source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35197 | ### 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.
If the price of a certain computer increased 30 percent from b dollars to 351 dollars, then 2b = | [
"540",
"570",
"619",
"649"
] | A | Before Price increase Price = b
After 30% Price increase Price = b+(30/100)*b= 1.3b = 351 (Given)
i.e. b= 351/1.3 = $270
i.e. 2b = 2*270 = 540
Answer: option A |
AQUA-RAT | AQUA-RAT-35198 | dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
_________________
-----------------------
tusharvk
Manager
Joined: 27 Oct 2008
Posts: 180
Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
The following is multiple choice question (with options) to answer.
Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 20 white marbles. How many red marbles could be in bag A? | [
"1",
"3",
"4",
"6"
] | C | 6 is the answer.
Bag A-
R:W:B = 2:6:9
Let W in bag A be 6K
Bab B -
R:W = 1:4
Let W in bag B be 4k
W = 20 = 6K+4k
k=2
Total Red's in bag A will be 2K = 4
C |
AQUA-RAT | AQUA-RAT-35199 | gravity, terminology, acceleration, notation
Title: What does dimensionless quantity 'number of $g$' mean? I am doing data analysis in which I found a quantity named "no. of $g$".
I don't know what it means or what is its usage.
Look at the image below.
I want to know the meaning and usage of "no. of $g$". The acceleration due to gravity near the Earth's surface is often denoted $g$.
Given any other acceleration $a$, we call $a/g$ the "number of $g$'s" because it is just the number you multiply by $g$ to get $a$.
For example, if we say an acceleration $a$ is "2 $g$'s" then that means
$$a = 2 \times g = 2 \times 9.8 \text{m}/\text{s}^2 = 19.6 \text{m}/\text{s}^2 \, .$$
The following is multiple choice question (with options) to answer.
On a scale that measures the intensity of a certain phenomenon, a reading of g+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of g. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3? | [
" 5",
" 50",
" 10^5",
" 5^10"
] | C | To solve this problem we need to examine the information in the first sentence. We are told that “a reading of g + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of g.”
Let’s practice this idea with some real numbers. Let’s say g is 2. This means that g + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.
Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.
Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.
We have found a pattern, which can be applied to the problem presented in the stem:
3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.
4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.
5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.
We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.
Answer C. |
AQUA-RAT | AQUA-RAT-35200 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
17 men take 21 days of 8 hours each to do a piece of work. How many days of 6 hours each would 21 women take to do the same. If 3 women do as much work as 2 men? | [
"34",
"87",
"30",
"99"
] | A | 3W = 2M
17M ------ 21 * 8 hours
21 W ------ x * 6 hours
14 M ------ x * 6
17 * 21 * 8 = 14 * x * 6
x = 34
Answer: A |
AQUA-RAT | AQUA-RAT-35201 | greatest common divisor
cehsu
104 views
Greatest Common Divisor
In mathematics, the greatest common divisor (gcd) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers. For example, the gcd of 8 and 12 is 4.
The greatest common divisor is also known as the greatest common factor (gcf), highest common factor (hcf), greatest common measure (gcm), or highest common divisor.
Check for the gcd
Write a getGCD function that takes two numbers as parameters and returns the gcd.
Sample Input and Output
Input: 8 12 Output: 4
Optimization
Check out Euclid's algorithm and see how much things speed up. How can you account for this in terms of time complexity?
See:
https://en.wikipedia.org/wiki/Euclidean_algorithm#Worked_example
https://en.wikipedia.org/wiki/Greatest_common_divisor#Complexity_of_Euclidean_method
Write the getGCD function
1
2
3
4
5
6
7
8
function getGCD (numOne, numTwo) {
}
module.exports = getGCD;
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Open Source Your Knowledge: become a Contributor and help others learn.
The following is multiple choice question (with options) to answer.
If n is a positive integer and the greatest common divisor of n and 16 is 4, and the greatest common divisor of n and 15 is 3, which of the following is the greatest common divisor of n and 96? | [
"20",
"22",
"24",
"12"
] | D | The greatest common divisor of n and 16 is 4 --> n is a multiple of 4 but not a multiple of 8.
The greatest common divisor of n and 15 is 3 --> n is a multiple of 3 but not a multiple of 5.
96=2*3*4*4 is divisible by 12. therefore the greatest common divisor is 12
D |
AQUA-RAT | AQUA-RAT-35202 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A is thrice efficient as B and C is twice as efficient as B. what is the ratio of number of days taken by A,B and C, when they work individually? | [
"2:6:3",
"2:6:9",
"2:6:6",
"2:6:1"
] | A | Answer: A) 2:6:3 |
AQUA-RAT | AQUA-RAT-35203 | ### 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, c are consecutive positive integers and a<b<c, which of the following must be true?
I. c-a=2
II. abc is an even integer
III. a-b is positive integer | [
"I",
"II",
"I and II",
"II and III"
] | C | If a, b, c are consequitive positive integers and a<b<c, which of the following must be true?
1. c-a=2
2. abc is an even integer
3. (a+b+c) is an integer
1.
a=a
b=a+1
c=a+2
c-a=a+2-a=2.
Correct
2. abc
a can be either even or odd
if a=even; abc=even;
if a=odd,b=even; abc=even
Rule:
multiple of any number of integers if atleast once multiplied by an even number will result in even.
even*even=even
even*odd=even
Correct
3. a-b = 1-2 = -1 => negative
Answer : C |
AQUA-RAT | AQUA-RAT-35204 | # How to calculate the area covered by any spherical rectangle?
Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a sphere with a radius $R$?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
• What do you mean by "rectangle"? Do you just want a quadrilateral, or some stronger condition? (Note that you cannot have four right angles in a quadrilateral on the sphere.) – Nick Matteo Mar 25 '15 at 13:16
• Yes, a quadrilateral having equal opposite sides (but not parallel) each as a great circle arc on a sphere. – Harish Chandra Rajpoot Mar 25 '15 at 13:21
• What do you mean by length and width? As the sides are not parallel, we don't have the usual idea of width and length. Do you simply mean the lengths of the sides? – robjohn Apr 4 '15 at 14:12
• Yes, length & width are simply the sides of the rectangle as great circles arcs on the spherical surface. – Harish Chandra Rajpoot May 28 '15 at 23:56
Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides $$w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2$$
The following is multiple choice question (with options) to answer.
The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 4900 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units? | [
"140 sq.units",
"280 sq.units",
"167sq.units",
"178sq.units"
] | B | Given that the area of the square = 4900 sq.units
=> Side of square = √4900 = 70 units
The radius of the circle = side of the square = 70 units Length of the rectangle = 2/5 * 70 = 28 units
Given that breadth = 10 units
Area of the rectangle = lb = 28 * 10 = 280 sq.units
Answer:B |
AQUA-RAT | AQUA-RAT-35205 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
In a Question paper there are 4 subjects. Each subject has has 5 question sets with each subject having one single question set that is the toughest of all. What is the total number of ways in which a candidate will not get all the four toughest question sets? | [
"19",
"85",
"120",
"624"
] | D | A candidate can get a question set in 5*5*5*5=5^4 number of ways (each aubject has 5 question sets and we have total of 4 subjects). Now, out of these cases there will be only one case when the candidate got all the four toughest question sets. Therefore the total number of ways in which a candidate will NOT get all the four toughest question set is is 5^4-1=624.
Answer: D. |
AQUA-RAT | AQUA-RAT-35206 | 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.
A card is randomly drawn from a deck of 52 cards. What is the probability getting an Ace or King or Queen? | [
"7/13",
"9/13",
"3/13",
"4/13"
] | C | Total number of cards = 52
Total number of Ace cards = 4
P(Ace) = 4/52=1/13
P(King) = 4/52=1/13
P(Queen) = 4/52=1/13
P(Ace or King or Queen) = P (Ace) + P (King)+ P(Queen)
=1/13+1/13+1/13=3/13
Answer C |
AQUA-RAT | AQUA-RAT-35207 | five -- is it 0.5 or five? -- a 0.5 kilogram block is placed on top of the polystyrene, and what fraction will be above the water now? Before, there is 90 percent above the water and now with this block on top what fraction will be above the water now? So the buoyant force is the weight of the fluid displaced, Archimedes principle, because the thing is floating that's going to equal the weight of the thing, so that's the mass of the polystyrene plus the mass of the block, times g. So we have this expression now, mass of the water displaced is the total mass of the block and polystyrene, and density of water is this as we've seen before. We can solve this for V w by multiplying by V w over density of water and we get the volume of water displaced is mass of the water divided by density of water. That's going to equal the fraction of the block that is submerged, times the total volume of the block. So we're solving for b here, the fraction submerged because this amount by which it's submerged, this is the volume submerged, is going to equal the volume of water displaced. So we have b then after we divide both sides by V p here, is the mass of the water divided by density of water times V p. Now the mass of the water is mass of the polystyrene block plus the mass of the block. Then we substitute for mass of the polystyrene which is density of polystyrene times its volume, and then we can divide this denominator into both terms on the top here. We get the ratio of the densities of polystyrene to water plus the mass of the block divided by density of water times volume of the polystyrene. This works out to 100 kilograms per cubic meter divided by 1000 kilograms per cubic mete plus 0.5 kilograms, mass of the block, divided by 1000 kilograms per cubic meter times 0.1 meter cubed, this is the volume of the polystyrene, this works out to 0.6. So 60 percent of the polystyrene block is submerged which means the percent above is 100 minus 60 which is 40 percent. That's 40 percent of the block, polystyrene, is above the water. Then if the fluid was not
The following is multiple choice question (with options) to answer.
What percent of 5.2 kg is 16 gms ? | [
"30",
"66",
"58",
"29"
] | A | Explanation:
Required percentage = (16/5200 * 100)% = 3/10% = 0.30%
Answer: A) .30% |
AQUA-RAT | AQUA-RAT-35208 | Joined: Oct 2011
Posts: 13,632
Thanks: 954
Quote:
Originally Posted by xoritos In a bakery, (1/4) of the muffins for sale are banana muffins, (2/5) are walnut muffins and the rest are chocolate muffins. There are 18 more chocolate muffins than banana muffins. How many muffins are there for sale altogether? Number of chocolate = c Number of walnut = w Number of banana = b
Total muffins = t
b = t/4
w = 2t/5
c = b+18
c = t/4 + 18 [1]
also:
c = t - b - w
c = t - t/4 - 2t/5 [2]
[1] = [2]
t/4 + 18 = t - t/4 - 2t/5
Solve for t
September 19th, 2018, 01:58 PM #4
Senior Member
Joined: May 2016
From: USA
Posts: 1,210
Thanks: 497
The following is multiple choice question (with options) to answer.
On rainy mornings, Mo drinks exactly N cups of hot chocolate (assume that N is an integer). On mornings that are not rainy, Mo drinks exactly 4 cups of tea. Last week Mo drank a total of 26 cups of tea and hot chocolate together. If during that week Mo drank 14 more tea cups than hot chocolate cups, then how many rainy days were there last week? | [
"2",
"3",
"4",
"5"
] | A | T= the number of cups of tea
C= the number of cups of hot chocolate
T+C = 26 T-C=14 -> T= 20. C=6.
Mo drinks 4 cups of tea a day then number of days that are not rainy = 20/4 = 5
So number of rainy days = 7-5 = 2
A is the answer. |
AQUA-RAT | AQUA-RAT-35209 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Carl can wash all the windows of his house in 7 hours. His wife Maggie can wash all the windows in 4 hours. How many hours will it take for both of them working together to wash all the windows? | [
"2",
"2 1/4",
"11 6/2",
"4 1/2"
] | C | Work hrs=AB/(A+B)= 28/11 =11 6/2
Answer is C |
AQUA-RAT | AQUA-RAT-35210 | => 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.
