds two objects need to move at in order to reach different destinations at the same time For example: On a graph there are two points. Each point has a different destination. The distance between Point A and its destination is 50, and the distance between Point B is 100. The points will begin moving at the same time at a constant speed until they both reach their destination. What speed does each point need to move in order to reach their destination at the same time as the other point if given a time, like 5 seconds. Intuitively, if one object travels twice the distance as another in the same amount of time, then first one is moving with twice the speed as the other.
You could show this mathematically by using the equations $$x_1=x_{01}+v_1t\\ x_2=x_{02}+v_2t$$ where $x_{01}$, $x_{02}$ are the initial positions of each object and $v_1$, $v_2$ are their respective speeds and $t$ is the time it takes both objects to reach their destinations.
Since they both start at the same position, then $x_{01}=x_{02}$ and so $$50=v_1t\\ 100=v_2t$$ since one travels a distance $50$ and the other $100$. Solving both these equations by dividing one into the other will give $$\frac{100}{50}=\frac{v_2 t}{v_1 t}$$ so that cancelling the $t$'s will give $$v_2=2v_1$$
The following is multiple choice question (with options) to answer.
A and B are 90 km apart they start to move each other simultaneously A at speed 10 and B at 5 km/hr if every hour they double their speed what is distance that A pass until he meet B? | [
"30 km",
"40 km",
"50 km",
"60 km"
] | D | in 2 hours a covers (10+20)=30 km
b covers( 5+10)=15 km, total covered=45
in the next hour they have to cover (90-45) km with the speeds 40km/hr and 20 km/hr respectively
let them take another t hrs to meet
so sum of the distance to be covered by both is 45km
(40*t)+(20*t)=45km
gives t= (3/4) hrs
we need to fid out the distance covered by a
which equals to =30+(40*(3/4))=60 km
ANSWER:D |
AQUA-RAT | AQUA-RAT-38278 | • You for $v_x=0$. What does that imply about $u_x,u_y,v_y$? – N. S. Jul 9 '14 at 12:58
• Ouch. So $u=c_1,v=c_2\quad (c_{1,2}\in\mathbb{C})$ ? but that's a constant function (not sure whether they asked for a linear function with incline $\neq 0$). Thanks. – user65985 Jul 9 '14 at 13:05
The following is multiple choice question (with options) to answer.
For all numbers u and v, the operation @ is defined by u@v = u^2 - uv. If xy ≠ 0, then which of the following can be equal to zero?
I. x@y
II. (xy)@y
III. x@(x + y) | [
"II",
"I and II",
"I and III",
"II and III"
] | B | u@v = u^2 - uv=u(u-v).... so u@v will be zero if u=v or u=0.. but u cannot be equal to 0.. as per Q, x and y can take any int value except 0...
now lets look at the choices..
when x=y, it will be 0... so ok...
when we put xy=y, it is possible when x=1 and y any integer... so ok again
when we put x=x+y.... only possibility when y=0 and it is given x and y cannot be 0....so not possible
only l and ll possible
answer: B |
AQUA-RAT | AQUA-RAT-38279 | Given: Pat bought 5 pounds of apples.
Target question: How many pounds of pears could Pat have bought for the same amount of money?
This is a good candidate for rephrasing the target question.
Let A = the price per pound of apples
Let P = the price per pound of pears
If Pat bought 5 pounds of apples, then 5A = the total amount that Pat spent
Pat then wants to spend her 5A dollars on pears
So, 5A/P = the number of pounds of pears Pat can buy with the 5A dollars
REPHRASED target question: What is the value of 5A/P?
Aside: Below, you'll find a video with tips on rephrasing the target question
Statement 1: One pound of pears costs 0.50\$ more than one pound of apples.
In other words, P = A + 0.5
Does this help us determine the value of 5A/P?
No.
Take 5A/P and replace P with A + 0.5 to get: 5A/P = 5A/(A + 0.5)
Since there's no way to determine the value of 5A/(A + 0.5) (aka 5A/P), we cannot answer the REPHRASED target question with certainty.
So, statement 1 is NOT SUFFICIENT
Statement 2: One pound of pears costs 3/2 times as much as one pound of apples.
In other words, P = (3/2)A or we can write P = 1.5A
Does this help us determine the value of 5A/P?
Yes!!
