source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36997 | Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240.
The following is multiple choice question (with options) to answer.
To create paint with a certain shade of gray, one must combine 2.75 liters of black paint with every one liter of white paint. Approximately how many liters of white paint must be combined with 350 liters of black paint to create the certain shade of gray? | [
"127.2",
"129.4",
"347.1",
"694.4"
] | A | Since the ratio of Black to White is 2.75 to 1 this implies that Amount of Black is going to be more than (approximately double) the amount of White. This implies Option C, D and E can not be correct.
Now between A and B:
Black:White = 2.75:1 Since the ratio is more than double, the amount of White Paint should be less than half of Black Paint. B is more than half so can not be the answer, A fits the criteria hence should be the correct answer. No calculation needed whatsoever. |
AQUA-RAT | AQUA-RAT-36998 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A pupil's marks were wrongly entered as 83 instead of 70. Due to the average marks for the class got increased by half. The number of pupils in the class is? | [
"18",
"82",
"26",
"27"
] | C | Let there be x pupils in the class.
Total increase in marks
= (x * 1/2) = x/2
x/2 = (83 - 70) => x/2
= 13 => x = 26.
Answer: C |
AQUA-RAT | AQUA-RAT-36999 | newtonian-mechanics, classical-mechanics
Title: Why is a $5-60 mph$ time slower than a $0-60 mph$ time for some automobiles? This doesn't make a lot of sense to me, from a physics 101 point of view. I've read a few blog entries on why this is, but none of them explain it well or are convincing. "something-something launch control. something-something computers." Nothing in physics terms or equations.
For instance, Car and Driver magazine tested the Porsche Macan GTS. The $x-60$ times are:
Rolling start, $5-60\; \mathrm{mph}: 5.4\;\mathrm{ s}$
$0-60\;\mathrm{mph}: 4.4\;\mathrm{s}$
That's a whole second - about $20$% faster from a dead stop than with some momentum - which seems rather huge.
edit:
here is the article for this particular example. But I've noticed this with many cars that are tested for $0-60$ and $5-60$ times.
Here is another example - an SUV.
Another example.
And finally, interesting, even for the Tesla Model S (EV) where power doesn't depend on engine RPM, $0-60$ is still slightly faster than $5-60.$ Ok, from the link given by @count_to_10, I think the answer is clear from this response:
You can launch from a dead stop at any RPM you want, whereas from 5
MPH it's assumed the car is already in gear at low RPMs.
When you start from a standstill, you can rev the engine to any RPM you like before throwing the clutch to engage the axle. Maybe you could match the static friction of the surface to achieve the maximum possible acceleration. When they start at 5 mph, another answer on that site makes it clear that they assume your RPMs are matched to your motion:
"What about rolling at 5mph and dropping the clutch like a regular
launch? Wouldn't that help?"
Yeah, but that's not how they test 5-60 or any other rolling
acceleration tests. That's the point of them: to test how much passing
power you have while already rolling, in gear without a clutch-drop.
So the engine has to move through the entire range of RPMs, which takes more power.
The following is multiple choice question (with options) to answer.
If a car at 60 km/hr instead of 50 km/hr,the car would have gone 20 km more. The actual distance traveled by the car is? | [
"100 km",
"58 km",
"60 km",
"90 km"
] | A | Let the actual distance traveled be x km. Then,
x/50 = (x + 20)/60
6x - 5x = 100 =>x = 100 km.
Answer :A |
AQUA-RAT | AQUA-RAT-37000 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4383
Location: India
GPA: 3.5
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Math Expert
Joined: 02 Sep 2009
Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A metal company's old machine makes bolts at a constant rate of 100 bolts per hour. The company's new machine makes bolts at a constant rate of 150 bolts per hour. If both machines start at the same time and continue making bolts simultaneously, how many minutes will it take the two machines to make a total of 500 bolts? | [
" 36",
" 72",
" 120",
" 144"
] | C | Old Machine
100 bolts in 60 mins
so,
5/3 bolts in 1 min
New Machine
150 bolts in 60 mins
so,
5/2 bolts in 1 min
together,
5/3 + 5/2 =
25/6 bolts in 1 min
so, for 500 bolts
500 * 6 / 25 = 120 mins
Ans C |
AQUA-RAT | AQUA-RAT-37001 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
Director
Joined: 13 Mar 2017
Posts: 703
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)
What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Intern
Joined: 03 Dec 2017
Posts: 18
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Which of the following is equal to 1(1/3)%? | [
"0.012/100",
"0.12/100",
"1.3/100",
"12/100"
] | C | This notation may be confusing for some, since it looks like we're multiplying 1 and 1/3
How about adding a space:Which of the following is equal to (1 1/3)%
(1 1/3)% = 1.3% = 1.3/100
Answer:
C |
AQUA-RAT | AQUA-RAT-37002 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A van takes 6 hours to cover a distance of 252 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time? | [
"28 Kmph",
"60 Kmph",
"70 Kmph",
"80 Kmph"
] | A | Time = 6
Distence = 252
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 252/9 = 28 Kmph
A |
AQUA-RAT | AQUA-RAT-37003 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
Two pipes A and B can separately empty a cistern in 45 min and 30 min respectively. There is a third pipe in the top of the cistern to fill it. If all the three pipes are simultaneously opened, then the cistern is full in 90 min. In how much time, the third pipe alone can fill the cistern? | [
"22.5 min",
"15 min",
"30 min",
"20 min"
] | A | Work done by the third pipe in 1 min = (1/45 + 1/30) - 1/60 = 4/90.
[+ve sign means filling]
The third pipe alone can fill the cistern in 90/4= 22.5 min.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37004 | # Random Gift Giving at a Party - Combinatorics Problem
Each of $$10$$ employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents?
Firstly, there are $$10^{10}$$ total ways to give the $$10$$ employees the $$10$$ presents. So this is our denominator.
My attempt was to consider the complement and consider the number of ways that either $$0$$ employees receive no presents (every employee gets a present) or $$1$$ employee receives no present.
Case 1: $$0$$ employees
There are $$10$$ employees and $$10$$ presents. So there are $$10^{10}$$ ways to give the presents.
Case 2: $$1$$ employee
Step 1: Decide which employee receives no presents: $$10$$ possibilities.
Step 2: Distribute the $$10$$ presents to the remaining $$9$$ employees: $$9^{10}$$ ways.
So the number of ways in which at least $$2$$ employees receive no presents is: $$1-(10^{10}+9^{10}$$).
So my final answer is: $$1-\displaystyle\frac{(10^{10}+9^{10})}{10^{10}}$$.
However, this answer does not match the answer in my textbook. Which is: $$1-\displaystyle\frac{10!-10\times 9 \times \frac{10!}{2!}}{10^{10}}$$
Where did my attempt go wrong and how can I correct it?
The following is multiple choice question (with options) to answer.
A company plans to award prizes to its top 3 salespeople, with the largest prize going to the top salesperson, the next-largest prize to the next salesperson, and a smaller prize to the third-ranking salesperson. If the company has 11 salespeople on staff, how many different arrangements of winners are possible? | [
"1,728",
"1,440",
"990",
"220"
] | C | [quote=Bunuel]A company plans to award prizes to its top 3 salespeople, with the largest prize going to the top salesperson, the next-largest prize to the next salesperson, and a smaller prize to the third-ranking salesperson. If the company has 11 salespeople on staff, how many different arrangements of winners are possible?
11*10*9
=990
Answer : C |
AQUA-RAT | AQUA-RAT-37005 | 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 1000 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is? | [
"288 m",
"256 m",
"1000 m",
"278 m"
] | C | Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (1000 + x)/60
= 65/3
x = 300 m.
Answer:C |
AQUA-RAT | AQUA-RAT-37006 | 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 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is? | [
"10.9",
"10.6",
"10.4",
"10.8"
] | D | Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.
Answer: D |
AQUA-RAT | AQUA-RAT-37007 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Tom reads at an average rate of 30 pages per hour, while Jan reads at an average rate of 38 pages per hour. If Tom starts reading a novel at 4:30, and Jan begins reading an identical copy of the same book at 5:18, at what time will they be reading the same page? | [
"7:48",
"8:18",
"8:48",
"9:18"
] | B | Since Tom reads an average of 1 page every 2 minutes, Tom will read 24 pages in the first 48 minutes. Jan can catch Tom at a rate of 8 pages per hour, so it will take 3 hours to catch Tom.
The answer is B. |
AQUA-RAT | AQUA-RAT-37008 | 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.
Car A leaves the airport at 9:00 am and travels at a constant rate of 30 km per hour. Car B leaves the airport at 9:30 am and travels in the same direction along the same highway at a constant rate of 40 km per hour. At what time will Car B be 40 km ahead of Car A? | [
"3:00 pm",
"3:30 pm",
"4:00 pm",
"4:30 pm"
] | A | At 9:30 am, Car A will be 15 km ahead of Car B.
Car B travels at a rate 10 km per hour faster than Car A.
Car B needs to travel 55 km more than Car A in order to be 40 km ahead.
This will take 55 km/10 km per hour=5.5 hours.
Car B will be 55 km ahead at 3:00 pm.
The answer is A. |
AQUA-RAT | AQUA-RAT-37009 | So on.
=====
Another thing to note:
(Assume only positive values)
$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$
So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$
That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.
But it doesn't go the other way!
$$\frac 12 < a < 4 \not \implies 1 < a < 2$$
Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$
[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)
• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06
The following is multiple choice question (with options) to answer.
If ab - c = a(b - c), which of the following must be true? | [
" a=0 and c=0",
" a=1/2 and b=2",
" b=1 and c=0",
" a=1 or c=0"
] | D | ab-c = a (b-c)
ab - c = ab - ac
c= ac
ac-c = 0
c(a-1) = 0
Either c = 0; or a = 1
D is the answer |
AQUA-RAT | AQUA-RAT-37010 | c, playing-cards
if( isSTR8( str8 ) ) return straight; // should be obvious
// determine highest and 2nd highest tallies of card face values
int hi1 = 0, hi2 = 0;
for( ; popV; popV >>= 4 ) { // finish search asap!
int pop = (int)(popV & 0xF);
if( hi1 < pop ) { hi2 = hi1; hi1 = pop; }
else if( hi2 < pop ) hi2 = pop;
}
switch( hi1 ) { // dispatch...
case 4: return fourOfKind;
case 3: return hi2 >= 2 ? fullhouse : threeOfKind;
case 2: return hi2 == 2 ? twoPair : onePair;
default: return highCard;
}
}
int *show( int *h ) { // called by test()
for( int i = 0; i < nCard; i++ )
printf( "%c%c ", "hcds"[h[i]/oneSuit], "A23456789XJQK"[h[i]%oneSuit] );
return h;
}
The following is multiple choice question (with options) to answer.
What is the difference between the place value and face value of 2 in the numeral 2578? | [
"2000",
"2002",
"1998",
"1990"
] | C | Difference between the place value and face value of 2 = 2000-2 = 1998
Answer is C |
AQUA-RAT | AQUA-RAT-37011 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
_________________
Intern
Joined: 26 May 2010
Posts: 10
Followers: 0
Kudos [?]: 33 [5] , given: 4
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 10 numbers is 23. If each number is increased by 4, what will the new average be? | [
"17",
"65",
"12",
"27"
] | D | Sum of the 10 numbers = 230
If each number is increased by 4, the total increase
=4 * 10 = 40
The new sum = 230 + 40 = 270 The new average
= 270/10 = 27.
