source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38397 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
The sum of four consecutive even numbers is 140. What would be the largest number? | [
"33",
"38",
"70",
"123"
] | B | Let the four consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1)
Their sum = 8x - 4 = 140 => x = 18
Smallest number is: 2(x +1) = 38.
Answer: B |
AQUA-RAT | AQUA-RAT-38398 | The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.
The following histogram displays the heights on the x -axis and relative frequency on the y -axis.
## Try it
The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5
11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5
12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14
Smallest value: 9
Largest value: 14
Convenient starting value: 9 – 0.05 = 8.95
Convenient ending value: 14 + 0.05 = 14.05
$\frac{14.05-8.95}{6}=0.85$
The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one.
The following is multiple choice question (with options) to answer.
Two people measure each other's height, the height of the taller person is H and the height of the other person is L. If the difference in their heights is equal to the average height, what is the Value of H/L | [
"1/3.",
"3.",
"2.",
"1/2."
] | B | Difference = Average
H - L = (H + L)/2,
Solving for H/L gives 3. A quick check, H be 18 and L be 6, 18 - 6 = (18 + 6)/2
B |
AQUA-RAT | AQUA-RAT-38399 | Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?
The Attempt at a Solution
Let ##B## denote the event that at least 2 kings are drawn, and ##A## the event that at least 1 king is drawn. Because ##B## is a strict subset of ##A##,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute ##P(A)##, ##P(A^c )## denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute ##P(B)##, ##P(B^c)## denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king ##P(1)## and the probability of not drawing a single king ##P(A^c)##.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$
The following is multiple choice question (with options) to answer.
From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is ? | [
"1/26",
"1/07",
"1/22",
"1/29"
] | A | Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13 = 1/26
Answer:A |
AQUA-RAT | AQUA-RAT-38400 | # 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 purchase price of an article is $48. In order to include 25% of cost for overhead and to provide $12 of net profit, the markup should be | [
"50%",
"25%",
"35%",
"40%"
] | A | Cost price of article = 48$
% of overhead cost = 25
Net profit = 12 $
We need to calculate % markup
Net profit as % of cost price = (12/48)*100 = 25%
Total markup should be = 25 + 25 = 50%
Answer A |
AQUA-RAT | AQUA-RAT-38401 | 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 men are going along a track rail in the opposite direction. One goods train crossed the first person in 20 sec. After 10 min the train crossed the other person who is coming in opposite direction in 18 sec .After the train has passed, when the two persons will meet? | [
"70 minutes",
"80 minutes",
"85 minutes",
"90 minutes"
] | D | Let us consider that speed of train , first man and second man are respectively t, f and s.
According to first given condition goods train crossed the first person moving in same direction in 20 sec.
so length of the will be 20(t-f)
similarly train crossed the second man in 18 sec.
so length of the train will be 18(t+s)
on comparing these two equations, we get 20(t-f) = 18(t+s)
or 2t = 20f + 18s
or t = 10f + 9s
now it is given that After 10 min the train crossed the other person who is coming in opposite direction so
if we consider this way of train as L then the next equation will be
L = 600t (here 600 is used for 10 minutes)
finally as asked in the question the time required to meet the two man after the train has passed will be given by
time = (L-600 f)/(f+s) {here 600f is used for the distance traveled by first man in 10 minutes}
= (600t-600f)/(f+s)
= [600(10f+9s)-600f]/(f+s)
= 600(10f+9s-f)/(f+s) = 600*9(f+s)/(f+s) = 600*9 seconds = 600*9/60 min
= 90 minutes
ANSWER:D |
AQUA-RAT | AQUA-RAT-38402 | Note: The author(s) of your text are calculating the probability that the top-ranked woman is in the $k$th position. For that to occur, we must select $k - 1$ of the five men to be ranked ahead of her and one of the five women to fill the $k$th position while choosing $k$ of the $10$ people. Hence, the answer in the text is equivalent to \begin{align*} P(X = 1) & = \frac{\binom{5}{0}\binom{5}{1}}{\binom{10}{1}} = \frac{1}{2}\\ P(X = 2) & = \frac{\binom{5}{1}\binom{5}{1}}{\binom{10}{2}} = \frac{5}{18}\\ P(X = 3) & = \frac{\binom{5}{2}\binom{5}{1}}{\binom{10}{3}} = \frac{5}{36}\\ P(X = 4) & = \frac{\binom{5}{3}\binom{5}{1}}{\binom{10}{4}} = \frac{5}{84}\\ P(X = 5) & = \frac{\binom{5}{4}\binom{5}{1}}{\binom{10}{5}} = \frac{5}{252}\\ P(X = 6) & = \frac{\binom{5}{5}\binom{5}{1}}{\binom{10}{6}} = \frac{1}{252} \end{align*}
The following is multiple choice question (with options) to answer.
If Ramola ranks 14th in a class of 26, what is her rank from the last? | [
"11th",
"13th",
"12th",
"14th"
] | B | when considered from last, 12 (i.e 26-14) students are ranked before her. so she is 13th rank from last.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38403 | # Probability of a certain ball drawn from one box given that other balls were drawn
Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
The following is multiple choice question (with options) to answer.
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green? | [
"2/3",
"8/21",
"3/7",
"9/22"
] | B | Explanation:
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
Answer: Option B |
AQUA-RAT | AQUA-RAT-38404 | # Probability Problem with $n$ keys
A woman has $n$ keys, one of which will open a door. a)If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her $k^{\mathrm{th}}$ try?
Attempt: On her first try, she will have the correct key with probability $\frac1n$. If this does not work, she will throw it away and on her second attempt, she will have the correct key with probability $\frac1{(n-1)}$. So on her $k^{\mathrm{th}}$ try, the probability is $\frac1{(n-(k-1))}$ This does not agree with my solutions.
b)The same as above but this time she does not discard the keys if they do not work.
Attempt: We want the probability on her $k^{\mathrm{th}}$ try. So we want to consider the probability that she must fail on her $k-1$ attempts. Since she keeps all her keys, the correct one is chosen with probability $\frac1n$ for each trial. So the desired probability is $(1-\frac{1}{n})^{k-1} (\frac1n)^k$. Again, does not agree with solutions.
I can't really see any mistake in my logic. Can anyone offer any advice? Many thanks
-
Ah, finally understand what that means now! Yeah, I am sort of new to this site. – CAF Nov 28 '12 at 22:37
Presumably the $(1/n)^k$ part at the end is a typo, you mean $1/n$. – André Nicolas Nov 28 '12 at 23:21
For $(a)$, probability that she will open on the first try is $\dfrac1n$. You have this right.
The following is multiple choice question (with options) to answer.
In a contest, a bowl contains 18 keys, only one of which will open a treasure chest. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize? | [
"1/18",
"1/9",
"5/36",
"1/4"
] | B | P(not winning) = 17/18*16/17 = 8/9
P(winning) = 1 - 8/9 = 1/9
The answer is B. |
AQUA-RAT | AQUA-RAT-38405 | 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 600 meter long train crosses a signal post in 40 seconds. How long will it take to cross a 7.2 kilometer long bridge, at the same speed? | [
"4 min",
"2 min",
"8 min",
"9 min"
] | C | S = 600/40 = 15 mps
S =7200/15 =480 sec =8 min.Answer:C |
AQUA-RAT | AQUA-RAT-38406 | # 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 wholesaler wishes to sell 100 pounds of mixed nuts at $2.50 a pound. She mixes peanuts worth $1.50 a pound with cashews worth $5.00 a pound. How many pounds of cashews must she use? | [
"40",
"45",
"50",
"55"
] | B | From the question stem we know that we need a mixture of 100 pounds of peanuts and cashews. If we represent peanuts as x and cashews as y, we get x + y = 100. Since the wholesaler wants to sell the mixture of 100 pounds @ $2.50, we can write this as: $2.5 * (x + y) = $1.5x + $4y
From the equation x + y = 100, we can rewrite y as y = 100 - x and substitute this into our equation to get:
$2.5 * (x + 100 - x) = $1.5x + $4(100 - x)
If you solve for x, you will get x = 60, and therefore y = 40. So the wholesaler must use 40 pounds of cashews.
You can substitute into the original equation to see that: $250 = $1.5(60) + $4(40)
Answer is B |
AQUA-RAT | AQUA-RAT-38407 | 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.
The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 50 kms. what was the total distance traveled in 12 hours? | [
"252 kms",
"152 kms",
"732 kms",
"752 kms"
] | C | Explanation:
Total distance travelled in 12 hours =(50+52+54+.....upto 12 terms)
This is an A.P with first term, a=50, number of terms,
n= 12,d=2.
Required distance = 12/2[2 x 50+{12-1) x 2]
=6(122)
= 732 kms.
Answer: C |
AQUA-RAT | AQUA-RAT-38408 | => 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.
workers decided to raise Rs.3 lacs by equal contribution from each. Had they contributed Rs.50 eachextra, the contribution would have been Rs.3.15 lacs . How many workers were they? | [
"300",
"230",
"500",
"560"
] | A | N * 50 = (315000 - 300000) = 15000
N = 15000 / 50
= 300
A |
AQUA-RAT | AQUA-RAT-38409 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Joe’s average (arithmetic mean) test score across 4 equally weighted tests was 60. He was allowed to drop his lowest score. After doing so, his average test score improved to 65. What is the lowest test score that was dropped? | [
"20",
"25",
"45",
"65"
] | C | The arithmetic mean of 4 equally weighted tests was 60. So what we can assume is that we have 4 test scores, each 60.
He dropped his lowest score and the avg went to 65. This means that the lowest score was not 60 and other three scores had given the lowest score 5 each to make it up to 60 too. When the lowest score was removed, the other 3 scores got their 5 back. So the lowest score was 3 *5 = 15 less than 60.
So the lowest score = 60 - 15 = 45
Answer (C) |
AQUA-RAT | AQUA-RAT-38410 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
Dhoni bought a 1-year, $40,000 certificate of deposit that paid interest at an annual rate of 6 percent compounded semiannually. What was the total amount of interest paid on this certificate at maturity? | [
"$2456",
"$2436",
"$2446",
"$2466"
] | B | 6 percent compounded semiannually --> 3% in 6 moths.
For the first 6 moths interest was 3% of $40,000, so $1200
For the next 6 moths interest was 3% of $41,200, so $1236
Total interest for 1 year was $1200+$1236=$2436
Answer : B |
AQUA-RAT | AQUA-RAT-38411 | 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 finish the job in 15 days.while A,B and C can finish it in 12 days . C alone will finish the job in | [
"40 days",
"30 days",
"60 days",
"70 days"
] | C | Explanation:
12 = (15 * x)/(15 + x)
180+12x=15x
3x=180
X=60
Answer: Option C |
AQUA-RAT | AQUA-RAT-38412 | 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.
10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work? | [
"6 days",
"7 days",
"8 days",
"9 days"
] | B | Explanation:
1 woman's 1 day's work = 1/70
1 Child's 1 day's work = 1/140
5 Women and 10 children 1 day work =
(5/70+10/140)=1/7
So 5 women and 10 children will finish the work in 7 days
Option B |
AQUA-RAT | AQUA-RAT-38413 | Your mistake is in 4th step. When you are taking $(-1+i)$ common from $R_2$. Then element at 2nd row and 3rd column is wrong I think.
• Sorry, I meant to multiple by $\frac{1}{-1+i}$. Already edited. – AndreGSalazar Apr 18 '17 at 16:58
• Then it becomes $\frac{i}{1-i}$. – Kanwaljit Singh Apr 18 '17 at 17:25
The following is multiple choice question (with options) to answer.
Adding two non multiples of R always results in a multiple of R; which of the following must be the value of R ? | [
"[A] two",
"[B] three",
"[C] five",
"[D] ten"
] | A | NON- multiples of 2 would always be ODD and sum of two odd would be EVEN.
so irrespective of value of non-multiples, sum of two them would always be even and thus multiple of 2..