If Sharon's weekly salary increased by 16 percent, she would earn $406 per week. If instead, her weekly salary were to decreased by 10 percent, how much would she earn per week? | [
"Rs.415",
"Rs.325",
"Rs.352",
"Rs.315"
] | D | Soln:-
(406/116)90 =315
In this case long division does not take much time.
(406/116)=3.5
3.5*90=Rs.315
ANSWER:D |
AQUA-RAT | AQUA-RAT-35211 | # Find the smallest positive number k
For any positive number $$n$$, let $$a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.
I have $$a_1=\sqrt 2$$ and $$a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N$$
Consider
$$a_1=\sqrt 2 \lt 2$$
$$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$$
$$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$$
Use Mathematical Induction, I conclude $$\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$$
Thus, $$3-a_{n+1}\gt1$$ and $$7-a_{n}\gt 5$$
Since $$k\in\mathbb N$$, I have $$\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$$
Hence, $$3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$$
Therefore $$k=5$$
Please check my solution, Is it correct?, Thank you
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 3:5:7. The largest number value is 70. Find difference between Smallest & largest number is? | [
"40",
"45",
"35",
"30"
] | A | == 3:5:7
Total parts = 15
= The largest number value is 70
= The largest number is = 7
= Then 7 parts -----> 70 ( 7 * 10 = 70 )
= smallest number = 3 & Largest number = 7
= Difference between smallest number & largest number is = 7 - 3 = 4
= Then 4 parts -----> 40 (4 * 10 = 40)
A |
AQUA-RAT | AQUA-RAT-35212 | 94
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 7 13 37 283 344887 cdot mbox{BIG}
392318858461667547739736838970286191634963299677388144641 = 3 7 cdot mbox{BIG}
94 GCD 21
95
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 31 37 21319 cdot mbox{BIG}
1569275433846670190958947355841530685282721029912780603393 = 7 151 32377 cdot mbox{BIG}
95 GCD 1 m is ODD
96
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 19 3169 8929 13249 52579 98641 333667 cdot mbox{BIG}
6277101735386680763835789423286894578616619782057578463233 = 3 19 37 73 109 433 577 1153 6337 38737 cdot mbox{BIG}
96 GCD 57
97
100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 37 1747 18043 cdot mbox{BIG}
25108406941546723055343157692989121989437950453043225952257 = 7 272959 cdot mbox{BIG}
97 GCD 1 m is ODD
98
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 7^3 13 37 43 127 1933 2689 63799 459691 cdot mbox{BIG}
100433627766186892221372630771639575307694744461798728007681 = 3 7^3 337 5419 748819 cdot mbox{BIG}
98 GCD 1029
99
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 = 3 757 55243 198397 cdot mbox{BIG}
401734511064747568885490523085924475930664863146446560428033 = 262657 cdot mbox{BIG}
99 GCD 1 m is ODD
The following is multiple choice question (with options) to answer.
106 × 106 + 94 × 94 = ? | [
"20072",
"20062",
"10072",
"20172"
] | A | Explanation :
(a + b)2 + (a − b)2 = 2(a2 + b2)
(Reference : Basic Algebraic Formulas)
1062 + 942 = (100 + 6)2 + (100 − 6)2 = 2(1002 + 62) = 2(10000 + 36) = 20072. Answer : Option A |
AQUA-RAT | AQUA-RAT-35213 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 110 m long is running with a speed of 40 km/h. In how many seconds will the train pass a man who is running at 4 km/h in the direction opposite to that in which the train is going? | [
"6",
"7",
"8",
"9"
] | D | The speed of the train relative to the man = 40 + 4 = 44 km/h.
44000 m/h * 1h/3600 s = (440/36) m/s
(110 m) / (440/36 m/s) = (110 * 36) / 440 = 36/4 = 9 seconds
The answer is D. |
AQUA-RAT | AQUA-RAT-35214 | Hence, the probability of selecting three balls of one color and two balls of a different color when five balls are selected from the $7 + 8 + 9 = 24$ balls in the bag is $$\frac{\dbinom{7}{3}\dbinom{8}{2} + \dbinom{7}{3}\dbinom{9}{2} + \dbinom{7}{2}\dbinom{8}{3} + \dbinom{7}{2}\dbinom{9}{3} + \dbinom{8}{3}\dbinom{9}{2} + \dbinom{8}{2}\dbinom{9}{3}}{\dbinom{24}{5}}$$
• I was wondering whether you could have any cases where you would draw 3 blue balls, 1 red, and 1 white ball? – user262291 Apr 9 '16 at 15:10
• That's a different question since choosing three of one color and two of another color when five balls are selected precludes the possibility of selecting three different colors. The probability of selecting three blue balls, one red ball, and one white ball in the first scenario is $$\frac{\binom{7}{3}\binom{7}{1}\binom{7}{1}}{\binom{21}{5}}$$ In the second scenario, the probability is $$\frac{\binom{9}{3}\binom{7}{1}\binom{8}{1}}{\binom{24}{5}}$$ The second scenario makes it clearer what I am counting, namely selections of three blue balls, one red ball, and one white ball. – N. F. Taussig Apr 9 '16 at 18:15
Whenever you can, relate the problem to a known category of familiar problems.
Here, you can treat it the way you compute poker probabilities:
We want $3-2$ of a kind, so numerator will be [Choose kinds] $\times$ [Choose balls from each kind]
The following is multiple choice question (with options) to answer.
A certain bag contains 100 balls — 50 white, 30 green, 8 yellow, 9 red, and 3 purple. If a ball is to be chosen at random, what is the probability that the ball will be neither red nor purple? | [
"0.9",
"0.75",
"0.6",
"0.88"
] | D | According to the stem the ball can be white, green or yellow, so the probability is (white + green + yellow)/(total) = (50 + 30 + 8)/100 = 88/100 = 0.88.
Answer is D |
AQUA-RAT | AQUA-RAT-35215 | 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 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time? | [
"40 Kmph",
"52 Kmph",
"60 Kmph",
"76 Kmph"
] | C | Time = 6
Distance = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
C) |
AQUA-RAT | AQUA-RAT-35216 | Thanks
• Your "40% overlap" would mean taht the surface overlaping is 40% of the surface of which rectangle ? – Evargalo Sep 28 '17 at 16:13
• I want to know if, say, 40% of the other rectangle is contained inside the self rectangle – John Lexus Sep 28 '17 at 16:14
For the equations, I will let the left of the first rectangle be $l_0$, the right be $r_0$, the top $t_0$ and the bottom $b_0$. The second rectangle is $l_1, r_1,$ etc. Their areas will be $A_0$ and $A_1$.
If the boxes don't overlap, obviously the percentage overlap is $0$.
If your boxes are found to be colliding, simply use this formula to calculate the area that is overlapping:
$$A_{overlap} = (\max(l_0, l_1)-\min(r_0, r_1))\cdot(\max(t_0, t_1)-\min(b_0, b_1)).$$
Now there are two ways to calculate a percentage error that could make sense for your explanation. If we're just checking the percentage within the first rectangle (or "self" in your program), the percent overlap is simple: $$P_{overlap} = \frac{A_{overlap}}{A_{self}}.$$
If you want the percentage to be equal whether it's calculated from either rectangle, the equation you're looking for is: $$\frac{A_{overlap}}{A_0+A_1-A_{overlap}}.$$
The following is multiple choice question (with options) to answer.
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is: | [
"32%",
"34%",
"42%",
"44%"
] | D | Explanation:
Let original length = x metres and original breadth = y metres.
Original area =xy m2New Length =120/100x=6/5x
New Breadth =120/100y=6/5y
=>New Area =6/5x∗6/5y
=>New Area =36/25xy
Area Difference=36/25xy−xy
=11/25xy
Increase%=Differnce/Actual∗100
=11xy/25∗1/xy∗100=44%
ANSWER IS D |
AQUA-RAT | AQUA-RAT-35217 | So, we can always stop when we reach the largest prime number smaller than the square root of the limit -- in this case, when we reach the largest prime number smaller than $\sqrt{200}$. (The square root, say $a$ has the property that $a\times a = 200$ so any other combination of factors must have one larger, one smaller).
$13\times 13 = 169 < 200$ and $17\times 17 = 289 > 200$, so we don't need to check as far as $17$ and D cannot be the answer, which leaves C.
Your reasoning is a bit "in reverse" when you are thinking about the 10. When you employed 2, the 10 & all of its multiples 20, 30, 40, et. al. were indeed eliminated. Note here that 25 was skipped. The fact that 5 is a factor of 10 makes no difference as that 5 never gets used. When actually using this method, its no problem at all to determine what number to use next: it's simply the next higher number not already marked out.
Doing the sieve with the primes in order, the first nontrivial multiple of $p$ to eliminate is $p^2$.
If $p$ is an odd prime, then $2p$ should have already been eliminated when you were crossing off the multiples of $2$. But $p^2$ is odd, not even, so it shouldn't have been eliminated at a prior step. Likewise, if $p$ is a prime greater than $3$, then you don't need to cross $2p$ and $3p$ off again because they should have already been crossed off.
Let's see what happens with D. You might forget to cross off $35$ if you skip ahead to $49$. And then at the end of the process you might come to the absurd conclusion that both $5$ and $25$, and possibly $35$ and $105$, are also prime, which is of course incorrect.
The following is multiple choice question (with options) to answer.
What is the greatest prime factor of 4^17 - 2^30? | [
"2",
"3",
"5",
"7"
] | C | 4^17 - 2^30
=(2^2)^17 - 2^30
= 2^34 - 2^30
= 2^30 (2^4 -1)
= 2^30 * (16-1)
= 2^30 * 15
= 2^30 * 3 * 5
The greatest prime factor is 5.
The answer is C. |
AQUA-RAT | AQUA-RAT-35218 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 10 hrs and pipe B can fill it in 8 hrs. If both the pipes are opened in the empty tank.There is an outlet pipe in 3/4 th of the tank. In how many hours will it be fill 3/4th of that tank? | [
"3 hr",
"3.10hr",
"3.15 hr",
"3.20 hr"
] | D | Part filled A in 1 hr= (1/10)
Part filled B in 1 hr= (1/8)
Part filled by (A+B) together in 1 hr=(1/10)+(1/8)=18/80
So, The tank will be full in 80/18 hrs.
Time taken to fill Exact 3/4th of the tank = (80/18) *(3/4) = 3.20 hrs
Answer : D |
AQUA-RAT | AQUA-RAT-35219 | agriculture
Title: What does "permanent field" mean in agriculture? I am reading a book that in a paragraph talks about the agricultural methods used in prehistoric Finland.
The further north and east, the more extensive the amount of
burn-beat cultivation, which was a far from primitive form of
agriculture. The yield was many times higher (twenty- to thirty-fold)
than on permanent fields (five- to ten-fold), and there were multiple
varieties of the technique
A history of Finland by Henrik Meinander.
One of them is burn-beating. Like I understand, in burn-beating people cut down the trees in the forests and burn the topsoil. This way they can use that soil for 3 to 6 years for cultivation.
The other method is permanent field. I have searched the internet and the result I got was "permanent crops", like here. In which case people planted trees once in a field and harvested them multiple times.
But in another research about prehistoric Finland it was saying:
The site of Orijärvi shows that permanent field cultivation, with
hulled barley as the main crop was conducted from approximately cal AD 600 onwards.
The following is multiple choice question (with options) to answer.
If a farmer wants to plough a farm field on time, he must plough 160 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially? | [
"600",
"490",
"448",
"435"
] | C | Let x be the number of days in the initial plan. Therefore, the whole field is 160â‹…x hectares. The farmer had to work for x+2 days, and he ploughed 85(x+2) hectares, leaving 40 hectares unploughed. Then we have the equation:
160x=85(x+2)+40
75x=210
x=2.8
So the farmer planned to have the work done in 6 days, and the area of the farm field is 160(2.8)=448 hectares
correct answer C |
AQUA-RAT | AQUA-RAT-35220 | # Linear Algebra Money Question
bleedblue1234
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # $10 bills A+B+C = 32 1A+5B+10C = 100 ## The Attempt at a Solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. ## Answers and Replies E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. azizlwl You can narrow the selection.$1 can only be in a group of 5.
