source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37597 | (1) $$a^2+b^2>16$$
Doesn't tell us about the value of a. a could be 2 or 10 or many other values.
(2) a=|b|+5
a is 5 or greater since |b| is at least 0. This tells us that 'a' does not lie between -4 and 4. Hence this statement is sufficient to answer the question.
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The following is multiple choice question (with options) to answer.
a is an integer greater than 9 but less than 21, b is an integer greater than 19 but less than 31, what is the range of a/b? | [
"1/4",
"1/2",
"2/3",
"1"
] | C | min value of a/b will be when b is highest and a is lowest ---> a=10 and b=30
So, a/b = 1/3
max value of a/b will be when b is lowest and a is highest ---> a=20 and b=20
So, a/b = 1
Range is 1-(1/3) = 2/3.
Answer should be C |
AQUA-RAT | AQUA-RAT-37598 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is? | [
"11",
"77",
"60",
"12"
] | C | Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(100 + 100)/8 = 3x => x = 25/3.
So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.
Answer:C |
AQUA-RAT | AQUA-RAT-37599 | 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 bottle contains a certain solution. In the bottled solution, the ratio of water to soap is 3:5, and the ratio of soap to salt is four times this ratio. The solution is poured into an open container, and after some time, the ratio of water to soap in the open container is quartered by water evaporation. At that time, what is the ratio of water to salt in the solution? | [
"9:100",
"100:9",
"12:100",
"12:20"
] | A | Water:soap = 3:5
Soap:Salt=12:20
=> For 12 soap, salt = 20
=> For 5 Soap, salt = (20/12)*5 = 100/12=25/3
So, water:soap:salt = 3:5:25/3 = 9:15:25
After open container, water:soap:salt = 2.25:15:25
So, water:salt = 2.25:25 = 9:100
ANSWER:A |
AQUA-RAT | AQUA-RAT-37600 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
A sum of 1200 lent at S.I at 10% will become twice in how many years? | [
"12 years",
"14 years",
"10 years",
"8 years"
] | C | Explanation:
To become twice means S.I should be as much as initial sum i.e, 1200
(1200 x 10 x T)/100 = 1200
T = 10 years
Answer: Option C |
AQUA-RAT | AQUA-RAT-37601 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
The banker's discount on a certain sum due 2 years hence is 11/10 of the true discount.The rate percent is | [
"11%",
"10%",
"5%",
"5.5%"
] | C | Solution
Let T.D be Rs 1.Then, B.D =Rs.11 / 10
= Rs.1.10.
Sum =Rs.(1.10 x 1/ 1.10-1)
= Rs.(110 / 10)
= Rs. 11.
S.I on Rs.11 for 2 years is Rs.1.10.
Rate =(100 x 1.10 / 11 x 2)%
= 5%
Answer C |
AQUA-RAT | AQUA-RAT-37602 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A is a working partner and B is a sleeping partner in the business. A puts in Rs.15000 and B Rs.25000, A receives 10% of the profit for managing the business the rest being divided in proportion of their capitals. Out of a total profit of Rs.9600, money received by A is? | [
"1797",
"2970",
"2782",
"4200"
] | D | 15:25 => 3:5
9600*10/100 = 960
9600 - 960 = 8640
8640*3/8 = 3240 + 960
= 4200
Answer:D |
AQUA-RAT | AQUA-RAT-37603 | # probability of the 3rd pick is a boy
10 students are in a class,
4 of these students are boys and 6 are girls
the teacher wanted to randomly select 3 students to represent the class in an event
what is the probability that the 3rd student is a boy?
Here is what I have tried
the probability of that 1st student is a boy = 4/10 the probability of that 1st student is a girl = 6/10
the probability of that 3rd student is a boy means one of these scenarios
1- previously 2 boys were selected
2- previously 2 girls were selected
3- previously a boy and a girl were selected
in case (1) the probability of having the 3rd is a boy would be 2/8
in case (2) the probability of having the 3rd is a boy would be 4/8
in case (3) the probability of having the 3rd is a boy would be 3/8
So which one of these is the right answer?!
• You can do it that way but there's no need to, each choice is like any other so the answer is $\frac 4{10}$. To do it the way you started, you need to weight each case by the probability of being in that scenario. – lulu Apr 28 at 22:50
• @lulu but when we select the 1st and 2nd student, number of students change and the distribution changed as well.. how come it is the same as the 1st pick? – asmgx Apr 28 at 22:53
• If you specify the first two choices then of course the answer changes, but if you don't specify them then there is no information from them, – lulu Apr 28 at 22:56
• – JMoravitz Apr 28 at 22:59
• If you were to simplify that god-awful expression however... it very simply results in $\frac{4}{10}$... which as many of the answers and comments in the linked question as well as other answers and comments here already will tell you makes sense as being "the third" person to be selected is not different enough from being "the first" person selected, so naturally the probability the third is a boy is going to be the same as the probability the first is a boy which is obviously $\frac{4}{10}$ with no tedious calculations required. – JMoravitz Apr 28 at 23:50
The following is multiple choice question (with options) to answer.
A teacher will pick a group of 3 students from a group of 7 students that includes Bart and Lisa. If one of all the possible 3-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa? | [
"1/7",
"3/14",
"1/4",
"1/3"
] | A | Probability = Favorable Outcomes / Total Outcomes
Total Outcomes= Total No. of ways of Picking Group of 3 out of 7 = 7C3 = 7! / (4! * 3!) = 35
Favorable Outcomes= Total No. of ways of Picking Group of 3 out of 7 such that B and L are always in the group (i.e. we only have to pick remaining two out of remaining 5 as B and L must be there is group) = 5C1 = 5
Hence,Probability=5/35=1/7
Answer: Option A |
AQUA-RAT | AQUA-RAT-37604 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, thereby increasing his average by 6 runs. His new average is : | [
"45 runs",
"46 runs",
"48 runs",
"49 runs"
] | C | 10(x-8) + 108 = 11x , x = 48 . This is the shortest way
or
10*6=60 runs for 10 innings and 108 in 11th innings then 108-60=48 runs
ANSWER:C |
AQUA-RAT | AQUA-RAT-37605 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief is spotted by a policeman from a distance of 175 meters. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken? | [
"350m",
"200m",
"400m",
"700m"
] | D | Relative speed of the policeman = (10-8) km/hr =2 km/hr.
Time taken by police man to cover
(175m/1000) x 1/2 hr = 7/80hr.
In 7/80 hrs, the thief covers a distance of 8 x 7/80 km = 7/10 km = 700 m
Answer is D. |
AQUA-RAT | AQUA-RAT-37606 | respectively. The 4th number is the addition of 2nd and 3rd number i.e. [4] The first few roots are 0, https://www.dur.ac.uk/bob.johnson/fibonacci/, https://maths.dur.ac.uk/~dma0rcj/PED/fib.pdf, https://home.att.net/~blair.kelly/mathematics/fibonacci/, https://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html. = − (−) Where = +, the golden ratio. Using equation (7), the definition of can be extended New York: Blaisdell, 1961. The Fibonacci Brousseau, A. The rest of the numbers are obtained by the sum of the previous two numbers in the series. 131-132). So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first -quite a task, even with a … The #1 tool for creating Demonstrations and anything technical. If and are two positive integers, then between and , there can "Fibonacci Resources." Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. "Fibonacci Numbers." 1 episode "Sabotage" Sequence--Part V." Fib. Oxford, England: Oxford University Press, 1966. New York: Wiley, 2001. which holds for arbitrary integers , , , , and with and from Johnson (2003) gives the very general identity. Guy (1990) notes the curious fact that for 10, 3, nombres remarquables. Math. Informatique 3, 36-57, 1991-1992. which corresponds to the decimal digits of A037917, A037918, Amer., 1985. of ways of picking a set (including the empty set) from A082116, A082117, https://mathworld.wolfram.com/FibonacciNumber.html. Freeman, pp. Middlesex, England: Fibonacci Sequence & Nature." The Fibonacci numbers obey the
The following is multiple choice question (with options) to answer.
What is the 4 digit number in which the 1st digit is 1/3 of the second, the 3rd is the sum of the 1st and 2nd, and the last is three times the second? | [
"1100",
"1200",
"1349",
"1430"
] | C | First digit is 1/3 second digit => The numbers can be 1 & 3, 2& 6, 3 & 9.
First + second = third => we can eliminate 3 & 9 since 3 + 9 = 12.
Last is 3 times the second => we can eliminate option 2 & 6 since 3 * 6 = 18.
Hence the number is 1349
C |
AQUA-RAT | AQUA-RAT-37607 | Maybe you should use a different approach.
It can be found that sets with the least number of elements are of the following type:
$$\{0,1,a_1,\dots ,a_m,\frac{n}{2}-a_m,\dots ,\frac{n}{2}-a_1,\frac{n}{2}-1,\frac{n}{2}\}$$
or
$$\{0,1,a_1,\dots ,a_m,\frac{n}{4},\frac{n}{2}-a_m,\dots ,\frac{n}{2}-a_1,\frac{n}{2}-1,\frac{n}{2}\}$$
example $$n \leq 20$$
elements can be found easily with this set
$$\{0,1,3,\dots ,\frac{n}{2}-1,\frac{n}{2}\}$$ number of elements $$2+\frac{n}{4}$$
then
$$\{0,1\}$$ for $$0 \leq n \leq 2$$
$$\{0,1,2\}$$ for $$3 \leq n \leq 4$$
$$\{0,1,3,4\}$$ for $$5 \leq n \leq 8$$
$$\{0,1,3,5,6\}$$ for $$9 \leq n \leq 12$$
$$\{0,1,3,5,7,8\}$$ for $$13 \leq n \leq 16$$
$$\{0,1,3,5,7,9,10\}$$ for $$17 \leq n \leq 20$$
example $$n >20$$
$$\{0,1,3,4,9,10,12,13\}$$ for $$21 \leq n \leq 26$$
The following is multiple choice question (with options) to answer.
Consider the set (0.34,0.304,0.034,0.43) The sum of the smallest and largest numbers
in the set is | [
"0.77",
"0.734",
"0.077",
"0.464"
] | D | Since each of the numbers in the set is between 0 and 1, then the tenths digit of each number
contributes more to its value than any of its other digits.
The largest tenths digit of the given numbers is 4, and so 0.43 is the largest number in the set.
The smallest tenths digit of the given numbers is 0, so 0.034 is the smallest number in the set.
