source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-34997 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 58? | [
"17",
"18",
"19",
"14"
] | A | Let the side of the square plot be a ft.
a2 = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 58 = Rs. 3944.
Answer: Option A |
AQUA-RAT | AQUA-RAT-34998 | # 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.
2 hours after train A leaves Lucknow a train B leaves the same stationtraveling in the same direction at an average speed of 36 km/hr. After traveling 6 hrsthe train B overtakes the train A. What is the average speed of the train A? | [
"22.5 km/hr",
"16 km/hr",
"27 km/hr",
"18 km/hr"
] | C | Explanation :
Total distance covered by B=36*6=216kmTotal time taken by A to cover same distance=2+6=8 hrsaverage speed of A=216/8=27 km/hr
Answer : C |
AQUA-RAT | AQUA-RAT-34999 | # 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.
The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit? | [
"2000",
"2887",
"2797",
"1172"
] | A | Explanation:
Let C.P. be Rs. x.
Then, (1920 - x)/x * 100 = (x - 1280)/x * 100
1920 - x = x - 1280
2x = 3200 => x = 1600
Required S.P. = 125 % of Rs. 1600 = 125/100 * 1600 = Rs. 2000.
Answer:A |
AQUA-RAT | AQUA-RAT-35000 | ### Show Tags
08 Feb 2012, 09:00
# of selections of books with no condition = 8C4 = 70
# of selections of books with no paperback book = 6C4 = 15
# of selections of books with at least one paperback book = 70 -15 = 55
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Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
20 May 2012, 03:36
3
The selection should at least contain one PAPERBACK book. There are a total of 2 PB and 6 HB books are available.
The combinations that at least one PB book will come out are: PHHH & PPHH
1. PHHH = 2C1 * 6C3 = 2 * (5*4) = 40
2. PPHH = 2C2 * 6C2 = 1 * ((6*5) / 2) = 15
In total 40+15 = 55 Ways
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Joined: 07 Feb 2011
Posts: 89
Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
25 Jan 2013, 07:20
Hmm....so when I did this I made an error that might have carried over from using combinatorics from probability. I found the total as 8C4 and to find the amount to subtract from it I had (6C4)(2C0). Why don't we multiply by the books not chosen and what's would it mean if we did?
I'm trying to understand a fundamental flaw I made here
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Joined: 02 Sep 2009
Posts: 52385
Re: There are 8 books in a shelf that consist of 2 paperback [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Thabo owns exactly 160 books, and each book is either paperback fiction, paperback nonfiction, or hardcover nonfiction. If he owns 20 more paperback nonfiction books than hardcover nonfiction books, and twice as many paperback fiction books as paperback nonfiction books, how many hardcover books nonfiction books does Thabo own? | [
"10",
"25",
"35",
"46"
] | B | I think we can use double-matrix method and solve using only one variable.
Our goal is to find the number of hardcover nonfiction books. Let that number be x. We are given that all 140 books are either paperback fiction, paperback nonfiction, or hardcover nonfiction. This implies that number of hardcover fiction books is 0.
Double-matrix: P = paperback; H = hardcover; F = fiction; NF = nonfiction
P H Total
F 2x+40 0
NF x+20 x
Total 3x+60 x 160
3x + 60 + x = 160
x = 25
Answer (B.) |
AQUA-RAT | AQUA-RAT-35001 | 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
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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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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
In a manufacturing plant, it takes 36 machines 4 hours of continuous work to fill 8 standard orders. At this rate, how many hours of continuous work by 72 machines are required to fill 24 standard orders? | [
"3",
"6",
"8",
"9"
] | B | the choices give away the answer..
36 machines take 4 hours to fill 8 standard orders..
in next eq we aredoubling the machines from 36 to 72, but thework is not doubling(only 1 1/2 times), = 4*48/72*24/8 = 6
Ans B |
AQUA-RAT | AQUA-RAT-35002 | It has taken a few days, but I believe I can at last answer my own question. The strategy is to show that, for every combination of the $(n,k)$ lock, there is a combination of the $(k-1,n+1)$ lock that needs the same number of turns to unlock. The argument begins by grouping lock combinations into equivalence classes in such a way that equivalent combinations require the same number of turns to open.
Assume throughout that the destination combination, that opens the lock, is $n$ zeros.
The first trick is one I used before: instead of looking at the positions of the dials, look at the differences between the positions of adjacent dials. On its own this doesn’t discard any information – the process is reversible – but it opens the door to a simplification: the order of the differences doesn’t matter. So we’ll say that two combinations are equivalent if they have the same multiset of differences, and we’ll write the differences as a nondecreasing sequence, to give a canonical form.
To motivate the next identification we’re going to make, let’s consider an example combination of the $(n=4, k=10)$-lock: 2593, which has differences 23447. The differences sum to $2k$, which means – as explained in my original blog post – that we can ignore the two largest differences and add up the others, so this combination takes $2+3+4 = 9$ turns to open. But, since the two largest differences didn’t even enter into the calculation, they could have been any pair of numbers that are both at least $4$ and sum to $11$. In particular they could have been $5$ and $6$ so that the differences were 23456. In this sense the combinations 2593 and 2594 are equivalent. We shall denote this equivalence class by the sequence (2,3,4), which we’ll call a lock sequence for the $(n, k)$-lock. Notice that the number of turns needed to open the lock is the sum of the terms of the lock sequence.
The following is multiple choice question (with options) to answer.
A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most? | [
"176",
"178",
"215",
"197"
] | C | Since each ring consists of six different letters, the total number of attempts possible with the three rings is
= 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts
= 216 - 1 = 215.
Answer: C |
AQUA-RAT | AQUA-RAT-35003 | If you role two dice, then there are 36 equally likely outcomes, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Of those, 11, the last of each of those six lines and all of the last, have at least one 6 so the probability that "one or both rolls is a 6" is 11/36. Your "13/36" is wrong because both of the "1/6" In your sum include "(6/6)". It should be 1/6+ 1/6- 1/36 where the 1/36 being subtracted is to take away one of the (6, 6)s.
The general rule for "Probability of A or B" is "Probability of A"+ "Probability of B" minus "Probability of A and B".
When you roll two dice, the probability the first die is even is 1/2, the probability the second die is 1/2, and the probability both are even is (1/2)(1/2)= 1/4 (the results of the two rolls are independent) so the probability that either one or both are even is 1/2+ 1/2- 1/4= 3/4.
The following is multiple choice question (with options) to answer.
If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is? | [
"1/98",
"1/2",
"1/6",
"1/2"
] | B | The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Answer: B |
AQUA-RAT | AQUA-RAT-35004 | problem was so easy...just plug and play !! Kudos [?]: 7 [0], given: 0 Manager Joined: 16 May 2011 Posts: 71 Kudos [?]: 18 [0], given: 2 Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:06 if we plug and play, why can't we test X = 1? then y/n.. Kudos [?]: 18 [0], given: 2 Manager Joined: 21 Feb 2012 Posts: 116 Kudos [?]: 148 [0], given: 15 Location: India Concentration: Finance, General Management GMAT 1: 600 Q49 V23 GPA: 3.8 WE: Information Technology (Computer Software) Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 24 Feb 2012, 10:46 The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even? 1) xz is even 2) y is even. Please explain. thanks.[/quote] Ans. let us take any 3 numbers, say x=3,y=18,z=54, or x=2,y=4,z=20 1)if xz is even then it means that either x or z is even,say that x is even, now there is no even number which is a factor of odd number, so z is definitely even, now if x is odd as in the above case, still then we can point out that z is even. 2)if y is even then it is clear that z will be even. Thus this question could be answered by any of the two questions. Kudos [?]: 148 [0], given: 15 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7670 Kudos [?]: 17334 [1], given: 232 Location: Pune, India Re: The positive integers x, y, and z are such that x is a [#permalink] ### Show Tags 27 Feb 2012, 05:20 1 This post received KUDOS Expert's post dchow23 wrote: if we plug and play, why
The following is multiple choice question (with options) to answer.
If x and y are positive integers such that x < y and z=x/y, which of the following must be true?
I. z>(x − 1)/(y + 1)
II. z<(x + 1)/(y + 1)
III. z>(x + 1)/(y - 1) | [
"I only",
"I and II",
"II and III",
"II only"
] | B | x < y --> Let x = 2 and y = 3 --> z = 2/3
I. z>(x − 1)/(y + 1) --> 2/3 > 1/4 --> True
II. z<(x + 1)/(y + 1) --> 2/3 < 3/4 --> True
III. z>(x + 1)/(y - 1) --> 2/3 > 3/2 --> False
Answer: B |
AQUA-RAT | AQUA-RAT-35005 | • $P_2$ 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.
The first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 50%?
I will see what is the quickest way to solve it then I will provide the explanation | [
"3",
"7",
"9",
"10"
] | A | The following approach might be the easiest one and less error prone.
We need on-time departure rate to be higher than 5/10, so it should be at least 6/11, which means that 6 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 6-3=3 subsequent flights need to depart on-time.
Answer: A |
AQUA-RAT | AQUA-RAT-35006 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
Find the remainder of the division (2^14)/7. | [
"1",
"2",
"3",
"4"
] | D | Find the pattern of the remainders after each power:
(2^1)/7 remainder 2
(2^2)/7 remainder 4
(2^3)/7 remainder 1 -->this is where the cycle ends
(2^4)/7 remainder 2 -->this is where the cycle begins again
(2^5)/7 remainder 4
Continuing the pattern to (2^14)/7 gives us a remainder of 4
Final Answer:
D) 4 |
AQUA-RAT | AQUA-RAT-35007 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
A number when divided by 296 leaves 75 as remainder. When the same number is divided by 37, the remainder will be: | [
"5",
"4",
"1",
"2"
] | C | Explanation:
Let x = 296q + 75
= (37 x 8q + 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.
Answer C |
AQUA-RAT | AQUA-RAT-35008 | 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.
Six years ago Rahul was P times as old as Sachin was. If Rahul is now 17 years old, how old is Sachin now in terms of P ? | [
"11/P + 6",
"P/11 +6",
"17 - P/6",
"17/P"
] | A | let's call Sachin age six years ago is x
so six years ago Rahul was: 17-6 = 11 years old
and six years ago Rahul was P times as old as Sachin was
=> 11/x = P
=> x = 11/P
=> Now Sachin's age will be : x+6 = 11/P+6
=> the answer is A |
AQUA-RAT | AQUA-RAT-35009 | # Evaluate the sum of the reciprocals
#### anemone
##### MHB POTW Director
Staff member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
#### mente oscura
##### Well-known member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
Hello.
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$
$$=qrs+prs+pqs+pqr=$$
$$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$
$$=-rrs-srs+pqs+pqr$$, (*)
$$(p+q)^3=-(r+s)^3$$
$$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$
$$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$
$$661+p^2q+pq^2=-r^2s-rs^2$$, (**)
For (*) and (**):
The following is multiple choice question (with options) to answer.
