source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36797 | 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.
Kuber borrowed rs 169 due in 2 years at 4%per annum compound interest.Find the Present worth? | [
"160.15",
"225",
"156.25",
"180.25"
] | C | Compound interest=[169*(1+(4/100))^2]
=[169*((104/100)^2)]
=(169*((26/25)^2))
=(169*25*25)/(26*26)
=Rs.156.25
Option C |
AQUA-RAT | AQUA-RAT-36798 | The operations above are perfectly reasonable. Let me show you the kind of situation that I think you were worried about, and you'll see why it's different from the situations above:
I take 4 exams and have an average score of 80. Then I take 2 more exams and on those 2, my average is 100. What's my overall average for the course?
The wrong way to do it is (80+100)/2 = 90. This is wrong because the averages were of different-sized groups. To get the correct answer, I know that on the first 4 exams I got 320 total points because when I divide 320 by 4, I get 80.
Similarly, for the last two exams, I must have gotten 200 points total. So for the six exams, I got 200+320 = 520 points and 520/6 = 86.66666 = my real grade average.
We saw this scenario last week: If we just average the two averages, the result can only be described as the average of the averages, not the average for the course. The latter has to be weighted, because for the course, each exam counts the same, not each of these two sets of exams.
As we saw last week, a weighted average can often be understood by breaking it down:
For your problem, the averages you are averaging are for the same period, so it works out. To convince you that it's true, let's just look at a situation where the averages came from 10 months of data.
Then in the first line of business, 500 people must have been hired, since 500/10 = 50. Similarly, 800 were hired in the other, since 800/10 = 80. Altogether, 1300 people were hired in the ten months, or 1300/10 = 130 per month, company-wide. Or if you're trying to get the average per line of work, 65 is right, since if each group had hired 65 people each month for 10 months, there would be 65*20 = 1300 total hires, so it works out.
Bottom line: always think about what you want, and what an average means, rather than use an average (or not!) unthinkingly.
## Using a weighted average
Averaging Averages
The following is multiple choice question (with options) to answer.
In three annual examinations, of which the aggregate marks of each was 500, a student secured
average marks 45% and 55% in the first and the second yearly examinations respectively. To secure 40% average total marks, it is necessary for him in third yearly examination to secure marks : | [
"100",
"350",
"400",
"450"
] | A | total marks:1500 for three exams
40% of 1500=600
first exam marks=45% of 500=225
second exam marks=55% of 500=275
let X be the third exam marks
225 + 275 + X =600
X=100
ANSWER:A |
AQUA-RAT | AQUA-RAT-36799 | $$\text{average rate of change}=\dfrac{45-35}{5}=\dfrac{10}{5}=2$$
On average, the temperature between 9 a.m. and 2 p.m. increased $$2^\circ F$$ per hour.
• How quickly was the temperature falling between 2 p.m. and 8 p.m.?
$$\text{average rate of change}=\dfrac{30-45}{6}=\dfrac{\text-15}{6}=\text-2.5$$
On average, the temperature between 2 p.m. and 8 p.m. dropped by $$2.5 ^\circ F$$ per hour.
In general, we can calculate the average rate of change of a function $$f$$, between input values $$a$$ and $$b$$, by dividing the difference in the outputs by the difference in the inputs.
$$\text{average rate of change}=\dfrac{f(b)-f(a)}{b-a}$$
If the two points on the graph of the function are $$(a, f(a))$$ and $$(b, f(b))$$, the average rate of change is the slope of the line that connects the two points.
The following is multiple choice question (with options) to answer.
The average temperature of the town in the first four days of a month was 58 degrees. The average for the second, third, fourth and fifth days was 59 degrees. If the temperatures of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day ? | [
"62 degrees",
"32 degrees",
"65 degrees",
"66 degrees"
] | B | Explanation:
Sum of temperatures on 1st, 2nd, 3rd and 4th days = (58 * 4) = 232 degrees ... (1)
Sum of temperatures on 2nd, 3rd, 4th and 5th days - (59 * 4) = 236 degrees ....(2)
Subtracting (1) From (2), we get :
Temp, on 5th day - Temp on 1st day = 4 degrees.
Let the temperatures on 1st and 5th days be 7x and 8x degrees respectively.
Then, 8x - 7x = 4 or x = 4. Answer: B
Temperature on the 5th day = 8x = 32 degrees. |
AQUA-RAT | AQUA-RAT-36800 | 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.
Sixty men can stitch 200 shirts in 30 days working 8 hours a day. In how many days can 45 men stitch 300 shirts working 6 hours a day? | [
"33",
"88",
"27",
"80"
] | D | We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D2 * 6) / 300
D2 = (60 * 30 * 8 * 300) / (200 * 45 * 6) => D2 = 80.
Answer:D |
AQUA-RAT | AQUA-RAT-36801 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Divide Rs. 800 among A, B and C so that A receives 1/3 as much as B and C together and B receives 2/3 as A and C together. A's share is? | [
"s.800",
"s.200",
"s.600",
"s.500"
] | B | A+B+C = 800
A = 1/3(B+C); B = 2/3(A+C)
A/(B+C) = 1/3
A = 1/4 * 800 => 200
ANSWER:B |
AQUA-RAT | AQUA-RAT-36802 | Math Expert
Joined: 02 Sep 2009
Posts: 47200
A is a prime number (A>2). If B = A^3, by how many different integers [#permalink]
### Show Tags
27 Sep 2017, 21:07
2
Huey002 wrote:
A is a prime number (A>2). If B = A^3, by how many different integers can B be equally divided?
(a) 3.
(b) 4.
(c) 5.
(d) 6.
(e) 7.
Finding the Number of Factors of an Integer
First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.
The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$
Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
According to above as $$b=a^3$$ and $$a$$ is a prime, then the number of factors of $$b$$ is $$3+1=4$$: $$1$$, $$a$$, $$a^2$$, $$a^3=b$$.
P.S. PLEASE NAME TOPICS PROPERLY. CHECK RULE 3 HERE: RULES OF POSTING. Thank you.
_________________
A is a prime number (A>2). If B = A^3, by how many different integers &nbs [#permalink] 27 Sep 2017, 21:07
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# Events & Promotions
The following is multiple choice question (with options) to answer.
In N is a positive integer less than 200, and 10N/60 is an integer, then N has how many different positive prime factors? | [
"2",
"3",
"5",
"6"
] | A | (A).
10n/60 must be an integer. => 1n/6 must be an integer. Hence n must be a multiple of 2*3.
=> n has 2 different prime integers. |
AQUA-RAT | AQUA-RAT-36803 | So on.
=====
Another thing to note:
(Assume only positive values)
$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$
So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$
That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.
But it doesn't go the other way!
$$\frac 12 < a < 4 \not \implies 1 < a < 2$$
Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$
[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)
• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06
The following is multiple choice question (with options) to answer.
On a certain number line, conditions are a<b<c<d<e and ab>0. Which of the following must be true?
I ab>0 II cd>0 III de>0 | [
"I&II only",
"II only",
"III only",
"I only"
] | D | Givena<b<c<d<e and abcde>0.
Suppose, for the product to be greater than 0, we can two variables as -ve or four variables can be -ve.
We are given : I ab>0 II cd>0 III de>0
case 1: a and b can be negative/positive and if they are... we get greater than 0.
case 2: we have cd , then c and d both can be negative/positive and if they are... we get greater than 0.
case 2: we have de , then c and d both can be negative/positive and if they are... we get greater than 0.
if cd both negative , then de both have to negative in order to get greater than 0. In order to abcde>0 , cde can't be negative else we get -ve value, then we need to have either a or b as negative then ab can't be greater than 0.
so cde have to be positive.
D |
AQUA-RAT | AQUA-RAT-36804 | ### Show Tags
11 Apr 2018, 09:13
2
hetmavani wrote:
In a school there are two sections A and B. When a few students from section B took transfer to section A the strength of section A increased by 22.5% and strength of B decreased by 30%. Now, what percentage of students in section A should be given a transfer to section B such that section A and section B have the same strengths?
A) $$\frac{100}{7}$$%
B) 20%
C) $$\frac{200}{7}$$%
D) 40%
E) $$\frac{300}{7}$$%
so 22.5% of A = 30% of B......
$$22.5*A = 30*B.......A = \frac{30B}{22.5} = \frac{4B}{3}$$..
Now let B = 300, so A = 400..
30% of 300 = 90..
so present strength - $$A= 400+90=490$$ and $$B = 300-90=210$$..
to equalize both should have $$\frac{300+400}{2}=350$$
so $$490-350 = 140$$ should move to B from A..
$$% = \frac{140}{490}*100 = \frac{200}{7}$$
C
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
GMAT online Tutor
In a school there are two sections A and B. When a few students from [#permalink] 11 Apr 2018, 09:13
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# Events & Promotions
The following is multiple choice question (with options) to answer.
If A got 80 marks and B got 60 marks, then what percent of A's mark is B's mark? | [
"60%",
"80%",
"65%",
"75%"
] | D | A's marks = 80 ; B's marks = 60.
Let x% of A = B => x/100 * 80 = 60
=> x = (60 * 100)/80 = 75
B's marks is 75% of A's marks.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36805 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
An entrepreneurship competition requires registering teams to have 3 team members, at least one of which must be a technology co-founder. If all team members must come from the auditorium during the meet and greet event which has 4 technologists and 7 businessmen, how many possible team submissions are possible? | [
"76",
"130",
"162",
"198"
] | B | We have 3 scenarios here:
1) 1 tech2 businessmen: 4C1 x 7C2 = 84
2) 2 tech1 businessman: 4C2 x 7C1 =42
3) 3 tech0 businessmen: 4C3 = 4
Total: 84+42+4 = 130
Answer: B |
AQUA-RAT | AQUA-RAT-36806 | instrumentation Calculating percentages can be calculated as a percentage is a number or that. By the symbol % '' or pct. number or ratio that represents a of! Sample size 1 values below and Click the calculate '' button to get one value ) Elongation! Provide any two values below and Click the calculate '' button to get the third value this by. } { l } \ ) percentage Elongation Example as percent '' simply!
The following is multiple choice question (with options) to answer.
2.09 can be expressed in terms of percentage as | [
"2.09%",
"20.9%",
"209%",
"0.209%"
] | C | Explanation:
While calculation in terms of percentage we need to multiply by 100, so
2.09 * 100 = 209
Answer: Option C |
AQUA-RAT | AQUA-RAT-36807 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man buys an article and sells it at a profit of 20%. If he had bought it at 20% less and sold it for Rs.75 less, he could have gained 25%. What is the cost price? | [
"256",
"375",
"287",
"255"
] | B | CP1 = 100 SP1 = 120
CP2 = 80 SP2 = 80 * (125/100) = 100
20 ----- 100
75 ----- ? => 375
Answer: B |
AQUA-RAT | AQUA-RAT-36808 | # Maximum and minimum value of $\int_0^1 f(x)dx$ given $|f'(x)|<2$
Let $$f:\mathbb R\to \mathbb R$$ be a differentiable function such that $$f(0)= 0$$ and $$f(1)= 1$$ and $$|f'(x)|<2 ~ \forall x \in \mathbb R$$, if $$a$$ and $$b$$ are real numbers such that the set of possible values of $$\displaystyle\int_0^1 f(x)dx$$ is the open interval $$(a,b)$$, then $$b-a$$ is: ?
Attempt:
$$I = \int_0^1 1.f(x) dx$$
$$\implies I = 1 - \int_0^1 xf'(x)dx$$ (Integration by parts)
$$-2 < f'(x) < 2$$
$$\implies -2x for $$x>0$$
$$\implies -1 < \int_0^1 xf'(x) dx < 1$$
Therefore $$I_{max} = 2$$ and $$I_{min} = 0$$
$$\implies b- a = 2$$ but answer given is $$b-a = \dfrac 3 4$$.
Please let me know my mistake, and the correct way to solve it.
• I'm guessing the problem statement says $f(x)$ is nonnegative? – user25959 Nov 2 '18 at 19:26
• @user25959 it does not. – Abcd Nov 2 '18 at 19:32
The following is multiple choice question (with options) to answer.
If the function f(x) is defined for all real numbers x as the maximum value of 2x + 4 and 12 + 4x, then for which one of the following values of x will f(x) actually equal 2x + 4 ? | [
"–3",
"–5",
"–6",
"–7"
] | B | Since f(x) defines maximum of 2x + 4 and 12 + 4x,
to find those x when 2x+4 has greater value,
2x + 4 > 12 + 4x
or x < -4
Only value that satisfy this is -5.
