source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36697 | mass, newtonian-gravity, statics
Title: How do you weigh a box on a scale whose limit is too low? As you will see I know nothing about physics and after being asked to solve a physics problem in a recent interview wanted to ask it of professionals and see what the response would be:
I have a set of domestic scales (actually just one scale) that weigh up to the maximum weight of 5kg and a large package that weighs more than 5kgs but less than 10kgs. How can I tell the exact weight using the inadequate scales.
The package is a long item akin to a steal girder but not in weight obviously.
I know that stack exchange has rules about this type of question so please treat it as light entertainment from someone in awe of your intellect. You put the package horizontally across three scales and add up the weight you see. If the center scale registers more than the limit, move the box, or put some pieces of paper under the scales which are registering a small weight to redistribute the weight.
A more physics-y question would be how to determine the weight when you have only one scale. This can be done by tilting the package at a small angle, by placing one end on a scale (perhaps propped up on something light, like your shoe, so that the weight momentum flow goes through the scale to the floor). To weigh this way requires knowing where the CM of the box is, and this can be determined by balancing it precisely on the edge of the scale (or on your finger, or on something else narrow).
Then the weight registered by the scale (minus the weight of the shoe) is the ratio of the length of the box to the distance from the end which is on the floor to the center of mass. If this is still too heavy, you can tilt the box up to 45 degrees, which will cut down the weight registered by the scale further by a factor of .707, which can be undone by multiplying by 1.414.
The following is multiple choice question (with options) to answer.
You have been given a physical balance and 7 weights of 45, 50, 53, 46, 52, 46 and 80 kgs. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 188 kgs. | [
"183",
"185",
"182",
"184"
] | B | 80 + 52 + 53 = 185
Answer: B |
AQUA-RAT | AQUA-RAT-36698 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A ? | [
"240",
"388",
"208",
"112"
] | A | The ratio of A & B investments = (3x8 + 2x4):(4x8 + 5x4)
=> 8:13
=> 8/21 x 630 = 240.
Answer: A |
AQUA-RAT | AQUA-RAT-36699 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many three letter words are formed using the letters of the word TIME? | [
"20",
"24",
"26",
"28"
] | B | Explanation:
The number of letters in the given word is four.
The number of three letter words that can be formed using these four letters is ⁴P₃ = 4 * 3 * 2 = 24.
Answer: Option B |
AQUA-RAT | AQUA-RAT-36700 | that can be factored with ease!
Use the fact that $73 \times 137=10001 = 10^4+1$.
Now, mark off the number in groups of four digits starting from the right, and add the four-digit groups together with alternating signs.
Applying the above rule, $1000027$ in groups of $4$ is $\underbrace{0100}$ $\underbrace{0027}$. Adding the groups with alternate signs gives $73$. Therefore, $1000027$ is divisble by $73$ and gives $13699$ as quotient.
For $13699$ apply the divisbility test by $7$ by marking of the digits in groups of $3s$ by starting from the right and adding together with alternate signs. Therefore, adding the groups $\underbrace{013}$ $\underbrace{699}$ with alternate signs gives $686$ which is divsible by $7$ and hence $13699$ is disvisble by $7$
$13699$ when divided by $7$ gives $1957$.
Now, $1957$ is $1900+57$ and hence is divisible by $19$ giving the quotient as $103$.
Combining them all gives $1000027 = 7\times19\times73\times103$
To begin with note that
$1000027 = 100^3 + 3^3 = (100+3)(100^2−3⋅100+32) = 103 \cdot 9709 = 103 \cdot (1001 \cdot 9 + 700) = 103 \cdot 7 \cdot (13 \cdot 11 \cdot 9+100) = 103 \cdot 7 \cdot 1387$
where we have used the well-known fact that $1001 = 7 \cdot 11 \cdot 13$
The following is multiple choice question (with options) to answer.
How many of the following numbers are divisible by 132?
264,396,462,792,968,2178,5184,6336 | [
"4",
"5",
"6",
"7"
] | A | Explanation:
132 = 4 x 3 x 11, So if the number is divisible by all three numbers 4,3 and 11,then the number is divisible by 132 also.
264 => 4,3,11(/)
396 => 4,3,11(/)
462 => 11,3
792 => 4,3,11(/)
968 => 11,4
2178 => 11,3
5184 => 3,4
6336 => 4,3,11(/)
Required number of numbers=4.
