source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35697 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
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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.
A tank is filled in 10 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank? | [
"30 hours",
"35 hours",
"70 hours",
"50 hours"
] | C | Let the pipe A can fill the tank in x hours
Then pipe B can fill the tank in x/2 hours and pipe C can fill the tank in x/4 hours
Part filled by pipe A in 1 hour = 1/x
Part filled by pipe B in 1 hour = 2/x
Part filled by pipe C in 1 hour = 4/x
Part filled by pipe A, pipe B and pipe C in 1 hour =( 1/x)+(2/x)+(4/x)=7/x
i.e., pipe A, pipe B and pipe C can fill the tank in x/7 hours
Given that pipe A, pipe B and pipe C can fill the tank in 10 hours
x/7=10
x=10×7=70 hours
Answer is C |
AQUA-RAT | AQUA-RAT-35698 | # Converting Repeating Decimal Numbers to Fractions
Is it possible to write any decimal number, with a repeating decimal part, and be able to convert it into the form n/d (where both n and d are natural numbers)?
I know rational numbers that are expressed in decimal notation will either terminate exactly (such as 1.25, which is the value 5/4), or repeat forever (such as 0.333..., which is the value 1/3).
So if I just come up with any random repeating decimal, like 2.175175175..., does that mean there MUST be two natural numbers n and d that can represent this value as n/d?
I'm just trying to get a better feel for rational numbers and decimals.
Yes, as long at the repeating decimal is a positive number. Here's how: Let x = .175175175... Then 1000x - x = 175. This implies 999x = 175 and we have .175175175... = 175/999.
Finally, 2.175175175... = 2 + 175/999 = 2173/999
Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.
Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .
Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$. So, $f=\dfrac{b}{10^p-1}$
Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.
The following is multiple choice question (with options) to answer.
The rational number for recurring decimal 0.125125.... is: | [
"125/990",
"125/999",
"125/900",
"12/999"
] | B | 0.125125...
= 0.125
= 125/999
Answer is B. |
AQUA-RAT | AQUA-RAT-35699 | time, sun
d represents the day of the year for 2000. For example d=7
would be January 7th, 2000.
If you're willing to lose some precision in exchange for
convenience, you can compute the day of the current year (instead
of counting all the way back to 2000), since the sun's declination
repeats yearly (roughly speaking).
To calculate the day of the year, remember that December 31st
(noon) of the previous year is day 0. This means day 1 is January
1st, and day 31 is January 31st. Day 31 is also "February 0", so if
you need a date in February, just add. February 19th, for example,
would be 31+19 or 50. Since February has 29 days this year, February
29th would be day 31+29 or day 60, which is also "March 0". However,
at our level of precision, it doesn't matter whether you could the
leap day or not: the results will be approximately the same.
The formula above is accurate to about 0.1 degrees for this
century. Since the sun's declination can change by as much as 0.4
degrees in a day, this amount of precision should suffice.
Technically, the formula above computes the sun's declination at
Greenwich noon for a given day, which is the time when most of the
world is observing the same day. Again, the inaccuracies from using
Greenwich noon (instead of the actual, as yet unknown, time) are
small enough to ignore for our purposes.
The following is multiple choice question (with options) to answer.
The 19th of September 1987 was Saturday. What day was the 20th of September 1990 if 1988 was a leap-year? | [
"Monday",
"Tuesday",
"Wednesday",
"Thursday"
] | D | There are 365 days in an ordinary year and 366 days in a leap-year.
365 divided by 7 (a week period) yields a remainder of 1, thus the same date after an ordinary year would fall on the next day of a week;
366 divided by 7 (a week period) yields a remainder of 2, thus the same date after a leap-year would fall two days after in a week;
Thus 3 years, 1988, 1989 and 1990, will accumulate 2+1+1=4 days, plus 1-day difference between 19th and 20th, which gives total of 5-day difference (5-day shift). Thus, 20th of September 1990 was Saturday+5 days=Thursday
Answer: D |
AQUA-RAT | AQUA-RAT-35700 | # 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 fruit seller sells mangoes at the rate of Rs.9 per kg and thereby loses 20%. At what price per kg, he should have sold them to make a profit of 5% | [
"Rs 8.81",
"Rs 9.81",
"Rs 10.81",
"Rs 11.81"
] | D | Explanation:
85 : 9 = 105 : x
x= (9×105/85)
= Rs 11.81
Option D |
AQUA-RAT | AQUA-RAT-35701 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
Length of a rectangular plot is 10 mtr more than its breadth. If the cost of fencin g the plot at 26.50 per meter is Rs. 5300, what is the length of the plot in mtr? | [
"46 m",
"55 m",
"58 m",
"78 m"
] | B | Let breadth = x metres.
Then, length = (x + 10) metres.
Perimeter = 5300/26.5 = 200 m.
2[(x + 10) + x] = 200
2x + 10 = 100
2x = 90
x = 45.
Hence, length = x + 10 = 55 m
B |
AQUA-RAT | AQUA-RAT-35702 | ### Show Tags
20 Nov 2016, 04:12
Whenever you get problems pertaining to 'loan' or 'borrowed amount' to be paid in equal installments, just apply the following formula, it is easy to remember:
Value of each equal annual installments = a
Rate of interest = r% p.a.
No. of installments per year = n
No. of years = t
Therefore, total no. of installments = n*t = N
Borrowed amount(or loan taken) = B
Then:
a[{100/(100+r)} + {100/(100+r)}^2 + ......... + {100/(100+r)}^N] = B
Now, in the above question(three equal installments to be paid in three months):
a[{100/110} + {100/110}^2 + {100/110}^3] = 1000
Solve the above equation and you will get 'a' as equal to 402.11
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Vipul Chhabra
Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GMAT 1: 730 Q49 V41
GPA: 3.5
### Show Tags
The following is multiple choice question (with options) to answer.
A car is purchased on hire-purchase. The cash price is $26 000 and the terms are a deposit of 10% of the price, then the balance to be paid off over 60 equal monthly installments. Interest is charged at 12% p.a. What is the monthly installment? | [
"$603",
"$624",
"$625",
"$626"
] | B | Explanation:
Cash price = $26 000
Deposit = 10% × $26 000 = $2600
Loan amount = $26000 − $2600
Number of payments = 60
= $23400
I=p*r*t/100
I=14040
Total amount = 23400 + 14040 = $37440
Regular payment = total amount /number of payments= 624
Answer: B |
AQUA-RAT | AQUA-RAT-35703 | 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.
Rahim and his uncle differ in their ages by 30 years. After 7 years, if the sum of their ages is 66, what will be the age of the uncle ? | [
"41",
"36",
"48",
"33"
] | A | A
41
Let uncle’s present age = x
Rahim's present age = y
y– x= 30 ...(i)
After 7 year
(x + 7) + (y+ 7) = 66
x + y + 14 = 66
x + y = 52 ...(ii)
combining (i) and (ii) we get
(x + y = 52) + (x – y = 30)
2x= 82
x= 41
Uncle's age is 41 |
AQUA-RAT | AQUA-RAT-35704 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 600. What is their H.C.F.? | [
"10",
"20",
"30",
"40"
] | A | Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x
60x = 600
x = 10
The numbers are (3 x 10), (4 x 10) and (5 x 10).
The H.C.F. is 10.
The answer is A. |
AQUA-RAT | AQUA-RAT-35705 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
As part of a game, six people each must secretly choose an integer between 1 and 6, inclusive. What is the approximate likelihood that all six people will choose different numbers? | [
"2%",
"12%",
"16%",
"20%"
] | A | Ans:a)
1st person has option no's- (1,2,3,4,5,6) - there fore probability of getting a no = 6c1/6c1 = 1
2nd person has option no's any five ,
he has to choose a no from five no's - there fore probability of getting a no = 5c1/6c1 = 5/6
3rd person has option no's any four ,
he has to choose a no from four no's -there fore probability of getting a no = 4c1/6c1 = 4/6
4th person has only one option - there fore probability of getting a no= 3c1/6c1 =3/6
5th person has only one option - there fore probability of getting a no= 2c1/6c1 = 2/6
6th person has only one option - there fore probability of getting a no= 1c1/6c1 = 1/6
=1*5/6*4/6*3/6*2/6*1/6 = 2% |
AQUA-RAT | AQUA-RAT-35706 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
2. The value of x + x(xx) when x = 2 is: | [
"5",
"10",
"8",
"15"
] | B | x + x(xx)
Put the value of x = 2 in the above expression we get,
2 + 2(22)
= 2 + 2(2 × 2)
= 2 + 2(4)
= 2 + 8
= 10
Answer : B |
AQUA-RAT | AQUA-RAT-35707 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average age of M boys is ‘b’ years and of ‘n’ girls ‘c’ years. The average age of all together is? | [
"(mb + nc)/ (m + m) years",
"(mb + nn)/ (m + n) years",
"(mb + nc)/ (m + n) years",
"(nb + nc)/ (m + n) years"
] | C | Explanation:
(mb + nc) / (m+ n)
Answer:C |
AQUA-RAT | AQUA-RAT-35708 | # How many ways to pair 6 chess players over 3 boards, disregarding seating arrangement.
The problem is how many chess pairs can I make from $6$ players, if it doesn't matter who gets white/black pieces, and it doesn't matter on which of the $3$ boards a pair is seated.
I have a possible solution which doesn't seem rigorous. Can someone tell me if 1) it's correct (the answer and logic) , 2) what is a different way to reason about it ? Seems very laborious to think of it the way I got there.
I started thinking about all the possible arrangements from $6$ people, and that's $6!=720$.
Now, if I think of the arrangement $A-B ; C-D ; E-F$, it's clear that within the $720$ total arrangements, I will have counted that same arrangement of pairs with each 'pair' seated on different boards $C-D ; A-B ; E-F$ etc.. for a total of $3!=6$ per arrangement. So if I divide by that, I will basically take each arrangement such as $A-B ; C-D ; E-F$ and count it only $1$ time instead of $6$ which is what I want $\to 720/6 = 120$.
So far I can think of the $120$ arrangements left as unique, fixed-position pairs. Meaning that for the arrangement of pairs $A-B ; C-D ; E-F$, I know I won't find the same pairs in different order.
I finally need to remove those arrangements where we have pairs swapped, since I don't care about who is playing white/black. I am still counting $A-B ; C-D ; E-F$ and $B-A ; C-D ; E-F$, $A-B ; D-C ; E-F$ etc.. Because each of those pairs can be in one of two states, $2 \cdot 2 \cdot 2 = 8$ gives me all possible arrangements where each pair swaps or doesn't. So if I divide by that number $120/8=15$ I should get the correct number.
The following is multiple choice question (with options) to answer.
