source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38997 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
How many unique positive odd integers less than 90 are equal to the product of a positive multiple of 5 and an odd number? | [
"4",
"9",
"11",
"12"
] | B | The Question basically asks how many positive odd integers less than 90 are odd multiples of 5
So we have 5,15,25,35,45,55,65,75,85
= 9
Ans B |
AQUA-RAT | AQUA-RAT-38998 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
If a number is formed by writing integers 1 to 150 one after another, like 12345...150. How many T digits does this integer consists of? | [
"150",
"339",
"342",
"359"
] | C | Number of single digit integers = 9
Number of 2 digit integers = 99 - 10 + 1 = 90
Number of 3 digit integers = 150 - 100 + 1 = 51
Number of digits in the integer T= 1*9 + 2*90 + 3*51 = 9 + 180 + 153 = 342
Answer: C |
AQUA-RAT | AQUA-RAT-38999 | $24 = 1^3 + 2^3 - 3^3 - 4^3 + 5^3 + 6^3 - 7^3 + 8^3 + 9^3 + 10^3 - 11^3 + 12^3 + 13^3 + 14^3 - 15^3 - 16^3$ (at least 7 other ways)
Each of the above representations is the shortest possible. The algorithm created a sequence of sets $S_n$ with $S_0 = \{0\}$ and $S_n = \{x+n^3, x-n^3 : x \in S_{n-1}\}$ for $n > 0$, so that $S_1 = \{1, -1\}$, $S_2 = \{9, -7, 7, -9\}$, etc. The algorithm also kept track of the sequence of signs used to arrive at each particular number.
• Nice work ! :D Thank's !!!
– r9m
Dec 24 '14 at 1:11
• @Unit: Do you suppose you can do the same for at least $1<N<100$ for $5$th powers? Kindly refer to my answer below. Dec 29 '14 at 3:54
• @TitoPiezasIII: Ahh, I just read this message. I will try, though it seems alexwlchan is almost there.
– Unit
Dec 29 '14 at 22:55
• @Unit: And it seems Zander even managed $0\leq N<10000$. The ingenuity of some people never ceases to amaze. Dec 29 '14 at 23:44
For 5th powers, there is an identity,
$$\sum\limits_{k=1}^{168}\pm(x+k)^5 = 480$$
analogous to the 3d powers mentioned by R. Millikan in his answer.
What remains to be shown is that all $0\leq N<240$ can be decomposed into sums of fifth powers. See this post.
(P.S. The number of addends can be reduced to just $m=168$.)
The following is multiple choice question (with options) to answer.
What is the smallest integer y for which 27^y> 3^24? | [
"7",
"8",
"9",
"10"
] | C | 27^y> 3^24
Converting into the same bases:
27^y> 27^8
Therefore for the equation to hold true, y> 8 or y= 9
Option C |
AQUA-RAT | AQUA-RAT-39000 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A tradesman by means of his false balance defrauds to the extent of 50%? in buying goods as well as by selling the goods. What percent does he gain on his outlay? | [
"120%",
"125%",
"150%",
"100%"
] | B | g% = 50 + 50 + (50*50)/100
= 125%
ANSWER:B |
AQUA-RAT | AQUA-RAT-39001 | # 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.
The ratio of two natural numbers is 5:6. If a certain number is added to both the numbers, the ratio becomes 7:8. If the larger number exceeds the smaller number by 10, find the number added? | [
"10",
"87",
"20",
"72"
] | C | Let the two numbers be 5x and 6x.
Let the number added to both so that their ratio becomes 7:8 be k.
(5x + k)/(6x + k) = 7/8
42x = 7k => k = 2x.
6x - 5x = 10 => x = 10
k = 2x = 20.
Answer:C |
AQUA-RAT | AQUA-RAT-39002 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is? | [
"11",
"16",
"77",
"88"
] | B | Ratio of times taken by Sakshi and Tanys = 125:100 = 5:4
Suppose Tanya takes x days to do the work.
5:4 :: 20:x => x= 16 days.
Hence, Tanya takes 16 days to complete the work.
Answer: B |
AQUA-RAT | AQUA-RAT-39003 | python, python-3.x
if seniors + carers >= int(40) or seniors <= int(1) or seniors >= int(37) or carers <= int(1) or (seniors >= int(24) and carers <= int(2)):
print("Enter Valid Info Please")
else:
print("Seniors going: " + repr(seniors) + " | Carers going: " + repr(carers))
if seniors >= int(1) and seniors <= int(16):
cost_coach = int(150)
cost_meal = int(14)
cost_ticket = int(21)
seats = int(16)
if seniors >= int(17) and seniors <= int(26):
cost_coach = int(190)
cost_meal = float(13.50)
cost_ticket = int(20)
seats = int(26)
if seniors >= int(27) and seniors <= int(36):
cost_coach = int(225)
cost_meal = int(13)
cost_ticket = int(19)
seats = int(36)
cost_coach_per_person = float(cost_coach / seniors)
rounded_cost_coach_per_person = float("%.2f" % cost_coach_per_person)
total_cost = (cost_meal * seniors) + (cost_ticket * seniors) + cost_coach
cost_per_person = cost_meal + cost_ticket + rounded_cost_coach_per_person
print("The total cost is $" + repr(total_cost))
print("The total cost per person is $" + repr(cost_per_person))
names_ppl_going = []
print("Logbook (Names and payment)")
n = int(1)
total_payment = float(0)
for x in range(seniors):
person_id = repr(n) + ")"
names_ppl_going.append(person_id)
name = input("[" + repr(n) + "] Enter a name: ")
name2 = name + ":"
names_ppl_going.append(name2)
The following is multiple choice question (with options) to answer.
If the charge of staying in a student youth hostel $18.00/day for the first week, and $15.00/day for each additional week, How much does it cost to stay for 23 days? | [
"$336",
"$289",
"$282",
"$274"
] | A | Total number of days of stay = 23
Charge of staying in first week = 18*7 = 126 $
Charge of staying for additional days =(23-7)*15 = 16*15 = 240 $
Total charge =126 +240 = 366$
Answer A |
AQUA-RAT | AQUA-RAT-39004 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
A worker makes a toy in every 2h. If he works for 80h, then how many toys will he make ? | [
"40",
"54",
"45",
"39"
] | A | Answer
Let number of toys be N.
More hours, More toys (Direct proportion)
2 : 80 : : 1 : N
⇒ N = 80/2 = 40 toys
Correct Option: A |
AQUA-RAT | AQUA-RAT-39005 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
In a school of 850 boys, 44% of Muslims, 28% Hindus, 10% Sikhs and the remaining of other communities. How many belonged to the other communities? | [
"120",
"125",
"153",
"200"
] | A | 44 + 28 + 10 = 82%
100 – 82 = 18%
850 * 18/100 = 153
ANSWER A |
AQUA-RAT | AQUA-RAT-39006 | # 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.
When a certain number X is divided by 61, the remainder is 24. What is the remainder when X is divided by 5? | [
"2",
"3",
"4",
"5"
] | C | When a certain number X is divided by 61, the remainder is 24. What is the remainder when X is divided by 5?
Putting a value say x = 24 we get remainder as 24 when divided by 61.
When 24 divided by 5 we get 4 as remainder.
C is the answer. |
AQUA-RAT | AQUA-RAT-39007 | ## 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 (2 to the x) - (2 to the (x-2)) = 3 (2 to the 11), what is the value of x? | [
"9",
"11",
"13",
"15"
] | C | (2 to the power x) - (2 to the power (x-2)) = 3 (2 to the power 11)
2^x - 2^(x-2) = 3. 2^11
Hence x = 13.
Answer is C |
AQUA-RAT | AQUA-RAT-39008 | # 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? | [
"2 2 1/8 kmph",
"2 3 1/2 kmph",
"4 2 1/2 kmph",
"2 2 1/2 kmph"
] | D | 100 --- 10 DS = 10
? ---- 1
75 ---- 15 US = 5
? ----- 1 S = (10 - 5)/2
= 2 2 1/2 kmph
Answer:D |
AQUA-RAT | AQUA-RAT-39009 | # 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.
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in scheme B? | [
"6400",
"2777",
"2666",
"2888"
] | A | Let the sum invested in scheme A be Rs. x and that in scheme B be Rs. (13900 - x). Then,
(x * 14 * 2)/100 + [(13900 - x) * 11 * 2]/100 = 3508
28x - 22x = 350800 - (13900 * 22)
6x = 45000 => x = 7500
So, sum invested in scheme B = (13900 - 7500) = Rs. 6400.
Answer: A |
AQUA-RAT | AQUA-RAT-39010 | Will give the value for two hits or fewer and two hits exactly respectively.
5. Oct 3, 2016
### RJLiberator
Allright, so now the answer to A is P_12(2)+P_12(1)+P_12(0) = 0.168+0.014+0.071 = 0.253
This makes more reasonable sense to me and I see how it makes sense with the additions of the probabilities as it relates to the question. Now I will consider part b.
6. Oct 3, 2016
### RJLiberator
Now that I got the hang of this, this is quite easy (I think).
So for part b, all I did was calculate the probabilities P_12(4) and P_12(3). I used my previous answer in A and summed up these values. I then took 1-this sum and that is the probability that he gets 5 hits or more.
This turned out to be 0.276 which is a reasonable number.
The following is multiple choice question (with options) to answer.
Three cannons are firing at a target. If their individual probabilities to hit the target are 0.9, 0.7, and 0.8 respectively, what is the probability that none of the cannons will hit the target after one round of fire?
prob. that all the cannons will hit the target = .06
prob. that none of the cannons will hit = 1 - .06 = .94 | [
"0.06",
"0.12",
"0.21",
"0.29"
] | A | The probability that eachdoesn'thit is: 0.1, 0.3, and 0.2. When we have multiple independent events, we multiply the probabilities:
.9*.7*.8 = 0.06.