A |
AQUA-RAT | AQUA-RAT-38414 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
Ashish and Rahul started walking from same point E. Ashish walked at the speed of 15 kmph to the east, and Rahul walked 25 kmph to south, at what time they meet 10 miles apart? | [
"15 min",
"16 min",
"17 min",
"18 min"
] | D | Consider 1 mile = 1.6 km
so 10 miles = 16 km
both are walking in at an angle 90 from same point so pythagorous theorem can be apply hence
(15x)^2 + (25x)^2 = 16^2
by solving and x hours are converted into min we got
approximate 18 min
ANSWER:D |
AQUA-RAT | AQUA-RAT-38415 | $\displaystyle \frac{3}{4} \times 0.04 = 0.03$
4. Outch ! Sorry, I'll think again about it !
5. Ok, so $\displaystyle P(A/R)=\frac{P(A \cap R)}{P\color{red}(R)}$
Plus $\displaystyle P(R/A)=\frac{P(A \cap R)}{P(A)} \Longleftrightarrow P(A \cap R)=P(R/A)P(A)$
Hence : $\displaystyle {\color{blue} P(A/R)=\frac{P(R/A)P(A)}{P(R)}}$
I misread and $\displaystyle {\color{blue} P(R/A)=.04}$, not $\displaystyle P(R \cap A)$
Let's calculate P(R).
The following is multiple choice question (with options) to answer.
Find the fourth proportional to 0.2,0.12 and 0.3? | [
"0.18",
"0.21",
"0.22",
"0.34"
] | A | Formula = Fourth propotional = (b × c)/a
A = 0.2 , B = 0.12 and C = 0.3
(0.12 × 0.3)/0.2
0.036/0.2 = 0.18
A |
AQUA-RAT | AQUA-RAT-38416 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
_________________
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Bottle R contains 100 capsules and costs $6.30. Bottle T contains 130 capsules and costs $2.99. What is the difference between the cost per capsule for bottle R and the cost per capsule for bottle T? | [
" $0.04",
" $0.12",
" $0.05",
" $0.03"
] | A | Cost per capsule in R is 6.30/100=0.063
Cost per capsule in T is 2.99/130=0.023
The difference is 0.004
The answer is A |
AQUA-RAT | AQUA-RAT-38417 | regular-languages, finite-automata, nondeterminism
Title: Minimal number of states for an NFA of all different words Given $\Sigma =\{0,1,@\}$, I am looking at a language $L=\{u@v | u,v\in \{0,1\}^k\wedge u\neq v\}$. So $u,v$ have only $0,1$s, same length $k$, yet are different.
Also, for me $k$ is a known constant. So the language ìs regular (at most $(2^k)\cdot (2^k-1)$ words).
I want to build the most efficient NFA for $L$. So far, I have been able to cone up with an NFA of $O(k^2)$ states, but I want to find one with $O(k)$ states. Of course one could argue that as $k$ is constant, its the same as $O(1)$, but assuming $k$ is a constant parameter - that is, given a value of $k$, I construct $L$ and an NFA for $L$ with $O(k)$ states.
My idea to use $k^2$ states:
A path for each index, as $u$ and $v$ of size $k$... they differ means there is at least one $1\leq i \leq k$ where they differ at that character.
Read '0' in $i$th character and '1' in $i+k+1$th character (due to '@'). Or other way.
In between make sure that $u$s length and $v$s length is $k$. Which characters you read in between? I don't care.
The following is multiple choice question (with options) to answer.
Using all the letters of the word "NOKA", how many words can be formed, which begin with N and end with A? | [
"8",
"2",
"9",
"3"
] | B | There are five letters in the given word.
Consider 4 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 2! ways.
The number of words = 2! = 2.
Answer:B |
AQUA-RAT | AQUA-RAT-38418 | 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.
The average age of 20 men in the class is 15.6 years. Five new men join and the new average becomes 16.56 years. What was the average age of five new men? | [
"15.5",
"15.4",
"15.25",
"20.4"
] | D | Total age of 20 men = 15.6 x 20 = 312
Now, total age of 25 men = 141.
Total age of five men added later = 414 - 312 = 202.
Hence, the total average of five men = 102/5 = 20.4
ANSWER:D |
AQUA-RAT | AQUA-RAT-38419 | If we take $F_1=F_2=1$, then we certainly have that $5|F_5=5$. Let's look at the Fibonacci numbers, reduced modulo $5$:
$F_1\equiv 1\\ F_2\equiv 1\\ F_3\equiv 2\\ F_4\equiv 3\\ F_5\equiv 0\\ F_6\equiv 3\\ F_7\equiv 3\\ F_8\equiv 1\\ F_9\equiv 4\equiv -1\\ F_{10}\equiv 0\\ F_{11}\equiv -1\\ F_{12}\equiv -1\\ F_{13}\equiv -2\\ F_{14}\equiv -3\\ F_{15}\equiv 0\\ F_{16}\equiv -3\\ F_{17}\equiv -3\\ F_{18}\equiv -1\\ F_{19}\equiv -4\equiv 1\\ F_{20}\equiv 0\\$
After this, the pattern begins to repeat, and it appears that $F_{n+20}\equiv -F_{n+10}\equiv F_n$ for all $n$. This, together with $F_5\equiv F_{10}\equiv 0\pmod 5$ is sufficient to establish your claim.
You don't have to define the Fibbonacci sequence recursively
You could say:
$F_n = \frac{\phi^n + \phi'^n}{\phi - \phi'}$
where $\phi = \frac {1 + \sqrt 5}{2}$ and $\phi' = \frac {1 - \sqrt 5}{2}$
The following is multiple choice question (with options) to answer.
If f(1, 1) = 5 and f(1, 5) = 26, what is the value of f(1, 10)? | [
"107",
"101",
"147",
"None of these"
] | B | Solution:
The function f(a, b) = a3 + b2
f(1, 2) therefore is = 1^3 + 2^2 = 1 + 4 = 5 and
f(1, 5) = 1^3 + 5^2 = 1 + 25 = 26
Therefore, f(1, 10) = 1^3 + 10^2 = 1 + 100 = 101
Answer B |
AQUA-RAT | AQUA-RAT-38420 | visualization
equating "relationship wealth" with "proportion of people of the opposite gender attracted"
the more a person is attractive the more they go for the top attractive people of the opposite gender,
Then for each gender the average number of likes received by level of attractiveness is calculated, for example:
In average a woman who has say 50% of attractiveness "likes" only 10% of the men, and we assume that she's going to like only the top 10% attractive men. Since this implies that the top 50% women are interested only in the top 10% men, it can be deduced by contrapositive that only the other 50% least attractive women can be interested by the 90% remaining (least attractive) men.
Another example from the other side: a man who has a level of 50% attractiveness likes all the women with more than 5% attractiveness in average. Again it is assumed that a woman who can attract a man in the top 50% will not be interested in the top bottom 50%, therefore the bottom 50% men can only "access" the bottom 5% of the women.
The whole graph is based on this assumption: if somebody at a particular level of attractiveness can attract the N% most attractive of the opposite gender, then anybody below this level of attractiveness is stuck with the remaining proportion. In other words it's not about how many people of the opposite gender one likes, it's about which level of attractiveness one can "afford" given their own level of attractiveness. That's what the graph shows: for example the 80% least attractive men can only afford the 22% least attractive women. By contrast the 78% most attractive women can afford to take their pick among the 20% top attractive men.
Beyond the caveats, my personal conclusion is: all you need is love... but can you afford it?
The following is multiple choice question (with options) to answer.
In a survey about potential presidential candidates A and B, 30% of the public likes A and 50% liked B.If the percentage of the public who like one candidate only is twice the percentage of the public who like both candidates, then what is the percentage of the public that liked neither. | [
"70%",
"55%",
"60%",
"65%"
] | A | T=100
A=30
B=50
Both=x
A only=30-x
B only=50-x
N=Neither
Given:
A only+B only=2*Both
30-x+50-x=2x
80=4x
x=20
T=A only+B only+Both+Neither
100=30-20+20-20+20+N
100=30+N
N=100-30=70
Ans:A |
AQUA-RAT | AQUA-RAT-38421 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2460? | [
"14299",
"27888",
"26687",
"14000"
] | D | 25*12: 30*12: 35*8
15:18:14
14/47 * 47000 = 14000 .Answer: D |
AQUA-RAT | AQUA-RAT-38422 | What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$
The following is multiple choice question (with options) to answer.
The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is | [
"338",
"140",
"287",
"277"
] | B | Explanation:
distance covered in 1 revolution = {\color{Blue} 2\Pi r} = 2 *(22/7) *20 = 880/7 cm
required no of revolutions = 17600 *(7/880) = 140
Answer: B) 140 |
AQUA-RAT | AQUA-RAT-38423 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 65 sq. cm. Find the length of the rectangle. | [
"12cm",
"14cm",
"16cm",
"18cm"
] | C | Explanation:
Let breadth = x. Then, length = 2x. Then,
(2x - 5) (x + 5) - 2x * x = 65 => 5x - 25 = 65 => x = 16.
Length of the rectangle = 16 cm.
Answer: Option C |
AQUA-RAT | AQUA-RAT-38424 | (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.
Suppose you want to buy three loaves of bread that cost $1.00 each
and a jar of peanut butter that costs $5. A jar of jelly is $2.75, but you
don’t need any jelly. You have $16. How much money will you have left
over? | [
"$1.50",
"$6",
"$8",
"$7"
] | C | The jelly is extra information. 16.00 – 3 x 1 – 5.00 =
16.00 – 3 – 5.00 = 8
You have $8 left.
correct answer C |
AQUA-RAT | AQUA-RAT-38425 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
30% people of a village in Sri Lanka died by bombardment, 25% of the remainder left the village on account of fear. If now the population is reduced to 6695, how much was it in the beginning? | [
"7000",
"6700",
"6695",
"7645"
] | C | X * (70/100) * (75/100) = 3515
X = 6695
Answer: C |
AQUA-RAT | AQUA-RAT-38426 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
A’s speed is 17/14 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat? | [
"1/14",
"1/17",
"3/14",
"3/17"
] | D | Let x be the fraction of the distance that B runs.
Let v be the speed at which B runs.
The time should be the same for both runners.
Time = D / (17v/14) = xD/v
(14/17)*D/v = x*D/v
x = 14/17
B should have a head start of 3/17 of the full distance.
The answer is D. |
AQUA-RAT | AQUA-RAT-38427 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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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.
If 5 machines can produce 20 units in 10 hours, how long would it take 20 machines to produce 200 units? | [
"50 hours",
"40 hours",
"25 hours",
"12 hours"
] | C | Here, we're told that 5 machines can produce 20 units in 10 hours....
That means that EACH machine WORKS for 10 hours apiece. Since there are 5 machines (and we're meant to assume that each machine does the same amount of work), then the 5 machines equally created the 20 units.
20 units/5 machines = 4 units are made by each machine every 10 hours
Now that we know how long it takes each machine to make 4 units, we can break THIS down further if we choose to...
10 hours/4 units = 2.5 hours per unit when 1 machine is working.
The prompt asks us how long would it take 20 machines to produce 200 units.
If 20 machines each work for 2.5 hours, then we'll have 20 units. Since 200 units is '10 times' 20, we need '10 times' more TIME.
(2.5 hours)(10 times) = 25 hours
Final Answer:
[Reveal]Spoiler:
C |
AQUA-RAT | AQUA-RAT-38428 | 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.
Two men can complete a piece of work in four days. Two women can complete the same work in eight days. Four boys can complete the same work in five days. If four men, eight women and 20 boys work together in how many days can the work be completed? | [
"1/2 day",
"1/8 day",
"1/3 day",
"8/2 day"
] | A | Two men take four days to complete the work four men would take (2 * 4)/4 = 2 days to complete it.
Similarly four women would take two days to complete it and 20 children would take one day to complete it.
All the three groups working togerther will complete 1/2 + 1/2 + 1/1 work in a day
= 2 times the unit work in a day.
They will take 1/2 a day to complete it working together.
Answer: A |
AQUA-RAT | AQUA-RAT-38429 | # Random Gift Giving at a Party - Combinatorics Problem
Each of $$10$$ employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents?
