source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39197 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
What number should replace the question mark?
105, 87, 70, 52, 35,? | [
"49",
"37",
"70",
"17"
] | D | D
17
The sequence progresses -18, -17, -18, -17. |
AQUA-RAT | AQUA-RAT-39198 | 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, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel ? | [
"650 meter",
"555 meter",
"500 meter",
"458 meter"
] | C | Explanation:
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
=>800+x=65/3∗60
=>800+x=20∗65=1300
=>x=1300−800=500
So the length of the tunnel is 500 meters.
Option C |
AQUA-RAT | AQUA-RAT-39199 | Show Tags
25 Nov 2010, 21:43
2
shrive555 wrote:
In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
20
40
216
720
729
Or out of 6 children, choose 3 in 6C3 ways = 20 ways.
Note: When you choose 3 children say, A, B and C are give them a red shirt, D, E and F get a green shirt. When you choose D, E and F and give them a red shirt, A, B and C automatically get the green shirts. So you do not need to multiply by 2! above.
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24 Feb 2012, 17:19
Approach 1:
1st Child: 6 has options
2nd Child: 5 has options…
Therefore, for all kids: 6 x 5 x 4 x 3 x 2 = 720 arrangements.
Since the reds are identical, we divide by 3!; Since the greens are identical, we divide by another 3!
So: in all, 720/[ 3! X 3! ] = 20 ways.
Approach 2 / MGMAT technique:
This is like anagramming RRRGGG. No of arrangements = 6! / 3! x 3! ways ==> 20.
Math Expert
Joined: 02 Sep 2009
Posts: 53771
Re: In how many different ways can 3 identical green shirts and [#permalink]
Show Tags
24 Feb 2012, 22:59
3
1
shrive555 wrote:
In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?
A. 20
B. 40
C. 216
D. 720
E. 729
1-2-3-4-5-6 (children)
B-B-B-G-G-G (shirts)
G-B-B-G-G-B
G-G-B-G-B-B
....
The following is multiple choice question (with options) to answer.
In how many different ways can 3 identical green boots and 3 identical red boots be distributed among 6 people such that each person receives a pair of boots? | [
"7209",
"4059659",
"2169",
"20"
] | D | How to solve:
1st Person: 6 has options
2nd Person: 5 has options…
Therefore, for all people: 6 x 5 x 4 x 3 x 2 = 720 arrangements.
Since the reds are identical, we divide by 3! Since the greens are identical, we divide by another 3!
So: in all, 720/[ 3! X 3! ] = 20 ways.
(e.g. Answer: D) |
AQUA-RAT | AQUA-RAT-39200 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How long does a train 110 m long travelling at 60 kmph takes to cross a bridge of 200 m in length? | [
"18.6 sec",
"14.9 sec",
"12.4 sec",
"16.8 sec"
] | A | A
18.6 sec
D = 110 + 200 = 310 m
S = 60 * 5/18 = 50/3
T = 310 * 3/50 = 18.6 sec
Answer is A |
AQUA-RAT | AQUA-RAT-39201 | "How does this help me?" you say
I'm very glad you asked
Let's think about $8$ as our 100% and we want to find 2 parts of 8, so using what we just learned we say that "2 parts of 8 is the same as $\frac{2}{8}$." To turn $\frac{2}{8}$ into a percentage we just need the denominator to equal 100 since earlier we said that 8 is our 100%
We could use an equation for this, so let's use an equation for this :D
We know that $8 \cdot \left(\text{some number}\right) = 100$ so;
$8 x = 100$
$\frac{1}{8} \cdot 8 x = 100 \cdot \frac{1}{8} = \frac{100}{8} = \frac{25}{2} = 12.5$
Now we know that if we multiply the denominator by 12.5 we will get 100 and if we multiply the numerator by 12.5 we will get a ratio with 100 as the denominator, so our percentage must be the numerator!
2/8*12.5/12.5 = 25/100 = 25%
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The following is multiple choice question (with options) to answer.
8 is 6% of a, and 6 is 8% of b. c equals b/a. What is the value of c? | [
"3/8",
"8/15",
"9/16",
"12/25"
] | C | 6a/100 = 8
a = 400/3
8b/100 = 6
b = 75
c = b/a = 75*3/ 400 = 9/16
The answer is C. |
AQUA-RAT | AQUA-RAT-39202 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
In a certain state, gasoline stations compute the price per gallon p, in dollars, charged at the pump by adding a 6 percent sales tax to the dealer's price per gallon d, in dollars, and then adding a gasoline tax of $0.18 per gallon. Which of the following gives the dealer's price per gallon d in terms of the price per gallon p charged at the pump? | [
"d = p/1.06 - 0.18",
"d= (p-0.18)/1.06",
"d = (p-0.06)/1.18",
"d = p-0.24"
] | B | Let dealers price (d) be 1. so adding 6% to dealers price is d+ 6% of d. i.e. 1 + 6% of 1 which is1 + 0.06. then add 0.18 to the value.
Now 1.06 + 0.18. this is now 1.24. you have the gasoline stations price (p) as 1.24 dollars.
Now sub 1.24 in the options to know which option gave you d = 1. d must equal 1 because you earlier picked 1 as the value of d in the question.
PS: always remember to start from E upwards.
Answer : B |
AQUA-RAT | AQUA-RAT-39203 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling 50 meters of cloth. I gain the selling price of 15 meters. Find the gain percent? | [
"42 6/8%",
"42 8/7%",
"72 6/7%",
"42 6/7%"
] | D | SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 --- 15 CP gain
100 --- ? => 42 6/7%
Answer: D |
AQUA-RAT | AQUA-RAT-39204 | 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.
If the product of the integers from 1 to n is divisible by 1029, what is the least possible value of n? | [
"7",
"14",
"21",
"28"
] | C | 1029 = 7 x 7 x 7 x 3
N must include at least 7, 2*7, and 3*7.
The answer is C. |
AQUA-RAT | AQUA-RAT-39205 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If b, x, y, and z are positive integers and (x)×(y)×(z)=b^2, which of the following could be the values of x, y, and z? | [
"3, 16, 25",
"9, 25, 24",
"2, 81, 32",
"2, 9, 16"
] | C | This is one of those questions where using the given options is fastest way to get the solution. We need to find a set where 3 numbers can be paired into 2 pairs of some numbers because we need a product which is a square number.
C fits the bill. 2*81*32 = 2*9*9*8*4 = 9*9*8*8
ans C |
AQUA-RAT | AQUA-RAT-39206 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A certain store sells all maps at one price and all books at another price. On Monday the store sold 9 maps and 6 books for a total of $20.00, and on Tuesday the store sold 9 maps and 8 books for a total of $30.00. At this store, how much less does a map sell for than a book? | [
" $0.25",
" $0.50",
" $0.75",
" $3.9"
] | D | 9x+6y=20
9x+8y=30
subtracting 1 from 2
-2y=-10
y=5
therefore
x=1.1
difference in price =3.9
D |
AQUA-RAT | AQUA-RAT-39207 | This online aptitude test on Compound Interest is useful for candidates preparing for banking exams - Bank PO, IBPS PO, SBI PO, RRB PO, RBI Assistant, LIC,SSC, MBA - MAT, XAT, CAT, NMAT, UPSC, NET etc. COMPOUND INTEREST TABLES 561 1 / 2 %Compound Interest Factors 1 2 Single Payment Uniform Payment Series Arithmetic Gradient Compound Present Sinking Capital Compound Present Gradient Gradient. Compound Interest Invest €500 that earns 10% interest each year for 3 years, where each interest payment is reinvested at the same rate: End of interest earned amount at end of period Year 1 50 550 = 500(1. In compound interest calculations interest earned, or due, for each period, is added to the principal. Fill in the table below by calculating the interest, new balance and total payment for each month for the two payment scenarios,$25. 1) Year 3 60. So we are sharing It here with you so that It can be helpful for the aspirants who are going to appear in upcoming exams like SSC CPO 2019, SSC CGL 2019, SSC CHSL 2019, SSC MTS 2020, RRB NTPC 2019, RRC Group D 2019 and other state exams. Feb 1, 2017 - Compound Interest worksheet with answer key (pdf). The stochastic calculus part of these notes is from my own book: Probabilistic Techniques in Analysis, Springer, New York, 1995. Compound interest is incredibly powerful. Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. The chart below from JP Morgan shows how one saver (Susan) who invests for only 10 years early in her career, ends up with more wealth than another saver. MONTHLY COMPOUNDING ANNUAL COMPOUNDING STRATEGY 3 VS. In the formula, A represents the final amount in the account after t years compounded 'n' times at interest rate 'r' with starting. As you can see, at the end of 10 years, you receive more than 50 percent more money in interest payments with compound interest ($15,939) than you do with simple interest ($10,000). Now, let's say you deposited the same amount of money on a bank for 2 years at 3% annual interest compounded annually. 2 : Nov 20,
The following is multiple choice question (with options) to answer.
The compound interest on Rs.2,000 at 10% per annum for 3 years payable half yearly is | [
"Rs.600",
"Rs.630",
"Rs.660",
"Rs.680"
] | D | amt=2000(1+r/2/100)^2n formulae as it is half yearlly
amt=2000(1+1/20)^4
amt=2680.19
ci= 2680.19-2000 = 680.00
ANSWER:D |
AQUA-RAT | AQUA-RAT-39208 | newtonian-mechanics, drag, relative-motion
Title: Relative motion-Acceleration My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.
Say a boat travels at 10 m/s in a water channel.
the water speed relative to ground is 0.
so the boat travels at 10 m/s relative to the ground.
now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
As after a while the boat speed relative to ground has become 0,
then from the ground-based observer's point of view, the boat has undergone a deceleration.
My question is;
Is this deceleration always necessarily equal to minus the water acceleration?
In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?
PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?
My question is; Is this deceleration always necessarily equal to minus the water acceleration?
The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:
$$f=kv\qquad \text{ for low speed}\\
f=kv^2\qquad \text{ for high speed}$$
where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.
The following is multiple choice question (with options) to answer.
A boat can move upstream at 35 kmph and downstream at 65 kmph, then the speed of the current is? | [
"15",
"8",
"9",
"12"
] | A | US = 35
DS = 65
M = (65- 35)/2
= 15
Answer:A |
AQUA-RAT | AQUA-RAT-39209 | Let P=[$(\sqrt{n}+0.7)^2$]
Second Hint
given $n \geq 1$, put n=1 gives P=2
n=2 gives P=4
n=3 gives P=5
n=4 gives P=7
n-5 gives P=8
n=6 gives P=9
n=7 gives P=11
Final Step
here missing number are
1,3,6,10,… which is following a certain pattern
1, 1+2, 3+3, 6+4, 10+5, 15+6, 21+7, 28+8, 36+9, 45+10, 55+11, 66+12.
so, $n_{12}$=78.
The following is multiple choice question (with options) to answer.
Find the missing number in the given sequence : 1,4,7,?,13,16,? | [
"10 & 19",
"8 & 17",
"9 & 18",
"11 & 20"
] | A | 1+3=4
4+3=7
7+3=10
10+3=13
13+3=16
16+3=19
ANSWER:A |
AQUA-RAT | AQUA-RAT-39210 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
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
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
In a manufacturing plant, it takes 36 machines 4 hours of continuous work to fill 4 standard orders. At this rate, how many hours of continuous work by 72 machines are required to fill 12 standard orders? | [
"3",
"6",
"8",
"9"
] | B | the choices give away the answer..
36 machines take 4 hours to fill 4 standard orders..
in next eq we aredoubling the machines from 36 to 72, but thework is not doubling(only 1 1/2 times), = 4*48/72*12/4 = 6
Ans B |
AQUA-RAT | AQUA-RAT-39211 | Example: Take $k=2002$, then $n=63$ and hence $2002$ lies in $(T_{63}-2002+1,63+2002-T_{63})=(15,49)$.
መለሰ
The following is multiple choice question (with options) to answer.
N is one of the numbers below. N is such that when multiplied by 0.75 gives 1. Which number is equal to N? | [
"1 1/2",
"1 1/3",
"5/3",
"3/2"
] | B | "N is such that when multiplied by 0.75 gives 1" is written mathematically as
N * 0.75 = 1
Solve for N
N = 1/0.75 = 100/75 = (75 + 25) / 75 = 75/75 + 25/75 = 1+1/3
correct answer B |
AQUA-RAT | AQUA-RAT-39212 | # Geometry
Two cylinders have the same volume. If the radius of cylinder I is 3 times the radius of cylinder II, then the height of cylinder II is how many times the height of cylinder I?