There are 9 players in a chess group, and each player plays each of the others once. Given that each game is played by two players, how many total games will be played? | [
"10",
"36",
"45",
"60"
] | B | 10 players are there.
two players play one game with one another.
so 9C2=9*8/2
=36
SO OPTION B is correct |
AQUA-RAT | AQUA-RAT-35709 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Find the simple interest on Rs. 5000 at 6 % per annum for the period from 5th Feb to 19th April, 2015. | [
"Rs. 40",
"Rs. 50",
"Rs. 60",
"Rs. 70"
] | C | Explanation:
Given:
1) Principal = Rs. 5000
2) Rate of interest = 6 %
3) Time = 5th Feb to 19th April, 2015
First find the time period 5th Feb to 19th April, 2015
Feb = 28 – 5 = 23 days
March = 31 days
April = 19 days
Total days = 23 + 31 + 19 = 73 days
Convert days into years, by dividing it by 365
Time = 73/365 = 1/5
Simple Interest = (P × R × T)/100
= [5000 × 6 × (1/5)]/100
=Rs.60
Simple Interest = Rs. 60
ANSWER IS C |
AQUA-RAT | AQUA-RAT-35710 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 12 seconds. Find the length of the train? | [
"150",
"872",
"287",
"200"
] | D | Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 12 = 200 meter
Answer:D |
AQUA-RAT | AQUA-RAT-35711 | fluid-dynamics, pressure, fluid-statics, potential-flow
Title: related to fluid mechanics and its velocity variation with respect to area of cross section of pipe through which it's flowing and height if water is drawn from a tank by two pipes of the same diameter and at the same depth such that one pipe ends at some height and the other reaches the ground in which pipe we can collect the water more faster? Not sure about the efficiency, since we don't need to spend any energy to collect the water, so, I assume you want to compare the speed of water collection.
If the horizontal sections of the two pipes had the same length, the speed of collection from the short pipe would be greater, since extending the length of a pipe downward cannot speed it up beyond what is already provided by the gravity.
In addition, provided that the water is collected at the end of a pipe (not at the ground level), collecting from the short pipe will eliminate the initial delay it would take for the water to reach the ground.
If the horizontal section of the short pipe was substantially longer than the horizontal section of the long pipe, more data would be needed to answer this question definitively.
The following is multiple choice question (with options) to answer.
One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in 4 hours. How long would it take to fill the pool if only the slower pipe is used? | [
"11.25",
"11.52",
"1.25",
"9"
] | D | Say the rate of the slower pipe is R pool/hour, then the rate of the faster pipe would be 1.25R=5R/4. Since when both pipes are opened, they fill the pool in four hours, then their combined rate is 1/4 pool/hour.
Thus we have that R + 5R/4 = 1/4 --> R = 1/9 pool/hour --> time is reciprocal of rate thus it's 9/1 =9 hours.
Answer: D. |
AQUA-RAT | AQUA-RAT-35712 | 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.
1000 boys have provisions for 15 days. If 200 more men join them, for how many days will the provisions last now? | [
"12",
"12.5",
"13",
"13.5"
] | B | Explanation:
1000*15 = 1200*x
x = 12.5
B |
AQUA-RAT | AQUA-RAT-35713 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
96 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part? | [
"88",
"66",
"42",
"11"
] | C | x/7 = y/9 => x:y = 7:9
7/16 * 96 = 42
Answer: C |
AQUA-RAT | AQUA-RAT-35714 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 160 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"266",
"288",
"245",
"215"
] | D | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (160 + X)/30 = 25/2 ==> 2(160 + X) = 750 ==> X
= 215 m.
Answer:D |
AQUA-RAT | AQUA-RAT-35715 | The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer $$n$$, can be determined with this formula:
$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where $$k$$ must be chosen such that $$5^k\leq{n}$$
Example:
How many zeros are in the end (after which no other digits follow) of 32!?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$. Notice that the denominators must be less than or equal to 32 also notice that we take into account only the quotient of division (that is $$\frac{32}{5}=6$$ not 6.4). Therefore, 32! has 7 trailing zeros.
The formula actually counts the number of factors 5 in $$n!$$, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Finding the powers of a prime number p, in the n!
The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
Example:
What is the power of 2 in 25!?
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$.
See this post to understand rationale behing this formulae
The following is multiple choice question (with options) to answer.
How many trailing zeros will be there after the rightmost non-zero digit in the value of 25! (factorial 25)? | [
"25",
"8",
"6",
"5"
] | C | Explanatory Answer
25! means factorial 25 whose value = 25 * 24 * 23 * 22 *....* 1
When a number that has 5 as its factor is multiplied by an even number, it results in a trailing zero.
(Product of 5 and 2 is 10 and any number when multiplied with 10 or a power of 10 will have one or as many zeroes as the power of 10 with which it has been multiplied)
In 25!, the following numbers have 5 as their factor: 5, 10, 15, 20, and 25.
25 is the square of 5 and hence it has two 5s in it.
In toto, it is equivalent of having six 5s.
There are at least 6 even numbers in 25!
Hence, the number 25! will have 6 trailing zeroes in it.
Choice C |
AQUA-RAT | AQUA-RAT-35716 | # Remainder on division with $22$
What is the remainder obtained when $$14^{16}$$ is divided with $$22$$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $$22$$, except the remainder?
How should I proceed?
• $14^{16} = (22 - 8)^{16}$ – ab123 Jul 3 '18 at 10:37
• $8^{16} = 64^8 = (66 - 2)^8$ – ab123 Jul 3 '18 at 10:41
• @ab123 Why not make an answer of that? – Arthur Jul 3 '18 at 10:41
You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
Since $$14^2 \equiv -2$$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$
or
$$14^2 = 22k -2$$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
The following is multiple choice question (with options) to answer.
A no.when divided by 142 gives a remainder 110, what remainder will be obtainedby dividingthe same no.14? | [
"16",
"17",
"18",
"19"
] | C | 142 + 110 = 252/14 = 18 (Remainder)
C |
AQUA-RAT | AQUA-RAT-35717 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
In a 100 m race between Rose and Diane, Rose lost to Diane by 0.75 m. If Diane was given a 7 m head start , how many meters more will it take before Rose overtakes Diane? | [
"12.91m",
"11.91m",
"12.5m",
"10.91m"
] | B | Distance Rose ran before completion of race---100 m-0.75m=99.25m
Distance gained on Diane over 99.25m---7m-0.75m=6.25
then Rose gains 99.25/6.25m = 1m on Kelly every 15.88 meters.
Therefore 15.88 divided by 0.75 of 1 meter.
ANSWER: B. 11.91m |
AQUA-RAT | AQUA-RAT-35718 | human-biology, genetics, hematology
Title: Possible genotypes for blood types? If I am blood type B, what are all the possible genotypes that could be expressed by my parents?
I think it might be 16 but I was reading online and saw this:
Similarly, someone who is blood type B could have a genotype of either
BB or BO.
So if someone can help me answer this, that would be great. Parent 1 and 2 have each 5 possible genotypes (OO, AO, BB, BO and AB).
Here a Punnett square with each possibilities. I highlighted the possible parent genotypes.
The total number of possible crosses is exactly 21. Note that here A = Ia, B = Ib and O = i.
OOxBB,OOxBO,OOxAB
AOxBB,AOxBO,AOxAB
BBxOO,BBxAO,BBxBB,BBxBO,BBxAB
BOxOO,BOxBB,BOxAO,BOxBO,BOxAB
ABxOO,ABxAO,ABxBB,ABxBO,ABxAB
This is starting from the information based on your blood type only (i.e. no information about your genotype).
Some background information. Antigen expressing alleles (here referred as Ia or A and Ib or B) are dominant. Not expressing an allele is notated i or O. Being of blood type O is when you don't express both alleles so it is a recessive trait (only possible genotype is ii or OO genotype).
The following is multiple choice question (with options) to answer.
The blood group of 200people is distributed as follows: 50 have A group, 65have B group, 70have O blood type and 15have type AB group. If a person from this group isselected at random, what is the probability that this person has O group type? | [
"0.35",
"0.5",
"1",
"1.23"
] | A | We construct a table of frequencies for the the blood groups as follows
group frequency
a 50
B 65
O 70
AB 15
We use the empirical formula of the probability
Frequency for O blood
P(E)= ________________________________________________
Total frequencies
= 70 / 200 = 0.35
A |
AQUA-RAT | AQUA-RAT-35719 | 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.
45 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30 persons, working 6 hours a day, complete the work? | [
"10",
"15",
"13",
"18"
] | B | Let the required number of days be x.
Less persons, More days (Indirect Proportion)
More working hours per day, Less days (Indirect Proportion)
Persons 30 : 45 :: 12 : x
Working hours/day 6 : 5
30 x 6 x x = 45 x 5 x 12
x = (45 x 5 x 12)/(30 x 6)
x = 15
ANSWER B |
AQUA-RAT | AQUA-RAT-35720 | ### 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 and B invests Rs.3500 and Rs.4000 respectively in a business. If A doubles his capital after 6 months. In what ratio should A and B divide that year's profit? | [
"21:5",
"21:16",
"21:2",
"9:9"
] | B | (3.5*6 + 7*6): (4*12)
63:48 =>
21:16
Answer:B |
AQUA-RAT | AQUA-RAT-35721 | Author Message
TAGS:
### Hide Tags
Manager
Joined: 26 Apr 2010
Posts: 122
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)
Followers: 2
Kudos [?]: 129 [0], given: 54
$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.
P,Q and R together earn Rs.1710 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day? | [
"s.40",
"s.60",
"s.90",
"s.100"
] | B | Explanation :
Amount Earned by P,Q and R in 1 day = 1710/9 = 190 ---(1)
Amount Earned by P and R in 1 day = 600/5 = 120 ---(2)
Amount Earned by Q and R in 1 day = 910/7 = 130 ---(3)
(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day
- Amount Earned by P,Q and R in 1 day = 120+130-190 = 60
=>Amount Earned by R in 1 day = 60
Answer : Option B |
AQUA-RAT | AQUA-RAT-35722 | Just to check each case: \begin{align*} f(S_T)=\begin{cases} 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + 0 + 0 = 3 & \text{if }S_T\leq 30, \\ 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + (30 - S_T) + 0 = 33- S_T & \text{if }30
I know @KeSchn already answered but hope this helps since this is how I usually do these. Of course, you can do this multiple ways but this gets to a correct answer relatively quickly.
Edit: Gordon's answer is definitely the way to go if you're comfortable with indicator functions. It does everything the graphical methods do without requiring any visualizing etc
The following is multiple choice question (with options) to answer.