Firstly, there are $$10^{10}$$ total ways to give the $$10$$ employees the $$10$$ presents. So this is our denominator.
My attempt was to consider the complement and consider the number of ways that either $$0$$ employees receive no presents (every employee gets a present) or $$1$$ employee receives no present.
Case 1: $$0$$ employees
There are $$10$$ employees and $$10$$ presents. So there are $$10^{10}$$ ways to give the presents.
Case 2: $$1$$ employee
Step 1: Decide which employee receives no presents: $$10$$ possibilities.
Step 2: Distribute the $$10$$ presents to the remaining $$9$$ employees: $$9^{10}$$ ways.
So the number of ways in which at least $$2$$ employees receive no presents is: $$1-(10^{10}+9^{10}$$).
So my final answer is: $$1-\displaystyle\frac{(10^{10}+9^{10})}{10^{10}}$$.
However, this answer does not match the answer in my textbook. Which is: $$1-\displaystyle\frac{10!-10\times 9 \times \frac{10!}{2!}}{10^{10}}$$
Where did my attempt go wrong and how can I correct it?
The following is multiple choice question (with options) to answer.
It is known that no more than 100 children will be attending a birthday. What is the smallest number of toffees that must be brought to the party so that each child receives the same number of toffees? | [
"15",
"320",
"400",
"150"
] | C | The number of children is not specified. We are just given that there are no more than 100 children.
If we assume that number of children is 75 then answer will be D.
If we assume that number of children is 100 then answer will be C because only 400 is divisible by 100. |
AQUA-RAT | AQUA-RAT-38430 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
A sum of money invested at C.I. amounts to Rs. 800 in 3 years to Rs. 840 in 4 years. The rate of interest per annum is? | [
"2 1/2 %",
"4%",
"5%",
"6 2/3 %"
] | C | S.I. on Rs. 800 for 1 year = (840 - 800) = Rs. 40 Rate = (100 * 40)/(800 * 1) = 5%
ANSWER:C |
AQUA-RAT | AQUA-RAT-38431 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
A vehicle has an insurance of 4/5 of its original value. Premium on the vehicle at the rate of 1.3% amounts to 910 Rs. What is the original value of the vehicle? | [
"87,508",
"87,502",
"87,500",
"87,509"
] | C | Explanation:
No explanation is available for this question!
ANSWER: C |
AQUA-RAT | AQUA-RAT-38432 | c#, parsing
\n20150805 0900| 6| 91| | 0| | 0| | 0| 0| 0.0| | 13.0| 10.0| 0| 0| |15.00| |
\n20150805 1000| 8| 84| | 0| | 0| | 0| 0| 0.0| | 13.0| 11.0|230| 8| |15.00| |
\n20150805 1100| 10|113| -RW |100| | 0| | 0|100| 0.1| RA| 13.0| 11.0|277| 11| |10.00| |
\n-------------+---+---+-----+---+-----+---+-----+---+---+----+---+-----+-----+---+---+---+-----+--+
\n20150805 1200| 10| 93| -RW |100| | 0| | 0|100| 0.1| RA| 13.0| 10.8|225| 9| | 7.00| |
\n-------------+---+---+-----+---+-----+---+-----+---+---+----+---+-----+-----+---+---+---+-----+--+
\n20150805 1300| 8|999| -RW |100| | 0| | 0| 33| 0.1| RA| 13.9| 11.4|163| 6| |15.00| |
\n20150805 1400| 8|999| -RW |100| | 0| | 0| 31| 0.1| RA| 14.4| 11.5|129| 7| |15.00| |
The following is multiple choice question (with options) to answer.
15.06 * 0.0000001 = ? | [
"15060000",
"0.001506",
"1.506e-06",
"0.1506"
] | C | Explanation:
Clearly after decimal 9 digits should be there.
Option C |
AQUA-RAT | AQUA-RAT-38433 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 20%, he is only able to save Rs. 220 per month. What is his monthly salary? | [
"5500",
"2999",
"2878",
"2990"
] | A | Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80*(20/100) = Rs. 96
Present Savings = 100 – 96 =Rs. 4
100 ------ 4
? --------- 220 => 5500
Answer: A |
AQUA-RAT | AQUA-RAT-38434 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
In how many ways can be 7 boys and 7 girls sit around circular table so that no two boys sit next to each other? | [
"(5!)^2",
"(6!)^2",
"6!7!",
"11!"
] | C | first fix one boy and place other 6 in alt seats so total ways is 6!
now place each girl between a pair of boys... total ways of seating arrangement of girls 7!
total is 6!*7!
ANS C |
AQUA-RAT | AQUA-RAT-38435 | # STRNO - Editorial
Setter: Kritagya Agarwal
Tester: Felipe Mota
Editorialist: Taranpreet Singh
Easy
Number theory
# PROBLEM:
Given two integers X and K, you need to determine whether there exists an integer A with exactly X factors and exactly K of them are prime numbers.
# QUICK EXPLANATION
A valid choice for A exists if the prime factorization of X has the number of terms at least K. So we just need to compute prime factorization of X.
# EXPLANATION
Let’s assume a valid A exists, with exactly K prime divisors.
The prime factorization of A would be like \prod_{i = 1}^K p_i^{a_i} where p_i are prime factors of A and a_i are the exponents. Then it is well known that the number of factors of A are \prod_{i = 1}^K (a_i+1). Hence we have X = \prod_{i = 1}^K (a_i+1) for a_i \geq 1, hence (a_i+1) \geq 2
So our problem is reduced to determining whether is it possible to write X as a product of K values greater than 1 or not.
For that, let us find the maximum number of values we can split X into such that each value is greater than 1. It is easy to prove that all values shall be prime (or we can further split that value). If the prime factorization of X is \prod r_i^{b_i}, then \sum b_i is the number of terms we can split X into.
For example, Consider X = 12 = 2^2*3 = 2*2*3. Hence we can decompose 12 into at most 3 values such that their product is X. However, if K < \sum b_i, we can merge the values till we get exactly K values. Suppose we have X = 12 and K = 2, so after writing 2, 2, 3, we can merge any two values, resulting in exactly two values.
So a valid A exists when the number of prime factors of X with repetition is at least K.
The following is multiple choice question (with options) to answer.
The expression x#y denotes the product of the consecutive multiples of 3 between x and y, inclusive. What is the sum of the exponents in the prime factorization of 21#39? | [
"17",
"18",
"19",
"20"
] | D | First, let's translate the expression 21#39, using the definition given:
21#39 = 21×24×27×30×33×36×39
We need the prime factorization of this product.
Let's factor out 3 from each multiple.
21#39 = 3^7(7×8×9×10×11×12×13)
Now let's replace each consecutive integer with its prime factorization:
21#39 = 3^7(7×2^3×3^2×(2×5)×11×(2^2×3)×13)
Let's group the prime bases:
21#39 = 2^6×3^10×5×7×11×13
The sum of the exponents is 6 + 10 + 1 + 1 + 1 + 1 = 20
The answer is D. |
AQUA-RAT | AQUA-RAT-38436 | ### 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.
John bought an old Vehicle for $42000. He spent $13000 on repairs and sold it for $64900. What is his profit percent? | [
"11%",
"15%",
"18%",
"17%"
] | C | C
18%
Total CP = Rs. 42000 + Rs. 13000 = Rs. 55000 and SP = Rs. 64900
Profit(%) = (64900 - 55000)/55000 * 100 = 18% |
AQUA-RAT | AQUA-RAT-38437 | s_1 :: Integer -> Integer
s_1 n
| n < 2 = error "s_1: n must be >= 2."
| otherwise = t_0 - t_1 - t_2 - t_3
where x_0 = floor ((fromIntegral n) / phi_r)
ff x = (x * (x + 1) * (x + 2)) div 6
(_, _, t_0) = fast_F m_k m_f x_0
t_1 = 2
t_2 = (sum_of_squares x_0) - (sum_of_squares 1)
t_3 = (ff x_0) - (ff 1)
m_f = M.insert 2 ((g 1) + (g 2)) $M.singleton 1 (g 1) m_k = M.insert 2 ((f 1) + (f 2))$ M.singleton 1 (f 1)
s :: Integer -> Integer
s n
| n < 2 = error "s: n must be >= 2."
| otherwise = (s_1 n) + (s_2 n)
The following is multiple choice question (with options) to answer.
If (2x + 1) times (x + 2) is an odd integer, then x must be: | [
"an odd integer",
"an even integer",
"a prime number",
"a composite number"
] | A | Solution:
(2x + 1) times (x + 2) means (2x + 1)(x + 2)
(2x + 1)(x + 2) = 2x^2+5x+2
For 2x^2+5x+2 to be odd, 5x+2 must be odd since 2x^2 is always even.
So 5x must be odd, hence x must be odd.
Therefore x is an odd integer.
Answer: A |
AQUA-RAT | AQUA-RAT-38438 | 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.
Anna left for city A from city B at 5.20 a.m. She traveled at the speed of 80 km/hr for 2 hrs 15 min. After that the speed was reduced to 60 km/hr. If the distance between two cities is 350 kms, at what time did Anna reach city A? | [
"10.77 a.m",
"10.85 a.m",
"10.27 a.m",
"10.25 a.m"
] | D | Distance covered in 2 hrs 15 min i.e., 2 1/4 hrs = 80 * 9/4 = 180 hrs.
Time taken to cover remaining distance = (350 - 180)/60 = 17/6 hrs
= 2 5/6 = 2 hrs 50 min
Total time taken = (2 hrs 15 min + 2 hrs 50 min) = 5 hrs 5 min.
So, Anna reached city A at 10.25 a.m.
Answer:D |
AQUA-RAT | AQUA-RAT-38439 | # Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$?
I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,
$$2^{10^k} \equiv 7 \pmod{9}$$
My proof is as follows:
Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$
Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that \begin{align*} 10^k + 2 &\equiv 0 \pmod{6} \\ 10^k &\equiv 4 \pmod{6} \,\, \text{ for all integers } k \geq 1 \end{align*}
Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$
The following is multiple choice question (with options) to answer.
How many different positive integers exist between 10^7 and 10^8, the sum of whose digits is equal to 2? | [
"6",
"7",
"5",
"8"
] | D | So, the numbers should be from 10,000,000 to 100,000,000
The following two cases are possible for the sum of the digits to be 2:
1. Two 1's and the rest are 0's:
10,000,001
10,000,010
10,000,100
10,001,000
10,010,000
10,100,000
11,000,000
7 numbers.
2. One 2 and the rest are 0's:
20,000,000
1 number.
Total = 8 numbers.
Answer: D |
AQUA-RAT | AQUA-RAT-38440 | In layman terms, this means if we assume there is a smallest number X where X>0 but there is no number such that X>Y>0, then X is the difference between 1 and 0.999... As all my opponent's formulas rely on the idea that 0.999... is a real number, he fails the idea when applied to hyperreal numbers.
The following is multiple choice question (with options) to answer.
