source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-33997 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
Three people have $36, $72, and $98, respectively. If they pool their money then redistribute it among them, what is the maximum value for the median amount of money? | [
"$103",
"$85",
"$98",
"$101"
] | A | Solution -
Total money distributed is $206. In order to maximize median, one person has to accept $0 and remaining two people share $103 each. Hence median is $103.
ANS A |
AQUA-RAT | AQUA-RAT-33998 | # how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
The following is multiple choice question (with options) to answer.
All the faces of cubes are painted with red colour.It is then cut into 64 equal small cubes.Find How many small cubes have only one face coloured ? | [
"4",
"8",
"16",
"24"
] | D | There are 64 small cubes,Hence one side side of the big cube is 3√64=4 cm
Number of small cubes having only one face coloured = (x - 2)2 x No. of faces
= (4 - 2)^2 x 6
= 24
Answer:D |
AQUA-RAT | AQUA-RAT-33999 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
If Juan takes 16 seconds to run y yards, how many seconds will it take him to run x yards at the same rate? | [
"16x/y",
"16y/x",
"x/(16y)",
"16/(xy)"
] | A | This problem is testing us on the Rate x Time = Distance relationship. This relationship also tells us that Rate = Distance/Time and Time = Distance/Rate.
Ultimately, we are looking for how many seconds it will take Juan to run x yards. Thus, the equation we’ll use is: Time = Distance/Rate. We know the distance is x yards, and we need to find Juan’s rate.
We can find Juan’s rate as follows: Rate = Distance/Time = y yards/16 seconds
Using that rate, we need to determine how long it takes him to run x yards at the same rate. So we have:
Time = Distance/Rate
Time = x yards/(y yards/16 seconds)
Time = (x yards) x (16seconds/y yards)
Time = 16x/y seconds
Answer A. |
AQUA-RAT | AQUA-RAT-34000 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
A man walks at a speed of 5 km/hr and runs at a speed of 7 km/hr. How much time will the man require to cover a distance of 10 1/2 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running? | [
"1 4/5 hours",
"2 hours",
"2 1/6 hours",
"2 1/3 hours"
] | A | Required time
= (5 1/4)/5 + (5 1/4)/7
= 1 4/5 hours.
Answer:A |
AQUA-RAT | AQUA-RAT-34001 | There will be one solution: name it $x_0$.
Find $y$ at $x_0$, and call it $y_0$.
Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$
The following is multiple choice question (with options) to answer.
A straight line in the xy- plane has y-intercept of 49 . On this line the x-coordinate of the point is 71 and y-coordinate is 12 then what is the slope of the line? | [
"0.5",
"0.51",
"-0.52",
"0.31"
] | C | Eq of line = Y = MX +C
C= 49
X=71
Y=12
Substitute given:
M = (Y-C) / X = (12 - 49) / 71 = -37/71 = -0.52
correct option is C |
AQUA-RAT | AQUA-RAT-34002 | homework-and-exercises, work, potential-energy
Title: Problem regarding finding work done This problem is given in my physics book. How much work will be done if anyone wants to stack up 12 bricks? Given that each brick is $10 cm$ high and each brick's mass is $2 kg$. ($g=9.8$)
Now in finding my answer I used the mass and height of all the bricks. But in my book they used the mass of all the bricks. But in using the height they used the height of only 11 bricks. So my result is $141.12 J$ and the book's result is $129.36 J$. Why is that? We are supposed to stack 12 bricks on one another. One brick is unmoved. The centre of mass of 11 bricks is 60 cm or 0.6m above initial level after stacking.
Now, the work done against the force of gravity is:
$W = mg \Delta h = 2(11)(9.8)(0.6) = 129.36J $
Another method is to calculate work done individually
$W = mg(\Delta h_1+\Delta h_2+\Delta h_3+\Delta h_4+\Delta h_5+\Delta h_6+\Delta h_7+\Delta h_8+\Delta h_9+\Delta h_{10}+\Delta h_{11})$
$\implies W = 2(9.8)(0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9+1+1.1)$
$\implies W = 129.36J$
The following is multiple choice question (with options) to answer.
Bob and Sue are building a brick wall. They can each lay fifty bricks per hour. If Bob takes a one hour break after every two hours of working, and Sue takes a one-hour break after every three hours of working, how long will it take them to finish the wall if they need to lay 600 bricks to finish the wall? | [
"6 hours",
"7 hours",
"8 hours",
"9 hours"
] | C | Bob works 2 hours out of every 3. Sue works 3 hours out of every 4. They can each lay 50 bricks per hour of work. Laying 600 bricks will take a total work of 600/50 = 12 hours. After 6 hours, Bob has done 4 hours of work and Sue has done 5 hours of work, for a total of 9 hours. After 7 hours, Bob has done 5 hours of work and Sue has done 6 hours of work, for a total of 11 hours. After 8 hours, Bob has done 6 hours of work and Sue has done 6 hours of work, for a total of 12 hours. Therefore, the answer is C. |
AQUA-RAT | AQUA-RAT-34003 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
If a clock strikes 12 in 33 seconds, it will strike 6 in how many seconds? | [
"33⁄2",
"15",
"12",
"22"
] | B | In order to strike 12, there are 11 intervals of equal time
= 33⁄11 = 3 seconds each
Therefore, to strike 6 it has 5 equal intervals, it requires 5 × 3 = 15 sec.
Answer B |
AQUA-RAT | AQUA-RAT-34004 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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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
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
What is the average (arithmetic mean) of the numbers 5, 6, 7, 7, 8, and 9? | [
" 4.2",
" 6.5",
" 6",
" 7"
] | D | {5, 6, 7, 7, 8, 9}={7-2,7-1, 7, 7,7+1,7+2} --> the average = 7.
Answer: D. |
AQUA-RAT | AQUA-RAT-34005 | ## 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.
(1/2) of a number is 3 more than the (1/6) of the same number? | [
"6",
"7",
"8",
"9"
] | D | Take a number as x.
(1/2) of a number is 3 more than the (1/6) of the same number.
i.e., x/2 = 3+(x/6)
=> x/3 = 3
=> x = 9
ANSWER:D |
AQUA-RAT | AQUA-RAT-34006 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern has a leak which would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold? | [
"480",
"287",
"289",
"270"
] | A | 1/x - 1/20 = -1/24
x = 120
120 * 4 = 480
Answer: A |
AQUA-RAT | AQUA-RAT-34007 | ## The nominal and effective rates are equivalent for annual compounding
Convert my salary to an equivalent hourly wage · Convert my hourly wage to an The number of compounding periods per year will affect the total interest earned Use this calculator to determine the effective annual yield on an investment. AssumptionsPart 1. Assumptions. Nominal/stated annual interest rate (0% to 40%). If the nominal interest rate is 8%, find the effective annual rate with quarterly compounding. Method 1: By Formula. m = 4, EAR = (1 + 0.08/4)4 - 1 = 0.0824 The nominal interest rate, also called annual percentage rate (APR), is simply the since interest is compounded monthly, the actual or effective interest rate is
The following is multiple choice question (with options) to answer.
A part of certain sum of money is invested at 9% per annum and the rest at 15% per annum, if the interest earned in each case for the same period is equal, then ratio of the sums invested is? | [
"4:2",
"4:8",
"5:3",
"4:0"
] | C | 15:9
= 5:3
Answer: C |
AQUA-RAT | AQUA-RAT-34008 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A can run 4 times as fast as B and gives B a start of 60 m. How long should the race course be so that A and B might reach in the same time? | [
"70 m",
"60 m",
"80 m",
"65 m"
] | C | Speed of A:Speed of B=4:1
means in a race of 4m A gains 3m.
then in a race of 60m he gains 60*(4/3)
I.e 80m
ANSWER:C |
AQUA-RAT | AQUA-RAT-34009 | 4. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
Psec/1bolt * 5bolts/1rev * 1rev/78.4in *...
The OP specified that the circumference of the wheel is 78.4in. Why do we need to include the radius in the conversion?
Last edited: Sep 8, 2015
5. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
Because I missed that little detail?
6. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
But the circumference of the wheel is irrelevant -- it's the circumference of the tire that matters. Perhaps the TS means the tire and not the wheel, but I'm not positive about that.
7. ### djsfantasi AAC Fanatic!
Apr 11, 2010
2,805
833
A circumference of 78.4" is a diameter of 10". Much more likely to be the diameter of a wheel.
Update: Helps to use the right equation...
Last edited: Sep 8, 2015
8. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
Or it's one of the new micro-cars that are becoming the snob-appeal rage.
9. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
As sensors go I like it. It would also be good for odometers too.
10. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
I think driveshaft sensors are probably better since you typically get several rotations of the shaft per rotation of the tire (and you can mount multiple magnets on the shaft to further improve resolution). It also tends to average out the distance traveled by the tires as you drive due to the natural behavior of the differential.
Jan 26, 2014
5
0
12. ### rjwheaton Thread Starter New Member
Jan 26, 2014
5
0
Thank you!
Not only does that answer my question, but it helps me understand how you did the calculations.
The following is multiple choice question (with options) to answer.
The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 250 resolutions? | [
"352 m",
"704 m",
"454 m",
"186 m"
] | A | In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 250 resolutions.
= 250 * 2 * 22/7 * 22.4
= 35200 cm
= 352 m
Answer: A |
AQUA-RAT | AQUA-RAT-34010 | (1) The median of set {x,−1,1,3,−x} is 0
(2) The median of set {x,−1,1,3,−x} is x2
Merging similar topics. Please refer to the solutions above.
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Re: What is the value of x? [#permalink]
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09 Jun 2017, 15:49
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: What is the value of x? [#permalink] 09 Jun 2017, 15:49
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The following is multiple choice question (with options) to answer.
If x is the median of the set {9/2, 10/3, 27/9, 21/5, x}, x could be | [
"16/5",
"17/5",
"28/7",
"30/7"
] | C | The median is the middle number once all the numbers are arranged in increasing/decreasing order.
We see that
10/3 = 3.something,
27/9 = 3.something
21/5 = 4.something
9/2 = 4.something
So x should greater than the smallest two numbers and smaller than the greatest two numbers. We can see that x = 4 is possible. (First look at the simplest option or the middle option since options are usually arranged in increasing/decreasing order)
Answer (C) |
AQUA-RAT | AQUA-RAT-34011 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
12, 38, 116, 350, 1052, ? | [
"5815",
"3815",
"3518",
"3158"
] | D | 12
12 × 3 + 2 = 38
38 × 3 + 2 = 116
116 × 3 + 2 = 350
350 × 3 + 2 = 1052
1052 × 3 + 2 = 3158
Answer is D. |
AQUA-RAT | AQUA-RAT-34012 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into a partnership. They invest Rs.40,000, Rs.80,000 and Rs.1,20,000 respectively. At the end of the first year, B withdraws Rs.40,000 while at the end of the second year, C withdraws Rs.80,000. In what ratio will the profit be shared at the end of 3 years? | [
"2 : 3 : 5",
"3 : 4 : 7",
"4 : 5 : 9",
"None of these"
] | B | Solution
A : B: C =(40000×36) : (80000×12+40000×24) : (120000×24+40000×12)
=144 :192 :336
= 3 : 4 : 7.
Answer B |
AQUA-RAT | AQUA-RAT-34013 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and B is closed after 3 minutes, how much further time would it take for A to fill the bucket? | [
"5min. 50sec.",
"5min. 30 sec.",
"6 min 36 sec.",
"6mins. 32 sec."
] | C | Explanation:
Part filled in 3 minutes =
3∗(1/12+1/15)=3∗9/60=9/20
Remaining part =1−9/20=11/20
=>1/12:11/20=1:X
=>X=6.6 mins.
=>X=6.6 mins.
So it will take further 6 mins 36 seconds to fill the bucket.