A. 12
B. 9
C. 6
D. 3
1. 👍 0
2. 👎 0
3. 👁 96
1. Cylinder 1 has nine times the base area of cylinder 2.
Volume is (base area) x (height)
For the volumes to be equal, cylinder 2 must have 9 times the height of cylinder 2.
1. 👍 0
2. 👎 0
posted by drwls
2. pi*r^2*h2 = pi*(3r)^2*h1.
pi*r^2*h2 = pi*9r^2*h1
Divide both sides by pi*r^2:
h2 = 9*h1.
1. 👍 0
2. 👎 0
posted by Henry
3. For the volumes to be equal, cylinder 2 must have 9 times the height of cylinder 1
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4. msh fahma 7aga
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## Similar Questions
1. ### physics
Consider two solid uniform cylinders that have the same mass and length, but different radii: the radius of cylinder A is much smaller than the radius of cylinder B. Rolling down the same incline, which one of the two cylinders
asked by rosa on November 18, 2013
2 cylinders are proportional the smaller cylinder has a radius of 4 centimeters, which is half as large as the radius of the larger cylinder. the volume of smaller cylinder =250 cubic cemtimeters. what is the volume of the larger
asked by brianna on December 14, 2008
3. ### math
2 cylinders are proportional. The smaller cylinder has a radius of 4 cm. which is half as large as the radius of the larger cylinder.The volume of the smaller cylinder is 250 cubic cm. What is the approximate volume of the larger
asked by brianna on December 14, 2008
4. ### College Chemistry
If the density of cylinder 1 is 3.55 , what is the density of cylinder 2? (The masses and volumes of two cylinders are measured. The mass of cylinder 1 is 1.35 times the mass of cylinder 2. The volume of cylinder 1 is 0.792 times
The following is multiple choice question (with options) to answer.
A cylinder of height h is 8/9 of water. When all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder, the new cylinder is 3/5 full. The height of the new cylinder is what percent of h? | [
"25%",
"50%",
"68%",
"80%"
] | C | Basically we can disregard the radius is 25% information, as we are only asked about the height of the original and the new cylinder.
This is becausethe new cylinder is 3/5 fullmeans the same as that it's height is 3/5.
Original cylinder 8/9
New cylinder 3/5
So 3/5/8/9 = 3/5 * 9/8 = 12/15 = 4/5 = 0.680 or 68%.
Answer C |
AQUA-RAT | AQUA-RAT-39213 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"10 m",
"16 m",
"245 m",
"19 m"
] | C | Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30
= 25/2 => 2(130 + x) = 750 => x
= 245 m.
Answer: C |
AQUA-RAT | AQUA-RAT-39214 | (A) 10
(B) 45
(C) 50
(D) 55
(E) 65
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
15 Nov 2017, 16:22
9
3
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
We can create the following equation:
Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things
150 = 0.6(150) + 0.5(150) + 0.3(150) - D - 2(5) + 5
150 = 90 + 75 + 45 - D - 10 + 5
150 = 205 - D
D = 55
_________________
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
29 May 2017, 05:24
8
5
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
This can be solved using Venn Diagram (refer the attachment)
AC = 60% of 150 = 90
Sunporch = 50% of 150 = 75
SP = 30% of 150 = 45
The following is multiple choice question (with options) to answer.
At a certain health club, 30 percent of the members use both the pool and sauna, but 35 percent of the members who use the pool do not use the sauna. What percent of the members of the health club use the pool? | [
"33 1/3%",
"51 23/27%",
"50%",
"62 1/2%"
] | B | P = pool S = sauna
given P+S = 30 then let only S be x and only P will be 100 - (30+x) = 70 -x
35% of (70-x) = x => 24.5 - 0.35x = x => x = 18 4/27% so only P = 70 -x = 51 23/27%
Answer B |
AQUA-RAT | AQUA-RAT-39215 | Arrange $2$ red books. Arrange the $5$ black books in any of the three positions as [bbb] [bb]. Take $3$ blue books and separate the adjacent bbb and bb. Remaining $3$ books can be arranged in remaining $5$ positions. So,
$\dbinom{3}{1}\dbinom{2}{1}\dbinom{5}{3}=60$
Arrange $2$ red books. Arrange the $5$ black books in any of the three positions as [bb] [bb] [b]. Take $2$ blue books and separate the adjacent bb and bb. Remaining $4$ books can be arranged in remaining $6$ positions. So,
$\dbinom{3}{2}\dbinom{6}{4}=45$
Total: $228$
Part (2): Find number of arrangements where blue books are apart and black books are apart but red books are together
Group the red books and consider as $1$ big red book.
Arrange $1$ big red book. Arrange the $5$ black books in any of the two positions as [bbbb] [b] . Take $3$ blue books and separate the adjacent bbbb. Remaining $3$ books can be arranged in remaining $4$ positions. So,
$\dbinom{2}{1}\dbinom{4}{3}=8$
Arrange $1$ big red book. Arrange the $5$ black books in any of the two positions as [bbb] [bb] . Take $3$ blue books and separate the adjacent bbbb. Remaining $3$ books can be arranged in remaining $4$ positions. So,
$\dbinom{2}{1}\dbinom{4}{3}=8$
Arrange $1$ big red book. Arrange the $5$ black books in any of the two positions as [bbbbb] . Take $4$ blue books and separate the adjacent bbbbb. Remaining $2$ books can be arranged in remaining $3$ positions. So,
$\dbinom{2}{1}\dbinom{3}{2}=6$
Total: $22$
From Part(1) and (2),
## $228-22=206$
The following is multiple choice question (with options) to answer.
15 binders can bind 1400 books in 21 days. How many binders will be required to bind 1600 books in 20 days? | [
"14",
"18",
"24",
"28"
] | B | Binders Books Days
15 1400 21
x 1600 20
x/15 = (1600/1400) * (21/20) => x = 18
ANSWER:B |
AQUA-RAT | AQUA-RAT-39216 | 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 after interchanging the position of the two digits is 54. What is the difference between the two digits of the number? | [
"4",
"6",
"3",
"Cannot be determined"
] | B | Let the two-digit no. be l0x + y.
Then, (10x + y) – (10y + x) = 36
or, 9(x – y) = 54
or, x – y = 6
Answer B |
AQUA-RAT | AQUA-RAT-39217 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
On dividing number by 357, we get 39 as remainder. On dividing the same number by 17, what will be the remainder? | [
"0",
"3",
"5",
"11"
] | C | Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.
x = 5k + 3
x2 = (5k + 3)2 = (25k2 + 30k + 9)
= 5(5k2 + 6k + 1) + 4
On dividing x2 by 5, we get 4 as remainder.
ANSWER:C |
AQUA-RAT | AQUA-RAT-39218 | Digital Sum Rule of Multiplication: The digital sum of the product of two numbers is equal to the digital sum of the product of the digital sums of the two numbers.
Example: The product of 129 and 35 is 4515.
Digital sum of 129 = 3 and digital sum 35 = 8. Product of the digital sums = 3 × 8 = 24 Digital sum = 6.
Digital sum of 4515 is = 4 + 5 + 1 + 5 = 15 = 1 + 5 = 6. Digital sum of the product of the digital sums =
digital sum of 24 = 6 Digital sum of the product (4515) = Digital sum of the product of the digital sums
(24) = 6
Applications of Digital Sum :
1.Rapid checking of calculations while multiplying numbers :
Suppose a student is trying to find the product 316 × 234 × 356, and he obtains the number 26525064. A
quick check will show that the digital sum of the product is 3. The digital sums of the individual numbers
(316, 234 and 356) are 1, 9, and 5. The digital sum of the product of the digital sum is 1 × 9 × 5 = 45 = 4 + 5 = 9. ⇒ the digital sum of the product of the digital sums (9) ≠ digital sum of the 26525064 (3). Hence,
the answer obtained by multiplication is not correct.
Note: Although the answer of multiplication will not be correct if the digital sum of the product of the digital
sums is not equal to digital sum of the product, but the reverse is not true i.e. the answer of multiplication
may or may not be correct if the digital sum of the product of the digital sums is equal to digital sum of
the product.
Suppose you have a question : ( Very absurd eg by me but learn the concept )
What is value is A+B+C+D+E?
$$878373 * 7738838339 * 827287 = E6D35C50B5547A87623729$$
HINT: Find the digital sum on LHS and equate it with that of RHS.
2.Determining if a number is a perfect square or not :
The digital sum of the numbers which are perfect squares will
always be 1, 4, 7, or 9.
The following is multiple choice question (with options) to answer.
In a two-digit number, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is: | [
"24",
"73",
"26",
"82"
] | A | Explanation:
Let the ten's digit be x. Then, unit's digit = x + 2. Number = 10x + (x + 2) = 11x + 2
Sum of digits = x + (x + 2) = 2x + 2
(11x + 2)(2x + 2) = 144
2x2 + 26x - 140 = 0
(x - 2)(11x + 35) = 0
x = 2
Hence, required number = 11x + 2 = 24.
Answer: A |
AQUA-RAT | AQUA-RAT-39219 | # Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?
A bag has:
$3$ red
$5$ black
$8$ green
marbles.
Total of 16 marbles.
You select a marble, and then another one right after. (without replacement).
What is the probability that $both$ are red?
Probability that first pick is red: $\frac{3}{16}$
Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)
Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$
How do I do this using combinations only?
• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59
Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$
Hope this helps you.
The following is multiple choice question (with options) to answer.
A box contains 4 red chips and 3 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors? | [
"1/2",
"12/21",
"7/12",
"2/3"
] | B | Total selection ways: 7C2=7!/3!4!=21
Selecting one blue chip out of two: 3C1=3!/2!1!=3
Selecting one red chip out of four: 4C1=4!/1!3!=4
Thus, (3C1*4C1)/7C2=(3*4)/21=12/21=B |
AQUA-RAT | AQUA-RAT-39220 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
894.7 – 573.07 – 95.007 = ? | [
"226.623",
"224.777",
"233.523",
"414.637"
] | A | Solution
Given expression = 894.7 - (573.07 + 95.007) = 894.7 - 668.077 = 226.623. Answer A |
AQUA-RAT | AQUA-RAT-39221 | kmph. com/2016/01/01/speed-distance-timeVideos, worksheets, 5-a-day and much more. If a car traveled 50 miles over the course of one hour then its average speed will be 50 mph. Distance, Time and Speed Word Problems | GMAT GRE Maths. Exercise: Problems on Speed Time and Distance. org/time-speed-distanceTime, Speed and Distance (popularly known as TSD) is an important topic for written round of placements for any company. sg//sol_04_distance_speed_and_time. Let the time taken to cover the distance downstream = t hrs. Distance Time Speed. Time is entered in minutes, speed in knots and distance in nautical miles (the same formula will work for statute miles and kilometres). Time = Distance / speed = 20/4 = 5 hours. Speed, Time, and Distance Worksheet. Equations: Acccceelleerraattiioonn = Final speed–Initial TTiimme = Final Speed–Initial Time Acceleration. When dealing with distance, rate and time, we always want to remember the nifty little formula, D = R x T, in which D stands for the distance, R stands for the rate (or speed), and T stands for the time. b. Acceleration – change in velocity over time. 4 Calculate speed, distance, and time I . Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. He drives 150 meters in 18 seconds. Distance (D) = Speed (S) × Time (T) X kmph = X × 5/18 m/s X m/s = X × 18/5 kmph. Aptitude Reasoning TRAINS DISTANCE SPEED TIME QUANTITATIVE APTITUDE . We define speed as distance divided by time. A train travels at a speed of 30mph and travel a distance of 240 miles. It can cross a pole in 10 seconds. The speeds are indicated in the rate column. Toon Train is traveling at the speed of 10 m/s at the top of a hill. If you run around the house randomly, and then end up back where you started, moving a total of 44 meters, what is your distance? Average speed for the entire trip is going to be equal to the total
The following is multiple choice question (with options) to answer.
A train when moves at an average speed of 60 kmph, reaches its destination on time. When its average speed becomes 45 kmph, then it reaches its destination 30 minutes late. Find the length of journey. | [
"70 km",
"76 km",
"85 km",
"90 km"
] | D | Explanation :
Solution: Difference between timings = 30 min. = 30/60 hr = 1/2 hr.
