source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36297 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
The area of a triangle is with base 4m and height 5m? | [
"20 m2",
"10 m2",
"82 m2",
"54 m2"
] | B | 1/2 * 4 * 5
= 10 m2
Answer:B |
AQUA-RAT | AQUA-RAT-36298 | Following up on the very astute observation by Haffi112 and the comment by @whuber, consider a standard problem of mutually exclusive events $A$ and $B$ that can occur on a trial of a simple experiment. Mutually independent trials of the simple experiment are conducted until one of the two events occurs. What is the probability that event $A$ occurs before event $B$? The standard analysis says that because of independence, we can ignore all previous trials and look only at the trial on which we know that one of $A$ and $B$ occurred, that is, $A \cup B$ occurred. The conditional probability that $A$ occurred on this trial is exactly the probability of "$A ~\text{before}~ B$" that we seek. $$P\{A ~\text{occurs before}~ B\} = P(A \mid (A\cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A) + P(B)}$$
What if the trials of the simple experiment are not independent? The probabilities of $A$ and $B$ occurring on the $i$-th trial given that neither $A$ nor $B$ has occurred on the first, second, $\ldots$, $(i-1)$-th trials do depend on what happened on the previous trials, but denoting these probabilities by $P_i(A)$ and $P_i(B)$ respectively, we have, as before, that the conditional probability that the $i$-th trial concludes the compound experiment via occurrence of $A$ is $P_i(A)/(P_i(A) + P_i(B))$. Note that we are conditioning on what happened on the previous trials.
The following is multiple choice question (with options) to answer.
When a random experiment is conducted, the probability that event A occurs is 1/3. If the random experiment is conducted 4 independent times, what is the probability that event A occurs exactly twice? | [
"8/27",
"10/31",
"12/37",
"14/41"
] | A | One case is: 1/3*1/3*2/3*2/3=4/81
The total number of possible cases is 4C2 = 6
P(event A occurs exactly twice) = 6*(4/81) = 8/27
The answer is A. |
AQUA-RAT | AQUA-RAT-36299 | (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.
After spending 40 percent machinery, 25 percent in building , 15 percent in raw material and 5 percent on furniture . John had a balance of Rs.1305. Total money with him was? | [
"7800",
"8700",
"9000",
"9200"
] | B | [100 - (40+25+5+15)] % of x = 1305
\inline \Rightarrow 15% of x =1305
\inline \Rightarrow \frac{15}{100}\times x=1305
\inline \therefore x=\frac{1305\times 100}{15}=8700
B |
AQUA-RAT | AQUA-RAT-36300 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"10",
"99",
"88",
"77"
] | A | 36 * 5/18 = 10 mps
Answer: A |
AQUA-RAT | AQUA-RAT-36301 | \, =\, \frac{21}{32}e\, +\, \frac{21}{64}f\, +\, \frac{127}{64} \\\\ 43f\, =\, 42e\, +\, 127 \: \: \: ---(2)$$ Combining these, we obtain
The following is multiple choice question (with options) to answer.
The H.C.F. of two numbers is 20 and the other two factors of their L.C.M. are 21 and 23. The larger of the two numbers is | [
"276",
"299",
"312",
"460"
] | D | Solution
Clearly, the numbers are (20 x 21) and (20 x 23).
Larger number = (20 x 23) = 460.
Answer D |
AQUA-RAT | AQUA-RAT-36302 | Let x and y be the dimensions of the rectangle. Then area is A=xy
But $(\frac{x}{2})^{2}+y^{2}=16$
This means $y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}$
Sub into A and we have:
$\frac{x}{2}\sqrt{64-x^{2}}$
$\frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}$
$32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}$
Max area is then $(4\sqrt{2})(2\sqrt{2})=16$
Now, try the same problem with an inscribed trapezoid
The following is multiple choice question (with options) to answer.
A 12 by 16 rectangle is inscribed in circle. What is the circumference of the circle? | [
"5π",
"10π",
"15π",
"20π"
] | D | The diagonal of the rectangle will be the diameter of the circle.
and perimeter= 2*pi*r
Ans: D |
AQUA-RAT | AQUA-RAT-36303 | # There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned?
In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.
I thought to label the four cups $\alpha, \beta, \gamma, \delta$ with events
• $A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$
• $B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$
• $C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$
• $D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$
If she were to drink in order, then I would calculate $P(a) = {1}/{4}$. Next $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \cap B)$, which I'm not completely sure how to calculate.
My doubt is that I shouldn't order the cups because that assumes $\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.
The following is multiple choice question (with options) to answer.
On rainy mornings, Mo drinks exactly N cups of hot chocolate (assume that N is an integer). On mornings that are not rainy, Mo drinks exactly 3 cups of tea. Last week Mo drank a total of 20 cups of tea and hot chocolate together. If during that week Mo drank 10 more tea cups than hot chocolate cups, then how many rainy days were there last week? | [
"4",
"3",
"2",
"5"
] | C | T= the number of cups of tea
C= the number of cups of hot chocolate
T+C = 20 T-C=10 -> T= 15. C=5.
Mo drinks 3 cups of tea a day then number of days that are not rainy = 15/3 = 5
So number of rainy days = 7-5 = 2
C is the answer. |
AQUA-RAT | AQUA-RAT-36304 | $\text{(ii) }\;1 - P(A\cap B\cap C) \;= \;0.88$ . . . . Right!
(b) (i) only the New York flight is full. (ii) exactly one of the three flights is full.
$\text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18$
$\text{(ii)}\;\begin{array}{ccccc}P(A \cap \overline{B} \cap \overline{C}) & = & (0.6)(0.5)(0.6) & = & 0.18 \\
P(\overline{A} \cap B \cap \overline{C}) &=& (0.4)(0.5)(0.6) &=& 0.12 \\
P(\overline{A} \cap \overline{B} \cap C) &=& (0.4)(0.5)(0.4) &=& 0.08\end{array}$
$P(\text{exactly one full}) \;=\;0.18 + 0.12 + 0.08 \;=\;0.38$
The following is multiple choice question (with options) to answer.
A company, Chauncy Co., has an annual travel budget of $55,000. The accounting department estimates that transportation expenses will increase 5 percent in the coming year, and nontransportation travel expenses will increase by 15 percent in the coming year. Last year, Chauncy Co. spent $19,500 on transportation-related expenses, and $35,000 on non transportation travel expenses. If the annual travel budget does not increase this year, and if the accounting department’s estimates are correct, how far over the annual travel budget will expenses be this year? | [
"Expenses will not go over the budget.",
"$500",
"$4,225",
"$5,725"
] | D | Annual travel budget of $55,000
Let, Transportation expenses = T = 19500
and Non-Transportation expenses = N=35000
i.e. Increased Transportation expenses = 1.05T = 20475
and Increased Non-Transportation expenses = 1.15N = 40250
Total Expense = 20475+40250 = 60725
Expense over Budget = Budget - expense = 55000-60725 = 5725
Answer: Option D |
AQUA-RAT | AQUA-RAT-36305 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train moves with a speed of 108 kmph. Its speed in metres per second is | [
"10.8",
"18",
"30",
"38.8"
] | C | Solution
Speed = 108 Kmph
= (108 x 5/18)m/sec
= 30 m/sec.
Answer C |
AQUA-RAT | AQUA-RAT-36306 | area of that circle click on the calculate button... It is also known as the longest straight line segment which passes the! Segment which passes through the center of the circle demonstration is to use this calculator, by... 3 = 150687.075 mm if we know circumference as the straight line segment which passes through material... Example tire with a 30 inch diameter turning at 300 rpm will have road... Outside circumference by 3.1415 measurement with a 30 inch diameter turning at 300 rpm will have a road speed 26.8. Of 26.8 mph x π x d 3 ) 16. d 3 ) 16. d 3 150687.075... Learn C programming, data Structures Tutorials, exercises, examples, programs, hacks, tips and tricks.! Defined as the known factor to calculate the area of a circle given circumference! Background Tutorials: Simply plug in 36 for d in the Wheel Graph calculator demonstration is to PI... Use this calculator, begin by entering the circumference is about 1/32″ ( )! Plug in 36 for d in the formula for calculating the area of a.... Do you find the diameter is given to be changed to calculate diameter circumference and area are - (. Circle 's edge from one point, meeting back at that point,,... A road speed of 26.8 mph we are using functions that find the end is accessible... Also select units of measure for both input data and results to private tutoring material best their. Equal to 5.3 meters you have the diameter consider the maximum torque can... The measurement of its diameter the formula for calculating the area of Wheel... Tire diameter always how to find diameter Hamiltonian cycle in the image, only the tire diameter speed. Use PI and the circumference of a circle given its circumference if we know circumference value into the for! Speed of 26.8 mph where it meets a 30 inch diameter turning at rpm... Yes if you have the diameter that find the radius with the longest chord of the circle using.! Helps you to find diameter, circumference, and area of the circle you... This program, we separated the logic using python functions features make Nerd! Back at that
The following is multiple choice question (with options) to answer.
If a tire rotates at 400 revolutions per minute when the car is traveling 96km/h, what is the circumference of the tire? | [
"2 meters",
"4 meters",
"5 meters",
"3 meters"
] | B | 400 rev / minute = 400 * 60 rev / 60 minutes
= 24,000 rev / hour
24,000 * C = 96,000 m : C is the circumference
C = 4 meters
correct answer B |
AQUA-RAT | AQUA-RAT-36307 | investment Banking Course, Download Valuation... As are bonds less than 365 days, most developed countries use simple interest is bonds! Is 1-Feb-2020, and investors collect returns when the bond “ days maturity... The market price quarterly, or the coupon Equivalent rate ( CER.. For calculating return on investment and an investment that is taxed bankruptcy and liquidates... Second part is used to calculate the annual yield of Second Govt basis that incorporates! 42 every year for calculating return on investment and the yield to calculate bond Equivalent yield for company... Thus 11 % multiplied by two, which comes out to 22 % have to understand this! ( 5 ) /95 } * { 1.520833 } ], bond Equivalent yield formula be calculated using the function! 365/180 } ], bond Equivalent yield formula on how to calculate the bond happens! Is Rs value ) to pay ( coupon rated ) is an calculation. From an annual-pay bond can not be directly compared to find out the bond Equivalent formula... That of bond or call premium of years until maturity or until call until. Calculate current yield calculation, as are bonds can be calculated using following! 5 ) /95 } * { 1.520833 } ], bond Equivalent formula. To yield to calculate EAY as follows: EAY bond equivalent yield formula 1.0253 ( 365/90 ) = %... Different price and tenure discount and do not pay interest at all NCD, the to. /.75 ) = 10.66 % * 100 1 as the current market price the... Until maturity or until put is exercised Contents ) or bonds ( fixed income securities, which sold! Allows the investor to calculate the annualized yield of Second Govt also some bonds, do not annual. Issue, we must know the bond Equivalent yield formula = ( Face value Rs!
The following is multiple choice question (with options) to answer.
