source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36197 | 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 500 meters long is running with a speed of 44 kmph. The time taken by it to cross a tunnel 230 meters long is? | [
"287 sec",
"288 sec",
"48 sec",
"61 sec"
] | D | D = 500 + 230 = 730
S = 44 * 5/18 = 12 mps
T = 730/12 = 61 sec
Answer: D |
AQUA-RAT | AQUA-RAT-36198 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A man rides at the rate of 50km/hr. But stops 20minutes to change horses at the end of every 25th kilometer. How long will he take to go a distance of 100 kilometers? | [
"3hr",
"6hr 30min",
"8hr 10min",
"7hr 20min"
] | A | speed of man = 50km/hr
number of rests = (100/25)-1 = 3
time taken for the man = (100/50)+3*(20/60) = 3 hr
Answer is A |
AQUA-RAT | AQUA-RAT-36199 | That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
VP
Joined: 07 Dec 2014
Posts: 1224
Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Consider there is an elevator and you are at the upside and coming down using elevator. You walk 20 steps and then you stop, then you reach to the ground in 10 minutes. If you walk 10 steps and then stop, then you reach to the ground in 20 minutes. What is the speed of the elevator and How many steps are there?? | [
"1 step/min",
"2 step/min",
"3 step/min",
"4 step/min"
] | A | Let e be the speed of the elevator, n be the no. of steps
n = 20 + 10*e = 10+20*e
e=1
n=30
No. of steps : 30
Speed : 1 step/min
ANSWER:A |
AQUA-RAT | AQUA-RAT-36200 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 44, the dividend is: | [
"5330",
"5336",
"4884",
"5345"
] | C | Divisor = (5 * 44) = 220
= 10 * Quotient = Divisor
=> Quotient = 220/10 = 22
Dividend = (Divisor * Quotient) + Remainder
Dividend = (220 * 22) + 42 = 4884.
C |
AQUA-RAT | AQUA-RAT-36201 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 42 minutes. The waste pipe can empty the full cistern in? | [
"8 min",
"7 min",
"5 min",
"9 min"
] | B | 1/10 + 1/15 - 1/x
= 1/42
x = 7
Answer: B |
AQUA-RAT | AQUA-RAT-36202 | Circular motion question
1. Nov 10, 2005
donjt81
This is the question...
A small wheel of radius 1.4cm drives a large wheel of radius 15cm by having their circumferences pressed together. If the small wheel turns at 407 rad/s, how fast does the larger one turn? Answer in rad/s
This is what I was thinking...
radius of smaller wheel = .014m
radius of larger wheel = .15m
circumference of smaller wheel = 2*pi*r = 2*3.14*.014 = .08792
angular velocity of smaller wheel (given) = 407 rad/s
angular velocity = circumference/time
time = circumference/angular velocity
=.08792/407 = .000216s
circumference of larger wheel = 2*pi*r = 2*3.14*.15 = .942
angular velocity = circumference/time
=.942/.000216 = 4361.11 rad/s
Does this approach look right?
2. Nov 10, 2005
BerryBoy
I disagree... By intuition, you can predict that the larger wheel is going to turn more slowly.. Try another approach..
Hint: Consider the fact that the speeds of circumferences are equal.
Eq: speed = w x r
w = angular velocity
r = radius
Does this help?
Sam
Last edited: Nov 10, 2005
3. Nov 10, 2005
donjt81
You are right... the larger wheel should go slower.
but since the speed of smaller wheel is 407 rad/s wont the larger wheel speed be the same?
so is the answer to the problem 407 rad/s for the larger wheel? but that doesnt make sense because the larger wheel is supposed to go slower...
I am confused...
4. Nov 10, 2005
BerryBoy
OK, so if the speed at the circumfrence is:
v = w x r (as I stated above).
If the wheels are in contact this speed is equal on both wheels (not the angular velocity). Therefore:
wsmall x rsmall = wlarge x rlarge
I can't give you anymore hints without doing it now.
The following is multiple choice question (with options) to answer.
The radius of the wheel of a bus is 175 cms and the speed of the bus is 66 km/h, then the r.p.m. (revolutions per minutes)of the wheel is | [
"100",
"250",
"300",
"330"
] | A | Radius of the wheel of bus = 175 cm. Then,
circumference of wheel = 2Ï€r = 350Ï€ = 1100 cm
Distance covered by bus in 1 minute
= 66â„60 × 1000 × 100 cms
Distance covered by one revolution of wheel
= circumference of wheel
= 1100 cm
∴ Revolutions per minute = 6600000/60×1100 = 100
Answer A |
AQUA-RAT | AQUA-RAT-36203 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
A candidate got 35% of the votes polled and he lost to his rival by 2250 votes. How many votes were cast? | [
"7500",
"2327",
"2997",
"2662"
] | A | 35%-----------L
65%-----------W
------------------
30%----------2250
100%---------? => 7500
Answer: A |
AQUA-RAT | AQUA-RAT-36204 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
12 men take 18 days to complete a job whereas 12 women in 18 days can complete 3⁄4 of the same job. How many days will 10 men and 8 women together take to complete the same job? | [
"6",
"13 1⁄2",
"12",
"Data inadequate"
] | B | 12 M × 18 = 12 W × 18 × 4⁄3
\ w = 3/4 M
10M + 8W = 10M + 8 × 3⁄4M = 16 M
\16 men can complete the same work
in 12×18/16=27/2=13 1/2days
Answer B |
AQUA-RAT | AQUA-RAT-36205 | ### Show Tags
04 May 2016, 10:03
1
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$
(A) $$9\sqrt{5}$$
(B) $$20\sqrt{5}$$
(C) $$41\sqrt{5}$$
(D) $$\sqrt{205}$$
(E) 100
Solution:
To solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.
√80 = √16 x √5 = 4√5
√125 = √25 x √5 = 5√5
Now we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.
4√5 + 5√5 = 9√5
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Intern
Joined: 01 Jan 2015
Posts: 31
Location: India
GPA: 3.71
WE: Consulting (Retail Banking)
### Show Tags
04 May 2016, 10:06
A
80=4*4*5
125=5*5*5
Root(80)=4Root(5)
Similarly 125
hence 9Root(5)
_________________
_________________
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3622
Location: India
GPA: 3.5
### Show Tags
04 May 2016, 10:19
Bunuel wrote:
$$\sqrt{80}+\sqrt{125} =$$
The following is multiple choice question (with options) to answer.
What is the square root of 121? | [
"11",
"12",
"16",
"18"
] | A | 11X11 = 121
ANSWER A |
AQUA-RAT | AQUA-RAT-36206 | MAT 142 3.4 Basics of Probability Theory
# MAT 142 3.4 Basics of Probability Theory
398.3k points
1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3
green candies and 4 orange candies. You select one at random. Find the probability that you
select one that is:
a. Brown
b. Green or orange
c. Not red
d. Yellow or blue
e. Yellow and blue
1. Shoppers at a local department store were asked to complete a survey of their shopping
experience. The results are shown in the table below:
** Satisfied with Service Not satisfied with Service** Totals
Made a purchase 130 494 624
Did not make a purchase 715 183 898
Totals 845 677 1522
__
a. What is the probability that a shopper selected at random made a purchase?
b. What are the odds that a shopper selected at random was satisfied with the service?
MAT 142
393k points
#### Oh Snap! This Answer is Locked
Thumbnail of first page
Excerpt from file: Math Tutorial MAT 142 Problem set Unit: Probability Topic: Basics of Probability Theory Directions: Solve the following problems. Please show your work and explain your reasoning. 1. A jar of M&Ms contains 12 brown candies, 4 yellow candies, 2 blue candies, 5 red candies, 3 green candies and 4
Filename: M1084.pdf
Filesize: 369.5K
Print Length: 2 Pages/Slides
Words: 292
J
0 points
#### Oh Snap! This Answer is Locked
Thumbnail of first page
Excerpt from file: Week4DQ#3DiscOps&ExtraordinaryItems Whyisitimportanttoreportdiscontinuedoperationsorextraordinaryitemsseparately fromincomefromcontinuingoperations?Isthismethodofreportingallowed?What concernsdoesthistypeofreportingcreate?Doestheaverageinvestorunderstandthe
Filename: ACC 280week4dqs.zip
Filesize: < 2 MB
Print Length: 1 Pages/Slides
Words: NA
The following is multiple choice question (with options) to answer.
A bag of P peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when Pm is divided by 18? | [
"6",
"4",
"2",
"8"
] | A | P = 9x + 6
m = 12y + 4
Pm = (9x + 6)*(12y + 4) = 108xy + 36x + 72y + 24
Remainder of nm/18 = (108xy + 36x + 72y + 24)/18
Observe that the first three terms are a multiple of 18
24 when divided by 18 leaves remainder 6
Hence mn/18 will leave remainder 6
A |
AQUA-RAT | AQUA-RAT-36207 | # Probability a 9-digit number has the digits 2,4, and 6 next to each other.
The integers $1,2,3,....,9$ are arraned (at random) in a row, resulting in a $9$-digit integer (without replacement). What is the probability that:
The result is even? $\frac49$ or $\frac{4(8!)}{9!}$
The result is divisible by $5$? The number must end in $0$ or $5$. Edit: As Andre pointed out we have no $0$. So, $\frac19$. or $\frac{(8!)}{9!}$
The digits $2, 4,$ and $6$ are next to each other (in any order)? The above two I have confidence in, it is this last one I'm a little confused on.
$9\choose 3$ ways to position $2,4,6$ in the 9-digit number and $3!$ ways they can be arranged. This doesn't appear to be right as after you divide by $9!$ you get $.13$% which seems unreasonably low. So, I thought to multiple by $6!$ to account for the number of ways the other $6$ numbers can be arranged. This gives you:
$9\choose 3$$3!6!/9!$
which equals $1$, and obviously isn't right. Any suggestions as to where I went wrong would be great.
-
You're right! Silly me. – Vincent Jun 16 '14 at 22:47
For $2,4,6$ next to each other, imagine choosing the $3$ spots that will be reserved for our three favoured guests. This can be done in $\binom{9}{3}$ equally likely ways.
The three spots can be in a row in $7$ ways, for the leftmost of the spots can be in any of the positions $1$ to $7$. Thus the required probability is $\frac{7}{\binom{9}{3}}$. This simplifies nicely.
The following is multiple choice question (with options) to answer.
1 boy forgot the last digit of a 7 digit telephone no. If he randomly dial thefinal 3 digits after correctly dialing the 1st four, then what is the chance of dialing the correct no.? | [
"2/989",
"3/456",
"1/1000",
"1/2314"
] | C | it is given that last 3 digits are randomly dialed
Then, each of the digit can be selected out of 10 digits in 10 ways. Hence, required probability
= 1/(10)3 = 1/1000
C |
AQUA-RAT | AQUA-RAT-36208 | b) prime c) even or prime? Homework Equations The Attempt at a Solution a) P(even) = 1/2 b) P(prime) = 9/16 c) c for confused Can someone please explain the theory behind answering Question c? Cheers. A series of ruby scripts for dice shenanigans. 9 Compute the expected value of the score when rolling two dice. The two probabilities are compared as a function of n in the graph below: (The final sentence in the question suggests that Melody's answer is what was being sought. a) A die is rolled, find the probability that the number obtained is greater than 4. 5 means the event A is equally likely to occur. Return to interactive exercise for conditions. For the first two dice, there are 3 3 = 9 favorable outcomes as shown here: For three dice, there are 3 3 3 = 27 favorable outcomes. Dice Probability Calculator App Calculate odds of sums and combinations for six-sided dice. The sum of the two dice values is a number from 2 through 12. It can do more than simple d20 rolls, like calculating average damage against a target's AC given a weapon. If you roll two dice, what is the probability that the sum of the two is odd? #N#As we can see in the chart above, there are 18 combinations that result in an odd sum. 40K Visual Dice Calculator (8th Ed) Please also check out the version for AoS. The probability of Dice 2 rolling a 1 is also 1/6. !Each boy at a school plays one of four sports. Usually the dice within one set are equal, but different from the kind of dice in the other set. I wrote it in Perl. There are 62=36. d) A card is drawn at random from a deck of cards. The dice probability calculator that the next throw is heads is 1 in 2. The critical step in calculating the odds in Sic Bo is to find the probability of any given total in the throw of three dice. There are still 36 different combinations, so: #N#Show Next Step. There is an equal probability of rolling each of the numbers 1-6. 129 Probability Probability of A = Number of outcomes for which A. Example: there are 5 marbles in a bag: 4 are. As you can see, 7 is the most common roll with two six-sided dice. Probability represents the
The following is multiple choice question (with options) to answer.