Answer: D |
AQUA-RAT | AQUA-RAT-37012 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"157 m",
"717 m",
"142 m",
"150 m"
] | D | Speed=(60 * 5/18) m/sec
= (50/3) m/sec Length of the train
= (Speed x Time) = (50/3 * 9) m
= 150 m.
Answer:D |
AQUA-RAT | AQUA-RAT-37013 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
How much water should be added to 14 liters of a 20%-solution of alcohol to reduce the concentration of alcohol in the solution by 75% ? | [
"25 liters",
"27 liters",
"26 liters",
"32 liters"
] | C | Let x ltr water to be added
2ltr alcohol to be represented as (20(1-3/4(new soln.=14+x)))
2=5%*(14+x)-------->x=26
Ans C |
AQUA-RAT | AQUA-RAT-37014 | # If $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$,then how to find $k$ such that $f(x + k) = f(x)$?
Let $f(x)$ be a function such that $f(x + 1) + f(x − 1) = f(x), \forall x \in \mathbb{R}$. Then for what value of $k$ is the relation $f(x + k) = f(x)$ necessarily true for every real $x$?
The answer/solution suggested in my module is like this: "this is a bit involved but can be proved that $k=6$". Could anybody explain me this?
-
Just add two consecutive relations: $$f(x+2)+f(x)=f(x+1)$$ $$f(x+3)+f(x+1)=f(x+2)$$ Then you'll get $$f(x+3)+f(x)=0$$ vor every real $x$. You have also $f(x+6)+f(x+3)=0$ for every real $x$, and therefore $f(x)=f(x+6)$ for every real $x$.
The following is multiple choice question (with options) to answer.
If f is a function defined for all k by f(k) = k^3 /16, what is f(2 k) in terms of f(k)? | [
"1/8 f(k)",
"8 f(k)",
"2 f(k)",
"10 f(k)"
] | B | f(k) = k^3 /16
f(2k) = (2k)^3 /16 =8 * k^3/16 = 8 x (k^3 /16) = 8 f(k).
So answer is B. |
AQUA-RAT | AQUA-RAT-37015 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Rs. 700 is divided among A, B, C so that A receives half as much as B and B half as much as C. Then C's share is | [
"500",
"400",
"300",
"1200"
] | B | C=x
B=x/2
A=x/4
A:B:C=1:2:4
C's Share Rs(4/7)*700 = 400
ANSWER B |
AQUA-RAT | AQUA-RAT-37016 | in the tool 's defaults in bond... Often misuse the term or use it inappropriately to gain an advantage in the equation this gives you$,. Value or face value of $100 calculate its return based on the market price a! ( YTM ) = [ ( face value/Bond price ) 1/Time period ] -1 the time that they.. Other details of the same risk & maturity Hey presto other details of the current yield an. Market value of 10 years and par value and periodic coupon payments to the holder a people... For calculating stocks and bonds & B bond will be calculated as follows: for bond a value/Bond ). Is 18.53 % relevance of the current yield involves two variables: annual cash and... Result by 100 to calculate realised yield of a bond formula bond ’ s yield, but is expressed as an annual.! And periodic coupon payments to the holder a promissory note the term or use it inappropriately to gain advantage! Calculations involving Money, credit, and bonds yield, or the return on the.... Bonds of the bond ’ s current price a promissory note in evaluating multiple bonds of the is. % interest has a maturity value of Rs Money, credit, and face of. Will be calculated as follows: for bond a credit, and face value impact relationship... ) statistic we 'll use the example, multiply 0.0477 by 100 to calculate its return on! For an example and compute the current yield for an example and compute the current of., credit, and bonds yield, but is expressed as an annual rate variables. Price result in bond yields - current yield formula can be seen in evaluating multiple bonds of the risk... Result in bond yields - current yield is: =YIELD ( C4, realised yield of a bond formula, C6, C7 C8. ( face value/Bond price ) 1/Time period ] -1 payments constitute the potential future cash flows into the... The amount of the same risk & maturity market value of Rs function of price, changes in result... In evaluating multiple bonds of the same risk & maturity bond refers to holder... Purchase price stocks and bonds yield, but is expressed as an rate... To maturity is considered a long-term bond yield, risk, return and more are very in! The relevance of the bond ’
The following is multiple choice question (with options) to answer.
Last year a certain bond yielded 5 percent of its face value in interest. If that interest was approximately 4 percent of the bonds selling price of $7,500, what is the bonds face value? | [
"$6,000",
"$6,750",
"$7,425",
"$7,500"
] | A | Interest = 0.05*face value = 0.04*7,500 --> face value = 0.04*7,500/0.05 = 6,000.
Answer: A. |
AQUA-RAT | AQUA-RAT-37017 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
X and Y invested in a business. They earned some profit which they divided in the ratio of 2 : 3. If X invested Rs.60,000. the amount invested by Y is | [
"Rs.45,000",
"Rs.50,000",
"Rs.60,000",
"Rs.90,000"
] | D | Solution
Suppose Y invested Rs.y
Then, 60000 /y = 2 / 3
‹=› y=(60000×3 / 2).
‹=› y=90000.
Answer D |
AQUA-RAT | AQUA-RAT-37018 | 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.
10 men and 15 women together can complete a work in 5 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work? | [
"127 days",
"150 days",
"177 days",
"187 days"
] | B | 1 man's 1 day work = 1/100
(10 men + 15 women)'s 1 day work = 1/5
15 women's 1 day work = (1/5 - 10/100) = 1/10
1 woman's 1 day work = 1/150
1 woman alone can complete the work in 150 days.
Answer:B |
AQUA-RAT | AQUA-RAT-37019 | 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.
The no. of girls in a class are seven times the no. of boys, which value cannever be the of total students? | [
"10",
"25",
"30",
"45"
] | C | Let the boys are X, then girls are 7X, total = X+7X = 8X
So it should be multiple of 8, 30 is not a multiple of 8.
C |
AQUA-RAT | AQUA-RAT-37020 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
Evaluate
1916−−−−√ | [
"116",
"115",
"114",
"113"
] | C | Explanation:
=2516−−−√
=25−−√16−−√
=54
=114
Answer: C |
AQUA-RAT | AQUA-RAT-37021 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains 135 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet? | [
"7.19",
"7.18",
"7.44",
"7.15"
] | C | T = (135 + 165)/ (80 + 65) * 18/5
T = 7.44
Answer: C |
AQUA-RAT | AQUA-RAT-37022 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
_________________
Number Properties | Algebra |Quant Workshop
Success Stories
Guillermo's Success Story | Carrie's Success Story
Ace GMAT quant
Articles and Question to reach Q51 | Question of the week
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities
Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2324
The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
If the price of a TV is first decreased by 20% and then increased by 40%, then the net change in the price will be : | [
"4% increase",
"12% increase",
"10% decrease",
"6% increase"
] | B | Explanation :
Solution: let the original price be Rs. 100.
New final price = 140% of(80% of 100) =Rs. 140/100 * 80/100 * 100 = Rs. 112.
.'. Increase = 12%
Answer : B |
AQUA-RAT | AQUA-RAT-37023 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
The sum of three consecutive numbers is 75. Name the numbers. | [
"24, 25, and 26.",
"34, 25, and 26.",
"29, 28, and 26.",
"44, 15, and 36."
] | A | The sum of three consecutive numbers is 75. Name the numbers.
Consecutive numbers are numbers in counting order. To solve problems of this type, let x equal
the first number. The second and third numbers can be expressed as x 1 and x 2.
Write an equation. x x 1 x 2 75
Solve. 3x 3 75
3x 72
x 24
Answer: The numbers are 24, 25, and 26.
Check: The numbers are consecutive, and their sum is 75.
correct answer A |
AQUA-RAT | AQUA-RAT-37024 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
The banker's gain on a bill due due 1 year hence at 12% per annum is Rs.6.6. The true discount is | [
"Rs.72",
"Rs.36",
"Rs.55",
"Rs.50"
] | C | Solution
T.D = [B.G x 100 / R x T]
= Rs.(6.6 x 100 / 12 x 1)
= Rs.55.
Answer C |
AQUA-RAT | AQUA-RAT-37025 | food
Title: Why might food go bad in an oxygen-free environment? I was recently watching a video from the International Space Station (Making a peanut butter sandwich in space), and I noticed he mentioned that a tortilla kept in an oxygen-free environment could last up to 18 months. Impressive, but it got me thinking: Why would some foods ever spoil if kept in a state where molds would not be able to grow? Would it be that other micro-organisms in things like the moisture present in the food eventually take their natural toll? Just a curious thought I had while watching through these awesome ISS videos.
Thanks for any answers! First of all, you assumed that this tortilla went bad after 18 month, not that austranaughts just decided to eat it?
More importantly, there is thing called anaerobs. Just like your muscle don't use oxygen in times of acute stress (force requirement), so there are organisms that don't care much about oxygen. As it happens in muscle, it happens in yeast during fermentation when sugars are turned into alcohols and released energy is used to propel molecular machinery. So, if there are sugar molecules on tortilla, there are still microbes that willing to prosper on it.
The following is multiple choice question (with options) to answer.
An orphanage had provision of food for 300 boys for 90 days. After 20 days, 50 boys left the orphanage. The number of days for which remaining food will last is : | [
"70",
"72",
"80",
"84"
] | D | Explanation :
Solution: After 20 days : 300 boys had food for 70 days.
Suppose 250 boys had food for x days. Now, less boys, more days (Indirect proportion)
.'. 250 : 300 : : 70 : x
=> 250*x = 300*70
=> x = 300*70/250 = 84.
Answer : D |
AQUA-RAT | AQUA-RAT-37026 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
The owner of a local jewelry store hired two watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave 1/2 of the diamonds he had then, and 5 more besides. He escaped with two diamonds. How many did he steal originally? | [
"50",
"36",
"25",
"None of these"
] | A | Explanation :
Since, thief escaped with 1 diamond,
Before 2nd watchman, he had ( 2 + 5 ) x 2 = 20 diamonds.
Before 1st watchman, he had ( 20 + 5 ) x 2 = 50 diamonds.
Answer : A |
AQUA-RAT | AQUA-RAT-37027 | organic-chemistry, biochemistry, heat, chemical-biology
It is a poorly phrased statement, but it is probably generally true. It would have been better to say something like, "heating tomatoes will increase the amount of all-trans-lycopene available for ingestion."
Here's a link to a full paper on the subject (click the "View" button near the top-right of the page). Basically two things are going on at the same time when lycopene containg food is heated. It is the naturally occurring, all-trans isomer of lycopene that is thought to be beneficial to the human body. Heating lycopene causes it to isomerize to a variety of cis-isomers. Heating in the presence of oxygen can also cause oxidation. Both of these processes (isomerization and oxidation) reduce the potency of the lycopene in the broth.
However, in parallel with these deleterious effects, heating also makes more lycopene available in the heated broth or extract. This is because heating disrupts cell walls and releases lycopene from the cells. The above link shows that, at least at moderate temperatures, the latter effect predominates and so more active all-trans-lycopene is made available upon heating tomatoes, despite the loss of some lycopene to isomerization and oxidation.
The following is multiple choice question (with options) to answer.