If 45-[28-{37-(15-*)}]= 55, then * is equal to: | [
"-29",
"-19",
"16",
"29"
] | C | 45-[28-{37-(15-*)}]= 55 => 45-[28-{37-15+*}]=55
45-[28-37+15-*]=55 => 45[43-37-*]=55
45-[6-*]=55 => 45-6+*=55
39+*=55 => *=55-39
= 16
ANSWER:C |
AQUA-RAT | AQUA-RAT-35723 | # 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 factory has a fixed cost of $40,000 a month, and a cost of $2.5 for every item produced. If the selling price of a single item is $5, what is the number of items must be sold monthly for the factory to cover its cost exactly? | [
"9,000",
"14,000",
"16,000",
"22,500"
] | C | selling price - Cost price = 2.5 so per unit profit =2.5
cost to recover =40000
no of items required = 40000/2.5 = 16000
Ans C |
AQUA-RAT | AQUA-RAT-35724 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
What will be the difference in simple and compound interest on 6000 after three years at the rate of 10 percent per annum? | [
"160",
"186",
"62",
"20"
] | B | For 3 years:
Diff.=Sum×(rate)2(300+rate)/(100)3
= 6000×10×10×310/100×100×100 = 186
Answer B |
AQUA-RAT | AQUA-RAT-35725 | that is y=???abc
#### Bacterius
##### Well-known member
MHB Math Helper
Re: find the value of x and y
$$x^2 \equiv x \pmod{1000} \tag{1}$$
[JUSTIFY]Clearly for $x$ coprime to $1000$, the only solutions are the trivial $x = 0$ and $x = 1$, so all nontrivial solutions for $x$ must be divisible by either $2$ or $5$, or both. This leaves us with about $600$ possible candidates. Now it should be clear that if a solution works for $1000$, it will also work for $100$, and for $10$. Let's try and find the solutions for $10$:[/JUSTIFY]
$$x^2 \equiv x \pmod{10} \tag{2}$$
[JUSTIFY]We have $5$ potential solutions here, namely $2, 4, 5, 6, 8$ (modulo $10$). Let's check each of them manually, some calculations show that only $5$ and $6$ work. So the solutions must have either $5$ or $6$ as a last digit. Let's now extend our search to $100$, armed with this information (which reduces the set of possible solutions considerably):[/JUSTIFY]
$$x^2 \equiv x \pmod{100} \tag{3}$$
[JUSTIFY]We know from the previous step that the potential solutions are $5, 6, 15, 16, \cdots$. A short exhaustive search tells us that only $25$ and $76$ work. Finally, we have narrowed down the search enough to look for the real solutions:[/JUSTIFY]
$$x^2 \equiv x \pmod{1000} \tag{4}$$
The only possible solutions are $25, 76, 125, 176, \cdots$. Again, trying those 20 potential solutions shows that only $376$ and $625$ work. Therefore, the solutions are $x \in \{ 0, \, 1, \, 376, \, 625 \}$. QED.
The following is multiple choice question (with options) to answer.
X and Y are both integers. If X/Y = 59.40, then what is the sum of all the possible two digit remainders of X/Y? | [
"450",
"616",
"672",
"900"
] | A | Remainder = 0.40 --> 40/100 --> Can be written as (40/4) / (100/4) = 10/25
So remainders can be 10, 20, 30, 40, ..... 90.
We need the sum of only 2 digit remainders --> 10 + 20 + 30 + 40 + 50 + 60 + 70+ 80 + 90 =450
Answer: A |
AQUA-RAT | AQUA-RAT-35726 | # 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 retailer buys a radio for Rs 232. His overhead expenses are Rs 15. He sellis the radio for Rs 300. The profit percent of the retailer is | [
"10%",
"50%",
"21.4%",
"52%"
] | C | Explanation:
cost price = (232 + 15) = 247 sell price = 300
gain = (53/247)*100 = 21.4%. Answer: C |
AQUA-RAT | AQUA-RAT-35727 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
Aman started a business investing Rs. 65,000. Rakhi joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business? | [
"10:105:112",
"78:105:112",
"72:105:112",
"74:105:112"
] | B | Explanation:
Aman : Rakhi : Sagar = (65000 * 36) : (105000 * 30) : (140000 * 24)
= 78:105:112
Answer: B |
AQUA-RAT | AQUA-RAT-35728 | Let us now make sure that our answer matches the textbook one $$\displaystyle\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right] + C = \frac{1}{24}\cos 6x - \frac{1}{16}\cos 4x - \frac{1}{8}\cos 2x + C$$ Indeed, observe that \begin{aligned} \cos 6x &= \cos \big(2(3x) \big) \\ & = \cos^2 3x - \sin^2 3x \\ & = 1 - 2 \sin^2 3x \\ & = 1 - 2 \big(3\sin x - 4\sin^3 x\big)^2 \\ & = 1 - 2 \big(9\sin^2 x - 24\sin^4 x + 16\sin^6 x\big) \\ & = 1 - 18\sin^2 x + 48 \sin^4 x - 32\sin^6 x \\ \cos 4x &= \cos \big(2(2x) \big) \\ & = \cos^2 2x - \sin^2 2x \\ & = 1 - 2 \sin^2 2x \\ & = 1 - 2 \big(2\sin x \cos x \big)^2 \\ & = 1 - 2 \big(4 \sin^2 x \cos^2 x\big) \\ & = 1 - 8\sin^2 x \left(1 - \sin^2 x \right) \\ & = 1 - 8\sin^2 x + 8 \sin^4 x \\ \cos 2x &= \cos^2 x - \sin^2 x \\ & = 1 - 2 \sin ^2 x \end{aligned}
The following is multiple choice question (with options) to answer.
Find the fourth proportional to 33.6 , 41.2 and 48.9? | [
"59.96",
"56.15",
"52.39",
"58.96"
] | A | Formula= Fourth proportional =(b x c)/a
a=33.6 , b=41.2 and c=48.9
(41.2 x 48.9)/33.6=59.96
A |
AQUA-RAT | AQUA-RAT-35729 | => 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.
Calculate the share of Y , if Rs. 2690 is divided among X, Y and Z in the ratio 5 : 7 : 9? | [
"890.7",
"826.7",
"895.7",
"896.7"
] | D | 5 + 7 + 9 =21
2690/21=128.1
Y's share =7*128.1
=896.7
ANSWER:D |
AQUA-RAT | AQUA-RAT-35730 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
The rate of increase of the price of sugar is observed to be three percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994, is Rs. 25 per kg. The inflation rate for the years 1994 and 1995 are expected to be 12% each. The expected price of sugar on January 1, 1996 would be | [
"33.06",
"34.1",
"34.2",
"24.6"
] | A | Explanation :
Increase in the price of sugar = (12+3)= 15%
Hence, price of the sugar on Jan 1, 1996
=> (25 * 115 * 115)/( 100 * 100 ) = Rs 33.06.
Answer : A |
AQUA-RAT | AQUA-RAT-35731 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The ratio between the sale price and the cost price of an article is 6:4. What is the ratio between the profit and the cost price of that article? | [
"23",
"1:2",
"2:5",
"3:5"
] | B | C.P. = Rs. 4x and S.P. = Rs. 6x.
Then, Gain = Rs. 2x
Required ratio = 2x : 4x = 1:2
B |
AQUA-RAT | AQUA-RAT-35732 | \begin{align} \E(Y) & = 1\cdot\Pr(Y=1) + \sum_{y=2}^\infty y\Pr(Y=y) \\[10pt] & = 1\cdot\Pr(X=0) + \underbrace{1\cdot\Pr(X=1)+ \sum_{y=2}^\infty y\Pr(X=y)}_\text{This is $\E(X)$} \\[10pt] & = 1\cdot e^{-1.2} + \E(X) = e^{-1.2}+1.2. \end{align}
• Thanks for mentioning this interesting approach. – d125q Jun 4 '15 at 23:37
$\lambda=0.6$ is the average number of fishes, that the fisherman is catching during one hour. $\lambda$ is the expected value of the poisson distribution. Thus in two hours he catches on average 1.2 fishes .
The probability, that he is not catching a fish during the first hour is $P(X=0)=e^{-0.6}\cdot \frac{0.6^0}{0!}=e^{-0.6}$
The probabiliy, that he is not catching a fish during the second hour is $P(Y=0)=e^{-0.6}\cdot \frac{0.6^0}{0!}=e^{-0.6}$.
$P(X=0 \cap Y=0)$=Probability, that the fisherman is not catching a fish during 2 hours.
$P(X=0 \cap Y=0)=P(X=0)\cdot P(Y=0|X=0)$.
$X$ and $Y$ are independent. The number of fishes, which are caught during the second hour does not depend on the number of fishes, which have been caught during the first hour.
The following is multiple choice question (with options) to answer.
40 men can catch 200 sharks in 20 days working 6 hours a day. In how many days 25 men can catch 300 sharks working 4 hours a day? | [
"30",
"34",
"24",
"20"
] | C | We have, M1D1H1/W1 = M2D2H2/W2
40*20*6/200 = 25*D2*4/300
D2= 40*20*6*300/200*25*4 = 24
ANSWER:C |
AQUA-RAT | AQUA-RAT-35733 | algorithms, optimization
but I still want all items to have 25% share.
Assuming amount can be decimal, how do I find the optimal way of rebalancing this portfolio back to 25% shares? How to determine how many of each item to sell so we won't have to buy it back again (sell everything and spend 25% on each item is not an option)?
What if I have extra $100 to spend while rebalancing?
Upd: what I mean by "optimal" is to keep the amount of transactions as less as possible, assuming each transaction costs something. Your problem specification is incomplete. What do you mean by optimal? Do you want to, say, minimize transactions costs? Do you want to make sure you do not cross the spread?
The original portfolio was worth \$800. The new portfolio is worth \$804. Assuming your actions do not change the market price, and assuming no transactions costs, that is a given.
You had \$200 allocated to each security. Now, you want \$201 allocated to each security:
| Name | Price | Amount | Notional |
|--------|-------|--------|----------|
| Item 1 | 220 | 1 | $220 |
| Item 2 | 130 | 2 | $260 |
| Item 3 | 17 | 4 | $68 |
| Item 4 | 32 | 8 | $256 |
For every security whose notional is above the desired amount, sell that many units:
Item 1: Sell 19/220 = 0.0863636363636364 units
Item 2: Sell 59/130 = 0.453846153846154 units
Item 4: Sell 55/32 = 1.71875 units
The following is multiple choice question (with options) to answer.
What is the CP of Rs 100 stock at 7 discount, with 1/5% brokerage? | [
"93.9",
"96.3",
"93.2",
"96.7"
] | C | Explanation:
Use the formula,
CP= 100 – discount + brokerage%
CP= 100-7+1/5
93.2
Thus the CP is Rs 93.2.
ANSWER: C |
AQUA-RAT | AQUA-RAT-35734 | # Seating couples around 2 tables
Here's my question and possible answer.
