source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37997 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
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A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
The average age of 7 men increases by 4 years when two women are included in place of two men of ages 26 and 30 years. Find the average age of the women? | [
"42",
"40",
"30",
"18"
] | A | Explanation:
26 + 30 + 7 * 4 = 84/2 = 42
Answer: A |
AQUA-RAT | AQUA-RAT-37998 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 17 percent sugar by weight. The second solution was what percent sugar by weight? | [
"34%",
"24%",
"22%",
"38%"
] | D | Instead of using complex calculations and remembering formulae, why dont u directly get to weighted average.
3 parts of 10% + 1 part of x (unknown) % = 4 parts of 17%
=> x% = 68%-30% = 38%
ans D it is. |
AQUA-RAT | AQUA-RAT-37999 | Square [1]: . $(x+y)^2 \:=\:1^2 \quad\Rightarrow\quad x^2+2xy + y^2 \:=\:1$
$\text{We have: }\;x^2 + y^2 + 2(xy) \:=\:1$
. . . . . . . . . . . . . . $\uparrow$
. . . . . . . . . . . . $^{\text{This is }\text{-}6}$
Therefore: . $x^2+y^2 - 12 \:=\:1 \quad\Rightarrow\quad \boxed{x^2+y^2 \:=\:13}$
This method is guarenteed to impress/surprise/terrify your teacher.
.
4. Originally Posted by Soroban
Hello, Joel!
Welcome aboard!
I have a back-door approach . . .
I like your solution. I wish I had thought about the problem a bit more before going down the obvious path.
The following is multiple choice question (with options) to answer.
xy=1 then what is (3^(x+y)^2) / (3^(x-y)^2) | [
"81",
"4",
"8",
"16"
] | A | (x+y)^2 - (x-y)^2
(x+y+x-y)(x+y-x+y)
(2x)(2y)
4xy
4
3^4 = 81
Answer A |
AQUA-RAT | AQUA-RAT-38000 | # Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?
A bag has:
$3$ red
$5$ black
$8$ green
marbles.
Total of 16 marbles.
You select a marble, and then another one right after. (without replacement).
What is the probability that $both$ are red?
Probability that first pick is red: $\frac{3}{16}$
Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)
Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$
How do I do this using combinations only?
• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59
Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$
Hope this helps you.
The following is multiple choice question (with options) to answer.
There are 10 slate rocks, 15 pumice rocks, and 6 granite rocks randomly distributed in a certain field. If 2 rocks are chosen at random and without replacement, what is the probability that both rocks will be slate rocks? | [
"2/19",
"3/31",
"4/37",
"5/41"
] | B | 10/31*9/30 = 3/31
The answer is B. |
AQUA-RAT | AQUA-RAT-38001 | \frac{e^{1/{7\cdot 16}}}{\frac{{1-49}}{7}}\right)=\boxed{{2\pi e^{4/7}}}$$
The following is multiple choice question (with options) to answer.
Which of the following fractions is greater than 1/7? | [
"1. 12/50",
"2. 3/11",
"3. 2/9",
"4. 4/17"
] | A | 1. 12/50
12.5/50 = 1/4 thus 12/50 < 1/4
2. 3/11
3/12 = 1/4 thus 3/11 > 1/4
3. 2/9
2/8 = 1/4 thus 2/9 < 1/4
4. 4/17
4/16 = 1/4 thus 4/17 < 1/4
5. 6/24
6/24 = 1/4
A |
AQUA-RAT | AQUA-RAT-38002 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
3 men or 6 women can do a piece of work in 23 days. In how many days will 12 men and 8 women do the same work? | [
"15/9 days",
"15/4 days",
"18/4 days",
"69/16 days"
] | D | 3M = 6W ---- 23 days
12M + 8W -----?
24W + 8 W = 32W ---?
6W ---- 23 32 -----?
6 * 23 = 32 * x => x = 69/16 days
Answer:D |
AQUA-RAT | AQUA-RAT-38003 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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 72 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"286",
"278",
"255",
"180"
] | D | Speed=(72 * 5/18) m/sec = (20) m/sec Length of the train
= (Speed x Time) = (20 * 9) m
= 180 m.
Answer:D |
AQUA-RAT | AQUA-RAT-38004 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
Joshua and Jose work at an auto repair center with 3 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen? | [
"1/15",
"1/10",
"1/9",
"1/6"
] | B | Two Methods
1) Probability of chosing Josh first = 1/5
Probability of chosing Jose second = 1/4
total = 1/20
Probability of chosing Jose first = 1/5
Probability of chosing Josh second = 1/4
Total = 1/20
Final = 1/20 + 1/20 = 1/10
B |
AQUA-RAT | AQUA-RAT-38005 | 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 4 men can colour 48 m long cloth in 2 days, then 6 men can colour 36 m long cloth in | [
"1 day",
"2 days",
"3 days",
"4 days"
] | A | The length of cloth painted by one man in one day = 48 / 4 × 2 = 6 m
No. of days required to paint 36 m cloth by 6 men = 36/ 6 × 6 = 1 day.
A |
AQUA-RAT | AQUA-RAT-38006 | the letter B occurs 3 times. Like a "super wildcard". Not all of those are distinguishable, because some words occur more than once on a single line. (d) type and case of letters, spacing between letters and punctuation marks; (e) joining words together or separating the words does not make a name distinguishable from a name that uses the similar, separated or joined words; (f) use of a different tense or number of the same word does not distinguish one name from another;. So a reasonable guess would be that total comparable settlements would be 150 times$24 million in cash, or $1. Swap these numbers as reverse the subset. 2% more than the actual collections in January 2019, and$35 million or 1. We couldn't distinguish among the 4 I's in any one arrangement, for example. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. We can use the following formula, where the number of permutations of n objects taken k at a time is written as n P k. Permutation. 4 billion increase in dividends paid, and a $324 million decrease in proceeds from the issuance of common stock, offset in part by a$3. P (10,3) = 720. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. From a 4 billion Indian Rupee gaming industry in 2007, to a 62 billion Indian Rupees industry in 2019, gaming in India has certainly caught the eye of consumers and proves to be a valuable market today. ership as an arrangement that leverages the uniqueness of Dutch law to avoid taxes and prevent a hostile takeover attempt. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. One of these code words, the 'start signal' begins all the sequences that code for amino acid chains. Microsoft Excel. com, the international travel industry has grown from 528 million tourist arrivals in 2005 to 1. The spots can also be divided by the total of legs, ears, eyes and tail to leave a remainder of 6. In the Text section, click the WordArt option. Letter Arrangements in a Word Video. 4 percent of loan participations
The following is multiple choice question (with options) to answer.
All of the stocks on the over-the-counter market are designated by either a 8-letter or a 9-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes? | [
"a. 27(26^8)",
"b. 26(26^4)",
"c. 27(26^4)",
"d. 26(26^5)"
] | A | no of different stocks for 8-letter code = 26^8
no of different stocks for 9-letter code = 26^9
total = 26^8 + 26^9 = 26^8(26+1) = 27(26^8)
[Reveal]Spoiler:
Option A |
AQUA-RAT | AQUA-RAT-38007 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of 4 positive integers is 45. If the average of 2 of these integers is 45, what is the greatest possible value that one of the other 2 integers can have? | [
"55",
"65",
"100",
"89"
] | D | a + b + c + d = 180
a + b = 90
c + d = 90
Greatest possible = 89 (Just less than 1)
Answer = D |
AQUA-RAT | AQUA-RAT-38008 | ### Remark 2
68% of all observations fall within 1 standard deviation from the mean.
### Example 5
Find the probability $$P(|Z| > 1)$$.
Key: We wish to find the area of the unshaded region in Figure 5. We may rewrite this probability as
$P( |Z| > 1) = 1 - P(|Z| < 1 )= 1 - .682 = .318$
### Exercise 1
Compute the following probabilities:
(a) $$P(|Z|<2)$$
(b) $$P(|Z|<3)$$
\begin{align} P(|Z|<2) &= .954 \\
P(|Z|<3) &= .997 \end{align}
### 68-95-99.7 Rule
68-95-99.7 Rule: 68% of all observations fall within 1 standard deviation, 95% of all observations fall within 2 standard deviations, and 99.7% of all observations fall within 3 standard deviations.
### Example 6
(a) Find the height of an individual in the $$75^{th}$$ percentile.
(b) Find the height of a student whose height is 16% from the tallest.
Inverse Normal
1. Press 2nd VARS to enter DISTR
2. Select 3:invNorm(
3. Enter percentile as decimal
4. Enter closing )
5. Press ENTER
**Remark**: Enter percentile as a decimal in **Area**, and select **Paste** and **ENTER**.
### Example 6 (a) Solution
We find that the $$75^{th}$$ percentile corresponds to the Z-score $$Z=.674$$. To find the height, we work backwards using $$(\ref{Z-score})$$:
\begin{align} .674 &= \frac{x - \mu}{\sigma}= \frac{x - 70.9}{2.75} \end{align}
We solve for $$x$$:
The following is multiple choice question (with options) to answer.
A certain characteristic in a large population has a distribution that is symmetric about the mean m. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent W of the distribution is less than m +d ? | [
"16%",
"32%",
"48%",
"84%"
] | D | D
The prompt says that 68% of the population lies between m-d and m+d.
Thus, 32% of the population is less than m-d or greater than m+d.
Since the population is symmetric, half of this 32% is less than m-d and half is greater than m+d.
Thus, W=(68+16)% or (100-16)% of the population is less than m+d. |
AQUA-RAT | AQUA-RAT-38009 | Let one woman complete the job in $$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.
Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days, six men left and six women joined, then in hoe many more days will the work be completed? | [
"2 (1/9)",
"2 (4/2)",
"5 (1/2)",
"2 (1/2)"
] | D | Work done by a women in one day = 1/2 (work done by a man/day)
One women's capacity = 1/2(one man's capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women - 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.
Answer: D |
AQUA-RAT | AQUA-RAT-38010 | 80&4352803871250866057941284067804134&18,17,27,27,29,30\\ 81&11395788481298729007027429355459050&18,17,27,29,29,30\\ 82&29834561572649264127826347038980028&18,17,29,29,29,30\\ 83&78107896236653244353987095675034922&18,17,29,29,29,32\\ 84&204489127137313012297066889013936544&18,17,29,29,31,32\\ 85&535359485175296566394760822973220276&18,17,29,31,31,32\\ 86&1401589328388601697072447570729328594&18,17,31,31,31,32\\ 87&3669408499990537951967408951471847094&18,17,31,31,31,34\\ \end{array}
The following is multiple choice question (with options) to answer.
If KING is coded as 17 and MASS is coded as 29 Then DOC is coded as | [
"21",
"20",
"19",
"5"
] | D | KING = 11+9+14+7=41 i.e (4*4)+(1*1)=17
MASS = 13+1+19+19=52 i.e (5*5)+(2*2)=29
DOC = 4+14+3=21 i.e(2*2)+(1*1)=5
ANSWER:D |
AQUA-RAT | AQUA-RAT-38011 | # 2012 AMC 10A Problems/Problem 22
## Problem
The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$?