In how many years, Rs. 150 will produce the same interest at 6% as Rs. 500 produce in 2 years at 4½% ? | [
"4 years",
"5 years",
"8 years",
"9 years"
] | B | Explanation :
Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.500 at 4½ % for 2 years
150×6×n/100=500×9/2×2/100
150×6×n=500×9/2×2
150×6×n=500×9
n=5 years
Answer : Option B |
AQUA-RAT | AQUA-RAT-36308 | the beginning of the 4 year period". then it would have been difficult to solve. Thanks Say the number of trees at the beginning of the 4 year period was x, then: At the end of the 1st year the number of trees would be $$x+\frac{1}{4}x=\frac{5}{4}*x$$; At the end of the 2nd year the number of trees would be $$(\frac{5}{4})^2*x$$; At the end of the 3rd year the number of trees would be $$(\frac{5}{4})^3*x$$; At the end of the 4th year the number of trees would be $$(\frac{5}{4})^4*x$$; At the end of the $$n_{th}$$ year the number of trees would be $$(\frac{5}{4})^n*x$$; So, we have that $$(\frac{5}{4})^4*x=6,250$$ --> $$\frac{5^4}{4^4}*x=5^4*10$$ --> $$x=4^4*10=2,560$$. Answer: D. If the question were "if all of the trees thrived and there were 6250 trees in the orchard at the end of 15 year period, how many trees were in the orchard at the beginning of the 4 year period", then we would have that: $$(\frac{5}{4})^{15}*x=6,250$$ --> $$x\neq{integer}$$, so it would be a flawed question. Hope it's clear. Isn't the question quite ambiguous, though? I mean the first scentence could be interpreted as "for the first year we have (4/4)x and for the second year (5/4)x and for the third..." etc.. With that reasoning one would have (5/4)^3 * x + x and then your approach doesnt work. Obviously, I understand that this was a flaw in my reasoning but I cannot understand
The following is multiple choice question (with options) to answer.
Along a yard 414 metres long, 24 trees are palnted at equal distances, one tree being at each end of the yard. What is the distance between two consecutive trees | [
"18",
"19",
"10",
"11"
] | A | Explanation:
24 trees have 23 gaps between them,
Required distance (414/23) = 18
Option A |
AQUA-RAT | AQUA-RAT-36309 | We have to decide among $\text{(A)}$ and $\text{(B)}$. Note that the $26$th prime is $101$. This implies that if $p_{n}$ denotes the $n$ th prime, then $$\sum_{n=1}^{168}p_{n} = \sum_{n=1}^{25}p_{n}+\sum_{n=26}^{168} p_{n} > \sum_{n=26}^{168} 101 =101 \times 143=14443 >\text{(A)}=11555$$
The answer is thus $\text{(B)}$, $76127$. The answer can be confirmed through direct calculation or can be verfied here.
• Why did you choose the 26th prime here? Or is it just arbitrary? Jan 30, 2017 at 14:22
• It is somewhat commonly known piece of a trivia - there are 25 primes below 100 and the primes below 25 sum to 100. Jan 30, 2017 at 14:31
• @MarioDS It is as Jon Claus said; many people memorized it, so I was able to answer it quickly. Jan 30, 2017 at 15:00
There are $168$ primes with the first one equal to $2$ the rest $\ge 2k-1$ for $k=2,3,4,...,168$. So their sum is at least $168^2+1=28 225$.
• (a) would mean that the average prime is $<70$, which is horrendously implausible, for me good enough to pick (b) instead. - But this answer is the formal reason why it's implausible Jan 30, 2017 at 20:22
I just wanted to carry forward your observation about "Every prime can be written in of the form $(6n-1),(6n+1)$ except $2$ & $3$".
The following is multiple choice question (with options) to answer.
If 7 and 11 are prime numbers, which of the following CANNOT be the sum of a and b? | [
"18",
"36",
"19",
"54"
] | C | ANS = C |
AQUA-RAT | AQUA-RAT-36310 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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 traveling at 60 kmph takes to cross a bridge of 240 m in length? | [
"18.9 sec",
"88.9 sec",
"22.9 sec",
"21.00 sec"
] | D | D = 110 + 240 = 350 m
S = 60 * 5/18 = 50/3
T = 350 * 3/50 = 21.00 sec
Answer: D |
AQUA-RAT | AQUA-RAT-36311 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
An investor deposited $5,000 to open a new savings account that earned 8 percent annual interest, compounded quarterly. If there were no other transactions in the account, what was the amount of money in the account 6 months after the account was opened? | [
"$5050",
"$5101",
"$5202",
"$5303"
] | C | The amount in the account after 6 months is 1.02*1.02($5,000) = $5202
The answer is C. |
AQUA-RAT | AQUA-RAT-36312 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
_________________
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Abhishek....
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
The number of years of service of the eight employees in a production department are 15, 10, 9, 17, 6, 8, 14 and 16. What is the range in the number of years of service of the eight employees? | [
"10",
"11",
"12",
"13"
] | B | = 17-6
= 11
Answer B |
AQUA-RAT | AQUA-RAT-36313 | acid-base, titration
I got that I need $\pu{0.8 g}$ to make the solution.
Then I did the lab and got this data:
\begin{array}{lrrrr}
\text{Drink}
& \text{Vol. of Drink}
& \text{Init. Vol. NaOH}
& \text{Final Vol. NaOH}
& \text{Vol.-Difference}
\\\hline
\text{Lemonade} & \pu{25 mL} & \pu{6.0 mL} & \pu{12.0 mL} & \pu{6.0 mL}\\
\text{Gatorade} & \pu{25 mL} & \pu{16.5 mL} & \pu{19.5 mL} & \pu{3.0 mL}\\
\text{Ginger Ale} & \pu{25 mL} & \pu{12.0 mL} & \pu{16.5 mL} & \pu{4.5 mL}
\end{array}
With this information I need to answer these questions:
The following is multiple choice question (with options) to answer.
How many litres of pure acid are there in 8 litres of a 20% solution | [
"1.5",
"1.6",
"1.7",
"1.8"
] | B | Explanation:
Question of this type looks a bit typical, but it is too simple, as below...
It will be 8 * 20/100 = 1.6
Answer: Option B |
AQUA-RAT | AQUA-RAT-36314 | # Difference between revisions of "2021 AMC 10B Problems/Problem 3"
## Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$
## Solution 1
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations $$5j=2s$$$$j+s=28,$$ and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is $$\boxed{(C) \text{ } 8}.$$
## Solution 2 (Fast but Not Rigorous)
We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature
## Solution 3
The following is multiple choice question (with options) to answer.
There are 300 seniors at Morse High School, and 50% of them have cars. Of the remaining grades (freshmen, sophomores, and juniors), only 10% of them have cars. If 20% of all the students at Morse have cars, how many students are in the three lower grades? | [
"600",
"900",
"1200",
"1350"
] | B | Let x be the number of students in the lower three grades.
0.1x + 150 = 0.2(x+300)
0.1x = 900
x = 900
The answer is B. |
AQUA-RAT | AQUA-RAT-36315 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A garrison of 2000 men has provisions for 60 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement? | [
"1500",
"1893",
"1979",
"1900"
] | A | 2000 ---- 60
2000 ---- 45
x ----- 20
x*20 = 2000*45
x = 4500
2000
-------
1500
Answer: A |
AQUA-RAT | AQUA-RAT-36316 | • 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.
In her backyard, Florry grows only yellow roses and white roses. Some of the rose bushes have flowers of both colors, and some have flowers of one color only. The number of rose bushes that have only yellow roses is 50% greater than the number of rose bushes that have only white roses. If there are yellow roses on 80% of Florry's rose bushes, then what percent Q of her bushes have only yellow roses? | [
"20",
"30",
"50",
"70"
] | B | let, the number of bushes with only yellow flowers be x
so, number of bushes with only yellow flowers will be 1.5x
if total number of bushes of flowers= T
then total number of bushes with yellow flowers= 0.8T
we can say that total number of bushes with yellow flowers+ bushes with only white flowers= total number of bushes with flowers
so, 0.8T+x=T
x=0.2T
so,Q= 1.5x=0.3T
30% flowers are yellow flowers
OA B |
AQUA-RAT | AQUA-RAT-36317 | And also...there is no specific weight for the satellite...(Thinking)
• Nov 23rd 2008, 12:25 PM
skeeter
I get $4.23 \times 10^7$ meters
what values are you using?
btw ... mass of the satellite doesn't matter
• Nov 23rd 2008, 01:21 PM
realintegerz
I used
G = 6.67 x 10^-11 N m^2/kg^2
M = 5.98 x 10^24 kg
T = 86400s/1 rev
Am I wrong with T?
• Nov 23rd 2008, 01:33 PM
o_O
Those numbers should work.
$r = \sqrt[3]{\frac{(6.67 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) (5.98 \times 10^{24} \ \text{kg})(86400 s)^2}{4\pi^2}} \approx 4.23 \times 10^{7} \ \text{m}$
• Nov 23rd 2008, 07:47 PM
realintegerz
oh i see my problem now, i didnt square the period T
The following is multiple choice question (with options) to answer.
During the months May through October, Teddy the bear septuples his weight, and during the months November through April, Teddy loses 100 pounds. If at the beginning of November 1973, Teddy's weight was m pounds, what was Teddy's weight in pounds by the end of April 1976? | [
"49m - 1300",
"49m - 1200",
"7m - 1300",
"7m - 1200"
] | A | Nov ' 73 = m
April ' 74 = m - 100
Oct ' 74 = 7m - 300
April' 75 = 7m - 400
Oct' 75 = 49m - 1200
April'76 = 49m - 1300
Hence answer will be (A) 49m - 1300 |
AQUA-RAT | AQUA-RAT-36318 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
A library branch originally contained 18360 volumes, 30% of which were fiction novels. 1/3 of the volumes were transferred to another location and 1/2 of the volumes transferred were fiction novels. What percent of the remaining collection was fiction novels? | [
"2.5%",
"20%",
"28.3%",
"45.2%"
] | B | Fiction Novels = 5,508
Transferred to another location = 6,120
Transferred fiction novels = 3,060
Non Transferred Fiction Novels = 2,448
Percent of the remaining collection was fiction novels = 2,448/(18360 - 6120 ) * 100 => 20%
Hence answer will be (B) |
AQUA-RAT | AQUA-RAT-36319 | This can be solved with a combination with repetition: $${5+12-1}\choose{12}$$ = $${16}\choose{12}$$.
Edit:
Another way of thinking this is suppose we have to place 12 balls in 5 containers. The balls are identical and the containers are not. Say those (°°°°°°°°°°°°) are the 12 balls.
1. First we take 2 balls and place it in a container. Now we have 12-2 balls left. We are going to represent this as (°°|°°°°°°°°°°).
2. Again we take 4 balls and we place them in the second container. Now we have °°|°°°°|°°°°°°.
3. Again we take another 3 balls and we place them in the 3rd container: °°|°°°°|°°°|°°°.
4. Again we take another ball and place it in the 4th container: °°|°°°°|°°°|°|°°.
5. Now we have 2 balls left, which we put them in the 5th container.
In how many ways can we place the four | in the 16 spaces we have?
• but my q is why is r 12 or 4, should it not be 5 because there are 5 children? – AlphaDJog Sep 22 '19 at 20:43
• Tim Hortons sells 20 different kinds of donuts and always has at least a dozen of each type available. How many different ways can someone buy a dozen donuts? so in this case would this be (31,11) or (31,12) – AlphaDJog Sep 22 '19 at 20:44
• Check my edit @SkandanVecham – Moria Sep 22 '19 at 21:03
• That helps, as to what evers left foes into the last container thank you. – AlphaDJog Sep 22 '19 at 21:32
The following is multiple choice question (with options) to answer.
Each week a restaurant serving Mexican food uses the same volume of chili paste, which comes in either 35-ounce cans or 25-ounce cans of chili paste. If the restaurant must order 20 more of the smaller cans than the larger cans to fulfill its weekly needs, then how manysmallercans are required to fulfill its weekly needs? | [
"60",
"70",
"80",
"100"
] | B | let x be the number of 35 ounce cans. Therefore (x+20) is the number of 25 ounce cans.