Homework Helper
This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.
Last edited by a moderator:
Homework Helper
Dearly Missed
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # \$10 bills
A+B+C = 32
1A+5B+10C = 100
## The Attempt at a Solution
The following is multiple choice question (with options) to answer.
A man has Rs. 176 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? | [
"33",
"60",
"75",
"90"
] | A | Let number of notes of each denomination be x.
Then x + 5x + 10x = 176
16x = 176
x = 11.
Hence, total number of notes = 3x = 33.
Answer: Option A |
AQUA-RAT | AQUA-RAT-35221 | =2.8/(0.2+1+2.8)
=2.8/4=0.7
Concept: Bayes’ Theorem
Is there an error in this question or solution?
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The following is multiple choice question (with options) to answer.
The number 0.8 is how much greater than 1/2 ? | [
"½",
"3/10",
"1/50",
"1/500"
] | B | let x be the difference
then
.8-1/2=x
8/10-1/2=x
x= 3/10
ans B |
AQUA-RAT | AQUA-RAT-35222 | Determine the value of each expression.
###### 7.
$4^{1/2}$
Solution
\begin{aligned}[t] 4^{1/2}\amp=\sqrt{4}\\ \amp=2 \end{aligned}
###### 8.
$27^{-1/3}$
Solution
\begin{aligned}[t] 27^{-1/3}\amp=\frac{1}{27^{1/3}}\\ \amp=\frac{1}{\sqrt[3]{27}}\\ \amp=\frac{1}{3} \end{aligned}
###### 9.
$\left(\frac{4}{9}\right)^{-1/2}$
Solution
\begin{aligned}[t] \left(\frac{4}{9}\right)^{-1/2}\amp=\left(\frac{9}{4}\right)^{1/2}\\ \amp=\sqrt{\frac{9}{4}}\\ \amp=\frac{3}{2} \end{aligned}
###### 10.
$8^{7/3}$
Solution
\begin{aligned}[t] 8^{7/3}\amp=(\sqrt[3]{8})^7\\ \amp=2^7\\ \amp=128 \end{aligned}
###### 11.
$100^{5/2}$
Solution
\begin{aligned}[t] 100^{5/2}\amp=(\sqrt{100})^5\\ \amp=10^5\\ \amp=100,000 \end{aligned}
###### 12.
$16^{-9/4}$
Solution
The following is multiple choice question (with options) to answer.
Find the value of x in each of the following equation: [(17.28/x) / (3.6*0.2)] = 2 | [
"12",
"14",
"16",
"18"
] | A | (17.28/x) = 2*3.6*0.2 x = (17.28/1.44) = (1728/14) = 12.
Answer is A. |
AQUA-RAT | AQUA-RAT-35223 | The game is "fair".
Playing repeatedly, you can expect to break even.
4. thanks all for the answers.
but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?
The following is multiple choice question (with options) to answer.
Three men sit at a table. One dollar is placed in front of each of them. When a buzzer sounds, the first man to touch the money in front of him keeps the money that is in front of him, while the other two receive nothing. If none of the three men touches the money in front of them, then the amount of money doubles for the next round. After three rounds of this activity, what is the maximum amount of money that any one man could receive? | [
"$4",
"$5",
"$10",
"$15"
] | A | The Optimized scenario
After 1st Round - $1. Let nobody touches the amount will double in next round.
After 2nd Round - $2. Let Man A touches it and get$2 or nobody touches
After 3rd Round - $2 Let the same Man A touches it and get another$2.( $4 in case nobody touches in 2nd round).
So, after 3rd round, the maximum amount a man receives is $4.
Answer (A) |
AQUA-RAT | AQUA-RAT-35224 | # If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
If we select a random integer number of the set $[1000000]$ what is the probability of the number selected contains the digit $5$?
My work:
We know the sample space $S:$"The set of number of 1 to 1000000" and $|S|=1000000$
Let $E$ the event such that $E:$"The set of number contains the digit 5 in $[1000000]$" We need calculate $|E|$.
I know in $[100]$ we have $5,15,25,35,45,50,51,52,53,54,55,56,57,58,59,65,75,85,95$ then we have 19 numbers contains the digit $5$ in the set $[100]$
Then in $[1000]-[500]$ we have 171 numbers have the digit 5. this implies [1000] have 271 number contains the digit 5.
. . .
Following the previous reasoning we have to $[10000]$ have 3439 number contains the digit 5.
Then, $[100000]$have 40951 number contains the digit 5.
Moreover, $[1000000]$ have 468559 number contain the digit 5.
In consequence the probability of we pick a digit contain the number 5 in the set $[1000000]$ is 0.468
Is correct this?
How else could obtain $|E|$?
Thanks
The following is multiple choice question (with options) to answer.
Raffle tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random will have a number with a hundreds digit of 3? | [
"2/5",
"2/7",
"33/83",
"51/250"
] | D | SOLUTION
The number of integers from 101 to 350, inclusive is 250, out of which 100 (from 200 to 299) will have a hundreds digit of 2. Thus the probability is 51/250.
Answer: D |
AQUA-RAT | AQUA-RAT-35225 | Your proof is correct, but you should elaborate on why $x^2>20+x$. This inequality is the same as $(x-5)(x+4)=x^2-x-20>0$ which is true when $x>5$.
Alternatively, just use $y=\frac{x^2-25}{2}$ to lead to a quick proof.
• can you possibly elaborate on this, I think I see why this works, however, my professor was not satified with the above - that is, I had not sufficiently shown why (in my proof) that 20+x is less than $x^2$. I can see why taking the above works, I just can't articulate it correctly I guess. – Ben Anderson Feb 8 '13 at 16:35
The following is multiple choice question (with options) to answer.
What is x if x + 2y = 20 and y = 5? | [
"A)10",
"B)8",
"C)6",
"D)4"
] | A | x = 20 - 2y
x = 20 - 10.
x = 10
Answer : A |
AQUA-RAT | AQUA-RAT-35226 | ### 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.
If x dollars is invested at 12.5 percent for one year and y dollars is invested at 9.5 percent for one year, the annual income from the 12.5 percent investment will exceed the annual income from the 8 percent investment by $60. If $2500 is the total amount invested, how much is invested at 9.5 percent? | [
"a. $1097.83",
"b. $1231.57",
"c. $1101.52",
"d. $1147.72"
] | D | 2 equations with 2 unknowns
12.5x / 100 - 9.5y / 100 = 60
and
x + y = 2500
Solving these 2 equations, x =1352.27 and y = 1147.72
Answer D. |
AQUA-RAT | AQUA-RAT-35227 | Statement 2: $$a>20$$. Nothing mentioned about $$b$$. Insufficient
Combining 1 & 2: $$a>20 =>2a>40$$. Add $$35$$ to both sides
$$2a+35>40+35 => a+b>75$$. Sufficient
Option C
Is the average (arithmetic mean) of a and b greater than 30? [#permalink] 04 Jan 2018, 11:07
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The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of a and b is 45 and the average of b and c is 80, what is the value of c − a? | [
"25",
"60",
"90",
"140"
] | B | -(a + b = 90)
b + c=160
c-a=60
B. 60 |
AQUA-RAT | AQUA-RAT-35228 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Rs. 6000 is lent out in two parts. One part is lent at 2% p.a simple interest and the other is lent at 10% p.a simple interest. The total interest at the end of one year was Rs. 450. Find the ratio of the amounts lent at the lower rate and higher rate of interest? | [
"5:11",
"5:6",
"5:2",
"5:8"
] | A | Let the amount lent at 2% be Rs. x
Amount lent at 10% is Rs. (6000 - x)
Total interest for one year on the two sums lent
= 2/100 x + 10/100 (6000 - x) = 600 - 3x/100
=> 600 - 3/100 x = 450 => x = 1875
Amount lent at 10% = 4125
Required ratio = 5:11
Answer:A |
AQUA-RAT | AQUA-RAT-35229 | are these 2's supposed to be squares?
RonL
7. Originally Posted by CaptainBlack
this last line should be:
=> x = (k/3) pi for k an integer
RonL
right, i forgot about the 1/3
The following is multiple choice question (with options) to answer.
If y/x is an integer, which of the following must also be an integer?
I. xy
II. x/y
III. x | [
"I alone",
"II alone",
"None of the above",
"I and III"
] | C | Lets take X = 1/6 and Y = 2/3
Then Y/X = 4 which is an integer.
But XY = 1/6 * 2/3 = 1/9 --> Not an integer.
X/Y = 1/6 divided by 2/3 = 1/4 --> Not an integer.
X alone is 1/6. Not an integer.
Hence C. None of the above. |
AQUA-RAT | AQUA-RAT-35230 | If you need a tester program that calculate permutation from index or viceversa, you can see here. It can be useful and it's easy to use. It's based on factoradic.
As example: it allows to calculate the correct index corresponding to the solution "2783905614" mentioned earlier Or obtain the 2,000,000th permutation of S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } )
It works up to 17 elements (max index = 355,687,428,096,000)
The following is multiple choice question (with options) to answer.
Evaluate permutation
7P7 | [
"5010",
"5020",
"5030",
"5040"
] | D | Explanation:
nPn=n!
7P7=7*6*5∗4∗3∗2∗1=5040
Option D |
AQUA-RAT | AQUA-RAT-35231 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference between a two-digit number and the number obtained by interchanging the digit is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1:2? | [
"7",
"8",
"6",
"5"
] | B | Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let the ten's and unit's digits be 2x and x respectively.
Then, (10 * 2x + x) - (10x + 2x) = 36
9x = 36
x = 4
Required difference = (2x + x) - (2x - x) = 2x = 8.
Answer: B |
AQUA-RAT | AQUA-RAT-35232 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
In a certain boys camp, 20% of the total boys are from school A and 30% of those study science. If there are 77 boys in the camp that are from school A but do not study science then what is the total number of boys in the camp? | [
"550",
"245",
"150",
"35"
] | A | Since 30% of the boys from school A study science, then 70% of the boys from school A do NOT study science and since 20% of the total number of boys are from school A, then 0.2*0.7 = 0.14, or 14% of the boys in the camp are from school A and do not study science.
We are told that this number equals to 77, so 0.14*{total} = 77 --> {total} = 550.
Answer: A. |
AQUA-RAT | AQUA-RAT-35233 | ### 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.
An analyst will recommend a combination of 2 industrial stocks, 2 transportation stocks, and 2 utility stocks. If the analyst can choose from 5 industrial stocks, 4 transportation stocks, and 3 utility stocks, how many different combinations of 6 stocks are possible?
-- | [
"12",
"19",
"60",
"180"
] | D | 5C2 * 4C2 * 3C2 = 10*6*3 = 180.
Answer: D. |
AQUA-RAT | AQUA-RAT-35234 | c, linux
if (fd_vk < 0) {
fail("could not initialize virtual keyboard\n");
}
set_vk_evbits();
set_vk_keybits();
struct uinput_user_dev dev;
memset(&dev, 0, sizeof(dev));
snprintf(dev.name, UINPUT_MAX_NAME_SIZE, vk_name);
dev.id.bustype = BUS_USB;
dev.id.vendor = 1;
dev.id.product = 1;
dev.id.version = 1;
if ((res = write(fd_vk, &dev, sizeof(dev)) < 0)) {
fail("could not write virtual keyboard data: %d\n", res);
}
if ((res = ioctl(fd_vk, UI_DEV_CREATE)) < 0) {
fail("could create virtual keyboard: %d\n", res);
}
}
/**
* Destroys the virtual keyboard and closes the file descriptor.