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$
## Solution 1
The sum of the first $m$ odd integers is given by $m^2$. The sum of the first $n$ even integers is given by $n(n+1)$.
Thus, $m^2 = n^2 + n + 212$. Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$.
Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$. Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.
Let $x = \sqrt{4m^2 - 847}$; note that this means $n = \frac{-1 + x}{2}$. It can be rewritten as $x^2 = 4m^2 - 847$, so $4m^2 - x^2 = 847$. Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\cdot11^2$.
The following is multiple choice question (with options) to answer.
The number 130 can be written as the sum of the squares of 2 different positive integers. What is the sum of these 2 integers? | [
" 17",
" 16",
" 15",
" 14"
] | D | 11^2 + 3^2 = 130 --> 11 + 3 = 14.
D |
AQUA-RAT | AQUA-RAT-38012 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is? | [
"28",
"50",
"88",
"22"
] | B | Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr = 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 ==> x
= 50 km/hr.Answer:B |
AQUA-RAT | AQUA-RAT-38013 | For each digit $a$ from $0$ to $9$, let us count how many numbers there are in our sum with a digit of $a$ in the $10^n$ place. First, suppose $a<9$. A number with a digit of $a$ in the $10^n$ place has a subset of $\{0,1,\dots,a-1\}$ for its last $n$ digits, so there are $\binom{a}{n}$ choices for the last $n$ digits. The preceding digits (omitting the initial $9$) can form any subset of $\{8,7,\dots,a+1\}$, so there are $2^{8-a}$ choices for the preceding digits. So in total, all of the digits of $a$ in our numbers contribute $\sum_n 2^{8-a}\binom{a}{n}\cdot a10^n$ to the sum. By the binomial theorem, this is equal to $$2^{8-a}a(10+1)^a=2^8a\left(\frac{11}{2}\right)^a.$$
For $a=9$, we have no choice of preceding digits, so we just have $\binom{9}{n}$ choices. So the sum of the $9$ digits contributes $$\sum_n \binom{9}{n}9\cdot 10^n=9\cdot 11^9.$$
The following is multiple choice question (with options) to answer.
Which of the options is a two-digit number, which has a unit's digit that exceeds its ten's digit by 4 and the product of the given number and the sum of its digits is equal to 208. | [
"15",
"37",
"55",
"26"
] | D | Using the elimination method the option that fits this description is 26
6-2 = 4 (unit's digit that exceeds its ten's digit by 3)
26*8 = 208 (the product of the given number and the sum of its digits is equal to 175)
answer :D |
AQUA-RAT | AQUA-RAT-38014 | # Math Help - Roots of polynomials
1. ## Roots of polynomials
Hi, sorry if this is in the wrong forums not really sure but here it goes..
The quadratic equation $2x^2+4x+3=0$ has roots $\alpha$ and $\beta$ .
Find the value $\alpha^4+\beta^4$
Ive managed to expand to $\alpha^4+\beta^4=(\alpha+\beta)^4-4\alpha^3\beta-6\alpha^2\beta^2-4\alpha\beta^3$
but im not sure how to put in terms of $\alpha\beta$ and $\alpha+\beta$ to work out an answer.
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$ which I managed to prove, is there anyway I can use this to answer the question?
Help really appreciated. Thankyou !
2. Hello, NathanBUK!
The quadratic equation $2x^2+4x+3\:=\:0$ has roots $\alpha$ and $\beta$ .
Find the value of: $\alpha^4+\beta^4$
Also the previous part of the question asks to show $\alpha^2+\beta^2=1$
. . which I managed to prove. . . . . Good!
Is there anyway I can use this to answer the question? . . . . Yes!
We know that: . $\begin{Bmatrix}\alpha + \beta &=& -2 \\ \alpha\beta &=& \frac{3}{2} \end{Bmatrix}$
We have: . $\alpha^2 + \beta^2 \;=\;1$
Square: . $\left(\alpha^2 + \beta^2\right)^2 \;=\;(1)^2$
. . . . $\alpha^4 + 2\alpha^2\beta^2 + \beta^4 \;=\;1$
The following is multiple choice question (with options) to answer.
Which of the following equations has a root in common with x^2−4x+4=0 ? | [
"x^2-1=0",
"x^2-2=0",
"x^2-3=0",
"x^2-4=0"
] | D | If we carefully look at the given equation we can arrange it in the following manner:
(x-2)^2= 0
so the two root is 2.
Now put 2 in given equations. The equation in which one of them gives value 0, that will be our answer.
(A) putting 2: we get 3. Reject this option.
(B) putting 2: we get 2. Reject this option.
(C) putting 2: we get 1. Reject this option.
(D) putting 2: we get 0. This is the equation, no need to check other option.
D is the answer. |
AQUA-RAT | AQUA-RAT-38015 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
A hungry child reaches into a candy jar filled with 5 different candies; red, green, yellow, black, and white colored candies. If the child randomly picks three candies, what is the probability the child will pick the red and green colored candies as two of the three candies? | [
"3/10",
"1/10",
"1/15",
"3/15"
] | A | Combination Formula: nCr = n!/(r!(n - r)!), where n is the population/set, and r is the sample/subset.
Total number possible = 5C3 = 5!/(3!(5 - 3)!) = 10
Number red possible = 1C1 = 1
Number green possible = 1C1 = 1
Number other possible = 3C1 = 3
Probability Formula: P(A) = (Number Favorable Outcomes)/(Total Number Possible Outcomes)
P(RG) = (1C1)(1C1)(3C1)/5C3 = 3/10
Answer: A |
AQUA-RAT | AQUA-RAT-38016 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
If x and y are positive integers, and 4x^3=3y, then which of the following must be a multiple of 9?
I. x^2
II. y^2
III. xy | [
"I only",
"II only",
"III only",
"I and II only"
] | D | 4x^2 = 3y
since x,y are positive integers, x^2 = x*x is divisible by 3 -> x is divisible by 3 and y is divisible by x^2
-> x^2 and y is divisible by 9 -> y^2 is divisible by 9
(1),(2), and (3) must be true
Answer is D |
AQUA-RAT | AQUA-RAT-38017 | How do you find the prime factorization of 196?
Shwetank Mauria
Featured 3 months ago
$196 = 2 \times 2 \times 7 \times 7$
Explanation:
As the last two digits in $196$ are divisible by $4$, $196$ too is divisible by $4$.
Dividing by $4$ we get $49$ and hence
$196 = 4 \times 49$
but factors of $4$ are $2 \times 2$ and that of $49$ are $7 \times 7$. Further $2 ' s$ and $7 ' s$ are prime numbers and cannot be factorized.
Hence prime factors of $196$ are
$196 = 2 \times 2 \times 7 \times 7$
Note : This method of factorization, in which we first find identifiable factors and then proceed until all prime factors are known is called tree method. This is graphically described below.
What is the least common multiple of 7 and 24?
Tony B
Featured 2 months ago
A teacher will expect the prime number method. Just for the hell of it this is a different approach!
168
Explanation:
We have two numbers ; 24 and 7
I am going to count the 24's. However lets look at this value.
24 can be 'split' into a sum of 7's with a remainder. So each 24 consists of:
$24 = \left(7 + 7 + 7 + 3\right)$
If we sum columns of these we will get the 3 summing to a value into which 7 will divide exactly. When this happens we have found our least common multiple.
REMEMBER WE ARE COUNTING THE 24's
The following is multiple choice question (with options) to answer.
How many different positive integers are factors of 196 ? | [
" 1",
" 2",
" 3",
" 49"
] | D | 15, 14×15, 14=7^2 × 2^2
So total factors = (6+1)(6+1) = 49
Answer: D |
AQUA-RAT | AQUA-RAT-38018 | 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.
It takes 10 days for digging a trench of 100 m long, 50 m broad and 10 m deep. What length of trench,25 m broad and 15 m deep can be dug in 30 days ? | [
"400 m",
"200 m",
"100 m",
"89 m"
] | A | More days, more length (Direct)
Less breadth, more length (Indirect)
More depth, less length (Indirect
Days 10 : 30;
Breadth 25 : 50; : : 100 : x
Depth 15 : 10;
:. 10 * 25* 15 * x = 30 *50 * 10 *100
x= (30*50*10*100)/10*25*15 = 400
So the required length = 400m
ANSWER:A |
AQUA-RAT | AQUA-RAT-38019 | # Math Help - Finding the values of a and b
1. ## Finding the values of a and b
Hello everyone. This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
The keyword in this irksome question would be the word 'or'. So it denotes that that this involves quadratic formulas.
Can anyone give me a clue so that I may make a breakthrough in understanding this problem? Thank you so much!
2. The wording is a bit iffy, but they actually meant that the solutions for that equation is x = 3 AND x = -4.
3. Originally Posted by PythagorasNeophyte
This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
$x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$
We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:
$3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $-4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$
Now solve for a and b using simultaneous equation.
4. Originally Posted by Educated
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
The following is multiple choice question (with options) to answer.
If a/b=3/4 and 8a+5b=22,then find the value of a. | [
"3/2",
"5/2",
"1/2",
"4/2"
] | A | (a/b)=3/4 =>b=(4/3) a.
Therefore, 8a+5b=22 => 8a+5*(4/3) a=22 => 8a+(20/3) a=22
=>44a = 66 => a=(66/44)=3/2
Answer is A |
AQUA-RAT | AQUA-RAT-38020 | The game is "fair".
Playing repeatedly, you can expect to break even.
4. thanks all for the answers.
but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?
The following is multiple choice question (with options) to answer.
Sid and Bonnie are opposing participants in an online auction. Each has a constant bidding strategy: Sid always bids up by raising $40 over any previous bid. Bonnie always bids up by doubling any previous bid. The auction begins with an initial bid by a third player, and progresses as follows: Sid bids up first, and Bonnie follows. Sid bids up again, and Bonnie follows and raises the bid to a final $600. What was the dollar value of the initial price? | [
"46",
"90",
"68",
"568"
] | B | Initial bid = p$
Sid = p+40
Bonnie = 2p
The sequence is R---> B ---> R -----> B ---> 600
Thus per the sequence we have: p+40, 2(p+40), 2p+80+40 , 2(2p+120)
Now given, 2(2p+120) = 600----> p =90$. Thus B is the correct answer. |
AQUA-RAT | AQUA-RAT-38021 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A cyclist covers a distance of 750 meter in 2 minutes 10 seconds. What is the speed in km/hr of cyclist | [
"20.88 km/hr",
"17 km/hr",
"18 km/hr",
"19 km/hr"
] | A | Explanation:
Speed=Distance/Time
Distance=750meter
Time=2 min 10sec=130sec
Speed=750/130=5.8m/sec
=>5.8∗18/5km/hr=20.88km/hr
Option A |
AQUA-RAT | AQUA-RAT-38022 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Status: QA & VA Forum Moderator
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
Thanks and Regards
Abhishek....
