source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37097 | Case 4: One digit is used four times, while two other digits are used once each.
Subcase 1: The leading digit is repeated.
We have nine ways of choosing the leading digit and $\binom{5}{3}$ ways of choosing the other three positions in which it appears. We have nine choices for the leftmost open position and eight choices for the remaining position. $$9 \cdot \binom{5}{3} \cdot 9 \cdot 8 = 6480$$
Subcase 2: The leading digit is not repeated.
We have nine ways of choosing the leading digit. We have nine ways of choosing the repeated digit and $\binom{5}{4}$ ways of selecting four of the five open positions in which to place it. We have eight ways of filling the remaining open position. $$9 \cdot 9 \cdot \binom{5}{4} \cdot 8 = 3240$$
That gives a total of $$9 + 405 + 81 + 810 + 405 + 6480 + 3240 = 11,430$$ excluded cases.
Hence, there are $$900,000 - 11,430 = 888,570$$ six-digit positive integers in which no digit appears more than three times.
This is neatly handled using exponential generating functions. Assuming first that we are allowed to have 0 as the first digit (e.g. we're talking about license plates or lock combinations): each of $10$ digits can occur up to $3$ times, and the order of the symbols matters. The answer is $$\left[\frac{x^6}{6!}\right] \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^{10} = 987,300.$$
The following is multiple choice question (with options) to answer.
A license plate in the country Kerrania consists of four digits followed by two letters. The letters A, B, and C are used only by government vehicles while the letters D through Z are used by non-government vehicles. Kerrania's intelligence agency has recently captured a message from the country Gonzalia indicating that an electronic transmitter has been installed in a Kerrania government vehicle with a license plate starting with 79. If it takes the police 12 minutes to inspect each vehicle, what is the probability that the police will find the transmitter within three hours? | [
" 18/79",
" 16/900",
" 1/25",
" 1/50"
] | B | Everything is correct except that you assumed the letters can't be repeated. It can be repeated.
AA
BB
CC
AB, AC, BC, BA, CA, CB.
Thus; total government vehicles = (10)^2*(3)^2 = 100*9 = 900
Vehicles inspected within 3 hours = 60*3/12 = 16
P = 16/900
Ans:B |
AQUA-RAT | AQUA-RAT-37098 | Transcript
TimeTranscript
00:00 - 00:59so this is the question 1 ka manufacture compile data that the indicated mileage decrease in the number of the miles between driven between recommended serving increased the manufacturer use the equation Y is equal to minus one upon 200 X + 35 to model the data based on the information how many miles per gallon could be expected if the 34900 miles between servicing this is the graph which had been drawn using the best fit line from the scatter plots and the line equation of the line is given out here and it is asking for thirty four thousand miles over the combined with the recommended servicing recommended servicing MI is the x-axis and gas mileage in the wire we have been given the value of the x-axis and we have to calculate simultaneous simultaneous value of Dubai actors using this equation so
01:00 - 01:59Y is equal to minus x upon 200 f-35 X equal to 34000 ok so be divided out here 34000 / actually 3448 3430 4030 430 400/200 35 - 8 - 6235 we have to target for 3434 100 and 200 + 35 is equal to - 1700 gets cancelled 2134 also gets cancelled 17 times its - 17 + 35 it is equal to 18
02:00 - 02:5997035 equal to 18 18 mile gal idhar answer 18 miles per gallon thank you
The following is multiple choice question (with options) to answer.
Dan’s car gets 32 miles per gallon. If gas costs $4/gallon, then how many miles can Dan’s car go on $54 of gas? | [
"236",
"354",
"432",
"512"
] | C | 54/4 = 13.5 gallons
13.5*32 = 432 miles
The answer is C. |
AQUA-RAT | AQUA-RAT-37099 | # Difference between revisions of "2019 AMC 10A Problems/Problem 23"
## Problem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
## Solution 1
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$
The following is multiple choice question (with options) to answer.
Two assembly line inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. If Lauren inspects every fifth widget, starting with the fifth, and Steven inspects every third, starting with the third, how many of the 95 widgets produced in the first hour of operation are not inspected by either inspector? | [
"50",
"43",
"45",
"55"
] | A | Widgets inspected by Lauren: ((95-5)/5)+1=18+1=19
Widgets inspected by Steven: ((93-3)/3)+1 =31+1 =32
Widgets inspected by both: ((90/15)+1 =6
Total : 19+32 -6=45
hence, widgets not inspected: 95-45=50
Option A |
AQUA-RAT | AQUA-RAT-37100 | # Math Help - Distance (Word Problem)
1. ## Distance (Word Problem)
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
2. Originally Posted by Beastkun
Hello I currently am stumped on a word problem that appeared on my homework, it is as follows...
"Driving from Dallas towards Memphis, JoJo averages 50mph. She figured that if she had averaged 60mph the driving time would have decreased by 3 hours. How far did she drive?"
I thought this would have been a simple D=R/T problem.. but somewhere I am not plugging the variables in correctly.. or setting the problem up the correct way. Please help.. this assignment is due in a few hours. Thanks!
1. Let d denote the covered distance and t the elapsed time. Then you know:
$\dfrac dt = 50$
and
$\dfrac{d}{t-3}=60$
2. Solve for d.
Spoiler:
You should come out with d = 900
3. I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
4. Originally Posted by Beastkun
I'm drawing a huge blank on how to make the two work together in one equation. The instructions note that I should set up and write a "rational equation", then solve AND answer the problem.
Unless I'm missing something.. I'm unsure of how I should incorporate the two to solve them.
Substitute d = 50t from the first equation into the second equation and solve for t and hence d.
5. What prove it is doing is simultaneous equations.
The following is multiple choice question (with options) to answer.
Bob bikes to school every day at a steady rate of a miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home? | [
"(a + y) / t",
"2(a + t) / ay",
"2ayt / (a + y)",
"2(a + y + t) / ay"
] | C | If we choose for d (distance): 10 miles, for a 10 and for y 5.
t would be 90 minutes or 1,5 hours.
If I try this for answer choice C it fits. |
AQUA-RAT | AQUA-RAT-37101 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A company has a job to prepare certain number of cans and there are three machines A, B and C for this job. A can complete the job in 2 days, B can complete the job in 5 days and C can complete the job in 7 days. How many days the company will it take to complete the job if all the machines are used simultaneously? | [
"71/59 days",
"70/59 days",
"3 days",
"12days"
] | B | Let the total number of cans to be prepared be 70.
The number of cans prepared by A in 1 day = 35.
The number of cans prepared by B in 1 day = 14.
The number of cans prepared by C in 1 day = 10.
Thus, the total number of cans that can be prepared by all the machines working simultaneously in a single day = 59.
Therefore, the number of days taken to complete the whole work = 70/59 days.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37102 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
An error 4% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? | [
"4.05%",
"4.02%",
"8.16%",
"3%"
] | C | Percentage error in calculated area
=(4+4+(4×4)/100)%
=8.16%
ANSWER:C |
AQUA-RAT | AQUA-RAT-37103 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of five numbers is 60. If one number is excluded, the average becomes 65. The excluded number is? | [
"40",
"6",
"58",
"35"
] | A | Excluded number
= (60 * 5) - (65 * 4)
= 300 - 260 = 40
Answer:A |
AQUA-RAT | AQUA-RAT-37104 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
Anil invested a sum of money at a certain rate of simple interest for a period of five years. Had he invested the sum for a period of eight years for the same rate, the total intrest earned by him would have been sixty percent more than the earlier interest amount. Find the rate of interest p.a. | [
"5352",
"5355",
"2789",
"2689"
] | B | Let the sum lent by Manoj to Ramu be Rs.P.
Amount gained by Manoj = P. 3.9 /100 = 3450.3.6/100
= Rs.824.85
27P = 62100
= 82485
P = (82485 + 62100)/27 = 3055 + 2300
= Rs.5355
Answer: B |
AQUA-RAT | AQUA-RAT-37105 | The answer is: 133,320.
List the $4! = 24$ permutations of the four digits
,. . and consider their sum.
. . $\begin{array}{cc}
&2468 \\ &2486 \\ &2648 \\ &2684 \\ &2846 \\ &2864 \\ &\vdots \\
&8246 \\ &8264 \\ &8426 \\ &8462 \\ &8624 \\ + & 8642 \\ \hline
\end{array}$
We find that each column has: six 2's, six 4's, six 6's, six 8's.
The total of each column is: . $6\!\cdot\!2 + 6\!\cdot\!4 + 6\!\cdot\!6 + 6\!\cdot\!8 \:=\:120$
Hence, the addition has the form:
. . $\begin{array}{cccccc}
&&& 1 & 2 & 0 \\ && 1 & 2 & 0 \\ & 1 & 2 & 0 \\ 1 & 2 & 0 \\ \hline 1 & 3 & 3 & 3 & 2 & 0 \end{array}$
Therefore, the sum is: . $133,\!320$
Edit: Plato beat me to it . . . *sigh*
.
The following is multiple choice question (with options) to answer.
What is the 100th digit of (3!*5! + 4!*3!)/2? | [
"4",
"3",
"2",
"7"
] | A | (3!*5! + 4!*3!)/2
=3!(5! + 4!)/2
=6 (120+24)/2
=432
100th digit of the above product will be equal to 4
Answer A |
AQUA-RAT | AQUA-RAT-37106 | One of you all sent a fairly interesting problem, so I thought I would work it out. The problem is I have a group of 30 people, so 30 people in a room. They're randomly selected 30 people. And the question is what is the probability that at least 2 people have the same birthday? This is kind of a fun question because that's the size of a lot of classrooms. What's the probability that at least someone in the classroom shares a birthday with someone else in the classroom? That's a good way to phrase as well. This is the same thing as saying, what is the probability that someone shares with at least someone else. They could share it with 2 other people or 4 other people in the birthday. And at first this problem seems really hard because there's a lot of circumstances that makes this true. I could have exactly 2 people have the same birthday. I could have exactly 3 people have the same birthday. I could have exactly 29 people have the same birthday and all of these make this true, so do I add the probability of each of those circumstances? And then add them up and then that becomes really hard. And then I would have to say, OK, whose birthdays and I comparing? And I would have to do combinations. It becomes a really difficult problem unless you make kind of one very simplifying take on the problem. This is the opposite of-- well let me draw the probability space. Let's say that this is all of the outcomes. Let me draw it with a thicker line. So let's say that's all of the outcomes of my probability space. So that's 100% of the outcomes. We want to know-- let me draw it in a color that won't be offensive to you. That doesn't look that great, but anyway. Let's say that this is the probability, this area right here-- and I don't know how big it really is, we'll figure it out. Let's say that this is the probability that someone shares a birthday with at least someone else. What's this area over here? What's this green area? Well, that means if these are all the cases where someone shares a birthday with someone else, these are all the area where no one shares a birthday with anyone. Or you could say, all 30 people have different birthdays. This is what
The following is multiple choice question (with options) to answer.