Option C |
AQUA-RAT | AQUA-RAT-34014 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
James took a 3-hour bike ride. In the second hour he traveled 6 miles, which was 20 percent farther than he traveled the first hour. If he traveled 25 percent farther in the third hour than he did in the second hour, how many miles did jose travel during the entire ride? | [
"54.0",
"54.9",
"18.5",
"57.0"
] | C | Let the distance travelled in the first hour be x. Thus, 1.2x = 6, x = 5. Now, the distance travelled in the 3rd hour = 6+1/4∗6=7.5 The only option ending with a 0.5 in the decimal place is C.
ANSWER:C |
AQUA-RAT | AQUA-RAT-34015 | mass, weight
I used two socks, a binder clip to secure the weights to it, a pen, some string, and multiple 20 fl oz ( = 1.3 lbs of water ) bottles, and tried to see if the response was hookian over several bottles. I got
1 bottle -- 1.5 distance units
2 bottles -- 2.8 distance units
3 bottles -- 3.9 distance units
I'm being vague about the distance units because I didn't actually have a ruler handy, so instead used the L scale on my sliderule to measure the extensions
Rendered as a plot we see:
While this looks decent, I'm troubled by the fact that it doesn't line up well with the zero point extension of the sock, also we were only able to take 3 measurements since with the addition of the 4th bottle, the binder clip gave way.
Regardless of the questions of hookian reliability of the socks, to weigh your nephew at around 30 lbs, you would need something like 25 nearly identical socks, or calibrate all of them individually, and figure out a way to reliably afix your nephew to the socks. Due to the impracticality of the method, I can't recommend this approach, though in the interest of science, and so that others need not follow in my footsteps, I've shared this failure here.
The following is multiple choice question (with options) to answer.
Dimitri weighs x pounds more than Allen weighs. Together, Allen and Dimitri weigh a total of 2y-y pounds. Which of the following represents Allen’s weight? | [
"y - x/2",
"2x - y/2",
"(y - x)/2",
"y - 2x"
] | C | These type of multi-variable story problems are usually perfect for TESTing VALUES.
Here, we're told that Dimitri weighs X pounds more than Allen weighs and that they weight a combined TOTAL of 2Y-Y or Y pounds.
IF....
Dimitri = 30
Allen = 20
X = 10
Y = 50
We're asked for Allen's weight, so we're looking for an answer that equals 20 when X = 10 and Y = 50.
Answer A: 50 - (10/2) = 45 NOT a match
Answer B: 20 - (50/2) = -5 NOT a match
Answer C: (50-10)/2 = 20 This IS a match
Answer D: 50 - 20 = 30 NOT a match
Answer E: 20 - 50 = -30 NOT a match
Final Answer:
C |
AQUA-RAT | AQUA-RAT-34016 | 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.
895.7 – 573.07 – 95.007 = ? | [
"227.623",
"224.777",
"233.523",
"414.637"
] | A | Solution
Given expression = 895.7 - (573.07 + 95.007) = 895.7 - 668.077 = 227.623. Answer A |
AQUA-RAT | AQUA-RAT-34017 | the largest and smallest integers in the group. This method should take three decimal arguments, and return the smallest of the three. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. It is similar to modulus – ‘%’ operator of C/C++ language. Note that this is not the modulo * operation (the result can be negative). Print its smallest integer divisor greater than 1. Find the least common multiple and greatest common divisor of the given three numbers : Find the least common multiple and the greatest common divisor of numbers 12, 24, 48, 192 and 288. LCM(Least Common Multiple) is the smallest positive number which is divisble by both the numbers. The macro below does the. During the reconstruction of the railway tracks 40 meter long pieces of rails were replaced by 15 meter long pieces. (For example: 7/3 = 3 and 10/2 = 5). Dynamically Calculate Prime Numbers. The smallest divisor (other than 1) of a composite number is a/an a) odd number b) even number c) prime number d) composite number. Thus the smallest number we can represent would be: 1E(-(MIN_EXP-digits-1)*4), eg, for digits=5, MIN_EXP=-32767, that would be 1e-131092. For example divisors of 6 are 1, 2, 3 and 6, so divisor_sum should return 12. gcd(a, b) = 5 3 · 7 2 = 6125. On a number line, we get hundredths by simply dividing each interval of one-tenth into 10 new parts. Example: The smallest weird number is 70. Write a Java program that computes the absolute value of the product of 2 numbers given by the user. Here dividend is 205 and divisor is 2 therefore remainder is 1. The last divisor will be the HCF of given numbers. In particular, we have added an entirely new Ch. The SMALL function will automatically ignore TRUE and FALSE values, so the result will be the nth smallest value from the set of actual numbers in the array. In this tutorial we will write couple of different Java programs to find out the GCD of two numbers. The first perfect number is 6, because
The following is multiple choice question (with options) to answer.
Which is smallest prime number? | [
"97",
"109",
"103",
"95"
] | A | Answer:A |
AQUA-RAT | AQUA-RAT-34018 | +0
# URGENT! I NEED HELP RIGHT AWAY!
0
261
1
+755
Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage. For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps. However, there are some amounts of postage he can't make exactly, such as 10 cents.
What is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps?
Explain how you know your answer is correct. (You should explain two things: why Larry can't make the amount of your answer, and why he CAN make any bigger amount.)
Jan 26, 2018
#1
+21199
+1
Larry has 4-cent stamps and 9-cent stamps, which he can combine to produce various amounts of postage.
For example, he can make 40 cents by using four 9-cent stamps and a 4-cent stamp, or by using ten 4-cent stamps.
However, there are some amounts of postage he can't make exactly, such as 10 cents.
What is the largest number of cents that Larry CANNOT make exactly from a combination of 4- and/or 9-cent stamps?
Explain how you know your answer is correct. (You should explain two things:
why he CAN make any bigger amount.)
The following is multiple choice question (with options) to answer.
The number of stamps that Kaye and Alberto had were in the ration of 5:3 respectively. After Kaye gave Alberto 10 of her stamps, the ration of the number of Kaye had to the number of Alberto had was 6:4. As a result of the gift, Kaye had how many more stamps than Alberto? | [
"80",
"85",
"65",
"55"
] | A | C
K1 = 5x
A1 = 3x
K2 = 5x - 10
A2 = 3x + 10
K2/A2 = 6/4
(5x-10)/(3x+10) = 6/4
4*(5x-10) =6*(3x+10)
20x - 40 = 18x + 60
2x = 100
x = 50
K2 = 5*50 - 10 = 240
A2 = 3*50 + 10 = 160
K2 - A2 = 80
Answer : A |
AQUA-RAT | AQUA-RAT-34019 | 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.
What distance will be covered by a bus moving at 54 kmph in 30 seconds? | [
"605",
"450",
"601",
"609"
] | B | 54 kmph = 54 * 5/18 = 15 mps
D = Speed * time = 15 * 30
= 450 m.
Answer: B |
AQUA-RAT | AQUA-RAT-34020 | We need to understand the compound interest formula: A = P(1 + r/n)^nt. A stands for the amount of money that has accumulated. P is the principal; that's the amount you start with. The r is the. is continuously, where interest is compounded essentially every second of every day for the entire term. This means 𝑛 is essentially infinite, and so we will use a different formula which contains the natural number 𝑒 to calculate the value of an investment. The formula for interest compounded continuously is 𝐴=𝑃𝑒𝑟𝑡. APY to APR Calculator : Enter the APY currently being earned (in percent): % Enter the number of compounding periods in a year ... If it’s compounded monthly: APY = (APR/12 + 1)^12 – 1. (where APR & APY are decimals; for 3% you’d put 0.03). But, always we have question about compounded continuously. To understand 'compounded continuously', let us consider the example given below. When we invest some money in a bank, it will grow continuously. That is, at any instant the balance is changing at a rate that equals 'r' (rate of interest per year) times the current balance. Formula for. The continuous compounding formula says A = Pe rt where 'r' is the rate of interest. For example, if the rate of interest is given to be 10% then we take r = 10/100 = 0.1. What Is e in Continuous Compounding Formula? 'e' in the continuous compounding formula is a mathematical constant and its value is approximately equal to 2.7183. In this section we cover compound interest and continuously compounded interest. Use the compound interest formula to solve the following. Example: If a $500 certificate of deposit earns 4 1/4% compounded monthly then how much will be accumulated at the end of a 3 year period?. Answer: At the end of 3 years the amount is$576.86. The formula for the principal plus interest is as follows: Total = Principal x e^(Interest x Years) Where: e – the exponential function, which is equal to 2.71828. Using Company ABC example above, the return on investment can be calculated as follows when using continuous compounding: = 10,000 x 2.71828^(0.05 x
The following is multiple choice question (with options) to answer.
Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit? | [
"Rs. 8600",
"Rs. 8620",
"Rs. 8820",
"Rs. 9820"
] | C | Explanation:
Amount
= Rs.[8000 x (1 + 5/100)2] = Rs. 8820.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-34021 | In how many ways can the nine letters $$\{A, B, C, D, E, F, G, H, I\}$$ be placed in the nine envelopes $$\{a, b, c, d, e, f, g, h, i\}$$ so that letter $$A$$ is placed in envelope $$b$$ and no letter is placed in the correct envelope?
If letter $$A$$ is placed in envelope $$b$$, there are two possibilities:
1. Letter $$B$$ is placed in envelope $$a$$. Then each of the seven letters not already placed has one prohibited envelope, namely its own envelope. Thus, there must be a derangement on the remaining seven envelopes. Hence, there are $$D_7$$ such cases.
2. Letter $$B$$ is not placed in envelope $$a$$. Then each of the eight letters not already placed has one prohibited envelope, its own envelope for each letter other than letter $$B$$ and envelope $$a$$ for letter $$B$$. Hence, there must be a derangement on the eight letters not already placed, so there are $$D_8$$ such cases.
Since these two cases are mutually exclusive, the number of ways the nine letters can be placed in the nine envelopes so that letter $$A$$ is placed in envelope $$b$$ and no letter is placed in the correct envelope is $$D_7 + D_8$$.
In how many ways can the nine letters $$\{A, B, C, D, E, F, G, H, I\}$$ be placed in the nine envelopes $$\{a, b, c, d, e, f, g, h, i\}$$ so that letter $$A$$ is placed in envelope $$b$$, letter $$B$$ is placed in envelope $$f$$, and no letter is placed in the correct envelope?
Again, there are two possibilities:
The following is multiple choice question (with options) to answer.
You have a bag of 9 letters: 3 Xs, 3 Ys and 3 Zs. You are given a box divided into 3 rows and 3 columns for a total of 9 areas. How many different E ways can you place one letter into each area such that there are no rows or columns with 2 or more of the same letter? | [
"5",
"6",
"9",
"12"
] | D | Consider one particular arrangement of the first row: XYZ, then we can construct only two boxes with so that no rows or columns have 2 or more of the same letter:
XYZ
YZX
ZXY
And:
XYZ
ZXY
YZX
Now, the first row itself can be arranged in 3!=6 ways (since there are three distinct letter), hence the total number of boxes possible E is 2*6=12.
Answer: D. |
AQUA-RAT | AQUA-RAT-34022 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
A store reported total sales of $385 million for February of this year. If the total sales for the same month last year was $320 million, approximately what was the percent increase in sales? | [
"2%",
"17%",
"20%",
"65%"
] | C | New Value – Old Value)/Old Value x 100
We are given:
February sales this year = 385 million
February sales last year = 320 million
We need to determine the percent increase between sales from last year to sales this year.
Thus, the new value = 385 million and the old value = 320 million. Let’s plug them into our percent change formula.
(New Value – Old Value)/Old Value x 100
[(385 – 320)/320] x 100
65/320 x 100
13/64 x 100 ≈ 13/65 x 100 ≈ 1/5 x 100 ≈ 20%.
The answer is C. |
AQUA-RAT | AQUA-RAT-34023 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\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.
Fence X is twice as long as fence Y, and fence Y is 2 feet shorter than fence Z. If 3 feet were added to each fence, which of the following must be true?