P, Q and R have Rs.7000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R? | [
"Rs.3000",
"Rs.3600",
"Rs.2400",
"Rs.2800"
] | D | Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(7000 - r) => 3r = 14000 - 2r
=> 5r = 14000 => r = 2800.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35010 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is? | [
"37.9",
"37.2",
"37.5",
"37.1"
] | C | Average speed = (2xy) /(x + y) km/hr
= (2 * 50 * 30) / (50 + 30) km/hr.
37.5 km/hr.
Answer: C |
AQUA-RAT | AQUA-RAT-35011 | # Maximum value of $ab+bc+ca$ given that $a+2b+c=4$
Question:
Find the maximum value of $ab+bc+ca$ from the equation,
$$a+2b+c=4$$
My method:
I tried making quadratic equation(in $b$) and then putting discriminant greater than or equal to $0$. It doesn't help as it yields a value greater than the answer.
Thanks in advance for the solution.
• Are the variables assumed to be positive? – Dr. Sonnhard Graubner Sep 7 '18 at 15:55
• No. It is not given. So no. Whenever they are positive, it gets mentioned in the question. – Love Invariants Sep 7 '18 at 15:56
• @LoveInvariants What is even nicer is that $4$ is the only positive integer that satisfies $$ab+bc+ca=a+2b+c\quad\ddot\smile$$ – TheSimpliFire Sep 7 '18 at 16:35
• @TheSimpliFire- Thats a very nice observation. 😊 – Love Invariants Sep 7 '18 at 16:53
Without using calculus:
Substituting $c=4-2b-a$, we get $$ab+bc+ca=ab+(a+b)(4-2b-a)=4(a+b)-(a+b)^2-b^2$$ and since $f(x)=4x-x^2=4-(x-2)^2$ has maximum at $(2,4)$, we have that $\max\{a+b\}=2$ so $$ab+bc+ca\le4-b^2$$ and the minimum value of $b$ is zero, yielding a maximum value of $4$.
• I was searching for something like this. Thanks. – Love Invariants Sep 7 '18 at 16:07
Hint: Plugging $$c=4-a-2b$$ into your given term we get
$$f(a,b)=-a^2+4a-2ab-2b^2+4b$$ and now differentiate this with respect to $a,b$
The following is multiple choice question (with options) to answer.
Let a,b,c,d,e,f be non-negative real numbers. Suppose that a + b + c + d + e + f = 1 and
ad + be + cf >= 1/18 . Find the maximum value of ab + bc + cd + de + ef + fa. | [
"7/36",
"6/36",
"5/36",
"4/36"
] | A | We partially factor
ab + bc + cd + de + ef + fa = a(b + f) + c(b + d) + e(d + f):
Notice that the terms in the parenthesis are very close to (b+d+f), except each part is missing
one term. We can add ad + cf + eb to this expression to get
a(b + f) + c(b + d)e(d + f) + ad + cf + eb
= a(b + d + f) + c(b + d + f) + e(b + d + f)
= (a + c + e)(b + d + f)
Let x = a+c+e and y = b+d+f. We have x+y = 1, so AM-GM gives us x+y
2
p
xy, hence
(a + c + e)(b + d + f) - 1
4 . Subtracting away ad + be + cf - 1
18 gives us
ab + bc + cd + de + ef + fa -
1/4
-1/18
=
7/36
where equality holds when a + c + e = b + d + f = 1
2 and ad + be + cf = 1/18 . We can verify that
equality holds when a = e = 1/9 , b = d = 1/4 ,c = 5/18, f = 0.
correct answer A |
AQUA-RAT | AQUA-RAT-35012 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
A basket contains 3 blue, 3 red and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted? | [
"2/21",
"3/25",
"1/6",
"9/28"
] | D | Probability of getting each color marble = (3C1*3C1*3C1)/9C3
= (3*3*3)/84
= 27/84 = 9/28
So answer is D |
AQUA-RAT | AQUA-RAT-35013 | 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.
You walk one mile to school every day. You leave home at the same time each
day, walk at a steady speed of 3 miles per hour, and arrive just as school begins. Today you
were distracted by the pleasant weather and walked the first half mile at a speed of only 2
miles per hour. At how many miles per hour must you run the last half mile in order to arrive
just as school begins today? | [
"4",
"6",
"8",
"10"
] | B | Recall the formula s = v *t, where in this case, s is distance (in miles) traveled at a
constant speed of v (miles per hour), for a time of t hours. The first statement indicates that
it usually takes you only 20 minutes to go to school (solve for t in 1 = 3 * t and change units)
Today, it took you 15 minutes to cover the first half mile (solve for t in 1=2 = 2 * t and
change units). You only have 5 more minutes to arrive (that is 1/12th of an hour), and need
to cover another half mile. Solve for v in the equation 1/2 = (1/12) *v.
correct answer B |
AQUA-RAT | AQUA-RAT-35014 | 2. How many ways can one choose three distinct characters to be placed into the forms from above.
3. Watch out for any duplication or overcounting.
There are two options (using Joe's suggestion in the comments):
• $\{a,b,c,c,c\}$. We can choose $3$ of the $5$ positions for the $c$'s, then there are two options for the $a$ and $b$ ($a$ first and then $b$ or the other way around). This gives $$2\binom{5}{3}=20$$ ways to write $a$, $b$, and $c$.
• $\{a,b,b,c,c\}$. We can choose $1$ of the $5$ positions for the $a$, and then $2$ of the remaining $4$ positions for the $b$'s. This gives $$5\binom{4}{2}=30$$ ways to write $a$, $b$, and $c$.
Now, there are $10\cdot 9\cdot 8=720$ ways to choose three digits for the positions $a$, $b$, and $c$. So, one might expect $$720\cdot (20+30)=36000$$ arrangements. The problem with this is that there is some overcounting since $abbcc$ and $accbb$ result in the same arrangement when the numbers in $b$ and $c$ are also flipped. This occurs for $a\leftrightarrow b$ in the first option and $b\leftrightarrow c$ in the second form. Therefore, we've overcounted by a factor of $2$ and there are $18000$ total arrangements. (Note that $720\cdot 30/2=10800$ matching @Green's approach).
The following is multiple choice question (with options) to answer.
In how many ways letters of the word LICENSING be arranged? | [
"362880",
"12540",
"65320",
"48962"
] | A | LICENSING has 9 letters, out of which I repeated 2 times, N repeated 2 times. Hence total ways = 10!/(2!*2!)= 362880 ways
A |
AQUA-RAT | AQUA-RAT-35015 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
A ratio between two numbers is 3 : 4 and their L.C.M. is 180. The first number is | [
"60",
"45",
"20",
"15"
] | B | Sol.
Let the required numbers be 3x and 4x. Then, their L.C.M. is 12x.
∴ 12x = 180⇔ x = 15. Hence, the first number is 45.
Answer B |
AQUA-RAT | AQUA-RAT-35016 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
A technician makes a round-trip to and from a certain service center by the same route. If the technician completes the drive to the center and then completes 50 percent of the drive from the center, what percent of the round-trip has the technician completed? | [
"70",
"65",
"60",
"75"
] | D | round trip means 2 trips i.e.to and fro. He has completed one i.e 50% completed. then he traveled another 50% of 50% i.e 25%. so he completed 50 +25 =75 % of total trip.
D |
AQUA-RAT | AQUA-RAT-35017 | 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.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is? | [
"4:9",
"4:3",
"4:5",
"4:1"
] | B | Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √16 : √9
= 4:3
Answer:B |
AQUA-RAT | AQUA-RAT-35018 | ### 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 shared profits in ratio of 3 : 7 : 6. Theypartnered for 14months, 8months and 7months respectively. What was he ratio of their investments ? | [
"12 : 49 : 64",
"20 : 49 : 64",
"10 : 19 : 64",
"20 : 29 : 64"
] | A | Simply multiply profit sharing ratio with investment ratio to get investment amount ratio.
Let X is the total investment
⇒14 x = 3
⇒ 8 x = 7
⇒ 7x = 8
⇒ Final investment ratio = 12 : 49 : 64
A |
AQUA-RAT | AQUA-RAT-35019 | 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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Abhishek....
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
Machine A can put caps on a set of bottles in 8 hours at a constant rate. If Machine A does the job for 6 hours and Machine B does the rest of the job, which works at 1/4 constant rate of Machine A. How long will it take for Machine B alone to do the rest of the job? | [
"4hrs",
"6hrs",
"8hrs",
"10hrs"
] | C | machine A will do 6/8 in 6 hrs, so A does 3/4 of the work ..
Therefore, B will do the remaining 1/4 th work alone..
As the speed of B is 1/4 rate of A, B will do the 1/4th work in same time that A takes to complete full job...
ans 8
C |
AQUA-RAT | AQUA-RAT-35020 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If x is an integer, then x(x - 1)(x - v) must be evenly divisible by three when v is any of the following values EXCEPT | [
"-4",
"-2",
"-1",
"2"
] | B | here's another approach
x(x - 1)(x - v)
all three are consecutive, so the product MUST be a multiple of 3
we don't know the value of v just yet ... so let's extend the series ... the extension itself reveals the answers
..(x-5)..(x-2)(x-1)x(x+1)..(x+4)..
we can see the possible values of v too from the series
v = 2 OR 2+3n [25]
v = -1 OR -1+3n [-1-4]
B i.e. -2 does not fit in any value of v
so B it is |
AQUA-RAT | AQUA-RAT-35021 | Combining the two cases:
4% of integers satisfy the condition.
Using the well-known summation formula, $$\sum_{i=1}^n i = \frac{(n+1)n}{2}$$, the question is equivalent to how often $$\sum_{i=1}^n i$$ is a multiple of $$50$$. To get a multiple of $$100$$, $$n$$ and $$n+1$$ together must contain two factors of $$5$$. Since $$n$$ and $$n+1$$ cannot both be a multiple of $$5$$, at least one of them is a multiple of $$25$$.
Now observe that
The following is multiple choice question (with options) to answer.
One fourth of one third of two fifth of a number is 35. What will be40% of that number | [
"A)140",
"B)420",
"C)180",
"D)200"
] | B | Explanation:
(1/4) * (1/3) * (2/5) * x = 35
then x = 35 * 30 = 1050
40% of 1050 = 420
Answer: Option B |
AQUA-RAT | AQUA-RAT-35022 | 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 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
An internet service provider charges $3.10 for the first 1/5 of a day plus $0.40 for each additional 1/5 of a day. What would this company charge for an internet usage that was 8 days long? | [
"$15.60",
"$16.00",
"$17.80",
"$18.70"
] | D | first 1/5 day = $3.10
rest of the days = 8 - (1/5) = 39/5
charge for the rest of the days = 39 *0.4 = 15.6
total charge = 3.10+15.6 = 18.7
Answer is D. |
AQUA-RAT | AQUA-RAT-35023 | acid-base, ph, titration
The calculation
There are two steps. You missed one (the one that involves the equilibrium). Let's work with equations A1 and A2 to work through the calculations.