Fresh tomato contains 60% lycopene and dry tomato contains 20% lycopene. How much dry tomato can be obtained from 100kg of fresh tomato? | [
"A)32kg",
"B)50kg",
"C)52kg",
"D)80kg"
] | B | Quantity of lycopene in 100 kg of fresh tomatoes = (100-60)% of 100kg = 40kg
Let the quantity of dry tomato obtained be x kg
then, (100-20)% of x = 40
(80/100)*x = 40
x = 50
correct option is B |
AQUA-RAT | AQUA-RAT-37028 | homework-and-exercises, reference-frames, angular-velocity
Title: Reference Frame and Angular Speed Related? I am given the following problem:
If an airplane propeller rotates at 2000 rev/min while the airplane flies at a speed of 480 km/h relative to the ground, what is the linear speed of a point on the tip of the propeller, at radius 1.5m, as seen by (a) the pilot and (b) an observer on the ground? The plane’s velocity is parallel to the propeller’s axis of rotation.
The following is multiple choice question (with options) to answer.
The end of a blade on an airplane propeller is 10 feet from the center. If the propeller spins at the rate of 1,980 revolutions per second, how many miles will the tip of the blade travel in one minute? (1 mile = 5,280 feet) | [
"200π",
"450π",
"300π",
"480π"
] | B | Distance traveled in 1 revolution = 2πr= 2π10/5280
Revolutions in one second= 1980
Revolutions in 60 seconds (one minute)= 1980*60
Total distance traveled= total revolutions *distance traveled in one revolution
1980*60 *2π10/5280= 450π
B is the answer |
AQUA-RAT | AQUA-RAT-37029 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 60? | [
"s.3944",
"s.3948",
"s.4080",
"s.3965"
] | C | Let the side of the square plot be a ft.
a2 = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 60
= Rs.4080.
Answer: C |
AQUA-RAT | AQUA-RAT-37030 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
When the smallest of 3 consecutive odd integers is added to four times the largest, it produces a result 728 more than 4times the middle integer. Find the numbers? | [
"650",
"678",
"698",
"728"
] | D | x + 4 (x + 4) = 728 + 4 (x + 2)
Solve for x and find all three numbers
x + 4 x + 16 = 728 + 4 x + 8
x = 720
x + 2 = 722
x + 4 = 724
Check: the smallest is added to four times the largest
720 + 4 * 724 = 3616
four times the middle
4 * 722 = 2888
3616 is more than 2888 by
3616 - 2888 = 728
D |
AQUA-RAT | AQUA-RAT-37031 | 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.
Man is 40% more efficient than Woman. How much time will they, working together, take to complete a job which 1 Man alone could have done in 20 days? | [
"11 days",
"25 days",
"22 days",
"11 2â„3 days"
] | D | Ratio of times taken by Man and Woman = 100 : 140 = 5 : 7.
Suppose Woman takes x days to do the work.
Then, 5 : 7 : : 20 : x ⇒ x = (20×7)/5 ⇒ x = 28
Man’s 1 day’s work = 1â„20; Woman’s 1 days work = 1â„28
(Man + Woman)’s 1 day’s work = (1â„20 + 1â„28) = 3â„35
∴ Man and Woman together can complete the job in 35â„3 days or 11 2â„3 days.
Answer D |
AQUA-RAT | AQUA-RAT-37032 | c#, performance, algorithm, recursion, combinatorics
Here is the prunning I came with. I create set of pieces (in total you have 2^18 = 265k combinations of pieces) where the sum of flips on the pieces match the number of flips for the row.
Is possible you can have a solution with the right amount of flips but there isnt possible place them on the board in a way to reach goal for all the row. That testing should be done in a later process rigth now just try to discard solution doesnt even have the right amount of flips.
Of course if you use 0 pieces you can't solve 1st row, and if you use all the pieces on 1st row you wont have pieces to solve the bottom rows.
So lets analyze 1st row: You need a solution with at least 6 flips or (6 + 3*x) flips. If you spend one extra flip on a sword that become crown and then need 2 more to return to sword.
One possible combination with 6 flips is {5,6,9}:
The following is multiple choice question (with options) to answer.
A standard Veggiematik machine can chop 36 carrots in 4 minutes. How many carrots can 5 standard Veggiematik machines chop in 6 minutes? | [
"36",
"54",
"108",
"270"
] | D | Direct Relationship:-
1 standard Veggiematik machine - 36 Carrots - 4 minutes,
1 standard Veggiematik machine - 9 Carrots - 1 minute,
Now
5 standard Veggiematik machine - ? Carrots - 6 minutes,
Hence = 9X5X6 =270 Carrots
Answer D |
AQUA-RAT | AQUA-RAT-37033 | $$3 = (321)(-18+41k) + (123)(47-107k) \text{ for any } k$$
so for instance we also get ($k=1$):
$$3 = (321)(23) + (123)(-60)$$
-
The following is multiple choice question (with options) to answer.
What is the place value of 3 in the numeral 3259 | [
"2000",
"3000",
"3500",
"4000"
] | D | Option 'D'
3 * 1000 = 3000 |
AQUA-RAT | AQUA-RAT-37034 | You can do it like this. Let's begin with the expression you have: $$7x-6y=5$$ Adding the term $-7x$ to both sides, they remain equal and become $$7x-6y-7x=5-7x$$ that is $$-6y=5-7x$$ Now let's multiply both sides of this equation by $-\dfrac{1}{6}$. We get: $$y=\dfrac{7}{6}x-\dfrac{5}{6}$$ which is in the form you wanted. By the way, the number $\dfrac{7}{6}$ is called the slope of the line, which is a measure of its inclination.
-
thanks dude, this is a brilliant answer – user1534664 Jul 25 '13 at 1:02
Can I also divide it by -6, instead of -1/6? – user1534664 Jul 25 '13 at 1:05
I understand why you used fractions now, because else you end up with alot of decimal places :) – user1534664 Jul 25 '13 at 1:07
:) thank you, I hope it was clear enough – Marra Jul 25 '13 at 1:39
Try simple and avoid teacher jargon, it's used for make your remember it, but in fact they are not associated with something real.
Try to solve it as a equation where the unknown is y
$\begin{array}{l} 7x - 6y = 5\\ - 6y = 5 - 7x\\ 6y = 7x - 5\\ y = \frac{{7x - 5}}{6} = \frac{7}{6}x - \frac{5}{6} \end{array}$
Then:
$f(x) = \frac{7}{6}x - \frac{5}{6}$
The following is multiple choice question (with options) to answer.
If 2x + y = 7 and x + 2y = 5, then xy/y = ? | [
" 1",
" 4/3",
" 3",
" 18/5"
] | C | 2*(x+2y = 5) equals 2x+4y=10
2x+4y=10
- 2x + y= 7
= 3y=3
Therefore Y = 1
Plug and solve...
2x + 1 = 7
2x=6
x=3
(3*1)/1
=3
C |
AQUA-RAT | AQUA-RAT-37035 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
Stay hungry, Stay foolish
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1059
Location: India
GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3400
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average of 10 consecutive integers is 15. Then, 9 is deducted from the first consecutive number, 8 is deducted from the second, 7 is deducted form the third, and so on until the last number which remains unchanged. What is the new average? | [
"10",
"10.5",
"11",
"11.5"
] | B | The total subtracted is (9+8+...+1) = (9*10) / 2 = 45
On average, each number will be reduced by 45/10 = 4.5
Therefore, the overall average will be reduced by 4.5
The answer is B. |
AQUA-RAT | AQUA-RAT-37036 | #### Opalg
##### MHB Oldtimer
Staff member
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
First, the minimum value of $xy$ must be positive, because if $xy\leqslant 0$ then $(x+y)z\geqslant 4$. So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and $$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$$ Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. But that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}$ is $8$ (occurring when $z^2 = 4$).
So we may assume that $xy>0$. Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$ Therefore $xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.
The following is multiple choice question (with options) to answer.
Find the value of x from the below equation: x^2−7x+10=0 | [
"5",
"3",
"6",
"7"
] | A | Here we need to find out a and b such that a + b = -7 and ab = +10
a = -5 and b = -2 satisfies the above condition.
Hence
x^2−7x+10=(x−5)(x−2)
x2−7x+10=(x−5)(x−2)
x^2−7x+10
=0
⇒(x−5)(x−2)
=0
x2−7x+10=0⇒(x−5)(x−2)=0
Step 3: Equate each factor to 0 and solve the equations
(x−5)(x−2)
=0
⇒(x−5)
=0or (x−2)=0
⇒x= 5 or 2
A |
AQUA-RAT | AQUA-RAT-37037 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
What is 0.1 percent of 12,356? | [
"0.12356",
"1.2356",
"12.356",
"0.012356"
] | C | Since, Percent=1/100, what=something(s), and is:=. We can write the question as s=0.1(1/100)12,356. The answer is 12.356. Hence, the correct answer is C. |
AQUA-RAT | AQUA-RAT-37038 | speed-of-light, measurements, si-units, metrology, length
Title: How is the Length of a Meter Physically Measured? I have two parts to this question.
First, I understand that the meter is defined as the distance light travels in 1/299,792,458 seconds. But how is this distance actually measured? The second is obviously from an atomic clock, but Wikipedia makes it appear that the distance is calculated by the counting of wavelengths. For the count of wavelength to be useful you must calculate the physical length of it, which is dependent on the speed of light, which is defined with meters and to increase the accuracy of the measurement, NIST recommends the use of a specific wavelength of laser (which is in meters). Does this not make the actual measurement of the length of the meter circular? Or does the fact that the speed of light is a constant defined in meter/second over come the appearance of the circular logic? The way I see the circular logic would be if the speed of light ever changed, the length of the meter would also change, which make it impossible for us to know the speed of light changed without referencing it back to an older physical object.
Further, the measurement is done in a vacuum but we can not actually create a perfect vacuum, so I assume there would be a pressure range allowed on the vacuum, and pressure measurements are also based on the definition of a meter (I would think this would further add circular logic to laboratory measurements performed in air and adjusted for refraction).
Second, how is this measurement for the meter actually used to calibrate physical objects? I.e. If I buy a meter stick that has been calibrated against the national standard, how do they actually compare the length of the stick vs the wavelength measurements? The length of a meter bar can be measured using a HeNe laser. The laser used is chosen because we are very good at stabilizing the frequencies it outputs. This means that if we can measure the frequencies before the testing, they will remain steady during the testing. A cesium clock can be used to determine the frequencies of light used with great precision, as the second is defined from said clocks.
Once you have a good measurement of frequency, you have a good measurement of wavelength (assuming a reasonable medium... a vacuum is best, as the speed of light is defined in a vacuum, thus no uncertainty). With a wavelength in hand, you can do interferometry.
The following is multiple choice question (with options) to answer.
The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is | [
"15 cm",
"25 cm",
"35 cm",
"42 cm"
] | C | Explanation:
Required Length = H.C.F of 700 cm, 385 cm and 1295 c
= 35 cm. Answer: C |
AQUA-RAT | AQUA-RAT-37039 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
St Charles Borromeo Liverpool Newsletter, Soft Skills Questions And Answers Pdf, Calphalon Classic Nonstick Stainless Steel 2-piece Fry Pan Set, Legendary Dragon Decks Price Guide, Toril Moi Feminist, Female, Feminine, Carlton Postcode Nsw, How To Build A Powerful Electric Motor From Scratch Pdf, Moca Poe Filter, Wow Console Commands, Is It Hard To Build An Acoustic Guitar, Yamaha Spare Parts Price List, Khao Soi Restaurant, 4 Pack - Lysol Concentrate Disinfectant, Original Scent 12 Oz, Hindrances To Fulfilling God's Purpose, Redmi Y2 4/64 Price In Bangladesh, How To Pronounce Photogenic, How To Shape Orecchiette, Purple Graphic Tee, Thistle Plants For The Garden, Naphthalene Is Acid Or Base, Prayer Points On Divine Touch, Beautiful Quilt Patterns, " />
The following is multiple choice question (with options) to answer.