I) X is twice as long as Y.
II) Y is 2 feet shorter than Z.
III) X is longer than Z. | [
"I only",
"II only",
"III only",
"I and II"
] | B | Length of Z = z
Length of Y = z-2
Length of X = 2z-4
Minimum possible length of fence Z is 2 feet, because if we take the value for length of z below 2, lengths of Z and Y would be negative which is not possible.
If Z = 2
then Y = 0 and X = 0
When we add 3 feet in each fence, we will get
Z= 5
Y=3
X=3
Only Statement II holds true with above values.
ANSWER:B |
AQUA-RAT | AQUA-RAT-34024 | With this relationship in mind, let’s take a look at statement 1.
## (1) The next time he prepares this dish, Malik will make half as many servings as he did the last time he prepared the dish.
Since we don’t know how many servings Malik made the last time, we still don’t know how many servings he will make the next time. So statement 1 is insufficient.
Let’s try the next statement.
## (2) Malik used 6 cups of pasta the last time he prepared this dish.
With this information, we can use the equation from our proportions before to find the number of serving Malik made the last time.
$4*6 = \frac{3}{2}s$
$24 = \frac{3}{2}s$
$24*\frac{2}{3} = s$
$s = 16$
Great! Now we know Malik prepared 16 servings of the dish last time. But, we still don’t know how many servings Malik is planning on preparing next time.
Insufficient.
## Let’s try both statements together:
Well, we know Malik made 16 servings last time from statement 2. We also know that the next time, Malik will make half as many servings as he did last time.
This means next time, Malik is planning on making 16/2 = 8 servings next time.
Since we have proportions relating servings to cups of pasta, we know we can find the number of cups of pasta Malik will need next time. So, we have all the information we need. Still, let’s finish off the question to make sure.
Plugging in 8 for the number of servings into our proportions equation, we have:
$4p = \frac{3}{2}*8$
$4p = 12$
$p = 3$
So, we know Malik will need 3 cups of pasta the next time he makes the dish. Both statements together were sufficient.
The following is multiple choice question (with options) to answer.
When Tom works alone he chops 4lb. salad in 3 minutes, and when Tammy works alone she chops 3 lb. salad in 4 minutes. They start working together, and after some time finish chopping 65 lb. of salad. Of those 80 lb., the salad quantity chopped by Tammy is what percent greater than the quantifying chopped by Tom?. | [
"125 %",
"100%",
"25%",
"225%"
] | A | Tom chops 8 lbs in 12 minutes
Tammy chopsWhen Tom works alone he chops 2 lb. salad in 3 minutes, and when Tammy works alone she chops 3 lb. salad in 2 minutes. They start working together, and after some time finish chopping 65 lb. of salad. Of those 80 lb., the salad quantity chopped by Tammy is what percent greater than the quantifying chopped by Tom?. 9 lbs in 6 minutes
So in the same amount of time, Tammy chops 125% more than Tom, since 9 is 125% greater than 4. So 125% is the answer.
Note that the actual time doesn't matter. If you multiply the time each work by x, you'll multiply the work each does by x, and 9x is still 125% greater than 4x.
Ans :A |
AQUA-RAT | AQUA-RAT-34025 | I understand your intuition, that multiplying by $100\% + p\%$ should be reversed by multiplying by $100\% - p\%$, but there are two ways to see that this won't work. The symbolic way is \begin{align} (100\% + p\%)(100\% - p\%) &= \left(1+\frac{p}{100}\right)\left(1-\frac{p}{100}\right) \\ &= 1 + \frac{p}{100} - \frac{p}{100} + \left(\frac{p}{100}\right)\left(\frac{-p}{100}\right) & \text{[by distributivity, or by FOILing]} \\ &= 1 - \frac{p^2}{10\,000} \text{,} \end{align} which isn't quite the same thing as $1$. The other way to see this is pictorially. When we multiply a quantity by $75\%$, we break it into four equal parts and discard one of the parts. When we multiply a quantity by $125\%$ we break it into four equal parts and add a fifth equal part. If we want to undo multiplying by $75\%$, we start with four equal parts and throw one away, but to get back to the original quantity, we have to add back a copy of one of the three remaining parts (that is, multiply by $100\% + 33.\overline{3}\%$). If we instead re-partition the three remaining parts into four parts (by erasing the divisions and drawing in new ones), we don't add back quite enough to recover the original quantity.
You're right about percentages being equivalent to decimals; they're also equivalent to fractions, and sometimes that's an easier way to think about them.
You can take a percentage and make a fraction out of it by putting it over 1. I'll do that with $125\%$:
The following is multiple choice question (with options) to answer.
$100 is divided amongst A,B and C so that A may get 1/4 as much as B and C together, B may get 3/5 as much as A and C together, then the share of A is | [
"$15",
"$20",
"$25",
"$18"
] | B | A:(B+C) = 1:4
A's share = 100*1/5 = $20
Answer is B |
AQUA-RAT | AQUA-RAT-34026 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 150 meters long completely crosses a 240 meters long bridge in 42 seconds. What is the speed of the train is? | [
"32",
"33",
"88",
"66"
] | B | S = (150 + 240)/42 = 390/42 * 18/5 = 33
Answer: B |
AQUA-RAT | AQUA-RAT-34027 | I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.
• If $$4$$ is in the first position, then the last position has $$4$$ possibilities to keep the number even ($$0$$, $$2$$, $$6$$, or $$8$$). In this case, you get the number of possibilities is: $$1\cdot 4\cdot 8\cdot 7\cdot 6=1344.$$ Here, the product order is "first, last, second, third fourth."
• If $$4$$ is in the last position, then the count above works. In this case, you get the number of possibilities is: $$1\cdot 8\cdot 8\cdot 7\cdot 6=2688.$$ Here, the product order is "last, first, second, third, fourth."
• If $$4$$ is in one of the middle positions, then the last position has $$4$$ possibilities. There are two possibilities here, either $$0$$ is the last position or it isn't.
• If $$0$$ is in the last position, then the number of possibilities is: $$3\cdot 1\cdot 8\cdot 7\cdot 6=1008.$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."
• If $$0$$ is not in the last position, then the number of possibilities is: $$3\cdot 3\cdot 7\cdot 7\cdot 6=2646$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."
Adding all of these up gives $$7686$$, as desired.
• I like this answer more than mine, it is much clearer. Thank you! – Student Nov 5 '18 at 17:47
The following is multiple choice question (with options) to answer.
Four of the five parts numbered (a), (b), (c), (d) and (e) are exactly equal. Which of the parts is not equal to the other four? The number of that part is the answer. | [
"16.80 × 4.50 + 4.4",
"1600 ÷ 40 + 16 × 2.5",
"5.5 × 8.4 + 34.6",
"1620 ÷ 20 – 1"
] | C | Others equal 80 whereas (c) equals 80.8.
Answer C |
AQUA-RAT | AQUA-RAT-34028 | # Thread: probability/combination flipping a coin.
1. ## probability/combination flipping a coin.
question is,
If you flip a coin 5 times what is the probability of getting 4 heads, and 1 tails?
total number of outcomes is $2^5$ because you flip a coin there are to possible outcomes 5 times.
Since there is no order it is a combination *this is where I am confused* I thought when you have two different things like,
"how many 4 card hands can be formed with 2 hearts and 2 diamonds" you do 13c2 x 13c2 because you have 13 of each and you are choosing two each.
but what I've seen searching online is when you flip a coin dice 5 times and want 4 heads it's 5c4=5? but what about the tails? shouldn't it be 5c4 x 5c1 because there are 5 heads and 5 tails? Also if this is the same with dice I'd like to know how to do that too.
thanks.
2. Originally Posted by brentwoodbc
question is,
If you flip a coin 5 times what is the probability of getting 4 heads, and 1 tails?
total number of outcomes is $2^5$ because you flip a coin there are to possible outcomes 5 times.
There are five ways to have 4 heads and one tail.
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
3. Originally Posted by Plato
There are five ways to have 4 heads and one tail.
THHHH
HTHHH
HHTHH
HHHTH
HHHHT
how would you do that if there where more, like 100 head's 125 tails.
I thought it would be 225C100 times 225C125 ?
but it seems to be 225C100?
Is dice the same thing? I dont see how this is different from having say 26 red cards and 5 black cards where you want a 6 card hand, it's 26c4 times 5c2
4. Originally Posted by brentwoodbc
how would you do that if there where more, like 100 head's 125 tails.
The probability of 100 heads and 125 tails is $\binom{225}{100}\left(\frac{1}{2}\right)^{225}$.
The following is multiple choice question (with options) to answer.
If a coin is flipped, the probability that the coin will land tails is 1/2. If the coin is flipped 3 times, what is the probability that it will land tails up on the first flip and not on the last 2 flips? | [
"1/8",
"1/4",
"1/3",
"1/2"
] | A | (1/2) * (1/2) * (1/2) = 1/8 Answer: A |
AQUA-RAT | AQUA-RAT-34029 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 5 a rupee. Find his net profit or loss percent? | [
"34 %",
"33 %",
"20 %",
"35 %"
] | C | The total number of mangoes bought by the shopkeeper be 30.
If he buys 6 a rupee, his CP = 5
He selling at 5 a rupee, his SP = 6
Profit = SP - CP = 6 - 5 = 1
Profit Percent = 1/5 * 100 = 20 %
Answer:C |
AQUA-RAT | AQUA-RAT-34030 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train? | [
"230 m",
"240 m",
"260 m",
"270 m"
] | D | 72 Kmph=72*5/18=20m/s
The train has to cover its length and the length of the platform
so the total distance =250+l
250+l=20*26
250+l=520
l=520-250
=270m
length of the train=270m
ANSWER:D |
AQUA-RAT | AQUA-RAT-34031 | # Chapter 7: Coordinate Geometry
1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.
|AB| = $$\sqrt{(x-5)^{2}+(y-1)^{2}}$$
= $$\sqrt{x^{2}+16-8x+y^{2}+4-4y}$$
|AC| = $$\sqrt{(x+2)^{2}+(y-5)^{2}}$$
= $$\sqrt{x^{2}+4+4x+y^{2}+16-8y}$$
Since, |AB| = |AC|
$$\Rightarrow$$ $$x^{2}+y^{2}-8x-4y+20$$ = $$x^{2}+y^{2}+4x-8y+20$$
$$\Rightarrow$$ 8y – 4y = 4x + 8x
$$\Rightarrow$$ y = 3x
2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)
Let the vertices of the $$\bigtriangleup$$ be P(0, 8), Q(0, 0) and R(6, 0)
∴ PQ = $$\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8$$
∴ QR = $$\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6$$
∴ RP = $$\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10$$
∴ Perimeter of $$\bigtriangleup$$ = 8 + 6 + 10 = 24 units
The following is multiple choice question (with options) to answer.
fill in ht blank:
(a) the point with coordinates (0,0) is called ____ of a rectangular coordinate system,
(b) to find the x-intercept of a line, we let ___ equal 0 and solve for ___; to find y- intercept , we let ____ equal 0 and solve for___ | [
"(a) the point with coordinates (0,0) is called axis of a rectangular coordinate system, (b) to find the x-intercept of a line, we let x equal 0 and solve for y ; to find y- intercept , we let y equal 0 and solve for y; to find y- intercept , we let x equal 0 and solve for x",
"(a) the point with coordinates (0,0... | C | (a) the point with coordinates (0,0) is called origin of a rectangular coordinate system, (b) to find the x-intercept of a line, we let y equal 0 and solve for x ; to find y- intercept , we let x equal 0 and solve for y
correct answer (C) |
AQUA-RAT | AQUA-RAT-34032 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
Income and expenditure of a person are in the ratio 5:4. If the income of the person is Rs.18000, then find his savings? | [
"Rs.3600",
"Rs.8600",
"Rs.3630",
"Rs.9600"
] | A | Let the income and the expenditure of the person be Rs.5x and Rs.4x respectively.