A cube with its sides numbered 1 through 6 is rolled twice, first landing on a and then landing on b. If any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6, what is the probability R that a + b is prime? | [
" 0",
" 1/12",
" 5/12",
" 7/18"
] | C | Total # of outcomes is 6*6=36;
Favorable outcomes:
a-b --> prime
1-1 --> 2;
1-2 --> 3;
2-1 --> 3;
1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;
1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;
6-5 --> 11;
5-6 --> 11.
Total of 15 favorable outcomes
R=15/36.
Answer: C. |
AQUA-RAT | AQUA-RAT-36209 | The number of passwords with neither an upper case letter nor a digit is $26 \cdot (62 - 26 - 10)^7 = 26 \cdot 26^7$.
Notice that if we subtract the number of passwords with no upper case letters and the number of passwords with no digits from $26 \cdot 62^7$, we will have subtracted those passwords with neither an upper case letter nor a digit twice, once when we subtracted the number of passwords with no upper case letters and once when we subtracted the number of passwords with no digits. Since we only want to subtract these numbers once, we must add them back. Hence, by the Inclusion-Exclusion Principle, the number of permissible passwords is $$26 \cdot 62^7 - 26 \cdot 52^7 - 26 \cdot 36^7 + 26 \cdot 26^7 = 63,003,474,293,760$$
• @JMoravitz Thank you. Clearly, you did a better job proofreading than I did. – N. F. Taussig Mar 24 '17 at 18:41
The following is multiple choice question (with options) to answer.
A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible? | [
"9! + 10!",
"2 x 10!",
"9! x 10!",
"19!"
] | B | If we choose the 10 different digits then they can be arranged (permutations) in 10! ways.
But the question asks at least 9 digits. So we have the possibility of choosing only 9 digits for the password ( but digit shouldn't repeat), so we can have a total of 10 different combinations and each combination can be arranged in 9! ways.
Therefore
10 x 9! + 10!
= 10! + 10!
= 2 x 10!
Therefore B |
AQUA-RAT | AQUA-RAT-36210 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
Two cars cover the same distance at the speed of 60 and 64 kmps respectively. Find the distance travelled by them if the slower car takes 1 hour more than the faster car. | [
"980 km",
"860 km",
"960 km",
"460 km"
] | B | B
60(x + 1) = 64x
X = 15
60 * 16 = 960 km |
AQUA-RAT | AQUA-RAT-36211 | # 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.
Sonika deposited Rs.14500 which amounted to Rs.12200 after 3 years at simple interest. Had the interest been 3% more. She would get how much? | [
"13505",
"12004",
"15003",
"14500"
] | A | (14500*3*3)/100 = 1305
12200
--------
13505
Answer: A |
AQUA-RAT | AQUA-RAT-36212 | 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 310 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long? | [
"40 sec",
"50 sec",
"44 sec",
"36 sec"
] | D | Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 310 + 140 = 450 m
Required time = 450 * 2/25 = 36 sec
Answer:D |
AQUA-RAT | AQUA-RAT-36213 | if you would like to know the stuff related arithmetic sequences and series, Please click here. But each term is from a sum of 41 and an arithmetic series. If his scores continued to increase at the same rate, what will be his score on his 9th quiz? Show all work. Revise calculations for arithmetic and geometric series Summary Arithmetic Geometric Constant difference = d OR Constant ratio = r OR Test Yourself Select the most correct answer from the options given. Secondly, the determinant of the arithmetic mean of two matrices can be considerably bigger than the determinants of the two matrices themselves, which does not make sense in some applications. Recursive and explicit equations for arithmetic and geometric. Practice with Arithmetic and Geometric Sequences. The NRICH Project aims to enrich the mathematical experiences of all learners. The first is to calculate any random element in the sequence (which mathematicians like to call the "nth" element), and the second is to find the sum of the geometric sequence up to the nth element. 1 #7,9,2331odd find the general (nth) term of the sequence. Find the 20th term in the arithmetic sequence for which _____ 4. What I want to do in this video is familiarize ourselves with a very common class of sequences. org READY ! Topic:!Distinguishing!betweenarithmetic!and!geometric!sequences!!. Marcus An unbiased forecast of the terminal value of a portfolio requires compounding of its initial lvalue ut its arithmetic mean return for the length of the investment period. Standard: CCSS. 2: Geometric Sequences pg. Chapter 13 Sequences and Series 251 (c) If S 2 = t 1 + t 2, what does S 2 represent? What does S n mean? Calculate S10, S20 and S50. Arithmetic And Geometric Sequences Worksheet Pdf in an understanding medium may be used to check students qualities and knowledge by answering questions. A gift to the children and math students of the world from the U. 7 Carry out a procedure to define a sequence recursively when given four or more consecutive terms of the sequence. Find the first 4 terms of the geometric sequence with a=-6 and r= -2/3 You will need to determine if the series is arithmetic. The terms arithmetic sequence, common difference, geometric sequence, and common ratio are then defined, examples are provided, and students respond to clarifying questions. symbol sequences by
The following is multiple choice question (with options) to answer.
An arithmetic Sequence is a number sequence which has a constant difference between terms. Considering this definition which of the following is an arithmetic sequence?
I. e^2, f^2, g^2, h^2, i^2
II. e-6, f-3, g-5, h-3, i-3
III. 2e, 2f, 2g, 2h,2i | [
"II and III",
"III only",
"I and III",
"I and II"
] | B | lets look at the three choices...
I. e^2, f^2, g^2, h^2, i^2
since the difference is not constant, it is not an arithmetic sequence...
II. e-6, f-3, g-5, h-3, i-3
since the difference is not constant, it is not an arithmetic sequence
III. 2e, 2f, 2g, 2h,2i
since the difference is constant it is an arithmetic sequence...
The answer B) |
AQUA-RAT | AQUA-RAT-36214 | # 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.
A person borrows Rs. 4000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6 p.a for 2 years. Find his gain in the transaction per year. | [
"100 rs",
"80 rs",
"160 rs",
"180 rs"
] | B | Gain in 2 years =
[(4000*6*2)/100]-[(4000*4*2)/100]
480-320=160
Gain in 1 year =(160/2)=80 rs
ANSWER:B |
AQUA-RAT | AQUA-RAT-36215 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular garden has a fence along three sides and a wall along the fourth side. The fenced side opposite the wall is twice the length of each of the other two fenced sides. If the area of the rectangular garden is 1800 square feet, what is the total length of the fence, in feet? | [
"60",
"100",
"120",
"200"
] | D | two sides EACH = x the third = 2x and the wall length is thus 2x too
x*2x = 2x^2 = 1800 ie x^2 = 900 ie x = 30
L = 60 W= 30
TOTAL LENGHT OF FENCE = 2*30+60 = 120
MY ANSWER IS C |
AQUA-RAT | AQUA-RAT-36216 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 40 natural numbers? | [
"20.5",
"22",
"30.5",
"35"
] | A | Sum of first 100 natural numbers = 40*41/2 = 820
Required average = 820/40 = 20.5
Answer is A |
AQUA-RAT | AQUA-RAT-36217 | 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.
Each of the 38 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 12 students sign up for the poetry club, 20 students for the history club, and 19 students for the writing club. If 3 students sign up for exactly two clubs, how many students sign up for all three clubs? | [
"2",
"3",
"4",
"5"
] | D | The total number in the three clubs is 12+20+19=51.
All 38 students signed up for at least one club.
3 of those students signed up for exactly one more club.
51 - 41 = 10 so 5 students must have signed up for exactly three clubs.
The answer is D. |
AQUA-RAT | AQUA-RAT-36218 | algorithms
Title: Divide numbers into fewest groups with each group's sum bounded We have an integer $K$ and an array of $N$ positive integers each smaller than $K$. How do we group the numbers into the minimum number of groups where the sum of each group is less than or equal to $K$? As comments suggested it is called the Bin Packing Problem.
The following is multiple choice question (with options) to answer.
A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is? | [
"31",
"42",
"50",
"22"
] | A | LCM = 30 => 30 + 1 = 31
ANSWER A |
AQUA-RAT | AQUA-RAT-36219 | Once again
Repetition helps too, so let’s recite: it starts with $289 = 17^2$, then continues by 102s: 391, 493. After that the twins 527, 529, followed by 629; then 667 and 697. Then two sets of twins each with its 99: 713, 731, 799; 841, 851, 899; then 901 to come after 899, and then the three sporadic values: 943, 961, 989!
Posted in arithmetic, computation, primes |
Quickly recognizing primes less than 1000: divisibility tests
I took a little hiatus from writing here since I attended the International Conference on Functional Programming, and since then have been catching up on teaching stuff and writing a bit on my other blog. I gave a talk at the conference which will probably be of interest to readers of this blog—I hope to write about it soon!
In any case, today I want to return to the problem of quickly recognizing small primes. In my previous post we considered “small” to mean “less than 100”. Today we’ll kick it up a notch and consider recognizing primes less than 1000. I want to start by considering some simple approaches and see how far we can push them. In future posts we’ll consider some fancier things.
First, some divisibility tests! We already know how to test for divisibility by $2$, $3$, and $5$. Let’s see rules for $7$, $11$, and $13$.
• To test for divisibility by $7$, take the last digit, chop it off, and subtract double that digit from the rest of the number. Keep doing this until you get something which obviously either is or isn’t divisible by $7$. For example, if we take $2952$, we first chop off the final 2; double it is 4, and subtracting 4 from $295$ leaves $291$. Subtracting twice $1$ from $29$ yields $27$, which is not divisible by $7$; hence neither is $2952$.
The following is multiple choice question (with options) to answer.
If k^2 is divisible by 240 what is the least possible value of integer K? | [
"60",
"30",
"12",
"90"
] | A | 240 can be written as (2^4)*3*5.
For k^2 to be divisible by 240 it should contain at least 2^4 and 3 and 5 in its factors. We can leave out option C because 12 doesnt have 5 as one of its factor. Now if we check for option B, 30 can be written as 2*3*5, hence 30^ 2 will have 2 as the maximum power of 2, so we can leave out this option too.
Option C is the right answer if we follow the same method as we followed for other two previous options.
60 = (2^2)*3*5; 60^2 = (2^4)*(3^2)*(5^2). So it shows that 60^2 is divisible by 240 and hence the answer.
Answer : A |
AQUA-RAT | AQUA-RAT-36220 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
If the cost price is 99% of selling price then what is the profit percentage. | [
"1",
"1.11",
"1.01",
"1.1"
] | C | selling price=Rs100 : then cost price=Rs 99:profit =Rs 1.
Profit={(1/99)*100}%=1.01%
Answer is C. |
AQUA-RAT | AQUA-RAT-36221 | ### Solution 1
Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$, and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$. Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$. $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$, the $70$ accounting for the difference between $2005$ and $44^2 = 1936$, inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$. Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$.
### Solution 2
Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$, where the $-19$ accounts for those numbers between $2005$ and $2024$.
Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$.
The following is multiple choice question (with options) to answer.
What is the sum of two consecutive even numbers, the difference of whose squares is 84? | [
"42",
"44",
"48",
"56"
] | A | Sol.
Let the numbers be x and x + 2.
then, (x + 2)² - x² = 84 ⇔ 4x + 4 = 84 ⇔ 4x = 80 ⇔ x=20.
∴ Required sum = x + (x + 2) = 2x + 2 = 42.
Answer A |
AQUA-RAT | AQUA-RAT-36222 | Assume that the set S has 7 elements. First, our two subsets can have 2 and 5 elements. Then, the formula to find number of subsets is. You can use Pascal's triangle to figure out how many subsets have no elements, one element, two elements and so on. Total subsets of with odd number of elements is: Oct 18, 2012 · Assume there are X_n subsets for [n] like that. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, C By having a factor of two and one of five, then the product of the elements of the set is just some multiple of $10$, precisely what we desire. Substituting for in, to get. My first inclination was to use something like RandomSample[Subsets[list, {25}], 1000], but the problem is the number of subsets of length 25 out of a 300 element set is way to big for the computer to deal with. A set containing 10 elements can have subsets which will contain 1 element, 3 elements, 5 elements, 7 elements, and 9 elements. Consider a set with one element: {A}. Thus, the required number of subsets Permutations and Combinations Questions & Answers : Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements. How many subsets does the set {apple, banana} have? It could have {apple}, or {banana}, and don't forget: the whole set: {apple, banana} the empty set: {} So a set with two elements has 4 subsets. Set A contains 34 elements and Set B contains 98 elements. asked by Elle on April 27, 2009; Sets. if they don't include n then there are X_(n-1) of those subsets if they include n then they can't include n-1 (no consecutive) so we have choices of n-2 numbers that is X_(n-2). There are ways to choose the 2-element subset. 1 :49. No distinction will be made between subsets except for their size. The objective is to find the number of subsets with more than two elements of a set having 100 elements. What you want is the
The following is multiple choice question (with options) to answer.