You had the correct intuition about what happens with the hydrochloric acid: When adding 25 mL ammonia solution to the hydrochloric acid solution, the reactants are present at stoichiometric ratio, and the only species that is left to influence the pH is ammonium cation. When adding 26 mL ammonia solution, however, the ammonia is in excess, so some of it turns into the ammonium cation, and some remains. Scenario (i) is called a weak acid in solution, scenario (ii) is called a buffer solution. We can ignore the chloride ion, it is a spectator ion.
The amount of hydrochloric acid present initially can be determined from initial concentration and volume:
$$n_{\mathrm{\ce{HCl},initial}} = c_{\mathrm{\ce{HCl},initial}} \cdot V_{\mathrm{\ce{initial}}}$$
$$\ \ \ =0.10\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.025\ \mathrm{L}$$
$$\ \ \ =2.5\times 10^{-3}\ \mathrm{mol}$$
The following is multiple choice question (with options) to answer.
The ratio of sodium chloride to ammonium in 100 kg of mixed ammonium normally used by three chemists is 1: 3. The amount of sodium chloride to be added to 100 kg of mixed ammonium to make the ratio 9 :25 is | [
"2 kg",
"6.625 kg",
"6.25 kg",
"6.35 kg"
] | A | sodium chloride in 100 kg = 1/4 x 100 = 25 kg
ammonium in 100 kg = 3 / 4 x 100 = 75 kg
Now 75 is 25 parts out of (25 + 9) = 34 parts.
Hence 9 parts = 9/25 x 75 = 27
Amount to be added = 27 - 25= 2 Kg
ANSWER:A |
AQUA-RAT | AQUA-RAT-35024 | 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.
There are 172 lights which are functional and each is controlled by a separate On/Off switch. Two children A and B start playing with the switches. A starts by pressing every third switch till he reaches the end. B, thereafter, presses every fifth switch till he too reaches the end. If all switches were in Off position at the beggining, How many lights are switched On by the end of this operation? | [
"83",
"85",
"87",
"79"
] | D | Editing my solution:
Number of switches = 172
Number of switches turned on by A: 3, 6, ... 171 = 57
Number of switches turned on by B: 5, 10, ....170 = 34
Few switches are turned on by A and later turned off by B: LCM(3,5) = 15x = 15, 30,....90 = 6.
Subtract the above 6 switches from both A and B as they are turned off.
Number of switches that are turned on = (57- 6) + (34 - 6) = 79
Answer: D |
AQUA-RAT | AQUA-RAT-35025 | general-relativity, spacetime, metric-tensor, coordinate-systems, matrix-elements
It has side lengths $a$ and $b$, and an angle $\theta$ between the two sides.
Then, according to the law of cosines,
the diagonal $\Delta s$ of this unit-parallelogram is given by
$$(\Delta s)^2=a^2+b^2+2ab\cos\theta$$
For a general parallelogram (with arbitrary coordinate
differences $\Delta x_1$ and $\Delta x_2$) its diagonal length $\Delta s$
is given by
$$(\Delta s)^2=a^2(\Delta x_1)^2+b^2(\Delta x_2)^2+2ab\cos\theta\ \Delta x_1\ \Delta x_2$$
That means the metric is
$$g_{\mu\nu}=\begin{bmatrix} g_{11} & g_{12} \\ g_{21} & g_{22} \end{bmatrix}
=\begin{bmatrix} a^2 & ab\cos\theta \\ ab\cos\theta & b^2 \end{bmatrix}$$
which gives you the geometric meanings of all the metric components.
Now you can repeat this reasoning in 3-dimensional space
(using a parallel-epiped instead of a parallelogram)
or in 4-dimensional space-time.
The diagonal metric components $g_{ii}$
give the square of the length of the unit-parallel-epiped
in $x_i$ direction.
And the off-diagonal metric components $g_{ij}$ are related
to the angle between the $x_i$ and $x_j$ direction.
$g_{ij}=0$ : they are orthogonal,
$g_{ij}>0$ : there is an acute angle ($<90°$) between them,
$g_{ij}<0$ : there is an obtuse angle ($>90°$) between them.
The following is multiple choice question (with options) to answer.
What is the measure of the angle W made by the diagonals of the any adjacent sides of a cube. | [
"30",
"45",
"60",
"75"
] | C | C.. 60 degrees
All the diagonals are equal. If we take 3 touching sides and connect their diagonals, we form an equilateral Triangle. Therefore, each angle would be 60.C |
AQUA-RAT | AQUA-RAT-35026 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. | [
"48.55",
"49",
"51",
"61"
] | A | Required average
= (50.25 x 16 + 45.15 x 8)/(16 + 8)
= (804 + 361.20)/24
= 1165.20/24
= 48.55
Answer is A |
AQUA-RAT | AQUA-RAT-35027 | • In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52
The following is multiple choice question (with options) to answer.
The sum of the heights of two high-rises is x feet. If the second high rise is 37 feet taller than the first, how tall will the first high rise be after they add an antenna with a height of z feet to the top? | [
"(x+z)/2 + 37",
"2x−(37+z)",
"(x−37)/2 + z",
"x/2 - 37 + z"
] | C | I will note h1 the height of high-rise 1 and h2 the height of high-rise 2. SO:
h1 + h2 = x
and h2 = h1 + 37 =>
Q: h1+ z = ?
h21+ h1 + 37 = x => 2h1= x-37 =? h1 = (x-37)/2
=> h1+ z = (x-37)/2 + z, CORRECT ANSWER C |
AQUA-RAT | AQUA-RAT-35028 | We have 8 members and a 3 slots. So total number of ways is 8*7*6, divided by 3! for double counts. 8*7*6/3!=56 .
Now what are the number of ways can we have two teams members on the committee?
We can pick any of the 8 for the first slot. The second slot is reserved for the team member of the first slot, so 1 ways. And the last slot can be any of the remaining 6.
So 8*1*6. But we have to account for double counts. Do we divide by 3! or 2!? This is were I am stuck. If we divide by 2!, we get 8*1*6/2=24
56-24=32, our answer. But I don't understand why we divide by 2!??
I hope I was a little more clearer. Thank you again for responding Bunuel!
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Re: Committee Combination Tough Problem [#permalink]
### Show Tags
02 Oct 2012, 04:36
Bunuel wrote:
alphabeta1234 wrote:
A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A)20
B)22
C)26
D)30
E)32
I'm not sure what are you exactly dong in your last approach.
Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.
The following is multiple choice question (with options) to answer.
In how many ways, a committee of 12 members can be selected from 8 men and 7 ladies, consisting of 7 men and 5 ladies? | [
"150",
"167",
"200",
"168"
] | D | (7 men out 8) and (5 ladies out of 7) are to be chosen
Required number of ways = 8C7*7C5 = 168
Answer is D |
AQUA-RAT | AQUA-RAT-35029 | III. All faces same There are 2 colors, so two ways for all faces to be the same.
Adding them up, we have a total of 20 ways to have four vertical faces the same color. The are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$
The following is multiple choice question (with options) to answer.
61 small identical cubes are used to form a large cube. How many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | [
"64",
"128",
"152",
"155"
] | D | 61 small cube will make a large cube with 4 cubes in each line i.e.
Adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line.
Total number of small cubes in new cube = 6^3 = 216
Extra cube required = 216 - 61 = 155
Hence, D is the answer. |
AQUA-RAT | AQUA-RAT-35030 | # conditional probability trick questions - drawing cards from a deck and the meaning of 'at least'
Suppose that a box contains one blue card and four red cards, which are labeled A, B, C, and D. Suppose also that two of these five cards are selected at random, without replacement.
a. If it is known that card A has been selected, what is the probability that both cards are red?
b. If it is known that at least one red card has been selected, what is the probability that both cards are red?
I have been assigned the problem above in a class. I am aware that the book considers the answer for A) to be 3/4, which is obvious. However, the answer to B) is claimed to be $$P(red_1)P(red_2|red_1) = 4/5 \cdot 3/4 = 3/5$$
The professor has not been able to explain why the answers to A and B are different in any way that makes sense to me. There is no way in which you cannot draw at least 1 red card in a draw of 2 cards. If you are told you have drawn at least 1 red card, then you have the same information that you did in A.
• The box has one blue card and 4 red cards? In this case wouldn't (b) premise - that is at least one red card was selected - always be true? It seems like you can never select two cards without at least one of them being red. – Karolis Koncevičius Oct 29 '14 at 18:19
• @Karolis Is that a problem? – whuber Oct 29 '14 at 18:24
• @whuber No, not a problem, sorry. I just thought maybe there was an error in formulation. Not used seeing Pr(something | whole_space) kind of questions. – Karolis Koncevičius Oct 29 '14 at 18:36
• To clarify, the red cards are labeled A, B, C, and D, and the blue card is unlabeled (or labeled something other than A, B, C, or D)? – Hao Ye Oct 29 '14 at 23:42
Look at the possible samples consistent with the information, each of which is equally likely:
The following is multiple choice question (with options) to answer.
A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 6 points if you must draw exactly 3 “low” cards? | [
"1",
"20",
"3",
"4"
] | B | Great question Ravih. This is a permutations problem (order matters) with repeating elements. Given thatlowcards are worth 1 pt andhigh cards2 pts, and you must draw 3 low cards, we know that you must also draw 1 high card. The formula for permutations problems with repeating elements isN!/A!B!...where N represents the number of elements in the group and A, B, etc. represent the number of times that repeating elements are repeated. Here there are 4 elements and thelowcard is repeated 3 times. As a result, the formula is:
5!/3! which represents (5*4*3*2*1)/(3*2*1) which simplifies to just 20, giving you answer B. |
AQUA-RAT | AQUA-RAT-35031 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 3 different offices? | [
"5",
"6",
"27",
"8"
] | C | Ans. Each employee can go into any of the two offices. Thus we have
=> 3 * 3 * 3= 27
Answer : C |
AQUA-RAT | AQUA-RAT-35032 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A person purchased a TV set for Rs. 16000 and a DVD player for Rs. 1800. He sold both the items together for Rs. 31150. What percentage of profit did he make? | [
"30%",
"75%",
"40%",
"70%"
] | B | The total CP = Rs. 16000 + Rs. 1800 = Rs. 17800 and SP = Rs. 31150
Profit(%) = (31150 - 17800)/17800* 100 = 75%
ANSWER:B |
AQUA-RAT | AQUA-RAT-35033 | So by my equation I get that they are of equal purity.
Kalyan.
4. ## Re: Purity
Hello, PASCALfan!
Two vessels contains equal amount of water and beer.
1 spoon beer is added to the water and mixed well.
From this mixture 1 spoon is added to the beer.
Which of the liquids is more pure?
This is the classic "Wine and Water" problem.
As long as the amount transferred each time remains the same,
. . the final concentrations are equal.
This is the basis of a stunning card trick.
You are seated at a small table across from your volunteer.
You show him a deck of cards, fanning them
. . so he can see they are all facing one way.
