source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35497 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
If the position of the first, sixth digits of the number 2796543018 are interchanged, similarly the positions of the the second and seventh digits are interchanged and so on, which of the following will be the left of seventh digit from the left end? | [
"1",
"2",
"6",
"7"
] | A | Explanation:
The new number formed is 4 3 0 1 8 2 7 9 6 5
The seventh digit from the left is 7, the 3rd digit from the left of 7 is 1
Answer: A) 1 |
AQUA-RAT | AQUA-RAT-35498 | Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
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Posts: 1835
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Re: Some part of a 50% solution of acid was replaced with an [#permalink]
### Show Tags
12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
The following is multiple choice question (with options) to answer.
A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 30% lemonade syrup? | [
"1.5",
"1.75",
"2.14",
"2.34"
] | C | Let the total solution is 150 L with 80 L water70 L syrup.
To make 30% syrup solution, the result solution must have 105 L syrup and 45 L syrup.
Therefore we are taking 25 L of syrup from initial solution and replacing with water.
using urinary method:
70 L syrup in 150 L solution
25 L syrup in 53.6 L solution
We started by multiplying 10
Now to get to the result we need to divide by 25 => amount of solution to be replaced with water = (53.6/25) = 2.14.
Correct option : C |
AQUA-RAT | AQUA-RAT-35499 | ### Which sequences will contain the number 1000 Charlie's way
If you look it at Charlie's way instead you can see that carrying on from where he started A2's sequence is the four times table just skipping out 1 number each time because it goes up in 8's so if you do 1000 divided by 4 =250 or divided by 8 just to make sure it's not a number that is skipped out it would =125 so A2=YES. A3's sequence is the 8 times table but skips out a numbers each time eg: 8-24(skips 16) because the sequence adds 16 each time so what you would do is 1000 divided by 8 = 125 = YES A4 is the 16 times table missing out a number each time so you would do 1000 divided by 16 which equals 62.5= NO A5 is the 32 times table missing out 1 each time so 1000 divided by 32 =31.25= NO A6 is the 64 times table missing out a number each time so do 1000 divided by 64 which equals 15.625=NO so out of the sequences shown only A2 and A3 have the number 1000
### Which sequences will contain the number 1000 Alison's way
If you do it Alison's way there is a bit of counting so if you think about what Alison said, each number is double the number in the row above.
To get to 1000 each sequence has to have the sequence above containing 500, which is less counting, and to make it even easier the row before that one has to have 250, and the one before that 125.
This way you can use the doubles as an advantage and do a lot less counting.
### A2 sequence
I was wrong on all three ways for sequence A2 as it does not have 1000 in its sequence as with Alison's way a1 doesn't reach 500 but 498 or 502 and with Bernard's way and Charlie's way it would equal 996 or 1004
### Persevering
Hi Sofia, well done for sticking at it with this problem even though you made a mistake initially, it's great to see you keeping going until you get the right answer!
### Math
The following is multiple choice question (with options) to answer.
Lucy is traveling from one end of a forest to the other. In order to find her way back, she is leaving morsels of bread in the following pattern: 2 morsels of wheat, 3 morsels of white, and 1 morsel of milk. The pattern repeats after she leaves the morsel of milk. If Lucy drops 2,000 morsels of bread, what are the last 3 morsels of bread that she drops? | [
"milk − wheat − wheat",
"wheat − wheat − white",
"white − milk − wheat",
"white − white − white"
] | A | In a single round Lucy drops 6 morsels.
Remainder (2000/6) = 2
Final action = 2 drops of Wheat --> Options B, C, D and E can be eliminated
Answer: A |
AQUA-RAT | AQUA-RAT-35500 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? | [
"3",
"4",
"5",
"7"
] | A | Area of the park = (60 x 40) m2 = 2400 m2.
Area of the lawn = 2109 m2.
Area of the crossroads = (2400 - 2109) m2 = 291 m2.
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
x2 - 100x + 291 = 0
(x - 97)(x - 3) = 0
x = 3.
A |
AQUA-RAT | AQUA-RAT-35501 | varies over another length-$$2\pi$$ interval the monotinicity division details are different, but any choice requires at least one such cut, succumbs to the above technique, and gets the same answer. Let's check another example from $$t_1=-\pi/2$$ to $$t_1=3\pi/2$$, giving two pieces (meeting at $$t=\pi/2$$) where the $$\pm$$ are $$+-$$:$$\int_{-1}^1\frac{2dt_2}{\sqrt{1-t_2^2}}-\int_1^{-1}\frac{2dt_2}{\sqrt{1-t_2^2}}\color{red}{=4\int_{-1}^1\frac{dt_2}{\sqrt{1-t_2^2}}}\color{blue}{=\left(\int_{-\pi/2}^{\pi/2}+\int_{\pi/2}^{3\pi/2}\right)2dt_1}.$$
The following is multiple choice question (with options) to answer.
If a continuous cable 10 meters long was marked off at intervals of 1/3 meter and also 1/4 meter, and then if the cable were cut at each mark, how many different size length of cable would result? | [
"1",
"2",
"3",
"4"
] | C | In this particular example length (10 meter) is just extraneous information.
The location of the marks for 1/3 meter would be 4/12, 8/12, 12/12, etc...
The location of the marks for 1/4 meter would be 3/12, 6/12, 9/12, 12/12, etc...
The distances between marks are 3/12, 1/12, and 2/12.
The answer is C. |
AQUA-RAT | AQUA-RAT-35502 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
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Joined: 04 Aug 2010
Posts: 322
Schools: Dartmouth College
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
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. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
The following is multiple choice question (with options) to answer.
Tap 'A' can fill the tank completely in 6 hrs while tap 'B' can empty it by 12 hrs. By mistake, the person forgot to close the tap 'B', As a result, both the taps, remained open. After 1 hrs, the person realized the mistake and immediately closed the tap 'B'. In how much time now onwards, would the tank be full? | [
"2.5 hours",
"4.5 hours",
"5.5 hours",
"1 hour"
] | C | Explanation :
Tap A can fill the tank completely in 6 hours
=> In 1 hour, Tap A can fill 1⁄6 of the tank
Tap B can empty the tank completely in 12 hours
=> In 1 hour, Tap B can empty 1⁄12 of the tank
i.e., In one hour, Tank A and B together can effectively fill 1⁄6 - 1⁄12 = 1⁄12 of the tank
=> In 1 hours, Tank A and B can effectively fill 1⁄12 × 1 = 1⁄12 of the tank.
Time taken to fill the remaining 1−(1/12) = 11/12 of the tank = (11/12)/(1/6) = 5.5 hours. Answer : Option C |
AQUA-RAT | AQUA-RAT-35503 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 149, what is the difference between the ages of Patrick and Monica? | [
"27",
"64",
"45",
"72"
] | B | Ages of P and Mi=3x+5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=47x=149
x=4
Difference if ages of Pa and Mo=25x-9x=16x=16*4=64
Answer B |
AQUA-RAT | AQUA-RAT-35504 | [1, 1, 5] -> [1, 2, 6] -> [1, 3, 7] -> [1, 4, 8] -> [1, 5, 9] -> ...
Is it my initial approach correct? How can I break it down in subproblems?
• Here's a hint: The final values don't matter, only the fact that they are equal. Think of what happens in terms of the relative differences of the elements in the array. Adding $i$ to all but one elements is equivalent to to decreasing $i$ from exactly one element -- the one you wouldn't add to. For example, in the first case your solution is equivalent to subtracting twice 2 from 5 to obtain [1,1,1]. – Vincenzo Jul 11 '19 at 11:28
• Thank you, you are right. In the second case, subtracting 1 from 3 and 5 from 7, you obtain [2, 2, 2, 2] in two steps. I'm going to try to code it. – Héctor Jul 11 '19 at 11:34
Increasing all elements but one is equivalent to decrease only that element. Find minimum value m, now for each element x you need to find minumum number of subtractions to go from x to m (change-making problem), since you can use greedy algorithm for coins (1, 2, 5), you can solve this problem in O(n).
• This will not always give the correct answer. Say the original array is [1, 5, 5]. Your algorithm would return 4 by decreasing the two 5 to 1. But one can actually do in 3 steps: decrease all of them to 0. – WhatsUp Oct 7 '19 at 1:38
The main idea is as in the answer by @izanbf1803.
However, there is a detail: after you find the minimum value m, you should go through the array again, keeping in mind that m, m - 1 and m - 2 could all be your final target. After getting the total number of steps for all three of them, just return the smallest one.
For example, if the array is [0, 0, 0], then the best final target is m = 0;
The following is multiple choice question (with options) to answer.
What is the least number to be subtracted from 11, 15, 21 and 30 each so that the resultant numbers become proportional? | [
"7",
"6",
"3",
"4"
] | C | Let the least number to be subtracted be 'x', then 11 - x, 15 - x, 21 - x and 30 - x are in proportion.
<=> (11 - x):(15 - x) = (21 - x):(30 -x)(21 - x)
From the options, when x = 3
=> 8 * 27 = 12 * 18 => then x = 3. => (11 - x)(30 - x) = (15 - x)(21 - x)
Answer: C |
AQUA-RAT | AQUA-RAT-35505 | # Ratio of angles in a triangle, given lengths of triangle's sides.
If I have a triangle $\,\triangle ABC,\,$ with sides of lengths $\,AB=6, \;BC=4, \;CA=5,\,$
then what can I know about the ratio of $\,\dfrac{\angle ACB}{\angle BAC}\,$?
-
Use en.wikipedia.org/wiki/Law_of_cosines, if Trigonometry is allowed – lab bhattacharjee Nov 12 '13 at 13:49
You can use the Law of Sines to find the ratio of the sines of your two angles:
$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, D \!$$
In your case, you'd have $$\dfrac 6{\sin(\angle ACB)} = \dfrac 4{\sin(\angle BAC)} \iff \dfrac {6}{4} = \dfrac{\sin(\angle ACB)}{\sin(\angle BAC)} = \frac 32$$
Alternatively, to compute the measures of your angles directly, use the Law of Cosines. $$c^2 = a^2 + b^2 - 2ab\cos\gamma\,\iff \cos \gamma = \dfrac{a^2 + b^2 - c^2}{2ab}$$
The following is multiple choice question (with options) to answer.
A triangle has three angles all with the same ratio to each other, 1:1:1. What is the angle measurement? | [
"45°",
"60°",
"75°",
"90°"
] | B | If the ratio is 1:1:1, and the total angle measurement is 180°, then the equation is
x + x + x = 180°
3x = 180°
x = 60°
Therefore the answer is 60° or B. |
AQUA-RAT | AQUA-RAT-35506 | ### 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.
Company C sells a line of 25 products with an average retail price of $1,200. If none of these products sells for less than $400, and exactly 12 of the products sell for less than $1,000, what is the greatest possible selling price of the most expensive product? | [
"10,000",
"11.0",
"12,000",
"13,200"
] | D | The average price of 25 products is $1,200 means that the total price of 25 products is 25*1,200=$30,000.
Next, since exactly 12 of the products sell for less than $1,000, then let's make these 12 items to be at $400 each (min possible).
Now, the remaining 12 items cannot be priced less than $1,000, thus the minimum possible price of each of these 12 items is $1,000.
Thus the minimum possible value of 24 products is 12*400+12*1,000=$16,800.
Therefore, the greatest possible selling price of the most expensive product is $30,000-$16,800=$13,200.
