source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35297 | ### Bonus Question
As for the Bonus question, it is also an interesting exercise. If we start with $$n=1600$$ fair coins and at each turn we eliminate those landing tails, we wonder, how many rounds on average we will have before running out of coins.
A first approximation would be to say that, on average, we would expect to lose half of the coins each time. After the first toss we will end up with around $$800$$, then $$400$$, and after $$6$$ round we would expect to have only around $$25$$ coins. Now we are expecting to have an odd number of coins but we could keep halving until we have close to a single coin. That is, after $$10$$ rounds, $$\dfrac{1600}{2^{10}}\approx 1.5625$$ and so that we may expect to have at least an eleventh round before running out of coins. But, is the answer then $$11$$, $$12$$, $$13$$,…?
Just like with the main puzzle, by rephrasing the problem in terms of flipping all the $$n$$ coins and just considering a different notion of success we can come up with a better solution. Let us focus on a single coin; we would keep flipping it as long as it keeps turning heads. Well, the number of times we will flip said coin can be modeled as a Geometric random variable where now the success is landing tails. We see then that we will run out of coins the last time where a given coin lands tails for the first time. This is just to say that we are interested in the maximum of a series of $$n$$ independent Geometric random variables with parameter $$p=0.5$$ or, more precisely, in its expected value.
The following is multiple choice question (with options) to answer.
A girl has exactly 6 quarters, 7 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter? | [
"1",
"2",
"3",
"4"
] | C | When the probability of NOT drawing a quarter is 50% or less, then the probability of drawing a quarter is at least 50%.
P(not drawing a quarter with 2 coins) = 17/23 * 16/22 = 136/253 > 50%
P(not drawing a quarter with 3 coins) = 136/253 * 15/21 = 680/1771 < 50%
The answer is C. |
AQUA-RAT | AQUA-RAT-35298 | ## 1 Answer
You also cannot succeed with $$8$$ tiles. Each of the tiles can only cover one of the squares marked with an $$\times$$: $$\begin{array}{|c|c|c|c|c|c|c|} \hline \times &\;\,&\;\,&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline \end{array}$$
Here is a solution with $$9$$ tiles:
• Wow that's so nice... and I am blind. – abc... Feb 23 at 3:53
• Very nice. For a slightly prettier solution with 9 tiles, you can start with a rotationally symmetric arrangement of 8 tiles covering all but the center square. Namely, put tiles on b1&c2, d2&e1, g2&f3, f4&g5, f7&e6, d6&c7, a6&b5, b4&a3; every square but d4 is covered. – bof Feb 23 at 5:40
The following is multiple choice question (with options) to answer.
A room 11 m 47 cm long and 7m 77 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor. | [
"636",
"640",
"647",
"651"
] | D | Explanation:
Area of the room = (1147 x 777) cm2.
Size of largest square tile = H.C.F. of 1147 cm and 777 cm = 37 cm.
Area of 1 tile = (37 x 37) cm2.
Number of tiles required =(1147×777)/(37×37)=651
Answer: Option D |
AQUA-RAT | AQUA-RAT-35299 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
The area of the floor of a room is 20 m2 that of a longer wall 15 m2 and of the shorter wall 12 m2, find the edge of the new cube? | [
"50 m^3",
"60 m^3",
"70 m^3",
"80 m^3"
] | B | Explanation:
lb = 20 ; lh = 15 ; fh = 12
(lbh)2 = 20 * 15 * 12 => lbh = 60 m^3
B) |
AQUA-RAT | AQUA-RAT-35300 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 300 m long train crosses a platform in 48 sec while it crosses a signal pole in 18 sec. What is the length of the platform? | [
"500",
"350",
"828",
"277"
] | A | Speed = 300/18 = 50/3 m/sec.
Let the length of the platform be x meters.
Then, (x + 300)/489 = 50/3
3x + 900 = 2400 => x = 500 m.
Answer: A |
AQUA-RAT | AQUA-RAT-35301 | Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240.
The following is multiple choice question (with options) to answer.
A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans, what is the least amount of paint T, in gallons, that must be purchased in order to measure out the portions needed for the mixture? | [
"2",
"2 1/2",
"3",
"3 1/2"
] | B | Given W:B = 3:5
That means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture.
But we want least amount of whiteblack paints for minimum of 2 gallons of mixture, so lets reduce keeping same ratio,
1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture, but we want only 2 gallons, lets further reduce
0.75: 1.25 gives 1+1.5 =2.5 gallons of mixture. This looks ok, but lets reduce further just to be sure
0.375: 0.625 gives 0.5 + 1 = 1.5 gallons of mixture, thats less than 2 gallons of mixture, so not acceptable.
So correct ans is 2.5 gallons. B |
AQUA-RAT | AQUA-RAT-35302 | 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 upstream at 25 kmph and downstream at 45 kmph, and then find the speed of the man in still water? | [
"65",
"86",
"35",
"78"
] | C | US = 25
DS = 45
M = (45 + 25)/2
= 35
Answer: C |
AQUA-RAT | AQUA-RAT-35303 | 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.
A mixture contains alcohol and water in the ratio 2:5. If 10liters of water is added to the mixture, the ratio becomes 2:7. Find the quality of alcohol in the given mixture. | [
"5",
"10",
"15",
"20"
] | B | Let the quantity of alcohol and water be 2x and 5x
2x/(5x+10) = 2/7
20x = 4(3x+5)
x=5
Quantity of alcohol = 2*5 = 10liters.
Answer is B |
AQUA-RAT | AQUA-RAT-35304 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A person crosses a 650 m long street in 5 minutes. What is his speed in km per hour? | [
"5.2 km/hr.",
"8.2 km/hr.",
"7.8 km/hr.",
"3.2 km/hr."
] | C | C
7.8 km/hr.
Speed=(600/5 * 60) m/sec
= 2.16 m/sec.
Converting m/sec to km/hr
=(2*18/5) km/hr
= 7.8 km/hr. |
AQUA-RAT | AQUA-RAT-35305 | 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 is 360 meter long is running at a speed of 72 km/hour. In what time will it pass a bridge of 140 meter length? | [
"11 seconds",
"38 seconds",
"40 seconds",
"25 seconds"
] | D | Speed = 72 Km/hr = 72*(5/18) m/sec = 20 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (1/20) = 25 seconds
Answer: D |
AQUA-RAT | AQUA-RAT-35306 | This function takes $2^4=16$ possible values.
Now, lets write $p(n,m)$ for the product of the digits from $n$th to $m$th. Note that $f$ are defined for this values since their prime factors are always among $2,3,5$ and $7$. If some of the values $$f(p(1,m))$$ is $(0,0,0,0)$ then $p(1,m)$ is a square. If not, the set $\{f(p(1,1)),f(p(1,2)),\ldots,f(p(1,16))\}$ has at most $15$ different elements. So there are two equal values, say $p(1,n)$ and $p(1,m)$. Then $$f(p(n+1,m))=f\left(\frac{p(1,m)}{p(1,n)}\right)=f(p(1,m))-f(p(1,n))=(0,0,0,0)$$ so $p(n+1,m)$ is a square, q.e.d.
The following is multiple choice question (with options) to answer.
What is the product of all the prime factors of 16? | [
"108",
"18",
"9",
"6"
] | D | The factors of 16 :
factors are 1,2,4.8,16 Ans - 2*2*2*2 = 16
Answer : D |
AQUA-RAT | AQUA-RAT-35307 | to calculate area, perimeter or volume of a figure. Find the area of the face we are looking directly at. A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other and joined at right angles. The Surface Area of a cuboid is the total of all the areas on each face added together. Then , Total surface Area of Cuboid = 2( lxb + bxh +hxl) square units. The injury is in the area of a small tarsal bone in the foot, the cuboid bone. Go to Surface Area or Volume. The cuboid bone is within the area of the mid-foot. Total surface area of cuboid is 1216 sq cm. The volume is found using the formula: Which is usually shortened to: It doesn't really matter which one is length, width or height, so long as you multiply all three together. For More Videos - Subscribe In this video, we will solve first two problem statements from class 9 NCERT, Exercise 13. Sol: Total area that can be painted = 9. Irregular Prism Volume Calculator - A prism has the same cross section all along its length. Write a Python program to calculate surface volume and area of a cylinder. 8 cm cubic and length of the two edges are 2 cm and 6 cm. You can change the Name, Class, Course, Date, Duration, etc. Some of the worksheets for this concept are Common 2 d and 3 d shapes, Geometric nets pack, Volume and surface area work, Surface area of solids, Surface area, Surface areas of prisms, Surface area of 3d shapes, Module mathematical reasoning. 6 x 3² = 54. So, total surface area of cuboid = 2(lb + bh + lb) = 2(5x × 4x + 4x × 2x + 5x × 2x) = 2(20x2 + 8x2 + 10x2) = 2(38x2) = 76x2 Total surface area of cuboid = 1216 cm2 76x2 = 1216cm2 x2 = 16 x = 4 Dimensions of cuboid are. Students learn how to find the surface area of cubes and cuboids using nets and 3D drawings. Submitted by IncludeHelp, on May 08, 2020. Answers included. It is also known as a right rectangular prism. A patient with cuboid
The following is multiple choice question (with options) to answer.
The surface of a cube is 24sq cm. Find its volume? | [
"8",
"5",
"7",
"3"
] | A | 6a2 = 24
a = 2 => a3 = 8cc
Answer: A |
AQUA-RAT | AQUA-RAT-35308 | 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.
Laura can paint 1/x of a certain room in 20 minutes. What fraction E of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates? | [
"1/(3x)",
"3x/(x-3)",
"(x – 3)/(3x)",
"x/(x-3)"
] | C | Options with variables are often done by plugging in numbers.
Both working together can paint the room in 1 hr so if their individual rates were equal, each would take 2 hours alone.
2 hours is 120 mins so in 20 mins each would complete E=120/20 = 1/6th of the room alone.
So if x = 6 (Laura completes 1/6th of the room in 20 mins), the correct option will give 1/6. (Joseph will also paint 1/6th of the room if their rates are same)
If you put x = 6 in the options, only option (C) will give 1/6
Answer (C) |
AQUA-RAT | AQUA-RAT-35309 | homework-and-exercises, kinematics, relative-motion
Title: Classical mechanics problem - relative velocity I'm having a hard time solving this problem concerning relative motion:
A hare runs with a constant speed $u$ on a straight line. A dog chases the hare by always running straight to it with speed $v$. Calculate the acceleration vector of the dog and its magnitude at the moment in which his velocity vector is perpendicular to the velocity vector of the hare. Suppose that at that instant the distance from the dog to the hare is equal to $d$.
The following is multiple choice question (with options) to answer.
A dog taken four every five leaps of hare but three leaps of the dog is equal to four leaps of the hare. Compare speed? | [
"16:15",
"17:15",
"18:15",
"19:15"
] | A | A dog taken four every five leaps of hare but three leaps of the dog is equal to four leaps of the hare.