The set S has 36 different subsets each of which contains exactly two elements. How many subsets of S could contain exactly seven elements each? | [
"81",
"63",
"54",
"36"
] | D | nC2 = 36
=> n*(n-1)/2 = 36 by middle term factor and n cannot be negative
=> n = 9
nC7 = 9C7 = 9!/7!*(9-7)!= 9*8*7!/7!*2 =36
So, Answer is D. |
AQUA-RAT | AQUA-RAT-36223 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The average of the marks of 12 students in a class is 36. If the marks of each student are doubled, find the new average? | [
"72",
"88",
"15",
"10"
] | A | Sum of the marks for the 12 students = 12 * 36 = 432. The marks of each student are doubled, the sum also will be doubled.
The new sum = 432 * 2 = 864. So, the new average = 864/12 = 72.
Answer: A |
AQUA-RAT | AQUA-RAT-36224 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
Car A runs at the speed of 65km/hr & reaches its destination in 8hr. Car B runs at the speed of 70 km/h & reaches its destination in 4h. What is the respective ratio of distances covered by Car A & Car B? | [
"11 : 5",
"11 : 8",
"13 : 7",
"15 : 7"
] | C | Sol. Distance travelled by Car A = 65 × 8 = 520 km
Distance travelled by Car B = 70 × 4 = 280 km
Ratio = 520/280 = 13 : 7
C |
AQUA-RAT | AQUA-RAT-36225 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
If the price of an article went up by 40%, then by what percent should it be brought down to bring it back to its original price? | [
"2/3%",
"28.57%",
"2/1%",
"1/3%"
] | B | Let the price of the article be Rs. 100.
40% of 100 = 40.
New price = 100 + 40 = Rs. 140
Required percentage = (140 - 100)/140 * 100
= 40/140 * 100= 28.57%.
Answer:B |
AQUA-RAT | AQUA-RAT-36226 | # If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?
It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
The following is multiple choice question (with options) to answer.
If n is the product of the integers from 1 to 13, inclusive, how many different prime factors greater than 1 does n have? | [
"Four",
"Five",
"Six",
"Seven"
] | C | n = 1*2*3*4*5*6*7*8*9*10*11*12*13
The prime factors of n are 2, 3, 5, 7, 11, and 13.
There are 6 prime factors.
The answer is C. |
AQUA-RAT | AQUA-RAT-36227 | For the LHS you want $4x\equiv x\pmod 3$. This is true because $3$ divides $4x-x=3x$. (Recall the definition of modulo: "$a\equiv b\pmod n$" means "$n$ divides $a-b$".)
Notationally, it is simpler to use bar notation:
$\overline{2x+11}=\overline{7}=\overline{1}\Longrightarrow$
$\overline{2x}+\overline{11}=\overline{1}\Longrightarrow$
$\overline{2x}=\overline{1}-\overline{11}=\overline{-10}$ $=\overline{2}\Longrightarrow$
$\overline{x}=\overline{1}$.
The following is multiple choice question (with options) to answer.
If m=9^(x−1), then in terms of m, 3^(4x−1) must be which of the following? | [
"m/3",
"9m",
"27m^2",
"m^2/3"
] | C | m = 9 ^ (X-1)
m = 3 ^ (2x-2)
m^2 = 3 ^ (4x-4)
27m^2 = 3 ^ (4x-1)
Answer C |
AQUA-RAT | AQUA-RAT-36228 | 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 1st 3 of 4 numbers is 16 and of the last 3 are 15. If the sum of the first and the last number is 13. What is the last numbers? | [
"4",
"7",
"5",
"9"
] | C | A + B + C = 48
B + C + D = 45
A + D = 13
A – D = 3
A + D = 13
2D = 10
D = 5
Answer:C |
AQUA-RAT | AQUA-RAT-36229 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
On a certain date, Pat invested $7,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 10 years will be $28,000, in how many years total will the total value of the investment plus interest increase to $56,000? | [
"15",
"16",
"18",
"20"
] | A | 28,000 = 7,000(1 + x)^10
4 = (1+x)^10 = 2^2
(1+x)^10 = ((1+x)^5)^2 = 2^2
Therefore, (1+x)^5 = 2
56,000 = 7,000(1 + x)^n
(1+x)^n = 8
(1+x)^n = 2^3
(1+x)^n = ((1+x)^5)^3 = (1+x)^15
Therefore, n = 15.
The answer is A. |
AQUA-RAT | AQUA-RAT-36230 | (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.
On a Saturday night, each of the rooms at a certain motel was rented for either $40 or $60. If 10 of the rooms that were rented for $60 had instead been rented for $40, then the total rent the motel charged for that night would have been reduced by 50 percent. What was the total rent the motel actually charged for that night ? | [
" $400",
" $800",
" $1,000",
" $1,600"
] | A | Let total rent the motel charge for all rooms =x
If 10 rooms that were rented for 60 $ had instead been rented for 40 $,
then total difference in prices = 20 $ * 10 = 200 $
Total rent the motel charged would have been reduced by 50 %
.5x = 200
=> x= 400
Answer A |
AQUA-RAT | AQUA-RAT-36231 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
It takes John 25 minutes to walk to the car park and 45 to drive to work. At what time should he get out of the house in order to get to work at 9:00 a.m.? | [
"6:50 a.m",
"5:50 a.m",
"4:50 a.m",
"7:50 a.m"
] | D | The time it takes John to get to work: time to walk to car park + time to drive
25 + 45 = 70 minutes = 1 hour and 10 minutes
John needs to get out of the house 1 hour and 10 minutes before 9:00 am at
9:00 - 1:10 = 7:50 a.m
correct answer D |
AQUA-RAT | AQUA-RAT-36232 | $\text{(ii) }\;1 - P(A\cap B\cap C) \;= \;0.88$ . . . . Right!
(b) (i) only the New York flight is full. (ii) exactly one of the three flights is full.
$\text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18$
$\text{(ii)}\;\begin{array}{ccccc}P(A \cap \overline{B} \cap \overline{C}) & = & (0.6)(0.5)(0.6) & = & 0.18 \\
P(\overline{A} \cap B \cap \overline{C}) &=& (0.4)(0.5)(0.6) &=& 0.12 \\
P(\overline{A} \cap \overline{B} \cap C) &=& (0.4)(0.5)(0.4) &=& 0.08\end{array}$
$P(\text{exactly one full}) \;=\;0.18 + 0.12 + 0.08 \;=\;0.38$
The following is multiple choice question (with options) to answer.
A, B and C have Rs.500 between them, A and C together have Rs.200 and B and C Rs.320. How much does C have? | [
"50",
"78",
"267",
"20"
] | D | A+B+C = 500
A+C = 200
B+C = 320
--------------
A+B+2C = 520
A+B+C = 500
----------------
C = 20
Answer: D |
AQUA-RAT | AQUA-RAT-36233 | 12n + 7 7, 19, 31, 43, 67, 79, 103, 127, 139, 151, ... A068229
12n + 11 11, 23, 47, 59, 71, 83, 107, 131, 167, 179, ... A068231
The following is multiple choice question (with options) to answer.
5n + 2 > 12 and 7n - 5 < 30; n must be between which numbers? | [
"1 and 8",
"2 and 6",
"0 and 9",
"2 and 5"
] | D | 5n > 10 --> n > 2
7n < 35 --> n < 5
2 < n < 5
Answer: D |
AQUA-RAT | AQUA-RAT-36234 | ## 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 -1 < x < 0, which of the following must be true?
I. x^5 < x^2
II. x^6 < 1 – x
III. x^7 < x^3 | [
"I,II,III only",
"I,II only",
"I,III only",
"II only"
] | A | Answer : A |
AQUA-RAT | AQUA-RAT-36235 | gears, mechanisms, homework
Title: Conical pulley system and transmission ratio (closed) I tried asking this on other SE forum but i think this is the correct one:
Basically our teacher has sent us quite a lot of exercises but I can't get past this one which says: a pulley cone consists of 3 pulleys of 150, 250 and 350 mm of diameter that link an identical but inverted cone. Determine the three possible transmission ratios. Our teacher doesn't explain very well and I don't know where to search. If someone knows something about this, anything would be helpful.
PD: if you think I didn't tried to solve it, in all the book there's no information about this type of mechanism and we don't see our teacher until the day we need to bring the homework. There's no more info. in the exercise sorry.
(ignore, i messed up the question here)Edit: by the way what I'm trying to look for here is for a formula about the transmission ratio, but I can't understand how to do the formula with this specific mechanism because I don't see how to get the speed of the pulleys, here's an example given by the book.
Edit 2: i finally came up with this answer:
This question defines a system of adjustable speed pulleys, of this type of mechanisms what we can see as a ratio is that the bigger the gear we choose the pulley will go slower, this means that if we choose the gears that are parallel to each other the speed will always be the same because we will always have gears that are proportional to each other.
350mm -> 150mm
250mm -> 250mm
150mm -> 350mm Your question is describing an adjustable speed pulley system. These are commonly found in bench drill presses. (See the link below.)
Figure 1. A 5-speed adjustable speed pulley system. Image source: Toolbox Buzz.
Since the centres of the shafts are fixed and the length of the belt is fixed this arrangement requires that the path of the belt on each "level" of the system must be a constant (because the belt isn't elastic). Therefore, as you increase the pulley circumference on one shaft you must decrease it on the other to keep the path length constant.
Question to get you thinking:
The following is multiple choice question (with options) to answer.
An automobile manufacturer offers a station wagon with either a 12-cylinder engine or a 10-cylinder engine and with either a manual transmission or an automatic transmission. A trailer hitch is also offered, but only on a station wagon with a 12-cylinder engine. How many combinations of the five options listed does the manufacturer offer for its station wagon? | [
" 44",
" 47",
" 48",
" 50"
] | C | 12-cylinder engine wagons = 2*2*2*2*2 = 32 (manual or automatic, with or without trailer);
10-cylinder engine wagons = 2*2*2*2 = 16 (manual or automatic).
Total = 32 + 16 = 48.
Answer:C. |
AQUA-RAT | AQUA-RAT-36236 | javascript, object-oriented, unit-conversion
All units are converted to a base unit type (kelvin and meters per second)
Usage as follows
Units.kmph = 100;
const mph = Units.mph;
Units.celsius = 38;
const f = Units.fahrenheit;
The following is multiple choice question (with options) to answer.
Express a speed of 60 kmph in meters per second? | [
"10 mps",
"16.67 mps",
"97.67 mps",
"17.67 mps"
] | B | 60 * 5/18
= 16.67 mps
Answer: B |
AQUA-RAT | AQUA-RAT-36237 | the letter B occurs 3 times. Like a "super wildcard". Not all of those are distinguishable, because some words occur more than once on a single line. (d) type and case of letters, spacing between letters and punctuation marks; (e) joining words together or separating the words does not make a name distinguishable from a name that uses the similar, separated or joined words; (f) use of a different tense or number of the same word does not distinguish one name from another;. So a reasonable guess would be that total comparable settlements would be 150 times$24 million in cash, or $1. Swap these numbers as reverse the subset. 2% more than the actual collections in January 2019, and$35 million or 1. We couldn't distinguish among the 4 I's in any one arrangement, for example. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. We can use the following formula, where the number of permutations of n objects taken k at a time is written as n P k. Permutation. 4 billion increase in dividends paid, and a $324 million decrease in proceeds from the issuance of common stock, offset in part by a$3. P (10,3) = 720. Explanation: The small letters are b, d, f, h, j, l, n, p, r, t, v, x, z. From a 4 billion Indian Rupee gaming industry in 2007, to a 62 billion Indian Rupees industry in 2019, gaming in India has certainly caught the eye of consumers and proves to be a valuable market today. ership as an arrangement that leverages the uniqueness of Dutch law to avoid taxes and prevent a hostile takeover attempt. How many distinguishable permutations of letters are possible in the word Tennessee? I understand this word has 9 letters, with 1-T, 4-E, 2-N, and 2-S. One of these code words, the 'start signal' begins all the sequences that code for amino acid chains. Microsoft Excel. com, the international travel industry has grown from 528 million tourist arrivals in 2005 to 1. The spots can also be divided by the total of legs, ears, eyes and tail to leave a remainder of 6. In the Text section, click the WordArt option. Letter Arrangements in a Word Video. 4 percent of loan participations
The following is multiple choice question (with options) to answer.