Answer: D. |
AQUA-RAT | AQUA-RAT-35507 | Once again
Repetition helps too, so let’s recite: it starts with $289 = 17^2$, then continues by 102s: 391, 493. After that the twins 527, 529, followed by 629; then 667 and 697. Then two sets of twins each with its 99: 713, 731, 799; 841, 851, 899; then 901 to come after 899, and then the three sporadic values: 943, 961, 989!
Posted in arithmetic, computation, primes |
Quickly recognizing primes less than 1000: divisibility tests
I took a little hiatus from writing here since I attended the International Conference on Functional Programming, and since then have been catching up on teaching stuff and writing a bit on my other blog. I gave a talk at the conference which will probably be of interest to readers of this blog—I hope to write about it soon!
In any case, today I want to return to the problem of quickly recognizing small primes. In my previous post we considered “small” to mean “less than 100”. Today we’ll kick it up a notch and consider recognizing primes less than 1000. I want to start by considering some simple approaches and see how far we can push them. In future posts we’ll consider some fancier things.
First, some divisibility tests! We already know how to test for divisibility by $2$, $3$, and $5$. Let’s see rules for $7$, $11$, and $13$.
• To test for divisibility by $7$, take the last digit, chop it off, and subtract double that digit from the rest of the number. Keep doing this until you get something which obviously either is or isn’t divisible by $7$. For example, if we take $2952$, we first chop off the final 2; double it is 4, and subtracting 4 from $295$ leaves $291$. Subtracting twice $1$ from $29$ yields $27$, which is not divisible by $7$; hence neither is $2952$.
The following is multiple choice question (with options) to answer.
What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ? | [
"1386",
"1276",
"1566",
"1466"
] | A | LCM of 12,14,18 and 22 = 2772
least no which will be exactly divisible by 12,14,18 and 22 = 2772
2772/2 = 1386
1386 is the number when doubled 2772
=>1386 is the least exactly divisible by 12,14,18 and 22.
ANSWER A |
AQUA-RAT | AQUA-RAT-35508 | the letter B occurs 3 times. Like a "super wildcard". Not all of those are distinguishable, because some words occur more than once on a single line. (d) type and case of letters, spacing between letters and punctuation marks; (e) joining words together or separating the words does not make a name distinguishable from a name that uses the similar, separated or joined words; (f) use of a different tense or number of the same word does not distinguish one name from another;. So a reasonable guess would be that total comparable settlements would be 150 times$24 million in cash, or $1. Swap these numbers as reverse the subset. 2% more than the actual collections in January 2019, and$35 million or 1. We couldn't distinguish among the 4 I's in any one arrangement, for example. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. We can use the following formula, where the number of permutations of n objects taken k at a time is written as n P k. Permutation. 4 billion increase in dividends paid, and a $324 million decrease in proceeds from the issuance of common stock, offset in part by a$3. P (10,3) = 720. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. From a 4 billion Indian Rupee gaming industry in 2007, to a 62 billion Indian Rupees industry in 2019, gaming in India has certainly caught the eye of consumers and proves to be a valuable market today. ership as an arrangement that leverages the uniqueness of Dutch law to avoid taxes and prevent a hostile takeover attempt. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. One of these code words, the 'start signal' begins all the sequences that code for amino acid chains. Microsoft Excel. com, the international travel industry has grown from 528 million tourist arrivals in 2005 to 1. The spots can also be divided by the total of legs, ears, eyes and tail to leave a remainder of 6. In the Text section, click the WordArt option. Letter Arrangements in a Word Video. 4 percent of loan participations
The following is multiple choice question (with options) to answer.
All of the stocks on the over-the-counter market are designated by either a 3-letter or a 4-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes? | [
"27(26^3)",
"26(26^4)",
"27(26^4)",
"26(26^5)"
] | A | Number of 4-letter codes: 26 * 26 * 26 = 26^3
Number of 5-letter codes: 26 * 26 * 26 * 26 = 26^4
Total Number of codes: 26^3 + 26^4 = 26^3 + 26*(26^3) = 27*(26^3)
Therefore, the answer is A: 27*(26^3). |
AQUA-RAT | AQUA-RAT-35509 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
_________________
# Jeffrey Miller
Jeff@TargetTestPrep.com
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
### Show Tags
07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
The following is multiple choice question (with options) to answer.
One dog tells the other that there are two dogs in front of me.The other one also shouts that he too had two behind him. How many are they? | [
"3",
"4",
"1",
"6"
] | A | Explanation:
Dog
Dog
Dog
So there are 3 dogs.
They are in circle.
Answer: A |
AQUA-RAT | AQUA-RAT-35510 | qp2u, wyb, poqr, 4pzm, hm, qi, 3b, 7y88v, hez7, 8up, yh4ju, gbgk, gb, jhh, caoa,
The following is multiple choice question (with options) to answer.
L29Q,L31S,P37U,R41W,T43U,? | [
"V43Y",
"U47Y",
"V47Z",
"V47Y"
] | D | L + Q = 29
L + S = 31
P + U = 37
R + W = 41
T + W = 43
29 , 31 , 37 , 41 , 43 ---> prime number number sequence
=> next number = 47
V + Y = 47 => V47Y
ANSWER:D |
AQUA-RAT | AQUA-RAT-35511 | 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 can do a work in 6 days , B can do a work in 8 days and C can do it in 12 days. B left work after 5 days. For how many number of days should A and C should work together to complete the remaining work ? | [
"1/2",
"2/3",
"3/4",
"4/3"
] | B | b work 1/8 * 5=5/8
remaining work =1-5/8=3/8
a and c work together =1/6+1/12=3/12=1/4
take remaining work is done in =8/3 *1/4=2/3 days
ANSWER:B |
AQUA-RAT | AQUA-RAT-35512 | 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.
An art gallery owner is hanging paintings for a new show. Of the eight paintings she has to choose from, she can only hang three on the main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings? | [
"18",
"30",
"64",
"6720"
] | D | Arrangement means permutations questions..
8P3 = 8!/3! = 6720
.
ANS option D is correct answer.. |
AQUA-RAT | AQUA-RAT-35513 | One of you all sent a fairly interesting problem, so I thought I would work it out. The problem is I have a group of 30 people, so 30 people in a room. They're randomly selected 30 people. And the question is what is the probability that at least 2 people have the same birthday? This is kind of a fun question because that's the size of a lot of classrooms. What's the probability that at least someone in the classroom shares a birthday with someone else in the classroom? That's a good way to phrase as well. This is the same thing as saying, what is the probability that someone shares with at least someone else. They could share it with 2 other people or 4 other people in the birthday. And at first this problem seems really hard because there's a lot of circumstances that makes this true. I could have exactly 2 people have the same birthday. I could have exactly 3 people have the same birthday. I could have exactly 29 people have the same birthday and all of these make this true, so do I add the probability of each of those circumstances? And then add them up and then that becomes really hard. And then I would have to say, OK, whose birthdays and I comparing? And I would have to do combinations. It becomes a really difficult problem unless you make kind of one very simplifying take on the problem. This is the opposite of-- well let me draw the probability space. Let's say that this is all of the outcomes. Let me draw it with a thicker line. So let's say that's all of the outcomes of my probability space. So that's 100% of the outcomes. We want to know-- let me draw it in a color that won't be offensive to you. That doesn't look that great, but anyway. Let's say that this is the probability, this area right here-- and I don't know how big it really is, we'll figure it out. Let's say that this is the probability that someone shares a birthday with at least someone else. What's this area over here? What's this green area? Well, that means if these are all the cases where someone shares a birthday with someone else, these are all the area where no one shares a birthday with anyone. Or you could say, all 30 people have different birthdays. This is what
The following is multiple choice question (with options) to answer.
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year? | [
"1",
"2",
"3",
"4"
] | C | (C) 3 , with 2 we will have exactly 50%, so for more than 50% should be 3. |
AQUA-RAT | AQUA-RAT-35514 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
X and Y invested in a business. They earned some profit which they divided in the ratio of 2 : 6. If X invested Rs.5,000. the amount invested by Y is | [
"Rs.45,000",
"Rs.50,000",
"Rs.60,000",
"Rs.15,000"
] | D | Solution
Suppose Y invested Rs.y
Then, 5000 /y = 2 / 6
‹=› y=(5000×6 / 2).
‹=› y=15000.
Answer D |
AQUA-RAT | AQUA-RAT-35515 | Now, since you know $2a=-3$, $2b=8$ yield the right-hand side above equal to $5$, then you know $a=-3/2$, $b=4$, $k=2$ work.
-
Great explanation, I think I've got it now. Thank you. – Paul Spinelli Feb 9 '13 at 22:32
Suppose that you knew an antiderivative $F(x)$ of $f(x)$, so that $F\,'(x)=f(x)$, and
$$\int_{-3}^8f(x)\,dx=F(8)-F(-3)\;.$$
Now you calculate
$$\int_a^bkf(2x)\,dx$$
by making the substitution $u=2x$, $du=2dx$:
$$\int_a^bkf(2x)\,dx=\frac{k}2\int_{2a}^{2b}f(u)\,du=\frac{k}2\big(F(2b)-F(2a)\big)\;.$$
Thus, you want $$\frac{k}2\big(F(2b)-F(2a)\big)=5\;.$$
Since we know almost nothing about $F$, it’s clear that we need to take $b=\frac82=4$ and $a=\frac{-3}2=-\frac32$, as you did. Then we’ll have $$\frac{k}2(5)=5\;,$$
which clearly requires that $k=2$.
-
The following is multiple choice question (with options) to answer.
If f(a)=2a, what is the value of (f(a-b)+f(a))/b ? | [
"4a/b",
"3a+4b/b",
"(4a-2b)/b",
"5a-6b*b"
] | C | f(a)=2a
f(a-b)= 2(a-b)
(f(a-b)+f(a))/b = (2(a-b)+2a)/b = (4a-2b)/b
Answer is C |
AQUA-RAT | AQUA-RAT-35516 | ### 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 factors does 36^2 have? | [
"2",
"8",
"24",
"25"
] | D | 36^2 = 2^4 3^4
total factors = (4+1)(4+1) = 25.
Answer is D. |
AQUA-RAT | AQUA-RAT-35517 | Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
-
M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
The following is multiple choice question (with options) to answer.
A number is doubled and 9 is added. If resultant is trebled, it becomes 75. What is that number | [
"8",
"10",
"12",
"14"
] | A | Explanation:
=> 3(2x+9) = 75
=> 2x+9 = 25
=> x = 8
Option A |
AQUA-RAT | AQUA-RAT-35518 | 7. wayki says:
Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).
I hate imperial so please allow me to convert it to metric.
0.07366 mm = 30.480m
Multiply all of this up by 1000 =
73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?
Those that come up with 8inches a mile are much closer to the truth.
8. wayki says:
Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.
9. Alex A says:
I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???
Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.
If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.
• mathscinotes says:
Thank you. Very nicely put.
mathscinotes
• Johnny Emerson Neeley says:
I think 18 miles is level at a 6' height😜😨, on a perfect sphere, 131,477,280 ft.
10. Steve says:
So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?
• mathscinotes says:
The following is multiple choice question (with options) to answer.