Answer: A) 4 |
AQUA-RAT | AQUA-RAT-36701 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
If the cost price of 12 pens is equal to the selling price of 9 pens, the gain percent is: | [
"80%",
"90%",
"33%",
"40%"
] | C | Explanation:
Let C.P. of each pen be Re. 1.
Then, C.P. of 9 pens = Rs. 9; S.P. of 9 pens = Rs. 12.
Gain % = 3/9 * 100 = 33%
Answer:C |
AQUA-RAT | AQUA-RAT-36702 | • In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52
The following is multiple choice question (with options) to answer.
The sum of the heights of two high-rises is x feet. If the first high rise is 36 feet taller than the second, how tall will the second high rise be after they add an antenna with a height of z feet to the top? | [
"(x+z)/2 + 36",
"2x−(37+z)",
"(x−36)/2 + z",
"x/2 - 37 + z"
] | C | I will note h1 the height of high-rise 1 and h2 the height of high-rise 2. SO:
h1 + h2 = x
and h1 = h2 + 36 =>
Q: h2 + z = ?
h2 + h2 + 36 = x => 2h2 = x-376=? h2 = (x-36)/2
=> h2 + z = (x-36)/2 + z, CORRECT ANSWER C |
AQUA-RAT | AQUA-RAT-36703 | To fill it completely we need 2 (jugs needed) + 5 (remaining jugs) = 7 (one carton's capacity)
Hence (B) 2 jugs
_________________
"Do not watch clock; Do what it does. KEEP GOING."
Intern
Joined: 02 Oct 2016
Posts: 23
Re: A worker carries jugs of liquid soap from a production line to a packi [#permalink]
### Show Tags
08 Apr 2018, 06:05
1
Since the worker, carried jugs of liquid soap(4 jugs/trip)
In 17 trips, he would carry 68 jugs of liquid soap.
Since he can fill 7 jugs in a carton, he will need to add 2 more jugs
such that he has 70 jugs and he can fill the jugs in 10 cartons.
Hence he needs 2 jugs(Option B) such that the last partially filled carton is full.
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Joined: 02 Mar 2017
Posts: 8
Location: United States
Schools: Cox '19, Jindal '19
GMAT 1: 660 Q55 V50
GPA: 3.65
A worker carries jugs of liquid soap from a production line to a packi [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
The cost of 2 liters of lube oil is $7. What is the cost of 5 liters of lube oil? | [
"17.5",
"20.5",
"24.0",
"28.5"
] | A | The cost of one liter of oil=7/2=3.5
The cost of five liters of oil = 3.5x5= $ 17.5
Answer: A |
AQUA-RAT | AQUA-RAT-36704 | algorithms, integers
If the prime factorization of $n+1$ is $p_1^{k_1},\ldots,p_t^{k_t}$, then $f(n)$ depends only on $(k_1,\ldots,k_t)$, and in fact only on the sorted version of this vector. Every number below $10^9$ has at most $29$ prime factors (with repetition), and since $p(29)=4565$, it seems feasible to compute $f$ (or rather $g$) for all of them, recursively. This solution could be faster if there were many different inputs; as it is, there are at most $10$.
It is also possible that this function, mapping partitions to the corresponding $g$, has an explicit analytic form. For example, $g(p^k) = 2^{k-1}$, $g(p_1\ldots p_t)$ is given by A000670, and $g(p_1^2 p_2\ldots p_t)$ is given by A005649 or A172109.
The following is multiple choice question (with options) to answer.
What is the greatest prime factor of 2^8 - 1? | [
"11",
"13",
"17",
"19"
] | C | 2^8-1 = (2^4-1)(2^4+1) = 15*17
The answer is C. |
AQUA-RAT | AQUA-RAT-36705 | (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.
There were 35 students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.84 per day while the average expenditure per head diminished by Re 1. What was the original expenditure of the mess? | [
"Rs 450",
"Rs 920",
"Rs 550",
"Rs.630"
] | D | Explanation :
Let the original average expenditure be Rs.x then,
42(x - 1) - 35x = 84 ? 7x = 126 ? x = 18
Therefore original expenditure = Rs.(35 18)=Rs.630.