Let the length of the journey be x km.
Then, x/45 - x/60 = 1/2
=> 4x -3x = 180/2
=> x = 90 km.
Answer : D |
AQUA-RAT | AQUA-RAT-39222 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A certain sum of money is divided among A, B and C so that for each Rs. A has, B has 65 paisa and C 40 paisa. If C's share is Rs.40, find the sum of money? | [
"782",
"276",
"727",
"205"
] | D | A:B:C = 100:65:40
= 20:13:8
8 ---- 40
41 ---- ? => Rs.205
Answer:D |
AQUA-RAT | AQUA-RAT-39223 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way? | [
"8 sec",
"7 sec",
"2 sec",
"6 sec"
] | D | T = 60/36 * 18/5
= 6 sec
Answer:D |
AQUA-RAT | AQUA-RAT-39224 | # 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.
Camel and carriage for Rs. 5000. He SP the camel at a gain of 20% and the carriage at a loss of 10%. If he gains 3% on the whole, then find the CP of the camel? | [
"Rs.2166.67",
"Rs.2224",
"Rs.2270.72",
"Rs.2145"
] | A | Now, in this numerical, there is no common loss and gain %.
Hence, solve it making equations.
Let cost price of camel be x.
As cost of camel and carriage = Rs 5000
Cost of carriage = Rs. (5000 – x)
After selling camel he gains 20% and on carriage a loss of 10%. But on the whole he gains 3%.
Therefore,
20% of x – 10 % of (5000 – x) = 3 % of 5000
20 × x – 10 × (5000 – x) = 3 × 5000
100 100 100
x – (5000 – x) = 150
5 10
10x – (5000 – x) × 10 = 150 × 10
5 10
2x-5000+x=1500
3x=1500+5000
x=2166.67
The cost of camel = Rs.2166.67
Option (A) is the correct answer |
AQUA-RAT | AQUA-RAT-39225 | 7. Originally Posted by HallsofIvy
Skeeters method, find the roots to the original equation, take the reciprocal, and create the new equation from that is the obvious and simplest method.
Here is a little more "sophisticated" method. Look at the general product (x- a)(x- b)= x^2- (a+b)x+ ab. If, instead, we have (x- 1/a)(x- 1/b) we get x^2- (1/a+ 1/b)x+ 1/ab. You can see that the last term, 1/ab, is just the reciprocal in the original equation, ab. Also 1/a+ 1/b= b/ab+ a/ab= (a+b)/ab so the coefficient of x in the last equation is the coeffient of x in the original equation divided by the last term.
Starting from 3x^2+2x-1=0, which is the same as x^2+ (2/3)x- 1/3= 0 (I divided both sides by 3), we can see that the equation having the reciprocals as roots is x^2+ (2/3)/(-1/3)x- 1(1/3)= x^2- 2x- 3= 0.
We can do the same kind of thing for "2a" and "2b". (x- 2a)(x- 2b)= x^2-(2a+2b)+ (2a)(2b)= x^2- 2(a+b)+ 4(ab)= 0. To create an equation that has 2 times the roots as its roots, multiply the coefficient of x by 2 and the final term by 4.
Whatever the roots of 4x^2-5x-2= 0, which is the same as x^2- (5/4)x- 1/2= 0, are, the roots of the equation x^2- (5/2)(2)x- (1/2)(4)= x^2- 5x- 2= 0 are twice that.
The following is multiple choice question (with options) to answer.
The roots of the equation axpower2 + bx + c =0 will be reciprocal if: | [
"a=b",
"a-bc",
"c=a",
"c=b"
] | C | based on the problem 1/k will be the other root of the given equation. root equation = c/a. k*1/k=c/a. c/a= 1 or a = c[multiplying on both sides].
the roots of the equation axsquared + bc + c= 0 will be reciprocal if a= c. correct answer: (C) |
AQUA-RAT | AQUA-RAT-39226 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
a plane travelled k miles in 1st 96 miles of flight time.if it completed remaining 300 miles of the trip in 1 minute what is its average speed in miles/hr for entire trip ? | [
"396/60+k",
"376/60+k",
"386/60+k",
"366/60+k"
] | A | k=96+300
tot distance= 396 miles
tot time= 1min + k
avg speed= 396/60+k
ANSWER:A |
AQUA-RAT | AQUA-RAT-39227 | # How to distribute $k$ prizes in $p$ student with each student would receive $(k-1)$ prizes at most?
A teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was $k$, and total number of kids was $p$. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive $(k – 1)$ prizes at the most. How many ways she could distribute the prizes?
Answer is $p^k-p$
If I understood the problem correctly, it is no-where stated that a student can get no prize at all, and the answer seems to be using this assumption,however I am a bit confused why the answer is not $p^{k-1}$?
If the restriction is that no student can get more than $(k-1)$ prizes then what is wrong with starting with $(k-1)$ distinct prizes and distributing those into $p$ distinct groups (students)?
-
Then what happens with the last prize? That one needs to be distributed too, right? But then it makes a difference if the first k-1 prizes went to a single student or more students... – N. S. Nov 6 '11 at 23:34
Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute $k$ prizes among $p$ students?".
For each prize, there are $p$ possibilities for the recipient of that prize. In total, that means there are $p \cdot p \cdot \dots \cdot p$ (with $k$ copies of $p$ in the product) possible ways to assign the prizes. This is the $p^k$ part.
Now let's impose the restriction that no student receives all the prizes (this is the same as saying no student receives more than $k-1$ prizes). Of the $p^k$ configurations we just enumerated, which ones are illegal under this new restriction? Exactly $p$ of them, since there is one illegal configuration for each student (namely, giving that student all the prizes).
Subtracting the illegal configurations from our previous enumeration, we get $p^k - p$ legal configurations.
The following is multiple choice question (with options) to answer.
In a function they are distributing noble prize. In how many ways can 2 prizes be distributed among 5 boys when a boy gets any no. of prizes? | [
"25",
"30",
"64",
"70"
] | A | Ans.(A)
Sol. In this case, repetitions are allowed, so all the two prizes can be given in 5 ways, i.e. (5 x 5) ways = 52 ways = 25 ways Or nr ways = 52 ways = 25 ways |
AQUA-RAT | AQUA-RAT-39228 | # 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 40 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream? | [
"20 kmph",
"13 kmph",
"65 kmph",
"55 kmph"
] | A | Speed of the boat in still water
= (40+80)/2
= 60 kmph. Speed of the stream
= (80-40)/2
= 20 kmph.
Answer: A |
AQUA-RAT | AQUA-RAT-39229 | algorithms, scheduling, algorithm-design
Title: Schedule two trains whose tracks overlap so they don't crash I've encountered scheduling problems in my algorithms class before like the type we use vertex cover to solve.
Recently I was asked this question and did not even know what algorithmic technique to use to answer it!
There are two trains, which run between stations and share a portion of track. For instance train 1 runs between stations A B C D E F G H and train 2 runs between stations I J K B C D E F L M so the trains share the track between B C D. At each end of the track the train turns around and goes the other way on its track. The trains go from station to station in 1 unit time. I need to prevent the trains from crashing head on on the shared track, I have the power to stop either train at any station and restart it again when I please.
How can I solve this problem without using a ton of if else clauses, what CS principles should I be using here?
edit
Some background, the first solution I wrote was just a simple check to see if the other train was on the shared track before sending the other train out. That seemed ad hoc to me and I'm sure it did to the company I interviewed for as well.
I figure there must be an algorithm or at least a school of thought for this type of problem, after all how do they build extensible systems for managing shared runway space at an airport, or managing traffic flow at stop lights.
My intuition is that this is a very introductory problem to a school of thought in computational problem solving I have not yet encountered.
I might be wrong, but could anybody clear the air for me? If you calculate the cycle, as there are only two trains and distance between every station is one, you could simply run them in order and appoint wait in some station to avoid crash.
If you have to implement some unfortunate events (first one stops for some reason, reschedule it to run on itd own time).
With two trains it will work without problems.
Otherwise use semaphores (it fits even literally ;)
There is no explicitly given synchronisation problem, so nasty trick works.
The following is multiple choice question (with options) to answer.
There are 12 intermediate stations between two places A and B. Find the number of ways in which a train can be made to stop at 4 of these intermediate stations so that no two stopping stations are consecutive? | [
"108",
"112",
"126",
"140"
] | C | Explanation :
Initially, let's remove the 4 stopping stations
Then we are left with 8 non-stopping stations (=12-4) as shown below
(non-stopping stations are marked as 1,2 ... 8)
Now there are 9 positions (as marked by * in the above figure) to place the 4 stopping stations
such that no two stopping stations are consecutive
This can be done in 9C4 ways
Hence, required number of ways = 9C4
=((9)(8)(7)(6)/(4)(3)(2)(1))=126
Answer C |
AQUA-RAT | AQUA-RAT-39230 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
If,
1 * 3 * 5 = 16
3 * 5 * 7 = 38
Then find,
5 * 6 * 9 =? | [
"65",
"59",
"72",
"80"
] | B | (9 * 6) + 5 = 59
B |
AQUA-RAT | AQUA-RAT-39231 | (b) Here, since the digits must strictly increase from left to right, consider two sub-cases:
(b1) If 0 is not included - Then, there are 9 digits, and for every choice of 4 digits from them, we have exactly one way to arrange them in strictly increasing order. So, there are 9C4 such numbers.
(b2) If 0 is included - Then, 0 will appear as the left most digit, and this will not be a four-digit number. Therefore, there are no such numbers at all.
The following is multiple choice question (with options) to answer.
Some numbers are very mysterious. They are all integers. They have more than one digit. If you multiply them with 2 their digits may shift ,but no new digit will include or no digit will disappear. Example: 12 is not such a number. Because if you multiply 2 with 12, it will become 24. And 24 includes a new digit 4 ,and fails to obtain the previous digit 1. Can you find one of them? | [
"138687",
"142587",
"144487",
"142667"
] | B | B
142587
142587*2=285174
These type of numbers are called cyclic numbers where a number multiplied by any digit the result will contain all the digit in the original number and no other digit.
cyclic numbers are generated by full reptend prime 7,17,19,23 etc.
where 1/7=0.142857……
1/17=0.058823529411764 |
AQUA-RAT | AQUA-RAT-39232 | # Project Euler #5 - Lowest common multiple of 1 through 20
This is my code. All comments welcome. Last time run it only took 335 milliseconds:
static bool CheckMultiples(int val)
{
for (int i = 1; i <= 20; i++)
{
if (val % i != 0)
return false;
}
return true;
}
static void Main(string[] args)
{
Stopwatch s = new Stopwatch();
s.Start();
int Num = 20;
while (!CheckMultiples(Num))
{
Num += 20;
}
s.Stop();
Console.WriteLine("The smallest number divisible by all numbers 1-20 is {0}.", Num);
Console.WriteLine("The time took is {0} milliseconds.", s.ElapsedMilliseconds);
}
There is a far more efficient approach to finding the least common multiple ($\operatorname{lcm}$) of a set of integers which, unlike your and rolfl's code, involves no trial-and-error. Recall that for any pair $(a, b)$ of natural numbers, we have $$\operatorname{lcm}(a, b) = \frac{ab}{\operatorname{gcd}(a, b)}.$$ This reduces the implementation of $\operatorname{lcm}$ to the implementation of $\operatorname{gcd}$, which is easily done using the simple and efficient Euclidean algorithm. Then, calculating the least common multiple of $\{1, 2, ..., 20\}$ is as simple as $\operatorname{reduce}(\operatorname{lcm}, \{1, 2, ..., 20\})$.
You can find a simple Python 3 implementation of this idea in ideone here. Even in a slow interpreted language like Python, this is several orders of magnitude faster than your or rolfl's code, running in about .04 milliseconds.
The following is multiple choice question (with options) to answer.
Find the lowest common multiple of 25, 35 and 50. | [
"350",
"420",
"510",
"320"
] | A | LCM=2*5*5*7=350.
Answer is A |
AQUA-RAT | AQUA-RAT-39233 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank? | [
"20 hours",
"15 hours",
"10 hours",
"12 hours"
] | A | Part filled by A in 1 hour = 1/36
Part filled by B in 1 hour = 1/45
Part filled by (A+B) in 1 hour = 1/36 + 1/45 = 9/180 = 1/20
Both the pipes together fill the tank in 20 hours
Answer is A |
AQUA-RAT | AQUA-RAT-39234 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article offering a discount of 5% and earned a profit of 23.5%. What would have been the percentage of profit earned if no discount was offered? | [
"60%",
"37%",
"30%",
"80%"
] | C | Let C.P. be Rs. 100.