Option: A |
AQUA-RAT | AQUA-RAT-39011 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can finish a piece of work in 5 days. B can do it in 10 days. They work together for 1 day and then A goes away. In how many days will B finish the work? | [
"8 days",
"5 days",
"6 days",
"7 days"
] | D | 1/5 + (1 + x)/10 = 1 => x = 7 days
Answer: D |
AQUA-RAT | AQUA-RAT-39012 | # arrangement
#### jacks
##### Well-known member
How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
#### Jameson
Staff member
How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
I am not 100% sure that I am right here but want to give it a shot.
The starting word has 9 letter with two A's. The number of choices for the first letter is 7 though, because I must be used at the end and M cannot be first, so 9-2=7 choices are left. Now write the number of possible letter choices for the second letter. Again it can't be I but this time it could be M, and we must note that a some letter (not I or M) has been chosen for the first spot. 9-1-1=7 again. Try to continue this process until the end of the word.
Lastly, there are two letters A which are identical so when they switch positions but everything else remains the same, the word is the same and we can't double count. To correct for this error you divide the answer you got above by 2!.
#### CaptainBlack
##### Well-known member
How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
If all the letters were different there would be $$8!$$ permutations which end with I, of which $$7!$$ start with M, so $$8!-7!=7.7!$$ would not start with M but end with I. However there is a repeated A so the final answer is half that.
CB
#### soroban
##### Well-known member
Hello, jacks!
I'll give it a try . . .
The following is multiple choice question (with options) to answer.
How many two letter words are formed using the letters of the word MICKY? | [
"28",
"21",
"26",
"20"
] | D | The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is 5P2
= 5 * 4 = 20.
Answer: D |
AQUA-RAT | AQUA-RAT-39013 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee? | [
"70",
"560",
"630",
"1,260"
] | B | Total number of ways to select at least one professor = Number of ways to select 3 people from 10 students and 7 professors - Number of ways to select 3 people from 10 student (i.e. without including any professor)
= 17C3 - 10C3 = 680 - 120 = 560
ANSWER:B |
AQUA-RAT | AQUA-RAT-39014 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 402 to 500 inclusive? | [
"14,550",
"18,550",
"22,550",
"26,550"
] | C | 2 + 4 + 6 + 8 +...+ 100 = 2550
402 + 404 + ... + 500 = 50(400) + (2+4+...+100) = 20,000 + 2550 = 22,550
The answer is C. |
AQUA-RAT | AQUA-RAT-39015 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B start a business, with A investing the total capital of Rs.5000, on the condition that B pays A interest @ 10% per annum on his half of the capital. A is a working partner and receives Rs.500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year? | [
"A)2500",
"B)3500",
"C)5000",
"D)6000"
] | B | Interest received by A from B = 10% of half of Rs.5000 = 10% * 2500 = 250.
Amount received by A per annum for being a working partner = 500 * 12 = Rs.1000.
Let 'P' be the part of the remaining profit that A receives as his share. Total income of A = (250 + 1000 + P)
Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally.
Income of A = Twice the income of B
(250 + 1000 + P) = 2(P)
P = 1250
Total profit = 2P + 1000
= 2*1250+ 1000 = 3500
Answer:B |
AQUA-RAT | AQUA-RAT-39016 | The operations above are perfectly reasonable. Let me show you the kind of situation that I think you were worried about, and you'll see why it's different from the situations above:
I take 4 exams and have an average score of 80. Then I take 2 more exams and on those 2, my average is 100. What's my overall average for the course?
The wrong way to do it is (80+100)/2 = 90. This is wrong because the averages were of different-sized groups. To get the correct answer, I know that on the first 4 exams I got 320 total points because when I divide 320 by 4, I get 80.
Similarly, for the last two exams, I must have gotten 200 points total. So for the six exams, I got 200+320 = 520 points and 520/6 = 86.66666 = my real grade average.
We saw this scenario last week: If we just average the two averages, the result can only be described as the average of the averages, not the average for the course. The latter has to be weighted, because for the course, each exam counts the same, not each of these two sets of exams.
As we saw last week, a weighted average can often be understood by breaking it down:
For your problem, the averages you are averaging are for the same period, so it works out. To convince you that it's true, let's just look at a situation where the averages came from 10 months of data.
Then in the first line of business, 500 people must have been hired, since 500/10 = 50. Similarly, 800 were hired in the other, since 800/10 = 80. Altogether, 1300 people were hired in the ten months, or 1300/10 = 130 per month, company-wide. Or if you're trying to get the average per line of work, 65 is right, since if each group had hired 65 people each month for 10 months, there would be 65*20 = 1300 total hires, so it works out.
Bottom line: always think about what you want, and what an average means, rather than use an average (or not!) unthinkingly.
## Using a weighted average
Averaging Averages
The following is multiple choice question (with options) to answer.
In three annual examinations, of which the aggregate marks of each was 500, a student secured
average marks 45% and 55% in the first and the second yearly examinations respectively. To secure 60% average total marks, it is necessary for him in third yearly examination to secure marks : | [
"300",
"350",
"400",
"450"
] | C | total marks:1500 for three exams
60% of 1500=900
first exam marks=45% of 500=225
second exam marks=55% of 500=275
let X be the third exam marks
225 + 275 + X =900
X=400
ANSWER:C |
AQUA-RAT | AQUA-RAT-39017 | The statement
"(the integer referred to by the decimal symbol 1) + (the integer referred to by the decimal symbol 1) = (the integer referred to by the decimal symbol 2)"
is indeed always true. Whether or not this is expressed in symbols as
"1 + 1 = 2"
depends on how you choose to represent integers.
The following is multiple choice question (with options) to answer.
If 1/a + 3/a + 4/a is an integer, which of the following must also be an integer? | [
"12/a",
"3/a",
"8/a",
"9/a"
] | C | 1/a + 3/a + 4/a
=8/a
Answer C |
AQUA-RAT | AQUA-RAT-39018 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How long does a train 110 m long running at the speed of 36 km/hr takes to cross a bridge 132 m length? | [
"82.1 sec",
"12.1 sec",
"24.2 sec",
"13.1 sec"
] | C | Speed = 36 * 5/18 = 10 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/10 = 24.2 sec.
Answer:C |
AQUA-RAT | AQUA-RAT-39019 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
What is the greatest number of three digits which is divisible by 5, 10, 15 ? | [
"985",
"970",
"955",
"990"
] | D | Greatest number of 3 digits = 999
LCM of 5, 10, 15 = 30
999 ÷ 30= 33, remainder = 9
Hence, greatest number of four digits which is divisible by 5,10,15
= 999 - 9 = 990
answer : D |
AQUA-RAT | AQUA-RAT-39020 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
In a race with 30 runners where 6 trophies will be given to the top 6 runners (the trophies are distinct: first place, second place, etc), how many ways can this be done? | [
"6^6 ways",
"8^9 ways",
"7^8 ways",
"8^7 ways"
] | A | 6 people can be prized with distinct prizes in 6^6 ways
ANSWER:A |
AQUA-RAT | AQUA-RAT-39021 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
At what time between 6 and 7 are the hands of the clock coincide? | [
"6:39",
"6:37",
"6:31",
"6:32"
] | D | For hour = 360/12 = 300300/hr
For Minute = full rotation = 36003600/hr
Let the line is 't' , for 6 = 6*30=18001800
then
30 t + 180=360 t
330t = 180
t = 180/330
t = 6/11 hr 6/11*60=360/11=32611611
Ans. D |
AQUA-RAT | AQUA-RAT-39022 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times? | [
"37/256",
"56/256",
"65/256",
"70/256"
] | A | More than 5 times tails = 6times +7times+8times = 8C6 + 8C7 + 8C8 = 37
- - - - - - - -
2 2 2 2 2 2 2 2
2^8 times total events and 37 events where tails side up .
So probability = 37/2^8 = 37/256 (Answer A) |
AQUA-RAT | AQUA-RAT-39023 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
A certain bakery baked a batch of 450 cookies one day. Of those, 320 contained nuts, 230 contained chocolate chips, and 85 contained neither nuts nor chocolate chips. What is the fewest possible number of cookies with both chocolate chips and nuts that would need to be added to that batch so that cookies with both nuts and chocolate chips represented more than 3/5 of all the cookies in the batch? | [
"166",
"213",
"413",
"438"
] | B | Cookies which have both nuts and chocolate chips = 185
Let fewest possible number of cookies with both chocolate chips and nuts that would need to be added to that
batch so that cookies with both nuts and chocolate chips represented more than 3/5 of all the cookies in the batch = x
(185+x)/(450+x) = 6/10
=>1850 + 10x = 2700 + 6x
=> 4x = 850
=> x = 212.5
Therefore x = 213
Answer B |
AQUA-RAT | AQUA-RAT-39024 | ### Show Tags
19 Aug 2015, 01:34
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Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
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Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The avg. age ofa group of 12 students is 20yers.If 4 more students join the group,the avg age increases by 1 year.The avg age of the new student is? | [
"22 years",
"23 years",
"24 years",
"25 years"
] | C | Total age of 12 students=12*20=240
If total age of 4 students=x
Then, (240+x)/(12+4) =(20+1), x=96
So average age of new students=96/4= 24 years
ANSWER:C |
AQUA-RAT | AQUA-RAT-39025 | 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 room has equal number of men and women. Eight women left the room, leaving twice as many men as women in the room . What was the total number of men and women present in the room initially? | [
"28",
"30",
"31",
"32"
] | D | Assume that initial number of men = initial number of women = x
2(x-8) = x
=> 2x - 16 = x
=> x = 16
Total number of men and women = 2x = 2 × 16 = 32
Answer is D. |
AQUA-RAT | AQUA-RAT-39026 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
In how many ways can you seat 4 people on a bench if one of them, Rohit, does not want to sit on the middle seat or at either end? | [
"720",
"6",
"2880",
"5040"
] | B | Since Rohit does not want to sit on the middle seat or at either end (3 chairs), then he can choose 1 chairs to sit. The remaining 3 people can sit in 3! ways. Thus the # of arrangements is 1*3! = 6.
