source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37197 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
An article is bought for Rs.600 and sold for Rs.1000, find the gain percent? | [
"73 1/3%",
"63 1/3%",
"33 1/3%",
"66.66%"
] | D | Explanation:
600 ---- 400
100 ---- ? => 66.66%
Answer:D |
AQUA-RAT | AQUA-RAT-37198 | the square root of 5. The set of the real numbers is denoted by the symbol \mathbb{R}.There are five subsets within the set of real numbers.Let’s go over each one of them. The square root of 25 is 5, which is a real number. Here is the answer to questions like: Square root of 0.25 or what is the square root of 0.25? Find x so that the sum of the lengths of the line segments AB and BC is as large as possible. . But you can also approximate the value of those square roots by hand, and sometimes you can rewrite the square root in a somewhat simpler form. Why don't libraries smell like bookstores? A square root is either positive or negative, but only positive numbers have real square roots. The 2nd root of 100, or 100 radical 2, or the square root of 100 is written as 100 2 = 100 = ± 10 . However, there is a number i defined as the square root of -1 in the complex numbers, so: sqrt (-25) = sqrt (25) • sqrt (-1) = 5 • i = 5i Get your answers by asking now. Question 673347: Explain why the square root of 25 belongs to 4 subsets of the real numbers , but the square root of 5 only belongs to 1 . The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. The principal square root function () = (usually just referred to as the "square root function") is a function that maps the set of nonnegative real numbers onto itself. For the numbers above, the square root was equal to an integer. The square root of x is rational if and only if x is a rational number that can be represented as a ratio of two perfect squares. Square root of 25 is a rational number because it is a terminating number, unlike Pi. When writing math, people often use sqrt (x) to mean the square root of x. Copyright © 2020 Multiply Media, LLC. the square root is not a whole number than you have to approximate the square root \pm \sqrt{3}= … If N is a positive real number, then n has two real values, with the same absolute value but with opposite signs. keep on with the discriminative to the different issues
The following is multiple choice question (with options) to answer.
The function f is defined by subtracting 25 from the square of a number and the function c is defined as the square root of one-half of a number. If c(f(x))=10, then which of the following is a possible value of x? | [
"-15",
"-5",
"0",
"5"
] | A | f(x) = x^2-25
c(x) = sqrt(x/2) not sqrt(x)/2, because the question clearly says its square root of (half of the number).
c(f(x)) = 10
c(x^2-25) =10
sqrt((x^2-25)/2) = 10
=> (x^2-25)/2 = 100
=>x^2 = 225
=> x = 15 or -15
Answer is A. |
AQUA-RAT | AQUA-RAT-37199 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
Seven years ago the average age of a family of 2 members was 25 years. A child has been born. due to Which the average age of the family is 23 years today. What is the present age of the child? | [
"7 years",
"5 years",
"3 years",
"8 years"
] | B | Explanation:
Average of the family of two numbers
7 years ago = 25
= (25+ 7) * 2 = 64 years
Average age of the family of three members at present = 23 years
= 23 * 3 = 69 YEARS
Age of child = 69 - 64 = 5 years
Answer: Option B |
AQUA-RAT | AQUA-RAT-37200 | Veritas Prep Reviews
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05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
A machine puts c caps on bottles in m minutes. How many hours will it take to put caps on b bottles | [
"60bm/c",
"bm/60c",
"bc/60m",
"60b/cm"
] | B | Explanation:
Substitute sensible numbers and work out the problem. Then change the numbers back to letters. For example if the machine puts 6 caps on bottles in 2 minutes, it will put 6 /2 caps on per minute, or (6 /2) x 60 caps per hour. Putting letters back this is 60c/m.If you divide the required number of caps (b) by the caps per hour you will get time taken in hours. This gives bm/60c
Answer: B |
AQUA-RAT | AQUA-RAT-37201 | # Write 100 as the sum of two positive integers
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.
Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then $100 = 44 + 56 = 4 \times 11 + 8 \times 7$.
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.
• I think you want to reword this 'Scalable algorithm to write N as the sum of two positive integers, for large N'
– smci
May 4, 2015 at 7:13
• This seems like a rather badly designed exercise since, looking at the answers, it's clear that just checking multiples of 7 and 11 is by far the simplest way of solving it. I once saw an exam question that made things much clearer by saying the equivalent of, "Proceed as if 100 is a very large number and you do not know that 100=44+56." May 4, 2015 at 21:07
From Bezout's Lemma, note that since $\gcd(7,11) = 1$, which divides $100$, there exists $x,y \in \mathbb{Z}$ such that $7x+11y=100$.
A candidate solution is $(x,y) = (8,4)$.
The following is multiple choice question (with options) to answer.
How many positive integers less than 100 are there such that they are multiples of 13 or multiples of 12 but not both? | [
"15",
"16",
"17",
"18"
] | A | for 13: 13...91=13*7 = 7
for 12: 12...96=12*8 =8
but there is no integer 13*12. so
N=(7)+(8)=15
A |
AQUA-RAT | AQUA-RAT-37202 | ### Show Tags
09 Jan 2010, 05:34
24
28
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?
A.54
B.432
C.2160
D.2916
E.148,824
i ve got it right.but this problem is very time consuming.can anyone suggest shorter method
We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.
Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).
9C1*6C*8C1*5C1=2160.
Answer: C.
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Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]
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25 Jan 2013, 13:03
12
7
Slightly different way of thinking:
On the 9x6 grid of possibilities, I can imagine a bunch of rectangles (with sides parallel to x and y axes). Each of these rectangles contains 4 triangles that fit the description of the question stem.
therefore:
Answer = ( # of Rectangles I can make on the grid) x 4
To create the rectangle, I need to pick 2 points on the x direction, and 2 points on the y direction. Therefore:
The following is multiple choice question (with options) to answer.
On the xy-coordinate plane, point A lies on the y-axis and point B lies on the x-axis. Points A, B, and C form a right triangle with a 90-degree angle at point C and the triangle has an area of 30. If AC is parallel to the x-axis, and BC is parallel to the y-axis, which of the following could be the coordinates of point C? | [
"(−2, 30)",
"(−4, 12)",
"(6, −9)",
"(9, −8)"
] | A | The area of the triangle formed will be:
1/2 * base * height =
1/2 * |(X-coordinate of point C)| * |(Y-coordinate of point C)| = 30
Thus |product of coordinates of point C| = 60
The answer is A. |
AQUA-RAT | AQUA-RAT-37203 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill cistern. How much time will be taken by A to fill the cistern separately? | [
"1 hr",
"2 hrs",
"6 hrs",
"8 hrs"
] | C | Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x2 - 2x - 24 = 0
(x - 6)(x + 4) = 0 => x = 6.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37204 | # How to distribute $k$ prizes in $p$ student with each student would receive $(k-1)$ prizes at most?
A teacher decided to encourage the kids by distributing prizes to them. Each of the prizes was different from the other. The total number of prize was $k$, and total number of kids was $p$. To encourage the kids, the number of prizes was more than the number of kids. But, the teacher imposed a restriction on herself that each kid would receive $(k – 1)$ prizes at the most. How many ways she could distribute the prizes?
Answer is $p^k-p$
If I understood the problem correctly, it is no-where stated that a student can get no prize at all, and the answer seems to be using this assumption,however I am a bit confused why the answer is not $p^{k-1}$?
If the restriction is that no student can get more than $(k-1)$ prizes then what is wrong with starting with $(k-1)$ distinct prizes and distributing those into $p$ distinct groups (students)?
-
Then what happens with the last prize? That one needs to be distributed too, right? But then it makes a difference if the first k-1 prizes went to a single student or more students... – N. S. Nov 6 '11 at 23:34
Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute $k$ prizes among $p$ students?".
For each prize, there are $p$ possibilities for the recipient of that prize. In total, that means there are $p \cdot p \cdot \dots \cdot p$ (with $k$ copies of $p$ in the product) possible ways to assign the prizes. This is the $p^k$ part.
Now let's impose the restriction that no student receives all the prizes (this is the same as saying no student receives more than $k-1$ prizes). Of the $p^k$ configurations we just enumerated, which ones are illegal under this new restriction? Exactly $p$ of them, since there is one illegal configuration for each student (namely, giving that student all the prizes).
Subtracting the illegal configurations from our previous enumeration, we get $p^k - p$ legal configurations.
The following is multiple choice question (with options) to answer.
what will be the highest no. of students in which 1001 pen and 910 pencil should be distributed so that everyone get equal no. of pen and pencils? | [
"91",
"910",
"1011",
"1911"
] | A | h.c.f of 910 and 1001=91
answer A |
AQUA-RAT | AQUA-RAT-37205 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
A began business with Rs.45000 and was joined afterwards by B with Rs.5400. When did B join if the profits at the end of the year were divided in the ratio of 2:1? | [
"8 months",
"4 months",
"7 months",
"9 months"
] | C | 45*12 : 54*x = 2: 1
x = 5
12 -5 = 7
Answer:C |
AQUA-RAT | AQUA-RAT-37206 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Two cans A & B contain 25% & 30 % of milk. If 6 parts of can A and 4 parts of can B acre mixed in a can C. What is the proportion of milk in can C. | [
"25%",
"26%",
"27%",
"28%"
] | C | suppose we take 10 lits of solution in tank A and B then
25% of in A is 2.5 lits and 30% in B is 3 lits milk
6 parts of A can have 1.5 lits of milk and 4 parts of B have 1.2 lts of milk so
totally 2.7 lits of milk in c i.e 27%.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37207 | This lets them bring a 15th bar and catch the train with less than 5 minutes to spare.
It is not possible to move all 16 bars.
Consider the total distance that each person moves, in each direction, carrying each possible number of bars. For example $$R_2^+$$ is the total distance that Rod moves forward while carrying 2 bars. We can set up a system of equations. Both Rod and Lia must move a net distance of $$1$$ mile forwards: $$R_0^+ - R_0^- + R_1^+ - R_1^- + R_2^+ - R_2^- = 1 \\ L_0^+ - L_0^- + L_1^+ - L_1^- = 1$$ Both of them must spend less than $$370$$ minutes moving (note we have taken the reciprocals of their speeds in minutes per mile): $$20(R_0^+ + R_0^-) + 30(R_1^+ + R_1^-) + 60(R_2^+ + R_2^-) \le 370 \\ 30(L_0^+ + L_0^-) + 40(L_1^+ + L_1^-) \le 370$$ Finally, the total net distance moved by all the bars must be $$16$$ miles forwards: $$R_1^+ - R_1^- + 2(R_2^+ - R_2^-) + L_1^+ - L_1^- = 16$$ And of course each distance must be nonnegative.
Next (thanks to RobPratt) we multiply these equations by $$\frac{1}{2}$$, $$\frac{3}{7}$$, $$\frac{1}{40}$$, $$\frac{1}{70}$$, and $$-1$$ respectively, and sum them, yielding:
The following is multiple choice question (with options) to answer.
A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is | [
"85",
"80",
"75",
"70"
] | C | Explanation:
Clearly, out of every 16 persons, there is one captain. So, number of captains (1200/16) = 75.
Answer: C |
AQUA-RAT | AQUA-RAT-37208 | 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 820 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is? | [
"510",
"540",
"480",
"520"
] | C | Speed = 78 * 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (820 + x)/60 = 65/3
x = 480 m.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37209 | 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.
Two trains are running in opposite directions with the same speed. If the length of each train is 300 m and they cross each other in 12 sec, then the speed of each train is? | [
"11",
"90",
"36",
"99"
] | B | Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (300 + 300)/12 => x = 25
Speed of each train = 25 m/sec.
= 25 * 18/5 =- 90 km/hr.