How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:
1) each couple must sit at the same table, and,
2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).
My solution:
Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)
$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)
$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$
$=2 * 8!$
Can someone verify this solution or provide the correct one?
First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$
Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4}$ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.
Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.
Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.
Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$
And again, if you do not differentiate between the tables, then divide this by $2$
The following is multiple choice question (with options) to answer.
For a fundraising dinner, a florist is asked to create flower arrangements for 8 tables. Each table can have one of the two types of bouquets available, one with a single type of flower or one with three different types of flowers. If the florist wants to make each table unique, what is the least number of types of flowers he needs? | [
"4",
"5",
"6",
"7"
] | A | Looking for (n + nC3) = 8
2 types of flower arrangements: some with 1 flower and some with 3 different flowers.
Therefore, total number of arrangements we could make from n different types of flowers = (n + nC3).
Since 4C3 = 4 (nC(n-1) always equals n combinations i.e. nC(n-1) = n), (n + nC3) = (4 + 4C3) = (4 + 4) = 8
Aim to get 8 distinct floral arrangements, 4 different types of flowers is the answer.
Answer: A |
AQUA-RAT | AQUA-RAT-35735 | 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 N is divided by T, the quotient is A and the remainder is V. Which of the following expressions is equal to N? | [
"ST",
"S + V",
"ST + A",
"T(S+V)"
] | C | Using the rule dividend = Quotient * divisor + remainder =>ST + A C is correct |
AQUA-RAT | AQUA-RAT-35736 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
2 hours after train A leaves Lucknow a train B leaves the same stationtraveling in the same direction at an average speed of 20 km/hr. After traveling 6 hrsthe train B overtakes the train A. What is the average speed of the train A? | [
"15 km/hr",
"16 km/hr",
"17 km/hr",
"18 km/hr"
] | A | Explanation :
Total distance covered by B=20*6=120kmTotal time taken by A to cover same distance=2+6=8 hrsaverage speed of A=120/8=15 km/hr
Answer : A |
AQUA-RAT | AQUA-RAT-35737 | Denote: $$y=ax, z=bx$$. Then: $$x+y+z=0 \iff x+ax+bx=0 \iff a=-1-b;\\ S={x\over y} + {y\over z} + {z\over x} = {y\over x} + {z\over y} + {x\over z} \iff \\ S={1\over a} + {a\over b} + b = a + {b\over a} + {1\over b} \iff \\ S=-{1\over 1+b} - {1+b\over b} + b = -1-b - {b\over 1+b} + {1\over b} \iff \\ \frac{(b+2)(2b+1)(b-1)}{b(1+b)}=0 \iff \\ b_{1,2,3}=\{\color{red}{-2},\color{green}{-\frac12},\color{blue}1\}; a_{1,2,3}=\{\color{red}{1},\color{green}{-\frac12},\color{blue}{-2}\}\\ S_1=\frac1a+\frac ab+b=\frac1{\color{red}{1}}+\frac{\color{red}1}{\color{red}{-2}}+(\color{red}{-2})=-\frac32.\\ S_2=\frac1a+\frac ab+b=\frac1{\color{green}{-\frac12}}+\frac{\color{green}{-\frac12}}{\color{green}{-\frac12}}+(\color{green}{-\frac12})=-\frac32.\\ S_3=\frac1a+\frac
The following is multiple choice question (with options) to answer.
If X:Y is 1:2 and Y:Z is 2:5 then X:Z is equal to | [
"2:5",
"3:5",
"1:5",
"4:5"
] | C | The two ratios given are having the same number 2 for Y in both the ratios.
Hence- X:Y = 1:2 Y:Z = 2:5
=> X:Z = 1:5
Answer C |
AQUA-RAT | AQUA-RAT-35738 | # 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.
man buys an article for 10% less than its value and sells it for 10% more than its value. His gain or loss percent is? | [
"12",
"76",
"26",
"20"
] | D | Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 - 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%
Answer: D |
AQUA-RAT | AQUA-RAT-35739 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
When Jessica withdrew $200 from her bank account, her account balance decreased by 2/5. If she deposits an amount equal to 1/3 of the remaining balance, what will be the final balance in her bank account? | [
"300",
"375",
"400",
"500"
] | C | As per the question 200=2a/5
thus- a which is the total amount =500
The amount thus left =300
She then deposited 1/3 of 300=100
total amount in her account =400
Answer C |
AQUA-RAT | AQUA-RAT-35740 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
The sum of all consecutive odd integers from −27 to 37, inclusive, is | [
"110",
"135",
"150",
"165"
] | D | The sum of the odd numbers from -27 to +27 is 0.
Let's add the remaining numbers.
29+31+33+35+37 = 5(33) = 165
The answer is D. |
AQUA-RAT | AQUA-RAT-35741 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern has a leak which would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold? | [
"480",
"289",
"279",
"276"
] | A | 1/x - 1/20 = -1/24
x = 120
120 * 4 = 480
Answer:A |
AQUA-RAT | AQUA-RAT-35742 | To solve more problems based on Heron’s … Question Bank Solutions 4046. Similarity to Heron's Formula. We can easily draw many more quadrilaterals and we can identify many around us. 2 + base. Important Solutions 2865. Chapters. where is the semiperimeter, or half of the triangle's perimeter. 4. Ex 12.2, 2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Program to find the angles of a quadrilateral in C++; Maximum area of rectangle possible with given perimeter in C++; Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts in C++; Find maximum volume of a cuboid from the given perimeter and area in C++; Maximum area rectangle by picking four sides from array in C++ side a: side b: side c: side d: sum of angles θ 1 +θ 2: area S . % Progress . Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. Sides of triangle are 12 cm, 13cm and 15cm. … or own an. Find its area. Now what I need is to implement a method get_area() which calculates the area of my quadrilateral, but I have no idea how. Watch Area and Perimeter of Quadrilaterals Videos tutorials for CBSE Class 8 Mathematics. Progress % Practice Now. Another construction of the point T A is to start at A and trace around triangle ABC half its perimeter, and similarly for T B and T C. Syllabus. Franchisee/Partner Enquiry (North) … Questionnaire. Change Equation Select to solve for a different unknown Scalene Triangle: No sides have equal length No angles are equal. A splitter of a triangle is a cevian (a segment from a vertex to the opposite side) that divides the perimeter into two equal lengths, this common length being called the semiperimeter of the triangle. It is classif The sides a n of regular inscribed polygons, where R is the radius of the circumscribed circle = a n = 2R sin 180 o /n; Area of a polygon of perimeter P and radius of in-circle … Properties of a quadrilateral. Find the Length
The following is multiple choice question (with options) to answer.
In the quadrilateral PQRS d=7 cm, h1=4.2 cm and
h2=2.1 cm. Find the area | [
"21",
"22",
"23",
"24"
] | B | area of quad.=1/2*any diagonal*(sum of 2 perpendiculars which is drawn on that diagona)
so 1/2*7*(4.2+2.1)=22.05
ANSWER:B |
AQUA-RAT | AQUA-RAT-35743 | 1. ## Number of streets
In one residantial area every street crosses over every other street and there are no three streets which crosses each other in same point.
How many streets there are in that residential area if there are 21 crosses?
2. Originally Posted by OReilly
In one residantial area every street crosses over every other street and there are no three streets which crosses each other in same point.
How many streets there are in that residential area if there are 21 crosses?
$7?$
RonL
3. Originally Posted by CaptainBlack
$7?$
RonL
Why 7?
4. Originally Posted by OReilly
Why 7?
Because if there are n lines and C(n) crossings, adding an extra line adds
n more cuts one for each existing line, so:
C(n+1)=C(n)+n.
C(1)=0,
C(2)=1+0
C(3)=2+1
C(4)=3+3
C(5)=4+6
C(6)=5+10
C(7)=6+15=21
RonL
5. Originally Posted by CaptainBlack
Because if there are n lines and C(n) crossings, adding an extra line adds
n more cuts one for each existing line, so:
C(n+1)=C(n)+n.
C(1)=0,
C(2)=1+0
C(3)=2+1
C(4)=3+3
C(5)=4+6
C(6)=5+10
C(7)=6+15=21
RonL
Quite right!
I have made a picture of how could these streets look like.
There is also second question.
How many there are areas which are edged with streets from every side?
I have counted from picture that there are 15, I must say that I can't think another way at the moment.
6. If you have $n$ lines than at most you can have.
$\frac{n(n-1)}{2}$ intersection points.
Thus,
$\frac{n(n-1)}{2}=21$
Thus,
$n^2-n+42=0$
Thus, $n=-6,7$
Disregard the negative which yields $n=7$
The following is multiple choice question (with options) to answer.
There are 200 cats in Cat-City. Out of the 200, 70 are street cats and the rest are domestic cats. 110 cats are gray, 50 out of the gray cats are NOT domestic ones. How many domestic cats are there which are not gray in Cat-City? | [
"90.",
"70.",
"50.",
"40."
] | B | STREET-CAT DOMESTIC-Cat
70 130
(50gray+20other) 60gray+70other
Answer:B |
AQUA-RAT | AQUA-RAT-35744 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Bhaman travelled for 15 hours. He covered the first half of the distance at 40 kmph and remaining half of the distance at 10 kmph. Find the distance travelled by Bhaman? | [
"240",
"230",
"260",
"220"
] | A | Let the distance travelled be x km.
Total time = (x/2)/40 + (x/2)/10 = 15 => x/80 + x/20 = 15 => (x + 4x)/80 = 15
=> x = 240 km
Answer:A |
AQUA-RAT | AQUA-RAT-35745 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 3025., the number of the member is the group is: | [
"55",
"67",
"77",
"87"
] | A | Money collected = (30.25 x 100) paise = 3025 paise.
Number of members =Square root of 3025 = 55.
Answer: Option A |
AQUA-RAT | AQUA-RAT-35746 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
What ratio must a shopkeeper mix Peas and Soybean of Rs.16 and Rs. 25/kg, As to obtain a mixture of Rs.21? | [
"10 : 7",
"9 : 8",
"4 : 5",
"13 : 11"
] | C | Correct option: (C)
Use rule of alligation, to determine the ratio
The required ratio of Soybean and Peas = 4 : 5 |
AQUA-RAT | AQUA-RAT-35747 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
Dick and Jane each saved $2,000 in 1989. In 1990 Dick saved 10 percent more than in 1989, and together he and Jane saved a total of $4,000. Approximately what percent less did Jane save in 1990 than in 1989? | [
"10%",
"20%",
"25%",
"30%"
] | A | 1990
Dick saved = $ 2200
Jane saved = $ 1800 (jane saved $200 less than she did the prior year)
Jane saved approximately $200/2000$ (10%) less in 1990
Answer: A |
AQUA-RAT | AQUA-RAT-35748 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the train? | [
"300 m",
"270 m",
"200 m",
"250 m"
] | B | speed = 72*5/18 = 20 m/sec
t = 26 sec
length => x
x+250 = 520
x= 270
ANSWER B |
AQUA-RAT | AQUA-RAT-35749 | algorithms, scheduling, algorithm-design
Title: Schedule two trains whose tracks overlap so they don't crash I've encountered scheduling problems in my algorithms class before like the type we use vertex cover to solve.