*/
static void destroy_vk(void) {
if (fd_vk <= 0) {
return;
}
int res;
if ((res = ioctl(fd_vk, UI_DEV_DESTROY)) < 0) {
close(fd_vk);
fail("could not destroy virtual keyboard: %d\n", res);
}
close(fd_vk);
}
/* Devices. */
static int fd_video;
static int fd_kb;
/**
* Opens and captures devices.
*/
static void capture_devices(void) {
int res;
if ((fd_video = open(fname_video, O_RDONLY)) < 0) {
fail("could not open video device %s for reading: %d\n",
fname_video, fd_video);
}
if ((res = ioctl(fd_video, EVIOCGRAB, 1)) < 0) {
fail("could not capture video device %s: %d\n", fname_video, res);
}
The following is multiple choice question (with options) to answer.
In a certain code VOLITION is written as UMIEUKRR, MARTINET is written as LYOPJPMX. How is DESTROYS written in this code? | [
"CCPPSQBW",
"CCPPSQBU",
"CCPPSQBV",
"CCPPSQBX"
] | A | as per the sequence the
ans is D-1 = C
E-2 = C
S-3=P
T-4=P
R+1=S
O+2=Q
Y+3=B
S+4=W
ans is CCPPSQBW
ANSWER:A |
AQUA-RAT | AQUA-RAT-35235 | # probability question on a customer
A very frustrated customer is trying to find an electronic receipt on their phone so they can return an item. The trouble is that the customer has three email accounts and can't remember which, one, account it was sent to. The customer assumes that there is an equal probability for each account. To add complexity the phone only has enough battery power to search one of the three email accounts
Unfortunately, the store says that this is the last day it will accept the return, and it is only 3 minutes till close, so there will not be any chance to get to a charger. The customer randomly decides to search one of the emails, without any bias.
Suppose, due to the organization and the various spam filters of the accounts, the chance of finding the receipt even if they were to search in the correct account is not a guarantee. The probability of finding it in account 1, assuming it was sent there is 62%, 54% if in account 2 and 56% for account 3.
Part (a) What is the probability that if the receipt is in account 2 that the customer will find it?
I think this one is (1/3)*(0.54) The third is for choosing the account and then times 0.54 is the prob for finding from the question. but the question seems like a conditional probability so I'm not sure.
Part (b) Calculate the probability the receipt was in account 2, if the search in account 2 is unsuccessful.
I think this one is ((1/3)(0.16))/((1/3)(0.31)+(1/3)(0.16)+(1/3)(0.49))
Part(c) What is the probability this person finds the email?
(1/3)(0.69)+(1/3)(0.84)+(1/3)(0.51) because we can either find it in account 1 or 2 or 3
Any help would be appreciated!
The following is multiple choice question (with options) to answer.
A manufacturer is using glass as the surface for the multi-touch screen of its smartphone. The glass on the manufactured phone has a 6% probability of not passing quality control tests. The quality control manager bundles the smartphone in groups of 10. If that bundle has any smartphone that does not pass the quality control test, the entire bundle of 10 is rejected.
What is the probability that a smartphone bundle that will be rejected by quality control? | [
" 0.25",
" .05^10",
" 1-0.94^10",
" 1-0.05^10"
] | C | Find the probability of the opposite event and subtract from 1.
The opposite event is that bundle will NOT be rejected by quality control, which will happen if all 10 phones pass the test, so P(all 10 phones pass test)=0.94^10.
P(at least one phone do not pass the test)=1- P(all 10 phones pass test)=1-0.94^10.
Answer: C. |
AQUA-RAT | AQUA-RAT-35236 | • I understand your answer completely now, thank you. – user103816 Dec 1 '14 at 15:25
• @user31782, great, glad to help. – Barry Cipra Dec 1 '14 at 15:31
$a'=ma+r, b'=nb +s\implies \dfrac{a'}{b'}=\dfrac{ma+r}{nb+s}$
\begin{align} \dfrac{a}{b}= \dfrac{ma+r}{nb+s} & \implies a(nb+s)=b(ma+r)\\ & \implies a\mid (ma+r)\land b\mid (nb+s) \\ & \implies r=s=0 \\ & \implies a'=ma, b'=nb \\ & \implies\dfrac{a}{b}= \dfrac{ma}{nb}\\ & \implies m=n\end{align}
The following is multiple choice question (with options) to answer.
If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of is : | [
"3 M^9",
"3 M^2",
"3 M^6",
"3 M^1"
] | B | Explanation:
We have : ( a + b + c) / 3 = M or (a + b + c) = 3M.
Now. .
Answer: B) 3 M^2 |
AQUA-RAT | AQUA-RAT-35237 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
There are 5 candidates in an election and 3 of them are tobe elected. A voter can cast any number of votes but notmore than three. The number of ways in which he can cast his vote is | [
"5",
"15",
"20",
"25"
] | D | Solution
The voter can cast one or two or three votes. So total number of ways in which he can cast his vote =5C1 + 5C2 + 5C3 = 5 + 10 + 10 = 25. Answer D |
AQUA-RAT | AQUA-RAT-35238 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
If $30,000 interest is invested in x percent simple annual interest for n years, which of the following represents the total amount of interest, in dollars, that will be earned by this investment in the n years? | [
"10,000(x^n)",
"30,000n(x/100)",
"10,000(1 + x/100)^n",
"30,000(x/100)^n"
] | B | Principal * no.of years * percentage of interest
Answer : B |
AQUA-RAT | AQUA-RAT-35239 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
Excluding stoppages, the speed of a bus is 50 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? | [
"7 min",
"6 min",
"9 min",
"10 min"
] | B | Due to stoppages, it covers 5 km less.
Time taken to cover 5 km =(5/50)x60 = 6 min
answer : B |
AQUA-RAT | AQUA-RAT-35240 | $$s=\frac{a_1+...+a_n}{n}$$.
If you want the average of $$a_1,...,a_n$$ and $$a_{n+1}$$, then $$s'=\frac{a_1+...+a_n+a_{n+1}}{n+1}=\frac{ns+a_{n+1}}{n+1} = \frac{(n+1)s+a_{n+1}}{n+1} - \frac{s}{n+1} = s + \frac{a_{n+1}-s}{n+1}$$
If you want the average of $$a_1,...,a_{n-1}$$ then $$s''=\frac{a_1+...+a_{n-1}}{n-1}=\frac{ns-a_n}{n-1}= \frac{(n-1)s-a_n}{n-1} + \frac{s}{n-1}=s+\frac{s-a_n}{n-1}$$.
• And $ns$ is the old sum, as you can derive from the first equation. Feb 16, 2011 at 13:21
• I'm tempted to downvote as ns is not explained in the answer Aug 22, 2014 at 20:32
• At first I commented that @501's answer was clearer but realized I didn't fully understand why the question worked. This answer, while denser, actually explains the result.
– jds
Dec 10, 2014 at 23:05
• @Celeritas $ns = n \cdot s$, $n$ and $s$ are defined quantities in the post. Jan 13, 2015 at 15:13
• This is the working answer. @501's answer doesn't work when increasing n by more than 1. Nov 16, 2017 at 4:26
To put it programmatically, and since the question was about how to both add and subtract:
The following is multiple choice question (with options) to answer.
If a = (1/6)b and c = 7a, then which of the following represents the average (arithmetic mean) of a, b, and c, in terms of a ? | [
" a + 4",
" (11/3)a",
" (13/3)a",
" (4 1/7)a"
] | C | Official Answer: C
The average of the three variables isa + b + c/3. However, we need to solve in terms of a, which means we must convert b and c into something in terms of a.
Were told that a =1/6b, which is equivalent to b = 6a. We can plug that in and simplify the average to:a + 6a + c/3
We also know that c = 7a, which we can plug directly into the average expression:
a + 6a + 7a/3
= 13a/3 = (13/3)a choiceC. |
AQUA-RAT | AQUA-RAT-35241 | 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 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 700 m long? | [
"180",
"190",
"130",
"120"
] | B | L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1900 m
T = D/S
T = 1900/10
T = 190 Sec
Answer: Option B |
AQUA-RAT | AQUA-RAT-35242 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are drawn at random, what is the probability that at least one is green? | [
"23/35",
"23/36",
"23/33",
"23/31"
] | A | Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that at least one green marble can be picked in the random draw of two marbles = Probability that one is green + Probability that both are green
= (⁶C₁ * ⁹C₁)/¹⁵C₂ + ⁶C₂/¹⁵C₂
= (6 * 9 * 2)/(15 * 14) + (6 * 5)/(15 * 14) = 36/70 + 1/7
= 46/70 = 23/35
Answer: A |
AQUA-RAT | AQUA-RAT-35243 | The book has 300 pages, 200 of which are in long chapters and 100 of which are in short chapters. If you pick a random page it is $\frac 23$ to be in a long chapter. So strategy 1 gives you a long chapter $\frac 23$ of the time. Strategy 2 gives you the chapter after a long one $\frac 23$ of the time. If the chapters alternate, strategy 2 will give you a short chapter $\frac 23$ of the time.
Thanks, but doesn't the book still have 200 pages from long and 100 pages from short chapters both ways? How come the probability of picking long chapters isn't $\frac{2}{3}$ for both strategies? In other words, doesn't that logic for strategy 1 also also apply to 2? – David Faux May 5 '12 at 0:04
The initial page selection is in fact $\frac 23$ to be in a long chapter either way. But the selected chapter in strategy 2 is not the chapter the selected page is in. Think of a 2 chapter book with the first chapter 200 pages and the second 100. $\frac 23$ of the time the page chosen will be in chapter 1. In strategy 1 you then read chapter 1. In strategy 2 you then read chapter 2. – Ross Millikan May 5 '12 at 0:08
Ah thanks. So if short and long chapters alternate, then a reader using strategy 2 flips forward to a short chapter if he/she lands in a long chapter. However, what if the chapters don't alternate and are ordered randomly? Would the probability of reading a short chapter for strategy 2 be less than $\frac{2}{3}$ because the reader will not always move forward to a short chapter after landing in a long chapter? – David Faux May 5 '12 at 0:23
It will depend upon the ordering of the chapters. If chapters 1-5 are long and 2-6 are short, you will have $\frac 2{15}$ to read any of 2-6 and $\frac 1{15}$ to read 1 or 7-10. So that is $\frac 9{15}$ to read a long and $\frac 6{15}$ to read a short. – Ross Millikan May 5 '12 at 0:34
The following is multiple choice question (with options) to answer.
Sophia finished 2/3 of a book. She calculated that she finished 90 more pages than she has yet to read. How long is her book? | [
"270",
"150",
"130",
"90"
] | A | Let x be the total number of pages in the book, then she finished 2/3*x pages.
Then she has x−2/3*x=1/3*x pages left.
2/3*x−1/3*x=90
1/3*x=90
x=270
So the book is 270 pages long.
Answer is A. |
AQUA-RAT | AQUA-RAT-35244 | # Remainder on division with $22$
What is the remainder obtained when $$14^{16}$$ is divided with $$22$$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $$22$$, except the remainder?
How should I proceed?
• $14^{16} = (22 - 8)^{16}$ – ab123 Jul 3 '18 at 10:37
• $8^{16} = 64^8 = (66 - 2)^8$ – ab123 Jul 3 '18 at 10:41
• @ab123 Why not make an answer of that? – Arthur Jul 3 '18 at 10:41
You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
Since $$14^2 \equiv -2$$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$
or
$$14^2 = 22k -2$$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
The following is multiple choice question (with options) to answer.
In a division, divident is 689, divisior is 36 and quotient is 19. Find the remainder. | [
"4",
"3",
"2",
"5"
] | D | Explanation:
689 = 36 x 19 + r
689 = 684 + r
r = 689 - 684 = 5
Answer: Option D |
AQUA-RAT | AQUA-RAT-35245 | 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 take twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 5 days. B can do the work alone in? | [
"19",
"12",
"11",
"30"
] | D | Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/5
6/x = 1/5 => x = 30
So, B takes 15 hours to finish the work.