Answer: B |
AQUA-RAT | AQUA-RAT-37210 | To solve more problems based on Heron’s … Question Bank Solutions 4046. Similarity to Heron's Formula. We can easily draw many more quadrilaterals and we can identify many around us. 2 + base. Important Solutions 2865. Chapters. where is the semiperimeter, or half of the triangle's perimeter. 4. Ex 12.2, 2 Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Program to find the angles of a quadrilateral in C++; Maximum area of rectangle possible with given perimeter in C++; Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts in C++; Find maximum volume of a cuboid from the given perimeter and area in C++; Maximum area rectangle by picking four sides from array in C++ side a: side b: side c: side d: sum of angles θ 1 +θ 2: area S . % Progress . Calculates the area and perimeter of a quadrilateral given four sides and two opposite angles. Sides of triangle are 12 cm, 13cm and 15cm. … or own an. Find its area. Now what I need is to implement a method get_area() which calculates the area of my quadrilateral, but I have no idea how. Watch Area and Perimeter of Quadrilaterals Videos tutorials for CBSE Class 8 Mathematics. Progress % Practice Now. Another construction of the point T A is to start at A and trace around triangle ABC half its perimeter, and similarly for T B and T C. Syllabus. Franchisee/Partner Enquiry (North) … Questionnaire. Change Equation Select to solve for a different unknown Scalene Triangle: No sides have equal length No angles are equal. A splitter of a triangle is a cevian (a segment from a vertex to the opposite side) that divides the perimeter into two equal lengths, this common length being called the semiperimeter of the triangle. It is classif The sides a n of regular inscribed polygons, where R is the radius of the circumscribed circle = a n = 2R sin 180 o /n; Area of a polygon of perimeter P and radius of in-circle … Properties of a quadrilateral. Find the Length
The following is multiple choice question (with options) to answer.
Side of a square is 10 cm. after joining the mid point of all sides makes a another inner square and this process goes to infinite.Find the sum of perimeter of all squares. | [
"333",
"278",
"27",
"8"
] | B | You can calculate the side of the small square in two ways. Take half of the side which is equal to 5 cm. Using Pythagorean theorem, x2=52+52x2=52+52 ⇒x2=50⇒x2=50 ⇒x=50−−√=52–√=102–√⇒x=50=52=102
In another way, we can equate the squares of the sides of the small square to square of the side which is equal to 10. ⇒x2+x2=102⇒x2+x2=102
⇒2x2=100⇒2x2=100⇒x=50−−√=52–√=102–√⇒x=50=52=102
So if you observe carefully, the side of the small square is 12–√12 part of the side of the bigger square.
So the side of the square inside the small square = 102–√×12–√=102=5102×12=102=5 and so on...
So areas of the perimeters = 4(10+102–√+102+...)4(10+102+102+...)
= 40(1+12–√+12+...)40(1+12+12+...)
The terms in the bracket are in GP withe common ratio of 12–√12
So Sum of the perimeters = 40⎛⎝⎜11−12–√⎞⎠⎟40(11−12) = 40(2–√2–√−1)
Answer:B |
AQUA-RAT | AQUA-RAT-37211 | $$(13)^2 = x^2 + y^2 + 2*30$$
$$x^2 + y^2 = 169 – 60 = 109$$
• Thus, the correct answer is Option D.
Method 2
• The sum of two numbers is 13 and their product is 30
• Since the sum is positive and even the product is positive we can conclude that both x and y should be positive.
o Keeping this in mind, let us focus on $$x*y = 30$$
$$x*y = 6*5 = 3*10 = 15*2$$
Out of the three cases above only $$x *y = 3* 10$$ gives us $$x + y = 3 + 10 = 13$$
o Thus, the value of $$x^2 + y^2 = 10^2 + 3^2 = 100 + 9 = 109$$
• And the correct answer is Option D.
Thanks,
Saquib
Quant Expert
e-GMAT
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Re: The sum of two numbers is 13, and their product is 30. What is the sum [#permalink] 26 Jul 2017, 08:41
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The following is multiple choice question (with options) to answer.
If x = 55 + 2q and y = 4q + 41, then for what value of q, x is equal to y ? | [
"-14",
"14",
"-7",
"7"
] | D | Explanation:
x = y <=> 55 + 2q = 4q + 41 <=> 2q = 14 <=> q = 7.
Answer: D |
AQUA-RAT | AQUA-RAT-37212 | ## Solution
Consider $f(2)$. The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$. Since $100\le (x+1)^2-x^2=2x+1$, this first happens at $x\ge \lfloor 99/2\rfloor = 50$. The perfect squares from here go: $2500, 2601, 2704, 2809\dots$. Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$. After the ones and tens make $100$, the hundreds place will go up by $2$, thus reaching our goal. Since $10^2=100$, the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$. The missing number is one less than the number of hundreds $(k=2)$ of $3600$, or $35$.
Now consider f(4). Instead of the difference between two squares needing to be $100$, the difference must now be $10000$. This first happens at $x\ge 5000$. After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$. This will take place at $\left(5000+100\right)^2=5100^2= 26010000$. Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.
The following is multiple choice question (with options) to answer.
Which of the following numbers is a perfect square? | [
"1266",
"1444",
"2022",
"4034"
] | B | perfect Square never ends with 2, 3, 7 and 8 so straight away you can eliminate 8122 and 2022.
38*38 = 1444
ANSWER:B |
AQUA-RAT | AQUA-RAT-37213 | a height. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . Preview. Using Properties of Parallelograms 1.40 B.4.58 C.3.69 D. Cannot be solved Parallelogram, triangles etc; The surface area and the volume of pyramids, prisms, cylinders and cones; About Mathplanet; SAT. By: Learn Zillion. Add to My Library . SURVEY . Substitute the area and the height h by their values in the above equation and solve for BC length of BC = 1000 / [ 15 sqrt (2) / 2] = 94.28 feet (rounded to two decimal places). 2X. If two lines are parallel, then its slopes will be equal. Our mission is to provide a free, world-class education to anyone, anywhere. The rules of a 30-60-90 are as follows: 60° x. x√3. You could do this length right over here as the base. If we have a quadrilateral where one pair and only one pair of sides are parallel then we have what is called a trapezoid. The formula is: Because the parallelogram has an angle of 60 degrees you can create a 30-60-90 triangle to find the height. Students are given a variety of parallelograms where the side lengths are algebraic expressions. Back to Geometry Calc Next to Interactive Parallelogram. So, x = 4 and y = 3. To find the area of a parallelogram, use the formula area = bh, where b is the length of the parallelogram and h is the height. So the area here is also the area here, is also base times height. It doesn't matter which one, as long as one of the angle's arms is the base. 7. It can be shown that opposite sides of parallelogram must be congruent , that opposite angles are also congruent , and that consecutive angles are supplementary . To find the missing coordinate of a parallelogram, we use one of the following methods. All courses. Designed with Geometer's Sketchpad in mind . A parallelogram with base b and height h can be divided into a trapezoid and a right triangle, and rearranged into a rectangle, as shown in the figure to the left. Find the altitude. Area of a parallelogram.
The following is multiple choice question (with options) to answer.
The area of a parallelogram is 72 cm^2 and its altitude is twice the corresponding base. What is the length of the base? | [
"1 cm",
"3 cm",
"6 cm",
"4 cm"
] | C | let base= x cm
height = 2x cm
area = x × 2x=2x^2
area=x × 2x=2x^2
Area is given as 72 cm^2
2x^2=72
x^2=36
x=6 cm
ANSWER:C |
AQUA-RAT | AQUA-RAT-37214 | If you role two dice, then there are 36 equally likely outcomes, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Of those, 11, the last of each of those six lines and all of the last, have at least one 6 so the probability that "one or both rolls is a 6" is 11/36. Your "13/36" is wrong because both of the "1/6" In your sum include "(6/6)". It should be 1/6+ 1/6- 1/36 where the 1/36 being subtracted is to take away one of the (6, 6)s.
The general rule for "Probability of A or B" is "Probability of A"+ "Probability of B" minus "Probability of A and B".
When you roll two dice, the probability the first die is even is 1/2, the probability the second die is 1/2, and the probability both are even is (1/2)(1/2)= 1/4 (the results of the two rolls are independent) so the probability that either one or both are even is 1/2+ 1/2- 1/4= 3/4.
The following is multiple choice question (with options) to answer.
If 2 dice as rolled, what is the probability of the total score being 2? | [
"1/12",
"5/12",
"2/3",
"1/3"
] | A | Using the rules of probability, the likelihood of rolling a 2 would be 1/12, as the total score of rolling a 2 is 1 and 1. Answer: A |
AQUA-RAT | AQUA-RAT-37215 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Rs. 5887 is divided between Shyam and Ram, such that Shyam's share at the end of 9 years is equal to Ram's share at the end of 11 years, compounded annually at the rate of 5%. Find the share of Shyam. | [
"3082",
"2887",
"3087",
"2887"
] | C | Explanation:
Shyam's share * (1+0.05)9 = Ram's share * (1 + 0.05)11
Shyam's share / Ram's share = (1 + 0.05)11 / (1+ 0.05)9 = (1+ 0.05)2 = 441/400
Therefore Shyam's share = (441/841) * 5887 = 3087
Answer: C) 3087 |
AQUA-RAT | AQUA-RAT-37216 | = \sqrt{2}.$$ Obviously $$q \neq 0$$. We can assume for convenience that $$p$$ and $$q$$ are both positive. We can actually iterate $$q$$ through several positive integers, choosing a $$p$$ that will get the division arbitrarily close to $$\sqrt{2}$$, but we can't possibly iterate $$q$$ through enough possibilities to be absolutely certain. The negation of $$A$$ is therefore much easier: we assume that the $$p$$ and $$q$$ we're looking for do exist. If the pair does exist, we will be able to deduce a lot of facts about these numbers. But if the pair does not exist, we should arrive at a contradiction. • I don't think your post addresses the question at all. There are existence theorems that cannot be proven without showing that its negation leads to contradiction. Feb 1, 2021 at 16:01 easy to follow but not exacting or rigorous answer... Yes as long as the contrapositive is constructed in such a way that the answer space is complete and you disprove the contrapositive in the general case. so picture a venn diagram that where there is clear lint between true and false and all the space is filled with one or the other with no overlaps. Then find a way to contradict false in such generality that you can cross out all of false. Well all you have left is true so accept it. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true". I think it's worth considering several cases. 1. If there is a contradiction between two unproven statements, that just means that they can't be true at the same time. You can't use the contradiction to directly prove or disprove either one. 2. If one of them appears particularly plausible, then that would be a good fit for your “big problem” description. But then what you have to do is conjecture that the plausible thing holds true, and prefix every statement you derive from this with “if the … conjecture holds, then …” or “assuming the …
The following is multiple choice question (with options) to answer.
If q and r are negative integers, which of the following statements must be true?
I. (-q)^r is positive.
II. (-q)^(-r) is positive.
III. q^(-r) is positive. | [
"None",
"II only",
"I and II",
"I and III"
] | C | Let's q = -2 and r = -3
I. (-q)^r = (2)^(-3) = 1/(2^3) --> Positive
II. (-q)^(-r) is positive. = 2^3 --> Positive
III. q^(-r) is positive. = -2^3 --> Negative
Answer: C |
AQUA-RAT | AQUA-RAT-37217 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can swim in still water at 4.5 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is? | [
"1.7",
"1.4",
"1.2",
"1.5"
] | D | M = 4.5
S = x
DS = 4.5 + x
US = 4.5 + x
4.5 + x = (4.5 - x)2
4.5 + x = 9 -2x
3x = 4.5
x = 1.5
Answer: D |
AQUA-RAT | AQUA-RAT-37218 | induction that the product of three consecutive integers is divisible by 6. Real numbers class 10. In mathematics, the least common multiple, also known as the lowest common multiple of two (or more) integers a and b, is the smallest positive integer that is divisible by both. Click 'show details' to verify your result. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Prove that n2-n is divisible by 2 for every positive integer n. Exercise: 2. For any positive integer n, prove that n3 – n is divisible by 6. The number is divisible by 6 means it must be divisible by 2 and 3. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. What are the two odd integers? 12. 111 is the smallest possible magic. Exercise 12. case (3) z is a multiple of three. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Some other very important questions from real numbers chapter 1 class 10. Prove that the product of any three consecutive positive integers is divisible by 6. Since n is a perfect square, n is congruent to 0 or 1 modulo 4. Solution for Determine whether the statement is true or false. 1, which is divisible by 9. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. : Therefore: n = 3p or 3p+1 or 3p+2, where p is some integer If n = 3p, then n is divisible by 3 If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3. Consider three consecutive integers, n, n + 1, and n+ 2. 7: given non-empty nite sets X and Y with jXj= jYj, a function X !Y is an injection if and only if it
The following is multiple choice question (with options) to answer.