Take 5A/P and replace P with 1.5A to get: 5A/P = 5A/1.5A = 5/1.5 = 3 1/3
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
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Re: Pat bought 5 pounds of apples. How many pounds of pears &nbs [#permalink] 26 Jul 2018, 13:24
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Pat bought 5 pounds of apples. How many pounds of pears
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The following is multiple choice question (with options) to answer.
Based on this year's costs, an orchard grower budgets P dollars for planing N new trees next year. If the average cost of planting each tree were to decrease 50% from this year's cost, then the greatest number of trees that the orchard grower could plant next year using P dollars would be: | [
"100% less than N",
"200% less than N",
"Equal to N",
"100% greater than N"
] | D | Correct answer is (D).
This year, the price of a tree is price1=P/N.
If this price decreases by 50% it becomes Price2=P/N*0.5
Then with P dollars, you can grow P/Price2 trees i.e. P/(P/N*0.5) i.e. N/0.5 i.e. 2*N
Which is 100% greater than N. |
AQUA-RAT | AQUA-RAT-38280 | The sum $$S$$ can be written in terms of the first number in the interval, $$a$$, and the length of the interval $$k$$. We have $$S = k(a + (k-1)/2)$$, or $$k$$ times the median of the sequence. For many $$a$$, one can pick a number c next to and different from the median, and c will be coprime to the integer which is either the median or twice the median (in case $$k$$ is even), and if c is also coprime to $$k$$, then we are done.
Before we look at two examples, let us note that for $$a+i$$ in the interval, $$(S,a+i) \gt 1$$ iff $$(k((k-1 - 2i)/2),a+i ) \gt 1$$. Thus we need to check that every number in the interval has (or avoids) a prime factor less than $$k$$. It will now be helpful to look at numbers near the median, and write $$j$$ for $$k$$ when $$k$$ is odd, and $$j$$ for $$k/2$$ when $$k$$ is even.
Example for odd $$k$$: $$3j$$ $$2j$$ $$j$$ $$0$$ $$j$$ $$2j$$ $$3j$$
Example for $$k$$ even: $$5j$$ $$3j$$ $$j$$ $$j$$ $$3j$$ $$5j$$
The following is multiple choice question (with options) to answer.
If S = {1, 2, 3, 4, 5, 6, 7}, how much less is the mean of the numbers in S than the median of the numbers in S? | [
"0",
"1",
"2",
"3"
] | A | Mean = (1 + 2 +3 + 4 + 5 + 6 + 7)/7 = 4
Median = 4
Difference = 4-4 = 0
Option A |
AQUA-RAT | AQUA-RAT-38281 | are these 2's supposed to be squares?
RonL
7. Originally Posted by CaptainBlack
this last line should be:
=> x = (k/3) pi for k an integer
RonL
right, i forgot about the 1/3
The following is multiple choice question (with options) to answer.
If x/y is an integer, which of the following must also be not an integer?
I. xy
II. y/x
III. x | [
"I alone",
"II alone",
"III alone",
"I, II and III"
] | D | Let's take X = 4/3 and Y = 2/3
Then X/Y = 2 which is an integer.
But XY = 4/3 * 2/3 = 8/9 --> Not an integer.
Y/X = 2/3 divided by 4/3 = 2/4 = 1/2 Not an integer.
X alone is 4/3. Not an integer.
Hence D. |
AQUA-RAT | AQUA-RAT-38282 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Chuck and Bob are not seated next to each other? | [
"24",
"48",
"72",
"74"
] | C | When the constraint on an arrangement says,Two people should not be seated together,we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit!
Let's see why....
These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.
Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
Answer :C |
AQUA-RAT | AQUA-RAT-38283 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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The following is multiple choice question (with options) to answer.
Bottle R contains 250 capsules and costs $8.25. Bottle T contains 130 capsules and costs $2.99. What is the difference between the cost per capsule for bottle R and the cost per capsule for bottle T? | [
" $0.25",
" $0.01",
" $0.05",
" $0.03"
] | B | Cost per capsule in R is 8.25/250=0.825/25=0.033
Cost per capsule in T is 2.99/130=0.023
The difference is 0.01
The answer is B |
AQUA-RAT | AQUA-RAT-38284 | # Chapter 7: Coordinate Geometry
1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.