Therefore, the sum of the smallest number in the set and the largest number in the set is
0.034 + 0.43 = 0.464
correct answer D |
AQUA-RAT | AQUA-RAT-37608 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
If the perimeter of a rectangular garden is 800 m, its length when its breadth is 100 m is? | [
"286 m",
"899 m",
"300 m",
"166 m"
] | C | 2(l + 100) = 800 => l
= 300 m
Answer: C |
AQUA-RAT | AQUA-RAT-37609 | # 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.
Bill has 75 cents in his pocket. He walks up to a food stand and asks how much for a hot dog. The vendor just raised the price of his hot dogs by 10% but believes Bill is down on his luck and so he sells the hot dog to Bill for 75 cents which is exactly 75% of his new price for hot dogs. What is the difference in the new price of hot dogs and the price the vendor sold a hot dog to Bill for? | [
"8.5%",
"17.5%",
"12.5%",
"14.5%"
] | B | Quantity X Rate = Price
1 X 1 = 1
0.75 X 1.10 = 0.825
Decrease in price = (0.175/1) × 100 = 17.5%
Answer = Option B |
AQUA-RAT | AQUA-RAT-37610 | Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:
0.04(total fish) = 50
This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:
Quote:
.04 (total) = tagged fish
We are missing two variables. We don't know the total fish and we don't know the tagged fish.
If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.
If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."
From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.
Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Lake A has a duck population of 30 females and 30 males. Lake B’s duck population consists of the same ratio of females to males. If 6 new female ducks are added to Lake B and no other ducks are added or removed, the new total duck population of Lake B could be which of the following? | [
"144",
"105",
"78",
"72"
] | D | Lake A has a duck population of 30 females and 36 males --> the ratio is F:M = 30:30 = 1:1.
Lake B’s duck population consists of the same ratio of females to males --> F:M = 1:1.
D |
AQUA-RAT | AQUA-RAT-37611 | # Find the remainder when $10^{400}$ is divided by 199?
I am trying to solve a problem
Find the remainder when the $10^{400}$ is divided by 199?
I tried it by breaking $10^{400}$ to $1000^{133}*10$ .
And when 1000 is divided by 199 remainder is 5.
So finally we have to find a remainder of :
$5^{133}*10$
But from here I could not find anything so that it can be reduced to smaller numbers.
How can I achieve this?
Is there is any special defined way to solve this type of problem where denominator is a big prime number?
• $10^{400}=1000^{133}\times10$, not $1000^{333}\times10$. – Gerry Myerson Oct 9 '12 at 4:53
• A standard beginning (for prime moduli) is to use the fact that if $p$ does not divide $a$, then $a^{p-1}\equiv 1\pmod{p}$. Thus $10^{198}\equiv 1\pmod{199}$. It follows that $10^{396}\equiv 1\pmod{199}$ and therefore $10^{400}\equiv 10^4\pmod{199}$. Now we have to calculate. In this case, there is a further shortcut, since $1000=(5)(199)+5\equiv 5\pmod{199}$. – André Nicolas Oct 9 '12 at 5:24
• – Martin Sleziak Jun 17 '16 at 8:22
You can use Fermat's little theorem. It states that if $n$ is prime then $a^n$ has the same remainder as $a$ when divided by $n$.
So, $10^{400} = 10^2 (10^{199})^2$. Since $10^{199}$ has remainder $10$ when divided by $199$, the remainder is therefore the same as the remainder of $10^4$ when divided by $199$. $10^4 = 10000 = 50*199 + 50$, so the remainder is $50$.
The following is multiple choice question (with options) to answer.
If n is a prime number greater than 4, what is the remainder when n^2 is divided by 12? | [
"0",
"2",
"1",
"3"
] | C | There are several algebraic ways to solve this question including the one under the spoiler. But the easiest way is as follows:since we can not have two correct answersjust pick a prime greater than 4, square it and see what would be the remainder upon division of it by 12.
n=7 --> n^2=49 --> remainder upon division 49 by 12 is 1.
Answer: C. |
AQUA-RAT | AQUA-RAT-37612 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A,B and C started a business by investing Rs.800/- , Rs.1,000/- and Rs.1,200/- respectively for two years.
Total profit is Rs.1000/-. Find the C's share? | [
"Rs. 600",
"Rs. 500",
"Rs. 400",
"Rs. 800"
] | C | Solution: A : B : C = (800 * 2) : (1,000 * 2) : (12,000 * 2) = 4 : 5 : 6.
So C's share = Rs. (1000 * 6/15) = Rs. 400.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37613 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of a train and that of a platform are equal. If with a speed of 72 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is? | [
"600",
"758",
"718",
"750"
] | A | Speed = [72 * 5/18] m/sec = 20 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 20 è x = 20 * 60 / 2
=600
Answer:A |
AQUA-RAT | AQUA-RAT-37614 | Kudos [?]: 4 [2] , given: 0
Re: Good set of PS 2 [#permalink] 19 Oct 2009, 08:02
2
KUDOS
Bunuel wrote:
4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
This one was solved incorrectly:
Days to finish the job for 10 people 110 days.
On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people).
Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.
I solved in a more easier way I think:
1) 10 man 110 days --> need for 1100 man.days
2) 55 days with 10 men --> 550 man.days
3) 40 days with 16 men --> 640 man.days
--> total man.days equals 1190 vs need for 1100 --> days of rain equals 90/16 max --> 5.625 --> rounded to 5
Senior Manager
Joined: 31 Aug 2009
Posts: 420
Location: Sydney, Australia
Followers: 6
Kudos [?]: 165 [1] , given: 20
The following is multiple choice question (with options) to answer.
Working in a South Side studio at a constant rate, Kanye can drop a full-length platinum LP in 5 weeks. Working at his own constant rate, Common can drop a full-length platinum LP in x weeks. If the two emcees work together at their independent rates, they can drop a full-length platinum compilation LP in 2 weeks. Assuming no efficiency is lost or gained from working together, how many weeks would it take Common, working alone, to drop a full-length platinum LP? | [
" 3 and 1/3 weeks",
" 3 weeks",
" 2 and 1/2 weeks",
" 2 and 1/3 weeks"
] | A | VERITAS PREPOFFICIAL SOLUTION:
Now, while your instinct may be to Go! and speed through your initial read of this rate problem, remember: slow motion (is) better than no motion. As you read each sentence, you should start jotting down variables and relationships so that by the time you get to the question mark you have actionable math on your noteboard and you don’t have to read the question all over again to get started. You should be thinking:
Working in a South Side studio at a constant rate, Kanye can drop a full-length platinum LP in 5 weeks.
Rate (K) = 1 album / 5 weeks
Working at his own constant rate, Common can drop a full-length platinum LP in x weeks.
Rate (C) = 1 album / x weeks
If the two emcees work together…
I’m adding these rates, so their combined rate is 1/5 + 1/x
…they can drop a full-length platinum compilation LP in 2 weeks.
And they’re giving me the combined rate of 1 album / 2 weeks, so 1/5 + 1/x = 1/2
Assuming no efficiency is lost or gained from working together, how many weeks would it take Common, working alone, to drop a full-length platinum LP?
I’m using that equation to solve for Common’s time, so I’m solving for x.
Now by this point, that slow motion has paid off – your equation is set, your variable is assigned, and you know what you’ve solving for. Your job is to solve for x, so:
1/5 + 1/x = 1/2, so let’s get the x term on its own:
1/x = 1/2 – 1/5. and we can combine the two numeric terms by finding a common denominator of 10:
1/x = 5/10 – 2/10
1/x = 3/10, and from here you have options but let’s cross multiply:
10 = 3x, so divide both sides by 3 to get x alone:
10/3 = x, and that doesn’t look like the answer choices so let’s convert to a mixed number: 3 and 1/3 (there’s that number again), for answer choice A. |
AQUA-RAT | AQUA-RAT-37615 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A contractor undertook to do a piece of work in 10 days. He employed certain number of laboures but 5 of them were absent from the very first day and the rest could finish the work in only 13 days. Find the number of men originally employed ? | [
"21.6",
"23.6",
"22.6",
"21.8"
] | A | Let the number of men originally employed be x.
10x = 13(x – 5)
or x = 21.6
Answer A |
AQUA-RAT | AQUA-RAT-37616 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio between the present ages of A and B is 5:3 respectively. The ratio between A's age 4 years ago and B's age 4 years hence is 1:1. What is the ratio between A's age 4 years hence and B's age 4 years ago? | [
"3:9",
"3:7",
"3:1",
"3:2"
] | C | Let the present ages of A and B be 5x and 3x years respectively.
Then, (5x - 4)/(3x + 4) = 1/1
2x = 8 => x = 4
Required ratio = (5x + 4):(3x - 4) = 24:8 = 3:1
Answer: C |
AQUA-RAT | AQUA-RAT-37617 | # 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.
If selling price is doubled, the profit triples. Find the profit percent. | [
"66%",
"120%",
"105%",
"100%"
] | D | Let C.P. be Rs. x and S.P. be Rs. y.
Then, 3(y - x) = (2y - x)
y = 2x.
Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.
Profit % = x/x x 100% =100%
Answer is D. |
AQUA-RAT | AQUA-RAT-37618 | 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 seconds and a platform 100 m long in 25 seconds. Its length is? | [
"886 m",
"787 m",
"876 m",
"150 m"
] | D | Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.
Answer:D |
AQUA-RAT | AQUA-RAT-37619 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
Veena ranks 79rd from the top in a class of 182. What is her rank from the bottom if 22 students have failed the examination? | [
"88",
"108",
"82",
"90"
] | C | total student=182
failed=22
paasd student=182-22=160
from bottom her rank is=160-79+1=82
ANSWER:C |
AQUA-RAT | AQUA-RAT-37620 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
From a group of 6boys&4girls a Committee of 4 persons is to be formed. In how many different ways can it be done, so that the committee has at least 2 boys? | [
"145",
"198",
"185",
"210"
] | C | The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
(i) 2m. 2w in 6C2 x 4 C1=6X5 - 2X1 X 3X3 2X1 =n an (ii) 3m. 1w in 6C3 x 4C1 =6X5X4 3X2X1 X4 _n - a ''''
(iii) 4m in 6C4 -2x1 = 15 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men. = 90 + 18 + 15 = 185
C |
AQUA-RAT | AQUA-RAT-37621 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
How many positive 5-digit integers have the odd sum of their digits? | [
"9*10^2",
"9*10^3",
"10^4",
"45*10^3"
] | D | We are looking at numbers between 10000 and 99999 both inclusive.
There are 90000 numbers.
Now for
10000 : sum of digits is odd;
10001 :sum of digits is even;
10002 : sum of digits is odd ; so on and so forth. So every alternate number is such that the sum of digit is odd.