The sum of three consecutive odd numbers is always divisible by:
I. 2 II. 3 III. 5 IV. 6 | [
"Only I",
"Only II",
"Only I and II",
"Only II and IV"
] | B | Let the three consecutive odd numbers be (2x+1), (2x+3) and (2x+5).
Their sum = (6x=9)= 3 (2x+3), which is always divisible by 3.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37219 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
Radha bought a watch for Rs. 144 and got a percentage of profit equal to the cost price of the watch. What is the cost price of the watch? | [
"Rs. 72",
"Rs. 78",
"Rs. 80",
"Rs. 90"
] | D | SP=144
CP=x
profit %=x
C.P.= (100/(100+gain%))*S.P.
x=(100/100+x) *144
x^2+100x=14400
x^2+180x-80x-14400=0
(x+180) (x-80)=0
x=-180
x=80
ANSWER:D |
AQUA-RAT | AQUA-RAT-37220 | PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.
PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.
IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
The following is multiple choice question (with options) to answer.
The average of first ten prime numbers which are odd is? | [
"15.9",
"15.7",
"15.1",
"15.8"
] | D | Sum of first 10 prime no. which are odd = 158
Average
= 158/10
= 15.8
Answer:D |
AQUA-RAT | AQUA-RAT-37221 | 4. As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!
$(2V/pi)=4r^3$ He divided each side by 2 to get...
$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?
You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.
What would you do now?
5. Originally Posted by SuperTyphoon
As great is this is, i have found an arithmetic error in the solution that has left me hanging once again!
$(2V/pi)=4r^3$ He divided each side by 2 to get...
$2r^3=(4V/pi)$ But wouldn't dividing by 2 give you $2r^3=(V/pi)$ ?
You would then get $r=\sqrt[3](V/2pi)$ thus $d=2r=2\sqrt[3](V/pi)$.
What would you do now?
I didn't divide through by two! I actually multiplied through by two and then skipped some steps. May I should have included the extra steps...
$\frac{2V}{\pi}=4r^3\implies {\color{red}2}\cdot\frac{2V}{\pi}={\color{red}2}\c dot 4r^3\implies \frac{4V}{\pi}=8r^3\implies \frac{4V}{\pi}=(2r)^3\implies 2r=\sqrt[3]{\frac{4V}{\pi}}$
Does this clarify things?
--Chris
6. Ahhhh yessss. That makes a big difference! Thanks for clearing that up. My calc teacher always discourages skipping steps for the same reason that happened here.
Thanks again!
The following is multiple choice question (with options) to answer.
In the formula V = 1/(2r)^2, if r is halved, then V is multiplied by? | [
"16",
"8",
"4",
"1/8"
] | C | Say r=6 => V1 = 1/144
when r=3; V2 = 1/36
V2 = 36*V1.
Answer : C |
AQUA-RAT | AQUA-RAT-37222 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
1000 men have provisions for 15 days. If 200 more men join them, for how many days will the provisions last now? | [
"12.8",
"12.6",
"12.5",
"12.2"
] | C | 1000*15 = 1200*x
x = 12.5.Answer: C |
AQUA-RAT | AQUA-RAT-37223 | You can use sample() and select specific probabilities for each integer. If you sum the product of the probabilities and the integers, you get the expected value of the distribution. So, if you have a mean value in mind, say $$k$$, you can solve the following equation: $$k = 1\times P(1) + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ You can arbitrarily choose two of the probabilities and solve for the third, which determines the fourth (because $$P(1)=1-(P(2)+P(3)+P(4))$$ because the probabilities must sum to $$1$$). For example, let $$k=2.3$$, $$P(4)=.1$$, and $$P(3)=.2$$. Then we have that $$k = 1 \times [1-(P(2)+P(3)+P(4)] + 2\times P(2) + 3\times P(3) + 4\times P(4)$$ $$2.3 = [1 - (P(2)+.1+.2)] + 2*P(2) + 3\times .2 + 4\times .1$$ $$2.3 = .7 + P(2) + .6 + .4$$ $$P(2)=.6$$ $$P(1)=1-(P(2)+P(3)+P(4)=1 - (.6+.1+.2)=.1$$
So you can run x <- sample(c(1, 2, 3, 4), 1e6, replace = TRUE, prob = c(.1, .6, .2, .1)) and mean(x) is approximately $$2.3$$
The following is multiple choice question (with options) to answer.
The value of x is to be randomly selected from the integers from 1 to 9, inclusive, and then substituted into the equation y = x^2 - 4x + 3. What is the probability that the value of y will be negative? | [
"1/9",
"2/9",
"1/3",
"4/9"
] | A | y will only be negative for x=2.
(We can check the values from 1 to 9 to be certain.)
P(y is negative) = 1/9
The answer is A. |
AQUA-RAT | AQUA-RAT-37224 | Let one woman complete the job in $$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.
Mangala completes a piece of work in 15 days, Raju completes the same work in 35 days. If both of them work together, then the number of days required to complete the work is | [
"6 days",
"10 1/2 days",
"8 1/2 days",
"9 days"
] | B | If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days.
That is, the required No. of days = 15 × 35/50 = 10 1/2 days.
B |
AQUA-RAT | AQUA-RAT-37225 | # whole numbers and division
Consider the whole number with one thousand digits that can be formed by writing the digits 2772 two hundred and fifty time in succession. Is it divisible by 9? Is it divisible by 11?
-
I answered yes to both because 2772 is divisible by 9 and 11. I am just trying to make sure this is correct! – SNS Feb 27 '12 at 23:19
If the sum of the digits of $n$ is divisible by 9, then (and only then) $n$ is divisible by 9. From this, your number is divisible by 9 (the sum of its digits is $250\cdot18$). I've forgotten the divisibility test for 11... – David Mitra Feb 27 '12 at 23:21
Can you show that your number is divisible by 2772? And can you finish up from there? – Gerry Myerson Feb 27 '12 at 23:26
2772/9=308 2772/11=252 I think that if the original number is divisible by 9 and 11 then if you continue adding the same numbers over and over it should always be divisible by 9 and 11. This is the part I want to double check on. – SNS Feb 27 '12 at 23:29
@DavidMitra If the alternating sum of the digits (add, subtract, add, subtract, etc.) is divisible by 11 then the number is divisible by 11. This is because $10^n\equiv (-1)^n\mod 10$. – Alex Becker Feb 27 '12 at 23:29
The answer is yes; but, in my opinion, you did not give enough information in your comment for a justification.
One way to show it is to use the divisibility tests for 9 and 11.
Let's call your number, obtained by writing "$2772$" two hundred and fifty times in succession, $y$.
A number $n$ is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $y$ is $250\cdot(2+7+7+2)=250(18)$, so $y$ is divisible by 9.
The following is multiple choice question (with options) to answer.
476 ** 0 is divisible by both 3 and 11.The non zero digits in the hundred's and ten's places are respectively: | [
"8 and 8",
"8 and 5",
"8 and 7",
"7 and 5"
] | D | Explanation:
Let the number be 476ab0
476ab0 is divisible by 3
=> 4 + 7 + 6 + a + b + 0 is divisible by 3
=> 17 + a + b is divisible by 3 ------------------------(i)
476ab0 is divisible by 11
[(4 + 6 + b) -(7 + a + 0)] is 0 or divisible by 11
=> [3 + (b - a)] is 0 or divisible by 11 --------------(ii)
Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation 1 and Equation 2.
if a=6 and b=2,
17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation(i).Hence this is not the answer
if a=8 and b=2,
17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- Meet equation(i)
[3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11---Does not meet equation(ii).Hence this is not the answer
if a=6 and b=5,
17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation (i) .Hence this is not the answer
if a=8 and b=5,
17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- Meet equation 1
[3 + (b - a)] = [3 + (5 - 8)] = 0 ---Meet equation 2
Since these values satisfies both equation 1 and equation 2, this is the answer
Answer: D) 8 and 5 |
AQUA-RAT | AQUA-RAT-37226 | ### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
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The following is multiple choice question (with options) to answer.
Sabrina is contemplating a job switch. She is thinking of leaving her job paying $85,000 per year to accept a sales job paying $45,000 per year plus 15 percent commission for each sale made. If each of her sales is for $2,000, what is the least number of sales she must make per year if she is not to lose money because of the job change? | [
"57",
"133",
"178",
"377"
] | B | Lets say she sales x items so her commission will be : x * 2000*15/100 = 300x
Salary difference is: 85000 - 45000 = 40000
So commission must at least be equal to salary difference i.e. x = 40000/300= 133 (Approx).
Hence answer is B |
AQUA-RAT | AQUA-RAT-37227 | 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 80 m in length crosses a telegraph post in 6 seconds. The speed of the train is? | [
"16 kmph",
"88 kmph",
"54 kmph",
"48 kmph"
] | D | S = 80/6 * 18/5
= 48 kmph
Answer: D |
AQUA-RAT | AQUA-RAT-37228 | (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.
At a garage sale, all of the items were sold at different prices. If the price of a radio sold at the garage sale was both the 14th highest price and the 21st lowest price among the prices of the items sold, how many items were sold at the garage sale? | [
"33",
"34",
"35",
"36"
] | B | There were 13 items sold at a higher price than the radio and 20 items sold at a lower price than the radio. Including the radio, there were 13 + 20 + 1 = 34 items sold.
The answer is B. |
AQUA-RAT | AQUA-RAT-37229 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes? | [
"399 km",
"162 km",
"277 km",
"887 km"
] | B | 12 m/s = 12 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 12 * 18/5 * 15/4 km = 162 km.Answer: B |
AQUA-RAT | AQUA-RAT-37230 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe A fills a swimming pool in 4 hours. Pipe B empties the pool in 6 hours. If pipe A was opened at 10:00 am and Pipe B at 11:00 am, at what time will the pool be full? | [
"22:00",
"20:00",
"21:00",
"18:00"
] | B | Pipe A fills the pool in 4 hrs.
1 hour's work : 1/4
Pipe B empties the pool in 6 hrs.
1 hour's work : 1/6
Together if they work, 1 hour's work = 1/4 -1/6 = 1/12
Given : Pipe A started at 10:00 a.m and Pipe B at 11:00 a.m
Pool filled after 1 hour by Pipe A : 1/4 or 3/12
After 11:00 a.m
Pool filled after 1 hour with both the pipes on : 1/12
Pool filled after 9 hours with both pipes on : 9/12
Pool filled in 1 hour + Pool filled in 9 hours = 3/12 +9/12 =1
Therefore, it takes 10 hrs to fill the pool
As Pipe A started at 10:00 a.m, pool is full at 20:00 hrs
Answer : B |
AQUA-RAT | AQUA-RAT-37231 | # 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.