0.0005?=0.01 | [
"5",
"0.05",
"0.5",
"50"
] | B | Explanation :
Required Answer = 0.0005/0.01 = 0.05/1 = 0.05. Answer : Option B |
AQUA-RAT | AQUA-RAT-38441 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
A toy store regularly sells all stock at a discount of 20% to 40%. If an additional 25% were deducted from the discount price during a special sale, what would be the lowest possible price of a toy costing $16 before any discount? | [
"$5.60",
"$7.20",
"$8.80",
"$9.60"
] | B | The lowest price would have to be calculated by finding the price of the $16 product after the initial 40% reduction, then the final price after 25% was taken off of the already-discounted price. So:
A 40% discount on a $16 product would = 60% of the original $16 price: $16 x .60 = $9.6
Next you have to find the price after an additional 25% had been discounted from the $9.6, which would be 75% of the $9.6: $9.6 x .75 = $7.20
The answer is B |
AQUA-RAT | AQUA-RAT-38442 | the area and volume of a sphere, using simple geometric arguments. The surface area is the surface part of a three-dimensional object. Find out what's the height of the cylinder, for us it's 9 cm. Volume of a cylinder is (to find half, divide by 2): Vcylinder = Abase * height In a cylinder, the base is a circle and the area of a circle can be found: Acircle = pi * radius2 After you find the. The area of some circle is known and equals 15. 809625 Cubic Inches The volume of a cylinder 5 inches in diameter and 6 inches high is 117. What is the formula for finding the volume of a cylinder? V = π r 2 h 3. The volume of the cylinder is the amount of space inside the cylinder. Visually Understanding Area of a Circle and Volume of a Cylinder. Then click Calculate. Don't forget the two end bits: Total Surface Area. 0021m^3 Please Help or try and Solve this for me, Thankyou Very Much. For the rectangular solid, the area of the base, B, is the area of the rectangular base, length × width. Volume of a Right Circular Cylinder. Volume of a cylinder. Pi is the ratio of a circle's circumference to its diameter: 3. Im doing Maths about 'Volume' one of the questions are finding the volume of a semi-cylinder. The volume of three cones is equal to the volume of one cylinder with the same base and height. The cylinder has two bases, the base has radius r. The length of each cylinder (h) is 15,000 m. Surface Area of a Cylinder= 2πr² + 2πrh. Surface area and volume of a cylinder. Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. Area of a Trapezoid. To find the volume of a cylinder we multiply the base area (which is a circle) and the height h. In this example, r and h are identical, so the volumes are πr 3 and 1 ⁄ 3 π r 3. Browse by Radius in Meters. For engine's, this would be: Swept
The following is multiple choice question (with options) to answer.
5 cylinders are built successively, the first is 4 meters tall and the following cylinders are each 4 meters taller than the previous. If the perimeter is 10pi m, what is the total surface area of the 5 cylinders, excluding the bases of each cylinder? | [
"725pi sq m",
"1025pi sq m",
"600pi sq m",
"125pi sq m"
] | A | s.a. = h x p + 2 x a
we need to excude the bottom area, so the formula becomes:
s.a. = h x p + a
p = 2 pi x r, so 10 pi = 2 pi x r
r = 5
since we have 5 cylinders with different heights the equation becomes
s.a. = ( 4 + 8 + 12 + 16 + 20 ) x 10pi + 5 x pi x r ^2
60 x 10 pi + 5 pi x 5^2
600 pi + 125 pi = 725 pi sq. m.
Answer is A |
AQUA-RAT | AQUA-RAT-38443 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
The length of a rectangular plot is 10 meters more than its width. If the cost of fencing the plot at $15.50 per meter is $682, what is the length of the plot in meters? | [
"10",
"12",
"14",
"16"
] | D | perimeter = 2L + 2W = 2L + 2(L-10) = 4L - 20
cost of fencing = (4L - 20) * 15.50 = 62L - 310 = 682
62L = 992
L = 16 meters
The answer is D. |
AQUA-RAT | AQUA-RAT-38444 | • I hope that the graphs would not converge on $50\%.$ When the $P_{ai}$ and $P_{bi}$ are independently uniformly distributed between $0$ and $1,$ they should converge to a value very close to $41\%.$ That is due to the fact that (a) both students have equal chances of coming out with a higher score and (b) there is about an $18\%$ chance that they will tie. – whuber Apr 19 '18 at 22:30
• It wasn't clear in my answer, but the the simulation used $P_{ai} = P_{bi} = 0.5.$ Not the probabilities 0.8 and 0.9 used in an earlier example. Does that clarify or am I still missing your point? – COOLBEANS Apr 19 '18 at 23:19
• You miss the point: When none of the $(P_{ai},P_{bi})$ pairs is $(0,1)$ or $(1,0),$ there is a positive chance that the students will get the same total score. When all those pairs are equal, as in your example, it therefore becomes impossible that one student's score will exceed the other's with a 50% chance. In your example the chance is only 41%. A close look at your plot suggests it is consistent with 41% but not with 50%. (The exact answer in your case is easy to obtain because all the probabilities are equal: 41% approximates $215955/524288.$) – whuber Apr 20 '18 at 13:40
The following is multiple choice question (with options) to answer.
In a particular course, only two grades are given all semester: the midterm exam, which counts for 25% of the semester grade, and the final exam, which counts for the remainder. Lindsey's semester grade was 88%. If the highest possible score on either exam is 100%, what is the lowest possible score P that Lindsey could have received on her midterm exam? | [
"52%",
"64%",
"72%",
"76%"
] | A | If the midterm counts for 25% of the total grade, then the final exam must count for the remaining 75%, meaning that in this weighted average problem the final is weighted 3 times as much as the midterm. Using the weighted average mapping strategy, then, and maximizing her final exam score so that you minimize her midterm score, you have:
100-------------88----------------------------x
And you know that the ratio is 3:1. Since the final exam score of 100 is weighted heavier, then the difference of 12 between 100 and 88 is the1part of the ratio, meaning that 3(12) is the 3 portion. That equals 36, so the midterm score (x) is 36 less than the weighted average of 88. 88 - 36 = 52, sothe correct answer is 52%. |
AQUA-RAT | AQUA-RAT-38445 | SIMPLE
$27^\frac{2}{3}=(3\cdot\ 3\cdot\ 3)^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=(3)^{3\cdot\frac{2}{3}}=**(3)^{\frac{6}{3}}**=3^2$
ACTUALLY. $27^2$. BECOMES $3^6$
The following is multiple choice question (with options) to answer.
If three eighth of a number is 141. What will be the approximately value of 32.08% of this number? | [
"90",
"120",
"160",
"60"
] | B | x * 3/8 = 141 => x= 376
376 * 32.08/100 = 120
ANSWER:B |
AQUA-RAT | AQUA-RAT-38446 | special-relativity, visible-light, reference-frames, rotational-dynamics, rotation
Title: Light’s Behavior in a Rotating Reference Frame Let’s say there is a laser on one side of a very large rotating table, and it’s beam is shining onto a target on the other side of this table. The target is equipped with a very sensitive buzzer that will sound if the laser moves off of the target. Here’s what I think will happen and why:
During acceleration the laser will move off of the target and the buzzer will sound. Once a constant speed is maintained the buzzer will continue to sound.
My logic for this is rotating frames are not inertia frames. They are always accelerating. Someone on the table could tell they are moving without looking beyond the table simply by placing a ball onto the table and watch it roll off. The Sagnac Effect influences my answer too. Rotating mirrors can move toward or away from a beam of light to shorten or lengthen it’s path.
The speed of light is constant. In order for the beam to stay on the target it would need to follow a curved path which is longer than a straight path.
Is my logic faulty??? Will the buzzer buzz???? Please set me straight. Thanks. As it will take time for the photons to reach the sensor after leaving the emitter, once the sensor starts moving away from where the emitter was originally pointed, the photons will no longer hit the same point on the sensor. So the buzzer will sound until the table stops spinning. This will happen in your spinning table situation, or in any situation where the sensor moves in any direction other than directly away from the emitter.
The following is multiple choice question (with options) to answer.
A searchlight on top of the watch-tower makes 1 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 20 seconds? | [
"1/4",
"1/3",
"1/2",
"2/3"
] | B | 1 revolutions per minute = 1 revolution every 60 seconds
So no matter what anybody appearing at the tower cannot stay in the dark for more than 60 seconds. This will be our total number of possibilities i.e the denominator.
P(man in dark for at least 20 seconds) = 1 - P (man in dark for max of 20 seconds) = 1 - 20/60 = 1 - 1/3 = 2/3
or the other way would be:
P(man in dark for at least 20 seconds) is like saying he can be in dark for 5,6,7...all the way to 60 seconds because that is the max. In this approach it would be 20/60 seconds = 1/3.
Answer is B |
AQUA-RAT | AQUA-RAT-38447 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
• 4 minutes ago
• 5 minutes ago
• 7 minutes ago
• 8 minutes ago
• 37 seconds ago
• A minute ago
• A minute ago
• 2 minutes ago
• 2 minutes ago
• 3 minutes ago
• 4 minutes ago
• 5 minutes ago
• 7 minutes ago
• 8 minutes ago
The following is multiple choice question (with options) to answer.
85% of a number is added to 24, the result is the same number. Find the number? | [
"330",
"267",
"278",
"160"
] | D | (85/100) * X + 24 = X
3X = 480
X = 160
Answer:D |
AQUA-RAT | AQUA-RAT-38448 | ## anonymous 5 years ago 100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!
1. anonymous
you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..
2. anonymous
$\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950$
3. anonymous
Could you solve it using factorials, please?
4. anonymous
hmm i found the answer as 9900
5. amistre64
if there are 3 people: 1,2 ; 1,3 ; 2,3 .... is that right?
6. anonymous
Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...
7. anonymous
yea but its harder to calculate this like that... so i think C(100,2) better way of solve
8. amistre64
5054 if we do that ..... add all the numbers from 1 to 100 :)
9. anonymous
No, all the numbers from 1 to 99.
10. amistre64
n(n+1) ------ :) 2
11. amistre64
ack .... 99 then lol
12. amistre64
wouldnt the last guy have noone to shake hands with?
13. anonymous
mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,
14. anonymous
Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.
15. anonymous
yea you right
16. amistre64
The following is multiple choice question (with options) to answer.
36 people {a1, a2… a36} meet and shake hands in a circular fashion. In other words, there are totally 36 handshakes involving the pairs, {a1, a2}, {a2, a3}, …, {a35, a36}, {a36, a1}. Then size of the smallest set of people such that the rest have shaken hands with at least one person in the set is | [
"76",
"55",
"44",
"12"
] | D | Ans: {a1, a2}, {a2, a3},{a3, a4}, {a4, a5},{a5, a6}, {a6, a7} …, {a35, a36}, {a36, a1}
From the above arrangement, If we separate a3, a6, a9, .....a36. Total 12 persons the reamining persons must have shaked hand with atleast one person. So answer is 12.
Answer:D |
AQUA-RAT | AQUA-RAT-38449 | # If $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$ = $$K(10)^9$$, then k is equal to
## Solution :
$$K(10)^9$$ = $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$
K = 1 + 2$$({11\over 10})$$ + 3$$({11\over 10})^2$$ + ….. + 10$$({11\over 10})^9$$ ……(i)
$$({11\over 10})$$K = 1$$({11\over 10})$$ + 2$$({11\over 10})^2$$ + 3$$({11\over 10})^3$$ + ….. + 10$$({11\over 10})^{10}$$ …..(ii)
On subtracting equation (ii) from (i), we get
K$$(1 – {11\over 10})$$ = 1 + $$({11\over 10})$$ + $$({11\over 10})^2$$ + …. + $$({11\over 10})^9$$ – 10$$({11\over 10})^{10}$$
$$\implies$$ K$$({10 – 11\over 10})$$ = $$1[({11\over 10})^{10} – 1]\over ({11\over 10} – 1)$$ – 10$$({11\over 10})^{10}$$
The following is multiple choice question (with options) to answer.
IF
6= I
7= E
8= I
9= I
10=E
THEN 11=? | [
"I",
"J",
"K",
"L"
] | D | IF
6= I... Second letter of ONE.
7= E.... Second letter of TWO.
8= I
9= I
10= E
THEN 11=L... Second letter of Eleven
ANSWER:D |
AQUA-RAT | AQUA-RAT-38450 | # How to prove $\sqrt {75025} + \sqrt {121393} + \sqrt {196418} + \sqrt{317811} \approx \sqrt {514229} + \sqrt {832040}$?
Let $$a = \sqrt {75025} + \sqrt {121393} + \sqrt {196418} + \sqrt{317811}$$ and $$b = \sqrt {514229} + \sqrt {832040}$$. By using SageMath we can see that $$a - b \approx 2.95301560981898 \cdot 10^{-9}$$ That means almost nine digits of accuracy!