If 1 kilometer is approximately 0.6 mile, which of the following best approximates the number of kilometers in 9 miles? | [
"27/5",
"26/5",
"23/5",
"22/5"
] | A | 1 km is approxmately equal to 0.6 miles
So 9 km = 9*0.6 = 5.4 miles . Multiple & divide by 10
i.e 5.4*10/10 = 54/10 = 27/5
Answer : A |
AQUA-RAT | AQUA-RAT-35519 | 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 400 meter long is running at a speed of 60 km/hour. In what time will it pass a signal post? | [
"28 seconds",
"30 seconds",
"40 seconds",
"24 seconds"
] | D | Speed = 60 Km/hr = 60*(5/18) m/sec = 150/9 m/sec
Total distance = 400 meter
Time = Distance/speed
= 400 * (9/150) = 24 seconds
Answer: D |
AQUA-RAT | AQUA-RAT-35520 | a professional dart player, has an 80% chance of hitting the bullseye on a dartboard with any throw. I tried using all the common factors for numbers 4, 9, and 13 but I cannot come up with a. Algebra 1 Common Core Answers Chapter 12 Data Analysis and Probability Exercise 12. Garrett throws a dart at a circular dart board. Some special discrete probability distributions Bernoulli random variable: It is a variable that has 2 possible outcomes: \success", or \fail-ure". Round your answer to the nearest hundredth. Slim, a professional dart player, has an 80% change of hitting the bullseye on a dartboard with any throw. pi * 10^2 = 100pi. P(white or black) 2. The power of Monte Carlo integration is that the error, given by the width of the distribution in the above plot, scales with the square root of the number of darts thrown, independent of the number of dimensions in the integral. What is the probability your dart will land closer to the center than closre to an edge?. Prerequisites. It is as if B has become the entire Sample space (i. And so we see that c = 65. For each of the five specific outcomes, you'll need to calculate the probability of achieving that outcome, the number of darts thrown to achieve that outcome, and the winnings for that outcome. pi * 20^2 = 400pi. Example: A dart is thrown at random onto a board that has the shape of a circle as shown below. The probability the dart will hit the. Give your answer as a fraction, decimal and percent. Let X and Y denote the x and y coordinates of the landing location of the dart. Find The Probability That The Dart Lands In The Shaded Circular Region. 05 otherwise. radius of circle A = 2 in. There is a very simple and very important rule relating P(A) and P(not A), linking the probability of any event happening with the probability of that same event not happening. Answer…… There is one “3” on a die and. The sum of the probabilities of all outcomes in a sample space is 1. If you throw 3 times, what is the probability that you will strike the bull's-eye all 3 times?. => 16 - 12. Geometric Probability : Example : Suppose you are considering the probability of hitting a target on a dart board. The
The following is multiple choice question (with options) to answer.
An archer is shooting arrows from various distances. She shoots at a target from 40 yards, 50 yards, and 60 yards. The probability that she hits within the bullseye from 40 yards is 4/5, the probability that she hits within the bullseye from 50 yards is 3/4, and the probability that she hits within the bullseye from 60 yards is 2/3. What is the probability that she hits the bullseye at least once in three shots, one from each distance listed. | [
"39/40",
"49/50",
"59/60",
"9/10"
] | C | The probability that she hits the bullseye at least once = 1 - (the probability that she misses the bullseye every shot)
The probability that she misses every shot = (prob that she misses from 40 yards)*(prob that she misses from 50 yards)*(prob that she misses from 60 yards)
Prob that she misses from 40 yards = 1 - 4/5 = 1/5; Prob that she misses from 50 yards = 1 - 3/4 = 1/4; Prob that she misses from 60 yards = 1 - 2/3 = 1/3
Thus the probability that she misses all shots = (1/5)*(1/4)*(1/3) = 1/60
Thus the probability that she makes at least one shot = 1 - (1/60) = 59/60
C |
AQUA-RAT | AQUA-RAT-35521 | 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.
How many seconds will a 600 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"25",
"36",
"40",
"45"
] | B | Relative speed of the train=63-3=60kmph=60*5/18=50/3 m/sec
t=600*3/50=36sec
ANSWER:B |
AQUA-RAT | AQUA-RAT-35522 | motor, hardware, dc
$$ P = \tau \times \omega$$
Mechanical power is the product of torque and angular velocity. We can calculate that for the given mechanical power of $800W$, with the target velocity of $18.45 rpm$ you will have $43.4 Nm$ of Torque available at the wheel axis. Now we have assumed $150mm$ wheel diameter, so the radius of the wheel is $75mm$. $43.4 Nm$ on a distance of $75mm$ produces $578.6 N$ of force. This is the maximum force that you can get at your wheels.
$40\%$ slope translates to $21.8 \deg$. From this we can deduce, that the sin component of your gravitational force cannot exceed $578 N$. At the limit
$$F= m * g * sin(21.8 \deg)$$
$ 578.6 N = m kg \times 9.81 m/s^2 \times 0.189$. So your absolute maximum weight from a static perspective is about $312 kg$.
Please note that this is the max weight you can move at $5km/h$ with a $1kW$ motor at a 40% slope on a perfectly flat terrain and it does not mean that with this weight your system can actually accelerate to this velocity. The more you are under this weight limit the more dynamic performance your system will have! Furthermore there might be some fictions forces to consider from the 140 cc engine that will further reduce the dynamic performance.
For the power capacity, if you put the two batteries in series, you get 24V and 18 Ah. This will mean that your batteries will be able to power your system under full load, for $18Ah / 43A = 0.41h$. So about 24 minutes of fun.
The following is multiple choice question (with options) to answer.
Fortunes, the latest SUV by Toyota Motors, consumes diesel at the rate of 1400×[1000x+x]1400×[1000x+x] litres per km, when driven at the speed of xx km per hour. If the cost of diesel is Rs 35 per litre and the driver is paid at the rate of Rs 125 per hour then find the approximate optimal speed (in km per hour) of Fortuner that will minimize the total cost of the round trip of 800 kms? | [
"34 km per hour",
"54 km per hour",
"49 km per hour",
"65 km per hour"
] | C | Given that the diesel consumption is at the rate
1400×[1000x+x]1400×[1000x+x]
Cost of diesel = Rs 35 per litre
Payment to the driver = Rs 125 per hour.
Also given that the SUV is driven at the speed of xx km per hour.
Total cost (c)(c)
=1400×[1000x+x]×800×35+125×800x=1400×[1000x+x]×800×35+125×800x
=70000x+70x+100000x=70000x+70x+100000x
Now differentiating both sides in the above equation with respect to xx.
dcdx=−170,000x2+70=0dcdx=−170,000x2+70=0
⇒ x=49 km per hour.
C |
AQUA-RAT | AQUA-RAT-35523 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
Find the annual income derived by investing $ 6800 in 40% stock at 136. | [
"550",
"2000",
"250",
"3000"
] | B | By investing $ 136, income obtained = $ 40.
By investing $ 6800, income obtained = $ [(40/136)*6800] = $ 2000.
Answer B. |
AQUA-RAT | AQUA-RAT-35524 | 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.
Youseff lives x blocks from his office. It takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work. It is takes him exactly 8 minutes more to walk to work than to ride his bike to work, then x equals? | [
"4",
"7",
"12",
"15"
] | C | Please follow posting guidelines, link is in my signatures.
As for your question, x/60 = blocks/time/block = block^2/time . This is not what you want. You are given x blocks and 60 seconds PER BLOCK. Thus you need to put it as 60*x to give you units of seconds as you are equating this to 480 (which is TIME in seconds.).
Thus the correct equation is : 60*x-20*x=480 ----> 40x=480--> x = 12.
Option C |
AQUA-RAT | AQUA-RAT-35525 | D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
The following is multiple choice question (with options) to answer.
At an examination in which full marks were 500. A got 10% less than B, B got 25% more than C and C got 20% less than D. If A got 360marks, what percentage of full marks was obtained by D? | [
"40%",
"44%",
"80%",
"90%"
] | C | Explanation:
A B C D
90 100 80 100
A D
90 ----- 100
360 ------ ? = 400
500 ------ 400
100 ------- ? => 80%
Answer: Option C |
AQUA-RAT | AQUA-RAT-35526 | Now all you have to do is to choose k so that –17+55k lies between 100 and 200.
Originally Posted by barhin
Question 2
Let X1 and X2 be two be two smallest positive integer for which the following statement is true 85x-12 is a multiple of 194 then X1+X2= (NB: The 1 and 2 in front of X1 and X2 are all subscripts X1 and X2)
For this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X – 12 = (multiple of 194). Then add suitable multiples of 194 to find the two smallest positive values for X.
3. Hello, barhin!
$\text{(1) Of all integer pairs }(x,y)\text{ that satisfy the equation: }\: 42x+55y\:=\:1$
. . $\text{only one such pair has }100
. . $\text{What is this value of }x\,?$
The problem can be solved without Euclid's algorithm.
. . But, of course, this solution takes much longer.
We have: . $42x + 55y \:=\:1$
$\text{Then: }\:x \:=\:\frac{1-55y}{42} \:=\:\frac{-42y + 1 - 13y}{42} \quad\Rightarrow\quad x \:=\:-y + \frac{1-13y}{42}$ .[1]
Since $x$ is an integer, $1-13y$ must be a multiple of 42.
. . $1-13y \:=\:42a \quad\Rightarrow\quad y \:=\:\frac{1-42a}{13} \quad\Rightarrow\quad y \:=\:-3a + \frac{1-3a}{13}$ .[2]
Since $y$ in an integer, $1-3a$ must be a multiple of 13.
The following is multiple choice question (with options) to answer.
For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 525 what is the least possible value of k? | [
"10",
"14",
"15",
"21"
] | A | 525 = 3*5*5*7
Thus k must include numbers at least up to the number 10 so that there are two appearances of 5.
The answer is A. |
AQUA-RAT | AQUA-RAT-35527 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
Three models (P, Q, and R) of cars are distributed among three showrooms. The number of cars in each showrooms must be equal and each model must be represented by at least one car in every showroom. There are 19 cars of Model P, 17 cars of Model Q, and 15 cars of Model R. What is the maximum number of cars of model P in any showroom ? | [
"17",
"16",
"15",
"14"
] | C | The total number of cars is 51. So each showroom has 17 cars(since the number of cars in each showrooms should be equal 51/3=17).
Moreover that the number of model P is maximum means that the numbers of model Q and R should be minimum. Since each model must be represented by at least one car in every showroom that minimum number should be 1.
So Maximum number of model P is 17-2= 15.
The answer is (C) |
AQUA-RAT | AQUA-RAT-35528 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 60 natural numbers? | [
"50.5",
"52",
"30.5",
"75"
] | C | Sum of first 100 natural numbers = 60*61/2 = 1830
Required average = 1830/60 = 30.5
Answer is C |
AQUA-RAT | AQUA-RAT-35529 | 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.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of B. | [
"240",
"288",
"277",
"390"
] | D | (3*8 + 2*4):(4*8 + 5*4)
8:13
13/21 * 630 = 390
Answer: D |
AQUA-RAT | AQUA-RAT-35530 | # Two screws are inspected and the first is found to be good. What is the probability that the second is also good
#### Ram123
##### New member
Question: All the screws in a machine come from the same factory but it is as likely to be from A as from factory B. The percentage of defective screws is 5% from A and 1% from B. Two screws are inspected and the first is found to be good. What is the probability that the second is also good?
----------------------------------------------------------------------------------------------------
My attempt:
G1: First screw is good. G2: Second screw is good
G1A: First screw comes from A. G1B: First screw comes from B
G2A: Second screw comes from A. G2B: Second screw comes from B
P(G1A)=1/2, P(G2A)=1/2, P(G1B)=1/2, P(G2B)=1/2
P(G1A $\cap$ G2A)=P(G1B $\cap$ G2B)=P(G1A $\cap$ G2B)=P(G1B $\cap$ G2A)=1/2*1/2=1/4
Now,
P(G1)=P(G1|G1A)P(G1A)+P(G1|G1B)P(G1B)
The following is multiple choice question (with options) to answer.