Tarun got 30% concession on the labelled price of an article and sold it for Rs. 8750 with 25% profit on the price he bought. What was the labelled price ? | [
"10000",
"22438",
"28789",
"10002"
] | A | Explanation:
Cost Price = = Rs. 7000. Let the labelled price be Rs. x.
Then,
= Rs.10000
Answer: A) 10000 |
AQUA-RAT | AQUA-RAT-37232 | Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have
\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}
And so we have our case where
\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}
The following is multiple choice question (with options) to answer.
A basketball player succeeds with 50% of her free throws. If she has a set of three free throws, what is the probability that she succeeds at least once? | [
"0.585",
"0.625",
"0.75",
"0.875"
] | D | P(missing all 3 free throws) = 0.5^3 = 0.125
P(succeeding at least once) = 1 - 0.125 = 0.875
The answer is D. |
AQUA-RAT | AQUA-RAT-37233 | • $$F$$ is a wide inclusion, i.e. $$F$$ is faithful and essentially surjective $$\Rightarrow$$
• $$F$$ dominant $$\Rightarrow$$ same conclusion as $$F$$ essentially surjective (since $$F$$ is essentially surjective on Cauchy completions and the Cauchy completion has the same presheaf category)
• (Assuming $$C$$ has finite limits) $$F$$ conservative $$\Leftrightarrow$$ the counit is a strong epimorphism
• $$F$$ separable $$\Rightarrow$$
• $$F$$ injective on objects $$\Rightarrow$$
• $$F$$ preserves limits $$\Rightarrow$$ $$F^L$$ exists (assuming $$C$$ is complete and "small enough")
• $$F$$ preserves colimits $$\Rightarrow$$ $$F^R$$ exists (assuming $$C$$ is cocomplete and "small enough")
• $$F$$ preserves $$\kappa$$-directed colimits $$\Rightarrow$$ $$F^L$$ preserves $$\kappa$$-presentable objects
The following is multiple choice question (with options) to answer.
Commodious means ‘spacious’, large, voluminous and extensive. | [
"22",
"28",
"277",
"199"
] | A | Answer:A |
AQUA-RAT | AQUA-RAT-37234 | # 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.
John started a business, investing Rs.18000. After 3 months and 4 months respectively, Rose and Tom joined him with capitals of 12000 and 9000. At the end of the year the total profit was Rs.4070. What is the difference between Rose ’s and Tom ’s share in the profit? | [
"s.340",
"s.362",
"s.370",
"s.382"
] | C | John : Rose : Tom
Ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8
= 6 : 3: 2
The difference between Rose’s and Tom’s share = 1 share
:.i.e. = Rs. 4070 × 1/11 = Rs.370.
C) |
AQUA-RAT | AQUA-RAT-37235 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
How many box's of length 40 Cm, width 30 Cm and height 10 Cm can be formed from a box of 6 m length, 4 m width and 2 m height. | [
"2000",
"5000",
"7000",
"4000"
] | D | (600×400×200)/(40×30×10) = 4000
Answer is D |
AQUA-RAT | AQUA-RAT-37236 | Time needed for both machine to produce 1500 units is given.
1/8 + 1/12
= 5/24
I hrs work done by both machine.
24/5 = 4.8.
Time needed to complete the entire work.
******* 1500 units must be considered as a single work. units are same for both machine. Therefore we don't need to do anything with units.
C is the correct answer.
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink]
### Show Tags
12 Aug 2019, 10:53
Bunuel wrote:
A new copy machine can run off 1,500 workbooks in 8 hours, while it takes on older copy machine 12 hours to do the same job. What is the total number of hours that it would take both copy machines working at the same time, but independently, to run off the 1,500 workbooks?
(A) 4.4
(B) 4.6
(C) 4.8
(D) 5
(E) 10
Let T be the number of hours to complete the job when both machines are running. We can create the following equation:
1500/8 * T + 1500/12 * T = 1500
T/8 + T/12 = 1
Let’s multiply each side by 24:
3T + 2T = 24
5T = 24
T = 4.8 hours
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink] 12 Aug 2019, 10:53
# A new copy machine can run off 1,500 workbooks in 8 hours, while it ta
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
The following is multiple choice question (with options) to answer.
One copy machine can make 30 copies a minute, and a second copy machine makes 15 copies a minute. If the two copiers work together, how long would it take them to make 900 copies? | [
"15 minutes",
"20 minutes",
"25 minutes",
"30 minutes"
] | B | Total work done by both machines in a minute = 30 + 15 = 45 copies
Total number of copies required = 900
Time = 900/45 = 20 mins
Answer B |
AQUA-RAT | AQUA-RAT-37237 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 41 kmph. For how many minutes does the bus stop per hour? | [
"7 min",
"8 min",
"14.4 min",
"10 min"
] | C | Due to stoppages, it covers 13 km less.
Time taken to cover 13 km =(13/54)x60 = 14.4 min
answer : C |
AQUA-RAT | AQUA-RAT-37238 | ## anonymous 5 years ago 100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!
1. anonymous
you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..
2. anonymous
$\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950$
3. anonymous
Could you solve it using factorials, please?
4. anonymous
hmm i found the answer as 9900
5. amistre64
if there are 3 people: 1,2 ; 1,3 ; 2,3 .... is that right?
6. anonymous
Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...
7. anonymous
yea but its harder to calculate this like that... so i think C(100,2) better way of solve
8. amistre64
5054 if we do that ..... add all the numbers from 1 to 100 :)
9. anonymous
No, all the numbers from 1 to 99.
10. amistre64
n(n+1) ------ :) 2
11. amistre64
ack .... 99 then lol
12. amistre64
wouldnt the last guy have noone to shake hands with?
13. anonymous
mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,
14. anonymous
Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.
15. anonymous
yea you right
16. amistre64
The following is multiple choice question (with options) to answer.
50 men shake hands with each other. Maximum no of handshakes without cyclic handshakes. | [
"190",
"200",
"1225",
"220"
] | C | total no. of handshakes = 49+48+47+...+3+2+1 = 19*(19+1)/2 = 1225
or, if there are n persons then no. of shakehands = nC2 = 50C2 = 1225
Answer : C |
AQUA-RAT | AQUA-RAT-37239 | 36&14,9,9,9,9,10\\ 37&16,9,9,9,9,10\\ 38&16,9,9,9,9,12\\ 39&16,9,9,11,9,12\\ 40&16,9,9,11,11,12\\ 41&16,9,11,11,11,12\\ 42&16,11,11,11,11,12\\ 43&18,11,11,11,11,12\\ 44&18,11,11,11,11,14\\ 45&18,11,11,13,11,14\\ 46&18,11,11,13,13,14\\ 47&18,11,13,13,13,14\\ 48&18,13,13,13,13,14\\ 49&18,13,13,13,13,16\\ 50&18,13,13,15,13,16\\ 51&18,13,13,15,15,16\\ 52&18,13,15,15,15,16\\ 53&18,15,15,15,15,16\\ 54&18,15,15,15,15,18\\ 55&18,15,15,15,17,18\\ 56&18,15,15,17,17,18\\ 57&18,15,17,17,17,18\\ 58&18,17,17,17,17,18\\ 59&18,17,17,17,17,20\\ 60&18,17,17,17,19,20\\ 61&18,17,17,19,19,20\\ 62&18,17,19,19,19,20\\ 63&18,17,19,19,19,22\\ 64&18,17,19,19,21,22\\ 65&18,17,19,21,21,22\\ 66&18,17,21,21,21,22\\ 67&18,17,21,21,21,24\\ 68&18,17,21,21,23,24\\
The following is multiple choice question (with options) to answer.
385, 462, 572, 396, 421, 671, 264 | [
"385",
"421",
"671",
"264"
] | B | Solution
In each number except 421,the middle digit is the sum of the other two. Answer B |
AQUA-RAT | AQUA-RAT-37240 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
An express traveled at an average speed of 120 km/hr, stopping for 4 min after every 85 kn. How long did it take to reach its destination 8400 km from the starting point? | [
"5 hrs 21 min",
"4 hrs 21 min",
"6 hrs 41 min",
"7 hrs 32 min"
] | D | D
7 hrs 32 min
Time taken to cover 840 km = 840/120 = 7 hrs.
Number of stoppages = 840/85 - 1 = 8
Total time of stoppages = 4 * 8= 32 min
Hence, total time taken = 7 hrs 32 min. |
AQUA-RAT | AQUA-RAT-37241 | + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 15+3x \\ 8. & 8+2x &-2 a_1 + 2 b& 3 a_1 -b & 14+3x \\ 8. & 7+2x &-2 a_1 + 2 b& 3 a_1 -b & 13+3x \end{array}$$
The following is multiple choice question (with options) to answer.
3 + 3 + 3 + 2 × 3^2 + 2 × 3^3 + 2 × 3^4 + 2 × 3^5 + 2 × 3^6 + 2 × 3^7 = | [
" 3^7",
" 3^8",
" 3^14",
" 3^28"
] | B | We have the sum of 9 terms. Now, if all terms were equal to the largest term 2*3^7 we would have: sum=9*(2*3^7)=2*3^9=~3^10, so the actual sum is less than 3^10 and more than 3^7 (option A) as the last term is already more than that. So the answer is clearly B.
Answer: B. |
AQUA-RAT | AQUA-RAT-37242 | 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-4)(b-2) =0, which of the following must be true ?
I. a=4 and b=2
II. if a is not = 4, then b=2
III. if a=4, then b is not=2 | [
"I only",
"II only",
"III only",
"I and II"
] | B | (a-4)(b-2)=0
So either a=4 or b=2 or both. It is 'or' not 'and', so it could be any of the three possibilities. Thus I and III are not necessarily true.
The answer is B. |
AQUA-RAT | AQUA-RAT-37243 | Time needed for both machine to produce 1500 units is given.
1/8 + 1/12
= 5/24
I hrs work done by both machine.
24/5 = 4.8.
Time needed to complete the entire work.
******* 1500 units must be considered as a single work. units are same for both machine. Therefore we don't need to do anything with units.
C is the correct answer.
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink]
### Show Tags
12 Aug 2019, 10:53
Bunuel wrote:
A new copy machine can run off 1,500 workbooks in 8 hours, while it takes on older copy machine 12 hours to do the same job. What is the total number of hours that it would take both copy machines working at the same time, but independently, to run off the 1,500 workbooks?
(A) 4.4
(B) 4.6
(C) 4.8
(D) 5
(E) 10
Let T be the number of hours to complete the job when both machines are running. We can create the following equation:
1500/8 * T + 1500/12 * T = 1500
T/8 + T/12 = 1
Let’s multiply each side by 24:
3T + 2T = 24
5T = 24
T = 4.8 hours
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Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink] 12 Aug 2019, 10:53
# A new copy machine can run off 1,500 workbooks in 8 hours, while it ta
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
The following is multiple choice question (with options) to answer.
Working alone, machine X can manufacture 1,200 nails in 12 hours. Working together, machines X and Y can manufacture 1,200 nails in 5 hours. How many hours does it take machine Y to manufacture 1,000 nails working alone? | [
"3 9/17",
"5 1/3",
"7",
"7 1/7"
] | D | rate of machine x = 1200/12
(1200/12 + y)5=1200
y=700/5
700/5 * t = 1000
t=7 1/7
D. |
AQUA-RAT | AQUA-RAT-37244 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Each shelf of a bookcase contained 11 books. If the librarian took out 33 books and rearranged the remaining books so that all shelves but the last one contained 7 books and that last shelf contained 6 books, how many shelves does the bookcase have? | [
"6",
"7",
"8",
"9"
] | C | Let x be the number of shelves.