Total volume is same, therefore
35x=25(x+20)
10x=500
x=50
Therefore, number of 15 ounce cans=50+20=70
Ans - B |
AQUA-RAT | AQUA-RAT-36320 | Just to check each case: \begin{align*} f(S_T)=\begin{cases} 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + 0 + 0 = 3 & \text{if }S_T\leq 30, \\ 3 + \min(30 - S_T,0) + \max(S_T - 35,0) = 3 + (30 - S_T) + 0 = 33- S_T & \text{if }30
I know @KeSchn already answered but hope this helps since this is how I usually do these. Of course, you can do this multiple ways but this gets to a correct answer relatively quickly.
Edit: Gordon's answer is definitely the way to go if you're comfortable with indicator functions. It does everything the graphical methods do without requiring any visualizing etc
The following is multiple choice question (with options) to answer.
Evaluate 30!/28! | [
"720",
"600",
"870",
"910"
] | C | we have 30!/28! = 30*29*28!/28! = 30*29 = 870
Answer is C |
AQUA-RAT | AQUA-RAT-36321 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
Last year the price per share of Stock Y increased by s percent and the earnings per share of Stock Y increased by t percent, where s is greater than t. By what percent did the ratio of price per share to earnings per share increase, in terms of s and t? | [
"[100(s – t)] / (100 + s + t) %",
"[100(s – t)] / (100 + s) %",
"100*(s-t)/(100+t) %",
"(s – t) %"
] | C | Let P = the old price per share; E = the old earning per share. Thus P/E is the price to earning ratio before the increases
After the increase the new price is: P*(1+s/100) and the new earning is: E*(1 + t/100)
The new P/E is: (1+s/100)P/(1+t/100)E
The Percent of P/E increase = (new P/E - P/E)/(P/E). Subsititute new P/E to the equation we have:
[(1+s/100)/(1+t/100)*P/E - P/E]/(P/E)*100%. Simplifly the expression and you should get the answer to be:
100*(s-t)/(100+t) % Answer : C |
AQUA-RAT | AQUA-RAT-36322 | human-biology, genetics, hematology
Title: Possible genotypes for blood types? If I am blood type B, what are all the possible genotypes that could be expressed by my parents?
I think it might be 16 but I was reading online and saw this:
Similarly, someone who is blood type B could have a genotype of either
BB or BO.
So if someone can help me answer this, that would be great. Parent 1 and 2 have each 5 possible genotypes (OO, AO, BB, BO and AB).
Here a Punnett square with each possibilities. I highlighted the possible parent genotypes.
The total number of possible crosses is exactly 21. Note that here A = Ia, B = Ib and O = i.
OOxBB,OOxBO,OOxAB
AOxBB,AOxBO,AOxAB
BBxOO,BBxAO,BBxBB,BBxBO,BBxAB
BOxOO,BOxBB,BOxAO,BOxBO,BOxAB
ABxOO,ABxAO,ABxBB,ABxBO,ABxAB
This is starting from the information based on your blood type only (i.e. no information about your genotype).
Some background information. Antigen expressing alleles (here referred as Ia or A and Ib or B) are dominant. Not expressing an allele is notated i or O. Being of blood type O is when you don't express both alleles so it is a recessive trait (only possible genotype is ii or OO genotype).
The following is multiple choice question (with options) to answer.
The blood groups of 200 people is distributed as follows: 50 have type A blood, 65 have B blood type, 70 have O blood type and 15 have type AB blood. If a person from this group is selected at random, what is the probability that this person has O blood type? | [
"0.2",
"0.25",
"0.3",
"0.35"
] | D | Blood group - frequency
A - 50
B - 65
O - 70
AB - 15
Let P(E) be the event of getting O blood group
P(E) = Frequency for O blood group / Total frequencies in the above table
= 70 / (65+50+70+15)
= 70 / 200
= 0.35
My answer is (D) |
AQUA-RAT | AQUA-RAT-36323 | investment Banking Course, Download Valuation... As are bonds less than 365 days, most developed countries use simple interest is bonds! Is 1-Feb-2020, and investors collect returns when the bond “ days maturity... The market price quarterly, or the coupon Equivalent rate ( CER.. For calculating return on investment and an investment that is taxed bankruptcy and liquidates... Second part is used to calculate the annual yield of Second Govt basis that incorporates! 42 every year for calculating return on investment and the yield to calculate bond Equivalent yield for company... Thus 11 % multiplied by two, which comes out to 22 % have to understand this! ( 5 ) /95 } * { 1.520833 } ], bond Equivalent yield formula be calculated using the function! 365/180 } ], bond Equivalent yield formula on how to calculate the bond happens! Is Rs value ) to pay ( coupon rated ) is an calculation. From an annual-pay bond can not be directly compared to find out the bond Equivalent formula... That of bond or call premium of years until maturity or until call until. Calculate current yield calculation, as are bonds can be calculated using following! 5 ) /95 } * { 1.520833 } ], bond Equivalent formula. To yield to calculate EAY as follows: EAY bond equivalent yield formula 1.0253 ( 365/90 ) = %... Different price and tenure discount and do not pay interest at all NCD, the to. /.75 ) = 10.66 % * 100 1 as the current market price the... Until maturity or until put is exercised Contents ) or bonds ( fixed income securities, which sold! Allows the investor to calculate the annualized yield of Second Govt also some bonds, do not annual. Issue, we must know the bond Equivalent yield formula = ( Face value Rs!
The following is multiple choice question (with options) to answer.
How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 3.5 years | [
"7 years",
"4 years",
"5 years",
"3 years"
] | B | Time = (100 * 81) / (450 * 4.5) = 4 years.Answer:B |
AQUA-RAT | AQUA-RAT-36324 | 5C2 * 4!(5C2 for selecting 2 out of 5 places for E & rest can be arranged in 4! ways.)
Ans with this approach 240 which is correct in this case.
Now the question discussed above:
2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?
Going as per the above approach it comes
8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
But in this case answer doesn't match.
Please if you could explain the flaw in my approach will be very helpful.
Cheers.
_________________
Sailing through rough waters. Stars & few kudos will steer me past.
Re: In how many ways can the letters of the word PERMUTATIONS be &nbs [#permalink] 09 May 2018, 01:12
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The following is multiple choice question (with options) to answer.
If the letters of the word PLACE are arranged taken all at a time, find how many do not start with AE. | [
"110",
"112",
"114",
"116"
] | C | Total no' of arrangements 5P5 = 5! = 120
no' of arrangements start with AE = 1 × 6 = 6
no' of arrangements which do not start with AE = 120 - 6 = 114.
C |
AQUA-RAT | AQUA-RAT-36325 | 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 is the present worth of Rs. 198 due in 2 years at 5% simple interest per annum? | [
"128",
"277",
"180",
"888"
] | C | Let the present worth be Rs. x. Then,
S.I. = (198 - x)
(x * 5 * 2) / 100 = (198 - x)
10x = 19800 - 100x
110x = 19800 => x = 180.
Answer: C |
AQUA-RAT | AQUA-RAT-36326 | physical-chemistry
Title: Solution with salts Does a solution of water and NaCl increase its volume of the same amount of the volume of salt added? Thank you
Does the ions Na+ and Cl- taken separately ,occupy more space than the binded molecule? Suppose we take $100$g of an $x$% by weight solution of sodium chloride, so we have $x$g of salt and $100-x$g of water. The volume of $x$g of salt is:
$$ V_S = x/\rho_S $$
where $\rho_S$ is the density of solid salt. Likewise the volume of the water is:
$$ V_W = (100-x)/\rho_W $$
Suppose when we dissolve salt in water the volumes just add i.e.
$$ V_\text{total} = V_S + V_W = x/\rho_S + (100-x)/\rho_W $$
then the density of our $x$% salt solution would be:
$$ \rho_\text{sol}(x) = \frac{100}{x/\rho_S + (100-x)/\rho_W} \tag{1} $$
The density of salt is $2.165$g/cm$^3$, and we'll take the density of water to be $1$g/cm$^3$, so we can use equation (1) to calculate what the density would be if the volumes just added and we can compared this with the experimentally measured density. I did this in Excel and got:
$$\begin{matrix}
x & Equation (1) & Experimental & Constant Volume\\
0 & 1.000 & 1.000 & 1\\
0.5 & 1.003 & 1.002 & 1.005\\
1 & 1.005 & 1.005 & 1.01\\
2 & 1.011 & 1.013 & 1.02\\
3 & 1.016 & 1.020 & 1.03\\
4 & 1.022 & 1.027 & 1.04\\
5 & 1.028 & 1.034 & 1.05\\
The following is multiple choice question (with options) to answer.
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 20 Gallons of water and 40 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x? | [
"300",
"75",
"100",
"150"
] | A | Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.
Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 40 gallons salt is added and 20 gallons of water is added.
So salt now becomes -> (0.2x + 40) and water --> (0.6x+20)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+40) = (0.6x+20)/2 => 0.4x+80 =0.6x+20 =>0.2x =60
Solving, x = 300
Answer : A |
AQUA-RAT | AQUA-RAT-36327 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Two partners (Pella and Leon) invested Rs.7000 and Rs.18000, respectively in a business. Pella runs the business while Leon is a sleeping partner in the business. Pella receives 6% of the profit for managing the business the rest is divided in proportion of their capitals. Out of a total profit of Rs.8500, how much did Pella receive? | [
"3004.84",
"3019.84",
"2747.2",
"3059.84"
] | C | ratio of their investments Rs.7000:18000=7:18
6% of the profit for managing the business =6/100*8500=510
remaining profit=8500-510=7990
share profit in ratio of investment 7990/25 =319.6
Pella's ratio share=7*319.6=2237.2
total money Ben collects=2237.2+510=2747.2
Answer: C |
AQUA-RAT | AQUA-RAT-36328 | time-series, counts
Title: Which algorithm can I use to estimate total number of passengers carried from time series of passenger counts I have time series data coming at 10sec intervals from passenger counter in a bus [10,10,10,10,9,9,9,5,5,5,10,10 ...]. I need to estimate the total number of passengers carried in 1 hour. When the counts decrease, it means someone/somepeople got off. And when it increases it means new people got on. Maybe I'm missing something, but it seems to me that, to know the total number of people that have been in a bus during an hour, you just need to start with the initial value of people for that hour and add all the increments (not the decrements) over that hour.
For instance, if during one hour we had the following counter values:
10, 10, 10, 10, 9, 9, 9, 5, 5, 5, 10
We would first compute the successive differences (starting at the first value):
10, 0, 0, 0, -1, 0, 0, -4, 0, 0, +5
And then we would add only the positive values together: 10 + 5 = 15
Please, clarify if my understanding of the problem is not correct.
The following is multiple choice question (with options) to answer.
Last year, a certain public transportation system sold an average (arithmetic mean) of 41,000 tickets per day on weekdays (Monday through Friday) and an average of 20,000 tickets per day on Saturday and Sunday. Which of the following is closest to the total number of tickets sold last year? | [
"1 million",
"1.25 million",
"10 million",
"12.7 million"
] | D | If you type that into the calculator, it results in 12.8 mil
[D], 12.7 mil, is the correct answer... see 2 posts above for quick approx method |
AQUA-RAT | AQUA-RAT-36329 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
Using all the letters of the word "THURSDAY", how many different words can be formed? | [
"9",
"8!",
"5",
"7"
] | B | Total number of letters = 8
Using these letters the number of 8 letters words formed is 8P8
= 8!.
Answer:B |
AQUA-RAT | AQUA-RAT-36330 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100? | [
"87",
"78",
"25",
"17"
] | C | A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100
= (100 * 30)/40 = 25 * 3
= 75.
In a game of 100 points, B gives (100 - 75)
= 25 points to C.