For time duration, when a dog takes 12 steps , hare takes 15 steps
but 12 steps of dog = 12*4/3 = 16 steps of hare
so ratio of speed of dog to speed of hare = 16:15
ANSWER:A |
AQUA-RAT | AQUA-RAT-35310 | --------------
Back to your question, after finding the 2 breakpoints, namely 0, and 1. We then, split it into 3 different parts:
1. x < 0:
When x < 0, x is negative, right? So |x| = -x
When x < 0, x - 1 is also negative, right? So |x - 1| = -(x - 1) = -x + 1
So, when x < 0, your function will be (without absolute signs) f(x) = |x| + |x - 1| = -x + (-x + 1) = -2x + 1
2. 0 <= x < 1: (pay attention to the <= sign, it should be less than or equal, not just less than, since you should consider x = 0 as well)
This can be worked out in almost the same manner.
3. x >= 1:
This part should be simple too, just follow my example. :)
Last edited: Sep 2, 2009
The following is multiple choice question (with options) to answer.
If x > 1, which of the following decreases as x decreases?
I.1/( x + x^2)
II. 1/(2x^2 − x)
III. (x + 1) | [
"I only",
"II only",
"III only",
"I and II only"
] | C | let x= 4 and 3 b old and new values
ST1: x= 4 -> 4+16 = 1/20
x=3-> 3+9 = 1/12 ...increased
ST2:x= 4-> 2*16-4 =1/ 28
x=3->2*9-3 = 1/15
Increased
St3: x=4-> 5
x=3->4
Ans: C |
AQUA-RAT | AQUA-RAT-35311 | CRGreathouse said:
If so, I might further guess that this is related to prime gaps, since 113 is the first prime before a record gap (the next prime is 113 + 14 = 127). But I can't get any further, even if both are right.
I don't think it is related to prime gaps in the usual sense , for example , the prime gap between 23 and 29 is 6 because 29 - 23 = 6 (I personally disagree with the definition but that's niether here nor there :mrgreen
But it may be related to prime gaps of another form which will become apparent once the rule for writing down 0's and 1's is discovered.
Note: The sequence consists entirely of 0's and 1's.
Similar threads
The following is multiple choice question (with options) to answer.
Two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line. For instance, 3 and 7 rhyme around 5. What integer Q between 1 and 20, inclusive, has the greatest number of distinct rhyming primes around it? | [
"Q=12",
"Q=15",
"Q=17",
"Q.18"
] | D | Since we are concerned with integers between 1 and 20, write down the primes till 40.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
(you should be very comfortable with the first few primes... )
2, 3, 5, 7, 11,12, 13, 17, 19, 23, 29, 31, 37 - Three pairs (11,13), (7,17), (5, 19)
2, 3, 5, 7, 11, 13,15,17, 19, 23, 29, 31, 37 - Three pairs (13, 17), (11, 19), (7, 23)
2, 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37 - Three pairs (11, 23), (5, 29), (3, 31)
2, 3, 5, 7, 11, 13, 17,18,19, 23, 29, 31, 37 - Four pairs (17, 19), (13, 23), (7, 29), (5, 31)
2, 3, 5, 7, 11, 13, 17, 19,20,23, 29, 31, 37 - definitely cannot be more than 4 since there are only 4 primes more than 20. So must be less than 4 pairs. Ignore.
Answer (D). |
AQUA-RAT | AQUA-RAT-35312 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man sells a horse for Rs.620 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse? | [
"227",
"299",
"740",
"860"
] | C | CP = SP + 1CP = SP - g
620 + x = 980 - 2x
3x = 360 => x = 120
CP = 620+ 120= 740
Answer: C |
AQUA-RAT | AQUA-RAT-35313 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
Logan buys 100 shares of par value Rs. 30 each, of a company, which pays an annual dividend of 12% at such a price that he gets 10% on his investment. Find the market value of a share. | [
"2",
"4",
"5",
"36"
] | D | Face value of each share = Rs.30
Total dividend received by Logan = 100×30×12/100 = Rs.360
Let market value of 100 shares = Rs.x
x×10/100=360
x = 3600
ie, Market value of 100 shares = Rs.3600
Hence, Market value of each share = Rs.36
Answer is D. |
AQUA-RAT | AQUA-RAT-35314 | # 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 4 a rupee and sells them at 3 a rupee. Find his net profit or loss percent? | [
"33 1/7 %",
"33 8/3 %",
"33 1/3 %",
"33 1/8 %"
] | C | The total number of mangoes bought by the shopkeeper be 12.
If he buys 4 a rupee, his CP = 3
He selling at 3 a rupee, his SP = 4
Profit = SP - CP = 4 - 3 = 1
Profit Percent = 1/3 * 100 = 33 1/3 %
Answer: C |
AQUA-RAT | AQUA-RAT-35315 | 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.
In a race, the speeds of A and B are in the ratio 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is? | [
"1 hr",
"2 hr",
"1.5 hr",
"2.5 hr"
] | B | Explanation :
Ratio of speeds = 3:4Distance remaining constant, the ratio of time taken = 4:3A takes 0.5 hours more than BHence time taken by A=4×0.5= 2 hour
Answer : B |
AQUA-RAT | AQUA-RAT-35316 | ### 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.
From the sale of sleeping bags, a retailer made a gross profit of 15% of the wholesale cost. If each sleeping bag was sold for $28, what was the wholesale cost per bag? | [
"3.0",
"3.36",
"24.34",
"25.0"
] | C | Cost price * 1.15 = Selling price --> Cost price * 1.15 = $28 --> Cost price = $24.34.
Answer: C. |
AQUA-RAT | AQUA-RAT-35317 | I believe that the answer is 0.18.
geno3141 Jul 12, 2017
#2
+1129
+1
Geno, you have the hippo but you are missing the potamus.
(Using subscripted variables makes it easier to present and keep track of the relations.)
$$\small \text {Let A1 = {pass exam 1}, A2 = {pass exam 2}, A3 = {pass exam 3}, and A = {passes all three exams}. }\\$$
$$\text{Because }\ A^c_2\subset A^c, \; A^c_2\cap Ac=A^c_2 \\ \mathbb{P}(A^c_2) = \mathbb{P}(A^c_2|A^c_1)\; \cdot\; \mathbb{P}(A^c_1 +\mathbb{P}(A^c_2|A_1) \; \cdot\; \mathbb{P}(A_1) \leftarrow \small \text{Apply Total Probability Formula}\\ = (0 \cdot 0.1) + (0.2 \cdot 0.9) = 0.18 \\ \text {Then }\\ \mathbb{P}(A^c_2|A_c) = \dfrac {\mathbb{P}(A^c_2 \cap A^c)} {\mathbb{P}(A^c)}\\ = \dfrac {0.18}{0.496} = 0.363$$
GA
GingerAle Jul 12, 2017
edited by GingerAle Jul 12, 2017
#3
+1
Rhinoceros !!!!!!!
noun, plural rhinoceroses (especially collectively) rhinoceros.
1.
any of several large, thick-skinned, perissodactyl mammals of the family Rhinocerotidae, of Africa and India, having one or two upright horns on the snout: all rhinoceroses are endangered.
Guest Jul 12, 2017
#4
+92888
0
Thanks guest,
Yep, Ginger sure got that one wrong :)
Gino is definitely affiliated with rhinos, not hippos :))
The following is multiple choice question (with options) to answer.
A wildlife preserve is being planned for 4,000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinoceroses is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population? | [
" 450",
" 330",
" 320",
" 310"
] | A | Number of rhinos = 4000
Watering area = 10,000 acres
Number of rhino to increase by 10 percent ,then number of rhino = 4400
Grazing area for a rhino = 100
Total grazing area for 4400 rhinos= 4400 * 100 = 4,40,000
Total area required for the wildlife preserve = 4,40,000 + 10,000 = 4,50,000 = 450 * 1000
Answer A |
AQUA-RAT | AQUA-RAT-35318 | The $$26$$ blocks consist of three groups: six of them are face-to-face with $$V_1$$, twelve of them are edge-touching-edge-only with $$V_1$$, and the remaining eight are vertex-touching-vertex-only with $$V_1$$.
The value of each $$I_f$$, $$I_e$$, and $$I_v$$ will be obtained using $$E(a,b,c)$$, successively building up starting from $$I_f$$.
It will involve a $$2$$-$$1$$-$$1$$ long box, a $$2$$-$$2$$-$$1$$ flat box, then finally the $$2$$-$$2$$-$$2$$ "double cube".
## Calculation Details
The following is multiple choice question (with options) to answer.
How many cubes of 10cm edge can be put in a cubical box of 1m edge | [
"3348",
"2888",
"1000",
"2877"
] | C | Explanation:
number of cubes=(100x100x100)/(10x10x10)=1000
Answer: C) 1000 |
AQUA-RAT | AQUA-RAT-35319 | # Difference between revisions of "2021 AMC 10B Problems/Problem 3"
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
## Solution 2 (Fast but Not Rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Solution 3
The following is multiple choice question (with options) to answer.
In a group of 26 junior high school students, 7 take French, 10 take Spanish, and 4 take both languages. The students taking both French and Spanish are not counted with the 7 taking French or the 10 taking Spanish. How many students are not taking either French or Spanish? | [
"7",
"6",
"5",
"4"
] | C | C
5
Add 7 + 10 + 4 to get 21.
Then subtract 21 from the total students ⇒ 26 – 21 = 5.
Answer is C |
AQUA-RAT | AQUA-RAT-35320 | If the year is $$19ab$$ then that has a value of $$1900 + 10a +b$$.
You can create two equations:
First $$P=2017-(1900+10a+b) \Rightarrow P=117-10a-b$$
Then $$P=1 + 9 + a + b \Rightarrow P=10+a+b$$
Combining those gives $$10+a+b=117-10a-b$$
$$2b=107-11a$$
$$b=\frac{107-11a}2$$
For this to work, $$a$$ must be odd.
$$a=9 \Rightarrow b=4$$
$$a=7 \Rightarrow b=15$$ - too large
$$a=5, 3, 1$$ will fail similarly.
$$a=9 \Rightarrow b=4$$
If the year is $$20ab$$ then that has a value of $$2000 + 10a +b$$.
You can create two equations:
First $$P=2017-(2000+10a+b) \Rightarrow P=17-10a-b$$
Then $$P=2 + 0 + a + b \Rightarrow P=2+a+b$$
Combining those gives $$2+a+b=17-10a-b$$
$$2b=15-11a$$
$$b=\frac{15-11a}2$$
For this to work, $$a$$ must be odd.
$$a=1 \Rightarrow b=2$$
$$a=3 \Rightarrow b<0$$ , similarly for $$a=5, 7, 9$$.
That gives you two possible answers: 1994 and 2012
• Thank you for your answer! I feel like this answer is the simplest and most easy to understand, so I have marked it correct. Jun 23 at 3:29
The following is multiple choice question (with options) to answer.