Answer: D |
AQUA-RAT | AQUA-RAT-35246 | Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4675
Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
14 Jun 2016, 17:21
5
KUDOS
Expert's post
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This post was
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AbdurRakib wrote:
The harvest yield from a certain apple orchard was 350 bushels of apples. If x of the trees in the orchard each yielded 10 bushel of apples, what fraction of the harvest yield was from these x trees?
A) $$\frac{x}{35}$$
B) 1–($$\frac{x}{35}$$)
C) 10x
D) 35–x
E) 350–10x
OG 2017 New Question
Dear AbdurRakib,
I'm happy to respond. This kind of word problem intimidates many people, but the actual calculation here is quite straightforward.
A fraction is a part over a whole. The whole is 350 bushels.
Of this set of x special trees, special for some unknown reason, each tree produced 10 bushels. That's 10x bushels from the lot of them. That's the part.
fraction = $$\frac{10x}{350}$$ = $$\frac{x}{35}$$
Does this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
15 Jun 2016, 00:41
Total yield by x trees = 10x
So, fraction of the harvest yield from these x trees = 10x/350 = x/35
Manager
Joined: 25 Jun 2016
Posts: 61
GMAT 1: 780 Q51 V46
Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
27 Jul 2016, 10:52
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KUDOS
Here are a couple of ways to solve this question:
Value Substitution:
Algebra:
The following is multiple choice question (with options) to answer.
Twenty percent of Country Y's yearly exports come from fruit exports. One-sixth of all fruit exports from Country Y are orange exports. If country Y generates $4.25 million from its orange exports, how much money does it generate from its yearly total of exports? | [
"$21.25m",
"$25.5m",
"$106.25m",
"$127.5m"
] | D | 2/10*1/6*(total) = 4.25
1/30*(total) = 4.25
(total) = 4.25*30 = 127.5
Answer: D. |
AQUA-RAT | AQUA-RAT-35247 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can do a work in 6 days. If A alone can do it in 15 days. In how many days can B alone do it? | [
"7",
"13",
"19",
"10"
] | D | 1/6 – 1/15 = 1/10
B can do the work in 10 days
Answer: D |
AQUA-RAT | AQUA-RAT-35248 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate got? | [
"55%",
"56%",
"57%",
"58%"
] | C | Total number of votes polled = (1136 + 7636 + 11628) = 20400
So, Required percentage = 11628/20400 * 100 = 57%
ANSWER : C |
AQUA-RAT | AQUA-RAT-35249 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is? | [
"50 m",
"150 m",
"200 m",
"300 m"
] | B | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35250 | 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.
Distances from you to Man united is written below.
CHELSEA and ARSENAL are 700 kms away
SPURS 1800 kms away
WOLVES 200 kms away
Based on the system , How far should it be to MANCITY ? | [
"1100 miles.",
"3100 miles.",
"2100 miles.",
"5100 miles."
] | C | Solution:
2100 miles.
Each vowel equivalent to 500 and each consonant equivalent to -200
Answer C |
AQUA-RAT | AQUA-RAT-35251 | 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 12 meters. What will be the distance covered by Amar during the time the car covers 1.6 km? | [
"6200 m",
"2400 m",
"300 m",
"4300 m"
] | B | B
2400 m
Distance covered by Amar = 18/1.2 (1.6km) = 3/8(1600) = 2400 m
Answer is B |
AQUA-RAT | AQUA-RAT-35252 | # Math Help - arithmetic sequence problem
1. ## arithmetic sequence problem
sry it's me with another question again. my test is coming up and my teacher is giving us all these challenge problems to prepare.
The sum of the first n terms in a certian arithmetic sequence is given by Sn = 3n^2 - n. Show that the nth term of the sequence is given by an = 6n - 4.
so i have this so far:
S_n = (n/2)(a1 + a_n) = 3n^2 - n
i then solved for (a1 + a_n) = 6n - 2
i also have the equation a_n = a1 + d(n-1)
i have tried solving for a_n, a1, and d using these equations but i'm not getting anywhere. Please help me.
2. A term in an arithmetic sequence can always be written a+ bn so you only need to find two values, a and b. For that you need two equations. Apply what you know about "the sum of the first n terms" to the sum of the first term only and the sum of the first two terms.
3. nth term of sequence can be find by $S_n - S_{n-1}$
Simple
$a_n = (3n^2 - n ) - (3(n-1)^2 - (n-1))$
= $3n^2 - n - 3n^2 - 3 +6n + n -1$
= $6n -4$
4. Hello, oblixps!
Yet another approach . . .
The sum of the first $n$ terms of an arithmetic sequence is given by: $S_n \:=\: 3n^2 - n$
Show that the $n^{th}$ term of the sequence is given by: $a_n \:= \:6n - 4$
The sum of the first $n$ terms of an arithmetic sequence: . $S_n \:=\:\frac{n}{2}\bigg[2a + (n-1)d\bigg]$
We will take $S_n \:=\:3n^2-n$ and hammer it into that form . . .
The following is multiple choice question (with options) to answer.
The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 40 terms of this sequence? | [
"(1+1/2) + 1/40",
"(1+1/2) – 1/42",
"(1+1/2) – 1/40",
"(1+1/2) – (1/41 +1/ 42)"
] | D | The answer would most certainly be[D]. But the question needs a slight modification.n>=1, since the answer does consider a1 under the sum.
The sequence is :
a1 = 1-1/3
a2 = 1/2 - 1/4
a3 = 1/3 - 1/5....
We can observe that the third term in the sequence cancels the negative term in the first. A similar approach can be seen on all the terms and we would be left with 1 + 1/2 from a1 and a2 along with -1/42 and -1/41 from a40 and a39 term which could not be cancelled.
Hence the sum = (1+1/2) – (1/41 +1/ 42)
Answer : D |
AQUA-RAT | AQUA-RAT-35253 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 120 meters long passes an electric pole in 12 seconds and another train of same length traveling in opposite direction in 8 seconds. The speed of the second train is | [
"60 Km",
"66 Km",
"72 Km",
"62 Km"
] | C | speed of first train=120/12=10 m/s
let speed of second train be x m/s. As they are approaching each other, relative speed is (10+x) m/s.
distance between each other is 120+120=240 m.
time of crossing each other is 8 sec
=>240/(10+x)=8
=>x=20m/s = 20*18/5 km/hr = 72 kmph
ANSWER:C |
AQUA-RAT | AQUA-RAT-35254 | Actually, there's an important detail that's hidden in this statement: "just chose one of their names at random". I assume "at random" means 50-50%.
This is the crux of the problem. If T is fired, under the assumption stated above, T will be told B or J with equal probability. On the other hand, conditional on T not being fired, there is probability $$\frac{5}{85}$$ (this is the probability B is fired conditional on T not being fired) that he will be told J and $$\frac{80}{85}$$ probability he will be told B. So if he is not fired, he is much more likely to be told B than J. Thus, being told J should make him think his odds of being fired are worse than if he was told B.
But note that if by "choosing one of their names at random" when T is fired, we allow for unequal probabilities, we could say that we will tell him J will probability $$\frac{5}{85}$$ and B otherwise. Note that this is still choosing one name at random (but I'm guessing not what the OP meant) and in this case, the odds of being told J is exactly the same whether he will be fired or not. Therefore, being told J will keep their job does not change his probability of being fired!
The following is multiple choice question (with options) to answer.
Jack and Jill work at a hospital with 3 other workers. For an internal review, 2 of the 5 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen? | [
" 1/3",
" 1/10",
" 1/15",
" 3/8"
] | B | 1/5C2=1/10.
Answer: B. |
AQUA-RAT | AQUA-RAT-35255 | 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 ratio of the present age of two brothers is 1:2 and 5 years back, the ratio was 1:3. What will be the ratio of their ages after 5 years? | [
"1:4",
"2:3",
"3:5",
"5:6"
] | C | Let the present ages of the two brothers be x and 2x years respectively.
Then, (x - 5)/(2x - 5) = 1/3
3(x - 5) = (2x - 5) => x = 10
Required ratio = (x + 5) : (2x + 5) = 15 : 25 = 3:5
ANSWER:C |
AQUA-RAT | AQUA-RAT-35256 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Albert borrowed a total of $ 6000 from Brian and Milton. He paid simple interest at the rate of 7 % Per yr and 9 % per yr respectively.At the end of three years he paid $ 1494 as total interest. What is the interest paid to Brian at the end of three years? | [
"491",
"481",
"471",
"441"
] | D | Let x be the amount borrowed form Brian. So amount borrowed form Milton = 6000-x.
1494 = 21/100x + 27/100(6000-x)
=>x = 2100.
Interest paid = 3*7/100*2100 = 441.
D |
AQUA-RAT | AQUA-RAT-35257 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 5000 natural numbers? | [
"2500",
"2500.5",
"2145.5",
"2005.5"
] | B | Sum of first 100 natural numbers = 5000*5001/2 = 12502500
Required average = 12502500/5000 = 2500.5
Answer is B |
AQUA-RAT | AQUA-RAT-35258 | 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.
A father was as old as his son's present at the time of your birth. If the father's age is 38 years now, the son's age five years back was? | [
"10 years",
"12 years",
"14 years",
"16 years"
] | C | Let the son's present age be x years. Then, (38 - x) = x
2x = 38.
x = 19.
Son's age 5 years back (19 - 5) = 14 years.
C |
AQUA-RAT | AQUA-RAT-35259 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
In a school of 850 boys, 44% of Muslims, 28% Hindus, 10% Sikhs and the remaining of other communities. How many belonged to the other communities? | [
"125",
"627",
"153",
"721"
] | C | 44 + 28 + 10 = 82%
100 – 82 = 18%
850 * 18/100 = 153
Answer:C |
AQUA-RAT | AQUA-RAT-35260 | 10.21, 9.82, 10.81, 10.3, 9.81, 11.48, 8.51, 9.55, 10.41, 12.17, 9.9, 9.07, 10.51, 10.26, 10.62, 10.84, 9.67, 9.75, 8.84, 9.85, 10.41, 9.18, 10.93, 11.41, 9.52]
The following is multiple choice question (with options) to answer.
8, 12, 18, 27, 40.5, (...) | [
"60",
"60.75",
"58",
"89"
] | B | 8
(8×3)÷ 2 = 12
(12×3)÷ 2 = 18
(18×3)÷ 2 = 27
(27×3)÷ 2 = 40.5
(40.5×3)÷ 2 = 60.75
Answer is B |
AQUA-RAT | AQUA-RAT-35261 | $x = \frac{- 4 \pm \sqrt{16 + 4}}{2}$
$x = \frac{- 4 \pm \sqrt{20}}{2}$
$x = \frac{- 4 \pm 2 \sqrt{5}}{2}$
$x = - 2 \pm \sqrt{5}$
Hope this helps!
The following is multiple choice question (with options) to answer.
If x≠4 and (x^2 - 16)/(2y) = (x - 4)/4, then in terms of y, x = ? | [
"(y - 8)/2",
"(y - 3)/2",
"y - 3",
"y - 6"
] | A | Since (x^2 – 16)=(x-4)(x+4) the original equation can be changed into 4*(x-4)(x+4) = 2y *(x-4).
By cancelling 2(x-3) (we can do it because x≠4) we get :
2*(x+4)=y.
So x=(y-8)/2. So the answer is A). |
AQUA-RAT | AQUA-RAT-35262 | tiling
Title: Minimizing number of MxM squared tiles in an infinite grid covered by any part of a shape Without restriction (e.g. continuity or convexity are not guaranteed), we're given an W x H raster-based shape that needs to be placed on an infinite tiling of M x M squares. The shape can placed at any integer positional offset within those squares.
The goal is to minimize the number of squares containing any part of that shape.