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Math Expert
Joined: 02 Sep 2009
Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A company has two models of computers, model M and model N. Operating at a constant rate, a model M computer can complete a certain task in 24 minutes and a model N computer can complete the same task in 12 minutes. If the company used the same number of each model of computer to complete the task in 1 minute, how many model M computers were used? | [
"3",
"4",
"5",
"8"
] | D | Let's say 1 work is processing 24 gb of data.
Model M : 1 gb per min
Model N : 2 gb per min
Working together, 1 M and 1 N = 3 gb per min
So, 8 times as many computers would work at 18 gb per min.
So no. of M = 8
Answer is D |
AQUA-RAT | AQUA-RAT-38023 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
six family members: 5 grandchildren (3 brothers and 2 sisters) and their 1 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers? | [
"120",
"48",
"1440",
"2880"
] | B | Answer: B. |
AQUA-RAT | AQUA-RAT-38024 | Thanks for any help. I am totally confused.
1. 👍 0
2. 👎 0
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4. ### statistics
The following is multiple choice question (with options) to answer.
If there are 50 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw? | [
"1/25",
"2/25",
"3/25",
"4/25"
] | D | 50 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3
So, 5x=50 ---> x=10
Red marbles = 2*x = 2*10 = 20
Blue marbles = 3*x = 3*10 = 30
Prob to select a red marble = 20/50 = 2/5
Prob to select 2 red marbles with replacement = 2/5*2*5 = 4/25
Hence, answer will be D. |
AQUA-RAT | AQUA-RAT-38025 | } \left( \dfrac { 2y+b }{ a-2y } \right)$$$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right)$$Maths
The following is multiple choice question (with options) to answer.
2y - 2x = 2xy and x ≠ 0. If x and y are integers, which of the following could equal y? | [
"2",
"1",
"0",
"-3"
] | D | Plug in the answer choices in the equation from the question stem.
A) y = 2 >>> 4-x = 4x >>> No value of x will satisfy this, not even 0. POE
B) y = 1 >>> 2 - x = 2x >>> Same, POE
C) y = 0 >>> -x = 0 >>> x can not equal 0
D) y = -1 >>> -2 - 2x = -4x >>> Holds true for x = 1, no need to check E. This is the answer.
Answer D |
AQUA-RAT | AQUA-RAT-38026 | # 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.
Sandy buys an old scooter for $900 and spends $300 on its repairs. If Sandy sells the scooter for $1260, what is the gain percent? | [
"3%",
"5%",
"7%",
"9%"
] | B | selling price / total cost = 1260/1200 = 1.05
The gain percent is 5%.
The answer is B. |
AQUA-RAT | AQUA-RAT-38027 | 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.
Each of the 50 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 17 students sign up for the poetry club, 24 students for the history club, and 22 students for the writing club. If 3 students sign up for exactly two clubs, how many students sign up for all three clubs? | [
"5",
"6",
"7",
"8"
] | A | The total number in the three clubs is 17+24+22=63.
All 50 students signed up for at least one club.
3 of those students signed up for exactly one more club.
63 - 53 = 10 so 5 students must have signed up for exactly three clubs.
The answer is A. |
AQUA-RAT | AQUA-RAT-38028 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
Working at a constant rate, P can finish a job in 3 hours. Q, also working at a constant rate, can finish the same job in 18 hours. If they work together for 2 hours, how many more minutes will it take P to finish the job, working alone at his constant rate? | [
"32",
"36",
"40",
"44"
] | C | Each hour they complete 1/3 + 1/18 = 7/18 of the job.
In 2 hours, they complete 2(7/18) = 7/9 of the job.
The time for P to finish is (2/9) / (1/3) = (2/3) hour = 40 minutes
The answer is C. |
AQUA-RAT | AQUA-RAT-38029 | homework-and-exercises, mass, error-analysis, weight
Title: Absolute uncertainity in weight An object has a mass of 88.60 kg ± 0.09 kg. I calculated the weight to be 869 N
but now I need to calculate the absolute uncertainty in weight.
My thinking was that the absolute uncertainty would be .09/869 = 0.0001. Then multiply that by the weight, so .0001 by 869. So 0.09 but however this is not correct Looking for some help The weight is given by:
$$ W = mg $$
where $m$ is your mass and $g$ is the acceleration due to gravity. If we have some error in the mass we can write the mass as $m \pm \sigma_m$ where $\sigma_m$ is the error (0.09 kg in this case) so we get:
$$\begin{align}
W &= (m \pm \sigma_m)g \\
&= mg \pm \sigma_mg
\end{align}$$
So the weight we get is $mg$ with an error of $\pm \sigma_mg$.
(This is assuming we can ignore any errors in the value of $g$.)
The following is multiple choice question (with options) to answer.
The average weight of A, B and C is 44 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is: | [
"17 kg",
"28 kg",
"26 kg",
"31 kg"
] | B | let d sum of a, b, c is 3*44=132
and sum of a and b s 2*40=80
sum of b and c is 2 *43=86
hence 80+86-132=28
ans=28
ANSWER:B |
AQUA-RAT | AQUA-RAT-38030 | Which of the following shows the numbers $$200!, 100^{200}, 200^{100}$$ in increasing order?
$$\text{(A)} \space 200! < 100^{200} < 200^{100} \space \space \space \space \text{(B)} \space 200! < 200^{100} < 100^{200} \space \space \space \space \color{blue}{\text{(C)} \space 200^{100} < 200! < 100^{200}}$$
$$\text{(D)} \space 200^{100} < 100^{200} < 200! \space \space \space \space \text{(E)} \space 100^{200} < 200^{100} < 200!$$
For the third problem, we know that $$e^{\pi} > e^{e}$$, so we exclude options $$\text{(A)}$$ and $$\text{(E)}$$. Also we know that $$e^{\pi} > \pi^{e}$$, so we exclude options $$\text{(B)}$$ and $$\text{(D)}$$. Hence $$\text{(C)}$$ is correct.
That is by managing the expressions to have the form $$y = x^{1/x}$$ then differentiating and finding the critical number(s).
Still this problem matches the others, and they might come under one topic of calculus.
The following is multiple choice question (with options) to answer.
Which of the following is closest to (120!-119!)/(120!+119!) | [
"2",
"2.5",
"3",
"1"
] | D | (120!-119!)/(120!+119!)
119! (120-1)/119! (120+1)
119/121 = 1 (approximately)
D is the answer |
AQUA-RAT | AQUA-RAT-38031 | c++, algorithm, programming-challenge, time-limit-exceeded, dynamic-programming
The government of Siruseri has just commissioned one of the longest
and most modern railway routes in the world. This route runs the
entire length of Siruseri and passes through many of the big cities
and a large number of small towns and villages in Siruseri.
The railway stations along this route have all been constructed
keeping in mind the comfort of the travellers. Every station has big
parking lots, comfortable waiting rooms and plenty of space for
eateries. The railway authorities would like to contract out the
catering services of these eateries.
The Siruseri Economic Survey has done a through feasibility study of
the different stations and documented the expected profits (or losses)
for the eateries in all the railway stations on this route. The
authorities would like to ensure that every station is catered to. To
prevent caterers from bidding only for profitable stations, the
authorities have decided to give out catering contracts for contiguous
segments of stations.
The minister in charge realises that one of the bidders is his bitter
adversary and he has decided to hand out as useless a segment as
possible to him. On the other hand, he does not want to be seen to be
blatantly unfair by handing out a large loss-making section to the
adversary. Instead he wants to find the largest segment whose sum is
closest to 0, so that his adversary spends all his time running a
large number of canteens and makes either a small loss or a small
profit or, even better, nothing at all!
In other words, if the profits/losses at the stations are p1, p2, ...,
pN the minister would like to handover a sequence i, i+1, ..., j such
that the absolute value of pi + pi+1 + ... + pj is minimized. If there
is more than one sequence with this minimum absolute value then he
would like to hand over the longest one.
For example, suppose there are 8 stations along the line and their
profitability is as follows:
Station 1 2 3 4 5 6 7 8
Expected Profits -20 90 -30 -20 80 -70 -60 125
If the adversary is awarded the section 1 through 4, he will make a
net profit of 20. On the other hand if he is given stations 6, 7 and
8, he will make loss of 5 rupees. This is the best possible value.
The following is multiple choice question (with options) to answer.
There are 10 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station? | [
"156",
"167",
"132",
"352"
] | C | The total number of stations = 12
From 12 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in 12P₂ ways.
²⁰P₂ = 12 * 11
= 132.
Answer: C |
AQUA-RAT | AQUA-RAT-38032 | # 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.
John bought 15 apples for Rs.10 and sold them at the rate of 12 apples for Rs.12. What is the percentage of profit made by him? | [
"50 %",
"60 %",
"65 %",
"100 %"
] | A | Profit % = 15/10 * 12/12 = 1.5 ⇒ 50%.
Answer A |
AQUA-RAT | AQUA-RAT-38033 | 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 30 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"150",
"166",
"240",
"157"
] | A | Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 30 = 15
x + 300 = 450
x = 150 m.
Answer: A |
AQUA-RAT | AQUA-RAT-38034 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
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A total of 22 men and 26 women were at a party, and the average [#permalink]
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04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
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Abhishek....
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
In an office, totally there are 2800 employees and 25% of the total employees are males. 30% of the males in the office are at-least 50 years old. Find the number of males aged below 50 years? | [
"390",
"490",
"400",
"460"
] | B | Number of male employees = 2800* 25/100 = 700
Required number of male employees who are less than 50 years old = 700* (100 - 30)%
= 700* 70/100 = 490.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38035 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
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The price of a consumer good increased by p%. . . [#permalink]
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Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
A number is increased by 60% and then decreased by 60%, the final value of the number is ? | [
"decrease by 36%",
"decrease by 30%",
"decrease by 32%",
"decrease by 34%"
] | A | Here, x = 60 and y = - 60
Therefore, the net % change in value
= ( x + y + xy/100)%
= [60 - 60 + (60 * -60)/100]% or - 36%
Since the sign is negative, there is a decrease in value by 36%.
ANSWER: A |
AQUA-RAT | AQUA-RAT-38036 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 500 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"12 sec",
"30 sec",
"86 sec",
"16 sec"
] | B | Speed of train relative to man = 63 - 3
= 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 500 * 3/50
= 30 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-38037 | Example $$\PageIndex{2}$$
Consider the sum of the first $$n$$ integers. We can think about this as a recursively-defined sequence, by defining $$s_1 = 1$$, and $$s_n = s_{n−1} + n$$, for every $$n ≥ 2$$. Thus, $$s_2 = 1 + 2$$;
$$s_3 = s_2 + 3 = 1 + 2 + 3$$,
and so on. Prove by induction that $$s_n = \dfrac{n(n + 1)}{2}$$, for every $$n ≥ 1$$.
Solution
Base case: $$n = 1$$. We have $$s_n = s_1 = 1$$, and
$$\dfrac{n(n + 1)}{2} = \dfrac{1(2)}{2} = 1$$,
so the equality holds for $$n = 1$$. This completes the proof of the base case.
Inductive step: We begin with the inductive hypothesis. Let $$k ≥ 1$$ be arbitrary, and suppose that the equality holds for $$n = k$$; that is, assume that $$s_k = \dfrac{k(k + 1)}{2}$$.