|AB| = $$\sqrt{(x-5)^{2}+(y-1)^{2}}$$
= $$\sqrt{x^{2}+16-8x+y^{2}+4-4y}$$
|AC| = $$\sqrt{(x+2)^{2}+(y-5)^{2}}$$
= $$\sqrt{x^{2}+4+4x+y^{2}+16-8y}$$
Since, |AB| = |AC|
$$\Rightarrow$$ $$x^{2}+y^{2}-8x-4y+20$$ = $$x^{2}+y^{2}+4x-8y+20$$
$$\Rightarrow$$ 8y – 4y = 4x + 8x
$$\Rightarrow$$ y = 3x
2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)
Let the vertices of the $$\bigtriangleup$$ be P(0, 8), Q(0, 0) and R(6, 0)
∴ PQ = $$\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8$$
∴ QR = $$\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6$$
∴ RP = $$\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10$$
∴ Perimeter of $$\bigtriangleup$$ = 8 + 6 + 10 = 24 units
The following is multiple choice question (with options) to answer.
In the rectangular coordinate system, what is the area of triangle CDE with points
C=(7,2)
D=(3,9)
E=(1,4) | [
"17",
"16.3",
"18",
"19"
] | A | Use the formula of the distance formula
square root (x2-x1)^2+(y2-y1)^2
Do this for each of the sides and then the value is given as A. |
AQUA-RAT | AQUA-RAT-38285 | # 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.
I bought two books; for Rs.460. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss? | [
"337.33",
"280.33",
"299.33",
"268.33"
] | D | x*(85/100) = (460 - x)119/100
x = 280
Answer: D |
AQUA-RAT | AQUA-RAT-38286 | kmph. com/2016/01/01/speed-distance-timeVideos, worksheets, 5-a-day and much more. If a car traveled 50 miles over the course of one hour then its average speed will be 50 mph. Distance, Time and Speed Word Problems | GMAT GRE Maths. Exercise: Problems on Speed Time and Distance. org/time-speed-distanceTime, Speed and Distance (popularly known as TSD) is an important topic for written round of placements for any company. sg//sol_04_distance_speed_and_time. Let the time taken to cover the distance downstream = t hrs. Distance Time Speed. Time is entered in minutes, speed in knots and distance in nautical miles (the same formula will work for statute miles and kilometres). Time = Distance / speed = 20/4 = 5 hours. Speed, Time, and Distance Worksheet. Equations: Acccceelleerraattiioonn = Final speed–Initial TTiimme = Final Speed–Initial Time Acceleration. When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time. b. Acceleration – change in velocity over time. 4 Calculate speed, distance, and time I . Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. He drives 150 meters in 18 seconds. Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. Aptitude Reasoning TRAINS DISTANCE SPEED TIME QUANTITATIVE APTITUDE . We define speed as distance divided by time. A train travels at a speed of 30mph and travel a distance of 240 miles. It can cross a pole in 10 seconds. The speeds are indicated in the rate column. Toon Train is traveling at the speed of 10 m/s at the top of a hill. If you run around the house randomly, and then end up back where you started, moving a total of 44 meters, what is your distance? Average speed for the entire trip is going to be equal to the total
The following is multiple choice question (with options) to answer.
The train travels at an average speed of 16km/h, to the top of the hill where the midpoint of the trip is. Going down hill, train travels at an average speed of 8km/h. Which of the following is the closest approximation of train's average speed, in kilometers per hour, for the round trip? | [
"9.6",
"10.6",
"4.3",
"2.5"
] | B | Since we're not given a distance to work with, we can TEST any distance that we like. Here, since the speeds are 16km/hour and 8 km/hour, using a multiple of both 16 and 8 would make the math easiest.
Let's set the Distance up the hill at 64 km.
So, going uphill, we have....
D = (R)(T)
64 = (16)(T)
64/16 = 4 = T
4 hours to go uphill
Going downhill, we have....
D = (R)(T)
60 = (20)(T)
64/8 = 8 = T
8 hours to go downhill
Total Distance = 120 km
Total Time = 4+8 = 12 hours
Average Speed = 128/12 = 10.6 km/hour
B |
AQUA-RAT | AQUA-RAT-38287 | c) $$\frac{1}{15,000}$$ percent
d) $$\frac{1}{4,000}$$ percent
e) $$\frac{1}{40}$$ percent
Let x = the desired percent
In other words: 2.5 million is x percent of 10 billion
In other words: 2,500,000 = x% of 10,000,000,000
In other words: 2,500,000 = (x/100)(10,000,000,000)
Divide both sides by 100,000 to get: 25 = (x/100)(100,000)
Simplify to get: 25 = 1000x
Divide both sides by 1000 to get: x = 25/1000 = 1/40
Cheers,
Brent
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Re: 2.5 million is what percent of 10 billion? [#permalink]
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19 Mar 2020, 07:56
Having some issues figuring out what I did wrong. ScottTargetTestPrep or JeffTargetTestPrep care to share any light?