(Exception for the above statement :
When it is 10009 the sum is even and for 10010 again the sum is even; But if you look at 10019 :sum is odd; 10020 : sum is odd
and this pattern continues so basically the number of odd sum of digits and even sum of digits are equal)
This means exactly half of the numbers will have odd sum of their digits. i.e 45000
Answer : D |
AQUA-RAT | AQUA-RAT-37622 | a bijection between X and Y if and only if both X and Y have the same number of elements. Loreaux, Jireh. Answer. Two simple properties that functions may have turn out to be exceptionally useful. If a and b are not equal, then f(a) ≠ f(b). For every y ∈ Y, there is x ∈ X such that f(x) = y How to check if function is onto - Method 1 In this method, we check for each and every element manually if it has unique image Check whether the following are onto? Then, at last we get our required function as f : Z → Z given by. Suppose that and . Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … f(x) = 0 if x ≤ 0 = x/2 if x > 0 & x is even = -(x+1)/2 if x > 0 & x is odd. Springer Science and Business Media. Sometimes functions that are injective are designated by an arrow with a barbed tail going between the domain and the range, like this f: X ↣ Y. Image 2 and image 5 thin yellow curve. Foundations of Topology: 2nd edition study guide. For example, 4 is 3 more than 1, but 1 is not an element of A so 4 isn't hit by the mapping. The function value at x = 1 is equal to the function value at x = 1. This function is an injection because every element in A maps to a different element in B. (the factorial function) where both sets A and B are the set of all positive integers (1, 2, 3...). ... Function example: Counting primes ... GVSUmath 2,146 views. There are special identity transformations for each of the basic operations. Why it's injective: Everything in set A matches to something in B because factorials only produce positive integers. HARD. Plus, the graph of any function that meets every vertical and horizontal line exactly once is a bijection. A one-one function is also called an Injective function. De nition 68. Let the extended function be f. For our example let f(x) = 0 if x is a negative integer. A function $f: R \rightarrow S$ is simply a unique
The following is multiple choice question (with options) to answer.
The function f is defined for all positive integers V by the following rule. f(V) is the number of positive integers each of which is less than V and has no positive factor in common with V other than 1. If Y is any prime, number then f(Y)= | [
"Y-1",
"Y-2",
"(Y+1)/2",
"(Y-1)/2"
] | A | If not the wording the question wouldn't be as tough as it is now. The GMAT often hides some simple concept in complicated way of delivering it.
This question for instance basically asks: how many positive integers are less than given prime number Y which have no common factor with Y except 1.
Well as Y is a prime, all positive numbers less than Y have no common factors with Y (except common factor 1). So there would be Y-1 such numbers (as we are looking number of integers less than Y).
For example: if Y=3 how many numbers are less than 3 having no common factors with 3: 1, 2, 3, --> 3-1=2.
Answer: A. |
AQUA-RAT | AQUA-RAT-37623 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
Ajay can ride 50km in 1 hour. In how many hours he can ride 1250km? | [
"10hrs",
"15hrs",
"20hrs",
"25hrs"
] | D | 1 hour he ride 50km
he ride 1250km in = 1250/50 * 1 = 25hours
Answer is D |
AQUA-RAT | AQUA-RAT-37624 | sql, sql-server, t-sql, stackexchange
Title: Popular questions by view count I made this query to create a graph of a user's popular questions and the view count on that question. It allows for a minimum of 500 views, and a score of 3.
DECLARE @allowed_min_views INT = 500;
DECLARE @allowed_min_score INT = 3;
DECLARE @user_id INT = ##UserId:int?-1##;
DECLARE @min_views INT = ##MinimumViews:int?500##;
DECLARE @min_score INT = ##MinimumScore:int?3##;
DECLARE @question INT = 1;
IF (@min_views < @allowed_min_views)
BEGIN
PRINT '@MinimumViews must be larger than 499.'
END
IF (@min_score < @allowed_min_score)
BEGIN
PRINT '@MinimumScore must be larger than 2.'
END
IF (@min_views >= @allowed_min_views AND @min_score >= @allowed_min_score)
BEGIN
SELECT
ViewCount
, Score
FROM Posts WHERE
PostTypeId = @question
AND OwnerUserId = @user_id
AND ViewCount >= @min_views
AND Score >= @min_score
ORDER BY ViewCount ASC;
END
Finally, here's some sample input (I'm using @Mat'sMug's user ID):
@user_id: 23788
@min_views: 500
@min_score: 7 Good things
You use good local variables, and you are consistent with your naming. The typical naming for T-SQL is using PascalCase, however there are no standards and snake_case or camelCase work just as good, as long as you are consistent (which you are).
You validate your values, although I am not quite sure why you chose 500 and 3 as arbitrary minimums (might be worth documenting).
The following is multiple choice question (with options) to answer.
To be considered for “movie of the year,” a film must appear in at least 1/4 of the top-10-movies lists submitted by the Cinematic Academy’s 795 members. What is the smallest number of top-10 lists a film can appear on and still be considered for “movie of the year”? | [
"191",
"199",
"193",
"212"
] | B | Total movies submitted are 795.
As per question we need to take 1/4 of 795 to be considered for top 10 movies = 198.75
approximate the value we 199.
IMO option B is the correct answer... |
AQUA-RAT | AQUA-RAT-37625 | # Probability of a certain ball drawn from one box given that other balls were drawn
Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
The following is multiple choice question (with options) to answer.
In a box, there are 10 red, 6 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? | [
"4/51",
"10/21",
"7/12",
"11/31"
] | B | Total number of balls = 10+6+5 = 21
E = event that the ball drawn is neither red nor green
= event that the ball drawn is red
n(E) = 10
P(E) = 10/21
Answer is B |
AQUA-RAT | AQUA-RAT-37626 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
John purchased some shirts and trousers for $1550. He paid $250 less for the shirts than he did for the trousers. If he bought 5 shirts and the cost of a shirt is $20 less than that of a trouser, how many trousers did he buy? | [
"4",
"5",
"6",
"7"
] | C | Given that the total purchase of two items cost 1550.
So the average purchase of one item will cost 1550/2 =775.
Its given as total shirt cost 250$ less. Hence Total Shirt cost = 775 - 125 and Total trouser cost = 775 + 125
5 shirts =650$ ==> One shirt =130$
One trouser = 130 + 20 =150$
Total trousers = 900 / 150 =6.
C |
AQUA-RAT | AQUA-RAT-37627 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
If (2^x)(3^y) = 324, where x and y are positive integers, then (2^x-1)(3^y-2) equals: | [
"16",
"18",
"48",
"96"
] | B | So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=2, y=4. now quickly 2^1(3^2)=18. Than answer should be number 2.
B |
AQUA-RAT | AQUA-RAT-37628 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
When positive integer a is divided by 6, the remainder is 1. Which of the following must be true?
I. a is a prime number
II. a is odd
III. a is divisible by 7 | [
"I only",
"II only",
"III only",
"I and II only"
] | B | a=6q+1
Examine II) put q=0,1,2,3....... we have a=1,7,13,...25 so a must be odd
Examine III) 14/6 has reminder 249/7 has reminder 1 so not always true
Examine I) a= 7 true but while 2 is prime, it does not give reminder 1
Answer: B |
AQUA-RAT | AQUA-RAT-37629 | rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron’s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 × 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The "base times height" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b×h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal ‘d1’ of a rhombus and a side ‘a’, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of […] where b is the
The following is multiple choice question (with options) to answer.
Find the area of a rhombus one side of which measures 20 cm and one diagonal is 23 cm. | [
"100cm2",
"150cm2",
"300cm2",
"368cm2"
] | D | Explanation: Let other diagonal = 2x cm.
Since diagonals of a rhombus bisect each other at right angles, we have:
(20)2 = (12)2 + (x)2 =>x =√(20)2 – (12)2= √256= 16 cm. _I
So, other diagonal = 32 cm.
Area of rhombus = (1/2) x (Product of diagonals) =(1/2× 23 x 32) cm2 = 368 cm2
Answer: Option D |
AQUA-RAT | AQUA-RAT-37630 | => 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.
Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck. | [
"$ 130",
"$ 145",
"$ 115",
"$ 120"
] | B | Let the 1st paycheck be x (integer).
Mrs. Rodger got a weekly raise of $ 145.
So after completing the 1st week she will get $ (x+145).
Similarly after completing the 2nd week she will get $ (x + 145) + $ 145.
= $ (x + 145 + 145)
= $ (x + 290)
So in this way end of every week her salary will increase by $ 145.
correct answer B |
AQUA-RAT | AQUA-RAT-37631 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The price of an article is cut by 10%. To restore it to the former value. The new price must be increased by ? | [
"10%",
"9 1/11",
"11 1/9",
"11%"
] | C | Answer
Let original price = Rs. 100.
Then, new price = Rs. 90.
∴ Increased on Rs. 90 = Rs. 10
Required increase % = (10 x 100) / 90 % = 111/9%
Correct Option: C |
AQUA-RAT | AQUA-RAT-37632 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A committee has 4 men and 7 women. What are the number of ways of selecting 3 men and 4 women from the given committee? | [
"110",
"120",
"130",
"140"
] | D | The number of ways to select three men and four women = 4C3 * 7C4 = 4 * 35 = 140
The answer is D. |
AQUA-RAT | AQUA-RAT-37633 | "Thursdays with Ron - Consolidated Verbal Master List - Updated"
Math Expert
Joined: 02 Sep 2009
Posts: 39609
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Dec 2016, 23:09
Nightfury14 wrote:
Bunuel wrote:
gdk800 wrote:
1) A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)
a. 48
b. 100
c. 120
d. 288
e. 600
[Reveal] Spoiler:
OA is B.
Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$;
So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.
Hi Bunuel
The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:
Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp
Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp
Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.
The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.
_________________
Intern
Joined: 18 May 2016
Posts: 20
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Jan 2017, 17:21
Hello,
The following is multiple choice question (with options) to answer.
Each of three charities in Novel Grove Estates has 8 persons serving on its board of directors. If exactly 4 persons serve on 3 boards each and each pair of charities has 5 persons in common on their boards of directors, then how many distinct persons serve on one or more boards of directors? | [
"8",
"13",
"16",
"24"
] | B | A intersect B intersect C = 4
A intersect B but not C = 1
B intersect C but not A = 1
C intersec A but not B = 1
A only is 2; B only is 2; C only is 2
Therefore, total 13 members serve on one or more boards of directors.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37634 | audio, modulation, frequency-modulation, fsk
well, "best" is always a reduction to a single set of optimization parameters (e.g. cost per bit, durability, ...) and isn't ever "universally true".