All of the stocks on the over-the-counter market are designated by either a 7-letter or a 7-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes? | [
"2(26^5)",
"26(26^4)",
"27(26^4)",
"2(26^7)"
] | D | Number of 4-letter codes: 26 * 26 * 26 * 26 * 26 * 26 * 26 = 26^7
Number of 5-letter codes: 26 * 26 * 26 * 26 * 26 * 26 * 26 = 26^7
Total Number of codes: 26^7 + 26^7 = 2*(26^7)
Therefore, the answer isD: 2*(26^7). |
AQUA-RAT | AQUA-RAT-36238 | P\left( S=18 \right) =\dfrac { 3 }{ P\left( T \right) } \\ P\left( S=21 \right) =\dfrac { 2 }{ P\left( T \right) } \\ Let\quad D\quad be\quad the\quad event\quad that\quad sum\quad is\quad divisible\quad by\quad 3\\ Then\quad P\left( D \right) =P\left( S=3 \right) +P\left( S=6 \right) +P\left( S=9 \right) +P\left( S=12 \right) +P\left( S=15 \right) +P\left( S=18 \right) +P\left( S=21 \right) \\ \Rightarrow P\left( D \right) =\dfrac { 22 }{ P\left( T \right) } =\dfrac { 22 }{ 66 } \\ \Rightarrow P\left( D \right) =\dfrac { 1 }{ 3 }$$Maths
The following is multiple choice question (with options) to answer.
If t = 20! + 17, then t is divisible by which of the following?
I. 15
II. 17
III. 19 | [
" None",
" I only",
" II only",
" I and II"
] | C | Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So t!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer. |
AQUA-RAT | AQUA-RAT-36239 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
What will come in place of the x in the following Number series? 6, 12, 21, x , 48 | [
"33",
"37",
"39",
"41"
] | A | (A)
The pattern is + 6, + 9, + 12, +15 ………..
So the missing term is = 21 + 12 = 33 |
AQUA-RAT | AQUA-RAT-36240 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 110 m long is running with a speed of 60 km/hr. In what time will it pass a man who is running at 6 km/hr in the direction opposite to that in which the train is going? | [
"2 sec",
"1 sec",
"6 sec",
"8 sec"
] | C | Explanation:
Speed of train relative to man = 60 + 6 = 66 km/hr.
= 66 * 5/18 = 55/3 m/sec.
Time taken to pass the men = 110 * 3/55 = 6 sec.
Answer: C |
AQUA-RAT | AQUA-RAT-36241 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1326
Location: Viet Nam
Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
### Show Tags
02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
If A lends Rs.25000 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is? | [
"1125",
"1225",
"1325",
"1145"
] | A | (25000*1.5*3)/100 => 1125
ANSWER:A |
AQUA-RAT | AQUA-RAT-36242 | # How to calculate the area covered by any spherical rectangle?
Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a sphere with a radius $R$?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
• What do you mean by "rectangle"? Do you just want a quadrilateral, or some stronger condition? (Note that you cannot have four right angles in a quadrilateral on the sphere.) – Nick Matteo Mar 25 '15 at 13:16
• Yes, a quadrilateral having equal opposite sides (but not parallel) each as a great circle arc on a sphere. – Harish Chandra Rajpoot Mar 25 '15 at 13:21
• What do you mean by length and width? As the sides are not parallel, we don't have the usual idea of width and length. Do you simply mean the lengths of the sides? – robjohn Apr 4 '15 at 14:12
• Yes, length & width are simply the sides of the rectangle as great circles arcs on the spherical surface. – Harish Chandra Rajpoot May 28 '15 at 23:56
Assume we are working on a sphere of radius $1$, or consider the lengths in radians and the areas in steradians.
Extend the sides of length $l$ until they meet. This results in a triangle with sides $$w,\quad\frac\pi2-\frac l2,\quad\frac\pi2-\frac l2$$
The following is multiple choice question (with options) to answer.
The length of a rectangle is two - fifths of the radius of a circle. The radius of the circle is equal to the side of the square, whose area is 1225 sq.units. What is the area (in sq.units) of the rectangle if the rectangle if the breadth is 10 units? | [
"140 sq.units",
"176 sq.units",
"675 sq.units",
"169 sq.units"
] | A | Given that the area of the square = 1225 sq.units
=> Side of square = √1225 = 35 units
The radius of the circle = side of the square = 35 units Length of the rectangle = 2/5 * 35 = 14 units
Given that breadth = 10 units
Area of the rectangle = lb = 14 * 10
= 140 sq.units
Answer:A |
AQUA-RAT | AQUA-RAT-36243 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet? | [
"10 am",
"12 am",
"10.30 am",
"12.30 am"
] | A | Assume both trains meet after x hours after 7 am
Distance covered by train starting from P in x hours = 20x km
Distance covered by train starting from Q in (x-1) hours = 25(x-1)
Total distance = 110
=> 20x + 25(x-1) = 110
=> 45x = 135
=> x= 3
Means, they meet after 3 hours after 7 am, ie, they meet at 10 am
Answer is A. |
AQUA-RAT | AQUA-RAT-36244 | # 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.
n is a whole number which when divided by 4 gives 3 as remainder. What will be the remainder when 2n is divided by 4 ? | [
"2",
"5",
"6",
"1"
] | A | Let n = 4q + 3. Then 2n = 8q + 6 = 4(2q + 1 ) + 2.
Thus, when 2n is divided by 4, the remainder is 2.
ANSWER A |
AQUA-RAT | AQUA-RAT-36245 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
If P represents the product of the first 13 positive integers, which of the following must be true?
I. P is an odd number
II. P is a multiple of 24
III. P is a multiple of 23 | [
"I only",
"II only",
"III only",
"None of the above"
] | B | The correct answer is B. |
AQUA-RAT | AQUA-RAT-36246 | 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.
In one alloy there is 12% chromium while in another alloy it is 8%. 15 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy. Find the percentage of chromium in the new alloy. | [
"9.4%",
"9.33%",
"9.6%",
"9.8%"
] | B | The amount of chromium in the new 15+30=45 kg alloy is 0.12*15+0.08*30=4.2 kg, so the percentage is 4.2/45*100=
9.33%.
Answer: B. |
AQUA-RAT | AQUA-RAT-36247 | Once again
Repetition helps too, so let’s recite: it starts with $289 = 17^2$, then continues by 102s: 391, 493. After that the twins 527, 529, followed by 629; then 667 and 697. Then two sets of twins each with its 99: 713, 731, 799; 841, 851, 899; then 901 to come after 899, and then the three sporadic values: 943, 961, 989!
Posted in arithmetic, computation, primes |
Quickly recognizing primes less than 1000: divisibility tests
I took a little hiatus from writing here since I attended the International Conference on Functional Programming, and since then have been catching up on teaching stuff and writing a bit on my other blog. I gave a talk at the conference which will probably be of interest to readers of this blog—I hope to write about it soon!
In any case, today I want to return to the problem of quickly recognizing small primes. In my previous post we considered “small” to mean “less than 100”. Today we’ll kick it up a notch and consider recognizing primes less than 1000. I want to start by considering some simple approaches and see how far we can push them. In future posts we’ll consider some fancier things.
First, some divisibility tests! We already know how to test for divisibility by $2$, $3$, and $5$. Let’s see rules for $7$, $11$, and $13$.
• To test for divisibility by $7$, take the last digit, chop it off, and subtract double that digit from the rest of the number. Keep doing this until you get something which obviously either is or isn’t divisible by $7$. For example, if we take $2952$, we first chop off the final 2; double it is 4, and subtracting 4 from $295$ leaves $291$. Subtracting twice $1$ from $29$ yields $27$, which is not divisible by $7$; hence neither is $2952$.
The following is multiple choice question (with options) to answer.
What is the largest number that will divide 90207, 232585 and 127986 without leaving a
remainder? | [
"257",
"905",
"351",
"498"
] | A | HCF is the best solution. bit its time taking.
see the numbers 90207 , 232585 , 127986.
choose 905. but its not possible because last digit must be 0 or 5.
choose 498. it is also not possible because last dight must be an even number.
only 2 left now.
its easy now to get the answer by division method
ans is 257.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36248 | # String probability (with conditional prob and combinations)
I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.
SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.
Q1. What is the probability that it contains exactly one letter that is a vowel?
A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)
Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?
A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))
SET2: How many distinct permutations of the letters in "letters"...
Q1. Begin with two vowels?
A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...
Q2. Begin with two e's or end with two t's?
A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.
Q3. Have the vowels together?
A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.
Thank you!
The following is multiple choice question (with options) to answer.
A single letter is selected at random from the word 'SIMS'. The probability that it is a vowel is..?? | [
"1/4",
"3/11",
"4/11",
"3/7"
] | A | total vowels = 1, total number of letters = 4
so probability = 1/4
ANSWER:A |
AQUA-RAT | AQUA-RAT-36249 | What you did is correct.
You can check by substitution: If $z=3-2i$ then
$$(z+2i)^2-9= ((3-2i)+2i)^2-9 = 3^2-9=0.$$
If $z=-3-2i$ then $$(z+2i)^2 - 9 = ((-3-2i) + 2i)^2 - 9 = (-3)^2 - 9 = 0.$$
-
The following is multiple choice question (with options) to answer.
If 2x + 3y = 59; 2y + z = 19 and x + 2z = 29, what is the value of x + y + z ? | [
"18",
"31",
"26",
"22"
] | B | On solving equation we get
x = 19, y = 7, z = 5
Answer B |
AQUA-RAT | AQUA-RAT-36250 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
If Sharon's weekly salary increased by 16 percent, she would earn $348 per week. If instead, her weekly salary were to increase by 10 percent, how much would she earn per week? | [
" $374",
" $330",
" $385",
" $392"
] | B | (348/116)110 =330
In this case long division does not take much time.
(348/116)=3
3*110=330(300+30)
Answer B |
AQUA-RAT | AQUA-RAT-36251 | algorithms
Title: How to calculate least common multiple (LCM) for two numbers with constants (shifts)? What's the most effective algorithm to calculate LCM for 2 numbers where each of them has their own "shift"?
Example:
We want to find LCM for 25 + 5 as c and 30 + 10 as c
Then for the first case, the sequence will look like this:
25*1+5, 25*2+5, 25*3+5, 25*4+5...
And the second one:
30*1+10, 30*2+10, 30*3+10, 30*4+10...
In this case, the result should be 130 - 5th member of the first sequence and 4th of the second one. Given integers $a,b \geq 1$ and $c,d \geq 0$, you are looking for the set of solutions of $ax + c = by + d$ over integers $x,y \geq 0$. We can assume without loss of generality that $d = 0$ (replace $c,d$ with $c-d,0$ or $0,d-c$) and that $(a,b,c) = 1$ (otherwise, divide everything by the GCD).
I claim that if there is any solution then necessarily $(a,b) = 1$. Indeed, suppose that $(a,b) = g > 1$. If $ax + c = by$ then $g \mid ax,by$ implies $g \mid c$, and so $(a,b,c) = g > 1$, contrary to assumption. Therefore $(a,b) = 1$.
If $ax + c = by$ then $ax + c \equiv 0 \pmod{b}$ and so $x \equiv -c/a \pmod{b}$ (note that we can divide by $a$ since $(a,b)=1$). Similarly, $y \equiv c/b \pmod{a}$. These equations essentially give us both the minimal solution and all solutions.
The following is multiple choice question (with options) to answer.
The HCF and LCM of two numbers m and n are respectively 8 and 200. If m + n = 84, then 1/m + 1/n is equal to | [
"1/35",
"3/50",
"5/37",
"2/35"
] | B | Answer
We have, m x n = 8 x 200 = 1600
∴ 1/m + 1/n = (m + n)/mn = 84/1600 = 3/50
Correct Option: B |
AQUA-RAT | AQUA-RAT-36252 | 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.
20 women can do a work in 9 days. After they have worked for 6 days. 6 more men join them. How many days will they take to complete the remaining work? | [
"4:87",
"4:3",
"4:2",
"4:7"
] | B | (20 * 16) women can complete the work in 1 day.
1 woman's 1 day work = 1/320
(16 * 15) men can complete the work in 1 day
1 man's 1 day work = 1/240
So, required ratio = 1/240 : 1/320 = 4:3.