If 9a - b = 10b + 65 = -12b - 2a, what is the value of 11a + 11b? | [
"-11",
"-5",
"0",
"5"
] | C | (i) 9a - 11b = 65
(ii) 2a + 22b = -65
Adding (i) and (ii):
11a + 11b = 0
The answer is C. |
AQUA-RAT | AQUA-RAT-35321 | • Possible duplicate of Chess Master Problem – Aqua Dec 21 '17 at 20:01
• – Aqua Dec 21 '17 at 20:04
• – Aqua Dec 21 '17 at 20:05
We define a sequence $a_1,a_2,\ldots$, by letting $a_i$ be the amount of games played on day $i$. It is given that for all $n\in\mathbb{N}$, we have: $$a_{7n+1}+a_{7n+2}+a_{7n+3}+a_{7n+4}+a_{7n+5}+a_{7n+6}+a_{7n+7}\le 12$$ Now, define $$b_k=\sum_{i=1}^{k}a_i$$ for all $k=1,2\ldots$. By the pigeonhole principle, we can find some $l>m$ such that: $$b_l\equiv b_m\pmod {20}$$ and $l-m\le 20$. This means that: $$\sum_{i=m+1}^{l}a_i\equiv 0\pmod {20}$$ and $$1\le\sum_{i=m+1}^{l}a_i<3\cdot 12=36$$ Since the chess player plays at least one game per day and $l>m$. We conclude that: $$\sum_{i=m+1}^{l}a_i=20$$ and we are done.
The following is multiple choice question (with options) to answer.
17 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played? | [
"154",
"184",
"240",
"272"
] | D | 2*17C2 = 2*136 = 272
The answer is D. |
AQUA-RAT | AQUA-RAT-35322 | 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 and B runs around a circular track. A beats B by one round or 10 minutes. In this race, they had completed 4 rounds. If the race was only of one round, find the A's time over the course: | [
"8.5 min",
"7.5 min",
"1.5 min",
"9.5 min"
] | B | Explanation:
B runs around the track in 10 min.
i.e , Speed of B = 10 min per round
A beats B by 1 round
Time taken by A to complete 4 rounds
= Time taken by B to complete 3 rounds
= 30 min
A's speed = 30/4 min per round
= 7.5 min per round
Hence, if the race is only of one round A's time over the course = 7 min 30 sec
Answer: B |
AQUA-RAT | AQUA-RAT-35323 | consecutive edges share any vertex with each other, nor can any two cycles be connected to each other by a path of consecutive edges. Path graphs can be characterized as connected graphs in which the degree of all but two vertices is 2 and the degree of the two remaining vertices is 1. A vertex may belong to no edge, in which case it is not joined to any other vertex. In other words, there is no specific direction to represent the edges. In mathematics, and more specifically in graph theory, a multigraph is a graph which is permitted to have multiple edges, that is, edges that have the same end nodes. Multiple edges , not allowed under the definition above, are two or more edges with both the same tail and the same head. This property can be extended to simple graphs and multigraphs to get simple directed or undirected simple graphs and directed or undirected multigraphs. The vertexes connect together by undirected arcs, which are edges without arrows. Reference: 1. This section focuses on "Tree" in Discrete Mathematics. The following are some of the more basic ways of defining graphs and related mathematical structures. Mary Star Mary Star. The order of a graph is its number of vertices |V|. The edge is said to joinx and y and to be incident on x and y. Typically, a graph is depicted in diagrammatic form as a set of dots or circles for the vertices, joined by lines or curves for the edges. Discrete Mathematics & Mathematical Reasoning Chapter 10: Graphs Kousha Etessami U. of Edinburgh, UK Kousha Etessami (U. of Edinburgh, UK) Discrete Mathematics (Chapter 6) 1 / 13 . Otherwise, it is called an infinite graph. Use your answers to determine the type of graph in Table 1 this graph is. The category of all graphs is the slice category Set ↓ D where D: Set → Set is the functor taking a set s to s × s. There are several operations that produce new graphs from initial ones, which might be classified into the following categories: In a hypergraph, an edge can join more than two vertices. The former type of graph is called an undirected graph while the latter type of graph is called a directed graph. The minimum degree is 5 and the same remarks apply to edges while... Represents a pictorial structure of a set of the only repeated vertices are adjacent if share! Such generalized graphs are the first one is the Difference between
The following is multiple choice question (with options) to answer.
A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph of 3 edges and 3 points. The degree of a point is the number of edges connected to it. For, example, a triangle is a graph with 3 points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any through sequence of edges. The number of edges 'e' in the graph must satisfy the condition | [
"11 ≤ e ≤ 66",
"10 ≤ e ≤ 66",
"11 ≤ e ≤ 65",
"0 ≤ e ≤ 11"
] | A | Explanation :
The least number of edges will be when one point is connected to each of the other 11 points, giving a total of 11 lines. One can move from any point to any other point via the common point.
The maximum edges will be when a line exists between any two points. Two points can be selected from 12 points in 12C2 i.e. 66 lines.
Answer : A |
AQUA-RAT | AQUA-RAT-35324 | ### Show Tags
30 Oct 2018, 10:20
Bunuel wrote:
When a company places a wholesale order on coffee cups with the company logo, they pay $692 for 80 cups. When Phil wants to buy just one such cup, it costs him$12.50. How much above the wholesale price per cup is he paying?
A. $0.55 B.$3.85
C. $7.35 D.$10.65
E. $14.25 The wholesale price is 692/80 =$8.65 per cup.
Thus, Phil pays 12.5 - 8.65 = \$3.85 above the wholesale price if he buys just one cup.
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
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Re: When a company places a wholesale order on coffee cups with the compan &nbs [#permalink] 30 Oct 2018, 10:20
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The following is multiple choice question (with options) to answer.
At the wholesale store you can buy an 8-pack of hot dogs for $1.55, a 20-pack for $3.05, and a 350-pack for $22.95. What is the greatest number of hot dogs you can buy at this store with $200? | [
"1,108",
"2,900",
"2,108",
"2,124"
] | B | We have $200 and we have to maximize the number of hot dogs that we can buy with this amount.
Let's try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money, which in this case is 350 for $22.95.
For the sake of calculation, let's take $23. 23x8 gives 184 ,i.e. a total of 350x8 = 2800 hot dogs . We are left with ~$16. Similarly, let's use $3 for calculation. We can buy 5 20-pack hot dogs (3x5), a total of 20x5 = 100 hot dogs. So we have 2900 hot dogs.
2108 looks far-fetched (since we are not likely to be left with > $1.55). Hence, (b) 2900 (ANSWER B) |
AQUA-RAT | AQUA-RAT-35325 | 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 telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job? | [
"15",
"14",
"12",
"10"
] | D | Explanation :
There are two equations to be formed 40m + 50f = 1000.
250m + 300f + 40 x 15m + 50 x 10f = A.
850m + 8000f = A.
m and f are the number of males and females. A is the amount paid by the employer.
Then the possible value of f = 8,9,10,11,12.
Iif f= 8, M =15.
If f = 9,10,11 then m will not be an integer while f =12 then m will be 10.
By putting f = 8 and m=15 , A = 18800.
when f =12 and m= 10 then A = 18100.
Hence, the number of males is 10.
Answer : D |
AQUA-RAT | AQUA-RAT-35326 | 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
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
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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.
A batsman makes a score of 100 runs in the 17th inning and thus increases his averages by 5. What is his average after 17th inning? | [
"39",
"35",
"20",
"40.5"
] | C | let the average after 16th inning =x
then total run after 16th inning=16x
then total run after 17th inning=16x+100
then average run after 17th inning=(16x+100)/17
(16x+100)/17=x+5 = > 16x+100 = 17x+85 => x =15
x=15;
average after 17th inning =15+5=20
ANSWER:C |
AQUA-RAT | AQUA-RAT-35327 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are traveling on a collision course. If train A is traveling at a speed of 350 mph and train B is traveling 28% slower, how much time will it take the trains to collide if the initial distance between the two is 2107 miles? | [
"Two hours and 30 minutes.",
"One hour and 10 minutes.",
"Two hours and 25 minutes.",
"Three hours and 5 minutes."
] | D | Answer is D. Three hours and 5 minutes.
Since they are racing towards each other, the speed is accumulative, ie. 350 mph + 350*0.72 = 602 (0.72 because the second train is 28% slower)
So time before collision is total distance / total speed = 2107/602 = 3 hours and 5 mins |
AQUA-RAT | AQUA-RAT-35328 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If three marbles are drawn what is the probability that one is yellow and two are red? | [
"1/5",
"2/47",
"3/8",
"12/455"
] | D | Explanation:
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When three marbles are drawn, the probability that one is yellow and two are red
= (²Câ‚)(â´Câ‚‚)/¹âµC₃
= (2 * 4 * 3 * 3 * 2)/(1 * 2 * 15 * 14 * 13) = 12/455.
D |
AQUA-RAT | AQUA-RAT-35329 | For an arbitrary natural number $N$, we get the inequality $$\sum_{i > k} \frac{N}{p_i} < \frac{N}{2}$$
Now, we call all the primes $p_1, p_2, \dots , p_k$ the small primes, and all the other primes the big primes.
Let $N_b$ denote the number of positive integers $n \leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.
We will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.
First, we estimate $N_b$ $$N_b \leq \sum_{i > k}\big\lfloor \frac{N}{p_i} \big\rfloor < \frac{N}{2}$$
Now, we look at $N_s$. We write every $n \leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \begin{align} b_n \leq \sqrt n \leq \sqrt N \\ \text{There are at most $\sqrt N$ different values of $b_n$} \\ \implies N_s \leq 2^k\sqrt N \end{align}
The following is multiple choice question (with options) to answer.
The product of the prime numbers between 0 and 20 is closest to... | [
"10^5",
"10^6",
"10^7",
"10^8"
] | C | Write out the primes 1 - 20
2 x 3 x 5 x 7 x 11 x 13 x 17 x 19
Round the numbers off in multiples of 5. Try to alternate rounding up and down
5 x 5 x 5 x 5 x 10 x 10 x 20 x 20
25 x 25 x 10 x 10 x 20 x 20
pull out 10 from each number
10^6 (2.5 x 2.5 x 2 x 2)
10^6 (5 x 4)
10^7 (2)
round off the 2.
answer is C |
AQUA-RAT | AQUA-RAT-35330 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
Ten years ago, the age of Anand was one-third the age of Bala at that time. The present age of Bala is 12 years more than the present age of Anand. Find the present age of Anand? | [
"22",
"16",
"27",
"187"
] | B | Let the present ages of Anand and Bala be 'a' and 'b' respectively.
a - 10 = 1/3 (b - 10) --- (1)
b = a + 12
Substituting b = a + 12 in first equation,
a - 10 = 1/3 (a + 2) => 3a - 30 = a + 2
=> 2a = 32 => a = 16.