Answer : D |
AQUA-RAT | AQUA-RAT-36706 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
The sum of the first 50 positive even integers is 2550. What is the sum T of even integers from 102 to 200 inclusive? | [
"5100",
"7550",
"10100",
"15500"
] | B | My solution is:
First 50 even integers:
2
4
6
8
<...>
Integers from 102 to 200
102
104
106
108
<...>
We notice that each integer from the second set is 100 more than the respective integer in the first set. Since we have 50 even integers from 102 to 200, then:
T=2550+(100*50)=7550.B |
AQUA-RAT | AQUA-RAT-36707 | # 2012 AMC 10A Problems/Problem 22
## Problem
The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$?
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$
## Solution 1
The sum of the first $m$ odd integers is given by $m^2$. The sum of the first $n$ even integers is given by $n(n+1)$.
Thus, $m^2 = n^2 + n + 212$. Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$.
Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$. Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.
Let $x = \sqrt{4m^2 - 847}$; note that this means $n = \frac{-1 + x}{2}$. It can be rewritten as $x^2 = 4m^2 - 847$, so $4m^2 - x^2 = 847$. Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\cdot11^2$.
The following is multiple choice question (with options) to answer.
What is the sum of the odd integers from 35 to 65, inclusive? | [
"800",
"550",
"555",
"600"
] | A | The mean is 50.
Sum=Mean(# of elements)
There are 16 odd numbers between 35-65 inclusive. 16*50=800
A |
AQUA-RAT | AQUA-RAT-36708 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
10% people of a village in Sri Lanka died by bombardment, 15% of the remainder left the village on account of fear. If now the population is reduced to 3213, how much was it in the beginning? | [
"A)3800",
"B)4200",
"C)4400",
"D)4500"
] | B | X * (90/100) * (85/100) = 3213
X = 4200
ANSWER:B |
AQUA-RAT | AQUA-RAT-36709 | newtonian-mechanics, momentum, acceleration, collision
Title: A problem regarding a fiction Recently I watched a fictional(I guess) video: a man is crossing the road while a truck accelerates towards him and a superhero flashes and saves his life by taking him out of the road. He was very fast, therefore only a flash could be seen. It seems like a silly incident though. After some time the question that occurred in my mind was 'if this happened in reality, could that man survive?'
This is my reasoning:
If the man had to collide with the truck it would create serious damages and this has been explained scientifically in many topics such as energy transferred to the body, force emitted, and so on. Also, I found that the acceleration of the vehicle performs a lower impact on the pedestrian. Nevertheless, everyone knows that more harm is done by a vehicle that goes with $60 \;\text{km/h}$ than $5 \;\text{km/h}$ at a collision. But in this scenario, the speed of the hero is exaggeratedly high(Find the video below). The speed of the truck is negligible in comparison. Thus the impulse on the man is massive when the hero catches him. Can a person endure such a great impulse? I heard that such a great change in momentum will disturb fluids in the person's body and feels extremely uncomfortable. And also can a person tolerate that acceleration? Thus the ridiculous thought that came to my mind was that the damage will be minor when the person had to be hit by the vehicle than is carried by the hero.
This seems to be a silly problem, but I am asking whether this could happen in the real world.
EDIT: Look for video here:- https://youtu.be/KJqhR2YSUXw
The following is multiple choice question (with options) to answer.
A flamboyant comic book villain has abducted a bystander! Captain Valiant is alerted to the crime and reaches the start of the villain's trail 30 minutes later. The villain is in a getaway car going 40 kmph, and Captain Valiant can fly at 50 kmph. How long will the bystander have to wait for rescue? | [
"1 hour",
"1 1/2 hours",
"2 hours",
"2 1/2 hours"
] | D | Distance covered by the villain in 1/2 hour = 20 km
Now, 10 km is compensated in 1 hour
Therefore 20 km will be compensated in 2 hours
Added to the half hour head start, the bystander has been kidnapped for 2 1/2 hours.
Correct Answer: D |
AQUA-RAT | AQUA-RAT-36710 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
Find the middle one when The sum of three consecutive even numbers is 36? | [
"11",
"12",
"14",
"16"
] | B | 3 consecutive numbers can be a - 1, a, a + 1
So sum of numbers = 3a = 36.
Hence a = 12.