The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked x? | [
"C",
"W",
"Y",
"I"
] | A | Explanation: If you fold the above picture at the dotted lines, X and C are opposite to each other.
Ans: A |
AQUA-RAT | AQUA-RAT-37040 | # Welcome to the LEAP Q&A Forum
Find your questions answered. Here.
## GMAT - Problem Solving - Geometry
LEAP Administrator 11 months ago
### The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. (more)
The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. If the area of the painting is 4,800 square inches, what is the width, in inches, of each of the four rectangles?
A. 15
B. 20
C. 25
D. 30
E. 40
from the figure
length of one single rectangle = 3w
width of the complete painting = w + 3w = 4w
area of the painting = l*total width = 3w*4w = 12w2
4800 = 12w2
w2 = 400
w = 20 inches
B is correct
LEAP Administrator 11 months ago
### The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be (more)
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
There are three faces of the rectangular box
6 x 8, 8 x 10 and 10 x 6
Case 1
If 6 x 8 is the base of the cylinder, 10 will be the height and 6 will be the diameter
Volume = πr2h = π32*10 = 90π cubic inch
Case 2
If 8 x 10 is the base of the cylinder, 6 will be the height and 8 will be the diameter
Volume = πr2h = π42*6 = 96π cubic inch
Case 3
If 10 x 6 is the base of the cylinder, 8 will be the height and 6 will be the diameter
Volume = πr2h = π32*8 = 72π cubic inch
The following is multiple choice question (with options) to answer.
The length of a rectangular floor is more than its breadth by 200%. If Rs.484 is required to paint the floor at the rate of Rs.3/sq m, what would be the length of the floor? | [
"12",
"18",
"20",
"22"
] | D | Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 484/3 = 161.33 sq m
l b = 161.33 i.e., l * l/3 = 161.33
l^2 = 484 => l = 22.
D |
AQUA-RAT | AQUA-RAT-37041 | factoring
Title: Number of digits in a binary product Assume i have 2 numbers in binary form (or, more precisely, assume to know the number of their digits, DF1, DF2):
101010101001010101010101010111111111111111111111010101
10101111111111111111010101
Is there a formula for the exact number of binary digits (DP) of the product?
DP = F (DF1, DF2)
[I need this, in a practical case, to size a target array] I assume you want to know the maximum possible bits that a product of two numbers might have. Under that assumption, let the first number $a$ have $m$ bits, and the second number $b$ have $n$ bits.
Then, $$a \leq 2^{m} - 1 $$
Similarly, $$b \leq 2^{n} - 1$$
Then, the product is:
$$a \cdot b \leq 2^{m+n} - 2^m - 2^n + 1$$
Now, both $2^m$ and $2^n$ are greater than $1$ (since $n, m > 0$).
So, $$a \cdot b < 2^{m+n} - 1$$
Therefore, at most $m+n$ bits are sufficient for the product of the two numbers. Of course, lesser bits might suffice. Also, the sign bit is not included in the computation, and each of $a$, $b$, and $a \cdot b$ might have a separate sign bit.
If you really need to optimize on space, you can check $a$ and $b$ for special cases. Some cases could be:
$a$ or $b$ is $0$: You need 1 bit
$a$ or $b$ is $1$: You need as many bits as the other number
One of the numbers is a power of $2$: It will contain exactly one $1$, say at position $i$ (LSB is position 0). Then, the multiplication is equivalent to left shifting the other number by $i$ bits.
The following is multiple choice question (with options) to answer.
Every digit of a number written in binary is either 0 or 1. To translate a number from binary, multiply the nth digit (reading from right to left) by 2^(n-1)
What is the largest prime number (written in binary) that is a factor of both 100000 and 1000000 ? | [
" 10",
" 11",
" 101",
" 1011"
] | A | Binary Divison can provide a quick answer if you are comfortable with it.
as option E is the biggest binary number we try with it first :
100010000/ 10=10000
1000100000/ 101 =100000
so answer is option is A |
AQUA-RAT | AQUA-RAT-37042 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg? | [
"2:1",
"1:2",
"1:5",
"5:1"
] | A | 1st variety rs 18 /kg---- 2nd variety rs 24/kg
mix cost 20/kg = (24-20) (20-18)
=4:2 = 2:1
ANSWER A |
AQUA-RAT | AQUA-RAT-37043 | 51. Trisaba
thanks for the help
52. dan815
|dw:1441062793507:dw|
53. dan815
this is the full break down
54. dan815
they will probably ask u a final question like the chance it will rain, when there is a prediction of rain
55. Trisaba
60%
56. Trisaba
bayes' theorem
57. Trisaba
thanks for helping dan
58. dan815
wait i got 36% lemme see
59. dan815
okay think about it like this out of every 97 days, 5% of those days say it will rain, and it wont rain out of every 3 days, 90% of those days say it will rain, and it will rain! what is the total number of days it will rain to not raining then?
60. Trisaba
61. dan815
0.03*0.9 : 0.05*0.97 this means that (0.03*0.9)/((0.03*0.9)+(0.05*0.97)) = chance it rains when it say it rains
62. Trisaba
60% is what i got
63. dan815
what did u say p(b) was
64. Trisaba
P(B) is .0755
65. Trisaba
look at my sheet posted pic
66. dan815
hard to read, what did u say P(B|A) was then
67. Trisaba
.9
68. Trisaba
$\frac{ 90 }{ 151 }$
69. Trisaba
it rounds to 60%
70. dan815
hmm i am not seeing anything wrong with the way im finding the answer, ill think about this more later but for now for every 100 days 3 days really rain 90% of 3 days is 2.7 so every 2.7 days out of the 3 days it rains, it will predict right now an additional bad predictions of 5% of 97 days is thrown in there 0.05*97=4.85 right preiction and rain = 2.7 total prediction of rainy days = 4.85+2.7 therefore 2.7/(4.85+2.7)= ~36%
71. Trisaba
|dw:1441064489549:dw|
72. Trisaba
The following is multiple choice question (with options) to answer.
At Chennai it rained as much on Tuesday as on all the others days of the week combined. If the average rainfall for the whole week was 3 cm. How much did it rain on Tuesday? | [
"2.625 cm",
"3 cm",
"10.5 cm",
"15 cm"
] | C | Total rainfall = 3 x 7 = 21 cm
Hence, rainfall received on Tuesday = 21/2 (as it rained as much on Tuesday as on all the others days of the week combined)
= 10.5 cm.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37044 | @Rajeshwar: Because when you divide $6k+1$ by $5$, the $5k$ part is already a multiple of $5$. Adding or subtracting multiples of five form a number does not change the remainder when you divide by $5$. E.g., $2$, $7=2+5$, $22=2+5\times 4$, $177 = 2 + 5\times35$, etc., all have the same remainder when divided by $5$. So the remainder you get when dividing $n=6k+1$ by $5$ is the same as the remainder you get when dividing $n-5k = k+1$ by $5$. – Arturo Magidin Jul 20 '12 at 2:31
The following is multiple choice question (with options) to answer.
If the number is decreased by 5 and divided by 7 the result is 7. What would be the result if 6 is subtracted and divided by 8? | [
"4",
"6",
"8",
"5"
] | B | Explanation:
Let the number be x. Then,
(x - 5)/7 = 7 => x - 5 = 49
x = 54
.: (x - 6)/8 = (54 - 6)/8 = 6
Answer: Option B |
AQUA-RAT | AQUA-RAT-37045 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Jaime earned enough money by selling seashells at 20 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold ? | [
"5",
"10",
"22",
"25"
] | C | Let's TEST Answer C: 22 seashells...
With 22 seashells, Jamie would have 22(20) =440 cents. This would allow him to buy 8 books for 440 cents total, so this is the correct answer.
C |
AQUA-RAT | AQUA-RAT-37046 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
A sum of money lent out at S.I. amounts to Rs. 720 after 2 years and to Rs. 1020 after a further period of 5 years. The sum is? | [
"776",
"267",
"269",
"600"
] | D | S.I for 5 years = (1020 - 720)
= Rs. 300.
S.I. for 2 years = 300/5 * 2
= Rs. 120.
Principal = (720 - 120)
= Rs.600.
Answer: D |
AQUA-RAT | AQUA-RAT-37047 | ## 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.
Which of the following values of x will satisfy the inequality x2 – x – 6 > 0 ? | [
"x < – 2 or x > 3",
"–2 < x < 3",
"–3 < x < 2",
"x > – 2 or x < 3"
] | A | x2 – x – 6 > 0
or, x2 – 3x + 2x – 6 > 0
or, x(x – 3) + 2(x – 3) > 0
or, (x – 3)(x + 2) > 0
or, x = 3 or – 2
∴ x < – 2 or x > 3
Answer A |
AQUA-RAT | AQUA-RAT-37048 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
Find the middle one when The sum of three consecutive even numbers is 36? | [
"11",
"12",
"15",
"16"
] | B | 3 consecutive numbers can be a - 1, a, a + 1
So sum of numbers = 3a = 36.
Hence a = 12.
B |
AQUA-RAT | AQUA-RAT-37049 | # Welcome to the LEAP Q&A Forum
Find your questions answered. Here.
## GMAT - Problem Solving - Geometry
LEAP Administrator 11 months ago
### The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. (more)
The figure shown above represents a modern painting that consists of four differently colored rectangles, each of which has length l and width w. If the area of the painting is 4,800 square inches, what is the width, in inches, of each of the four rectangles?
A. 15
B. 20
C. 25
D. 30
E. 40
from the figure
length of one single rectangle = 3w
width of the complete painting = w + 3w = 4w
area of the painting = l*total width = 3w*4w = 12w2
4800 = 12w2
w2 = 400
w = 20 inches
B is correct
LEAP Administrator 11 months ago
### The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be (more)
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has the maximum volume?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
There are three faces of the rectangular box
6 x 8, 8 x 10 and 10 x 6
Case 1
If 6 x 8 is the base of the cylinder, 10 will be the height and 6 will be the diameter
Volume = πr2h = π32*10 = 90π cubic inch
Case 2
If 8 x 10 is the base of the cylinder, 6 will be the height and 8 will be the diameter
Volume = πr2h = π42*6 = 96π cubic inch
Case 3
If 10 x 6 is the base of the cylinder, 8 will be the height and 6 will be the diameter
Volume = πr2h = π32*8 = 72π cubic inch
The following is multiple choice question (with options) to answer.