step 1] Set up equation
2,500,000 = $$\frac{x}{100} * 10,000,000,000$$
I simplified the equation as follows
2,500,000 = 25 x $$10^5$$
10,000,000,000 = 10 x $$10^9$$
25 x $$10^5$$ = $$\frac{x}{10^2} * 10^9$$
Then
25 x $$10^5$$ = x * $$10^{11}$$
$$\frac{25 x 10^5 }{10^{11} }$$ = x
25 x $$10^ {-6}$$ = x
Which is equivalent to
$$\frac{25}{1,000,000}$$ = x
The following is multiple choice question (with options) to answer.
If 25% of x is 15 less than 15% of 1500, then x is? | [
"657",
"765",
"865",
"840"
] | D | 25% of x = x/4 ; 15% of 1500
= 15/100 * 1500 = 225
Given that, x/4 = 225 - 15
=> x/4 = 210 => x = 840.
Answer:D |
AQUA-RAT | AQUA-RAT-38288 | (1\times3\times7\times9)^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon\equiv 9^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon$, where $\epsilon$ is either $1$, $3$ or $9$ depending on $n$ the last digit of $n$. These two reduction rules together with the modular reduction of $n\underset{5}{!}$ cut the size of $n$ in half, and repeated application will bring the size of $n$ down fairly fast. We have:
The following is multiple choice question (with options) to answer.
What is the least number of square tiles required to pave the floor of a room
15 m 17 cm long and 9m 2 cm broad? | [
"814",
"802",
"836",
"900"
] | A | L = 15 M 17 CM =1517 CM
B = 9 M 2 CM = 902 CM^2
AREA = 1517 * 902 CM^2
FIND OUT HCF = 41
AREA OF SQUARE = 41 * 41 = 1517 * 902 / 41 * 41 = 37 * 22
=407 * 2 =814
ANSWER A |
AQUA-RAT | AQUA-RAT-38289 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
2 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work? | [
"12/9 days",
"30/11 days",
"82/8 days",
"22/76 days"
] | B | 2M = 6W ---- 20 days
12M + 8W -----?
36W + 8 W = 44W ---?
6W ---- 20 44 -----?
6 * 20 = 44 * x => x = 30/11 days
Answer:B |
AQUA-RAT | AQUA-RAT-38290 | 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 contractor undertakes to do a job within 100 days and hires 10 people to do it. After 20 days, he realizes that one fourth of the work is done so he fires 2 people. In how many more days Q will the work get over? | [
"Q=60",
"Q=70",
"Q=75",
"80"
] | C | We can also use the concept of man-days here
100 days -->10men so the job includes 100*10=1000 man-days
After 20 days
1/4 of Job is completed so 1/4 X 1000 man-days=250 man-days Job is done
Now the Balance Job=1000-250=750 man-days worth of Job
Since 2 men are fired so B/L men=8
Therefore Total no. of days of Job=750 man-day/8 days = 375/4=94 days (approx.)
Now since this is total and Ques. is asking for additional no. of days,
So 94-20=74 days
The nearest approx. to answer is 75
Ans: C (75 days) |
AQUA-RAT | AQUA-RAT-38291 | WAIT!
S=11!
Because we have 11! ways of arranging the 11 items.
Sorry about that.
5. Hello, sweeetcaroline!
Edit: Plato is absolutely right . . . *blush*
6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order.
What is the probability that no two neighboring beads are the same color?
There are: . ${11\choose6,4,1} \:=\:2310$ possible orders.
Place the 6 Red beads in a row, leaving a space between them.
. . $\begin{array}{ccccccccccccc} R & \_ & R & \_ & R & \_ & R & \_& R & \_ & R\end{array}$
Place the Blue bead is any of the 5 spaces: . $5$ choices.
Drop the 4 White beads in the remaining 4 spaces: . $1$ way.
Hence, there are: . $5\cdot1 \,=\,5$ ways.
Therefore, the probability is: . $\frac{5}{2310} \;\;=\;\;\frac{1}{462}$
6. Originally Posted by sweeetcaroline
the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
Can someone give me a line of these beads in which no two neighboring beads are the same color other than the five listed below?