Then, S.P. = Rs. 123.50
Let marked price be Rs. x. Then, 95/100 x = 123.50
x = 12350/95 = Rs. 130
Now, S.P. = Rs. 130, C.P. = Rs. 100
Profit % = 30%.
Answer & Answer:C |
AQUA-RAT | AQUA-RAT-39235 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Two mechanics were working on your car. One can complete the given job in 5 hours, but the new guy takes eight hours. They worked together for the first two hours, but then the first guy left to help another mechanic on a different job. How long will it take the new guy to finish your car? | [
"7/4",
"4/3",
"14/5",
"10/3"
] | C | Rate(1)=1/5
Rate(2)=1/8
Combined = 13/40
Work done in 2 days=13/20
Work left = 7/20
Rate * Time = Work left
1/8 * Time = 7/20
Time=14/5
C |
AQUA-RAT | AQUA-RAT-39236 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
An optometrist charges $150 per pair for soft contact lenses and $85 per pair for hard contact lenses. Last week she sold 5 more pairs of soft lenses than hard lenses. If her total sales for pairs of contact lenses last week were $1,455, what was the total number of pairs of contact lenses that she sold? | [
" 11",
" 13",
" 15",
" 17"
] | A | (x+5)*150 +x*85=1455
=>x=3
total lens=3+(3+5)= 11
Answer A |
AQUA-RAT | AQUA-RAT-39237 | 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.
What would be the compound interest on an amount of Rs.5500 at the rate of 5 p.c. per annum after 2 years? | [
"610.25",
"573.85",
"563.75",
"420.5"
] | C | CI=P[(1+(R/100))^T-1]
=5500*[(1+(5/100))^2-1]
=5500*[(1.05)^2-1]
=5500*0.1025
=Rs.563.75
Option C |
AQUA-RAT | AQUA-RAT-39238 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
Eight percent of the programmers in a startup company weigh 200 pounds or more. Twenty-five percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less. What percent of the programmers in the startup company weigh between 100 and 200 pounds? | [
"15%",
"20%",
"25%",
"69%"
] | D | Initially 92% and 8% split
80% is further divided as 25% and 75%
Q is asking about that 75%
let total be '100' then that 75% is (3/4)∗92
so, the required % is [(3/4)∗92/100]∗100 = 69%
ANSWER:D |
AQUA-RAT | AQUA-RAT-39239 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
A water tank is one-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how many minutes will it take to empty or fill the tank completely? | [
"3",
"6",
"9",
"12"
] | A | The combined rate of filling/emptying the tank = 1/10 - 1/6 = -1/15
Since the rate is negative, the tank will be emptied.
A full tank would take 15 minutes to empty.
Since the tank is only one-fifth full, the time is (1/5) * 15 = 3 minutes
The answer is A. |
AQUA-RAT | AQUA-RAT-39240 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The price of 3 pants and 6 t-shirts is Rs. 2250. With the same money one can buy 1 pant and 12 t-shirts. If one wants to buy 8 t-shirts, how much shall she have to pay ? | [
"1300",
"1100",
"1000",
"1200"
] | D | Let the price of a pant and a t-shirt be Rs. x and Rs. y respectively.
Then, 3x + 6y = 2250 .... (i)
and x + 12y = 2250 .... (ii)
Divide equation (i) by 3, we get the below equation.
= x + 2y = 750. --- (iii)
Now subtract (iii) from (ii)
x + 12y = 2250 (-)
x + 2y = 750
----------------
10y = 1500
----------------
y= 150
cost of 8 t-shirts = 8*150 = 1200
Answer : D |
AQUA-RAT | AQUA-RAT-39241 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
_________________
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The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
When the price of an article was reduced by 30% its sale increased by 80%. What was the net effect on the sale? | [
"26% increase",
"44% decrease",
"60% increase",
"66% increase"
] | A | if n items are sold for $p each, revenue is $np. If we reduce the price by 30%, the new price is 0.7p. If we increase the number sold by 80%, the new number sold is 1.8n. So the new revenue is (0.7p)(1.8n) = 1.26np, which is 1.26 times the old revenue, so is 26% greater.
ANSWER:A |
AQUA-RAT | AQUA-RAT-39242 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
The average expenditure of a labourer for 10 months was 85 and he fell into debt. In the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30. His monthly income is | [
"180",
"100",
"112",
"110"
] | C | Income of 10 months = (10 × 85) – debt
= 850 – debt
Income of the man for next 4 months
= 4 × 60 + debt + 30
= 270 + debt
∴ Income of 10 months = 1120
Average monthly income = 1120 ÷ 10 = 112
Answer C |
AQUA-RAT | AQUA-RAT-39243 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
How many terms are there in 2, 4, 8, 16,..., 512? | [
"14",
"11",
"12",
"9"
] | D | 2, 4, 8, 16, ..., 512 is a G.P with a =2
and r =4/2 =2
Let the number of terms be n. Then
2 x 2 ^n-1 = 512
or 2^n-1 = 256 = 2^8
Thus n - 1 =8
n= 9
ANSWER:D |
AQUA-RAT | AQUA-RAT-39244 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours? | [
"233",
"552",
"376",
"287"
] | B | Given that distance travelled in 1st hour = 35 km
and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
...
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term, a=35, number of terms,n = 12 and common difference, d=2.
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . . is called an Arithmetic Progression(AP)where a is the first term and d is the common difference of the AP
Sum of the first n terms of an Arithmetic Progression(AP),Sn=n2[2a+(n−1)d]where n = number of terms
Hence, [35+37+39+... (12 terms)]=S12=122[2×35+(12−1)2]=6[70+22]=6×92=552
Hence the total distance travelled = 552 km
Answer :B |
AQUA-RAT | AQUA-RAT-39245 | circle, ellipse, trapezoid, cube, sphere, cylinder and cone. 2πrh=πrl [r is radius of. A cylinder is a solid figure, with a circular or oval base or cross section and straight and parallel sides. ' and find homework help for other Math questions. Derive the formula for the surface area of a cone of radius r and height h. As we can seem the ratio is 2/3. volume = Pi * radius 2 * length. Similarly, the volume of a cube is V =L*L*L. This is within the range provided by the "64 to 74%" rule of thumb. Grade 8 » Geometry » Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. The volume of this triangular pyramid is 252 cm3. Cylinder: V = π R 2 L where R is the radius of its base and L the length of it. The Volume Formula of a Sphere. That is, Dm (the dioptric power at any meridian of a cylindric lens) is equal to D (the maximum power of the cylinder) multiplied by the sine squared of the given angle. When solving problems about volume of cones and cylinders, you highlight the base and the height. Tape together as shown. 01 mL pretty reliably. The surface area of an open ended cylinder (as shown) is 2 RL If the cylinder has caps on the ends, the surface area is 2 RL+2 R 2; The volume of a cylinder is R 2 L Note that =3. MORE PRACTICE : F. Volume of cone = (1/3)πr2h Volume of hemisphere = (2/3)πr3 Volume of cylinder = πr2h Given :-the cone, hemisphere and cylinder have equal base and same height. Car load volume to move storage. A similar figure is the (circular) cylinder, which has two congruent circular bases and a tube-shaped body, as shown below. Circle and sphere are both round in shape but whereas a circle is a figure, a sphere is an object. Available in clear color. When used in a classroom setting, the task could be supplemented by questions that ask students to thinking about the relationship between volume and liquid capacity. We can use the relationship between the volume of a cone and a cylinder, both conceptually and computationally, to solve real-world problems. The volume of a
The following is multiple choice question (with options) to answer.
The volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2. What is the length of the wire? | [
"2:5",
"2:6",
"2:3",
"2:4"
] | A | The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r12h1/r22h2 => 1/10 = (1)2h1/(2)2h2
=> h1/h2 = 2/5
i.e. h1 : h2
=2:5
Answer: A |
AQUA-RAT | AQUA-RAT-39246 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
In a group of 6 boys&4 girls a Committee of 4 persons is to be formed. In how many different ways can it be done so that the committee has at least 1girl? | [
"120",
"130",
"150",
"195"
] | D | The committee of 4 persons is to be so formed that it has at least 1 woman The different ways that we can choose to form such a committee are:
(i) lw. 3 m in t 6X5X4' x 6C3 = 4x — — 80 3X2X1
x 6c2 =42:: x 26:: = 90 (ii) 2w. 2 m in °C2 (iii) 3w. 1 m in 4C3 x 6C1 = 4 x 6 = 24 (iv) 4w in 6C4 = 1 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least one woman. = 80 + 90 + 24 + 1 = 195
D |
AQUA-RAT | AQUA-RAT-39247 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When positive integer r is divided by 13, the remainder is 2. When n is divided by 8, the remainder is 5. How many such values are less than 180? | [
" 0",
" 1",
" 2",
" 3"
] | B | the equation that can be formed r is 13x+2=8y+5..
13x-3=8y...
as we can see x can take only odd values as the RHS will always be even..
Also x can take values till 13 as 13*14>180..
now we have to substitue x as 1,3,5,7,9,11,13...
once we find 7 fitting in , any other value need not be checked as every 4th value will give us answer so next value will be 15..
ans 1.. B |
AQUA-RAT | AQUA-RAT-39248 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
Torry has submitted 2/5 of his homework assignments, and he received an average grade of 75 for those assignments. If he wishes to receive an average grade of 105 for all his homework assignments, the average grade for Torry's remaining homework assignments must be what percent greater than the average grade for the assignments he has already submitted? | [
"15%",
"20%",
"25%",
"66 2/3%"
] | D | 0.4*75 + 0.6x = 105
30 + 0.6x=107
0.6x = 75
x= 125
125/75= 1.66666...
Ans:D |
AQUA-RAT | AQUA-RAT-39249 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A block of wood has dimensions 10cm x 10cm x 50cm. The block is painted red and then cut evenly at the 25 cm mark, parallel to the sides, to form two rectangular solids of equal volume. What percentage of the surface area of each of the new solids is not painted red? | [
"5.5%",
"8.3%",
"11.6%",
"14.2%"
] | B | The area of each half is 100+4(250)+100 = 1200
The area that is not painted is 100.
The fraction that is not painted is 100/1200 = 1/12 = 8.3%
The answer is B. |
AQUA-RAT | AQUA-RAT-39250 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
A board 7ft. 9 inches long is divided into 3 equal parts . What is the length of each part? | [
"2 FT 7 IN",
"1 FT 7 IN",
"2 FT 6 IN",
"3 FT 7 IN"
] | A | Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.
Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches
ANSWER A |
AQUA-RAT | AQUA-RAT-39251 | 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.
In an office, 20 percent of the workers have at least 5 years of service, and a total of 16 workers have at least 10 years of service. If 90 percent of the workers have fewer than 10 years of service, how many of the workers have at least 5 but fewer than 10 years of service? | [
" 32",
" 64",
" 50",
" 144"
] | A | (10/100)Workers = 16 = > number of workers = 160
(20/100)*Workers = x+16 = > x = 32
Answer A |
AQUA-RAT | AQUA-RAT-39252 | Pre-checking the number $A$
Very quickly, 7 is found to be a factor of $A$. We have $A=7 \times B$ where $B$ is:
$B=$ 9636514689378269172619627962314350702683338048951
We then try to find small factors of $B$. The number 283 is found to be a factor of $B$. Then we have $A=7 \times 283 \times C$ where $C$ is the following 47-digit number:
$C=$ 34051288655046887535758402693690285168492360597
By tagging a 6 in front of the 49-digit factor of the 9th Fermat number, we obtain a composite number. The prime or composite status for the number $A$ is settled. Now the next question: is the remaining factor $C$ prime or composite?
___________________________________________________________________
Pre-checking the number $C$
We perform the two pre-checks as described above. In searching for small factors, we find that $C$ is not divisible by all prime numbers less than 200,000. So if $C$ has a factor, it will have to be a large one. Or it could be that $C$ is prime. Instead of continuing to search for factors, we can do the second check. We calculate $2^{C-1} \ (\text{mod} \ C)$ and find that:
$2^{C-1} \equiv t \ (\text{mod} \ C)$
where $t=$ 7426390354305013563302537374271875139618265902
Since $t$ clearly is not 1, this tells us that $C$ is composite.