I can see, for example, that "large" is already a relative term, and for a small office, the optimum solution for backing up "large" amounts of data is a simple hard drive, or a hard drive array.
For a company, backup tapes might be better, depending on how often they need their data back. (Tapes are inherently pretty slow and can't be accessed at "random" points)
So I figured I can store a relatively large amount of data on a cassette tape.
Uh, you might be thinking of a Music Casette, right? Although that's magnetic tape, too, it's definitely not the same tape your first sentence referred to: It's meant to store an analog audio signal with low audible distortion for playback in a least-cost cassette player, not for digital data with low probability of bit error in a computer system.
Also, Music Cassettes are a technology from 1963 (small updates afterwards). Trying to use them for the amounts of data modern computers (even arduinos) deal with sounds like you're complaining your ox cart doesn't do 100 km/h on the autobahn.
But after reading up about it for a bit it turns out that they can store very small amounts of data. With baud rates varying between 300 to 2400 something between ~200KB to ~1.5MB can be stored on a 90 minute (2x45min) standard cassette tape.
Well, so that's a lot of data for when music-cassette-style things were last used with computers (the 1980s).
Also, where do these data rates drop from? That sounds like you're basing your analysis on 1980's technology.
These guys can store 90 minutes of audio. Even if we assume the analog audio quality on them was equivalent of 32Kbps that's about 21MB of data.
The following is multiple choice question (with options) to answer.
Dropbox charges $3.10 for first 1/5 of a month for 500 GB storage, plus $0.40 for each additional 1/5 of a month. What would Dropbox charge for a data storage of 500 GB for 8 months ? | [
"$15.60",
"$16.00",
"$17.80",
"$18.70"
] | D | first 1/5 month charge = $3.10
rest of the months = 8 - (1/5) = 39/5
charge for the rest of the months = 39 *0.4 = 15.6
total charge = 3.10+15.6 = 18.7
Answer is D. |
AQUA-RAT | AQUA-RAT-37635 | the acceptance volume is the volume of water that the bladder is designed to hold, which is smaller than the overall tank size. The volume flow, Q = ∫ u(2π r)dr, can be estimated by a numerical quadrature formula such as Simpson’s rule. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308. This process is carried out in a structure called sedimentation tank or settling tank. Indirect tank heating uses a heat transfer medium to apply the heat to the tank. Motor fuel 60 to 250 gallons (Primary or alternate fuel system tanks for autos, trucks, buses. It is defined as the ratio of pressure stress to volumetric strain. The barometric formula is used in both ISA and USSA to model how the pressure and density of air changes with altitude. When the substance in the tank is near the mid-fill level, each inch of depth corresponds to a much larger volume that when the substance is near the top or bottom of the tank. We manufacture our tanks to UL-142, UL-2085, API-650, and other industry-proven standards for Above Ground Storage of Flammable and Combustible Liquids. calculate the volume of ellipsoidal head or dish 2 1. Major Axis of an Ellipse. Stated as a mathematical formula, taking into consideration the total volume of a pressure tank, it looks like this: Drawdown = P1V / P2 – P1V / P3 where, P1 is the pre-charge pressure. Inspiratory Reserve Volume (IRV). The problem can be generalized to other cones and n-sided pyramids but for the moment consider the right. Many times, this formula will use the height of the prism, or depth (d), rather than the length (l), though you may see either abbreviation. We use cookies to improve your navigation experience. Volume (V) = π * R 2 * ((4/3) * R + H) Volume (V) = 3. This means that the normal line at this point is a vertical line. Conservation. por | set 15, 2020 | Sem categoria | 0 Comentários | set 15, 2020 | Sem categoria | 0 Comentários. When a sphere is falling through a fluid it is completely submerged, so there is only
The following is multiple choice question (with options) to answer.
A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is: | [
"9",
"1",
"2",
"4"
] | B | Answer: B) 1 dm |
AQUA-RAT | AQUA-RAT-37636 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
A box contains ELEVEN bulbs out of which 5 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good? | [
"325/330",
"325/420",
"369/430",
"546/558"
] | A | Required probability
= 1-5C4/11C4
= 1 - 5/330
= 325/330
Answer:A |
AQUA-RAT | AQUA-RAT-37637 | Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240.
The following is multiple choice question (with options) to answer.
A certain gallery is hanging a series of 6 paintings. All the paintings will be exhibited in a row along a single wall. Exactly 2 of the paintings are on panel, the remainder are on canvas. In how many ways can the paintings be exhibited if the works on panel must be the second and sixth in the row?
OE | [
"240",
"200",
"122",
"48"
] | D | We need to consider them separately because the question specifies:
1. Panel paintings have to be placed only at 2nd and 6th place.
2. Implies that Canvas has to be placed 1,3,4,5th place.
So we have two separatebucketsto consider.
So panels can be ordered in 2!
Canvas can be ordered in 4!
since relative position of panels to canvas is fixed (Panels 2nd and 6th) the answer is 2! x 4! = 48
D |
AQUA-RAT | AQUA-RAT-37638 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream? | [
"76 kmph",
"17 kmph",
"14 kmph",
"18 kmph"
] | C | The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1
= 3:1
Speed of the stream = 42/3 = 14 kmph.
Answer: C |
AQUA-RAT | AQUA-RAT-37639 | Is the blue area greater than the red area?
Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?
Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}
Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.
Is this true? If so, is there another way of proving the assertion?
Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.
Source:
The following is multiple choice question (with options) to answer.
The roof of an apartment building is rectangular and its length is 3 times longer than its width. If the area of the roof is 588 feet squared, what is the difference between the length and the width of the roof? | [
"28.",
"40.",
"42.",
"44."
] | A | Answer is A: 28
Let w be the width , so length is 3w. Therefore : w*3w = 588, solving for, w = 14 , so 3w-w = 2w = 2*14 = 28 |
AQUA-RAT | AQUA-RAT-37640 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
What value will replace the question mark in the following equation ? 5172.49 + 378.352 + ? = 9318.678 | [
"3767.836",
"3677.863",
"3767.863",
"3767.683"
] | A | Let 5172.49 + 378.352 + x = 9318.678
Then , x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836
Answer is A. |
AQUA-RAT | AQUA-RAT-37641 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
If n is a prime number greater than 11, what is the remainder when n^2 is divided by 12 ? | [
"0",
"1",
"2",
"3"
] | B | There are several algebraic ways to solve this question, but the easiest way is as follows:since we cannot have two correct answers just pick a prime greater than 11, square it and see what would be the remainder upon division of it by 12.
n=17 --> n^2=289 --> remainder upon division 289 by 12 is 1.
Answer: B. |
AQUA-RAT | AQUA-RAT-37642 | 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 piece of work in 6 days and A alone can do it in 12 days. In how many days can B alone can do it? | [
"11 days",
"15 days",
"12 days",
"21 days"
] | C | Explanation:
A and B can do work 1/6 in 1 day
A alone can do 1/12 work in 1 day
B alone can do (1/6 -1/12) =1/12 work in 1 day
=> complete work can be done in 12 days by B
Answer: Option C |
AQUA-RAT | AQUA-RAT-37643 | # 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 is the smallest positive perfect square that is divisible by 9, 21, and 49? | [
"225",
"324",
"441",
"529"
] | C | The number needs to be divisible by 3^2, 3*7, and 7^2.
The smallest such perfect square is 3^2*7^2 = 441
The answer is C. |
AQUA-RAT | AQUA-RAT-37644 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
A box measuring 35 inches long by 20 inches wide by 10 inches deep is to be filled entirely with identical cubes. No space is to be left unfilled. What is the smallest number of cubes that can accomplish this objective? | [
"17",
"18",
"54",
"56"
] | D | least number of cubes will be required when the cubes that could fit in are biggest.
5 is the biggest number that could divide all three, 35, 20 and 10.
Thus side of cube must be 5, and total number of cubes = 35/5 * 20/5*10/5 = 56
Ans D it is. |
AQUA-RAT | AQUA-RAT-37645 | Slope = m = y 2 y 1 x 2 x 1 Slope = m = y 2 y 1 x 2 x 1 size 12{"Slope"=m= { {y rSub { size 8{2} } - y rSub { size 8{1} } } over {x rSub { size 8{2} } - x rSub { size 8{1} } } } } {}
### Example 6
#### Problem 1
Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.
### Example 7
#### Problem 1
Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.
### Example 8
#### Problem 1
Graph the line that passes through the point (1, 2) and has slope 3434 size 12{ - { {3} over {4} } } {} .
### Example 9
#### Problem 1
Find the slope of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
### Example 10
#### Problem 1
Find the slope of the line y=3x+2y=3x+2 size 12{y=3x+2} {}.
### Example 11
#### Problem 1
Determine the slope and y-intercept of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
## Determining the Equation of a Line
### Section Overview
In this section, you will learn to:
1. Find an equation of a line if a point and the slope are given.
2. Find an equation of a line if two points are given.
So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on it, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.
An equation of a line can be written in two forms, the slope-intercept form or the standard form.
The following is multiple choice question (with options) to answer.