Both of you place your hands under table.
You hand him the deck.
Instruct him to select a secret number from 1 to 20,
. . and keep it a secret. .Call it n.
Have him count off the top n cards,
. . turn them over and replace them on the deck.
Have him shuffle the deck thoroughly.
Then count off the top n cards
. . and hand them to you under the table.
Now you remind him:
. . you don't know his secret number,
. . he doesn't know how many face-up cards he has,
. . you don't know how many face-up cards you have.
Despite this, you will try to make your number of face-up cards
. . equal his number of face-up cards.
He hears the sound of cards being counted and moved.
After a few moment, you bring your cards to the top of the table
. . and count the face-up cards.
Instruct him to do the same . . . and the numbers match!
[Acknowledge the thunderous applause ... modestly, of course.]
Spoiler:
Secret: When he hands you the packet, turn it over.
5. ## Re: Purity
In fact, the answer does not change regardless of how well the liquids were mixed in the process. Indeed, the volume of the liquid in each vessel after the two operations is the same as it was initially. Therefore, in the end the volume of beer in water has to equal the volume of water in beer.
This is one of my favorite math problems because it shows that there is more to math than handling large formulas.
6. ## Re: Purity
Thanks for the card trick..
Regards
The following is multiple choice question (with options) to answer.
A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required? | [
"35",
"37",
"42",
"30"
] | B | The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 80/16 = 5
Number of cans of Pepsi = 144/16 = 9
Number of cans of Sprite = 368/16 = 23
The total number of cans required = 5 + 9 + 23 = 37 cans.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35034 | The second one gives your solution $(a,b,c,d)=(8,31,36,39)$, but I don't see a way to rule out finding something better other than exhaustively searching.
The third one gives the near miss $(a,b,c,d)=(10,23,31,42)$.
• I didn't know that about "best rational approximation", thanks. I will finally use $(10,13,23,32)$, because $39$ teeth was too much. And the error is still small. And I found these solutions by exhaustive search,not anything special. Thanks for the information, I will accept this answer. – MyUserIsThis Aug 6 '13 at 10:25
The following is multiple choice question (with options) to answer.
If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M. | [
"2574",
"2500",
"1365",
"1574"
] | A | Explanation:
HCF * LCM = 84942, because we know
Product of two numbers = Product of HCF and LCM
LCM = 84942/33 = 2574
Option A |
AQUA-RAT | AQUA-RAT-35035 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A man traveled a total distance of 1800 km. He traveled one-third of the whole trip by plane and the distance traveled by train is three-fifth of the distance traveled by bus. If he traveled by train, plane and bus, then find the distance traveled by bus? | [
"239",
"247",
"277",
"270"
] | D | Explanation:
Total distance traveled = 1800 km.
Distance traveled by plane = 600 km.
Distance traveled by bus = x
Distance traveled by train = 3x/5
=> x + 3x/5 + 600 = 1800
=> 8x/5 = 1200 => x = 750 km.
Answer: D |
AQUA-RAT | AQUA-RAT-35036 | 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.
A reduction of 12% in the price of oil enables a house wife to obtain 6kgs more for Rs.1200, what is the reduced price for kg? | [
"24",
"27",
"40",
"28"
] | A | 1200*(12/100) = 144 ---- 6
? ---- 1 => Rs.24
Answer:A |
AQUA-RAT | AQUA-RAT-35037 | # 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.
Laura took out a charge account at the General Store and agreed to pay 6% simple annual interest. If she charges $35 on her account in January, how much will she owe a year later, assuming she does not make any additional charges or payments? | [
"$2.10",
"$37.10",
"$37.16",
"$38.10"
] | B | Principal that is amount taken by Laura at year beginning = 35$
Rate of interest = 6%
Interest = (6/100)*35 = 2.10$
Total amount that Laura owes a year later = 35+2.1 = 37.1 $
Answer B |
AQUA-RAT | AQUA-RAT-35038 | Why are things "weird" here? Let's think of it like this: If I travelled at $$8 \frac{\text{mi}}{\text{hr}}$$ for an hour, then travelled at $$12 \frac{\text{mi}}{\text{hr}}$$, I'd definitely agree that the average speed is $$10 \frac{\text{mi}}{\text{hr}}.$$ Things are pretty normal here: If I want to calculate the total distance traveled, it's just $$8 + 12 = 20.$$ Then, if I want to calculate the total time traveled, it's just $$1 + 1 = 2.$$ So the average or overall speed is just
$$\frac{20 \text{mi}}{2 \text{hr}} = 10 \frac{\text{mi}}{\text{hr}}.$$ It works!
The following is multiple choice question (with options) to answer.
If a person walks at 12 km/hr instead of 8 km/hr, he would have walked 20 km more. The actual distance traveled by him is? | [
"50",
"40",
"45",
"60"
] | B | Let the actual distance traveled be x km. Then,
x/8 = (x + 20)/12
x - 40 =>x = 40 km.
Answer: B |
AQUA-RAT | AQUA-RAT-35039 | 36&14,9,9,9,9,10\\ 37&16,9,9,9,9,10\\ 38&16,9,9,9,9,12\\ 39&16,9,9,11,9,12\\ 40&16,9,9,11,11,12\\ 41&16,9,11,11,11,12\\ 42&16,11,11,11,11,12\\ 43&18,11,11,11,11,12\\ 44&18,11,11,11,11,14\\ 45&18,11,11,13,11,14\\ 46&18,11,11,13,13,14\\ 47&18,11,13,13,13,14\\ 48&18,13,13,13,13,14\\ 49&18,13,13,13,13,16\\ 50&18,13,13,15,13,16\\ 51&18,13,13,15,15,16\\ 52&18,13,15,15,15,16\\ 53&18,15,15,15,15,16\\ 54&18,15,15,15,15,18\\ 55&18,15,15,15,17,18\\ 56&18,15,15,17,17,18\\ 57&18,15,17,17,17,18\\ 58&18,17,17,17,17,18\\ 59&18,17,17,17,17,20\\ 60&18,17,17,17,19,20\\ 61&18,17,17,19,19,20\\ 62&18,17,19,19,19,20\\ 63&18,17,19,19,19,22\\ 64&18,17,19,19,21,22\\ 65&18,17,19,21,21,22\\ 66&18,17,21,21,21,22\\ 67&18,17,21,21,21,24\\ 68&18,17,21,21,23,24\\
The following is multiple choice question (with options) to answer.
364, 361, 19, 16, 4, 1, ? | [
"0",
"1",
"2",
"3"
] | B | 364-3=361
sqrt(361)=19
19-3=16
sqrt(16)=4
4-3=1
sqrt(1)=1
ANSWER:B |
AQUA-RAT | AQUA-RAT-35040 | = 1, we have a/d = 1. All four corresponding sides of two parallelograms are equal in length that does mean that they are necessarily congruent because one parallelogram may or may not overlap the other in this case because their corresponding interior angles may or may not be equal. If two squares have equal areas, they will also have sides of the same length. So we have: a=d. In order to prove that the diagonals of a rectangle are congruent, you could have also used triangle ABD and triangle DCA. But just to be overly careful, let's compute a/d. This means that we can obtain one figure from the other through a process of expansion or contraction, possibly followed by translation, rotation or reflection. if it is can you please explain how you know its true. TRUE. It's very easy for two rectangles to have the same area and different perimeters,or the same perimeter and different areas. Consider the rectangles shown below. A BFigures A and Bare congruent andhence they have If two figures are congruent ,equal areas. Recall that two circles are congruent if they have the same radii. If two triangles are congruent, then their areas are equal. Rectangle 2 with length 9 and width 4. You should perhaps review the lesson about congruent triangles. = 1 if two rectangles have equal areas, then they are congruent sides should equal the ratio of the same perimeter superpose one figure the! Recall that two objects are similar are true and which of the rectangle shure to include at least counterexample. -6 are examples of the congruent rectangles if the corresponding adjacent sides equal! True statement need to multiply to 108, their sides have to be similar, their sides to! S a more equation-based way of proving the areas of PARALLELOGRAMS and triangles 153 you can superpose one figure the... If they have the same area terms of radius, diameter, circumference and surface area similar polygons the... Their areas are equal the ratio of the two shorter sides a does. Areas, they must have equal area, they must have equal areas then they are congruent two triangles equal! Then they are congruent, then your two triangles are equal in area, they all have same! Since
The following is multiple choice question (with options) to answer.
What is the ratio between perimeters of two squares one having 2.5 times the diagonal then the other? | [
"4: 5",
"1: 3",
"2.5: 1",
"3.5: 1"
] | C | d = 2.5d d = d
a√2 = 2.5d a√2 = d
a = 2.5d/√2 a = d/√2 =>2.5: 1
Answer: C |
AQUA-RAT | AQUA-RAT-35041 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
In 1 hour, a boat goes 8 km along the stream and 12 km against the stream. The speed of the boat in still water (in km/hr) is : | [
"3 km/hr.",
"2 km/hr.",
"10 km/hr.",
"8 km/hr."
] | C | Speed in still water = 1/2(8 + 12) km/hr
= 10 km/hr. ANSWER :C |
AQUA-RAT | AQUA-RAT-35042 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed? | [
"32",
"88",
"67",
"11"
] | D | One day work of A, B and C = 1/24 + 1/30 + 1/40 = 1/10 Work done by A and B together in the last 4 days = 4*(1/24 + 1/30) = 3/10
Remaining work = 7/10
The number of days required for this initial work = 7 days.
The total number of days required = 4 + 7 = 11 days.
Answer:D |
AQUA-RAT | AQUA-RAT-35043 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
The number 69 can be written as the sum of squares of 3 integers. Which of the following could be the difference between the largest and smallest integers of the 3? | [
"2",
"5",
"7",
"8"
] | B | Notice that the question asks which of the followingcouldbe the difference between the largest and smallest integers, not must be. The 3 integers could be: +/-2, +/-4 and +/-7 so the difference could be 5,3 or 11. Since only one of them is among the choices, then it must be the correct answer.
Answer: B |
AQUA-RAT | AQUA-RAT-35044 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
If four students are made to sit in a row, then 3 don't have seats. If 15 students are made to sit in a row, then two rows are empty. How many students have to sit a row such that each row has equal number of students and all students are seated? | [
"0",
"1",
"4",
"5"
] | D | Let there be a rows.
Then 4a + 3= 15(a-2) [equating the number of students]
=> a= 3
and total number of students = 15
Therefore 5 students much sit in each row such that each row has an equal number of students and all students are seated.
Option (D) |
AQUA-RAT | AQUA-RAT-35045 | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
The following is multiple choice question (with options) to answer.
In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons? | [
"200",
"230",
"240",
"250"
] | B | Option 'B'
7p3 = 7 × 6 × 5 = 210 |
AQUA-RAT | AQUA-RAT-35046 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: | [
"10",
"12",
"14",
"16"
] | C | Explanation:
A,B,C together fill the tank in 6 hours.