Answer is (B), |
AQUA-RAT | AQUA-RAT-36809 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A truck driver drove for 2 days. On the second day, he drove 3 hours longer and at an average speed of 15 miles per hour faster than he drove on the first day. If he drove a total of 1,020 miles and spent 22 hours driving during the 2 days, what was his average speed on the first day, in miles per hour? | [
"25.0",
"30.6",
"37.8",
"40.2"
] | C | Day 1
2t+3=22
t=9.5
Day 2
t+3
9.5+3=12.5
9.5r+12.5(r+15)=1020
r=37.8
Answer:C |
AQUA-RAT | AQUA-RAT-36810 | solar-system, amateur-observing, asteroids, near-earth-object
Is there a distinction between NEOs and near-Earth asteroids? Is there a difference?
What (actually) defines an Aten-class near-earth asteroid?
Has Hubble ever been used to try to image a near Earth asteroid?
Is the passage of three asteroids near Earth today just coincidental?
https://space.meta.stackexchange.com/q/1459/12102 The Near-Earth close approches website shows close approaches to the Earth by near-Earth objects (NEOs). The table showing all close encounters indicates the absolute magnitude.
The data can be exported to a CSV file to estimate the apparent magnitude for each object, using the following equation.
$$
m = H + 5 \log_{10} \bigg( \frac{d_{BS}d_{BO}}{d_0^2} \bigg) - q(\alpha)
$$
where $H$ is the absolute magnitude, $m$ is the apparent magnitude, d_{*} are the distances between the objects and $a(\alpha)$ is the reflected light. $q(\alpha)$ is a number between 0 and 1.
I only want to know what happens when the object is closest to the Earth, so I use the approximation that the distance from the Sun to the NEO is 1AU.
$q(\alpha)$ is complicated to compute, so I just compute $m$ using $q=0$ and $q=1$. This leads to
min value $ = H + 5 \log (d_{BO}) - 1 < m < H + 5 \log (d_{BO}) = $ max value
with $d_{BO}$ the distance between the Earth and the NEO expressed in astronomical units (AU).
The server is unhappy when I try to get the entire database, so I limited my export to the objects that come reasonably close to Earth (d<0.05 AU), with no time limit.
Among these 24588 objects, 4 have a maximal magnitude less than 6, and 16 have a minimal magnitude less than 6. So between 1900 and 2200, no more than 16 NEOs are visible by the naked eye.
The following is multiple choice question (with options) to answer.
The speeds of three asteroids were compared. Asteroids X-13 and Y-14 were observed for identical durations, while asteroid Z-15 was observed for 2 seconds longer. During its period of observation, asteroid Y-14 traveled three times the distance X-13 traveled, and therefore Y-14 was found to be faster than X-13 by 6000 kilometers per second. Asteroid Z-15 had an identical speed as that of X-13, but because Z-15 was observed for a longer period, it traveled five times the distance X-13 traveled during X-13's inspection. Asteroid X-13 traveled how many kilometers during its observation? | [
"500",
"1,600/3",
"1,000",
"1,500"
] | D | X13: (t, d, s)
Y14: (t, 3d, s+6000mi/hour)
Z15: (t+2 seconds, s, 5d)
d=?
Distance = Speed*Time
x13: d = s*t
x14: 3d = (s+6000)*t ===> 3d = ts+6000t
z15: 5d = s*(t+2t) ===> 5d = st+2st ===> 5d - 2st = st
3d = 5d - 2st + 6000t
-2d = -2st + 6000t
2d = 2st - 6000t
d = st - 3000t
x13: d = s*t
st - 3000t = s*t
s - 3000 = s
-1500 = s
I got to this point and couldn't go any further. This seems like a problem where I can set up individual d=r*t formulas and solve but it appears that's not the case. For future reference how would I know not to waste my time setting up this problem in the aforementioned way? Thanks!!!
The distance of Z15 is equal to five times the distance of X13 (we established that x13 is the baseline and thus, it's measurements are d, s, t)
S(T+2) = 5(S*T)What clues would I have to know to set up the equation in this fashion? Is it because I am better off setting two identical distances together?
ST+2S = 5ST
T+2 = 5T
2=4T
t= 1/2
We are looking for distance (d=s*t) so we need to solve for speed now that we have time.
Speed y14 - speed x13
Speed = d/t
3d/t - d/t = 6000 (remember, t is the same because both asteroids were observed for the same amount of time)
2d = 6000
2 = 3000
d=s*t
d=3000*(1/2)
d=1500
ANSWER:D |
AQUA-RAT | AQUA-RAT-36811 | ### Show Tags
30 Jun 2018, 18:53
12/(.6x)=12/x+4
20=12+4x
x=2
24/2
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
04 Jul 2018, 18:25
1
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost: pq = 12 q = 12/p After the orange’s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is: (0.6p)(q + 4) = 12 0.6pq + 2.4p = 12 Substituting for q, we have: 0.6p(12/p) + 2.4p = 12 7.2 + 2.4p = 12 2.4p = 4.8 p = 2 That is, each orange is$2. So for \$24, we can buy 24/2 = 12 oranges.
_________________
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Re: When the price of oranges is lowered by 40%, 4 more oranges &nbs [#permalink] 04 Jul 2018, 18:25
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The following is multiple choice question (with options) to answer.
A reduction of 20% in the price of salt enables a lady to obtain 10kgs more for Rs.400, find the original price per kg? | [
"s.7",
"s.9",
"s.10",
"s.11"
] | C | 100*(20/100) = 20 --- 10
? --- 1 => Rs.2
400 --- 80
? --- 2 => Rs.10
Answer:C |
AQUA-RAT | AQUA-RAT-36812 | They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
The true discount on a bill of Rs. 2560 is Rs. 360. What is the banker's discount? | [
"Rs. 432",
"Rs. 422",
"Rs. 419",
"Rs. 442"
] | C | Explanation :
F = Rs. 2560
TD = Rs. 360
PW = F - TD = 2560 - 360 = Rs. 2200
True Discount is the Simple Interest on the present value for unexpired time
=>Simple Interest on Rs. 2200 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest on the face value of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
=360/2200 × 2560 = 0.16 × 2560 = Rs. 419 Answer : Option C |
AQUA-RAT | AQUA-RAT-36813 | population-biology, population-dynamics
Title: Why will world population keep growing if all women have only 2.1 children Why is it that, even if we were going to immediately agree that every women will not have more than 2.1 children on average, the world population would continue to grow for another 60 years? I've learned that a TFR of 2.1 means the population is at 0 growth rate. I guessed the answer was the population-lag effect, but since we are assuming that ALL women immediately have only 2.1 children on average, wouldn't there be a 0 population growth immediately and not after 60 years? Thanks in advance! As you mention, the population-lag effect is responsible for this. From The Wikipedia article on TFR: http://en.wikipedia.org/wiki/Total_fertility_rate#Population-lag_effect
A population that has recently dropped below replacement-level
fertility will continue to grow, because the recent high fertility
produced large numbers of young couples who would now be in their
childbearing years.
Thus, even if the current TFR implies long-term stability, the recent history of TFR values will continue to affect birth rates, and thus population growth/decrease, in the future.
Imagine that you have a stable population, with static birth and death rates. Now imagine that the birth rate during a single year for some reason doubles before dropping back to normal the next year, thus transiently increasing TFR. Now, it is easy to see that when the newly born, larger-than-normal generation reaches reproductive age, the population will increase again because the birth rate will increase in proportion with the size of the generation, while the death rate is unaffected until the enlarged generation reaches old age and starts dying off. Later, the population will decrease to its long-term stable level, which will be larger than before the TFR spike.
The following is multiple choice question (with options) to answer.
Statements: Population increase coupled with depleting resources is going to be the scenario of many developing countries in days to come. Conclusions: 1) The population of developing countries will not continue to increase in future. 2) It will be very difficult for the governments of developing countries to provide its people decent quality of life. | [
"Only conclusion I follows",
"Only conclusion II follows",
"Either I or II follows",
"Neither I nor II follows"
] | B | The fact given in I is quite contrary to the given statement. So, I does not follow. II mentions the direct implications of the state discussed in the statement. Thus, II follows.
B |
AQUA-RAT | AQUA-RAT-36814 | that can be factored with ease!
Use the fact that $73 \times 137=10001 = 10^4+1$.
Now, mark off the number in groups of four digits starting from the right, and add the four-digit groups together with alternating signs.
Applying the above rule, $1000027$ in groups of $4$ is $\underbrace{0100}$ $\underbrace{0027}$. Adding the groups with alternate signs gives $73$. Therefore, $1000027$ is divisble by $73$ and gives $13699$ as quotient.
For $13699$ apply the divisbility test by $7$ by marking of the digits in groups of $3s$ by starting from the right and adding together with alternate signs. Therefore, adding the groups $\underbrace{013}$ $\underbrace{699}$ with alternate signs gives $686$ which is divsible by $7$ and hence $13699$ is disvisble by $7$
$13699$ when divided by $7$ gives $1957$.
Now, $1957$ is $1900+57$ and hence is divisible by $19$ giving the quotient as $103$.
Combining them all gives $1000027 = 7\times19\times73\times103$
To begin with note that
$1000027 = 100^3 + 3^3 = (100+3)(100^2−3⋅100+32) = 103 \cdot 9709 = 103 \cdot (1001 \cdot 9 + 700) = 103 \cdot 7 \cdot (13 \cdot 11 \cdot 9+100) = 103 \cdot 7 \cdot 1387$
where we have used the well-known fact that $1001 = 7 \cdot 11 \cdot 13$
The following is multiple choice question (with options) to answer.
Which number need to add to 172835 to get a number exactly divisible by 136? | [
"27",
"25",
"26",
"21"
] | D | 172835 / 136 = 1270 and reminder = 115.
136-115 = 21
So, the next number divisible by 115 is 21 places in front of 172835
Which means 21 + 172835 =172856
21 should be added to 172835
D |
AQUA-RAT | AQUA-RAT-36815 | # 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 speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream? | [
"17 kmph",
"40 kmph",
"16 kmph",
"18 kmph"
] | B | Speed downstream
= 60 + 20 = 80 kmph
Speed upstream
= 60 - 20
= 40 kmph
Answer:B |
AQUA-RAT | AQUA-RAT-36816 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
The market value of a certain machine decreased by 30 percent of its purchase price each year. If the machine was purchased in 1982 for its market value of $10,000, what was its market value two years later? | [
" $8,000",
" $4,000",
" $3,200",
" $2,400"
] | B | B.
Market value in 1982= $ 10000
Market value in 1983 = $ 10000- ($ 10000 x 30/100) = 10000-3000= $ 7000
Market value in 1984 = Market value in 1983 - (30 % of $10000)= 7000-3000 = $4000 |
AQUA-RAT | AQUA-RAT-36817 | 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.
Ayesha's father was 40 years of age when she was born while her mother was 32 years old when her brother four years younger to her was born. What is the difference between the ages of her parents? | [
"12 years",
"14 years",
"16 years",
"7 years"
] | A | Explanation:
Mother's age when Ayesha's brother was born = 32 years.
Father's age when Ayesha's brother was born = (40 + 4) = 44 years.
Required difference = (44 - 32) = 12 years.
Answer: Option A |
AQUA-RAT | AQUA-RAT-36818 | 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.
12 men take 18 days to complete a job whereas 12 women in 18 days can complete 5/6 of the same job. How many days will 10 men and 8 women together take to complete the same job? | [
"6",
"13 1â„2",
"13",
"Data inadequate"
] | C | 12 M × 18 = 12 W × 18 × 6/5
\ w = 5/6 M
10M + 8W = 10M + 8 × 5/6M = 16 2/3 M
\16 2/3men can complete the same work
in 12×18/16 2/3=13 days
Answer C |
AQUA-RAT | AQUA-RAT-36819 | tiling
Title: Minimizing number of MxM squared tiles in an infinite grid covered by any part of a shape Without restriction (e.g. continuity or convexity are not guaranteed), we're given an W x H raster-based shape that needs to be placed on an infinite tiling of M x M squares. The shape can placed at any integer positional offset within those squares.
The goal is to minimize the number of squares containing any part of that shape.
For example, this shape (shown in orange) can be placed in a way that occupies seven squares on a tiling of 10 x 10:
However, a more optimal placement is an offset of (+2, 0) which occupies six squares by leaving the central bottom square unoccupied:
Can the brute-force solution (O(W * H * M * M)) be improved upon? How / why not? I found an algorithm. The TL;DR is that we can efficiently precompute a parallel image that indicates whether each position in the image contains any occupied pixels if we were to start a subgrid at that position (going down and to the right, chosen arbitrarily), and determining which collection of those positions produces the minimum total number of occupied subgrids can then be done in simple iteration.