B |
AQUA-RAT | AQUA-RAT-36711 | Equilateral Triangle Calculator. Find the Length of the Side and Perimeter of an Equilateral Triangle Whose Height is √ 3 Cm. The height of an equilateral triangle of side ' a ' is given by. The special right triangle gives side ratios of , , and . A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe Thus these are properties that are unique to equilateral triangles, and knowing that any one of them is true directly implies that we have an equilateral triangle. Originally Answered: The side length of an equilateral triangle is 6 cm. Solution: According to height of equilateral triangle formula 4.1. s = … To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles. improve our educational resources. Varsity Tutors LLC An equilateral triangle also has three equal 60 degrees angles. Area of equilateral triangle = 4 3 a 2 Also, area = 2 1 × base × height = 2 1 a × height … OwlCalculator.com. Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. How to find the height of an equilateral triangle. An equilateral … The height of an equilateral triangle having each side 12cm, is (a) 6√2 cm (b) 6√3m (c) 3√6m (d) 6√6m. Therefore, the value of X will be twice the value of the height: What is the height of an equilateral triangle with a side length of 8 in? What is the triangle's height ? ... Now apply the Pythagorean theorem to get the height (h) or the length of the line you see in red. = 5.2 units approx. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Take √(3) = 1.732 And so to do it, we remember that the area of a triangle is equal to 1/2 base times height. A triangle ABC that has the sides a, b, c,
The following is multiple choice question (with options) to answer.
What is the are of an equilateral triangle of side 18 cm? | [
"66√3 cm2",
"74√3 cm2",
"81√3 cm2",
"64√5 cm2"
] | C | Area of an equilateral triangle = √3/4 S2
If S = 18, Area of triangle = √3/4 * 18 * 18 = 81√3 cm2;
Answer:C |
AQUA-RAT | AQUA-RAT-36712 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invested Rs.6300, Rs.4200 and Rs.10500 respectively, in a partnership business. Find the share of A in profit of Rs.12100 after a year? | [
"3630",
"2899",
"277",
"2870"
] | A | 6300:4200:10500
3:2:5
3/10 * 12100 = 3630
Answer:A |
AQUA-RAT | AQUA-RAT-36713 | +0
The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.
0
401
4
The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.
Guest May 27, 2015
#2
+18829
+13
The lines 3x+y=1 and 5x-y=15 intersect at the center of circle O. If the circle is tangent to the y-axis, find the equation of the circle.
The following is multiple choice question (with options) to answer.
Find the value of X, Y by solving the below equations
X + Y = 15
X - Y = 5 | [
"10, 5",
"8, 7",
"9, 6",
"11, 4"
] | A | X + Y = 15 ---(I)
X - Y = 5 -----(II) by adding (I) and (II)
------------
2X = 20 ==> X = 20/2 = 10
By Replacing the value of X in (I) we get 10 + Y = 15 ==>
Y = 15 -10 = 5.
So, X = 10, Y = 5
Answer A) 10, 5 |
AQUA-RAT | AQUA-RAT-36714 | given that $$f(x)$$ has local maximum at $$x=1$$ and local minimum at $$x=0$$ and $$2$$i.e $$f^{\prime}(x)=0$$ when $$x=0,1,2$$$$\Rightarrow$$ $$f^{\prime}(1)=8+5b+6c=0$$ and $$f^{\prime}(2)=4+5b+12c=0$$solving above two equations for $$b$$ and $$c$$ gives$$b=\displaystyle\dfrac{-12}{5}$$ and $$c=\displaystyle\dfrac{2}{3}$$$$\therefore$$ $$\displaystyle f(x)=2x^4-\dfrac{12}{5}x^5+\dfrac{2}{3}x^6$$$$\displaystyle 5f(3) = 5(162-\dfrac{12}{5}3^5+\dfrac{2}{3}3^6) = 324$$Hence, $$5f(3) = 324$$Mathematics
The following is multiple choice question (with options) to answer.
Find the maximum value of the function f(x)=x−5 if x is a number between -5 and 13. | [
"6",
"7",
"8",
"9"
] | C | Solution:
Since f(x) is a linear function whose slope is 1, a positive number, it is strictly increasing for all x. Therefore its maximal value is reached for the largest value of x, x=13 and f(x)=13-5=8.
Answer C |
AQUA-RAT | AQUA-RAT-36715 | Thanks once again for the perfectly clear answer. You have understood the problem well. I think that I can work things out myself now.
• All times are GMT -7. The time now is 09:21 AM.
The following is multiple choice question (with options) to answer.