Recently I was asked this question and did not even know what algorithmic technique to use to answer it!
There are two trains, which run between stations and share a portion of track. For instance train 1 runs between stations A B C D E F G H and train 2 runs between stations I J K B C D E F L M so the trains share the track between B C D. At each end of the track the train turns around and goes the other way on its track. The trains go from station to station in 1 unit time. I need to prevent the trains from crashing head on on the shared track, I have the power to stop either train at any station and restart it again when I please.
How can I solve this problem without using a ton of if else clauses, what CS principles should I be using here?
edit
Some background, the first solution I wrote was just a simple check to see if the other train was on the shared track before sending the other train out. That seemed ad hoc to me and I'm sure it did to the company I interviewed for as well.
I figure there must be an algorithm or at least a school of thought for this type of problem, after all how do they build extensible systems for managing shared runway space at an airport, or managing traffic flow at stop lights.
My intuition is that this is a very introductory problem to a school of thought in computational problem solving I have not yet encountered.
I might be wrong, but could anybody clear the air for me? If you calculate the cycle, as there are only two trains and distance between every station is one, you could simply run them in order and appoint wait in some station to avoid crash.
If you have to implement some unfortunate events (first one stops for some reason, reschedule it to run on itd own time).
With two trains it will work without problems.
Otherwise use semaphores (it fits even literally ;)
There is no explicitly given synchronisation problem, so nasty trick works.
The following is multiple choice question (with options) to answer.
X can lay railway track between two station in 16; Y can do the same work in 20 days,both are working together for 2 days after Y left the job then how many days X requires to complete the remaining work ? Total number of days taken to complete the whole work? | [
"7 Days and 9 days",
"5.89 Days and 7.89 days",
"8.26 Days and 10.26 days",
"7.75 Days and 9.75 days"
] | D | First take LCM of (16,20) = 160; Total work = 160; X's one day work = 160/16 =10; Y's one day work =160/20 = 8; One day work of both X+Y= 10+8 = 18; Total work completed in 2 days = 18*2=36; remaining work is = 160 - 36 =124; remaining work completed by X in = 124/16= 7.75 days.Total number of days taken to complete the whole work = 2+7.75 = 9.75 days.
Answer =D |
AQUA-RAT | AQUA-RAT-35750 | # 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.
By selling an article at Rs.600, a profit of 25% is made. Find its cost price? | [
"228",
"200",
"279",
"480"
] | D | SP = 600
CP = (SP)*[100/(100+P)]
= 600 * [100/(100+25)]
= 600 * [100/125] = Rs.480.Answer:D |
AQUA-RAT | AQUA-RAT-35751 | # Chapter 7: Coordinate Geometry
1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.
|AB| = $$\sqrt{(x-5)^{2}+(y-1)^{2}}$$
= $$\sqrt{x^{2}+16-8x+y^{2}+4-4y}$$
|AC| = $$\sqrt{(x+2)^{2}+(y-5)^{2}}$$
= $$\sqrt{x^{2}+4+4x+y^{2}+16-8y}$$
Since, |AB| = |AC|
$$\Rightarrow$$ $$x^{2}+y^{2}-8x-4y+20$$ = $$x^{2}+y^{2}+4x-8y+20$$
$$\Rightarrow$$ 8y – 4y = 4x + 8x
$$\Rightarrow$$ y = 3x
2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)
Let the vertices of the $$\bigtriangleup$$ be P(0, 8), Q(0, 0) and R(6, 0)
∴ PQ = $$\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8$$
∴ QR = $$\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6$$
∴ RP = $$\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10$$
∴ Perimeter of $$\bigtriangleup$$ = 8 + 6 + 10 = 24 units
The following is multiple choice question (with options) to answer.
If two sides of a triangle have lengths 4 and 7, which of the following could be the perimeter of the triangle?
I. 9
II. 15
III. 19 | [
"None",
"I only",
"II only",
"II and III only"
] | D | Let x be the length of the third side.
7-4 < x < 7+4
3 < x < 11
14 < perimeter < 22
The answer is D. |
AQUA-RAT | AQUA-RAT-35752 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat? | [
"188m",
"278m",
"887m",
"100m"
] | D | A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.
Answer: D |
AQUA-RAT | AQUA-RAT-35753 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two trains start simultaneously from opposite ends of a 140-km route and travel toward each other on parallel tracks. Train X, traveling at a constant rate, completes the 140-km trip in 4 hours. Train Y, travelling at a constant rate, completes the 140-km trip in 3 hours. How many kilometers had Train X traveled when it met Train Y? | [
"60",
"64",
"68",
"72"
] | A | If the two trains cover a total distance D, then Train X travels (3/7)*D while Train Y travels (4/7)*D. If the trains travel 140 km to the meeting point, then Train X travels (3/7)*140 = 60 km.
The answer is A. |
AQUA-RAT | AQUA-RAT-35754 | Write down the circumference formula. The area, diameter and circumference will be calculated. Practice. Radius Of A Circle Formula. For example: enter the radius and press 'Calculate'. We know the area of the pizza is 220 square inches. Formula for calculating radius of a inscribed circle of a rhombus if given height ( r ) : radius of a circle inscribed in a rhombus : = Digit 2 1 2 4 6 10 F first two years of college and save thousands off your degree. In the figure above, drag the orange dot around and see that the radius is always constant at any point on the circle. Preview; Assign Practice; Preview. Notice that the radius is the same length at any point around the circle. % How to Do Your Best on Every College Test. The radius is 7.5 inches. a. Enter any single value and the other three will be calculated.For example: enter the radius and press 'Calculate'. Diameter Which is the circle's 'width'. Solution: Formula for circumference of circle is = 2 π r. Substitute π = 22/7 and r = 14. A great time-saver for these calculations is a little … imaginable degree, area of The area of a circle is: π times the Radius squared: A = π r 2. or, when you know the Diameter: A = (π /4) × D 2. or, when you know the Circumference: A = C 2 / 4 π. See diameter of a circle For many geometric figures, the radius has a well-defined relationship with other measures of the figure. The nature of the radius makes it a powerful building block for understanding many other measurements about a circle, for example its diameter, its circumference, its area and even its volume (if you're dealing with a three-dimensional circle, … just create an account. You will also learn how to find the radius of a circle using three different formulas based on diameter, circumference, and area. Enter the radius, diameter, circumference or area of a Circle to find the other three. A radius can be a line from any point on the circumference to the center of the circle. Circumference of Circle is the distance all the way around the circle. The radius of a circle is a line from the center to its edge Learning Outcomes.
The following is multiple choice question (with options) to answer.
The area of a circle is subtracted from its diameter, and the circumference is then added to this total, the result is 0. What is the radius of the circle? | [
"2*pi",
"pi",
"1",
"2"
] | D | The equation is; diameter - Area + Circumference = d - A + C = d - pi * r^2 + 2 * pi * r = 2r - pi * r(r + 2).
By plugging in the answers we can test the answers quickly; then, 2 is the only possible answer.
Answer: D |
AQUA-RAT | AQUA-RAT-35755 | $\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, 30,000 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? | [
"104,000",
"60,000",
"80,000",
"90,000"
] | A | The distance traveled by each tyre:
4/5 * 1, 30, 000km = 104,000 km. A |
AQUA-RAT | AQUA-RAT-35756 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
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07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
The following is multiple choice question (with options) to answer.
In a group of ducks and cows, the total number of legs are 34 more than twice the no. of heads. Find the total no.of buffaloes. | [
"11",
"12",
"14",
"17"
] | D | Let the number of buffaloes be x and the number of ducks be y
=> 4x + 2y = 2 (x + y) + 34
=> 2x = 34 => x = 17
D |
AQUA-RAT | AQUA-RAT-35757 | Location: Cambridge, MA
Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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08 Jul 2012, 18:46
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catty2004 wrote:
92. If a, b, c, and d, are positive numbers, is a/b < c/d?
1) 0 < (c-a) / (d-b)
Hi catty,
We're looking for whether a/b < c/d. Fortunately, we're told a useful bit of info in the question stem. All four terms are positive. That's very important with inequalities, because it means that we can multiply and divide without having to worry about the direction of the inequality signs. In this case, we could rephrase the question to whether ad < bc by cross-multiplying. This will be useful laters.
Statement 1) is not useful, however. (c-a) and (d-b) could both be positive or negative; that means that when me multiply to get rid of a term, we might or might not have to flip the terms. Since any of the variables could be greater or less than any of the other variables, this statement is insufficient.
Statement 2) gives us exactly what we want. Here, with no subtraction, everything stays positive. That means we can divide out (ad/bc) from both sides without flipping the inequality. We get ad/bc < 1, and can cross-multiply to get ad < bc. That answers our question with a definite yes, so it's sufficient and the answer is (B)
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Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink]
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The following is multiple choice question (with options) to answer.
The positive number x is q percent greater than the positive number b, which is p percent less than x itself. If a is increased by p percent, and the result is then decreased by q percent to produce a positive number c, which of the following could be true?
I. c > x
II. c = x
III. c < x | [
"I only",
"II only",
"III only",
"I and II only"
] | C | Let q = 10% then p = 100/11 %
let b = 100 then x = 110
after increasing x by p and decreasing by q we get c= 108
therefore c<x
C it is answer |
AQUA-RAT | AQUA-RAT-35758 | Author Message
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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 certain company had a total annual expenditure of 2.55∗10^7 on employee salaries last year. If the company employed 425 people, what was the average employee salary? | [
"$60,000",
"$25,000",
"$35,000",
"$40,000"
] | A | Given: Total annual expenditure of 2.55∗10^7 on employee salaries
Total employees = 425
Observe that 425*6 = 2550
Therefore try to bring the numerator in terms of 2550
Average salary = (2550*10^4) /425 = 6*10^4 = 60,000
Option A |
AQUA-RAT | AQUA-RAT-35759 | 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.
In a class of students, the ratio of girls to boys is 1:2. If 3 more girls joined the class, the ratio would be 3:4. How many students are in the class? | [
"16",
"18",
"20",
"22"
] | B | Let x be the number of students in the class.