=> Part filled by pipes A,B,C together in 1 hr =1/6
All these pipes are open for only 2 hours and then C is closed.
Remaining part =1−1/3=2/3
This remaining part of 2/3 is filled by pipes A and B in 7 hours
Therefore, part filled by pipes A and B in 1 hr
=(2/3)/7=2/21
Part filled by pipe C in 1 hr
=(1/6-2/21)=7-4/42 =3/42 =1/14
i.e., C alone can fill the tank in 14 hours.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35047 | ### 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 sells his goods 50% cheaper than B but 50% dearer than C. The cheapest is? | [
"33.3",
"33.6",
"33.0",
"33.4"
] | C | Let B = 100
A = 50
C * (150/100) = 50
3C = 100
C = 33.3 then 'C' Cheapest
Answer:C |
AQUA-RAT | AQUA-RAT-35048 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A class of students consists of 6 senior and 2 junior students. If a class council is created with 3 senior and 1 junior student, in how many ways can the class council be formed? | [
"12",
"20",
"40",
"57"
] | C | Choose 3 senior from 5 senior and choose 1 junior from 3 junior:
3C6 * 1C2 = 20*2 = 40
Ans: C |
AQUA-RAT | AQUA-RAT-35049 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
The 100-milliliter solution of sugar and water is 15% sugar. How much water (in milliliters) must be added to make a solution that is 8% sugar? | [
"72.5",
"77.5",
"82.5",
"87.5"
] | D | In the original solution the amount of sugar is 0.15*100=15 ml.
Let the volume of the new solution be V.
0.08V=15 ml.
V=187.5 ml.
The amount of water we need to add is 187.5 - 100 = 87.5 ml.
The answer is D. |
AQUA-RAT | AQUA-RAT-35050 | 5. Originally Posted by ANDS!
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
Thanks.
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
Why aren't these included in the answer given in the post above?
6. Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
7. Soroban only considered these 2 cases:
Originally Posted by Soroban
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
The following is multiple choice question (with options) to answer.
An office supply store stocks three sizes of notebooks, each in five colors: blue, green, yellow, red, or pink. The store packs the notebooks in packages that contain either 4 notebooks of the same size and the same color or 4 notebooks of the same size and of 4 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible? | [
"18",
"24",
"30",
"45"
] | C | First let's consider the small notebooks.
There are 5 ways to choose notebooks of the same color.
The number of ways to choose four notebooks with different colors is 5C4=5.
There are 10 different packages we can make with the small notebooks.
We can use the same process to find 10 different packages with medium and large notebooks.
The total number of different packages is 30.
The answer is C. |
AQUA-RAT | AQUA-RAT-35051 | Since these 6 possibilities are exhaustive, we have $$(1+2+3+4+5+6)q/6 = 1 \implies q = {6 \over 21} = {1 \over 3.5}$$ as I originally suspected; in particular, this implies that $$P(X_\tau = M + j) = (6-j)q/6 = (6-j)/21,$$ agreeing with Mike's answer.
The following is multiple choice question (with options) to answer.
If p/q = 2/7 , then 2p+q= ? | [
"11",
"14",
"13",
"15"
] | A | let p=2, q=7 then 2*2+7=11
so 2p+q=11.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35052 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A company produces computers with 7 different speeds, and the ratio of each speed to the next one is a fixed ratio. If the fastest computer is 5.4 GHz, while slowest speed is 0.2 GHz, what is the speed of the second slowest computer? | [
"(3^1/2) (0.2)",
"(27^1/7) (0.2)",
"(5.2/7) + 0.2",
"(0.2)^2"
] | A | 1st speed=0.2
Let x be ratio,then 1st/2nd=x
0.2/2nd =x------>2nd=0.2/x------(a)
.
.
.
7th=0.2/x^6=5.4
x^6=0.2/5.4=1/27----->1/3^3
x=1 / 3^(3*1/6)------->1/3^1/2
therefore 2nd speed(from (a)) =0.2/(1/3^1/2)-------->3^1/2*0.2
Ans A |
AQUA-RAT | AQUA-RAT-35053 | ### 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 is a working partner and B is a sleeping partner in the business. A puts in Rs.15000 and B Rs.25000, A receives 10% of the profit for managing the business the rest being divided in proportion of their capitals. Out of a total profit of Rs.9600, money received by A is? | [
"Rs.3240",
"Rs.3600",
"Rs.3800",
"Rs.4200"
] | D | 15:25 => 3:5
9600*10/100 = 960
9600 - 960 = 8640
8640*3/8 = 3240 + 960
= 4200
ANSWER:D |
AQUA-RAT | AQUA-RAT-35054 | Note that $-169 + ( 4 \times 52 ) = 39$, so $-169 \equiv 39 \mod 52$.
Try this yourself for the second question.
• how did you get the four? – Nicole Oct 13 '14 at 23:36
• @Nicole Four is the number of times you need to add 52 to -169 to get a positive number. Note that adding any lesser multiple of 52 will leave you with a negative number. – Nick Oct 13 '14 at 23:38
The following is multiple choice question (with options) to answer.
The product of two numbers is 266 and their difference is 5. What is the bigger number ? | [
"13",
"15",
"19",
"24"
] | C | Explanation:
Let the two numbers be A and B, here A > B
AB = 266
B = 266/A -----------------(I)
Given,
A – B = 5 ----------- (II)
Substitute from (I) in (II), we get
A – 266/A = 5
A2 – 5A + 266 = 0
(A – 19)(A – 14) = 0
Therefore , A = 19 or A = 14
Hence, bigger number = A = 19
ANSWER: C |
AQUA-RAT | AQUA-RAT-35055 | # Conditional Probability of rainfall
1. Apr 26, 2012
### TranscendArcu
1. The problem statement, all variables and given/known data
3. The attempt at a solution
a) P(Pickwick has no umbrella | it rains) = $\frac{\frac{1}{3}\frac{1}{3}}{\frac{1}{2}} = \frac{2}{9}$, which is the answer according to my answer key.
b) For part b we have:
There is a rain forecast which means he will bring the umbrella. The probability that it won't rain is 1/3.
There is a non-rain forecast which means he brings the umbrella with a probability of 1/3 and it will not rain with a prob of 2/3.
P(Pickwick has umbrella | no rain) = $\frac{1}{3} + \frac{1}{3}\frac{2}{3} = \frac{5}{9}$. But the answer is apparently 5/12. What have I done incorrectly here?
2. Apr 26, 2012
### MaxManus
I got the same as you at b) and I cant see why it is not correct.That's no guarantee for that you are correct though
3. Apr 26, 2012
### Ray Vickson
The following is multiple choice question (with options) to answer.
The probability of rain showers in Barcelona on any given day is 0.3. What is the probability that it will rain on exactly one out of three straight days in Barcelona? | [
"0.441",
"0.072",
"0.432",
"0.72"
] | A | IMO This is just for first day! It can rain on 2nd day or 3 rd day (and not rain on other days) in 3 consecutive days!
Hence,
0.3*0.7*0.7 + 0.7*0.3*0.7 + 0.7*0.7*0.3 = 0.441
Option - A |
AQUA-RAT | AQUA-RAT-35056 | I'll apply it to approximations for pi.
• 256/81 = 3 13/81 = 3 + 1/7 + 1/57 + 1/10773
• 22/7 = 3 1/7 = 3 + 1/7
• 223/71 = 3 10/71 = 3 + 1/8 + 1/64 + 1/4544
• 355/113 = 3 16/113 = 3 + 1/8 + 1/61 + 1/5014 + 1/27649202 + 1/1911195900442808
256/81 is from an approximation of the area of a circle: ((8*diameter)/9)^2
3. gives several reduction algorithms. Here are some of those that fit the Rhind papyrus's numbers:
A simple one for odd x is 2/x = 2/(x+1) + 2/x*(x+1))
A fancier one is 2/x = 1/a + (2*a-x)/(a*x) where a is some number with lots of divisors.
A fraction a/b where b has lots of divisors can be resolved by looking for some sum of divisors that adds up to a. For power-of-2 divisors, that is always possible, from a binary representation. For numbers with powers of higher primes, that is more difficult, but it is sometimes possible.
The number 6 = 2*3 and its proper divisors are 1, 2, and 3. One can form 1, 2, 3, 4 = 1+3, 5 = 2+3.
But prime numbers have only one proper divisor, so that is only possible for 2. That is also true of odd numbers more generally. One cannot represent 2 with the proper divisors of an odd number greater than 1.
There are also even numbers where that is not possible, like 10 = 2*5, with proper divisors 1, 2, 5. One can form 1, 2, 5, 6 = 1+5, 7=2+5, but not 3, 4, 8, 9.
2/(x*y) = 1/(a*x) + 1/(a*x*y) where a = (x+1)/2
The following is multiple choice question (with options) to answer.
How many odd, positive divisors does 640 have? | [
"6",
"8",
"12",
"15"
] | C | Make a prime factorization of a number: 540=2^2*3^3*5 --> get rid of powers of 2 as they give even factors --> you'll have 3^3*5 which has (3+1)(1+1)=12 factors.
Another example: 60=2^2*3*5 it has (2+1)(1+1)(1+1)=12 factors out of which (1+1)(1+1)=4 are odd: 1, 3, 5 and 15 the same # of odd factors as 60/2^2=15 has.
Answer: C. |
AQUA-RAT | AQUA-RAT-35057 | ## 35.
The Correct Answer is (1/29) — The probability of choosing one student wanting to enter finance is 6/30. The probability of choosing another student wanting to enter finance is then 5/29. The probability of both of these events occurring is therefore (6/30)(5/29) = 30/870, which can be reduced to 1/29.
## 36.
The Correct Answer is (40) — Since the small triangle and the entire triangle share two angles (the one on the right, and the right angle), they must share all three angles, and therefore are similar triangles. You can use the Pythagorean Theorem on the bigger triangle to find that the length of the horizontal side is the square root of 1002 - 602, which is 80. (This is a special 3-4-5 triangle with the side lengths multiplied by 20). Since the ratio of the vertical side to the horizontal side of the big triangle is 60/80 = 3/4, this must be the same as the corresponding ratio in the small triangle. Since the vertical side has length 30, the length marked x must have length 40.
## 37.
The Correct Answer is (10) — There are 6 full cages of mice, which means there are 6 × 5 = 30 mice. If we let n be the number of male mice, then the number of female mice is twice that at 2n. There are 30 mice in total, so n + 2n = 30. Solving for n gives you n = 10. Since n is the number of male mice, there are 10 male mice.
## 38.
The Correct Answer is (17) — The reduced cost of maintaining each cage is 0.5($1.25) =$0.625 per cage per day. The student needs 21 cages to house 102 mice, so her daily cost is $0.625 × 21 =$13.125. She has a budget of $225, so she can afford to maintain the cages for$225/\$13.215 = 17.14 days. Rounding this to the nearest whole day gives you 17 days.
The following is multiple choice question (with options) to answer.