To investigate any particular reason why these surprising digits of accuracy come I considered the function $$f(x)= \sqrt{x +317811} + \sqrt{x + 196418} + \sqrt{x + 121393} + \sqrt{x + 75025} -\sqrt{x + 832040} - \sqrt{x + 514229}$$
The Taylor series of $$f$$ around $$x =0$$ with approximate coefficients looks like $$f(x) = f(0) + 0.00403020948350145x -2.13362649294736 \cdot 10^{-8}\frac12 x^2 + O(x^3)$$
If $$\alpha$$ is a root of the equation $$f(x) = 0$$ where $$\alpha$$ is very close to $$0$$ (definitely there is a root between $$-1$$ and $$0$$) then $$0 = f(\alpha) = f(0) + 0.00403020948350145 \alpha -2.13362649294736 \cdot 10^{-8}\frac12 \alpha^2 +\text{higher error terms}$$
Of course, we can use computer programs to find a bound on $$\alpha$$ but the whole process is not mathematically elegant.
The following is multiple choice question (with options) to answer.
We all know that square root of number 121 is 11.
But do you know what si the square root of the number "12345678987654321" ? | [
"222222222",
"111111111",
"333333333",
"777777777"
] | B | B
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer) |
AQUA-RAT | AQUA-RAT-38451 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Joe went on a diet 3 months ago when he weighed 222 pounds. If he now weighs 198 pounds and continues to lose at the same average monthly rate, in approximately how many months will he weigh 170 pounds? | [
" 3",
" 3.5",
" 4",
" 4.5"
] | B | 222 - 198 = 24 pounds lost in 3 months
24/3 = 8, so Joe is losing weight at a rate of 8 pounds per month.
...in approximately how many months will he weigh 170 pounds?
A simple approach is to just list the weights.
Now: 198 lbs
In 1 month: 190 lbs
In 2 months: 182 lbs
In 3 months: 174 lbs
In 4 months: 166 lbs
Since 170 pounds is halfway between 174 and 166, the correct answer must be 3.5 months.
Answer: B |
AQUA-RAT | AQUA-RAT-38452 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of first 10 even numbers is? | [
"17",
"199",
"16",
"11"
] | D | Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
Answer:D |
AQUA-RAT | AQUA-RAT-38453 | # 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.
Look at this series: 23, 22, 24, 23, 25, 24, ... What number should come next? | [
"20",
"26",
"28",
"21"
] | B | In this simple alternating subtraction and addition series; 1 is subtracted, then 2 is added, and so on.
23-1 = 22
22+2 = 24
24-1 = 23
23+2 = 25
25-1 = 24
24+2 = 26
Answer is B |
AQUA-RAT | AQUA-RAT-38454 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, which A and B together can do it in 2 hours. How long will C alone take to do it? | [
"12hours",
"10hours",
"6hours",
"8hours"
] | A | A's 1 hour work = 1/4;
(B + C)'s 1 hour work = 1/3;
(A + B)'s 1 hour work = 1/2
(A + B + C)'s 1 hour work = (1/4 + 1/3) = 7/12
C's 1 hour work = (7/12 - 1/2) = 1/12
C alone will take 12 hours to do the work.
Answer:A |
AQUA-RAT | AQUA-RAT-38455 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
If the sum of three consecutive positive odd integers is A, then the sum of the next three consecutive odd integers in terms of A is: | [
"A+3",
"A+9",
"A+18",
"2A + 3"
] | C | In Case of Consecutive Integers or Integers in Arithmetic Progression Mean = Median
I.e. Median = A/3 = Mean = Second Integer
First Integer = A/3 - 2
Second Integer = A/3
Third Integer = A/3 + 2
i.e.
Fourth Integer = A/3 + 4
Fifth Integer = A/3 + 6
Sixth Integer = A/3 + 8
Now Mean of next 3 Integers = Median = A/3 + 6
i.e. Sum of Next 3 integers = (A/3 + 6)*3 = A+18
Answer: option C |
AQUA-RAT | AQUA-RAT-38456 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of 35 numbers is 25. If each number is multiplied by 5, find the new average? | [
"125",
"777",
"255",
"199"
] | A | Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.Answer: A |
AQUA-RAT | AQUA-RAT-38457 | Since we have $21$ terms taking $20$ possible values, there are some $0 \le i < j \le 20$ such that $S_i = S_j$. It follows that the total number of hours of study between days $i+1$ and $j$ (inclusive) is a multiple of $20$.
If it is not exactly equal to $20$ hours, then it must be at least $40$ hours. However, this is over a span of at most $20$ days. Dividing this into three periods of at most $7$ days (say the first week, second week, and third week), by averaging we find that she must have worked at least $14$ hours during one of the weeks, which is not allowed.
Thus she must actually have studied exactly $20$ hours between days $i+1$ and $j$.
• Yes, that's a little cleaner as route to Wiley's lemma.Thanks. – Joffan Jul 17 '16 at 23:05
• @Joffan, agreed. It's indeed cleaner to use 20 "holes" with 21 "pigeons" rather than separating $S_l=S_i+20$ as an individual case in my proof. Thank you, Shagnik. – Wiley Jul 18 '16 at 3:59
The proof consists of two parts.
• Part I: Prove that a period of $20$ days is enough such that there must exist some period of consecutive days during which totally $20$ hours are spent on studying.
• Part II: A counterexample which shows that $19$ days are not enough is presented.
Proof of Part I
The following is multiple choice question (with options) to answer.
In an intercollegiate competition that lasted for 3 days, 175 students took part on day 1, 210 on day 2 and 150 on day 3. If 80 took part on day 1 and day 2 and 70 took part on day 2 and day 3 and 20 took part on all three days, how many students took part only on day 1? | [
"25",
"45",
"55",
"70"
] | B | Day 1&2 = 80; Only Day 1&2 (80-20) = 60,
Day 2&3 = 70; Only Day 2&3 (70-20) = 50,
Only Day 1 = 175 - (60+50+20) = 45
Answer:B |
AQUA-RAT | AQUA-RAT-38458 | => 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.
Two employees X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week? | [
"Rs.200",
"Rs.220",
"Rs.250",
"Rs.300"
] | C | Let the amount paid to X per week = x
and the amount paid to Y per week = y
Then x + y = 550
But x = 120% of y = 120y/100 = 12y/10
∴12y/10 + y = 550
⇒ y[12/10 + 1] = 550
⇒ 22y/10 = 550
⇒ 22y = 5500
⇒ y = 5500/22 = 500/2 = Rs.250
C |
AQUA-RAT | AQUA-RAT-38459 | 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.
The monthly incomes of A and B are in the ratio 5 : 2. B's monthly income is 12% more than C's monthly income. If C's monthly income is Rs. 16000, then find the annual income of A? | [
"Rs. 537600",
"Rs. 180000",
"Rs. 201600",
"Rs. 504000"
] | A | B's monthly income = 16000 * 112/100 = Rs. 17920
B's monthly income = 2 parts ----> Rs. 17920
A's monthly income = 5 parts = 5/2 * 17920 = Rs. 44800
A's annual income = Rs. 44800 * 12 = Rs. 537600
ANSWER:A |
AQUA-RAT | AQUA-RAT-38460 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two stations A and B are 110 km apart on a straight line. One train starts from A at 4 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet? | [
"15 a.m",
"10 a.m",
"7 a.m",
"02 a.m"
] | C | Suppose they meet x hours after 4 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
45x = 135
x = 3.
So, they meet at 7 a.m.
Answer:C |
AQUA-RAT | AQUA-RAT-38461 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A dishonest dealer professes to sell his goods at Cost Price but still gets 25% profit by using a false weight. What weight does he substitute for a kilogram? | [
"800 1/7 grams",
"800 grams",
"800 8/3 grams",
"883 1/3 grams"
] | B | If the cost price is Rs.100, then to get a profit of 25%, the selling price should be Rs.125.
If 125kg are to be sold, and the dealer gives only 100kg, to get a profit of 25%.
How many grams he has to give instead of one kilogram(1000 gm).
125 gm ------ 100 gm
1000 gm ------ ?
(1000 * 100)/125 =800 grams.
Answer:B |
AQUA-RAT | AQUA-RAT-38462 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man's speed with the current is 21 km/hr and the speed of the current is 4.3 km/hr. The man's speed against the current is | [
"9.2",
"10.3",
"11.5",
"12.4"
] | D | Man's rate in still water = (21 - 4.3) km/hr = 16.7 km/hr.
Man's rate against the current = (16.7 - 4.3) km/hr = 12.4 km/hr.
Answer:D |
AQUA-RAT | AQUA-RAT-38463 | ### Exercise 23
x+2y=10x+2y=10 size 12{x+2y="10"} {}
(35)
y+z=5y+z=5 size 12{y+z=5} {}
(36)
z+w=3z+w=3 size 12{z+w=3} {}
(37)
x+w=5x+w=5 size 12{x+w=5} {}
(38)
### Exercise 24
x+w=6x+w=6 size 12{x+w=6} {}
(39)
2x+y+w=162x+y+w=16 size 12{2x+y+w="16"} {}
(40)
x2z=0x2z=0 size 12{x - 2z=0} {}
(41)
z+w=5z+w=5 size 12{z+w=5} {}
(42)
## SYSTEMS OF LINEAR EQUATIONS – SPECIAL CASES
Solve the following inconsistent or dependent systems by using the Gauss-Jordan method.
### Exercise 25
2x+6y=82x+6y=8 size 12{2x+6y=8} {}
(43)
x+3y=4x+3y=4 size 12{x+3y=4} {}
(44)
### Exercise 26
The sum of the digits of a two digit number is 9. The sum of the number and the number obtained by interchanging the digits is 99. Find the number.
### Exercise 27
2xy=102xy=10 size 12{2x - y="10"} {}
(46)
4x+2y=154x+2y=15 size 12{ - 4x+2y="15"} {}
(47)
### Exercise 28
The following is multiple choice question (with options) to answer.
The sum of two numbers is 21 and the sum of their squares is 527. Find the product of the numbers. | [
"48",
"41",
"46",
"43"
] | D | Let a and b be the two numbers
(a+b)^2=a^2+2ab+b^2
Given (a+b)=21 a^2+b^2=527
So, 21^2=527+2ab
2ab=527-441
2ab=86
ab=43
Ans D |
AQUA-RAT | AQUA-RAT-38464 | ### 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 a, b, and c are consecutive positive integers and a > b > c, which of the following must be true?
I. c - a = 2
II. abc is an odd integer.
III. (a + b + c)/3 is an integer. | [
" I only",
" II only",
" I and III only",
" II and III only"
] | C | Since a, b, and c are consecutive positive integers and a < b < c, then c = a + 2, from which it follows that c - a = 2. So, I is true.
Next, out of 3 consecutive integers at least 1 must be even, thus abc=even. II is false.
Finally, since b = a + 1, and c = a + 2, then (a + b + c)/3 = (a + a + 1 + a + 2)/3 = a + 1 = integer. III is true as well.
Answer: C. |
AQUA-RAT | AQUA-RAT-38465 | area of that circle click on the calculate button... It is also known as the longest straight line segment which passes the! Segment which passes through the center of the circle demonstration is to use this calculator, by... 3 = 150687.075 mm if we know circumference as the straight line segment which passes through material... Example tire with a 30 inch diameter turning at 300 rpm will have road... Outside circumference by 3.1415 measurement with a 30 inch diameter turning at 300 rpm will have a road speed 26.8. Of 26.8 mph x π x d 3 ) 16. d 3 ) 16. d 3 150687.075... Learn C programming, data Structures Tutorials, exercises, examples, programs, hacks, tips and tricks.! Defined as the known factor to calculate the area of a circle given circumference! Background Tutorials: Simply plug in 36 for d in the Wheel Graph calculator demonstration is to PI... Use this calculator, begin by entering the circumference is about 1/32″ ( )! Plug in 36 for d in the formula for calculating the area of a.... Do you find the diameter is given to be changed to calculate diameter circumference and area are - (. Circle 's edge from one point, meeting back at that point,,... A road speed of 26.8 mph we are using functions that find the end is accessible... Also select units of measure for both input data and results to private tutoring material best their. Equal to 5.3 meters you have the diameter consider the maximum torque can... The measurement of its diameter the formula for calculating the area of Wheel... Tire diameter always how to find diameter Hamiltonian cycle in the image, only the tire diameter speed. Use PI and the circumference of a circle given its circumference if we know circumference value into the for! Speed of 26.8 mph where it meets a 30 inch diameter turning at rpm... Yes if you have the diameter that find the radius with the longest chord of the circle using.! Helps you to find diameter, circumference, and area of the circle you... This program, we separated the logic using python functions features make Nerd! Back at that
The following is multiple choice question (with options) to answer.