Compute a "row occupancy" grid which extends to the "left" of the image by M additional pixels. Each pixel in this grid is TRUE if any pixel within M pixels to the right of it (including itself) is nonempty. This can be done in O(HW) time because you can iterate each row backwards, keeping track of how long it's been since the last time you saw an occupied pixel
Compute a "grid occupancy" grid, which is computed similarly. This grid extends "above" the top of the image by M additional pixels as well as to the left by M additional pixels. The algorithm is the same as the prior step, except applied to the row occupancy grid - count from the bottom up, keeping track of the last time you saw an occupied row, marking TRUE when the last occupied row was within the last M pixels.
The following is multiple choice question (with options) to answer.
Rectangular tile each of size 25cm by 65cm must be laid horizontally on a rectangular floor of size 150cm by 390cm,such that the tiles do not overlap and they are placed with edges jutting against each other on all edges. A tile can be placed in any orientation so long as its edges are parallel to the edges of floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is: | [
"50",
"40",
"25",
"36"
] | D | Area of tile = 25*65 = 1625
Area of floor = 150*390 = 58500
No of tiles = 58500/1625 = 36
So, the no of tile = 36
ANSWER:D |
AQUA-RAT | AQUA-RAT-36820 | + b_y^2b_x^2 + b_y^4 + b_y^2 d^2\\ - d^2 a_x^2 - d^2 a_y^2 + d^2 b_x^2 -3 d^2 b_y^2 + d^4\\ = 0$$
The following is multiple choice question (with options) to answer.
Simplify (5^y + 5^y + 5^y + 5^y + 5^y) (4^y + 4^y + 4^y + 4^y) | [
" 4^4y * 3^3y",
" 20^y+1",
" 16^y + 9^y",
" 12^y"
] | B | ans B
(5^y + 5^y + 5^y + 5^y + 5^y) (4^y + 4^y + 4^y + 4^y)
(5*5^y)(4*4^y)
20*20^y=20^(y+1).. |
AQUA-RAT | AQUA-RAT-36821 | Since we have $21$ terms taking $20$ possible values, there are some $0 \le i < j \le 20$ such that $S_i = S_j$. It follows that the total number of hours of study between days $i+1$ and $j$ (inclusive) is a multiple of $20$.
If it is not exactly equal to $20$ hours, then it must be at least $40$ hours. However, this is over a span of at most $20$ days. Dividing this into three periods of at most $7$ days (say the first week, second week, and third week), by averaging we find that she must have worked at least $14$ hours during one of the weeks, which is not allowed.
Thus she must actually have studied exactly $20$ hours between days $i+1$ and $j$.
• Yes, that's a little cleaner as route to Wiley's lemma.Thanks. – Joffan Jul 17 '16 at 23:05
• @Joffan, agreed. It's indeed cleaner to use 20 "holes" with 21 "pigeons" rather than separating $S_l=S_i+20$ as an individual case in my proof. Thank you, Shagnik. – Wiley Jul 18 '16 at 3:59
The proof consists of two parts.
• Part I: Prove that a period of $20$ days is enough such that there must exist some period of consecutive days during which totally $20$ hours are spent on studying.
• Part II: A counterexample which shows that $19$ days are not enough is presented.
Proof of Part I
The following is multiple choice question (with options) to answer.
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is: | [
"250",
"276",
"280",
"285"
] | D | Since the month begins with sunday,to there will be five sundays in the month
Average required = (510x5 + 240x25 /30) = 8550/30 =285
Answer: Option D |
AQUA-RAT | AQUA-RAT-36822 | python, performance, primes
Title: Express a number as sum of 2 primes
Problem: Given an even number (greater than 2), return two prime
numbers whose sum will be equal to given number.
NOTE 1: A solution will always exist. (Goldbach's conjecture)
Note 2: If there are more than one solutions possible, return the
lexicographically smaller solution.
This solution becomes slower for higher order numbers. How can I optimise this code?
import math
# Function to create a list of all prime numbers uptil given number
def list_prime(A):
primes = [] # initialise an empty list of primes
bool_a = [True for i in range(2,A+3)] # creates boolean list of index
for i in range(2, int(math.sqrt(A))+1):
if bool_a[i]:
for j in range( i * i, A +1 , i): # eliminate all multiples of i*i+i, i*i+2i and so on uptil A+1
if bool_a[j]:
bool_a[j] = False;
for i in range(2,A): # the index of reamaining bools are prime numbers
if bool_a[i]:
primes.append(i)
return primes
#Function that returns two prime numbers whose sum is equal to the given number
def prime_sum(A):
solution_set = []
for i in (list_prime(A)):
for j in (list_prime(A)[::-1]):
if i + j == A:
solution_set.append((i,j))
break
return min(solution_set) Using the Sieve of Eratosthenes seems like a reasonable way to generate a large number of primes.
One thing that you could do is directly return your bitset representation of the primes, rather than copying it to a condensed list. This would save a bit of time in the short term, and be helpful for the next step.
for i in (list_prime(A)):
for j in (list_prime(A)[::-1]):
The following is multiple choice question (with options) to answer.
Find a sum for 1st 6 prime number's? | [
"41",
"28",
"30",
"34"
] | A | Required sum = (2 + 3 + 5 + 7 + 11 + 13) = 41
Note: 1 is not a prime number
Option A |
AQUA-RAT | AQUA-RAT-36823 | earth
Well, lets look at the values returned compared to days in a non-leap year:
Month N1 Day Diff
1. 30 31 -1
2. 61 59 +2
3. 91 90 +1
4. 122 120 +2
5. 152 151 +1
6. 183 181 +2
7. 213 212 +1
8. 244 243 +1
9. 275 273 +2
10. 305 304 +1
11. 336 334 +2
12. 366 365 +1
A bit messy, right? Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:
Month N1 Day Diff
1. 31 31 0
2. 61 59 +2
3. 92 90 +2
4. 122 120 +2
5. 153 151 +2
6. 183 181 +2
7. 214 212 +2
8. 245 243 +2
9. 275 273 +2
10. 306 304 +2
11. 336 334 +2
12. --- 365 ---
From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later. This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:
N = N1 - (N2 * N3) + day - 30
The following is multiple choice question (with options) to answer.
The average salary of a person for the months of January, February, March and April is Rs.8000 and that for the months February, March, April and May is Rs.8500. If his salary for the month of May is Rs.6500, find his salary for the month of January? | [
"4500",
"5000",
"6000",
"6500"
] | A | Sum of the salaries of the person for the months of January, February, March and April = 4 * 8000 = 32000 ----(1)
Sum of the salaries of the person for the months of February, March, April and May = 4 * 8500 = 34000 ----(2)
(2)-(1) i.e. May - Jan = 2000
Salary of May is Rs.6500
Salary of January = Rs.4500
ANSWER A |
AQUA-RAT | AQUA-RAT-36824 | I want to draw your attention to the expression $$(x_i - \overline{x})$$ in the numerator. Ignore the summation notation for the purposes of this answer. Note, we decrement the sample mean (i.e., $$\overline{x}$$) from each realization of $$x_{i}$$. If we were to increase or decrease each $$x_{i}$$ by the same amount, the mean will change. However, the distance of each realization of $$x_{i}$$ from that central tendency remains the same. In other words, each deviation from the mean is the same.
Subtracting each $$x_{i}$$ from a constant (e.g., $$100 - x_{i}$$) flips the vector of values to its mirror image, then slides it along the number line by a constant amount. Suppose the first realization of $$x_{1}$$ = 20 and $$\overline{x}$$ = 4. In keeping with the foregoing expression,
$$(20 - 4) = 16,$$
which is the deviation from the sample mean. Now, assume we don't move along the number line just yet and we simply put a negative sign in front of each $$x_{i}$$, such that we have $$-(x_{i})$$; this flips the sign of the mean as well. The first realization, $$x_{1}$$, is now $$-20$$. Again, substituting $$-20$$ into the expression,
$$(-20-(-4)) =-16,$$
which is the same number of units away from the mean. Note, squared deviations result in a positive number, so the numerator remains the same. You can run this code in R which builds upon @knrumsey's insightful response, but breaks it down further:
x <- rnorm(10000, 30, 5) # Simulates 10,000 random deviates from the normal distribution
The following is multiple choice question (with options) to answer.
The mean of 50 observations is 200. But later he found that there is decrements of 15 from each observations. what is the the updated mean is ? | [
"170",
"185",
"190",
"196"
] | B | 185
Answer is B |
AQUA-RAT | AQUA-RAT-36825 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
Set S consists of integers {6, 7, 10, 12, 15}. If integer n is included in the set, the average (arithmetic mean) of set S will increase by 40%. What is the value of integer n? | [
"26",
"30",
"34",
"38"
] | C | The average of the numbers in set S is 10.
If we increase the mean by 40%, the new mean is 14.
Thus, on average, 5 numbers increase by 4.
Therefore n = 14+20 = 34
The answer is C. |
AQUA-RAT | AQUA-RAT-36826 | A. 216
B. 504
C. 3,024
D. 3,600
E. 7,371
The number of 2-digit codes is 9 x 9 = 9^2, the number of 3-digit codes is 9 x 9 x 9 = 9^3, and the number of 4-digit codes is 9 x 9 x 9 x 9 = 9^4, so the total number of codes is:
9^2 + 9^3 + 9^4 = 9^2(1 + 9 + 9^2) = 81(91) = 7371
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Re: Each item sold at a market is labelled with a 2-, 3-, or 4- digit code &nbs [#permalink] 14 Nov 2017, 06:20
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A local bank that has 9 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 9 codes? | [
"3",
"4",
"5",
"6"
] | A | Pick any two integer.
Integers: 12
Code: 11, 12, 21, 22 = 4 Codes
Add one more integer: 3
13, 31, 33, 23, 32 = 5 Codes
Total = 9 Codes. Enough. Answer: A |
AQUA-RAT | AQUA-RAT-36827 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
The area of square ABCD above is 12. The shaded region is the intersection of the square and a circular region centered at C. What is the area of the shaded region? | [
"18−9∗π",
"12−3∗π",
"9+9/4∗π",
"9+9/2∗π"
] | B | =Square - Quarter circle
=12- pi*12/4
=12- pi*3
ANSWER:B |
AQUA-RAT | AQUA-RAT-36828 | feature a number of tools to solve geometry problems. Each part is called semi circle. A rectangle is one of the many fundamental shapes you'll see in math. Round your answer to the nearest hundredth. A square has sides of length 12 cm. To make the pies, roll the pastry on a floured surface into a large rectangle, about 12 inches by 24-plus inches. Learn how to calculate perimeter and area for various shapes. The distance around the outside of a semicircle is generally called its perimeter. It is evident that if we make a large number of cuts, the figure formed will approximate a rectangle whose length is equal to one-half of the circumference and whose width is equal to the radius. Fold the entire piece of fabric in half to create a large square. Area of a circle intuition. Requires knowledge of Conic Sections. make up Remember, by o shape. Creating a design with pavers requires planning. Repeat the same method the other side. where r is the radius of the circle. The word ‘area’ stands for the space occupied by a flat object or figure. To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Draw and cut out the wing shape from the flap and insert into the slit. 1165 1225 112. Calculations at a semicircle. 28 square cm. asked by ian on August 29, 2016. How to Find the Area of a Semicircle To find the area of a semi-circle, you need to know the formula for the area of a circle. Semi-circles are drawn on and as diameters. Determine the diameter of the circle you intend to make. 4 in L, Base 6in - 4694841. in Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 251 16. Practice: Area of a circle. In the article below, we provide the semicircle definition and explain how to find the perimeter and area of a semicircle. What is the objective function in terms of the base of the rectangle, x? (Type an expression. The Gimp provides the Rectangle, Ellipse and Lasso selection tools to help you cut out specific parts of a photo or illustration to keep, eliminate or edit. Area
The following is multiple choice question (with options) to answer.
The parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 14 cm. Find the circumference of a semicircle whose diameter is equal to the side of the square. (Round off your answer to two decimal places) | [
"23.57",
"23.54",
"23.50",
"23.55"
] | A | Let the side of the square be a cm.