From a nine-member dance group, five will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the nine members, what is the probability that both will be chosen to volunteer? | [
"a) 1/21",
"b) 35/2",
"c) 4/21",
"d) 2/7"
] | B | Total number of ways to choose 5 out of 9 = 9C5 = 144.
Number of ways to choose 2 (any but Kori and Jason) out of 7 (excluding Kori and Jason) = 7C2 = 2520.
P = favorable/total = 2520/144 = 35/2.
Answer: B. |
AQUA-RAT | AQUA-RAT-37107 | => $$\frac{74}{100} \times x$$ = 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.
A began business with Rs.45000 and was joined afterwards by B with Rs.5400. When did B join if the profits at the end of the year were divided in the ratio of 2:1? | [
"6",
"9",
"7",
"5"
] | C | 45*12 : 54*x = 2: 1
x = 5
12 -5 = 7
Answer: C |
AQUA-RAT | AQUA-RAT-37108 | So:
W_2(L,F) = 0, If 2F+1 > L
W_2(L,F) = L - 2F, If 2F+1 <= L
We can compute W_2(L), for all valid F, by summing W_2(L,F) for F ranging from 1 to (L-1)/2:
...skipping the algebra...
If L is odd, W_2(L) = (L^2 - 2L + 1) / 4
If L is even, W_2(L) = (L^2 - 2L) / 4
Now, we can use this result in calculating the three-number problem:
If we have a first number F, then our second number, S, and third number, T, must satisfy: F < S < T F + S + T <= L
Assuming we have chosen an F small enough that there are S and T that can satisfy F + S + T <= L, then
W_3(L,F) = W_2(L - 3F)
W_3(L,F) = ((L-3F)^2-2*(L-3F)+1)/4, if 3F-L is odd
W_3(L,F) = ((L-3F)^2-2*(L-3F))/4, if 3F-L is even
Our upper limit for F is F + (F + 1) + (F + 2) <= L or F <= (L - 3) / 3
Finally, we can compute W_3(L) for all valid F, by summing W_3(L,F) for F ranging from 1 to I = Int((L-3)/3), where INT() rounds down to the nearest integer.
...skipping even more algebra...
If (L-3)/3 is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I)/8
The following is multiple choice question (with options) to answer.
W is an even integer greater than 300,000 and smaller than 1,000,000. How many numbers can W be? | [
"300,000",
"349,999",
"350,000",
"399,999"
] | B | W 1,000,000-300,000=700,000 integers
700,000/2= 350,000 even integers.
350,000+1 inclusive. But since 1,000,000 and 300,000 are not included.
350,001-2=349,999
B |
AQUA-RAT | AQUA-RAT-37109 | 5C2 * 4!(5C2 for selecting 2 out of 5 places for E & rest can be arranged in 4! ways.)
Ans with this approach 240 which is correct in this case.
Now the question discussed above:
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
Going as per the above approach it comes
8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
But in this case answer doesn't match.
Please if you could explain the flaw in my approach will be very helpful.
Cheers.
_________________
Sailing through rough waters. Stars & few kudos will steer me past.
Re: In how many ways can the letters of the word PERMUTATIONS be &nbs [#permalink] 09 May 2018, 01:12
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The following is multiple choice question (with options) to answer.
If the letters of the word PLACE are arranged taken all at a time, find how many do not start with AE | [
"110",
"114",
"118",
"120"
] | B | Total no'of arrangements 5P5 = 5! = 120
no'of arrangements start with AE = 1 × 6 = 6
no'of arrangements which do not start with AE = 120 - 6 = 114.
B) |
AQUA-RAT | AQUA-RAT-37110 | python, game, python-2.x
m_write(players)
# Tally up the votes for each player
votes = [p.vote for p in players]
# Shuffle to eliminate max() bias
random.shuffle(votes)
# The most voted player is lynched, with ties broken randomly
lynched = max(votes, key=votes.count)
if lynched is not None:
log('The town has killed {}!'.format(lynched.name))
log('They were {}.'.format(lynched.get_role()))
for p in players:
p.add_message("""\
The town has killed {}!
They were {}.""".format(lynched.name, lynched.get_role))
players, mafia, cop, doctor = kill(lynched, players, mafia, cop, doctor)
else:
log('The town opted to lynch no one today.')
for p in players:
p.add_message('The town opted to lynch no one today.')
m_write(players)
for p in players:
execute(p)
p.vote = None
# Don't go to night if a win condition's been met.
if not mafia or (len(players) - len(mafia)) <= len(mafia):
break
# Day ends, night begins
# MAFIA NIGHT ACTION
# Each mafioso votes for a victim. The most voted player is then killed,
# unless saved that night by the doctor.
for m in mafia:
m.add_message('It is night. Vote for a victim.')
m_write(mafia)
victim_votes = []
for m in mafia:
try:
execute(m)
victim_votes.append(name_to_player[m_read(m)])
log('{} votes to kill {}.'.format(m.name, victim_votes[-1].name))
except:
# Vote to kill no one on invalid input
victim_votes.append(None)
log(m.name + ' votes to kill no one.')
# Shuffle to eliminate max() bias
random.shuffle(victim_votes)
The following is multiple choice question (with options) to answer.
At a recent small town election for mayor a total of 967 votes were cast for the four candidates, the winner exceeding his opponents by 53, 79 and 105 votes, respectively. How many votes were cast for the candidate in fourth place? | [
"134",
"178",
"196",
"166"
] | C | The number of votes the winning candidate received was
967 + 53 + 79 + 105 /4= 301.
The fourth place received 301 – 105 = 196.
C |
AQUA-RAT | AQUA-RAT-37111 | 5. ## Re: How to solve for c, given the roots of the equation?
Originally Posted by bartholomew
How do I find the value of c for the question below?
The roots of the equation $x^2 + 6x + c = 0$ are k and k-1. Find the value of c.
Here is a third way.
The sum of the roots: $k+(k-1)=-6.$
The product of the roots: $k(k-1)=c.$
The following is multiple choice question (with options) to answer.
Find the value of x from the below equation? : x^2+5x+4 = 0 | [
"1",
"-3",
"3",
"2"
] | B | a = 1, b = 5, c = 4
x1,2 = (-5 ± √(5^2 - 4×1×4)) / (2×1) = (-5 ± √(25-16)) / 2 = (-5 ± 3) / 2
x1 = (-5 + 3)/2 = -2/2 = -1
x2 = (-5 - 1)/2 = -6/2 = -3
B |
AQUA-RAT | AQUA-RAT-37112 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. 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.
In a sequence of 39 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence? | [
"2",
"-2",
"0",
"18"
] | D | Which term is the greatest? The first or the last? It is given to you that every term is 7 less than the previous term. Hence as you go on, your terms keep becoming smaller and smaller. The first term is the greatest term.
An = 281 + (39- 1)*(-7)
An = 281 - 266 = 15
D |
AQUA-RAT | AQUA-RAT-37113 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Arun borrowed a certain sum from Manish at a certain rate of simple interest for 2 years. He lent this sum to Sunil at the same rate of interest compounded annually for the same period. At the end of two years, he received Rs. 2400 as compound interest but paid Rs. 2000 only as simple interest. Find the rate of interest. | [
"20%",
"10%",
"40%",
"50%"
] | C | Explanation:
Let the sum be x
Simple interest on x for 2 years = Rs.2000
Simple interest = PRT/100
2000 = x × R × 2100
⇒ xR = 100000 --- (1)
Compound Interest on x for 2 years = 2400
P(1+R/100)T−P=2400
x(1+R/100)2 − x = 2400
x(1+2R/100 + R2/10000) − x = 2400
x(2R/100 + R2/10000) = 2400
2xR/100 + xR2/10000 = 2400 --- (2)
Substituting the value of xR from (1) in (2) ,we get
(2 × 100000)/100 + (100000×R)/10000 = 2400
2000 + 10R = 2400
10R = 400R = 40%
Answer: Option C |
AQUA-RAT | AQUA-RAT-37114 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man sells a horse for Rs.600 and loses something, if he had sold it for Rs.720 , his gain would have been double the former loss. Find the cost price of the horse? | [
"27",
"98",
"27",
"40"
] | D | CP = SP + 1CP = SP - g
600 + x = 720 - 2x
3x = 120 => x = 40
Answer:D |
AQUA-RAT | AQUA-RAT-37115 | 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.
In what ratio must a grocer mix rice worth Rs.60 a kg and Rs.65 a kg.So that by selling the mixture at Rs. 68.20 a kg, He may gain 10%? | [
"1:2",
"3:2",
"4:5",
"1:3"
] | B | S.P of 1kg mix = RS.68.20,
Gain = 10% C.P of 1kg mix =Rs.(100/110 x 68.20) =Rs.62 = 3 : 2
ANSWER B |
AQUA-RAT | AQUA-RAT-37116 | ### 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 enter into partnership with capitals in the ratio 3 : 4. At the end of 10 months A withdraws,and the profits now are divided in the ratio of 5 : 4. Find how long B remained in the business? | [
"9 months",
"8 months",
"7 months",
"6 months"
] | D | Initially A’s investment = 3x and B’s investment = 4x
Let B remain in the business for ‘n’ months.
⇒ 3x × 10 : 4x × n = 5 : 4
∴ 3x × 10 × 4 = 4x × n × 5
⇒ n = 6
Answer D |
AQUA-RAT | AQUA-RAT-37117 | # 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.
A group of hikers is planning a trip that will take them up a mountain using one route and back down using another route. They plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up, but the time each route will take is the same. If they will go up the mountain at a rate of 6 miles per day and it will take them two days, how many miles long is the route down the mountain? | [
"16",
"17",
"18",
"19"
] | C | On the way down, the rate is 1.5*6 = 9 miles per day.
The distance of the route down the mountain is 2*9 = 18 miles.
The answer is C. |
AQUA-RAT | AQUA-RAT-37118 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
Peter is a trail runner who decides to take a day off work to run up and down a local mountain. He runs uphill at an average speed of 4 miles per hour and returns along the same route at an average speed of 6 miles per hour. Of the following, which is the closest to his average speed, in miles per hour, for the trip up and down the mountain? | [
" 4.8",
" 5.8",
" 6.0",
" 6.3"
] | A | to calculate average of 2 speeds a and b when distance is constant
Formula - 2ab/(a+b)
Here Peter's uphill speed = 4 mph
Peter's downhill speed = 6 mph
2*4*6/(4+6) = 48/10 = 4.8
Correct answer - A |
AQUA-RAT | AQUA-RAT-37119 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
Calculate the different number of ways 7 boys and 2 girls can sit on a bench? | [
"362881",
"362880",
"311880",
"362280"
] | B | npn = n!
9p9 = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362880
B |
AQUA-RAT | AQUA-RAT-37120 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
The population of a town increased from 1,75,000 to 2,10,000 in a decade. The average percent increase of population per year is | [
"2%",
"5%",
"6%",
"8.75%"
] | A | Solution
Increase in 10 years = (210000 - 175000)
= 35000.