The length of a rectangular floor is more than its breadth by 200%. If Rs. 324 is required to paint the floor at the rate of Rs. 3 per sq m, then what would be the length of the floor? | [
"16",
"17",
"18",
"19"
] | C | Let the length and the breadth of the floor be l m and b m respectively.
l = b + 200% of b = l + 2b = 3b
Area of the floor = 324/3 = 108 sq m
l b = 108 i.e., l * l/3 = 108
l2 = 324 => l = 18.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37050 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
What will be the compound interest on a sum of Rs. 20,000 after 3 years at the rate of 12% p.a.? | [
"s:8098.56",
"s:10123.29",
"s:10123.20",
"s:10123.28"
] | A | Amount
= [20000 * (1 + 12/100)3]
= 20000 * 28/25 * 28/25 * 28/25
= Rs. 28098.56
C.I. = (28098.56 - 20000)
= Rs: 8098.56
Answer: A |
AQUA-RAT | AQUA-RAT-37051 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
Find large number from below question The difference of two numbers is 1200. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder | [
"1437",
"1250",
"1540",
"1600"
] | A | Let the smaller number be x. Then larger number = (x + 1365).
x + 1200 = 6x + 15
5x = 1185
x = 237
Large number = 237+1365 = 1437
A |
AQUA-RAT | AQUA-RAT-37052 | we need to know definitions... They survived choices in the wrong spot: you still haven ’ t answered the question... Out how to restore/save my reputation the process of trying to answer this I... © 2021 Stack Exchange is a typical Bayes Theorem problem, it ’ s the probability of a student the! Kountz, Ashwini Miryala, Kyle Scarlett, Zachary Zell conditional probability for multiple random variables 4 0! Will finish the test to measure what the student both knows the answer being correct and knowing the answer 0.5. 3 or 4 choices for each question without the table helps me see more clearly we! Because if you know the answer but are not correct Stuck in wrong. Fida uses K ’ to mean “ not K ” of choices in the of! ) / ( 1/4 ) = P ( K∩C ) = 0 — you work., M = multiple choice examination has 5 questions and move 1/12 to right! Level and filesystem for a large storage server knew the answer, you will guess one of subject! That where you put 1/12, and conditional probability is a bit more confusing for.. The subject obtain a particular one are free and with help function teaching... Properties of conditional probabilities and solve problems pen if it is entirely forgivable not conditional probability multiple choice questions seen. A question and answer site for people studying math at any level and filesystem for a large server. Is the probability they knew the answer terms of service, privacy policy and policy... Has 5 questions on some of the subject options randomly authors use “ ~K or. Way I thought about P ( C ’ ∩ K ), which know. Saw when I read this problem was a reversal of information how many choices there are options... Into your RSS reader will get the question in order to work any! A recent question about probability has ties to Venn diagrams, tables, and Miriam symbol some! Completed a multiple-choice test with four options for each question has only one true answer multiple-choice questions are ;! Result of your quiz after you finish the novel that I am reading no! Design / logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa codes and datasets: them... Can ’ t be true professionals in related fields to buy them of choices! ( bleeding '', he who fears will
The following is multiple choice question (with options) to answer.
----------------YES---------NO----UNSURE Subject M----500--------200-----100 Subject R----400--------100-----300 A total of 800 students were asked whether they found two subjects, M and R, interesting. Each answer was either yes or no or unsure, and the numbers of students who gave these answers are listed in the table above. If 190 students answered yes only for subject M, how many of the students did not answer yes for either subject? | [
"100",
"210",
"300",
"400"
] | B | Since 190 students answered yes only for subject M, then the remaining 310 students who answered yes for subject M, also answered yes for subject R. So, 310 students answered yes for both subjects.
If 310 students answered yes for both subjects, then 400-310=90 students answered yes only for subject R.
So, we have that:
200 students answered yes only for subject M;
90 students answered yes only for subject R;
300 students answered yes for both subjects;
Therefore 800-(200+90+300)=210 students did not answer yes for either subject.
Answer: B. |
AQUA-RAT | AQUA-RAT-37053 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A person purchased a TV set for Rs. 12000 and a DVD player for Rs. 6250. He sold both the items together for Rs. 31150. What percentage of profit did he make? | [
"16.68%",
"87.68%",
"70.68%",
"17.68%"
] | C | The total CP = Rs. 12000 + Rs. 6250 = Rs. 18250 and SP
= Rs. 31150
Profit(%) = (31150 - 18250)/18250 * 100
= 70.68%
Answer:C |
AQUA-RAT | AQUA-RAT-37054 | 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.
One years ago,Promila was four times as old as her daughter Sakshi.Six years hence,Promila’s age will excepted her daughter’s age by 9 years.The ratio of the present ages of Promila and her daughter is | [
"9 : 2",
"11 : 3",
"12 : 5",
"13 : 4"
] | D | Solution
Let the ages of Promila and Sakshi 1 year ago be 4x and x years respectively.
Then,[(4x +1)+6]-[(x + 4)+ 4]=64 ⇔ 4x = 48 ⇔ x =12.
∴ Required ratio =(4x +1 ) : (x + 1) : 13 : 4. Answer D |
AQUA-RAT | AQUA-RAT-37055 | ### Show Tags
20 Nov 2016, 04:12
Whenever you get problems pertaining to 'loan' or 'borrowed amount' to be paid in equal installments, just apply the following formula, it is easy to remember:
Value of each equal annual installments = a
Rate of interest = r% p.a.
No. of installments per year = n
No. of years = t
Therefore, total no. of installments = n*t = N
Borrowed amount(or loan taken) = B
Then:
a[{100/(100+r)} + {100/(100+r)}^2 + ......... + {100/(100+r)}^N] = B
Now, in the above question(three equal installments to be paid in three months):
a[{100/110} + {100/110}^2 + {100/110}^3] = 1000
Solve the above equation and you will get 'a' as equal to 402.11
_________________
Before getting into the options, have an idea of what you seek
Thanks & Regards,
Vipul Chhabra
Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5
### Show Tags
The following is multiple choice question (with options) to answer.
We made a down payment of $80 and borrowed the balance on a new violin which cost $400. The balance with interest was paid in 23 monthly payments of $14 each and a final payment of $22. The amount of interest paid was what percent of the amount borrowed? | [
"5.5%",
"6.5%",
"7.5%",
"8.5%"
] | C | We borrowed $320.
To pay back this loan, we paid 23*$14 + $22=$344.
The interest was $344-$320=$24.
The percentage was $24/$320=0.075=7.5%.
The answer is C. |
AQUA-RAT | AQUA-RAT-37056 | ### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
_________________
"Please hit +1 Kudos if you like this post"
_________________
Manish
The following is multiple choice question (with options) to answer.
Sandy purchased 20 dozen toys at the rate of $144 per dozen. Sandy sold each toy at the rate of $12.60. What was the percentage profit? | [
"3%",
"5%",
"7%",
"9%"
] | B | A dozen toys cost $144, so each toy cost $12.
selling price / cost price = 12.60 / 12 = 1.05
The percentage profit is 5%.
The answer is B. |
AQUA-RAT | AQUA-RAT-37057 | 5. Originally Posted by ANDS!
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
Thanks.
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
Why aren't these included in the answer given in the post above?
6. Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
7. Soroban only considered these 2 cases:
Originally Posted by Soroban
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
The following is multiple choice question (with options) to answer.
At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using ten colors? | [
"21",
"35",
"105",
"120"
] | D | The prompt tells us that substance is identified by an UNORDERED combination of 3 colors and that no chemical may be assigned the same 3 colors as any other chemical.
As an example, a chemical with the colors A/B/C means that the following combinations CANNOT be used by any other chemical:
ABC
ACB
BAC
BCA
CAB
CBA
By extension, the first part of your calculation would be the number of possible substances that could be identified with 10 colors: 10c3 = 120 possible substances
D |
AQUA-RAT | AQUA-RAT-37058 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
Find the simple interest on Rs.450 for 8 months at 5 paisa per month? | [
"277",
"270",
"180",
"266"
] | C | I = (450*8*5)/100 = 180
Answer: C |
AQUA-RAT | AQUA-RAT-37059 | ### 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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
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.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
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.
A solution of 66 litres contains milk and water in the ratio 7:x. If four litres of water is added to the solution, the ratio becomes 3:2, find the value of x? | [
"8",
"5",
"3",
"4"
] | D | Total new quantity = original sol + water =66+4 =70
New ratio = 3:2, New quantity of milk =3/5*70 = 42 Lit,
New quantity of water = 2/5*70 = 28 Lit
Water present initially = (28-4) = 24 Lit
Ratio = 42/24 =7/4 There for x = 4
ANSWER:D |
AQUA-RAT | AQUA-RAT-37060 | Its reversal is 100 × [10 − (A − C)] + 10 × 9 + 1 × [(A − C) − 1]. The sum is thus 101 × [(A − C) − 1] + 20 × 9 + 101 × [10 − (A − C)] = 101 × [(A − C) − 1 + 10 − (A − C)] + 20 × 9 = 101 × [−1 + 10] + 180 = 1089.
However, I don’t particularly care for the succinct explanation, and so I’d prefer to give my audience the following explanation. Let’s write our original three-digit number as $ABC$, which of course stands for $100 \times A + 10 \times B + C$. Then, when I reverse the digits, the new three-digit number will be $CBA$, or $100 \times C + 10 \times B + A$.
Of course, because the first number is bigger than the second number, this means that the first hundreds digit is bigger than the second hundreds digit. This means that the first ones digit has to be less than the second ones digit. In other words, when we subtract, we have to borrow from the tens place. However, the tens digits are the same for both numbers. That means that I have to borrow from the hundreds place also.
I’ll illustrate this for both subtraction problems:
Now I’ll subtract. The hundreds digit will be $A - 1 - C$. The tens digit will be $9 + B - B$, or simply $9$. Finally, the ones digit will be $10 + C - A$. This is a little hard to write on a board, so I’ll add some dotted lines to separate the hundreds digits from the tens digit from the ones digit:
The next step is to reverse the digits and add:
I’ll begin with the ones digit:
$(10 + C - A) + (A - 1 - C) = 10 - 1 = 9$.
No matter what, the ones digit is a 9.
Continuing with the tens digits, I get $9 + 9 = 18$. I’ll write down $8$ and carry the $1$ to the next column.
Finally, adding the hundreds digits (and the extra $1$), I get
The following is multiple choice question (with options) to answer.
The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number? | [
"11",
"28",
"156",
"44"
] | B | Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
Answer: B |
AQUA-RAT | AQUA-RAT-37061 | 7. wayki says:
Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).
I hate imperial so please allow me to convert it to metric.
0.07366 mm = 30.480m
Multiply all of this up by 1000 =
73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?
Those that come up with 8inches a mile are much closer to the truth.
8. wayki says:
Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.
9. Alex A says:
I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???
Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.
If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.
• mathscinotes says:
Thank you. Very nicely put.
mathscinotes
• Johnny Emerson Neeley says:
I think 18 miles is level at a 6' height😜😨, on a perfect sphere, 131,477,280 ft.
10. Steve says:
So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?
• mathscinotes says:
The following is multiple choice question (with options) to answer.
On a map, 1.5 inches represent 24 miles. How many miles approximately is the distance if you measured 44 centimeters assuming that 1-inch is 2.54 centimeters? | [
"174.2",
"277",
"288.1",
"296"
] | B | 1.5 inch = 2.54*1.5 cm.
So, 2.54*1.5 represents 24 miles.
So for 44 cm.:
44/(2.54*1.5) = x / 24 ---> x = 24*44/(3.81) = 277
Answer will be B. |
AQUA-RAT | AQUA-RAT-37062 | In a bag there are 10 Black balls, 8 White balls and 5 Red balls.