A certain animal shelter has 28 cats and 46 dogs. How many dogs must be adopted and taken away from the animal shelter so that 70 percent of the animals in the shelter will be cats? | [
"6",
"12",
"28",
"34"
] | D | (28 + 46 - x)*0.7 = 28
x = 34.
Answer: D. |
AQUA-RAT | AQUA-RAT-35058 | A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Another weighted average approach, expressed a bit differently.
Track on milk. We know the desired concentration of milk in the resultant mixture.
Milk is a fraction (or percentage or concentration) of all three mixtures of milk and water. The weighted average formula accounts for water by way of volume.
This formula is easy (concentration can be a percentage or a fraction):
$$(Concentration_{A})(Vol_{A}) + (Concentration_{B})(Vol_{B})$$
$$=(Concentration_{A+B})(Vol_{A+B})$$
Let A = # of gallons of A (volume)
1) Use ratios and desired percentage to find the concentration of milk in A, B, and end mixture
(With ratios, remember to find $$\frac{part}{whole}$$)
In mixture A, $$\frac{Milk}{Water}=\frac{2}{5}$$
2 parts milk, 5 parts water, total parts = 7
So in A, milk is $$\frac{2parts}{7parts}=\frac{2}{7}$$
In the second mixture, B, milk is what fraction?
$$\frac{M}{W}=\frac{5}{4}$$
B, concentration of milk $$=\frac{5}{4+5}=\frac{5}{9}$$
Resultant mixture, desired concentration =
40% milk $$=\frac{40}{100}=\frac{2}{5}$$
The volume of B is 90 gallons.
The volume of the resultant mixture is (A + B).
What is the volume of A?
2) Use weighted average to find the unknown volume of A (steps can be combined)
$$\frac{2}{7}A + \frac{5}{9}(90)=\frac{2}{5}(A+90)$$
$$\frac{2}{7}A + 50=\frac{2}{5}A + \frac{2}{5}(90)$$
The following is multiple choice question (with options) to answer.
Three grades of milk are 1 percent, 2 percent and 5 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 5 percent grade are mixed to give x+y+z gallons of a 3 percent grade, what is x in terms of y and z? | [
"y + 3z",
"(y +z) / 4",
"2y + 3z",
"z - 0.5 y"
] | D | Based on the given information, the following equation could be dervied: 1% * x + 2% * y + 5% * z = 3% * (x+y+z) Then use algebra to figure out 2%*x = -1%*y + 2%*z. Thus, x = z - 0.5 y
D |
AQUA-RAT | AQUA-RAT-35059 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work? | [
"56",
"28",
"45",
"24"
] | D | D
24
Let the required number of days be x.
Less men, More days (Indirect Proportion)
27 : 36 :: 18 : x <-> 27 x x = 36 x 18
x = (36 x 18)/27
x = 24 |
AQUA-RAT | AQUA-RAT-35060 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A person decided to build a house in 100days. He employed 100men in the beginning and 100 more after 50 days and completed the construction in stipulated time. If he had not employed the additional men , how many days behind schedule would it have been finished? | [
"40",
"80",
"50",
"60"
] | C | 200 men do the rest of the work in 100-50 = 50days
100 men can do the rest of the work in 50*200/100 = 100days
Required number of days = 100-50 = 50 days
Answer is C |
AQUA-RAT | AQUA-RAT-35061 | Hint: if $$8$$ is the largest single digit factor than doesn't it make sense that the largest such three digit number would be in the $$8$$ hundreds.
Now, that was the hard way. Were there any handy mathematical observations you might have used to make this easier? Would knowing that $$24 = 2^3*3$$ is the unique prime factorization of $$24$$ have helped you find the ways to find three term factorizations? Would knowing there are $$k!$$ ways to arrange $$k$$ objects have helped? What if some of the objects were indistinguishable?
$$xyz= 2^3 .3$$
powers of $$2$$ can be divided among $$x,y,z$$ in
$$^{3+(3-1)} \ C _{3-1}= 10$$ ways
and power of $$3$$ can be divided into $$^{1+(3-1)} \ C _{3-1}= 3$$ ways
thus giving total $$10\times 3 =30$$ways
but here we also counted arrangements like $$2 \times 1\times 12$$ and $$1\times 1\times 24$$ which violates 3 digit number policy so, by subtracting these i.e, $$\left(3! + \dfrac{3!}{2!}\right)=9$$ ways
we get
total number of $$3$$ digit numbers having product $$4 != (30-9)=21$$
and out of all such $$3$$ digit numbers largest is $$8 3 1$$
The following is multiple choice question (with options) to answer.
The number 243 has been divided into three parts in such a way that one third of the first part, fourth of the second part and half of the third part are equal. Determine the largest part. | [
"108",
"276",
"297",
"266"
] | A | Explanation:
It is given that 243 has been divided into 3 numbers x y and z for instance.
Second condition mentioned in the question is:
=> x/3 = y/4 =z/2
Clearly the biggest number is y. So let us solve w.r.t. y.
=> x = 3y/4
=> x =3z/2
=> z = y/2
=> x+y+z = 243
=> y/2 + 3y/4 + y = 243
9y/4 = 243
=> 243*4/9
=> 108
ANSWER: A |
AQUA-RAT | AQUA-RAT-35062 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle? | [
"60 cm2",
"30 cm2",
"45 cm2",
"15 cm2"
] | B | The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle = 6 * 5 = 30 cm2
14. The area of a square is 40
ANSWER:B |
AQUA-RAT | AQUA-RAT-35063 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangule, whose length and breadth are in the ratio of 6 : 5. What is the area of the rectangle? | [
"87 cm2",
"30 cm2",
"76 cm2",
"26 cm2"
] | B | The circumference of the circle is equal to the permeter of the rectangle.
Let l = 6x and b = 5x 2(6x + 5x) = 2 * 22/7 * 3.5
=> x = 1
Therefore l = 6 cm and b = 5 cm Area of the rectangle
= 6 * 5
= 30 cm2
Answer: B |
AQUA-RAT | AQUA-RAT-35064 | triangle is a kind of acute triangle, and is always equilateral. An obtuse triangle is one where one of the angles is greater than 90 degrees. A(6,8) B(-1,4) C(5,4) is the obtuse triangle … FAQ. The given 96º angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. A triangle is a polygon made up of 3 sides and 3 angles.. We can classify triangles according to the length of their sides. This is one of the three types of triangles, based on sides.. We are going to discuss here its definition, formulas for perimeter and area and its properties. Angles measures more than 90° degrees are also of different lengths 2. a triangle with one angle! Are always all different sides all of…, as ∠A measures more than but. If all sides of a triangle with more than 90° degrees pair because a triangle are equal 5,4 is. There are are equal—each one measures 60 degrees, the triangle ABC is obtuse... Acute, obtuse, acute, or right angles most general scalene:! Note: it is called an isosceles or scalene triangle are equal 5,4. An obtuse-angled triangle is a triangle in which the number is shown as the product of two factors can! Insert > Shapes and select the Freeform tool and obtuse by moving it 4 spaces to the angle... Sum of all the angles are the same size, imagine two angles equal—each! All different to also be an equilateral triangle: if all sides are equal, then it is obtuse... Of all the three angles are acute and so the triangle may be right, obtuse, acute, right. Is the obtuse triangle can be an equilateral triangle: No sides have equal length No angles are ;! The other elements of a scalene triangle but it can ’ t draw than! Triangle has one angle measuring more than 90 degrees the sum of all the sides of a is! Which has at least one angle scalene obtuse triangle has measurement greater than 90° - all three are... Select the Freeform tool an acute, obtuse or right angle of new, high-quality pictures added every day scalene. Collinear vertices ) is the same i )
The following is multiple choice question (with options) to answer.
How many obtuse triangle can be made when one of the side of the triangle is taken between the 1st natural number and 8. | [
"1",
"2",
"3",
"4"
] | C | 3 triangles can be formed....because we have to take the property that sum of two sides of a triangle should be greater than the third side
ANSWER:C |
AQUA-RAT | AQUA-RAT-35065 | # 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.
John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in John's account 6 months after the account was opened? | [
"$10,100",
"$10,101",
"$10,200",
"$10,201"
] | C | john amount is compounded quarterly so the formula is A=P*(1+R/100)^N her p=principle R=rate of interest N=no of terms
rate of interest is 4% for a yr
so for quarter R=4/4=1%
john receive amt after 6 months so there are two quarter
so N=2
Amount A=P(1+R/100)^2
=10000(1+1/100)^2
=10000(1+0.01)^2
=1020
ANSWER:C |
AQUA-RAT | AQUA-RAT-35066 | 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 least value must be assigned to * so that the number 63576*2 is divisible by 8? | [
"1",
"2",
"3",
"4"
] | C | Sol.
The given number is divisible by 8, if the number 6x2 is divisible by 8.
Clearly, the least value of x is 3.
Answer C |
AQUA-RAT | AQUA-RAT-35067 | A group of friends rent a beach house and decide to split the cost of the rent and food. Four friends pay $170 each. Five friends pay$162 each. Six friends pay 157. If nine people were to share the expense, how much would each pay? Let’s look at this in a table. n\begin{align*}n\end{align*} (number of friends) t\begin{align*}t\end{align*} (share of expense) 4 170 5 162 6 157 9 ??? As the number of friends gets larger, the cost per person gets smaller. This is an example of inverse variation. An inverse variation function has the form f(x)=kx\begin{align*}f(x)=\frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is called the constant of variation and must be a counting number and x0\begin{align*}x \neq 0\end{align*}. To show an inverse variation relationship, use either of the phrases: • Is inversely proportional to • Varies inversely as Example 1: Find the constant of variation of the beach house situation. Solution: Use the inverse variation equation to find k\begin{align*}k\end{align*}, the constant of variation. Solve for k:y170170×4k=kx=k4=k4×4=680\begin{align*}&& y &= \frac{k}{x}\\ && 170 &= \frac{k}{4}\\ \text{Solve for} \ k: && 170 \times 4 &= \frac{k}{4} \times 4\\ && k &= 680\end{align*} You can use this information to determine the amount of expense per person if nine people split the cost. yy=680x=6809=75.56\begin{align*}y &= \frac{680}{x}\\ y &= \frac{680}{9}=75.56\end{align*} If nine people split the expense, each would pay75.56.
Using a graphing calculator, look at a graph of this situation.
The following is multiple choice question (with options) to answer.
The total cost of a vacation was divided among 3 people. If the total cost of the vacation had been divided equally among 4 people, the cost per person would have been $40 less. What was the total cost cost of the vacation? | [
"$200",
"$300",
"$480",
"$500"
] | C | C for cost.
P price per person.
C= 3*P
C=4*P-160
Substituting the value of P from the first equation onto the second we get P = 160.