Answer:B |
AQUA-RAT | AQUA-RAT-35331 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A shipment of 1500 heads of cabbage, each of which was approximately the same size, was purchased for $900. The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head. The following day the rest were sold at a price equal to 10% less than the price each head sold for on the day before. What was the total gross profit on this shipment. | [
"$100",
"b $115",
"c $187.5",
"d $130.5"
] | C | Purchase cost = 900/1500= $ 0.6
25% above the cost per head = (125/100)*0.6= $ 0.75
2/3 of the heads were sold means 2/3*1500 =1000 ,Selling amount for 1000 cabbage = 1000 * 0.75 = $ 750
Remaining were sold at a price equal to 10% less than the price each head sold for on the day before i.e , (90/100)*0.75 = $ 0.675
Hence selling cost for the remaining cabbage = 500 * 0.675 = $ 337.5
Total selling cost = 750 + 337.5 = $ 1087.5
Gross profit = $ 1087.5 - $ 900 = $ 187.5
Hence Answer is C. |
AQUA-RAT | AQUA-RAT-35332 | owned for a long time ) / ( $110 –$ 100 average return the! Heavily on key aspects, such as the value of money. [ 10 ] very little.... Capital employed ) measured in different currencies muni bonds and interest rates in the form of project. Find information on government bonds yields, muni bonds and interest rates in the first of! Be higher than the cost of funds: the amount invested externe bronnen en zijn misschien niet nauwkeurig returns of! Money manager controls cash flows it all goes according to plan! ) share purchased at point! And add one to the equation, requiring some interpretation to determine the success of project. Of 20,000 USD divided by 100,000 USD, which equals 20 percent returns... Well connected to the percentage can be calculated work on a project, product or advertising campaign and guide... That this does not apply to interest rates or yields where there no. Return measures the increase in size of an asset you have an asset you have an you! Exclude accrued fees from valuations funds and brokerage firms are now providing personalized account returns on 's., i.e., optimized returns the quarter-end stock price appropriate one ÷ $100 ) ÷ initial value )$. After year 1 of objective helps your website fulfill your needs rate at which items. Be profitable the holding period return growth rate … Get updated data about US Treasuries pattern is not to! 'S how basic earnings and gains/losses work on a mutual fund the single period last... Be earned on a mutual fund ( 1/ investment ’ s term in years is called the of. Expenses ( gross profit rate ) the price often changes a great deal, the return of: the invested... It will earn more interest in the cash flow example below, the IRR a... This period. ) = ( $dividend ) / ($ stock price increased by 10 % rate return... And add one to the right is an example of a stock for $50 per share as in...$ dividend ) / ( \$ stock price a percentage of the reinvestment/ex-dividend.. 3.02 % for consecutive periods measured in different currencies other words, the dollar returns the. Interested to know how the investment 's purchase price negative initial value ) ÷ initial
The following is multiple choice question (with options) to answer.
Anil invested a capital of Rs.5000 and Sunil invested Rs.10000. Anil invested for a time period which is double than the time period for which Sushil invested. How will a profit of Rs.4000 be divided between them? | [
"1000, 3000",
"3000,1000",
"2500,1500",
"2000,2000"
] | D | suppose time duration for sunil is 'x' and thus time duration for anil will be '2x'
So their investment ratio will be 5000*2x:10000*x= 1:1
Therefore profit 4000 will be divided into 1:1 ratio i.e. 2000 & 2000
ANSWER:D |
AQUA-RAT | AQUA-RAT-35333 | 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 passes a station platform in 32 seconds and a man standing on the platform in 12 seconds. If the speed of the train is 72 km/hr, what is the length of the platform? | [
"340",
"360",
"380",
"400"
] | D | The speed of the train is 72 * 5/18 m/s = 20 m/s
The length of the train is 20 x 12 = 240 m
platform + 240 = 20 * 32
platform = 640 - 240 = 400
The answer is D. |
AQUA-RAT | AQUA-RAT-35334 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Two persons A and B can complete a piece of work in 30 days and 45 days respectively. If they work together, what part of the work will be completed in 5 days? | [
"5/18",
"1/6",
"1/4",
"1/9"
] | A | A's one day's work = 1/30
B's one day's work = 1/45
(A + B)'s one day's work = 1/30 + 1/45 = 1/18
The part of the work completed in 5 days = 5 (1/18) = 5/18.
ANSWER A |
AQUA-RAT | AQUA-RAT-35335 | => 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.
In store A there are 20 pairs of pants for every 40 store B has. The price ratio between the pants in store B and the pants in store A is 3:4. If all the pants were sold in both places until the stock ran out, what is the ratio between the total amount stores A earned to the total amount store B earned? | [
"3:16.",
"2:3.",
"1:3.",
"3:4."
] | B | 1st statement : ratio of pants
Store A : Store B
20x : 40x
X:2X
Price :
4y:3y
Total revenue
4xy : 6xy
2:3
Answer is B |
AQUA-RAT | AQUA-RAT-35336 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 15 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one? | [
"78",
"60",
"26",
"23"
] | B | Relative speed = 45 + 15 = 60 km/hr.
60 * 5/18 =50/3 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 3/50 = 60 sec.
Answer:B |
AQUA-RAT | AQUA-RAT-35337 | javascript, jquery, e-commerce
You could also consider that droping the quantity to zero is the same than removing the item. This is easy with a switch and no break in the 'minus' and 'quantity == 0' case:
switch(button) {
case 'minus':
if (quantity > 1) {
quantity--;
updBasketItems = parseInt(basketItems) - 1;
updTotalPrice = parseFloat(basketTotal) - parseFloat(itemPrice);
updItemPrice = parseFloat(subTotal) - parseFloat(itemPrice);
break;
}
case 'delete':
updTotalPrice = parseFloat(basketTotal) - parseFloat(subTotal);
updBasketItems = parseInt(basketItems) - parseInt(quantity);
$(this).parent(".item-row").fadeOut('slow',function(){
$(this).remove();
});
break;
case 'plus':
quantity++;
updBasketItems = parseInt(basketItems) + 1;
updTotalPrice = parseFloat(basketTotal) + parseFloat(itemPrice);
updItemPrice = parseFloat(itemPrice) + parseFloat(subTotal);
break;
}
The following is multiple choice question (with options) to answer.
The cost of an item was reduced from its usual selling price of $23.99 to $17.99. Which of the following is closest to the percent decrease in the price of the item? | [
"17%",
"21%",
"24%",
"25%"
] | D | The price decreased by 23.99 - 17.99 = $6.
$6 is a bit more than 25% of original price of $24.
Answer: D. |
AQUA-RAT | AQUA-RAT-35338 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
In a bag,there are coins of 25 p,10 p and 5 p in the ratio of 1 : 2 : 3.If there are Rs.30 in all,how many 5 p coins are there ? | [
"50",
"100",
"150",
"200"
] | C | Solution
Let the number of 25 p,10 p and 5 p coins be x,2x and 3x respectively.
Then,sum of their values = Rs.(250x/100 +10x2x100 +5x3x/100) = Rs.60x/100
∴60/100 = 30 x =30x100/60 = 50.
Hence,the number of 5 p coins = (3x50) = 150
Answer C |
AQUA-RAT | AQUA-RAT-35339 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
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 $$0.3*1$$.
The following is multiple choice question (with options) to answer.
If 20 liters of chemical X are added to 80 liters of a mixture that is 5% chemical X and 95% chemical Y, then what percentage of the resulting mixture is chemical X? | [
"24%",
"26%",
"28%",
"30%"
] | A | The amount of chemical X in the solution is 20 +0.05(80) = 24 liters.
24 liters / 100 liters = 24%
The answer is A. |
AQUA-RAT | AQUA-RAT-35340 | ### 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 gets 3 times as much money as B gets, B gets only Rs.25 more then what C gets. The three gets Rs.575 in all. Find the share of B? | [
"130",
"120",
"218",
"140"
] | B | A+B+C = 575
A = 3B
3B+B+B-25 = 575
5B = 600
B = 120
Answer: B |
AQUA-RAT | AQUA-RAT-35341 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
How many liters of water must be added to 18 liters of milk and water containing 10% water to make it 40% water? | [
"6",
"7",
"8",
"9"
] | D | By rule of alligation:
40% - 10% = 30%
100% - 40% = 60%
Quantity of pure water : Quantity of the mixture = 1 : 2
There are 18 liters of mixture, so we need to add 9 liters of pure water.
The answer is D. |
AQUA-RAT | AQUA-RAT-35342 | 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.
In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran? | [
"70 kg",
"69 kg",
"61 kg",
"67 kg"
] | D | Explanation:
Let Kiran's weight = x. Then
According to Kiran, 65 < x < 72 ----(equation 1)
According to brother, 60 < x < 70 ----(equation 2)
According to mother, x ≤≤ 68 ----(equation 3)
Given that equation 1,equation 2 and equation 3 are correct. By combining these equations, we can write as
65<x≤68
i.e., x = 66 or 67 or 68
Average of different probable weights of Kiran = 66+67+683=67
Answer: Option D |
AQUA-RAT | AQUA-RAT-35343 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of six numbers is 3.75. The average of two of them is 3.4, while the average of the other two is 3.85. What is the average of the remaining two numbers ? | [
"4.7",
"4.6",
"4.0",
"4.1"
] | C | Explanation:
Sum of the remaining two numbers = (3.75 * 6) - [(3.4 * 2) + (3.85 * 2)]
= 22.50 - (6.8 + 7.7) = 22.50 - 14.5 = 8.
Required average = (8 / 2) = .
Answer: C) 4.0 |
AQUA-RAT | AQUA-RAT-35344 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
A committee of 5 persons is to be formed from 3 company secretaries, 4 economists and 1 chartered accountant. What is the probability that each of the three professions is represented in the committee? | [
"1/28",
"3/28",
"60/14",
"4/28"
] | C | 3/8*4/7*1/6*5! = 3/14. We are multiplying by 5! because we can choose secretary,economist, and accountant is several way:
Answer: C. |
AQUA-RAT | AQUA-RAT-35345 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A batsman scored 100 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? | [
"45(4/11) %",
"40 %",
"45(5/11) %",
"44(5/11) %"
] | B | Explanation :
Total runs scored = 100
Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60
Total runs scored by running between the wickets = 100 - 60 = 40
Required % = (40/100) × 100 = 40%
Answer : Option B |
AQUA-RAT | AQUA-RAT-35346 | 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 10:30 am. They shall first cross each other at ? | [
"7:15 am",
"7:30 am",
"10: 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 10:30 + 12 mins = 10:42
Option C |
AQUA-RAT | AQUA-RAT-35347 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
Maria bought 10 notebooks and 5 pens costing 2 dollars each.How much did Maria pay? | [
"$10",
"$20",
"$30",
"$40"
] | C | 2 × (10 + 5) = 2 × 10 + 2 × 5 = 20 + 10 = 30 dollars
correct answer is C)$30 |
AQUA-RAT | AQUA-RAT-35348 | # Number of triples of divisors who are relatively prime as a triple
Given a number $n \in \mathbb{N}$, define $a(n)=\{(d_1,d_2,d_3): d_i|n,\ \gcd(d_1,d_2,d_3)=1,\ 1 \leq d_1 \leq d_2 \leq d_3\}$. What is $|a(n)|$?