Three balls are chosen at random and one is found to be Black.
Find the probability that remaining two are white.
. $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$
I see it as a Conditional Probability problem . . .
Given that at least one ball is Black,
. . find the probability that we have one Black and two White balls.
Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$
There are ${23\choose3} = 1771$ possible ways to choose 3 balls.
To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways.
. . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$
The opposite of "at least 1 Black" is "NO Blacks".
There are: . ${13\choose3} = 286$ ways to choose no Blacks.
So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls.
. . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$
The following is multiple choice question (with options) to answer.
A bag contains 6 white and 4 black balls .2 balls are drawn at random. find the probability that they are of same colour. | [
"71/9",
"7/15",
"7/18",
"4/17"
] | B | let S be the sample space
Then n(S)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45
Let E=event of getting both balls of same colour
Then n(E)=no of ways(2 balls out of six) or(2 balls out of 4)
=(6c2+4c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21
P(E)=n(E)/n(S)=21/45=7/15
Option B |
AQUA-RAT | AQUA-RAT-37063 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man swims downstream 30 km and upstream 18 km taking 3 hours each time, what is the speed of the man in still water? | [
"5",
"8",
"9",
"2"
] | B | 30 --- 3 DS = 10
? ---- 1
18 ---- 3 US = 6
? ---- 1 M = ?
M = (10 + 6)/2 = 8
Answer: B |
AQUA-RAT | AQUA-RAT-37064 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
find the number, difference between number and its 3/5 is 60. | [
"150",
"153",
"154",
"155"
] | A | Explanation:
Let the number = x,
Then, x-(3/5)x = 60,
=> (2/5)x = 60 => 2x = 60*5,
=> x = 150
Answer: Option A |
AQUA-RAT | AQUA-RAT-37065 | The following is an example of a method that you could use to figure this out. The method is called bisection. It is not the fastest, but it is clear how to extract digits from it because it has explicit error bounds.
You know $1^2<2$. You know $2^2>2$. So $1<\sqrt{2}<2$ so its first decimal digit is $1$.
You check $1.5^2>2$. You check $1.25^2<2$. You check $1.375^2<2$. You check $1.4375^2>2$. You check $1.40625^2<2$. Now you know $1.40625<\sqrt{2}<1.4375$ so you know the first two digits are $1.4$.
You continue: you check $1.421875^2>2$. You check $1.4140625^2<2$. You check $1.41796875^2>2$. So $1.4140625<\sqrt{2}<1.41796875$, so you know the first three decimal digits now.
You can keep going; at each time you know $\sqrt{2}$ is in between two numbers getting closer together, so as soon as those numbers have a new digit in common, you know that digit of $\sqrt{2}$. On average it takes $\log_2(10) \approx 3.3$ steps to get a new correct decimal digit.
At the cost of slightly more iterations, you can make calculations easier (if you're doing it by hand) by rounding the lower bound down and/or the upper bound up. For instance back in the second paragraph of iterations you could have said $1.4<\sqrt{2}<1.44$ and then continued, obtaining $1.41<\sqrt{2}<1.415$ at the end of the third paragraph.
The following is multiple choice question (with options) to answer.
What is the decimal equivalent of (1/4)^1? | [
"0.0016",
"0.0625",
"0.16",
"0.25"
] | D | (1/4)= 0.25
Answer : D |
AQUA-RAT | AQUA-RAT-37066 | 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.
Albert father was 48 years of age when she was born while her mother was 46 years old when her brother 2 years younger to her was born. What is the difference between the ages of her parents? | [
"2",
"4",
"6",
"8"
] | B | Mother's age when Albert's brother was born = 46years
Father's age when Albert's brother was born = 48+2 = 50years
Required difference = 50-46 = 4 years
Answer is B |
AQUA-RAT | AQUA-RAT-37067 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
The contents of a certain box consist of 20 apples and 30 oranges. How many oranges must be added to the box so that exactly 20% of the pieces of fruit in the box will be apples? | [
"50",
"55",
"75",
"85"
] | A | apple = (apple + orange + x)*0.2
20 = (30 + 20 + x)*0.2
x = 50
Answer: A |
AQUA-RAT | AQUA-RAT-37068 | ## Effective interest rate of 12 compounded quarterly
Rate (EAR) from a stated nominal or annual interest rate and compounding frequency. APY Calculator to Calculate Annual Percentage Yield from a Stated Nominal Interest Rate For example, if one saving institution offers an annual interest rate of 1% compounded annually, whereas APY = (1 + .04875/12 )12 – 1. This words out to a 12% interest rate. However, since interest is compounded monthly, the actual or effective interest rate is higher because interest in the current
8 Sep 2014 But loan interest is almost never compounded annually! To convert a nominal interest rate to an effective interest rate, we have to pay close daily and annual compounding is a lot bigger at 12%/yr interest than at 4%/yr. 5 Feb 2019 It is likely to be either monthly, quarterly, or annually. Locate the stated interest rate in the loan documents. Enter the compounding period and Where r is the interest rate per period in decimal form so R = r * 100 and, i is the effective interest rate in decimal form so I = i * 100. P is the rate per compounding period where P = R/m. The effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other).
## (b) The annual interest rate is 2.4%, and the number of interest periods is 12. Table 3 shows the effects of interest rates (compounded quarterly) on the Effective Rate of Interest Formula If interest is compounded m times per year, then.
The following is multiple choice question (with options) to answer.
A certain sum amounts to Rs.1725 in 3 years and Rs.1875 in 5 years. Find the rate % per annum? | [
"8%",
"5%",
"2%",
"3%"
] | B | 3 --- 1725
5 --- 1875
--------------
2 --- 150
N = 1 I = 75 R = ?
P = 1725 - 225 = 1500
75 = (1500*1*R)/100
R = 5%
Answer: B |
AQUA-RAT | AQUA-RAT-37069 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2013, What is the value of n? | [
"22",
"38",
"62",
"122"
] | D | Same logic but with application of factorials.
2*4*6*8*...*n = 2(1*2*3*4*...n/2) = 2*(n/2)!
Now we have: 2*(n/2)!/2013=2*(n/2)!/(3×11×61)
For the expression 2*(n/2)! To be divisible by 2013 min value of n/2 should be 61.
So n/2=61 => n=122
Answer D. |
AQUA-RAT | AQUA-RAT-37070 | To get the required result in the book you first have to calculate the amount of the equivalent annual payment. The formula for payments at the beginning of every month is
$C_1=12\cdot r+\frac{\color{blue}{13}\cdot r\cdot i}{2}$
In case of payments at the end of every month $\color{blue}{13}$ has to be replaced by $11$.
$C_1=12\cdot 200+\frac{\color{blue}{13}\cdot 200\cdot 0.045}{2}=2458.5$
To get the Future value after 10 years we use the formula for annual payments.
$C_{10}=2458.5\cdot \frac{1-1.045^{10}}{1-1.045}=30210.56$
But in general I wouldn´t say that your method is worse then the method above. Your result differs from my result about $0.015\%$ only.
The following is multiple choice question (with options) to answer.
Find the sum lend at C.I. at 5 p.c per annum will amount to Rs.650 in 2 years? | [
"221",
"287",
"400",
"589"
] | D | Explanation:
650 = P(21/20)2
P = 589.56
Answer:D |
AQUA-RAT | AQUA-RAT-37071 | program. Determine the radius Measure the distance from the center of the circle of which the semicircle is part to its edge. *Note: Model is a radical function with limited domain. Example 2 Determine the area of the largest rectangle that can be inscribed in a circle of radius 4. Note: The rectangle and the "bumpy edged shape" made by the sectors are not an exact match. 5 m, find the area of the playground. 3137085/2)^2 = 100. formule o f area of a circle but bas ic rectangle formula helped her to find out the area. Area of Semi-Circle = 1 ⁄ 2 * π *r 2. Here, the circle is cut into 8 equal parts. Mike Henderson was best known as a committed tennis coach and a devoted follower of Christ. a) Create a rectangle by cutting off the right-angled triangle and moving it. Apart from these, some software also feature options to calculate diagonal, height, length, breadth, etc. Formulas, explanations, and graphs for each calculation. A composite area consisting of the rectangle, semicircle, and a triangular cutout is shown (Part A figure). The second moment of area, also known as area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. 99 for a replacement flag or$49 for a complete kit, purchasing these promotional flags is a must. Volume of Half Cylinder Calculator. Area of a Circle. Area of rectangle is the region covered by the rectangle in a two-dimensional plane. With the Line tool (), draw two lines: one that divides the outer circle in half and one that divides the inner circle that you created with the Offset tool. Selections are so vast and varied, the experience can be overwhelming. Adjust and Print Full Scale Circle Divider Templates- Setting Circles Drag Diameter slider to size. We cut circles into small sectors and arranged them into a form as close to a rectangle as possible. asked by ian on August 29, 2016. Creating a design with pavers requires planning. Learn how to make memes, slideshows, vision boards, book covers and photo collages! Check out beautiful hex color codes for coral, periwinkle, emerald green, royal blue, and teal. The Rhino Reflex Salon Mats feature 1" thick sponge and are available
The following is multiple choice question (with options) to answer.
In May, the groundskeeper at Spring Lake Golf Club built a circular green with an area of 90π square feet. In August, the groundskeeper doubled the distance from the center of the green to the edge of the green. What is the total area of the renovated green? | [
"360π",
"400π",
"450π",
"500π"
] | A | Area = πR^2, so doubling the radius results in an area that is 4 times the original area.
4(90π) = 360π
The answer is A. |
AQUA-RAT | AQUA-RAT-37072 | # smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2|<0.001$$
How to proceed? Thanks in advance.
• You are almost done ! – Khosrotash Aug 11 '17 at 14:49
$$|2[1-(-\frac{1}2)^n]-2|<0.001\\ |-2(\frac{-1}2)^n|<0.001\\ +2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\ +2.\frac{1}{2^n}<\frac{1}{1000}\\ \frac{2^n}{2}>1000\\ 2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
The following is multiple choice question (with options) to answer.
Given that 0.0010010 x 10^q > 10^3, and q is an integer, which of the following is the smallest possible value of q? | [
"3",
"4",
"5",
"6"
] | D | 0.0010010 x 10^q > 10^3, and q is an integer
This will be true only when LHS will be 1001
so the smallest possible value of q to make LHS > 10^3 is q=6
ANSWER:D |
AQUA-RAT | AQUA-RAT-37073 | # Find area of the triangle ABC, given the coordinates of vertices in plane
$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.
I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.
I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.
I do not understand how this answer is achieved.
• Every heard of the Shoelace formula? Oct 2 '16 at 15:34
• Another way would also be to draw a rectangle around the triangle and subtract each section not included in the triangle, but you would need the graph for that. Oct 2 '16 at 15:35
• Fine comments above. Still another way is to exploit en.wikipedia.org/wiki/Pick%27s_theorem Oct 2 '16 at 15:36
• OP; stop changing it. it's fine. Oct 2 '16 at 15:39
• In order to not to waste your efforts, you may just apply Heron's formula to the computed side lengths, too. Oct 2 '16 at 15:44
Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area. Let's consider the approach suomynonA suggested in the comments. Consider the figure below.
We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$. I will leave the details of the calculations to you.
Follow-through
As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.
The following is multiple choice question (with options) to answer.