(1/3)*x=(3/7)(x+3)-3
7x=9x+27-63
2x=36
x=18
The answer is B. |
AQUA-RAT | AQUA-RAT-35760 | box there are nine fair coins and one two-headed coin. One of two coins is selected at random and tossed three times. Tossing a Coin A balanced coin is tossed three times. One is a two-headed coin ( having head on both faces ), another is a biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tails 40% of the times. They are: HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT. find the probability of getting a result of an odd number of heads in 1/2 of the cases. A coin flip: A fair coin is tossed three times. 5 Probability of getting a tail = 0. In a short trial, heads may easily come up on every flip. There are three ways that can happen. The outcomes of the three tosses are recorded. There are many other kinds of situations, however, where the probability of an event is not independent but dependent — that is, where the. sum of three times a. Probability of Coin Tosses. One Head : 160 times c. 3 10 Illowsky et al. enter your value ans - 5/16. There were five heads and three tails in the eight tosses. The probability of rolling a six on one die is 1/6. Subjective probability is more connected to the original meaning of the word probable for example, because simply calculating the probability of any horse winning There are three types of probability problems that occur in mathematics. Poker night 2 steam achievement manager. Solution: n=20, p=0. 5 Number of heads$\leq$Number of tails. When three coins are tossed at random, what is the probability of getting : (i) 3 heads ? (ii) 2 heads Best Answer for Silver Coins, Round, Sloven A fair coin is tossed 5 times. Let E = event of getting exactly 3 heads. The Questions and Answers of A coin is tossed three times. The answer would be a cumulative probability. If$X\$ is the number of heads. A consecutive streak or a run can happen in random. what's the probability of getting three heads? You can put this solution on YOUR website! a fair coin is tossed 3 times, 1. A coin has a probability of 0. So let's start again with a fair coin. The chances of getting a girl should be the same whether or
The following is multiple choice question (with options) to answer.
A fair coin is tossed 3 times. What is the probability of getting at least 2 heads? | [
"3/4",
"3/2",
"1/4",
"1/2"
] | D | Let's find the probability of 2 heads , 3 heads
P(HHH)=((1/2)^3=1/8.
P(HHT)=(3!/2!)*(1/2)^3=3/8
Total Probablity = 1/8 + 3/8
=1/2
Answer(D) |
AQUA-RAT | AQUA-RAT-35761 | # If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term. - Mathematics
Sum
If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.
#### Solution
Let the first term, common difference
And number of terms of the AP: 9, 7, 5,…. are
a1, d1 and n1, respectively
i.e., first term (a1) = 9
And common difference (d1) = 7 – 9 = –2
⇒ T"'"_(n_1) = a_1 + (n_1 - 1)d_1
⇒ T"'"_(n_1) =9 + (n_1 - 1)(-2)
⇒ T"'"_(n_1) = 9 - 2n_1 + 2
⇒ T"'"_(n_1) = 11 - 2n_1 ......(i)
∵ nth term of an AP, Tn = a + (n – 1)d
Let the first term, common difference and the number of terms of the AP: 24, 21, 18, … are a2, d2 and n2, respectively
i.e., first term, (a2) = 24
And common difference (d2) = 21 – 24 = – 3
∴ Its nth term, T"'"_(n_2) = a_2 + (n_2 - 1)d_2
⇒ T"'"_(n_2) = 24 + (n_2 - 1)(-3)
⇒ T"'"_(n_2) = 24 - 3n_2 + 3
⇒ T"'"_(n_2) = 27 - 3n_2 .....(ii)
Now, by given condition,
nth terms of the both APs are same
i.e., 11 - 2n_1 = 27 - 3n_2 ......[From equations (i) and (ii)]
The following is multiple choice question (with options) to answer.
Which one of the following is the sum of the numbers 1, 3, 5, 7, 9…. up to nth number? | [
"N2 + 1",
"N2 - 1",
"N(N + 1)/2",
"N2"
] | D | Explanation:
N2
Answer: D |
AQUA-RAT | AQUA-RAT-35762 | # 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 costs a dealer Rs.50,000.Dealer raised price by Rs.10,000 and then deducted 4% of new price.
What is the profit/loss %? | [
"14",
"15",
"15.2%",
"16"
] | C | Ans: Let P be the percentage of price raised=(10000/500000) * 100 =20%
Discount %=4%
profit %=(20-4- 20*4/100 ) =15.2%
Answer C |
AQUA-RAT | AQUA-RAT-35763 | # A positive integer (in decimal notation) is divisible by 11 $\iff$ …
(I am aware there are similar questions on the forum)
## What is the Question?
A positive integer (in decimal notation) is divisible by $11$ if and only if the difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by $11$.
For example consider the integer 7096276.
The sum of the even positioned digits is $0+7+6=13.$ The sum of the odd positioned digits is $7+9+2+6=24.$ The difference is $24-13=11$, which is divisible by 11.
Hence 7096276 is divisible by 11.
## (a)
Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11.
## (b)
Show that divisibility statement is true for three-digit integers $abc$. Hint: $100 = 99+1$.
## What I've Done?
I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But...the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing.
I need some help on the modular arithmetic on these.
For example... Is 77 divisible by 11? $$7+7 = 14 \equiv ...$$
I don't know what to do next.
Thanks very much, and I need help on both (a) and (b).
In order to apply the divisibility rule, you have to distinguish between odd and even positioned digits. (It doesn't matter how you count.)
Example: In 77 the first position is odd, the second even, so you would have to calculate $7-7=0$, which is divisible by 11.
The following is multiple choice question (with options) to answer.
If 'c' is a positive integer exactly divisible by 6 or 5 but not divisible by 12.What could possibly be the value of 'c' ? | [
"30",
"60",
"120",
"24"
] | A | 60 and 120 are both divisible by 6 and 5 but also by 12. so they are not the right answer.
24 and 72 are both clearly not divisible by 5(not correct)
30 is both divisible by 6 and 5 but not by 12.
answer : (A) 30 |
AQUA-RAT | AQUA-RAT-35764 | 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.
What number continues the sequence?
25, 50, 27, 46, 31, 38, 39, ? | [
"11",
"33",
"52",
"22"
] | D | D
22
There are two alternate sequences, the first increases by 2, 4, 8 etc and the second decreases by 4, 8, 16 etc. |
AQUA-RAT | AQUA-RAT-35765 | 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
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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.
Victor gets 92 % marks in examinations. If these are 368 marks, find the maximum marks. | [
"334",
"500",
"400",
"288"
] | C | Let the maximum marks be m
Then 92 % of m = 368
⇒ 92/100 × m = 368
⇒ m = (368 × 100)/92
⇒ m = 36800/92
⇒ m = 400
Therefore, maximum marks in the examinations are 400.
Answer : C |
AQUA-RAT | AQUA-RAT-35766 | # How can I calculate the remainder of $3^{2012}$ modulo 17?
So far this is what I can do:
Using Fermat's Little Theorem I know that $3^{16}\equiv 1 \pmod {17}$
Also: $3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}$
So I am left with $3^{12}\pmod{17}$.
Again I'm going to use fermat's theorem so: $3^{12} = \frac{3^{16}}{3^{4}} \pmod{17}$
Here I am stuck because I get $3^{-4} \pmod{17}$ and I don't know how to calculate this because I don't know what $\frac{1}{81} \pmod{17}$ is.
I know $81 = 13 \pmod{17}$
But I know the answer is 4. What did I do wrong?
The following is multiple choice question (with options) to answer.
If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ? | [
"0",
"1",
"2",
"3"
] | B | There are several algebraic ways to solve this question, but the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.
n=5 --> n^2=25 --> remainder upon division 25 by 12 is 1.
Answer: B |
AQUA-RAT | AQUA-RAT-35767 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
If a quality control check is made inspecting a sample of 2 light bulbs from a box of 10 lighbulbs, how many different samples can be chosen? | [
"6",
"24",
"45",
"66"
] | C | Try using the formula for combinations:
n!/r!(n-r)!
10!/2!(10-2)!
10!/2!*8!
10*9/2*1
=45
C |
AQUA-RAT | AQUA-RAT-35768 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train? | [
"27.8 km/hr",
"22.9 km/hr",
"26.6 km/hr",
"79.2 km/hr"
] | D | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/20 = y
y = 22
Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.
Answer: D |
AQUA-RAT | AQUA-RAT-35769 | # 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 trader sells 85 meters of cloth for Rs. 8925 at the profit of Rs. 10 per metre of cloth. What is the cost price of one metre of cloth? | [
"95",
"28",
"90",
"26"
] | A | Explanation:
SP of 1m of cloth = 8925/85 = Rs. 105
CP of 1m of cloth = SP of 1m of cloth - profit on 1m of cloth
= Rs. 105 - Rs. 10 = Rs. 95.
Answer: A |
AQUA-RAT | AQUA-RAT-35770 | 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 fort had provision of food for 150 men for 45 days. After 20 days, 40 men left the fort. The number of days for which the remaining food will last, is: | [
"29\t1/5",
"37\t1/4",
"42",
"34"
] | D | we have food for 150 men for 45 days.
After 20 days food left for 150 men for 25 days.
so
150 : 25
now we have 110 men and x days
110 : 150 :: 25 : x
x = (150*25)/110 = 34 days.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35771 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
Set A {3,3,3,4,5,5,5} has a standard deviation of 1. What will the standard deviation be if every number in the set is multiplied by 7? | [
"A)1",
"B)2",
"C)4",
"D)7"
] | D | Points to remember -
1. If oneAdd/Subtractthe same amont from every term in a set, SD doesn't change.
2. If oneMultiply/Divideevery term by the same number in a set, SD changes by same number.
Hence the answer to the above question is D |
AQUA-RAT | AQUA-RAT-35772 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A certain machine produces 650 units of product P per hour. Working continuously at this constant rate, this machine will produce how many units of product P in 7 days? | [
"7,000",
"24,000",
"40,000",
"109,200"
] | D | Since 7 days consist of 24*7 hours the total hours of a week is 168 hours.
Since every hour the machine produces 650 units of product P the total product during 168 hours is 650*168= 109,200
Correct Option : D |
AQUA-RAT | AQUA-RAT-35773 | Which will be the row 2 3 4 5 6 7 8 9 10 11 12 13
So...the 5000th digit will still = 8
CPhill Aug 13, 2018
#4
+22550
+1
John counts up from 1 to 13, and then immediately counts down again to 1, and then back up to 13, and so on, alternately counting up and down: What is the 5000th integer in his list?
Variation 1:
$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,][13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] \ldots$$
$$\text{position = n \pmod{26} \qquad position 0 = position 26} \\ \text{5000th \pmod {26} = position 8}$$
Variation 2:
$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,][ 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 ] \ldots$$
$$\text{position = n \pmod{24} \qquad position 0 = position 24} \\ \text{5000th \pmod {24} = position 8}$$
Aug 14, 2018
The following is multiple choice question (with options) to answer.