A line that passes through (–1, –4) and (4, k) has a slope = k. What is the value of k? | [
" 3/4",
" 1",
" 4/3",
" 2"
] | B | Slope = (y2-y1)/(x2-x1)
=> k = (k+4)/(4+1)
=> 5k = k+4
=> k =1
Ans B it is! |
AQUA-RAT | AQUA-RAT-37646 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? | [
"520",
"720",
"890",
"870"
] | B | The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
ANSWER B 72 |
AQUA-RAT | AQUA-RAT-37647 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A, B and C started a business with capitals of Rs. 8000, Rs. 10000 and Rs. 12000 respectively. At the end of the year, the profit share of B is Rs. 1500. The difference between the profit shares of A and C is? | [
"Rs.628",
"Rs.629",
"Rs.629",
"Rs.600"
] | D | Ratio of investments of A, B and C is 8000 : 10000 : 12000 = 4 : 5 : 6
And also given that, profit share of B is Rs. 1500
=> 5 parts out of 15 parts is Rs. 1500
Now, required difference is 6 - 4 = 2 parts
Required difference = 2/5 (1500) = Rs.600'
Answer: D |
AQUA-RAT | AQUA-RAT-37648 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
For a race a distance of 224 meters can be covered by P in 14 seconds and Q in 32 seconds. By what distance does P defeat Q eventually? | [
"232m",
"288m",
"324m",
"231m"
] | B | Explanation:
This is a simple speed time problem. Given conditions:
=>Speed of P= 224/14 = 16m/s
=>Speed of Q=224/32 = 7m/s
=>Difference in time taken = 18 seconds
Therefore, Distance covered by P in that time= 16m/s x 18 seconds = 288 metres
ANSWER: B |
AQUA-RAT | AQUA-RAT-37649 | # 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 list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is? | [
"1",
"4",
"5",
"6"
] | B | 65*(90/100)*((100-x)/100) = 56.16
x = 4%
Answer: B |
AQUA-RAT | AQUA-RAT-37650 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
A sum of money invested at C.I. amounts to Rs. 800 in 3 years to Rs. 840 in 4 years. The rate of interest per annum is? | [
"3",
"2",
"5",
"7"
] | C | Explanation:
S.I. on Rs. 800 for 1 year = (840 - 800) = Rs. 40 Rate = (100 * 40)/(800 * 1) = 5%
Answer:C |
AQUA-RAT | AQUA-RAT-37651 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds does a train 120 meters long, traveling at 54 km/h, take to completely cross a bridge of 180 meters in length? | [
"18",
"20",
"22",
"24"
] | B | 54 km/h = 54000/3600 = 15 m/s
Time = 300 / 15 = 20 seconds
The answer is B. |
AQUA-RAT | AQUA-RAT-37652 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If x + 5 > 2 and x - 3 < 6, the value of x must be between which of the following pairs of numbers? | [
" -3 and 10",
" -3 and 9",
" 2 and 7",
" 3 and 4"
] | B | Let’s isolate x in each inequality.
x + 5 > 2
x > -3
Next we simplify x – 3 < 7.
x – 3 < 6
x < 9
We know that x is greater than -3 and less than 9.
The answer is B. |
AQUA-RAT | AQUA-RAT-37653 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
In a stable there are men and horses. In all, there are 22 heads and 72 feet. How many men and how many horses are in the stable? | [
"8 men and 14 horses.",
"7 men and 14 horses.",
"6 men and 14 horses.",
"5 men and 14 horses."
] | A | 8 men and 14 horses.
Let M = men and H = horses. We can come up with 2 equations.
M + H = 22
2M + 4H= 72
Solving the 2 equations will give 8 men and 14 horses.
Answer A |
AQUA-RAT | AQUA-RAT-37654 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"120 metres",
"180 metres",
"324 metres",
"150 metres"
] | D | Speed=(60 * 5/18) m/sec = (50/3) m/sec Length of the train = (Speed x Time) = (50/3 * 9) m = 150 m.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37655 | # Is $\frac{200!}{(10!)^{20} \cdot 19!}$ an integer or not?
A friend of mine asked me to prove that $$\frac{200!}{(10!)^{20}}$$ is an integer
I used a basic example in which I assumed that there are $200$ objects places in $20$ boxes (which means that effectively there are $10$ objects in one box). One more condition that I adopted was that the boxes are distinguishable but the items within each box are not. Now the number of permutations possible for such an arrangement are : $$\frac{200!}{\underbrace{10! \cdot 10! \cdot 10!\cdots 10!}_{\text{20 times}}}$$ $$\Rightarrow \frac{200!}{(10!) ^{20}}$$
Since these are just ways of arranging, we can be pretty sure that this number is an integer.
Then he made the problem more complex by adding a $19!$ in the denominator, thus making the problem: Is $$\frac{200!}{(10!)^{20} \cdot 19!}$$ an integer or not?
The $19!$ in the denominator seemed to be pretty odd and hence I couldn't find any intuitive way to determine the thing. Can anybody please help me with the problem?
The following is multiple choice question (with options) to answer.
How many of the positive factors of 19 are not factors of 29? | [
"0",
"1",
"2",
"3"
] | B | Factors of 19 - 1, 19
factors of 29- 1, 29
Comparing both, we have three factors of 19 which are not factors of 29- 19,
The answer is B |
AQUA-RAT | AQUA-RAT-37656 | 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 130 m long crosses a platform with a speed of 70 kmph in 20 sec; find the length of the platform? | [
"150",
"145",
"165",
"162"
] | D | D = 130 + x
T = 20
S = 130+x/15 * 18/5 = 70 kmph.
=> (130+x) * 18 = 5250
=> 130 + x = 292 => x= 162 m
Answer: D |
AQUA-RAT | AQUA-RAT-37657 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper forced to sell at cost price, uses a 850 grams weight for a kilogram. What is his gain percent? | [
"17.65%",
"9%",
"11.11%",
"12 %"
] | A | Shopkeeper sells 850g instead of 1000g.
So, his gain = 1000 - 850 = 150g.
Thus, %Gain = (150 *100)/850 = 17.65%.
answer : OPTION A |
AQUA-RAT | AQUA-RAT-37658 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A store sells chairs and tables. If the price of 2 chairs and 1 table is 60% of the price of 1 chair and 2 tables, and the price of 1 table and 1 chair is $72, what is the price, in dollars, of 1 table? (Assume that every chair has the same price and every table has the same price.) | [
"55",
"84",
"63",
"75"
] | C | LET C= CHAIR ; T =TABLE
2C+1T = 0.6 (1C + 2T) OR C( 2-0.6) = 1.2T -1T OR 1.4C = 0.2 T THEREFORE C= 0.2/1.4 T = 1/7 T
IC +1T =72 OR 1/7T + 1T =60 THEREFORE T= 72*7/8 =63
C |
AQUA-RAT | AQUA-RAT-37659 | # 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.
Raja sold two cycles, each for Rs.990. If he made 10% profit on the first and 10% loss on the second, what is the total cost of both cycles? | [
"2000",
"6700",
"3400",
"7800"
] | A | A
(10*10)/100 = 1%loss
100 --- 99
? --- 1980 => Rs.2000 |
AQUA-RAT | AQUA-RAT-37660 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
90 percent of the members of a study group are women, and 60 percent of those women are lawyers. If one member of the study group is to be selected at random, what is the probability that the member selected is a woman lawyer? | [
"0.34",
"0.44",
"0.54",
"0.64"
] | C | Say there are 100 people in that group, then there would be 0.9*0.60*100=54 women lawyers, which means that the probability that the member selected is a woman lawyer is favorable/total=54/100.
Answer: C |
AQUA-RAT | AQUA-RAT-37661 | As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $$5!=5\times 4\times 3\times 2\times 1=120$$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $$3!=3\times 2\times 1=6$$ ways to do, making (again) a total of $$(5!)\times (3!)=120\times 6=720$$ ways of arranging them.
-
+1 for the graphics! :-) – joriki Jul 8 '12 at 19:53
Thanks!$\text{}$ – Zev Chonoles Jul 8 '12 at 20:00
Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
-
thanks joriki, but not quiet clear about what you say – user1419170 Jul 8 '12 at 19:49
@user1419170: Could you be more specific? What part isn't clear to you? – joriki Jul 8 '12 at 19:49
this part Treat the group of three boys as an additional girl – user1419170 Jul 8 '12 at 19:51
@user1419170: See Brian's and Saurabh's answers, which use the same idea but phrase it differently. – joriki Jul 8 '12 at 19:53
The following is multiple choice question (with options) to answer.
In a group of 5 boys and 5 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be selected? | [
"100",
"110",
"179",
"205"
] | D | Number of ways to select four children is 10C4 = 210
Number of ways to choose only girls is 5C4 = 5
Number of ways that at least one boy is selected is 210 - 5 = 205
The answer is D. |
AQUA-RAT | AQUA-RAT-37662 | # 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.
Ms. Lopez deposits $ 140 in an account that pays 20% interest, compounded semiannually. How much money will there be in the account at the end of one year? | [
"$118.00",
"$120.00",
"$169.40",
"$122.00"
] | C | Using Formula - A = P(1+r/n)^nt
Given
P=140
n=2
t=1
r=0.2
Substituting value in formula
A = 140(1+0.2/2)^2
A= 169.40$=C |
AQUA-RAT | AQUA-RAT-37663 | So by my equation I get that they are of equal purity.
Kalyan.
4. ## Re: Purity
Hello, PASCALfan!
Two vessels contains equal amount of water and beer.
1 spoon beer is added to the water and mixed well.
From this mixture 1 spoon is added to the beer.
Which of the liquids is more pure?
This is the classic "Wine and Water" problem.
As long as the amount transferred each time remains the same,
. . the final concentrations are equal.
This is the basis of a stunning card trick.
You are seated at a small table across from your volunteer.
You show him a deck of cards, fanning them
. . so he can see they are all facing one way.
Both of you place your hands under table.
You hand him the deck.
Instruct him to select a secret number from 1 to 20,
. . and keep it a secret. .Call it n.
Have him count off the top n cards,
. . turn them over and replace them on the deck.
Have him shuffle the deck thoroughly.
Then count off the top n cards
. . and hand them to you under the table.
Now you remind him:
. . you don't know his secret number,
. . he doesn't know how many face-up cards he has,
. . you don't know how many face-up cards you have.
Despite this, you will try to make your number of face-up cards
. . equal his number of face-up cards.
He hears the sound of cards being counted and moved.
After a few moment, you bring your cards to the top of the table
. . and count the face-up cards.
Instruct him to do the same . . . and the numbers match!
[Acknowledge the thunderous applause ... modestly, of course.]
Spoiler:
Secret: When he hands you the packet, turn it over.
5. ## Re: Purity
In fact, the answer does not change regardless of how well the liquids were mixed in the process. Indeed, the volume of the liquid in each vessel after the two operations is the same as it was initially. Therefore, in the end the volume of beer in water has to equal the volume of water in beer.
This is one of my favorite math problems because it shows that there is more to math than handling large formulas.
6. ## Re: Purity
Thanks for the card trick..
Regards
The following is multiple choice question (with options) to answer.
Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q, the ratio of alcohol and water in the resulting mixture is? | [
"19:1",
"19:4",
"19:8",
"19:5"
] | D | Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed
= 4.75 : 1.25 = 19:5.
Answer: D |
AQUA-RAT | AQUA-RAT-37664 | 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 get? | [
"45%",
"57%",
"34%",
"90%"
] | B | B
57%
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 x 100)% = 57%. |
AQUA-RAT | AQUA-RAT-37665 | # How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
My try:
Let $A_2, A_3, A_5, A_7$ be the set of numbers between 1 and 1,000 that are divisible by 2, 3, 5, and 7 respectively. I used the inclusion-exclusion formula for $|A_2\cup A_3\cup A_5\cup A_7|= |A_2|+|A_3|+|A_5|+|A_7|-|A_2\cap A_3|-|A_2\cap A_5|-|A_2\cap A_7|-|A_3\cap A_5|-|A_3\cap A_7|-|A_5\cap A_7|+|A_2\cap A_3\cap A_5|+|A_2\cap A_3\cap A_7|+|A_2\cap A_5\cap A_7|+|A_3\cap A_5\cap A_7|-|A_2\cap A_3\cap A_5\cap A_7| = 500+333+200+142-166-100-71-66-47-28+33+23+14+9-4 = 772$
And the result I received was - 772.