The following is multiple choice question (with options) to answer.
a and b are positive integers less than or equal to 8. If a and b are assembled into the six-digit number ababab, which of the following must be a factor of ababab? | [
"3",
"4",
"5",
"6"
] | B | 484848 when divided by 4 gives the result 121212
CONCEPT: Rule of Divisibility of 4 isIf sum of the digits of the Number is divisible by 4 then the number will be divisible by 2
SUm of the digits of No. which is divisible by 4 hence the Number 121212 will be divisible by 4
B |
AQUA-RAT | AQUA-RAT-39253 | • Any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.
• A Pythagorean triple consists of three positive integers $$a$$, $$b$$, and $$c$$, such that $$a^2 + b^2 = c^2$$. Such a triple is commonly written $$(a, b, c)$$, and a well-known example is $$(3, 4, 5)$$. If $$(a, b, c)$$ is a Pythagorean triple, then so is $$(ka, kb, kc)$$ for any positive integer $$k$$. There are 16 primitive Pythagorean triples with c ≤ 100:
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).
• A right triangle where the angles are 30°, 60°, and 90°.
This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio $$1 : \sqrt{3}: 2$$.
Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
• A right triangle where the angles are 45°, 45°, and 90°.
The following is multiple choice question (with options) to answer.
A right triangle has sides of a, b, and 9, respectively, where a and b are both integers. What is the value of (a + b)? | [
"64",
"49",
"96",
"81"
] | D | LET a= hypotenuse , b =base ,9 = perpendicular . therefore a^2 -b^2 =9^2 or (a+b)(a-b) = 81
a+b =81/a-b ' a-b cannot be zero ..therefore a+ b =81 where a-b is equal to 1
D |
AQUA-RAT | AQUA-RAT-39254 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way? | [
"10 sec",
"8 sec",
"5 sec",
"6 sec"
] | D | T = 60/36 * 18/5 = 6 sec
ANSWER D |
AQUA-RAT | AQUA-RAT-39255 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
In the graduating class of a certain college, 48 percent of the students are male and 52 percent are female. In this class 40 percent of the male and 30 percent of the female students are 25 years old or older. If one student in the class is randomly selected, approximately what is the probability that he or she will be less than 25 years old? | [
"A)0.9",
"B)0.6",
"C)0.45",
"D)0.3"
] | B | Percent of students who are 25 years old or older is 0.4*48+0.3*52=~34, so percent of people who are less than 25 years old is 100-34=66.
Answer: B. |
AQUA-RAT | AQUA-RAT-39256 | So, we can always stop when we reach the largest prime number smaller than the square root of the limit -- in this case, when we reach the largest prime number smaller than $\sqrt{200}$. (The square root, say $a$ has the property that $a\times a = 200$ so any other combination of factors must have one larger, one smaller).
$13\times 13 = 169 < 200$ and $17\times 17 = 289 > 200$, so we don't need to check as far as $17$ and D cannot be the answer, which leaves C.
Your reasoning is a bit "in reverse" when you are thinking about the 10. When you employed 2, the 10 & all of its multiples 20, 30, 40, et. al. were indeed eliminated. Note here that 25 was skipped. The fact that 5 is a factor of 10 makes no difference as that 5 never gets used. When actually using this method, its no problem at all to determine what number to use next: it's simply the next higher number not already marked out.
Doing the sieve with the primes in order, the first nontrivial multiple of $p$ to eliminate is $p^2$.
If $p$ is an odd prime, then $2p$ should have already been eliminated when you were crossing off the multiples of $2$. But $p^2$ is odd, not even, so it shouldn't have been eliminated at a prior step. Likewise, if $p$ is a prime greater than $3$, then you don't need to cross $2p$ and $3p$ off again because they should have already been crossed off.
Let's see what happens with D. You might forget to cross off $35$ if you skip ahead to $49$. And then at the end of the process you might come to the absurd conclusion that both $5$ and $25$, and possibly $35$ and $105$, are also prime, which is of course incorrect.
The following is multiple choice question (with options) to answer.
How many integers V are prime numbers in the range 200 < V < 220? | [
" 1",
" 2",
" 3",
" 4"
] | A | My friend, every single odd number greater than can be written either as 4V+1 or as 4V+3. If you divide any odd number by 4, you will get a remainder of either 1 or 3. That's not a rule unique to prime numbers at all.
The 6V+1 or 6V-1 rule is basically every odd number that is not divisible by three, so it narrows the search a little.
Here's how I thought about the problem. First, eliminate all the even numbers and the odd multiples of 5 in that range. That leaves us with:
{201, 203, 207, 209, 211, 213, 217, 219}
Eliminate the four multiples of 3. Notice that 21 is a multiple of 3, so 210 is also a multiple of 3. If we add or subtract 3 or 9, we get more multiples of three. When we eliminate those, we are left with.
{203, 209, 211, 217}
Now, notice that a cool thing about this range is that 210 is also a multiple 7 (again, because 21 is a multiple of 7). This means that
210 - 7 = 203
210 + 7 = 217
Those two numbers are also multiples of 7, so eliminate them from the list. Now, we are left with
{209, 211}.
We've already checked all the prime numbers less than 10, so we know that neither of these numbers is divisible by anything less than 10. We have to check 11 now. We know that 22 is a multiple of 11, so 220 is also a multiple of 11. This means that
220 - 11 = 209
is also a multiple of 11. We can eliminate this from the list also.
That leaves us with just 211. There's no zero option in the question, so this must be a prime number.
Answer = (A) |
AQUA-RAT | AQUA-RAT-39257 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
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The following is multiple choice question (with options) to answer.
Carol spends 1/3 of her savings on a stereo and 1/3 less than she spent on the stereo for a television. What fraction of her savings did she spend on the stereo and television? | [
"1/4",
"2/7",
"5/9",
"1/2"
] | C | Total Savings = S
Amount spent on stereo = (1/3)S
Amount spent on television = (1-1/3)(1/3)S = (2/3)*(1/3)*S = (2/9)S
(Stereo + TV)/Total Savings = S(1/3 + 2/9)/S = 5/9
Answer: C |
AQUA-RAT | AQUA-RAT-39258 | Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$
-
Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring: \begin{align} \left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2 &=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\ &=8+2\sqrt{16-12}\\[6pt] &=12 \end{align} Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
-
Write as $\sqrt{4+2\sqrt{3}} = a+b\sqrt{3}$. Now square both sides, equate real and radical part. This gives two equations in $a$ and $b$. Now eliminate $a$, solve for $b$. Goes perfect. Same for the other term.
-
The following is multiple choice question (with options) to answer.
Find √? /13 = 4 ? | [
"76",
"5776",
"304",
"2704"
] | D | Answer
Let √N/13= 4
Then √N = 13 x 4 = 52
∴ N = 52 x 52 = 2704.
Correct Option: D |
AQUA-RAT | AQUA-RAT-39259 | # Probability - analyzing "randomness" of data
Forgive me, I am a probability novice and am looking for a little guidance. My question is based on real-world data. I have obscured it a bit for confidentiality reasons but the spirit of the question is the same. Okay, here is the setup: Suppose that 100 students from 37 different schools have applied to take part in a math camp. We are told that 188 students will be chosen randomly from the 3700 total applications. Suppose the following number of students are selected from each school:
1. 7
2. 8
3. 1
4. 5
5. 11
6. 3
7. 6
8. 15
9. 3
10. 7
11. 43
12. 1
13. 1
14. 2
15. 1
16. 23
17. 4
18. 3
19. 5
20. 5
21. 6
22. 2
23. 16
24. 1
25. 1
26. 2
(Schools 27-37 have 0 students selected)
Now, I am suspicious about the large number of students chosen from School 11 (43 students), so I wish to analyze this data to determine the likelihood that the applications were randomly selected. Mathematically I believe this equates to determining whether or not the data follows a normal distribution.
My attempt at the solution is the following. Since there are 37 different schools, and each school submitted the same number of applications, I would expect 188/37 ~ 5 students chosen from each school (i.e. this is the mean of my random variable). I would like to determine a range such that - if the students were randomly selected - "there is a 99% probability that the number of students chosen from each school would be between x and y" (so that I can see whether 43 falls into this range). However I am unsure what to use for the standard deviation.
• You could attempt an analytic solution, but why not try a computer simulation first? Dec 15 '15 at 0:28
• Such a huge deviation assures you that the selection wasn't random and there was human factor involved. It IS possible that their random number generator got royally messed up though.
– A.S.
Dec 15 '15 at 2:48
The following is multiple choice question (with options) to answer.
As part of her MBA program, Karen applied for three different Spring Break outreach projects, each of which selects its students by a random lottery of its applicants. If the probability of her being accepted to each individual project is 50%, what is the probability that Karen will be accepted to at least one project? | [
"1/2",
"3/4",
"5/8",
"7/8"
] | D | Since the probability of Karen's being accepted to each individual project is 50%, the probability of her not being accepted to each project is 50%. The probability that she will be accepted to none is, therefore, (1/2)*(1/2)*(1/2)=1/8.
P(Karen will be accepted to at least one project) = 1 - (1/8) = 7/8.
The answer is D. |
AQUA-RAT | AQUA-RAT-39260 | • I am also stuck at this point. If the first rose is $Red$ then the Bride enters the Church and in that case the probability is $\frac{10}{20}$. But now come the cases when the first rose is $White$: $WRR$, $WWRRR$, $WRWRR$ and so on. No matter what, if the first rose is $White$, the last two roses must be $Red$. And the total number of roses required ($\leqslant 20$) to enter the church is $Odd$, where the number of $Red$ roses will never exceed that of the $White$ roses but once when the Bride finally enters the church. Leaves me in doldrums, though. – JackT Oct 18 '17 at 7:42
• @Maths_student Actually, if you have got $x_1$ Red Roses and $y_1$ White roses, then you must get $y_1-x_1+1$ more red roses to enter the church. Or, rather, $x_2-y_2$ should be $y_1-x_1+1$. – MalayTheDynamo Oct 18 '17 at 7:49
• Your edit is incorrect. If she takes a red rose on the first try, she enters, so the probability has to be at least $1/2$ as you said. She can also enter the church if she initially takes a white rose, then takes two red roses in a row. Clearly, the probability is much greater than $1/20$. – N. F. Taussig Oct 18 '17 at 9:34
The following is multiple choice question (with options) to answer.
At the end of the day, February 14th, a florist had 120 roses left in his shop, all of which were red, white or pink in color and either long or short-stemmed. A third of the roses were short-stemmed, 15 of which were white and 15 of which were pink. The percentage of pink roses that were short-stemmed equaled the percentage of red roses that were short-stemmed. If none of the long-stemmed roses were white, what percentage of the long-stemmed roses were red? | [
"20%",
"25%",
"62.5%",
"75%"
] | C | R + W + P = 120
S + L = 120
1/3 * 120 = 40
Short-Stemmed White = 15
Short-Stemmed Pink = 15
=> Short-Stemmed Red = 10
15/P =10/R
=> R = 2R/3
So Total Long Stemmed = 80
And Long Stemmed Red + Long Stemmed Pink = 80
So Long Stemmed Red/Long Stemmed = ?
Total White = 20 (As no Long stemmed white)
=> R + 2R/3 + 20 = 120
=> 5R = 300 and R = 60
Long Stemmed R = 60 - 10 = 50
So Long Stemmed Red/R = 50/80 = 62.5%
Answer - C |
AQUA-RAT | AQUA-RAT-39261 | Thus, it’s possible to make it into the second round with just $3$ wins.
With only $2$ wins, however, it’s impossible to get to the second round. The top three finishers cannot have more than $18$ wins altogether (either $7+6+5$ or $6+6+6$). That leaves $10$ wins amongst the remaining $5$ teams, so either each of the bottom $5$ teams has $2$ wins, or one of them has at least $3$ wins. In neither case is a team with just $2$ wins assured of getting into the second round.