In a circuit board factory, all circuit boards that pass a verification process are certified. Every board that fails the verification process is indeed faulty, but 1/8 of those that pass are also faulty.
Approximately how many faulty circuit boards exist in a group of 2,400 circuit boards where 64 fail inspection? | [
"72",
"356",
"200",
"256"
] | B | Total of 2,400 boards. All that fail verification are indeed faulty. So the 64 are indeed faulty. 1/8 those that pass are also faulty.
From the 2,400 we know 64 fail. So 2,336 must pass. Of these 1/8 are faulty. 2,336 divided by 8 gives you 292.
What one must do now is to add to the 292 which were not detected the actually detected faulty ones, namely the 64.
Total faulty: 356.
Answer: B |
AQUA-RAT | AQUA-RAT-35531 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A, B and C start at the same time in the direction to run around a circular stadium. A completes a round in 252 seconds and C in 198 seconds, all starting at the same point. after what time will they meet again at the starting point ? | [
"26 minutes 18 seconds",
"42 minutes 36 seconds",
"45 minutes",
"46 minutes 12 seconds"
] | D | Solution
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e, 46 minutes 12 seconds. Answer D |
AQUA-RAT | AQUA-RAT-35532 | 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.
Diana is painting statues. She has 7/16 of a gallon of paint remaining. Each statue requires 1/16 gallon of paint. How many statues can she paint? | [
"7",
"20",
"28",
"14"
] | A | number of statues=all the paint÷amount used per statue
=7/16 ÷ 1/16
=7/16*16/1
=7/1
=7
Answer is A. |
AQUA-RAT | AQUA-RAT-35533 | 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 can do a piece of work in 30 days; B can do the same in 30 days. A started alone but left the work after 10 days, then B worked at it for 10 days. C finished the remaining work in 10 days. C alone can do the whole work in? | [
"30 days",
"65 days",
"86 days",
"45 days"
] | A | 10/30 + 10/30 + 10/x = 1
x =30 days
Answer:A |
AQUA-RAT | AQUA-RAT-35534 | ## [1] 22.13
### Changing the Budget
A broader social question is whether the budget is being set at an appropriate level. One way to examine this is to look at how the QALY outcome changes as the budget changes.
1. What is the rate of change of $$f_A(x_A)$$ with respect to $$x_A$$? Of course, this depends on the value of $$x_A$$, so evaluate the derivative at the optimal expenditure.
2. What is the rate of change of $$f_B(x_B)$$ with respect to $$x_B$$? Again, evaluate this at the optimal expenditure (keeping in mind that the expenditure on B is $$50-x_A$$).
3. Using the chain rule, simplify the expression $$f_A(x_A) +f_B(50-x_A)$$ and show that, if you are at the optimal values of $$x_A$$ and $$x_B$$, it must be the case that $$\frac{\partial}{\partial x_A} f_A( x_A) = \frac{\partial}{\partial x_B} f_B (x_B)$$
This suggests another way to look at the optimum, plotting out the
difference between the derivatives to find inputs $$x_A$$ and $$x_B$$
where they are equal.
dfA = D(fA(xA) ~ xA)
dfB = D(fB(xB) ~ xB)
plotFun(dfA(xA) - dfB(xB) ~ xA & xB, xA.lim = range(0, 50), xB.lim = range(0,
50))
There are many pairs of values $$(x_A, x_B)$$ on this graph where the two derivatives are equal. Find several. Then explain how the optimal value given the budget constraint $$x_A + x_B = 50$$ corresponds to just one of these.
\begin{AnswerText}
The optimal point is at the intersection of the constraint (shown in red) and the set of points where the two derivatives are equal.
The following is multiple choice question (with options) to answer.
In the new budget , the price of kerosene oil rose by 25%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase ? | [
"10",
"20",
"30",
"40"
] | B | Reduction in consumption = [((R/(100+R))*100]%
=[(25/125)*100]%=20%.
Answer is B. |
AQUA-RAT | AQUA-RAT-35535 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
At present, the ratio between the ages of Arun and Deepak is 4:3. After 7 years, Arun's age will be 19 years. What is the age of Deepak at present? | [
"16",
"14",
"12",
"9"
] | D | Let the present ages of Arun and Deepak be 4x and 3x years respectively.
Then, 4x + 7 = 19 => x = 12
Deepak's age = 3x = 9 years.
Answer: D |
AQUA-RAT | AQUA-RAT-35536 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
The area of a square field 3136 sq m, if the length of cost of drawing barbed wire 3 m around the field at the rate of Rs.1.10 per meter. Two gates of 1 m width each are to be left for entrance. What is the total cost? | [
"399",
"272",
"732.6",
"277"
] | C | Answer: Option C
Explanation:
a2 = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.1 = 732.6 Answer: C |
AQUA-RAT | AQUA-RAT-35537 | type-theory, types-and-programming-languages
&\quad\quad\;\;if\: 0 \:then \: 0 \:else \: 0\}
\end{align}$$
So $S_2$ contains 3+9+27= 39 elements.
Let $i= 2$, we will get $S_3=S_{2+1}=\cdots$, which contains 3+39*3+39^3=59439 elements.
The following is multiple choice question (with options) to answer.
If W is the set of all the integers between 69 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers? | [
"20",
"32",
"33",
"34"
] | A | Multiples of 2 from 69 to 99=Multiples of 2 from 1 to 99-Multiples of 2 from 1 to 68=[99/2]-[68/2]=49-34=15
Multiples of 3 from 69 to 99=Multiples of 3 from 1 to 99-Multiples of 3 from 1 to 68=[99/3]-[68/3]=33-23=10
Multiples of2 and 3 bothi.e.6 from 69 to 99=Multiples of 6 from 1 to 99-Multiples of 6 from 1 to 68=[99/6]-[68/6]=16-11=5
These 8 Numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3
i.e. Total Numbers in W = 15 + 10 - 5 = 20
Answer Option A |
AQUA-RAT | AQUA-RAT-35538 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
Joshua and Jose work at an auto repair center with 2 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen? | [
"1/15",
"1/12",
"1/9",
"1/6"
] | D | Two Methods
1) Probability of chosing Josh first = 1/4
Probability of chosing Jose second = 1/3
total = 1/12
Probability of chosing Jose first = 1/4
Probability of chosing Josh second = 1/3
Total = 1/12
Final = 1/12 + 1/12 = 1/6
D |
AQUA-RAT | AQUA-RAT-35539 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
The average of 7 numbers is 25. If each number be multiplied by 5. Find the average of new set of numbers? | [
"A)110",
"B)122",
"C)120",
"D)125"
] | D | Explanation:
Average of new numbers = 25 * 5 = 125
Answer: Option D |
AQUA-RAT | AQUA-RAT-35540 | The last line shows we are guaranteed to have a solution if $n (\log_2 13 - \log_2 12) \geq 1$. (Any interval of length at least $1$ contains at least one integer.) This says there is a choice of $m$ for all $n \geq \dfrac{1}{\log_2 13 - \log_2 12} = 8.659\dots$. An exhaustive search will find a solution for $n = 9$ and may have found one for a smaller $n$. Since $9$ is not a very big number, exhaustive search is feasible.
But we can do better. Note that $2^3 = 8 < 12 < 13 < 16 = 2^4$, so $m/n \in (3,4)$. Write $m = 3n + m'$ so that $0 < m' < n$ and then \begin{align*} 0.584 \ldots = \log_2 12/8 < \dfrac{m'}{n} < \log_2 13/8 = 0.700\dots \end{align*} From here, we only need to check which $n$, when multiplied by $\log_2 13/8$ have a new integer part, then check to see whether $\log_2 12/8$ was left far enough behind. (A way to do this efficiently is to use Bresenham's line algorithm for finding where the integer part increments.) We get this table: \begin{align*} n && \lfloor \log_2 &13/8 \rfloor & \lfloor \log_2 &12/8 \rfloor \\ 1 && 0 && 0 \\ 2 && 1 && 1 \\ 3 && 2 && 1 \end{align*} and we're done. We read that $n=3$ allows $m'=2$ and then $m = 3\cdot 3 + 2 = 11$.
The following is multiple choice question (with options) to answer.
Solve 2log53 X log9x + 1 = log53 | [
"3/5",
"4/5",
"4",
"4/6"
] | A | Solution:
We can rewrite the above equation in the below format
=> log532 X log9x = log53 - 1
=> log59 X log9x = log53 - log55
=> log59 X log9x = log5(3/5)
By Using the change of base rule in Left side we get
=> log5x = log5(3/5)
The value of X = 3/5
ANSWER IS A |
AQUA-RAT | AQUA-RAT-35541 | to 6. A chord is a straight line joining any two parts of the circumference. If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. To find the perimeter, P, of a semicircle, you need half of the circle's circumference, plus the semicircle's diameter: P = 1 2 ( 2 π r ) + d The 1 2 and 2 cancel each other out, so you can simplify to get this perimeter of a semicircle formula. Calculate the area of one surface of the table mat. o Pi (π) Objectives. Area of a Semicircle In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle’s radius. If someone could please help with the following question? What is the formula to calculate the distance perpendicular from the section line to the nearest edge of the circumference at any point X along the section line? Thanks in advance. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. 1 foot ≈ 0. Its unit length is a portion of the circumference. The area of a circle is. This is the perimeter of the semi-circle!. perimeter (circumference) of a semicircle: pi times radius. A segment is the shape formed between the chord and the arc. A major arc is an arc that is larger than a semicircle. 14(15) Replace d with 15. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. Find the circumference of the quad-rant with radius 4. Shapes and Figures. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Please find teh circumference of the basket so Mrs. !Shown below is a compound shape made from a rectangle and semi-circle. When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its
The following is multiple choice question (with options) to answer.
The perimeter of a semi circle is 144 cm then the radius is? | [
"22",
"28",
"98",
"37"
] | B | 36/7 r = 144
=> r
= 28
Answer: B |
AQUA-RAT | AQUA-RAT-35542 | units, geometry, dimensional-analysis
Title: Physical representation of volume to surface area I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.
My question is: what is the physical representation of length in this ratio? For the case of a sphere the ratio you found is:
$$ \frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3} $$
We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus.
One approach is then to ask "what is a function divided by its derivative". This is really similar to the area to perimeter ratio of a circle.
$$ \frac{A}{P} = \frac{ \pi R^2}{ 2 \pi R} = \frac{R}{2} $$
Of course you see the "2" because of the value of the exponent, which comes from there being two dimensions, just like the sphere. So now we have explained part of the answer, which is that the linear dimension is divided by the number of dimensions. This is still unsatisfactory because we have no clear sense of how we should define this particular "characteristic length".
One attempt at resolution of this problem would be to test the idea for a square-cube system.
$$ \frac{V}{S} = \frac{ R^3}{ 6 R^2} = \frac{R}{6} $$
$$ \frac{A}{P} = \frac{ R^2}{4 R} = \frac{R}{4} $$
The following is multiple choice question (with options) to answer.
The volumes of two cubes are in the ratio 27: 125, what shall be the ratio of their surface areas? | [
"9:22",
"9:20",
"9:25",
"9:26"
] | C | a13 : a23 = 27 : 125
a1 : a2 = 3 : 5
6 a12 : 6 a22
a12 : a22 = 9:25
Answer:C |
AQUA-RAT | AQUA-RAT-35543 | A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:
(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:
99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)
Therefore, the probability is: 24,000 / 90,000 ==4 / 15
Feb 4, 2023
The following is multiple choice question (with options) to answer.