There were similar questions asked about pairs rather than triples here: Number of pairs of nontrivial relatively prime divisors and here: Number of Relatively Prime Factors but the addition of the third component seems to make the arguments used in these two questions obsolete.
An example:
Let $n=p$ for some prime $p$. Then the divisors of $n$ are $\{1,p\}$, and the triples of those divisors whom are relatively prime (order does not matter) as a triple are $\{(1,1,1),(1,1,p),(1,p,p)\}$. Thus we see that for any prime, the answer is $|a(n)|=3$.
Similarly, (its not hard to check) for $n=p^2$, we have $|a(n)|=6$.
In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.
Also for $n=pq$ where $p$ and $q$ are distinct primes, $|a(n)|=13$ (again, not hard to check).
This leads me to believe that like in the other answers for the two referenced questions, the answer may be found using some nice combinatorics, but if looked at just right from a Number Theoretic perspective, may be a multiplicative function or composition of multiplicative functions, based on how the answers seem to only depend on the powers of the primes in the prime factorization of the number.
The following is multiple choice question (with options) to answer.
If x = 3d, where d is a prime number greater than 2, how many different positive even divisors does x have, including x? | [
"0",
"1",
"2",
"3"
] | A | d is prime greater than 2 thus d is odd
3d is odd
Thus no even factors
Ans:A |
AQUA-RAT | AQUA-RAT-35349 | 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.
Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the job in 6 days, working together.
For how many days have each of them worked and what percentage of the job have each of them completed? | [
"5 days and 40%",
"6 days and 30%",
"6 days and 40%",
"7 days and 40%"
] | C | First we will find the daily productivity of every worker. If we consider the whole job as unit (1), Elizabeth does 115 of the job per day and Tony does 75% of 115, i.e.
75100⋅115=120. Suppose that Tony worked alone for x days. Then he finished x20 of the total job alone. Working together for 6 days, the two workers finished 6⋅(115+120)=6⋅760=710 of the job.
The sum of x20 and 710 gives us the whole job, i.e. 1. So we get the equation:
x20+710=1
x20=310
x=6. Tony worked for 6 + 6 = 12 days and Elizabeth worked for 6 days. The part of job done is 12⋅120=60100=60% for Tony, and 6⋅115=40100=40% for Elizabeth.
Answer is C. |
AQUA-RAT | AQUA-RAT-35350 | ### Show Tags
15 Aug 2010, 03:11
Bunuel wrote:
praveengmat wrote:
How many factors does 36^2 have?
A 2
B 8
C 24
D 25
E 26
Finding the Number of Factors of an Integer:
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
Back to the original question:
How many factors does 36^2 have?
$$36^2=(2^2*3^2)^2=2^4*3^4$$ --> # of factors $$(4+1)*(4+1)=25$$.
Or another way: 36^2 is a perfect square, # of factors of perfect square is always odd (as perfect square has even powers of its primes and when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only odd answer in answer choices is 25.
Hope it helps.
Thanks a ton !!.. loved the approach !
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Re: Help: Factors problem !! [#permalink]
### Show Tags
14 Oct 2010, 13:58
1
Factors of a perfect square can be derived by using prime factorization and then using the formula to find perfect square's factors.
The following is multiple choice question (with options) to answer.
How many different positive integers are factors of 36 ? | [
" 4",
" 6",
" 7",
" 9"
] | D | 6×6=3^2 × 2^2
So total factors = (2+1)(2+1) = 9
Answer: D |
AQUA-RAT | AQUA-RAT-35351 | ### 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.
A departmental store charges the commission of 18 percent on the first sale of items worth $50,000 and then an additional 10 percent on any sale price that exceeds the initial $50,000. If the department store earned $24,000 in commissions for the sale of a series of diamonds, what was the sale price of the diamond set? | [
" $215,000",
" $365,000",
" $115,000",
" $200,000"
] | D | 24,000 = 0.18 *(50000) + 0.10 (x - 50000)
24000 - 9000 = 0.10 (x-50000)
150000 = x - 50000
x = 2,00,000
Answer : D |
AQUA-RAT | AQUA-RAT-35352 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
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The following is multiple choice question (with options) to answer.
Fill in the missing number:
18 14 ? 11 10.5 10.25 | [
"10.1",
"12",
"10.2",
"10.5"
] | B | 12
Divide each number by 2 and then add 5 to it.
ANSWER B |
AQUA-RAT | AQUA-RAT-35353 | 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.
P, Q and R can do a piece of work in 20, 30 and 60 days respectively. In how many days can P do the work if he is assisted by Q and R on every third day? | [
"15 DAYS",
"12 DAYS",
"10 DAYS",
"18 DAYS"
] | A | P's 2 day's work = 1/20*2=1/10
P+Q+R)'S 1 DAY'S WORK = 1/20+1/30+1/60 = 6/60 = 1/10
WORK IN DAYS = 1/5
NOW 1/5 DONE IN 3 DAYS
WHOLE WORK WILL BE DONE IN(3*5) = 15 DAYS
ANSWER A |
AQUA-RAT | AQUA-RAT-35354 | 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 300 m long passed a pole in 14 sec. How long will it take to pass a platform 540 m long? | [
"12 sec",
"59 sec",
"40 sec",
"45 sec"
] | C | Speed = 300/14
= 21 m/sec.
Required time = (300 + 540)/21
= 40 sec.
Answer: C |
AQUA-RAT | AQUA-RAT-35355 | Maybe the examiner didn't think through your method.
From 10,000 to 99,999 the numbers divisible by 5 end in 0 or 5.
The first (most significant) digit can be any of nine from 1 to 9.
The 2nd digit can be any of 10.
The 3rd digit can be any of 10.
The 4th digit can be any of 10.
The 5th digit can be either of 2.
That's 9(10)(10)(10)2=18,000
The following is multiple choice question (with options) to answer.
Which is greater than 99? | [
"9",
"99",
"9.9",
"999"
] | D | 999 greater than 99.
Answer: D |
AQUA-RAT | AQUA-RAT-35356 | induction that the product of three consecutive integers is divisible by 6. Real numbers class 10. In mathematics, the least common multiple, also known as the lowest common multiple of two (or more) integers a and b, is the smallest positive integer that is divisible by both. Click 'show details' to verify your result. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Prove that n2-n is divisible by 2 for every positive integer n. Exercise: 2. For any positive integer n, prove that n3 – n is divisible by 6. The number is divisible by 6 means it must be divisible by 2 and 3. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. What are the two odd integers? 12. 111 is the smallest possible magic. Exercise 12. case (3) z is a multiple of three. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Some other very important questions from real numbers chapter 1 class 10. Prove that the product of any three consecutive positive integers is divisible by 6. Since n is a perfect square, n is congruent to 0 or 1 modulo 4. Solution for Determine whether the statement is true or false. 1, which is divisible by 9. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. : Therefore: n = 3p or 3p+1 or 3p+2, where p is some integer If n = 3p, then n is divisible by 3 If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3. Consider three consecutive integers, n, n + 1, and n+ 2. 7: given non-empty nite sets X and Y with jXj= jYj, a function X !Y is an injection if and only if it
The following is multiple choice question (with options) to answer.
The least common multiple of positive integer q and 3-digit integer n is 690. If n is not divisible by 3 and q is not divisible by 2, what is the value of n? | [
"115",
"230",
"460",
"575"
] | B | The LCM of n and q is 690 = 2*3*5*23.
q is not divisible by 2, thus 2 goes to n
n is not divisible by 3, thus 3 goes to q.
From above:
n must be divisible by 2 and not divisible by 3: n = 2*... In order n to be a 3-digit number it must take all other primes too: n = 2*5*23 = 230.
Answer: B. |
AQUA-RAT | AQUA-RAT-35357 | • $\frac{999,999}{7}=142857$. You used $1,000,000-1$ instead. But I don't understand how from $2\cdot 999,999 = 2\times \frac {1,000,000 -1}7 = 1,999,998$ we go to $285714$ via replacing $2$ with the formula for $100$.
– Jim
Sep 22 at 19:55
• If $100 = 7\times 14 + 2$ then $2 = 100 - 7\times 14$. So $2\times \frac {999,999}7 = (100-7\times 14)\times \frac {999999}7 = (100 - 7\times 14)\times \frac {10^6 -1}7$ Now just expand $(100-7\times 14)\times \frac{10^6 - 1}7 = 100\times \frac {10^6-1}7 - 7\times 14\times \frac {10^6-1}7= 100\times \frac {10^6-1}7 +-14\times (10^6 -1) = 100\times \frac {10^6-1}7 - 14\times 10^6 + 14$. ... Now $100\times {10^6-1}7$ will add two zeros to $142857$ to get $14285700$. And $-14\times 10^6=-14000000$ removes the leading $14$ to get $285700$ and $+14$ adds it back to the other side: $285714$. Sep 22 at 20:06
• $1999998 = 2\cdot 999,999 \ne 2\cdot \frac {999999}7=285714$. Why did you do anything with $2\times 999999$ that was not part of the question and never part of my solution? Sep 22 at 20:08
The following is multiple choice question (with options) to answer.
Find the value of 72516 x 9999 = m? | [
"345434667",
"246465757",
"465767867",
"725087484"
] | D | 72516 x 9999 = 72516 x (10000 - 1)
= 72516 x 10000 - 72516 x 1
= 725160000 - 72516
= 725087484
D |
AQUA-RAT | AQUA-RAT-35358 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A rectangular rug with side lengths of 4 feet and 6 feet is placed on a square floor that has an area of 64 square feet. If the surface of the rug does not extend beyond the area of the floor, what fraction of the area of the floor is not covered by the rug? | [
"A.3/16",
"1/4",
"C.1/2",
"5/8"
] | D | Area of the rectangular rug = 4*6 = 24
Fraction not covered by the rug = (Total Area - Rug area) / Total Area
= (64 - 24)/64
= 5/8=D |
AQUA-RAT | AQUA-RAT-35359 | # For dinner, $n$ people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so ...
Problem:
For dinner, $$n$$ ($$n \geq 4$$) people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so that Ana and Ivan do not sit next to each other and at least one of them sits next to Mark? (Note: the round table implies seating arrangements that differ only in rotation.)
My attempt:
If I have $$n$$ people sitting around circular table, the number of different arrangements are $$(n-1)!$$.
If I have $$2$$ people Mark and Ana number of arrangements that they can sit next to each other is $$2 \cdot (n-2)!$$. So the number of arrangements that Mark sit next to Ivan is also $$2 \cdot (n-2)!$$, and sitting next to Ana also $$2 \cdot (n-2)!$$.