Answer:B |
AQUA-RAT | AQUA-RAT-36253 | sam, vcf, gatk
@SQ SN:chr14_GL000225v1_random LN:211173
@SQ SN:chr14_KI270722v1_random LN:194050
@SQ SN:chr14_GL000194v1_random LN:191469
@SQ SN:chr14_KI270723v1_random LN:38115
@SQ SN:chr14_KI270724v1_random LN:39555
@SQ SN:chr14_KI270725v1_random LN:172810
@SQ SN:chr14_KI270726v1_random LN:43739
@SQ SN:chr15_KI270727v1_random LN:448248
@SQ SN:chr16_KI270728v1_random LN:1872759
@SQ SN:chr17_GL000205v2_random LN:185591
@SQ SN:chr17_KI270729v1_random LN:280839
@SQ SN:chr17_KI270730v1_random LN:112551
@SQ SN:chr22_KI270731v1_random LN:150754
@SQ SN:chr22_KI270732v1_random LN:41543
@SQ SN:chr22_KI270733v1_random LN:179772
@SQ SN:chr22_KI270734v1_random LN:165050
@SQ SN:chr22_KI270735v1_random LN:42811
@SQ SN:chr22_KI270736v1_random LN:181920
@SQ SN:chr22_KI270737v1_random LN:103838
@SQ SN:chr22_KI270738v1_random LN:99375
The following is multiple choice question (with options) to answer.
Shipment ------No. of Defective Chips/shipment---Total Chips in shipment
S2 -------------- 5 ------------------------ ----------------------12000
S3 -------------- 6 ----------------------------------------------- 18000
S4 -------------- 4 ----------------------------------------------- 16000
A computer chip manufacturer expects the ratio of the number of defective chips to be total number of chips in all future shipments equal to the corresponding ratio for shipmemts S1,S2,S3 and S4 comined as shown in the table above. What is the expected number J of defective chips in a shipment of 60000 chips? | [
"14",
"20",
"22",
"24"
] | B | I agree with your solution = 20. But the question is:
There are different combination to get 60,000 chips. For example: 1*S3 + 2*S4 + 2*S2. In this way, we ship 60,000 chips with only 6 + 4*2 + 2*2 = 18 defective chips, better than the average of 20.
The question is to find the expected number J of defective chips, i guess it assume the minimum #, therefore it might not be 20. |
AQUA-RAT | AQUA-RAT-36254 | 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 passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"227",
"268",
"240",
"889"
] | C | Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Answer:C |
AQUA-RAT | AQUA-RAT-36255 | That is why an answer of $$14/27$$ should not surprise you. Now for the calculation:
There are 14 possibilities for the first child (gender and day of week) and 14 possibilities for the second. However one of them must be a girl born on a Saturday, so there are 27 valid combinations. 14 of them have the sibling being a boy so the probability of a boy is $$\frac{14}{27}$$.
The following is multiple choice question (with options) to answer.
Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl. | [
"25/39",
"25/32",
"25/22",
"25/34"
] | D | Explanation:
The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.
So the required probability:
Answer: D) 25/34 |
AQUA-RAT | AQUA-RAT-36256 | Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4675
Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
14 Jun 2016, 17:21
5
KUDOS
Expert's post
3
This post was
BOOKMARKED
AbdurRakib wrote:
The harvest yield from a certain apple orchard was 350 bushels of apples. If x of the trees in the orchard each yielded 10 bushel of apples, what fraction of the harvest yield was from these x trees?
A) $$\frac{x}{35}$$
B) 1–($$\frac{x}{35}$$)
C) 10x
D) 35–x
E) 350–10x
OG 2017 New Question
Dear AbdurRakib,
I'm happy to respond. This kind of word problem intimidates many people, but the actual calculation here is quite straightforward.
A fraction is a part over a whole. The whole is 350 bushels.
Of this set of x special trees, special for some unknown reason, each tree produced 10 bushels. That's 10x bushels from the lot of them. That's the part.
fraction = $$\frac{10x}{350}$$ = $$\frac{x}{35}$$
Does this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern
Joined: 01 May 2015
Posts: 44
Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
15 Jun 2016, 00:41
Total yield by x trees = 10x
So, fraction of the harvest yield from these x trees = 10x/350 = x/35
Manager
Joined: 25 Jun 2016
Posts: 61
GMAT 1: 780 Q51 V46
Re: The harvest yield from a certain apple orchard was 350 bushels of appl [#permalink]
### Show Tags
27 Jul 2016, 10:52
2
KUDOS
Here are a couple of ways to solve this question:
Value Substitution:
Algebra:
The following is multiple choice question (with options) to answer.
Harper's Apple Farm began expansion of it's apple harvest in 1900. From 1900 to 1950, the number of apples harvested by Harper's Apple Farm increased by 25%. From 1900 to 2000, the number of apples harvested increased by 350%. What was the percentage increase in the number of apples harvested from the year 1950 to the year 2000? | [
" 100%",
" 167%",
" 260%",
" 250%"
] | C | Let x be the number of apples harvested in 1900; then, x + 0.25x = 1.25x is the number of apples harvested for 1900-1950, and x + 3.5x = 4.5x is the number of apples harvested for 1900-2000.
Now, percentage increase in the number of apples harvested from the year 1950 to the year 2000 is (4.5x - 1.25x)/1.25x = 2.6 = 260%
Answer is C |
AQUA-RAT | AQUA-RAT-36257 | SIMPLE
$27^\frac{2}{3}=(3\cdot\ 3\cdot\ 3)^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=(3)^{3\cdot\frac{2}{3}}=**(3)^{\frac{6}{3}}**=3^2$
ACTUALLY. $27^2$. BECOMES $3^6$
The following is multiple choice question (with options) to answer.
If k is a non-negative integer and 30^k is a divisor of 929,260 then 3^k - k^3 = | [
"0",
"1",
"45",
"130"
] | B | 9+2+9+2+6+0 = 28, so this number is not divisible by 3 and thus not divisible by 30.
Therefore, k=0
3^k - k^3 =1-0=1
The answer is B. |
AQUA-RAT | AQUA-RAT-36258 | algorithms, integers
If the prime factorization of $n+1$ is $p_1^{k_1},\ldots,p_t^{k_t}$, then $f(n)$ depends only on $(k_1,\ldots,k_t)$, and in fact only on the sorted version of this vector. Every number below $10^9$ has at most $29$ prime factors (with repetition), and since $p(29)=4565$, it seems feasible to compute $f$ (or rather $g$) for all of them, recursively. This solution could be faster if there were many different inputs; as it is, there are at most $10$.
It is also possible that this function, mapping partitions to the corresponding $g$, has an explicit analytic form. For example, $g(p^k) = 2^{k-1}$, $g(p_1\ldots p_t)$ is given by A000670, and $g(p_1^2 p_2\ldots p_t)$ is given by A005649 or A172109.
The following is multiple choice question (with options) to answer.
If integer k is equal to the sum of all even multiples of 5 between 300 and 610, what is the greatest prime factor of k? | [
"32",
"7",
"11",
"13"
] | A | if we break down what the stem is asking what is the sum of all mult of 10 between 300 and 610.
using arithmetic progression to find n : 610 = 300 + (n - 1) 10
310+ 10 = 10n
320 =10n => n = 32
the sum would be: 32* mean
mean = [610 + 300] / 2 = 455
32*455
A |
AQUA-RAT | AQUA-RAT-36259 | Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:
0.04(total fish) = 50
This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:
Quote:
.04 (total) = tagged fish
We are missing two variables. We don't know the total fish and we don't know the tagged fish.
If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.
If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."
From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.
Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
CEO
Joined: 12 Sep 2015
Posts: 2705
Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
In a certain pond, 30 fish were caught, tagged, and returned to the pond. A few days later, 30 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond, what`s the approximate number of fish in the pond? | [
"400",
"750",
"1250",
"2500"
] | B | If x is total number of fish in the pond :
4 = 30/x * 100
=> x = 750
So answer is B |
AQUA-RAT | AQUA-RAT-36260 | mass, weight
Title: Mass versus Weight What are the difference between mass and weight? I keep getting confused in my physics class, and I am in 8th grade. Thank you in advance. Mass is a measure of how much matter (atoms) make up an object. Weight is a force that results from that quantity of matter accelerating in a gravitational field. Weight = mass*acceleration due to gravity. Weight can change depending on what gravitational field you are in, mass cannot change.
The following is multiple choice question (with options) to answer.
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 15 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.) | [
"25 pounds",
"46 pounds",
"150 pounds",
"100 pounds"
] | C | Since only water evaporates, then the weight of pulp (non-water) in grapes and raisins is the same. Thus 0.08*{weight of grapes}=0.8*{weight of raisins} --> 0.08x = 0.8*15 --> x = 150.
Answer: C. |
AQUA-RAT | AQUA-RAT-36261 | $y = x$ if $x \geq -1$.
It also means
$y = -(x + 1) - 1$ if $x + 1 < 0$, i.e. $x < -1$
$y = -x - 2$ if $x < -1$.
So all the points which satisfy these two linear equations will be your solution.
The following is multiple choice question (with options) to answer.
If x = 1 - q and y = 2q + 1, then for what value of q, x is equal to y ? | [
"0",
"1",
"-1",
"-2"
] | A | Explanation:
x = y <=> 1 - q = 2q + 1 <=> 3q = 0 <=> q = 0.
Answer: A |
AQUA-RAT | AQUA-RAT-36262 | # conditional probability trick questions - drawing cards from a deck and the meaning of 'at least'
Suppose that a box contains one blue card and four red cards, which are labeled A, B, C, and D. Suppose also that two of these five cards are selected at random, without replacement.
a. If it is known that card A has been selected, what is the probability that both cards are red?
b. If it is known that at least one red card has been selected, what is the probability that both cards are red?
I have been assigned the problem above in a class. I am aware that the book considers the answer for A) to be 3/4, which is obvious. However, the answer to B) is claimed to be $$P(red_1)P(red_2|red_1) = 4/5 \cdot 3/4 = 3/5$$
The professor has not been able to explain why the answers to A and B are different in any way that makes sense to me. There is no way in which you cannot draw at least 1 red card in a draw of 2 cards. If you are told you have drawn at least 1 red card, then you have the same information that you did in A.
• The box has one blue card and 4 red cards? In this case wouldn't (b) premise - that is at least one red card was selected - always be true? It seems like you can never select two cards without at least one of them being red. – Karolis Koncevičius Oct 29 '14 at 18:19
• @Karolis Is that a problem? – whuber Oct 29 '14 at 18:24
• @whuber No, not a problem, sorry. I just thought maybe there was an error in formulation. Not used seeing Pr(something | whole_space) kind of questions. – Karolis Koncevičius Oct 29 '14 at 18:36
• To clarify, the red cards are labeled A, B, C, and D, and the blue card is unlabeled (or labeled something other than A, B, C, or D)? – Hao Ye Oct 29 '14 at 23:42
Look at the possible samples consistent with the information, each of which is equally likely:
The following is multiple choice question (with options) to answer.
In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and multiples the integer on the card by the next larger integer. If each possible product is between 29 and 200, then the least and greatest integers on the card could be | [
"3 and 15",
"3 and 20",
"5 and 13",
"4 and 14"
] | C | There child draws one number then multiplies that number by the number above it. The number must be more than 29 and less than 200.
x * (x+1)
Answer gives us choices of 3,4 and 5 (JUST ABOVE 29)
3*(3+1) =12 <- WRONG
4*(4+1) =20 <- WRONG
5*(5+1) =30 <- CORRECT AS IT IS ABOVE 29
Answer gives us choices 13,14,15 and 20 (BELOW 200)
Immediately by knowing squares you should be able to rule out 15 (225) and 20 (400). And common sense dictates that as 14^2 is 196 that 14*(14+1) is above 200.
13*(13+1)=182 <- CORRECT
Answer is C) 5 and 13 |
AQUA-RAT | AQUA-RAT-36263 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The price of a consumer good increased by pp% during 20122012 and decreased by 1212% during 20132013. If no other change took place in the price of the good and the price of the good at the end of 20132013 was 1010% higher than the price at the beginning of 20122012, what was the value of pp? | [
"−2%",
"2%",
"22%",
"25%"
] | D | As per question=> Price was simple 10 percent greater
Hence x[1+10/100] must be the final price.
Equating the two we get
=> x[110/100]=x[1+p/100][88/100]
=> 44p+4400=5500
=> 44p=1100
=> p=1100/44=> 100/4=> 25.
So p must be 25
ANSWER:D |
AQUA-RAT | AQUA-RAT-36264 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | [
"4",
"8",
"16",
"17"
] | B | Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
Answer:B |
AQUA-RAT | AQUA-RAT-36265 | When morning comes and all $N$ pirates wake up, they see the greatly diminished pile and a happy fat monkey.
The question is: how many bananas could there have been in the original pile for this to occur?