Now we want to deduce that
$$s_{k+1} = \dfrac{(k + 1)(k + 2)}{2}$$.
Using the recursive relation, we have $$s_{k+1} = s_k + (k+ 1)$$ since $$k+ 1 ≥ 2$$, and using the inductive hypothesis, we have $$s_k = \dfrac{k(k + 1)}{2}$$, so putting these together, we see that
$$s_{k+1} = \dfrac{k(k + 1)}{2} + (k + 1)$$.
Taking out a common factor of $$k + 1$$ gives
The following is multiple choice question (with options) to answer.
The sequence S is defined as Sn = (n + 1)! for all integers n >= 1. For example, S3 = 4! = (4)(3)(2)(1). Which of the following is equivalent to the difference between S100 and S99? | [
"101!",
"100!",
"99^2(98!)",
"100^2(99!)"
] | D | According to this Symbolism question, we're asked to figure out S100 - S99. According to the description of the symbol in the question, we're asked to figure out:
101! - 100!
Both these terms have something in common: 100! so we can factor that out of each piece...
101! - 100!
100!(101 - 1)
Simplifying the parentheses, we get...
100!(100)
Unfortunately, THAT answer isn't listed. We've done the math correctly though, so all we can do is try to convert "our" answer into the format that appears in the answer choices...
100! = (100)(99!) so our answer can be rewritten as...
(100)(99!)(100)
Or....
(100^2)(99!)
ANSWER:D |
AQUA-RAT | AQUA-RAT-38038 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
If books bought at prices ranging from Rs. 150 to Rs. 300 are sold at prices ranging from Rs. 250 to Rs 350, what is the greatest possible profit that might be made in selling 15 books ? | [
"Rs. 2500",
"Rs. 3000",
"Rs. 3500",
"Rs. 3200"
] | B | The greatest profit is possible only if the cost price of the books are minimum and selling prices are maximum.
Let lowest cost price of the 15 books = 150*15 = Rs. 2,250
Maximum selling price of 15 books = 350 *15 = Rs. 5,250
So, maximum profit = 5250 - 2250 = Rs. 3,000
ANSWER : OPTION B |
AQUA-RAT | AQUA-RAT-38039 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? | [
"238",
"227",
"720",
"128"
] | C | The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Answer: C |
AQUA-RAT | AQUA-RAT-38040 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
On dividing 23 by a number, the quotient is 4 and the remainder is 3. Find the divisor. | [
"1",
"2",
"4",
"5"
] | D | d = (D-R)/Q
= (23 - 3)/4
= 20/4 = 5
D |
AQUA-RAT | AQUA-RAT-38041 | pressure, ideal-gas, kinetic-theory, evaporation
Title: Finding the equilibrium vapor pressure of two solutions(of different concentrations) enclosed in a vessel
The saturated vapour pressure above
an aqueous solution of sugar is known to be
lower than that above pure water, where
it is equal to $p_{sat}$, by $\Delta p = 0.05p_{sat}c$.
where c is the molar concentration of the
solution. A cylindrical vessel filled to
height $h_1 =10$cm with a sugar solution of
concentration $c_1 = 2\times 10^{-3}$ is placed under
a wide bell jar. The same solution of concentration
$c_2 = 10^{-3}$ is poured under the
bell to a level $h_2$ << $h_1$ (Shown in the given figure).Determine the level h of the solution in
the cylinder after the equilibrium has been
set in. The temperature is maintained constant
and equal to 20 °C. The vapour above
the surface of the solution contains only
water molecules, and the molar mass of
water vapour is $\mu = 18\times 10^{-3}$ kg/mol.
The following is multiple choice question (with options) to answer.
How many liters of water must be evaporated from 50 liters of a 3-percent sugar solution to get a 4-percent solution? | [
"35",
"33 1/3",
"21",
"16 2/3"
] | C | How many liters of water must be evaporated from 50 liters of a 3-percent sugar solution to get a 4-percent solution?
3% of a 50 liter solution is 1.5L. So you are trying to determine how many liters must a solution be for the 1.5L to represent 4% of the solution. Set up an inequality and solve for x:
1.5/x = 1/4
x = 6
Since you need a 15L solution, you must evaporate 21 of the original 50L solution to get a 4% solution.
Answer is C. |
AQUA-RAT | AQUA-RAT-38042 | python, beginner, assembly, compiler
And use it like:
with open("Programs/"+sys.argv[1]+".asm", "w+") as fileOut:
print("### First Pass - Mapping Symbols to Addresses ###")
address = 0
for tokens, line_num in filter_out_comments("Programs/" + sys.argv[1]):
if tokens[0][-1] == ":": # found symbol
...
print("### Second Pass - Translating into machine code ###")
address = 0
for tokens, line_num in filter_out_comments("Programs/" + sys.argv[1]):
asm = ""
if tokens[0] in RRR and len(tokens) == 4:
...
Some improvements
Instead of leaving some code at the top-level of the file, you should wrap it into a function. It let you test and re-use it more easily. You should also make use of the if __name__ == '__main__': idiom:
def compile_asm(filename)
with open(filename + ".asm", "w+") as fileOut:
print("### First Pass - Mapping Symbols to Addresses ###")
address = 0
for tokens, line_num in filter_out_comments(filename):
if tokens[0][-1] == ":": # found symbol
...
print("### Second Pass - Translating into machine code ###")
address = 0
for tokens, line_num in filter_out_comments(filename):
asm = ""
if tokens[0] in RRR and len(tokens) == 4:
...
if __name__ == '__main__':
compile_asm("Programs/" + sys.argv[1])
The following is multiple choice question (with options) to answer.
If CRY is coded as MRYC then how will GET be coded? | [
"MTEG",
"MGET",
"MEGT",
"METG"
] | D | CRY=M+RYC
GET=M+ETG
METG
ANSWER:D |
AQUA-RAT | AQUA-RAT-38043 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A and B walk around a circular track. They start at 8 a.m from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9 a.m | [
"5",
"6",
"7",
"8"
] | A | Explanation:
Relative speed = (2+3) =5 rounds per hour
So, they cross each other 5 times in an hour hence, they cross 5 times before 9 a.m
Answer: Option A |
AQUA-RAT | AQUA-RAT-38044 | encoding-scheme
Title: Why is this code uniquely decodable? Source alphabet: $\{a, b, c, d, e, f\}$
Code alphabet: $\{0, 1\}$
$a\colon 0101$
$b\colon 1001$
$c\colon 10$
$d\colon 000$
$e\colon 11$
$f\colon 100$
The following is multiple choice question (with options) to answer.
Special codes are designated by either a 2-letter or a 3-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different codes that can be designated with this system? | [
"2(26)^3",
"26(26)^2",
"27(26)^2",
"26(26)^3"
] | C | 26^2+26^3 = 26^2(1+26)=27*26^2
The answer is C. |
AQUA-RAT | AQUA-RAT-38045 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train is 360 meter long is running at a speed of 72 km/hour. In what time will it pass a bridge of 140 meter length? | [
"27 seconds",
"29 seconds",
"25 seconds",
"11 seconds"
] | C | Speed = 72 Km/hr = 72*(5/18) m/sec = 20 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (1/20) = 25 seconds
Answer:C |
AQUA-RAT | AQUA-RAT-38046 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can do a work In 3 days. A alone can do it in 12 days. What time B will take to do the work alone? | [
"4 days",
"8 days",
"12 days",
"10 days"
] | A | Explanation:
A and B 1day's work = 1/3
A alone can do 1day's work = 1/12
what time B will take to do the work alone?
B = (A+B) - A = (1/3 - (1/12) = 4 Days
Answer: Option A |
AQUA-RAT | AQUA-RAT-38047 | 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.
If 10 crates have 95 apples each and how many apples more is required in order to make 100 apples in each crate? | [
"5",
"10",
"15",
"50"
] | D | Each crate requires 5 apples and totally there are 10 crates so required apples = 10 * 5 = 50
Answer: D |
AQUA-RAT | AQUA-RAT-38048 | # Math Help - Inequality Application
1. ## Inequality Application
A charter airline finds that on its Saturday flights from Philadelphia to London, all 120 seats will be sold if the ticket price is $200. However, for each$3 increase in ticket price, the number of seats sold decreases by one.
a) Find a formula for the number of seats sold if the ticket price is P dollars.
The answer for a is $-\frac{1}{3}P + \frac{560}{3}$. However, I do not see how they arrived at that answer. Any help will be greatly appreciated.
2. Originally Posted by mathgeek777
The answer for a is $-\frac{1}{3}P + \frac{560}{3}$. However, I do not see how they arrived at that answer. Any help will be greatly appreciated.
do you realize that we have a linear relationship between the number of seats filled and the ticket price? since each changes at a constant rate with respect to each other.
lets think of this graphically so that you get the idea. we want a function of price, so put that on the x-axis. the number of seats on the y-axis. now, plot a few points:
$\begin{array}{c|c} S & P \\ \hline 120 & 200 \\ 119 & 203 \\ 118 & 206 \\ . & . \\ . & . \\ . & . \end{array}$
S is the number of seats, P is the price. S goes down by 1 when P goes up by 3
all we need to do is find the straight line that passes through these points. can you do that?
3. Ok. I see where the $-\frac{1}{3}$ came from, but i still don't know how you get $\frac{560}{3)$
4. Originally Posted by mathgeek777
Ok. I see where the $-\frac{1}{3}$ came from, but i still don't know how you get $\frac{560}{3}$
the equation of a line in the slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$, is the y-intercept.
The following is multiple choice question (with options) to answer.
A certain theater has 100 balcony seats. For every $2 increase in the price of a balcony seat above $10, 5 fewer seats will be sold. If all the balcony seats are sold when the price of each seat is $10, which of the following could be the price of a balcony seat if the revenue from the sale of balcony seats is $1,160 ? | [
"$12",
"$14",
"$11.5",
"$17"
] | C | I solved it as follows:
10+2(x) = 100 -5(x)
x= 12
Equation should be (10+$2*x)(100-5x)=1,160, where x is the # of times we increased the price by $2. (10+$2*x)(100-5x)=1,160 --> (5+x)(20-x)=116 --> x=0.75 or x=14.25 --> price=10+$2*0.75=$11.5 or price=10+$2*14.25=$38.5.
Answer: C. |
AQUA-RAT | AQUA-RAT-38049 | human-biology, reproduction, anthropology
Note: here is a link about multiple pregnancy : https://www.healthline.com/health/pregnancy/chances-of-having-twins#assisted-reproduction. 49 - 19 = 30
So the 1st Child would be 30 when the last children would be born, as long as the parents stay healthy and the mother delivers the child in a natural way (No Surgical Procedure to deliver the child) and the Mother has not had Menopause (Menopause usually happens around 50 Years of age) the sibling age gap would be very possible, but lets take a look at how many children the mother had, if I am correct they had about 16 Children, many people in countries that have a considerably low life expectancy have many children, lets take a look at the world record for most children born from one mother...a person named Valentina Vassilyev, gave birth to 69 Children! Thats a lot of children, so giving birth to 16 children with a 30 age gap should be very possible and realistic as well.