I would appreciate if you could confirm my method and result, and I'd be happy to see a different, more elegant approach.
• It should be “2 , 3, 5 or 7”. – user371838 Jan 1 '18 at 13:01
• I am afraid there may be an elegant solution if you insist with and, and not so, if it is or. – user371838 Jan 1 '18 at 13:02
• It should be or. My apologies – Moshe King Jan 1 '18 at 13:03
• I haven't checked the numbers, but your inclusion-exclusion argument should work. – Paul Aljabar Jan 1 '18 at 13:10
The following is multiple choice question (with options) to answer.
How many numbers between 100 and 630 are divisible by 2, 3, and 7 together? | [
"112",
"77",
"267",
"13"
] | D | Explanation:
As the division is by 2, 3, 7 together, the numbers are to be divisible by: 2*3*7 = 42
The limits are 100 and 600
The first number divisible is 42*3 = 126
To find out the last number divisible by 42 within 630:
630/42 = 13
Hence, 42*15 = 630 is the last number divisible by 42 within 630
Hence, total numbers divisible by 2, 3, 7 together are (15 – 2) = 13
ANSWER: D |
AQUA-RAT | AQUA-RAT-37666 | genetics, entomology
Title: Why do ladybugs have a different number of points on their backs Everytime I see a ladybug I ask myself this question.
Why does every ladybug have a different amount of points on its back? Is it because of its age? Or because of its genes? Is it inheritable? The spots on the back of Ladybugs over the surface is defense mechanism to avoid predators. The spots come in different shapes and different numbers. Some say that those dots tell us their age. Since some ladybugs have 24 spots which means its age would be 24 years and that is not at all possible. So this is a popular misconception running around us.
But the real truth about the number of spots on the ladybugs' back is:
The number of spots on a ladybug does have significance. The spots and other markings do help you identify the species of ladybug. Some species have no spots at all. The record-holder for most spots is the 24-spot ladybug ( Subcoccinella 24-punctata), which has 24 spots, of course. Ladybugs aren't always red with black spots, either. The twice-stabbed ladybug ( Chilocorus stigma) is black with two red spots.
So now we know that the number of spots on the back of Ladybugs will help us to identify what species it belongs to. To give an insight into different species of them you can refer to this Identifying Ladybugs.
The above image shows different species of Ladybugs where you can see the varying number of spots in each bugs' back. More information can be found here Stripes on Ladybugs.
The following is multiple choice question (with options) to answer.
Jack and Jill collect ladybugs. Jack only collects the ones with 2 spots,
and Jill only collects the ones with 7 spots. Jack has 5 more ladybugs
than Jill. The total number of spots found on all of their ladybugs is 100.
How many ladybugs do they have in their combined collection? | [
"17",
"21",
"23",
"25"
] | D | If x is the number of ladybugs that Jack has and y is the number of ladybugs that Jill has, then
x − y = 5 and 2x + 7y = 100. Multiplying the first of these equations by 5 and the second by 2
and adding gives 9(x + y) = 225. Hence, x + y = 25.
correct answer D |
AQUA-RAT | AQUA-RAT-37667 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Two persons A and B can complete a piece of work in 30 days and 60 days respectively. If they work together, what part of the work will be completed in 10 days? | [
"1/8",
"1/3",
"1/6",
"1/2"
] | D | A's one day's work = 1/30
B's one day's work = 1/60
(A + B)'s one day's work = 1/30 + 1/60 = 1/20
The part of the work completed in 10 days = 10 (1/20) = 1/2.
Answer:D |
AQUA-RAT | AQUA-RAT-37668 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $200 before any discount? | [
" $112.00",
" $145.60",
" $163.80",
" $182.00"
] | A | Since the question is essentially just about multiplication, you can do the various mathstepsin a variety of ways (depending on whichever method you find easiest).
We're told that the first discount is 10% to 30%, inclusive. We're told that the next discount is 20% off of the DISCOUNTED price....
We're told to MAXIMIZE the discount (thus, 30% off the original price and then 20% off of the discounted price). Thatmathcan be written in a number of different ways (fractions, decimals, etc.):
30% off = (1 - .3) = (1 - 30/100) = (.7) and the same can be done with the 20% additional discount...
The final price of an item that originally cost $200 would be.....
($200)(.7)(.8) =
($200)(.56)=112
Final Answer:
A |
AQUA-RAT | AQUA-RAT-37669 | => 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 $348 per week. If instead, her weekly salary were to increase by 15 percent, how much would she earn per week? | [
" $374",
" $382",
" $345",
" $392"
] | C | (348/116)115 =345
In this case long division does not take much time.
(348/116)=3
3*115=345(300+45)
Answer C |
AQUA-RAT | AQUA-RAT-37670 | Thus the result is $2$ as I exclaimed before.
The faster way is to determine the parity of $b+e$ first.
Since $b+e=3+2=5$ is odd, the result is the one’s digit of $a+f+5=6+1+5$, and hence the result is $2$.
If you don’t know the formula then you would complete the tree diagram like this:
and clearly it takes more time to get the result without the trick.
Modulo $10$
Here is a tip to use the formula $a+f+5(b+e)$.
If the numbers are relatively small, then just compute it and find the one’s digit.
But remember that you can always do modulo $10$ computation anytime. (Considering the one’s digit or the remainder of division by $10$.)
For example, if $a=11, f=29, b=17, e=104$ are given, I first find the one’s digits of them and can assume that $a=1, f=9, b=7, e=4$.
Then $a+f=10$ and the one’s digit is $0$. So I can safely ignore the $a+f$ part in the formula.
Next, $b+e=11$ and the one’s digit is $1$.
(Or you could first determine the parity of $b+e$.)
Thus by the formula, the result is $5$.
Do Practice and Have some Fun!
Practice how to use the formula several times and get used to it.
The following is multiple choice question (with options) to answer.
If a * b = a + b/ab , find the value of 5 * (5 * -2) : | [
"-3",
"-10",
"-1.66",
"3/5"
] | B | (5 * -2) = (5 x (-2))/(5 + (-2)) = -10/3
So, 5 * (5 * -2) = 5 * (-10/3) = (5 * (-10/3))/(5 + (-10/3))
= (-50/3) * (3/5)
= -10.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37671 | You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
09 Mar 2018, 03:18
HarveyKlaus wrote:
Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
Question:
The following is multiple choice question (with options) to answer.
If x>y>0, which of the following must be true:
I. x^2>y^2
II. x^3 > y^3
III. x<y | [
"I and II",
"II only",
"III only",
"II and III"
] | A | Assuming x and y to be integers
X=3,Y=2
X^2=3^2=9
Y^2=2^2=4
So I is true
Now lets look at option III because if option 3 is true E is the answer or else A
As per question X>Y>0 Hence |X| >Y
So OA=A |
AQUA-RAT | AQUA-RAT-37672 | 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 and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last? | [
"5 days",
"10 days",
"11 days",
"20 days"
] | B | Work done by X in 4 days =1/20 x 4 = 1/5
Remaining work = (1 -1/5) = 4/5
(X + Y)'s 1 day's work = 1/20+1/12 = 8/60 = 2/15
Now, 2/15 work is done by X and Y in 1 day.
So,4/5work will be done by X and Y in 15/2 x 4/5= 6 days.
Hence, total time taken = (6 + 4) days = 10 days.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37673 | # 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.
Krishan and Nandan jointly started a business. Krishan invested three times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was? | [
"28,029",
"28,199",
"28,299",
"28,000"
] | D | Explanation:
3:1
2:1
------
6:1
1 ----- 4000
7 ----- ? => Rs.28,000
Answer: D |
AQUA-RAT | AQUA-RAT-37674 | 6. Hello, Migotek84!
Here's a very primitive method . . .
Compute: . $17^{-1}\text{ (mod 23)}$
Let: . $x \:\equiv\:17^{-1}\text{ (mod 23)}$
Multiply by 17: . $17x \:\equiv\:1\text{ (mod 23)}$
Then: . $17x \:=\:23a + 1\:\text{ for some integer }a.$
$\text{Solve for }x\!:\;\;x \:=\:\dfrac{23a+1}{17} \:=\:a + \dfrac{6a+1}{17}$ .[1]
Since $x$ is an integer, $6a+1$ is a multiple of 17.
. . $6a+1 \:=\:17b\:\text{ for some integer }b.$
$\text{Solve for }a\!:\;\;a \:=\:\dfrac{17b-1}{6} \:=\:2b + \dfrac{5b-1}{6}$ .[2]
Since $a$ is an integer, $5b-1$ is a multiple of 6.
. . $5b-1 \:=\:6c\:\text{ for some integer }c.$
$\text{Solve for }b\!:\;\;b \:=\:\dfrac{6c+1}{5} \:=\:c + \dfrac{c+1}{5}$ .[3]
Since $b$ is an integer, $c+1$ is a multiple of 5.
. . The first time this happens is when . $c = 4$
Substitute into [3]: . $b \:=\:4 + \frac{4+1}{5} \quad\Rightarrow\quad b \:=\:5$
The following is multiple choice question (with options) to answer.
What is the units digit of 23^3 * 17^2 * 39^2? | [
"9",
"7",
"5",
"3"
] | D | The units digit of 23^3 is the units digit of 3*3*3 = 27 which is 7.
The units digit of 17^2 is the units digit of 7*7 = 49 which is 9.
The units digit of 39^2 is the units digit of 9*9 = 81 which is 1.
The units digit of 7*9*1 = 63 is 3.
The answer is D. |
AQUA-RAT | AQUA-RAT-37675 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 362, 482, 551, 263, 344, 284 | [
"362",
"482",
"551",
"344"
] | D | Explanation :
In all numbers except 344, the product of first and third digits is the middle digit.
Answer : Option D |
AQUA-RAT | AQUA-RAT-37676 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
The ratio, by weight, of coffee to sugar to water in a pot of coffee is 3:1:15. The ratio is then altered so that the ratio of coffee to sugar is halved while the ratio of coffee to water is doubled. If the altered pot of coffee contains 12 grams of sugar, how many grams of water does it contain after alteration? | [
"35",
"40",
"45",
"50"
] | C | The new ratio of coffee to sugar is 1.5:1
The new ratio of coffee to water is 6:15.