Set S consists of numbers 2, 3, 6, 48, and 164. Number K is computed by multiplying one random number from set S by one of the first 10 non-negative integers, also selected at random. If Z=6^K, what is the probability that 678,463 is not a multiple of Z? | [
"10%",
"25%",
"50%",
"90%"
] | D | Probability of any random number being picked from set S = 1
Probability of kk not being 0 = 9/10 ( as there are total of 10 ways to pick up k and 9 ways for k not being 0)
Since it's an AND event , we will multiply the probabilities of both the events.
Hence total probability = 1∗9/10=90%. Answer:D |
AQUA-RAT | AQUA-RAT-35544 | Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers.
When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.
## Related Chapters
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The following is multiple choice question (with options) to answer.
HCF and LCM two numbers are 12 and 396 respectively. If one of the numbers is 36, then the other number is? | [
"36",
"66",
"132",
"264"
] | C | 12 * 396 = 36 * x
x = 132
ANSWER:C |
AQUA-RAT | AQUA-RAT-35545 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 8 pumps work to empty the tank in 1 day? | [
"6",
"10",
"11",
"12"
] | A | 3 pumps take 16 hrs total (8 Hrs a day)
If 1 pump will be working then, it will need 16*3=48 hrs
1 pump need 48 Hrs
If I contribute 8 pumps then
48/8=6 hrs.
answer :A |
AQUA-RAT | AQUA-RAT-35546 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
A bus leaves Los Angeles at 9:00 a.m. traveling north at 50 miles per hour. At 11:00 a.m. a plane leaves Los Angeles traveling north at 200 miles per hour. At what time will the plane overtake the bus? | [
"11:40 a.m.",
"12:15 p.m.",
"10:00 a.m.",
"11:25 a.m."
] | A | At 11:00am, the bus is 100 miles ahead of plane
The plane travels at 200 mph, which gains 150 mph on bus
100/250=2/5 hours = 40 minutes
11:00 + 0:40 = 11:40
The plane will overtake bus at 11:40am
Answer:A |
AQUA-RAT | AQUA-RAT-35547 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
What is the smallest no. that should be added to 31110 to make it exactly divisible by 9? | [
"1",
"3",
"5",
"6"
] | B | If a number is divisible by 9, the sum of its digits must be a multiple of 9.
Here, 3+1+1+1+0=6, the next multiple of 9 is 9.
3 must be added to 31110 to make it divisible by 9
B |
AQUA-RAT | AQUA-RAT-35548 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
An bus covers a certain distance at a speed of 240 kmph in 5 hours. To cover the samedistance in 1hr, it must travel at a speed of? | [
"600 km/hr",
"720 km/hr",
"730 km/hr",
"750 km/hr"
] | B | Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 hours as 5/3 hours]
Required speed = 1200 x 3 km/hr = 720 km/hr.
B |
AQUA-RAT | AQUA-RAT-35549 | 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$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 130 m long and traveling at 45 km/hr can cross in 30 sec is? | [
"300",
"230",
"245",
"250"
] | C | Speed = 45 * 5/18 = 25/2 m/sec.
Time = 30 sec
Let the length of bridge be x meters.
Then, (130 + x)/30 = 25/2
x = 245 m.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35550 | mechanical-engineering, structural-engineering, mathematics, experimental-physics
Title: Calculations of a sphere I apologize if the question is very basic, I admit I'm not the best at math and physics.
I am doing an experiment with a sphere, the measurements are as follows:
The sphere has a diameter of 330 mm (13 inches)
D1 measures 75mm
I want D2 to be attached to D1 and both ends of D2 to be attached to the sphere. D1 must be attached to one end of the sphere. I want to know how long D2 should be. How can I do that calculation?
Update: First, consider a circle with same diameter as the sphere, with the distance $d_2$ being the projection of $D_2$ onto the plane and assuming $d_1=D_1$ relative to your figure:
The radius of the circle is equal to the radius of the sphere, i.e. $r=165$mm.
We get a planar measure for $d_2$ that is
$$d_2=2\sqrt{r^2-(r-d_1)^2}$$
To get the length $D_2$ in 3D, we multiply $d_2$ by $\pi/2$
$$D_2=\pi\sqrt{r^2-(r-d_1)^2}$$
The following is multiple choice question (with options) to answer.
The curved surface of a sphere is 64 π cm2. Find its radius? | [
"9",
"8",
"4",
"5"
] | C | 4 πr2 = 64 => r = 4
Answer:C |
AQUA-RAT | AQUA-RAT-35551 | What is the probability that a spade is drawn from a pack of 50 cards with 11 spades? $11/50$.
$$\mathsf P(A\mid B\cap C) ~{= \dfrac{\mathsf P(A\cap B\cap C)}{\mathsf P(B\cap C)} \\= \dfrac{\binom{13}3/\binom{52}3}{\binom{13}2/\binom{52}{2}} \\ = \dfrac{11}{50}}$$
Reason: $\{B\cap C\}$ is the event of drawing two spades when drawing two cards from the full deck . That the position of these cards is second and third is irrelevant.
The following is multiple choice question (with options) to answer.
A card is drawn at random from a pack of 52 cards. What is the probability that it is neither a spade nor a jack? | [
"9/13",
"4/13",
"2/13",
"6/13"
] | A | There are 13 spades including a jack and 3 more jacks of other suits.
Probability of getting spade or a jack:
= (13 + 3)/52
= 16/52
= 4/13
Therefore, the probability of getting neither a spade nor a jack:
= 1 − 4/13
= 9/13
ANSWER:A |
AQUA-RAT | AQUA-RAT-35552 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
At a contest with 3000 participants, 1/2 of the people are aged 8 to 14. Next year, the number of people aged 8 to 14 will increase by 1/6. After this change, what percentage of the total 3000 people will the 8- to 14-year-olds represent? | [
"53.33%",
"50%",
"58.33%",
"55.33%"
] | C | I just wanted to mention a couple of things here:
* this is a pure ratio question; the number 3000 is completely irrelevant, and you can ignore it if you like. When we increase something by 1/6, we are multiplying it by 1 + 1/6 = 7/6, so the answer here must be (1/2)*(7/6) = 7/12 = 58.33 %.
Answer : C |
AQUA-RAT | AQUA-RAT-35553 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
Strawberries are purchased for $5/lb at the Farmer's Market. 4/5 of the strawberries will be used for making jam (1/5 of strawberries are rotten and thrown out). The strawberries are crushed and 1 lb of strawberries produces 2 cups of strawberry juice which is mixed with sugar to make the jam. If 3 cups of strawberry juice can make one jar of jam, how much money is needed to buy strawberries to make x number of jars of jam, in dollars? | [
"2x",
"75x/8",
"8x/15",
"2x/3"
] | B | Initial cost for 1 lb = $5, but ONLY 4/5 is used.
So the cost of 4/5 lb = $5, or 1 lb = $25/4,
and 1 lb of strawberries produces 2 cups of strawberry juice,so 2 cups of strawberry juice costs $25/4, but 3 cups are needed for 1 jar, so cost of 3 cups = $(25/4)(3/2) = $75/8.
If 1 jar costs $75/8, x number of jars will cost $75x/8..
Answer is B |
AQUA-RAT | AQUA-RAT-35554 | Let x and y be the dimensions of the rectangle. Then area is A=xy
But $(\frac{x}{2})^{2}+y^{2}=16$
This means $y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}$
Sub into A and we have:
$\frac{x}{2}\sqrt{64-x^{2}}$
$\frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}$
$32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}$
Max area is then $(4\sqrt{2})(2\sqrt{2})=16$
Now, try the same problem with an inscribed trapezoid
The following is multiple choice question (with options) to answer.
The parameter of a square is double the perimeter of a rectangle. The area of the rectangle is 480 sq cm. Find the area of the square? | [
"398",
"379",
"237",
"480"
] | D | Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.
Answer: D |
AQUA-RAT | AQUA-RAT-35555 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The tax on a commodity is diminished by 20% but its consumption is increased by 10%. Find the decrease percent in the revenue derived from it? | [
"62%",
"18%",
"72%",
"12%"
] | D | 100 * 100 = 10000
80 * 110 = 8800
10000------- 1200
100 ------- ? = 12%
Answer: D |
AQUA-RAT | AQUA-RAT-35556 | # Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube.
Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube. What I have so far: we express the prime factorizations of $35$ and $7$ as $5\cdot7$ and $7$, respectively. Then $n$ must be of the form $n=5^{\alpha}7^{\beta}$. Thus we see that $$35n=5^{\alpha+1}7^{\beta+1}$$ and $$\frac{n}{7}=5^{\alpha}7^{\beta-1}.$$ From here we proceed to see that if $35n$ is a perfect square, then $\alpha+1$ and $\beta+1$ are even, thus $\alpha$ and $\beta$ themselves must be odd, and so, $\alpha\equiv b\equiv1\bmod2$. Clearly, $n$ must be divisible by $7$ for $n/7$ to be an integral quantity. (This is where I'm stuck..)
I know $\alpha$ and $\beta-1$ are odd and that $n$ must be divisible by 7, however, I'm not sure what to fill in for the question marks below based on these conditions.
$$\alpha\equiv?\bmod3\quad\text{and}\quad\beta\equiv?\bmod3$$
Once I know why and how to fill in those question marks, I know it's a matter of solving a system of congruences via the chinese remainder theorem to finish the problem.
The following is multiple choice question (with options) to answer.
If N is a positive integer and 14N/60 is an integer. What is the smallest Value of N for which N has exactly four different prime factors.? | [
"30",
"60",
"180",
"210"
] | D | 14N/60 tells us that N must be multiple of 30.
Or N has at least 2 *3 * 5 in it.
To find out the smallest value of N such that it has 4 different prime factors, we will multiple 30 with next smallest prime factor which is 7 here.
So, Least value of N is 30 *7 =210 .
ANSWER:D |
AQUA-RAT | AQUA-RAT-35557 | ### Question 3
Multiple choice!
The decimal $0.23$ equals:
(A) $\dfrac {23}{10}$
(B) $\dfrac {23}{100}$
(C) $\dfrac {23}{1000}$
(D) $\dfrac {23}{10000}$
### Question 4
The decimal $0.0409$ equals:
(A) $\dfrac {409}{100}$
(B) $\dfrac {409}{1000}$
(C) $\dfrac {409}{10000}$
(D) $\dfrac {409}{100000}$
## SIMPLE FRACTIONS AS DECIMALS
Some decimals give fractions that simplify further.
For example,
$0.5 = \dfrac{5}{10} = \dfrac{1}{2}$
and
$0.04 = \dfrac{4}{100} = \dfrac{1}{25}$.
Conversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.
For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and
$\dfrac{13}{20} = \dfrac{13 \times 5}{20 \times 5} = \dfrac{65}{100} = 0.65$.
What fractions (in simplest terms) do the following decimals represent?
$0.05$, $0.2$, $0.8$, $0.004$
### Question 6
Write each of the following fractions as a decimal.
$\dfrac {2} {5}$, $\dfrac {1} {25}$, $\dfrac {1} {20}$, $\dfrac {1} {200}$, $\dfrac {2} {2500}$
### Question 7
MULTIPLE CHOICE!
The decimal $0.050$ equals
The following is multiple choice question (with options) to answer.