• yes.....it is fixs
– josf
Nov 2 '19 at 14:49
• "... seating arrangements that differ only in rotation... " are what? Considered to be the same, or different? From the context, I assume they should be considered to be the same, but the language isn't clear. Nov 11 '19 at 16:28
Method 1: Seat Mark. We will use him as our reference point.
Only Ana sits next to Mark: She can be seated in two ways, to his left or to his right. That leaves $$n - 2$$ seats. Since Ivan cannot sit next to Ana or Mark, he may be seated in $$n - 4$$ ways. The remaining $$n - 3$$ people can be seated in the remaining $$n - 3$$ seats in $$(n - 3)!$$ ways as we proceed clockwise around the table relative to Mark. Hence, there are $$2(n - 4)(n - 3)!$$ such arrangements.
The following is multiple choice question (with options) to answer.
There are 10 reading spots in a room. Each reading spot has a round table. Each round table has 4 chair. If different no of persons are sitting at each reading spot. And if there are 10 persons inside the room then how many reading spots donot have atleast a single reader. | [
"5",
"6",
"8",
"9"
] | B | 10 reading spots and 4 chairs each
total 10 persons
different no. of seating must
so
1st-1person
2nd-2person
3rd-3person
4th-4person
by above way 10 peoples are seated
so empty chairs = 10chairs - 4chairs = ans: 6chairs
ANSWER:B |
AQUA-RAT | AQUA-RAT-35360 | So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
VP
Joined: 07 Dec 2014
Posts: 1128
Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A fruit seller sells 8 oranges at a cost price of 9. The profit percent is | [
"12 1⁄2",
"11 1⁄9",
"5 15⁄17",
"8 2⁄3"
] | A | Let C.P. of one orange = 1
Then C.P. of 8 oranges = 8
S.P of 8 oranges = 9
∴ Gain % = 9−8/8×100=100/8=12 1/2%
Answer A |
AQUA-RAT | AQUA-RAT-35361 | # 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.
By selling an article at Rs.300, a profit of 25% is made. Find its cost price? | [
"s.486",
"s.455",
"s.240",
"s.480"
] | C | SP = 300
CP = (SP)*[100/(100+P)]
= 300* [100/(100+25)]
= 300* [100/125]
= Rs.240
Answer: C |
AQUA-RAT | AQUA-RAT-35362 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
A and B start a business with Rs.5000 and Rs.15000 respectively. How should they share their profits at the end of one year? | [
"1:3",
"3:4",
"2:5",
"3:7"
] | A | They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
5000 : 15000 => 1:3
ANSWER:A |
AQUA-RAT | AQUA-RAT-35363 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
P , Q, R enter into a partnership & their share are in the ratio 1/2 : 1/3 : 1/4 , after two months , P withdraws half of the capitals & after 10 months , a profit of Rs 378 is divided among them . What is P's share? | [
"114",
"126",
"134",
"144"
] | B | Explanation :
The ratio of their initial investment = 1/2 : 1/3 : 1/4
= 6 : 4: 3
Let's take the initial investment of P, Q and R as 6x, 4x and 3x respectively
A:B:C = (6x * 2 + 3x * 10) : 4x*12 : 3x*12
= (12+30) : 4*12 : 3*12
=(4+10) : 4*4 : 12
= 14 : 16 : 12
= 7 : 8 : 6
P's share = 378 * (7/21) = 126. Answer : Option B |
AQUA-RAT | AQUA-RAT-35364 | inorganic-chemistry, solubility
$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)
So then
$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$
Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!
Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility. The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's. Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals. See here, under "Applications".
The following is multiple choice question (with options) to answer.
Out of the total production of iron from hematite, an ore of Iron, 20% of the ore gets wasted, and out of the remaining iron, only 25% is pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000 kg, then the quantity of hematite mined from that mine in the year is | [
"5,00,000 kg",
"4,00,000 kg",
"4,50,000 kg",
"None of these"
] | B | Solution: Let 100 kg of hematite be obtained then 20% of it get wasted that means 80 kg of ore remains.
Pure iron = 25% of remaining ore = 80*25/100 = 20 kg.
20 kg pure Iron is obtained from 100 of hematite.
1 kg pure Iron is obtained from = 100/20 hematite;
Then, 80000 kg pure Iron is obtained from = (100/20)*80000 = 400000 kg hematite.
Answer: Option B |
AQUA-RAT | AQUA-RAT-35365 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
3 gentlemen and 3 gentlewomen are candidates for 2vacancies. A voter has to vote for 2candidates. In how many ways can one cast his vote? | [
"11",
"12",
"15",
"16"
] | C | There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,
= 6C2 = 6!/2!*4!
= 15.
C |
AQUA-RAT | AQUA-RAT-35366 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
If the largest three digit number is subtracted from the smallest five digit number, then the remainder is | [
"9000",
"9001",
"90001",
"9009"
] | B | Sol.
Required number = (10000 - 999) = 9001.
Answer B |
AQUA-RAT | AQUA-RAT-35367 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average weight of a group of boys is 30 kg. After a boy of weight 39 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ? | [
"4",
"8",
"5",
"2"
] | B | Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 39 kg joins the group, total weight of boys = 30x + 39
So 30x + 39 + 31(x + 1) = > x = 8.
Answer: B |
AQUA-RAT | AQUA-RAT-35368 | Sort by:
1)If r is an odd prime, let it be of the form 2n+1 where n is a positive integer. If we put this value of r in our equation, we get 2(n+1)q-2np=(2n+1)^2. As L.H.S. is even and R.H.S is odd we won't have any solutions for p,q,r in this case. So r=2. We get 3q-p=4 which means q=(p+4)/3. We can see that q can be an integer when p=2,5,11,17,23,29,41,47,53,59,71,..... After which p+q will exceed 111. So when p is 71 then q is not prime, when p is 59 q is 21 which is again not a prime. But when p is 53, q is 19 which is indeed a prime. So our largest solution set for p,q is 53 and 19 and r=2. So largest possible value of pqr is 53×19×2=2014.
- 4 years, 6 months ago
Great solution! You would get 7 out of 7. Only 1 improvement is to make your proof look neater (paragraphing, spacing, etc.)
- 4 years, 6 months ago
$\Large{{ 5 }^{ a }+{ b }^{ 2 }={ 3 }^{ c }\\ \Longrightarrow \frac { { 5 }^{ a }+{ b }^{ 2 } }{ 3 } ={ 3 }^{ c-1 }}$
Since $3^{c-1}$ is an integer. We have
${ 5 }^{ a }+{ b }^{ 2 }\equiv 0\quad \left( \mod 3 \right)$
The following is multiple choice question (with options) to answer.
P is a prime number. If 6p is between 15 and 95, which of the following T can be a value of 7p + 2? | [
"T=15",
"T=79",
"121",
"65"
] | B | [3,5,7,11,13] are prime numbers that fits this equation 15< 6X <95. Afterwards subtract 2 from all the answer n see if it's divisible by 7. Therefore A and E are out. Leaves out with B C D.
B =11
C =17
D = 9
Only 11 is the set of prime numbers that fit between 15 and 95. Therefore 79 is the answer.B |
AQUA-RAT | AQUA-RAT-35369 | 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
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Kudos [?]: 79 [1] , given: 10
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.
A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3.Find his average after 17th inning? | [
"19",
"29",
"39",
"49"
] | C | let the average after 17th inning =x
Then average after 16th inning=(x-3)
Therefore 16(x-3)+87=17x
Therefore x=39
ANSWER:C |
AQUA-RAT | AQUA-RAT-35370 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
### GMAT/MBA Expert
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
Nine men went to a hotel. 8 of them spent 3 each over their meals and the ninth spent 2 more than the average expenditure of all the nine. The total money spent by all of them was | [
"26",
"40",
"29.25",
"27"
] | C | Let the average expenditure of all the ninte be x
Then, 3 × 8 + x + 2 = 9x
⇒ x = 3.25
∴ Total money spent = 9x = 9 × 3.25 = 29.25
Answer C |
AQUA-RAT | AQUA-RAT-35371 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. How far is the place? | [
"2.86",
"2.88",
"2.89",
"2.834"
] | B | M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 - 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
Answer: B |
AQUA-RAT | AQUA-RAT-35372 | Note: The author(s) of your text are calculating the probability that the top-ranked woman is in the $k$th position. For that to occur, we must select $k - 1$ of the five men to be ranked ahead of her and one of the five women to fill the $k$th position while choosing $k$ of the $10$ people. Hence, the answer in the text is equivalent to \begin{align*} P(X = 1) & = \frac{\binom{5}{0}\binom{5}{1}}{\binom{10}{1}} = \frac{1}{2}\\ P(X = 2) & = \frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{5}{18}\\ P(X = 3) & = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{5}{36}\\ P(X = 4) & = \frac{\binom{5}{3}\binom{5}{1}}{\binom{10}{4}} = \frac{5}{84}\\ P(X = 5) & = \frac{\binom{5}{4}\binom{5}{1}}{\binom{10}{5}} = \frac{5}{252}\\ P(X = 6) & = \frac{\binom{5}{5}\binom{5}{1}}{\binom{10}{6}} = \frac{1}{252} \end{align*}
The following is multiple choice question (with options) to answer.
The population of a town is 45000; 5/9th of them are males and the rest females 40% of the males are married. What is the percentage of married females? | [
"56%",
"50%",
"57%",
"10%"
] | B | Male = 45,000* 5/9 = 25,000
Female = 45,000* 4/9 = 20,000
Married Male = 25,000* 40/100 = 10,000
Married Female = 10,000
20,000 ------------ 10,000
100 ------------- ? => 50%
Answer: B |
AQUA-RAT | AQUA-RAT-35373 | # Homework Help: Probability of choosing stale donuts out of 24
Tags:
1. Sep 2, 2016
### TheSodesa
1. The problem statement, all variables and given/known data
There are 6 stale donuts in a set of 24. What is the probability of:
a) there being no stale donuts in a sample of 10?
b) there being 3 stale donuts in a sample of 10?
c) What is the chance of a stale doughnut being found?
2. Relevant equations
$N \, permutations = N!$
3. The attempt at a solution
Let $X$ denote the number of stale donuts in a set of 10.
a) I used the idea of permutations like so:
$$P(X = 0) = \frac{\frac{18!}{8!}}{\frac{24!}{15!}} \approx 0.335$$
This was incorrect, according to the testing software
b) Here I followed the same idea:
$$P(X=3) = \frac{\frac{18!}{11!} \times \frac{6!}{3!}}{\frac{24!}{15!}} \approx 0.041$$
c) This is just the complement of part $a$:
$$P(X \, is \, at \, least \, 1) = 1 - 0.061 = 0.665$$
Parts $a$ and $b$ (and $c$ as a consequence of $a$ being wrong) are apparently wrong and I'm not sure whats wrong with my reasoning.