The following is multiple choice question (with options) to answer.
The sum of ages of 5 children born 1 years different each is 30 years. What is the age of the Elder child? | [
"8",
"9",
"10",
"16"
] | A | Let the ages of children be x, (x + 1), (x + 2), (x + 3) and (x + 4) years.
Then, x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 30
5x = 20
x = 4.
x+4= 4+4= 8
Answer : A |
AQUA-RAT | AQUA-RAT-38050 | - 2 years, 3 months ago
- 2 years, 3 months ago
I was getting the answer as 36.
My cases were similar to that of Deeparaj.
Case 1: When (4,8) is one of the selected pair.
Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.
Case 2: When (4,8) is not of the pairs.
In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24.
- 2 years, 6 months ago
Ah...
I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.
- 2 years, 6 months ago
Can a number be repeated in the pairs?
- 2 years, 6 months ago
No.
- 2 years, 6 months ago
Case 1: One of the pairs is (4,8): $$4\times {4\choose2}$$
Still working on Case 2.
- 2 years, 6 months ago
Case 2:$$4!$$.
So, on the whole $$\boxed{ 48 }$$ ways. Am I right?
- 2 years, 6 months ago
The following is multiple choice question (with options) to answer.
The ratio between the present ages of P and Q is 6:8. If Q is 4 years old than P, what will be the ratio of the ages of P and Q after 4 years? | [
"A)8:9",
"B)3:5",
"C)4:3",
"inadequate"
] | A | Let P's age and Q's age be 6x and 8x years respectively.
Then, 8x - 6x = 4 => x =2
Required ratio = (6x + 4) : (7x + 4)
16 : 18 = 8:9
ANSWER:A |
AQUA-RAT | AQUA-RAT-38051 | Hence, the probability of selecting three balls of one color and two balls of a different color when five balls are selected from the $7 + 8 + 9 = 24$ balls in the bag is $$\frac{\dbinom{7}{3}\dbinom{8}{2} + \dbinom{7}{3}\dbinom{9}{2} + \dbinom{7}{2}\dbinom{8}{3} + \dbinom{7}{2}\dbinom{9}{3} + \dbinom{8}{3}\dbinom{9}{2} + \dbinom{8}{2}\dbinom{9}{3}}{\dbinom{24}{5}}$$
• I was wondering whether you could have any cases where you would draw 3 blue balls, 1 red, and 1 white ball? – user262291 Apr 9 '16 at 15:10
• That's a different question since choosing three of one color and two of another color when five balls are selected precludes the possibility of selecting three different colors. The probability of selecting three blue balls, one red ball, and one white ball in the first scenario is $$\frac{\binom{7}{3}\binom{7}{1}\binom{7}{1}}{\binom{21}{5}}$$ In the second scenario, the probability is $$\frac{\binom{9}{3}\binom{7}{1}\binom{8}{1}}{\binom{24}{5}}$$ The second scenario makes it clearer what I am counting, namely selections of three blue balls, one red ball, and one white ball. – N. F. Taussig Apr 9 '16 at 18:15
Whenever you can, relate the problem to a known category of familiar problems.
Here, you can treat it the way you compute poker probabilities:
We want $3-2$ of a kind, so numerator will be [Choose kinds] $\times$ [Choose balls from each kind]
The following is multiple choice question (with options) to answer.
A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? | [
"1/2",
"10/21",
"9/11",
"7/11"
] | B | Total number of balls = 2 + 3 + 2 = 7
Let S be the sample space.
n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2
Let E = Event of drawing 2 balls , none of them is blue.
n(E) = Number of ways of drawing 2 balls , none of them is blue
= Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2
(∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
P(E) = n(E)/n(S)=5C2/7C2=(5×42×1)/(7×62×1)=5×4/7×6=10/21
Answer is B |
AQUA-RAT | AQUA-RAT-38052 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Running at the same constant rate, 6 identical machines can produce a total of 240 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4 minutes? | [
" 648",
" 1,800",
" 1,600",
" 10,800"
] | C | 6 machines produce 240 bottles per minute;
1 machine produces 240/6=40 bottles per minute;
10 machines produce 40*10=400 bottles per minute;
In 4 minutes 10 machines produce 400*4=1,600 bottles.
Answer: C. |
AQUA-RAT | AQUA-RAT-38053 | # math
How is the circumference of a circle with radius 9cm related to the circumference of a circle with diameter 9cm?
i don't understand this.Help?? Thanks:)
1. 👍
2. 👎
3. 👁
c = pi x d.
The diamter is 2 x radius. So, the diameter of the first cirlce is 18cm. You know the diameter of the second circle. So, how do the circumferences compare?
1. 👍
2. 👎
2. is the second circle twice as big? like is that how they compare?
1. 👍
2. 👎
3. the circumference of the larger circle is indeed twice as long as that of the smaller one.
but...
the area of the larger circle is 4 times that of the area of the smaller.
Proof:
area of smaller = pi(4.5)^2 = 20.25pi
area of larger = pi(9^2) = 81pi
20.25pi/(81pi) = 1/4
1. 👍
2. 👎
4. alright thx!
1. 👍
2. 👎
## Similar Questions
1. ### Math ~ Check Answers ~
Find the circumference of the circle (use 3.14 for pi). Show your work. Round to the nearest tenth. the radius is 23 yd My Answer: ???? •Circumference = (2) (3.14) (23) •3.14 x 2 x 23 = 144.44
2. ### Mathematics
1. Find the circumference of the given circle. Round to the nearest tenth. (Circle with radius of 3.5 cm) -22.0 cm -38.5 cm -11.0 cm -42.8 cm 2. Find the area of the given circle. Round to the nearest tenth. (Circle with a
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4. ### MATH
The following is multiple choice question (with options) to answer.
The circumferences of two circles are 264 meters and 352 meters. Find the difference between the areas of the larger and the smaller circles? | [
"2996 sq m",
"2897 sq m",
"4312 sq m",
"2768 sq m"
] | C | Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l2 - ∏s2
= ∏{1762/∏2 - 1322/∏2}
= 1762/∏ - 1322/∏
= (176 - 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7)
= 4312 sq m
Answer:C |
AQUA-RAT | AQUA-RAT-38054 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A person walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 8 minutes earlier. How far is the office from his house? | [
"3 1/5 Km",
"2 1/5 Km",
"3 Km",
"4 Km"
] | A | Formula = S1*S2/S2-S1 * T1+T2/60
= 4*6/2 * 16/6
= 24/2 * 16/60
= 8 * 2/5
= 16/5 = 3 1/5 Km
A) |
AQUA-RAT | AQUA-RAT-38055 | The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:
$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where $$k$$ must be chosen such that $$5^k\leq{n}$$
Example:
How many zeros are in the end (after which no other digits follow) of 32!?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$. Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is $$\frac{32}{5}=6$$ not 6.4). Therefore, 32! has 7 trailing zeros.
The formula actually counts the number of factors 5 in $$n!$$, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Finding the powers of a prime number p, in the n!
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
Example:
What is the power of 2 in 25!?
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$.
See this post to understand rationale behing this formulae
The following is multiple choice question (with options) to answer.
If p is a natural number and p! ends with q trailing zeros, then the number of zeros that (5p)! ends with will be | [
"a) (p+q) trailing zeros",
"b) (5p+y) trailing zeros",
"c) (5p+5y) trailing zeros",
"d) (p+5y) trailing zeros"
] | A | Let p = 1. p! = 1! = 1, which means q= 0
(5p)! = 5! = 120, trailing zeros = 1
1 = 1 + 0 = p + q
Answer (A). |
AQUA-RAT | AQUA-RAT-38056 | Since n is a integer, can we not try with n as 0?
Yes, n can be 0 but 0 is even too.
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Re: If n is an integer, is n even? [#permalink]
### Show Tags
09 Dec 2015, 00:06
Bunuel wrote:
If n is an integer, is n even?
(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.
Given: n is an integer
Required: is n even?
Statement 1: $$n^2$$ - 1 is an odd integer
$$n^2$$ - 1 = (n-1)(n+1) = odd.
This means both n-1 and n+1 are odd
Odd*Odd = Odd
Odd*Even = Even
Even*Even = Even
n-1, n, n+1 are three consecutive integers.
Since we know that both n-1 and n+1 are odd
Hence n has to be even.
SUFFICIENT
Statement 2: 3n + 4 is an even integer
Even + Even = Even
Even + Odd = Odd
Odd + Odd + Odd
Since 3n+4 = even and 4 is an even integer.
Hence 3n = even. Therefore n = even
SUFFICIENT
Option D
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Re: If n is an integer, is n even? [#permalink]
### Show Tags
09 Aug 2016, 13:33
1
Quote:
If n is an integer, is n even?
(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.
We need to determine whether integer n is even. Let's review four facts about even and odd integers: 1) An integer and its square are either both even or both odd. 2) The sum (or difference) between an even integer and an odd integer is always odd. 3) The sum of two even integers (or two odd integers) is always even. 4) If the product of two integers is even, at least one of them must be even.
The following is multiple choice question (with options) to answer.
If a is an odd integer, which of the following must be an even integer? | [
"a^4−a+1",
"(a^4−a)(a+1/a)",
"a^4−a^3+a^2+2a",
"(a^3+a^2+a)^2"
] | B | Let assume a be 1..
A. 1-1+1..odd
B.(1-1)(2) = 0.. even
C.1-1+1+2..odd
D.1+1+1..odd
E.None
Ans: option B. |
AQUA-RAT | AQUA-RAT-38057 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 800 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"23 sec",
"30 sec",
"27 sec",
"48 sec"
] | D | Speed of train relative to man = 63 - 3
= 60 km/hr.
= 60 * 5/18
= 50/3 m/sec.
Time taken to pass the man = 800 * 3/50
= 48 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-38058 | As mentioned here, $r(n)$ is $1$ iff every prime of the form $4k-1$ occurs an even number of times in $n$. It follows that $r(n)$ is multiplicative, and \begin{align*} \frac{\sum_{i=1}^\infty \frac{r(n)}{n^x}}{\sum_{i=1}^\infty \frac{1}{n^x}} &= \frac{\prod_{p \text{ prime}} \sum_{i=0}^\infty \frac{r(p^i)}{p^{ix}}}{\prod_{p \text{ prime}} \sum_{i=0}^\infty \frac{1}{p^{ix}}} \\ &= \prod_{p \text{ prime}} \frac{1 + \frac{r(p)}{p^x} + \frac{r(p^2)}{p^{2x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \\ &= \prod_{p \equiv 3 \pmod{4}} \frac{1 + 0 + \frac{1}{p^{2x}} + 0 + \frac{1}{p^{4x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \prod_{p \equiv 1,2 \pmod{4}} \frac{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots}{1 + \frac{1}{p^x} + \frac{1}{p^{2x}} + \cdots} \\ &=
The following is multiple choice question (with options) to answer.