The area of a triangle will be when a = 1m, b = 2m, c = 7m, a, b, c being lengths of respective sides? | [
"3",
"6",
"5",
"9"
] | C | S = (1 + 2 + 7)/2
= 5 Answer:C |
AQUA-RAT | AQUA-RAT-37074 | ### Show Tags
23 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?
Aman
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
24 Jun 2019, 03:25
pkloeti wrote:
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.
Does that approach hold up in general? Bunuel VeritasPrepKarishma
Thanks a lot for the feedback!
I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
_________________
Karishma
Veritas Prep GMAT Instructor
Intern
Joined: 12 Feb 2018
Posts: 9
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
04 Aug 2019, 10:54
i tried like this:
Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85
The following is multiple choice question (with options) to answer.
0.6 of a number equals 0.09 of another number. The ratio of the number is: | [
"2:3",
"1:15",
"20:3",
"3:20"
] | D | Explanation :
Let 0.6 of x be equal to 0.09 of y:
i.e. 0.6x = 0.09y
=> x/y = 0.09/0.6 = 3/20
Answer : D |
AQUA-RAT | AQUA-RAT-37075 | You can reduce the labour a little by writing
$\dfrac{1}{x(x-1)^3(x-2)^2} = \dfrac{(1-x+x^2)+x(x-1)}{x(x-1)^3(x-2)^2}$
$=\dfrac{x(x-2)^2-(x-1)^3}{x(x-1)^3(x-2)^2} + \dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3} - \dfrac{1}{x(x-2)^2}+\dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3}- \dfrac{1}{x(x-2)^2}+ \dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}+2 \left[\dfrac{1}{x-1} - \dfrac{1}{x-2} \right]$
If you wish you can further decompose $\dfrac{1}{x(x-2)^2}= \dfrac{1}{(x-2)^2}+\dfrac{1}{2} \left[\dfrac{1}{x} - \dfrac{1}{x-2} \right]$
This expression can be readily integrated.
The following is multiple choice question (with options) to answer.
If Rs. 782 be divided into three parts, proportional to 1/2: 2/3: 3/4, then the first part is: | [
"202",
"203",
"204",
"205"
] | D | (1/2+2/3+3/4)x=782
x=782*12/23=408
fist part is: (1/2)x=408/2=204
ANSWER:D |
AQUA-RAT | AQUA-RAT-37076 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
The average marks of a class of 24 students is 40 and that of another class of 50 students is 60. Find the average marks of all the students? | [
"52.2",
"59.5",
"52.8",
"53.5"
] | D | Sum of the marks for the class of 24 students = 24 * 40 = 960
Sum of the marks for the class of 50 students = 50 * 60 = 3000
Sum of the marks for the class of 74 students =
960 + 3000 = 3960
Average marks of all the students = 3960/74
= 53.5
Answer:D |
AQUA-RAT | AQUA-RAT-37077 | Show Tags
14 Sep 2018, 17:42
SW4 wrote:
How many consecutive zeros will appear at the end of 43! if that number is expanded out to its final result?
A) 12
B) 11
C) 10
D) 9
E) 8
To determine the number of trailing zeros in a number, we need to determine the number of 5- and-2 pairs within the prime factorization of that number. Each 5-and-2 pair creates a 10, and each 10 creates an additional zero.
Since we know there are fewer 5s in 43! than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.
To determine the number of 5s within 43!, we can use the following shortcut in which we divide 43 by 5, then divide the quotient of 43/5 by 5 and continue this process until we no longer get a nonzero quotient.
43/5 = 8 (we can ignore the remainder)
8/5 = 1 (we can ignore the remainder)
Since 1/5 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 5 within 43!.
Thus, there are 8 + 1 = 9 factors of 5 within 43! This means we have nine 5-and-2 pairs, so there are 9 consecutive zeros at the end of 43! when it is expanded to its final result.
_________________
Scott Woodbury-Stewart
Founder and CEO
Scott@TargetTestPrep.com
122 Reviews
5-star rated online GMAT quant
self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.
Re: How many consecutive zeros will appear at the end of 43! if that numbe [#permalink] 14 Sep 2018, 17:42
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If in a certain sequence of consecutive multiples of 50, the median is 725, and the greatest term is 950, how many terms that are smaller than 725 are there in the sequence? | [
"6",
"5",
"8",
"12"
] | B | Since the median is 725 we know there must be a even number of integers because 50 is not a multiple of 725.
So the list around 725 must go. 600 650 700(725) 750 800 850 900 950
Since we know there are 5 numbers greater than 725 then there must be 5 numbers less then 725.
answer:B |
AQUA-RAT | AQUA-RAT-37078 | java, xml, swing
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/12 00:00:00</date>
<riddle>ipso lorem 7</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/19 00:00:00</date>
<riddle>ipso lorem 8</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/03/26 00:00:00</date>
<riddle>ipso lorem 9</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/02 00:00:00</date>
<riddle>ipso lorem 10</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/09 00:00:00</date>
<riddle>ipso lorem 11</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/16 00:00:00</date>
<riddle>ipso lorem 12</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/23 00:00:00</date>
<riddle>ipso lorem 13</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
<week>
<date>2012/04/30 00:00:00</date>
<riddle>ipso lorem 14</riddle>
<lastWeekSolution>I'm Going to get this</lastWeekSolution>
</week>
</weeks>
The following is multiple choice question (with options) to answer.
What will be the day of the week 15th August, 2010? | [
"Tuesday",
"Saturday",
"Monday",
"Sunday"
] | D | 15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.
Jan. Feb. Mar. Apr. May. Jun. Jul. Aug.
(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days
227 days = (32 weeks + 3 days) 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 = 0 odd days.
Given day is Sunday
Answer is D. |
AQUA-RAT | AQUA-RAT-37079 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
Find the area of the square, one of whose diagonals is 3.8 m long ? | [
"7.22m^2",
"3.86m^2",
"8.96m^2",
"2.68m^2"
] | A | Area of the square= 1/2(diagonal)^2
=(1/2*3.8*3.8)m^2
= 7.22m^2
Answer (A) |
AQUA-RAT | AQUA-RAT-37080 | They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. Prove that the product of three consecutive positive integer is divisible by 6. gl/9WZjCW prove that the product of three consecutive positive integers is divisible by 6. Therefore, n = 3p or 3p + 1 or 3p + 2 , where p is some integer. 1 3 + 2 3 + 3 3 +. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. By the laws of divisibility, anything divisible by 2 and 3 is divisible by 6. Whenever a number is divided by 3 , the remainder obtained is either 0,1 or 2. Find the smallest number that, when. Essentially, it says that we can divide by a number that is relatively prime to. Let the three consecutive positive integers be n , n + 1 and n + 2. Let n be a positive integer. 1 Consecutive integers with 2p divisors. If A and B are set of multiples of 2 and 3 respectively, then show that A = B and A∪B. Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. The array contains integers in the range [1. Prove that one of any three consecutive positive integers must be divisible by 3. ← Prev Question Next Question →. Btw jayshay - if you said 7n, 7n+1 and 7n+2 then your 'proof' would effectively be proving that the product of 3 consecutive integers is a multiple of 7. We have to prove this for any arbitrary k ∈Z, so fix such a k. (Examples: Prove the sum of 3 consecutive odd integers is divisible by 3. Prove: The product of any three consecutive integers is divisible by 6; the product of any four consecutive integers is divisible by 24; the product of any five consecutive integers is divisible by 120. If A =40, B =60 and
The following is multiple choice question (with options) to answer.
If x is the sum of six consecutive integers, then x is divisible by which of the following: I. 6 II. 5
III. 7 | [
"I only",
"II only",
"III only",
"I,III only"
] | B | We're told that X is the SUM of 6 CONSECUTIVE INTEGERS. We're asked what X is divisible by....
Let's TEST VALUES....
IF we use the 6 consecutive integers: 5, 6, 7,8,9 and 10, then the sum = 45.
45 is divisible by 5
45 is NOT divisible by 6
45 is NOT divisible by 7
There's only one answer that 'fits' with these facts. Answer : B |
AQUA-RAT | AQUA-RAT-37081 | 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.
Which number is the odd one out ?
9654 4832 5945 7642 7963 8216 3647 | [
"3647",
"9654",
"5945",
"7963"
] | A | The product of first two digits in a number equals the last two digits, except in 3647. Hence, it is the odd one.
The answer is 3647.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37082 | 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 490 m long, running with a speed of 63 km/hr will pass a tree in? | [
"25",
"26",
"27",
"28"
] | D | Speed = 63 * 5/18 = 35/2 m/sec
Time taken = 490 * 2/35 = 28 sec
Answer: Option D |
AQUA-RAT | AQUA-RAT-37083 | # Runner's High (Speed)
I find the following mind-boggling.
Suppose that runner $$R_1$$ runs distance $$[0,d_1]$$ with average speed $$v_1$$. Runner $$R_2$$ runs $$[0,d_2]$$ with $$d_2>d_1$$ and with average speed $$v_2 > v_1$$. I would have thought that by some application of the intermediate value theorem we can find a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$. This is not necessarily so!
Question. What is the smallest value of $$C\in\mathbb{R}$$ with $$C>1$$ and the following property?
Whenever $$d_2>d_1$$, and $$R_2$$ runs $$[0,d_2]$$ with average speed $$Cv_1$$, then there is a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$.
The following is multiple choice question (with options) to answer.
In a race of length Lmetres, Johnson beats Lewis by M metres and Greene by N metres, By how many metres does Lewis beat Greene in the same race ? (M<N) | [
"L(L-N) / L-M",
"(N-M)*L/(L-M)",
"L-N",
"M-N"
] | B | the ans is quite simple N-M but since the ans is not given, the best way to 'find' the correct option is to plug in the numbers. For my work, i took the following numbers:
L = 100
M = 10
N =20.
So basically, L beat G by 10 meters. Now whatever options comes close to the figure 10 is the right ans..
looking the the opion C to E, one can make out that they are incorrect
Out of which, i choose B. |
AQUA-RAT | AQUA-RAT-37084 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
Ratio between Rahul and Deepak is 4:3, After 6 Years Rahul age will be 38 years. What is Deepak present age? | [
"22",
"24",
"77",
"266"
] | B | Present age is 4x and 3x,
=> 4x + 6 = 38 => x = 8
So Deepak age is = 3(8) = 24
Answer: B |
AQUA-RAT | AQUA-RAT-37085 | 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.
The dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 7 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each? | [
"s.4529",
"s.4586",
"s.4597",
"s.6342"
] | D | Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 7
=Rs.6342
Answer: D |
AQUA-RAT | AQUA-RAT-37086 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
A man is 27 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is? | [
"11",
"25",
"27",
"22"
] | B | Let the son's present age be x years.
Then, man's present age = (x + 27) years.
(x + 27) + 2 = 2(x + 2) x + 29 = 2x + 4 => x = 25.
Answer: B |
AQUA-RAT | AQUA-RAT-37087 | ## Extra 01
How many arrangements can be made using the letters from the word COURAGE? What if the arrangements must contain a vowel in the beginning?
• $$4 \times 6!$$
## Extra Problem 02
How many arrangements are possible using the words
• EYE
• CARAVAN?
## 3(a)
There are (p+q) items, of which p items are homogeneous and q items are heterogeneous. How many arrangements are possible?
## 2(j)
There are 10 letters, of which some are homogeneous while others are heterogeneous. The letters can be arranged in 30240 ways. How many homogeneous letters are there?