If a tire rotates at 400 revolutions per minute when the car is traveling 96km/h, what is the circumference of the tire? | [
"7 meters",
"9 meters",
"4 meters",
"5 meters"
] | C | 400 rev / minute = 400 * 60 rev / 60 minutes
= 24,000 rev / hour
24,000 * C = 96,000 m : C is the circumference
C = 4 meters
correct answer C |
AQUA-RAT | AQUA-RAT-38466 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
Approximately what percent of the employees with 5 or more years of employment are NOT taking either workshop? | [
"86%",
"64%",
"50%",
"14%"
] | B | Total = n(A) + n(B) - Both + None
None = Total - n(A) - n(B) + Both
Calculate "none" for each case
None(5+ yrs Line) = 200 - 50 - 40 + 30 = 140
None(5+ yrs Staff) = 160 - 40 - 50 + 20 = 90
Employees not taking either workshop/Total number of employees = (140 + 90)/(200 + 160) = 230/360 = About 64%
ANSWER:B |
AQUA-RAT | AQUA-RAT-38467 | 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 undertake to do a piece of work for $600. A alone can do it in 6days while B alone can do it in 8days. With the help of C, they finish it in 3days. Find the share of A? | [
"$100",
"$150",
"$300",
"$250"
] | C | C's 1day work = (1/3)-(1/6 + 1/8) = 1/24
A:B:C= 1/6 : 1/8: 1/24 = 4:3:1
A's share = 600*4/8 = $300
Answer is C |
AQUA-RAT | AQUA-RAT-38468 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
The true discount on a bill due 9 months hence at 16% per annum is Rs. 171.The amount of the bill is | [
"Rs. 1386",
"Rs. 1764",
"Rs. 1425",
"Rs. 2268"
] | C | Solution
32.5
Let P.W. be Rs. x.Then,S.I. on Rs.x at 16% for 9 months =Rs.171.
∴ x 16 x9/ 12x 1/100}= 171 or x = 1425.
∴ P.W. = Rs. 1425.
Answer C |
AQUA-RAT | AQUA-RAT-38469 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If the two-digit integers P and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of P and N? | [
"181",
"165",
"121",
"99"
] | A | Since the question asks for the answer that CANNOT be the sum of P and N, and the answers are numbers, we can use a combination of TESTing VALUES and TESTing THE ANSWERS to eliminate the possible values and find the answer to the question.
We're told that P and N are two-digit positive integers and have the SAME DIGITS but in REVERSE ORDER. We're asked which of the 5 answers CANNOT be the SUM of P and N.
Let's start with the 'easiest' answer first:
44. Can we get to 44 in the manner described?
Yes, if the numbers are 13 and 31.....13+31 = 44. Eliminate Answer E
Now let's work through the rest of the list....
Can we get to 99 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 18 and 81.....18+81 = 99. Eliminate Answer D
Can we get to 121 in the manner described?
Yes, there are several ways to do it. For example, if the numbers are 38 and 83.....38+83 = 121. Eliminate Answer C
Can we get to 165 in the manner described?
Yes, there are a couple of ways to do it. For example, if the numbers are 78 and 87.....78+87 = 165. Eliminate Answer B
There's only one answer left....
A |
AQUA-RAT | AQUA-RAT-38470 | Use $\int_a^b f(x)dx=\int_a^nf(a+b-x)dx$ and $\int_0^{2a}f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$
The following is multiple choice question (with options) to answer.
If f(a)=a^2, what is the value of (f(a+b)−f(b))/b ? | [
"2a+a^2",
"b",
"2a",
"2a + b"
] | A | start with f(a+b) before start calculations: f(a+b)=(a+b)^2= a^2+2ab+b^2
a^2+2ab+b^2 - b^2/b = (simplify a^2)
2ab+a^2/b = (cancel b's)
= 2a+a^2
Ans : A |
AQUA-RAT | AQUA-RAT-38471 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A can run 22.5 m while B runs 25 m. In a kilometre race B beats A by: | [
"120 meter",
"300 meter",
"100 meter",
"200 meter"
] | C | when B runs 25 m, A runs 22.5 m
=> When B runs 1000 metre, A runs 22.5/25 x 1000 = 900m.
=> When B runs 1 kilometre, A runs 900 m
Hence, in a kilometre race, B beats A by
ANSWER:C
(1 kilometre - 900 metre) = (1000 metre - 900 metre) = 100 metre |
AQUA-RAT | AQUA-RAT-38472 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
The sum of the present ages of two persons A and B is 54. If the age of A is twice that of B, find the sum of their ages 5 years hence? | [
"50",
"64",
"70",
"80"
] | B | A + B = 54, A = 2B
2B + B = 54 => B = 18 then A = 36.
5 years, their ages will be 41 and 23.
Sum of their ages = 41 + 23 = 64.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38473 | 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. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. If they start at 8 a.m. from the same point in opposite directions, how many times shall they cross each other before 9.30 a.m.? | [
"5",
"6",
"7",
"8"
] | C | Explanation:
Relative speed = Speed of A + Speed of B (∴ they walk in opposite directions)
=2+3=5 rounds per hour
Therefore, they cross each other 5 times in 1 hour and 2 times in 12 hour
Time duration from 8 a.m. to 9.30 a.m. =1.5 hour
Hence they cross each other 7 times before 9.30 a.m.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-38474 | => 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.
Sam’s car was fined when he gave Joe and Peter a ride, so they decided to help Sam pay the fine. Joe paid $10 more than 1/4 of the fine and Peter paid $10 less than 1/3 of the fine, leaving pay $25 less than 1/2 the fine to complete the payment. What fraction of the fine did Sam pay? | [
"$120",
"$125",
"$130",
"$135"
] | B | Call the fine F. Joe paid (1/4)F + 4 and Peter paid (1/3)F – 4, leaving (1/2)F – 5 left. If we add those three up, they should add up to F.
F = [(1/4)F + 10] + [(1/3)F – 10] + [(1/2)F – 25]
F = (1/4)F + (1/3)F + (1/2)F – 25
Multiply all terms by 12 to clear the fractions.
12F = 3F + 4F + 6F – 300
12F = 13 F – 300
–F = – 300
F = 300
Well, if the fine cost $300, then Sam paid the part not covered by Joe or Peter. Half the fine is $150, and Sam paid $25 less than this: $125
Answer = B |
AQUA-RAT | AQUA-RAT-38475 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
Mary built a model ship. She spent 10 hours constructing the ship and 5 hours painting it, for what fraction of the total time did she spend painting the ship? | [
"1/3",
"1/5",
"1/4",
"2/3"
] | A | We don't need to get into calculations for solving this question. We can use the concept of weighted averages.
The average time for the entire project is 15 hours; so, she spent a lesser duration of time painting the ship than constructing it.
5---(5---5)
This shows that you can divide the entire project into 3 equal parts. Thus, 1/3 of the journey was spent painting.
Answer: A |
AQUA-RAT | AQUA-RAT-38476 | # Homework Help: A somewhat simple Probabilty problem
1. Dec 16, 2011
### Whitishcube
1. The problem statement, all variables and given/known data
Suppose two teams are playing a set of 5 matches until one team wins three games (or best of five).
Considering the possible orderings for the winning team, in how many ways could this series end?
2. Relevant equations
3. The attempt at a solution
So I think I have the solution, I just would like my logic to be checked in this.
If we think of the case where the winning team gets three wins first, we can think of the
last two games as losses, so we can essentially think of the number of possible outcomes
as the number of permutations of the set {W, W, W, L, L} (W is win, L is loss). This ends up being the multinomial coefficient:
$$\left( \begin{array}{c} 5\\ 3,2 \end{array} \right)= \frac{5!}{2! 3!} = 10,$$
so there are 10 possible outcomes.
Is this correct? Probability is a strange feel compared to most of the other math I have encountered...
2. Dec 16, 2011
### BruceW
Your answer would be correct if they always played all 5 games. But this isn't true, because they will stop as soon as the winner has won 3 games. So you have over-counted. You've got to think of a different way of counting the possible outcomes.
3. Dec 16, 2011
### Whitishcube
Hmm any pointers besides that? The only things I can think of are permutations and combinations. If they don't always play 5 games though, how can I use these methods here?
4. Dec 16, 2011
### dacruick
technically you could use the logic that they play all 5 games and subtract from that the amount of ways that 5 games wouldn't be played. You know that a minimum of 3 have to be played.
5. Dec 16, 2011
### Whitishcube
So you're saying I should take the possible number of ways the 5 games could be played, which is 10 ways, and subtract the number of games where they win before all 5 games are played?
The following is multiple choice question (with options) to answer.
A and B participate in a seven-person footrace on the basketball court during All-Star Weekend. If all seven contestants finish (including Charles Barkley) and there are no ties, how many different arrangements of finishes are there in which A defeats B ? | [
" 5040",
" 2520",
" 720",
" 120"
] | B | PRINCETON REVIEW OFFICIAL SOLUTION:
A beats B? Carmelo beats B? No one said anything about specific results, especially not on a question that includes so many people.
Many questions on the GMAT can be overwhelming in appearance so first thing…Relax. This question looks nasty, but it’s pretty straightforward, requiring two things: the basic first calculation and then some additional work to solve the problem. It would be extremely painful to try to calculate this whole question in one fell swoop. So don’t.
Calculate out all possibilities first to get your bearings. That is, how many different arrangements, period? Since the order of arrangement again matters, this is another permutation. And since we’re placing all seven runners, this calculation is…
7 x 6 x 5 x 4 x 3 x 2 x 1, cash and prizes worth 5040 arrangements.
Now, what to do? Again, some nice deep breathing to control your nerves – relax and think. These 5040 arrangements are a comprehensive list; every possible seven-person order of finish has been counted. So, we have accounted for every time A beats B AND every time LeBron beats Carmelo. Doesn’t it make sense that, for every time Carmelo beats LeBron, there is exactly one other instance in which LeBron beats Carmelo? Or more simply, doesn’t it make sense that, if these 5040 represents everything, that Carmelo beats LeBron exactly 50% of the time and vice versa?
The thought takes time. You need to have the presence to recognize that grinding your way in, around, and through this question could take roughly forever. Stepping back and recognizing that this is simply a 50/50 proposition and applying that thought to the 5040 you cranked out gives you 2520. The answer is B. |
AQUA-RAT | AQUA-RAT-38477 | 3. 0725. This calculator shows the steps and work to convert a fraction to a decimal number. Write the repeating part as the numerator of the fraction. Okay, so you can use division to see if a fraction has a repeating decimal, but how do you convert a repeating decimal into a fraction? Changing a repeating decimal in which all numbers after the decimal repeat, such as zero point one repeating, zero point two repeating, etc. Ask students to describe each piece as a fraction (1/8), a decimal (0. $$\frac{decimal}{1}$$ Step 2: For Every number after the decimal point multiply by 10 for both top and bottom (i. 125 Arranging decimal numbers by size When comparing decimal numbers and arranging them in order it is usually easiest to line up the numbers vertically with the decimal points in a vertical line. Convert fractions to repeating decimals Grade 5 Decimals Worksheet Convert to decimals, round to 3 digits if necessary. 5 ), this percentage is simpler to convert than was the previous one. Here is a simple online Recurring Decimal to Fraction Calculator to convert from Repeating Decimal to Fraction. to Fractions Focus 4 - Learning Goal #1: Students will know that there are numbers that are not rational, and approximate them with rational numbers. Writing Terminating Decimals as Fractions. 3 2 1 0. Example: Convert 0. 7025 = 3 + . 444444…. Name: _____ Converting Fractions, Decimals, and Percents fraction decimal percent a. 5555555555 Step 2: After examination, the repeating digit is 5 Step 3: To place the repeating digit ( 5 ) to the left of the decimal point, you need to move the decimal point 1 repeating part of the number is immediately next to the decimal. Converting decimals to fractions To convert a decimal to a fraction we rst split it into the whole number part added to the decimal part Example: 3. Convert decimal to fraction changing 1. Fractions as decimals. Decide how many numbers are repeating. org and *. Math Worksheets Fractions as Decimals Fractions as Decimals. org This file derived from G7-M2-TE-1. Use repeating DECIMALS INTO. To convert fractions to decimals, look at the fraction as a division problem. Students should be able to convert fractions to decimals, decimals
The following is multiple choice question (with options) to answer.