11x - 33 = 7(x-1) + 6
4x = 32
x = 8
The answer is C. |
AQUA-RAT | AQUA-RAT-37245 | java, beginner, collections
if(input.equals("stdtogrpRemove")){
System.out.println("Enter group to found: ");
groupToFound = sc.nextLine();
System.out.println("Enter student surname to remove: ");
studentSurname = sc.nextLine();
while (groupIT.hasNext()){
Group group = groupIT.next();
String strInt = Integer.toString(group.getGroupNumb());
if (groupToFound.equals(strInt) || studentSurname.equals(group.getStudList())) {
group.setStudList(null, null);
}
}
}
if (input.equals("findstdfromgrp")){
System.out.println("Enter group to found: ");
groupToFound = sc.nextLine();
System.out.println("Enter student surname to remove: ");
studentSurname = sc.nextLine();
while (groupIT.hasNext()){
Group group = groupIT.next();
String strInt = Integer.toString(group.getGroupNumb());
if (groupToFound.equals(strInt) || studentSurname.equals(group.getStudList())) {
System.out.println("Student exist in list");
}
}
}
if (input.equals("findstdfromsbj")){
}
if (input.equals("findstdfromtch")){
}
}
The following is multiple choice question (with options) to answer.
Find the one which does not belong to that group? | [
"16",
"36",
"21",
"64"
] | C | Explanation:
16, 36, 64 and 4 are perfect squares but not 21.
Answer:C |
AQUA-RAT | AQUA-RAT-37246 | water, mixtures, salt
Title: Why water volume went up by almost the same amount after adding salt? Seems like it's a well-known fact that adding salt to water will not raise the water volume by much (e.g.: Why there is no change in water level when salt is added? )
I've done this experiment at home and saw unexpected results:
Setup:
Start with 500 ml of tap water in measuring cup (500 grams)
Measure 50 ml of iodized salt in measuring cup (80 grams)
Add salt to water, stir until dissolved
Result: around 540 ml (give or take) of salt water in measuring cup weighing 580 grams
Weight was measured with kitchen scale, volume measure by eye, but ~40 ml increase in total volume seems way too much.
Followed up by replacing salt with sand, volume went up by about the same amount.
Video: https://www.youtube.com/watch?v=rLTn5jKqipU
What am I missing here?
p.s.: the commenter pointed out that my salt volume is off, I should have looked up the density to go by weight rather than by measuring cup notches. Going by weight, I've added 37 ml or salt, making the result event closer. Volumes were judged by eye in food measuring cups, not exactly the proper lab equipment.
You have water volume.
Calculate salt volume from its mass and density.
Calculate their total volume before dissolution.
Calculate solution mass concentration in mass %.
From tabulated data or online calculators, obtain its density.
Calculate solution volume from its total mass and density.
Subtract from it the total volume of separate water and salt.
You have the volume difference.
The following is multiple choice question (with options) to answer.
How many kg of pure salt must be added to 30kg of 2% solution of salt and water to increase it to a 10% solution? | [
"8/3",
"9/7",
"7/4",
"5/2"
] | A | Amount salt in 30kg solution = 2*30/100 = 0.6 kg
Let x kg of pure salt be added
then(0.6+x)/(30+x)=10/100
60+100x=300+10x
90x = 240
x = 8/3
Answer is A |
AQUA-RAT | AQUA-RAT-37247 | # Sort of a challenge
#### topsquark
##### Well-known member
MHB Math Helper
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
-Dan
#### Bacterius
##### Well-known member
MHB Math Helper
Well the Earth's radius is $6371 ~ \text{km}$, so the original rope has a length of $2 \pi \times 6371 ~ \text{km}$. If we add $6 ~ \text{m} = 6 \times 10^{-3} ~ \text{km}$ to this rope, its new radius is:
$$\frac{2 \pi \times 6371 + 6 \times 10^{-3}}{2 \pi} \approx 6371.0009 ~ \text{km}$$
So the rope now "floats" about $0.0009 ~ \text{km} = 90 ~ \text{cm}$ above the ground. To be exact, $95.5 ~ \text{cm}$ (this is $\frac{6}{2 \pi} ~ \text{m}$).
Wait, what? My mind is blown
#### MarkFL
Staff member
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
The following is multiple choice question (with options) to answer.
The dimensions of a field are 10 m by 10 m. A pit 10 m long, 5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field. What will be the rise in the height of field as a result of this operation ? | [
"2 m",
"3 m",
"4 m",
"5 m"
] | C | The volume of the earth removed is 10*5*4 = 200 m^3.
The remaining area of the field is 10*10 - 10*5 = 50 m^2.
200 m^3 of the earth evenly spread over the area of 50 m^2 will rise the height by (height) = (volume)/(area) = 200/50 = 4 m.
Answer: C |
AQUA-RAT | AQUA-RAT-37248 | 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.
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 6000, the number of valid votes that the other candidate got, was: | [
"2800",
"2700",
"2160",
"2200"
] | C | C
Number of valid votes = 80% of 6000 = 4800.
Valid votes polled by other candidate = 45% of 4800
= (45/100 x 4800) = 2160. |
AQUA-RAT | AQUA-RAT-37249 | We need to understand the compound interest formula: A = P(1 + r/n)^nt. A stands for the amount of money that has accumulated. P is the principal; that's the amount you start with. The r is the. is continuously, where interest is compounded essentially every second of every day for the entire term. This means 𝑛 is essentially infinite, and so we will use a different formula which contains the natural number 𝑒 to calculate the value of an investment. The formula for interest compounded continuously is 𝐴=𝑃𝑒𝑟𝑡. APY to APR Calculator : Enter the APY currently being earned (in percent): % Enter the number of compounding periods in a year ... If it’s compounded monthly: APY = (APR/12 + 1)^12 – 1. (where APR & APY are decimals; for 3% you’d put 0.03). But, always we have question about compounded continuously. To understand 'compounded continuously', let us consider the example given below. When we invest some money in a bank, it will grow continuously. That is, at any instant the balance is changing at a rate that equals 'r' (rate of interest per year) times the current balance. Formula for. The continuous compounding formula says A = Pe rt where 'r' is the rate of interest. For example, if the rate of interest is given to be 10% then we take r = 10/100 = 0.1. What Is e in Continuous Compounding Formula? 'e' in the continuous compounding formula is a mathematical constant and its value is approximately equal to 2.7183. In this section we cover compound interest and continuously compounded interest. Use the compound interest formula to solve the following. Example: If a $500 certificate of deposit earns 4 1/4% compounded monthly then how much will be accumulated at the end of a 3 year period?. Answer: At the end of 3 years the amount is$576.86. The formula for the principal plus interest is as follows: Total = Principal x e^(Interest x Years) Where: e – the exponential function, which is equal to 2.71828. Using Company ABC example above, the return on investment can be calculated as follows when using continuous compounding: = 10,000 x 2.71828^(0.05 x
The following is multiple choice question (with options) to answer.
Albert invested amount of 8000 in a fixed deposit for 2 years at compound interest rate of 5 % per annum. How much Albert will get on the maturity of the fixed deposit | [
"Rs. 8510",
"Rs. 8620",
"Rs. 8730",
"Rs. 8820"
] | D | Explanation:
=>(8000×(1+5/100)2)
=>8000×21/20×21/20
=>8820
Option D |
AQUA-RAT | AQUA-RAT-37250 | newtonian-mechanics, drag, relative-motion
Title: Relative motion-Acceleration My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.
Say a boat travels at 10 m/s in a water channel.
the water speed relative to ground is 0.
so the boat travels at 10 m/s relative to the ground.
now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
As after a while the boat speed relative to ground has become 0,
then from the ground-based observer's point of view, the boat has undergone a deceleration.
My question is;
Is this deceleration always necessarily equal to minus the water acceleration?
In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?
PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?
My question is; Is this deceleration always necessarily equal to minus the water acceleration?
The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:
$$f=kv\qquad \text{ for low speed}\\
f=kv^2\qquad \text{ for high speed}$$
where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.
The following is multiple choice question (with options) to answer.
The Speed of a boat in still water is 25 kmph. If it can travel 10 km upstream in 1 hr, What time it would take to travel the same distance downstream? | [
"26 minutes",
"23 minutes",
"15 minutes",
"19 minutes"
] | C | Speed of boat in still water = 25 km/hr
Speed upstream = 10⁄1 = 10 km/hr
Speed of the stream = (25-10) = 15 km/hr
Speed downstream = (25+15) = 40 km/hr
Time taken to travel 10 km downstream = 10/40 hours=(10×60)/40= 15 minutes
Answer is C |
AQUA-RAT | AQUA-RAT-37251 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What is the smallest number which, when increased by 5, is divisible by 7, 8, and 24? | [
"148",
"153",
"158",
"163"
] | D | LCM(7,8,24)=24x7=168
So the least divisible number is 168, and the number we are looking for is 168-5=163.
The answer is D. |
AQUA-RAT | AQUA-RAT-37252 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
Ravi and Kavi start a business by investing ₹ 5000 and ₹ 72000, respectively. Find the ratio of their profits at the end of year. | [
"2 : 72",
"5 : 72",
"7 : 72",
"1 : 72"
] | B | Ratio of profit = Ratio of investments
= 5000 : 72000 = 5 : 72
Answer : B |
AQUA-RAT | AQUA-RAT-37253 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group? | [
"22 years",
"30 years",
"32 years",
"44 years"
] | C | The total age of 10 students with average 20 = 200
The total age of 12 students with average 22 = 264
The total age of 2 new students = 64
Their average =32
ANSWER:C |
AQUA-RAT | AQUA-RAT-37254 | that is y=???abc
#### Bacterius
##### Well-known member
MHB Math Helper
Re: find the value of x and y
$$x^2 \equiv x \pmod{1000} \tag{1}$$
[JUSTIFY]Clearly for $x$ coprime to $1000$, the only solutions are the trivial $x = 0$ and $x = 1$, so all nontrivial solutions for $x$ must be divisible by either $2$ or $5$, or both. This leaves us with about $600$ possible candidates. Now it should be clear that if a solution works for $1000$, it will also work for $100$, and for $10$. Let's try and find the solutions for $10$:[/JUSTIFY]
$$x^2 \equiv x \pmod{10} \tag{2}$$
[JUSTIFY]We have $5$ potential solutions here, namely $2, 4, 5, 6, 8$ (modulo $10$). Let's check each of them manually, some calculations show that only $5$ and $6$ work. So the solutions must have either $5$ or $6$ as a last digit. Let's now extend our search to $100$, armed with this information (which reduces the set of possible solutions considerably):[/JUSTIFY]
$$x^2 \equiv x \pmod{100} \tag{3}$$
[JUSTIFY]We know from the previous step that the potential solutions are $5, 6, 15, 16, \cdots$. A short exhaustive search tells us that only $25$ and $76$ work. Finally, we have narrowed down the search enough to look for the real solutions:[/JUSTIFY]
$$x^2 \equiv x \pmod{1000} \tag{4}$$
The only possible solutions are $25, 76, 125, 176, \cdots$. Again, trying those 20 potential solutions shows that only $376$ and $625$ work. Therefore, the solutions are $x \in \{ 0, \, 1, \, 376, \, 625 \}$. QED.
The following is multiple choice question (with options) to answer.
If x 2 – y 2 = 101, find the value of x 2 + y 2 ,
given that x and y are natural numbers. | [
"5100",
"5101",
"5102",
"5103"
] | B | x^2 – y^2 = 101
=> (x+y)*(x-y) = 101*1
x+y = 101
x-y = 1
so, x = 51, y = 50
(x^2 + y^2) = 51^2 + 50^2 = 5101
ANSWER:B |
AQUA-RAT | AQUA-RAT-37255 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
Virginia, Adrienne, and Dennis have taught history for a combined total of 93 years. If Virginia has taught for 9 more years than Adrienne and for 9 fewer years than Dennis, for how many years has Dennis taught? | [
" 23",
" 32",
" 40",
" 41"
] | C | Let number of years taught by Virginia = V
number of years taught by Adrienne = A
number of years taught by Dennis = D
V+A+D = 96
V= A+9
=>A = V-9
V= D-9
=> A = (D-9)-9 = D-18
D-9 + D-18 + D = 93
=> 3D = 93 + 27 = 120
=> D = 40
Answer C |
AQUA-RAT | AQUA-RAT-37256 | Your best bet in a situation like this, where there are relatively few numbers between $50$ and $100$ which are divisible by $7$, is to just list them, and count the number of entries on the list. (Of course, you'd like a more efficient means if the interval over which multiples of a given number span is very very large!) Note: When computing the number of integers between $50$ and $100$ that are divisible by both $7$ and $11$, there is only one such number: $7 \times 11 = 77$.