Increase% = (35000/175000×100)%
= 20%.
Required average = (20/10)%
= 2%.
Answer A |
AQUA-RAT | AQUA-RAT-37121 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
Mysoon collects glass ornaments. Ten more than 1/6 of the ornaments in her collection are handmade, and 1/2 of the handmade ornaments are antiques. If 1/6 of the ornaments in her collection are handmade antiques, how many ornaments are in her collection? | [
"36",
"60",
"108",
"60"
] | D | The number of ornaments = a
Ten more than 1/6 of the ornaments in her collection are handmade => Handmade = 10+a/6
1/2 of the handmade ornaments are antiques => Handmade ornaments = 1/2*(10+a/6) = 5 + a/12
1/6 of the ornaments in her collection are handmade antiques => Handmade ornaments = a/6
=> 5 + a/12 = a/6 => a = 60
Ans: D |
AQUA-RAT | AQUA-RAT-37122 | the largest and smallest integers in the group. This method should take three decimal arguments, and return the smallest of the three. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. It is similar to modulus – ‘%’ operator of C/C++ language. Note that this is not the modulo * operation (the result can be negative). Print its smallest integer divisor greater than 1. Find the least common multiple and greatest common divisor of the given three numbers : Find the least common multiple and the greatest common divisor of numbers 12, 24, 48, 192 and 288. LCM(Least Common Multiple) is the smallest positive number which is divisble by both the numbers. The macro below does the. During the reconstruction of the railway tracks 40 meter long pieces of rails were replaced by 15 meter long pieces. (For example: 7/3 = 3 and 10/2 = 5). Dynamically Calculate Prime Numbers. The smallest divisor (other than 1) of a composite number is a/an a) odd number b) even number c) prime number d) composite number. Thus the smallest number we can represent would be: 1E(-(MIN_EXP-digits-1)*4), eg, for digits=5, MIN_EXP=-32767, that would be 1e-131092. For example divisors of 6 are 1, 2, 3 and 6, so divisor_sum should return 12. gcd(a, b) = 5 3 · 7 2 = 6125. On a number line, we get hundredths by simply dividing each interval of one-tenth into 10 new parts. Example: The smallest weird number is 70. Write a Java program that computes the absolute value of the product of 2 numbers given by the user. Here dividend is 205 and divisor is 2 therefore remainder is 1. The last divisor will be the HCF of given numbers. In particular, we have added an entirely new Ch. The SMALL function will automatically ignore TRUE and FALSE values, so the result will be the nth smallest value from the set of actual numbers in the array. In this tutorial we will write couple of different Java programs to find out the GCD of two numbers. The first perfect number is 6, because
The following is multiple choice question (with options) to answer.
The smallest 4 digit prime number is? | [
"A)101",
"B)1003",
"C)109",
"D)113"
] | B | The smallest 3-digit number is 1000, which is divisible by 2.
1000 is not a prime number.
1003< 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
1003 is a prime number.
Hence 1003 is the smallest 4-digit prime number.
B) |
AQUA-RAT | AQUA-RAT-37123 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19, 29 and 34 respectively is? | [
"1994",
"1494",
"1349",
"1496"
] | C | LCM = 1400
1400 - 6 = 1394
ANSWER:C |
AQUA-RAT | AQUA-RAT-37124 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
a bank was to give a loan out to a newly married couple for 800,000 with no interest for 2 years and raise the price to 900,000 after 2 years. what is the interest rate? | [
"17%",
"13%",
"12%",
"7%"
] | D | for 0 years = (900,000- 800,000) = 100,000.
for 2 years = (100,000 x 2)/2= 100,000
Principal = (800,000 - 100,000) = 700,000.
Hence, rate = (100 x 100,000)/(700,000x2) = 7% ANSWER: D |
AQUA-RAT | AQUA-RAT-37125 | homework-and-exercises, kinematics
Title: Calculating how long does a hare remain stationary in a race I'm solving book exercises - there's a classic problem involving a hare and a tortoise. This is what I did, but apparently the final answer is wrong.
Basically it's a 1000m race, both animals have a constant speed (hare at 8m/s and tortoise at 0.2m/s). The hare runs 800m and then stops to tease the tortoise. Then, at some point in the future, the hare resumes and both animals finish at the same time.
(a) How far is the tortoise from the finish line when the hare
resumes the race?
Well, since both animals finish at the same time, I just need to find out how long it takes the hare to finish the remaining 200m and then check how much distance could the tortoise cover in that time.
So
$$\frac{200\text{m}}{8\text{m/s}} = 25\text{s}$$
Meaning that the tortoise could cover barely
$$25\text{s}\cdot0.2\text{m/s} = 5\text{m}$$
This appears to be correct according to the book.
(b) For how long in time was the hare stationary?
The following is multiple choice question (with options) to answer.
A horse chases a pony 8 hours after the pony runs. Horse takes 12 hours to reach the pony. If the average speed of the horse is 320 kmph, what s the average speed of the pony? | [
"182kmph",
"192kmph",
"193kmph",
"196kmph"
] | B | pony take 20 hours and horse take 12 hours...then Distance chased by them is 320*12.so speed of pony is (320*12)/20=192kmph.
ANSWER IS B |
AQUA-RAT | AQUA-RAT-37126 | investment Banking Course, Download Valuation... As are bonds less than 365 days, most developed countries use simple interest is bonds! Is 1-Feb-2020, and investors collect returns when the bond “ days maturity... The market price quarterly, or the coupon Equivalent rate ( CER.. For calculating return on investment and an investment that is taxed bankruptcy and liquidates... Second part is used to calculate the annual yield of Second Govt basis that incorporates! 42 every year for calculating return on investment and the yield to calculate bond Equivalent yield for company... Thus 11 % multiplied by two, which comes out to 22 % have to understand this! ( 5 ) /95 } * { 1.520833 } ], bond Equivalent yield formula be calculated using the function! 365/180 } ], bond Equivalent yield formula on how to calculate the bond happens! Is Rs value ) to pay ( coupon rated ) is an calculation. From an annual-pay bond can not be directly compared to find out the bond Equivalent formula... That of bond or call premium of years until maturity or until call until. Calculate current yield calculation, as are bonds can be calculated using following! 5 ) /95 } * { 1.520833 } ], bond Equivalent formula. To yield to calculate EAY as follows: EAY bond equivalent yield formula 1.0253 ( 365/90 ) = %... Different price and tenure discount and do not pay interest at all NCD, the to. /.75 ) = 10.66 % * 100 1 as the current market price the... Until maturity or until put is exercised Contents ) or bonds ( fixed income securities, which sold! Allows the investor to calculate the annualized yield of Second Govt also some bonds, do not annual. Issue, we must know the bond Equivalent yield formula = ( Face value Rs!
The following is multiple choice question (with options) to answer.
How much time will take for an amount of Rs. 480 to yield Rs. 81 as interest at 4.5% per annum of simple interest? | [
"8 years",
"4 years",
"3 years 9 months",
"9 years"
] | C | Time = (100 * 81) / (480 * 4.5)
= 3 years 9 months
Answer: C |
AQUA-RAT | AQUA-RAT-37127 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invests Rs.2000, Rs.3000 and Rs.4000 in a business. After one year A removed his money; B and C continued the business for one more year. If the net profit after 2 years be Rs.3200, then A's share in the profit is? | [
"388",
"299",
"266",
"400"
] | D | 2*12 : 3*12 : 4*24
1: 3: 4
1/8 * 3200 = 400.Answer: D |
AQUA-RAT | AQUA-RAT-37128 | algorithm, graph, f#
77 112,3500 160,105 189,5702 191,5135 124,8896 198,5081 19,7013 73,2438 63,5873 129,2337 11,815 133,2481 192,561 32,1689 50,5220 87,7040 25,2729 65,4069 106,9161 153,4483 56,7923 172,4771 13,8638
78 10,3771 68,7924 12,5753 30,5469 158,6367 122,6207 27,3066 116,2732 41,63 72,5839 161,6310 4,8058 104,1377 83,3955 29,8190 98,6603 154,8423 137,1910 135,6919 73,212 145,7244
79 141,7918 101,3205 165,3768 96,3059 119,4117 152,6519 57,2330 42,9404 166,8726 161,8395 30,503 89,5169 134,5792 117,9043 129,7314 43,5052 109,9677 58,2365 44,9988 167,820 193,7737 194,5784
80 36,2032 84,4645 1,982 115,1417 151,6728 112,5208 51,7589 152,9606 113,917 18,5252 121,2257 75,2187 57,8722 133,7217 179,7729 119,108 66,2902 40,9657 97,7213 172,7715 89,7224 19,62 46,7968 21,9884
81 115,2608 197,5540 97,8866 101,4493 64,9744 11,7299 71,1711 109,2519 136,1409 39,7400 75,2609 142,424 141,4032 183,3061 184,4485 95,7627 5,2469 143,9810 65,3054 89,6124
The following is multiple choice question (with options) to answer.
195,383,575,763,955,___ | [
"1153",
"1143",
"1243",
"1343"
] | B | 19*1 = 19 5
19*2 = 38 3
19*3 = 57 5
19*4 = 76 3
19*5 = 95 5
19*6 = 114 3
so ans=1143
ANSWER:B |
AQUA-RAT | AQUA-RAT-37129 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
Find the number of ways of arranging the letters of the word "MATERIAL" such that all the vowels in the word are to come together? | [
"1446",
"1440",
"1444",
"1441"
] | B | In the word, "MATERIAL" there are three vowels A, I, E.
If all the vowels are together, the arrangement is MTRL'AAEI'.
Consider AAEI as one unit. The arrangement is as follows.
M T R L A A E I
The above 5 items can be arranged in 5! ways and AAEI can be arranged among themselves in 4!/2! ways.
Number of required ways of arranging the above letters = 5! * 4!/2!
= (120 * 24)/2 = 1440 ways.
Answer:B |
AQUA-RAT | AQUA-RAT-37130 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
In a manufacturing plant, it takes 36 machines 8 hours of continuous work to fill 8 standard orders. At this rate, how many hours of continuous work by 72 machines are required to fill 12 standard orders? | [
"3",
"6",
"8",
"9"
] | B | the choices give away the answer..
36 machines take 4 hours to fill 8 standard orders..
in next eq we aredoubling the machines from 36 to 72, but thework is not doubling(only 1 1/2 times), = 8*36/72*12/8 = 6
Ans B |
AQUA-RAT | AQUA-RAT-37131 | \$7,382.94 3.2% 01.09.08 \$19.6879
\$7,402.63 3.2% 01.10.08 \$19.7404
\$7,422.37 3.2% 01.11.08 \$19.7930
\$7,442.17 3.2% 01.12.08 \$19.8458
The following is multiple choice question (with options) to answer.