The new ratio of coffee to sugar to water is 6:4:15.
If there are 12 grams of sugar, then there are (15/4)*12=45 grams of water.
The answer is C. |
AQUA-RAT | AQUA-RAT-37677 | # Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have $$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$ There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B The answer for this question should be $0.35$, I am not sure where I did wrong.
I just figured maybe I should do this instead
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?
• Possible duplicate of Expected values of a dice game with a 30-sided die and a 20-sided die. – samjoe Mar 29 '17 at 6:00
• this is a different question – szd116 Mar 29 '17 at 6:05
• No, look properly. The first part of the question is same as yours, just it asks the probability of A winning. – samjoe Mar 29 '17 at 6:11
• I read it, but still couldn't figure out this problem – szd116 Mar 29 '17 at 6:13
• Look the answer to first part is 7/20 right? Now look at answer to that question. – samjoe Mar 29 '17 at 6:23
The following is multiple choice question (with options) to answer.
In a certain game of dice, the player’s score is determined as a sum of four throws of a single die. The player with the highest score wins the round. If more than one player has the highest score, the winnings of the round are divided equally among these players. If Jim plays this game against 26 other players, what is the probability of the minimum score that will guarantee Jim some monetary payoff? | [
"41/50",
"1/221",
"1/1296",
"1/84"
] | C | Toguaranteethat Jim will get some monetary payoff he must score the maximum score of 6+6+6+6=24, because if he gets even one less than that so 23, someone can get 24 and Jim will get nothing.
P(24)=1/6^4=1/1296.
Answer: C. |
AQUA-RAT | AQUA-RAT-37678 | beautiful black bare babes
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I have a list of prices where I am trying to calculate the change in percentage of each number. I calculated the differences with prices = [30.4, 32.5, 31.7, 31.2, 32.7, 34.1, 35.8, 37.8, 36.3... Stack Overflow. ... Calculating change in percentage between two numbers (Python) Ask Question Asked 10 years, 1 month ago. Modified 2 years, 7 months. Step 1: Find the difference between the two numbers, i.e a - b. Step 2: Then, find the average of two numbers, i.e (a+b)/2. Step 3: Take the ratio of the difference and the average. Step 4: Multiply the fraction obtained by 100 and.
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Calculating the percent change between two given quantities is quite an easy process. When the initial or old value and final or new values of a quantity are known, percent change formula.
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The formula for % change between two numbers. The simple formula is shown below. = (A2/A1)-1. In this example, cell A2 contains the new number and cell A1 contains the original number. The formula will return a.
Calculating percentage change between 2 numbers. .
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Thus to calculate the percentage increase we will follow two steps: Step-1: Calculate the difference i.e. increase between the two numbers. i.e. Increase = New Number - Original Number. Step-2: Divide the increased value by the original. In computing the growth or decline of a variable, you can quickly use this percentage change calculator to find the percentage increase or decrease in the value of two numbers. How to use our FREE Percent Change Calculator It is very simple, easy and quick to use! Step 1: Simply fill in the initial and new values in the provided boxes. Calculating percentage change between 2 numbers. Percentage Difference Formula: Percentage difference equals the absolute value of the change in value, divided by the average of the 2 numbers, all multiplied by 100. We then append the percent sign, %, to designate the % difference..
The following is multiple choice question (with options) to answer.
If each side of a square is increased by 25%, find the percentage change in its area? | [
"28.88",
"56.25",
"22.89",
"29.18"
] | B | let each side of the square be a , then area = a x a
As given that The side is increased by 25%, then
New side = =
New area =
increased area=
Increase %= % = 56.25%
Answer: B |
AQUA-RAT | AQUA-RAT-37679 | 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 piece of work can be done by 6 men and 5 women in 6 days or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in how many days ? | [
"3 days",
"4 days",
"5 days",
"6 days"
] | A | Explanation:
To calculate the answer we need to get 1 man per day work and 1 woman per day work.
Let 1 man 1 day work =x
and 1 woman 1 days work = y.
=> 6x+5y = 1/6
and 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90
(9 men + 15 women)'s 1 days work =
(9/54) + (15/90) = 1/3
9 men and 15 women will finish the work in 3 days
Option A |
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### Show Tags
05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
Machine E and machine B are each used to manufacture 660 sprockets. It takes machine E 10 hours longer to produce 660 sprockets than machine B. Machine B produces 10 percent more sprockets per hour than machine E. How many sprockets per hour does machine E produces? | [
"6",
"6.6",
"60",
"100"
] | A | I think the correct answer is A.
Machine E produces at a speed of E sp/hour and B at a speed of B sp/hour.
so, 660/E=(660/B)+10 and E=1,1 E--->1,1*660=660+11E--->A=6, so answer A is correct |
AQUA-RAT | AQUA-RAT-37681 | Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:37
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
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05 Oct 2010, 12:53
Bunuel wrote:
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!
this might clear my entire probability confusion i hope...
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The following is multiple choice question (with options) to answer.
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible? | [
"144",
"152",
"90",
"168"
] | C | The first digit can be filled in 6 ways
For second digit , it can either be 0 or 1
Case 1 -
If second digit is 1 ,Third digit can take 8 values
number of codes = 6 * 1 * 8 = 48
Case 2 -
If second digit is 0,Third digit can take 7 values ( Third digit can't be zero)
number of codes = 6 * 1 * 7= 42
Total number of codes = 90
Answer C |
AQUA-RAT | AQUA-RAT-37682 | $\frac {0.367}{0.978} = 0.375.$
13. ## Re: Probability with a "known"
Another way to do this is with combinations (since you don't care about the order that you pick the fruit). You will get the same answer either way. With combinations, you can just divide the number of outcomes. So, if we figure out the number of outcomes with (at least two red AND at least one red or green) divided by the number of outcomes with at least one apple (red or green), that will give us the same probability.
Just as ebaines did, we want the number of outcomes with exactly 0 red and exactly 1 red, then take the compliment (total number of outcomes minus outcomes that give us only 0 or 1 red). For exactly 0 red, that is $\binom{24}{5}$ outcomes.
For exactly 1 red, that is $\binom{24}{4}\binom{8}{1}$ outcomes (note: since we are doing combinations, we don't care which of the apples is red). So, there are $\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}$ outcomes with at least 2 red.
For at least one apple (red or green), we want the number of outcomes with 0 apples and take the compliment: $\binom{32}{5} - \binom{16}{5}$.
So, the probability that you get at least 2 red apples given that you pick at least one apple (red or green):
$\dfrac{\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}}{\binom{32}{5} - \binom{16}{5}} = \dfrac{73,864}{197,008} = \dfrac{1319}{3518} \approx 0.375$
14. ## Re: Probability with a "known"
The following is multiple choice question (with options) to answer.
A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is -. | [
"25/42",
"9/20",
"10/23",
"41/42"
] | A | Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37683 | # ONEORTWO - Editorial
Author: utkarsh_25dec
Testers: tabr, iceknight1093
Editorialist: iceknight1093
TBD
None
# PROBLEM:
There are two piles of stones: the first with X, the second with Y.
Chef and Chefina alternate moves, with Chef moving first.
• Chef can either remove one stone from both piles or remove two stones from the first pile.
• Chefina can either remove one stone from both piles or remove two stones from the second pile.
Under optimal play, who wins?
# EXPLANATION:
Note that Chef always removes at least one stone from the first pile, and Chefina always removes at least one from the second.
So, ideally Chef always wants there to be stones remaining on the first pile, while Chefina always wants there to be stones remaining on the second pile.
Intuitively, this should tell you that both players will use the “remove one stone from both piles” move as much as possible.
This fixes the set of moves of the game, and so it can be easily simulated to see who wins.
A little bit of casework should tell you that:
• If X \geq Y+2, Chef will always win.
• If X = Y+1, Chef will win if X is even and lose otherwise.
• If X = Y, Chef will win if X is odd and lose otherwise.
• If X = Y-1, Chef will win if X is odd and lose otherwise.
• If X \leq Y-2, Chef will always lose.
This can be simplified into:
• If X \geq Y+2, Chef wins
• If X \leq Y-2, Chef loses
• if |X-Y| \leq 1, Chef wins if and only if \min(X, Y) is odd
In fact, this intuition is correct, and the above cases solve the problem!
Detailed proof
Intuition is great to have, but can’t always be relied on. It’s good to get into the habit of proving your solutions.
Let’s prove the above cases by constructing a strategy for them.
The following is multiple choice question (with options) to answer.
Bilal and Ahmed play the following game with n sticks on a table. Each must remove 1,2,3,4, or 5 sticks at a time on alternate turns, and no stick that is removed is put back on the table. The one who removes the last stick (or sticks) from the table wins. If Bilal goes first, which of the following is a value of n such that Ahmed can always win no matter how Bilal plays? | [
"7",
"10",
"11",
"12"
] | D | If the number of sticks on a table is a multiple of 6, then the second player will win in any case (well if the player is smart enough).
Consider n=6, no matter how many stick will be removed by the first player (1, 2, 3 ,4 or 5), the rest (5, 4, 3, 2, or 1) can be removed by the second one.
The same for n=12: no matter how many sticks will be removed by the first player 1, 2, 3 ,4 or 5, the second one can remove 5, 4, 3, 2, or 1 so that to leave 6 sticks on the table and we are back to the case we discussed above.
Answer: D. |
AQUA-RAT | AQUA-RAT-37684 | ## Circular Combination
1. Arrange 8 dancers in circular fashion.
2. Use 8 pearls in a band to make a necklace.
3. Arrange 8 science and 7 arts students circularly so no two arts students are together.
• 7!
• $$\frac{(8-1)!}{2}$$
• $$^8P_7 \times \space 7!$$
# Combination
## Concept
A, B, C
In how many ways can we select 2?
• AB, AC, BC
• When making teams $$AB \equiv BA$$
## Formulae And Notation
Formulae (Don’t Memorize!)
• $$^nC_r = {n! \over r!(n-r)!}$$
• $$^nC_r \times ^rP_r = ^nP_r = ^nP_r \times r!$$ (Permutation-combination relationship)
• $$^{n+1}C_r = ^nC_r + ^nC_{r-1}$$
• $$^nC_r=?$$ (from above, expanding twice up to n-2)
• $$^nC_r = ^nC_{n-r}$$
Notation
• $$^nC_r \equiv$$ $$n \choose r$$
• Select 5 people from 6 $$\rightarrow ^6C_5 =$$ $$6 \choose 5$$
## Expaniosn of $$^nC_r$$
The following is multiple choice question (with options) to answer.