Parameter of the rectangle = 2(16 + 14) = 60 cm Parameter of the square = 60 cm
i.e. 4a = 60
A = 15
Diameter of the semicircle = 15 cm
Circimference of the semicircle
= 1/2(∏)(15)
= 1/2(22/7)(15) = 330/14 = 23.57 cm to two decimal places
Answer: A |
AQUA-RAT | AQUA-RAT-36829 | 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:1,3,10,21,64,129,356,777 | [
"3",
"10",
"21",
"64"
] | D | 1=1
3=3
10=1+0=1
21=2+1=3
64=6+4=10=1+0=1
129=1+2+9=12=1+2=3
356=3+5+6=14=1+4=5( wrong one.. it should be 1)
777=7+7+7=21=2+1=3
Alternative 1 and 3 series..
ANSWER:D |
AQUA-RAT | AQUA-RAT-36830 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
In a party attended by 12 persons, each clinch their glass with every other. How many glass clinches? | [
"51",
"52",
"53",
"66"
] | D | Total no. of person = 12
Total no. of glass clinches = n(n-1)/2
=12*11/2
= 66
ANSWER:D |
AQUA-RAT | AQUA-RAT-36831 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In an exam, Amar scored 64 percent, Bhavan scored 36 percent and Chetan 44 percent. The maximum score awarded in the exam is 800. Find the average mark scored by all the three boys? | [
"384",
"826",
"207",
"269"
] | A | Average mark scored by all the three boys =
[64/100 (800) + 36/100 (800) + 44/100 (800)] / 3
= 384
Answer: A |
AQUA-RAT | AQUA-RAT-36832 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
A group of 55 adults and 70 children go for trekking. If there is meal for either 70 adults or 90 children and if 28 adults have their meal, find the total number of children that can be catered with the remaining food. | [
"42",
"54",
"90",
"70"
] | B | Explanation:
As there is meal for 70 adults and 28 have their meal, the meal left can be catered to 42 adults.
Now, 70 adults = 90 children
7 adults = 9 children
Therefore, 42 adults = 54 children
Hence, the meal can be catered to 54 children
ANSWER B |
AQUA-RAT | AQUA-RAT-36833 | # Math Help - Questions Involving Z-Score, Can you please check my answer?
1. The lifespan of a certain brand of light bulbs in a photographic machine is normally distributed with a mean of 210 hours and a standard deviation on 50 hours.
a) Determine the z-score of a light bulb that has a lifespan of exactly 124 hours?
z= x - mean / standard deviation
z= 124-210 / 50
z= -86 / 50
z= -1.72
Z-Score is -1.72
b) What is the probability that a randomly chosen light bulb would have a lifespan of less than 180 hours? (Not sure if correct)
z= x- mean / Standard Deviation
z= 180 - 210 / 50
z= -0.60
P(x<180) = P (z<-0.60) = 0.2258
So there is a 0.2258 (22.58%) chance the lifespan is less than 180 hours.
c) What is the probability that a randomly chosen light bulb would have a lifespan of between 200 and 250 hours? (Not sure if right, italic/bold part I'm a little iffy about)
z= x - mean / standard deviation
z= 200-210 / 50
z= -0.2
z= x - mean / standard deviation
z= 250-210 / 50
z= 0.8
P(200<z<250) = p(-0.2<z<-0.8) = P(z<0.8) - p(z<-0.2) = 0.7881 - 0.4207 = 0.3674
There's a 0.3674 (36.74%) chance the lifespan is between 200 and 250.
2. ## Re: Questions Involving Z-Score, Can you please check my answer?
A) Correct
B) p = 0.27425
C) Correct
3. ## Re: Questions Involving Z-Score, Can you please check my answer?
A) Correct
B) p = 0.27425
C) Correct
Can you tell me how you get 0.27425? On the Z-score table, the Z-score 0.6 has a probability of 0.2258? Or am I reading it wrong?
4. ## Re: Questions Involving Z-Score, Can you please check my answer?
The following is multiple choice question (with options) to answer.
Life expectancy is defined by the formula 2SB/G, where S = shoe size, B = average monthly electric bill in dollars, and G = GMAT score. If Melvin's GMAT score is twice his monthly electric bill, and his life expectancy is 70, what is his shoe size? | [
"75",
"70",
"65",
"40"
] | B | Solution -
Given that, G = 2B and Life expectancy(L) = 70
L=2SB/G -> 70 = 2SB/2B --> S=70.
ANS B |
AQUA-RAT | AQUA-RAT-36834 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Posts: 140
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4? | [
"56",
"86",
"25",
"76"
] | D | Average after 11 innings = 36
Required number of runs
= (36 * 11) - (32 * 10)
= 396 - 320
= 76.
Answer:D |
AQUA-RAT | AQUA-RAT-36835 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
For all integers a and b, a%b = (a + b)(a - b). If 8%x = 15, then which of the following could be a value of x? | [
"-7",
"-6",
"2",
"3"
] | A | 8%x = (8+x)(8-x)
64-x^2 = 15
x^2 = 49
x=-7,7
Ans: A |
AQUA-RAT | AQUA-RAT-36836 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
1+2+2^2+2^3+2^4+2^5+2^6+2^7=？ | [
"(2^4-1)(2^4+1)",
"2^6+1",
"2^5-1",
"2^5+1"
] | A | From 1+2+2^2+2^3+2^4+2^5+2^6+2^7=1(2^8-1)/(2-1)=2^8-1=(2^4-1)(2^4+1), the correct answer is A. |
AQUA-RAT | AQUA-RAT-36837 | # Thread: probability of 52 card deck
1. ## probability of 52 card deck
I have a few simple probability questions regarding draws from a deck of cards.
1) If two cards are drawn face down, what is the probability that the second card is an ace?
2) If it is known the first card draw is an ace, how would that change the answer to (1)?
3) What is the probability that 2 randomly drawn cards are both aces?
4) If two cards are drawn from a deck, how many different combinations of the two cards are possible if the order is not considered and if the order is considered?
For (1), I think the answer is 4/51, but do I need to include the probability that the first card is not an ace?
For (2), I think it's 3/51.
For (3), I think it's: 4/52 * 3/51 = 1/221
For (4) - no idea
2. 1. You didn't learn anything about the 1st card since it's face down. So this is like choosing 1 card from 52, your probability is 4/52.
2. Correct
3. Correct
4. Order doesn't count - 52 choose 2.
Order does count - 52 x 51
3. Hello, chemekydie!
1) If two cards are drawn from a standard deck,
what is the probability that the second card is an ace?
I'll do this the Long Way.
There are 4 Aces and 48 Others.
We must consider both possibilities:
[1] The first card is an Ace: . $p(\text{1st Ace}) \:=\:\frac{4}{52}$
. . The second card is an Ace: . $P(\text{2nd Ace}) \:=\:\frac{3}{51}$
. . Hence: . $P(\text{Ace, then Ace}) \:=\:\frac{4}{52}\cdot\frac{3}{51} \:=\:\frac{12}{2652}$
[2] The first card is not an Ace: . $P(\text{1st not-Ace}) \:=\:\frac{48}{52}$
The following is multiple choice question (with options) to answer.
One card is drawn from a pack of 52 cards. What is the probability that the card drawn is a face card? | [
"4/13",
"5/16",
"9/17",
"2/5"
] | A | Out of 52 cards there will be 16 face cards are there.
probability of getting a face card = 16/52 = 4/13
Answer is A |
AQUA-RAT | AQUA-RAT-36838 | => 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.
The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 18000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is | [
"20",
"42",
"22",
"23"
] | B | Sol.
Let the total number of workers be x. Then,
8000x = (18000 × 7) + 6000 ( x – 7)
‹=› 2000x = 84000
‹=› x = 42.
Answer B |
AQUA-RAT | AQUA-RAT-36839 | # Combination Question
#### Swazination
##### New member
How many combinations can I make if I have 4 letters A,G,C,T and I want to make groups of 5. Yes, you can repeat the letters.
#### Denis
##### Senior Member
From AAAAA to TGCAA
or (if replaced by digits 1 to 4):
from 1111 to 43211
Yes?
#### ksdhart2
##### Full Member
Combinations/Probability problems of this sort almost always come down to pattern recognition. In fact, learning how to find a pattern and extrapolate what comes next will help you greatly in nearly every aspect of mathematics. So let's try using those skills and see what we can come up with for this problem. Let's start with a much simpler version and build up as we go along. Suppose we had two letters (A and G) and one slot. How many combinations would there be? Well, obviously the answer is two. That's not very interesting nor does it seem helpful just yet, but we'll keep in mind all the same.
Now suppose we had two letters and two slots. How many combinations would there be then? Thinking about it, we can force the first slot to be A, then there's two possibilities for the second slot. Similarly, if we force the first slot to be G, we have two additional combinations. So we have two possibilities for the first slot, and for each of those two possibilities, we have two possibilities for the second slot. That gives us a total of 2 * 2 = 4 = 22 combinations.
Now suppose we had two letters and three slots. How many combinations would there be then? I see that we can arbitrarily fix the first two slots and then let the third one vary. Since we already know that there's four possible ways to arrange the first two slots, and for each of those four possibilities, there's two choices for the third slot, that gives us a total of 4 * 2 = 8 = 23 combinations. Are you seeing a pattern? How many combinations would there be with four slots? Five slots? n slots?
The following is multiple choice question (with options) to answer.
Find the number of selections that can be made taking 4 letters from the word "ENTRANCE". | [
"70",
"36",
"35",
"72"
] | B | We have 8 letters from which 6 are unique.
Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.
Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.
15+10+10+1=36
Answer: B. |
AQUA-RAT | AQUA-RAT-36840 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
The average of 30 numbers is 25. If each number is multiplied by 5, find the new average? | [
"115",
"125",
"135",
"145"
] | B | Sum of the 30 numbers = 30 * 25 = 750
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.
ANSWER:B |
AQUA-RAT | AQUA-RAT-36841 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\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.
During a clearance sale, a retailer discounted the original price of its TVs by 25% for the first two weeks of the month, then for the remainder of the month further reduced the price by taking 20% off the sale price. For those who purchased TVs during the last week of the month, what percent of the original price did they have to pay? | [
"40%",
"45%",
"55%",
"60%"
] | D | As it's a percentage problem, start with $100.
The first discount is 25% of $100, so the sales prices is 75% of $100 = $75.
The second discount is 20% of $75. Move the decimal once to the left and double it = $7.5*2 = $15.
The final sales price is $75-$15 = $60, which is 60% of $100.
The correct answer is D |
AQUA-RAT | AQUA-RAT-36842 | I'm glad I now know what the Pigeonhole Principal is after having heard it so many times. :)
There being $14$ distinct numbers$\in N\gt 13$, simply apply the Division Algorithm to get the remainders $0-12$ for $13$ of those numbers. So produce set $\{13q_0 + r_0,13q_1 + r_1,...,13q_{12}+r_{12}\}$. Remainders $r_1-r_{12}$ are the numbers $0$ through $12$ and are distinct. if any were the same the difference would be divisible by $13$ and done. now the $14th$ number must have the same remainder as one of the other $13$ and so the difference is divisible by $13$. I actually stumbled on this in an induction proof. Induction on the range of numbers/greatest number $n$.
Base case $n=14$:must have all numbers $1$ through $14$. $14-1=13$.
Assume true for $n$. For $n+1$ can apply what is said above. If all $14$ chosen from set through range $n$ done/true, by inductive hypothesis.
Otherwise if $n+1$ chosen have set/subset $\{\{i_1,i_2,...,i_{13}\},n+1\}$ and can apply what is said above. That is, if none of $\{i_1,i_2,...,i_{13}\}$ have same remainder mod $13$ they cover the entire range of remainders mod $13$ and therefore $n+1$ mod $13$ must be same as one of $\{i_1,i_2,...,i_{13}\}$mod $13$. This completes the induction. (if same mod $13$ difference between the two is divisible by $13$)
Induction is not needed for this problem, though.
The following is multiple choice question (with options) to answer.
Find the greatest number which is such that when794, 858 and 1351 are divided by it, the remainders are all same. | [
"1",
"5",
"6",
"7"
] | A | Given, the remainders are same i.e. differences of that numbers are exactly divisible. Hence, you have to find HCF (x–y, y–z, z–x). 858–794 = 64; 1351–794 = 557; 1351–858 = 493. HCF of (64, 557, 493) = 1.
Answer:A |
AQUA-RAT | AQUA-RAT-36843 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
Find the annual income derived by investing $ 6800 in 50% stock at 136. | [
"550",
"2500",
"250",
"3000"
] | B | By investing $ 136, income obtained = $ 50.
By investing $ 6800, income obtained = $ [(50/136)*6800] = $ 2500.