A sum of money deposited at C.I. amounts to Rs.6000 in 6 years and to Rs.7500 in 7 years. Find the rate percent? | [
"25%",
"20%",
"15%",
"10%"
] | A | 6000 --- 1500
100 --- ? => 25%
Answer: A |
AQUA-RAT | AQUA-RAT-37132 | 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.
In what time will a train 120 m long cross an electric pole, it its speed be 144 km/hr? | [
"3 sec",
"2.9 sec",
"3.5 sec",
"7.5 sec"
] | A | Speed = 144 * 5/18 = 40 m/sec
Time taken = 120/40
= 3 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-37133 | mechanical-engineering, gears, machine-design
Title: Velocity or Speed Ratios in Gears and Gear Trains I have seen the different definitions of the velocity ratio for Gears and Gear trains. I have referred a book called Theory of machines by S. S. Ratan. In that book the definitions are as following:
For gears, velocity ratio is ratio of angular velocity of follower to that of the driving gear.
For gear train, speed ratio is given as the ratio of speed of driving to that of the driven shaft.
Why are these two definitions different (inverse of each other) as I think Velocity ratio and speed ratio are the same thing. Also if there is a difference, it will be helpful if a bit explanation is provided. TL;DR IMHO, this is something that think has its roots to older times, where the strength design of gears required tables and charts
Like solarMike said, they are essentially the same thing. I don't think many people nowadays pay too much attention to this detail.
IMHO, the reason for the existence stems from the different need. Like you stated, (some authors) use (note that I'll start the opposite way):
For gear train, speed ratio is given as the ratio of speed of driving to that of the driven shaft.
In that case, the big picture is important. I.e the entire drivetrain, and more specifically its kinematics. In that case the important is to define the angular velocities and torque on the various stages of the drivetrain (or maybe just at the end). Therefore, it that case, the speed ratio focuses on that.
For gears, velocity ratio is ratio of angular velocity of follower to that of the driving gear.
The following is multiple choice question (with options) to answer.
The speeds of three motor bikes are in the ratio 6 : 5 : 4. The ratio between the time taken by them to
travel the same distance is : | [
"10 : 12 : 15",
"12 : 10 : 8",
"15 : 12: 10",
"10 : 15 : 12"
] | A | Expl : Ratio of time taken : 1/6 :1/5 : 1/4 = 10 : 12 : 15
Answer: A |
AQUA-RAT | AQUA-RAT-37134 | 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.
A sum of money at simple interest amounts to $ 780 in 3 years and to $930 in 4 years. The sum is: | [
"$153",
"$698",
"$330",
"$549"
] | C | C
$330
S.I. for 1 year = $(930 - 780) = $150.
S.I. for 3 years = $(150 x 3) = $450.
Principal = $(780 - 450) = $330. |
AQUA-RAT | AQUA-RAT-37135 | First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.
Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$
Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$
And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$
So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
The following is multiple choice question (with options) to answer.
A student chose a number, multiplied it by 2, then subtracted 138 from the result and got 102. What was the number he chose? | [
"240",
"120",
"150",
"200"
] | B | Let x be the number he chose, then
2⋅x−138=102
2x=240
x=120
So answer is B |
AQUA-RAT | AQUA-RAT-37136 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
In how many ways can the letters {U, R, R, R, R} be arranged? | [
"5",
"20",
"120",
"720"
] | A | My answer : A
There is 5! ways to arrange those letters in total, however letter U repeats 1! times and letter R repeats 4! times
=> the way to arrange without repeating = 5!/ [1!x4!] = 5 |
AQUA-RAT | AQUA-RAT-37137 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A local elementary school has 1049 students and nine teachers, How many students would have to be added to the school to make sure each teacher has the same amount of students in their classroom? | [
"4",
"6",
"5",
"1"
] | A | Using the rule on the divisibility of nine, we add the digits of 1049 together to get 14. Since we are doing addition, the next number that is divisible by nine would be 18. Therefore, our answer 4(option A) |
AQUA-RAT | AQUA-RAT-37138 | ### 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 entered into a partnership investing Rs.25000 and Rs.30000 respectively. After 4 months C also joined the business with an investment of Rs.35000. What is the share of C in an annual profit of Rs.47000? | [
"14028",
"14038",
"14079",
"14000"
] | D | 25*12: 30*12: 35*8
15:18:14
14/47 * 47000 = 14000
Answer:D |
AQUA-RAT | AQUA-RAT-37139 | 1.
The Correct Answer is (D) — To solve this question, isolate for x:
\small&space;\begin{align*}&space;3x&space;-&space;4&space;&=&space;-\frac{1}{2x}&space;\\&space;3x&space;+\frac{1}{2x}&space;&=&space;4&space;\\&space;\frac{7}{2}x&space;&=&space;4&space;\\&space;x&space;&=&space;\frac{8}{7}&space;\end{align*}
2.
To factor a quadratic where the second-order term has a coefficient of 1, find two numbers that multiply up to the constant, –7, and add up to the first-order term’s coefficient, 6. These two numbers are 7 and –1, so you can rewrite the quadratic equation as follows:
x2 + 6x – 7 = (x + 7)(x – 1)
To find the solutions—or roots of the resulting quadratic function—set the equation equal to zero. You can see that this happens when x = –7 or x = 1. (F) is the only answer choice that lists one of these options.
If you chose (H) or (J), you probably forgot that a root is the negative of one of those two numbers you found, because it’s a value of x that makes one of the bracketed expressions equal to zero.
3.
The Correct Answer is (B) — You want to avoid writing out the entire sequence up to the 21st term, thankfully, you can use the formula for an arithmetic sequence, an = a1 + d(n – 1) , to find the nth term algebraically. Simply set a1 = 11, d = –3, and n = 21 for the 21st term in a sequence where the first term is 11 and it decreases by 3 each term, and solve:
The following is multiple choice question (with options) to answer.
If the roots of a quadratic equation are 20 and -7, then find the equation? | [
"x2 + 13x - 140 = 0",
"x2 - 13x + 140 = 0",
"x2 - 13x - 140 = 0",
"x2 + 13x + 140 = 0"
] | C | Explanation:
Any quadratic equation is of the form
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-37140 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. 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.
38, 87, 22,76, 31,47, 13, 82
Which of the following numbers is greater than three-fourth of the numbers but less than one-fourth of the numbers in the list above? | [
"a- 55",
"b- 68",
"c- 79",
"d- 81"
] | C | Important:- Arrange the numbers in ascending order first.
13, 22, 31, 38, 47, 76, 82, 87
3/4th of the number list represents- 76 (6th number)
remaining 1/4th list represents 82 (7 th number)
79 fits in between above 2
Answer:- C |
AQUA-RAT | AQUA-RAT-37141 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder . | [
"1638",
"1863",
"1683",
"1836"
] | C | L.C.M. of 5,6,7,8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3)=1683
Answer is C. |
AQUA-RAT | AQUA-RAT-37142 | # 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.
Tom and John traveled in the same direction along the equal route at their constant speed rates of 15 km per hour and 10 km per hour, respectively. After 15 minutes Tom passed John, Tom reaches a certain Gas station, how many T minutes it takes John to reach the station? | [
"5 min",
"6 min",
"7 and 1/2 min",
"8 min"
] | C | Since the question states “after 15 minutes”, we can say Tom traveled 15/4km for 15 minutes as he can travel 15km per hour. Hence, using the same logic, we can say John traveled 10/4km as he travels 10km per hour. So, John has to travel (15/4)-(10/4)km=5/4km more. Since John’s speed is 10km/hour, which means 1km/6minutes. As he has to travel 5/4km more, it is going to take him 6(5/4) minutes. Hence, T=6(5/4)=15/2 minutes. The correct answer is C. |
AQUA-RAT | AQUA-RAT-37143 | 60 - 24 = 36.
I can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help
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Posts: 129
The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
### Show Tags
Updated on: 07 Aug 2019, 22:03
We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)
Number of ways the 5 letters can be arranged is $$\frac{5!}{2!}$$ = 60 ways
To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].
Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.
Originally posted by Sayon on 01 Aug 2019, 21:03.
Last edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.
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Joined: 12 Apr 2019
Posts: 9
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
How many 1/4s are there in 37 1/2? | [
"150",
"450",
"500",
"650"
] | A | Required number = (75/2)/(1/4)
= (75/2 x 4/1)
= 150.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37144 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
When a certain number X is divided by 52, the remainder is 19. What is the remainder when X is divided by 7? | [
"2",
"3",
"4",
"5"
] | D | When a certain number X is divided by 52, the remainder is 19. What is the remainder when X is divided by 7?
Putting a value say x = 19 we get remainder as 19 when divided by 52.
When 19 divided by 7 we get 5 as remainder.
D is the answer. |
AQUA-RAT | AQUA-RAT-37145 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A cycle is bought for Rs.800 and sold for Rs.1080, find the gain percent? | [
"22",
"20",
"35",
"88"
] | C | 800 ---- 180
100 ---- ? => 35%
Answer:C |
AQUA-RAT | AQUA-RAT-37146 | The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a?
(1) The cost of producing 500 units is $15.833. (2) The cost of producing 1,000 units is twice the cost of producing 500 units. Originally posted by inakihernandez on 24 Apr 2016, 10:20. Last edited by Vyshak on 24 Apr 2016, 11:36, edited 1 time in total. Edited the url ##### Most Helpful Expert Reply Math Expert Joined: 02 Aug 2009 Posts: 6958 The cost C, in dollars of producing x units of a certain commodity is [#permalink] ### Show Tags 24 Apr 2016, 10:58 11 4 inakihernandez wrote: The cost C, in dollars of producing x units of a certain commodity is given by the formula C(x) = 10.000 + bx + ax^2 where a, and b are constants. What is the value of a? (1) The cost of producing 500 units is$15.833.
(2) The cost of producing 1,000 units is twice the cost of producing 500 units.
Hi
the equation has got two constants a and b..
lets see the statements
The following is multiple choice question (with options) to answer.
The cost of producing x tools by a company is given by
C(x) = 2500 x + 5500 (in $)
a) What is the cost of 100 tools? | [
"222500 $",
"125800 $",
"225900 $",
"255500 $"
] | D | Solution
C(100) = 2500*100 + 5500 = 255500 $
Answer D |
AQUA-RAT | AQUA-RAT-37147 | # What is the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ is an integer?
Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit.
I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any pattern. I coudn't find any pattern but I eventually solved the problem to be $n=6$, but is there any easier more practical approach to this problem?
-
I've added LaTeX formatting to your question; did I interpret your meaning correctly? – Zev Chonoles Dec 1 '11 at 6:09
Yes it is perfectly portrayed. Thanks Zev. :) – Kelly Rocks Dec 1 '11 at 6:17
The following is multiple choice question (with options) to answer.
If n is an integer, then the least possible value of |31 - 4n| is? | [
"0",
"1",
"2",
"3"
] | B | |31 - 4n| represents the distance between 31 and 4n on the number line. Now, the distance will be minimized when 4n, which is multiple of 4, is closest to 31. Multiple of 4 which is closest to 31 is 32 (for n = 8),
so the least distance is 1: |31 - 32| = 1.