If, however, the question asked how many such numbers are divisible by $7$ OR $11$, then you'd need to list/count
(1) those divisible by $7$,
(2) those divisible by $11$;
(3) those divisible by both $7$ and $11$: any multiple of $7 \times 11$ in the given range;
Then add the number of entries on list (1) to those on (2), and then subtract the number of entries on (3) from that sum. (Otherwise, without subtracting, you'd be counting, e.g. 77, twice, since it would appear on both (1) and (2).)
In general, if the lower bound of the interval under consideration is $L$, and the upper bound of the interval under consideration is $U$, then the smallest multiple of a positive integer $n$ that is larger than or equal to $L$ is $L/n$, and if the result is not an integer, round up to the next integer (say $k_{min}$). On the other hand, the largest multiple of $n$ that is less than or equal to $U$ is $U/n$, but if the result is not an integer, round down (drop off the remainder), say $k_{max}$. Then, count all multiples of $n$ such that $L \leq nk_{min} < nk_i < nk_{max} \leq U$, where $nk_i$ are all the multiples of $n$ between the least and greatest integers in the given interval [$L, U$].
-
The following is multiple choice question (with options) to answer.
How many ODD integers n, where 100 <= n <= 200, are divisible neither by seven nor by nine? | [
"37",
"38",
"39",
"40"
] | C | Total Integers from 100 to 200 inclusive = 101 (51 even and 50 odd)
Total Odd Integers from 100 to 200 = 50
Total No. that are multiple of 7 from 1-200 = 200/7 = 28 (i.e. 14 even and 14 odd)
Total No. that are multiple of 7 from 1-100 = 100/7 = 14 (i.e. 7 even and 7 odd)
Total No. that are ODD multiple of 7 from 100-200 = 14-7 = 7
Total No. that are multiple of 9 from 1-200 = 200/9 = 22 (i.e. 11 even and 11 odd)
Total No. that are multiple of 9 from 1-100 = 100/9 = 11 (i.e. 5 even and 6 odd)
Total No. that are ODD multiple of 9 from 100-200 = 11-6 = 5
Total No. that are odd multiple of 7 and 9 will be multiple of 63 and count of such numbers from 100-200 = 1
So Total No. that are odd multiple of 7 and/or 9 from 100-200 = 7+5-1 = 11
So Total No. that are NOT odd multiple of 7 and/or 9 from 100-200 = 50-11 = 39
Answer: Option C |
AQUA-RAT | AQUA-RAT-37257 | # Project Euler #5 - Lowest common multiple of 1 through 20
This is my code. All comments welcome. Last time run it only took 335 milliseconds:
static bool CheckMultiples(int val)
{
for (int i = 1; i <= 20; i++)
{
if (val % i != 0)
return false;
}
return true;
}
static void Main(string[] args)
{
Stopwatch s = new Stopwatch();
s.Start();
int Num = 20;
while (!CheckMultiples(Num))
{
Num += 20;
}
s.Stop();
Console.WriteLine("The smallest number divisible by all numbers 1-20 is {0}.", Num);
Console.WriteLine("The time took is {0} milliseconds.", s.ElapsedMilliseconds);
}
There is a far more efficient approach to finding the least common multiple ($\operatorname{lcm}$) of a set of integers which, unlike your and rolfl's code, involves no trial-and-error. Recall that for any pair $(a, b)$ of natural numbers, we have $$\operatorname{lcm}(a, b) = \frac{ab}{\operatorname{gcd}(a, b)}.$$ This reduces the implementation of $\operatorname{lcm}$ to the implementation of $\operatorname{gcd}$, which is easily done using the simple and efficient Euclidean algorithm. Then, calculating the least common multiple of $\{1, 2, ..., 20\}$ is as simple as $\operatorname{reduce}(\operatorname{lcm}, \{1, 2, ..., 20\})$.
You can find a simple Python 3 implementation of this idea in ideone here. Even in a slow interpreted language like Python, this is several orders of magnitude faster than your or rolfl's code, running in about .04 milliseconds.
The following is multiple choice question (with options) to answer.
Find the lowest common multiple of 15, 25 and 35. | [
"360",
"420",
"510",
"525"
] | D | LCM=3*5*5*7=525.
Answer is D |
AQUA-RAT | AQUA-RAT-37258 | As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $$5!=5\times 4\times 3\times 2\times 1=120$$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $$3!=3\times 2\times 1=6$$ ways to do, making (again) a total of $$(5!)\times (3!)=120\times 6=720$$ ways of arranging them.
-
+1 for the graphics! :-) – joriki Jul 8 '12 at 19:53
Thanks!$\text{}$ – Zev Chonoles Jul 8 '12 at 20:00
Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
-
thanks joriki, but not quiet clear about what you say – user1419170 Jul 8 '12 at 19:49
@user1419170: Could you be more specific? What part isn't clear to you? – joriki Jul 8 '12 at 19:49
this part Treat the group of three boys as an additional girl – user1419170 Jul 8 '12 at 19:51
@user1419170: See Brian's and Saurabh's answers, which use the same idea but phrase it differently. – joriki Jul 8 '12 at 19:53
The following is multiple choice question (with options) to answer.
In a class there are 20 boys and 25 girls. In how many ways can a boy and a girl be selected? | [
"288",
"500",
"277",
"266"
] | B | We can select one boy from 20 boys in 20 ways.
We select one girl from 25 girls in 25 ways
We select a boy and girl in 20 * 25 ways i.e.,
= 500 ways.
Answer: B |
AQUA-RAT | AQUA-RAT-37259 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
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Posts: 140
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
The average of runs of a cricket player of 10 innings was 35. How many runs must he make in his next innings so as to increase his average of runs by 4 ? | [
"76",
"79",
"85",
"87"
] | B | Explanation:
Average = total runs / no.of innings = 35
So, total = Average x no.of innings = 35 x 10 = 350.
Now increase in avg = 4runs. So, new avg = 35+4 = 39runs
Total runs = new avg x new no. of innings = 39 x 11 = 429
Runs made in the 11th inning = 429 - 350 = 79 Answer: B |
AQUA-RAT | AQUA-RAT-37260 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
3 years ago the average age of a family of 5 members was 17 years. With the birth of a new baby, the average age of six members remains the same even today. Find the age of the new baby. | [
"1 year",
"2 years",
"1 1⁄2 years",
"cannot be determined"
] | B | Sum of present ages of the six members
= (17 × 6) years = 102 years.
Sum of present ages of the 5 members (excluding baby)
= 5 × (17 + 3) years = 100 years.
∴ Age of the baby = 102 – 100 = 2 years
Answer B |
AQUA-RAT | AQUA-RAT-37261 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A boat covers a distance of 30 kms in 2 ½ hours running down streams. While returning, it covers the same distance in 3 ¾ hours. What is the speed of the boat in kmph? | [
"12",
"8",
"14",
"10"
] | D | b+s=30*2/5
b-s=30*4/15
solving ths we get value of b
b=10
ANSWER:D |
AQUA-RAT | AQUA-RAT-37262 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"245",
"248",
"240",
"246"
] | C | Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37263 | \$7,382.94 3.2% 01.09.08 \$19.6879
\$7,402.63 3.2% 01.10.08 \$19.7404
\$7,422.37 3.2% 01.11.08 \$19.7930
\$7,442.17 3.2% 01.12.08 \$19.8458
The following is multiple choice question (with options) to answer.
A sum was put at simple interest at certain rate for 3 years. Had it been put at 1% higher rate it would have fetched Rs. 63 more. The sum is: A. Rs. 2,400 B. Rs. 2,100 C. Rs. 2,200 D.Rs. 2,480 | [
"2000",
"2100",
"2200",
"2300"
] | B | 1 percent for 3 years= 63
1 percent for 1 year = 21
=> 100 percent = 2100
ANSWER:B |
AQUA-RAT | AQUA-RAT-37264 | # 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.
Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B. | [
"2:1",
"1:2",
"1:1",
"4:3"
] | D | Explanation:
5% of A + 4% of B
=2/3(6% of A + 8% of B)
A/B=4/3
A : B = 4 : 3
ANSWER IS D |
AQUA-RAT | AQUA-RAT-37265 | For each of these, $\sum_{n=0}^{\infty} \frac{G_n}{ 10^n} =\frac1{1-\frac1{10}-\frac1{100}-...-\frac1{10^{m}}} =\frac{10^m}{10^m-10^{m-1}-...-1} =\frac{10^m}{8...(m-1 \ 8s)9}$.
The following is multiple choice question (with options) to answer.
What is the sum of all the multiples of 10 between 0 and 85? | [
"500",
"620",
"450",
"360"
] | D | The multiples of 10 between 0 and 85 are 10, 20, 30, 40, 50, 60, 70, 80. If these are all added together, the result is 360.
Final Answer:
D |
AQUA-RAT | AQUA-RAT-37266 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
Which number can we add to 897326 to get a number exactly divisible by 456? | [
"58",
"66",
"74",
"82"
] | D | 897326 / 456 = 1967 with a remainder of 374.
We need to add 456 - 374 = 82
The answer is D. |
AQUA-RAT | AQUA-RAT-37267 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
A, B and C started a business A's investment is 2 times of B's investment is 5 times of C's investment. Find the profit of B in the total profit of Rs. 16000. | [
"Rs. 25600",
"Rs. 5000",
"Rs. 7200",
"Rs. 8000"
] | B | Explanation:
A = 2B and B = 5C
=> A : B = 2:1 and B : C = 5 : 1
=> A : B : C = 10 : 5 : 1
Ratio of capitals of A, B and C = 10 : 5 : 1
B's share = Rs.[(5/16) × 16000] = Rs. 5000
Answer: Option B |
AQUA-RAT | AQUA-RAT-37268 | 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 many seconds does a train 130 meters long, traveling at 72 km/h, take to completely cross a bridge of 210 meters in length? | [
"13",
"15",
"17",
"19"
] | C | 72 km/h = 72000/3600 = 20 m/s
Time = 340 / 20 = 17 seconds
The answer is C. |
AQUA-RAT | AQUA-RAT-37269 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
Kudos [?]: 24 [2], given: 0
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Posts: 149
Kudos [?]: 121 [0], given: 24
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of a group of 14 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group? | [
"22 years",
"30 years",
"36 years",
"32 years"
] | C | Answer
The average age of a group of 14 students is 20.
Therefore, the sum of the ages of all 14 of them = 14 * 20 = 280
When two new students join the group, the average age increases by 2. New average = 22.
Now, there are 16 students.
Therefore, the sum of the ages of all 16 of them = 16 * 22 = 352
Therefore, the sum of the ages of the two new students who joined = 352 - 280 = 72
And the average age of each of the two new students = 72/2 = 36 years.
Answer C |
AQUA-RAT | AQUA-RAT-37270 | # 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.