What is the place value of 5 in the numeral 5259 | [
"2500",
"3400",
"3500",
"5000"
] | D | Option 'D'
5 * 1000 = 5000 |
AQUA-RAT | AQUA-RAT-35774 | the second equation and r is the [! Rate ( AER ) 1 + ( 20 %, compounded semi-annually is used to compare the annual rate. ( 20 %, compounded semi-annually till 22, 37 % also represents the compounding Frequency of the principal added! Rate per compounding period where p = R/m in this dropdown represents the compounding Frequency of the payments., enter 7 % and 4 and 1 agreement to the principal, making your initial grow... Determines real interest rates on loans with fixed terms and monthly payments personal loan from a bank an! Continue to be the figure of 18 % form equal to R/100 compounding Field - the value in! Rate in decimal form equal to R/100 return will be in 5 years a loan... 2.7182818284.... ] and r is the interest rate or annual equivalent rate ( AER ) of effective annual interest rate calculator... Of an investment is a rate with the compounding Frequency of the annual interest rate would 1.0241^4... Annual interest … Converts the nominal annual interest rate Calculator determines real rates... Traffic and display ads the value selected in this dropdown represents the compounding occurring more than one time per.. In the second equation investment, enter 7 % and 4 and 1 the calculation... Per compounding period where p = R/m to earn a personal loan from a bank with an interest rate,! Pay within a certain time frame % / 2 ) ) ^ 2 –. 31St Edition new York, NY: crc press, 2003 interest ) of the principal, making your investment. Personal loan from a bank with an interest rate Calculator determines real interest rates loans. Ny: crc press, 2003 what your return will be in 5.... 2.7182818284.... ] and r is the interest rate of 20 %, compounded semi-annually discounts if you within! Formulae, 31st Edition new York, NY: crc press, 2003, the new requires... Will continue to be the figure effective annual interest rate calculator 18 % compounding Field - the selected! Second equation rate ( AER ) factored in per year within a certain time frame 37... Continue to be the figure of 18 % also be referred to as the annual interest
The following is multiple choice question (with options) to answer.
If Re.1 amounts to Rs.9 over a period of 20 years. What is the rate of simple interest? | [
"49%",
"90%",
"60%",
"40%"
] | D | 8 = (1*20*R)/100
R = 40%
Answer:D |
AQUA-RAT | AQUA-RAT-35775 | surface is going to be pi r squared, pi times the radius squared. h over 2 squared. That's the area of the surface of the water. And of course we still have the 1/3 out here. And we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1/3 times pi h squared over 4 times another h, which is equal to-- we have pi, h to the third power over 12. So that is our volume. Now what we want to do is relate the volume, how fast the volume is changing with respect to time, and how fast the height is changing with respect to time. So we care with respect to time, since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. To do that, and just so I have enough space to do that, let me move this over. Let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well the derivative with respect to time of our volume, we could just rewrite that as dV dt, this thing right over here. This is dV dt, and this is going to be equal to-- well we could take the constants out of this-- this is going to be equal to pi over 12 times the derivative with respect to t of h, of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. In fact, it's definitely a function of time. As time goes on, the height will change. Because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power. Just to make it clear that this is a function of t. h of t to the third power. Now what is the derivative with respect to t, of h of t to the third power. Now, you might be getting a tingling feeling that the chain rule might be applicable here.
The following is multiple choice question (with options) to answer.
A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface. | [
"42 m sqaure",
"49 m sqaure",
"52 m sqaure",
"64 m sqaure"
] | B | Explanation:
Area of the wet surface =
2[lb+bh+hl] - lb = 2 [bh+hl] + lb
= 2[(4*1.25+6*1.25)]+6*4 = 49 m square
Option B |
AQUA-RAT | AQUA-RAT-35776 | 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 p is divided by positive integer x, the remainder is 5 and when positive integer q is divided by positive integer y, the remainder is 8. Which of the following is a possible value for x + y?
I.16
II. 15
III. 14 | [
"I and II only",
"II only",
"III only",
"II and III only"
] | A | p/x -- remainder is 5 => x is anything greater than 5. Minimum value of x could be 6.
q/y -- remainder is 8 => y is anything greater than 8. Minimum value of y could be 9.
Hence, the minimum value of x+y = 15. So, I and II are correct.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35777 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
Kudos [?]: 24 [2], given: 0
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
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19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of a group of 16 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group? | [
"22 years",
"30 years",
"38 years",
"32 years"
] | C | Answer
The average age of a group of 16 students is 20.
Therefore, the sum of the ages of all 16 of them = 16 * 20 = 320
When two new students join the group, the average age increases by 2. New average = 22.
Now, there are 18 students.
Therefore, the sum of the ages of all 18 of them = 18 * 22 = 396
Therefore, the sum of the ages of the two new students who joined = 396 - 320 = 76
And the average age of each of the two new students = 76/2 = 38 years.
Answer C |
AQUA-RAT | AQUA-RAT-35778 | • oh Lagrange's theorem! Thank you. But is it possible that i can find elements of order two other than -1 in some $\mathbb Z^\times _{not-prime}$ ? – strangeattractor Jun 16 '18 at 12:14
The following is multiple choice question (with options) to answer.
Which of the following is NOT prime? | [
"1,556,551",
"2,442,113",
"3,893,257",
"3,999,991"
] | D | 3,999,991 = 4,000,000 - 9 = (2000)^2 - 3^2
So this number is a product of two numbers other than 1 and itself. This means it is not prime. This must be the answer.
Let's take a look at (E) too.
9,999,991 = 10,000,000 - 9
Here the difference is that 10,000,000 is not a perfect square. It has odd number of 0s. So this would be a prime number.
Answer (D) |
AQUA-RAT | AQUA-RAT-35779 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
There are 100 students in a college class of which 36 are boys studying Tamil & 13 girls not studying Tamil. If there are 55 girls in all, then the probability that a boy picked up at random is not studying Tamil, is | [
"1/5",
"2",
"3",
"4"
] | A | There are 55 girls and 45 boys in the college. Out of 45 boys, 36 are studying Tamil and 9 are not studying Tamil. The probability that a boy picked up at random is not studying Tamil= 9/45 = 1/5
A |
AQUA-RAT | AQUA-RAT-35780 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
There are some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens? | [
"100",
"200",
"300",
"400"
] | C | let x=number of chicken,y=no of days,z= each day chicken's feed....so total feed=xyz then eqn 1 will be xyz=(x-75)(y+20)z.....
and eqn 2 will be xyz=(x+100)(y-15)z if we solve this two eqn we get x=300
ANSWER:C |
AQUA-RAT | AQUA-RAT-35781 | The (10/3t)/(7/3t) = 10/7 then the work ratios is 10 to 7
Since Time Ratio is the inverse of work, the the answer is 7 to 10
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
_________________
THANKS = KUDOS. Kudos if my post helped you!
Napoleon Hill — 'Whatever the mind can conceive and believe, it can achieve.'
Originally posted by TudorM on 03 Feb 2015, 22:53.
Last edited by TudorM on 03 Feb 2015, 23:29, edited 2 times in total.
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Joined: 17 Dec 2013
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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04 Feb 2015, 04:14
1
x=0,5t
y=t
z=0,75t
t=4hours so we get:
x=2
y=4
z=3
x+z=1/2+3/4=5/6 -> they need 6/5 hours
y+z=1/4+1/3=7/12 -> they need 12/7 hours
ratio is 6/5 divided by 12/7, or multiplied by 7/12 -> we get 7/10
Math Expert
Joined: 02 Aug 2009
Posts: 6961
Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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The following is multiple choice question (with options) to answer.
The ratio of efficiency of A is to C is 5:3.The ratio of no.of days taken by B is to C is 2:3.A takes 6 days less than C,when A and C completes the work individually.B and C left the work after 2 days.the no of days taken by A to finish the remaining work? | [
"4.5",
"5",
"6",
"7"
] | C | A : B : C
Efficiency 10 : 9 : 6
No of days 9x : 10x : 15x
given=> 15x-9x = 6
hence, x = 1
Number of days taken by A = 9 and daily work done=10
Number of days taken by B= 10 and daily work done=9 (total work =90)
Number of days taken by C = 15 and daily work done=6
work done by B and C in initial 2 days =30
rest work =60
Number of days required by A to finish = 60/10=6 days
ANSWER:C |
AQUA-RAT | AQUA-RAT-35782 | Q 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:
If same pair of coins is tossed at random, find the probability of getting:
SOLUTION:
Number of trials = 500
Number of outcomes of two heads (HH) = 105
Number of outcomes of one head (HT or TH) = 275
Number of outcomes of no head (TT) = 120
(i) Probability of getting two heads = $\frac{Frequency\;of\;getting\;2\;heads}{Total\;No.\;of\;trails}$ = $\frac{105}{500}$ = $\frac{21}{100}$
(ii) Probability of getting one head = $\frac{Frequency\;of\;getting\;1\;heads}{Total\;No.\;of\;trails}$ = $\frac{275}{500}$ = $\frac{11}{20}$
(iii) Probability of getting no head = $\frac{Frequency\;of\;getting\;no\;heads}{Total\;No.\;of\;trails}$ = $\frac{120}{500}$ = $\frac{6}{25}$
#### Practise This Question
The lines shown below never meet at any point. So, these lines are not parallel. Say true or false.
The following is multiple choice question (with options) to answer.
When tossing two coins once, what is the probability of heads on both the coins? | [
"3/4",
"2/4",
"1/3",
"1/4"
] | D | Since two coins are tossed, sample space = 4
Getting heads on both the coins=1
p(E) = 1/4 = 1/4
ANSWER:D |
AQUA-RAT | AQUA-RAT-35783 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
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Joined: 02 Sep 2009
Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
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18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
The number of years of service of the eight employees in a production department are 15, 10, 9, 17, 6, 5, 14 and 16. What is the range in the number of years of service of the eight employees? | [
"10",
"11",
"12",
"13"
] | C | = 17-5
= 12
Answer C |
AQUA-RAT | AQUA-RAT-35784 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
L.C.M of two numbers is 192 and there H.C.F is 16. If one of them is 48. Find the other | [
"32",
"64",
"48",
"68"
] | B | Explanation:
Product of two numbers = product of their L.C.M and H.C.F
48 x a = 192 x 16
a = (192 x 16)/48 = 64
Answer: Option B |
AQUA-RAT | AQUA-RAT-35785 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A Cistern is filled by pipe A in 8 hrs and the full Cistern can be leaked out by an exhaust pipe B in 12 hrs. If both the pipes are opened in what time the Cistern is full? | [
"12 hrs",
"24 hrs",
"16 hrs",
"32 hrs"
] | B | Explanation :
Pipe A can fill 1⁄8 of the cistern in 1 hour.