For example, this shape (shown in orange) can be placed in a way that occupies seven squares on a tiling of 10 x 10:
However, a more optimal placement is an offset of (+2, 0) which occupies six squares by leaving the central bottom square unoccupied:
Can the brute-force solution (O(W * H * M * M)) be improved upon? How / why not? I found an algorithm. The TL;DR is that we can efficiently precompute a parallel image that indicates whether each position in the image contains any occupied pixels if we were to start a subgrid at that position (going down and to the right, chosen arbitrarily), and determining which collection of those positions produces the minimum total number of occupied subgrids can then be done in simple iteration.
Compute a "row occupancy" grid which extends to the "left" of the image by M additional pixels. Each pixel in this grid is TRUE if any pixel within M pixels to the right of it (including itself) is nonempty. This can be done in O(HW) time because you can iterate each row backwards, keeping track of how long it's been since the last time you saw an occupied pixel
Compute a "grid occupancy" grid, which is computed similarly. This grid extends "above" the top of the image by M additional pixels as well as to the left by M additional pixels. The algorithm is the same as the prior step, except applied to the row occupancy grid - count from the bottom up, keeping track of the last time you saw an occupied row, marking TRUE when the last occupied row was within the last M pixels.
The following is multiple choice question (with options) to answer.
Rectangular tile each of size 50cm by 40cm must be laid horizontally on a rectangular floor of size 120cm by 150cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is: | [
"9",
"15",
"12",
"11"
] | A | Area of tile = 50*40 = 2000
Area of floor = 120*150= 18000
No of tiles = 18000/2000 =9
So, the no of tile = 9
ANSWER:A |
AQUA-RAT | AQUA-RAT-35263 | \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{r,s} \ . \hspace{1cm} (4)$$ Note that we have also changed the order of the product operations over $i$ and $j$ here. Now the average over the $s_j$ can be performed easily: for each $j$, $$\left\langle \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{s_j} = \frac1 2 \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right) \ ,$$ which is independent of $j$, so that $$\Big\langle P(Ms=0) \Big\rangle_M = 2^{-n} \left\langle \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right)^n \right\rangle_{r} \ . \hspace{2cm} (5)$$ To evaluate the right hand side of $(5)$ we expand the $n^{\rm th}$ power, $$\Big\langle P(Ms=0) \Big\rangle_M = 2^{-n} \sum_{k=0}^n \begin{pmatrix} {n} \\ {k} \end{pmatrix} 2^{-km} \prod_{i=1}^m \left\langle \Big( \omega^{r_i} + \omega^{-r_i} \Big)^k \right\rangle_{r} \ , \hspace{2cm} (6)$$ and likewise the $k^{\rm th}$ power
The following is multiple choice question (with options) to answer.
A salt manufacturing company produced a total of 5000 tonnes of salt in January of a particular year. Starting from February its production increased by 100 tonnes every month over the previous months until the end of the year. Find its average monthly production for that year? | [
"2989",
"2765",
"5550",
"2989"
] | C | Total production of salt by the company in that year
= 5000 + 5100 + 5200 + .... + 6100 = 66600.
Average monthly production of salt for that year
= 66600/12
= 5550.
Answer:C |
AQUA-RAT | AQUA-RAT-35264 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II, and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage Q of those asked favoured more than one of the 3 proposals. | [
" 10",
" 12",
" 17",
" 22"
] | C | Bunuel, my answer for exactly 2 people was 17 and this was my approach:
100%=(A+B+C)-(AnB+AnC+BnC)-5%+22% which leads me to
Q=100%=(50+30+20)-(at least 2 people)-5%+22%.C |
AQUA-RAT | AQUA-RAT-35265 | # 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 owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8339 for a computer table, then what was the cost price of the computer table? | [
"Rs. 5725",
"Rs. 5275",
"Rs. 6275",
"Rs. 6725"
] | D | CP = SP * (100/(100 + profit%))
= 8339(100/124) = Rs. 6725.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35266 | homework-and-exercises, fluid-statics, bernoulli-equation
You ignore the area of the hole.
Rate of water going in: $r'= \frac{0,2cm^3}{s}$
Rate of water going out: $r=1mm^2*\sqrt{2gh}$
Set the two rates equal and solve for $h$
$\frac{0,2cm^3}{s} = 1mm^2*\sqrt{2gh}$
$\frac{0,2cm^3}{s*0,01cm^2} = \sqrt{2gh}$
$\frac{20cm}{s} = \sqrt{2gh}$
The units work out correctly
I will leave the rest to you
Hint square both sides
Solving for the $t$ in $h(t)$ would be a more complex
The following is multiple choice question (with options) to answer.
A river 2m deep and 45 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is? | [
"4500 M3",
"4580 M3",
"9000 M3",
"4900 M3"
] | C | Explanation:
(6000 * 2 * 45)/60 = 9000 M3
Answer: Option C |
AQUA-RAT | AQUA-RAT-35267 | Kudos [?]: 53125 [5] , given: 8043
Re: problem solving question on ratios [#permalink] 16 Dec 2010, 13:47
5
KUDOS
Expert's post
2
This post was
BOOKMARKED
spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134
Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$
$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.
$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).
The following is multiple choice question (with options) to answer.
The present ratio of students to teachers at a certain school is 45 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers? | [
"4",
"8",
"10",
"12"
] | A | We are given that the ratio of students to teacher is 45 to 1. We can rewrite this using variable multipliers.
students : teachers = 45x : x
We are next given that student enrollment increases by 50 and the number of teachers increases by 5. With this change the new ratio becomes 25 to 1. We can put all this into an equation:
Students/Teachers 25/1 = (45x + 50)/(x + 5)
If we cross multiply we have:
25(x + 5) = 45x + 50
25x + 125 = 45x + 50
3.75 = x
x~4
Since x is the present number of teachers, currently there are 4 teachers.
Answer A. |
AQUA-RAT | AQUA-RAT-35268 | But suppose we weren't done. The table would continue \begin{align*} 4 && 2 && - \\ 5 && 3 && 2 \\ 6 && 4 && 3 \\ 7 && 4 && - \\ 8 && 5 && 4 \\ 9 && 6 && 5 \end{align*} where we stop when we hit $n=9$, which is guaranteed to have a solution. Notice, there is no reason to find the value in the 3rd column when the integer part in the second column has not increased.
Note that we can actually calculate the multiples of $\log_2 13/8 = 0.700 \dots$ "by eye", so the only work in this particular problem is the integer part of the multiples of $\log_2 12/8$.
• Calculation of log to base 2 is easy in floating point. A crude approximation is to write a dot, then the mantissa portion of the floating pint representation. If you have a decimal number between 1 and 2, just subtract 1. It is exact for 1 and 2, and a bit low at the middle of the interval. For more on logs, see Doerfler's wonderful book. – richard1941 Jul 12 '17 at 0:26
$12 \lt 2^{m/n} \lt 13$
$\log_2 12 \lt \frac{m}{n} \lt \log_2 13$
$[3; 1, 1, 2, 2, \dots] \lt \frac{m}{n} \lt [3; 1, 2, 2, 1, \dots]$
The continued fractions match up to $[3; 1]$. Since $[3; 1, 1]$ and $[3; 1, 2]$ are underestimations of $\log_2 12$ and $\log_2 13$, we take $[3; 1, 1]$ (the "smaller" one lexicographically) and add $1$ to the last number (round "up"), so the answer is $[3; 1, 2]$.
The following is multiple choice question (with options) to answer.
If log 27 = 1.431, then the value of log 9 is : | [
"0.934",
"0.945",
"0.954",
"0.958"
] | C | Solution
Log 27 = 1.431 ⇒ log (33) = 1.431 ⇒ 3 log 3 = 1.431
⇒ log 3 = 0.477
∴ log 9 = log (32) = 2 log 3 = (2 × 0.477) = 0.954. Answer C |
AQUA-RAT | AQUA-RAT-35269 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Ram purchased a TV and a DVD Player for Rs. 25000 and Rs. 5000 respectively. He sold the DVD Player at a loss of 4 percent and TV at a profit of 10 percent. Overall he make a.? | [
"2200",
"2500",
"2800",
"2300"
] | D | Let the SP of the TV and the DVD Player be Rs. r and Rs. m respectively.
r = 25000(1 + 10/100) = 25000 + 2500
m = 5000(1 - 4/100) = 5000 - 200
Total SP - Total CP = r + m - (25000 + 5000) = 2500 - 200 = Rs. 2300
As this is positive, an overall profit of Rs. 2300 was made.
Answer: D |
AQUA-RAT | AQUA-RAT-35270 | P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2 P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2
(1)
Show whether or not the pair {A,B}{A,B} is independent.
### Solution
P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc)P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc), and P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc)P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc).
PA = 0.4*0.7 + 0.3*0.3
PA = 0.3700
PB = 0.6*0.7 + 0.2*0.3
PB = 0.4800
PA*PB
ans = 0.1776
PAB = 0.4*0.6*0.7 + 0.3*0.2*0.3
PAB = 0.1860 % PAB not equal PA*PB; not independent
## Exercise 2
Suppose {A1,A2,A3} ci |C{A1,A2,A3} ci |C and ci |Cc ci |Cc, with P(C)=0.4P(C)=0.4, and
The following is multiple choice question (with options) to answer.
A = 10% of x, B = 10% of y, C = 10% of x + 10% of y. On the basis of the above equalities, what is true in the following? | [
"A is equal to B",
"A is greater than B",
"B is greater than A",
"Relation cannot be established between A and B"
] | D | The given information gives no indication regarding the comparison of x and y.
Answer D |
AQUA-RAT | AQUA-RAT-35271 | 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
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
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
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
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 after 100 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used? | [
"180 grams",
"200 grams",
"250 grams",
"350 grams"
] | B | The 24% alcohol solution decreases by 1/3 once the 100 grams of water is added so 100/.08 =1250*.16% (2/3 of 24%) is 200 grams, thus answer is B. |
AQUA-RAT | AQUA-RAT-35272 | 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.
In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m? | [
"127.6",
"127.5",
"127.9",
"127.8"
] | B | When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = (900 * 700)/800 = 6300/8 = 787.5 m.
In a race of 1000 m, A beats C by (1000 - 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = (600 * 212.5)/1000
= 127.5 m.
Answer: B |
AQUA-RAT | AQUA-RAT-35273 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected? | [
"80 ways",
"87 ways",
"30 ways",
"28 ways"
] | C | The number of ways to select three men and two women such that one man and one woman are always selected = Number of ways selecting two men and one woman from men and five women
= ⁴C₂ * ⁵C₁ = (4 * 3)/(2 * 1) * 5
= 30 ways.
Answer: C |
AQUA-RAT | AQUA-RAT-35274 | 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}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
X completes a work in 12 days and Y complete the same work in 24 days. If both of them work together, then the number of days required to complete the work will be ? | [
"8 days",
"10 days",
"12 days",
"20 days"
] | B | If X can complete a work in x days and Y can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 12 × 24/ 36 = 8 days.