Half of 3 percent written as decimal is | [
"5",
"0.5",
"0.05",
"0.015"
] | D | Explanation:
It will be 1/2(3%) = 1/2(3/100) = 3/200 = 0.015
Answer: Option D |
AQUA-RAT | AQUA-RAT-38478 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
Legendary Member
Posts: 1892
Joined: 14 Oct 2017
Followed by:3 members
### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Legendary Member
Posts: 2663
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
Jane and Thomas are among the 6 people from which a committee of 3 people is to be selected. How many different possible committees of 3 people can be selected from these 6 people if at least one of either Jane or Thomas is to be selected? | [
"12",
"14",
"16",
"18"
] | C | The total number of ways to choose 3 people from 6 is 6C3 = 20.
The number of committees without Jane or Thomas is 4C3 = 4.
There are 20-4 = 16 possible committees which include Jane and/or Thomas.
The answer is C. |
AQUA-RAT | AQUA-RAT-38479 | \begin{align} \dbinom{6}{2} & \longleftrightarrow \dbinom{6}{6-2} \\[8pt] AB & \longleftrightarrow CDEF \\ AC & \longleftrightarrow BDEF \\ AD & \longleftrightarrow BCEF \\ AE & \longleftrightarrow BCDF \\ AF & \longleftrightarrow BCDE \\ BC & \longleftrightarrow ADEF \\ BD & \longleftrightarrow ACEF \\ BE & \longleftrightarrow ACDF \\ BF & \longleftrightarrow ACDE \\ CD & \longleftrightarrow ABEF \\ CE & \longleftrightarrow ABDF \\ CF & \longleftrightarrow ABDE \\ DE & \longleftrightarrow ABCF \\ DF & \longleftrightarrow ABCE \\ EF & \longleftrightarrow ABCD \end{align} There are exactly as many ways to choose $2$ out of $6$ as to choose $6-2$ out of $6$ because each way of choosing $2$ out of $6$ has a corresponding way of choosing $6-2$ out of $6$ and vice-versa.
• NOTE TO FUTURE USERS: This combined with TheSparkThatThought's answer and Ned's comment at the original question will guide you to the way someone must think. Jun 27 '14 at 23:12
Consider a collection of $n$ objects. Choosing $k$ of them to place into a set is equivalent to choosing $n-k$ to leave out.
Edit: Consider a high school dodgeball game with a red team and a blue team. There are $n$ total students, and the blue team has $k$ students. Since every student plays, there are $n-k$ students on the red team.
Because the PE teacher is biased, he lets the blue team pick all of their players first. They have $\binom{n}{k}$ ways to do this. After that, the red team has no choices. They must pick all of the remaining $n-k$ students.
The following is multiple choice question (with options) to answer.
In a class of 37 students 26 play football and play 20 long tennis, if 17 play above, many play neither? | [
"6",
"8",
"10",
"12"
] | B | 26 + 20 - 17 = 29
37 - 29 = 8 play neither
Answer is B |
AQUA-RAT | AQUA-RAT-38480 | 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.
P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 35 days to complete the same work. then Q alone can do it in | [
"20",
"22",
"25",
"28"
] | D | Work done by P and Q in 1 day = 1/10
Work done by R in 1 day = 1/35
Work done by P, Q and R in 1 day = 1/10 + 1/35 =9/70
But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 9/70
=> Work done by P in 1 day = 9/140
=> Work done by Q and R in 1 day = 9/140
Hence work done by Q in 1 day = 9/140– 1/35 = 1/28
So Q alone can do the work in 28 days
Answer is D. |
AQUA-RAT | AQUA-RAT-38481 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
6 + 12 + 18 + 5 x 6^2 + 5 x 6^3 + 5 x 6^4 + 5 x 6^5 + 5 x 6^6 = | [
" 6^7",
" 6^10",
" 6^12",
" 6^16"
] | A | Let's see that 6 + 12 + 18 is 36 or 6^2, then: 6^2 + 5 x 6^2 becomes 6 x 6^2 that is 6^3. So we follow with the next element: 6^3 + 5 x 6^3 is equal to 6 x 6^3 that is 6^4. Then is assumed right before 5 x 6^6: 6^6 + 5 x 6^6 becomes 6 x 6^6 that is 6^7. Thus, the correct answer is the option A, 6^7. |
AQUA-RAT | AQUA-RAT-38482 | kmph. com/2016/01/01/speed-distance-timeVideos, worksheets, 5-a-day and much more. If a car traveled 50 miles over the course of one hour then its average speed will be 50 mph. Distance, Time and Speed Word Problems | GMAT GRE Maths. Exercise: Problems on Speed Time and Distance. org/time-speed-distanceTime, Speed and Distance (popularly known as TSD) is an important topic for written round of placements for any company. sg//sol_04_distance_speed_and_time. Let the time taken to cover the distance downstream = t hrs. Distance Time Speed. Time is entered in minutes, speed in knots and distance in nautical miles (the same formula will work for statute miles and kilometres). Time = Distance / speed = 20/4 = 5 hours. Speed, Time, and Distance Worksheet. Equations: Acccceelleerraattiioonn = Final speed–Initial TTiimme = Final Speed–Initial Time Acceleration. When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time. b. Acceleration – change in velocity over time. 4 Calculate speed, distance, and time I . Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. He drives 150 meters in 18 seconds. Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. Aptitude Reasoning TRAINS DISTANCE SPEED TIME QUANTITATIVE APTITUDE . We define speed as distance divided by time. A train travels at a speed of 30mph and travel a distance of 240 miles. It can cross a pole in 10 seconds. The speeds are indicated in the rate column. Toon Train is traveling at the speed of 10 m/s at the top of a hill. If you run around the house randomly, and then end up back where you started, moving a total of 44 meters, what is your distance? Average speed for the entire trip is going to be equal to the total
The following is multiple choice question (with options) to answer.
A car takes 6 hours to cover a distance of 180 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time? | [
"50 Kmph",
"20 Kmph",
"65 Kmph",
"70 Kmph"
] | B | Time = 6
Distence = 180
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 180/9 = 60 Kmph
B) |
AQUA-RAT | AQUA-RAT-38483 | I don't see it...
• May 21st 2010, 12:36 PM
wonderboy1953
Hint
Look for cycles.
• May 21st 2010, 03:22 PM
Soroban
Hello, pollardrho06!
Quote:
Find the number of positive integers not exceeding 1000
that are divisible by 3 but not by 4.
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3.
But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
• May 21st 2010, 03:46 PM
pollardrho06
Quote:
Originally Posted by Soroban
Hello, pollardrho06!
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3.
But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
Wow!! The greatest integer function!! Gr8!! Thanks!!
The following is multiple choice question (with options) to answer.
How many positive integers less than 70 have a reminder 01 when divided by 3? | [
"13",
"24",
"35",
"16"
] | B | 1 also gives the remainder of 1 when divided by 3. So, there are total of 24 numbers.
Answer: B. |
AQUA-RAT | AQUA-RAT-38484 | $$T = \frac {60}{160 + w} + \frac {60}{160-w} = \frac {60(160-w) + 60(160 + w)} {(160 + w)(160 - w)} = \frac {9600 - 60w + 9600 + 60w}{(160 + w)(160 - w)}= \frac {19200}{(160 + w)(160 - w)} = \frac {19200}{25600 +160w-160w-w^2}=\frac {19200}{25600-w^2}$$
For this scenario of finding round trip time where you have 2 islands, your wind direction is always parallel to your flight path, and the wind is always in the same direction you could reduce it to this formula.
r=rate (in this instance 160 km/h) w=wind speed (assuming here is in units km/h) d=km between the 2 islands
$$T = \frac {d(r-w) + d(r +w)}{(r+w)(r-w)} = \frac {dr+dr} {r^2-w^2} = \frac {2dr}{r^2-w^2}$$
If we take the problem with no wind.
r= 160 km/h w=wind speed (assuming here is in units km/h) d=60km $$\frac {(2)(60)(160)}{160^2-0^2} = \frac {19200}{25600} = \frac {3}{4}$$ of an hour
Problem with 30 km/h wind speed r= 160 km/h w=30km/h d=60km $$\frac {(2)(60)(160)}{160^2-30^2} = \frac {19200}{25600-900}= \frac {19200}{24700} = \frac {192}{247}$$ of an hour = ~$$0.777$$ of an hour
The following is multiple choice question (with options) to answer.
A plane flies 720 km with the wind and 630 km against the wind in the same length of time. If the speed of the wind is 15 km/h, what is the speed of the plane in still air? | [
"205 km/h",
"215 km/h",
"225 km/h",
"235 km/h"
] | C | The speed of the plane in still air = x km/h
The speed of the wind is 15 km/h
Speed with the wind = (x + 15) km/h
Speed against the wind = (x – 15) km/h
Time = Distance/ Speed
720 / (x+15) = 630 / (x-15)
720(x-15) = 630(x+15)
72x - 1080 = 63x + 945
9x = 2025
x = 225
Therefore, the speed of the plane in still air is 225 km/h.
The answer is C. |
AQUA-RAT | AQUA-RAT-38485 | 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.
Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km? | [
"328",
"288",
"2800",
"600"
] | D | Distance covered by Amar = 18/4.8 (1.6km)
= 3/8(1600)
= 600 m
Answer:D |
AQUA-RAT | AQUA-RAT-38486 | # Find the least next N-digit number with the same sum of digits.
Given a number of N-digits A, I want to find the next least N-digit number B having the same sum of digits as A, if such a number exists. The original number A can start with a 0. For ex: A-> 111 then B-> 120, A->09999 B-> 18999, A->999 then B-> doesn't exist.
You get the required number by adding $9$, $90$, $900$ etc to $A$, depending on the digits of $A$.
First Case If $A$ does not end in a row of $9$s find the first (starting at the units end) non-zero digit. Write a $9$ under that digit and $0$s under all digits to the right of it. Add the two and you get $B$.
Example: $A=3450$. The first non-zero digit is the 5 so we write a $9$ under that and a $0$ to its right and add:
\begin{align} 3450\\ 90\\ \hline 3540 \end{align}
There is a problem if the digit to the left of the chosen non-zero digit is a $9$. In this case we write a $9$ under that $9$ and $0$s to its right. And if there are several $9$ we put our $9$ under the highest one.
Example: $A=3950$. The first non-zero digit is the 5 but there is a $9$ to its left. We write a $9$ under that $9$ instead and $0$s to its right and add:
\begin{align} 3950\\ 900\\ \hline 4850 \end{align}
Second case If $A$ does end in a row of $9$s write a $9$ under the highest of the row of $9$s and $0$s under all digits to the right of it. Add the two and you get $B$. As you say, if $A$ is entirely $9$s there is no solution.
The following is multiple choice question (with options) to answer.
The sum of the digits of a two digit number is 7. When the digits are reversed then the number is decreased by 9. Then find the number | [
"43",
"34",
"44",
"35"
] | B | Let the original no be 10x+y
Sum of the digits x+y=7
When the digits are reversed then it is 10y+x
Since it is decreased by 9 it is written as (10y+x)-(10x+y)=9 solving this we get y=4
Substitute in x+y=7 we get x=3
so the number is 34 correct ans is B |
AQUA-RAT | AQUA-RAT-38487 | (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.
Craig had 200 apples. He lost 20 apples and sold 43 apples. A customer returned one apple. How many apples did he have left? | [
"137",
"138",
"135",
"140"
] | B | For lost, use negative. For gain, use positive.