Bill has d animals, 16 of which were sold at a profit of 10% each while the rest were sold at a loss of 20% each. Overall, Bill made neither a profit nor a loss. Which of the following is equal to d? | [
"6",
"12",
"18",
"24"
] | D | Total d animals, suppose each has price of x - total price = dx
now 16 were sold at profit of 10%, making equation - - 16*1.1x
now d-16 were sold at loss of 20%, making eqn --- (d-16)*0.8x
Now he made no profit or loss so,
dx = 16*1.1x + (d-16)*0.8x
solving for d, we get d=24
Ans D |
AQUA-RAT | AQUA-RAT-37271 | 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 which has 300 m long, is running 80 kmph. In what time will it cross a person moving at 16 kmph in same direction ? | [
"25 sec",
"35 sec",
"15 sec",
"45 sec"
] | C | Time taken to cross a moving person = length of train/ relative speed
Time taken = 300/((80-16) (5/18) = 300/ 64*(5/18) = 300/20= 15 sec
Answer: C |
AQUA-RAT | AQUA-RAT-37272 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely? | [
"16 sec",
"87 sec",
"12 sec",
"18 sec"
] | C | D = 250 m + 250 m = 500 m
RS = 80 + 70 = 150 * 5/18 = 125/3
T = 500 * 3/125 = 12 sec
Answer: C |
AQUA-RAT | AQUA-RAT-37273 | # 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.
What profit percent is made by selling an article at a certain price, if by selling at 2/3rd of that price, there would be a loss of 20%? | [
"20%",
"90%",
"60%",
"40%"
] | A | SP2 = 2/3 SP1
CP = 100
SP2 = 80
2/3 SP1 = 80
SP1 = 120
100 --- 20 => 20%
Answer: A |
AQUA-RAT | AQUA-RAT-37274 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 140 metres long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"876 m",
"157 m",
"235 m",
"156 m"
] | C | Speed = [45 X 5/18] m/sec
= [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (140 + x)/30 = 25/2
=> 2(140 + x) = 750 => x
= 235 m.
Answer: C |
AQUA-RAT | AQUA-RAT-37275 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief is noticed by a policeman from a distance of 225 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes ? | [
"50 meter",
"100 meter",
"110 meter",
"125 meter"
] | D | Explanation:
Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr
Distance covered in 6 minutes =
1/60∗6=1/10=100meters
So distance between them after 6 minutes = 225 - 100
= 125 meters
Option D |
AQUA-RAT | AQUA-RAT-37276 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A person can row at 9 kmph and still water. He takes 4 1/2 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 1 kmph? | [
"28",
"20",
"998",
"21"
] | B | Let the distance between A and B be x km.
Total time = x/(9 + 1) + x/(9 - 1) = 4.5
=> x/10 + x/8 = 9/2 => (4x + 5x)/40 = 9/2 => x = 20 km. Answer: B |
AQUA-RAT | AQUA-RAT-37277 | In solving many kinematic problems for gears, there is a key rule to key in mind: When two gears are meshing, the tangential velocities of both gears at the point of meshing are equal.
(Also, as said in EMiller's answer, gear C seems to be the smaller orange gear, and gear D the larger yellow gear, contrary to the diagram's labelling).
This rule will help provide the equations we need to solve this problem. Note that there are 2 pairs of meshing gears (A meshes with C and B meshes with D). The points of meshing for each pair are $P$ and $Q$ respectively. Therefore, this gives us 2 equations to work with. It is then a matter of relating the tangential velocities at the point of meshing with the velocity of the centre of gear C and D. I will define the following terms:
Let $v_{a,p}, v_{c,p}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $v_{b,q}, v_{d,q}$ be the tangential velocities of gears A and C, respectively, at point P.
Let $V$ be the velocity of the centre of gear C and D.
(The following diagrams represents gears A and C, and then B and D, as circles that are in pure rolling with one another:)
According to the rule above, we get the following two equations:
$$v_{a,p}=v_{c,p} \qquad (Eq.1)$$ $$v_{b,q}=v_{d,q} \qquad (Eq.2)$$
$v_{a,p}$ and $v_{b,q}$ are simple enough to determine since the centres of gears A and B don't move:
$$v_{a,p}=\omega_a R_a = 0.4\text{m/s} \qquad v_{b,q}=\omega_b R_b = 0.4\text{m/s}$$
The following is multiple choice question (with options) to answer.
A wheel that has 6 cogs is meshed with a larger wheel of 14 cogs. If the smaller wheel has made 35 revolutions, what will be the number of revolutions made by the larger wheel? | [
"15",
"12",
"21",
"9"
] | A | Explanation:
Let the number of revolutions made by the larger wheel be x
More cogs, less revolutions (Indirect proportion)
Hence we can write as
(cogs) 6 : 14 :: x : 35
⇒6×35=14 × x
⇒x = 15
Answer: Option A |
AQUA-RAT | AQUA-RAT-37278 | Thus,
200 x 0.3 = 60 received student loans
200 x 0.4 = 80 received scholarships
We are trying to determine what percent of those surveyed said that they had received neither student loans nor scholarships.
Let’s fill all this information into a table. Note that each row sums to create a row total, and each column sums to create a column total. These totals also sum to give us the grand total, designated by 200 at the bottom right of the table.
Statement One Alone:
25 percent of those surveyed said that they had received scholarships but no loans.
Using statement one we can determine the number of students who received scholarships but no loans.
200 x 0.25 = 50 students who received scholarships but no loans.
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.
Statement Two Alone:
We are given that 50 percent of those surveyed who said they had received loans also said that they had received scholarships. From the given information we know that 60 students received loans; thus, we can determine the number of these 60 students who also received scholarships.
60 x 0.5 = 30 students who received loans who also received scholarships
We can fill the above information into our table.
Thus, the percent of those surveyed who said that they had received neither student loans nor scholarships is (90/200) x 100 = 45%. Statement two is sufficient to answer the question.
_________________
# Jeffrey Miller
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Joined: 16 May 2017
Posts: 1
Re: In a survey of 200 college graduates, 30 percent said they [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
48% of respondents to a survey consisted of those who used both M and N and those who used neither M nor N. 46% of respondents did not use M. 42% used N. What percent of respondents used only N? | [
"19.5%",
"21.5%",
"20%",
"18.5%"
] | C | From the question, we can summarize the information as follows:
N Not N
M X
Not M Ans y 46
42 100
Then we can set a system of 3 equations 3 unknowns described below:
X + Y = 48
X + Ans = 42
Y + Ans = 46
Sum up all three equations to get 2X + 2Y + 2Ans = 136
Divide both sides by 2 to get X + Y + Ans = 68
Replace X + Y with 48 to get 48 + Ans = 68
Substract 48 from both sides to get Ans = 68 - 48 = 20
The answer must be C. |
AQUA-RAT | AQUA-RAT-37279 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Zayn is a trail runner who decides to take a day off work to run up and down a local mountain. He runs uphill at an average speed of 6 miles per hour and returns along the same route at an average speed of 8 miles per hour. Of the following, which is the closest to his average speed, in miles per hour, for the trip up and down the mountain? | [
" 2.4",
" 5.8",
" 6.0",
" 6.3"
] | A | to calculate average of 2 speeds a and b when distance is constant
Formula - 2ab/(a+b)
Here Zayn's uphill speed = 6 mph
Zayn's downhill speed = 8 mph
2*6*8/(6+8) = 96/40 = 2.4
Correct answer - A |
AQUA-RAT | AQUA-RAT-37280 | # 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.
WHEN A PERSON SELLS TWO ITEMS $600 each, ONE AT A GAIN OF 20% AND OTHER AT A LOSS OF 20%.THEN THE SELLER INCURES A LOSS OR GAIN OF: | [
"1% loss",
"no loss no gain",
"4% loss",
"2% gain"
] | C | When two items are sold at the same price, one at a profit of a% and other at a loss of a% (which means the cost price of one item was lower than the selling price and the cost price of the other item was higher than the selling price), there will always be a loss of (a^2)/100 %. e.g. Here a = 20, so loss % = (20)^2/100 % = 4%.
Answer : C |
AQUA-RAT | AQUA-RAT-37281 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What is the smallest number which when diminished by 8, is divisible by 9, 6, 12 and 18? | [
"50",
"45",
"44",
"60"
] | C | Explanation:
Required number = LCM of (9, 6, 12 and 18) + 8
= 36 + 8 = 44
Answer: Option C |
AQUA-RAT | AQUA-RAT-37282 | We can take leading digits common, reducing a single digit from each number
$S_n=1((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))+2((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))...9(1\times1...\times1)+9(1\times1...2)...+9(9\times9...\times9))$
Now we can take $\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change
$S_n=(\sum_{n=1}^{9})(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$
Using the lemma we can get
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$
If you apply it again and again
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\boxed{45^n}$ Hence proved
Challenge:
• I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$
Note by Zakir Husain
3 days, 14 hours ago
The following is multiple choice question (with options) to answer.
Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H | [
"153",
"150",
"137",
"123"
] | D | let the number be x, now x is halved => x/2..the question asks what is x?
x(number) + x/2 (half of that number) = something
=> 3x/2 = something
=> x = 2 (something) / 3
so the answer choice must be divisible by 3
eliminate C, E right away
check the rest
D satisfies
3x/ 2 = 123
=> x = 82
Verify. 82 + 41 = 123
Answer is D |
AQUA-RAT | AQUA-RAT-37283 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Aarti can do a piece of work in 6 days. In how many days will she complete three time of work of same type? | [
"6 days",
"18 days",
"21 days",
"3 days"
] | B | We have the important relation, More work, More time (days)
A piece of work can be done in 6 days.
Three times of work of same type can be done in 6 x 3
= 18 days
Answer B |
AQUA-RAT | AQUA-RAT-37284 | ### 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 man invested Rs. 4940 in Rs. 10 shares quoted at Rs. 9.50. If the rate of dividend be 14%, his annual income is | [
"Rs.728",
"Rs.648",
"Rs.720",
"Rs.622"
] | A | Explanation :
Market Value of a share = Rs.9.50
Investment = Rs.4940
Number of shares = 4940/9.50 = 520
Face Value of a share = Rs.10
dividend = 14%
dividend per share = (10×14)100 = Rs. 1.4
His annual income = 520 × 1.4 = Rs.728
Answer : Option A |
AQUA-RAT | AQUA-RAT-37285 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
Twenty percent of the programmers in a startup company weigh 200 pounds or more. Twenty-five percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less. What percent of the programmers in the startup company weigh between 100 and 200 pounds? | [
"15%",
"20%",
"25%",
"60%"
] | D | Initially 80% and 20% split
80% is further divided as 25% and 75%
Q is asking about that 75%
let total be '100' then that 75% is (3/4)∗80
so, the required % is [(3/4)∗80/100]∗100 = 60%
ANSWER:D |
AQUA-RAT | AQUA-RAT-37286 | # For dinner, $n$ people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so ...
Problem:
For dinner, $$n$$ ($$n \geq 4$$) people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so that Ana and Ivan do not sit next to each other and at least one of them sits next to Mark? (Note: the round table implies seating arrangements that differ only in rotation.)
My attempt:
If I have $$n$$ people sitting around circular table, the number of different arrangements are $$(n-1)!$$.
If I have $$2$$ people Mark and Ana number of arrangements that they can sit next to each other is $$2 \cdot (n-2)!$$. So the number of arrangements that Mark sit next to Ivan is also $$2 \cdot (n-2)!$$, and sitting next to Ana also $$2 \cdot (n-2)!$$.
• yes.....it is fixs
– josf
Nov 2 '19 at 14:49
• "... seating arrangements that differ only in rotation... " are what? Considered to be the same, or different? From the context, I assume they should be considered to be the same, but the language isn't clear. Nov 11 '19 at 16:28
Method 1: Seat Mark. We will use him as our reference point.