Answer B. |
AQUA-RAT | AQUA-RAT-36844 | $$15 + 2\sqrt{10} + 2\sqrt{14} + 2\sqrt{35} > 19 + 2\sqrt{34}$$
$$2\sqrt{10} + 2\sqrt{14} + 2\sqrt{35} > 4 + 2\sqrt{34}$$
I don't think the number of surds is decreased. We still have five surds (although one is simply $$\sqrt{16} = 4$$ and the rest have a factor of $$2$$ out). I can't see a way to continue it; how would you do it?
- 4 years, 8 months ago
I'm more concerned with looking at the sum of many terms. With few terms, there could be various tricks involved that could help us slightly reduce the number. In that sense, 5/6 was chosen arbitrarily.
It is also not immediately clear that 6 cannot work. There could be some kind of algebraic manipulation that reduces the number.
Staff - 4 years, 8 months ago
569
- 2 years, 7 months ago
The following is multiple choice question (with options) to answer.
How many even multiples of 15 are there between 149 and 301? | [
"5",
"6",
"9",
"10"
] | B | 150 = 10*15
300 = 20*15
The even multiples are 15 multiplied by 10, 12, 14, 16, 18, and 20 for a total of 6.
The answer is B. |
AQUA-RAT | AQUA-RAT-36845 | 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 team P of 20 engineers can complete work or task in 32 days. Another team Q of 16 engineers can complete same task in 30 days.Then the ratio of working capacity of 1 member of P to the 1 member of Q is
a. 3 : 2 | [
"22",
"34",
"77",
"29"
] | B | Explanation:
Let the capacity of an engineer in P = x units, and in Q = y units.
Working capacity of P = x × 32 × 20
Working capacity of Q = y × 16 × 30
As the total work is same, we equate the above equations.
⇒ x × 32 × 20 = y × 16 × 30
⇒ xy=16×3032×20=34
Answer: B |
AQUA-RAT | AQUA-RAT-36846 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
8 gentlemen and 3 gentlewomen are candidates for 2vacancies. A voter has to vote for 2 candidates. In how many ways can one cast his vote? | [
"51",
"52",
"55",
"56"
] | C | There are 11 candidates and a voter has to vote for any two of them.
So, the required number of ways is,
= 11C2 = 55.
C |
AQUA-RAT | AQUA-RAT-36847 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in. | [
"88 days",
"27 days",
"78 days",
"60 days"
] | D | A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days
Answer: D |
AQUA-RAT | AQUA-RAT-36848 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
What is the least common multiple of 8,12,32,14 | [
"672",
"682",
"328",
"964"
] | A | The given numbers are 8,12,32,14
thus the l.c.m is 2*2*2*1*3*4*7=672
the answer is A |
AQUA-RAT | AQUA-RAT-36849 | 12n + 7 7, 19, 31, 43, 67, 79, 103, 127, 139, 151, ... A068229
12n + 11 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, ... A068231
The following is multiple choice question (with options) to answer.
Look at this series: 70, 71, 76, __, 81, 86, 70, 91, ... What number should fill the blank? | [
"60",
"70",
"82",
"77"
] | B | B
70
In this series, 5 is added to the previous number; the number 70 is inserted as every third number. |
AQUA-RAT | AQUA-RAT-36850 | Voiceover: Easily the most quoted number people give you when they're publicizing information about their credit cards is the APR. I think you might guess or you might already know that it stands for annual percentage rate. What I want to do in this video is to understand a little bit more detail in what they actually mean by the annual percentage rate and do a little bit math to get the real or the mathematically or the effective annual percentage rate. I was actually just browsing the web and I saw some credit card that had an annual percentage rate of 22.9% annual percentage rate, but then right next to it, they say that we have 0.06274% daily periodic rate, which, to me, this right here tells me that they compound the interest on your credit card balance on a daily basis and this is the amount that they compound. Where do they get these numbers from? If you just take .06274 and multiply by 365 days in a year, you should get this 22.9. Let's see if we get that. Of course this is percentage, so this is a percentage here and this is a percent here. Let me get out my trusty calculator and see if that is what they get. If I take .06274 - Remember, this is a percent, but I'll just ignore the percent sign, so as a decimal, I would actually add two more zeros here, but .06274 x 365 is equal to, right on the money, 22.9%. You say, "Hey, Sal, what's wrong with that? "They're charging me .06274% per day, "they're going to do that for 365 days a year, "so that gives me 22.9%." My reply to you is that they're compounding on a daily basis. They're compounding this number on a daily basis, so if you were to give them $100 and if you didn't have to pay some type of a minimum balance and you just let that$100 ride for a year, you wouldn't just owe them $122.9. They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as
The following is multiple choice question (with options) to answer.
The credit card and a global payment processing companies have been suffering losses for some time now. A well known company recently announced its quarterly results. According to the results, the revenue fell to $42.0 billion from $69.0 billion, a year ago. By what percent did the revenue fall? | [
"20.8",
"30.4",
"31.8",
"39.1"
] | D | $69-$42=27$
(27/69)*100=39.13%
ANSWER:D |
AQUA-RAT | AQUA-RAT-36851 | Your assistance. This is the actual change, where the original value is subtracted from the new value. Calculate The Discount Percentage Between Two Numbers. So, while this result might be what contact centres wants to see, it does not represent the facts. For example, the percent difference between 30% and 50% is 20%. Different relations between two numbers. com's Numbers to Ratio Calculator is an online basic math function tool to find the quantitative relationship or ratio between two or three given numbers or to reduce the ratio to its lowest terms. # Function to calculate the maximum difference between two elements in a. The Percentage Difference Calculator has 3 ways to calculate the differences between two numbers. 4 day ago Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Use the calculator below to analyze the results of a difference in sample means hypothesis test. Example, total=1,100 and you need to find percent that equals to 100. A relative delta compares the difference between two numbers, A and B, as a percentage of one of the numbers. With delayed retirement credits , a person can receive his or her largest benefit by retiring at age 70. Just Now Percentage Difference Formula. Sales have been poor and the owner decided to mark down each item to$15 to speed up the sales. percentagedifferencecalculator. The percentage diference between 10 and 12 is 18. Determine the absolute difference between two numbers: 15 - 25 = -10; Take the average of those two figures: (15 + 25)/ 2 = 20; Calculate the difference by dividing it by the average: 10/20 = 0. Hi Mahmoud, yes, I was tried that method, it seems to be not working. While most students find percentages to be an easier topic than one such as combinatorics, some individuals initially trip on the difference between a percent change and a percent of a number. com/7344/c-program-to-calculate-percentage-difference-between-2-numbers/Given two numbers, write a C program to calculate percentage differe. Tips: Put numbers in as you like, and the result will automatically be generated. Step 4: Convert that to a percentage (by multiplying by 100 and adding a "%" sign). How To Calculate The Percentage Difference Between Two. Then take this number times 100%, resulting in 40%. The Percent Change Calculator finds
The following is multiple choice question (with options) to answer.
The difference between the value of a number increased by 12.5% and the value of the original number decreased by 25% is 30. What is the original number Q? | [
"60",
"80",
"40",
"120"
] | B | (1 + 1/8)x - (1 - 1/4)x = 30
(9/8)x - (3/4)x = 30
x = 80=Q
Answer: B |
AQUA-RAT | AQUA-RAT-36852 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half ? | [
"15 min",
"20 min",
"27.5 min",
"30 min"
] | D | et total time consumed in filling the tank is n.
(A+B)'s 1 minute work is 1/60 + 1/40 and (A+B) will take half of total time ie
(n/2)(1/60 + 1/40)----------(1)
and rest half will be filled by B only in half of total time means (n/2)(1/40)--(2)
(1)+(2)
(n/2)(1/60 + 1/40) + (n/2)(1/40)=1
n=30
ANSWER:D |
AQUA-RAT | AQUA-RAT-36853 | To get the required result in the book you first have to calculate the amount of the equivalent annual payment. The formula for payments at the beginning of every month is
$C_1=12\cdot r+\frac{\color{blue}{13}\cdot r\cdot i}{2}$
In case of payments at the end of every month $\color{blue}{13}$ has to be replaced by $11$.
$C_1=12\cdot 200+\frac{\color{blue}{13}\cdot 200\cdot 0.045}{2}=2458.5$
To get the Future value after 10 years we use the formula for annual payments.
$C_{10}=2458.5\cdot \frac{1-1.045^{10}}{1-1.045}=30210.56$
But in general I wouldn´t say that your method is worse then the method above. Your result differs from my result about $0.015\%$ only.
The following is multiple choice question (with options) to answer.
What annual payment dischargea debit of Rs.12900, due in 4yrs.At 5% rate? | [
"2000",
"1300",
"3000",
"4300"
] | C | A.P.=(200X12900)/[4(200+5X3)]=3000 Ans
Alternative
100+105+110+115=12900
430=12900
100=12900/(430)X100=3000
C |
AQUA-RAT | AQUA-RAT-36854 | $\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.
Bullock likes to keep a spare tyre in his car every time. On a certain day, he travels 1, 00,000 km and just to make the most of all the tyres, he changes the tyres between his journey such that each tyre runs the same distance.
What is the distance traveled by each tyre? | [
"50,000 km.",
"60,000 km.",
"70,000 km.",
"80,000 km."
] | D | Solution:
80,000.
Explanation:
The distance traveled by each tyre:
4/5 * 1, 00, 000km = 80,000 km.
Answer D |
AQUA-RAT | AQUA-RAT-36855 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
One-third of 1206 is what percent of 400 ? | [
"3",
"30",
"300",
"100.5"
] | D | Answer
Let one-third of 1206 is N% of 400.
∵ 1206/3 = (N x 400)/100
∴ N = (402 x 100) / 400 = 100.5
Correct Option: D |
AQUA-RAT | AQUA-RAT-36856 | Hint: $7\equiv1\mod3$, and $7\equiv-1\mod8$, and $24=3\cdot8$.
• How do I use $24 = 3 \cdot 8$? Is there a fact about modular arithmetic that I'm unaware of here? – Newb Mar 7 '14 at 18:03
• The Chinese remainder theorem! – Álvaro Lozano-Robledo Mar 7 '14 at 18:07
• @Newb: What number in between $0$ and $23$ gives the remainder $1$ when divided through both $3$ and $8$ ? :-) – Lucian Mar 7 '14 at 18:09
Also $$7^{n + 2} = \pars{24p + \mu}7^{2} = 24p\times 49 + 48\mu + \mu = \pars{49p + 2\mu}\times 24 + \mu\,,\ \left\vert% \begin{array}{rcl} \mbox{Also,}&& \\ n = 0 & \imp & \mu = 1 \\ n = 1 & \imp & \mu = 7 \end{array}\right.$$
Then, $$7^{n} \mod 24 =\left\lbrace \begin{array}{rcl} 1 & \mbox{if} & n\ \mbox{is}\ even \\ 7 & \mbox{if} & n\ \mbox{is}\ odd \end{array}\right.$$
Then $$\color{#00f}{\large 7^{1000}\mod 24 = 1}$$
The following is multiple choice question (with options) to answer.
When 24 is divided by the positive integer i, the remainder is 4. Which of the following statements about i must be true?
I. i is even
II. i is a multiple of 5
III. i is a factor of 20 | [
"III only",
"I and II only",
"I and III only",
"II and III only"
] | D | 24= m*i + 4
m*i=20
put values of m
m=1 i=20
m=2 i=10
m=4 i=5
I. not true
II. true
III. true
D. correct |
AQUA-RAT | AQUA-RAT-36857 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
In ten years, David will be four times as old as Aaron. Twenty years ago, David was twice as old as Ellen. If David is nine years older than Ellen, how old is Aaron? | [
"1.5–5",
"6–10",
"11–15",
"16–20"
] | A | Let David's present age be 'd', Aaron's present age be 'a' and Ellen's present age be 'e'
In ten years, David will be four times as old as Aaron --> d+10 = 4(a+10)
Twenty years ago, David was twice as old as Ellen --> d-20 = 2(e-20)
David is seven years older than Ellen --> d = e + 9
e+9-20 = 2e-40
e-11 = 2e-40
e = 29
d = 36
46 = 4a + 40
a = 1.5
Answer: A |
AQUA-RAT | AQUA-RAT-36858 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A certain industrial machine requires 15 quarts (480 ounces) of oil every day, and it is given a mixture of oil A and oil B. Oil A consists of 5% special additives while oil B consists of 8% special additives. If the machine needs exactly 40 ounces of the special additives to ensure smooth operation, then how much of oil A, in ounces, should be in the mixture? | [
"a.\tA – 53.33",
"b.\tB – 280",
"c. C – 290",
"d.\tD – 300"
] | A | 5%*x + 8% *(480-x)=40
1.6 = 3 %*x
x=53.33
A |
AQUA-RAT | AQUA-RAT-36859 | algorithms, integers
If the prime factorization of $n+1$ is $p_1^{k_1},\ldots,p_t^{k_t}$, then $f(n)$ depends only on $(k_1,\ldots,k_t)$, and in fact only on the sorted version of this vector. Every number below $10^9$ has at most $29$ prime factors (with repetition), and since $p(29)=4565$, it seems feasible to compute $f$ (or rather $g$) for all of them, recursively. This solution could be faster if there were many different inputs; as it is, there are at most $10$.