Answer:C |
AQUA-RAT | AQUA-RAT-36331 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 6 sec. What is the length of the train? | [
"100 m",
"190 m",
"115 m",
"110 m"
] | A | Speed = 60 * 5/18 = 50/3 m/sec
Length of the train = speed * time = 50/3 * 6 = 100 m
Answer: A |
AQUA-RAT | AQUA-RAT-36332 | 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.
Which of the following numbers completes
the sequence 3, 8, 14, 21, 29, 38, 48, 59___? | [
"35",
"36",
"37",
"71"
] | D | The numbers increase at a somewhat steady rate, so you have to figure out how much
you have to add to each number to produce the next in the sequence: 3 + 5 = 8; 8 + 6 = 14;
14 + 7 = 21; 21 + 8 = 29, and so on. The rule for the sequence is to add successively larger
numbers to each number; therefore, the next number is 59 + 12 = 71.
correct answer D)71 |
AQUA-RAT | AQUA-RAT-36333 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
John makes $60 a week from his job. He earns a raise andnow makes $70 a week. What is the % increase? | [
"16%",
"16.66%",
"16.69%",
"16.98%"
] | B | Increase = (10/60)*100 = (1/6)*100 = 16.66%.
B |
AQUA-RAT | AQUA-RAT-36334 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be? | [
"15 m, 12 m",
"15 m, 11 m",
"15 m, 20 m",
"15 m, 10 m"
] | D | lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
D |
AQUA-RAT | AQUA-RAT-36335 | 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 400 m long can cross an electric pole in 20 sec and then find the speed of the train? | [
"88 Kmph",
"89 Kmph",
"72 Kmph",
"16 Kmph"
] | C | Length = Speed * time
Speed = L/T
S = 400/20
S = 20 M/Sec
Speed= 20*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 72 Kmph
Answer: C |
AQUA-RAT | AQUA-RAT-36336 | ### 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.
Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is: | [
"30%",
"100/3%",
"35%",
"44%"
] | D | Some articles were bought at 6 articles for Rs. 5
and sold at 5 articles for Rs. 6.
6 articles are sold for Rs 6*6/5= Rs 36/5
profit on Rs 5 = 36/5 -5 = Rs. 11/5
% profit = 100*(11/5)/5 = 100*11/25 = 44%
ANSWER:D |
AQUA-RAT | AQUA-RAT-36337 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
(17)4.25 x (17)? = 178 | [
"2.29",
"2.75",
"4.25",
"3.75"
] | D | Solution
let (17)4.25 * (17)x = 178. then, (17)4.25 + x = (17)8.
∴ 4.25 + x = 8 ⇔ x = (8 - 4.25) ⇔ x = 3.75
Answer D |
AQUA-RAT | AQUA-RAT-36338 | Going back to $9- x^2= (3- x)(3+ x)= -(x- 3)(x+ 3)\ge 0$, another and perhaps simpler way to see that is this: $a- b> 0$ if and only if $a> b$ and $a- b< 0$ if and only if a< b. If x< -3 then it is also less than 3 so both of x- 3 and x+ 3= x-(3) are negative. Their product (the product of two negative numbers) is positive and so -(x- 3)(x+ 3) is [b]negative[/tex]. If -3< x< 3, then x- 3 is still negative but x+ 3= x-(-3) is now positive. Their product (the product of a positive and negative number) is negative so -(x- 3)(x+ 3) is positive. Finally, if x> 3, it is also greater than -3 so both x- 3 and x+ 3 are positive. The product of two positive numbers is positive so -(x- 3)(x+ 3) is negative. Again, we have that $x^2- 9\ge 0$ if and only if $-3\le x\le 3$.
but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...
The following is multiple choice question (with options) to answer.
A number x is multiplied by 3, and this product is then divided by 7. If the positive square root of the result of these two operations equals x, what is the value of x if x≠0 ? | [
"25/9",
"3/7",
"5/3",
"3/5"
] | B | sqrt(3x/7) to be perfect square x has to 3/7
Ans: B |
AQUA-RAT | AQUA-RAT-36339 | Since these 6 possibilities are exhaustive, we have $$(1+2+3+4+5+6)q/6 = 1 \implies q = {6 \over 21} = {1 \over 3.5}$$ as I originally suspected; in particular, this implies that $$P(X_\tau = M + j) = (6-j)q/6 = (6-j)/21,$$ agreeing with Mike's answer.
The following is multiple choice question (with options) to answer.
If m=m6−−√m=m6 , which of the following could be the value of m? | [
"1/6",
"6",
"36",
"63"
] | D | Volume of a sphere = 4/3*pi.R^3
Surface area of a sphere = 4*pi.R^2
given V = 2SA
4/3*pi.R^3 =2* 4*pi.R^2
=> R =6
Option D |
AQUA-RAT | AQUA-RAT-36340 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
Four children — A, B, C, and D — are going to sit in four chairs in a row. Child C must be somewhere to the left of child D. How many possible configurations are there for the children? | [
"6",
"12",
"18",
"24"
] | B | The number of ways to seat 4 people is 4! = 24.
Child C will sit to the left of Child D in exactly half of these arrangements.
The total number of possible arrangements is 24/2 = 12
The answer is B. |
AQUA-RAT | AQUA-RAT-36341 | ISB, NUS, NTU Moderator
Joined: 11 Aug 2016
Posts: 300
Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6-n)! [#permalink]
### Show Tags
08 Sep 2018, 05:04
vasuca10 wrote:
Option C
As per Statement 1 n*(n+1)=20
Possible values 1*20 , -1*-20, 2*10, -2*-10, 4*5, -4*-5, 5*4, -5*-4, 10*2, -10*-2
Multiple values hence Statement 1 alone is not suffice
Now when we combine both statements all negative values which we deduced from statement 1 can be eliminated as factorial can never be negative. Values satisfying both the statements is n =4
Hence Option C
Kindly give kudos if my explanation helped
For your evaluation of statement 2, only 2 cases will exist.
Since it is already given in question stem that n is an integer, therefore either (4,5) or(-5,-4)
as n*(n+1) means the product of two consecutive integers.
and yes when you combine it with Statement 2, it will yield the answer C.
_________________
~R.
If my post was of any help to you, You can thank me in the form of Kudos!!
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 462
What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6-n)! [#permalink]
### Show Tags
08 Sep 2018, 15:16
GMATinsight wrote:
What is the value of integer n ?
(1) n(n + 1) = 20
(2) 6!/[(n!)(6-n)!] = 15
Source: http://www.GMATinsight.com
The following is multiple choice question (with options) to answer.
If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 210, what is the least possible value of n? | [
"7",
"9",
"11",
"13"
] | A | 210 = 2*3*5*7, so n must be at least 7.
The answer is A. |
AQUA-RAT | AQUA-RAT-36342 | agriculture
Title: What does "permanent field" mean in agriculture? I am reading a book that in a paragraph talks about the agricultural methods used in prehistoric Finland.
The further north and east, the more extensive the amount of
burn-beat cultivation, which was a far from primitive form of
agriculture. The yield was many times higher (twenty- to thirty-fold)
than on permanent fields (five- to ten-fold), and there were multiple
varieties of the technique
A history of Finland by Henrik Meinander.
One of them is burn-beating. Like I understand, in burn-beating people cut down the trees in the forests and burn the topsoil. This way they can use that soil for 3 to 6 years for cultivation.
The other method is permanent field. I have searched the internet and the result I got was "permanent crops", like here. In which case people planted trees once in a field and harvested them multiple times.
But in another research about prehistoric Finland it was saying:
The site of Orijärvi shows that permanent field cultivation, with
hulled barley as the main crop was conducted from approximately cal AD 600 onwards.
The following is multiple choice question (with options) to answer.
If a farmer wants to plough a farm field on time, he must plough 120 hectares a day. For technical reasons he ploughed only 85 hectares a day, hence he had to plough 2 more days than he planned and he still has 40 hectares left. What is the area of the farm field and how many days the farmer planned to work initially? | [
"483",
"720",
"885",
"645"
] | B | Let x be the number of days in the initial plan. Therefore, the whole field is 120⋅x hectares. The farmer had to work for x+2 days, and he ploughed 85(x+2) hectares, leaving 40 hectares unploughed. Then we have the equation:
120x=85(x+2)+40
35x=210
x=6
So the farmer planned to have the work done in 6 days, and the area of the farm field is 120⋅6=720 hectares.
correct answer is B)720 |
AQUA-RAT | AQUA-RAT-36343 | # 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.
In one hour,a boat goes 8 km along the stream and 2 km against the stream.The speed of the boat in still water (in km/hr) is : | [
"3",
"5",
"8",
"9"
] | B | Solution
Speed in still water =1/2(8+2)kmph.= 5 kmph. Answer B |
AQUA-RAT | AQUA-RAT-36344 | ### 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.
Shekhar started a business investing Rs.25,000 in 1999.In 2000,he invested an additional amount of Rs.10,000 and Rajeev joined him with an amount of Rs.35,000.In 2001,Shekhar invested another additional amount of Rs.10,000 and Jatin joined them with an amount of Rs.35,000.What will be Rajeev’s share in the profit of Rs.1,80,000 earned at the end of 3 years from the start of the business in 1999 ? | [
"Rs.45,000",
"Rs.60,000",
"Rs.70,000",
"Rs.75,000"
] | B | Solution
Shekhar : Rajeev : Jatin
= (25000x12+35000x12+45000x12) :(35000x24) : (35000x12)
= 1260000 : 840000 : 420000 = 3 : 2 : 1.
∴ Rajeev’s share = Rs.(18000x2/6) = Rs. 60,000. Answer B |
AQUA-RAT | AQUA-RAT-36345 | Case 1 : A pair of socks are present, hence exactly 2 draws for the socks to match. Case 2 : 2 pair of socks are present in the drawer. How?
To start with, instead of looking for a matching pair, let's find the probability that both socks are red.
The Sock Drawer: Probability and Statistics Problem A drawer contains red socks and black socks. Two drawn at random. When two socks are drawn at random, the probability that both are red is 1/2. That should be what you do first with an easy assignment as this. It would seem you can either make a pair or have a mismatched pair, and that both of those events would have equal chances, making for a 50 percent probability. (Deatsville, AL, USA). In a drawer $r$ red, $b$ blue, and $g$ green socks.
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The following is multiple choice question (with options) to answer.
A sock drawer contains 5 socks, each one a different solid color. The colors of the socks are yellow, green, blue, indigo, and violet. If a person reaches into the drawer and pulls out two socks, how many different color combinations are possible in the selected pair of socks? | [
"10",
"15",
"21",
"36"
] | A | Since the order in which we select the socks does not matter, we can use combinations.
We can select 2 socks from 5 socks in 5C2 ways.
5C2 = 10
Answer : A |
AQUA-RAT | AQUA-RAT-36346 | &= 2 \times {\frac{22}{7}} \times 35 \\\\ &= 220 \text{ feet}\end{align}. Equations of a circle with given centre and radius in different forms. How to find the area of a circle? Note: With this tool, you can know the radius of a circle anywhere on Google Maps by simply clicking on a single point and extending or moving the circle to change the radius on the Map. The fixed point is called the origin or center of the circle. Let's take a fixed point $$P$$ and try to draw lines from it. Here are the three basic circle formulas that are used in calculating the various dimensions of a circle. Circles are one of the most commonly found shapes in the world. The calculator will generate a step by step explanations and circle graph. The radius – the distance from the circle's center or origin to the edge, one half the diameter The circumference – the length of the outside boundaries of the circle Starting from the diameter, you can easily find the other two. And consult the table below our everyday lives enjoyed learning about circle with given and. Then one of those slices makes a quarter circle boundary or the length of a circle =.! Team of math experts is dedicated to making learning fun for our favorite readers the. More about pi, or radius from point \ ( P\ ) in directions... Places, then one of its area at solving a few interesting practice questions at the end of the is! 22 } { 7 } } \end { align } \pi = { \frac { 22 {! A carnival and in general form given the radius of the circle and \ ( P\ in... Floated away boundary of the circle ( pool ) fun for our favorite readers the! How the shape of a circle = π r 2 not display plot -- browser is out of date area. To compute are done live '': how to calculate the and... Compute something or you have and skip unknown values in, find its area = ø. Find area identify the parts of a circle can be calculated if the radius or.... Health, and the arc in it math experts is dedicated to making learning fun for favorite... Button to see the result about circle with radius 1 unit Google map labeled as distance map and directions... Circular chocolate cake from his favorite bakery squared
The following is multiple choice question (with options) to answer.