The decimal 0.1 is how many times greater than the decimal (0.01)^4? | [
"10^5",
"10^6",
"10^7",
"10^8"
] | C | 0.1 = 10^-1
(0.01)^4 = (10^-2)^4 = 10^-8
10^7 * 10^-8 = 10^-1
The answer is C. |
AQUA-RAT | AQUA-RAT-35558 | 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-35559 | 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 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"177 m",
"176 m",
"240 m",
"187 m"
] | C | Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Answer: C |
AQUA-RAT | AQUA-RAT-35560 | dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
_________________
-----------------------
tusharvk
Manager
Joined: 27 Oct 2008
Posts: 180
Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
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 picked at random, what is the probability that they are all blue? | [
"1/455",
"1/458",
"1/453",
"1/456"
] | A | Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that all the three marbles picked at random are blue
= ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13)
= 1/455
Answer: A |
AQUA-RAT | AQUA-RAT-35561 | # Finding total after percentage has been used?
Tried my best with the title. Ok, earlier while I was on break from work (I have low level math, and want to be more fond of mathematics)
I work retail, and 20% of taxes are taken out, and I am wanting to find out how much I made before 20% is taken out.
So I did some scribbling, and this is without googling so please be gentle. but lets say I make 2000 dollars per paycheck, so to see how much money I get after taxes I do 2000 * .8 which gives me 1600. I can also do 2000 * .2 which gives me 400 and then do 2000 – 400 (imo I don’t like this way since I have a extra unnecessary step)
Anyways, I was pondering how to reverse that to have the answer be 2000, and what I did was 1600 * 1.20 (120%) which gave me 1920, and I thought that is odd so I did 1600 * 1.25 and answer was 2000 exact.
My question is why did I have to add extra 5% (25%) to get my answer? I am sure I did something wrong, and I fairly confident the formula I am using is a big no no.
edit; wow thank you all for your detailed answers. I am starting to like mathematics more and more.
#### Solutions Collecting From Web of "Finding total after percentage has been used?"
If $20\%$ of your pay is removed for taxes, then you retain $80\%$ of it. Thus, the amount you receive after taxes is
$$\text{net} = 80\%~\text{of gross} = \frac{80}{100}~\text{gross} = \frac{4}{5}~\text{gross}$$
To determine your gross pay from your net pay, you must multiply your net pay by $5/4$ since
$$\text{gross} = \frac{5}{4} \cdot \frac{4}{5}~\text{gross} = \frac{5}{4}~\text{net}$$
The following is multiple choice question (with options) to answer.
Last year Elaine spent 20% of her annual earnings on rent. This year she earned 20% more than last year and she spent 30% of her annual earnings on rent. The amount she spent on rent this year is what percent of the amount spent on rent last year? | [
" 185.5",
" 180.0",
" 167.5",
" 172.5"
] | B | For this it is easiest to use simple numbers. Let's assume that Elaine's annual earnings last year were $100.
She would've spent $20 of this on rent.
This year she earned 20% more, or $120
She would've spent 30% of this on rent, or $36
Do $36/$20
This will give you 180%
B is the correct answer. |
AQUA-RAT | AQUA-RAT-35562 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train? | [
"150m",
"157m",
"750m",
"850m"
] | A | Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
Answer:A: |
AQUA-RAT | AQUA-RAT-35563 | c#, performance, beginner, programming-challenge
Run it, iterating through all divisibles found and all first digits of 10-digit number:
int[] primes = { 2, 3, 5, 7, 11, 13, 17 };
var divisibles = GetDivisibles(2);
for (int i = 1; i < primes.Length; i++)
{
divisibles = JoinDivisibles(divisibles, GetDivisibles(primes[i]));
}
long sum = 0;
foreach (int divisible in divisibles)
{
for (long i = 1; i <= 9; i++)
{
long number = i * 1000000000 + divisible;
if (IsPandigital(number))
sum += number;
}
}
Console.WriteLine(sum);
The following is multiple choice question (with options) to answer.
Calculate the average of all the numbers between 10 and 86 which are divisible by 9. | [
"49.5",
"41.5",
"45.5",
"40.5"
] | A | Explanation:
numbers divisible by 9 are 18,27,36,45,54,63,72,81,
Average = (18+27+36+45+54+63+72+81,) / 8 = 396/8 = 49.5
ANSWER: A |
AQUA-RAT | AQUA-RAT-35564 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two stations P and Q are 200 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet? | [
"10 am",
"12 noon",
"10.30 am",
"12.30 am"
] | B | Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 200
=> 20x + 25(x-1) = 200
=> 45x = 225
=> x= 5
Means, they meet after 5 hours after 7 am, ie, they meet at 12 noon
Answer is B. |
AQUA-RAT | AQUA-RAT-35565 | Cost of building the base in is
\begin{align}70\times ~\left( lb \right) & =70\left( 4 \right) \\ & =280 \end{align}
Cost of building the walls in is
\begin{align} 2h\left( l~+~b \right)~\times ~45&=2\times 2\left( 2+2 \right)\times 45 \\ & =720 \end{align}
Required total cost is is
$280 + 720 = 1000$
Thus, the total cost of the tank will be ₹ $$1000.$$
## Chapter 6 Ex.6.ME Question 10
The sum of the perimeter of a circle and square is $$k$$, where $$k$$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
### Solution
Let $$r$$ be the radius of the circle and $$a$$ be the side of the square.
Then, we have:
\begin{align}&2\pi r + 4a = k\\ &\Rightarrow a = \frac{{k - 2\pi r}}{4}\end{align}
The sum of the areas of the circle and the square $$\left( A \right)$$ is given by,
\begin{align}A &= \pi {r^2} + {a^2}\\ &= \pi {r^2} + \frac{{{{\left( {k - 2\pi r} \right)}^2}}}{{16}}\end{align}
Hence,
\begin{align}\frac{{dA}}{{dr}}& = 2\pi r + \frac{{2\left( {k - 2\pi r} \right)\left( { - 2\pi } \right)}}{{16}}\\& = 2\pi r - \frac{{\pi \left( {k - 2\pi r} \right)}}{4}\end{align}
Now,
The following is multiple choice question (with options) to answer.
A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 55 paise per sq m is | [
"Rs. 209.20",
"Rs. 309.20",
"Rs. 409.20",
"Rs. 509.20"
] | C | Explanation:
Area to be plastered = [2(l + b) × h] + (l × b)
= [2(25 + 12) × 6] + (25 × 12) = 744 sq m
Cost of plastering = 744 × (55/100) = Rs. 409.20 Answer: C |
AQUA-RAT | AQUA-RAT-35566 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
Working at a constant rate, P can finish a job in 3 hours. Q, also working at a constant rate, can finish the same job in 15 hours. If they work together for 2 hours, how many more minutes will it take P to finish the job, working alone at his constant rate? | [
"30",
"36",
"40",
"45"
] | B | Each hour they complete 1/3 + 1/15 = 2/5 of the job.
In 2 hours, they complete 2(2/5) = 4/5 of the job.
The time for P to finish is (1/5) / (1/3) = (3/5) hour = 36 minutes
The answer is B. |
AQUA-RAT | AQUA-RAT-35567 | This has been offset by payments, whose value at the time of the $$(k)$$-th payment is
$$\displaystyle P[(1+j)^{k-1} + (1+j)^{k-2} + \cdots + 1] = P\left[\frac{(1+j)^k - 1}{(1 + j) - 1}\right] = P\left[\frac{(1+j)^k - 1}{j}\right].$$
This means that the loan balance, immediately after your $$(k)$$-th payment is
$$\displaystyle L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right].$$
During the period between the payments $$(k)$$ and $$(k+1)$$, the interest on this loan balance is
$$\displaystyle \left\{L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right]\right\} \times j$$
$$\displaystyle = \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}.$$
Therefore, the principal reduction for payment $$(k+1)$$ is
$$\displaystyle P - \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}$$
$$\displaystyle = P - L(1 + j)^k j + P\left[(1+j)^k - 1\right]$$
Using equation (1) above, this equals
$$\displaystyle = P - P \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right](1 + j)^k j + P\left[(1+j)^k - 1\right]$$
The following is multiple choice question (with options) to answer.
A money lender lends Rs.2000 for six months at 20% p.a. rate. It the interest is reckoned quarterly than find the amount given after time limit? | [
"2205",
"2229",
"2205",
"2984"
] | A | A = 2000(21/20)2
= 2205
Answer:A |
AQUA-RAT | AQUA-RAT-35568 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
Find the least number of five digits which is exactly divisible by 12 and 18? | [
"A)1080",
"B)10080",
"C)10025",
"D)11080"
] | B | The smallest five digit numbers are
10025,10080,11080
10025 is not divisible by 12
10080 is divisible by both 12 and 18
ANSWER:B |
AQUA-RAT | AQUA-RAT-35569 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 22,140, then what is B's share in the profit? | [
"Rs. 2000",
"Rs. 3100",
"Rs. 4100",
"Rs. 5200"
] | C | Explanation :
Ratio of the initial investment = 7/2 : 4/3 : 6/5
= 105 : 40 : 36
From this ratio, we can assume that actual initial investments of A, B and C
are 105x, 40x and 36x respectively
A increases his share 50% after 4 months. Hence the ratio of their investments =
(105x * 4) + (105x * 150/100 * 8) : 40x * 12 : 36x : 12
= 105 + (105 * 3/2 * 2) : 40*3 : 36 * 3
= 105 * 4 : 40 *3 : 36 * 3
= 35 * 4 : 40 : 36
= 35 : 10 : 9
B's share = total profit * (10/54) = 22,140 * 10/54 = 4100. Answer : Option C |
AQUA-RAT | AQUA-RAT-35570 | rust, rational-numbers
#[derive(Debug)]
pub struct Fraction {
numerator: i64,
denominator: i64,
}
impl Fraction {
/// Creates a new fraction with the given numerator and denominator
/// Panics if given a denominator of 0
pub fn new(numerator: i64, denominator: i64) -> Self {
if denominator == 0 { panic!("Tried to create a fraction with a denominator of 0!") }
if denominator < 0 {
Self { numerator: -numerator, denominator: -denominator }
} else {
Self { numerator, denominator }
}
}
/// Returns a new Fraction that is equal to this one, but simplified
pub fn reduce(&self) -> Self {
// Use absolute value because negatives
let gcd = gcd(self.numerator.abs(), self.denominator.abs());
Self {
numerator: (self.numerator / gcd),
denominator: (self.denominator / gcd),
}
}
/// Returns a decimal equivalent to this Fraction
pub fn to_decimal(&self) -> f64 {
self.numerator as f64/ self.denominator as f64
}
}
impl fmt::Display for Fraction {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
let temp = self.reduce();
if temp.denominator == 1 {
write!(f, "{}", temp.numerator)
} else {
write!(f, "{}/{}", temp.numerator, temp.denominator)
}
}
}
The following is multiple choice question (with options) to answer.
Which number, when found in the denominator of any given fraction, gives a terminating decimal? | [
"75",
"80",
"85",
"90"
] | B | 75 = 3 * 5 * 5; 80 = 2^4 * 5; 85 = 5 * 17; 90 = 2 * 3^2 * 5; 95 = 5 * 19.
Only Option B has 2 and 5 as factors, hence terminating.
Option B. |
AQUA-RAT | AQUA-RAT-35571 | # 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.
an article with cost price of 180 is sold at 15% profit. what is the selling price? | [
"198",
"200",
"204",
"207"
] | D | sp=1.15*180=207
ANSWER:D |
AQUA-RAT | AQUA-RAT-35572 | 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.