Let, $$m = \text{number of homogeneous items}$$
• n(arrangements) = 30240 = $$\frac {10!}{m!}$$
• $$m! = \frac{10!}{30240}=120$$
• m = 5
## 2(k)
A library has 8 copies of one book, 3 copies of another two books each, 5 copies of another two books each and single copy of 10 books. In how many ways can they be arranged?
Total books = $$1 \times 8+3 \times 2+5 \times 2 + 8 \times 1 + 10$$ = 42
• n(arrangements) = $$\frac{42}{8!(3!)^2(5!)^2}$$
## 2(l)
A man has one white, two red, and three green flags; how many different signals can he produce, each containing five flags and one above another?
Flags: W = 2, R = 2, G = 3, Total = 7
Answer
## 2 (m)
A man has one white, two red, and three green flags. How many different signals can he make, if he uses five flags, one above another?
## 3(a)
How many different arragnements can be made using the letters of the word ENGINEERING? In how many of them do the three E’s stand together? In how many do the E’s stand first?
i
ii
iii
## 3(b)
In how many ways can the letters of the word CHITTAGONG be arranged, so that all vowels are together?
The following is multiple choice question (with options) to answer.
In how many different ways can the letters of the word ‘EVER’ be arranged? | [
"12",
"280",
"300",
"310"
] | A | Number of ways = 4!/2! = 4X3X2X1/2X1 = 12
A |
AQUA-RAT | AQUA-RAT-37088 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
The average of numbers 0.54204, 0.54203, 0.54202 and 0.54201 is ? | [
"0.54202",
"0.54204",
"0.542022",
"0.542025"
] | D | Answer
Average = (0.54204 + 0.54203 + 0.54202 + 0.54201)/ 4
=2.1681/ 4
=0.542025
Correct Option: D |
AQUA-RAT | AQUA-RAT-37089 | P.S. I am very new to this site and I know similar questions have been asked and it'd have been better to comment over there but it's not letting me comment due to low reputation. Sorry :)
• Those answers are both correct. They are the same. Sep 16 at 2:23
• 90C4 * 10 / 100C5 is equal to 90P4 * 10 / 100P5. Either way of counting gives the same answer. Sep 16 at 2:25
• yes I do realize that both answers in the end come out to be correct but that's only because the denominators cancel out. But it's technically incorrect to use combination instead of permutation and I am trying to understand why. And I still don't know what the solution for the bulbs question should be. To me it just seems like 9C5 but if order is important in Beatles songs question, it should be important here as well in which case it'd be 9P5. Sep 16 at 2:27
• "but its technically incorrect to use combination instead of permutation" Says who? For probability scenarios we get to choose what sample space we use to describe the scenario. It is a choice. So long as we chose appropriately, there can be multiple choices. Sep 16 at 2:40
• Umm okay so, in words, probability for scenario 1, is total ways in which the first Beatles song is the 5th song / total ways of choosing 5 from 100 What I just said is permutation or combination? Why? That's what I am trying to understand as in, even though they both yield the same answer, what's the difference conceptually? Can anyone put the two solutions in actual words highlighting the conceptual difference please? I am sorry I am not able to exactly construct the question I have Sep 16 at 2:47
Let $$S = \displaystyle \frac{\binom{90}{4} \times \binom{10}{1}}{\binom{100}{5}}.$$
The following is multiple choice question (with options) to answer.
Jill’s compact disc player randomly plays a song, so that no song is repeated before the entire album is played. If Bill plays a disc with 18 songs, what are the chances that the third song he hears will be his favorite? | [
"1/18",
"1/12",
"1/11",
"3/14"
] | A | The player plays the songs on the album at random, so there are 18 songs that can be the third one played. Only 1 of these ways will result in Bill's favorite song being the third one played.
Total outcomes ==> 18
Total positive outcomes ==> 1
Correct answer is A) 1/18 |
AQUA-RAT | AQUA-RAT-37090 | thermodynamics, temperature
Case 2:Milk is added at the beginning of the experiment
The only change from case 1 would be that the initial temperature of the solution would now be $T_0 - \Delta T$ instead of $T_0$.
So the solution fot $T$ will be :$T(t) = T_{env} + (T_0 -\Delta T- T_{env}) e^{-k t}$
Therefore after time $\tau$ the temperature of the solution will be:
$T_2(\tau) = T_{env} + (T_0 -\Delta T- T_{env}) e^{-k t}$
So finally we have,$T_2(\tau)-T_1(\tau)=\Delta T (1-e^{-k t})$
Now for all $t>0$,$(1-e^{-k t})$ is always positive.
So $T_2(\tau)>T_1(\tau)$ always.
Moral of the story:"If you want hot tea,add milk in the beginning!"
Note:Here we assumed Newton's Law of Cooling was valid which was somewhat a simplistic assumption which may not be true in the real world.
The following is multiple choice question (with options) to answer.
A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 4 cups of tea are prepared, 2 in one way and 2 in other. Find the different possible ways of presenting these 4 cups to the expert. | [
"25",
"24",
"30",
"34"
] | B | Solution:
Since, there are 2 cups of each kind, prepared with milk or tea leaves added first, are identical hence, total number of different people ways of presenting the cups to the expert is,
[4!/(2!x2!)]= 24
Answer: Option B |
AQUA-RAT | AQUA-RAT-37091 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B started a business in partnership investing Rs.20,000 and Rs.15,000 respectively. After six months, C joined them with Rs.20,000. What will be A's share in the total profit of Rs.26,000 earned at the end of 2 years from the starting of the business? | [
"Rs.7500",
"Rs.9000",
"Rs.9500",
"Rs.10,400"
] | D | Solution
A : B : C =(20000×24):(15000×24):(20000×18)
=4 :3 :3
A's share =Rs.(26000×4/10)
= Rs.10400.
Answer D |
AQUA-RAT | AQUA-RAT-37092 | # probability of rolling a 6
Select a die from a bag containing 2 dice, one die has 6 on all faces, and the other is a fair sided die.
Choosing one die at random, roll it, and get a 6. If you roll the same die, what is the probability that the next roll is also a 6?
Would this be (0.5)(1) + (.5)(1/6) = 0.5833, as picking one of the two dice is equally likely and the probability of obtaining the result is what follows.
• What you have is indeed the probability of throwing a 6 if each for is equally likely to be the chosen one. However, the fact that the first throw is a 6 gives you information that makes it more likely that the die you have is the fudged die. – Arthur Oct 30 '19 at 16:33
• A priori, the probability of choosing the fair die is $1/2$, but after you've rolled a $6$, it's rather more likely that you picked the biased die, isn't it? So you need to update the probabilities. – saulspatz Oct 30 '19 at 16:34
We are told that the first chosen die gave us a $$6$$, let this event be $$A$$.
Let $$B$$ be the event that the chosen die has only $$6$$.
$$P(B|A)=\frac{P(B)P(A|B)}{P(A)}=\frac{\frac12}{\frac12+\frac12\cdot \frac16}=\frac67$$
Let $$C$$ be the event that $$6$$ is observed again.
$$P(C|A)=P(C|BA)P(B|A)+P(C|B^CA)P(B^c|A)=\frac67 + \frac{1}{6}\frac{1}{7}=\frac{37}{42}$$
The following is multiple choice question (with options) to answer.
What is the probability of rolling a nine with two, fair, six-sided dice? | [
"1/6",
"1/9",
"2/13",
"1/12"
] | B | There are 4 ways to roll a nine: 3 and 6, 6 and 3, 4 and 5, 5 and 4 with two six-sided dice. There are 6 times 6 = 36 ways to roll two dice. The probability of rolling a 9 is thus 4/36 = 1/9. B |
AQUA-RAT | AQUA-RAT-37093 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
In Mike's opinion, his weight is greater than 65kg but leas than 72 kg. His brother does not agree with Mike and he thinks that Mike's weight is greater than 60kg but less than 70kg. His mother's view is that his weight cannot be greater than 68 kg. If all of them are correct in their estimation, what is the average of different probable weights of Mike? | [
"54.3 kg",
"55.5 kg",
"66.5 kg",
"67.8 kg"
] | D | Let Mike's weight be X kg.
According to Mike, 65 < X < 72.
According to Mike's brother, 60 < X < 70.
According to Mike's mother, X < 68.
The values satisfying all the above conditions are 66 and 67.
Required average = (66 + 67) / 2 = 66.5 kg
D |
AQUA-RAT | AQUA-RAT-37094 | # If $(a+b)(a+c)(b+c)=8abc$ prove $a=b=c$
For positive numbers $$a,b,c$$ we have $$(a+b)(a+c)(b+c)=8abc$$. Prove $$a=b=c$$
I tried expanding the expression. after simplifying we have,
$$a^2b+ab^2+b^2c+ca^2+ac^2+bc^2=6abc$$ But not sure how to continue.
I also noticed that we have,
$$(a+b)(a+c)(b+c)=(2a)(2b)(2c)$$ $$(a+b)+(a+c)+(b+c)=(2a)+(2b)+(2c)$$ But I don't know if it helps.
• Have you tried AM-GM inequality...? Dec 14, 2021 at 14:12
• Dec 14, 2021 at 14:27
By AM-GM, $$a+b \geqslant 2\sqrt{ab}$$ with equality if and only if $$a=b$$. Multiplying together the three similar inequalities we get $$(a+b)(b+c)(c+a) \geqslant 8abc$$ with equality if and only if $$a=b=c$$.
Continuing from where you left, you have: $$b(a-c)^2+a(b-c)^2+ b^2c+ca^2-2abc=0$$,which is same as $$b(a-c)^2+a(b-c)^2+c(b-a)^2=0$$ So you now have sum of three non-negative numbers equal to $$0$$. Can you take it from here?
The following is multiple choice question (with options) to answer.
If a + b + c = 13, then find the ab + bc + ca : | [
"22",
"99",
"50",
"29"
] | C | Answer: C) 50 |
AQUA-RAT | AQUA-RAT-37095 | 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.
In order to complete a reading assignment on time, Terry planned to read 100 pages per day. However, she read only 75 pages per day at first, leaving 625 pages to be read during the last 6 days before the assignment was to be completed. How many days in all did Terry have to complete the assignment on time? | [
"8",
"7",
"9",
"10"
] | B | D=100*X - according to the plan, assignment to be completed on time reading 100 pages per day for next X days. But, Terry's plans changed so she read as follow:
75 pages for first Y days and 625 pages for last 6 days, we get these equations:
75*Y+625=100*X
X-Y=6 --------->>X planned number of days, Y - actually used reading 75 pages per day and 6 leftover days used to complete a lump 625 pages
From above we find that X=Y+6 and 75Y+625=100Y+600 or 25Y=25 --->>>>> Y=1, hence X=7
Answer : B |
AQUA-RAT | AQUA-RAT-37096 | \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain
The following is multiple choice question (with options) to answer.
If a*b*c=130, b*c*d = 65, c*d*e=1000 and d*e*f=250 the (a*f)/(c*d) = ? | [
"1/2",
"1/4",
"3/4",
"2/3"
] | A | Explanation :
a∗b∗c/b∗c∗d= 130/65 => a/d = 2
d∗e∗f/c∗d∗e= 250/1000 => f/c = 14
a/d* f/c = 2 * 1/4 = 1/2
Answer : A |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.