• Is this for all sets of possible bananas? – awesomepi May 22 '14 at 15:58
• ...The number of possible bananas is limited depending on N, N can be any natural number (except 1). – kaine May 22 '14 at 16:27
• The way I remember it, concludes with the pirates waking up, finding a greatly diminished pile of bananas, each taking their share of $x$ bananas, and feeding the final banana to the monkey. That would give you a starting pint, but I'm not sure it's correct. – SQB May 23 '14 at 7:01
• @SQB I did not remember it ending that way but that could be my bad memory. Mathematically though, I am pretty sure that would be the same as N+1 pirates so Ross's answer still applies. – kaine May 23 '14 at 12:52
• @kaine it is not the same, since N+1 pirates would always divide by (N+1) here it is always a division by N, only it happens N+1 times with the waking up part. – Falco Aug 6 '14 at 11:22
The basic idea is to work backwards.
The last pirate must have found $N+1$ bananas, because he had to find enough bananas remaining for at there to be least 1 banana in the pile left for each pirate. He took 1, fed 1 to the monkey, and left $N-1$. That $N+1$ means that the next to last pirate found $\frac N{N-1}\cdot (N+1)+1$ and so on.
A cute trick is to recognize that there could have been
$-(N-1)$ bananas. Each pirate gives the monkey $1$, takes $-1$, and leaves the pile the same size as before. Since we need to divide it by $N$ for each pirate, the next solution is higher by $N^N$, so the minimal positive solution is $N^N-N+1$ bananas.
The following is multiple choice question (with options) to answer.
On the independence day, bananas were be equally distributed among the children in a school so that each child would get two bananas. On the particular day 380 children were absent and as a result each child got two extra bananas. Find the actual number of children in the school? | [
"237",
"287",
"760",
"287"
] | C | Explanation:
Let the number of children in the school be x. Since each child gets 2 bananas, total number of bananas = 2x.
2x/(x - 380) = 2 + 2(extra)
=> 2x - 760 = x => x = 760.
Answer: C |
AQUA-RAT | AQUA-RAT-36266 | Hence, if we eliminate at most $13$ sets of $2$ numbers, then each difference of the $2$ numbers from the remaining $105-13=92$ is distinct. However, the difference of $2$ numbers is either $1,2,\cdots, 88,89(=99-10)$, which is a contradiction. QED
Added : Byron Schmuland and Ross Millikan independently show eleven 2-digit numbers so that every pair has a different sum. So, we now know that the minimum $n_{\text{min}}$ of $n$ has to satisfy $\color{red}{12\le n_{\text{min}}\le 15}$.
The following is multiple choice question (with options) to answer.
The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number ? | [
"69",
"78",
"96",
"Cannot be determined"
] | D | Solution
Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x = y =3 or y = x = 3.
Solving x + y = 15 and x - y = 3, we get : x = 9, y = 6.
Solving x + y = 15 and y - x = 3, we get : x = 6, y = 9.
So, the number is either 96 or 69. Hence, the number cannot be determined
Answer D |
AQUA-RAT | AQUA-RAT-36267 | ### Show Tags
19 May 2015, 12:37
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 878
Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
### Show Tags
20 May 2015, 03:32
BrainLab wrote:
we have here 4 (4 years) successive increases of 25% or *1,25 --> 1,25^4 * X = 6250, X = 2560
See MGMAT (Percents) for detailed explanation of such question types.....
Dear BrainLab
Perfect logic but for easier calculation, you may want to work with ratio here (1/4 increase per annum) instead of percentages (25% increase per annum). Both convey the same thing but the equation
$$(\frac{5}{4})^4*X = 6250$$
will take lesser time to solve (especially if you know that $$5^4 = 625$$) than $$(1.25)^4*X = 6250$$
Hope this was useful!
Japinder
_________________
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Math Expert
Joined: 02 Sep 2009
Posts: 44373
Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink]
### Show Tags
18 Jan 2016, 23:36
Expert's post
1
This post was
BOOKMARKED
Re: Each year for 4 years, a farmer increased the number of trees in a [#permalink] 18 Jan 2016, 23:36
The following is multiple choice question (with options) to answer.
How much is 25% of 40 is greater than 4/5 of 25? | [
"10",
"776",
"66",
"12"
] | A | (25/100) * 40 – (4/5) * 25
10 - 20 = 10
Answer: A |
AQUA-RAT | AQUA-RAT-36268 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The average age of 15 students of a class is 15 years. Out of these, the average age of 4 students is 14 years and that of the other 9 students is 16 years. Tee age of the 15th student is? | [
"20years",
"22years",
"24years",
"25years"
] | D | Age of the 15th student
=[15 * 15 - (14 * 4 + 16 * 9)]
= (225 - 200) = 25 years.
Answer:D |
AQUA-RAT | AQUA-RAT-36269 | Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink] 21 Apr 2017, 03:36
Go to page 1 2 Next [ 25 posts ]
Similar topics Replies Last post
Similar
Topics:
16 A, B, C and D are positive integers such that A/B = C/D. Is C divisibl 7 16 Mar 2016, 06:12
5 For non–zero integers a, b, c and d, is ab/cd positive? 8 05 Sep 2016, 15:05
If a, b, c, and d are positive, is a/b > c/d? 1 04 May 2017, 19:50
If a, b, c, and d are positive numbers, and a/b = c/d, what is the val 2 28 Jul 2016, 14:34
8 If a, b, c, and d are positive numbers and a/b < c/d , which of the fo 7 27 Mar 2017, 22:14
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If a,b are positive integers and a/b=87.5 which of the following could be the value of b? | [
"1. 25",
"2. 26",
"3. 27",
"4. 28"
] | B | Given a/b = 87.75.
this is 87 and 1/2 = 175/2
Since a and b are integers we know that whatever is the value of b, it must be divisible by 2 completely(to make a an integer)
Ans:B |
AQUA-RAT | AQUA-RAT-36270 | +0
# Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the
0
45
3
+338
Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?
supermanaccz Nov 3, 2017
#2
+91051
+2
Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?
I like this question.
I will assume Tell goes first becasue William gets to go first all the rest of the time so it is Tell's turn to go first this time.
The prob that the Tell wins on his first shot is 0.5
The prob that the William misses on his 1st shot and Tell hits on his second shot is 0.5*0.5*0.5
The prob that the William misses on his 1st and 2nd shots and Tell hits on his 3rd shot is 0.5*0.5*0.5*0.5*0.5
etc
The prob that Tell wins is the limiting sum of a GP with a=0.5 and r=0.5^2
So
\begin{align} P(Tell\; wins) &= 0.5+0.5^3+0.5^5+.....\\ &=\frac{0.5}{1-0.5^2}\\ &=\frac{1}{2}*\frac{4}{3}\\ &=\frac{2}{3} \end{align}
Melody Nov 3, 2017
Sort:
#1
0
The chance of the one that shoots first winning is 50%.
Guest Nov 3, 2017
#2
+91051
+2
Both William and Tell hit their target half the time. They decide to fight a duel in which they exchange shots until one is hit. What is the chance that the one that shoots first will win?
I like this question.
I will assume Tell goes first becasue William gets to go first all the rest of the time so it is Tell's turn to go first this time.
The following is multiple choice question (with options) to answer.
Three archers each have an equal chance of hitting a target, and if only two of the three shoot the likelihood of the two hitting the target is 4/9 . What is the likelihood of all three men missing the target? | [
"1/27",
"16/81",
"8/27",
"19/27"
] | A | Let the probability of an archer hitting a target be xx. Then the probability of two hitting the target will be P=x∗x=49P=x∗x=49 --> x=23x=23, so the probability of an archer missing the target will be P=1−23=13P=1−23=13.
The probability of all three men missing the target will be P=(13)3=127P=(13)3=127.
Answer: A. |
AQUA-RAT | AQUA-RAT-36271 | Author Message
TAGS:
### Hide Tags
Manager
Joined: 26 Apr 2010
Posts: 122
Concentration: Strategy, Entrepreneurship
Schools: Fuqua '14 (M)
Followers: 2
Kudos [?]: 129 [0], given: 54
$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
An analysis of the monthly incentives received by 5 salesmen : The mean and median of the incentives is $7000. The only mode among the observations is $12,000. Incentives paid to each salesman were in full thousands. What is the difference between Q the highest and the lowest incentive received by the 5 salesmen in the month? | [
"$4000",
"$5000",
"$9000",
"$11,000"
] | D | Break down the question in to steps:
Step #1: incentives received by 5 salesmen -->abcde
Step #2: mean and median of the incentives is $7000: via number properties --> total incentives = 7,000 * 5 = 35, 000ab7,000de
Step #3: only mode among the observations is $12,000: mode is the value that appears most often in a set of data. Therefore 12,000 must occur more then once and since 12,000 is larger than the mean (7,000) the most it can occur is twice in our list. If we place the numbers in ascending order we haveab7,00012,00012,000
Step #4: What is the difference between the highest and the lowest incentive:
- The total paid has to equal mean * # of numbers = 7,000 * 5 = 35 000.
- The three values we have so far (12,000 12,000 and 7, 000) equal 31,000.
- Therefore the first two numbers( ab) must equal 4, 000
- There is only one mode which is 12,000, therefore 4,000 must be equal to a set of two numbers that are not the and a < b
if a = 1,000 b = 3,000 --> difference Q between the highest and the lowest incentive = 12,000 - 1,000 =11, 000
no other options for a or b to fit the criteria
Final List:1,0003,0007,00012,00012,000
Answer D |
AQUA-RAT | AQUA-RAT-36272 | ### Show Tags
27 Oct 2016, 16:26
3
GMATPrepNow wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?
A) 7.5
B) 10
C) 12
D) 12.5
E) 15
We are given that Bill and Ted each competed in a 240-mile bike race. We are also given that Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Both Bill and Ted had distance of 240 miles. If we let Ted’s speed = r, we can let Bill’s speed = r - 5. We can use the above information to determine the time of Bill and Ted in terms of variable r.
Since time = distance/rate, Ted’s time = 240/r and Bill’s time = 240/(r - 5).
Since Ted completed the race 4 hours sooner than Bill did, we can create the following equation:
240/r + 4 = 240/(r - 5)
To eliminate the denominators of the fractions we can multiply the entire equation by r(r-5) and we have:
240(r - 5) + 4[r(r - 5)] = 240r
240r - 1,200 + 4r^2 - 20r = 240r
4r^2 - 20r - 1,200 = 0
r^2 - 5r - 300 = 0
(r - 20)(r + 15) = 0
r = 20 or r = -15
Since r must be positive, r = 20.
Thus, Bill’s rate = 20 - 5 = 15 mph.
_________________
# Scott Woodbury-Stewart
Founder and CEO
Scott@TargetTestPrep.com
214 REVIEWS
5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE
NOW WITH GMAT VERBAL (BETA)
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4958
GMAT 1: 770 Q49 V46
Re: Bill and Ted each competed in a 240-mile bike race. [#permalink]
The following is multiple choice question (with options) to answer.
Two dogsled teams raced across a 300-mile course in Wyoming. Team A finished the course in 3 fewer hours than did team B. If team A’s average speed was 5 miles per hour greater than that of team B, what was team B’s average speed, in miles per hour? | [
"12",
"15",
"18",
"20"
] | D | Speed of B= S
Speed of A= S+5
Time taken by B=300/S Time Taken by A= 300/(S+5)
Difference in their time is 3 hrs.
300/S - 3 = 300/(S+5)
Plug in the values from the option
No need of choosing 12 & 18 it won't satisfy
a) if we take 15 then
T(B) = 300/15 =20
T(A)= 300/20 =15. Hence from this option difference comes out to be 5 which is not correct
b) if we take 20 then
T(B) = 300/20 = 15
T(A)= 300/25 = 12
Difference in timing is 3 hence Option D is correct |
AQUA-RAT | AQUA-RAT-36273 | algorithms, integers
If the prime factorization of $n+1$ is $p_1^{k_1},\ldots,p_t^{k_t}$, then $f(n)$ depends only on $(k_1,\ldots,k_t)$, and in fact only on the sorted version of this vector. Every number below $10^9$ has at most $29$ prime factors (with repetition), and since $p(29)=4565$, it seems feasible to compute $f$ (or rather $g$) for all of them, recursively. This solution could be faster if there were many different inputs; as it is, there are at most $10$.
It is also possible that this function, mapping partitions to the corresponding $g$, has an explicit analytic form. For example, $g(p^k) = 2^{k-1}$, $g(p_1\ldots p_t)$ is given by A000670, and $g(p_1^2 p_2\ldots p_t)$ is given by A005649 or A172109.
The following is multiple choice question (with options) to answer.
If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have? | [
"Four",
"Five",
"Six",
"Seven"
] | A | n=8!, so it has 4 prime factors: 2, 3, 5, and 7.
Answer: A. |
AQUA-RAT | AQUA-RAT-36274 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest? | [
"3%",
"4%",
"5%",
"6%"
] | D | S.I. = Rs. (15500 - 12500) = Rs. 3000.