Answer: B. |
AQUA-RAT | AQUA-RAT-39027 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
The annual cost of owning and operating a car, C dollars, is a linear function of the distance, d kilometers, it is driven. | [
"0.3d + 1609",
"0.3d + 16089",
"0.3d + 1600",
"0.3d + 1602"
] | C | c = md + b
The cost is $ 4600 for 10 000 km
4600 = m (10 000) + b
10 000 m + b = 4600
b = -10 000 m + 4600 ------- (i)
The cost is $ 9100 for 25 000 km
9100 = m (25 000) + b
9100 = 25 000 m + b
25 000 m + b = 9100
b = -25 000 m + 9100 --------- (ii)
From equation (i) and (ii) we get;
-10 000 m + 4600 = -25 000 m + 9100
-10 000 m + 25 000 m = 9100 – 4600
15 000 m = 4500
m = 4500/15 000
m = 3/10
m = 0.3
Now, put the value of m = 0.3 in equation (i) we get;
b = -10 000 (0.3) + 4600
b = -3000 + 4600
b = 1600
Answer: m = 0.3 and b = 1600
(b) c = md + b
m = 0.3
b = 1600
c = 0.3d + 1600
Answer: C = 0.3d + 1600 |
AQUA-RAT | AQUA-RAT-39028 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Linda and Angela contract to paint a neighbor's house. Even though Linda spends 80% more time painting the house than Angela, each receives a payment of m dollars when the work is completed. If Angela decides to pay Linda n dollars so that they would have received the same compensation per hour worked, what is n in terms of m ? | [
"1/2 m",
"1/3 m",
"1/4 m",
"1/5 m"
] | A | After the final transaction:
Angela received $(m-n) for t hours;
Linda received $(m+n) for 1.5t hours;
We are told that after that transaction they received the same compensation per hour: (m-n)/t=(m+n)/1.5t --> 1.5m-1.5n=m+n -->m=5n --> n=m/2.
Answer: A. |
AQUA-RAT | AQUA-RAT-39029 | # Mean and Median in a Classic River Crossing Problem
Consider the following classic problem:
Four people on the west side of a river wish to use their single boat to get to the east side of a river. Each boat ride can hold at most two people, and the time it takes to get across will be the time preferred by the slower occupant. The time preferences are: $1, 2, 5$ and $10$ minutes. What is the minimal amount of time in which you can get all four people across the river, where an eastbound trip must have two occupants and a westbound trip must have one occupant?
Answer: $17$ minutes. (Though many mistakenly believe it is $19$ minutes.)
Question 1: If you look at all possible ways of ferrying these four people across the river, subject to the above constraints, what would the average (mean and median) times be?
Question 2: What if you replace the time preferences with $x_1, x_2, x_3$ and $x_4$ minutes?
Question 3: What if you have time preferences $x_1, \ldots, x_n$ minutes for $n$ people, respectively?
Probably the easiest way to broach Question 1 would be to write a quick program to compute the answer, and perhaps doing this for several different time preferences would give some insight into the mean and median in the general four person case. I'm not quite sure how I would start thinking about the general $n$ person case; perhaps by solving it for $n = 1, 2, 3, 4$.
Answers (even partial ones) to any or all of my questions would be greatly appreciated!
The following is multiple choice question (with options) to answer.
The toll for crossing a certain bridge is $0.65 each crossing. Drivers who frequently use the bridge may instead purchase a sticker each month for $13.00 and then pay only $0.30 each crossing during that month. If a particular driver will cross the bridge twice on each of x days next month and will not cross the bridge on any other day, what is the least value of x for which this driver can save money by using the sticker? | [
"14",
"15",
"16",
"28"
] | C | Option #1: $0.75/crossing....Cross twice a day = $1.5/day
Option #2: $0.30/crossing....Cross twice a day = $0.6/day + $13 one time charge.
If we go down the list of possible answers, you can quickly see that 14 days will not be worth purchasing the sticker. 1.5x14 (21) is cheaper than 0.6x14 +13 (21.4)...it's pretty close so let's see if one more day will make it worth it... If we raise the number of days to 15, the sticker option looks like a better deal...1.5x15 (22.5) vs 0.6x15 + 13 (22). Answer: C |
AQUA-RAT | AQUA-RAT-39030 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many 4-letter word with or without meaning can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not allowed | [
"5040",
"1254",
"4510",
"1203"
] | A | 'LOGARITHMS' contains 10 different letters
Required number of words
= Number of arrangements of 100 letters taking
4 at a time
= 10P4
= 10 * 9 * 8 * 7
= 5040
Ans: A |
AQUA-RAT | AQUA-RAT-39031 | (ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED.
The following is multiple choice question (with options) to answer.
If n is a 9-digit positive integer, all of whose digits are the same, which of the following must be true?
I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 10 | [
"I and II only",
"I only",
"I and III only",
"II and III only"
] | A | Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3
Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.
Rules for divisibility by 9: The sum of the digits should a multiple of 9
Consider no 11111111...9 times = The sum 9*1=27----> divisbible by 3,9
consider number to be 222....9 times, then sum of the no. 9*2=18 divisible by 3,9
So why so because when you sum the numbers either you can add the digits 9 times or multiply the digit *9..
Note that since 9 is divisble by 9 and 3 and thus the sum of the nos will be divisible by all the nos.
A |
AQUA-RAT | AQUA-RAT-39032 | # 2d kinematics
1. Sep 13, 2013
### Jabababa
1. The problem statement, all variables and given/known data
A diver jumps off a 3.00 meter high diving board with an initial velocity of 1.75 m/s vertically upward.
a) What is the diver's veolcity when she reaches the water? (assume the surface of the water is 3.00 meters below the board.)
b) What is the highest point the diver reaches above the water?
2. Relevant equations
3. The attempt at a solution
a) Since we know the height, initial velocity and acceleration, we use V^2= Vo^2 + 2ad
v=square root of vo^2 + 2ad
v = square root of 1.75^2 + 2(9.80)(3)
v= 7.86527 m/s = 7.87m/s
b)we use the same equation but this time we only calculate the diver going up, the maximum height of the the jump.
V^2 = Vo^2 + 2ad
d= V^2-Vo^2/2a
d= Vo^2/2a
d= 0.3125m
The maximum height = diver's jump + the 3.00m below the diving board.
so the answer is 3.3125m is the highest point the diver reaches above the water.
Can anyone check see if what i did is right? Thank you.
2. Sep 13, 2013
### Staff: Mentor
A useful check for this is that the velocity she hits the water should equal what she would have if she fell in freefall starting with zero velocity from the max height attained.
Your answer to (a) looks correct.
3. Sep 13, 2013
### Jabababa
thank you NascentOxygen
Is b) correct as well? The logic behind it seems correct to me. Finding the max height the diver jumps plus the 3.00m below the diving board seems reasonable to me.
4. Sep 13, 2013
### Tanya Sharma
I am not sure what you have done in part b) .
For maximum height above diving board use the relation v2=u2+2as where v=0,u=1.75m/s,a=-9.8 m/s2.
5. Sep 13, 2013
### Jabababa
The following is multiple choice question (with options) to answer.
In a diving competition, each diver has a 25% chance of a perfect dive. The first perfect dive of the competition, but no subsequent dives, will receive a perfect score. If Janet is the third diver to dive, what is her chance of receiving a perfect score? (Assume that each diver can perform only one dive per turn.) | [
"1/5",
"1/15",
"4/25",
"9/64"
] | D | Probability = Favourable Outcomes / Total Outcomes = 1 - (Unfavourable Outcomes / Total Outcomes)
Given:Probability of Perfect dive = 0.25 = 25/100 = 1/4
i.e. Probability of Non-Perfect dive = 0.75 = 75/100 =3/4
The probability of Janet to dive and get a perfect score depends on that other other previous two dives must be Imperfect
Method-1:
i.e. Probability of First two being Imperfect and Third being Perfect dive = (3/4)*(3/4)*(1/4) =9/64
D |
AQUA-RAT | AQUA-RAT-39033 | 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.
If 40 men take 15 days to to complete a job, in how many days can 25 men finish that work? | [
"24",
"25",
"26",
"27"
] | A | Ans. 24 days |
AQUA-RAT | AQUA-RAT-39034 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains of equal are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is? | [
"50 m",
"72 m",
"80 m",
"82 m"
] | A | Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 - 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.
ANSWER:A |
AQUA-RAT | AQUA-RAT-39035 | ##### Well-Known Member
Subscriber
Actually, it looks like the discount rate is a risk-free rate not necessarily equal to a LIBOR of the correct period length.
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi Brian,
In many countries such as Thailand, the Treasury yield curve is illiquid across maturities. On the other hand, the LIBOR/Swap Yield curve is liquid across maturities worldwide. The LIBOR swap zero curve is obtained by bootstrapping and is almost risk-free. That is why it is used to discount the Cash flows to obtain the value of the FRA. The Fixed rate cannot be used for discounting because FRA's are OTC - the fixed rate is unique to that particular FRA.
Thanks!
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Thanks Nicole - appreciate it
Jayanthi
#### brian.field
##### Well-Known Member
Subscriber
Thanks for your message Jayanthi. While I agree with what you are saying, generally, it isn't really addressing my philosophical question.
Consider today at time t=0 and an FRA that requires a fixed payment of 5% and a floating payment of 1YLIBOR from t=3 to t=4.
At t=0, we do not know that actual 1YLIBOR rate that will apply in 3 years but we do know the 1Y forward 1YLIBOR rate payable in 3 years. We establish the FRA at t=0, and at this time, the fixed rate is set so that the FRA's value is 0 at t=0. By the time we get to t=3, the FRA may not equal 0.
My point is that at t=3, we will know that value of the FRA at t=4 and it is customary to discount the t=4 value back to t=3 with a risk-free rate. I thought that I read that this discount rate was the appropriate tenor LIBOR.
My question is as follows:
The following is multiple choice question (with options) to answer.
The banker's discount on Rs. 1600 at 15% per annum is the same as true discount on Rs. 1680 for the same time and at the same rate. The time is: | [
"1 months",
"4 months",
"7 months",
"9 months"
] | B | S.I. on Rs. 1600 = T.D. on Rs. 1680.
Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
Time = 100 x 80 year = 1 year = 4 months.
1600 x 15 3
View Answer Discuss in Forum Workspac
Answer:B |
AQUA-RAT | AQUA-RAT-39036 | \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain
The following is multiple choice question (with options) to answer.
The average weight of D, E and F is 42 kg. If the average weight of D and E be 35 kg and that of E and F be 41 kg, what is the weight of E? | [
"23 Kg",
"24 Kg",
"25 Kg",
"26 Kg"
] | D | Let the weight of D, E and F are a,b and c respectively.
Average weight of D,E and F = 42
a + b + c = 42 × 3 = 126 --- equation(1)
Average weight of D and E = 35
a + b = 35 × 2 = 70 --- equation(2)
Average weight of E and F = 41
b + c = 41 × 2 = 82 --- equation(3)
equation(2) + equation(3) - equation(1)
=> a + b + b + c - (a + b + c) = 70+82 - 126
=> b = 152-126 = 26
Weight of B = 26 Kg
ANSWER:D |
AQUA-RAT | AQUA-RAT-39037 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day? | [
"48 min. past 19",
"48 min. past 12",
"48 min. past 10",
"48 min. past 11"
] | A | Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
Answer: A) 48 min. past 12. |
AQUA-RAT | AQUA-RAT-39038 | The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.
As Martin Schulz pointed out in the comments (now deleted),
9165784320 is divisible by 7
• I have 1385679240 & 7165932840 neither divisible by 7 – DeNel Oct 27 '18 at 0:10
• My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that. – DeNel Oct 27 '18 at 0:11
• "The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4. – elias Oct 27 '18 at 2:08
• the correct wording would be the last 2 digits must be divisible by 4 and the last 3 digits must be divisible by 8. In fact we just need to check for divisibility by 8, since that implies divisibility by 4 as well – phuclv Oct 27 '18 at 6:24
This is a good problem to attack by computer:
It turns out that there are 7344 solutions, including 1056 to the bonus.
So I thought, how far can we go?
And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.
Code used (Python 3 IDLE):
The following is multiple choice question (with options) to answer.
If the number 67925*4 is completely divisible by 8, then the smallest whole number in place of * will be? | [
"1",
"2",
"0",
"4"
] | C | The number 5x4 must be divisible by 8.
x=0 as 504 is divisible by 8.
Correct Option : C |
AQUA-RAT | AQUA-RAT-39039 | • $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 40%? | [
"4",
"6",
"5",
"2"
] | D | We need on-time departure rate to be higher than 4/10, so it should be at least 5/11, which means that 5 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 5-3=2 subsequent flights need to depart on-time.
Answer: D |
AQUA-RAT | AQUA-RAT-39040 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
Pam and Stanley packed several boxes with reams of paper. While both packed, Pam packed 40% of the boxes. After Pam stopped, Stanley packed the same number of boxes that he had packed while working with Pam. What is the ratio of the number of boxes Pam packed to the number of boxes Stanley packed? | [
"1 to 4",
"1 to 3",
"2 to 6",
"3 to 4"
] | C | Correct Answer: C
Solution: C. We know that when Pam and Stanley were both working, the ratio was 2 boxes by Pam to 3 boxes by Stanley. We also know that Stanley continued working after Pam stopped. He packed as many boxes alone as he had packed whlie working with Pam, effectively doubling his number of boxes. Thus, the ratio of Pam's boxes to Stanley's boxes is 2 to 6. Answer C is correct. |
AQUA-RAT | AQUA-RAT-39041 | the first question correctly: 1/5 For that 1/5 of the time when the first question has been guessed correctly, the second question could be guessed correctly 1/5 of the time. Instead, Microsoft is adding more questions of new types. It's probably the biggest struggle students have with multiple choice questions. Each question is multiple choice with 4 possible answers, only one of which is correct. Accel Math 6/7 -Probability Study Guide True/False Indicate whether the statement is true or false. Ever got stuck on multiple-choice questions in a test where you were not certain about the answer? Or, you had to rush through the last few questions due to paucity of time. 20 22) A test consists of 10 multiple choice questions, each with five possible answers, one of which is correct. My assumption is that you would choose A after reading the question and the answers (and considering them). You're right. What is the probability of getting exactly 4 correct answers? a multiple-choice test in which each question has 4 choices only one of which is correct. Probability of Two Events Occurring Together: Independent. Ruth Carver 1,219 views. (a) If student goes into exam completely unprepared and guesses on all 10, what is the probability of getting at least 5 answers right?. Let K be the event that you know the answer, C the event that you get the answer correct. 05 MC) The teacher of a seventh-grade class flipped a coin 100 times. A test contains 3 multiple choice questions. For additional insight into and help with multiple choice question you can check out Albert’s AP Biology practice questions. Those who perform well get a more difficult second testlet, while those who do not perform well receive a second medium difficulty testlet. Sex-linked recessive. What is the. That gives 0. Explore the combined power of descriptive statistics with challenging AP® Statistics probability questions. Test and improve your knowledge of Data Analysis, Probability & Statistics with fun multiple choice exams you can take online with Study. What is the probability of getting 5 multiple-choice questions answered correctly, if for each question the probability of answering it correctly is 1/3. If you write 50% below the answers on the test paper, you would be marked wrong. A Stats 10 test has 4 multiple choice questions with four choices with one correct answer each. As per new pattern of CBSE there are 10 questons of MCQ from
The following is multiple choice question (with options) to answer.
If 85 percent of the test takers taking an old paper and pencil GMAT exam answered the first question on a given math section correctly, and 65 percent of the test takers answered the second question correctly, and 5 percent of the test takers answered neither question correctly, what percent answered both correctly? | [
"60 %",
"55 %",
"70%",
"75%"
] | B | {Total} = {First correctly} + {Second correctly} - {Both correctly} + {Neither correctly}
100 = 85 + 65 - {Both correctly} + 5
{Both correctly} = 55.
Answer: B. |
AQUA-RAT | AQUA-RAT-39042 | # Math Help - Integers
1. ## Integers
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
2. ## Re: Integers
Originally Posted by fsiwaju
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
1. Set $j = k + 2m, m\in \mathbb{N}$. Then (j - k) is even.
2. Check all 5 terms, replacing j by k + 2m. For instance:
$j + 2k = k + 2m + 2k = 3k + 2m~\implies~$ ..... j+2k is only even if k is even because then j is even too.
...
$jk + j = (k+2m)k + (k + 2m) = (k + 2m)(k + 1)~\implies~$..... (jk + j) is even if k is even because then the 1st bracket would be even too; (jk + j) is even if k is odd because then the 2nd bracket would be even too. A product is even if one factor is even.
3. ## Re: Integers
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
if j - k is even, then j and k are both even or both odd.
4. ## Re: Integers
if j - k is even either j and k are both odd, or j and k are both even.
The following is multiple choice question (with options) to answer.
If x and j are integers and 2x–j= 11, then 4x+ j CANNOT be | [
" –5",
" 1",
" 13",
" 17"
] | D | 2x-j=11....j=2x-11
4x+j=4x+2x-11=6x-11
6x-11=-5...x=1
6x-11=1... x=2
6x-11=13...x=4
6x-11=17..X is not integer
6x-11=551..X is not integer
I think the choice E is 55 not 551. Otherwise both DE CANNOT be solution=D |
AQUA-RAT | AQUA-RAT-39043 | The correct answer is 4. The chart below shows the remainder calculated vs. the remainder that is the answer (4). The values that don't match up are once again in red. Please note that I removed 3 from the table because it was a root of the equation, so the results weren't useful for my purposes.
In summary, does the remainder theorem not work for smaller values or does the method of my textbook not work for smaller values. If the latter, what is a better method that is still feasible under a short time limit (SAT 2 Subject Test).
Edit:I realized the remainder theorem is working (I'm getting the correct remainder), so it isn't the remainder theorem that is the problem. What explains the difference between the actual remainder and the remainder predicted by the answer of the textbook?
• @Masacroso Your comment made me realize that the issue doesn't lie with the remainder theorem, but with the solution/method of the textbook. What is the reason for this difference? – Eric Wiener Dec 18 '16 at 4:15
• Don't know that I like that textbook's advice. In both cases it is far easier to actually calculate the remainder (in any of a number of ways) than try to "guess" it. In the latter case, the remainder of any polynomial $P(x)$ divided by $x-3$ is always $P(3)$. – dxiv Dec 18 '16 at 4:18
• Wow, the textbook's solution is really terrible. It happens that their method works as long as you choose $x$ large enough (and the polynomials have integer coefficients and the polynomial you're dividing by is monic), but they haven't given any justification that the $x$ they have chosen is large enough in this particular problem. – Eric Wofsey Dec 18 '16 at 4:22
The following is multiple choice question (with options) to answer.
x and y are positive integers. When x is divided by 9, the remainder is 2, and when x is divided by 7, the remainder is 4. When y is divided by 11, the remainder is 3, and when y is divided by 13, the remainder is 12. What is the least possible value of y - x? | [
"12",
"13",
"14",
"15"
] | C | When x is divided by 9, the remainder is 2: So, the possible values of x are: 2, 11, 21, 29, etc.
When x is divided by 7, the remainder is 4: So, the possible values of x are: 4,11,18,... STOP. Since both lists include 11, the smallest possible value of x is 11.
When y is divided by 11, the remainder is 3: So, the possible values of y are: 3, 14, 25, etc.