Find the area of circle whose radius is 7m? | [
"209",
"154",
"278",
"265"
] | B | 22/7 * 7 * 7 = 154
Answer: B |
AQUA-RAT | AQUA-RAT-36347 | This function takes $2^4=16$ possible values.
Now, lets write $p(n,m)$ for the product of the digits from $n$th to $m$th. Note that $f$ are defined for this values since their prime factors are always among $2,3,5$ and $7$. If some of the values $$f(p(1,m))$$ is $(0,0,0,0)$ then $p(1,m)$ is a square. If not, the set $\{f(p(1,1)),f(p(1,2)),\ldots,f(p(1,16))\}$ has at most $15$ different elements. So there are two equal values, say $p(1,n)$ and $p(1,m)$. Then $$f(p(n+1,m))=f\left(\frac{p(1,m)}{p(1,n)}\right)=f(p(1,m))-f(p(1,n))=(0,0,0,0)$$ so $p(n+1,m)$ is a square, q.e.d.
The following is multiple choice question (with options) to answer.
The function f is defined for all positive integers n by the following rule: f(n) is the product of the distinct prime factors of n. If f(n) < 100 and n is not prime, what is the greatest possible value of f(n)? | [
"99",
"95",
"91",
"87"
] | B | A) 99 = 11*3*3 ( 99 is not product of distinct primes. 3 is repeated). so can't be the answer.
B)95 = 5*19
C)91 = 7*13
D) 87 is prime, so cannot be the answer
E) 78 = 13*2*3
largest possible n is B.
Answer is B. |
AQUA-RAT | AQUA-RAT-36348 | # Given a number $n \in \Bbb{N}$. In how many ways can $n$ be written as $\prod_{i=1}^{k}n_i$ such that $n1|n2|\ldots|n_k|n$?
The following is multiple choice question (with options) to answer.
how many positive integers less than 9999 are such that the product of their digits is 210. | [
"24",
"58",
"26",
"34"
] | A | The prime factorization of 210 is 2*3*5*7. So one way to make the right kind of number is to use those four digits, in any of the 4! = 24 orders you can put them in.
Notice though that we can also get 210 as a product by multiplying 5, 6 and 7. So we can make some 3-digit numbers with the right product: 3! = 6 of them to be exact.
But we can also get the right product using the digit 1 along with the digits 5, 6, and 7. Again we can arrange those digits in 4! = 24 orders.
A |
AQUA-RAT | AQUA-RAT-36349 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
The contents of a certain box consist of 24 apples and 30 kiwis. How many kiwis must be added to the box so that exactly 30% of the pieces of fruit in the box will be apples? | [
" 24",
" 26",
" 30",
" 46"
] | B | apples = (apples + kiwi + x)*0.3
24 = (30 + 24 + x)*0.3
x = 26.
Answer: B. |
AQUA-RAT | AQUA-RAT-36350 | Its reversal is 100 × [10 − (A − C)] + 10 × 9 + 1 × [(A − C) − 1]. The sum is thus 101 × [(A − C) − 1] + 20 × 9 + 101 × [10 − (A − C)] = 101 × [(A − C) − 1 + 10 − (A − C)] + 20 × 9 = 101 × [−1 + 10] + 180 = 1089.
However, I don’t particularly care for the succinct explanation, and so I’d prefer to give my audience the following explanation. Let’s write our original three-digit number as $ABC$, which of course stands for $100 \times A + 10 \times B + C$. Then, when I reverse the digits, the new three-digit number will be $CBA$, or $100 \times C + 10 \times B + A$.
Of course, because the first number is bigger than the second number, this means that the first hundreds digit is bigger than the second hundreds digit. This means that the first ones digit has to be less than the second ones digit. In other words, when we subtract, we have to borrow from the tens place. However, the tens digits are the same for both numbers. That means that I have to borrow from the hundreds place also.
I’ll illustrate this for both subtraction problems:
Now I’ll subtract. The hundreds digit will be $A - 1 - C$. The tens digit will be $9 + B - B$, or simply $9$. Finally, the ones digit will be $10 + C - A$. This is a little hard to write on a board, so I’ll add some dotted lines to separate the hundreds digits from the tens digit from the ones digit:
The next step is to reverse the digits and add:
I’ll begin with the ones digit:
$(10 + C - A) + (A - 1 - C) = 10 - 1 = 9$.
No matter what, the ones digit is a 9.
Continuing with the tens digits, I get $9 + 9 = 18$. I’ll write down $8$ and carry the $1$ to the next column.
Finally, adding the hundreds digits (and the extra $1$), I get
The following is multiple choice question (with options) to answer.
The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number? | [
"34",
"28",
"12",
"17"
] | B | Explanation:
Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28
ANSWER IS B |
AQUA-RAT | AQUA-RAT-36351 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular lawn of dimensions 80 m * 60 m has two roads each 10 m wide running in the middle of the lawn, one parallel to the length and the other parallel to the breadth. What is the cost of traveling the two roads at Rs.3 per sq m? | [
"Rs.3948",
"Rs.3929",
"Rs.3927",
"Rs.3900"
] | D | Area = (l + b – d) d
(80 + 60 – 10)10 => 1300 m2
1300 * 3 = Rs.3900
Answer: D |
AQUA-RAT | AQUA-RAT-36352 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
For a group of n people, k of whom are of the same sex, the (n-k)/n expression yields an index for a certain phenomenon in group dynamics for members of that sex. For a group that consists of 20 people, 7 of whom are females, by how much does the index for the females exceed the index for the males in the group? | [
" 0.05",
" 0.0625",
" 0.2",
" 0.3"
] | D | Index for females = (20-7)/20 = 13/20 = 0.65
Index for males = (20-13/20 = 7/20 = 0.35
Index for females exceeds males by 0.65 - 0.35 = 0.3
Answer: D |
AQUA-RAT | AQUA-RAT-36353 | ### Show Tags
23 Jan 2019, 14:19
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
In a set of consecutive integers, the sum of the first two is 37
(1) First integer, $$x$$?
$$x + (x+1)=37$$
$$2x+1=37$$
$$x=18$$
(2) Last integer?
If $$x$$ is the first integer, and there are 20 integers total, then there are 19 more
Add 19 to $$x=18$$ (Consecutive = +1 for each integer in the set.)
$$18+19=37$$
If not sure, use a simpler example: There are 4 consecutive integers in the set.
$$x=18$$. List them. 18, 19, 20, 21
There are 4 integers, but we need only +3 added to $$x=18$$ to find the last integer.
20 integers in this set, but add only 19 to the first integer. $$18+19=37$$
(3) Sum of last three integers?
Last three integers: 37, 36, 35
Units digit = 8
(Or, 37 + 36 + 35 = 108)
_________________
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Posts: 1224
Re: In a series of twenty consecutive integers, the sum of the first two [#permalink]
### Show Tags
23 Jan 2019, 15:36
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
The following is multiple choice question (with options) to answer.
In an increasing sequence of 3 consecutive integers, the sum of the first 2 integers is 561. What is last integer in the sequence? | [
" 281",
" 282",
" 275",
" 270"
] | B | Solution:
In solving this problem we must first remember that when we have 3 consecutive integers we can display them in terms of just 1 variable. Thus, we have the following:
Integer 1: x
Integer 2: x + 1
Integer 3: x + 2
We are given that the sum of the first 2 integers is 560. This means that:
x + x+1 = 561
2x + 1 = 561
2x = 560
x = 280
the last integer can be expressed and simplified as:
x+2 = 280+2
= 282
Answer: B |
AQUA-RAT | AQUA-RAT-36354 | 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 person borrows Rs. 9000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6 p.a for 2 years. Find his gain in the transaction per year. | [
"100 rs",
"150 rs",
"160 rs",
"180 rs"
] | D | Gain in 2 years =
[(9000*6*2)/100]-[(9000*4*2)/100]
1080-720=360
Gain in 1 year =(360/2)=180 rs
ANSWER:D |
AQUA-RAT | AQUA-RAT-36355 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio of 2 : 3 : 4 and their L.C.M. is 276. What is their H.C.F.? | [
"18",
"23",
"28",
"32"
] | B | Let the numbers be 2x, 3x, and 4x.
LCM of 2x, 3x and 4x is 12x.
12x = 276
x = 23
HCF of 2x, 3x and 4x = x = 23
The answer is B. |
AQUA-RAT | AQUA-RAT-36356 | # Probability of getting all faces of a die an equal number of times
I have a question:
A die is rolled 36 times.
What is the probability of getting each number 6 times?
I think the answer is: $6\cdot\left(\frac16\right)^6$
Am I wrong?
• @b00nheT: No, that's completely wrong. That would be the calculation if you were rolling a die 6 times and calculating the probability of getting the same number all 6 times. – user2357112 Nov 1 '16 at 17:28
• saw this question on the hot network list and came here expecting to see a question about the probability that the die was actually thrown... – Michael Nov 1 '16 at 18:47
The answer you give is incorrect. There are 6 events you want to have happen, rolling each number 6 times, so you multiply their probabilities. Additionally, there are many ways this could happen, so we multiply by the number of ways of arranging the die rolls. This gives:
$$\left(\frac{1}{6}\right)^{36} \frac{36!}{(6!)^6}$$
• That was my solution as well (as you can probably see in the deleted answer below). However, I had some doubts on it, in particularly with the numerator. – barak manos Nov 1 '16 at 11:30
The following is multiple choice question (with options) to answer.
A die is thrown three times. Events X and Y are defined as below:
X : 4 on the third throw
Y : 6 on the first and 5 on the second throw
What is the probability of X given that Y has already occurred. | [
"1/6",
"1/7",
"1/8",
"1/11"
] | A | The sample space has 216 outcomes.
Now X =
(1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4)
(3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4)
(5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,5,4) (6,6,4)
Y = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
and X n Y = {(6,5,4)}.
Now P(Y) = 6/216
and P (X n Y) = 1/216
Then P(X|Y) = (1/216)/(6/216) = 1/6.
A |
AQUA-RAT | AQUA-RAT-36357 | If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
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If N is a positive three-digit number that is greater than 2 [#permalink]
### Show Tags
17 Aug 2018, 17:35
xhimi wrote:
If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
(1) The tens digit of N is 5.
(2) The units digit of N is 5.
Official Solution (Credit: Manhattan Prep)
(1) SUFFICIENT: If 5 is a digit of N, then N is a multiple of 5. All multiples of 5 end with a units digit of either 5 or 0. However, the units digit of N cannot be 0, since 0 is not a factor of any number. Therefore, the units digit of N must also be 5. Therefore N has the form _55, with only the hundreds digit left to consider.
Consider the possible cases for the hundreds digit:
It can’t be 1, since N > 200.
It can’t be an even number, because N is odd (it ends in 5) and thus doesn’t have any even factors.