Income, 5x = 18000 => x = 3600
Savings = Income - expenditure = 5x - 4x = x
So, savings = Rs.3600
Answer: A |
AQUA-RAT | AQUA-RAT-34033 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
The area of a triangle is with base 12m and height 5m? | [
"88 m2",
"30 m2",
"66 m2",
"77 m2"
] | B | 1/2 * 12 * 5 = 30 m2
Answer: B |
AQUA-RAT | AQUA-RAT-34034 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
The average score of a cricketer in 2 matches is 40 and in other 3 matches is 10. Then find the average score in all the 5 matches? | [
"25",
"22",
"30",
"35"
] | B | Average in 5 matches = (2*40 + 3*10)/2+3 = 80+30 / 5 = 110/5 = 22
Answer is B |
AQUA-RAT | AQUA-RAT-34035 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 147, what is the difference between the ages of Patrick and Monica? | [
"48",
"47",
"46",
"49"
] | A | let their ages are 9x,15x and 25x respectively.
difference between the ages of Patrick and Monica = 25x-9x=16x where
49x=147 or x=3
so diff in ages = 16x=16*3=48 yrs
ANSWER:A |
AQUA-RAT | AQUA-RAT-34036 | Is the blue area greater than the red area?
Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?
Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}
Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.
Is this true? If so, is there another way of proving the assertion?
Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.
Source:
The following is multiple choice question (with options) to answer.
The roof of an apartment building is rectangular and its length is 3 times longer than its width. If the area of the roof is 768 feet squared, what is the difference between the length and the width of the roof? | [
"32.",
"40.",
"42.",
"44."
] | A | Answer is A : 42
Let w be the width , so length is 3w. Therefore : w*4w = 768, solving for, w = 16 , so 3w-w = 2w = 2*16 = 32 |
AQUA-RAT | AQUA-RAT-34037 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
In a group of 6 boys&4 girls a Committee of 4 persons is to be formed. In how many different ways can it be done so that the committee has at least 1girl? | [
"180",
"185",
"190",
"195"
] | D | The committee of 4 persons is to be so formed that it has at least 1 woman The different ways that we can choose to form such a committee are:
(i) lw. 3 m in t 6X5X4' x 6C3 = 4x — — 80 3X2X1
x 6c2 =42:: x 26:: = 90 (ii) 2w. 2 m in °C2 (iii) 3w. 1 m in 4C3 x 6C1 = 4 x 6 = 24 (iv) 4w in 6C4 = 1 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least one woman. = 80 + 90 + 24 + 1 = 195
D |
AQUA-RAT | AQUA-RAT-34038 | 1111121, 1111202, 1111211, 1112002, 1120002, 1120012, 1120102, 1121002, 1121102, 1122002, 1200002, 1200012, 1200102, 1200112, 1200202, 1201002, 1201012, 1202002, 1210002, 1210102, 1210202, 1211002, 1212002, 1220002, 1220102, 2000002, 2000012, 2000022, 2000102, 2000112, 2000122, 2000202, 2000212, 2001002, 2001012, 2001022, 2001102, 2001112, 2001122, 2001202, 2001212, 2010012, 2010022, 2011012, 2020012, 2020022,$ and their inverses.
The following is multiple choice question (with options) to answer.
The unit digit in the product (124 * 812 * 816 * 467) is: | [
"2",
"7",
"6",
"8"
] | C | Explanation:
Unit digit in the given product = Unit Digit in (4*2*6*7) = 6
ANSWER: C |
AQUA-RAT | AQUA-RAT-34039 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
A man is 24 years older than his son. In two years, his age will be twice the age of his son. What is the present age of his son? | [
"21",
"20",
"24",
"22"
] | D | Let present age of the son =x years
Then, present age the man =(x+24) years
Given that, in 2 years, man's age will be twice the age of his son
⇒(x+24)+2=2(x+2)⇒x=22
Answer is D. |
AQUA-RAT | AQUA-RAT-34040 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A building contractor employs 20 male, 15 female and 5 child workers. To a male worker he pays Rs.25 per day, to a female worker Rs.20 per day and a child worker Rs.8 per day. The average wage per day paid by the contractor is? | [
"24",
"21",
"20",
"22"
] | B | 20 15 5
25 20 8
500 + 300 + 40 = 840/40 = 21
Answer: B |
AQUA-RAT | AQUA-RAT-34041 | = 407.So it is not an Armstrong number. Call that n. Then take every digit in the number and raise it to the n power. Answer Save. equals the number itself. Recent Posts. These numbers posses a unique dimension, but unfortunately they are not of any practical use. We need to calculate the sum of cube of each digit. A good illustration of Armstrong’s number is 407. 8 Answers. All one digit numbers are Armstrong numbers. 4150 = 4^5 + 1^5+ 5^5 + 0^5. Step 8.repeat steps 4 to 6 until number (n)!=0. For example, 8208 has 4 digits, and 8208 = 8^4 + 2^4 + 0^4 + 8^4, so 8208 is an Armstrong number. For example, 407 is given as input. sum of the digits in the number. Explore Armstrong numbers, identify all Armstrong numbers less than 1000, and investigate a recursive sequence. Armstrong number: The number is said to be an Armstrong number if the sum of the cubes of each digits is equal to the number itself. Here we have taken power 3 because the number 370 is 3 digit number. For example, 370 is an Armstrong number since 370 = 3*3*3 + 7*7*7 + 0*0*0. An Armstrong number, also known as narcissistic number, is a number that is equal to the sum of the cubes of its own digits. In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself. An Armstrong number (AKA Plus Perfect number, or narcissistic number) is a number which is equal to its sum of n-th power of the digits, where n is the number of digits of the number.. For example, 153 has 3 digits, and 153 = 1^3 + 5^3 + 3^3, so 153 is an Armstrong number. Do you all know the meaning of an armstrong number? 1 decade ago. Pranil. 6 years ago. Example 2: Check Armstrong for N digits Number. An Armstrong number is an n-digit base b number such that the sum of its (base b) digits raised to the power n is the number itself. #To check Armstrong number for 3 digits. An armstrong Number of N digits is an integre such that the sum of its every
The following is multiple choice question (with options) to answer.
I purchased a superb bike and like my bike my number plate is also superb.
The number plate is a five digit number plate and its water image is 78633 more than the number plate.
Also all digits are unique.
Whats is my number plate number ? | [
"10935",
"10968",
"23984",
"13725"
] | B | B
10968
Water image is 89601 |
AQUA-RAT | AQUA-RAT-34042 | 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.
Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 2.5 times the amount of the second solution? | [
"62",
"65",
"68",
"70"
] | B | SOL:
Lets assume that 25L of sol1 was mixed with 10L of sol2.
Alcohol in Sol1: 25*3/4 = 18.75L
Alcohol in Sol2: 10*2/5 = 4L
Total Alcohol in the mixed solution of 35L = 18.75 + 4 = 22.75L
Alcohol concentration in the new sol: (22.75/35)*100 = 65%
ANSWER:B |
AQUA-RAT | AQUA-RAT-34043 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can finish a work in 12 days while A alone can do the same work in 20 days. In how many days B alone will complete the work? | [
"76 days",
"30 days",
"98 days",
"31 days"
] | B | B = 1/12 – 1/20 = 2/60 = 1/30=> 30 days
Answer: B |
AQUA-RAT | AQUA-RAT-34044 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Mitali, a student of VIIIth standard, ranks eighth when put in class VIII A, ranks fifth when put in class VIII B, and ranks sixth when put in class VIII C. What will be Mitali's rank if students of all the three sections are put together? | [
"17th",
"16th",
"15th",
"Cannot be determined"
] | A | 7+4+5=16
16 students have higher rank than her..
So answer is 17th
ANSWER:A |
AQUA-RAT | AQUA-RAT-34045 | evolution, zoology, anatomy, species
Title: Examples of animals with 12-28 legs? Many commonly known animals' limbs usually number between 0 and 10. For example, a non-exhaustive list:
snakes have 0
Members of Bipedidae have 2 legs. Birds and humans have 2 legs (but 4 limbs)
Most mammals, reptiles, amphibians have 4 legs
Echinoderms (e.g., sea stars) typically have 5 legs.
Insects typically have 6 legs
Octopi and arachnids have 8 legs
decapods (e.g., crabs) have 10 legs
....But I can't really think of many examples of animals containing more legs until you reach 30+ legs in centipedes and millipedes. Some millipedes even have as many as 750 legs! The lone example I am aware of, the sunflower sea star, typically has 16-24 (though up to 40) limbs.
So my question is: what are some examples of animals with 12-28 legs? As a couple of counterexamples, species in the classes Symphyla (Pseudocentipedes) and Pauropoda within Myriapoda have 8-11 and 12 leg pairs respectively, so between 16 to 24 legs (sometimes with one or two leg pair stronlgy reduced in size).
(species in Symphyla, from wikipedia)
Another common and species-rich group with 14 walking legs (7 leg pairs) is Isopoda.
(Isopod, picture from wikipedia)
You also need to define 'legs' for the discussion to be meaningful. As you say, decapods have 10 legs on their thoracic segments (thoracic appendages), but they can also have appendages on their abdomens (Pleopods/swimming legs), which will place many decapods in the 10-20 leg range.
(Decapod abdominal appendages/legs in yellow, from wikipedia)
So overall, in Arthropoda, having 12-28 legs doesn't seem all that uncommon. There are probably other Arthropod groups besides those mentioned here that also have leg counts in this range.
However, for a general account, the most likely answer (if there is indeed a relative lack of 12-28 legged animals) is probably evolutionary contingencies and strongly conservative body plans within organism groups.
The following is multiple choice question (with options) to answer.
A group of people, cats, and birds has seventy legs, thirty heads, and twenty
tails. How many cats are among this group? (Assume all birds have two legs and a tail.) | [
"0",
"5",
"10",
"15"
] | B | Write P, C, and B for the number of people, cats, and birds respectively, so that
we have
2P + 4C + 2B = 70; P + C + B = 30; C + B = 20:
Comparing the last two equations we obtain P = 10, and the rst equation simplies to
4C + 2B = 50, or 2C + B = 25. Therefore,
C = (2C + B) / (C + B) = 25 / 5 = 5:
Alternatively,
C = 1
2
2P + 4C + 2B
P + C + B
= 35 - 30 = 5:
correct answer B |
AQUA-RAT | AQUA-RAT-34046 | 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.
3 men and 10 boys finish a job in 9 days, 7 men and 7 boys finish it in 8 days. 9 men and 9 boys shall finish it in how many days? | [
"18",
"09",
"07",
"19"
] | B | 3 M + 10 B ----- 09 days
7 M + 7 B ------- 8 days
9 M + 9 B -------?
21 M + 70 B = 63 M +63 B
7 B= 42 M => 6 M = 1 B =>1 B =6 M
3 M + 60 M = 63 M ---- 9 days
9 M + 54 M = 63 M -----? => 9 days
Answer: B |
AQUA-RAT | AQUA-RAT-34047 | 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 can row his boat with the stream at 16 km/h and against the stream in 4 km/h. The man's rate is? | [
"1 kmph",
"3 kmph",
"6 kmph",
"7 kmph"
] | C | DS = 16
US = 4
S = ?
S = (16 - 4)/2 = 6 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-34048 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Sam's dad is 4 times older than his son. In 22 years from now, however, he'll only be twice his son's age. How old will Sam be 22 years from now? | [
"22 years",
"33 years",
"44 years",
"55 years"
] | B | At the present,
Sam's age is x.
Sam's father's age is y= 4x.
In 22 years from now,
Sam's age will be x+22.
Sam's father's age will be y+22=2 (x+22).