Answer: B. |
AQUA-RAT | AQUA-RAT-37148 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a job in 10days and B in 20days. If they work on it together for 4 days, then the fraction of the work that is left is? | [
"2/5",
"8/15",
"3/11",
"1/12"
] | A | A's 1 day work = 1/10
B's 1day work = 1/20
A+B 1day work = 1/10 + 1/20 = 3/20
A+B 4days work = 3/20*4 = 3/5
Remaining work = 1 - 3/5 = 2/5
Answer is A |
AQUA-RAT | AQUA-RAT-37149 | |x - a| = b
x is a distance 'b' away from 'a' on either side on the number line.
-4 < x < 8
Mid point of -4 and 8 is 2. Both -4 and 8 are 6 away from 2.
So the absolute value form will be |x - 2| < 6
For more, check: http://www.veritasprep.com/blog/2011/01 ... edore-did/
_________________
Karishma
Veritas Prep GMAT Instructor
##### General Discussion
Intern
Joined: 09 Jul 2016
Posts: 22
Re: Which of the following inequalities is equivalent to −4 < x < 8? [#permalink]
### Show Tags
22 Aug 2016, 02:45
2
We know that |x| < a means -a < x < a, where Sum of lower limit of x (i.e -a) and the upper limit of x (i.e a), is 0
Given is, -4 < x < 8, let's say by adding y to this inequality we will get into the above format
-4+y < x+y < 8+y
Now, to move this into the mod format, we need to have (-4+y) + (8+y) = 0 => y = -2
Thus, -6< x-2 < 6 => |x-2| < 6.
Senior Manager
Joined: 23 Apr 2015
Posts: 271
Location: United States
WE: Engineering (Consulting)
Re: Which of the following inequalities is equivalent to −4 < x < 8? [#permalink]
### Show Tags
22 Aug 2016, 19:28
2
2
Bunuel wrote:
Which of the following inequalities is equivalent to −4 < x < 8?
A. |x - 1| < 7
B. |x + 2| < 6
C. |x + 3| < 5
D. |x - 2| < 6
E. None of the above
Given −4 < x < 8,
Now try to get into the mod form,
subtract two from both sides of the equation, (graphically the line remains the same)
The following is multiple choice question (with options) to answer.
Which of the following inequalities is equivalent to −4 < x < 8? | [
"|x - 1| < 7",
"|x + 2| < 6",
"|x + 3| < 5",
"|x - 2| < 6"
] | D | We know that |x| < a means -a < x < a, where Sum of lower limit of x (i.e -a) and the upper limit of x (i.e a), is 0
Given is, -4 < x < 8, let's say by adding y to this inequality we will get into the above format
-4+y < x+y < 8+y
Now, to move this into the mod format, we need to have (-4+y) + (8+y) = 0 => y = -2
Thus, -6< x-2 < 6 => |x-2| < 6.
Hence, answer is D |
AQUA-RAT | AQUA-RAT-37150 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If (x - 1)^2 = 324, which of the following could be the value of x - 3? | [
"17",
"15",
"13",
"-20"
] | D | (x - 1)^2 = 324
(x - 1) = 18 or -18
x = 19 or -17
x - 3 = 16 or -20
The answer is D. |
AQUA-RAT | AQUA-RAT-37151 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
The average of five numbers is 27. If one number is excluded, the average becomes 20. The excluded number is? | [
"52",
"53",
"54",
"55"
] | D | Excluded number = (27 * 5) - (20 * 4)
= 135 - 80
= 55.
Answer: D |
AQUA-RAT | AQUA-RAT-37152 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of the four numbers K, 2K + 3, 3K – 5 and 5K + 1 is 118, what is the value of K? | [
" 11",
" 15 3/4",
" 22",
" 118"
] | D | K + 2K +3 + 3K - 5 + 5K +1 = 11K -1
(11K -1)/4 = 118
11K = 118 * 4 +1 = 472 +1 = 473
K = 473 / 11 = 43.
Answer D. |
AQUA-RAT | AQUA-RAT-37153 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
A boy was asked to multiply a number by 22. He instead multiplied the number by 44 and got the answer 308 more than the correct answer. What was the number to be multiplied? | [
"10",
"12",
"08",
"14"
] | D | Let the number be x
22x + 308 = 44x
=> 44x - 22x = 308
=> 22x = 308
=> x = 308/22 = 154/11 = 14
Answer is D. |
AQUA-RAT | AQUA-RAT-37154 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
The average weight of 2 person's increases by 4.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person? | [
"74 kg",
"76.5 kg",
"85 kg",
"78"
] | A | Explanation:
Total weight increased = (2 x 4.5) kg = 9 kg.
Weight of new person = (65 + 9) kg = 74 kg.
Answer: A |
AQUA-RAT | AQUA-RAT-37155 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A pair of articles was bought for $720 at a discount of 10%. What must be the marked price of each of the article? | [
"$300",
"$500",
"$350",
"$400"
] | D | S.P. of each of the article = 720/2 = $360
Let M.P = $x
90% of x = 360
x = 360*100/90 =$400
Answer is D |
AQUA-RAT | AQUA-RAT-37156 | 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. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
A pipe can filled a tank in 40 minutes and another pipe can empty it in 1hour. If the tank is already half full and both the taps are opened together, then the tank is filled in how many minutes? | [
"25min",
"20min",
"15min",
"10min"
] | A | T = 1/2 (25*50 / 50-25) = 25 minutes
Tank is filled in 25 minutes
Answer is A |
AQUA-RAT | AQUA-RAT-37157 | # 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.
Two people P Q start a race in a circular track in opposite way different but constant speed. First they meet 900m cw from the starting pt. Then they meet 800m ccw from the starting pt. What's the circumference of the circle. | [
"2600",
"2700",
"2800",
"2900"
] | A | when they first met 900m cw from the starting pt.
second time, they will meet at 1800 m cw from starting point which is 800 m CCW from starting point.
so circumference of circle = 1800+800 = 2600 m
ANSWER:A |
AQUA-RAT | AQUA-RAT-37158 | # Ratio of angles in a triangle, given lengths of triangle's sides.
If I have a triangle $\,\triangle ABC,\,$ with sides of lengths $\,AB=6, \;BC=4, \;CA=5,\,$
then what can I know about the ratio of $\,\dfrac{\angle ACB}{\angle BAC}\,$?
-
Use en.wikipedia.org/wiki/Law_of_cosines, if Trigonometry is allowed – lab bhattacharjee Nov 12 '13 at 13:49
You can use the Law of Sines to find the ratio of the sines of your two angles:
$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, D \!$$
In your case, you'd have $$\dfrac 6{\sin(\angle ACB)} = \dfrac 4{\sin(\angle BAC)} \iff \dfrac {6}{4} = \dfrac{\sin(\angle ACB)}{\sin(\angle BAC)} = \frac 32$$
Alternatively, to compute the measures of your angles directly, use the Law of Cosines. $$c^2 = a^2 + b^2 - 2ab\cos\gamma\,\iff \cos \gamma = \dfrac{a^2 + b^2 - c^2}{2ab}$$
The following is multiple choice question (with options) to answer.
In a triangle, the ratio of two angles is 7:2, and the third angle is the difference between the first two angles. What is the smallest angle, in degrees? | [
"90/11",
"120/7",
"150/11",
"180/7"
] | D | the three angles are 7x, 2x, and (7x-2x)=5x for some value x (to create the ratio)
the three angles of a triangle add up to 180, so 7x+2x+5x = 180
14x = 180
x=180/14
x = 90/7
the smallest angle is 2x = 2*(90/7) = 180/7
D |
AQUA-RAT | AQUA-RAT-37159 | ### 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.
Ravi and sunil are partners in a business. Ravi invests Rs.12,000 for 4 months and Sunil invested Rs.10000 for 5 months then after one year ratio of their profits will be | [
"1:3",
"2:3",
"24:25",
"20:22"
] | C | =(12000*4):(10000*5)
=48000:50000
=24:25
ANSWER:C |
AQUA-RAT | AQUA-RAT-37160 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A train covers a distance in 50 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 50min will be. | [
"60 km/h",
"55 km/h",
"48 km/h",
"70 km/h"
] | C | Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=50/60hr=5/6hr
New speed = 40* 6/5 kmph= 48kmph
Answer : C |
AQUA-RAT | AQUA-RAT-37161 | solubility, alloy, teaching-lab
What other such situations are there is chemistry; when would one actually need to do such simultaneous calculations? You may also give the weight of a sample made of an alloy $\ce{Zn-Mg}$. Then you dip it into some diluted $\ce{HCl}$. Both metals will react simultaneously according to $$\ce{Zn + 2 HCl -> H_2 + ZnCl_2}$$ and $$\ce{Mg + 2 HCl -> H_2 + MgCl_2}$$ This will produce an important amount of $\ce{H_2}$ gas. You can measure its volume. So you have two measured data, the original mass and the volume of gas. It is enough for calculating the proportion of $\ce{Zn}$ and $\ce{Mg}$ in the original sample.
Example : Take a mixture of $0.1$ mol $\ce{Zn}$ + $0.3$ mol $\ce{Mg}$. Of course the molar masses of $\ce{Zn}$ and $\ce{Mg}$ are respectively $65.39$ g/mol and $24.32$ g/mol. The total mass is : $6.54$ g + $7.29$ g = $13.83$ g. Suppose the temperature and the pressure of the gas is such that $1$ mole gas occupies $24$ L. So $0.1$ mol $\ce{Zn}$ will produce $0.1$ mol $\ce{H_2}$ which occupies $2.4$ L gas. Then $0.3$ mol $\ce{Mg}$ will produce $0.3$ mol $\ce{H_2}$, or $7.2$ L gas. The total volume of $\ce{H_2}$ is $2.4$ L + $7.2$ L = $9.6$ L.
The following is multiple choice question (with options) to answer.
135 kg of an alloy A is mixed with 145 kg of alloy B. If alloy A has lead and tin in the ratio 3:5 and alloy B has tin and copper in the ratio 2:3, then the amount of tin in the new alloy is? | [
"100.6kg",
"142kg",
"135kg",
"110.8kg"
] | B | Quantity of tin in 135kg of A = 135*5/8 = 84kg
Quantity of tin in 145kg of B = 145*2/5 = 58kg
Quantity of tin in the new alloy = 84+58 = 142kg
Answer is B |
AQUA-RAT | AQUA-RAT-37162 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Carrie likes to buy t-shirts at the local clothing store. They cost $9.65 each. One day, she bought 12 t-shirts. How much money did she spend? | [
"$115.8",
"$248.75",
"$200",
"$171.6"
] | A | $9.65*12=$115.8. Answer is A. |
AQUA-RAT | AQUA-RAT-37163 | => 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.