B) |
AQUA-RAT | AQUA-RAT-35275 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
How many possible integer values are there for x if |7x - 3| < 6 ? | [
"One",
"Two",
"Three",
"Four"
] | B | Solution: |7x-3| < 6
let 7x=a therefore we have |a-3| < 6 ==> read this as origin is at +3 and we have to move +6 to the right and -6 to the left
(the less than sign represents that the a must be within boundaries )
(3-6)----------3----------(3+6)
now, we have -3<a<9
but a =7x ==> -3<7x<9
dividing all values by +4 we have -0.46 <x < 1,2
Now question says Integer values (not rational ) therefore we have 0,1
Hence 2 B |
AQUA-RAT | AQUA-RAT-35276 | Let one woman complete the job in $$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 alone can complete a work in 16 days and B alone can do in 12 days. StarƟng with A, they work on
alternate days. The total work will be completed in | [
"1314",
"1312",
"1334",
"1344"
] | C | ExplanaƟon:
A's 1 day work = 1/16
B's 1 day work = 1/12
As they are working on alternate day's
So their 2 days work = (1/16)+(1/12)
= 7/48
[here is a small technique, Total work done will be 1, right, then mulƟply numerator Ɵll denominator, as
7*6 = 42, 7*7 = 49, as 7*7 is more than 48, so we will consider 7*6, means 6 pairs ]
Work done in 6 pairs = 6*(7/48) = 7/8
Remaining work = 1-7/8 = 1/8
On 13th day it will A turn,
then remaining work = (1/8)-(1/16) = 1/16
On 14th day it is B turn,
1/12 work done by B in 1 day
1/16 work will be done in (12*1/16) = 3/4 day
So total days =
1334
Answer: C |
AQUA-RAT | AQUA-RAT-35277 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
A Jar of marble contains 6 yellow and 4 blue marbles, and no other marbles. If four are pulled out of the jar, one after the other, without being replaced, what is the probability that exactly two of the four marbles will be yellow and two blue.
A
B
C
D 9 | [
"1/10",
"1/7",
"1/6",
"3/7"
] | D | 4 marbles out of the given 10 are pulled out of the jar, one after the other, without being replaced
The number of possible combinations (total events) = 10C4
We need to find the probability that exactly two of the four marbles will be yellow(out of 6) and two blue( out of 4).
Here the order in which the marbles are removed isn't asked. So no permutation reqd.
The number of possible combinations to such a specific event = 6C2 X 4C2
The Probability of the specific event = [ 6C2 X 4C2 ] / 10C4
= 15 X 6 / 210
= 3/7
Correct Option : D |
AQUA-RAT | AQUA-RAT-35278 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
You collect baseball cards. Suppose you start out with 15. Maria takes half of one more than the number of baseball cards you have. Since you're nice, you give Peter 1 baseball card. Since his father makes baseball cards, Paul decides to triple your baseball cards. How many baseball cards do you have at the end? | [
"17",
"18",
"19",
"20"
] | B | Solution
Start with 15 baseball cards.
Maria takes half of one more than the number of baseball cards you have. So Maria takes half of 15 + 1 which is 8,
so you're left with 15 - 8 = 7.
Peter takes 1 baseball card from you: 7 - 1 = 6 baseball cards.
Paul triples the number of baseball cards you have: 6 × 3 = 18 baseball cards.
So you have 18 at the end.
Correct answer : B |
AQUA-RAT | AQUA-RAT-35279 | inorganic-chemistry, aqueous-solution, solubility
Title: Solutions of several salts What happens when you have a solution of a salt, and then try to dissolve another salt in that solution? I know about the common ion-effect etc, but how do you solve problems like this:
How much AgCl can you solve per L in a 3.8g/L NaCl solution? To solve a problem like this, we need to know the solubility product constant $K_\mathrm{sp}$ for $\ce{AgCl}$, where at equilibrium,
$$K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]=1.8\times 10^{-10}\ \mathrm{M^2}$$
This relationship stipulates that under equilibrium conditions at all times, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ must satisfy this relationship.
If the solution contains just $\ce{AgCl}$, we can determine the solubility. The concentrations of both ions must be equal based on the chemical formula.
$$\begin{array}{lcl}
x&=&[\ce{Ag+}]=[\ce{Cl-}]\\
x^2 &=& 1.8\times 10^{-10} \\
x &=& \sqrt{1.8\times 10^{-10}}=1.342...\times 10^{-5}
\end{array}$$
We have $1.342\times 10^{-5}\ \mathrm{mol/L}=1.92\times 10^{-3}\ \mathrm{g/L}$ of $\ce{AgCl}$ in soltion.
The following is multiple choice question (with options) to answer.
A tank contains 10,000 gallons of a solution that is 3 percent sodium chloride by volume. If 4,000 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride? | [
" 1.25%",
" 5%",
" 6.25%",
" 6.67%"
] | B | The remaining solution will be approximately what percent sodium chloride?means:what percent of the remaining solution is sodium chloride. Now, since the remaining solution is 10,000-4,000=6,000 gallons and sodium chloride is 300 gallons (3% of initial solution of 10,000 gallons) then sodium chloride is 300/6,000*100=5% of the remaining solution of 6,000 gallons.
Answer: B. |
AQUA-RAT | AQUA-RAT-35280 | human-biology, reproduction, human-genetics
Title: Very frequent multiple births in humans 18th century Feodor Vassilyev is said to have had children by two wives, each of whom only ever had twins, triplets or quadruplets. His first wife has 16 sets of twins, 7 of triplets and 4 of quads; his second had 6 sets of twins and 2 of triplets. Is there any known plausible biological explanation for this, or do we have to dismiss it as a fabrication?
I could understand a woman's body being unusually susceptible to multiple births. I can't find information on whether these women tended to have monozygotic or polyzygotic offspring, but neither option seems unviable to me. However, since it's unlikely two of Feodor's sexual partners would share such a trait, one would think to attribute it to him. Presumably there would have to be a mechanism by which paternal DNA can trigger embryo fissions, in which case I imagine the offspring would be polyzygotic. Risk factors for dizygotic twinning are related to multiple follicular development, and include maternal family history, ethnicity, geography, maternal parity, maternal age, and, of course, use of assisted reproductive technology. There may be a genetic component to monozygotic twinning as well, but that rate is fairly consistent across populations. Other risk factors, such as diet and supplementation, have been proposed, but the data are less robust. There are some interesting studies demonstrating geographic clusters of twinning, but these tend to be demographic clusters that are associated with other risk factors.
Though a higher risk of twinning can be transmitted from a father to his daughters, the father's family history of twinning is not a significant risk factor (for his own children to be twins). To clarify -- if a man has a family history of multiple births, his children are no more likely to be twins than the general population, but his daughers are more likely to give birth to twins. This study is one example of the studies that have shown a significant independent association between maternal family history and twinning, with no significant independent association with paternal family history.
The following is multiple choice question (with options) to answer.
In certain year in Country C, x sets of twins and y sets of triplets were born. If there were z total babies born in Country C in this year, and x and y were both greater than 0, which of the following represents the fraction T of all babies born who were NOT part of a set of twins or triplets? | [
"z - 2x - 3y",
"(2x+3y)/z",
"(x+y)/z",
"1 - (2x+3y)/z"
] | D | X set of twins implies 2x babies of twin category.
Y set of triplets means 3y babies of triplet category.
Let K be the babies other than twin or triplet category.
Total babies = z = 2x+3y+k.
or K=Z-(2x+3y)
=> Fraction of babies other than twin or triplet categories = K/Z = Z-(2x+3y)/Z.
T= 1- (2x+3y)/Z. Hence answer is D. |
AQUA-RAT | AQUA-RAT-35281 | => 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.
Each writer for the local newspaper is paid as follows: a dollars for each of the first s stories each month, and a + b dollars for each story thereafter, where a > b. How many more dollars will a writer who submits s+ a stories in a month earn than a writer who submits s+ b stories? | [
"(a – b)( a + b + s)",
"a – b",
"a^2 – b^2",
"s( a – b)"
] | C | total earned for s+a stories =a + a(a+b)
total earned for s+b stories =a+ b(a+b)
difference = a+a(a+b) -a -b(a+b)
=(a+b)(a-b) =a^2-b^2
answer is C |
AQUA-RAT | AQUA-RAT-35282 | 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.
In a race of 600 metres, A can beat B by 60 metres and in a race of 500 metres, B can beat C by 50 metres. By how many metres will A beat C in a race of 400 metres? | [
"76 metres",
"80 metres",
"70 metres",
"84 metres"
] | A | A runs B runs C runs
600 metres race 600m 540 m
500 metres race 500 m 450m
Combing ratio A runs B runs C runs
300metres - 2700meters - 2430metres
Unitary A runs B runs C runs
Method 400mtres - 360 metres - 324 metres
? A beats C by 400-324 = 76 metres.
Answer: A. |
AQUA-RAT | AQUA-RAT-35283 | investment Banking Course, Download Valuation... As are bonds less than 365 days, most developed countries use simple interest is bonds! Is 1-Feb-2020, and investors collect returns when the bond “ days maturity... The market price quarterly, or the coupon Equivalent rate ( CER.. For calculating return on investment and an investment that is taxed bankruptcy and liquidates... Second part is used to calculate the annual yield of Second Govt basis that incorporates! 42 every year for calculating return on investment and the yield to calculate bond Equivalent yield for company... Thus 11 % multiplied by two, which comes out to 22 % have to understand this! ( 5 ) /95 } * { 1.520833 } ], bond Equivalent yield formula be calculated using the function! 365/180 } ], bond Equivalent yield formula on how to calculate the bond happens! Is Rs value ) to pay ( coupon rated ) is an calculation. From an annual-pay bond can not be directly compared to find out the bond Equivalent formula... That of bond or call premium of years until maturity or until call until. Calculate current yield calculation, as are bonds can be calculated using following! 5 ) /95 } * { 1.520833 } ], bond Equivalent formula. To yield to calculate EAY as follows: EAY bond equivalent yield formula 1.0253 ( 365/90 ) = %... Different price and tenure discount and do not pay interest at all NCD, the to. /.75 ) = 10.66 % * 100 1 as the current market price the... Until maturity or until put is exercised Contents ) or bonds ( fixed income securities, which sold! Allows the investor to calculate the annualized yield of Second Govt also some bonds, do not annual. Issue, we must know the bond Equivalent yield formula = ( Face value Rs!
The following is multiple choice question (with options) to answer.
What is the rate percent when the simple interest on Rs.720 amount to Rs.180 in 4 Years? | [
"6.25",
"7.25",
"8.25",
"9.25"
] | A | 180 = (720*4*R)/100
R = 6.25%
Answer: A |
AQUA-RAT | AQUA-RAT-35284 | # Difference between revisions of "2021 AMC 10B Problems/Problem 3"
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
## Solution 2 (Fast but Not Rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Solution 3
The following is multiple choice question (with options) to answer.
In a sample of 800 high school students in which all students are either freshmen, sophomores, juniors, or seniors, 22 percent are juniors and 75 percent are not sophomores. If there are 160 seniors, how many more freshmen than sophomores are there among the sample of students? | [
"42",
"48",
"64",
"76"
] | C | 200 are sophomores.
The number of freshmen is 600 - 160 - 0.22(800) = 264
The answer is C. |
AQUA-RAT | AQUA-RAT-35285 | 2 } \right) \\ \\ \therefore \quad y=\sqrt { \frac { { b }^{ 2 } }{ { a }^{ 2 } } } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } } \\ \\ \therefore \quad y=\frac { b }{ a } \cdot \sqrt { { a }^{ 2 }-{ x }^{ 2 } }$
The following is multiple choice question (with options) to answer.
Given: b2 = ac and ax=by=cz.
Determine the value of y. | [
"2xz/(x+z)",
"8xz/(x+z)",
"3xz/(x+z)",
"7xz/(x+z)"
] | A | Explanation:
No explanation is available for this question!
ANSWER: A |
AQUA-RAT | AQUA-RAT-35286 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo? | [
"12",
"51",
"13",
"15"
] | D | Two students can be selected from 6 in 6C2 =15 ways.
Therefore, the teacher goes to the zoo 15 times.
Ans: D |
AQUA-RAT | AQUA-RAT-35287 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.10000? | [
"3310",
"1879",
"2779",
"2672"
] | A | A = 10000(11/10)^3
= 13310
= 10000
----------
3310
Answer: A |
AQUA-RAT | AQUA-RAT-35288 | # Combinatorics using averages
How many solutions exist for $$x+2y+4z=100$$ in non-negative integers?