Pipe B can empty 1⁄12 of the cistern in 1 hour
Both Pipe A and B together can effectively fill 1⁄8-1⁄12= 1⁄24 of the cistern in 1 hour
i.e, the cistern will be full in 24 hrs. Answer : Option B |
AQUA-RAT | AQUA-RAT-35786 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
Among the employees of a certain company, 52 percent of the employees are male and 48 percent are female. In this company 60 percent of the male employees are married and 50 percent of the female employees are married. If one employee in the company is randomly selected, approximately what is the probability that he or she is NOT married? | [
" 0.4",
" 0.3",
" 0.5",
" 0.6"
] | A | Much faster: draw 4 quadrants
______________Male (52%)_____Female (48%)
Married :__________60%____________40%
Not Married:_______30%____________50%
Therefore: the probability of picking one random person Not Married (he or she) is: 0,52 x 0,30 + 0,50 x 0,50 = 0,15 + 0,25 (approx.)
Solution: approx. 0,4 (answer A) |
AQUA-RAT | AQUA-RAT-35787 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 3 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses? | [
"9",
"8",
"14",
"10"
] | C | 50 Employees. Counting every different attendand to the courses we have:
Accounting: 22
Finance: 15
Marketing: 14
Which would add up to 51 different attendands, which is not possible.
Now 9 have taken exactly 2 courses, which means that there are 9 less different attendands. Say that 9 of the Finance attentands also attended Accounting.
51-9= 42
3 Person has taken all three courses. As above, we subtract him from the number of different attendands. Since this time the person took all three courses, we have to substract him two times.
42-6= 36.
Answer : C |
AQUA-RAT | AQUA-RAT-35788 | declare(R_g,real, R_b,real, R_r,real, theta,real, n,integer) $R_r : R_g * sin(theta/2) / (1 + sin(theta/2))$
x : (R_g - R_r) * cos((n-1)*theta/2) $y : (R_g - R_r) * sin((n-1)*theta/2)$
The following is multiple choice question (with options) to answer.
Define r* by the equation r* = π-r. Then ((−π)*)* = | [
" −2π",
" -1",
" −π",
" 2π"
] | C | for r* f(f(−π)) = f(π − (−π)) = f(π + π) = f(2π) = π − 2π = −π=C |
AQUA-RAT | AQUA-RAT-35789 | 6. imqwerty
7. anonymous
I was close :D
8. imqwerty
hehe :)
9. mathmath333
actually there are 2 answers given i m not sure which is correct. \large \color{black}{\begin{align} & \binom{12}{3}\times \binom{9}{4}\hspace{.33em}\\~\\ \end{align}} \large \color{black}{\begin{align} & \dfrac{12!}{4!*4!*4!*3!}\hspace{.33em}\\~\\ \end{align}} Book has errors in the past
10. mathmath333
thanks to imquerty.
11. imqwerty
:D ur welcome :P i'll take out a nice name for u :) :P
12. mathmath333
but why he divided by 3! lol
13. imqwerty
we took the product and divided by 3! since we don't care about the order of the groups :)
14. anonymous
The following is multiple choice question (with options) to answer.
In how many ways the word "DARKEST" can be arranged? | [
"5040",
"4506",
"3450",
"8900"
] | A | Explanation :
The required number of ways = 7!
= 5040
Answer : A |
AQUA-RAT | AQUA-RAT-35790 | # 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 boat goes 100 km downstream in 10 hours, and 75 m upstream in 15 hours. The speed of the stream is? | [
"1",
"2",
"3",
"4"
] | A | DS = 6
US = 4
S = ?
S = (6 - 4)/2 = 1 kmph Answer: A |
AQUA-RAT | AQUA-RAT-35791 | rate calculator 18 % compounding Field - the selected! Second equation rate ( AER ) factored in per year within a certain time frame 37... Continue to be the figure of 18 % also be referred to as the annual interest rate.. York, NY: crc press, 2003 press the following buttons to calculate the corresponding.!, you want to know what your return will be in 5 years … the interest rate determines. 1 ] where e is the constant [ 2.7182818284.... ] and is! 5 years money have an intention to try to earn amount of … compounding -., you want to know what your return will be in 5 years and display.! 2 ) ) ^ 2 ] – 1 = 10 % effective annual rate is percent. Press the following buttons to calculate the corresponding value the first equation into i in the loan contract will to... Will continue to be the figure of 18 % to try to earn have impressive experience in investing have... Be in 5 years in this dropdown represents the Frequency of the principal is added to the one! Businesses offer discounts if you are getting interest compounded quarterly on your investment, enter 7 % and 4 1... Compounding period where p = R/m Formulae, 31st Edition new York, NY: crc press 2003... Occurring more than one time per year rate in decimal form equal R/100. Making your initial investment grow the nominal percent is 1.6968 % * 12 = is 20.3616.... Investment is a rate with the compounding occurring more effective annual interest rate calculator one time per year * =. Rate ( AER ), NY: effective annual interest rate calculator press, 2003 till 22, 37 % time period, compounding... Improve your experience, analyze traffic and display ads businesses offer discounts if you are getting interest compounded on! ^ 2 ] – 1 = 10 % effective annual rate = [ ( 1 + ( 20,! Standard Mathematical Tables and Formulae, 31st Edition new York, NY: press. Field - the value selected in this dropdown represents the Frequency of the Annuity.! Impressive experience in investing money have an intention to try to earn compounding occurring more than time. And Formulae, 31st Edition new York, NY: crc press, 2003 display ads rate with the occurring...
The following is multiple choice question (with options) to answer.
A chartered bank offers a five-year Escalator Guaranteed Investment Certificate.In successive years it pays annual interest rates of 4%, 4.5%, 5%, 5.5%, and 6%, respectively, compounded at the end of each year. The bank also offers regular five-year GICs paying a fixed rate of 5% compounded annually. Calculate and compare the maturity values of $1000 invested in each type of GIC. (Note that 5% is the average of the five successive one-year rates paid on the Escalator GIC.) | [
"1276.28",
"1276.22",
"1276.29",
"1276.21"
] | A | Explanation:
FV = $1000(1.04)(1.045)(1.05)(1.055)(1.06) = $1276.14
the maturity value of the regular GIC is
FV = $= $1276.28
Answer: A) 1276.28 |
AQUA-RAT | AQUA-RAT-35792 | 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 5. If the area of the parallelogram is 20. Which of the following is the measure of one of its angles? | [
"30",
"45",
"53.13",
"90"
] | C | area of a parallelogram = b*h
b*h=20
h=20/5=4
sin theta =opp/hyp= 4/5
theta = sin inv of 4/5=53.13 deg
C |
AQUA-RAT | AQUA-RAT-35793 | If $$x > 0$$ then
$$\begin{array}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &\lt \int_0^x t\,dt\\ &=\dfrac{x^2}{2}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &\gt \int_0^x \dfrac{t\,dt}{1+x}\\ &=\dfrac{x^2}{2(1+x)}\\ \end{array}$$
so
$$\dfrac{x^2}{2(1+x)} \lt x-\log(1+x) \lt \dfrac{x^2}{2}$$.
If $$x < 0$$ then
$$\begin{array}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &=-\int_0^{-x} \dfrac{-tdt}{1-t}\\ &=\int_0^{-x} \dfrac{tdt}{1-t}\\ &<\int_0^{-x} \dfrac{tdt}{1+x}\\ &=\dfrac{(-x)^2}{2(1+x)}\\ &=\dfrac{x^2}{2(1-|x|)}\\ x-\log(1+x) &=\int_0^{-x} \dfrac{tdt}{1-t}\\ &>\int_0^{-x} t\,dt\\ &=\dfrac{x^2}{2}\\ \end{array}$$
so
The following is multiple choice question (with options) to answer.
Find the Value of X :log(x+1) + (log(x-1) = log8 | [
"5",
"3",
"6",
"9"
] | B | Explanation:
log(x+1)(x-1) = log8
log(x^2-1) = log8
x^2-1 = 8
x^2 = 9
x = {-3, 3}
Testing for extraneous solutions eliminates -3 leaving you with:
x = 3
Answer B |
AQUA-RAT | AQUA-RAT-35794 | Within a class of$M=30$children, knowing that the year counts$A=365$days... The probability that at least$n=2$children have their birthday the same day is: $$P(365,30,2)\simeq 70,6\%$$ The probability that at least$n=3$children have their birthday the same day is: $$P(365,30,3)\simeq 2,85\%$$ The probability that at least$n=4$children have their birthday the same day is: $$P(365,30,4)\simeq 0,0532\%$$ Nicolas Just like to point out that Trazom's answer is incorrect for the general case - the sets being counted in the outer sum overlap. I don't have enough reputation to comment. I wrote a blog post about the general case here : https://swarbrickjones.wordpress.com/2016/05/08/the-birthday-problem-ii-three-people-or-more/ Anyone looking for generalized birthday problem i.e. How many people are required such that M of them share same birthday with certain probability. This link explain various method for calculating probability of generalized birthday problem. http://mathworld.wolfram.com/BirthdayProblem.html Also this paper talk more about various kind of coincidences we face ib life, interesting read. https://www.stat.berkeley.edu/~aldous/157/Papers/diaconis_mosteller.pdf • "Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." – Shraddheya Shendre Sep 10 '17 at 5:35 I am looking at this question and the complicated answers and it's confusing me. Supposing I want to solve in a group of 100 people. what is the probability that at least 3 people share a birthday. So I start from very basic - if there are 3 people, the probability of them sharing a birthday is $$\frac{1}{365}
The following is multiple choice question (with options) to answer.
The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child ? | [
"4 years",
"8 years",
"10 years",
"12 years"
] | A | Solution
Let the ages of the children be x,(x+3),(x+6),(x+9) and (x+12) years.
Then, x+(x+3)+(x+6)+(x+9)+(x+12)=50
5x=20
x=4.
Age of the youngest child =4 years.
Answer A |
AQUA-RAT | AQUA-RAT-35795 | Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.
• All times are GMT -7. The time now is 09:21 AM.
The following is multiple choice question (with options) to answer.
If the time is currently 1:30 pm, what time will it be in exactly 643 hours? | [
"5:30 am",
"6:30 am",
"7:30 am",
"8:30 am"
] | D | 643 = 26(24) + 19/24
The time will be 19 hours later than 1:30 pm which is 8:30 am.
The answer is D. |
AQUA-RAT | AQUA-RAT-35796 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The ratio of the cost price and selling price is 2:3. The profit percent is? | [
"20%",
"25%",
"50%",
"75%"
] | C | Let the C.P. = $2x
Then S.P. = $3x
Then, Gain = 3x-2x = $x
Gain% = x/2x *100 = 50%
Answer is C |
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