When y is divided by 13, the remainder is 12: So, the possible values of y are: 12, 25, ...STOP. Since both lists include 25, the smallest possible value of y is 25
Since the smallest possible values of x and y are 11 and 25 respectively, the smallest possible value of y - x is 14. So,C is the correct answer to the original question. |
AQUA-RAT | AQUA-RAT-39044 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
Aryan runs at a speed of 40 metre/minute. Rahul follows him after an interval of 5 minutes and runs at a speed of 50 metre/minute. Rahul's dog runs at a speed of 60 metre/minute and start along with Rahul. The dog reaches Aryan and then comes back to Rahul, and continues to do so till Rahul reaches Aryan. What is the total distance covered by the dog? | [
"1100 meters",
"1200 meters",
"1250 meters",
"1300 meters"
] | B | Let X = Minutes
And Y = Aryan runs speed = 40 meter per X
And Z = Rahul runs speed = 50 meter per X (i.e., 200 + 50x)
And D = Total distance covered by the dog = X* 60meters
Y = Z
40x = 200 + 50x
200 = 10x
x = 200/10
x = 20
So, by 20minutes, both Aryan and Rahul meets.(Rahul reaches Aryan.)
So, dog covered distance, D = X * 60 meters = 20*60 = 1200 meters
ANSWER:B |
AQUA-RAT | AQUA-RAT-39045 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Andrew purchased 11kg of grapes at the rate of 98 per kg and 7 kg of mangoes at the rate of 50 per kg. How much amount did he pay to the shopkeeper? | [
"1000",
"1055",
"1428",
"1500"
] | C | Cost of 11 kg grapes = 98 × 11 = 1078.
Cost of 7 kg of mangoes = 50 × 7 = 350.
Total cost he has to pay = 1078 + 350= 1428
C |
AQUA-RAT | AQUA-RAT-39046 | 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.
Amith was 3 times as old as Harsh 6 years back. Amith will be 5/3 times as old as Harsh 6 years hence. How much old is Harsh today? | [
"12",
"11",
"15",
"16"
] | A | A
12
Let Harsh’s age 6 years back = A
Then, Amith’s age 6 years back = 32.
Then 5/3 (A + 6 + 6) = (3A + 6 + 6)
5(A + 12) = 3 (3A + 12)
4A = 24
A = 6
Harsh’s age today = (A + 6) years
= 12 years. |
AQUA-RAT | AQUA-RAT-39047 | I got y = 1/2
6. Originally Posted by magentarita
You got the right answer. But how did you get y = 1?
I got y = 1/2
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
7. Originally Posted by magentarita
ok but none the choices given for the minimum point is (-2, 1/2).
(2,33)
(2,17)
(-2,-15)
(-2,1)
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.
Reason: I forgot to divid y by 2.
Remember orginal question: y =( 2x^2 ) + 8x +9
(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side
(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side
y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side
y = a [ ( x-h)^2 ] + k
where a=2, h=-2, k=1 sorry
8. I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.
From $y=ax^2+bx+c$, you want to convert to $y=a(x-h)^2+k$, where $(h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
$y=2x^2+8x+9$ Factor 2 out of first 2 terms
$y=2(x^2+4x)+9$ Next, complete the square in parentheses.
$y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side
$y=2(x+2)^2+1$ Now, we're in vertex form.
The following is multiple choice question (with options) to answer.
If (x/y)=(9/5), find the value (x^2+y^2)/(x^2-y^2) | [
"53/28",
"59/11",
"51/77",
"41/11"
] | A | = (x^2+y^2)/(x^2-y^2) = ( x^2 /y^2+ 1)/ ( x^2 /y^2-1) = [(9/5)^2+1] / [(9/5)^2-1]
= [(81/25)+1] / [(81/25)-1] = 53/28
Answer is A. |
AQUA-RAT | AQUA-RAT-39048 | all three = neither =5 = 10/3 % of 150
working formula :-
Total = AC + sunporch + swimming pool - Exactly two - 2*all three + neither
100 = 60 + 50 + 30- Exactly two - 2*10/3 + 10/3
100 =140 - Exactly two - 10/3
Exactly two = 40 - 10/3 = (120-10)/3 = 110/3 %
110/3 % of 150 = 110*150/300 =55
Ans D
Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1317
Location: Malaysia
Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
28 Jun 2017, 05:54
1
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
Hi Bunuel,
Why formula (1) is incorrect? https://gmatclub.com/forum/advanced-overlapping-sets-problems-144260.html
(i) In term of percentage
Total = A + B + C − (sum of 2−group overlaps) + (all three) + Neither
100 = 60 + 50 + 30 - (x) + (5/150)(100) + (5/150)(100)
100 = 140 + (20/3) - x
x = 40 + (20/3)
x = 140/3%
(x/150)(100) = (140/3)
x = (140/300)(150)
x = 70
(ii) In term of number
Total = A + B + C − (sum of 2−group overlaps) + (all three) + Neither
150 = 90 + 75 + 45 - x + 5 + 5
x = 70
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
The following is multiple choice question (with options) to answer.
A motor pool has 300 vehicles of which 40 percent are trucks. 20 percent of all the vehicles in the motor pool are diesel, including 15 trucks. What percent of the motor pool is composed of vehicles that are neither trucks nor diesel? | [
"45%",
"90%",
"65%",
"55%"
] | A | Trucks = 40% of 300 = 120
Other Vehicles (Excluding Trucks) = 300-120 = 180
Diesel Vehicle = 20% of 300 = 60 (Including 15 Trucks)
Other Diesel Vehicles (Excluding Trucks) = 60-15 = 45
Trucks that are NOT diesel = 120 - 15 = 105
Other Vehicles that are NOT diesel Vehicles = 180-45 = 135
Required % = (135/300)*100 = 45%
Answer: option A |
AQUA-RAT | AQUA-RAT-39049 | 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.
P is three times as fast as Q and working together, they can complete a work in 9 days. In how many days can Q alone complete the work? | [
"16 days",
"17 days",
"19 days",
"12 days"
] | D | P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 9 days, which means Q can do the work in 48 days.
Hence, P can do the work in 12 days.
Answer:D |
AQUA-RAT | AQUA-RAT-39050 | = 1/9, i. Like area, surface area is also expressed in square units. The surface area and the volume of a cylinder have been known since ancient times. As the cube gets bigger, the volume increases much more rapidly than the surface area, because the volume increases as the cube of the linear dimension, but the surface area increases as the square. Let r , in centimeter be the radius of base of cylinder and h, in centimeter be height of cylinder. Enjoy the calculator! Buy a comprehensive geometric formulas ebook. Here's a quick example: Compute the flux of the vector field through the piece of the cylinder of radius 3, centered on the z -axis, with and. We are learning to: Calculate the surface area of a cylinder. I really want to find the surface area of just the side of the cylinder, not the top and bottom Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Since the area of a circle is and the area of a rectangle is lw, we can make a formula for the surface area of a cylinder:. Surface Area = Areas of top and bottom +Area of the side. The area of one square face A face = a ⋅ a = a 2. Specify the cylinder radius ( r ) and length ( L ), the liquid depth ( h ), and optionally select unit of length, to calculate volume of the liquid. Surface Area of a Cube. A cylinder which has slant height l l l and circular flat surfaces with radius r r r has a total surface area equal to 2 π r (r + l) 2 \pi r (r + l ) 2 π r (r + l). Embeddable Player. For a right circular cylinder of radius r and height h, the lateral area is the area of the side surface of the cylinder: A = 2π rh. 56 cm 2 Area of top of the cylinder=πr 2 =1. Example 6: A cube whose sides are 10. Work out the surface area of a cylinder. Problem: Consider a cylinder with a radius of 3 meters and a height of 6 meters. When the two ends are directly aligned on each other it is a Right Cylinder otherwise it is an Oblique Cylinder: Surface Area of a Cylinder. In our example, 3. Your email is safe
The following is multiple choice question (with options) to answer.
A larger cube has 64 cubic inch as a volume and in the cube there are 64 smaller cubes such that their volume is 1 cubic inch. What is the difference between the surface areas’ sum of the 64 smaller cubes and the surface area of the larger cube, in square inch? | [
"54",
"64",
"81",
"288"
] | D | Volume of larger cube = 64 = 4^3
Side of larger cube = 4
Volume of smaller cube = 1 --> Side of smaller cube = 1
Surface area of larger cube = 6 * 4^2 = 96
Surface area of 27 smaller cubes = 64 * 6 * 1 = 384
Difference = 384 - 96 = 288
Answer: D |
AQUA-RAT | AQUA-RAT-39051 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
Calculate the largest 5 digit number which is exactly divisible by 77? | [
"99111",
"99946",
"99869",
"99792"
] | B | Largest 4 digit number is 99999
After doing 99999 ÷ 77 we get remainder 53
Hence largest 5 digit number exactly divisible by 77 = 99999 - 53 = 99946
B |
AQUA-RAT | AQUA-RAT-39052 | evolution, zoology, anatomy, species
Title: Examples of animals with 12-28 legs? Many commonly known animals' limbs usually number between 0 and 10. For example, a non-exhaustive list:
snakes have 0
Members of Bipedidae have 2 legs. Birds and humans have 2 legs (but 4 limbs)
Most mammals, reptiles, amphibians have 4 legs
Echinoderms (e.g., sea stars) typically have 5 legs.
Insects typically have 6 legs
Octopi and arachnids have 8 legs
decapods (e.g., crabs) have 10 legs
....But I can't really think of many examples of animals containing more legs until you reach 30+ legs in centipedes and millipedes. Some millipedes even have as many as 750 legs! The lone example I am aware of, the sunflower sea star, typically has 16-24 (though up to 40) limbs.
So my question is: what are some examples of animals with 12-28 legs? As a couple of counterexamples, species in the classes Symphyla (Pseudocentipedes) and Pauropoda within Myriapoda have 8-11 and 12 leg pairs respectively, so between 16 to 24 legs (sometimes with one or two leg pair stronlgy reduced in size).
(species in Symphyla, from wikipedia)
Another common and species-rich group with 14 walking legs (7 leg pairs) is Isopoda.
(Isopod, picture from wikipedia)
You also need to define 'legs' for the discussion to be meaningful. As you say, decapods have 10 legs on their thoracic segments (thoracic appendages), but they can also have appendages on their abdomens (Pleopods/swimming legs), which will place many decapods in the 10-20 leg range.
(Decapod abdominal appendages/legs in yellow, from wikipedia)
So overall, in Arthropoda, having 12-28 legs doesn't seem all that uncommon. There are probably other Arthropod groups besides those mentioned here that also have leg counts in this range.