It is also possible that this function, mapping partitions to the corresponding $g$, has an explicit analytic form. For example, $g(p^k) = 2^{k-1}$, $g(p_1\ldots p_t)$ is given by A000670, and $g(p_1^2 p_2\ldots p_t)$ is given by A005649 or A172109.
The following is multiple choice question (with options) to answer.
How many positive integers y less than 30 have no common prime factor with 30? | [
"5",
"6",
"7",
"8"
] | D | y=30=2*3*5. So, the number must be less than 30 and not have primes 2, 3, or 5.
This means that the number could be: 1, 7, 11, 13, 17, 19, 23, or 29. Total of 8 numbers.
Answer: D. |
AQUA-RAT | AQUA-RAT-36860 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
A man covered a certain distance at a speed of 8 Kmph and returned at a speed of 4 Kmph. Find the average speed of whole journey? | [
"5 2/3 Kmph",
"5 1/3 Kmph",
"5 4/3 Kmph",
"6 1/3 Kmph"
] | B | Option 'B' |
AQUA-RAT | AQUA-RAT-36861 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
Look at this series: 21, 9, 21, 11, 21, 13, 21, ... What number should come next? | [
"14",
"15",
"21",
"23"
] | B | Explanation:
In this alternating repetition series, the random number 21 is interpolated every other number into an otherwise simple addition series that increases by 2, beginning with the number 9.
Answer is B |
AQUA-RAT | AQUA-RAT-36862 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
The number of bacteria in a petri dish increased by 50 percent every 2 hours. If there were 108 million bacteria in the dish at 2: 00 p.m., at what time were there 32 million bacteria in the dish? | [
"6: 00 p.m.",
"8: 00 p.m.",
"6: 00 a.m",
"8: 00 a.m."
] | D | Given:
he number of bacteria in a petri dish increased by 50 percent every 2 hours
There were 108 million bacteria in the dish at 2: 00 p.m
Since the bacteria is increasing by 50% every 2 hours, this means that the number is multiplied by 1.5 every 2 hours.
Required: At what time were there 32 million bacteria in the dish?
Assume that there were n intervals of 2 hours between 32 million and 108 million.
32 * (1.5)^n = 108
(3/2)^n = 108/32 = 27/8 = (3/2)^3
Hence n = 3 intervals.
Each interval was of 2 hours. Therefore, the number of hours = 6
Subtracting 6 hours from 2 PM, we get 8 AM as the answer.
Option D |
AQUA-RAT | AQUA-RAT-36863 | 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.
When positive integer n is divided by 3, the remainder is 1. When n is divided by 5, the remainder is 4. What is the smallest positive integer p, such that (n + p) is a multiple of 11? | [
"1",
"2",
"5",
"7"
] | D | When positive integer n is divided by 3, the remainder is 1 i.e., n=3x+1
values of n can be one of {1, 4, 7, 10, 13, 16, 19, 22..............49, 52, 59..................}
Similarly,
When n is divided by 5, the remainder is 5..i.e., n=5y+4
values of n can be one of {4, 9, 14, 19,...}
combining both the sets we get
n={4,19, 52, ...........}
What is the smallest positive integer p, such that (n + p) is a multiple of 11 or 11x
in case of n=4 p=7
so for min value of p, we take min value of n.
D is the answer. |
AQUA-RAT | AQUA-RAT-36864 | 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 garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement? | [
"1888",
"2766",
"2999",
"1900"
] | D | 2000 ---- 54
2000 ---- 39
x ----- 20
x*20 = 2000*39
x = 3900
2000
-------
1900
Answer: D |
AQUA-RAT | AQUA-RAT-36865 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
Due to construction, the speed limit along an 10-mile section of highway is reduced from 55 miles per hour to 25 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | [
"A) 6.24",
"B) 13.1",
"C) 10",
"D) 15"
] | B | Old time in minutes to cross 10 miles stretch = 10*60/55 = 10*12/11 = 10.9
New time in minutes to cross 10 miles stretch = 10*60/25 = 10*12/5 = 24
Time difference = 13.1
Ans:B |
AQUA-RAT | AQUA-RAT-36866 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
• 4 minutes ago
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• A minute ago
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The following is multiple choice question (with options) to answer.
By doing some calculation, 2+3=18, 3+5=42, 6+2=48 and find 5+6=?? | [
"74",
"70",
"72",
"76"
] | C | Given 2+3=18==> (2^2+3^2)+(2+3)=18
3+5=42==> (3^2+5^2)+(3+5)=42
6+2=48==> (6^2+2^2)+(6+2)=48
So 5+6=72==> (5^2+6^2)+(5+6)=72
ANSWER:C |
AQUA-RAT | AQUA-RAT-36867 | A group of friends rent a beach house and decide to split the cost of the rent and food. Four friends pay $170 each. Five friends pay$162 each. Six friends pay 157. If nine people were to share the expense, how much would each pay? Let’s look at this in a table. n\begin{align*}n\end{align*} (number of friends) t\begin{align*}t\end{align*} (share of expense) 4 170 5 162 6 157 9 ??? As the number of friends gets larger, the cost per person gets smaller. This is an example of inverse variation. An inverse variation function has the form f(x)=kx\begin{align*}f(x)=\frac{k}{x}\end{align*}, where k\begin{align*}k\end{align*} is called the constant of variation and must be a counting number and x0\begin{align*}x \neq 0\end{align*}. To show an inverse variation relationship, use either of the phrases: • Is inversely proportional to • Varies inversely as Example 1: Find the constant of variation of the beach house situation. Solution: Use the inverse variation equation to find k\begin{align*}k\end{align*}, the constant of variation. Solve for k:y170170×4k=kx=k4=k4×4=680\begin{align*}&& y &= \frac{k}{x}\\ && 170 &= \frac{k}{4}\\ \text{Solve for} \ k: && 170 \times 4 &= \frac{k}{4} \times 4\\ && k &= 680\end{align*} You can use this information to determine the amount of expense per person if nine people split the cost. yy=680x=6809=75.56\begin{align*}y &= \frac{680}{x}\\ y &= \frac{680}{9}=75.56\end{align*} If nine people split the expense, each would pay75.56.
Using a graphing calculator, look at a graph of this situation.
The following is multiple choice question (with options) to answer.
Average of money that group of 4 friends pay for rent each month is $800.After one persons rent is increased by 20% the new mean is $880.What was original rent of friend whose rent is increased? | [
"800",
"900",
"1600",
"1100"
] | C | 0.2X = 4(880-800)
0.2X = 320
X = 1600
Answer C |
AQUA-RAT | AQUA-RAT-36868 | meteorology, climate-change, gas, pollution
Title: Regarding various types of atmospheric pollution Does all the car pollution (from about 150 million cars at least in the U.S. and a lot more in all of North America and the rest of the world) all the smoke-stack pollution of various factories and all the Airline pollution running day after day have a deleterious and damaging effect on the general atmosphere and, over time, the climate?
Given all the observed pollution that China has caused itself and some of the resulting weird weather events there this certainly seems to be evidence of the damaging effects of car and factory pollution. Has anyone calculated how much exhaust from cars is produced in one day on average in a 'moderate' sized city?
Of course it seems with all the increased oil production in the U.S. and elsewhere we, human beings are going to keep are love-affair with gas-powered cars for the next 200 or 300 years. That is if we don't use up all the oil and gas in the ground before then. As a USA resident, the EPA is the best place to start when wondering about the emissions inventory of atmospheric pollutants or pollutant precursors that affect the National Ambient Air Quality Standards (e.g. Particulate Matter, Carbon Monoxide, Sulfur Dioxide, Lead, Nitrogen Oxides, Volatile Organic Compounds). The EPA compiles a comprehensive emissions inventory of all criteria pollutants at the county level which is available in the National Emissions Inventory (compiled once every 3 years). You can see the summary of your county at http://www.epa.gov/air/emissions/where.htm. As for the effects of atmospheric pollution, it is important to consider the lifetime of said pollutants in the atmosphere in order to put their environmental impacts into perspective. For instance, the air pollutants covered by the National Ambient Air Quality Standards have immediate health effects when high concentrations are breathed in regularly. Both animals and plants are adversely affected by these irritating and sometimes toxic chemicals, but these pollutants are also reactive and do not last long in the atmosphere unless they are constantly being replenished (e.g. daily traffic). Air quality also impacts critical nitrogen loads on ecosystems and possible production of acid rain.
The following is multiple choice question (with options) to answer.
Worldwide production of motor vehicles was 3.9 million vehicles in 1946 and 65.4 million in 1987. Of the following, which is closest to the average (arithmetic mean) annual increase, in millions, in worldwide production of motor vehicles during this period? | [
"0.08",
"1.0",
"1.1",
"1.5"
] | D | Change in production of motor vehicles = 65.4 - 3.9 = 61.5 million in 41 years (1987 - 1946)
average increase per year = 61.5/41 = 1.5 million (apporx.)
Answer (D) |
AQUA-RAT | AQUA-RAT-36869 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
If C + D = 15 and C and D are any numbers, which of the following is a possible value for 8C + 5D? | [
"99",
"29",
"0",
"09"
] | C | c + d = 15
8c + 5d = 3c + 5C + 5d = 5(c+d) + 3C = 5(15) + 3C=75 + 3C= c=-25
Now 3C is a multiple of 3, so the right answer should be 75 + multiple of 3.
c+d=15=-25+d=15=d=40
So, 0 is the right answer
Hence, C is the right answer. |
AQUA-RAT | AQUA-RAT-36870 | ### $\operatorname{Var}(X)$
Here, we will just apply EVvE's Law:
\begin{align} \operatorname{Var}(X) &= \mathbb{E}\left( \operatorname{Var}(X|N) \right) + \operatorname{Var}\left( \mathbb{E}(X|N) \right) \quad \text{and working on each part} \\\\ \mathbb{E}\left( \operatorname{Var}(X|N) \right) &= \mathbb{E}(N \sigma^2) \quad \text{treating } N \text{ as fixed, independence} \\ &= \sigma^2 \, \mathbb{E}(N) \\ \\ \operatorname{Var}\left( \mathbb{E}(X|N) \right) &= \operatorname{Var}(\mu \, N) \quad \text{} \\ &= \mu^2 \, \operatorname{Var}(N) \\\\ \Rightarrow \operatorname{Var}(X) &= \boxed{ \sigma^2 \, \mathbb{E}(N) + \mu^2 \, \operatorname{Var}(N) } \end{align}
Idiot-checking that result, we confirm that
• $\mathbb{E}(X) = \mu \, \mathbb{E}(N)$ means that the average total sales is the average per-customer spending amount times the average number of customers
• $\sigma^2$ is in terms of money-squared; $\mu$ is in terms of money, so $\operatorname{Var}(X)$ is in terms of money-squared
• both $\mathbb{E}(N)$ and $\operatorname{Var}(N)$ are dimensionless values ("people" are not units in this case)
• if we knew the MGF of each $X_j$, then we could condition on $N$ to treat it as fixed in order to get the MGF of $X$ (multiply by $n$)
## Statistical Inequality
The following is multiple choice question (with options) to answer.