If we take both equations,
y=4x, and
y+22=2 (x+22),
then
y+22=4x+22=2 (x+22)
4x+22=2x+44
4x-2x=44-22
2x=22
x=11
This is, however, Sam's current age. Since the question is how old he'll be in 22 years from now, the answer 11+22=33.
ANSWER: B |
AQUA-RAT | AQUA-RAT-34049 | vertex, (-3,-4). (b) Find the exact maximum or minimum value of f, and compare it with your answer to part (a). These are all quadratic equations in disguise:. Because the object is simply falling from the maximum height, Vo is 0 and so it's not a quadratic equation. Meaning of quadratic function. The vertex of a parabola is an extreme point of a quadratic function and in general, it is known as maximum or minimum of a parabola. The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a parabola. This calculus tutorial will demonstrate how linearization works, and the way to apply it into an issue. Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. Rather, it is in the shape of a flattened (or weighted) catenary, which is the shape we see if we hang a chain that is thin in the middle between two fixed points. The discriminant is -7. Section 4-4 : Finding Absolute Extrema. I have included following : Vertex of the parabola Axis of symmetry Graph opens Focus Directrix Latus rectum equation Latus rectum length maximum or minimum value at vertex Domain Range X intercepts y intercept Added two more functions. Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: − b ± √ b 2 − 4 a c. Using the first term of the quadratic formula the general form of a parabola is: y = ax² + bx + x. The parabola equation is y=9 -x^2. Based on the quadratic function shown in the table below, which of the following is the range of this function? (2) (4) y £11 For Problems 3 — 5, use tables on your calculator to help you investigate these functions. If you like the worksheet you can print it straight from your browser. The maximum value is 1 since that is the hightes point, or the greatest y-value, of the graph X Advertisement. Therefore, a quadratic function may have one, two, or zero roots. Best mix for a given depth. For a quadratic Bezier, this is actually quite simple. This Parabola
The following is multiple choice question (with options) to answer.
A(4, w^2) is the (x, y) coordinate of point located on the parabola Y = X^2 + 9. What is the value of w? | [
"3.",
"4.",
"5.",
"6."
] | C | y=x^2 +9
w^2=4^2+9
w^2=25
W=5
Answer C |
AQUA-RAT | AQUA-RAT-34050 | ### 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.
Arun, Kamal and Vinay invested Rs.8000, Rs.4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months, there was a gain of Rs. 4005, then what will be the share of Kamal? | [
"10000 and 15000",
"15000 and 10000",
"5000 and 20000",
"20000 and 5000"
] | A | As both A and B invest the same amounts, the ratio of their profits at the end of the year is equal to the ratio of the time periods for which they have invested.
Thus, the required ratio of their profits = A : B = 8 : 12 = 2 : 3.
Hence, share of A in the total profit = 2/5x25000 = Rs.10000
Similarly, share of B in the total profit =3/5x25000 = Rs.15000 answer :A |
AQUA-RAT | AQUA-RAT-34051 | The remaining case is that there are two $7's$ and the $a_1\lt 7$. From the range, $a_7=a_1+15$. Thus, from the mean, $a_1$, $a_7$ and the other undetermined number (denoted by $b$) add up to $70-2\times 7-9-10=37$. Thus, $b+2a_1+15=37$ or $b+2a_1=22$. Note $b\leq a_7=a_1+15$. Thus, $22=b+2a_1\leq3a_1+15$ or $a_1>2$. Noting that $a_1\lt 7$, the possible combinations of $(a_1,b)$ are $(3,16)$, $(4,14)$, $(5,12)$, and $(6,10)$. The last combination is disallowed as $7$ is the mode. Thus, the possible sequences are: $(3,7,7,9,10,16,18)$, $(4,7,7,9,10,14,19)$, and $(5,7,7,9,10,12,20)$.
### Solution 5
This is by N. N. Taleb:
Let $X=\left\{x_{(1)},x_{(2)},x_{(3)},x_{(4)},x_{(5)},x_{(6)},x_{(7)}\right\}$ ordered such that $x_{(1)} \leq x_{(2)} \ldots \leq x_{(7)}$.
We have $x_{(7)}=x_{(1)}+15$, $x_{(4)}=9$, and we need to solve for:
The following is multiple choice question (with options) to answer.
If a(a + 6) = 7 and b(b + 6) = 7, where a ≠ b, then a + b = | [
" −48",
" −6",
" 2",
" 46"
] | B | a(a + 6) = 7
=> we have a = 1 or -7
also b(b + 6) = 7
=> b = 1 or -7
given a ≠ b
1) when a =1, b= -7 and a + b = -6
1) when a = -7, b= 1 and a + b = -6
Answer choice B |
AQUA-RAT | AQUA-RAT-34052 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
How many positive even integers less than 100 contain digits 4 or 7? | [
"16",
"17",
"18",
"19"
] | C | One digit numbers: 4
Two digit numbers:
4 at tens place: 40,42,44,46,48
7 at tens place: 70,72,74,76,78
4 at units place: 14,24,34,54,64,84,94 (duplicates removed)
If 7 is at units place, the number cant be even
Total:1+5+5+7= 18
Answer C |
AQUA-RAT | AQUA-RAT-34053 | • +1 though irritated that you just scooped me! Apr 20, 2016 at 17:16
• Thanks for the answer. I have made an appropriate system of equations and solved it but my results differ from those from the answers in my textbook. I got $3x^3-2x^2+3x+1$ (confirmed by WolframAlpha: m.wolframalpha.com/input/?i=%28x^2-x-3%29%28ax%2Bb%29%2B13x-2%3D%28x^2-2x%2B5%29%28cx%2Bd%29-1-7x&x=0&y=0 ) but according to the textbook the answer is $3x^3-5x^2+6x+4$ could you please verify if I have done this correctly? Apr 20, 2016 at 18:03
• It's easy enough to check. The remainders on dividing $3x^3 - 2x^2 + 3 x + 1$ by $x^2-x-3$ and $x^2 -2x+5$ are $13x+4$ and $-4x-19$ respectively. So you must have done something wrong. The textbook's answers are correct. Apr 20, 2016 at 19:46
The following is multiple choice question (with options) to answer.
If m=9^(x−1), then in terms of m, 3^(4x−2) must be which of the following? | [
"m/3",
"9m",
"9m^2",
"m^2/3"
] | C | 3(4x−2)=3^2(2x−1)=9(2x−1)=9∗9(2x−2)=9∗(9(x−1))2=9m
Answer: C. |
AQUA-RAT | AQUA-RAT-34054 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Dan's age after 18 years will be 5 times his age 6 years ago. What is the present age of Dan? | [
"8",
"10",
"12",
"14"
] | C | Let Dan's present age be x.
x+18 = 5(x-6)
4x = 48
x = 12
The answer is C. |
AQUA-RAT | AQUA-RAT-34055 | Remark. The lemma shows that the least common multiple is not just "least" in terms of size. It's also "least" in the sense that it divides every other common multiple.
Theorem. Let m and n be positive integers. Then
Proof. I'll prove that each side is greater than or equal to the other side.
Note that and are integers. Thus,
This shows that is a multiple of m and a multiple of n. Therefore, it's a common multiple of m and n, so it must be greater than or equal to the least common multiple. Hence,
Next, is a multiple of n, so for some s. Then
(Why is an integer? Well, is a common multiple of m and n, so by the previous lemma .)
Similarly, is a multiple of m, so for some t. Then
In other words, is a common divisor of m and n. Therefore, it must be less than the greatest common divisor:
The two inequalities I've proved show that .
Example. Verify that if and .
, , and
Proposition. The element has order in .
Proof.
The first component is 0, since it's divisible by m; the second component is 0, since it's divisible by n. Hence, .
Next, I must show that is the smallest positive multiple of which equals the identity. Suppose , so . Consider the first components. in means that ; likewise, the second components show that . Since k is a common multiple of m and n, it must be greater than or equal to the least common multiple : that is, . This proves that is the order of .
Example. Find the order of in . Find the order of .
The element has order .
On the other hand, the element has order . Since has order 30, the group is cyclic; in fact, .
Remark. More generally, consider , and suppose has order in . (The 's need not be cyclic.) Then has order .
Corollary. is cyclic of order if and only if .
Note: In the next proof, " " may mean either the ordered pair or the greatest common divisor of a and b. You'll have to read carefully and determine the meaning from the context.
Proof. If , then . Thus, the order of is . But has order , so generates the group. Hence, is cyclic.
The following is multiple choice question (with options) to answer.
Which of the following CANNOT be the least common multiple of two positive integers x and y | [
"xy",
"x",
"y",
"y - x^2"
] | D | The least common multiple of two positive integers cannot be less than either of them. Therefore, since y - x^2 is less than x, it cannot be the LCM of a x and y.
Answer: D |
AQUA-RAT | AQUA-RAT-34056 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A car after covering ½ of a journey of 100 km develops engine trouble and later travels at ½ of its original speed. As a result, it arrives 2 hours late than its normal time. What is the normal speed of the car is? | [
"287",
"27",
"28",
"25"
] | D | [50/x + 50/(x/2)] – 100/x = 2
x = 25
Answer: D |
AQUA-RAT | AQUA-RAT-34057 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Albert is 2 times Mary’s age and 4 times as old as Betty. Mary is 14 years younger than Albert. How old is Betty? | [
"5",
"7",
"10",
"15"
] | B | A = 2M = M+14
M = 14
A = 28
A = 4B, and so B = 7
The answer is B. |
AQUA-RAT | AQUA-RAT-34058 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many words can be formed by using all letters of the word “PACKHORSE” | [
"40320",
"44000",
"42340",
"56000"
] | A | The word PACKHORSE contains 8 different letters.
Required number of words = 8p8 = 8! = (8x7x6x5x4x3x2x1) = 40320.
Answer A. |
AQUA-RAT | AQUA-RAT-34059 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
Manager
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is | [
"165 runs",
"170 runs",
"172 runs",
"174 runs"
] | D | Solution
Let the highest score be x. Then, lowest score =(x-172)
= [(50 x 40)-(x+(x-172)]
= 38 x 48
2x= 2000+172-1824
2x=348
x= 174.
Answer D |
AQUA-RAT | AQUA-RAT-34060 | python-3.x, programming-challenge, integer
Title: Calculate the cost of Biryani lessons Problem:
According to a recent survey, Biryani is the most ordered food. Chef
wants to learn how to make world-class Biryani from a MasterChef. The
chef will be required to attend the MasterChef's classes for X weeks
and the cost of classes per week is Y coins. What is the total amount
of money that the Chef will have to pay?
Input Format:
The first line of input will contain an integer T — the number of test
cases.
The first and only line of each test case contains two space-separated
integers.
for _ in range(int(input())):
x,y = map(int, input().split())
print(x*y)
I tried to solve this simple calculation problem with fewer declared variables. Is there any way to optimize the above code? Your code is concise, readable, and idiomatic. It works as you intended. It's small but nice snippet of code.
You could probably turn it into a one-liner, but it would get rather complex and readability would suffer, implying issues with maintainability, re-usability and reviewing.
Regarding style, I have two minor remarks:
The indentation is unusual, PEP 8 and recommends 4 spaces for indents, not 8. But this might as well be an issue with question formatting.
The variable names you chose, x and y, are undescriptive. I would prefer weeks and cost_per_week, although your approach of using the names in the problem description is understandable.
The following is multiple choice question (with options) to answer.
Alok ordered 16 chapatis, 5 plates of rice, 7 plates of mixed vegetable and 6 ice-cream cups. The cost of each chapati is Rs.6, that of each plate of rice is Rs.45 and that of mixed vegetable is Rs.70. The amount that Alok paid the cashier was Rs.1015. Find the cost of each ice-cream cup? | [
"34",
"76",
"29",
"12"
] | A | Explanation:
Let the cost of each ice-cream cup be Rs.x
16(6) + 5(45) + 7(70) + 6(x) = 1015
96 + 225 + 490 + 6x = 1015
6x = 204 => x = 34.