The author, Martin Erickson, in his book, Aha! Solutions, published by MAA, gives the following brief solution :
There are $$26$$ choices for $$z$$, namely, all integers from $$0$$ to $$25$$. Among these choices, the average value of $$4z$$ is $$50$$. So, on average, $$x+2y=50$$. In this equation, there are $$26$$ choices for $$y$$, namely, all integers from $$0$$ to $$25$$. The value of $$x$$ is determined by the value of $$y$$. Hence, altogether there are $$26^2=676$$ solutions to the original equation.
Can somebody explain to me why this mindblowing solution works using averages?
• Just because you got the numerical answer, doesn't necessarily mean that the solution must have a correct reasoning. Sep 12 '21 at 19:23
• @CalvinLin Is there no reasonable explanation for given solution? Sep 12 '21 at 19:26
• I didn't say that. There could be. But this requires more of an explanation/details than just what i written. Sep 12 '21 at 19:27
• @CalvinLin Unfortunately the author gives only this much solution to above problem. Sep 12 '21 at 19:28
There are $$26$$ choices for $$z$$ in $$\{0,1,\ldots,25\}$$.
Given $$z$$, we have $$x+2y=100-4z$$, which leaves $$\frac{1}{2}(100-4z)+1=51-2z$$ choices for $$y$$ in $$\{0,1,\ldots,\frac{1}{2}(100-4z)\}$$. Having fixed $$y$$, we must have $$x=100-4z-2y$$.
The following is multiple choice question (with options) to answer.
If the average of 6x and 8y is greater than 100, and x is twice y, what is the least integer value of x ? | [
"22",
"20",
"21",
"23"
] | B | Substitution can be used in the following way:
Always start with the equation: x = 2y. It is more straight forward to manage as compared to the inequality.
Substitute y = x/2, not the other way because you need to find the minimum value of x. So you can get rid of y.
Now go on to the inequality. So 8y = 8x/2 = 4x
Now average of 6x and 4x is greater than 100. Average of 6x and 4x is 5x. So,
5x > 100
x > 20
Answer : B |
AQUA-RAT | AQUA-RAT-35289 | 8 Since g(X)>f(X) we may conclude that g is always greater than f.
3. Mar 13, 2010
### Werg22
Re: x=2^x
Much simpler is to write 2^x - x = 0, and take the derivative of LHS: ln(2)2^x - 1. On [0, ∞), this is clearly strictly positive, so the function 2^x - x is strictly increasing on that interval. At x = 0, 2^x - x = 1, so it never attains 0 on [0, ∞) (it only gets larger than 1). It also can never attain 0 on (-∞, 0), so there is no solution.
4. Mar 13, 2010
### arildno
Re: x=2^x
Most definitely not!
At x=0, for example, ln(2)-1 is negative.
5. Mar 13, 2010
### Werg22
Re: x=2^x
Ah, then I retract my solution.
6. Mar 14, 2010
### Dragonfall
Re: x=2^x
Ah, thanks for the replies!
7. Mar 15, 2010
### uart
Re: x=2^x
This question could be turned into a nice little exercise for introductory calculus.
Q. Find the largest value of "a" for which the equation, $a^x = x$ has a real solution? It has a fairly cute solution.
a = e^(1/e), with the solution occurring at x = e.
The following is multiple choice question (with options) to answer.
What is the greatest value of positive integer x such that 2^x is a factor of 100^40? | [
"80",
"90",
"95",
"105"
] | A | Put in simple words, we need to find the highest power of 2 in 100^40
100 = 2^2*5^2
Therefore 100^40 = (2^2*5^2)^40 = 2^80*5^80
Answer : A |
AQUA-RAT | AQUA-RAT-35290 | Equilateral Triangle Calculator. Find the Length of the Side and Perimeter of an Equilateral Triangle Whose Height is √ 3 Cm. The height of an equilateral triangle of side ' a ' is given by. The special right triangle gives side ratios of , , and . A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe Thus these are properties that are unique to equilateral triangles, and knowing that any one of them is true directly implies that we have an equilateral triangle. Originally Answered: The side length of an equilateral triangle is 6 cm. Solution: According to height of equilateral triangle formula 4.1. s = … To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles. improve our educational resources. Varsity Tutors LLC An equilateral triangle also has three equal 60 degrees angles. Area of equilateral triangle = 4 3 a 2 Also, area = 2 1 × base × height = 2 1 a × height … OwlCalculator.com. Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. How to find the height of an equilateral triangle. An equilateral … The height of an equilateral triangle having each side 12cm, is (a) 6√2 cm (b) 6√3m (c) 3√6m (d) 6√6m. Therefore, the value of X will be twice the value of the height: What is the height of an equilateral triangle with a side length of 8 in? What is the triangle's height ? ... Now apply the Pythagorean theorem to get the height (h) or the length of the line you see in red. = 5.2 units approx. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Take √(3) = 1.732 And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. A triangle ABC that has the sides a, b, c,
The following is multiple choice question (with options) to answer.
The area of an equilateral triangle is subtracted from its base, and the perimeter is then added to this total, the result is 4. What is the height of the equilateral triangle if its base is 2? | [
"2",
"3",
"4",
"5"
] | C | An eqilateral triangle means all sides are equal; therefore, base = side. The equation is; base - Area + Perimeter = b - A + P = b -1/2bh + 3b, and b = 2. So, 2 - 2/2h + 6 = 4, and 8 - h = 4.
By plugging in the answers we can test the answers quickly; then, 4 is the only possible answer.
Answer: C |
AQUA-RAT | AQUA-RAT-35291 | # related rates help!!!
• November 15th 2007, 04:53 PM
singh1030
related rates help!!!
At noon, ship A is 125 km east of ship B. Ship A is sailing west at 35 http://maple-ta.nss.udel.edu:8080/tm...pogppajdcb.gif, and ship B is sailing north at 35 http://maple-ta.nss.udel.edu:8080/tm...pogppajdcb.gif. How fast is the distance between the ships changing at 2:00 P.M. in http://maple-ta.nss.udel.edu:8080/tm...pogppajdcb.gif?
• November 15th 2007, 05:35 PM
Soroban
Hello, singh!
Quote:
At noon, ship A is 125 km east of ship B.
Ship A is sailing west at 35 km/hr, and ship B is sailing north at 35 km/hr.
How fast is the distance between the ships changing at 2:00 P.M.?
I hope you made a sketch . . .
Code:
B * | \ | \ | \ 35t | \ x | \ | \ | \ * - - - - - - - * - - - - - - * Q 125-35t A 35t P : - - - - - - 125 - - - - - - :
At noon, ship A is at point P.
In $t$ hours, it has moved 35t km to point A.
Then: $AP \:=\:x$, and $QA \:=\:125-35t$
At noon, ship B is at point Q.
In $t$ hours, it has moved 35t km to point B.
Their distance is: . $x \;=\;\sqrt{(125-35t)^2 + (35t)^2}$
So we have: . $x \;=\;\left(2450t^2 - 8750t + 15,625)^{\frac{1}{2}}$
Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
The following is multiple choice question (with options) to answer.
At noon, ship A is 100km west of ship B. Ship A is sailing east at 35 km/hr and ship B is sailing north at 25km/hr. How fast is the distance between the ships changing at 4.00pm. | [
"35.21 km/hr.",
"36.21 km/hr.",
"37.21 km/hr.",
"39.21 km/hr."
] | B | Let the position of ship B at noon be at the origin
=>ship A has x-coordinate = -100km and B has x-coordinate =0km
At 4.00 p.m., ship A is at x = -100+4*35=40km
and B is at x = 0 and y = 4 * 25 = 100 km
Distance between them at 4.00 p.m. = √[(100)^2 + (40)^2] = 20√(29) km
Distance between the ships,
s^2 = x^2 + y^2
=> 2s ds/dt = 2x dx/dt + 2y dy/dt
=> rate of change of distance between the ships, ds/dt
= (x dx/dt + y dy/dt) / s
= [40 * 35 + 100 * 25] / [20√(29)] km/hr
= (1400 + 2500) / [20√(29)] km/hr
= 195/√(29) km/hr
≈ 36.21 km/hr.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35292 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
A boy wants to go abroad by boat and return by flight. He has a choice of 3 different boat to go and 4 flight to return. In how many ways, can the boy perform his journey? | [
"20",
"40",
"18",
"12"
] | D | Ans.(D)
Sol. Number of choices to go by ship = 3 Number of choices to return by airline = 4 From, fundamental principle, total number of ways of performing the journey = 3 x 4 = 12 ways |
AQUA-RAT | AQUA-RAT-35293 | there are 60 total minutes. So if the input is like hour = 12 and min := 30, then the result will be 165°. 8. 7. Formulas for Clock 1. The distance between the 2 and the 3 on the clock is 30°. (A) 100 0 (B) 80 0 (C) 120 0 (D) 200 0. The hour hand is in #3 [which is actually 15 minutes.] Angle 2 = 60°. University of California-Berkeley, Bachelor in Arts, Cellular and Molecular Biology. I understand this working all the way up until the '2n-1' part, where n is a positive integer. Write a program to determine the angle between the hands of a clock. If Varsity Tutors takes action in response to ø = 30h – 11/2 m -----when minute hand is in first 1/2 (between 12 to 6) 54 = 30 × 7 – 11/2 m. 11/2 m = 156. m = 28.36 min = 28 4/11 min = 21.8 s. at 7 hr 28 4/11 min. $${\text{192}}{\frac{1}{2}^ \circ }$$ C. 195º. M = Minute. View Answer. This problem is know as Clock angle problem where we need to find angle between hands of an analog clock at a given time. A clock is started at noon. B. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such If the required angle is A, then A = 30*H - (11*M/2) where H= hour hand, M= Minute hand. Angle Between Hands of a Clock. The angle between the two hands must be 10. Step 2: Press the "Calculate" button. as Boston University, Bachelor of Science, Business Administration and Management. It happens due to only one such incident between 12 and 1'o clock. In one hour, they will form two right angles and in 12 hours there are only 22 right angles. In each hour the minute hand covers more distance of 55 min in comparison to the hour hand. ... Find the angle between the minute hand and hour hand of a clock when the time is 7.20? Suppose we have two numbers, hour and minutes. on or linked-to by the Website infringes your
The following is multiple choice question (with options) to answer.
At 6′o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12′o clock | [
"47",
"76",
"28",
"66"
] | D | Explanation:
For ticking 6 times, there are 5 intervals.
Each interval has time duration of 30/5 = 6 secs
At 12 o'clock, there are 11 intervals,
So total time for 11 intervals = 11 × 6 = 66 secs.
Answer: D |
AQUA-RAT | AQUA-RAT-35294 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What should be the least number to be added to the 2496 number to make it divisible by 5? | [
"12",
"17",
"18",
"4"
] | D | Answer : 4
Option : D |
AQUA-RAT | AQUA-RAT-35295 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
The interest on a certain deposit at 4.5% p.a. is Rs. 202.50 in one year. How much will the additional interest in one year be on the same deposit at 5% p.a ? | [
"Rs. 20.25",
"Rs. 22.50",
"Rs. 25",
"Rs. 42.75"
] | B | Solution
S.I=Rs.202.50, R=4.5%, T= 1 Year
principal = Rs (100*202.50 / 4.5*1)= Rs.4500
Now P =Rs.4500, R =5%, T= 1 year
S.I=(4500*5*1/100)= Rs.225
Different in interest = Rs. ( 225- 202.50)= Rs.22.50
Answer B |
AQUA-RAT | AQUA-RAT-35296 | => 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.
Sheila works 8 hours per day on Monday, Wednesday and Friday, and 6 hours per day on Tuesday and Thursday. She does not work on Saturday and Sunday. She earns $396 per week. How much does she earn in dollars per hour? | [
"11",
"10",
"9",
"8"
] | A | Let Sheila earn x dollars per hour
So, on Monday, Wednesday and Friday , she earns 8x each
And, on Tuesday and Thursday, she earns 6x each
In total , over the week she should earn, 3(8x) + 2(6x) = 36x
She earns $396 per week
36x =396
x =11
Correct Option : A |
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