If the consumer price index for a sample of goods and services purchased in Dallas rose from 100 at the end of 1967 to x at the end of 1985, what was the average T(arithmetic mean) annual increase in the index over this period? | [
" (x + 100)/18",
" x/18",
" (100 − x)/18",
" (x − 100)/18"
] | D | CPI in end of 1967 = 100
CPI in end of 1985 = x
Number of years = 18
Average annual increase in CPI T= (x-100)/18
Answer D |
AQUA-RAT | AQUA-RAT-36871 | angles, diagonals, height, perimeter and area of parallelograms. Tags: Question 9 . Free Parallelogram Sides & Angles Calculator - Calculate sides, angles of an parallelogram step-by-step This website uses cookies to ensure you get the best experience. Sum of adjacent angles of a parallelogram is equal to 180 degrees. A parallelogram is a two-dimensional geometrical shape, whose sides are parallel to each other. Each diagonal of a parallelogram separates it into two congruent triangles. Parallelogram Problems - length, width, perimeter, area Solve the following. The area of this parallelogram, or well it used to be this parallelogram, before I moved that triangle from the left to the right, is also going to be the base times the height. These unique features make Virtual Nerd a viable alternative to private tutoring. 4 2/3 is one side 7 1/3 and the other side is this i dont know what to do (i) Using slope (ii) Using midpoint formula (iii) Using section formula. Opposite sides are equal in length and opposite angles are equal in measure. Classify Ratios Using a Decision Tree. Classify Types . A parallelogram where all angles are right angles is a rectangle! Popular pages @ mathwarehouse.com . The figure is a parallelogram. Substitute in to solve for x. In a parallelogram, the diagonals bisect each other, so you can set the labeled segments equal to one another and then solve for . Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? is the answer A. Problem 3 In the parallelogram below, BB' is the angle bisector of angle B and CC' is the angle bisector of angle C. Find the lengths x and y if the length of BC is equal to 10 meters. 4. Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there! How to solve problems on the parallelogram sides measures - Examples Problem 1 Find the perimeter of the parallelogram, if its sides are 7 cm and 12 cm long. A parallelogram is a quadrilateral with opposite sides parallel. Area
The following is multiple choice question (with options) to answer.
Each side of a certain parallelogram has length 8. If the area of the parallelogram is 24. Which of the following is the measure of one of its angles? | [
"22",
"45",
"60",
"90"
] | A | area of a parallelogram = b*h
b*h=24
h=24/8=3
sin theta =opp/hyp = 3/8
theta = sin inv of 3/8 = 22 deg
A |
AQUA-RAT | AQUA-RAT-36872 | python, optimization, programming-challenge
Hint 3:
Think of the number 145 given in the problem; since you know that this works, there's no need to check numbers that are a permutations of its digits: 154, 415, 451, 514, 541. The trick here is to look at the problem the other way around: instead of finding numbers whose Sum of Factorials of Digits equals the original number, find an ordered sequence of digits whose Sum of Factorials can be split into a sequence of digits, sorted, and that compares equal to the original sequence.
The following is multiple choice question (with options) to answer.
Find out the wrong number in the given sequence of numbers.
52, 51, 48, 43, 34, 27, 16 | [
"22",
"34",
"77",
"99"
] | B | Explanation:
Subtract 1, 3, 5, 7, 9, 11 from successive numbers.
So, 34 is wrong.
Answer: B) 34 |
AQUA-RAT | AQUA-RAT-36873 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Two pipes P and Q can fill a tank in 5 hours and 10 hours respectively. If both pipes are opened simultaneously, how much time will be taken to fill the tank? | [
"4 hours 20 min",
"5 hours 49 min",
"3 hours 20 min",
"3 hours 22 min"
] | C | Explanation:
Part filled by P in 1 hour = 1/5
Part filled by Q in 1 hour = 1/10
Part filled by (P + Q) in 1 hour = ( 1/5 + 1/10) = (3/10)
Time taken to fill the tank is (10/3) = 10/3*60 = 200 mins = 3 hrs 20 mins
ANSWER C |
AQUA-RAT | AQUA-RAT-36874 | # Find Smallest Positive Integer
#### anemone
##### MHB POTW Director
Staff member
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
#### Wilmer
##### In Memoriam
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Unfortunately, got no exotic formula for you
#### anemone
##### MHB POTW Director
Staff member
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Hi Wilmer,
Thanks for participating and thanks for the compliment to this problem as well.
Yes, 3987 is the answer to this problem but...
Unfortunately, got no exotic formula for you
30 minutes in the corner, please...
#### Wilmer
##### In Memoriam
30 minutes in the corner, please...
No fair! You only asked: "Find the smallest positive integer [FONT=MathJax_Math-italic]n" [/FONT]
#### anemone
The following is multiple choice question (with options) to answer.
What is the smallest positive integer x such that 450x is the cube of a positive integer? | [
"2",
"15",
"30",
"60"
] | C | 450 = 2 x 3^2 x 5^2 now we need two 2s, one 3 and one 5 to make it perfect cube.
So x= 2^2 x 3 x 5 = 60.
Answer is C. |
AQUA-RAT | AQUA-RAT-36875 | (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.
There are 35 students in a hostel. If the no. of students increases by 10, the expenses of the mess increase by Rs.42/day while the average expenditure per head diminishes by Rs1. Find the original expenditure of the mess? | [
"304.5",
"340.9",
"342.9",
"345.9"
] | A | Suppose the average join the mess , total expenditure = 35x + 42
Now, the average expenditure = (35x + 42)/ (35 + 10) = x – 1
or, 35x + 42 = 45x – 45
or, x = 8.7
Thus the original expenditure of the mess = 35 x 8.7 = Rs. 304.5
A |
AQUA-RAT | AQUA-RAT-36876 | # 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 car mechanic purchased four old cars for Rs. 1 lakh. He spent total 2 lakh in the maintenance and repairing of these four cars. what is the average sale price of the rest three cars to get 50% total profit if he has already sold one of the four cars at Rs. 1.2 lakh? | [
"1.8 lakh",
"1.1 lakh",
"1.9 lakh",
"8.1 lakh"
] | B | Explanation:
Total cost of 4 cars = 1+2 = 3 lakh
Total SP of 4 cars = 3 x 1.5 = 4.5 lakh
SP of 1 car = 1.2 lakh
SP of rest 3 cars = 4.5 - 1.2 = 3.3 lakh
Average SP of all the 3 cars = 1.1 lakh
Answer: B) 1.1 lakh |
AQUA-RAT | AQUA-RAT-36877 | # Conditional probability exercise
Charlotte87
I have some problems getting conditional probability right... Does this look like it should?
## Homework Statement
Assume that there are bags of tulip bulbs in the basement, ant that they contain 25 bulbs each. yellow bags contain 20 yellow tulips and 5 red tuplips, and red bags contain 15 red and 10 yellow tulips. 60% of the bags in the basement are yellow, the others are red. One bulb is chosen at random from a random bag in the basement, and then planted
a) what is the probabilit that the tulip turns out yellow?
b) given that the tulip turns out yellow, what is the probability that it came from a yellow bag?
## Homework Equations
Let RB be redbag, YB yellowbag, RT red tulip and YT yellow tulip. Then as far as I can read from this exercise I have the following information:
P(YB)=0.6
P(RB)=0.4
P(YT|YB)=20/25=4/5
P(RT|YB)=1/5
P(YT|RB)=10/25=2/5
P(RT|RB)=3/5
## The Attempt at a Solution
a) $P(YT)=P(YB\cap YT)+P(RB\cap YT)=(0.6*4/5)+(0.4*2/5)=16/25$
b) $P(YB|YT)=P(YB \cap YT)/P(YT) = (0,6*4/5)/(16/25)=3/4$
It is particularly this last one I am unsure about.
Homework Helper
Gold Member
Staff Emeritus
Homework Helper
I have some problems getting conditional probability right... Does this look like it should?
## Homework Statement
Assume that there are bags of tulip bulbs in the basement, ant that they contain 25 bulbs each. yellow bags contain 20 yellow tulips and 5 red tuplips, and red bags contain 15 red and 10 yellow tulips. 60% of the bags in the basement are yellow, the others are red. One bulb is chosen at random from a random bag in the basement, and then planted
The following is multiple choice question (with options) to answer.
For an agricultural experiment, 300 seeds were planted in one plot and 200 were planted in a second plot. If exactly 20 percent of the seeds in the first plot germinated and exactly 35 percent of the seeds in the second plot germinated, what percent of the total number of seeds germinated? | [
"12%",
"26%",
"29%",
"30%"
] | B | In the first plot 20% of 300 seeds germinated, so 0.20 x 300 = 60 seeds germinated.
In the second plot, 35% of 200 seeds germinated, so 0.35 x 200 = 70 seeds germinated.
Since 60 + 70 = 130 seeds germinated out of a total of 300 + 200 = 500 seeds, the percent of seeds that germinated is (130/500) x 100%, or 26%.
Answer: B. |
AQUA-RAT | AQUA-RAT-36878 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 10 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15). Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day)? | [
"5:00",
"5:34",
"5:42",
"6:00"
] | D | C1 loses 15 minutes every hour. So after 60 minutes have passed, C1 displays that 60-15 = 45 minutes have passed.
C2 gains 15 minutes for every 60 minutes displayed on C1. Thus, the time displayed on C2 is 75/60 = 5/4 the time displayed on C1. So after 60 minutes have passed, C2 displays the passing of (5/4 * 45) minutes.
C3 loses 20 minutes for every 60 minutes displayed on C2. Thus, the time displayed on C3 is 40/60 = 2/3 the time displayed on C2. So after 60 minutes have passed, C3 displays the passing of (2/3 * 5/4 * 45) minutes.
C4 gains 20 minutes for every 60 minutes displayed on C3. Thus, the time displayed on C4 is 80/60 = 4/3 the time displayed on clock 3. So after 60 minutes have passed, C4 displays the passing of 4/3 * 2/3 * 5/4 * 45 = 50 minutes.
C4 loses 10 minutes every hour.
In 6 hours, C4 will lose 6*10 = 60 minutes = 1 hour.
Since the correct time after 6 hours will be 6pm, C4 will show a time of 6-1 = 6pm.
The correct answer is D. |
AQUA-RAT | AQUA-RAT-36879 | $8xy^3+8x^2-8x^3y-8y^2=8(xy^3-y^2+x^2-x^3y)$
Now, $xy^3-y^2+x^2-x^3y=y^2(xy-1)-x^2(xy-1)=(xy-1)(y^2-x^2)=(xy-1)(y+x)(y-x)$
-
I don't think there is a general method.
However spotting obvious factors - like $8$ - and grouping factors of the same degree together so $x^2$ and $y^2$ have degree $2$ and $xy^3$ and $x^3y$ have degree 4 - is a useful step to have in mind.
-
$$8xy^3+8x^2-8x^3y-8y^2= 8xy^3-8x^3y+8x^2-8y^2$$
$$= 8xy(y^2-x^2)-8(y^2-x^2)= (8xy-8)(y^2-x^2)$$
$$= 8(1-xy)(x^2-y^2).$$
-
The following is multiple choice question (with options) to answer.
Factor 3x3 - x2y +6x2y - 2xy2 + 3xy2 - y3 = | [
"(3x - 2y)(x + y)",
"(3x - y)(x + y)(x - y)",
"(3x - y)(x + y)2",
"(3x - y)(x2 + y2)"
] | C | Solution:
3x3 - x2y + 6x2y - 2xy2 + 3xy2 - y3=
x2(3x - y) + 2xy(3x - y) + y2(3x-y) =
(3x - y)(x2 + 2xy + y2)=
(3x - y)(x + y)2
Answer C |
AQUA-RAT | AQUA-RAT-36880 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The maximum number of students among them 1234 pens and 874 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is: | [
"91",
"2",
"1001",
"1911"
] | B | Explanation:
Required number of students = H.C.F of 1234 and 874 = 2. Answer: B |
AQUA-RAT | AQUA-RAT-36881 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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$$.Physics
The following is multiple choice question (with options) to answer.
The two trains of lengths 400 m, 600 m respectively, running at same directions. The faster train can cross the slower train in 180 sec, the speed of the slower train is 48 km. then find the speed of the faster train? | [
"58 Kmph",
"68 Kmph",
"78 Kmph",
"55 Kmph"
] | B | Length of the two trains = 600m + 400m
Speed of the first train = X
Speed of the second train= 48 Kmph
1000/X - 48 = 180
1000/x - 48 * 5/18 = 180
50 = 9X - 120
X = 68 Kmph
ANSWER:B |
AQUA-RAT | AQUA-RAT-36882 | • but the balls are being replaced. how is this possible? – minatozaki Dec 15 '17 at 2:39
• The first ball you select can be any of the four colors. Then you replace it and you must select one of the other three balls next, etc. – Remy Dec 15 '17 at 2:40
• @minatozaki If the balls were not replaced, the second ball would be one of the remaining three balls with probability $\frac33,$ etc., and the answer would be $\frac44\cdot\frac33\cdot\frac22\cdot\frac11 = 1$; that is, if you don't replace, then for certain you will get four different colors. – David K Dec 15 '17 at 12:56
We could do this by counting the number of ways to draw four balls and the number of ways to draw four balls without getting any duplicates. There are $4!$ ways to not get a duplicate as every drawing can be thought of as an ordering and if we don't allow duplicates then we have a permutation. There are $4^4$ different possible drawings as replacement is allowed, this gives us $$\frac{4!}{4^4} = \frac{3}{32}$$
• If there is a -1 is there a reason, anything you'd like explained or improved? – Benji Altman Dec 15 '17 at 2:52
• Someone down-voted my answer too without explanation. – Remy Dec 15 '17 at 2:52
• +1, But one thing to be careful of with this method - it only works when all possible combinations of draws are equally likely. However, in this case, that obviously holds. – Paul Sinclair Dec 15 '17 at 17:45
The probability of drawing $4$ different balls is the product of the probabilities of drawing a new ball on all $4$ draws.