However, for a general account, the most likely answer (if there is indeed a relative lack of 12-28 legged animals) is probably evolutionary contingencies and strongly conservative body plans within organism groups.
The following is multiple choice question (with options) to answer.
There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 580. The number of hares in the zoo is? | [
"17",
"26",
"90",
"27"
] | C | 200*2 = 400
580
-----
180
1----2
?----180 = 90
Answer: C |
AQUA-RAT | AQUA-RAT-39053 | As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $$5!=5\times 4\times 3\times 2\times 1=120$$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $$3!=3\times 2\times 1=6$$ ways to do, making (again) a total of $$(5!)\times (3!)=120\times 6=720$$ ways of arranging them.
-
+1 for the graphics! :-) – joriki Jul 8 '12 at 19:53
Thanks!$\text{}$ – Zev Chonoles Jul 8 '12 at 20:00
Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
-
thanks joriki, but not quiet clear about what you say – user1419170 Jul 8 '12 at 19:49
@user1419170: Could you be more specific? What part isn't clear to you? – joriki Jul 8 '12 at 19:49
this part Treat the group of three boys as an additional girl – user1419170 Jul 8 '12 at 19:51
@user1419170: See Brian's and Saurabh's answers, which use the same idea but phrase it differently. – joriki Jul 8 '12 at 19:53
The following is multiple choice question (with options) to answer.
In a group of 5 boys and 3 girls, four children are to be selected. In how many different ways can they be selected such that at least one girl is included? | [
"65",
"50",
"80",
"35"
] | A | The total number of ways to choose 4 children is 8C4 = 70
The number of ways to choose only boys is 5C4 = 5
The number of groups which include at least one girl is 65.
The answer is A. |
AQUA-RAT | AQUA-RAT-39054 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
The value of a 10.5% stock, in which an income of Rs.756 is derived by investing Rs.9000, brokerage being% is : | [
"Rs. 118",
"Rs. 120",
"Rs. 124.75",
"Rs. 131"
] | C | For an income of Rs.756, investment = Rs.9000
For an income of Rs., investment = = Rs.125
For a Rs.100 stock, investment = Rs.125.
Market value of Rs. 100 stock = = Rs. 124.75
C |
AQUA-RAT | AQUA-RAT-39055 | # How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
The following is multiple choice question (with options) to answer.
A university cafeteria offers 1 flavors of pizza - pepperoni. If a customer has an option (but not the obligation) to add extra cheese, mushrooms or both to any kind of pizza, how many different pizza varieties are available ? | [
"4",
"8",
"12",
"1"
] | D | 1 flavours * 1 choices = 1C1*1C1 = 1*1=1=D |
AQUA-RAT | AQUA-RAT-39056 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\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.
The price of a coat in a certain store is $500. If the price of the coat is to be reduced by $350, by what percent is the price to be reduced? | [
"70%",
"15%",
"20%",
"25%"
] | A | price of a coat in a certain store = $500
the price of the coat is to be reduced by $350
% change = (Final Value - Initial Value)*100 / Initial Value
% Reduction= (Reduction in Price)*100 / Initial Value
i.e. % Reduction= (350)*100 / 500 = 70%
Answer: Option A |
AQUA-RAT | AQUA-RAT-39057 | 5. Originally Posted by Nath
"I have 2 children. One is female. What is the probability the other is also female?"
This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.
I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.
I'm interested to get some more opinions...
Here is an alternative "logical" way to consider this.
There are 4 possibilities in regard to the children being boy or girl.
The probability of having a boy and a girl is twice the probability of having 2 girls,
and is also twice the probability of having 2 boys.
(i) 1st and 2nd children are girls
(ii) 1st child is a girl and the 2nd is a boy
(iii) 1st child is a boy and the 2nd is a girl
(iv) 1st and 2nd children are boys
GG
GB
BG
BB
One is female.
This reduces to...
GG
GB
BG
3 cases of equal probability and in only 1 of these cases is the other child also a girl.
Hence the probability of the 2nd child also being a girl is 1/3.
Also note that in 2 of these 3 cases, the 2nd child is a boy,
therefore, if one of the children is a girl, the probability that the other is a boy is 2/3.
6. ## Solutions
This was my solution, but alas my friend disagrees:
I have 2 children, 1 is female. This gives four possible scenarios:
1. The female has a younger sister
2. The female has an older sister
3. The female has a younger brother
4. The female has an older brother
Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
---
My friend, with the PhD in Maths thinks the following:
The following is multiple choice question (with options) to answer.
A team consists of 4 boys and 3 girls. If you pick two persons at the same time, what's the probability that one person is boy and other one is girl ? | [
"2/7",
"5/7",
"4/7",
"3/7"
] | C | P(1st b, 2nd g) = 4/7*3/6 = 4/14;
P(1st g, 2nd b) = 3/7*4/6 = 4/14.
P = 4/14 + 4/14 = 4/7.
Answer: C. |
AQUA-RAT | AQUA-RAT-39058 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A certain article of clothing was discounted during a special sale to 2/5 of its original retail price. When the clothing didn't sell, it was discounted even further to 1/2 of its original retail price during a second sale. By what percent did the price of this article of clothing decrease from the first sale to the second sale? | [
"50%",
"33.33%",
"25%",
"16.66%"
] | D | Say the original retail price of the item was $200.
The price after the first sale = 3/5 * $200 = $120.
The price after the second sale = 1/2 * $200 = $100.
The percent change from the first sale to the second = (120 - 100) / 120 = 1/3 = 16.66%.
Answer: D. |
AQUA-RAT | AQUA-RAT-39059 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 48, the dividend is: | [
"5808",
"5336",
"5343",
"5345"
] | A | Divisor = (5 * 48) = 240
= 10 * Quotient = Divisor
=> Quotient = 240/10 = 24
Dividend = (Divisor * Quotient) + Remainder
Dividend = (240 * 24) + 48 = 5808.
A |
AQUA-RAT | AQUA-RAT-39060 | # Probability of a certain ball drawn from one box given that other balls were drawn
Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
The following is multiple choice question (with options) to answer.
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? | [
"8/21",
"7/11",
"5/27",
"11/32"
] | A | Total number of balls = 8+7+6 = 21
E = event that the ball drawn is neither red nor green
= event that the ball drawn is red
n(E) = 8
P(E) = 8/21
Answer is A |
AQUA-RAT | AQUA-RAT-39061 | 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.
After working for 6 days, David was joined by Moore. Together they completed the remaining job in 3 days. How many days will it take both of them to complete the entire job, given that it would have taken David 12 days to complete the job alone? | [
"7",
"4",
"6",
"2"
] | C | Explanation:
David and Moore complete half work in 3 days
=> they can complete whole work in 6 days
Answer: Option C |
AQUA-RAT | AQUA-RAT-39062 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 seconds and a man standing on the platform in 25 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"877 m",
"165 m",
"167 m",
"887 m"
] | B | Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train
= (15 x 25)m = 375 m. Let the length of the platform be x meters. Then, (x + 375)/36 = 15
==> x + 375 = 540 ==> x
= 165 m.
Answer: B |
AQUA-RAT | AQUA-RAT-39063 | # Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
The following is multiple choice question (with options) to answer.
In a rectangular coordinate system, what is the area of a rectangle whose vertices have the coordinates (-8, 1), (1, 1), (1, -7 and (-8, -7)? | [
"144",
"36",
"72",
"56"
] | C | Length of side 1= 8+1=9
Length of side 2= 7+1= 8
Area of rectangle= 9*8= 72
C is the answer |
AQUA-RAT | AQUA-RAT-39064 | dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
_________________
-----------------------
tusharvk
Manager
Joined: 27 Oct 2008
Posts: 180
Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
The following is multiple choice question (with options) to answer.
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow? | [
"2/29",
"2/17",
"2/21",
"2/28"
] | C | Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles. Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 * 2)/(15 * 14) = 1/35
Probability that both are yellow = ²C₂/¹⁵C₂ = (2 * 1)/(15 * 14) = 1/105
Probability that one blue and other is yellow = (³C₁ * ²C₁)/¹⁵C₂ = (2 * 3 * 2)/(15 * 14) = 2/35
Required probability = 1/35 + 1/105 + 2/35
= 3/35 + 1/105 = 1/35(3 + 1/3)
= 10/(3 * 35) = 2/21
Answer: C |
AQUA-RAT | AQUA-RAT-39065 | java
Title: Computing quantity-based discounts I'm a CS professor and provided this code as a small piece of a refactoring assignment:
private double getTotalPrice(int quantity, double itemPrice) {
if (quantity == 13) {
return itemPrice * 12; // Baker's dozen
}
else if (quantity >= 6) {
return itemPrice * quantity * .95;
} else {
return itemPrice * quantity;
}
}
The following is multiple choice question (with options) to answer.
A discount of 20 percent on an order of goods followed by a discount of 20 percent amounts to | [
"the same as one 45 percent discount",
"the same as one 40 percent discount",
"the same as one 36 percent discount",
"the same as one 30 percent discount"
] | C | Say initial order = 100
Discount 20% = 20
Value = 100 - 20 = 80
Again discount 20% =80∗20100=16=80∗20100=16
Value = 80-16 = 64
Total discount = 20+16 = 36
Answer = C) the same as one 36 percent discount |
AQUA-RAT | AQUA-RAT-39066 | # A simple riddle related to addition of odd numbers
I'm not sure if this type of question can be asked here, but if it can then here goes:
Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.
PS : The inception of this question is a result of a random discussion that I was having during the break hour :)
-
Hint: adding an odd number of odd numbers will give ... (is this a real question ?) – Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that – 403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even algebra.com/algebra/homework/word/numbers/… True? – 403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. – Raymond Manzoni May 25 '13 at 8:36
A direct approach:
Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like $${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$ (the integers $a,b,\ldots,i$ may or may not be the same). Grouping things together, this is equal to $$2(a+b+c+d+e+f+g+h+i+4)+1.$$ Thus, the result is odd.