Rate =((100 x 3000)/(12500 x 4))% = 6%
Answer :D |
AQUA-RAT | AQUA-RAT-36275 | # conditional probability trick questions - drawing cards from a deck and the meaning of 'at least'
Suppose that a box contains one blue card and four red cards, which are labeled A, B, C, and D. Suppose also that two of these five cards are selected at random, without replacement.
a. If it is known that card A has been selected, what is the probability that both cards are red?
b. If it is known that at least one red card has been selected, what is the probability that both cards are red?
I have been assigned the problem above in a class. I am aware that the book considers the answer for A) to be 3/4, which is obvious. However, the answer to B) is claimed to be $$P(red_1)P(red_2|red_1) = 4/5 \cdot 3/4 = 3/5$$
The professor has not been able to explain why the answers to A and B are different in any way that makes sense to me. There is no way in which you cannot draw at least 1 red card in a draw of 2 cards. If you are told you have drawn at least 1 red card, then you have the same information that you did in A.
• The box has one blue card and 4 red cards? In this case wouldn't (b) premise - that is at least one red card was selected - always be true? It seems like you can never select two cards without at least one of them being red. – Karolis Koncevičius Oct 29 '14 at 18:19
• @Karolis Is that a problem? – whuber Oct 29 '14 at 18:24
• @whuber No, not a problem, sorry. I just thought maybe there was an error in formulation. Not used seeing Pr(something | whole_space) kind of questions. – Karolis Koncevičius Oct 29 '14 at 18:36
• To clarify, the red cards are labeled A, B, C, and D, and the blue card is unlabeled (or labeled something other than A, B, C, or D)? – Hao Ye Oct 29 '14 at 23:42
Look at the possible samples consistent with the information, each of which is equally likely:
The following is multiple choice question (with options) to answer.
Ten cards numbered 1 to 10 are placed in a box and then one card is drawn randomly. If it is known that the number on the drawn card is more than 4, what is the probability that it is an odd number? | [
"1/2",
"3/10",
"7/10",
"5/6"
] | A | There are 6 numbers higher than 4 and 3 of them are odd.
The probability is 3/6=1/2.
The answer is A. |
AQUA-RAT | AQUA-RAT-36276 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
Three coins are tossed. What is the probability of getting at most two tails? | [
"5/8",
"6/8",
"7/8",
"3/8"
] | C | Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)
Hence, total number of outcomes possible when 3 coins are tossed, n(S) = 2 × 2 × 2 = 8
(∵ i.e., S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH})
E = event of getting at most two Tails = {TTH, THT, HTT, THH, HTH, HHT, HHH}
Hence, n(E) = 7
P(E) = n(E)/n(S)=7/8
Answer is C. |
AQUA-RAT | AQUA-RAT-36277 | • #### 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 lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m? | [
"Rs.3988",
"Rs.3900",
"Rs.3228",
"Rs.3922"
] | B | Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3
= Rs.3900
Answer:B |
AQUA-RAT | AQUA-RAT-36278 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
if the ratio of two number is 3:4 and LCM of the number is 180 then what is the number. | [
"15",
"20",
"25",
"30"
] | A | product of two no = lcm *hcf
3x*4x=180*x
x=15
ANSWER:A |
AQUA-RAT | AQUA-RAT-36279 | (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.
1. Internet: $80, with no sales tax, plus $10 shipping fee
2. Store X: $70, a 10% discount, plus 30% tax on discounted price
3. Store Y: $90, no tax
4. Store Z: $70, a 30% sales tax, and $10 rebate after tax
Isaac can purchase a certain item in four different ways, as shown in the table. What is the lowest price for which he can purchase this item? | [
"88.1",
"81.0",
"81.9",
"81.5"
] | B | Option 2 = (70 - 7) + 0.3 * 63= 81.90
Option 4 = 70 + 0.3 * 91 - 10
= 81
So the OA is correct - B |
AQUA-RAT | AQUA-RAT-36280 | # Find the value of the series $\frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \frac{32}{1025} + …$ to 20 terms.
Find the value of the series $$\frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \frac{32}{1025} + ...$$ to 20 terms.
The numerator seems to follow a clearly defined pattern here: it is an AP with common difference of 8. However, I cannot seem to find a pattern for the denominator here. The ratios are $$13, 5, 3.15$$ etc. whereas the differences are $$60, 260, 700$$. Even the double differences are not seemingly following a pattern. One last thing I noticed was that $$5 = 2^2+1$$, $$65 = 8^2+1$$, $$325 = 18^2+1$$, $$1025 = 32^2+1$$. The numbers $$2,6,18,32$$ are again not in an obvious pattern.
What is the answer to the problem? Finding the nth term is also good enough, as I should be able to take it from there.
Note: I read this question, and the topic behind these kind of problems seems to involve the calculus of finite differences. If someone could also point to some resources regarding this topic, I would read up more on this as well.
The following is multiple choice question (with options) to answer.
In each series, look for the degree and direction of change between the numbers. In other words, do the numbers increase or decrease, and by how much?
Look at this series: 88, 3, 80, 12, __, 21 ... What number is missing? | [
"72",
"78",
"68",
"30"
] | A | A
72
This is an alternating addition and subtraction series. In the first pattern, 8 is subtracted from each number to arrive at the next. In the second, 9 is added to each number to arrive at the next. |
AQUA-RAT | AQUA-RAT-36281 | = 4 52. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Two fair dice are rolled and the sum of the points is noted. For example: 1 roll: 5/6 (83. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. and only one way to roll a 12 (6-6, or boxcars). EXPERIMENTAL PROBABILITIES Simulate rolling two dice 120 times. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Two dice are tossed. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Sample space S = {H,T} and n(s) = 2. That intuition is wrong. When you roll a pair of dice there are 36 possible outcomes. To find the probability we use the mutually exclusive probability formula P(A) + P(B). A sum less than or equal to 4. Isn’t that kind of cool?. Two dice are tossed. So the probability of getting a sum of 4 is 3/36 or 1/12. 10 5 13 ! Find the probability distribution. So 1/36 is part of the. Sum of Two Dice. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Rolling two dice. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. The fundamental counting principle tells us there are 6*6=36 ways to roll two dice, all of them equally likely if the dice are fair. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Let B be the event - The sum
The following is multiple choice question (with options) to answer.
What is the probability of getting a sum 9 from two throws of a dice? | [
"1/5",
"1/9",
"1/7",
"1/12"
] | B | In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
P(E) = n(E)/n(S) = 4/36 = 1/9 .
Answer : B. |
AQUA-RAT | AQUA-RAT-36282 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
Machine–A produces 40% of the total output and Machine-B produces 60% of the total output. An average of nine units out of a thousand goods manufactured by Machine-A and one unit of 150 units produced by Machine-B prove to be defective. What is the probability that a unit chosen at random from the total daily output of the factory is defective? | [
"0.76",
"0.076",
"0.0076",
"0.00076"
] | C | Let total Production be 10000 units....
A produces 4000 units and 36 units are defective
B produces 6000 units and 40 units are defective
So, Out of total 10,000 units 76 units are defective...
So the required probability = 76/10000 => 0.0076
Answer will be (C) |
AQUA-RAT | AQUA-RAT-36283 | They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. Prove that the product of three consecutive positive integer is divisible by 6. gl/9WZjCW prove that the product of three consecutive positive integers is divisible by 6. Therefore, n = 3p or 3p + 1 or 3p + 2 , where p is some integer. 1 3 + 2 3 + 3 3 +. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. By the laws of divisibility, anything divisible by 2 and 3 is divisible by 6. Whenever a number is divided by 3 , the remainder obtained is either 0,1 or 2. Find the smallest number that, when. Essentially, it says that we can divide by a number that is relatively prime to. Let the three consecutive positive integers be n , n + 1 and n + 2. Let n be a positive integer. 1 Consecutive integers with 2p divisors. If A and B are set of multiples of 2 and 3 respectively, then show that A = B and A∪B. Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. The array contains integers in the range [1. Prove that one of any three consecutive positive integers must be divisible by 3. ← Prev Question Next Question →. Btw jayshay - if you said 7n, 7n+1 and 7n+2 then your 'proof' would effectively be proving that the product of 3 consecutive integers is a multiple of 7. We have to prove this for any arbitrary k ∈Z, so fix such a k. (Examples: Prove the sum of 3 consecutive odd integers is divisible by 3. Prove: The product of any three consecutive integers is divisible by 6; the product of any four consecutive integers is divisible by 24; the product of any five consecutive integers is divisible by 120. If A =40, B =60 and
The following is multiple choice question (with options) to answer.
The sum of three integers is 23. The largest integer is 3 times the middle integer, and the smallest integer is 12 less than the largest integer. What is the product of the three integers? | [
"125",
"150",
"175",
"225"
] | D | Let x be the middle integer. Then the largest is 3x and the smallest is 3x - 12.
3x - 12 + x + 3x = 23
x = 5
The three integers are 3, 5, and 15.
The product is 3*5*15 = 225
The answer is D. |
AQUA-RAT | AQUA-RAT-36284 | # Expected Number of Single Socks when Matching Socks
Whenever I go through the big pile of socks that just went through the laundry, and have to find the matching pairs, I usually do this like I am a simple automaton:
I randomly pick a sock, and see if it matches any of the single socks I picked out earlier and that haven't found a match yet. If there is a match, I will fold the two socks together and put them in the 'done' pile, otherwise I will add the single sock to the 'no match yet' pile of single socks, and pick out another random sock.
So, as I was doing this last night, I started thinking about this, and figured that the following would be true: The 'no match yet' pile can be expected to slowly grow, up to some point somewhere in the 'middle' of the process, after which the pile will gradually shrink, and eventually go down back to $0$. In fact, my intuition is that the expected number of loose socks as a function of the number of socks picked so far, is a symmetric function, with the maximum being when I have picked half of the socks.
So, my questions are:
With $n$ pairs of socks, what is the expected number of loose socks that are in my 'no match yet' pile after having picked $k$ socks?
Is it true that this function is a symmetric function, and that the maximum is for $k=n$? (if so, I figure there must be a conceptual way of looking at the problem that makes this immediately clear, without using any formulas ... what is that way? Is it just that I can think of reversing the process?)
Of course, this is all assuming there are $n$ pairs of socks total, and that there are no single socks in the original pile, and while this is something that never seems to apply to the pile of socks coming through my actual laundry, let's assume for the sake of mathematical simplicity that there really just are $n$ pairs of socks.
The following is multiple choice question (with options) to answer.
Andy has 20 pairs of matched gloves. If he loses 13 individual gloves, which of the following is NOT a possible number of matched pairs he has left? | [
"9",
"11",
"13",
"14"
] | D | We can solve this is second. The question isSelect an answer choice where there is no possibility of having a number as a matched pair
Case1: If 13 gloves are gone the maximum probability of losing maximum gloves is 13 pairs of gloves out of 20. This means i m having 7 pairs of gloves. [Minimum number of gloves i can have is 7 pairs]
Case2: If 13 gloves are gone the maximum probability of losing minimum gloves is 13/2 = 6.5 means i can lose 7 glove. in this scenario i will be having 13 gloves.
Hence10 < Available Glove Pair < 14
any given time I cant possess 14 glove pair for the above mentioned scenario. And thats the answer. D. |
AQUA-RAT | AQUA-RAT-36285 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
There are a lot of houses such that the numbers of their doorplates are odd numbers and the first number of the doorplates is 445, the last number of the doorplates is 705. How many houses are there? | [
"141",
"111",
"121",
"131"
] | D | So it starts from 445 and goes like 447, 449,...... 705. and both first and last number are inclusive. Since every other number is odd, it's just 1/2 of the numbers and since it starts with an odd and ends with an odd inclusive add one to the result.
i.e., ([705-445][/2]+1 = 131
Ans D |
AQUA-RAT | AQUA-RAT-36286 | There are only finitely many cases where the factors add up to 72. And these are the numbers 30, 46, 51, 55 and 71. You may want to check what happens to the sum of the reciprocal of their factors.
Now, for the general case, let $1 = d_1 < d_2 < \cdots < d_k = n$ be the factors of $n$. Then $\dfrac{1}{d_1} = \dfrac{d_k}{n}$, and in general $\dfrac{1}{d_i} = \dfrac{d_{k-i+1}}{n}$. Thus,
$\dfrac{1}{d_1} + \cdots \dfrac{1}{d_k} = \dfrac{d_k + d_{k-1} + \cdots + d_1}{n} = \dfrac{\sigma(n)}{n}$,
where $\sigma(n)$ is the sum of factors of $n$.
-
+1 for nice detailed way. ;-) – S. Snape Feb 19 '13 at 15:25
There's no need to find $n$. Note that for every $d$ dividing $n$, $\frac{n}{d}$ is also a divisor of $n$. Therefore the sum of all $d_i$ is equal to the sum of all $\frac{n}{d_i}$, which means...