Plugging in the value of P in the first equation, we get C= 480. Which leads us to answer choice C |
AQUA-RAT | AQUA-RAT-35068 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A boat goes 100 km downstream in 10 hours, and 75 m upstream in 15 hours. The speed of the stream is? | [
"27 1/2 kmph",
"22 1/2 kmph",
"32 1/2 kmph",
"22 1/1 kmph"
] | B | 100 --- 10 DS = 10
? ---- 1
75 ---- 15 US = 5
? ----- 1 S = (10 - 5)/2
= 22 1/2 kmph
Answer: B |
AQUA-RAT | AQUA-RAT-35069 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Winners will be announced the following day at 11 AM Pacific/1 PM Eastern Time.
If A, X, Y, and Z are unique nonzero digits in the equation:
XY
*YX
____
AYZ
And Y > X, the 2-digit number XY = ? | [
"13",
"21",
"22",
"24"
] | C | The reasoning is as follows:-
B) can't be answer because Y<X
D) can't be answer because it XY x YX will give 4 digit number. Thus wrong
E) can't be answer because it XY x YX will give 4 digit number. Thus wrong
A) can't be answer because it XY x YX will give 3 digit number but the middle digit will not be equal to Y. Thus wrong
C) is the correct answer because Y>xwill give 2 digit numbermiddle digit is Y
C |
AQUA-RAT | AQUA-RAT-35070 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If x and y are positive odd integers, then which of the following must also be an odd integer?
I. x^(y+1)
II. x(y+1)
III. (y+1)^(x-1) + 1 | [
"I only",
"II only",
"III only",
"I and III"
] | A | I. x^(y+1)>> (odd)^even=odd
II. x(y+1)>>(odd)*even=even
III. (y+1)^(x-1) + 1>>(even)^even+1>>even+1>> odd if x is not equal to 1 and if x=1 then its (even)^0+1=1+!=2 even.
answer A |
AQUA-RAT | AQUA-RAT-35071 | # 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.
Sonika deposited Rs.8000 which amounted to Rs.9200 after 3 years at simple interest. Had the interest been 0.5% more. She would get how much? | [
"9320",
"96288",
"26667",
"1662"
] | A | (8000*3*0.5)/100 = 120
9200
--------
9320
Answer:A |
AQUA-RAT | AQUA-RAT-35072 | 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.
Car A runs at the speed of 100km/hr & reaches its destination in 5hr. Car B runs at the speed of 80 km/h & reaches its destination in 6h. What is the respective ratio of distances covered by Car A & Car B? | [
"11 : 5",
"11 : 8",
"13 : 7",
"25 : 24"
] | D | Sol. Distance travelled by Car A = 100 × 5 = 500 km
Distance travelled by Car B = 80 × 6 = 480 km
Ratio = 500/480 = 25 : 24
Answer : D |
AQUA-RAT | AQUA-RAT-35073 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Sarah is driving to the airport. After driving at 40 miles per hour for one hour, she realizes that if she continues at that same average rate she will be an hour late for her flight. She then travels 50 miles per hour for the rest of the trip, and arrives 30 minutes before her flight departs. How many miles did she drive in all? | [
"140",
"340",
"210",
"245"
] | B | After driving at 40 miles per hourfor one hour, this distance left to cover is d-40. Say this distance is x miles.
Now, we know that the difference in time between covering this distance at 40 miles per hour and 50 miles per hour is 1+ 1/2 = 3/2 hours.
So, we have that x/40 - x/50 = 3/2 --> 5x/200 - 4x/200 = 3/2 --> x/200 = 3/2 --> x = 300.
Total distance = x + 40 = 340 miles.
Answer: B |
AQUA-RAT | AQUA-RAT-35074 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 187, 264, 386, 473, 682, 781 | [
"386",
"187",
"781",
"682"
] | A | Explanation :
In all numbers except 386, the middle digit is the sum of other two digits.
Answer : Option A |
AQUA-RAT | AQUA-RAT-35075 | 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 speed of a train is 110 kmph. What is the distance covered by it in 11 minutes? | [
"15 kmph",
"11 kmph",
"88 kmph",
"20 kmph"
] | D | 110 * 11/60
= 20 kmph
Answer:D |
AQUA-RAT | AQUA-RAT-35076 | 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 sec and a man standing on the platform in 20 sec. If the speed of the train is 54 km/hr. What is the length of the platform? | [
"288",
"240",
"88",
"66"
] | B | Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 20 = 300 m.
Let the length of the platform be x m . Then,
(x + 300)/36 = 15 => x = 240 m.
Answer: B |
AQUA-RAT | AQUA-RAT-35077 | This can be solved using Venn Diagram (refer the attachment)
AC = 60% of 150 = 90
Sunporch = 50% of 150 = 75
SP = 30% of 150 = 45
Using the venn diagram we can form the following equations
a+b+c+d+e+f+g+h = 150
a+b+c+d+f+g = 140 ----- (1) (as e = h = 5)
Taking 1 circle at a time
AC = a+b+d = 85 ---- (2) .. (as e = 5)
SunPorch = c+b+f = 70 ---- (3) .. (as e = 5)
SP = g+d+f = 40 ---- (4) .. (as e = 5)
equation (2) + (3) + (4)
a+c+g+ 2(d+b+f) = 195
subtracting this with equation (1)
d+b+f = 195 - 140 = 55
Attachments
Ven.png [ 6.81 KiB | Viewed 14100 times ]
##### General Discussion
Manager
Joined: 17 May 2015
Posts: 245
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
29 May 2017, 02:31
VP
Joined: 05 Mar 2015
Posts: 1000
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
29 May 2017, 07:23
2
1
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
all three = neither =5 = 10/3 % of 150
working formula :-
Total = AC + sunporch + swimming pool - Exactly two - 2*all three + neither
The following is multiple choice question (with options) to answer.
150 is what percent of 60 ? | [
" 250%",
" 20%",
" 50%",
" 200%"
] | A | 60*x=150 --> x=2.5 -->2.5 expressed as percent is 250%.
Answer: A. |
AQUA-RAT | AQUA-RAT-35078 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man swims downstream 72 km and upstream 45 km taking 9 hours each time; what is the speed of the current? | [
"2.5",
"1.5",
"1.7",
"1.2"
] | B | 72 --- 9 DS = 8
? ---- 1
45 ---- 9 US = 5
? ---- 1 S = ?
S = (8 - 5)/2
= 1.5
Answer: B |
AQUA-RAT | AQUA-RAT-35079 | ### 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.
UBA Capital recently bought Brand new vehicles for office use. UBA capital only went for Toyota and Honda and bought more of Toyota than Honda at the ratio of 8:2. If 80% of the Toyota bought and 20% of the Honda bought were SUV’s. How many SUV’s did UBA capital buy in the aforementioned purchase? | [
"66%",
"64%",
"68%",
"69%"
] | C | let total no of Vehicles bought be 100, Toyota 80 and Honda 20, so total number of SUV's bought for Toyota and Honda respectively 80* 80/100=64 and 20* 20/100=4
so total 68 SUV's were bought out of 100 Vehicles bought..so required % is 68%
ANSWER:C |
AQUA-RAT | AQUA-RAT-35080 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
When Rahul was born, his father was 32 years older than his brother and his mother was 27 years older than his sister. If Rahul's brother is 6 years older than him and his mother is 3 years younger than his father, how old was Rahul's sister when he was born ? | [
"10 years",
"9 years",
"8 years",
"7 years"
] | C | When Rahul was born, his brother's age = 6 years; his father's age = (6 + 32) years = 38 years,
his mother's age = (38 - 3) years = 35 years;
his sister's age = (35 - 27) years = 8 years.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35081 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"176 m",
"187 m",
"175 m",
"150 m"
] | D | Speed=(60 * 5/18) m/sec
= (50/3) m/sec Length of the train
= (Speed x Time)
= (50/3 * 9) m
= 150 m.
Answer: D |
AQUA-RAT | AQUA-RAT-35082 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A man covers a distance of 180 Km at 72 Kmph and next 190km at 88 Kmph. what is his average speed For his whole journey of 370 Km? | [
"77 Kmph",
"77.15 Kmph",
"77.25 Kmph",
"79 Kmph"
] | C | Formula = 2*F.S*S.P/F.S+S.P
= 2*72*88/88+72
=79.2 Kmph
Option 'C' |
AQUA-RAT | AQUA-RAT-35083 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In a college the ratio of the numbers of boys to the girls is 8:5. If there are 160 girls, the total number of students in the college is? | [
"562",
"356",
"452",
"416"
] | D | Let the number of boys and girls be 8x and 5x
Then , 5x = 160
x= 32
Total number of students = 13x = 13*32 = 416
Answer is D |
AQUA-RAT | AQUA-RAT-35084 | = 4 52. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Two fair dice are rolled and the sum of the points is noted. For example: 1 roll: 5/6 (83. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. and only one way to roll a 12 (6-6, or boxcars). EXPERIMENTAL PROBABILITIES Simulate rolling two dice 120 times. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Two dice are tossed. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Sample space S = {H,T} and n(s) = 2. That intuition is wrong. When you roll a pair of dice there are 36 possible outcomes. To find the probability we use the mutually exclusive probability formula P(A) + P(B). A sum less than or equal to 4. Isn’t that kind of cool?. Two dice are tossed. So the probability of getting a sum of 4 is 3/36 or 1/12. 10 5 13 ! Find the probability distribution. So 1/36 is part of the. Sum of Two Dice. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Rolling two dice. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. The fundamental counting principle tells us there are 6*6=36 ways to roll two dice, all of them equally likely if the dice are fair. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Let B be the event - The sum
The following is multiple choice question (with options) to answer.
Two dice are tossed once. What is the probability of getting an even number on the first die or a total sum of 10 on the two dice? | [
"5/9",
"11/18",
"19/36",
"23/36"
] | C | There are 36 possible outcomes for two dice.
The number of outcomes with an even number on the first die is 3*6=18.
The number of outcomes with a sum of 10 is 3.
We need to subtract the outcomes that have both an even number on the first die and a sum of 10 because we have counted those outcomes twice. This number of outcomes is 2.
The number of outcomes with either an even number on the first die or a sum of 10 is 18+3-2=19.
The probability is 19/36.
The answer is C. |
AQUA-RAT | AQUA-RAT-35085 | 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.
Three friends A, B and C run around a circular track of length 120 metres at speeds of 5 m/s, 7 m/sec and 15 m/sec, starting simultaneously from the same point and in the same direction. How often will the three of them meet ? | [
"Every 60 seconds",
"Every 120 seconds",
"Every 30 seconds",
"None of these"
] | A | Explanation :
A and B will meet for the first time in:
=>(Circumference of track/relative speed) seconds
=>120/2 = 60 seconds.
This also means that A and B will continue meeting each other every 60 seconds.
When B and C will meet for the first time.
B and C will meet for the first time in 120/8=15 seconds.
This also means that they will meet every 15 seconds after they meet for the first time i.e. A and B meet every 60 seconds and multiples of 60 seconds and B and C meet every 15 seconds and multiples of 15 seconds.
The common multiples to both these time, will be when A and B and B and C will meet i.e. when A, B and C will meet.