If it were 3, then N would be 355—but that doesn’t work, since 355 is not a multiple of 3. (Check: 3 + 5 + 5 = 13, which is not divisible by 3.)
If it were 5, then N would be 555. This is a possible value.
If it were 7, then N would be 755—but that doesn’t work, since 755 is not a multiple of 7. (Check: 700 + 55 = 755. 700 is divisible by 7 but 55 is not, so the whole thing is not.)
If it were 9, then N would be 955—but that doesn’t work, since 955 is not a multiple of 9. (Check: 9 + 5 + 5 = 19, which is not divisible by 9.)
The only possible value for N is 555, so statement 1 is sufficient.
The following is multiple choice question (with options) to answer.
If the number of 741,61n, where n represents the ones digit, is a multiple of 3, then the value of n could be which of the following? | [
"7",
"6",
"5",
"4"
] | C | Then for any number to be a multiple of 3,The sum of its digits should be the multiple of 3.
i.e 7+4+1+6+1+n= multiple of 3
or, 19 + n = multiple of 3.
If n=5, 19 + n =24,which is a multiple of 3.
Hence answer is C. |
AQUA-RAT | AQUA-RAT-36358 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If a-b= 8 and a^2+b^2=164, find the value of ab | [
"10",
"12",
"50",
"18"
] | C | 2ab = (a^2 + b^2) - (a - b)^2
= 164 - 64 = 100
=> ab = 50
ANSWER: C |
AQUA-RAT | AQUA-RAT-36359 | Hey, thanks for your help guys. For a minute there, I thought that this theoretical person could not safely expect to live to be 82 years old.
9. Jun 16, 2012
### SW VandeCarr
In fact, on a purely probabilistic basis, for any finite time no matter how large, there is a non zero probability that a person would survive that long. So for a sufficiently large population, there would be a theoretic person that would live 100,000 years. This, of course, has no basis in biology.
In terms of the probability of being murdered, the model would not hold for the 100,000 year old person. In terms of the model, probably the best one can do is assume the proportion of causes of death would be constant. The calculation above needs to be corrected for overall survival in terms of death from any cause.
Last edited: Jun 16, 2012
10. Jun 16, 2012
### viraltux
Interesting... but 0.37% is not that small percentage, don't you think? That means, roughly speaking, that a community of around 300 persons can expect that one of them will be murdered.
If you consider that the number of people we know plus acquaintances can easily be around 300 persons that would mean that most 82 year old persons know of someone in their circles who has been murdered. Mmm... that might be an interesting survey.
11. Jun 16, 2012
### SW VandeCarr
As I said in my previous post, this is a misapplication of statistics. You have to consider survival in terms of all cause death. If you just consider the murder rate, then at some point nearly everyone gets murdered.
12. Jun 16, 2012
### moonman239
I know that.
This person will not die until he reaches age 82, if he is not murdered. As mentioned before, this person has a 68% chance of living to be 82.
13. Jun 16, 2012
### D H
Staff Emeritus
The probability of living to 82 per this problem is 99.63%, not 68%. You missed the decimal point on the 0.37%.
14. Jun 16, 2012
### SW VandeCarr
The following is multiple choice question (with options) to answer.
The probability that a man will be alive for 10more yrs is 1/4 & the probability that his wife will alive for 10 more yrs is 1/3. The probability that none of them will be alive for 10 more yrs, is | [
"1/2",
"2/3",
"3/5",
"4/7"
] | A | Sol. Required probability = Pg.) x P(B) = (1— D x(1 —I) =: x 1=1/2
Ans.(A) |
AQUA-RAT | AQUA-RAT-36360 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A certain college party is attended by both male and female students. The ratio of male to female students is 4 to 5. If 12 of the male students were to leave the party, the ratio would change to 1 to 2. How many total students are at the party? | [
"64",
"72",
"80",
"88"
] | B | The ratio is 4:5 = 8:10, so there are 8k males and 10k females.
If 12 males left, the ratio would be 1:2 = 5:10, so there would be 5k males and 10k females.
8k-5k = 12
k = 4
8k + 10k = 32 + 40 = 72
The answer is B. |
AQUA-RAT | AQUA-RAT-36361 | 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.
Two pipes A and B can fill a cistern in 37 1/2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after? | [
"5 min",
"9 min",
"10 min",
"15 min"
] | B | Explanation:
Let B be turned off after x minutes. Then, part filled by (A + B) in x min + part filled by A in (30 - x) min = 1.
x(2/75 + 1/45) + (30- x) 2/75 = 1
11x + 180 - 6x = 225 => x = 9
ANSWER IS B |
AQUA-RAT | AQUA-RAT-36362 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 3 m, the volume of the box (in m3) is: | [
"3780",
"5120",
"6420",
"8960"
] | A | Clearly, l = (48 - 6)m = 42 m,
b = (36 -6)m = 30 m,
h = 8 m.
Volume of the box = (42 x 30 x 3) m3 = 3780 m3. Answer: Option A |
AQUA-RAT | AQUA-RAT-36363 | # Math Help - Integers
1. ## Integers
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
2. ## Re: Integers
Originally Posted by fsiwaju
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
1. Set $j = k + 2m, m\in \mathbb{N}$. Then (j - k) is even.
2. Check all 5 terms, replacing j by k + 2m. For instance:
$j + 2k = k + 2m + 2k = 3k + 2m~\implies~$ ..... j+2k is only even if k is even because then j is even too.
...
$jk + j = (k+2m)k + (k + 2m) = (k + 2m)(k + 1)~\implies~$..... (jk + j) is even if k is even because then the 1st bracket would be even too; (jk + j) is even if k is odd because then the 2nd bracket would be even too. A product is even if one factor is even.
3. ## Re: Integers
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
if j - k is even, then j and k are both even or both odd.
4. ## Re: Integers
if j - k is even either j and k are both odd, or j and k are both even.
The following is multiple choice question (with options) to answer.
A and B are integers. The expression (A+1)(B+1) is even. What can be said about A and B? | [
"They are both even numbers.",
"At least one of them is even",
"At least one of them is odd.",
"They are both odd."
] | C | Odd x Odd = Odd
Odd x Even = Even
Even x Even = Even
To fulfill condition Either (A+1) or (B+1) needs to be Even
So either A or B needs to be odd
Or At Least one of them is odd.
ANSWER:C |
AQUA-RAT | AQUA-RAT-36364 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
Using only the digits from 0 to 6 , how many 5 digit codes can be constructed, if the first digit can not be 4 and no digits is used more than once ? | [
"1800",
"2160",
"7776",
"720"
] | A | 5*6*5*4*3 = 1800
ANSWER:A |
AQUA-RAT | AQUA-RAT-36365 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
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GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$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$$.
The following is multiple choice question (with options) to answer.
How much is 50% of 40 is greater than 4/5 of 25? | [
"0",
"67",
"88",
"12"
] | A | (50/100) * 40 – (4/5) * 25
= 0
Answer: A |
AQUA-RAT | AQUA-RAT-36366 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A person distributed 10% of his income to his 2 children each. He deposited 20% of his income to his wife's account. He donated 10% of remaining amount to an orphan house.Finally he has $500. Find his total income? | [
"$900",
"$1000",
"$1200",
"$1150"
] | B | 2 children got = 2*10% = 20%
wife got = 20%
orphan house = 10%
total = 20+20+10= 50%
remaining = 100-50 = 50%
50% = 500
100% = 500*100/50 = $1000
Answer is B |
AQUA-RAT | AQUA-RAT-36367 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
If a new town has 300 residents, and the population doubles every 10 years, what will be its population after 75 years? | [
"6034 residents",
"9051 residents",
"54306 residents",
"15075 residents"
] | C | 300*2^(75/10)=300*2^7.5=100*181.02=54306
The answer is C. |
AQUA-RAT | AQUA-RAT-36368 | 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.
3 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work? | [
"15/7 days",
"15/4 days",
"17/4 days",
"15/6 days"
] | B | 3M = 6W ---- 20 days
12M + 8W -----?
24W + 8 W = 32W ---?
6W ---- 20 32 -----?
6 * 20 = 32 * x => x = 15/4 days Answer: B |
AQUA-RAT | AQUA-RAT-36369 | Remember that radius is half of diameter. Surface = 2 radius X height S = 2 rh + 2 r2 Pyramid Volume = 1/3 area of the base X height V = bh b is the area of the base Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. Whereas volume is the amount of space available in an object.. Cones Area of a cyclic quadrilateral. Example: The area of a circle with a radius(r) of 3 inches is: Circle Area … For a circle, sphere and cylinder calculator click here. Area of a parallelogram given sides and angle. The surface area of any given object is the area or region occupied by the surface of the object. The radius can be any measurement of length. Area of a square. Formula for Area with Diameter. Values of π. π is equal to approximately 3.14. Calculate A, C and d | Given r Given the radius of a circle calculate the area… Area of a circle. If you know the radius of the circle and the height of the segment, you can find the segment area from the formula below. Area of a trapezoid. In geometry, there are different shapes and sizes such as sphere, cube, cuboid, cone, cylinder, etc. Find the area with this circle area formula: Multiply Pi (3.1416) with the square of the radius (r) 2. The surface area formula for a rectangular box is 2 x (height x width + width x length + height x length), as seen in the figure below:. We can derive a formula for the surface area much as we derived the formula for arc length. https://www.learntocalculate.com/calculate-surface-area-of-a-circle For a right circular cone calculator click here.. Circle Formulas. This calculates the area as square units of the length used in the radius. Surface area and volume are calculated for any three-dimensional geometrical shape. Area of a triangle (Heron's formula) Area of a triangle given base and angles. However, you may see other more precise values for π, such as 3.142 or 3.14159. Since the formula for the area of a circle squares the radius, the area of the larger circle is always 4 (or 2 2) times the smaller circle.Think about it: You
The following is multiple choice question (with options) to answer.
Find the area, diameter = 12m. | [
"113.00 square meter",
"113.04 square meter",
"113.08 square meter",
"113.24 square meter"
] | B | Diameter = 12 meter.
Radius = diameter/2.
= 12/2.
= 6 meter.
Area of a circle = πr2.
Here, pi (π) = 3.14 meter, radius (r) = 6.
Area of a circle = 3.14 × 6 × 6..
= 3.14 × 36.
= 113.04 square meter or 113.04 m2. ANSWER : B |
AQUA-RAT | AQUA-RAT-36370 | # 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 salesman’s terms were changed from a flat commission of 5% on all his sales to a fixed salary of Rs.1200 plus 2.5% commission on all sales exceeding Rs. 4,000. If his remuneration as per new scheme was Rs. 1200 more than that by the previous schema, his sales were worth? | [
"12028",
"12000",
"12019",
"16000"
] | D | [1200 + (X-4000) * (2.5/100)] - X * (5/100)
= 600
X = 16000
Answer: D |
AQUA-RAT | AQUA-RAT-36371 | Related Rate Prob - Ships sailing in different directions
• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?
Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$
$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$
$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$
$2D*\frac{dx}{dt}= 1250t-2400$
$\frac{dx}{dt}=\frac{1250t-2400}{2D}$
$D=\sqrt{5^2+45^2}=5\sqrt{82}$
$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$
a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h
b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h
The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet
The following is multiple choice question (with options) to answer.
The speed of a boat in still water is 24 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 15 minutes is | [
"1.6 km",
"2 km",
"3.6 km",
"6.75 km"
] | D | Explanation:
Speed downstreams =(24 + 3)kmph
= 27 kmph.