200+-20+-43+1
correct answer is B)138 |
AQUA-RAT | AQUA-RAT-38488 | You can use sample() and select specific probabilities for each integer. If you sum the product of the probabilities and the integers, you get the expected value of the distribution. So, if you have a mean value in mind, say $$k$$, you can solve the following equation: $$k = 1\times P(1) + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ You can arbitrarily choose two of the probabilities and solve for the third, which determines the fourth (because $$P(1)=1-(P(2)+P(3)+P(4))$$ because the probabilities must sum to $$1$$). For example, let $$k=2.3$$, $$P(4)=.1$$, and $$P(3)=.2$$. Then we have that $$k = 1 \times [1-(P(2)+P(3)+P(4)] + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ $$2.3 = [1 - (P(2)+.1+.2)] + 2*P(2) + 3\times .2 + 4\times .1$$ $$2.3 = .7 + P(2) + .6 + .4$$ $$P(2)=.6$$ $$P(1)=1-(P(2)+P(3)+P(4)=1 - (.6+.1+.2)=.1$$
So you can run x <- sample(c(1, 2, 3, 4), 1e6, replace = TRUE, prob = c(.1, .6, .2, .1)) and mean(x) is approximately $$2.3$$
The following is multiple choice question (with options) to answer.
The value of x is to be randomly selected from the integers from 1 to 8, inclusive, and then substituted into the equation y = x^2 - 4x + 3. What is the probability that the value of y will be negative? | [
"1/4",
"1/8",
"1/2",
"3/8"
] | B | y will only be negative for x=2.
(We can check the values from 1 to 8 to be certain.)
P(y is negative) = 1/8
The answer is B. |
AQUA-RAT | AQUA-RAT-38489 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man buy a book in Rs50 & sale it Rs70. What is the rate of profit ??? | [
"10%",
"40%",
"30%",
"25%"
] | B | cp=50
sp=70
profit=70-50=20
%=20/50*100=40%
ANSWER:B |
AQUA-RAT | AQUA-RAT-38490 | ### Show Tags
05 Sep 2017, 11:01
2
In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hoursat 60 mph. What was his average speed for the whole trip?
(A) 50
(B) 53.5
(C) 55
(D) 56
(E) 57.5
The total distance traveled from City A to City B is 50*1 + 60*3 = 230
Since John drove for 4 hours to travel this distance,
the average speed of the whole trip is $$\frac{Distance}{Time} = \frac{230}{4}$$ = 57.5(Option E)
_________________
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Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink]
### Show Tags
17 Jan 2019, 03:38
How does this work using the harmonic mean?
$$\frac{1/50+3/60}{4}$$=$$\frac{7/100}{4}$$=57.1
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink] 17 Jan 2019, 03:38
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# In traveling from city A to city B, John drove for 1 hour at 50 mph an
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The following is multiple choice question (with options) to answer.
John starts his journey from A to B and Tom starts his journey from B to A at the same time.After passing each other they complete their journeys in 3 1/3 and 4 4/5 hours respectively.Find the speed of Tom if John traveled at 12 miles/hr. | [
"a) 8.33 m/hr",
"b) 9 m/hr",
"c) 10 m/hr",
"d) 20 m/hr"
] | C | Your calculations look perfect, but I think they answer a different question from the one the question designer intended. I don't find the wording of this question to be good at all, but I'm pretty sure what they mean is this: from the point in time at which John and Tom meet, it takes a *further* 3 1/3 hours for John to complete his journey, and a *further* 4 4/5 hours for Tom to complete his journey. So if it takes them t hours to meet, John will end up traveling for a total of t + 10/3 hours, and Tom a total of t + 24/5 hours. That's a more difficult problem than the one you solved. I don't have time to write up a solution at the moment, but I can later if that would be helpful.
C) |
AQUA-RAT | AQUA-RAT-38491 | life easy!, 456/789 ) thirty seconds fraction converted in to decimals an percent shown! + 1/3 = ( 1×3+1×2 ) / ( 2×3 ) = 5/6 learn to add, subtract,,... Use the calculator provided returns fraction inputs in both cases, fractions are presented their. 'S a division as 1 3/2 easily subtract mixed fractions and mixed fractions, select operators! Answers are provided in simplified, mixed fractions, select arithmetic operators drop-downs! To a mixed number calculator ' study math with us and make sure that Mathematics easy. Division and comparison of two fractions … multiply mixed numbers with space fraction represents the relationship between two ;! Cases, fractions, including comparison in this case, you can enter to! \\ ( { 5 \over 4 } \\ ) multiply, and press calculate change a mixed fraction calculator solve... Integers, decimals, and press calculate factor ( GCF ) see the greatest common factor calculator 1 ) mixed. A complete package of utilities that will make your calculation easy by this mixed fraction ( called., \\ ( { 11 \over 4 mixed fraction calculator \\ ) operation you want to perform math operations on proper., input values in this case, you will need to divide fractions are described below hand can. To change a mixed number to a mixed fraction ( also called number. Negative fractions ( e.g., 456/789 ), Compare and order fractions multiply mixed numbers find! Want to use the slash “ / ” between the whole number is part. Fractions online math solver with step-by-step solutions of whole number, be sure to leave a space between whole. Is that it supports two kinds of fraction math problems it ’ s simplest form than you! Fraction math problems terms use our simplify fractions calculator, simplify, Compare Mix or Compound fractions a! { 11 \over 4 } \\ ) simplifies the result fraction converted in to decimals percent... Find the addition and subtraction of mixed fraction calculator to do the four operations... By hand you can easily subtract mixed fractions online five and three thirty seconds 2 or fractions. Represents the relationship between two integers ; practically it 's a division
The following is multiple choice question (with options) to answer.
find the number, difference between number and its 3/5 is 50. | [
"120",
"123",
"124",
"125"
] | D | Explanation:
Let the number = x,
Then, x-(3/5)x = 50,
=> (2/5)x = 50 => 2x = 50*5,
=> x = 125
Answer: Option D |
AQUA-RAT | AQUA-RAT-38492 | Hope it helps.
Hi Bunuel,
Can we say that the sum of the numbers will be 1111 to 4444 ? since the repetition of the digits are allowed. I think this range would cover all the numbers. Then we can simply use the formula for arithmetic mean to calculate the sum ? Please correct my thinking if wrong.
Yes, the numbers range from 1111 to 4444: 1111, 1112, 1113, 1114, 1121, 1121, ..., 4444. But the numbers are NOT evenly spaced so you cannot get the sum by multiplying the mean by the number of terms.
Does this make sense?[/quote]
Ah yes , got it thanks.
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Joined: 18 Sep 2014
Posts: 223
Re: What is the sum of all 4-digit numbers that can be formed [#permalink]
### Show Tags
27 Mar 2015, 02:35
4
dimitri92 wrote:
I am sorry I don't have the OA. But I think it is solvable without the OA
What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?
As each digit of four digit number (abcd) can take 4 options (1, 2, 3, 4) total there can be 4^4 numbers.
Obviously each digit (a, b, c, and d) will take the value of 1, 2, 3, 4 equal number of times, so each digit will take the value of 1, 2, 3, 4 - $$\frac{4^4}{4}=4^3=64$$ times: units digit will take the values of 1, 2, 3, 4 - 64 times and the same with tens, hundreds, thousands digits.
So the sum would be 64*(1+2+3+4)+64*10*(1+2+3+4)+64*100*(1+2+3+4)+64*1000*(1+2+3+4)=64*10*(1+10+100+1000)=711040.
Actually there is the direct formula for this kind of problems. Of course it's better to understand the concept, then to memorize the formula but in case someone is interested here it is:
The following is multiple choice question (with options) to answer.
A four digit number is such that the product of all of its digits is 126. The sum of all the digits is equal to the 2 digit number formed by using thousands digit and tens digit (Thousand digit in tens place & ten digit in units place) which in turn is equal to 19. Then difference of units and thousand place of the number is, given that this difference is positive. | [
"1692",
"1792",
"1795",
"1892"
] | B | the number is 1792.
Let the number be represented as 1000x+100y+10z+t
acc to ques,
xyzt=126
x + y + z + t = 10x + z = 19
so, number in thousands and tens place are 1 and 9 respectively.
the remaining factor is 14 which is 7*2
So the number is 1792.
ANSWER:B |
AQUA-RAT | AQUA-RAT-38493 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
In a group of hats consisting of only blue hats, green hats, and purple hats, the ratio of blue hats to green hats to purple hats is 7:4:12. If there are a total of 115 hats in this group, how many of these hats are not blue? | [
"28",
"42",
"48",
"80"
] | D | Since the hats Blue, Green and Purple are in ratio of 7:4:12...The total no. of balls will be
7x+4x+12x=115 or 23x=115...here 7x,4x and 12x represent hats of each type
23x=115 so x=5...Blue hats =7*5=35...Not blue will be 115-35=80.
Ans D |
AQUA-RAT | AQUA-RAT-38494 | Example 2
In a large population of adults, 45% have a post secondary degree.
If people are selected at random from this population,
a) what is the probability that the third person selected is the first one that has a post secondary degree?
b) what is the probability that the first person with a post secondary degree is randomly selected on or before the 4th selection?
Solution to Example 2
a)
Let "having post secondary degree" be a "success". If a person from this population is selected at random, the probability of "having post secondary degree" is $p = 45\% = 0.45$ and "not having post secondary degree" (failure) is $1 - p = 1 - 0.45 = 0.55$
Selecting a person from a large population is a trial and these trials may be assumed to be independent. This is a geometric probability problem. Hence
$P(X = 3) = (1-0.45)^2 (0.45) = 0.1361$.
b)
On or before the 4th is selected means either the first, second, third or fourth person. The probability may be written as
$P(X \le 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
Substitute by the formula $P(X = x) = (1 - 0.45)^{x-1} 0.45$ to write
$P(X \le 4) = (1 - 0.45)^{1-1} 0.45 + (1 - 0.45)^{2-1} 0.45 + (1 - 0.45)^{3-1} 0.45 + (1 - 0.45)^{4-1} 0.45 = 0.9085$
## Sums of the terms of a Geometric sequence
The following is multiple choice question (with options) to answer.
In country Z, 10% of the people do not have a university diploma but have the job of their choice, and 15% of the people who do not have the job of their choice have a university diploma. If 40% of the people have the job of their choice, what percent of the people have a university diploma? | [
"39%",
"45%",
"55%",
"65%"
] | A | Setting up a matrix is how I solve this one.
Diploma No Diploma Totals
Job of Choice w/Diploma
Job of Choice w/o Diploma = 10%
Job of Choice Total = 40%
Not Job of Choice with Diploma =.15X
Not Job of Choice w/o Diploma= .85X
Total Not Job of Choice = X
Total with Diploma
Total without Diploma
Total citizen = 100
If 40% of people have their job of choice, then 60% of people do NOT have their job of choice. 15% of 60% = 9%. We can also see that 30% of the people have their job of choice and a diploma (40%-10%=30%). 30% + 9% = 39%. Therefore 39% of the people in Country Z have a diploma.
Ans A |
AQUA-RAT | AQUA-RAT-38495 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
if a person has to divide equally $13011 among all his 12 people then how much minimum should he keep with himself so that he divides equally among all 12 people in whole numbers? | [
"$1",
"$2",
"$3",
"$4"
] | C | He would need to keep $ 3 to himself so he can give $ 1084 to all 12 people so option C is correct one |
AQUA-RAT | AQUA-RAT-38496 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
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Joined: 06 Sep 2013
Posts: 1647
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Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
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If a portion of a half water/half alcohol mix is replaced [#permalink]
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02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
The following is multiple choice question (with options) to answer.
if 50% of (x-y) = 40% of (x+y) then what percent of x is y? | [
"2.5%",
"11%",
"5%",
"15%"
] | B | 50% of (x-y)=40% of(x+y)
(50/100)(x-y)=(40/100)(x+y)
5(x-y)=4(x+y)
x=9y
x=9y
therefore required percentage
=((y/x) X 100)% = ((y/9y) X 100) =11%
Answer is B. |
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