A man leaves a point P at 6 a.m. and reaches the point Q at 10 a.m. another man leaves the point give at 8 a.m. and reaches the point P at 12 noon. At what time do they meet? | [
"6",
"5",
"4",
"9"
] | C | 9 a.m.
Answer: C |
AQUA-RAT | AQUA-RAT-36716 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
Tough and Tricky questions: Percents.
Over the course of a year, a certain microbrewery increased its beer output by 60 percent. At the same time, it decreased its total working hours by 30 percent. By what percent did this factory increase its output per hour? | [
"228.5",
"228",
"229.5",
"220"
] | B | Lets assume the initial production was 100 litres of beer for 100 hr.
With the 60% increase the total amount of beer production will be 160 litres and with 30 % decrease in total hours will be reduced to 70 hr.
100hr ----> 100 lts
1hr -----> 1 lts
70hr -----> 160 lts
1hr -----> 2.28 lts
Total Increase in production for 1 hr = 228%
Answer B |
AQUA-RAT | AQUA-RAT-36717 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A dishonest dealer claims to sell a product at its cost price. He uses a counterfeit weight which is 20% less than the real weight. Further greed overtook him and he added 25% impurities to the product. Find the net profit percentage of the dealer? | [
"44%",
"40%",
"50%",
"56.25%"
] | D | The dealer uses weight which is 20% less than the real weight. or (1- 1/5) or 4/5 of real weight.
It means that he is selling $4 worth of product for $5.
The dealer then further added 25% impurities to the product.
It means that he is selling $5 worth of product for $6.25.
So his profit is $6.25-$4 = $2.25
and his profit percent is
(2.25/4)*100 = 56.25%
Answer:- D |
AQUA-RAT | AQUA-RAT-36718 | # 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.
The C.P of 10 pens is equal to the S.P of 12 pens. Find his gain % or loss%? | [
"16 2/7%",
"16 2/6%",
"16 2/3%",
"16 7/3%"
] | C | 10 CP = 12 SP
12 --- 2 CP loss
100 --- ? => 16 2/3%
Answer: C |
AQUA-RAT | AQUA-RAT-36719 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
At 15:00 there were 20 students in the computer lab. At 15:03 and every three minutes after that, 4 students entered the lab. If at 15:10 and every ten minutes after that 8 students left the lab, how many students were in the computer lab at 15:44 ? | [
"7",
"14",
"25",
"27"
] | D | Initial no of students + 3 * (1 + No of possible 3 minute intervals between 15:03 and 15:44) -
8 *(1 + No of possible 10 minute intervals between 15:10 and 15:44)
20 + 3*14 -8 * 4 = 27
D |
AQUA-RAT | AQUA-RAT-36720 | # What is the maximum number of circles in proximity to a given point.
The title maybe a bit obscure so I'll try my best to explain the problem here.
Below is the Picture that I'll take help from.
Say I have a circle A of Radius R now if I take a point C inside this circle then obviously the center A will lie inside the circle originating from C with same radius R. Now I want to know, what is the maximum number of centers of circles(each of radius R) that can be there in the circle originating from C.
Instead of point C, no other center could be inside any other circle. To extend the example, if I have the circle with center B(Radius R), with center just outside the circle A, this center will also lie in the circle originating from C. Hence so long, I have the number to be Two, i.e. in circle from C, both centers A and B will lie.
So what is the maximum number?
The following is multiple choice question (with options) to answer.
Each of the 43 points is placed either inside or on the surface of a perfect sphere. If 16% or fewer of the points touch the surface, what is the maximum number of segments which, if connected from those points to form chords, could be the diameter of the sphere? | [
"3",
"11",
"13",
"23"
] | A | Maximum number of points on the surface is 16%*43 = 6.88 ... or 6 since it has to be an integer
Now note that if two points form a diameter, they cannot be part of any other diameter.
So in the best case we can pair up the points
We have 6 points, so at best we can form 3 pairs (6).
So, answer is (A) |
AQUA-RAT | AQUA-RAT-36721 | BUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point
$$\\=4.4615\times 10^{-1} \times 10^{15}\\\\ =4.4615\times 10^{14}\qquad \mbox{Correct to 4 significant figures}\\\\$$
So does that all make sense now?
Melody Nov 21, 2014
#2
+91465
0
That was a good question anon
Melody Nov 21, 2014
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The following is multiple choice question (with options) to answer.