Anup start a bike at 6am. and rides it at 60kmph. Raghu started at 7pm with his car at 120kmph. When will Raghu overtake Anup? | [
"8:00am",
"9:40am",
"10:00am",
"12:00pm"
] | A | Suppose bike is overtaken by car xhrs after 6am
Distance covered by the bike in x hrs = distance covered by the car in x-1 hr
60x = 120(x-1)
60x = 120
x = 2 hrs
Required time = 6+2 = 8:00am
Answer is A |
AQUA-RAT | AQUA-RAT-35573 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas? | [
"81:121",
"81:131",
"81:145",
"81:167"
] | A | Ratio of the sides
= ³√729 : ³√1331 = 9 : 11
Ratio of surface areas
= 92 : 112 = 81:121
Answer:A |
AQUA-RAT | AQUA-RAT-35574 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
The sum of three consecutive multiples of 3 is 99. What is the largest number? | [
"36",
"33",
"30",
"27"
] | A | Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 99
9x = 90
x = 10
Largest number = 3x + 6 = 36
Answer : A |
AQUA-RAT | AQUA-RAT-35575 | filters
a6 = 128 + (224 *w2 * w1 * w3) - (56 *w2 * w1*w1 * w3) + (448 *w2 * w1 * w4) - (112 *w2 * w1*w1 * w4) + (448* w1 * w3 * w4) + (224 *w2 * w3 * w4) - (640 *w1) - (320 *w2) - (320 *w3) - (640 *w4) + (64 *w2 * w1) + (112 *w2 * w1*w1) + (32 *w1*w1) - (224 *w2 * w1 * w3 * w4) - (168 *w2 * w1*w1 * w3 * w4) + (64 *w1 * w3) + (32 *w2 * w3) + (112 *w1*w1 * w3) + (128 *w1 * w4) + (64 *w2 * w4) + (64 *w3 * w4) + (224 *w1*w1 * w4) - (112 *w1*w1 * w3 * w4) + (32 *w4*w4) + (224 *w4*w4 * w1) - (56 *w4*w4 * w1*w1) - (168 *w4*w4 * w1 * w2 * w3) + (210 *w4*w4 * w1*w1 * w2 * w3) - (112 *w4*w4 * w1 * w2) - (84 *w4*w4 * w1*w1 * w2) + (112 *w4*w4 * w2) + (112 *w4*w4 * w3) - (112 *w4*w4 * w1 * w3) - (84 *w4*w4 * w1*w1 * w3) - (56 *w4*w4 * w2 * w3)
The following is multiple choice question (with options) to answer.
Find the average of the series : 4555,250,355,450,150 | [
"129",
"1152",
"267",
"207"
] | B | Average = (4555+250+355+450+150)/5
= 5760/5
= 1152
Answer: B |
AQUA-RAT | AQUA-RAT-35576 | ### Show Tags
04 Aug 2019, 03:54
generis wrote:
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?
A. 4
B. 1
C. 0
D. -1
E. -4
Don't remove brackets and solve. The question asks for the logic behind absolute value.
$$y$$ is a positive integer
$$|x| < 5 − y$$
LHS is nonnegative - it is positive or 0.
Least value for x is ZERO
Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.
The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0
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Why can't x be a negative integer, in that case it should be -1 i.e. option D.
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
08 Aug 2019, 18:54
I don't understand why x can't be negative
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Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
If x is a positive integer, then the least value of x for which x! is divisible by 1,000,0000 is?
Can someone please explain intuitively what the question is asking? | [
"5",
"9",
"25",
"15"
] | C | In order x! to be divisible by 1,000, it should have at least 3 trailing zeros. A trailing 0 in factorial of a number is produced by 2 and 5 in it: 2*5 = 10. So, we need 10 to be in x! at least in power of 3.
5! = 120 has 1 trailing zeros.
10! will have 2 trailing zeros.
15! will have 3 trailing zeros.
20! will have 5 trailing zeros.
25! will have 7 trailing zeros. C |
AQUA-RAT | AQUA-RAT-35577 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
population is 20000. Pop increases by 10% every year, then the pop after 3 years is? | [
"26630",
"26640",
"36620",
"26620"
] | D | population after 1st year = 20000*10/100 = 2000 ===> 20000+2000 = 22000
population after 2nd year = 22000*10/100 = 2200 ===> 22000+2200 = 24200
population after 3rd year = 24200*10/100 = 2420 ===> 24200+2420 = 26620
ANSWER:D |
AQUA-RAT | AQUA-RAT-35578 | 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$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is: | [
"288",
"66",
"245",
"77"
] | C | Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.Answer: C |
AQUA-RAT | AQUA-RAT-35579 | # Generalisation of $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$
The following is multiple choice question (with options) to answer.
The value of (4 3) + 2 is | [
"33",
"10",
"14",
"24"
] | C | Evaluating, (4 3) + 2 = 12 + 2 = 14.
correct answer C |
AQUA-RAT | AQUA-RAT-35580 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train crosses a platform of 150 m in 15 sec, same train crosses another platform of length 250 m in 20 sec. then find the length of the train? | [
"150",
"878",
"277",
"992"
] | A | Length of the train be ‘X’
X + 150/15 = X + 250/20
4X + 600 = 3X + 750
X = 150m
Answer: A |
AQUA-RAT | AQUA-RAT-35581 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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 ,125 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is | [
"250 m",
"245 m",
"235 m",
"220 m"
] | A | Explanation:
Assume the length of the bridge = x meter
Total distance covered = 125+x meter
total time taken = 30s
speed = Total distance covered /total time taken = (125+x)/30 m/s
=> 45 × (10/36) = (125+x)/30
=> 45 × 10 × 30 /36 = 125+x
=> 45 × 10 × 10 / 12 = 125+x
=> 15 × 10 × 10 / 4 = 125+x
=> 15 × 25 = 125+x = 375
=> x = 375-125 =250
Answer: Option A |
AQUA-RAT | AQUA-RAT-35582 | ### 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.
Three partners A,B and C invest Rs.18000, Rs.22500 and Rs.27000 respectively in a business. If total profit is Rs.18750 then B's share is (in Rs.) | [
"10000",
"6250",
"5000",
"7500"
] | B | 18000:22500:27000
4:5:6
B's share=18750*5/15=6250
ANSWER:B |
AQUA-RAT | AQUA-RAT-35583 | $$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games
The following is multiple choice question (with options) to answer.
If the probability that the Chicago Bears win the Super Bowl is 1/20 and the probability that the Chicago Cubs win the World Series is 1/7, what is the probability that either the Bears win the Super Bowl or the Cub win the World Series (but not both)? | [
"2/7",
"3/16",
"4/25",
"5/28"
] | D | 19/20*1/7 + 1/20*6/7 = 25/140 = 5/28
The answer is D. |
AQUA-RAT | AQUA-RAT-35584 | Voiceover: Easily the most quoted number people give you when they're publicizing information about their credit cards is the APR. I think you might guess or you might already know that it stands for annual percentage rate. What I want to do in this video is to understand a little bit more detail in what they actually mean by the annual percentage rate and do a little bit math to get the real or the mathematically or the effective annual percentage rate. I was actually just browsing the web and I saw some credit card that had an annual percentage rate of 22.9% annual percentage rate, but then right next to it, they say that we have 0.06274% daily periodic rate, which, to me, this right here tells me that they compound the interest on your credit card balance on a daily basis and this is the amount that they compound. Where do they get these numbers from? If you just take .06274 and multiply by 365 days in a year, you should get this 22.9. Let's see if we get that. Of course this is percentage, so this is a percentage here and this is a percent here. Let me get out my trusty calculator and see if that is what they get. If I take .06274 - Remember, this is a percent, but I'll just ignore the percent sign, so as a decimal, I would actually add two more zeros here, but .06274 x 365 is equal to, right on the money, 22.9%. You say, "Hey, Sal, what's wrong with that? "They're charging me .06274% per day, "they're going to do that for 365 days a year, "so that gives me 22.9%." My reply to you is that they're compounding on a daily basis. They're compounding this number on a daily basis, so if you were to give them $100 and if you didn't have to pay some type of a minimum balance and you just let that$100 ride for a year, you wouldn't just owe them $122.9. They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as
The following is multiple choice question (with options) to answer.
The credit card and a global payment processing companies have been suffering losses for some time now. A well known company recently announced its quarterly results. According to the results, the revenue fell to $52.0 billion from $69.0 billion, a year ago. By what percent did the revenue fall? | [
"20.8",
"24.6",
"31.8",
"32.5"
] | B | $69-$52=17$
(17/69)*100=24.6%
ANSWER:B |
AQUA-RAT | AQUA-RAT-35585 | orbit, the-sun, earth
The CSPICE libraries assume that the Gregorian
calendar reformation occurred on 4 October 1582
(meaning the day after 4 October 1582 was 15 October
1582). Looking at these lines:
SOL -13191695511.794357 A.D. 1581-12-11 20:07:27
EQU -13183992131.003845 A.D. 1582-03-10 23:57:07
SOL -13175951250.170158 A.D. 1582-06-12 01:31:48
EQU -13167875920.223862 A.D. 1582-09-13 12:40:38
SOL -13160138634.917915 A.D. 1582-12-22 01:55:23
EQU -13152434815.793312 A.D. 1583-03-21 05:52:23
SOL -13144394485.035870 A.D. 1583-06-22 07:17:53
EQU -13136319216.460808 A.D. 1583-09-23 18:25:42
you can see that the dates of the solstices/equinoxes jump
ahead 10 days per the reformation.
The CSPICE libraries use the Julian calendar prior to
4 October 1582. In reality, the Julian calendar was
introduced in 46 BCE:
https://en.wikipedia.org/wiki/Julian_calendar
Prior to 46 BCE, there were other calendar systems in
use, but the CSPICE libraries assume the Julian
calendar goes back indefinitely:
https://en.wikipedia.org/wiki/Proleptic_Julian_calendar
My calculations go back to 13201 BCE (the limits of DE431,
the ephemeris I'm using), and it's possible human beings
weren't even using calendars regularly at the time: quoting
"https://en.wikipedia.org/wiki/History_of_calendars#Prehistory":
A mesolithic arrangement of twelve pits and an arc found in
Warren Field, Aberdeenshire, Scotland, dated to roughly
10,000 years ago, has been described as a lunar calendar
and dubbed the "world's oldest known calendar" in 2013.
Notes:
The following is multiple choice question (with options) to answer.
What day of the week was 1 January 1701 | [
"Tuesday",
"Monday",
"Friday",
"Saturday"
] | D | Explanation:
1 Jan 1701 = (1700 years + 1st Jan 1701)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1700
= Number of odd days in 100 years
= 5
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
1st Jan 1701 = 1 odd day
Total number of odd days = (0 + 5 + 1) = 6
6 odd days = Saturday
Hence 1 January 1701 is Saturday.
Answer: Option D |
AQUA-RAT | AQUA-RAT-35586 | # 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.