There is some ambiguity in the situation in which each of the bottom $5$ teams wins $2$ matches. (This can happen, e.g., if teams $1,2$ and $3$ all beat each of teams $4,5,6,7$, and $8$, team $4$ beats teams $5$ and $6$, team $5$ beats teams $6$ and $7$, team $6$ beats teams $7$ and $8$, team $7$ beats teams $8$ and $4$, and team $8$ beats teams $4$ and $5$.) In this case the rules as given in the question don’t specify what happens. Some of the possibilities are: that only the top three teams go to the second round; that there is a playoff for the fourth position; that the fourth position is chosen randomly; or that the fourth position is decided by some tie-breaker like goal differential. As long as four teams always move on to the second round, it’s still possible for a team with just $2$ wins in the first round to move on, but if that does happen, other teams with $2$ wins fail to move on.
The following is multiple choice question (with options) to answer.
Prizes totaling $24,000 were awarded unequally between 3 contestants. Which of the following choices could be the highest prize? | [
"(a) $5,000",
"(b) $7,000",
"(c) $15,000",
"(d) $25,000"
] | C | The highest prize cannot possibly be less than 1/3rd of 24,000 (because in this case the sum of 3 prizes would be less than 24,000) and cannot be more than 24,000.
Answer: C. |
AQUA-RAT | AQUA-RAT-39262 | 36&14,9,9,9,9,10\\ 37&16,9,9,9,9,10\\ 38&16,9,9,9,9,12\\ 39&16,9,9,11,9,12\\ 40&16,9,9,11,11,12\\ 41&16,9,11,11,11,12\\ 42&16,11,11,11,11,12\\ 43&18,11,11,11,11,12\\ 44&18,11,11,11,11,14\\ 45&18,11,11,13,11,14\\ 46&18,11,11,13,13,14\\ 47&18,11,13,13,13,14\\ 48&18,13,13,13,13,14\\ 49&18,13,13,13,13,16\\ 50&18,13,13,15,13,16\\ 51&18,13,13,15,15,16\\ 52&18,13,15,15,15,16\\ 53&18,15,15,15,15,16\\ 54&18,15,15,15,15,18\\ 55&18,15,15,15,17,18\\ 56&18,15,15,17,17,18\\ 57&18,15,17,17,17,18\\ 58&18,17,17,17,17,18\\ 59&18,17,17,17,17,20\\ 60&18,17,17,17,19,20\\ 61&18,17,17,19,19,20\\ 62&18,17,19,19,19,20\\ 63&18,17,19,19,19,22\\ 64&18,17,19,19,21,22\\ 65&18,17,19,21,21,22\\ 66&18,17,21,21,21,22\\ 67&18,17,21,21,21,24\\ 68&18,17,21,21,23,24\\
The following is multiple choice question (with options) to answer.
385, 462, 572, 396, 429, 671, 264 | [
"385",
"429",
"671",
"264"
] | B | Solution
In each number except 429,the middle digit is the sum of the other two. Answer B |
AQUA-RAT | AQUA-RAT-39263 | # 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 sum of money becomes double itself in 8 years at simple interest. How many times will it become 24 years at the same rate? | [
"2",
"4",
"6",
"8"
] | C | P ---- 2P ---- 8 years
6P ---- 24 years Answer: C |
AQUA-RAT | AQUA-RAT-39264 | 713, 731, 799, 841, 851, 899
Next come a pair of twins, a 99, then another pair of twins, then another 99! 713 and 731 are twins because they have the same digits, with the 1 and 3 reversed. 731 we have already seen: it is from the same 17-sequence as 527 and 629. 713 is $729 - 16 = 27^2 - 4^2 = (27-4)\cdot (27+4)$. 799 is again from the 17-sequence beginning with 289.
841 and 851 are twins because they have the same digits except the 4 and the 5, which are consecutive. 841 is $29^2$, and 851 is $900 - 49 = 30^2 - 7^2 = (30 - 7) \cdot (30 + 7)$. Finally we have 899 which is $900 - 1 = 30^2 - 1^2 = (30 - 1) \cdot (30 + 1)$.
901, 943, 961, 989
I haven’t thought of a nice story to tell about these—I think of the last three as sort of “sporadic”, but there’s only three of them so it’s not that hard. Someone else could probably come up with a nice mnemonic.
901 in any case is not too hard to remember because it’s a twin with 899, and it’s also the end of the 17-sequence that started with 289.
943 is $23 \cdot 41$. It’s $1024 - 81 = 32^2 - 9^2$, but unlike some of the other differences of squares I’ve highlighted, I doubt this will actually help me remember it.
961 is $31^2$. I think it’s cute that $169$, $196$, and $961$ are all perfect squares.
Last but not least, $989 = 23 \cdot 43$. If you happen to know that $33^2 = 1089$ (I sure don’t!), then this is easy to remember as $33^2 - 10^2 = (33-10) \cdot (33+10)$.
Once again
The following is multiple choice question (with options) to answer.
What is the difference between the place values of two sevens in the numeral 54179759 ? | [
"699990",
"99990",
"99980",
"69300"
] | D | Explanation:
Required Difference
= 70000 - 700 = 69300
Answer is D |
AQUA-RAT | AQUA-RAT-39265 | 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.
Out of a total of 1,000 employees at a certain corporation, 52 percent are female and 40 percent of these females work in research. If 40 percent of the total number of employees work in research, how many male employees do NOT work in research? | [
" 520",
" 480",
" 392",
" 288"
] | D | total number of female employees =52% =520
female employees working in research =(2/5)*520=208
Total no of employees working in research =40% =400
Total male employees =48%=480
male employees working in research =400-208=192
male employees not working in research =480-192=288
Answer D |
AQUA-RAT | AQUA-RAT-39266 | $$12=2^2*3$$, so we should calculate how many 2-s and 3-s are in $$18!$$.
Calculating 2-s: $$\frac{18}{2}+\frac{18}{2^2}+\frac{18}{2^3}+\frac{18}{2^4}=9+4+2+1=16$$. So the power of $$2$$ (the highest power) in prime factorization of $$18!$$ is $$16$$.
Calculating 3-s: $$\frac{18}{3}+\frac{18}{3^2}=6+2=8$$. So the power of $$3$$ (the highest power) in prime factorization of $$18!$$ is $$8$$.
Now as $$12=2^2*3$$ we need twice as many 2-s as 3-s. $$18!=2^{16}*3^8*a=(2^2)^8*3^8*a=(2^2*3)^8*a=12^8*a$$. So $$18!=12^8*a$$ --> $$x=8$$.
2. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors.
Let's take your example $$36$$. $$36=2^2*3^2$$, the number of factors of $$36$$ is $$(2+1)(2+1)=9$$: 1, 2, 3, 4, 6, 9, 12, 18, 36.
1, 3, 9 - THREE ODD factors - "ODD number of Odd-factors";
2, 4, 6, 12, 18, 36 - SIX EVEN factors - "EVEN number of Even-factors".
Perfect square always has even number of powers of prime factors
The following is multiple choice question (with options) to answer.
What is the greatest power that 9 can be raised to so that the resulting number is a factor of 18!? | [
"2",
"3",
"4",
"5"
] | A | the number of 9s in 18!= 18/9=2..
ans A |
AQUA-RAT | AQUA-RAT-39267 | 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 145 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is? | [
"13 km/hr",
"57.2 km/hr",
"17 km/hr",
"18 km/hr"
] | B | Speed of the train relative to man = (145/10) m/sec
= (14.5) m/sec. [(14.5) * (18/5)] km/hr
= 52.2 km/hr. Let the speed of the train be x km/hr. Then, relative speed
= (x - 5) km/hr. x - 5 = 52.2 ==> x
= 57.2 km/hr.
Answer:B |
AQUA-RAT | AQUA-RAT-39268 | # Conceptual reason for why the volume of an ocahedron is four times the volume of a tetrahedron
The image below shows that a regular octahedron can be scaled by a factor of $$2$$ (resulting in a $$2^3$$ factor in volume) and decomposed as six octahedra and eight tetrahedra.
If $$V_o$$ and $$V_t$$ respectively represent the volumes of a regular octahedron and a regular tetrahedron with the same edge lengths, then $$2^3V_o = 6V_o + 8V_t,$$ and solving for $$V_o$$ yields $$V_o = 4V_t$$.
Image from Wikipedia
Is there a conceptual reason why the volume of an octahedron is $$4$$ times the volume of a tetrahedron that doesn't rely on a decomposition like this? For example, is there a way that you can chop up four tetrahedra to fit them into an octahedron?
Equally useful, is there some nice way to see that a square-based pyramid has twice the volume of a tetrahedron? Perhaps integrating as slices of equilateral triangles vs slices of squares?
### A higher dimensional analog.
A "nice to have" quality of the answer would be if it generalizes to the higher dimensional case. If $$V_o^{(n)}$$ and $$V_t^{(n)}$$ denote the (hyper)volumes of the $$n$$-dimensional cross-polytope and $$n$$-dimensional simplex respectively, then
$$V_o^{(n)} = \frac{\sqrt{2^n}}{n!} \text{ and } V_t^{(n)} = \frac{\sqrt{n+1}}{n!\sqrt{2^n}} \text { with ratio } \frac{V_o^{(n)}}{V_t^{(n)}} = \frac{2^n}{\sqrt{n+1}}.$$
Is there a conceptual reason why this relationship is "nice"?
The following is multiple choice question (with options) to answer.
What is the volume of an tetrahedron with edge length of 1? | [
"3.11545466",
"0.76678113",
"0.11785113",
"2.11785113"
] | C | The answer is sqr(2)/12 = 0.11785113
The following can be derived using the Pythagorean theorem:
The height of a face is 3.5/2.
The area of any face is 3.5/4.
The distance from any corner to the center of a joining face is 1/3.5.
The height of the tetrahedron is (2/3).5
The area is 1/3*base*height =
(1/3) * 3.5/4 * (2/3).5 = 21/2/12.
The above cube has sides of length 1/sqr(2). The area formed by the diagonals along four of the faces as shown in the diagram is a tetrahedrom. Left over are four pyramids. The base of each pyramid is half of the side of the cube and the height is the height of the cube. Thus the area of each pyramid is: 1/2*(1/sqr(2))2 * 1/sqr(2) * 1/3 = (12*sqr(2))-1
The total area of the cube is (1/sqr(2))3 = (2*sqr(2))-1
The area of the tetrahedrom is thus (2*sqr(2))-1 - 4*(12*sqr(2))-1 =
3/(6*sqr(2)) - 2/(6*sqr(2)) = 1/(6*sqr(2)) = sqr(2)/12
correct answer C |
AQUA-RAT | AQUA-RAT-39269 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
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Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
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Concentration: General Management, Technology
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Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
A can contains a mixture of liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liter T of liquid A was contained by the can initially? | [
"10",
"20",
"21",
"25"
] | C | As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total
If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4
Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4
Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.
Hence C is the correct answer.
A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.
Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) T= 7*3 = 21.C |
AQUA-RAT | AQUA-RAT-39270 | Hint:
$$\frac{2\sqrt x-3x+x^2}{\sqrt x}=2-3\sqrt x+x^{3/2}$$
$\dfrac{-3x}{\sqrt{x}} = -3\sqrt{x}$, and $\dfrac{x^2}{\sqrt{x}} = x^{\frac{3}{2}}$
The following is multiple choice question (with options) to answer.
If x^2=x+2, then x^3=? | [
"3x+2",
"3x-2",
"2x+1",
"2x-3"
] | A | Given X^2 = X+2
X^3 = X*X^2 = x*(X+2)-- Substituted from above.
= X^2 + 2X = X+2+2X = 3X+2.
Hence A. |
AQUA-RAT | AQUA-RAT-39271 | # 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.
In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit | [
"70%",
"80%",
"90%",
"100%"
] | A | Explanation:
Let C.P.= Rs. 100.
Then, Profit = Rs. 320,
S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 - 125) = Rs. 295
Required percentage = (295/420) * 100
= 70%(approx)
Option A |
AQUA-RAT | AQUA-RAT-39272 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
A tank is filled in TEN hours by three pipes A, B and C. Pipe A is twice as fast as pipe B, and B is twice as fast as C. How much time will pipe B alone take to fill the tank? | [
"56 hours",
"28 hours",
"35 hours",
"66 hours"
] | C | 1/A + 1/B + 1/C = 1/10 (Given)
Also given that A = 2B and B = 2C
=> 1/2B + 1/B + 2/B = 1/10
=> (1 + 2 + 4)/2B = 1/10
=> 2B/7 = 10
=> B = 35 hours.