Last edited: Sep 2, 2016
2. Sep 2, 2016
### TheSodesa
I was completely wrong. I was supposed to use combinations instead:
a) $$P(X = 0) = \frac{18 \choose 10}{24 \choose 10} =39/1748 \approx 0.22$$
b) $$P(X = 3) = \frac{{18 \choose 7} \times {6 \choose 3}}{{24 \choose 10}} =1560/4807 \approx 0.325$$
The following is multiple choice question (with options) to answer.
Halle, Julia and Drew have 5 donuts to share. If one of them can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed. | [
"20",
"21",
"22",
"23"
] | B | these are the no of ways:
0 2 3 - 3! ways
2 2 1 - 3!/2 ways
5 0 0 - 3!/2 ways
4 1 0 - 3! ways
1 1 3 - 3!/2 ways
So total no of ways = 21
ANS:B |
AQUA-RAT | AQUA-RAT-35374 | Each of your methods gives you $1.496 \times 10^9 ~\rm{m}$ - accurate to 4 significant digits. That's better than you should expect.
The accuracy of your result is only ever as good as the accuracy of the inputs. Some of the data you have is given to 5 significant digits - that will give you the "more accurate" results.
I recommend that you learn about error propagation. There are millions of resources online for this - a relatively basic introduction can be found here
• Sorry for the late reply! I have been really busy with the project and school. I actually totally forgot about error propagation and simply kept the numbers intact without thinking!
– user156610
May 30, 2017 at 9:01
$a = 149598023$
$e = 0.0167086$
$p = a(1-e^2)$
$p = a(0.99972082268604) = 149,556,258.62576513369892$
$b = \frac{p}{ \sqrt{1-e^2}}$
$= \frac{p} {0.99986040159916324190990908617052}$
$= 149,577,139.35522085944399314340085$
$b = \sqrt{p*a}$ = 149,577,139.35522085944399314340085
$pa = rmax * rmin$
$22,373,320,617,691,160.86368910923516 = rmax * rmin$
$rmax = \frac{p}{(1-e)} = \frac{p}{0.9832914} =152,097,596.5270978$
$rmin = \frac{p}{(1+e)} = \frac{p}{1.0167086} =147,098,449.4729022$
The following is multiple choice question (with options) to answer.
In expressing a length 810472 km as nearly as possible with three significant digits, find the percentage error. | [
"0.034",
"0.024",
"0.352",
"0.063"
] | A | Error = (81.5 – 81.472)km = 0.028.
Required percentage = [(0.028/81.472)*100]% = 0.034%.
Answer A. |
AQUA-RAT | AQUA-RAT-35375 | # How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
The following is multiple choice question (with options) to answer.
If there are eight varieties of ice cream to be tasted and judged in a competition, what is the probability that black currant is tasted before Vanilla and Vanilla is tasted before chocolate, although not necessarily consecutively? | [
"1/8!",
"3!/8!",
"5!/8!",
"1/3!"
] | D | 8 flavors are too much. Lets try out only 3 flavors --> Black Currant, Vanilla, Chocalate
We can get Black Currant before Vanilla and Vanilla before Chocolate in only 1 way.
Probability = Possible/Total = 1/3!
Now lets consider 4 flavors --> BC, Van, Choc and X
Possibilities --> 1) BC, X, Van, Choc
2) X, BC, Van, Choc
3) BC, Van, Choc, X
4) BC, Van, X, Choc
Probability = 4/24 = 1/3!
Since the probability remains the same for both the cases, the probability will mostly remain true for the 8 varieties too.
Answer: D |
AQUA-RAT | AQUA-RAT-35376 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
A picnic attracts 200 persons. There are 20 more men than women, and 20 more adults than children. How many men are at this picnic? | [
"240",
"65",
"110",
"130"
] | B | Adult + Children = 200
Let, Children = y
then, Adult = y+20
i.e. y+(y+20) = 200
i.e. y = 900
i.e. Adult = 90+20 = 110
Adults include only Men and Women
i.e. Men + Women = 110
Let women, W = x
then Men, M = x+20
i.e. x+(x+20) = 2x+20 = 110
i.e. x = 45
i.e. Men, M = 45+20 = 65
Answer: Option B |
AQUA-RAT | AQUA-RAT-35377 | to two sides: And the diagonals "p" and "q" of a rhombus bisect each other at right angles. Yes; opposite sides are congruent. Example 1: Find the area and perimeter of the square whose side length is 4 meters. A trapezoid is a 4-sided figure with a single set of parallel sides. This selection focuses on what the area of a rectangular object (like a room) means, and how it ’ s measured. Side "a" is 4 cm longer than side "b". A kite is cyclic if and only if it has two right angles. 6 Solve real - world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. 79, and 168. Construct a line parallel to a given line and a given distance from it. Example: Prove that the following four points will form a rectangle when connected in order. The length of the base and the height are the only two numbers you need to know when calculating the area of any triangle. During the four-month closure, the museum stepped up its virtual presence, with daily Facebook posts and some virtual. Discover the area formula for trapezoids, and explore alternative methods for calculating the area of a trapezoid. Depending on the number of equal sides, triangles may be classified as: Scalene: A triangle with no equal sides or angles, i. the length of c can be determined as: c = √ a 2 + b 2 = √ 3 2 +4 2 = √ 25 = 5. To find the area of a rectangle, multiply the length by the width. First and fore most a Trapezoid is a quadrilateral which is a 4 sided plane shape. Area of the rectangle = 4 × 2 = 8 cm 2 We need 8 one-centimetre squares to make a rectangle 4 cm long and 2 cm wide. The sum of the four straight angles is 720° and the sum of the four interior angles is 360°, so the sum of the four exterior angles is 360°. The quadrilateral area calculator is the best choice! Choose the option with your given parameters. Construct a circle of a given radius tangent to two lines through a point. If you have a rectangle with two sides 17 cm and two sides 34 cm, the
The following is multiple choice question (with options) to answer.
The cross-section of a cannel is a trapezium in shape. If the cannel is 6 m wide at the top and 4 m wide at the bottom and the area of cross-section is 10290 sq m, the depth of cannel is? | [
"258",
"260",
"262",
"264"
] | A | 1/2 * d (6+4)
= 10290
d = 258
Answer: A |
AQUA-RAT | AQUA-RAT-35378 | # How many numbers made out of 5 distinct digits contain 4 and are even
This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.
I distinguished some cases (not sure if I distinguished in a 'smart' way)
• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.
The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.
This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?
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.
The following is multiple choice question (with options) to answer.
How many four-digit numbers that do not contain the digits 2, 3, 4, or 6 are there? | [
"1025",
"1080",
"2096",
"3040"
] | B | The 1st digit can be filled up by the numbers: {1,5,7,8,9} = 5 ways
The 2nd digit can be filled up by the numbers: {0,1,5,7,8,9} = 6 ways
The 3rd digit can be filled up by the numbers: {0,1,5,7,8,9} = 6 ways
The 4th digit can be filled up by the numbers: {0,1,5,7,8,9} = 6 ways
The total number of such four-digit numbers is 6*7*7*7 = 1080
The answer is B. |
AQUA-RAT | AQUA-RAT-35379 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
John walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 8 minutes earlier. How far is the office from his house? | [
"3 1/5 Km",
"3 2/5 Km",
"3 3/5 Km",
"3 5/7 Km"
] | A | Formula = S1*S2/S2-S1 * T1+T2/60
= 4*6/2 * 16/6
= 24/2 * 16/60
= 8 * 2/5
= 16/5 = 3 1/5 Km
A) |
AQUA-RAT | AQUA-RAT-35380 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
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PS Forum Moderator
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Posts: 1059
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GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
Calculate the average of first 16 even numbers is? | [
"16",
"15",
"12",
"17"
] | D | Explanation:
Sum of 10 even numbers = 16 * 17 = 272
Average = 272/16 = 17
Answer: Option D |
AQUA-RAT | AQUA-RAT-35381 | # Math Help - Ratio of inscribed and circumscribed triangles
1. ## Ratio of inscribed and circumscribed triangles
An equilateral triangle is inscribed in a circle, and the circle is inscribed in another equilateral triangle. Find the ratio of the area of the smaller triangle to that of the larger triangle.
Im getting 1:3 but the answer is 1:4.
Thank You
2. Hello, spred!
An equilateral triangle is inscribed in a circle,
and the circle is inscribed in another equilateral triangle.
Find the ratio of the area of the smaller triangle to that of the larger triangle.
Code:
A
D ∆ - - - - - - - * ∆ * - - - - - - - ∆ E
\ * / \ * /
\ * / \ * /
\ * / \ * /
\ / \ /
\ * / \ * /
\ * / ∆ O \ * /
\* / | \ */
\ / | \ /
C ∆ - - - * - - - ∆ B
* D /
\* */
\ * * * /
\ /
\ /
\ /
\ /
\ /
∆
F
The center of the circle is $O.$
The inscribed triangle is $ABC.$
The circumscribed triangle is $D{E}F.$
$O$ is the centroid of $\Delta ABC$ and $\Delta D{E}F.$
Draw segment $OB.$
In right triangle $ODB,\,\text{ let }r = OD.$
. . Then the altitude $AD$ of $\Delta ABC$ is $3r.$
Since $\angle ODB = 30^o\!:\;\;OB = 2r$
. . Then the altitude $DB$ of $\Delta D{E}F$ is $6r$
The ratio of the altitudes is: . $3r: 6r \:=\: 1: 2$
Therefore, the ratio of the area is: . $1^2:2^2 \;=\;1:4$
The following is multiple choice question (with options) to answer.
A square and an equilateral triangle have the same perimeter. What is the ratio of the area of the circle circumscribing the square to the area of the circle inscribed in the triangle? | [
"9:8",
"15:8",
"27:8",
"8:9"
] | C | let x be side of square
perimeter of square=4x=perimeter of triangle=3*side of triangle
so side of eq. triangle=(4/3)*x
diameter of circle circumscribing the square=sqrt(2)*x
area of circle circumscribing the square=pi*(sqrt(2)*x)^2/4=(pi/2)*x^2 ----(1)
to find radius of the circle inscribed in the triangle
area of triangle=r*s=sqrt(3)/4 * (4x/3)^2
now s=(4/3)*x+(4/3)*x+(4/3)*x/2=2x
so sqrt(3)/4 * (4x/3)^2=r*2x gives
r={2/3*(3^1/2)}*x
area of the circle inscribed in the triangle=pi*[{2/3*(3^1/2)}*x]^2
=pi*(4/27)*x^2 -------(2)
so reqd ratio= eqn(1)/eqn(2)
=[(pi/2)*x^2]/[pi*(4/27)*x^2]=27/8
so reqd ratio=27:8
ANSWER:C |
AQUA-RAT | AQUA-RAT-35382 | feature a number of tools to solve geometry problems. Each part is called semi circle. A rectangle is one of the many fundamental shapes you'll see in math. Round your answer to the nearest hundredth. A square has sides of length 12 cm. To make the pies, roll the pastry on a floured surface into a large rectangle, about 12 inches by 24-plus inches. Learn how to calculate perimeter and area for various shapes. The distance around the outside of a semicircle is generally called its perimeter. It is evident that if we make a large number of cuts, the figure formed will approximate a rectangle whose length is equal to one-half of the circumference and whose width is equal to the radius. Fold the entire piece of fabric in half to create a large square. Area of a circle intuition. Requires knowledge of Conic Sections. make up Remember, by o shape. Creating a design with pavers requires planning. Repeat the same method the other side. where r is the radius of the circle. The word ‘area’ stands for the space occupied by a flat object or figure. To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Draw and cut out the wing shape from the flap and insert into the slit. 1165 1225 112. Calculations at a semicircle. 28 square cm. asked by ian on August 29, 2016. How to Find the Area of a Semicircle To find the area of a semi-circle, you need to know the formula for the area of a circle. Semi-circles are drawn on and as diameters. Determine the diameter of the circle you intend to make. 4 in L, Base 6in - 4694841. in Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 251 16. Practice: Area of a circle. In the article below, we provide the semicircle definition and explain how to find the perimeter and area of a semicircle. What is the objective function in terms of the base of the rectangle, x? (Type an expression. The Gimp provides the Rectangle, Ellipse and Lasso selection tools to help you cut out specific parts of a photo or illustration to keep, eliminate or edit. Area
The following is multiple choice question (with options) to answer.