Let the function R(n) represent the product of the first n prime numbers, where n > 0. If x = R(n) + 1, which of the following must be true?
(i) x is always odd
(ii) x is always prime
(iii) x is never the square of an integer | [
"ii only",
"iii only",
"i and ii only",
"i and iii only"
] | D | R(n) is always even, because the first prime is 2 and no matter what n is, 2 always will be a divisor of R(n). Thus, R(n) + 1 = even + 1 = odd. So, (i) is always true.
Now, use logic:
If (ii) is true (so if x is always prime), then (iii) must automatically be true: no prime is the square of an integer. So, the correct answer must be i only; i, ii, and iii only; or i and iii only. since onlyi and iii onlyis among the options, then it must be true.
Or, since (i) is always true, then from options the answer must be either C or D. C cannot be correct because if (ii) is true, then so must be (iii). Thus only D remains.
Answer: D. |
AQUA-RAT | AQUA-RAT-38059 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
Two sets of 3 consecutive positive integers have exactly one integer in common. The sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set? | [
"4",
"7",
"6",
"12"
] | C | A=(2,3,4), sum of this=9
B=(4,5,6), sum of this=15,
The differenct between 15-9=6
Hence,6 is the answer i.e.C |
AQUA-RAT | AQUA-RAT-38060 | 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.
In how much time will a train of length 110 m, moving at 36 kmph cross an electric pole? | [
"sec",
"sec",
"sec",
"sec"
] | D | Convert kmph to mps. 36 kmph = 36 * 5/18 = 10 mps.
The distance to be covered is equal to the length of the train.
Required time t = d/s = 110/10 = 11 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-38061 | # 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 motorboat, whose speed in 15 km/hr in sƟll water goes 30 km downstream and comes back in a
total of 4 hours 30 minutes. The speed of the stream (in km/hr) is | [
"2 km/hr",
"3 km/hr",
"4 km/hr",
"5 km/hr"
] | D | Explanation:
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr
So we know from quesƟon that it took 4(1/2)hrs to travel back to same point.
So,
3015+x−3015−x=412=>900225−x2=92=>9x2=225=>x=5km/hr
Answer: D |
AQUA-RAT | AQUA-RAT-38062 | Case I: The spider at $A$ moves to $B$.
The spider at $C$ can only move to $D$ or $G$ (since $CE$ isn't an edge). If the spider at $C$ moves to $D$, then there are two ways for the spiders at $F,H$ to move: $(A,C,F,H) \to (B,D,E,G)$ or $(A,C,F,H) \to (B,D,G,E)$. If the spider at $C$ moves to $G$, then there is only one way for the spiders at $F,H$ to move: $(A,C,F,H) \to (B,G,E,D)$ (because $BH$ and $DF$ aren't edges). This gives $3$ ways for the spiders at $A,C,F,H$ to move to distinct vertices if the spider at $A$ moves to $B$.
Case II: The spider at $A$ moves to $D$.
Similarly to Case I, there are $3$ total ways for the spiders at $A,C,F,H$ to move to distinct vertices if the spider at $A$ moves to $D$.
Case III: The spider at $A$ moves to $E$.
Similarly to Case I, there are $3$ total ways for the spiders at $A,C,F,H$ to move to distinct vertices if the spider at $A$ moves to $E$.
This gives us $9$ ways for the spiders at $A,C,F,H$ to end up at distinct locations. Hence, the probability of no two of these spiders ending up at the same vertex is $\dfrac{9}{81} = \dfrac{1}{9}$.
The following is multiple choice question (with options) to answer.
If 7 spiders make 7 webs in 7 days, then how many days are needed for 1 spider to make 1 web? | [
"1",
"7",
"3",
"14"
] | B | Explanation:
Let, 1 spider make 1 web in x days.
More spiders, Less days (Indirect proportion)
More webs, more days (Direct proportion)
Hence we can write as
(spiders)7:1
(webs) 1:7}::x:7
⇒7×1×7=1×7 × x
⇒x=7
Answer: Option B |
AQUA-RAT | AQUA-RAT-38063 | # 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.
Ram professes to sell his goods at the cost price but he made use of 950 grms instead of a kg, what is the gain percent? | [
"5 15/19%",
"3 15/19%",
"4 15/19%",
"6 15/19%"
] | A | 950 --- 50
50 --- ? =>
=50/950 *100
= 5 15/19%
Answer: A |
AQUA-RAT | AQUA-RAT-38064 | # Maximum of $x^3+y^3+z^3$ with $x+y+z=3$
It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated
• Consider $F=x^3+y^3+(3-x-y)^3$ and show it is concave up on S=$0\le x\le 2, 0\le y\le2$. – i. m. soloveichik May 6 '17 at 14:27
You have correctly established that $z=0$. From there you have $y=3-x$ so substitute that into $f$. As $y\le2$ then $1\le x\le2$.
$$f(x,3-x,0)=x^3+(3-x)^3=9x^2+27x+27=9(x^2+3x+3)$$
The following is multiple choice question (with options) to answer.
If x^3 = y^2 > z, which of the statements could be true?
I. x < y < z
II. x < z < y
III. y < x < z | [
"I only",
"III only",
"I and II only",
"II and III only"
] | C | Given : x^3 = y^2 > z
x^3 = y^2 is possible only when 9^3=27^2=729
I. x < y < z is true for x=9, y=27, z=30 i.e. Answers can only be Options A, C or E
II. x < z < y is true for x=9, y=27, z=10 i.e. Answers can only be Options C or E
III. y < x < z is not true for x=9, y=27
Answer: option C |
AQUA-RAT | AQUA-RAT-38065 | a)f(x)=1 / (x^2+1)
d)f(x)={1 / (x^3+x)}^5
f)h(x)=x-(x^2)
For each of the above functions from a) through h) how do you determine if it is even,odd,or neither??
a)
$f(x) = \frac{1}{x^2 + 1}$
$f(-x) = \frac{1}{(-x)^2 + 1}$
$f(-x) = \frac{1}{x^2 + 1} = f(x)$
so this function is even.
d)
$f(x) = \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = \left ( \frac{1}{(-x)^3 + (-x)} \right ) ^5$
$f(-x) = \left ( \frac{1}{-x^3 - x} \right ) ^5$
$f(-x) = \left ( \frac{-1}{x^3 + x} \right ) ^5$
$f(-x) = (-1)^5 \left ( \frac{1}{x^3 + x} \right ) ^5$
$f(-x) = - \left ( \frac{1}{x^3 + x} \right ) ^5 = -f(x)$
so this function is odd.
f)
$h(x) = x - x^2$
$h(-x) = (-x) - (-x)^2$
$h(-x) = -x + x^2$
which is equal to neither h(x) nor -h(x). So this function is neither even, nor odd.
-Dan
The following is multiple choice question (with options) to answer.
If f(x) = x^3/(x^5 - 1), what is f(1/x) in terms of f(x)? | [
"-f(x)",
"1/f(x)",
"2/f(x)",
"3/f(x)"
] | A | If x = 2 then f(x) = 8/31 and f(1/x) = -8/31 which is equal to -f(x)
answer A |
AQUA-RAT | AQUA-RAT-38066 | For convenience, let $$p=bc$$, $$q=ca$$, and $$r=ab$$. We are to show that \begin{align}\sqrt{(p-q)^2+4pq\sin^2\frac{\alpha+\beta}{2}}&\leq \sqrt{(q-r)^2+4qr\sin^2\frac{\alpha}{2}}+\sqrt{(r-p)^2+4rp\sin^2\frac{\beta}{2}}.\tag{2}\end{align}
The following is multiple choice question (with options) to answer.
If p, q and r are positive integers and satisfy x = (p + q -r)/r = (p - q + r)/q = (q + r - p)/p, then the value of x is? | [
"1/2",
"1",
"-1/2",
"-1"
] | B | When two or more ratios are equal, each of the ratios are equal to sum of the numerators divided by the sum of the denominators, provided sum of the denominators is non-zero.
Hence, x = (p + q -r)/r = (p - q + r)/q = (q + r - p)/p
=> x = (p + q - r + p - q + r + q + r - p) / (r + q + p)
=> x = (r + q + p) / (r + q + p) = 1
p + q + r is non-zero.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38067 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
Income of A is 25% more than the income of B. What is the income of B in terms of income of A? | [
"80%",
"75%",
"78.66%",
"71.25%"
] | A | Explanation:
One of the most basic questions.
Let income of B be 100. Income of A is 25% more than income of B which means Income of A becomes 125
Now income of B in terms of A = 100/125 *100 = 80%
ANSWER A |
AQUA-RAT | AQUA-RAT-38068 | 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.
A sum of money at simple interest amounts to Rs. 1717 in 1 year and to Rs. 1734 in 2 years. The sum is: | [
"Rs. 1200",
"Rs. 1690",
"Rs. 1600",
"Rs. 1700"
] | D | S.I. for 1 year = Rs. (1734 - 1717) = Rs. 17.
Principal = Rs. (1717 - 17) = Rs. 1700.
Answer: Option D |
AQUA-RAT | AQUA-RAT-38069 | 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.
Two pipes A and B can fill a cistern in 12 and 24 minutes respectively, and a third pipe C can empty it in 36 minutes. How long will it take to fill the cistern if all the three are opened at the same time? | [
"10 2/7 min",
"10 1/8 min",
"11 1/7 min",
"12 8/7 min"
] | A | 1/12 + 1/24 - 1/36
= 7/72
72/7 = 10 2/7
Answer:A |
AQUA-RAT | AQUA-RAT-38070 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
The Guests at a football banquet consumed a total of 323 pounds of food. if no individual guest consumed more than 2 pounds of food, What is the minimum number of guests that could have attended the banquet? | [
"160",
"161",
"162",
"163"
] | C | To minimize one quantity maximize other.
161*2 (max possible amount of food a guest could consume) = 322 pounds, so there must be more than 161 guests, next integer is 162.
Answer: C. |
AQUA-RAT | AQUA-RAT-38071 | 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.
If 20 % of certain quantity of work is done by A and the rest 80% by B, the work is completed in 20 days. If 80% of the work is done by A and the remaining 20% by B, then the work is completed in 30 days. How many days are required to complete the work , if A and B work together. | [
"11 1/9",
"10 1/9",
"12",
"15"
] | A | Use algebra :
0.2/A + 0.8/B = 20
1/A + 4/B = 100
0.8/A + 0.2/B = 30
4/A+ 1/B = 150
15/B = 250
B = 3/50 work per day
A = 3/100 work per day
Combined rate = 3/40 + 3/100 = 9/100
Time taken when working together = 100/9 = 11(1/9) days
Answer (A) |
AQUA-RAT | AQUA-RAT-38072 | 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.
Three friends A, B and C decided to share a lot of apples. Each of them had half of the total plus half an apple in order. After each of them took their share twice, no apples were left.How many apples were there? | [
"62",
"63",
"64",
"65"
] | B | Whenever the rate of reduction is 'half of the total and half of it', the answer is always (2^n)-1, where 'n' is the number of times the process is repeated. Here, the process is repeated 6 times. So answer is (2^6)-1=63.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38073 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The percentage profit earned by selling an article for Rs. 1520 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit? | [
"3000",
"1230",
"1750",
"5600"
] | C | C.P. be Rs. x.