Answer: C |
AQUA-RAT | AQUA-RAT-39273 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is : | [
"16000",
"18000",
"20000",
"22000"
] | C | Explanation:
Number of bricks =Courtyard area /1 brick area
=(2500×1600 /20×10)=20000
Option C |
AQUA-RAT | AQUA-RAT-39274 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
In what time will a railway train 40 m long moving at the rate of 36 kmph pass a telegraph post on its way? | [
"3",
"5",
"9",
"4"
] | D | T = 40/36 * 18/5 = 4 sec
Answer:D |
AQUA-RAT | AQUA-RAT-39275 | First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.
Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$
Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$
And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$
So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
The following is multiple choice question (with options) to answer.
There are between 70 and 80 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection? | [
" 71",
" 73",
" 76",
" 77"
] | D | Let number of cards = x
If the cards are counted 3 at a time , there are 2 left over-
x= 3p+2
x can take values 71 , 77
If the cards are counted 4 at a time , there is 1 left over
x= 4q+1
x can take values 73, 77
Therefore , x = 77
Answer D |
AQUA-RAT | AQUA-RAT-39276 | 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.4800 is invested at a compound interest for three years, the rate of interest being 10% p.a., 20% p.a. and 25% p.a. for the 1st, 2nd and the 3rd years respectively. Find the interest received at the end of the three years. | [
"3388",
"3120",
"2177",
"1987"
] | B | Let A be the amount received at the end of the three years.
A = 4800[1 + 10/100][1 + 20/100][1 + 25/100]
A = (4800 * 11 * 6 * 5)/(10 * 5 * 4)
A = Rs.7920
So the interest = 7920 - 4800 = Rs.3120
Answer: B |
AQUA-RAT | AQUA-RAT-39277 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
Sum of two numbers is 30. Two times of the first exceeds by 5 from the three times of the other. Then the numbers will be? | [
"A)5",
"B)9",
"C)11",
"D)13"
] | C | Explanation:
x + y = 30
2x – 3y = 5
x = 19 y = 11
C) |
AQUA-RAT | AQUA-RAT-39278 | homework-and-exercises, kinematics
Title: Which approach can I use to find the total distance travelled by the car? I am really struggling with the following problem:
A car starts from rest with constant acceleration and moves in a
straight line. At some point the car passes two marked road sections
of length $L$ in the time $T$ and $\frac{T}{2}$. What is the
acceleration, the velocity $v_L$ at the end of the first section and
the total distance travelled until the end of the first section.
I am not sure if I am setting up the equations properly but here is what I got so far:
I tried to draw a diagram
The following is multiple choice question (with options) to answer.
On her annual road trip to visit her family in Seal Beach, California, Traci stopped to rest after she traveled 1⁄2 of the total distance and again after she traveled 1⁄4 of the distance remaining between her first stop and her destination. She then drove the remaining 200 miles and arrived safely at her destination. What was the total distance, in miles, from Traci’s starting point to Seal Beach? | [
" 250",
" 300",
" 800",
" 400"
] | C | Let D = total distance
Traci traveled 1/2 = D/2
i.e. remaining distance = D/2
She traveled 1/4 th of D/2 = D/8
Thus:
D = (D/2) + (D/8) + 200
D = 800
ANSWER: C |
AQUA-RAT | AQUA-RAT-39279 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
(3108 + 6160) / 28 | [
"380",
"350",
"331",
"310"
] | C | Explanation:
As per BODMAS rule, first we will solve the equation in bracket then we will go for division
= (9268)/28 = 331
Option C |
AQUA-RAT | AQUA-RAT-39280 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
_________________
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Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
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The following is multiple choice question (with options) to answer.
A pipe takes a hours to fill the tank. But because of a leakage it took 3 times of its original time. Find the time taken by the leakage to empty the tank | [
"50 min",
"60 min",
"90 min",
"80 min"
] | C | pipe a can do a work 60 min.
lets leakage time is x;
then
1/60 -1/x=1/180
x=90 min
ANSWER:C |
AQUA-RAT | AQUA-RAT-39281 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If 12 men can reap 120 acres of land in 36 days, how many acres of land can 54 men reap in 54 days? | [
"267",
"766",
"810",
"868"
] | C | 12 men 120 acres 36 days
54 men ? 54 days
120 * 54/12 * 54/36
10 * 54 * 3/2
54 * 15 = 810
Answer:C |
AQUA-RAT | AQUA-RAT-39282 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can complete work in 4 days. A alone starts working and leaves it after working for 3 days completing only half of the work. In how many days it can be completed if the remaining job is undertaken by B? | [
"4",
"8",
"5",
"6"
] | D | Explanation:
(A+B) one day work =1/4
now A does half of the work in 3 days so A can complete the whole work in 6 days
A’s one day work =1/6
B’s one day work=1/4 - 1/6= 1/12
B alone can complete the work in 12 days so half of the work in 6 days
Answer: Option D |
AQUA-RAT | AQUA-RAT-39283 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
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Re: Mixture problem-Can someone explain this [#permalink]
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02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water. What is the ratio of alcohol to water in the resulting solution? | [
"1:2",
"3:7",
"3: 5",
"4:7"
] | A | Container 1 has 12 ounces in the ratio 1:2
or,
x+2x=12 gives x(alcohol)=4 and remaining water =8
container 2 has 9 ounces in the ratio 1:2
or,
x+2x=9 gives x(alcohol)=3 and remaining water =6
mixing both we have alcohol=4+3 and water =8+6
ratio thus alcohol/water =7/14=1/2
Answer A |
AQUA-RAT | AQUA-RAT-39284 | Alice speaks the truth with probability 3/4 and Bob speaks the truth with probability 2/3. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is 3 while Bob says (to Carl) the number is 1. Find the probability that the number is actually 1.
UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ({1,2,⋯,6} - {The number that actually showed up}). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.
My attempt:
Case 1: Number 1 showed up.
Chance of all this happening = $\frac{1}{6}\cdot \frac{2}{3}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)=\frac{1}{180}$
Case 2: Number 3 showed up
Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{120}$
Case 3: Other number showed up
Chance of all this happening = $\frac{4}{6}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{450}$
So, total = $\overline{)\frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}$
Is my attempt correct? If not, how to do this problem?
You can still ask an expert for help
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The following is multiple choice question (with options) to answer.
In a class, 30% of the students speaks truth, 20% speaks lie and 10% speaks both. If a student is selected at random, what is the probability that he has speak truth or lie? | [
"1/4",
"2/3",
"3/5",
"2/5"
] | D | D) 2/5 |
AQUA-RAT | AQUA-RAT-39285 | of numbers in a data set from its mean value and can be represented using the sigma symbol (σ). deviation, standard error, sample variance). Next, add all the squared numbers together, and divide the sum by n minus 1, where n equals how many numbers are in your data set. The formula you'll type into the empty cell is =STDEV.P( ) where "P" stands for "Population". Type in the box above, of your data set up to 5000 data points ( ). It easy to quickly discover the mean, variance and population standard deviation while. Value, subtract the mean it measures the dispersion is the average temperature for the first time the., sum and other important statistical numbers like standard deviation calculator supports both continuous binomial. Their height in centimeters tool that allows researchers to measure the reliability statistical. Histogram, bars group numbers into ranges histogram chart to show the sample that can be read on Inquirer.net Manileno.com! Browser ( not on our server, it remains private ) square the result that same population would have... Have high volatility ( low SD ) is: basically, the higher the standard deviation will more! Select the numbers are dispersed at a height of the data falls within each section by! =Stdev.S ( ) here go through everything line by line you 'll instead in... Mean calculator will find standard deviation present counting measures such as the sample mode data button! The right place, commas or new lines above shows that only 4.6 % of data! A range of 1 standard deviation calculator to help you solve your statistical questions to standards... -1Σ from the mean the wider the dispersion standard deviation histogram calculator the mathematical average of all pixel intensities, so median. Statistical questions or remember √Variance, what is variance hypothesis, it can also use Excel Google. Be the entire population or from a sample calculation for the data falls within a of! 2 standard deviations are generally close to the mean, median, mode, and so on majority! To show the sample mode ; you can retrieve these and use them other! Continuous and binomial data points are too many to interpret, understand or remember with this calculator a is... By ⦠you can copy and paste from Excel data observations are on either
The following is multiple choice question (with options) to answer.
If a certain sample of data has a mean of 21.0 and a standard deviation of 3.0, which of the following values is more than 2.5 standard deviations from the mean? | [
"12.0",
"13.5",
"17.0",
"23.5"
] | B | Value ismore than 2.5SDfrom the mean means that the distance between the mean and the value must be more than 2.5*SD=7.5. So the value must be either less than 21-7.5=13.5or more than 21+7.5=28.5.
Answer: B. |
AQUA-RAT | AQUA-RAT-39286 | Difficulty:
65% (hard)
Question Stats:
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Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in? | [
"3 9/19 hrs",
"3 9/77 hrs",
"3 9/17 hrs",
"3 9/57 hrs"
] | C | Net part filled in 1 hour = 1/5 + 1/6 - 1/12 = 17/60
The tank will be full in 60/17 hrs, i.e., 3 9/17 hrs.Answer: C |
AQUA-RAT | AQUA-RAT-39287 | # Find the remainder of $3^{333}$ divided by $100$
Find the remainder of $$3^{333}$$ divided by $$100$$
So I can find that $$100=2^2\cdot 5^2$$
Then I want to find $$3^{333}$$ mod $$4$$ and mod $$25$$ and use chinese remainder theorem to find a solution mod $$100$$.
I can find that $$3^{333}\equiv (-1)$$ mod $$4$$
But then $$3^{333}=((3^3)^3)^{37}\equiv (27^3)^{37}\equiv (2^3)^{37}\equiv 8^{37}$$ mod $$25$$
But I cannot find $$8^{37}$$ mod $$25$$
• Mar 20 '20 at 20:57
• @WhatsUp I can see I can reduce it to $3^{320}\times 3^{13}\equiv 3^{13}$ mod $100$, but not sure what I do with $3^{13}$ Mar 20 '20 at 21:01
• You used the first link I gave. Now use the second one. Mar 20 '20 at 21:02
• $3^{20}=3 486 784 401\equiv 1 \text{ mod }100$. Just a little hint :)
– user759746
Mar 20 '20 at 21:03
• Hint $\large\ 3^{333} = 3(-1+10)^{166} =\,\ldots$ (Binomial Theorem, only first two terms survive $\bmod 100$) Mar 20 '20 at 21:11
$$3^{333}$$ being divided by 100.
$$=(3^9)^{37}$$
$$=19683^{37}$$
$$=(19700-17)^{37}$$
$$\equiv -17^{37} \quad \bmod 100$$
The following is multiple choice question (with options) to answer.
What is the remainder when 1024*1047*1050*1053 is divided by 33? | [
"3",
"27",
"30",
"21"
] | A | take the remainder from each of 1024/33, 1047/33 and so on..
1024/33 gives remainder = 1
1047/33 gives remainder = 24
1050/33 gives remainder = 27
1053/33 gives remainder = 30
the net remainder is the product of above individual remainders. i.e =1*24*27*30
break them into pairs 1*24/33 gives remainder 24
and 27*30/33 gives remainder 18
so 24*18/33 gives remainder 3.
A |
AQUA-RAT | AQUA-RAT-39288 | # Prove that the set of positive rational values that are less than $\sqrt{2}$ has no maximum value
Its been a while since I've written a proof and would appreciate some feedback on this one.
Question: Given the set of rational positive values, $\{q | q \in \mathbb{Q} \wedge 0 \lt q \lt \sqrt{2}\}$, show that there is no maximum value for $q \lt \sqrt{2}$
Response
Given $x=\sqrt{2}$, suppose that $x$ can be defined as a positive rational number such that $x$ is composed of positive numbers $p, q$ such that $x=\frac{p}{q}$ where $q \neq 0$ and that $p,q$ are simplified to the lowest possible terms.
It follows that $2=\frac{p^{2}}{q{^2}}$ or $p^{2} = 2 \cdot q{^2}$. Therefore, $p^{2}$ must be an even number as it is the product of some $n$ and an even number. As a result, $p$ is an even number because otherwiese, $p^{2}$ would be odd.
If $p$ is an even number, then $p=2n$ for some number $n$.