$RWRWRWRWRBR$
$RWRWRWRBRWR$
$RWRWRWBWRWR$
$RWRBRWRWRWR$
$RBRWRWRWRWR$
The following is multiple choice question (with options) to answer.
R-R-R-G-G-G-G-Y-Y-Y-B-R-R-R-G-G-G-G-Y-Y-Y-B… B-R-R
The preceding is a representation of the different colored beads on a string. The beads follow a repeating pattern and the colors Red, Green, Yellow, and Blue are represented by R, G, Y, and B respectively. Which of the following is a possible number of beads in the missing section of the string represented above? | [
"64",
"65",
"66",
"67"
] | B | The repeated pattern R-R-R-G-G-G-G-Y-Y-Y-B has 11 beads.
But the missing section includes R-R-R-G-G-G-G-Y-Y-Y at the end.
Thus the number of beads in the missing section has the form 11k + 10.
The answer is B. |
AQUA-RAT | AQUA-RAT-38292 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If 20 men can build a wall 66 metres long in 10 days, what length of a similar can be built by 86 men in 8 days? | [
"227.04 mtrs",
"378.4 mtrs",
"478.4 mtrs",
"488.4 mtrs"
] | A | If 20 men can build a wall 66 metres long in 10 days,
length of a similar wall that can be built by 86 men in 8 days = (66*86*8)/(10*20) = 227.04 mtrs
ANSWER:A |
AQUA-RAT | AQUA-RAT-38293 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains of equal are running on parallel lines in the same direction at 49 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is? | [
"50",
"88",
"65",
"55"
] | C | Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 49 - 36 = 13 km/hr.
= 13 * 5/18 = 65/18 m/sec.
2x/36 = 65/18 => x = 65.
Answer:C |
AQUA-RAT | AQUA-RAT-38294 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
### GMAT/MBA Expert
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\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.
Three coworkers P , W , R meet for Dinner. P had 5 sandwiches, W had 3 and R had none. Both PW share their sandwiches with R such that each one got the same amount. If R paid $ 8 to PW, how much of $ 8 should P get? Assume both sandwiches and Dollars can be split. | [
"a) 5$",
"b) 3 $",
"c) 7 $",
"d) 4 $"
] | C | we got total 8 sandwiches ....we are dividing equally in 3 people.
per person: 8/3
initially
p:8
W:3
r:0
now since r ate 8/3 so he is paying 8 dollars for 8/3 sandwiches.
now p also ate 8/3 of his 5 sandwiches = so he gave 5-8/3 to r ==>7/3 to r
ok we have 8 dollars for 8/3 sandwich
therefore for 7/3 sandwich we will get 7 dollars.
hence p should get 7 dollars.
hence C |
AQUA-RAT | AQUA-RAT-38295 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A company’s four cars running 10 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 25% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days? | [
"1200 lt",
"1555 lt",
"1674 lt",
"1728 lt"
] | C | Given that
4 cars running 10 hrs a day consume 1200 lts. of fuel in 10 days.
1 car consumption per hour per day = 1200 /4 *10 *10 = 3 litre
Now question say new car consume 20% less fuel than the company’s four cars = 75/100 of 3 = 2.25 ( 25 percent less than 3)
Hence we calculate total consumption for next 6 days, the company will need to run 5 new cars for 12 hrs = 2.25 *12 *6*5 =810
similarly = old 4 car consumption for next 6 days for 12 hrs = 3*6*12*4 = 864
hence total is = 810+864 = 1674 lt
Ans is C . |
AQUA-RAT | AQUA-RAT-38296 | ### Show Tags
09 Oct 2015, 14:30
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
I found an almost identical question with the same question stem, but different statements.
Here it is: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Manager
Joined: 12 Sep 2015
Posts: 80
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8803
Location: Pune, India
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
### Show Tags
08 Apr 2016, 21:31
MrSobe17 wrote:
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
z can take any value in that case. Think of a case in which z = 12.
$$5^2 * 3 * 2^2 = 3 * x^y$$
Here x = 10
_________________
The following is multiple choice question (with options) to answer.
For integers x, y, and z, if (3^x) (4^y) (5^z) = 52,428,800,000 and x + y + z = 17, what is the value of xy/z? | [
"0",
"2",
"4",
"6"
] | A | The number 52,428,800,000 is not divisible by 3. (We can verify this by adding the digits of the number to see that the sum is not a multiple of 3.)
Thus x = 0 and xy/z = 0.
The answer is A. |
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