Answer:A |
AQUA-RAT | AQUA-RAT-34061 | homework-and-exercises, fluid-statics, bernoulli-equation
You ignore the area of the hole.
Rate of water going in: $r'= \frac{0,2cm^3}{s}$
Rate of water going out: $r=1mm^2*\sqrt{2gh}$
Set the two rates equal and solve for $h$
$\frac{0,2cm^3}{s} = 1mm^2*\sqrt{2gh}$
$\frac{0,2cm^3}{s*0,01cm^2} = \sqrt{2gh}$
$\frac{20cm}{s} = \sqrt{2gh}$
The units work out correctly
I will leave the rest to you
Hint square both sides
Solving for the $t$ in $h(t)$ would be a more complex
The following is multiple choice question (with options) to answer.
A river 2m deep and 45 m wide is flowing at the rate of 4 kmph the amount of water that runs into the sea per minute is? | [
"6000 M3",
"4580 M3",
"18500 M3",
"4900 M3"
] | A | Explanation:
(4000 * 2 * 45)/60 = 6000 M3
Answer: Option A |
AQUA-RAT | AQUA-RAT-34062 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 6:30 am. They shall first cross each other at ? | [
"7:15 am",
"7:30 am",
"6: 42 am",
"7:50 am"
] | C | Explanation:
Relative speed between two = 6-1 = 5 round per hour
They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.
So 6:30 + 12 mins = 6:42
Option C |
AQUA-RAT | AQUA-RAT-34063 | 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.
At what rate percent on simple interest will Rs.750 amount to Rs.950 in 5 years? | [
"3.33%",
"5.93%",
"4.33%",
"5.33%"
] | D | 200 = (750*5*R)/100
R = 5.33%
ANSWER:D |
AQUA-RAT | AQUA-RAT-34064 | 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.
In a partnership between A, B and C. A's capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C's share is Rs.130, what is B's capital? | [
"Rs.3250",
"Rs.6250",
"Rs.10250",
"Rs.11750"
] | A | 200 + 130 = 330
800 - 330 = 470
200 ---- 5000
470 ---- ? => 11750
ANSWER:A |
AQUA-RAT | AQUA-RAT-34065 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank in gallons is? | [
"100",
"110",
"120",
"140"
] | C | Let the total time taken by both the pipes be 120 minutes to fill the tank.
Therefore, the number of gallons filled by both the tanks in one minute = (6 + 5) = 11 gallons.
At the same time a waste pipe is also opened which can empty 3 gallons per minute.
Hence, the number of gallons filled in one minute = (11 - 3) = 8 gallons.
Therefore, the number of gallons filled when all the three pipes are opened for 15 minutes = 8 x 15 = 120 gallons.
ANSWER:C |
AQUA-RAT | AQUA-RAT-34066 | 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?
159
248
951
357
122
582 | [
"951",
"246",
"325",
"147"
] | A | A
951
In the rest there is the same difference between each digit, eg: 9(–4) 5 (–4)1 |
AQUA-RAT | AQUA-RAT-34067 | ### Show Tags
07 Nov 2012, 05:44
breakit wrote:
Bunuel wrote:
breakit wrote:
Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)
Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622
Hope it helps.
Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)
[color=#ff0000] $$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> $$(10-w)(9-w)<9$$ --> $$w>6$$,
$$\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}$$ --> cross-multiply (multiply both parts by 10*9): $$(10-w)(10-w-1)<9$$ --> $$(10-w)(9-w)<9$$.
Hope it's clear.
_________________
Intern
Joined: 09 Sep 2012
Posts: 23
Schools: LBS '14, IMD '16
Re: If 2 different representatives are to be selected at random [#permalink]
### Show Tags
13 Mar 2013, 16:44
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.
This is extreme value problem
for p > 1/2 , p1 * p2 > 1/2 ie p1 or p2 > 1/4
The following is multiple choice question (with options) to answer.
150 college students were asked in a survey if they preferred Windows or Mac brand computers. 40 Students claimed that they preferred Mac to Windows brand computers. One third as many of the students who preferred Mac to Windows, equally preferred both brands. 40 of the students had no preference. How many of the students in the survey preferred Windows to Mac brand computers? | [
"25",
"40",
"70",
"60"
] | C | 150 = 40(Mac) + x(Window) + 40(both) => x=70
ANSWER:C |
AQUA-RAT | AQUA-RAT-34068 | # Is it linear or exponential growth
Arithmetic and Geometric progressions.
### Is it linear or exponential growth
The question is :
In 2018, the population in China was 1,427,647,786.
According to data, China’s population growth rate is 0.59% per year. What will the population be in China in 2021?
A) 8,423,121
B) 1,436,070,908
C) 1,444,543,726
D) 1,452,917,152
E) 1,453,142,111
The book answer is (E). But I think it is linear growth because it is only 3 years.
My solution is : The population in 2021 = 1427647786 + 142647786(.0059)3=1452917152 . which is in choice (D) not (E).
-------------
Please , I need help. What is the mistake in my solution.
Thank you very much.
Guest
### Re: Is it linear or exponential growth
"But I think it is linear growth because it is only three years."
I have no idea why you think whether growth is linear or exponential depends on how long a time you are dealing with!
"China's population growth rate is 0.59%"
Since the growth rate is a constant, the growth must be exponential!
The population in 2018 was 1,427,647,786.
Since the population increased by 0.57% the next year, 2019, the population is
(1.0057)(1,427,647,786)= 1,435,785,378
The next year, 2020, the population is
(1.0057)(1435785378)= 1,443,969,355
The next year, 2021, the population is
(1.0057)(1,443,969,355)= 1,452,199,980
That's closer to D than to E but it certainly is NOT linear!
Guest
### Re: Is it linear or exponential growth
Since I cannot edit on this board I have to repeat that post but with the correct 0.59% rather than 0.57%.
The following is multiple choice question (with options) to answer.
In one year, the population, of a village increased by 20% and in the next year, it decreased by 20%. If at the end of 2nd year, the population was 9600, what was it in the beginning? | [
"10000",
"8000",
"1988",
"1277"
] | A | x*120/100 * 80/100 = 9600
X*0.96=9600
X=9600/0.96
=> 10000
Answer:A |
AQUA-RAT | AQUA-RAT-34069 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
The price of a T.V. set worth Rs. 8000 is to be paid in 20 installments of Rs. 750 each. If the rate of interest be 6% per annum, and the first installment be paid at the time of purchase, then the value of the last installment covering the interest as well will be? | [
"7250",
"7200",
"7820",
"6000"
] | A | Money paid in cash = Rs. 750
Balance payment = (8000- 1000) = Rs. 7250
Answer:A |
AQUA-RAT | AQUA-RAT-34070 | ### 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.
In a mixture of 60 liters, the ratio of milk and water is 2:1. What amount of water must be added to make the ratio of milk and water as 1:2? | [
"20 liters",
"30 liters",
"60 liters",
"29 liters"
] | C | 2:1 --- 60
1:2
2:4
---------------
3
3 ----------- 60
3 ----------- ? => 60 liters
Answer:C |
AQUA-RAT | AQUA-RAT-34071 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man swims downstream 72 km and upstream 45 km taking 6 hours each time; what is the speed of the current? | [
"2.28",
"2.24",
"2.27",
"2.25"
] | D | 72 --- 6 DS = 12
? ---- 1
45 ---- 6 US = 7.5
? ---- 1 S = ?
S = (12 - 7.5)/2 = 2.25
Answer:D |
AQUA-RAT | AQUA-RAT-34072 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A book store bought copies of a new book by a popular author, in anticipation of robust sales. The store bought 400 copies from their supplier, each copy at wholesale price z. The store sold the first 150 copies in the first week at 80% more than z, and then over the next month, sold a 100 more at 20% more than z. Finally, to clear shelf space, the store sold the remaining copies to a bargain retailer at 40% less than z. What was the bookstore’s net percent profit or loss on the entire lot of 400 books? | [
" 30% loss",
" 10% loss",
" 10% profit",
" 20% profit"
] | D | [[z(150)(1.8)+z(100)(1.2)+z(150)(0.6)]/400z] - 1
[50[(3)(1.8) + (2)(1.2) + (3)(0.6)] / (50)(8)] - 1
[(5.4 + 2.4 + 1.8)/8] - [8/8]
+1.6/8
+20%
answer is D |
AQUA-RAT | AQUA-RAT-34073 | Alternatively, the lcm of 54 and 72 can be found using the prime factorization of 54 and 72:
• The prime factorization of 54 is: 2 x 3 x 3 x 3
• The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(54,54) = 216
In any case, the easiest way to compute the lcm of two numbers like 54 and 72 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 54,72. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 54 and 72
What is the least common multiple of 54 and 72 used for? Answer: It is helpful for adding and subtracting fractions like 1/54 and 1/72. Just multiply the dividends and divisors by 4 and 3, respectively, such that the divisors have the value of 216, the lcm of 54 and 72.
$\frac{1}{54} + \frac{1}{72} = \frac{4}{216} + \frac{3}{216} = \frac{7}{216}$. $\hspace{30px}\frac{1}{54} – \frac{1}{72} = \frac{4}{216} – \frac{3}{216} = \frac{1}{216}$.
## Properties of LCM of 54 and 72
The most important properties of the lcm(54,72) are:
• Commutative property: lcm(54,72) = lcm(72,54)
• Associative property: lcm(54,72,n) = lcm(lcm(72,54),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The following is multiple choice question (with options) to answer.
54 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780. The bigger part is : | [
"33",
"34",
"26",
"28"
] | B | Explanation:
Let the two parts be (54 - x) and x.
Then, 10 (54 - x) + 22x = 780
=> 12x = 240
=> x = 20.
Bigger part = (54 - x) = 34.
Answer: B) 34 |
AQUA-RAT | AQUA-RAT-34074 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in still water is 40kmph and the speed of the current is 10kmph. Find the speed downstream and upstream? | [
"50,30 kmph",
"40,20 kmph",
"29,25 kmph",
"26,20 kmph"
] | A | Speed downstream = 40 + 10
= 50 kmph
Speed upstream = 40 - 10
= 30 kmph
Answer: A |
AQUA-RAT | AQUA-RAT-34075 | 0.104768, 0.0591745, 0.0475319, 0.0452583}, {0.0510719, 0.0599374,
0.0730602, 0.0975814, 0.101289, 0.0691997, 0.0498054, 0.044892,
0.043122}, {0.0460517, 0.0567025, 0.0574044, 0.0587778, 0.0537118,
0.0487221, 0.0474098, 0.0413977, 0.04477}}
The following is multiple choice question (with options) to answer.
0.04 x 0.0162 is equal to: | [
"6.48 x 10(power -4)",
"6.84 x 10(power -4)",
"4.68 x 10(power -4)",
"5.48 x 10(power -4)"
] | A | 4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10(power -4)
Answer is A. |
AQUA-RAT | AQUA-RAT-34076 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
The average salary/head of all theworkers in a workshop is Rs.850, if the average salary/head of 7technician is Rs.1000 and the average salary/head of the rest is Rs.780, the total no. of workersin the work-shop is ? | [
"18",
"19",
"22",
"26"
] | C | Let the total number of workers be y.
So sum of salary for all workers = sum of salary of 7 technician + sum of salary for other y -7 workers.
7 x 1000 + 780(y -7) = 850 y
⇒ 7000 + 780y - 5460 = 850y
⇒ 70y = 1540
∴ y = 22
So total number of workers = 22
C |
AQUA-RAT | AQUA-RAT-34077 | Time needed for both machine to produce 1500 units is given.
1/8 + 1/12
= 5/24
I hrs work done by both machine.
24/5 = 4.8.
Time needed to complete the entire work.
******* 1500 units must be considered as a single work. units are same for both machine. Therefore we don't need to do anything with units.
C is the correct answer.