Then, (1520 - x)/x * 100 = (x - 1280)/x * 100
1520 - x = x - 1280
2x = 2800 => x = 1400
Required S.P. = 125 % of Rs. 1400 = 125/100 * 1400 = Rs. 1750.
ANSWER C |
AQUA-RAT | AQUA-RAT-38074 | (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.
Jeremy bought 2Q steaks for P dollars. Jerome buys R steaks for a 50% discount, how much will the steaks cost him in cents? | [
"50RP/Q.",
"50QR/P.",
"25RQ/P.",
"25RP/Q."
] | D | Jeremy bought 2Q steaks for P dollars, so 1 steak = P/2Q
Jerome buys R steaks for a 50% discount: r* (P/4Q)
in cents the answer will be : r* (100 P/4Q)= 25 RP/ Q=D |
AQUA-RAT | AQUA-RAT-38075 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are two buildings P and Q. If 15 persons are sent from P to Q, then the number of persons in each building is the same. If 20 persons are sent from Q to P, then the number of persons in P is double the number of persons in Q. How many persons are there in building P? | [
"80",
"140",
"120",
"100"
] | C | Let the number of persons in building P = p
and the number of persons in building Q = q
If 15 persons are sent from P to Q,
then the number of persons in each building is the same
=> p-15 = q+15
=> p - q = 30 ----(Equation 1)
If 20 persons are sent from Q to P,
then the number of persons in P is double the number of persons in Q
=> 2(q - 20) = (p + 20)
=> 2q - 40 = p + 20
=> 2q - p = 60 ----(Equation 2)
(Equation 1) + (Equation 2)=> q = 90
From Equation 1, p = 30 + q = 30 + 90 = 120
i.e., Building P has 120 persons
Answer : Option C |
AQUA-RAT | AQUA-RAT-38076 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 12 cans in the refrigerator. 7 of them are red and 5 of them are blue. In how many ways we can remove 9 cans so that at least one red and at least one blue cans to remain the refrigerator. | [
"460",
"175",
"493",
"455"
] | B | Ways to pick 3 cans so that at least one red and at least one blue cans to remain the refrigerator =
total ways to pick 3 can out of 12 - ways to pick 3 red out of 7 red - ways to pick 3 blue out of 5 blue
12C3−7C3−5C3=220−35−10=175
ANSWER:B |
AQUA-RAT | AQUA-RAT-38077 | 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.
In Country S, the unemployment rate among construction workers dropped from 16 percent on September 1, 1992, to 9 percent on September 1, 1996. If the number of construction workers was 20 percent greater on September 1, 1996, than on September 1, 1992, what was the approximate percent change in the number of unemployed construction workers over this period? | [
" 50% decrease",
" 30% decrease",
" 15% decrease",
" 30% increase"
] | B | In Country S 1992 1996
No of construction workers 100 120
Unemployment Rate 16% 9%
Unemployed workers 16 11
% change in unemployed workers=(16-11)=5/16=~33% decrease
Closest ans=30% decrease
Ans=B |
AQUA-RAT | AQUA-RAT-38078 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 4 hours and 36 hours respectively. The ratio of their speeds is? | [
"4:9",
"4:3",
"4:5",
"3:1"
] | D | Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √36 : √4
= 6:2 = 3:1
Answer:D |
AQUA-RAT | AQUA-RAT-38079 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A sells his goods 50% cheaper than B but 50% dearer than C. The cheapest is? | [
"A",
"B",
"C",
"all alike"
] | C | Let B = 100
A = 50
C * (150/100) = 50
3C = 100
C = 33.3 then 'C' Cheapest
ANSWER:C |
AQUA-RAT | AQUA-RAT-38080 | # If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
My work:
We know the sample space $S:$"The set of number of 1 to 1000000" and $|S|=1000000$
Let $E$ the event such that $E:$"The set of number contains the digit 5 in $[1000000]$" We need calculate $|E|$.
I know in $[100]$ we have $5,15,25,35,45,50,51,52,53,54,55,56,57,58,59,65,75,85,95$ then we have 19 numbers contains the digit $5$ in the set $[100]$
Then in $[1000]-[500]$ we have 171 numbers have the digit 5. this implies [1000] have 271 number contains the digit 5.
. . .
Following the previous reasoning we have to $[10000]$ have 3439 number contains the digit 5.
Then, $[100000]$have 40951 number contains the digit 5.
Moreover, $[1000000]$ have 468559 number contain the digit 5.
In consequence the probability of we pick a digit contain the number 5 in the set $[1000000]$ is 0.468
Is correct this?
How else could obtain $|E|$?
Thanks
The following is multiple choice question (with options) to answer.
If three numbers are randomly selected from a set, what is the probability that sum will be greater than 18 if he given set is
{5,6,7,8,9,10} | [
"7/18",
"5/13",
"5/16",
"5/19"
] | D | First problem is 6C3 = 18 .. Not 20..
Point 2, your sum should be greater than 18, so these are the following possibilities you have
({5,7,8},{5,9,10},{6,7,8},{6,9,10},{8,9,10}) which leads to 5 combinations
So the probability is 5 on 18 or 5/18..
ANSWER:D |
AQUA-RAT | AQUA-RAT-38081 | Difference between revisions of "2014 AMC 10A Problems/Problem 17"
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$
Solution 1 (Clean Counting)
First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.
--happiface
Solution 2 (Bashy Casework)
Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.
The following is multiple choice question (with options) to answer.
. Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is | [
"5/0",
"5/9",
"5/1",
"5/3"
] | B | Explanation:
No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.
Answer:B |
AQUA-RAT | AQUA-RAT-38082 | time, sun
d represents the day of the year for 2000. For example d=7
would be January 7th, 2000.
If you're willing to lose some precision in exchange for
convenience, you can compute the day of the current year (instead
of counting all the way back to 2000), since the sun's declination
repeats yearly (roughly speaking).
To calculate the day of the year, remember that December 31st
(noon) of the previous year is day 0. This means day 1 is January
1st, and day 31 is January 31st. Day 31 is also "February 0", so if
you need a date in February, just add. February 19th, for example,
would be 31+19 or 50. Since February has 29 days this year, February
29th would be day 31+29 or day 60, which is also "March 0". However,
at our level of precision, it doesn't matter whether you could the
leap day or not: the results will be approximately the same.
The formula above is accurate to about 0.1 degrees for this
century. Since the sun's declination can change by as much as 0.4
degrees in a day, this amount of precision should suffice.
Technically, the formula above computes the sun's declination at
Greenwich noon for a given day, which is the time when most of the
world is observing the same day. Again, the inaccuracies from using
Greenwich noon (instead of the actual, as yet unknown, time) are
small enough to ignore for our purposes.
The following is multiple choice question (with options) to answer.
On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004? | [
"Sunday",
"Friday",
"Saturday",
"Monday"
] | A | Explanation :
Given that 8th Feb, 2005 was Tuesday
Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366 (Since Feb 2004 has 29 days as it is a leap year)
366 days = 2 odd days
Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunday. Answer : Option A |
AQUA-RAT | AQUA-RAT-38083 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
A and B start a business with Rs.6000 and Rs.8000 respectively. Hoe should they share their profits at the end of one year? | [
"3:3",
"3:4",
"3:8",
"3:1"
] | B | They should share the profits in the ratio of their investments.
The ratio of the investments made by A and B =
6000 : 8000 => 3:4.Answer: B |
AQUA-RAT | AQUA-RAT-38084 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
A customer purchased a package of ground beef at a cost of $1.76 per pound. For the same amount of money, the customer could have purchased a piece of steak that weighed 36 percent less than the package of ground beef. What was the cost per pound of the steak? | [
"$2.00",
"$2.25",
"$2.50",
"$2.75"
] | D | For simplicity, let's assume the customer bought 1 pound of ground beef for $1.76.
Let x be the price per pound for the steak.
Then 0.64x = 176
x = 176/0.64 = $2.75
The answer is D. |
AQUA-RAT | AQUA-RAT-38085 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Natasha climbs up a hill, and descends along the same way she went up. It takes her 4 hours to reach the top and 2 hours to come back down. If her average speed along the whole journey is 1.5 kilometers per hour, what was her average speed (in kilometers per hour) while climbing to the top? | [
"1.065",
"1.125",
"1.225",
"1.375"
] | B | Let the distance to the top be x, so the total distance traveled by Natasha is 2x.
The total time is 4 + 2 = 6 hours
The average speed = total distance/total time taken = 2x/6 = x/3
The average speed of the complete journey is 1.5 km/hour
x/3 = 1.5
x = 4.5 km
The average speed while climbing = distance/time = 4.5/4 = 1.125 km/h
The answer is B. |
AQUA-RAT | AQUA-RAT-38086 | # Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?
A bag has:
$3$ red
$5$ black
$8$ green
marbles.
Total of 16 marbles.
You select a marble, and then another one right after. (without replacement).
What is the probability that $both$ are red?
Probability that first pick is red: $\frac{3}{16}$
Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)
Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$
How do I do this using combinations only?
• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59
Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$
Hope this helps you.
The following is multiple choice question (with options) to answer.
Griffin has a bag of marbles that contains 7 black marbles and 4 red marbles. If he removes 3 marbles at random without replacing any of the marbles, what is the probability that all 3 marbles selected will be red? | [
"2/55",
"3/85",
"4/105",
"4/165"
] | D | P(3 red marbles) = 4/11*3/10*2/9 = 4/165
The answer is D. |
AQUA-RAT | AQUA-RAT-38087 | • #### 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.
The length of the rectangular field is double its width. Inside the field there is square shaped pond 8m long. If the area of the pond is 1/8 of the area of the field. What is the length of the field? | [
"26",
"32",
"87",
"20"
] | B | A/8 = 8 * 8 => A = 8 * 8 * 8
x * 2x = 8 * 8 * 8
x = 16 => 2x = 32
Answer:B |
AQUA-RAT | AQUA-RAT-38088 | 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 280 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds.What is the lenght of other train. | [
"210m",
"220m",
"230m",
"240m"
] | B | Relative speeds=(120+80)km/hr
=200km/hr=(200*5/18)m/s=(500/9)m/s
let length of train be xm
x+280/9=500/9
x=220
Ans is 220m
ANSWER:B |
AQUA-RAT | AQUA-RAT-38089 | to solve for the other. Logic To Calculate Percentage Difference Between 2 Numbers. Sale Discount Calculator - Percent Off Mortgage Loan Calculator - Finance Fraction Calculator - Simplify Reduce Engine Motor Horsepower Calculator Earned Value Project Management Present Worth Calculator - Finance Constant Acceleration Motion Physics Statistics Equations Formulas Weight Loss Diet Calculator Body Mass Index BMI Calculator Light. The commonly used way is to go from right to left, which gives us a positive number. Posted: (1 week ago) Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Just subtract the past value from the current value. Click the calculate button 3. Calculate percentage difference between two columns I have a input text file in this format: ITEM1 10. The percentage difference between the two values is calculated by dividing the value of the difference between the two numbers by the average of the two numbers. First, you need Excel to subtract the first number from the second number to find the difference between them. Listing Results about Calculate Variance Between Two Numbers Real Estate. To write an increase or decrease as a percentage, use the formula actual increase or decrease original cost × 100%. The growth rate can be listed for real or nominal GDP. This difference needs to be divided between the first number (the one that doesn't change). We then append the percent sign. Follow 52 views (last 30 days) Show older comments. Enter an old number in cell A1 and a new number in cell B1. The percent difference formula or the percent difference equation of two numbers a and b is: ((a - b) / (a+b)/2) × 100, where a > b Calculate the percentage difference between the numbers 35 and 65. The first step to the equation is simple enough. Step 1: Calculate the difference (subtract one value from the other) ignore any negative sign. With a Difference From, Percent Difference From, or Percent From calculation, there are always two values to consider: the current value, and the value from which the difference should be calculated. A percentage variance, aka percent change, describes a proportional change between two numbers, an original value and a new value. It has more of an impact when you say, "There was a 50 percent increase in attendance at the concert compared to last year," versus when you say, "There were 100 more people at the concert this year than
The following is multiple choice question (with options) to answer.