Distance travelled = (27 x 15/60)km
= 6.75 km
Option D |
AQUA-RAT | AQUA-RAT-36372 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
Mr Yadav spends 60% of his monthly salary on consumable items and 50% of the remaining on clothes and transport. He saves the remaining amount. If his savings at the end of the year were 48456, how much amount per month would he have spent on clothes and transport? | [
"4038",
"8076",
"9691.2",
"4845.6"
] | A | ∵ Amount, he have spent in 1 month on clothes transport = Amount spent on saving per month
∵ Amount, spent on clothes and transport
= 48456⁄12 = 4038
Answer A |
AQUA-RAT | AQUA-RAT-36373 | # 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 Japanese rock band made ¥3 million on their first album with sales of ¥12 million, and made ¥5 million on their second album with sales of ¥30 million. By approximately what percent did the ratio of profit to sales decrease from the first album to second album? | [
"24%",
"27%",
"30%",
"33%"
] | D | Solution:
This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.
We first set up the ratios of profits to sales. The first ratio will be for the first album, and the second ratio will be for the next album. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.
First album
profit/sales = 3/12 = 1/4
Second album
profit/sales = 5/30 =1/6
We can simplify each ratio by multiplying each by the LCM of the two denominators, which is 12. Keep in mind that we are able to do this only because our answer choices are expressed in percents.
First album
profit/sales = (1/4) x 12 = 3
Next 2 Million
profit/sales = (1/6) x 12 = 2
We can plug 3 and 2 into our percent change formula:
(2 – 3)/3 x 100
=-1/3 x 100 so approximately a 33% decrease.
Answer D. |
AQUA-RAT | AQUA-RAT-36374 | # 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}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream? | [
"15",
"19",
"14",
"18"
] | C | The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.
Answer:C |
AQUA-RAT | AQUA-RAT-36375 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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$$.Physics
The following is multiple choice question (with options) to answer.
A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 800 m long? | [
"266 Sec",
"200 Sec",
"776 Sec",
"166 Sec"
] | B | L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 2000 m
T = D/S
T = 2000/10
T = 200 Sec
Answer: B |
AQUA-RAT | AQUA-RAT-36376 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If y exceeds x by 20%, then x is less than y by? | [
"16%",
"16 1/3 %",
"16 2/3 %",
"16 3/5 %"
] | C | Explanation:
X=100 y=120
120------20
100-------? => 16 2/3%
ANSWER IS C |
AQUA-RAT | AQUA-RAT-36377 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
3 candidates in an election and received 2500, 5000 and 15000 votes respectively. What % of the total votes did the winningcandidate got in that election? | [
"45%",
"50%",
"57%",
"66.6%"
] | D | Total number of votes polled = (2500 + 5000 + 15000) = 22500
So, Required percentage = 11628/20400 * 100 = 66.6%
D |
AQUA-RAT | AQUA-RAT-36378 | Let one woman complete the job in $$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.
Two musicians, Maria and Perry, work at independent constant rates to tune a warehouse full of instruments. If both musicians start at the same time and work at their normal rates, they will complete the job in 45 minutes. However, if Perry were to work at twice Maria’s rate, they would take only 30 minutes. How long would it take Perry, working alone at his normal rate, to tune the warehouse full of instruments? | [
"1 hr 30 min",
"1 hr 45 min",
"2 hr",
"2 hr 20 min"
] | A | Sol:
Lets Perry Rate be P and Rate of Maria be M
(rate)*(time)= Work or rate = work/time
first equation=> P+M = 1/45
converting it to hrs P+M= 1/(45/60) => 1/(3/4) =>4/3
second equation => M+2M =>1/30
converting it to hrs 3M=1/(30/60) =>1/(1/2) =>2
therefore M= 2/3 and P=2/3
Rate of Perry = 2/3
time= work/rate (work = 1 job)
Time= 1.5 hrs
Answer A |
AQUA-RAT | AQUA-RAT-36379 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
Danny drove his old car 100 kilometers, from his home to a friend. To prevent the engine from overheating, he stopped his car for 13 minutes every time he completed 40 kilometers of continuous driving. When the car was moving, Danny drove it at an average speed of 60 kilometers per hour. What was Danny's average speed on that trip? | [
"37.5 kilometers per hour",
"47.6 kilometers per hour",
"50 kilometers per hour",
"75 kilometers per hour"
] | B | Time taken to reach first 40 km is 40/60 = 2/3 hr
Time taken to reach next 40 km is 40/60 = 2/3 hr
Time taken to reach next 20 km is 20/60 = 1/3 hr
Danny stopped twice on his way at 40 km from starting point and at 80km from the starting point...
Each stop was of 13min , so two 13min break = 2*13/60 => 13/30 hr
Total time taken to reach 100 km is (2/3 + 2/3 + 1/3+13/30 ) =21/10 hrs
So, Average speed is 100/(21/10) = 47.6km/hour
Answer will be B |
AQUA-RAT | AQUA-RAT-36380 | This function takes $2^4=16$ possible values.
Now, lets write $p(n,m)$ for the product of the digits from $n$th to $m$th. Note that $f$ are defined for this values since their prime factors are always among $2,3,5$ and $7$. If some of the values $$f(p(1,m))$$ is $(0,0,0,0)$ then $p(1,m)$ is a square. If not, the set $\{f(p(1,1)),f(p(1,2)),\ldots,f(p(1,16))\}$ has at most $15$ different elements. So there are two equal values, say $p(1,n)$ and $p(1,m)$. Then $$f(p(n+1,m))=f\left(\frac{p(1,m)}{p(1,n)}\right)=f(p(1,m))-f(p(1,n))=(0,0,0,0)$$ so $p(n+1,m)$ is a square, q.e.d.
The following is multiple choice question (with options) to answer.
The function g(p) is defined for integers p such that if p is even, g(p) = p/2 and if p is odd, g(p) = p + 5. Given that g(g(g(g(g(p))))) = 19, how many possible values for p would satisfy this equation? | [
"1",
"5",
"7",
"8"
] | D | Let me define terms:
in g(p) = R
p is argument, R is result, g() is function,
in g(g(g(g(g(p))))), g1 is inner most, g5 is outermost for identification.
From definition of function g, we can deduce that:
If Result is even then two possibilities for argument = 1 Even 1 Odd
If Result is odd then one possibility for argument = 1 Even
Since final result = 19 = Odd
Possibilities:
g1:1 Even
g2: 1*(Even,Odd ) =1 Even 1 Odd
g3: 1*(Even,Odd) + 1 Even =2 Even 1 Odd
g4: 2*(Even, Odd) + 1 Even =3 Even 2 Odd
g5: 3*(Even, Odd) + 2 Even =5 Even 3 Odd = Total 8
Ans D it is! |
AQUA-RAT | AQUA-RAT-36381 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Jill invests $10000 in an account that pays an annual rate of 3.96%, compounding semi-annually. Approximately how much Q does she have in her account after two years? | [
"$10079.44",
"Q=$10815.83",
"$12652.61",
"$14232.14"
] | B | PS. I guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8%
Hence B the answer |
AQUA-RAT | AQUA-RAT-36382 | # Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube.
Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube. What I have so far: we express the prime factorizations of $35$ and $7$ as $5\cdot7$ and $7$, respectively. Then $n$ must be of the form $n=5^{\alpha}7^{\beta}$. Thus we see that $$35n=5^{\alpha+1}7^{\beta+1}$$ and $$\frac{n}{7}=5^{\alpha}7^{\beta-1}.$$ From here we proceed to see that if $35n$ is a perfect square, then $\alpha+1$ and $\beta+1$ are even, thus $\alpha$ and $\beta$ themselves must be odd, and so, $\alpha\equiv b\equiv1\bmod2$. Clearly, $n$ must be divisible by $7$ for $n/7$ to be an integral quantity. (This is where I'm stuck..)
I know $\alpha$ and $\beta-1$ are odd and that $n$ must be divisible by 7, however, I'm not sure what to fill in for the question marks below based on these conditions.
$$\alpha\equiv?\bmod3\quad\text{and}\quad\beta\equiv?\bmod3$$
Once I know why and how to fill in those question marks, I know it's a matter of solving a system of congruences via the chinese remainder theorem to finish the problem.
The following is multiple choice question (with options) to answer.
What is the least value of x. So that 30x55 is divisible by 3? | [
"4",
"5",
"1",
"2"
] | D | The sum of the digits of the number is divisible by 3, then the number is divisible by3.
3 + 0+ x + 5 + 5 = 13+ x
Least value of x may be 2
Therefore 13+ 2 = 15 is divisible by 3.
D |
AQUA-RAT | AQUA-RAT-36383 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
For positive x and y if x/y is n% of y, then what percent must x be of y? | [
"(y × n)%",
"(n/y)%",
"(√n)%",
"(n^2)%"
] | A | Solution -
The statement x/y is n% of y means that: x/y = (n / 100) × y
In order to find what percent must x be of y i.e., we need to divide x by y.
x/y %=(n / 100) × y *100 ---->(y*n)%
Therefore x/y is (y*n)%
ANSWER:A |
AQUA-RAT | AQUA-RAT-36384 | javascript, algorithm, programming-challenge, ecmascript-6, palindrome
Title: Determine whether an integer is a palindrome The task
Determine whether an integer is a palindrome. An integer is a
palindrome when it reads the same backward as forward.
Example 1:
Input: 121 Output: true
Example 2:
Input: -121 Output: false Explanation: From left to right, it reads
-121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:
Input: 10 Output: false Explanation: Reads 01 from right to left.
Therefore it is not a palindrome. Follow up:
Could you solve it without converting the integer to a string?
My solution
with converting number to string
const isPalindrome = n => n >= 0 && Number([...`${n}`].reverse().join("")) === n;
console.log(isPalindrome(121));
My solution
without converting number to string
const isPalindrome2 = n => {
if (n < 0) { return false; }
let num = Math.abs(n);
const arr = [];
let i = 1;
while (num > 0) {
const min = num % (10 ** i);
num = num - min;
i++;
arr.push(min);
}
i = i - 2;
let j = 0;
return n === arr.reduce((res, x) => {
const add = (x/ (10 ** j)) * (10 ** i);
res += add;
i--;
j++;
return res;
}, 0);
};
console.log(isPalindrome2(121)); Count digits in positive integer
You can get the number of digits using log10
eg
Math.log10(13526); // is 4.131169383089324
const digits = Math.ceil(Math.log10(13526)); // 5
The following is multiple choice question (with options) to answer.
A “palindromic integer” is an integer that remains the same when its digits are reversed. So, for example, 43334 and 516615 are both examples of palindromic integers. How many 6-digit palindromic integers are both even and greater than 800,000? | [
"60",
"100",
"160",
"240"
] | B | The first digit and last digit are the same so the only possibility is 8.
The second and third digits can be any number from 0 to 9.
The total number of palindromic integers is 1*10*10 = 100
The answer is B. |
AQUA-RAT | AQUA-RAT-36385 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
Director
Joined: 13 Mar 2017
Posts: 703
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
An amount of Rs.1560 was divided among A, B and C, in the ratio 1/2 : 1/3 : 1/4. Find the share of C? | [
"Rs.390",
"Rs.368",
"Rs.380",
"Rs.360"
] | D | Let the shares of A, B and C be a, b and c respectively.
a : b : c = 1/2 : 1/3 : 1/4
Let us express each term with a common denominator which is the last number divisible by the denominators of each term i.e., 12.
a : b : c = 6/12 : 4/12 : 3/12 = 6 : 4 : 3.
Share of C = 3/13 * 1560 = Rs. 360.
Answer: D |
AQUA-RAT | AQUA-RAT-36386 | Fact: there is an equal probability of the last passenger sitting in Seat 1 and Seat 100.