The following is multiple choice question (with options) to answer.
In the set of positive integers from 1 to 90, what is the sum of all the odd multiples of 5? | [
"180",
"245",
"320",
"405"
] | D | reduce 1 - 90
5 - 15 - 25 - 35 - 45--55--65--75--85 are valid multiples. Add them --> 405
D |
AQUA-RAT | AQUA-RAT-39067 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 9 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city? | [
"14",
"16",
"21",
"24"
] | D | I treat such problems as work ones.
Work=Rate*Time
Mileage(m)=Rate(mpg)*Gallons(g)
X gallons is a full tank
{462=RX
{336=(R-9)X solve for R, R=33
33-9=24 mpg
D |
AQUA-RAT | AQUA-RAT-39068 | 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.
A large tanker can be filled by two pipes A and B in 60 and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half? | [
"15 min",
"20 min",
"27.5 min",
"30 min"
] | D | Part filled by (A + B) in 1 minute = (1/60 + 1/40) = 1/24
Suppose the tank is filled in x minutes.
Then, x/2(1/24 + 1/40) = 1
x/2 * 1/15 = 1 => x = 30 min
ANSWER:D |
AQUA-RAT | AQUA-RAT-39069 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
The ratio of the volumes of two cubes is 729 : 1000. What is the ratio of their total surface areas? | [
"81 : 100",
"9 : 11",
"729 : 1331",
"27 : 121"
] | A | Ratio of the sides = ³√729 : ³√1000 = 9 : 10
Ratio of surface areas = 9^2 : 10^2 = 81 : 100
ANSWER:A |
AQUA-RAT | AQUA-RAT-39070 | 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$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 21 seconds. What is the length of the train? | [
"350 m",
"278 m",
"876 m",
"150 m"
] | A | Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train
= (Speed x Time)
= (50/3 * 21) m
= 350 m.
Answer: A |
AQUA-RAT | AQUA-RAT-39071 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\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 20 percent of the drive from the center, what percent of the round-trip has the technician completed? | [
"5%",
"10%",
"60%",
"40%"
] | C | The complete round trip consists of driving to the service center and then back home again.
So, once the technician drives to the service center he/she has already competed 50% of the entire trip.
Since the technician completes a portion of the trip back home,the correct answer must be greater than 50%
Only one answer choice works.
Answer:
C |
AQUA-RAT | AQUA-RAT-39072 | # 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.
An amount of Rs.95000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 3/4 %, then the amount invested in each share was? | [
"s. 52500; Rs. 47500",
"s. 59375; Rs. 35625",
"s. 72500; Rs. 27500",
"s. 82500; Rs. 17500"
] | B | Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (95000 - x). Then,
(x * 9 * 1)/100 + [(95000 - x) * 11 * 1]/100 = (95000 * 39/4 * 1/100)
(9x + 1045000 - 11x)/100 = 39000/4 = 18525/2
x = 59375
Sum invested at 9% = Rs. 59375
Sum invested at 11% = Rs. (95000 - 59375) = Rs. 35625.
ANSWER:B |
AQUA-RAT | AQUA-RAT-39073 | Since 1000 = 999 + 1, 100 = 99 + 1, and 10 = 9 + 1, this equation can be rewritten
$$999 d_{1}+d_{1}+99 d_{2}+d_{2}+9 d_{3}+d_{3}+d_{4}=3 k$$
Rearranging gives
$$d_{1}+d_{2}+d_{3}+d_{4}=3 k-999 d_{1}-99 d_{2}-9 d_{3}$$
$$=3 k-3\left(333 d_{1}\right)-3\left(33 d_{2}\right)-3\left(3 d_{3}\right)$$
We can now factor a 3 from the right side to get
$$d_{1}+d_{2}+d_{3}+d_{4}=3\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$
since $$\left(k-333 d_{1}-33 d_{2}-d_{3}\right)$$ is an integer, we have shown that $$d_{1}+d_{2}+d_{4}+d_{4}$$ is divisible by 3.
(2) Assume $$d_{1}+d_{2}+d_{3}+d_{4}$$ is divisible by $$3 .$$ Consider the number $$d_{1} d_{2} d_{3} d_{4} .$$ As remarked above,
$$d_{1} d_{2} d_{3} d_{4}=d_{1} \times 1000+d_{2} \times 100+d_{3} \times 10+d_{4}$$
so
The following is multiple choice question (with options) to answer.
If r = 199,999 and s = 991,999, which of the following is the units digit of r^5 + s^4? | [
"0",
"1",
"2",
"8"
] | A | The exponents of 9 cycle between 9 (odd exponents) and 1 (even exponents).
Then the sum of r^5+s^4 will have the units digit of 9+1 = 10 as a units digit.
The answer is A. |
AQUA-RAT | AQUA-RAT-39074 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
At present, the ratio between the ages of Arun and Deepak is 4:5. After 5 years, Arun's age will be 25 years. What is the age of Deepak at present? | [
"18",
"20",
"25",
"23"
] | C | Let the present ages of Arun and Deepak be 4x and 3x years respectively.
Then, 4x + 5 = 25 => x = 5
Deepak's age = 5x = 25 years.
Answer: C |
AQUA-RAT | AQUA-RAT-39075 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Mr. Kramer, the losing candidate in a two-candidate election, received 942,568 votes, which was exactly 47 percent of all votes cast. Approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast? | [
"10%",
"6%",
"15%",
"17%"
] | B | Let me try a simpler one.
Lets assume that candidate got 47% votes and total votes is 100.
Candidate won = 47
Remaining = 53
To get 50%, candidate requires 3 votes from 100 which is 3% and 3 votes from 53.
3/53 = .056= 5.6%
Which is approx 6%. Hence the answer is B. |
AQUA-RAT | AQUA-RAT-39076 | = 0 \\ a^7+ b^7 + c^7 = -7mk^2$$ Then your identity reads $$\left( -\frac{3m}{3} \right) \left( -\frac{7mk^2}{7} \right) = \left( -\frac{5mk}{5} \right)^2 \\ \left( -m \right) \left( -mk^2 \right) = (mk)^2$$
The following is multiple choice question (with options) to answer.
Find the value of c from (8)^3 x 9^3 ÷ 679 = c. | [
"249.7",
"529.7",
"549.7",
"594.7"
] | C | Given Exp. =(8)^3 x 9^3 ÷ 679 = c
= 512 x 729 ÷ 679
= 549.7
C |
AQUA-RAT | AQUA-RAT-39077 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 46 kmph respectively. In how much time will they cross each other, if they are running in the same direction? | [
"72 sec",
"210 sec",
"192 sec",
"252 sec"
] | B | Solution
Relative Speed = (46 - 40 ) Kmph
= 6 kmph
=(6 x 5 / 18)m/sec
= (30 / 18 ) m/sec
Time taken = (350 x 18 / 30) sec
= 210 sec.
Answer B |
AQUA-RAT | AQUA-RAT-39078 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many 3-digit even numbers are possible such that if one of the digits is 4, the next/succeeding digit to it should be 7? | [
"305",
"5",
"365",
"405"
] | B | 470, 472, 474, 476, and 478, so total 5. Hence Option B. |
AQUA-RAT | AQUA-RAT-39079 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
What is the next number of the following sequence
10, 7, 12, 10, 14, __ ? | [
"13",
"14",
"15",
"16"
] | A | combine two series:
1st series is:10 12 14
10+2=12
12+2=14
14+2=16
2nd series is:7 10
means its 7+3=10
10+3=13
ANSWER:A |
AQUA-RAT | AQUA-RAT-39080 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular grass field is 75 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m? | [
"1350",
"277",
"299",
"266"
] | A | Area = (l + b +2d) 2d
= (75 + 55 +2.5 * 2) 2*2.5 => 675
675 * 2 = Rs.1350
Answer:A |
AQUA-RAT | AQUA-RAT-39081 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many different number of ways 7 men and 2 women can sit on a shopa which can accommodate persons? | [
"A)160",
"B)170",
"C)190",
"D)15120"
] | D | 9p5 = 9 x 8 x 7 × 6 × 5 = 15120
Option 'D' |
AQUA-RAT | AQUA-RAT-39082 | # 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 C.P of 15 books is equal to the S.P of 18 books. Find his gain% or loss%? | [
"16 2/3%",
"16 2/8%",
"16 7/3%",
"16 5/3%"
] | A | 15 CP = 18 SP
18 --- 3 CP loss
100 --- ? => 16 2/3% loss
Answer: A |
AQUA-RAT | AQUA-RAT-39083 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
If the L.C.M of two numbers is 750 and their product is 18750, find the H.C.F of the numbers | [
"23",
"28",
"48",
"25"
] | D | Explanation:
H.C.F = (Product of the numbers) / (Their L.C.M) = 18750/750 = 25.
Answer: D |
AQUA-RAT | AQUA-RAT-39084 | # If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term. - Mathematics
Sum
If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.
#### Solution
Let the first term, common difference
And number of terms of the AP: 9, 7, 5,…. are
a1, d1 and n1, respectively
i.e., first term (a1) = 9
And common difference (d1) = 7 – 9 = –2
⇒ T"'"_(n_1) = a_1 + (n_1 - 1)d_1
⇒ T"'"_(n_1) =9 + (n_1 - 1)(-2)
⇒ T"'"_(n_1) = 9 - 2n_1 + 2
⇒ T"'"_(n_1) = 11 - 2n_1 ......(i)
∵ nth term of an AP, Tn = a + (n – 1)d
Let the first term, common difference and the number of terms of the AP: 24, 21, 18, … are a2, d2 and n2, respectively
i.e., first term, (a2) = 24
And common difference (d2) = 21 – 24 = – 3
∴ Its nth term, T"'"_(n_2) = a_2 + (n_2 - 1)d_2
⇒ T"'"_(n_2) = 24 + (n_2 - 1)(-3)
⇒ T"'"_(n_2) = 24 - 3n_2 + 3
⇒ T"'"_(n_2) = 27 - 3n_2 .....(ii)
Now, by given condition,
nth terms of the both APs are same
i.e., 11 - 2n_1 = 27 - 3n_2 ......[From equations (i) and (ii)]
The following is multiple choice question (with options) to answer.