The first draw yields a new ball, guaranteed: $$P(\text{ball 1 new})=1$$
For the second draw, there are $3$ possible new balls and $4$ total balls, so: $$P(\text{ball 2 new})=\frac34$$
The following is multiple choice question (with options) to answer.
20 balls are numbered 1 to 20. A ball is drawn and then another ball is drawn without replacement. What is the probability that both balls have even numbers? | [
"7/30",
"3/14",
"9/38",
"4/21"
] | C | P(1st ball is even) = 10/20
P(2nd ball is also even) = 9/19
P(both balls are even) = 10/20 * 9/19 = 9/38
The answer is C. |
AQUA-RAT | AQUA-RAT-36883 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row downstream at the rate of 10 Km/hr and upstream at 4 Km/hr. Find man's rate in still water and the rate of current? | [
"9,6",
"7,3",
"9,3",
"6,6"
] | B | Explanation:
Rate of still water = 1/2 (10 + 4) = 7 Km/hr
Rate of current = 1/2 (10-4) = 3 Km/hr
Answer: Option B |
AQUA-RAT | AQUA-RAT-36884 | By brute force, here are the possible sequences.
(8,0,...),(7,1,0,...),(6,2,0,...),(6,1,1,0,...),(5,3,0,...),(5,2,1,0,...)(5,1,1,1,0,...),(4,4,0,...),(4,3,1,0...),(4,2,2,0,...),(4,2,1,1,0,...),(4,1,1,1,1,0,...),(3,3,2,0,...),(3,3,1,1,0,...),(3,2,2,1,0,...),(3,2,1,1,1,0,...),(3,1,1,1,1,1,0,...),(2,2,2,2,0,...),(2,2,2,1,1,0,...),(2,2,1,1,1,1,0,...),(2,1,1,1,1,1,1,0,...),(1,1,1,1,1,1,1,1,0,...)
As you can see, there are 22 of them. I'm not sure how to generalize this result at the moment, perhaps someone who is more familiar with the problem will come around with a counting method. Perhaps a recurrence relation.
Looking for a pattern at the sequence of numbers for this problem with 1,2,3,... number of objects: I get the amounts: 1,2,3,5,7,11,15,22.
By google searching I find http://mathworld.wolfram.com/PartitionFunctionP.html which has a great deal of history about the problem and methods of calculating the sequence.
The following is multiple choice question (with options) to answer.
What is the next number of the following sequence
10,8,18,16,34,32,.... | [
"44",
"66",
"55",
"77"
] | B | 10
10-2=8
8+10=18
18-2=16
16+18=34
34-2=32
32+34=66
answer will be 66
ANSWER:B |
AQUA-RAT | AQUA-RAT-36885 | leaves at 9:10 from one station and its speed is 90 km/h, what time does it get to the next station? You may speak with a member of our customer support team by calling 1-800-876-1799. Every word problem has an unknown number. When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Select it and click on the button to choose it. Linear inequalities word problems. A car traveled 281 miles in 4 hours 41 minutes. Solution. Find the equilibrium price. Given : The total cost of 80 units of the product is $22000. Explain to students that you can find the rate (or speed) that someone is traveling if you know the distance and time that she traveled. Using a few simple formulas and a bit of logic can help students quickly calculate answers to seemingly intractable problems. Solution Let x be the first number. Solution Let y be the second number x / y = 5 / 1 x + y = 18 Using x / y = 5 / 1, we get x = 5y after doing cross multiplication Replacing x = 5y into x + y = 18, we get 5y + y = 18 6y = 18 y = 3 For a certain commodity, the demand equation giving demand "d" in kg, for a price "p" in dollars per kg. Difference between a number and its positive square root is 12. Related articles: The 4 Steps to Solving Word Problems. If the …, solve word problem Not rated yetIf a number is added to the numerator of 11/64 and twice the number is added to the denominator of 11/64, the resulting fraction is equivalent to 1/5. How many years …, Price of 3 pens without discount Not rated yetA stationery store sells a dozen ballpoints pens for$3.84, which represents a 20% discount from the price charged when a dozen pens are bought individually. A park charges $10 for adults and$5 for kids. Do you have some pictures or graphics to add? If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. Multi-Step All Operations Word Problems These Word Problems worksheets will produce word problems
The following is multiple choice question (with options) to answer.
In a fuel station the service costs $1.75 per car, every liter of fuel costs 0.65$. Assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty, how much money total will it cost to fuel all cars? | [
"320$",
"380$",
"420$",
"450$"
] | D | Total Cost = ( 1.75*12 ) + ( 0.65 * 12 * 55 ) = 21 + 429 => 450
Hence answer will be (D) 450 |
AQUA-RAT | AQUA-RAT-36886 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two cyclist start on a circular track from a given point but in opposite direction with speeds of 7m/s and 8m/s. If the circumference of the circle is 360meters, after what time will they meet at the starting point? | [
"20sec",
"15sec",
"30sec",
"24sec"
] | D | They meet every 360/7+8 = 24sec
Answer is D |
AQUA-RAT | AQUA-RAT-36887 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it? | [
"12",
"9",
"8",
"6"
] | A | A's 1 hour's work = 1 ;
4
(B + C)'s 1 hour's work = 1 ;
3
(A + C)'s 1 hour's work = 1 .
2
(A + B + C)'s 1 hour's work = ( 1 + 1 ) = 7 .
4 3 12
B's 1 hour's work = ( 7 - 1 ) = 1 .
12 2 12
Therefore A alone will take 12 hours to do the work. |
AQUA-RAT | AQUA-RAT-36888 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 120 km downstream? | [
"60/18",
"30/29",
"60/12",
"60/13"
] | D | Speed downstream = 20 + 6 = 26 kmph.
Time required to cover 60 km downstream = d/s = 120/26
= 60/13 hours.
Answer:D |
AQUA-RAT | AQUA-RAT-36889 | #### Solution
0424, R=.60.40, C=0.8.20, The value=0.80424 size 12{ left [ matrix { 0 {} # - 4 {} ## - 2 {} # 4{} } right ]} {}, R=.60.40 size 12{R= left [ matrix { "." 6 {} # 0 {} # "." 4 {} # 0{} } right ]} {}, C=0.8.20 size 12{C= left [ matrix { 0 {} ## "." 8 {} ## "." 2 {} ## 0 } right ]} {}, The value=0.8 size 12{"The value"= - 0 "." 8} {}
(8)
### Exercise 14
1324120513241205 size 12{ left [ matrix { - 1 {} # 3 {} # 2 {} # 4 {} ## 1 {} # 2 {} # 0 {} # 5{} } right ]} {}
(9)
### Exercise 15
5113101284015380551131012840153805 size 12{ left [ matrix { - 5 {} # - 1 {} # - 1 {} # 3 {} ## - "10" {} # 1 {} # 2 {} # - 8 {} ## 4 {} # 0 {} # 1 {} # 5 {} ## 3 {} # - 8 {} # 0 {} # 5{} } right ]} {}
(10)
#### Solution
The following is multiple choice question (with options) to answer.
Given that 268x74= 19432, find the value of 2.68x.74. | [
"1.9432",
"1.0025",
"1.5693",
"1.0266"
] | A | Solution
Sum of decimals places =(2+2)
= 4.
Therefore, = 2.68×.74
= 1.9432
Answer A |
AQUA-RAT | AQUA-RAT-36890 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
There is a truck driver starts from one place to pick another person at
another place. The driver goes to that place by 6 AM and then picking him
he returns back.one day the other person woke a bit early and started
walking towards the first place.one the way he meets truck driver and joins
him and both went back to the first place. This day the other person came
20 min early to the time he usually comes to the first place?Tell the time
when the driver met the second person that day? | [
"5.10 AM",
"5.20 AM",
"5.40 AM",
"5.50 AM"
] | D | Given that the first person reaches the first place by 6.00 AM
But,one day the second person reached the place 20 min early.
and met the first person on the way ad reached the stop at 6.00 AM.
Therfore,They would have met at 5.50 AM.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36891 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
The length of a rectangular plot is thrice its width. If the area of the rectangular plot is 675 sq meters, then what is the width (in meters) of the rectangular plot? | [
"12",
"15",
"18",
"21"
] | B | Area = L*W = 3W^2 = 675
W^2 = 225
W = 15
The answer is B. |
AQUA-RAT | AQUA-RAT-36892 | # Subsets of $\{1,2 \dots n\}$ with no consecutive integers
How many subsets with cardinality k of $$\{1, 2, \dots n\}$$ contain no consecutive integers? I know that there are $$F_{n+2}$$ subsets of $$\{1, 2, \dots n\}$$ with no consecutive integers, but I do not know how to go about finding the number for a given $$k$$.
This is equivalent to chooosing a sequence of $$k$$ ones and $$n-k$$ zeroes with no adjacent ones. An example with $$n=8$$ and $$k=3$$ is $$00101001, \text{ corresponding to the set }\{3,5,8\}$$ To choose such a sequence, start with a string of $$n-k$$ zeroes, with $$n-k-1$$ spaces between the zeroes, plus two extra spaces before and after, for $$n-k+1$$ spaces total: $$\;\_\; 0\;\_\;0\;\_\;0\;\_\;0\;\_\;0\;\_\qquad,\text{ with 8-3+1=6 gaps}.$$ Each of the $$k$$ $$1$$'s goes into exactly one gap. We need to choose $$k$$ of these gaps to put a $$1$$ in. This can be done in $$\binom{n-k+1}{k}\text{ ways.}$$
Without loss of generality, let your $$k$$ elements from a selected subset be $$x_1
The following is multiple choice question (with options) to answer.
List K consists of 14 consecutive integers. If -4 is the least integer in list K, what is the range of the positive integers in list K? | [
"5",
"6",
"7",
"8"
] | D | Answer = D = 8
If least = -4, then largest = 9
Range = 9 - 1 = 8 |
AQUA-RAT | AQUA-RAT-36893 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.200 when the rate to profit rises from 5 to 7 percent. Find the Capital of B? | [
"18",
"99",
"77",
"66"
] | A | 1500 ---- 270
100 ---- ? => 18%
Answer: A |
AQUA-RAT | AQUA-RAT-36894 | Author Message
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Manager
Joined: 26 Apr 2010
Posts: 122
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
A professional athlete was offered a three-year contract to play with Team K that provided for an annual salary of $120,000 in the first year, an increase in annual salary of 20% over the previous year for the next two years, and a bonus of $50,000 on signing. Team L offered a three-year contract providing for an annual salary of $150,000 in the first year, an increase in annual salary of 10% over the previous year for the next two years, and no signing bonus. If he accepts the offer of Team L and fulfills the three-year contract terms, the athlete will receive how much more money by choosing Team L over Team K ? | [
"$3,500",
"$5,000",
"$2,500",
"$9,700"
] | D | Team K's contract = $120,000 + $120,000*1.2 + $120,000*1.2*1.2 + $50,000 = $486,800
Team L's contract = $150,000 + $150,000*1.1 + $150,000*1.1*1.1 = $496,500
The difference = $9,700.
Answer: D. |
AQUA-RAT | AQUA-RAT-36895 | # 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.
At what price must an article costing Rs.47.50 be marked in order that after deducting 5% from the list price. It may be sold at a profit of 25% on the cost price? | [
"62.5",
"62.3",
"62.7",
"62.2"
] | A | CP = 47.50
SP = 47.50*(125/100)
= 59.375
MP*(95/100)
= 59.375
MP = 62.5
Answer:A |
AQUA-RAT | AQUA-RAT-36896 | beautiful black bare babes
## 550sx graphics
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 6%, find the percentage change in its area? | [
"10.68%",
"8.72%",
"9.28%",
"12.36%"
] | D | let each side of the square be a , then area = a x a
New side = 106a / 100 = 53a /50
New area =(53a x 53a) / (50 x 50) = (2809a²/2500)
increased area== (2809a²/2500) - a²
Increase %= [(309a²/2500 ) x (1/a² ) x 100]% = 12.36%
answer :D |
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