A paper is in a square form whose one side is 20 cm. Two semi circles are drawn on its opposites as diameters. If these semi circles are cut down what is the area of the remaining paper? | [
"8.79",
"8.77",
"8.75",
"8.72"
] | C | (5 * 3.5)/2 = 8.75
Answer: C |
AQUA-RAT | AQUA-RAT-35383 | 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 660 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is? | [
"640",
"277",
"500",
"297"
] | A | Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (660 + x)/60 = 65/3
x = 640 m.
Answer: A |
AQUA-RAT | AQUA-RAT-35384 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Arjun and Bhishma are running on a circular track of length 600m. Speed of Arjun is 30 m/s and that of Bhishma is 20 m/s .They start fro the same point at the same time in the same direction. When will they meet again for the first time? | [
"22",
"60",
"27",
"21"
] | B | Actually Arjun has to make a lead of 600 m, because when Arjun will be 600m ahead of Bhishma, they will be together again as a person when completes the total length it starts retracing he same path and thus Arjun and Bhishma can be together again.
Since , they make a difference of 10 m in 1 second. so, he will create 600m difference in 60 second.
ANSWER=B |
AQUA-RAT | AQUA-RAT-35385 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
"A" and "B" run around a circular track starting from the same point simultaneously in the same direction. "A" meets "B" for the first time when "A" is exactly in the middle of his 5th round. If "A" is faster than "B" and take 70 seconds to complete 1 lap, how long will B take to complete 1 lap? | [
"90 seconds",
"54.44 seconds",
"110 seconds",
"63 seconds"
] | A | A is faster than B. Every second he increases distance between A and B. They will meet for the first time again when he increases distance between them by one full circle. (Say the lap is of 100m. He keeps increasing distance between them. When he is 90 m ahead of him, it is the same as 10 m behind him because they are moving in a circle. Finally when he is 100 m ahead of him, he is exactly 1 lap ahead and hence both are at the same point).
A takes 70 secs for one full lap. So he covers 4.5 laps in exactly 70*4.5 = 315 seconds.
In 315 secs, B completes 3.5 laps. So for each lap, he takes 315/3.5 = 90 secs
Answer (A) |
AQUA-RAT | AQUA-RAT-35386 | 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 16 kmph in still water.it takes him thrice as long to row up as to row down the river.find the rate of stream | [
"8 km/hr.",
"6 km/hr.",
"9 km/hr.",
"15 km/hr."
] | A | Let man’s rate upstream be x kmph.then ,his rate downstream=3xkmph.
So,2x=16 or x=8.
Rate upstream=8 km/hr,rate downstream=24 km/hr.
Hence,rate of stream=1/2(24-8)km/hr=8 km/hr.
Answer is A. |
AQUA-RAT | AQUA-RAT-35387 | 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 7a = 2b = 15, then 42ab = | [
"75",
"125",
"675",
"1100"
] | C | 7a * 2b = 15 * 15 = 225
14 ab = 225
i.e. 42 ab = 675
Answer: Option C |
AQUA-RAT | AQUA-RAT-35388 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
6,9,15,27,51, (...) | [
"44",
"86",
"12",
"99"
] | D | 6
6× 2 - 3 = 9
9 × 2 - 3= 15
15 × 2 - 3 = 27
27 × 2 - 3 = 51
51 × 2 - 3 = 99
Answer is D |
AQUA-RAT | AQUA-RAT-35389 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post. If the length of each train be 120 m, in what time will they cross other travelling in opposite direction? | [
"77 sec",
"12 sec",
"55 sec",
"43 sec"
] | B | Speed of the first train = 120/10 = 12 m/sec.
Speed of the second train = 120/5 = 8 m/sec.
Relative speed = 12 + 8 = 20 m/sec.
Required time = (120 + 120)/20 = 12 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-35390 | # find the number of possible positive integer solutions when the inequality is $a \times b \times c \lt 180$
As most of you know there is a classical question in elementary combinatorics such that if $$a \times b \times c = 180$$ , then how many possible positive integer solution are there for the equation $$?$$
The solution is easy such that $$180=2^2 \times 3^2 \times 5^1$$ and so , for $$a=2^{x_1} \times 3^{y_1} \times 5^{z_1}$$ , $$b=2^{x_2} \times 3^{y_2} \times 5^{z_2}$$ , $$c=2^{x_3} \times 3^{y_3} \times 5^{z_3}$$ .
Then: $$x_1+x_2+x_3=2$$ where $$x_i \geq0$$ , and $$y_1+y_2+y_3=2$$ where $$y_i \geq0$$ and $$z_1+z_2+z_3=1$$ where $$z_i \geq0$$.
So , $$C (4,2) \times C(4,2) \times C(3,1)=108$$.
Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $$a \times b \times c \lt 180$$ instead of $$a \times b \times c = 180$$
The following is multiple choice question (with options) to answer.
For how many unique pairs of nonnegative integers {a, b} is the equation a^2 - b^2 = 99 true? | [
"1",
"3",
"5",
"7"
] | B | Answer C
(a+b)(a-b)=99
3 cases For (a+b), (a-b)
99, 1
33, 3
11, 9
Answer B |
AQUA-RAT | AQUA-RAT-35391 | thermodynamics, energy, home-experiment
Note 3: Peak hours tend to be during the day (this may reverse if we get wide-spread solar energy), so keeping the house cool during the day would cost even more. You pretty much have it right. We have two scenarios:
1 - Leave air conditioning on all day. Say that outside temperature is 90 degrees and inside temperature is 70 degrees. Then, all day long, the air conditioning has to remove any heat that gets into the house continuously. Heat transfer depends on the difference in temperature between outside and inside the house. In this case, heat transfer per unit time is equal to a constant K times 20 degrees. Over a whole day of D units of time, the total energy removed by the air conditioner is $20KD$ This assumes that convection (ie air draft) can be neglected.
2 - Stop the air conditioning during the day. If you stop the air conditioning, the flow of heat in the house initially will be the same. However, as time goes by, the inside temperature will increase. This will gradually reduce heat flow as it is proportional to the difference in temperatures. If the house is small enough and badly insulated, you could even get to a point where the inside temperature is equal to the outside temperature, at which point no more heat enters the house. When you get back from work, or if you have a timer a bit before, air conditioning is restarted. You now have to remove all the accumulated heat. However, since less heat entered the house over the duration of the day than if you had kept the air conditioning on, you have less work to do to cool down the house. Yes, the heat will have accumulated in insulation, furniture, walls, etc, but it doesn't matter. There is less heat total to remove.
The following is multiple choice question (with options) to answer.
An air-conditioning unit costs $470. On December there was a discount for Christmas of 16%. Six months later, the holiday season was over so the company raised the price of the air-conditioning by 12%. How much will an air-conditioning unit cost in November? | [
"$442",
"$470",
"$472",
"$484"
] | A | if its previous November(before discount) then price is $470.
but if its November of next year then
16% discount on $470=470(1-16/100)=$394.8
again a corrected raised price of 12% over $394.8=394.8(1+12/100)=442.176~$4442
Ans A |
AQUA-RAT | AQUA-RAT-35392 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Molly can do a task in 15 days while Sandy can do the task in 10 days. How many days will it take to complete the task if both work together? | [
"6",
"7",
"7.5",
"8"
] | A | Each day, Molly can do 1/15 of the task.
Each day, Sandy can do 1/10 of the task.
Each day, they can do 1/15 + 1/10 = 1/6 of the task.
The task will be completed in 6 days.
The answer is A. |
AQUA-RAT | AQUA-RAT-35393 | P\left( S=18 \right) =\dfrac { 3 }{ P\left( T \right) } \\ P\left( S=21 \right) =\dfrac { 2 }{ P\left( T \right) } \\ Let\quad D\quad be\quad the\quad event\quad that\quad sum\quad is\quad divisible\quad by\quad 3\\ Then\quad P\left( D \right) =P\left( S=3 \right) +P\left( S=6 \right) +P\left( S=9 \right) +P\left( S=12 \right) +P\left( S=15 \right) +P\left( S=18 \right) +P\left( S=21 \right) \\ \Rightarrow P\left( D \right) =\dfrac { 22 }{ P\left( T \right) } =\dfrac { 22 }{ 66 } \\ \Rightarrow P\left( D \right) =\dfrac { 1 }{ 3 }$$Maths
The following is multiple choice question (with options) to answer.
If p and q are the two digits of the number 653pq such that this number is divisible by 80, then p+q is equal to : | [
"2",
"3",
"4",
"6"
] | A | Since 653xy is divisible by 5 as well as 2, so y = 0.
Now, 653x0 must be divisible by 8.
So, 3x0 must be divisible by 8. This happens when x = 2
x + y = (2 + 0) = 2.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35394 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
A and B start a business with Rs.6000 and Rs.2000 respectively. Hoe should they share their profits at the end of one year? | [
"3:5",
"3:4",
"3:9",
"3:1"
] | D | They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
6000 : 2000 => 3:1,Answer: D |
AQUA-RAT | AQUA-RAT-35395 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
A, B and C have Rs.1000 between them, A and C together have Rs.700 and B and C Rs.600. How much does C have? | [
"100",
"200",
"300",
"400"
] | C | A+B+C = 1000
A+C = 700
B+C = 600
--------------
A+B+2C = 1300
A+B+C = 1000
----------------
C =300
ANSWER:C |
AQUA-RAT | AQUA-RAT-35396 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Smith can do a work in 15 days while Johnson can do it in 25 days. How long will they take if both work together? | [
"3 4/9 days",
"8 4/9 days",
"9 3/8 days",
"6 3/8 days"
] | C | Explanation:
1/15 + 1/25 = 8/75
75/8 = 9 3/8 days
Answer: C |
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