An American cafeteria offers 3 flavors of pizza - chicken, Hawaiian and vegetarian. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ? | [
"4",
"8",
"12",
"9"
] | D | 3 flavours * 3 choices = 3C1*3C1 = 3*3=9=D |
AQUA-RAT | AQUA-RAT-35587 | solution: solution to get the median: the mean compared to sets a mean questions with solutions b Question! To new CBSE exam Pattern, MCQ questions for Class 10to help you are presented 6 ICSE.. A- a ; each being equal to the common difference we will help you sum must 12! Step 3: the given mean of a set of numbers deep understanding of maths.... Mean μ ' and the new standard deviation σ MCQ questions for Class 10to help.... To get the mean compared to sets a, a, b and C. a given data set you. Is 8 of all 5 numbers about Harmonic progression formula for nth term, sum of 5! Multiply all data values by a constant k and Calculate the mean been designed to test for understanding! Μ ' and the new standard deviation σ understanding of maths concepts b – =. Averages problem is a vey easy Question so the sequence is 2, 3 8... Mean is also called average of 56, 41, 59, 52, 42 and.... ' and the new standard deviation σ exercises on calculating the mean of given sets and word problems on mean. Set, you may need to determine the possible values of the 3 numbers in. Aggarwal Solutions Class 10 Chapter 9 - Benefits the workers earn a salary of Rs 6 to get the for. Set has a mean μ ' and the new standard deviation of the page 7 +11 ) /5 5.6! Using the average of the given classes will give the sum is 18 Divide by! Their sum must be 12, so you have x+y+z=12 ⇒ a = A- a ; each being equal the. To sets a, b are in A.P x in the middle solution: Question 21 properties... One of the 5 numbers using the average of 56, 41, 59, 52, and..., a, b and C. a given data set a = 3 if r1, r2 r3... B are in A.P a ) Calculate the mean example 2: Compute sum of set! So you have x+y+z=12 a complete solution 2 # the mean questions with solutions mean (... Lower part of the 20 people and Harmonic mean the minimum number 1! Between the first and second must be equal to the difference second and third, giving the x-y=y-z! =
The following is multiple choice question (with options) to answer.
Set S has a mean of 10 and a standard deviation of 1.5. We are going to add two additional numbers to Set S. Which pair of numbers would decrease the standard deviation the most? | [
"{2, 10}",
"{16, 16}",
"{10, 18}",
"{9, 11}"
] | D | answer B,C, and E mean is not 10 therefore not accepted . D mean is 10 but S.D is one unit. therefore accepted
answer is D mean is 10 and S.D decreases minimum.
D |
AQUA-RAT | AQUA-RAT-35588 | to 6. A chord is a straight line joining any two parts of the circumference. If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. To find the perimeter, P, of a semicircle, you need half of the circle's circumference, plus the semicircle's diameter: P = 1 2 ( 2 π r ) + d The 1 2 and 2 cancel each other out, so you can simplify to get this perimeter of a semicircle formula. Calculate the area of one surface of the table mat. o Pi (π) Objectives. Area of a Semicircle In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle’s radius. If someone could please help with the following question? What is the formula to calculate the distance perpendicular from the section line to the nearest edge of the circumference at any point X along the section line? Thanks in advance. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. 1 foot ≈ 0. Its unit length is a portion of the circumference. The area of a circle is. This is the perimeter of the semi-circle!. perimeter (circumference) of a semicircle: pi times radius. A segment is the shape formed between the chord and the arc. A major arc is an arc that is larger than a semicircle. 14(15) Replace d with 15. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. Find the circumference of the quad-rant with radius 4. Shapes and Figures. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Please find teh circumference of the basket so Mrs. !Shown below is a compound shape made from a rectangle and semi-circle. When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its
The following is multiple choice question (with options) to answer.
A semicircle has a radius of 14. What is the approximate perimeter of the semicircle? | [
"72",
"76",
"80",
"84"
] | A | The perimeter of a circle is 2*pi*r.
The perimeter of a semicircle is 2*pi*r/2 + 2r = pi*r + 2r
The perimeter is pi*14 + 2*14 which is about 72.
The answer is A. |
AQUA-RAT | AQUA-RAT-35589 | 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.
15 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would 21 women take to do the same. If 3 women do as much work as 2 men? | [
"13",
"65",
"30",
"88"
] | C | 3W = 2M
15M ------ 21 * 8 hours
21 W ------ x * 6 hours
14 M ------ x * 6
15 * 21 * 8 = 14 * x * 6
x = 30. Answer: C |
AQUA-RAT | AQUA-RAT-35590 | # Largest integer that can't be represented as a non-negative linear combination of $m, n = mn - m - n$? Why?
This seemingly simple question has really stumped me:
How do I prove that the largest integer that can't be represented with a non-negative linear combination of the integers $m, n$ is $mn - m - n$, assuming they are coprime?
I got as far as this, but now I can't figure out where to go:
$mx + ny = k$, where $k = mn - m - n + c$, for some $c > 0$
$\Rightarrow m(x + 1) + n(y + 1) = mn + c$
If I could only prove this must have a non-negative solution for $x$ and $y$, I'd be done... but I'm kind of stuck.
Any ideas?
-
I don't understand the question. Are you searching within $m.\mathbb{Z}+n.\mathbb{Z}=gcd(m,n).\mathbb{Z}$? Do you assume both $m$ and $n$ to be positive? – Olivier Bégassat Sep 23 '11 at 14:44
@OlivierBégassat: No, I'm searching for solutions in the natural numbers, not just the integers. – Mehrdad Sep 23 '11 at 14:49
@HenningMakholm: Ah, right, my bad; I forgot to mention that. Fixed, thanks. – Mehrdad Sep 23 '11 at 14:54
So you need to prove (a) $nm-m-n$ is not a non-negative linear combination ($mn$ itself fails to be a positive combination), and (b) that every $k \ge (n-1)(m-1)$ is a non-negative linear combination. Is there one of these halves you have already figured out? – Henning Makholm Sep 23 '11 at 15:10
@DJC: Nah, the original question also said positive linear combination, but only in 1 place. No worries. :) – Mehrdad Sep 23 '11 at 18:13
Here's an alternative (but perhaps more pedestrian) proof:
The following is multiple choice question (with options) to answer.
If the two-digit integers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of M and N? | [
"181",
"165",
"121",
"99"
] | A | M=10x+y
N=10y+x
M+N=11x+11y=11(x+y)
In other words the answer is a multiple of 11.
Now the question becomes " which of the following is NOT a multiple of 11?"
Answer -181
ANSWER:A |
AQUA-RAT | AQUA-RAT-35591 | 1. Four $1$'s. Every column and row contains exactly one $1$, so the number of possible arrangements equals: $$4! = 24$$
2. Six $1$'s. One column contains three $1$'s, the three others contain one $1$. Once the former $1$'s have been placed in the grid, either all three remaining $1$'s are placed in the empty row and columns, or one $1$ is placed in the empty row and columns, and two $1$'s are placed in one of the non-empty rows. The number of possible arrangements thus equals: $${4 \choose 1}{4 \choose 3}\bigg(1+{3 \choose 1}{3 \choose 2}\bigg) = 4 \cdot 4 \cdot (1 + 3 \cdot 3) = 160$$
3. Eight $1$'s. Two columns contain three $1$'s and two columns contain one $1$. The six $1$'s must be distributed over all rows, otherwise it is impossible to have the three rows contain an odd number of $1$'s. Once these $1$'s have been placed in the grid, the remaining two $1$'s must be placed in the rows which contain two $1$'s. As such, the number of possible arrangements equals: $${4 \choose 2}{4 \choose 3}{3 \choose 2}{2 \choose 1} = 6 \cdot 4 \cdot 3 \cdot 2 = 144$$
The total number of arrangements thus equals:
$$24 + 160 + 144 + 160 + 24 = 512$$
The following is multiple choice question (with options) to answer.
How many terms are there in 2,4,8,16……1024? | [
"10",
"12",
"14",
"16"
] | A | Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.
Let the number of terms be n . Then
2 x 2^n-1 =1024 or 2^n-1=512 = 2^9.
n-1=9 or n=10
Answer is A. |
AQUA-RAT | AQUA-RAT-35592 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
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.
The following is multiple choice question (with options) to answer.
Rs 50000 is divided into two parts One part is given to a person with 10% interest and another part is given to a person with 20 % interest. At the end of first year he gets profit 8000 Find money given by 10%? | [
"20000",
"40000",
"50000",
"60000"
] | A | let first parrt is x and second part is y then x+y=50000----------eq1
total profit=profit on x +profit on y
8000=(x*10*1)/100 + (y*20*1)/100
80000=x+2y-----------------------------------eq2
80000=50000+y
so y=30000 then x =50000-30000=20000
first part =20000
ANSWER:A |
AQUA-RAT | AQUA-RAT-35593 | 2) If we draw NO number '2' card then $$p_2=\frac{28^{\underline{8}}}{30^{\underline{8}}}$$
3) If we draw NO number '1' card by the $5$th draw AND NO number '2' card then, similarly to case 1), $$p_3=\frac{1}{30^{\underline{8}}}\left(26^{\underline{8}}+(3+3)\cdot 26^{\underline{7}}+3\cdot 2\cdot 26^{\underline{6}}\right)$$
Hence, the desired probability is $$p=1-(p_1+p_2-p_3)=211/1566\approx 0.134738.$$
The following is multiple choice question (with options) to answer.
Seven cards numbered from 1 to 7 are placed in an empty bowl. First one card is drawn and then put back into the bowl, and then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 9, what is the probability that one of the two cards drawn is numbered 6? | [
"1/3",
"2/3",
"1/5",
"2/5"
] | A | There are 6 ways to get a total sum of 9.
2 of these ways include the number 6.
The probability that one of the cards is a 6 is 2/6 = 1/3
The answer is A. |
AQUA-RAT | AQUA-RAT-35594 | 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 bicyclist travels uphill from town A to town B for 2 hours at an average speed of 5 miles per hour and returns along the same road at an average speed of 6 miles per hour. What is the bicyclist’s average speed for the round trip, in miles per hour? | [
"(a) 60/11",
"(b) 5",
"(c) 26/5",
"(d) 27/5"
] | A | from the uphill we can find out the distance = 5*2 = 10
average speed = total distance/total time
= 20/ (2+ (10/6)) = 60/11
Answer is A. |
AQUA-RAT | AQUA-RAT-35595 | Equilateral Triangle Calculator. Find the Length of the Side and Perimeter of an Equilateral Triangle Whose Height is √ 3 Cm. The height of an equilateral triangle of side ' a ' is given by. The special right triangle gives side ratios of , , and . A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe Thus these are properties that are unique to equilateral triangles, and knowing that any one of them is true directly implies that we have an equilateral triangle. Originally Answered: The side length of an equilateral triangle is 6 cm. Solution: According to height of equilateral triangle formula 4.1. s = … To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles. improve our educational resources. Varsity Tutors LLC An equilateral triangle also has three equal 60 degrees angles. Area of equilateral triangle = 4 3 a 2 Also, area = 2 1 × base × height = 2 1 a × height … OwlCalculator.com. Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. How to find the height of an equilateral triangle. An equilateral … The height of an equilateral triangle having each side 12cm, is (a) 6√2 cm (b) 6√3m (c) 3√6m (d) 6√6m. Therefore, the value of X will be twice the value of the height: What is the height of an equilateral triangle with a side length of 8 in? What is the triangle's height ? ... Now apply the Pythagorean theorem to get the height (h) or the length of the line you see in red. = 5.2 units approx. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Take √(3) = 1.732 And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. A triangle ABC that has the sides a, b, c,
The following is multiple choice question (with options) to answer.
the perimeter of an equilateral triangle is 45.if one of the side is the side of an isoceles triangle of perimeter 40 then what is the lenght of base of isoceles triangle. | [
"10 units",
"20 units",
"30 units",
"40 units"
] | A | base of isoceles triangle is 40-15-15= 10 units.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35596 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
What annual payment will discharge a debt of Rs. 1090 due in 2 years at the rate of 5% compound interest? | [
"993.2",
"586.21",
"534.33",
"543.33"
] | B | Explanation:
Let each installment be Rs. x. Then,
x/(1 + 5/100) + x/(1 + 5/100)2 = 1090
820x + 1090 * 441
x = 586.21
So, value of each installment = Rs. 586.21
Answer: Option B |
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