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink]
### Show Tags
12 Aug 2019, 10:53
Bunuel wrote:
A new copy machine can run off 1,500 workbooks in 8 hours, while it takes on older copy machine 12 hours to do the same job. What is the total number of hours that it would take both copy machines working at the same time, but independently, to run off the 1,500 workbooks?
(A) 4.4
(B) 4.6
(C) 4.8
(D) 5
(E) 10
Let T be the number of hours to complete the job when both machines are running. We can create the following equation:
1500/8 * T + 1500/12 * T = 1500
T/8 + T/12 = 1
Let’s multiply each side by 24:
3T + 2T = 24
5T = 24
T = 4.8 hours
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink] 12 Aug 2019, 10:53
# A new copy machine can run off 1,500 workbooks in 8 hours, while it ta
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
The following is multiple choice question (with options) to answer.
Two identical machines have the ability to produce both nuts and bolts. However, it takes 1 second to produce a bolt but 2 seconds to produce a nut. What is the fastest the two machines, working together, can produce 800 nuts and 800 bolts? | [
"1,200 seconds",
"1,500 seconds",
"1,750 seconds",
"2,000 seconds"
] | A | To minimize the amount of time, we should make both machines work all the time while producing 800 nuts and 800 bolts.
Two machines to produce 800 bolts will need (800*1)/2 = 400 seconds.
Two machines to produce 800 nuts will need (800*2)/2 = 8000 seconds.
Total = 400 + 800 = 1,200.
Answer: A. |
AQUA-RAT | AQUA-RAT-34078 | ## 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.
There are 7 orators A, B, C, D, E, F and G. In how many ways can the arrangements be made so that A always comes before B and B always comes before C. | [
"7! / 3!",
"8! / 6!",
"5! x 3!",
"8! / (5! x 3!)"
] | A | Select any three places for A, B and C. They need no arrangement amongst themselves as A would always come before B and B would always come before C.
The remaining 4 people have to be arranged in 4 places.
Thus, 7C3 x 4! = 7!x4!/(4!x3!)=7!/3!
ANSWER:A |
AQUA-RAT | AQUA-RAT-34079 | Actually.. while writing this I tested to see what the difference is for rational cases like $$\sqrt{4}$$ and $$\sqrt{9}$$. Yes, now I understand it. It wasn't the evenness of $$\sqrt{2}$$'s example that was special, it was the fact that it was divisible by 2 (the same thing I know ).
4. Mar 13, 2010
### CompuChip
Yes, in this case it is the same.
But in the general case $\sqrt{n}$, you want to say something about the divisibility by $n$.
(Note that in passing you have also shown that the square of an even number is always divisible by not just 2, but by 4).
5. Mar 13, 2010
### icystrike
If you add an additional information that gcd(p,q)=1 , which also means the greatest common divisor of p and q is 1 . However for $$\sqrt{2}$$ you can draw the conclusion that the gcd(p,q) is atleast 2 and for $$\sqrt{3}$$ gcd(p,q) is atleast 3. In fact you can prove a generalised one analogously which states the $$\sqrt{prime}$$ will lead you to gcd(p,q) is atleast the prime number , yet definition of prime states it cannot be 1 , will leads to the irrationality of all prime.
The following is multiple choice question (with options) to answer.
If q=4p where p is a prime number greater than 2, how many different positive even divisors does q have, including q? | [
"two",
"three",
"four",
"six"
] | C | q=2, 2, p (p is in the prime factorization since were told it is a prime number in the question)
2
4 (because of 2*2)
2p (because of 2 * p)
4p (because of 2 * 2 * p)=C |
AQUA-RAT | AQUA-RAT-34080 | 4x 8x 8x 8x 8x 4x 8x 8x 4x
?????? ?????? ?????? ?????? ?????? ?????? ?????? ?????? ??????
?----- ?----- ?----- ?----- ?----- ?----- ?----- ?---?? ?---??
?-K-K- ?-K-K- ?-K-K- ?-K--K ?-K--K ?-K--K ?-K--K ?-K--- ?-K---
?----- ?----- ?----- ?----- ?----- ?----- ?----- ?---K- ?----K
?-K-?? ??-K-? ?---?? ?-K-?? ??-K-? ?---?? ??---? ?----- ??----
?---?? ??---? ?-K-?? ?---?? ??---? ?-K-?? ??-K-? ?-K-?? ??-K-?
Observe that in each case the bottom-right square is now a choice independent of the remaining unplaced kings, and the spaces available for the remaining unplaced kings fit a mere 3 patterns:
96x 64x 4x
?????? ?????? ??????
? ? ? ??
? ? ?
? ? ?
? ?? ??
? ?? ??
Calculate the number of placements for each of these patterns (which is a straight combinatorics-on-words problem) and you just need to apply the appropriate multiplicities.
The following is multiple choice question (with options) to answer.
1398 x 1398 | [
"1954404",
"1981709",
"18362619",
"2031719"
] | A | 1398 x 1398 = (1398)2
= (1400 - 2)2
= (1400)2 + (2)2 - (2 x 1400 x 2)
= 1954404.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34081 | beginner, reinventing-the-wheel, rust
So format_sign would print a + or - and with or without the space and then format_constant will just print the number and format_coefficient can return "" if the number is exactly one.
Sorry, I don't have a rust compiler at hand.
The following is multiple choice question (with options) to answer.
If * stands for +, / stands for * and + stands for /, then which of the following equations is TRUE? | [
"7*14/7+35-14=-4.2",
"5+7*7-5/14=24",
"14*5/7+35-14=-10.1",
"5/7*7+35-14=4"
] | A | 7*14/7+35-14 will become
7+14*7/35 - 14 = -7 + 14/5 = -21/5 = -4.2
ANSWER:A |
AQUA-RAT | AQUA-RAT-34082 | "Thursdays with Ron - Consolidated Verbal Master List - Updated"
Math Expert
Joined: 02 Sep 2009
Posts: 39609
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Dec 2016, 23:09
Nightfury14 wrote:
Bunuel wrote:
gdk800 wrote:
1) A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)
a. 48
b. 100
c. 120
d. 288
e. 600
[Reveal] Spoiler:
OA is B.
Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$;
So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.
Hi Bunuel
The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:
Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp
Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp
Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.
The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.
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Joined: 18 May 2016
Posts: 20
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Jan 2017, 17:21
Hello,
The following is multiple choice question (with options) to answer.
A firm has 3 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner? (2 groups are considered different if at least one group member is different) | [
"64",
"72",
"84",
"96"
] | A | The total number of ways to form a group of 3 is 9C3 = 84
The number of groups without a senior partner is 6C3 = 20
The number of groups which include a senior partner is 84-20 = 64
The answer is A. |
AQUA-RAT | AQUA-RAT-34083 | # Math Help - Distance (Word Problem)
1. ## Distance (Word Problem)
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
2. Originally Posted by Beastkun
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
1. Let d denote the covered distance and t the elapsed time. Then you know:
$\dfrac dt = 50$
and
$\dfrac{d}{t-3}=60$
2. Solve for d.
Spoiler:
You should come out with d = 900
3. I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
4. Originally Posted by Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
Substitute d = 50t from the first equation into the second equation and solve for t and hence d.
5. What prove it is doing is simultaneous equations.
The following is multiple choice question (with options) to answer.
Peter and Rhea lives some distance apart from each other, separated by a straight road. They both take out their cars and start driving at the same time with the same speed towards each other's home.
After some time, their cars meet at 500 miles away from Peter's house. They keep driving and reach at each other's home. Without stopping by, they turn back and start driving again. This time, they meet 300 miles away from each other's home.
What is the distance between their houses? | [
"1200 Miles.",
"110 Miles.",
"1300 Miles.",
"1400 Miles."
] | A | Solution:
Let the distance between Peter's (hereafter referred as P) and Rhea's (hereafter referred as R) be D.
When P and R meet for the first time, P has traveled 500 miles.
In that case, R has traveled (D - 500) miles.
=> Speed of P / Speed of R = 500 / (D - 500) ...... (1)
When P and R meet for the second time, P has traveled (D - 500) + 300 miles
Now, R has traveled (500 + D - 300)
=> Speed of P / Speed of R = 5 / (D - 5) ..... (2)
Equating (1) and (2)
500 / (D - 500) = (D - 2) / (D + 2)
D = 1200 Miles.
Answer A |
AQUA-RAT | AQUA-RAT-34084 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
A type of extra-large SUV averages 12.2 miles per gallon (mpg) on the highway, but only 7.6 mpg in the city. What is the maximum distance, in miles, that this SUV could be driven on 20 gallons of gasoline? | [
"190",
"284.6",
"300",
"244"
] | D | so 12.2 * 20 = 244..
IMO option D is correct answer.. |
AQUA-RAT | AQUA-RAT-34085 | WAIT!
S=11!
Because we have 11! ways of arranging the 11 items.
Sorry about that.
5. Hello, sweeetcaroline!
Edit: Plato is absolutely right . . . *blush*
6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order.
What is the probability that no two neighboring beads are the same color?
There are: . ${11\choose6,4,1} \:=\:2310$ possible orders.
Place the 6 Red beads in a row, leaving a space between them.
. . $\begin{array}{ccccccccccccc} R & \_ & R & \_ & R & \_ & R & \_& R & \_ & R\end{array}$
Place the Blue bead is any of the 5 spaces: . $5$ choices.
Drop the 4 White beads in the remaining 4 spaces: . $1$ way.
Hence, there are: . $5\cdot1 \,=\,5$ ways.
Therefore, the probability is: . $\frac{5}{2310} \;\;=\;\;\frac{1}{462}$
6. Originally Posted by sweeetcaroline
the problem: 6 red beads, 4 white beads, and 1 blue bead are placed in a line in random order. What is the probability that no two neighboring beads are the same color?
Can someone give me a line of these beads in which no two neighboring beads are the same color other than the five listed below?
$RWRWRWRWRBR$
$RWRWRWRBRWR$
$RWRWRWBWRWR$
$RWRBRWRWRWR$
$RBRWRWRWRWR$
The following is multiple choice question (with options) to answer.
R-R-R-G-G-G-Y-Y-B-R-R-R-G-G-G-Y-Y-B… B-R-R
The preceding is a representation of the different colored beads on a string. The beads follow a repeating pattern and the colors Red, Green, Yellow, and Blue are represented by R, G, Y, and B respectively. Which of the following is a possible number of beads in the missing section of the string represented above? | [
"62",
"63",
"64",
"65"
] | A | The repeated pattern R-R-R-G-G-G-Y-Y-B has 9 beads.
But the missing section includes R-R-R-G-G-G-Y-Y at the end.
Thus the number of beads in the missing section has the form 9k + 8.
The answer is A. |
AQUA-RAT | AQUA-RAT-34086 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
If 4 pounds of dried apricots that cost x dollars per pound are mixed with 2 pounds of prunes that cost y dollars per pound, what is the cost, in dollars, per pound of the mixture? | [
"(4x + 2y)/5",
"(4x + 2y)/(x + y)",
"(4x + 2y)/(xy)",
"5(4x + 2y)"
] | A | Total Cost = Weight(in pounds) * Price/pound;
To find Total cost/pound, Divide by total pounds.
Cost of Dried apricots = 4x;
Cost of prunes = 2y;
Cost per pound = (4x + 2y)/5;
Ans is (A). |
AQUA-RAT | AQUA-RAT-34087 | java, algorithm, knapsack-problem
Title: Stampcalculator - Given a set of stamps, what combinations are there to reach a certain amount? Background
My mother has a hobby of buying and reselling books via online trading sites. After a price is agreed on, the books have to be put into an envelope for mailing. On this envelope, stamps have to be applied to pay for postage (mailing fee). My mother buys stamps in bulk at various prices, varying from 75 to 90% of face value. As the postage prices change every year, new stamp combinations have to be calculated that will satisfy the postage fee with a minimum amount of hassle.