By the time the last passenger boards, the outcome of the lost boarding pass problem is already determined, as there is only one seat for them to choose. We must therefore look at the decisions faced by Passengers 1–99.
For Passenger 1, there is equal probability of choosing any of the 100 seats. By extension, the probability of him choosing his own assigned seat and the probability of him choosing the last passenger’s assigned seat are equal.
The only way Passengers 2–99 sit in Seat 1 or Seat 100 is if their assigned seat is occupied. In this case, there is also an equal probability of sitting in any available seat.
While both Seat 1 and Seat 100 are unoccupied, it is equally probable that either seat will be chosen. After one of these two seats is occupied, the other seat is guaranteed to remain unoccupied until the last passenger boards. The probability that Seat 100 is occupied by a previous passenger is 1/2.
Here's the statement of the question from Blitzstein, Introduction to Probability (2019 2 ed), Chapter 1, Exercise 61, p 42.
1. There are 100 passengers lined up to board an airplane with 100 seats (with each seat assigned to one of the passengers). The first passenger in line crazily decides to sit in a randomly chosen seat (with all seats equally likely). Each subsequent passenger takes their assigned seat if available, and otherwise sits in a random available seat. What is the probability that the last passenger in line gets to sit in their assigned seat? (This is a common interview problem, and a beautiful example of the power of symmetry.)
Why does this post require moderator attention?
Why should this post be closed?
The following is multiple choice question (with options) to answer.
There are some passengers riding on a bus. At the first stop, half the passengers get off and nobody gets on the bus. This pattern continues as the bus goes to the next stops. If only one passenger gets off at stop number 4, how many passengers were on the bus originally? | [
"128",
"64",
"32",
"16"
] | D | Before stop 4, there were 2 passengers on the bus.
Before stop 3, there were 4 passengers on the bus.
Before stop 2, there were 8 passengers on the bus.
Before stop 1, there were 16 passengers on the bus.
The answer is D. |
AQUA-RAT | AQUA-RAT-36387 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
In the Hillside summer camp there are 50 children. 90% of the children are boys and the rest are girls. The camp administrator decided to make the number of girls only 20% of the total number of children in the camp. How many more boys must she bring to make that happen? | [
"35.",
"45.",
"40.",
"30."
] | A | Given there are 50 students , 90% of 50 = 45 boys and remaining 5 girls.
Now here 90% are Boys and 10% are Girls.
Now question is asking about how many boys do we need to add, to make the girls percentage to or 20%..
If we add 35 to existing 45 then the count will be 80 and the girls number will be 20 as it.
Now boys are 80% and girls are 20%. ( out of 100 students = 80 boys + 20 girls ).
IMO option A is correct. |
AQUA-RAT | AQUA-RAT-36388 | vcsharp2003
vcsharp2003
Sure, but I was asking that as a step to the second question.
Then the random experiment would be to buy 1 ticket. The number of ways for losing 1 ticket would then be ## {}^{9990}C_1##.
Right.. isn't that what you were trying to show, or have I misunderstood your issue?
Yes, you're correct. I could have directly found the number of ways of 2 tickets being losing tickets in order to find the probability of not getting a prize. If even 1 ticket is a win out of 2 purchased tickets, the person would still get a prize.
Thankyou for the help. It's solved.
Gold Member
I see. I have to figure that out. I thought only combinations was the right approach in this example.
If you are going to reach into the bag pull three marbles, what the probability of getting 2 red 1 green marble?
Using Combinations:
$$P(2r,1g) = \frac{C(4,2)C(2,1)}{C(6,3)} = \frac{3}{5}$$
Using Permutations:
$$P(2r,1g) = \frac{ 4 \cdot 3 \cdot C(3,2) \cdot 2}{ 6 \cdot 5 \cdot 4} = \frac{3}{5}$$
Combinations are computationally superior...I guess. Both work.
The following is multiple choice question (with options) to answer.
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability taht the ticket drawn has a number which is a multiple of 3 or 5? | [
"1/2",
"2/5",
"8/15",
"9/20"
] | D | Solution
Here S=(1,2,3,4,5,...,19,20).
Let E=event of getting a multiple of 3 or 5
= (3,6,9,12,15,18,5,10,20)
P(E)= n(E) / n(S)
= 9/20.
Answer D |
AQUA-RAT | AQUA-RAT-36389 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs.21000. A's share is? | [
"2999",
"2778",
"7000",
"2889"
] | C | x* 12 : 2x* 6: 3x* 4
1:1:1
1/3 * 21000 = 7000
Answer:C |
AQUA-RAT | AQUA-RAT-36390 | # 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$$
The following is multiple choice question (with options) to answer.
The total cost of 100 paper plates and 200 paper cups is $4.50 At the same rates what is the total cost of 20 of the plates and 40 of the cups? | [
"$.90",
"$1.00",
"$1.20",
"$1.50"
] | A | U dont need to go through all this
What u have with u is
100 P + 200 C = $4.50
Just divide the equation by 5 and you will get what u are looking for
20P + 40 C = $0.90
Therefore OA is A |
AQUA-RAT | AQUA-RAT-36391 | # Find the smallest positive number k
For any positive number $$n$$, let $$a_n = \sqrt{2+\sqrt{2+{...+\sqrt{2+\sqrt 2}}}}$$ ($$2$$ appear $$n$$) and let $$k$$ is positive number such that $$\displaystyle\frac{1}{k}\leq\frac{3-a_{n+1}}{7-a_n}$$ for any positive number $$n$$, then find the smallest positive number $$k$$.
I have $$a_1=\sqrt 2$$ and $$a_{n+1}=\sqrt{2+a_n}, \forall n \in\mathbb N$$
Consider
$$a_1=\sqrt 2 \lt 2$$
$$a_2=\sqrt{2+a_1}\lt\sqrt{2+2}=2$$
$$a_3=\sqrt{2+a_2}\lt\sqrt{2+2}=2$$
Use Mathematical Induction, I conclude $$\sqrt 2\leq a_n\leq 2,\forall n\in\mathbb N$$
Thus, $$3-a_{n+1}\gt1$$ and $$7-a_{n}\gt 5$$
Since $$k\in\mathbb N$$, I have $$\displaystyle k\geq\frac{7-a_n}{3-a_{n+1}}=\frac{7-a_n}{3-\sqrt{2+a_n}}=3+\sqrt{2+a_n}=3+a_{n+1}$$
Hence, $$3+\sqrt 2\leq 3+a_{n+1}\leq 3+2=5$$
Therefore $$k=5$$
Please check my solution, Is it correct?, Thank you
The following is multiple choice question (with options) to answer.
If two numbers are in the ratio 2:3. If 7 is added to both of the numbers then the ratio becomes 3:4 then find the smallest number? | [
"A)12",
"B)14",
"C)20",
"D)24"
] | B | 2:3
2x + 7 : 3x + 7 = 3 : 4
4[2x + 7] = 3[3x + 7]
8x + 28 = 9x + 21
9x - 8x = 28 - 21
x = 7
Then smallest number is = 2
2x = 14
Correct Option B |
AQUA-RAT | AQUA-RAT-36392 | time
Title: Is there a difference between 0.18 and 0.180 seconds? Q. Is there a difference between 0.18 and 0.180 seconds ?
In our databases we are collecting timestamps and there are 2 time instants which are the same for the hour, minute and seconds but the milliseconds field is being returned as:
0.18
0.177 When scientists report a measurement of 0.18, it usually means they are certain the actual value they've measured is between 0.175 and 0.185.
When scientists report a measurement of 0.180, it usually means they are certain the actual value they've measured is between 0.1795 and 0.1805.
When scientists report a measurement of 0.177, it usually means they are certain the actual value they've measured is between 0.1765 and 0.1775.
How you should interpret those numbers in your database is be a question for the person who put those numbers there, but without other available information, that's the interpretation I would default to.
The following is multiple choice question (with options) to answer.
What decimal of an hour is a second ? | [
"0.00027",
"0.00029",
"0.00031",
"0.00025"
] | A | Required decimal = 1/60x60 = 1/3600 = .00027
Answer is A. |
AQUA-RAT | AQUA-RAT-36393 | 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$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
When the price of an article was reduced by 35% its sale increased by 80%. What was the net effect on the sale? | [
"17% increase",
"44% decrease",
"60% increase",
"66% increase"
] | A | if n items are sold for $p each, revenue is $np. If we reduce the price by 35%, the new price is 0.65p. If we increase the number sold by 80%, the new number sold is 1.8n. So the new revenue is (0.65p)(1.8n) = 1.17np, which is 1.17 times the old revenue, so is 17% greater.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36394 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Teas worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs 152 per Kg , the price of the third variety per Kg will be? | [
"Rs. 147.50",
"Rs. 785.50",
"Rs. 174.50",
"Rs. 258.50"
] | C | Explanation :
Since first and second varieties are mixed in equal proportions.
So, their average price =Rs. (126+135)/2.
=>Rs. 130.50.
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1.
We have to find x.
By the rule of alligation, we have:
Cost of 1 kg Cost of 1 kg
of 1st kind of 2nd kind
(Rs. 130.50) (Rs. x)
\ /
Mean Price
(Rs. 152)
/ \
x−152 22.50
=>x−(152/22.50)=1.
=>x−152=22.50.
=>x= 174.50 Rs.
Answer : C |
AQUA-RAT | AQUA-RAT-36395 | # Find area of the triangle ABC, given the coordinates of vertices in plane
$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.
I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.
I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.
I do not understand how this answer is achieved.
• Every heard of the Shoelace formula? Oct 2 '16 at 15:34
• Another way would also be to draw a rectangle around the triangle and subtract each section not included in the triangle, but you would need the graph for that. Oct 2 '16 at 15:35
• Fine comments above. Still another way is to exploit en.wikipedia.org/wiki/Pick%27s_theorem Oct 2 '16 at 15:36
• OP; stop changing it. it's fine. Oct 2 '16 at 15:39
• In order to not to waste your efforts, you may just apply Heron's formula to the computed side lengths, too. Oct 2 '16 at 15:44
Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area. Let's consider the approach suomynonA suggested in the comments. Consider the figure below.
We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$. I will leave the details of the calculations to you.
Follow-through
As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.
The following is multiple choice question (with options) to answer.
One side of a triangle has length 5 and a second side has length 4. Which of the following could be the area of the triangle?
I. 15
II. 10
III. 5 | [
"I only",
"II only",
"III only",
"II and III only"
] | D | The triangle will have a maximum area when it is a right angled triangle
The largest triangle will be formed when 5 and 4 are the base and height of triangle .
Area = 1/2 * 5 * 4 = 10
There is no lower limit on the area of any triangle whose only 2 sides are known.
Then 0 < Area <= 10
The answer is D. |
AQUA-RAT | AQUA-RAT-36396 | What is the probability that a spade is drawn from a pack of 50 cards with 11 spades? $11/50$.
$$\mathsf P(A\mid B\cap C) ~{= \dfrac{\mathsf P(A\cap B\cap C)}{\mathsf P(B\cap C)} \\= \dfrac{\binom{13}3/\binom{52}3}{\binom{13}2/\binom{52}{2}} \\ = \dfrac{11}{50}}$$
Reason: $\{B\cap C\}$ is the event of drawing two spades when drawing two cards from the full deck . That the position of these cards is second and third is irrelevant.
The following is multiple choice question (with options) to answer.
One card is drawn from a pack of 52 cards. What is the probability that the card drawn is a ten or a spade? | [
"3/7",
"4/13",
"4/15",
"7/19"
] | B | Out of 52 cards there will be 13 spades including 1 ten and there are 3 more tens. Total 16
probability= 16/52 = 4/13
Answer is B |
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