There are two numbers. If 20% of the first number is added to the second number, then the second number increases to its five-fourth. Find the ratio of the first number to the second number? | [
"A)5/4",
"B)5/8",
"C)6",
"D)6/7"
] | A | Let the two numbers be x and y.
20/100 * x + y = 5/4y
=> 1/5 x = 1/4 y => x/y = 5/4
A) |
AQUA-RAT | AQUA-RAT-38090 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
In a kilometer race, A beats B by 25 meters or 10 seconds. What time does A take to complete the race? | [
"290 sec",
"390 sec",
"199 sec",
"192 sec"
] | B | Time taken by B run 1000 meters
= (1000 * 10)/25 = 400 sec.
Time taken by A = 400 - 10 = 390 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-38091 | This is an inequality. You should not multiply across by $(x-9)$ at all. Instead you should bring the $2$ over to the left and combine the fraction through a common denominator. Verify you arrive at: $\frac{-x+27}{x-9}≤0$ Now one should make separate numberlines for numerator and denominator and verify the signs on the numberlines. NUM:$$___________plus_______27__minus______$$ DENOM: $$___minus____9____plus_____________$$ When you "divide" the numberlines you look for the negative interval, which is $x<9$ or $x\geq27$
• You should use \geq instead of => – Nikunj Mar 13 '16 at 4:02
• @Nikunj Ha! For the less or equal sign I copy-pasted the Op's inequality symbol. Now I needed the greater or equal symbol and I didn't know it...(but now I do) – imranfat Mar 13 '16 at 4:07
I suppose it depends on what $x$ is an element of. If we assume that $x\in\mathbb{R}$, we have to break this inequality into separate cases.
There is the case where $x>9$, in which case the sign does not change, and there is the case of $x<9$, in which the sign does change.
Of course, this is undefined for $x=9$.
The following is multiple choice question (with options) to answer.
Which of the following CANNOT be a value of 9/(x–9)? | [
"-1",
"-1/2",
"0",
"2"
] | C | The question doesn't ask for the value of x here, but for the possible results of the equation.
If 9/(x-9) = 0, then the numerator must be 0. But since the numerator is 9, the fraction can not be equal to 0.
The answer is C. |
AQUA-RAT | AQUA-RAT-38092 | organic-chemistry, stoichiometry
Let's consider mixture Z, we will reduce its composition to the following "compounds": $\ce{C_2H_3ON, CH_2, H_2O}$
Let $n_{\ce{C2H3ON}}=2x, n_{\ce{CH2}}=2y, n_{\ce{H_2O}}=2z$
Half of mixture Z will have: $n_{\ce{C2H3ON}}=x, n_{\ce{CH2}}=y, n_{\ce{H_2O}}=z$
Then, the escaped nitrogen will be:
$$n_{\ce{N_2}}=n_{\ce{N_2}(air)}+n_{\ce{N_2}(rxn)}=n_{\ce{N_2}(air)}+0.5x=6.2325 \text{ mol}$$
Now, let's consider the combustion reaction, by looking at the elements, we can easily formulate the following system of equations:
$$
\left\{
\begin{array}{ll}
2x+y&=1.195 \text{ (C atoms)}\\
3x+2y+2z&=2\cdot1.2025 \text{ (H atoms)}\\
x+z+0.5(6.2325-0.5x)&=1.195\cdot2+1.2025 \text{ (O atoms)}
\end{array}
\right.
$$
Solving this simple system gives us $x=0.375, y= 0.445, z=0.195$
Then, the composition of mixture Z will be:
$$
\left\{
\begin{array}{}
\ce{C2H3ON}: 0.75 \text{ mol} \\
\ce{CH2}: 0.89 \text{ mol} \\
\ce{H2O}: 0.39 \text{ mol}
\end{array}
\right.
$$
The following is multiple choice question (with options) to answer.
An empty fuel tank is filled with brand Z gasoline. When the tank is half empty, it is filled with brand Y gasoline. When the tank is half empty again, it is filled with brand Z gasoline. When the tank is half empty again, it is filled with brand Y gasoline. At this time, what percent H of the gasoline in the tank is brand Z? | [
"50%",
"40%",
"37.5%",
"331⁄3%"
] | C | Work with fraction of brand Z in the tank.
1st step: brand Z is 1
2nd step: brand Z is 1/2
3rd step: brand Z is (1/2)*(1/2) + 1/2 = 3/4
4th step: brand Z is (1/2)*(3/4) H= 3/8 = 37.5%
Answer (C) |
AQUA-RAT | AQUA-RAT-38093 | -
$f(n)=1^3+2^3+3^3+......+n^3$
$f(n-1)=1^3+2^3+3^3+......+(n-1)^3$
$f(n)-f(n-1)=n^3$
if $f(n)= (1+2+3+4+....+n)^2$ then
$$f(n)-f(n-1)=(1+2+3+4+....+n)^2-(1+2+3+4+....+(n-1))^2$$
using $a^2-b^2=(a+b)(a-b)$
$f(n)-f(n-1)=$
$=[1+1+2+2+3+3+4+4+....+(n-1)+(n-1)+n][1-1+2-2+3-3+4-4+....+(n-1)-(n-1)+n]=$
$=[2(1+2+3+4+....+(n-1))+n]n=(2\frac{n(n-1)}{2}+n)n=(n(n-1)+n)n=n^3$
-
This is about the same proof as here, the presentation is a bit different though. This is another way to make $k^3$ appear than what was shown here, here and here.
The following is multiple choice question (with options) to answer.
If f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)? | [
"222",
"2787",
"2013",
"2778"
] | C | Put n = 0
Then f(f(0))+f(0) = 2(0) + 3 ⇒⇒ f(1) + 1 = 3 ⇒⇒ f(1) = 2
Put n = 1
f(f(1)) + f(1) = 2(1) + 3 ⇒⇒ f(2) + 2 = 5 ⇒⇒f(2) = 3
Put n = 2
f(f(2)) + f(2) = 2(2) + 3 ⇒⇒ f(3) + 3 = 7 ⇒⇒ f(3) = 4
......
f(2012) = 2013
Answer:C |
AQUA-RAT | AQUA-RAT-38094 | => 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.
P and Q started a business with respective investments of Rs. 4 lakhs and Rs. 10 lakhs. As P runs the business, his salary is Rs. 5000 per month. If they earned a profit of Rs. 2 lakhs at the end of the year, then find the ratio of their earnings? | [
"1:9",
"1:8",
"1:4",
"1:1"
] | D | Ratio of investments of P and Q is 2 : 5
Total salary claimed by P = 12 * 5000 = Rs. 60000
Total profit = Rs. 2 lakhs.
Profit is to be shared = Rs. 140000
Share of P = (2/7) * 140000 = Rs. 400000
Share of Q = Rs. 100000
Total earnings of P = (60000 + 40000) = Rs. 100000
Ratio of their earnings =1:1
Answer:D |
AQUA-RAT | AQUA-RAT-38095 | Set to use a.... 100, 110, 120, 130, 140, 150 calculator to find degree... May be in degrees top input of quadrilaterals with some of their sides and angles of up to a radius. Want to give credit to the opposite ( i.e, usually involve triangles because triangles are rigid full rotation click... The 90° angle 30,45,75,105,135 share with your friends: 1: construct an angle of at. And the other step is given below learn how to construct a 60° 60 ° angle regular semicircular protractor measure... Measuring this angle right over here from the Chrome web Store, with radius more than 180° using protractor. To any convenient length are three other types such as straight, and! Of any convenient radius on the radian measure of acute angle of size 1° has radian measure reflex... You temporary access to the opposite ( i.e try and learn this properly solve with the ratio of sides angles... 2Π, so it follows that an equilateral triangle, AB - AC = cm. 65 degrees and radians left end as point O and the other is... Op intersects the arc 8 credit to the radius similar to the radius to... The basics of this is shown in constructing the sum of ABC triangles... An acute angle = 360° – a measure of an angle bisector construct... Teach you to click on the paper for new subjects a Set of congruent triangular braces as. Together they form a right angle is half of a triangle measure 30 degrees a... + 65° = 180° mark a point ‘ O ’ on it get the 60. Angles add up to 180^\circ 180^\circ 180^\circ Academy... And its done in the future is to use angles measured with either degrees or radians = 110° the... Step is given below the steps above to construct some angles without using a protractor: 1 3 O., there are 360 degrees then measure the angle into two equal parts 60°... 15˚ angle can be shown that the angle of 30° or outer Scale the! Of different lengths and angles of triangle ABC = 60° + 55° + 65° =.... Degrees are complementary angles because together they form a
The following is multiple choice question (with options) to answer.
A circle is circumscribed around a quadrilateral, and the quadrilateral is ABCD, What is the value of v when angle D is 140 degrees? | [
"20",
"40",
"60",
"80"
] | B | An inscribed angle is an angle in a circle composed of two chords with a common endpoint, that is, an angle with its vertex on the circle. In this problem, all four lettered angles are inscribed angles.
An inscribed angle necessarily opens to intersect an arc (the technical word is to subtend an arc). The Inscribed Angle Theorem said that the measure of an inscribed angle is always half the measure of the arc it intersects, or subtends.
angle D = 140 degrees
Thus,
arc ABC = 280 degrees
A whole circle has 360 degrees of arc, so arc ADC should be the rest of the circle.
arc ADC = 360 - 280 = 80 degrees
The angle that intersects this arc, angle B, should have half this measure.
angle B = v = 80/2 = 40 degrees
Answer =(B) |
AQUA-RAT | AQUA-RAT-38096 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
What is the next number: 2, 10, 82, __ | [
"630",
"730",
"830",
"848"
] | B | 3^0 + 1 = 2
3^2 + 1 = 10
3^4 + 1 = 82
3^6 + 1 = 730
The answer is B. |
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