On a certain scale of intensity, each increment of 10 in magnitude represents a tenfold increase in intensity. On this scale, an intensity corresponding to a magnitude of 160 is how many times an intensity corresponding to a magnitude of 125? | [
"40",
"100",
"400",
"1000"
] | D | Increase of 40 in magnitude corresponds to 10^4 increase in intensity:
If intensity for 125 is x then for 135 it'll be 10*x, for 145 it'll be 10*10*x=10^2*x, for 155 it'll be 10*10*10*x=10^3*x and for 160 it'll be 10*10*10*10*x=10^3*x.
Answer: D. |
AQUA-RAT | AQUA-RAT-36722 | 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 together, printer M and printer N would finish the task in 24 minutes. Printer M alone would finish the task in 60 minutes. How many pages does the task contain if printer N prints 5 pages a minute more than printer M? | [
"400",
"450",
"600",
"1200"
] | C | 24*M + 24*N = x pages
in 24 mins printer M will print = 24/60 * x pages = 2/5*x pages
thus in 24 mins printer printer N will print x - 2/5*x = 3/5*x pages
Also it is given that printer N prints 5 more pages per min that printer M. In 24 mins printer N will print 120 more pages than printer M
thus 3/5*x - 2/5*x = 120 => x = 600 pages
C) |
AQUA-RAT | AQUA-RAT-36723 | S = 1-1+1-1+1-1....
Notice that if I add another copy of the sum 1-1+1-1+1-1.... at the beginning of the
sum, I get exactly the same sum. It is still 1-1+1-1+1-1....
Thus,
S + S = S
2S = S
And since you "proved" S = ½
It must be the case that
2S = ½
Thus, S = ¼
And, of course, I can add as many sets of S to each other as a like and they will still be the same sum, they will still be 1-1+1-1+1-1.... So let us say that I added S to itself 999 times, giving me 1000S = S.
Since S = ½ and S = ¼
And since S = 1000S
Then,
1000S = ½. Thus, S = 1/2000
And 1000S = ¼. Thus S =1/4000
This is obviously absurd and self-contradictory. Thus, we know that the math is wrong. It Is NOT the case that S=½ nor any of the other values we could come up with. It is not the case that the associative property holds for this particular series. And, for that matter, it does not hold that S + S = 2S.
Since it is possible to have multiple contradictory sums for S, it must be the case that S fails to exist.
Thus when S fails to exist, it is possible to get various nonsensical and contradictory solutions.
Basic mathematical operations all require that S exists, if S does not exist the operations can still produce "answers" but they will be nonsense.
The following is multiple choice question (with options) to answer.
If 'a' and 'b' are non-zero numbers such that their sum is ten times the product, what is the value of 1/a + 1/b? | [
"2",
"5",
"10",
"15"
] | C | a+b = 10ab
1/a + 1/b = (b + a)/(ab) = 10.
The answer is C. |
AQUA-RAT | AQUA-RAT-36724 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two friends plan to walk along a 22-km trail, starting at opposite ends of the trail at the same time. If Friend P's rate is 20% faster than Friend Q's, how many kilometers will Friend P have walked when they pass each other? | [
"10",
"11",
"12",
"13"
] | C | If Q complete x kilometers, then P completes 1.2x kilometers.
x + 1.2x = 22
2.2x=22
x = 10
Then P will have have walked 1.2*10=12 km.
The answer is C. |
AQUA-RAT | AQUA-RAT-36725 | 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.
1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of the girls pass. The percentage of the total who failed is ? | [
"63.3%",
"52.4%",
"81.2%",
"75.4%"
] | A | Total number of students = 1100+700 = 1800
Number of students passed = (42% of 1100+30% of 700) = 462+210 = 672
number of failures = 1128*100/1800 = 190/3%= 63.3%
Answer is A |
AQUA-RAT | AQUA-RAT-36726 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 10 numbers is 23. If each number is increased by 4, what will the new average be? | [
"36",
"27",
"72",
"29"
] | B | Sum of the 10 numbers = 230
If each number is increased by 4, the total increase =
4 * 10 = 40
The new sum = 230 + 40 = 270 The new average = 270/10
= 27.