The common multiple of 60 and 15 will be 60,120,180 etc. i.e. they will meet every 60 seconds.
Answer : A |
AQUA-RAT | AQUA-RAT-35086 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
The speed of a car is 100 km in the first hour and 60 km in the second hour. What is the average speed of the car? | [
"12",
"75",
"80",
"54"
] | C | S = (100 + 60)/2
= 80 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-35087 | The most basic test one could come up with works easily: clearly with the given hypotheses $p$ must be relatively prime with $24$ (which is really all that matters about $p$) and checking the squares of all $8$ elements of $(\Bbb Z/24\Bbb Z)^\times$ one finds$~1$ each time. I just wanted the obvious to be said.
There are numerous ways to save part of the the work of this computation (using the Chinese remainder theorem $(\Bbb Z/24\Bbb Z)^\times\cong(\Bbb Z/3\Bbb Z)^\times\times(\Bbb Z/8\Bbb Z)^\times$ is one obvious way) but given the tiny amount of work the check is to begin with, there is no real need to elaborate on such savings.
We know for every positive integer $n$ , $\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}6$ . If $p>3$ is a prime then $\dfrac{p-1}2$ is a positive integer , so $$\sum_{k=1}^{\dfrac{p-1}2} k^2=\dfrac{\dfrac{p-1}2\bigg(\dfrac{p-1}2+1\bigg)\Bigg(2\bigg(\dfrac{p-1}2\bigg)+1\Bigg)}6=\dfrac {(p-1)(p+1)p}{24}=\dfrac {p(p^2-1)}{24}$$ ,
so $\dfrac {p(p^2-1)}{24}$ is a sum of some positive integers , and hence an integer i.e. $24$ divides $p(p^2-1)$ , but
also $24=3 ×2^3$ and since $p>3$ is a prime , so $g.c.d (p,24)=1$ , hence $24$ divides $p^2-1$
The following is multiple choice question (with options) to answer.
If the sum of two numbers is 24 and the sum of their squares is 400, then the product of the numbers is | [
"40",
"44",
"80",
"88"
] | D | according to the given conditions x+y = 24
and x^2+y^2 = 400
now (x+y)^2 = x^2+y^2+2xy
so 24^2 = 400 + 2xy
so xy = 176/2 = 88
ANSWER:D |
AQUA-RAT | AQUA-RAT-35088 | algorithms, natural-language-processing, counting, mathematical-analysis
Goal is to implement this as a function in a programming language like Python or PHP and to use the (personalised) result for creating an output like "this article can be read in approximately 12-14 minutes".
It would be nice if such information is publicly available. There is a similar question, so I will add a bit over the answer.
There are too many factors independent of language but solely based on individual reading abilities that such estimate would not be possible.
I assume perfect text without spelling errors and unclear grammar.
Giving it a try:
People can read 20, 50, 200, 2000... pages per hour, it depends on reading skills - only prescreening test will give you answer to that: give 2 pages, measure time, or use eye tracker and do the same.
This depends on language used, but the reading speed is measured in native language, so the differences are in the next point as for language (or decrease the speed for non-native by... proficiency at that language).
People use Cohorts imagine that the dictionary forms trie, and the speed of recognizing given word depends on amount of words with the same prefix.
Cognitive Psychology: A Student's Handbook, fifth edition by Michael W. Eysenck,Mark T. Keane, page 347.
For example: "Av..." What are possible words?
Average, Avalanche... Avengers if someone seen the movie, comics or ads.
"Ave.." - word on it's own, Average fits, Avengers also.
If person reads about statistics, "average" is already good choice, about Roman Empire - Ave it is, about movies played - Avengers.
But reading continues to the next syllabe.
It depends on mood, recent readings, coherence of text, familiarity with topic, recent topics seen (not only read).
About topic: reading novel is faster than technical text. Technical text for someone proficient will not decrease reading speed.
So the task given is important, if someone is checked for understanding text, reading speed decreases.
The language itself is not a factor, you should not take mean reading speed but median of your target group.
Yes age is factor, but not that usable as it gives fraction of other factors.
With dyslexia the decrease also vary per person, $30%$ decrease is very optimistic case.
The following is multiple choice question (with options) to answer.
If Rosé reads at a constant rate of 3 pages every 5 minutes, how many seconds will it take her to read N pages? | [
"100",
"2N",
"5/2*N",
"24N"
] | A | Rose would read 1 page in 5/3 min
Rose would read N page in (5/3)*N min i.e. (5/3)*N*60 Seconds = 100N Seconds.
Option A is the correct Answer. |
AQUA-RAT | AQUA-RAT-35089 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 70 kg. What is the weight of the new person? | [
"90",
"65",
"85",
"95"
] | A | Total increase in weight = 8 × 2.5 = 20
If x is the weight of the new person, total increase in weight = x−70
=> 20 = x - 70
=> x = 20 + 70 = 90
Answer is A. |
AQUA-RAT | AQUA-RAT-35090 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
When Rahul was born, his father was 30 years older than his brother and his mother was 24 years older than his sister. If Rahul's brother is 6 years older than him and his mother is 3 years younger than his father, how old was Rahul's sister when he was born ? | [
"10 years",
"7 years",
"8 years",
"9 years"
] | D | When Rahul was born, his brother's age = 6 years; his father's age = (6 + 30) years = 36 years,
his mother's age = (36 - 3) years = 33 years;
his sister's age = (33 - 24) years = 9 years.
Answer: Option D |
AQUA-RAT | AQUA-RAT-35091 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Arun and Tarun can do a work in 10 days.After 4 days tarun went to his village.How many days are required to complete the remaining work by Arun alone.Arun can do the work alone in 50 days. | [
"16 days.",
"17 days.",
"18 days.",
"30 days."
] | D | They together completed 4/10 work in 4 days.
balance 6/10 work will be completed by Arun alone in 50*6/10 = 30 days.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35092 | # Identifying Prime Numbers from 507 to 10647
#### ErikHall
##### New member
So lets say you have the Numbers (2677; 6191 and 1091), how would you go about the check that 2677 and 1091 are Prime ?
The conditions are:
- You dont have a calculator
- You just have a Pen
Is there a clever way to do it except looking at the Last Digit ?
Thanks for the Help !
Also small fun fact, if you use the Google Calculator and divide by Zero, it says "Infinity". Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
#### lookagain
##### Senior Member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
#### lev888
##### Full Member
Is there a clever way to do it except looking at the Last Digit ?
Could you explain the last digit method?
#### ErikHall
##### New member
Could you explain the last digit method?
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
#### ErikHall
##### New member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
#### Subhotosh Khan
##### Super Moderator
Staff member
Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
Incorrect.
The following is multiple choice question (with options) to answer.
Which of the following is a prime number ? | [
"33",
"81",
"97",
"93"
] | C | C
97
Clearly, 97 is a prime number. |
AQUA-RAT | AQUA-RAT-35093 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 40, what is the compound interest on the same sum at the rate and for the same time? | [
"s. 41.00",
"s. 51.22",
"s. 51.219",
"s. 41.18"
] | A | Explanation:
Sum = (40 * 100) / (2 * 5) = Rs. 400
Amount = [400 * (1 + 5/100)2] = Rs. 441
C.I. = (441 - 400) = Rs. 41
Answer:A |
AQUA-RAT | AQUA-RAT-35094 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.20000? | [
"6620",
"3378",
"2768",
"2999"
] | A | A = 20000(11/10)3
= 26620
= 20000
----------
6620
Answer: A |
AQUA-RAT | AQUA-RAT-35095 | Since $$^{2n}x \ge \mathrm{e}^{-1}$$ for all $$0 < x < \mathrm{e}^{-\mathrm{e}}$$, we have $$I_4(n) := \int_{\frac35\mathrm{e}^{-\mathrm{e}}}^{\mathrm{e}^{-\mathrm{e}}} \frac{1}{^{2n}x}\mathrm{d} x \le \int_{\frac35\mathrm{e}^{-\mathrm{e}}}^{\mathrm{e}^{-\mathrm{e}}} \frac{1}{x^{x^{1/\mathrm{e}}}}\mathrm{d} x < 0.0715.$$
One can use Mathematical Induction to prove that $$^{2n} x \ge \frac34$$ for all $$0 < x < \frac35 \mathrm{e}^{-\mathrm{e}}$$. We have $$I_5(n) := \int_0^{\frac35 \mathrm{e}^{-\mathrm{e}}} \frac{1}{^{2n}x}\mathrm{d} x \le \int_0^{\frac35 \mathrm{e}^{-\mathrm{e}}} \frac{1}{x^{x^{3/4}}}\mathrm{d} x < 0.0485.$$
Thus, $$I(n) = I_1(n) + I_2(n) + I_3(n) + I_4(n) + I_5(n) < 2$$.
The following is multiple choice question (with options) to answer.
(1/5)^e * (1/4)^18 = 1/(2(10)^35). What is e? | [
"17",
"18",
"34",
"35"
] | D | We need to arrive at a common base. -->
(5)^(-e) * 2^(-36) = 2^(-36)*5^(-35)
5^(-e)=5^(-35)
-e=-35
e=35=D |
AQUA-RAT | AQUA-RAT-35096 | 2. ### math
list all the 3 digit numbers that fit these clues. the hundreds digits is less than 3. the tens digit is less than 2. the ones digit is greater than 7
3. ### Math
Think of a five-digit number composed of odd numbers. The thousands digit is two less than the ten thousands digit but two more than the hundreds digit. The tens digit is two more than ones digit which is four less than the
4. ### Math (Confused)
How many international direct-dialing numbers are possible if each number consists of a four-digit area code (the first digit of which must be nonzero) and a five-digit telephone numbers (the first digit must be nonzero)? a.
1. ### math
how many positive 4-digit numbers are there with an even digit in the hundreds position and an odd digit in the tens position? a. 10,000 b. 5,040 c. 2,500 d. 2,250
2. ### math
Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3-digit integer to the original 4-digit
3. ### Math
7 digit number no repetition of numbers digit 5 in thousands place, greatest digit in the millions place, digit in the hundred thousands place is twice the digit in the hundreds place, digit in the hundreds place is twice the
4. ### math
the hundred thousands digit of a six-digit even numbers is 3 more than the thousand digit,which is twice the ones digit.give at least four numbers that satisfy the given condition.
The following is multiple choice question (with options) to answer.
An integer is said to be “diverse” if no two of its digits are the same. For example, 327 is “diverse” but 404 is not. How many “diverse” two digit numbers are there ? | [
"70",
"72",
"81",
"90"
] | C | A two digit number has to be formed from digits 0 to 9 such that unit's digit and ten's digit are not same.
The unit's digit can have 10 values from 0 to 9.
Ten's digit can have 9 values from 1 to 9
So total number of two digit numbers will be 9 x 10 = 90
But there are 9 numbers which are not diverse.
11,22,33,44,55,66,77,88,99
so, the number of diverse two digit numbers will be 90-9 = 81
Answer:- C |
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