There are five different models that are to appear in a fashion show. Two are from Europe, two are from South America, and two are from North America. If all the models from the same continent are to stand next to each other, how many ways can the fashion show organizer arrange the models? | [
"96",
"48",
"64",
"24"
] | A | Since we have 3 continental pairs (EU, SA, NA), these 3 pairs have 3*2*1 = 6 Combinations. Within each pair, you have however 2 different ways to put them together for each of the pair (2*2*2*2= 16). So we have 6*16 =96.
Please correct me if im wrong.
Answer A. |
AQUA-RAT | AQUA-RAT-37685 | 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.
81, 64, 27, 16... | [
"12",
"4",
"9",
"3"
] | C | 81/3 = 27
64/4 = 16
27/3 = 9
ANSWER: C |
AQUA-RAT | AQUA-RAT-37686 | are two constants of integration. a) Total cost when output is 4 units. Marginal revenue is the derivative of total revenue with respect to demand. Derivatives in Physics • In physics, the derivative of the displacement of a moving body with respect to time is the velocity of the body, and the derivative of velocity W.R.T time is acceleration. To calculate the marginal average cost, we need to first calculate the average cost, where TC(Q) is the total cost to produce Q units. Take, for example, a total cost function, TC: For a given value of Q, say Q=10, we can interpret this function as telling us that: when we produce 10 units of this good, the total cost is 190. It adjusts along with iso-quant and the output remains constant. The Second derivative is negative so you see that anything greater than 50 would make the First derivative less than 0 and the First derivative is the Marginal Change in Total Revenue. Particular attention is paid to the distinction between horizontal and vertical differentiation as well as to the related issues of product quality and durability. The welfare-enhancing benefit of product differentiation is the greater variety of products available to consumers, which comes at the cost of a higher average total cost of production. MIT OpenCourseWare 115,327 views For a function z = f(x, y, .. , u) the total differential is defined as Each of the terms represents a partial differential. Partial Differentiation: you take into account only the effect of one variable, say K, on Y. Secondly, the product at hand is differentiated from these alternatives and therefore captures differentiation value. Determine the marginal cost, marginal revenue, and marginal profit at x = 100 widgets.. Chain Rule and Total Differentials 1. The above equation is known as total differentiation of function f(x). Take the total cost formula of TC = 50 + 6Q and divide the right side to get average total costs. Marginal function in economics is defined as the change in total function due to a one unit change in the independent variable. To get average total cost at a specific point, substitute for the Q. Application of differentiation in business optimization problems are also covered. (dy/dx) measures the rate of change of y with respect to x. 2. In Economics and commerce we come across many such variables where one variable is a function of … Horizontal differentiation is less strong for simple products, such as erasers,
The following is multiple choice question (with options) to answer.
Marginal cost is the cost of increasing the quantity produced (or purchased) by one unit. If the fixed cost for n products is $12,000 and the marginal cost is $200, and the total cost is $16,000, what is the value of n? | [
" 20",
" 50",
" 60",
" 80"
] | A | Total cost for n products = fixed cost for n products + n*marginal cost --> $16,000 = $12,000 + n * $200 --> n = 20.
Answer: A. |
AQUA-RAT | AQUA-RAT-37687 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 125 m long passes a man, running at 8 km/hr in the same direction in which the train is going, in 10 sec. The speed of the train is? | [
"53",
"50",
"60",
"79"
] | A | Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 45 km/hr
Let the speed of the train be x km/hr. Then, relative speed = (x - 8) km/hr.
x - 8 = 45 => x = 53 km/hr.
Answer: Option A |
AQUA-RAT | AQUA-RAT-37688 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shop owner sells 110 eraser and gains SP of 22 metres. Find the gain %? | [
"25%",
"35%",
"27%",
"23%"
] | A | Gain=22 oranges
Total 110 oranges
CP 110-22=88
(22/88)*100
=25%
Ans A |
AQUA-RAT | AQUA-RAT-37689 | Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
-
M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
The following is multiple choice question (with options) to answer.
What number is 75 more than three-fourth of itself? | [
"138",
"225",
"300",
"324"
] | C | 3/4x+75= x
That means 75=1/4x
x= 75*4= 300
C is the answer |
AQUA-RAT | AQUA-RAT-37690 | # 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 cash difference between the selling prices of an article at a profit of 4% and 6% is Rs 3. The ratio of two selling prices is | [
"51:52",
"52:53",
"53:54",
"54:55"
] | B | Explanation:
Let the Cost price of article is Rs. x
Required ratio =
104% of x/ 106% of x
=104/106=52/53=52:53
Option B |
AQUA-RAT | AQUA-RAT-37691 | Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink] 21 Apr 2017, 03:36
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The following is multiple choice question (with options) to answer.
If c and d are both negative and cd < d^2, which of the following must be true? | [
"c < d < c^2 < d^2",
"c < d < d^2 < c^2",
"d < c < c^2 < d^2",
"c^2 < d^2 < d < c"
] | C | Because cd <d^2 and both are negative, I thought c < d. So I crossed off answers c), d), and e).
And because c < d , c^2 < d^2
ans C |
AQUA-RAT | AQUA-RAT-37692 | 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 ages of Ashley and Mary are in the ratio 4:7. The sum of their ages is 22. Find the age of Mary. | [
"11yrs",
"12yrs",
"13yrs",
"14yrs"
] | D | a:m=4:7
let the ages of a & m be x &22-x
x/22-x=4/7
x=8
a=8yrs
m=14yrs
ANSWER:D |
AQUA-RAT | AQUA-RAT-37693 | 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.
How much time will take for an amount of Rs. 400 to yield Rs. 81 as interest at 4.5% per annum of simple interest? | [
"4 1/2 years",
"6 1/2years",
"7 1/2years",
"12 1/2years"
] | A | Time = (100 * 81) / (400 * 4.5) = 4 1/2years
ANSWER:A |
AQUA-RAT | AQUA-RAT-37694 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
Kudos [?]: 24 [2], given: 0
Manager
Joined: 21 Jan 2015
Posts: 149
Kudos [?]: 121 [0], given: 24
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
After replacing an old member by a new member, it was found that the average age of five numbers of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member? | [
"2 years",
"4 years",
"8 years",
"15 years"
] | D | Sol.
Age decrease = (5 × 3) years = 15 years.
So, the required difference = 15 years.
Answer D |
AQUA-RAT | AQUA-RAT-37695 | Since n is a integer, can we not try with n as 0?
Yes, n can be 0 but 0 is even too.
_________________
Senior Manager
Joined: 20 Aug 2015
Posts: 386
Location: India
GMAT 1: 760 Q50 V44
Re: If n is an integer, is n even? [#permalink]
### Show Tags
09 Dec 2015, 00:06
Bunuel wrote:
If n is an integer, is n even?
(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.
Given: n is an integer
Required: is n even?
Statement 1: $$n^2$$ - 1 is an odd integer
$$n^2$$ - 1 = (n-1)(n+1) = odd.
This means both n-1 and n+1 are odd
Odd*Odd = Odd
Odd*Even = Even
Even*Even = Even
n-1, n, n+1 are three consecutive integers.
Since we know that both n-1 and n+1 are odd
Hence n has to be even.
SUFFICIENT
Statement 2: 3n + 4 is an even integer
Even + Even = Even
Even + Odd = Odd
Odd + Odd + Odd
Since 3n+4 = even and 4 is an even integer.
Hence 3n = even. Therefore n = even
SUFFICIENT
Option D
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Re: If n is an integer, is n even? [#permalink]
### Show Tags
09 Aug 2016, 13:33
1
Quote:
If n is an integer, is n even?
(1) n^2 - 1 is an odd integer.
(2) 3n + 4 is an even integer.
We need to determine whether integer n is even. Let's review four facts about even and odd integers: 1) An integer and its square are either both even or both odd. 2) The sum (or difference) between an even integer and an odd integer is always odd. 3) The sum of two even integers (or two odd integers) is always even. 4) If the product of two integers is even, at least one of them must be even.
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Functions. Let a be a positive integer. Let n#a equal n^(2a) if a is odd and n^(3a) if a is even. Then (2#3) + (2#3) – (3#3) is equal to | [
"64",
"82",
"128",
"729"
] | D | Answer:
3^6 + 3^6 - 3^6 = 3^6 = 729
ANs D |
AQUA-RAT | AQUA-RAT-37696 | in with angle 1 degrees, Area of a Parallelogram diagonal-e 83 m diagonal-f 73 ft with angle 0.5 radians, Area of a Parallelogram diagonal-e 75 in diagonal-f 22 yd with angle 130 degrees, Area of a Parallelogram diagonal-e 63 ft diagonal-f 47 m with angle 12 degrees, Area of a Parallelogram diagonal-e 23 cm diagonal-f 17 in with angle 0.8 radians, Area of a Parallelogram diagonal-e 60 ft diagonal-f 25 m with angle 2.6 radians. Math operations to check the side length and opposite angles are equal in measure free! Grade 5 through grade 8 have been included here obtained in the above steps to calculate area... B ) between any two sides of a parallelogram are equal ( a equal! Magnitude of cross-vector products for two adjacent sides are a, b and a b. Here are the length of a parallelogram with a and b is equal with b.! A simple quadrilateral with two pairs of parallel sides the above steps to calculate of! Each other grade 5 through grade 8 have been included here useful for everyone save. Q and sides are parallel to each other at the center of the parallelogram a. Area are listed here, 1 graphs for each calculation Width::... Parallelogram sides are in equal length, corner angles or angles or any others always! Results: area of a parallelogram is base x height an enormous range of area of a parallelogram easier. Are using base and height as in the details below: you need two measurements calculate! 11 cm magnitude of cross-vector products for two adjacent sides, one corner angle from the question angle. Is 12 cm and angle is 75° 2: a parallelogram = b×h square units = 4 × =..., find the area of a parallelogram calculator is a free online calculator with step by solution. Calculate ” to get your area of a parallelogram = base × height square units inside polygon! 3, then your area of a parallelogram is a perfect example of the parallelogram area a... Are corner angles, diagonals of a parallelogram = 20 cm 2 are 3 cm, find the distance its. Is given by area of a parallelogram = 20 sq.cm to build a
The following is multiple choice question (with options) to answer.
What is the area of a square field whose diagonal of length 20 m? | [
"300 sq m",
"250 sq m",
"200 sq m",
"400 sq m"
] | C | d2/2 = (20 * 20)/2 = 200
ANSWER:C |
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