On Thursday Mabel handled 90 transactions. Anthony handled 10% more transactions than Mabel, Cal handled 2/3rds of the transactions that Anthony handled, and Jade handled 14 more transactions than Cal. How much transactions did Jade handled? | [
"80",
"81",
"82",
"83"
] | A | Solution:
Mabel handled 90 transactions
Anthony handled 10% more transactions than Mabel
Anthony = 90 + 90 × 10%
= 90 + 90 × 0.10
= 90 + 9
= 99
Cal handled 2/3rds of the transactions than Anthony handled
Cal = 2/3 × 99
= 66
Jade handled 14 more transactions than Cal.
Jade = 66 + 14
= 80
Jade handled = 80 transactions.
Answer: A |
AQUA-RAT | AQUA-RAT-37164 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 70 sq. cm. Find the length of the rectangle. | [
"12cm",
"14cm",
"16cm",
"18cm"
] | D | Explanation:
Let breadth = x. Then, length = 2x. Then,
(2x - 5) (x + 5) - 2x * x = 70 => 5x - 25 = 70 => x = 18.
Length of the rectangle = 18 cm.
Answer: Option D |
AQUA-RAT | AQUA-RAT-37165 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman in his 11th inning makes a score of 69 and their by increasing his average by 1. What is his average after the 11th inning? | [
"58",
"59",
"68",
"69"
] | B | 10x + 69 = 11(x + 1)
x = 58 + 1 = 59
Answer:B |
AQUA-RAT | AQUA-RAT-37166 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man sells a horse for Rs.800 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse? | [
"27",
"98",
"27",
"60"
] | D | CP = SP + 1CP = SP - g
800 + x = 980 - 2x
3x = 180 => x = 60
Answer:D |
AQUA-RAT | AQUA-RAT-37167 | This leaves ten admissible configurations, and for each there are $(3!)^2=36$ ways of permuting the boys and girls, for a total of 360 ways.
• ok, but where exatcly am i going wrong? – Anvit Dec 10 '17 at 11:21
• @AnvitGarg Note that you have considered the case that boys and girls should sit in alternate positions. But, the question says that a boy can be adjacent to atleast one girl. – Rohan Dec 10 '17 at 11:22
• @AnvitGarg You overlooked permutations such as $b_1g_1g_2b_2b_3g_3$ in which two or more girls are consecutive. – N. F. Taussig Dec 10 '17 at 11:22
The following is multiple choice question (with options) to answer.
In how many different number of ways 7 boys and 4 girls can sit on a shopa which can accommodate 3 persons? | [
"800",
"900",
"200",
"990"
] | D | Option 'D'
11p3 = 11x10x9 = 990 |
AQUA-RAT | AQUA-RAT-37168 | Therefore, $$p_{11} = 0.5$$ is the probability of being sunny tomorrow, given that it is sunny today.
You would want to use your record to test the MC assumption.
Example 2:
Recall the weather pattern MC in example 1. You are planning a two-day holiday to begin in seven days, i.e., you are away on day $$7$$ and $$8$$. A travel insurance deal will pay you \$2500 if it rains on both days, nothing if not, and the premium is \$100. Should you buy this insurance if it is sunny today?
One way to make a decision would be to compare the expected pay-out with the premium. The actual return is
$$R = \begin{cases} 2500 & \text{if} \ X_7 = X_8 = 3 \\ 0 & \text{otherwise}, \end{cases}$$
where $$X_n$$ is the weather state on day $$n$$. Counting today as day $$0$$, the expected return is
\begin{align} E(R) &= 2500 \times P(X_7 = X_8 = 3 \vert X_0 = 1) \\ &= 2500P(X_8 = 3 \vert X_7 = 3, X_0 = 1)P(X_7 = 3 \vert X_0 = 1) \\ &= 2500P(X_8 = 3 \vert X_7 = 3) p^{(7)}_{13} \\ &= 2500p_{33}p^{(7)}_{13} \end{align}
Evaluation of the one number $$p^{(7)}_{13}$$ requires evaluation of $$\mathcal{P}^7$$, where $$\mathcal{P}^{(n)}$$ is the $$n$$-step transition matrix.
You will find that $$p^{(7)}_{13} = 0.2101$$, and $$p_{33} = 0.1$$, so $$E(R) = 52.52$$.
The following is multiple choice question (with options) to answer.
Sun Life Insurance company issues standard, preferred and ultra-preferred policies. Among the company's policy holders of a certain age, 50 are standard with the probability of 0.01 dying in the next year, 30 are preferred with a probability of 0.008 of dying in the next year and 20 are ultra-preferred with a probability of 0.007 of dying in the next year. If a policy holder of that age dies in the next year, what is the probability of the decreased being a preferred policy holder? | [
"0.1591",
"0.2727",
"0.375",
"0.2767"
] | B | Explanation :
The percentage of three different types of policy holders and the corresponding probability of dying in the next 1 year are as follow :
Type, Standard, Preferred, Ultra-Preferred
Percentage , 50 , 30, 20
Probability, 0.01 , 0.008, 0.007
The expected number of deaths among all the policy holders of the given age [say P] during the next year.
=>T×(50×0.01/100)+(30×0.008/100)+(20×0.007/100).
=>T×(0.88/100).
Where T= Total number of policy holder of age P
If any of these policy holders (who die during the next year) is picked at random, the probability that he is a preferred policy holder is :-
=>((30x0.88xT)/100)/ (Tx(0.88/100)).
=>24/88.
=>3/11.
=>0.2727.
Answer : B |
AQUA-RAT | AQUA-RAT-37169 | 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 100 meters long completely crosses a 300 meters long bridge in 45 seconds. What is the speed of the train is? | [
"25 kmph",
"32 kmph",
"40 kmph",
"38 kmph"
] | B | S = (100 + 300)/45 = 400/45 * 18/5 = 32 kmph
answer :B |
AQUA-RAT | AQUA-RAT-37170 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
Nine friends are planning to attend a concert. If another friend also goes with them, they will get a group discount on their tickets. If the new total price of the tickets with the group discount is the same as the original total price of the tickets without the group discount, how much is the discount? | [
"5%",
"10%",
"20%",
"25%"
] | B | Let x be the original price of one ticket.
The total original cost is 9x.
The new cost is 10y, where y is the discounted price of one ticket.
9x = 10y
y = 9x/10 = 0.9x which is a discount of 10%.
The answer is B. |
AQUA-RAT | AQUA-RAT-37171 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A certain sum of money is divided among A, B and C so that for each Rs. A has, B has 65 paisa and C 40 paisa. If C's share is Rs.40, find the sum of money? | [
"228",
"267",
"911",
"205"
] | D | A:B:C = 100:65:40
= 20:13:8
8 ---- 40
41 ---- ? => Rs.205
Answer: D |
AQUA-RAT | AQUA-RAT-37172 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? | [
"2.91 m",
"3 m",
"5.82 m",
"None of these"
] | B | Explanation:
Area of the park = (60 x 40) m2 = 2400 m2
Area of the lawn = 2109 m2
Area of the crossroads = (2400 - 2109) = 291
Let the width of the road be x metres. Then,
60x + 40x - x2 = 291
=> x2 - 100x + 291 = 0
=> (x - 97)(x - 3) = 0
=> x = 3
Answer: B |
AQUA-RAT | AQUA-RAT-37173 | c#, parsing
\n20150805 0900| 6| 91| | 0| | 0| | 0| 0| 0.0| | 13.0| 10.0| 0| 0| |15.00| |
\n20150805 1000| 8| 84| | 0| | 0| | 0| 0| 0.0| | 13.0| 11.0|230| 8| |15.00| |
\n20150805 1100| 10|113| -RW |100| | 0| | 0|100| 0.1| RA| 13.0| 11.0|277| 11| |10.00| |
\n-------------+---+---+-----+---+-----+---+-----+---+---+----+---+-----+-----+---+---+---+-----+--+
\n20150805 1200| 10| 93| -RW |100| | 0| | 0|100| 0.1| RA| 13.0| 10.8|225| 9| | 7.00| |
\n-------------+---+---+-----+---+-----+---+-----+---+---+----+---+-----+-----+---+---+---+-----+--+
\n20150805 1300| 8|999| -RW |100| | 0| | 0| 33| 0.1| RA| 13.9| 11.4|163| 6| |15.00| |
\n20150805 1400| 8|999| -RW |100| | 0| | 0| 31| 0.1| RA| 14.4| 11.5|129| 7| |15.00| |
The following is multiple choice question (with options) to answer.
15.06 * 0.000001 = ? | [
"15060000",
"0.001506",
"0.01506",
"1.506e-05"
] | D | Explanation:
Clearly after decimal 8 digits should be there.
Option D |
AQUA-RAT | AQUA-RAT-37174 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed? | [
"720",
"420",
"450",
"350"
] | A | Explanation:
The word 'LOGARITHMS' has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters =10X9X8
=720
ANSWER IS A |
AQUA-RAT | AQUA-RAT-37175 | also have a clearance (play) vertical to tooth depth. 14 * 5 2 * ((4/3) * 5 +10) Volume (V) = 1309 m 3. The Tank Capacity Calculator below allows you to type in your desired tank diameter and height and provides an estimated volume by gallon amount. White & blue vertical tanks are translucent for convenient product level viewing. Calculations are performed for: - Vertical tank - Horizontal tank - Rectangular tank - Ecliptic tank - Tanks with conical bottom, flat bottom, Torispherical head, Elliptical head, Hemispherical head. The reference point of the water level in the tank used to calculate the water rising level was made to the bell mouth top side for the vertical overflow outlet type as shown in Figure 4, and made a pipe inside bottom for the horizontal overflow outlet type. • if k should be negative, the vertical stretch or shrink is. Hence, we can describe the concentration of salt in the tank by concentration of salt = S 100 kg/l. User-defined axis scale type using formulas. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. If you have a complex tank, or need something special created please don't hesitate contact us for more info. The radius BO of the hemisphere (as well as of the cone) = ( ½) × 4 cm = 2 cm. , the Volumes of complicated shapes can be calculated with integral calculus if a formula exists for the In other systems the conversion is not trivial; the capacity of a vehicle's fuel tank is rarely stated in cubic. This Excel spreadsheet helps you calculate the liquid volume in partially filled horizontal tanks. 000233 D3 ( D = ID in inches ) ( DR = ID ). Nonsymmetrical Vertical Curves G1 & G2 Tangent Grades in percent A The absolute of the Algebraic difference in grades in percent BVC Beginning of Vertical Curve EVC End of Vertical Curve VPI Vertical Point of Intersection l1 Length of first section of vertical curve l2 Length of second section of vertical curve L Length of vertical curve. Volume formula of a rectangular prism Volume of a right circular cone is equal to one third of the product of the area of its base times the height. Adding air manually is not the best way to ensure proper. Internal lining material is natural rubber with thickness 4. There are two main methods to tank
The following is multiple choice question (with options) to answer.
A fish tank is half full of water. If 10 gallons of water were added, the tank would be 7/8 full. What is the capacity of the tank (in gallons)? | [
"11 3/7",
"22 3/7",
"26 2/3",
"30 3/8"
] | C | Solution -
Let capacity of tank be X gallons.