Substituting $p=2n$ into the original equation:
$$2= \frac{(2n)^{2}}{q^{2}}$$ $$2= \frac{(4n^{2}}{q^{2}}$$ $$2q^{2} = 4n^{2}$$ $$q^{2} = 2n^{2}$$
Therefore, $q^{2}$ is an even number, which makes $q$ even as well. This is a contradiction as $p, q$ are defined to be simplified to the lowest possible terms, which would not be possible if $p, q$ were even. Therefore, $\sqrt{2}$ must be an irrational number.
The following is multiple choice question (with options) to answer.
Q and R are two-digit positive integers that have the same digits but in reverse order. If the positive difference between Q and R is less than 20, what is the greatest possible value of Q minus R? | [
"15",
"16",
"17",
"18"
] | D | A two-digit integer "ab" can be expressed algebraically as 10a+b.
Q-R=(10a+b)-(10b+a)=9(a-b)<20.
The greatest multiple of 9 which is less than 20 is 18.
The answer is D. |
AQUA-RAT | AQUA-RAT-39289 | Remark. The lemma shows that the least common multiple is not just "least" in terms of size. It's also "least" in the sense that it divides every other common multiple.
Theorem. Let m and n be positive integers. Then
Proof. I'll prove that each side is greater than or equal to the other side.
Note that and are integers. Thus,
This shows that is a multiple of m and a multiple of n. Therefore, it's a common multiple of m and n, so it must be greater than or equal to the least common multiple. Hence,
Next, is a multiple of n, so for some s. Then
(Why is an integer? Well, is a common multiple of m and n, so by the previous lemma .)
Similarly, is a multiple of m, so for some t. Then
In other words, is a common divisor of m and n. Therefore, it must be less than the greatest common divisor:
The two inequalities I've proved show that .
Example. Verify that if and .
, , and
Proposition. The element has order in .
Proof.
The first component is 0, since it's divisible by m; the second component is 0, since it's divisible by n. Hence, .
Next, I must show that is the smallest positive multiple of which equals the identity. Suppose , so . Consider the first components. in means that ; likewise, the second components show that . Since k is a common multiple of m and n, it must be greater than or equal to the least common multiple : that is, . This proves that is the order of .
Example. Find the order of in . Find the order of .
The element has order .
On the other hand, the element has order . Since has order 30, the group is cyclic; in fact, .
Remark. More generally, consider , and suppose has order in . (The 's need not be cyclic.) Then has order .
Corollary. is cyclic of order if and only if .
Note: In the next proof, " " may mean either the ordered pair or the greatest common divisor of a and b. You'll have to read carefully and determine the meaning from the context.
Proof. If , then . Thus, the order of is . But has order , so generates the group. Hence, is cyclic.
The following is multiple choice question (with options) to answer.
Which of the following CANNOT be the least common multiple of two positive integers x and y | [
"xy",
"x",
"y",
"xy - y"
] | D | The least common multiple of two positive integers cannot be less than either of them. Therefore, since xy - y is less than x, it cannot be the LCM of a x and y.
Answer: D |
AQUA-RAT | AQUA-RAT-39290 | the subarray [5,2], leaving us with [6,3] with sum 9. The sum of the digits in 21 is 3. As 59 is not wholly divisible by 3 the question is invalid. I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n. Example 2 List all the factors of 32. The difference of the sum of odd and even places is " 5″. The difference between the number and the sum are 99a + 9b, which is divisible by three and nine. 2] And use our formula for the sum of the natural numbers: [6. Next: Write a program in C++ to find LCM of any two numbers using HCF. Sn=n/2{a+an] Sn=165150 &Sn=123200. Because any number which follows the formula 9n + 3 or 9n + 6 violates the statement. The sum of those 8 digits is not divisible by three so the largest possible integer must use no more than 7 of them (since 3, 6 and 9 would be eliminated). number divisible by 9 is : 117. In the given series of numbers how many 7's are there which are immediately followed by 9 and immediately not preceeded by 5 ?. Given an array of random integers, find subarray such that sum of elements in the element is divisible by k. C Program to print the numbers which are not divisible by 2, 3 and 5. The divisibility criteria are a roundabout way to know if a number is divisible by another without directly doing the calculation. N numbers A number K Output Format A number representing the count of subarrays whose sum is divisible by K. 1, −1, n and − n are known as the trivial divisors of n. First, we used the For loop to iterate from 1 to maximum value (Here, number = 5). Numbers that are the sum of two squares. Test whether 8 is divisible by the other numbers 8 is not divisible by 3 completely. And the only thee digit number which fits the bill is 290. Step-by-step explanation: * Lets explain how to solve the problem - The number is divisible by 6 if it divisible by 2 and 3 - Any even number
The following is multiple choice question (with options) to answer.
The integer x is divisible by both 9 and 36. Which of the following must be an integer? | [
"x/16",
"x/36",
"x/25",
"x/46"
] | B | Prime factorization of 9 = 3^2
Prime factorization of 36= 3^2*2^2
LCM of 9 and 36 = 3^2 * 2^2 = 36
Therefore x/36 must be an integer
Answer B |
AQUA-RAT | AQUA-RAT-39291 | what I want to do is express the stuff in the parentheses as a sum of a perfect square and then some number over here. And I have x squared minus 4x. If I wanted this to be a perfect square, it would be a perfect square if I had a positive 4 over here. If I had a positive 4 over there, then this would be a perfect square. It would be x minus 2 squared. And I got the 4, because I said, well, I want whatever half of this number is, so half of negative 4 is negative 2. Let me square it. That'll give me a positive 4 right there. But I can't just add a 4 willy-nilly to one side of an equation. I either have to add it to the other side or I would have to then just subtract it. So here I haven't changed equation. I added 4 and then I subtracted 4. I just added zero to this little expression here, so it didn't change it. But what it does allow me to do is express this part right here as a perfect square. x squared minus 4x plus 4 is x minus 2 squared. It is x minus 2 squared. And then you have this negative 2 out front multiplying everything, and then you have a negative 4 minus negative 4, minus 8, just like that. So you have y is equal to negative 2 times this entire thing, and now we can multiply out the negative 2 again. So we can distribute it. Y is equal to negative 2 times x minus 2 squared. And then negative 2 times negative 8 is plus 16. Now, all I did is algebraically arrange this equation. But what this allows us to do is think about what the maximum or minimum point of this equation is. So let's just explore this a little bit. This quantity right here, x minus 2 squared, if you're squaring anything, this is always going to be a positive quantity. That right there is always positive. But it's being multiplied by a negative number. So if you look at the larger context, if you look at the always positive multiplied by the negative 2, that's going to be always negative. And the more positive that this number becomes when you multiply it by a negative, the more negative this entire expression becomes. So if you think about it, this is going to be a downward-opening
The following is multiple choice question (with options) to answer.
The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to: | [
"0.02",
"0.2",
"0.04",
"0.4"
] | C | Given expression = (11.98)^2 + (0.02)^2 + 11.98 * x.
For the given expression to be a perfect square, we must have
11.98 * x = 2 * 11.98 * 0.02 or x = 0.04
Answer = C |
AQUA-RAT | AQUA-RAT-39292 | • #### 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.
The perimeter of a rectangular yard is completely surrounded by a fence that measures 18 meters. What is the length of the yard if the area of the yard is 20 meters squared? | [
"8",
"2",
"7",
"5"
] | D | Perimeter of rectangular yard = 2(l + b) = 18 --> l + b = 9
Area = l * b = 20
b = 9 - l
l(9 - l) = 20
29l - l^2 = 20
l^2 - 9l + 20 = 0
Upon simplifying we get l = 5 or 4. Only 5 is there in the answer choice.
Answer: D |
AQUA-RAT | AQUA-RAT-39293 | 7. wayki says:
Here you said there is a 2.9 thousandths of an inch curvature for each 100 feet of horizontal distance.....(heheh).
I hate imperial so please allow me to convert it to metric.
0.07366 mm = 30.480m
Multiply all of this up by 1000 =
73mm fall for every 30,000km. Are you mad? One would have nearly gone around the whole circumference by then - for what a 73mm fall in curve? When the diamerter is ?
Those that come up with 8inches a mile are much closer to the truth.
8. wayki says:
Correction: "When the diameter is"........12,742km? The observor has curved through thousands of km not less than 1 centremeter you madman.
9. Alex A says:
I can assure you the author is not mad. You, on the other hand, I am not so sure about. You take 1 number from the post, and then completely miss the point of the post (was that deliberate?) and abuse the number in the most absolutely ridiculous way possible to draw a completely wrong conclusion. Then you delude yourself into thinking that is evidence the author is mad???
Your most amusing part is that you seem to suggest that "8 inches a mile" is about correct. You should try applying your same abuse to this number, and you will again assume the author is mad. Trust me, the author is not the madman.
If you want to understand this, you should read the section titled "Conclusion" and pay particular attention to "square-law relationship" and "What the contractor did was erroneously assume that the deviation varied linearly with distance". A mistake you made as well.
• mathscinotes says:
Thank you. Very nicely put.
mathscinotes
• Johnny Emerson Neeley says:
I think 18 miles is level at a 6' height😜😨, on a perfect sphere, 131,477,280 ft.
10. Steve says:
So, this begs the question: At what point would the curvature of the earth "swamp out" the instrument error?
• mathscinotes says:
The following is multiple choice question (with options) to answer.
On a map, 1 inch represents 28 miles. How many inches would be necessary to represent a distance of 383.6 miles? | [
"5.2",
"7.4",
"13.7",
"21.2"
] | C | 28 miles represented by 1 inch
1 mile represented by (1/28) inch
383.6 miles represented by (1/28) * 383.6
approximately = 14
We should use estimation here as the answer options are not close .(28*10 =280 and 28*20 = 560 , hence we select the only option between 10 and 20 )
Answer C |
AQUA-RAT | AQUA-RAT-39294 | ## 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 -3 < x < 7 and -6 < y < 3, which of the following specifies all the possible values of xy? | [
"-42 < xy < 21",
"-42 < xy < 18",
"-28 < xy < 18",
"-24 < xy < 21"
] | B | The least value of xy is a bit more than 7*(-6)=-42 and the largest value of xy is a bit less than (-3)*(-6)=18.
Therefore -42 < xy < 18.
Answer: B. |
AQUA-RAT | AQUA-RAT-39295 | I have to design a laboratory activity to answer the question, "What is the relationship between the diameter of a circle and the area of the circle?" My teacher said to use a previous activity as a templete to help me. For the
6. ### Math
Two perpendicular diamters(cutting each other at right angles) of a circle cut the circumference at four(4) points. A square is formed by joining the 4 points. If the circumference of the circle is 132cm, find the area of the
7. ### Algebra
The Circumference and area of a circle of radius r are givin by 2 [pie] r and [pie] r[2], respectively use 3.14 for the constant [pie] A. What is the circumference of a circle with a radius of 2 m? B.What is the area of a circle
8. ### Geometry
A circle with a radius of 1/2 ft is dilated by a scale factor of 8. Which statements about the new circle are true? Check all that apply. A.The length of the new radius will be 4 feet. B.The length of the new radius will be 32
9. ### Math
23.What is the approximate circumference of a circle with a radius of 6 centimeters? A.12 B.18 C.24 D.36 D 24.What is the length of the diameter if the radius is 15? A.5 B.15 C.30 D.45 C 27.What is the circumference of the circle
10. ### math
suppose you copy the circle using a size factor of 150%.what will bet he radius,diameter,circumference,and the area of the image? radius will be 1.5 times as big diameter will be 1.5 times as big but the formula for area will be
11. ### area of circle
My circle has a radius of 8ft. What is the distance of the circle? What "distance" are you talking about? Diameter? Circumference? What is the area of a circle with the radius of 32 cm????
More Similar Questions
The following is multiple choice question (with options) to answer.
If the area of circle is 616 sq cm then its circumference? | [
"76",
"55",
"88",
"21"
] | C | the area of circle =pie* r^2 = 616 => r = 14
2 * 22/7 * 14 = 88
ANSWER:C |
AQUA-RAT | AQUA-RAT-39296 | (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.
Company X provides bottled water to its employees in 5 liter bottles, $5 each. At least how many 5L bottles must Company X buy monthly, so that new contract with $40 fixed monthly rate and $2 for 20L bottle each paid off? (Assume that no other costs apply) | [
" 4",
" 9",
" 12",
" 20"
] | B | let the no. of 5 liter bottles be x, so the no. of 20l bottles will be x/4 (to equate the vol.) since the total cost will be equal , 5x= 40+2*x/4 so x= 8.88 or 9. Answer is (B). |
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