-
The sum of divisors function is usually written with a letter $\sigma,$ so they are asking about $\sigma(n) = 72.$ Now, for a prime number $p,$ we do get $\sigma(p) = 1 + p,$ so in particular $\sigma(71) = 72.$ And that leads to at least one value for the sum of the reciprocals of the divisors.
We know that $\sigma(n) \geq n+1.$ So, the target 72 demands that $n \leq 71.$ Find out what other numbers solve $\sigma(n) = 72.$
-
The following is multiple choice question (with options) to answer.
The sum q of prime numbers that are greater than 60 but less than 70 is | [
" 67",
" 128",
" 191",
" 197"
] | B | A prime number is a number that has only two factors: 1 and itself. Therefore, a prime number is divisible by two numbers only.
Let's list the numbers from 61 to 69.
61, 62, 63, 64, 65, 66, 67, 68, 69
Immediately we can eliminate the EVEN NUMBERS because they are divisible by 2 and thus are not prime.
We are now left with: 61, 63, 65, 67, 69
We can next eliminate 65 because 65 is a multiple of 5.
We are now left with 61, 63, 67, 69.
To eliminate any remaining values, we would look at those that are multiples of 3. If you don’t know an easy way to do this, just start with a number that is an obvious multiple of 3, such as 60, and then keep adding 3.
We see that 60, 63, 66, 69 are all multiples of 3 and therefore are not prime.
Thus, we can eliminate 63 and 69 from the list because they are not prime.
Finally, we are left with 61 and 67, and we must determine whether they are divisible by 7. They are not, and therefore they must be both prime. Thus, the sum q of 61 and 67 is 128.
Answer B. |
AQUA-RAT | AQUA-RAT-36287 | units, geometry, dimensional-analysis
Title: Physical representation of volume to surface area I was looking at this XKCD what-if question (the gas mileage part), and started to wonder about the concept of unit cancellation. If we have a shape and try to figure out the ratio between the volume and the surface area, the result is a length. For example, a sphere of radius 10cm has the volume of $\approx 4118 cm^3$ and an area of $\approx 1256 cm^2$. Therefore, the volume : surface area is $\approx 3.3 cm$.
My question is: what is the physical representation of length in this ratio? For the case of a sphere the ratio you found is:
$$ \frac{V}{S} = \frac{ \frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3} $$
We can actually pass off the volume as being the integral of the surface area here. That's passable when you check the calculus.
One approach is then to ask "what is a function divided by its derivative". This is really similar to the area to perimeter ratio of a circle.
$$ \frac{A}{P} = \frac{ \pi R^2}{ 2 \pi R} = \frac{R}{2} $$
Of course you see the "2" because of the value of the exponent, which comes from there being two dimensions, just like the sphere. So now we have explained part of the answer, which is that the linear dimension is divided by the number of dimensions. This is still unsatisfactory because we have no clear sense of how we should define this particular "characteristic length".
One attempt at resolution of this problem would be to test the idea for a square-cube system.
$$ \frac{V}{S} = \frac{ R^3}{ 6 R^2} = \frac{R}{6} $$
$$ \frac{A}{P} = \frac{ R^2}{4 R} = \frac{R}{4} $$
The following is multiple choice question (with options) to answer.
What is the ratio E of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled? | [
"1/4",
"3/8",
"1/2",
"3/5"
] | D | One side surface area of a cube = x*x = x^2
Total 6 sides = 6x^2
As for the rectangular, Height (H) and Width (W) are same as Cube, x. Only Length = 2x.
L x H = 2x * x = 2x^2 ----> 4 sides = 2x^2 * 4 = 8x^2
W * H = x * x = x^2 ------> 2 sides = x^2 * 2 = 2x^2
Total 6 sides = 8x^2 + 2x^2 = 10x^2
Ratio of cube area to rectangular area E= 6x^2 / 10x^2 ----> 6/10 ----> 3/5 (D) |
AQUA-RAT | AQUA-RAT-36288 | # Is it possible to solve for $a, b \in \mathbb{N}$?
I need to solve the following equation so that both $a$ and $b$ are natural numbers.
$$ab - 2a = 2b$$
I must also prove that the solutions found are the only ones possible.
Is it possible to do so, and if yes how can I do it? Can this be solved as some type of diophantine equation?
• $ab=2(a+b)$ means $a\mid 2b$ and $b\mid 2a$. No Diophantine equations needed. – Dietrich Burde Feb 23 '15 at 16:34
Hint:
$$ab-2a=2b\iff (a-2)(b-2)=4$$
Hint $\ b\neq 2\,\Rightarrow\ a = \dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$
$(0,0)$ is a solution.
It can't be that $b = 1$, neither can it be that $b = 2$. Assume $b > 2$.
Considering the equation modulo $b - 2$, we get that $b - 2$ divides $4$. We conclude that $b - 2$ is either $1$, or $2$, or $4$. So $b = 3$ or $b = 4$ or $b = 6$. It happens that all of these give solutions.
Therefore the solutions are $(0,0)$, and the pairs for which $b \in \{3,4,6\}$.
The following is multiple choice question (with options) to answer.
Given a + b = 1, find the value of 2a + 2b. Two solutions are presented below. Only
one is correct, even though both yield the correct answer. | [
"3",
"5",
"4",
"2"
] | D | Because a + b = 1,
2a + 2b = 2(a + b) = 2 × 1 = 2.
correct answer D |
AQUA-RAT | AQUA-RAT-36289 | 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 crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train? | [
"271",
"180",
"188",
"127"
] | B | Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
Answer:B |
AQUA-RAT | AQUA-RAT-36290 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in upstream is 50 kmph and the speed of the boat downstream is 90 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"70,20 kmph",
"78,10 kmph",
"70,70 kmph",
"78,10 kmph"
] | A | Speed of the boat in still water
= (50+90)/2
= 70 kmph. Speed of the stream
= (90-50)/2
= 20 kmph.
Answer:A |
AQUA-RAT | AQUA-RAT-36291 | So: $$a_{16}=2^{16-1}=2^{15}$$, $$a_{17}=2^{17-1}=2^{16}$$ and $$a_{18}=2^{18-1}=2^{17}$$ --> $$2^{15}+2^{16}+2^{17}=2^{15}(1+2+4)=7*2^{15}$$.
_________________
Manager
Joined: 06 Feb 2010
Posts: 149
Schools: University of Dhaka - Class of 2010
GPA: 3.63
WE: Business Development (Consumer Products)
### Show Tags
23 Oct 2010, 05:02
In the sequence 1, 2 , 4, 8, 16, 32,............each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
A) 2^18
B) 3(2^17)
C) 7(2^16)
D) 3(2^16)
E) 7(2^15)
What is the easiest formula for this math?
_________________
Practice Makes a Man Perfect. Practice. Practice. Practice......Perfectly
Critical Reasoning: http://gmatclub.com/forum/best-critical-reasoning-shortcuts-notes-tips-91280.html
Collections of MGMAT CAT: http://gmatclub.com/forum/collections-of-mgmat-cat-math-152750.html
MGMAT SC SUMMARY: http://gmatclub.com/forum/mgmat-sc-summary-of-fourth-edition-152753.html
Sentence Correction: http://gmatclub.com/forum/sentence-correction-strategies-and-notes-91218.html
Arithmatic & Algebra: http://gmatclub.com/forum/arithmatic-algebra-93678.html
The following is multiple choice question (with options) to answer.
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 17th, and 18th terms in the sequence? | [
"2^18",
"3(2^16)",
"7(2^16)",
"3(2^16)"
] | B | the sequence comes out to be 2^0,2^1,2^2,2^3 and so on...
17th term 2^16
18th term 2^17
adding all three
we get
(2^16)+(2^16)*2
so answer is 2^16(1+2)=>3(2^16)
answer B |
AQUA-RAT | AQUA-RAT-36292 | (1) xz is even
(2) y is even.
Given:
x is a factor of y --> $$y=mx$$, for some non-zero integer $$m$$;
y is a factor of z --> $$z=ny$$, for some non-zero integer $$n$$;
So, $$z=mnx$$.
Question: is z even? Note that $$z$$ will be even if either $$x$$ or $$y$$ is even
(1) $$xz$$ even --> either $$z$$ even, so the answer is directly YES or $$x$$ is even (or both). But if $$x$$ is even and as $$z=mnx$$ then z must be even too (one of the multiples of z is even, so z is even too). Sufficient.
(2) $$y$$ even --> as $$z=ny$$ then as one of the multiples of z even --> z even. Sufficient.
_________________
Kudos [?]: 128579 [6], given: 12180
SVP
Joined: 29 Aug 2007
Posts: 2472
Kudos [?]: 841 [1], given: 19
### Show Tags
19 Oct 2009, 23:18
1
KUDOS
amitgovin wrote:
The positive integers x, y, and z are such that x is a factor of y and y is a factor of z. Is z even?
1) xz is even
2) y is even.
y/x = k where k is an integer.
y = xk ....................i
z/y = m where m is an integer.
z = ym = xkm .....................ii
If a factor is even, then the source of the factor must be even.
The following is multiple choice question (with options) to answer.
If x is an odd integer and y and z are even integers, which of the following CANNOT be an integer? | [
"y/z",
"x/y",
"z/x",
"yx/z"
] | B | x = Odd
y = Even
z = Even
Checking Options:
A. y/z Even/Even may be an Integer e.g. 4/2=2 Hence Incorrect Option
B. x/y Odd/Even CAN NEVER be an Integer Hence Correct Option
Answer : B |
AQUA-RAT | AQUA-RAT-36293 | 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.
Train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds. The length of train P is three-fourths the length of train Q. What is the ratio of the speed of train P to that of train Q? | [
"15:8",
"15:5",
"15:9",
"15:2"
] | A | Given that train P crosses a pole in 30 seconds and train Q crosses the same pole in one minute and 15 seconds.
Let the length of train P be LP and that of train Q be LQ
given that LP = 3/4 LQ
As the train P and Q crosses the pole in 30 seconds and 75 seconds respectively,
=> Speed of train P = VP = LP/30
Speed of train Q = VQ = LQ/75
LP = 3/4 LQ
=> VP = 3/4 LQ/(30) = LQ/40
Ratio of their speeds = VP : VQ
= LQ/40 : LQ/75 => 1/40 : 1/75 = 15:8
Answer:A |
AQUA-RAT | AQUA-RAT-36294 | $$L(100) \pmod{20} = \color{red}7$$
The remainder when the 100th term of the sequence is divided by 20 is 7
May 23, 2021
edited by heureka May 23, 2021
edited by heureka May 23, 2021
The following is multiple choice question (with options) to answer.
Find the greatest number which leaves the same remainder when it divides 20, 40 and 90. | [
"18",
"8",
"12",
"10"
] | D | 90 - 40 = 50
40 - 20 = 20
90 - 20 = 70
The H.C.F of 20, 50 and 70 is 10.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36295 | 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
VP
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
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4383
Location: India
GPA: 3.5
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 )
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Math Expert
Joined: 02 Sep 2009
Posts: 52917
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.
It takes one machine 5 hours to complete a production order and another machine 3 hours to complete the same order. How many hours would it take both amhcines working simultaneously at their respective rates to complete the order? | [
"a) 7/12",
"b) 1 1/2",
"c) 2 2/9",
"d) 3 1/2"
] | C | RT = W
Given:
R = 1/5 + 1/4 = 9/20
T = ?
W = 1
T = 1 / (9/20) = 20/9 = 2 2/9
Answer: C |
AQUA-RAT | AQUA-RAT-36296 | ## Solution
Consider $f(2)$. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$. Since $100\le (x+1)^2-x^2=2x+1$, this first happens at $x\ge \lfloor 99/2\rfloor = 50$. The perfect squares from here go: $2500, 2601, 2704, 2809\dots$. Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$. After the ones and tens make $100$, the hundreds place will go up by $2$, thus reaching our goal. Since $10^2=100$, the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$. The missing number is one less than the number of hundreds $(k=2)$ of $3600$, or $35$.
Now consider f(4). Instead of the difference between two squares needing to be $100$, the difference must now be $10000$. This first happens at $x\ge 5000$. After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$. This will take place at $\left(5000+100\right)^2=5100^2= 26010000$. Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
The following is multiple choice question (with options) to answer.
What is the least number to be subtracted from 990 to make it a perfect square? | [
"100",
"90",
"30",
"71"
] | B | The numbers less than 990 and are squares of certain number is 900.
The least number that should be subtracted from 990 to make it perfect square = 990 - 900 = 90.
ANSWER:B |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.