Water level increased to 7/8 from 1/2 when 10 gallons of water added ---> (7/8 - 1/2)*X = 10
X = 80/3 = 26 2/3.
ANS C. |
AQUA-RAT | AQUA-RAT-37176 | ##### Well-Known Member
Subscriber
Actually, it looks like the discount rate is a risk-free rate not necessarily equal to a LIBOR of the correct period length.
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Hi Brian,
In many countries such as Thailand, the Treasury yield curve is illiquid across maturities. On the other hand, the LIBOR/Swap Yield curve is liquid across maturities worldwide. The LIBOR swap zero curve is obtained by bootstrapping and is almost risk-free. That is why it is used to discount the Cash flows to obtain the value of the FRA. The Fixed rate cannot be used for discounting because FRA's are OTC - the fixed rate is unique to that particular FRA.
Thanks!
Jayanthi
#### Dr. Jayanthi Sankaran
##### Well-Known Member
Thanks Nicole - appreciate it
Jayanthi
#### brian.field
##### Well-Known Member
Subscriber
Thanks for your message Jayanthi. While I agree with what you are saying, generally, it isn't really addressing my philosophical question.
Consider today at time t=0 and an FRA that requires a fixed payment of 5% and a floating payment of 1YLIBOR from t=3 to t=4.
At t=0, we do not know that actual 1YLIBOR rate that will apply in 3 years but we do know the 1Y forward 1YLIBOR rate payable in 3 years. We establish the FRA at t=0, and at this time, the fixed rate is set so that the FRA's value is 0 at t=0. By the time we get to t=3, the FRA may not equal 0.
My point is that at t=3, we will know that value of the FRA at t=4 and it is customary to discount the t=4 value back to t=3 with a risk-free rate. I thought that I read that this discount rate was the appropriate tenor LIBOR.
My question is as follows:
The following is multiple choice question (with options) to answer.
The banker's discount on a bill due 4 months hence at 15% is Rs. 420. The true discount is: | [
"400",
"360",
"480",
"320"
] | A | T.D. = B.D. x 100 / 100 + (R x T)
= Rs.(420 x 100 / 100 + (15 x 1/3))
= Rs. (420 x 100 / 105)
= Rs. 400.
Correct Answer is A. |
AQUA-RAT | AQUA-RAT-37177 | => 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.
A worker is paid a regular rate of Rs.20 for completing a survey. The worker completes 50 surveys per week. For any survey involving the use of her cellphone, she is paid at a rate of that is 10% higher than her regular rate. If she completed 35 surveys involving the use of her cellphone, how much did she get that week? | [
"570",
"670",
"1070",
"470"
] | C | amount earned by using her cell phone = 35 * 22 = 770
amount earned as usual = 15 * 20 = 300
Total = 770+300 = 1070
Answer : C |
AQUA-RAT | AQUA-RAT-37178 | So total 6 ways we can assign project to three groups.
So total number of ways we can do the required is 15 Ways of distributing people in 3 groups X 6 ways of distributing 3 projects among the three formed groups.
= 90
Hope this helps
Probus
_________________
Probus
~You Just Can't beat the person who never gives up~ Babe Ruth
Manager
Joined: 16 Jul 2016
Posts: 54
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
05 Mar 2019, 19:03
Suppose the six students are 1,2,3,4,5,6
Now student 1 HAS to be paired with someone else. He/she can be paired with 2,3,4,5, or 6. So that's 5 possibilities.
Once that has occurred (without loss of generality suppose 1 pairs with 2. 3 can then pair with 4,5, or 6. That's 3 possibilities. Only one pair is left. 5x3=15 groups of 3.
For each group there are 3x2x1 different arrangements of the topics. 15x6=90.
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Joined: 22 Oct 2017
Posts: 19
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:35
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
Can someone please explain to me why the red part of 15*3!, because I really can't figure out the reasoning behind the permutation of the assignments,
guaranteed kudos for accurate responses
Thanks
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Joined: 04 Aug 2010
Posts: 546
Schools: Dartmouth College
Re: Six students are equally divided into 3 groups, then, the three groups [#permalink]
### Show Tags
06 Mar 2019, 03:57
arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?
(A) 30
(B) 60
(C) 90
(D) 180
(E) 540
The following is multiple choice question (with options) to answer.
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements Q are possible? | [
" 30",
" 60",
" 90",
" 180"
] | C | 90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks.
But now you can order the 3 tasks in 3! ways. T1 T2 T3 or T2 T1 T3.... etc etc.
I was confused between 90 and 540 but since question used the wordarrangementsdecided to go with complete arrangements Q including the order of tasks.
could you explain the highlighted step... i'm getting 90 = 15 * 3!
suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z
one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90
But now you can fruther decide which task you want to perform first X Y or Z..=C |
AQUA-RAT | AQUA-RAT-37179 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman in his 19th inning makes a score of 100 and their by increasing his average by 2. What is his average after the 19th inning? | [
"58",
"60",
"62",
"64"
] | D | 18x + 100 = 19(x + 2)
x = 62 + 2 = 64
Answer:D |
AQUA-RAT | AQUA-RAT-37180 | ## 35.
The Correct Answer is (1/29) — The probability of choosing one student wanting to enter finance is 6/30. The probability of choosing another student wanting to enter finance is then 5/29. The probability of both of these events occurring is therefore (6/30)(5/29) = 30/870, which can be reduced to 1/29.
## 36.
The Correct Answer is (40) — Since the small triangle and the entire triangle share two angles (the one on the right, and the right angle), they must share all three angles, and therefore are similar triangles. You can use the Pythagorean Theorem on the bigger triangle to find that the length of the horizontal side is the square root of 1002 - 602, which is 80. (This is a special 3-4-5 triangle with the side lengths multiplied by 20). Since the ratio of the vertical side to the horizontal side of the big triangle is 60/80 = 3/4, this must be the same as the corresponding ratio in the small triangle. Since the vertical side has length 30, the length marked x must have length 40.
## 37.
The Correct Answer is (10) — There are 6 full cages of mice, which means there are 6 × 5 = 30 mice. If we let n be the number of male mice, then the number of female mice is twice that at 2n. There are 30 mice in total, so n + 2n = 30. Solving for n gives you n = 10. Since n is the number of male mice, there are 10 male mice.
## 38.
The Correct Answer is (17) — The reduced cost of maintaining each cage is 0.5($1.25) =$0.625 per cage per day. The student needs 21 cages to house 102 mice, so her daily cost is $0.625 × 21 =$13.125. She has a budget of $225, so she can afford to maintain the cages for$225/\$13.215 = 17.14 days. Rounding this to the nearest whole day gives you 17 days.
The following is multiple choice question (with options) to answer.
A certain lab experiments with white and brown mice only. In one experiment, 2/3 of the mice are white. If there are 13 brown mice in the experiment, how many mice in total are in the experiment? | [
"39",
"33",
"26",
"21"
] | A | Let total number of mice = M
Number of white mice = 2/3 M
Number of brown mice = 1/3 M = 13
=> M = 39
Answer A |
AQUA-RAT | AQUA-RAT-37181 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
If u r genius solve it:-
40 * 14 = 11
30 * 13 = 12
20 * 12 = 6
10 * 11 = ? | [
"1",
"2",
"3",
"4"
] | B | B
2
Logic :
40*14=560=5+6+0=11
30*13=390=3+9+0=12
20*12=240=2+4+0=6
So 10*11=110=1+1+0=2 |
AQUA-RAT | AQUA-RAT-37182 | Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have
\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}
And so we have our case where
\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}
The following is multiple choice question (with options) to answer.
A team of eight entered for a shooting competition. The best marks man scored 85 points. If he had scored 92 points, the average scores for. The team would have been 84. How many points altogether did the team score? | [
"288",
"665",
"168",
"127"
] | B | Explanation:
8 * 84 = 672 – 7 = 665
Answer:B |
AQUA-RAT | AQUA-RAT-37183 | You've apparently made two mistakes: (1) you neglected to add $1$ to the difference when counting the integers in an inclusive range, and (2) you neglected to add the final term N(P ∩ Q ∩ R), which is $30$. Correcting these will give the answer $(450 + 300 + 180 - 150 - 60 - 90 + 30 = 660)$.
You cannot just use the number of numbers in the range to determine how many are multiples of something. For example, there are two even numbers in the range $[2..4]$ but only one even number in the range $[3..5]$. That is why Andre's comment tells you to identify the first and last multiple in the range, which you can then use to determine how many there are.
The following is multiple choice question (with options) to answer.
How many integers are there between 55 and 108, inclusive? | [
"51",
"54",
"56",
"57"
] | B | I guess the easiest way to answer this is -
Say you have two positive integers, x and y where y > x
Then the number of integers between x and y is given by - (y - x) + 1
In this case, it's (108 - 55) + 1 =54. Option B |
AQUA-RAT | AQUA-RAT-37184 | We have to consider 4 cases totally..
1) + +
2) + -
3) - +
4) - -
If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.
Regards,
Mustu
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Re: Inequalities, Is |X| < 1 ? [#permalink]
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22 Jul 2011, 14:34
So does (2) implies that x IS NOT = 3? Hence we can use x =1/3 as our value?
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Re: Inequalities, Is |X| < 1 ? [#permalink]
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22 Jul 2011, 18:48
kanishk wrote:
So does (2) implies that x IS NOT = 3? Hence we can use x =1/3 as our value?
Exactly. So 1) gives us two options, but we don't know which x is. 2) tells us that x does not equal 3, so we know that x = 1/3
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Re: Inequalities, Is |X| < 1 ? [#permalink]
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24 Jul 2011, 13:28
kanishk wrote:
So does (2) implies that x IS NOT = 3? Hence we can use x =1/3 as our value?
Yes, you got it right
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Re: Inequalities, Is |X| < 1 ? [#permalink]
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31 Jul 2011, 20:33
Thanks 144144
This is a wonderful method
statement 1 :
(X+1)^2 = 4 (x-1)^2
this translated to a quadratic equation
The following is multiple choice question (with options) to answer.
3 of the 4 expressions (1), (2), (3) & (4) given below are exactly equal . Which of the expression is not equal to the other four expressions? | [
"(A + B)2 - 4AB",
"(A – B)2 + 4AB",
"A2 + B2 - 4AB + 2AB",
"A2 – B2 + 2B(B – A)"
] | B | 1) (A + B)2 – 4AB = A2 + 2AB + B2 - 4AB = A2 + B2 - 2AB
2) (A – B)2 + 4AB = A2 – 2AB + B2 + 4AB = A2+ B2 + 2AB
3) A2 +B2 – 4AB + 2AB = A2 + B2 - 2AB
4) A2 – B2 + 2B(B – A) = A2 – B2 + 2B2 – 2AB = A2 + B2 – 2AB
Therefore, (1) = (3) = (4) ≠ (2)
Therefore, expression (2) is wrong
ANSWER IS B |
AQUA-RAT | AQUA-RAT-37185 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
P is three times as fast as Q and working together, they can complete a work in 36 days. In how many days can Q alone complete the work? | [
"16",
"48",
"88",
"55"
] | B | P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 36 days, which means Q can do the work in 144 days.