On January 1, 2004 two new societies, S1 and S2, are formed, each with n members. On the first day of each subsequent month, S1 adds b members while S2 multiplies its current number of members by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r? | [
"2",
"9",
"1.8",
"1.7"
] | A | Explanation :
Jan 1 – S1 and S2 has n members. Till july 2nd, there are 6 more months, and hence 6 more additions.
In S1, b members are added each month. So after 6 months, S1 has n + 6b.
In S2, r members are multiplied each month. So after 6 months, S2 has n * r6.
After 6 months, both are same, hence
=> n + 6b = nr6. ----------------- (1)
=> b = 10.5 n. (given)
=> 6b = 63n hence.
Substituting in (1)
So, 64n = nr6. Cancelling n on both sides, we have r6 = 64 or r = 2.
Answer : A |
AQUA-RAT | AQUA-RAT-39085 | ### 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 are entered into a partnership. A invested Rs.6500 for 6 months, B invested Rs.8400 for 5 months and C invested for Rs.10000 for 3 months. A is a working partner and gets 5% of the total profit for the same. Find the share of C in a total profit of Rs.7400. | [
"1999",
"1900",
"19077",
"19055"
] | B | 65 * 6 : 84 * 5 : 100 * 3
26:28:20
C share = 74000 * 95/100 = 7030 * 20/74 => 1900
Answer: B |
AQUA-RAT | AQUA-RAT-39086 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Seven different books (A, B, C, D,E,F and G) are to be arranged on a shelf. Books A , B and C are to be arranged first and second starting from the right of the shelf. The number of different orders in which books D,E, F and G may be arranged is | [
"4!",
"15!",
"16!",
"28!"
] | A | Solution
Since books A,B and C are arranged first and second, only books D, E,F and G will change order. Therefore it an arrangement problem involving 4 items and the number of different order is given by
4!
Answer A |
AQUA-RAT | AQUA-RAT-39087 | 18. freckles
so... notice that every term on top has a common factor of h and the denominator also has a common factor h
19. freckles
$f'(x)=\lim_{h \rightarrow 0} \frac{h(2ax+ah+b)}{h(1)}$
20. haleyelizabeth2017
Oh!
21. freckles
we can write as: $f'(x)=\lim_{h \rightarrow 0} \frac{h}{h} \cdot \frac{2ax+ah+b}{1}$
22. haleyelizabeth2017
which the h's cancel, right?
23. freckles
yep and that whole h on the bottom was the thing that was giving us problems but guess what?
24. freckles
it is no longer there because we canceled that mean h guy on bottom
25. haleyelizabeth2017
:)
26. freckles
so we can plug in 0 now
27. freckles
for h that is
28. haleyelizabeth2017
and it leaves us with 2ax+b?
29. freckles
$f'(x) =\lim_{h \rightarrow 0} (2ax+ah+b)=2ax+a(0)+b=2ax+0+b=2ax+b$ you are right
30. haleyelizabeth2017
Is that it?
31. freckles
$f(x)=ax^2+bx+c \text{ gives us } f'(x)=2ax+b$
32. freckles
yes now you can find the derivative of any quadratic
33. freckles
you actually have a formula for it now we just found a formula for it above
34. haleyelizabeth2017
Awesome! Didn't know it was that simple...the video I watched on it made it seem so confusing lol
35. haleyelizabeth2017
So there is never a c...
36. freckles
The following is multiple choice question (with options) to answer.
The H.C.F. of two numbers is 18 and the other two factors of their L.C.M. are 11 and 15. The larger of the two numbers is: | [
"270",
"300",
"299",
"322"
] | A | the numbers are (18 x 11) and (18 x 15).
Larger number = (18 x 15) = 270.
ANSWER :A |
AQUA-RAT | AQUA-RAT-39088 | 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.
Pipe A fills a swimming pool in 4 hours. Pipe B empties the pool in 6 hours. If pipe A was opened at 9:00 am and Pipe B at 10:00 am, at what time will the pool be full? | [
"19:00",
"17:00",
"16:00",
"18:00"
] | A | Pipe A fills the pool in 4 hrs.
1 hour's work : 1/4
Pipe B empties the pool in 6 hrs.
1 hour's work : 1/6
Together if they work, 1 hour's work = 1/4 -1/6 = 1/12
Given : Pipe A started at 9:00 a.m and Pipe B at 10:00 a.m
Pool filled after 1 hour by Pipe A : 1/4 or 3/12
After 10:00 a.m
Pool filled after 1 hour with both the pipes on : 1/12
Pool filled after 9 hours with both pipes on : 9/12
Pool filled in 1 hour + Pool filled in 9 hours = 3/12 +9/12 =1
Therefore, it takes 10 hrs to fill the pool
As Pipe A started at 9:00 a.m, pool is full at 19:00 hrs
Answer : A |
AQUA-RAT | AQUA-RAT-39089 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
Sum of ages of two friends is 25 while difference is 3 then their ages are | [
"15 years, 17 years",
"11 years, 14 years",
"16 years, 20 years",
"10 years, 16 years"
] | B | 11,14
ANSWER:B |
AQUA-RAT | AQUA-RAT-39090 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
The average marks of 10 students in a class is 100. But a student mark is wrongly noted as 50 instead of 10 then find the correct average marks? | [
"A)78",
"B)82",
"C)96",
"D)91"
] | C | correct avg marks = 100+(10-50)/10
avg = 100-4 = 96
Answer is C |
AQUA-RAT | AQUA-RAT-39091 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
The difference between the ages of two persons is 10 years. Fifteen years ago, the elder one was twice as old as the younger one. The present age of the elder person is | [
"25 years",
"30 years",
"35 years",
"40 years"
] | C | Sol.
Let their ages of x years and (x + 10) years respectively.
Then, (x + 10) - 15 = 2(x - 15)
⇔ x - 5 = 2x - 30
⇔ x = 25.
∴ Present age of the elder person = (x + 10)
= 35 years.
Answer C |
AQUA-RAT | AQUA-RAT-39092 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
How many times does the minute hand go round in a day? | [
"21",
"20",
"12",
"26"
] | C | C
12
period = 65 minutes/60 minutes = 1.08.
The number of cycles per hour then = 1/period = 1/1.08 = .923.
The number of cycles in 24 hours, then = (0.923)(24) = 22.153. That's how you compute how many times the minute hand catches up the hour hand in 24 hours (round to 22) |
AQUA-RAT | AQUA-RAT-39093 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
If a man buys 20 lollipops for $90 and sold them for $2 dollars determine his loss. | [
"$ 40.",
"$ 50.",
"$ 60.",
"$ 70."
] | B | Solution:
Cost of 20 lollipops = $90
Sold each lollipop for $2
So he sold 20 lollipops for $(20 × 2) = $40
Loss = $(90 – 40) = $50
Therefore, loss = $ 50.
Answer B |
AQUA-RAT | AQUA-RAT-39094 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A committee has 3 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee? | [
"127",
"60",
"882",
"272"
] | B | The number of ways to select two men and three women
= 3C₂ * ⁶C₃
= (3 *2 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 60
Answer: B |
AQUA-RAT | AQUA-RAT-39095 | # 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.
In shop contains 500 articles each of a cost of $10. He sold them at a price of $12. If the shop keeper has a order of 200 articles, how much profit he will get? | [
"200",
"2000",
"1500",
"400"
] | D | Total money after selling articles = 200*12 = 2400
cost price = 200*10 = 2000
profit = 400
correct option is D |
AQUA-RAT | AQUA-RAT-39096 | Originally Posted by dumluck
That's facinating so $1 + 3 + 5 + 7 + 9 + 11 = 6^2?$
• Mar 5th 2011, 10:31 AM
dumluck
Quote:
Originally Posted by HallsofIvy
The sum of all integers from 1 to n is, as you noted in your first post, $\frac{n(n+1)}{2}$. Notice that the sum of even numbers, 2+ 4+ 6+ 8+ ...= 2(1+ 2+ 3+ 4+ ...) so that the sum of all even integers from 2 to 2n is 2 times the sum of all numbers from 1 to n: The sum of all even numbers from 1 to 2n is $n(n+ 1)$. The sum of odd numbers, from 1 to 2n+ 1, is the sum of all integers from 1 to 2n+1 minus the sum of all even numbers from 2 to 2n:
$\frac{2n(2n+1)}{2}- n(n+1)= \frac{2n(2n+1)}{2}- \frac{2n(n+1)}{2}= \frac{2n(2n+1- (n+1))}{2}$
and that gives the very simple formula Wilmer indicates.
thanks, let me try something here so I can further my understanding.
Let's say we are looking for the even numbers between 1 - 11
2+ 4+ 6+ 8+ 10 ...= 2(1+ 2+ 3+ 4+ 5 + 6 + 7 + 8 + 9 + 10) :
This gives us...
2+4+6+8+10 = 30
1+2+3+4+5+6+7+8+9+10 = 55 * 2 = 110.
so 30 != 110. So this is what is causing me some confusion.
1 3 5 7 9
1 4 9 16 25
The pattern here is as soraban states (i.e) $X^2$ but if we are looking for the odd integers in a range of 200 that is not starting at 0 would that not be quite difficult?
say odd integers from 800 - 1000.
The following is multiple choice question (with options) to answer.
For all even integers n, h(n) is defined to be the sum of the even integers between 6 and n, inclusive. What is the value of h(18)/h(10) ? | [
" 1.8",
" 4",
" 6",
" 18"
] | B | CONCEPT: When terms are in Arithmetic Progression (A.P.) i.e. terms are equally spaced then
Mean = Median =(First+Last)/2
and Sum = Mean*Number of terms
h(18) = [(6+18)/2]*7 = 84
h(10) = (6+10)/2]*3 = 24
h(18)/h(10) = (84) / (24) ~ 4
Answer : B |
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