Only Ana sits next to Mark: She can be seated in two ways, to his left or to his right. That leaves $$n - 2$$ seats. Since Ivan cannot sit next to Ana or Mark, he may be seated in $$n - 4$$ ways. The remaining $$n - 3$$ people can be seated in the remaining $$n - 3$$ seats in $$(n - 3)!$$ ways as we proceed clockwise around the table relative to Mark. Hence, there are $$2(n - 4)(n - 3)!$$ such arrangements.
The following is multiple choice question (with options) to answer.
At a dinner party, 4 people are to be seated around a circular table. Two seating arrangement are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group? | [
"5",
"10",
"6",
"32"
] | C | Has to be C.
Simple formula: There are (n - 1)! ways to arrange n distinct objects in a circle
hence (4-1)! = 3!= 6 |
AQUA-RAT | AQUA-RAT-37287 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
In an office in Singapore there are 60% female employees. 50 % of all the male employees are computer literate. If there are total 62% employees computer literate out of total 1100 employees, then the no. of female employees who are computer literate ? | [
"462",
"674",
"672",
"960"
] | A | Solution: Total employees,
= 1100
Female employees, 60% of 1100.
= (60 *1100) /100 = 660.
Then male employees,
= 440
50 % of male are computer literate,
= 220 male computer literate.
62 % of total employees are computer literate,
= (62 *1100) /100 = 682 computer literate.
Thus, Female computer literate = 682 - 220
= 462.
Answer: Option A |
AQUA-RAT | AQUA-RAT-37288 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $215 million, which represented 10 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars? | [
"17,010",
"2,150",
"1,890",
"1,701"
] | B | 10% ---- 215 millions
for 100% => ( 215 * 100% )/10% = 2150.
Option B. |
AQUA-RAT | AQUA-RAT-37289 | dominion wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
scenario 1: 1 red, 1 blue
scenario 2: 2 blue
total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
3c1*7c2+7c3/10c3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
7c2*3c1/10c3
_________________
-----------------------
tusharvk
Manager
Joined: 27 Oct 2008
Posts: 180
Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
27 Sep 2009, 01:53
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Soln: (7C2 + 7C1*3C1)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Soln: 3C2/10C2
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Soln: (7C2*3C1 + 7C3)/10C3
The following is multiple choice question (with options) to answer.
A box contains 3 blue marbles, 2 red, 6 green marbles and 2 yellow marbles. If three marbles are picked at random, what is the probability that they are all blue? | [
"1/286",
"2/455",
"1/91",
"4/455"
] | A | Given that there are three blue marbles, two red marbles, six green marbles and one yellow marbles.
Probability that all the three marbles picked at random are blue = 3C₃/13C₃ = (1 * 3 * 2 * 1)/(13 * 12 * 11) = 1/286
ANSWER:A |
AQUA-RAT | AQUA-RAT-37290 | that the distribution is normal ( outliers included ) by! Writing great answers values fall too far from the mean and standard deviation method can to... Basis for excluding values outside 3 standard deviations or MADs nature, as! A word or phrase to be possible the rest of a boxplot with! Our tips on writing great answers a bad way to detect ''.. Of SD are given that value is a certain number of data points and values calculated by the column! Values of SD are given 24 inches tall third quartile 1.5 IQR rule method that isnât affected. Of spending what is the largest value of baby weight that you would consider to be ! Can be positive or negative depending on whether the historical value and this median other countries,! Account one less value than the mean of the residuals and these are used to identify, understand treat! Why is n't standard deviation or variance with median deviation and the standard deviation of the values are â¦. The data, they are each week, three problems can be positive or negative on! Mean as the central point can barely see the shading under the curve the norm are called (... Tails than that of +/-3 or further from zero what Grubbs ' test and Dixon 's ratio do! Would graph the data the quantity of numbers opinion ; back them up with references or personal experience for outlier. Design / logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa other... Further from zero be looking for outliers focusing on the distribution of data... 'S a bit painful to decide which one, it 's a substantive one ) from the first.... ) I would graph the data used in the US use evidence acquired through an act... Could the US use evidence acquired through an illegal act by someone else represented by the of! The research, I 'd check further e. g. lengths of products.... Not be rejected just because it is unusually extreme data fit can begin “! N'T reason it out, well, are arbitrary rules any better ( 1.5 * 83 higher. Less than the majority of your flowers grew about 8-12 inches, so theyâre now about 32-36 tall... Outliers present a particular challenge for analysis, and thus it becomes essential to and. Square the resulting number identify, understand and treat these values,
The following is multiple choice question (with options) to answer.
If a sample of data has mean of 20 and SD of 3, which of the following values is more than 2 SDs from the mean? | [
"a. 12",
"b. 14.5",
"c. 17",
"d. 23.5"
] | A | Value ismore than 2SDfrom the mean means that the distance between the mean and the value must be more than 2SD=6. So the value either <20-6=14 or >20+6=26.
Answer: A. |
AQUA-RAT | AQUA-RAT-37291 | Math Help - Counting
1. Counting
The problem
How many positive integers from 100-999 are divisible by 7.
I want to confirm if my approach is the best approach in finding the answer.
The smallest integer that can be divided by 7 is
105+112+119+126+133+140+147+154+161+168+175+182+18 9+196
there are 14 numbers divisible by 7 from the range 100-200 so from 100-900 would be 9*14 = 126 numbers there is still 99 numbers unaccounted for
but the answer in my textbook states 128 numbers in total. How is that possible?
2. Originally Posted by ff4930
The problem
How many positive integers from 100-999 are divisible by 7.
I want to confirm if my approach is the best approach in finding the answer.
The smallest integer that can be divided by 7 is
105+112+119+126+133+140+147+154+161+168+175+182+18 9+196
there are 14 numbers divisible by 7 from the range 100-200 so from 100-900 would be 9*14 = 126 numbers there is still 99 numbers unaccounted for
but the answer in my textbook states 128 numbers in total. How is that possible?
The first number in the range divisible by 7 is 105, then every 7th number is
divisible by 7, and so there are a total of floor[(999-105)/7] +1 =128 such
numbers.
RonL
3. Here is another way using the floor function.
$\left\lfloor {\frac{{999}}{7}} \right\rfloor - \left\lfloor {\frac{{99}}{7}} \right\rfloor = 128$
4. Thank You guys for the reply, I was doing it the long way =/.
but one thing can I do [999-100]/7 I still got the same answer if I rounded down
There are 1 more thing that is confusing me.
same problem as before but finding the integers that are divisible by 3 or 4
I found all the numbers divisble by 3 which is 300 and divisble by 4 which is 225, but the answer is 450.
The following is multiple choice question (with options) to answer.
The lowest number which should be added to 11390 so that the sum is exactly divisible by 7, 9, 10 and 12 is: | [
"1010",
"2210",
"1250",
"1210"
] | D | L.C.M. of 7, 9, 10 and 12 = 1260.
On dividing 11390 by 1260, the remainder is 50.
Number to be added = (1260 - 50) = 1210. Answer: Option 'D' |
AQUA-RAT | AQUA-RAT-37292 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A group consists of 8 men, 6 women and 4 children. In how many ways can 1 man , 2 woman and 2 child selected from the given group? | [
"780",
"750",
"720",
"620"
] | C | One man, two women and two child can be selected in 8C1 * 6C2 * 4C2 ways
8*6*5/2*1*4*3/2*1
= 720 ways
Answer : C |
AQUA-RAT | AQUA-RAT-37293 | substitute our 3rd conclusion:
$1 = x \oplus y \oplus w = 0 \oplus y \oplus 0 = 0 \oplus y = y \Longrightarrow y = 1$
-
The following is multiple choice question (with options) to answer.
For all real numbers v, an operation is defined by the equation v* = v - v/3. If (v*)* = 16, then v= | [
"30",
"33",
"36",
"39"
] | C | (v*)* = (v - v/3) - (v - v/3) / 3
16 = 2v/3 - 2v/9 = 4v/9
v = 36
The answer is C. |
AQUA-RAT | AQUA-RAT-37294 | 267 views
In how many rearrangements of the letters of the word SCINTILLATING will no two 'I's appear together?
1. ${ }^{11} \mathrm{C}_{3} * 13!$
2. $\frac{10 !}{2 ! * 2 ! * 2 !}$
3. ${ }^{11} \mathrm{C}_{3} * 3 ! * 10 !$
4. ${ }^{11} \mathrm{C}_{3} \frac{10 !}{2 ! * 2 ! * 2 !}$
Step $1$ of solving this question: Segregate the word into 'I's and letters that are not 'I's:
SCINTILLATING is a $13$ letter word comprising $3$ Is, $2$ Ls, $2 \mathrm{Ns}, 2$ Ts and once each of $\mathrm{S}, \mathrm{C}, \mathrm{A}$, and $\mathrm{G}$.
The condition given is that no two 'I's should appear together. The same condition can be reworded as "there should be at least one letter in between any two 'I's The way to achieve this condition is to place the remaining $10$ letters in such a way that there is exactly one gap between any two of these letters.
By this action we are ensuring that not more than one 'I' can be placed in the gap - making sure that no two 'I's appear together.
It will appear like this: $$_{-}\mathrm{S}_{-} \mathrm{C}_{-}\mathrm{N}_{-}\mathrm{T}_{-}\mathrm{L}_{-} \mathrm{L}_{-} \mathrm{A}_{-} \mathrm{T}_{-}\mathrm{N}_{-} \mathrm{G}_{-}$$ There are a total of $11$ places where the 'I's can be placed including the one before the first letter and the one after the last letter - and no two 'I's will appear together.
The following is multiple choice question (with options) to answer.
how many such letters pairs r in word SMUGGLER having same no. of letters left between them in the word as they have in the series | [
"1",
"2",
"3",
"4"
] | B | 2 letter pair.S-U,G-E.
ANSWER:B |
AQUA-RAT | AQUA-RAT-37295 | Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
### Show Tags
16 Apr 2017, 07:45
1
KUDOS
Top Contributor
1
This post was
BOOKMARKED
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Another approach.
Take the task of arranging the 5 letters and break it into stages.
Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways
IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.
Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)
[Reveal] Spoiler:
D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.
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The following is multiple choice question (with options) to answer.
How many words can be formed by using all letters of the word “MANFUL” | [
"720",
"240",
"750",
"860"
] | A | The word MANFUL contains 6 different letters.
Required number of words = 6p6 = 6! = (5x4x3x2x1) = 720.
Answer A. |
AQUA-RAT | AQUA-RAT-37296 | # Sum and Product of n-positive integers
If I have $n$-positive integers, and I compute their sum and product, is there any different group of $n$-positive integers that will have the same sum and product?
For example, if $a,...,z$ denote 26 positive integers, and we define:
\begin{align} a+b+c+d+....+z &= \text{Sum} \\ a \cdot b \cdot c \cdot d \cdot .... \cdot z &= \text{Product} \end{align}
Is there any way I can get the same Sum and Product from a different group of 26 (in this example) positive integers?
EDIT:
A friend of mine pointed out that knowing that we have a group of 3 that works, we can show that it works for all positive groups of $n$ integers.
For Example: $\{3,3,10 \}$ and $\{2,5,9 \}$ both yield Sum $=16$ and Product $=90$.
Now we can just continually add a number (let's say 1) as the next integer to get multiple solutions for $n =4,5,6,...$.
Explicitly, $\{ 3,3,10,1 \}$ and $\{ 2,5,9,1 \}$ both give Sum$=17$ and Product$=90$.
The following is multiple choice question (with options) to answer.
If the sum of two positive integers is 20 and the difference of their squares is 40, what is the product of the two integers? | [
"108",
"99",
"128",
"135"
] | B | Let the 2 positive numbers x and y
x+ y = 20 -- 1
x^2 - y^2 = 40
=> (x+y)(x-y)=40 -- 2
Using equation 1 in 2 , we get
=> x-y = 2 -- 3
Solving equation 1 and 3 , we get
x= 11
y= 9
Product = 11*9 = 99
Answer B |
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