Hence, P can do the work in 48 days.
Answer:B |
AQUA-RAT | AQUA-RAT-37186 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
Difference between two numbers is 5, six times of the smaller lacks by 8 from the four times of the greater. Find the numbers? | [
"5,1",
"7,3",
"9,8",
"11,6"
] | D | Explanation:
x – y = 5
4x – 6y = 8
x = 11 y = 6
D) |
AQUA-RAT | AQUA-RAT-37187 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day? | [
"12",
"13",
"15",
"16"
] | C | 2 day work = 1/20*2=1/10
A+B+C 1 day work = 1/10
work done in 3 days = 1/5
1/5 work is done in 3 days
whole work will done in (3*5)=15 days |
AQUA-RAT | AQUA-RAT-37188 | Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
SVP
Joined: 30 Apr 2008
Posts: 1835
Location: Oklahoma City
Schools: Hard Knocks
Re: Some part of a 50% solution of acid was replaced with an [#permalink]
### Show Tags
12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
The following is multiple choice question (with options) to answer.
Some of 10%-intensity red paint is replaced with 20% solution of red paint such that the new paint intensity is 15%. What fraction of the original paint was replaced? | [
"2/5",
"2/3",
"1/2",
"1/3"
] | C | Let total paint = 1
Let amount replaced = x
10 (1-x) + 20x = 15
x = 1/2
ANSWER:C |
AQUA-RAT | AQUA-RAT-37189 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Solve x2 – 7x = 0 | [
"0,7",
"1.7",
"2.7",
"3,7"
] | A | This quadratic factors easily: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved
x2 – 7x = 0
x(x – 7) = 0
x = 0 or x – 7 = 0
x = 0 or x = 7
The solution is x = 0, 7
ANSWER A |
AQUA-RAT | AQUA-RAT-37190 | - 7 years, 2 months ago
solution for qn 5 (i dont know if it is correct)
$first\quad lets\quad consider\quad the\quad equation\quad \\ { e }^{ i\pi }=-1\\ therefore,\quad i\pi =\log _{ e }{ (-1) } \\ =>\quad i\pi =\log _{ e }{ { i }^{ 2 } } \\ =>\log _{ e }{ i } =\frac { i\pi }{ 2 } \\ now\quad lets\quad get\quad back\quad to\quad the\quad question.\\ let\quad ,\quad { i }^{ i }=t\\ =>\quad \log _{ e }{ t } =i\log _{ e }{ i } =i(\frac { i\pi }{ 2 } )=-\frac { \pi }{ 2 } \\ therefore,\quad t={ i }^{ i }={ e }^{ -\frac { \pi }{ 2 } }=0.2078795$
- 7 years, 2 months ago
Its correct..Check out my note on the same.
- 7 years, 2 months ago
great note!
- 7 years, 2 months ago
SOLUTION #4
Using the Law of Cosines, we have $388 = 61 + 153 - 2 \sqrt{61} \cdot \sqrt{153} \cdot \cos{\alpha} \leftrightarrow \cos{\alpha} = - \frac{29}{\sqrt{1037}}$.
By the unit circle definition, $\sin^2 {\alpha} + \frac{841}{1037} = 1 \leftrightarrow \sin{\alpha} = \frac{14}{\sqrt{1037}}$. Note $0 < \alpha < \pi$.
The following is multiple choice question (with options) to answer.
The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages? | [
"30, 40",
"30, 46",
"30, 29",
"31, 40"
] | A | Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X - 5):(4X - 5) = 5:7
X = 35 - 25 => X = 10.
Their present ages are: 30, 40.
Answer: A |
AQUA-RAT | AQUA-RAT-37191 | ### 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.
Three business people wish to invest in a new company. Each person is willing to pay one third of the total investment. After careful calculations, they realize that each of them would pay $2,200 less if they could find two more equal investors. How much is the total investment in the new business? | [
"$11,000",
"$50,000",
"$16,500",
"$ 6,600"
] | C | Total Investment between 5: (x/5)
Total investment including 2200 less between 3 people(x-(2200*3))/3
Set both eq. equal to each other.
16,500
Answer C) |
AQUA-RAT | AQUA-RAT-37192 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100? | [
"15",
"77",
"25",
"18"
] | C | A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.
In a game of 100 points, B gives (100 - 75) = 25 points to C.
Answer:C |
AQUA-RAT | AQUA-RAT-37193 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
A car dealer has only blue and white cars in his show room. He has 24 blue cars and 30 white cars. He wants to have 60% more blue cars then white. How many whites cars does he have to sell to achieve his sales goal? | [
"12",
"13",
"14",
"15"
] | C | We can use simple arithmetic to solve this problem by testing our answers.
Lets test B:
If he sold 13 white cars:
24 blue cars and 17 white cars for a total of 41 cars
24/41=58.5% not quite our answer
Lets test C
If he sold 14 white cars
24 blue cars and 16 white cars for a total of 40 cars
24/40=60%
Our answer is C |
AQUA-RAT | AQUA-RAT-37194 | 2 + 7x = 3 + 11y.
I solved for y and got:
$y = \frac{7}{11}x - \frac{1}{11}$.
So I'm looking for an integer value of x that gives me an integer value for 7 (since all the numbers in both sequences are integers). I found the first one by trial and error. x=8 gives y=5. So the point (8,5) is on my line, and gives 58 for both 2 + 7x and 3 + 11y. (Points in which both coordinates are integers are called "lattice points" - there may be an easier way to find them, but I don't know it. Essentially, that's what we're looking for here - lattice points on our line.)
So 58 is in both sequences. Now, noting that the slope of our line is 7/11, we can increase our y value by 7 and our x value by 11 (since slope is change in y / change in x) to get the next value of x and y that gives integer coordinates: (19, 12). Plugging x = 19 and y = 12 into the appropriate equations gives 135 in both, so that's the next term.
Keep using slope to "jump" from each integer-coordinate point to the next. Use x and y, plug them into your equations, and you should be good to go:
(8,5), (19,12), (30, 19), (41, 26), ...
Hope this makes sense. I kind of figured it out while writing this, so it's a little... disorganized.
• May 22nd 2008, 01:17 PM
Mathstud28
Quote:
Originally Posted by Mathnasium
If I'm reading your original post correctly, you don't want the first ten terms of each sequence, right? You want the first ten terms that are COMMON to BOTH sequences. If so:
Here's what I did - it's slightly better than listing out every term.
The first sequence's terms can be described by 2 + 7x.
The second sequence's terms can be described by 3 + 11y.
I want to know when these are equal, so:
2 + 7x = 3 + 11y.
I solved for y and got:
The following is multiple choice question (with options) to answer.
if x:y = 1:5, then find the value of
(7x+3y):(2x+y) | [
"14:5",
"15:5",
"16:5",
"22:7"
] | D | Explanation:
let x = 1k and y = 5k, so
=7(k)+3(5k)/ 2(k)+1(5k)
=22k/7k
=22:7
Option D |
AQUA-RAT | AQUA-RAT-37195 | If $k = 5$ then $m$ may be $1....9$. Total: $9 + 7+ ..... + 1 = 25$ ways.
If $k = 6$ then $m$ may be $1....10$. Total: $10 + 25 = 35$ ways.
If $k = 7$ then $m$ may be $1..... 10$. Total: $10 + 10 + 25 = 45$ ways.
If $k = 8 .... 10$ then $m$ may be $1.... 10$. Total $10 + 10 + ..... + 10 + 25$ ways.
Total number of ways: $1 + 3 + 5+ 7 + +9 + 10+10+10 + 10+10 = 75$
For $n$ even we will have
$(1 + 3 + 5 + ..... + (2*(\frac n2)-1) + (n + n + n + .....n) =$
$(\frac n2)^2 + \frac n2*n = \frac 34 n^2$.
If $n$ is odd we will have.
$(1 + 3 + 5 + ..... + (2*(\frac {n-1}2) - 1) + (n+n+n .... +n) =$
$(\frac {n-1{2)^2 + \frac {n+1} 2*n = \frac 14(3n^2 + 1)$.
The following is multiple choice question (with options) to answer.
If the arithmetic mean of seventy five numbers is calculated, it is 35. If each number is increased by 5, then mean of new number is? | [
"43",
"40",
"87",
"77"
] | B | A.M. of 75 numbers = 35
Sum of 75 numbers = 75 * 35 = 2625
Total increase = 75 * 5 = 375
Increased sum = 2625 + 375 = 3000
Increased average = 3000/75 = 40.
Answer:B |
AQUA-RAT | AQUA-RAT-37196 | # Algebra
posted by Angela
If 3 is subtracted from the numerator of a fraction, the value of the fraction is 1/2. If 6 is added to the denominator of the ORIGINAL fraction, the value of the fraction is 1/2. What is the original fraction?
1. Damon
(x-3)/y = .5
x/(y+6) = .5
-----------------
.5 y = x - 3
.5 y + 3 = x
these two equations are the same, choose any old x (infinite number of solutions)
if x = 13
.5 y = 10 so y = 20
x/y = 13/20
check that
(13 -3)/20 = 1/2 sure enough
(13)/(26) = 1/2 sure enough
2. Angela
OK, THANK YOU. But I still do not understand why x=13. that is exactly what my text uses, as well. How did you know to use 13 for x?
3. Angela
I just reworked it using your method and subsituted as you did. Then I used x=15 and got 15/24 for my final answer. However text says the correct answer is 13/20, does not say infinite number. I am thinking this is a book error! Anyway, thank you!
4. Damon
I just used 13 because I wanted easy numbers
trying x = 15 now
.5 y = 12
y = 24
so
15/24 yes
15-3/24 = 1/2 yes
15/30 = 1/2 yes
so 15/24 is just as good as 13/20
## Similar Questions
1. ### algebra
The denominator of a fraction is 12 more than the numerator. If 16 is added to the numerator and 16 is subtracted from the denominator, the value of the resulting fraction is equal to 2/1. Find the original fraction
2. ### Maths
When the numerator and denominator of a fraction are each increased by 5, the value of the fraction becomes 3/5 When the numerator and denominator of that same fraction are each decreased by 5, the fraction is then 1/5 Find the original …
3. ### Algebra
The following is multiple choice question (with options) to answer.
The denominator of a fraction is 6 greater than the numerator. If the numerator and the denominator are increased by 1, the resulting fraction is equal to 4â„5. What is the value of the original fraction? | [
"17/23",
"23/29",
"29/35",
"31/37"
] | B | Let the numerator be x.
Then the denominator is x+6.
x+1 / x+7 = 4/5.
5x+5=4x+28.
x=23.
The original fraction is 23/29.
The answer is B. |
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