My mother is not all that good with calculations. And having to manually recalculate what stamp combinations are best to get to the postage required for a 250 - 500 gram package takes a lot of time. Therefore, she calculates a few options via trial and error - sometimes overshooting the postage amount by a few cents - and just uses those.
Some stamps are preferred to be used, whereas others aren't, due to the cost not always being the face value of a stamp. Asking my mother about the stamp prices individually would have taken too much time, so I just made a program that generates a list of options. It requires recompilation when you need to add different stamps or want to get to a different limit, but that's okay. The whole point of this was to not spend too much time on it (the goal is to save time!) - which is why I wrote it in the total time of 1 hour (from idea to solution).
Problem description
Given a target amount, a set of stamp denominations, and a maximum amount of usable stamps, print a list of stamp combinations that will get to the target amount.
Programmed in 1 hour. I have placed everything in one file to ensure that it can run in Ideone, so that she could theoretically make the changes herself, if needed.
How it works
Using a list of stamps, make partial combinations that are at or below the target amount. For each combination, make a new combinations via duplicating the current combination and adding a stamp with a value equal to or below the last added stamp. Combinations that go past the goal are discarded. Combinations that meet the goal are added to a solutions list.
The following is multiple choice question (with options) to answer.
Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought ? | [
"10",
"9",
"12",
"8"
] | A | Explanation :
At least two stamps of each type were ordered initially.
So, Rs.2 ( 5 + 2 + 1 ) = Rs.16 have been spent.
That leaves ( 20 – 16 ) = Rs.4.
In this Rs.4, three more stamps of one rupee were given, thus accounting for Rs.19 in all.
Since, one more rupee remains, it means that one more stamps of Rs.1 were bought initially.
So the total number of stamps is 2(of Rs.5) + 2(of Rs.2) + 6(of Rs.1). i.e 10.
Answer : A |
AQUA-RAT | AQUA-RAT-34088 | # Probability of getting at least one head given there at least two heads. (solution verification)
A coin with probability, $p$, of coming up heads is flipped three times (the flips are independent). What is the probability of getting at least one head given there are at least two heads?
My attempt at a solution: First I realize that there are eight possible outcomes: TTT, HHH, THH, HHT, HTH, TTH, THT, HTT. Now consider: \begin{align*} &P(1~\text{head} \mid \text{at least $2$ heads}) + P(2~\text{heads} \mid \text{at least $2$ heads})+ P(3~\text{heads} \mid \text{at least $2$ heads})\\ & \quad = \frac{P(1~\text{head}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})} + \frac{P(2~\text{heads}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})}\\ & \qquad + \frac{P(3~\text{heads}) \cap P(\text{at least $2$ heads})}{P(\text{at least $2$ heads})}\\ & \quad = 0 + \frac{P(2~\text{heads})}{P(\text{at least $2$ heads})} + \frac{P(3~\text{heads})}{P(\text{at least $2$ heads})}\\ & \quad = \frac{3p^2(1-p)}{3p^2(1-p)+ p^3} + \frac{p^3}{3p^2(1-p)+ p^3} \end{align*}
The following is multiple choice question (with options) to answer.
Three unbiased coins are tossed. What is the probability of getting at most two heads? | [
"3/4",
"1/4",
"7/8",
"3/8"
] | C | Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) = n(E) = 7 .
n(S) 8
Answer : C. |
AQUA-RAT | AQUA-RAT-34089 | c a b c u v c
7 5 8 7 1 2 7 = 7
13 7 15 13 1 3 13 = 13
19 16 21 19 2 3 19 = 19
31 11 35 31 1 5 31 = 31
37 33 40 37 3 4 37 = 37
43 13 48 43 1 6 43 = 43
49 39 55 49 3 5 49 = 7^2
61 56 65 61 4 5 61 = 61
67 32 77 67 2 7 67 = 67
73 17 80 73 1 8 73 = 73
79 51 91 79 3 7 79 = 79
91 19 99 91 1 9 91 = 7 * 13
91 85 96 91 5 6 91 = 7 * 13
97 57 112 97 3 8 97 = 97
103 40 117 103 2 9 103 = 103
109 95 119 109 5 7 109 = 109
127 120 133 127 6 7 127 = 127
133 23 143 133 1 11 133 = 7 * 19
133 88 153 133 4 9 133 = 7 * 19
139 69 160 139 3 10 139 = 139
151 115 171 151 5 9 151 = 151
157 25 168 157 1 12 157 = 157
163 75 187 163 3 11 163 = 163
169 161 176 169 7 8 169 = 13^2
181 104 209 181 4 11 181 = 181
193 175 207 193 7 9 193 = 193
199 56 221 199 2 13 199 = 199
211 29 224 211 1 14 211 = 211
217 208 225 217 8 9 217 = 7 * 31
217 87 247 217 3 13 217 = 7 * 31
223 168 253 223 6 11 223 = 223
229 145 264 229 5 12 229 = 229
241 31 255 241 1 15 241 = 241
247 203 275 247 7 11 247 = 13 * 19
247 93 280 247 3 14 247 = 13 * 19
259 155 299 259 5 13 259 = 7 * 37
259 64 285 259 2 15 259 = 7 * 37
271 261 280 271 9 10 271 = 271
277 217 312 277 7 12 277 = 277
283 192 325 283 6 13 283 = 283
301 136 345 301 4 15 301 = 7 * 43
301 279 319 301 9 11 301 = 7 * 43
307 35 323 307 1 17 307 = 307
313 105 352 313 3 16 313 = 313
331 320 341 331 10 11 331 = 331
337 272 377 337 8 13 337 = 337
343 37 360 343 1 18 343 = 7^3
349 111 391 349 3 17 349 = 349
361 185 416 361 5 16 361 = 19^2
The following is multiple choice question (with options) to answer.
What number has a 5:1 ratio to the number 13? | [
"22",
"50",
"88",
"65"
] | D | 5:1 = x: 13
x = 65
Answer: D |
AQUA-RAT | AQUA-RAT-34090 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most? | [
"19",
"20",
"1",
"0"
] | A | Avg of 20 nos = 0
sum of 20 numbers/20 = 0
sum of 20 nos = 0
=>19 positive nos
ANSWER A |
AQUA-RAT | AQUA-RAT-34091 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively. The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports. If the number of boys participating in the sports is 15, then how many girls are there in the class ? | [
"20",
"25",
"30",
"Data inadequate"
] | C | In a class of 60 students, the number of boys and girls participating in the annual sports is in the ratio 3 : 2 respectively.
say 3x and 2x respectively.
If the number of boys participating in the sports is 15,no of girls participating in sports = 10.
Total students participating out of 60 in class = 15+10=25
Total students not participating out of 60 in class = 60-25=35
The number of girls not participating in the sports is 5 more than the number of boys not participating in the sports.
so out of 45 not participating in sports, 20 are girls and 15 are boys.
No of girls in class = 20+10=30
ANSWER:C |
AQUA-RAT | AQUA-RAT-34092 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Lisa earns a profit of $3 on each ribbon that she sells, and $5 on each bracelet. If at the end of the day Lisa has earned $120 and she has sold at least 10 pieces of each item. At least how many bracelets she has sold? | [
"2",
"5",
"9",
"12"
] | D | Since Lisa has sold at least 10 pieces for each item, let us say Lisa sells (10 + r) number of ribbons and (10 + b) number of bracelets, her profit for the day = 3(10 + r) + 5(10 + b)
=> 3(10 + r) + 5(10 + b) = 120
=> 30 + 3r + 50 + 5b = 120
=> 3r + 5b = 120 - 30 - 50 = 40
=> 5b = 40 - 3r
=> b = 8 - (3/5)r
To minimize b, we need to maximize r and it will be a multiple of 5.
r = 5, b = 8 - (3/5)5 = 5
r = 10, b = 8 - (3/5)10 = 2
Beyond that b will become -ve. So Lisa has sold at least 10 + 2 = 12 bracelets.
Answer: D |
AQUA-RAT | AQUA-RAT-34093 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
The monthly incomes of A and B are in the ratio 5 : 2. B's monthly income is 12% more than C's monthly income. If C's monthly income is Rs. 14000, then find the annual income of A? | [
"Rs. 420000",
"Rs. 180000",
"Rs. 201600",
"Rs. 470400"
] | D | B's monthly income = 14000 * 112/100 = Rs. 15680
B's monthly income = 2 parts ----> Rs. 15680
A's monthly income = 5 parts = 5/2 * 15680 = Rs. 39200
A's annual income = Rs. 39200 * 12 = Rs. 470400
ANSWER:D |
AQUA-RAT | AQUA-RAT-34094 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
If Bill can buy 3 pairs of jeans and 2 shirts for $69 or 2 pairs of jeans and 3 shirts for $61, how much does one shirt cost? | [
"$9",
"$12",
"$13.20",
"$15"
] | A | 3J + 2S = 69
2J + 3S = 61
----------------
5J + 5S = 130 ----(divide by 5)---> J + S = 26
3J + 2S = J + 2(J + S) = J + 52 = 69 ---> J = 17
3*17 + 2S = 69
51 + 2S = 69
2S = 18
S = 9
Answer: A |
AQUA-RAT | AQUA-RAT-34095 | 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.
Every student in a room is either a junior or a senior. There is at least one junior and at least one senior in the room. If 2/3 of the juniors is equal to 1/2 of the seniors, what fraction of the students in the room are juniors? | [
"3/20",
"1/3",
"3/7",
"2/7"
] | C | Let total number of juniors= J
total number of seniors =S
(2/3) J = (1/2) S
=> S = 4/3 J
Total number of students = J+S = (7/3) J
Fraction of the students in the room are juniors = J/(J+S) = J/[(7/3) J]
=3/7
Answer C |
AQUA-RAT | AQUA-RAT-34096 | Since 2m+1 is greater than P1, x1 is positive. An even number minus an odd plus an odd equals an even, and half an even number is an integer. Hence, x1 is a solution to S.
In the second case, where 2m+1 is less than P1, let’s try N = 2m+1 as a solution. This yields: $x_1 = (\frac{2S}{2^{m+1}} – 2^{m+1} + 1)/2 \\ S = 2^m\cdot P_1 \cdot S_2 \implies \\ x_1 = (\frac{2 \cdot 2^m \cdot P_1 \cdot S_2}{2^{m+1}} – 2^{m+1} + 1)/2 \\ = (P_1 \cdot S_2 – 2^{m+1} + 1)/2$
Since P1 is greater than 2m+1, x1 is again positive. P1 and S2 are both odd, hence P1S2 is odd. An odd number minus an even plus an odd is an even number, and half of an even number is an integer. Hence, x1 is a solution to S.
This proves that S has a solution R for all values that are not powers of 2; it also illustrates why 23•16 requires 23 numbers (23 < 32) while 23•8 requires only 16 (16 < 23). As we would expect, 592 (37•16) requires 32 numbers, not 37 (3+4+...+34); 1184 (37•32) requires 37 (14+15+...+50). All of this proves the original hypothesis: A positive integer S is equal to the sum of consecutive positive integers iff it has odd prime factors.
## 1 Comment
1. paulhartzer
It occurred to me later that this provides a direct proof all odd numbers have N=2 solutions: All odd numbers are of the form 2^0•P_1•S_2. Since 2^(0+1) = 2 will always be lower than P_1, N=2^(0+1)=2 will always provide a solution.
The following is multiple choice question (with options) to answer.
If x is a positive integer, which of the following must be odd? | [
"x+1",
"x^2+x",
"x^2+x+5",
"x^2−1"
] | C | A. X+1 = can be odd or even. Since O + O =E or E + O = O
B. X^2 + X = X(X+1). Since from the above derivation we already know the term X+1 can be Odd or Even, directly substitute here. X(ODD) = Even (When X is Even) or X(EVEN) = EVEN [When X is ODD]
C. Here's the answer. Since we know the term x^2 + X can always take a EVEN number, EVEN + 5 = ODD
Hence C. |
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