Answer:B |
AQUA-RAT | AQUA-RAT-36727 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 18, 16, 12, 24, 11, 34, 46 | [
"16",
"46",
"16",
"11"
] | D | Explanation :
11 is the only odd number in the given series. Answer : Option D |
AQUA-RAT | AQUA-RAT-36728 | population-biology, population-dynamics
Title: Why will world population keep growing if all women have only 2.1 children Why is it that, even if we were going to immediately agree that every women will not have more than 2.1 children on average, the world population would continue to grow for another 60 years? I've learned that a TFR of 2.1 means the population is at 0 growth rate. I guessed the answer was the population-lag effect, but since we are assuming that ALL women immediately have only 2.1 children on average, wouldn't there be a 0 population growth immediately and not after 60 years? Thanks in advance! As you mention, the population-lag effect is responsible for this. From The Wikipedia article on TFR: http://en.wikipedia.org/wiki/Total_fertility_rate#Population-lag_effect
A population that has recently dropped below replacement-level
fertility will continue to grow, because the recent high fertility
produced large numbers of young couples who would now be in their
childbearing years.
Thus, even if the current TFR implies long-term stability, the recent history of TFR values will continue to affect birth rates, and thus population growth/decrease, in the future.
Imagine that you have a stable population, with static birth and death rates. Now imagine that the birth rate during a single year for some reason doubles before dropping back to normal the next year, thus transiently increasing TFR. Now, it is easy to see that when the newly born, larger-than-normal generation reaches reproductive age, the population will increase again because the birth rate will increase in proportion with the size of the generation, while the death rate is unaffected until the enlarged generation reaches old age and starts dying off. Later, the population will decrease to its long-term stable level, which will be larger than before the TFR spike.
The following is multiple choice question (with options) to answer.
Out of a world population of approximately 6.6 billion, 1.2 billion people live in the richer countries of Europe, North America, Japan and Oceania and is growing at the rate of 0.25% per year, while the other 5.4 billion people live in the lees developed countries and is growing at the rate of 1.5%. What will be the world population in 5 years if we assume that these rates of increase will stay constant for the next 5 years. (round answer to 3 significant digits) | [
"7.02 billion.",
"7.03 billion.",
"7.04 billion.",
"7.05 billion."
] | B | Solution to Problem 12:
Let us first calculate the population PR in 5 years in the richer countries
PR = (1.2 + 0.25% * 1.2) = 1.2(1 + 0.25%) after one year
PR = 1.2(1 + 0.25%) + 0.25% * 1.2(1 + 0.25%)
= 1.2(1 + 0.25%) 2after two years
Continue with the above and after 5 years, PR will be
PR = 1.2(1 + 0.25%) 5 after 5 years
Similar calculations can be used to find the population PL in less developed countries after 5 years.
PL = 5.4(1 + 1.5%) 5 after 5 years
The world population P after 5 years will be
P = PR + PL = 1.2(1 + 0.25%) 5 + 5.4(1 + 1.5%) 5 = 7.03 billion.
Answer B |
AQUA-RAT | AQUA-RAT-36729 | ### 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.
A right triangle AWC has to be constructed in the xy-plane so that the right angle is at A and AW is parallel to x axis. The coordinates of A, W and C are integers and satisfy the inequalities -1 ≤ x ≤ 7 and 1 ≤ y ≤ 7. How many different triangles can be constructed with these properties? | [
"63",
"336",
"567",
"3024"
] | D | All the cordinates are integer hence,
possible AW values are
{-1,0}, {-1,1} ....{-1,7}: 8 ways
....
{7,-1}, {7,0} ......{7,6}: 8 ways
9*8 ways = 72
for AC values can be
{1,2}, {1,3} .... {1,7} : 6ways
......
{7,1}, {7,2} ..... {7,6} : 6 ways
7*6 = 42
Total = 72 * 42 = 3024 hence D |
AQUA-RAT | AQUA-RAT-36730 | ### Show Tags
19 Aug 2015, 01:34
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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]
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19 Aug 2015, 01:55
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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.
After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago.What is the difference between the ages of the replaced and the new member? | [
"2 years",
"4 years",
"8 years",
"15 years"
] | D | the age 5 members will be 13,14,15,16 and 17
3 years before there age was 10,11,12,13 and 14 hence the average is 12(10+11+12+13+14/5=60/5=12).
now if the oldest man 17 is replaced by 2 the the average is become same as three years ago the are .
as (13+14+15+16+2)/5 = 60/5=12.
hence the difference between the ages of the replaced and the new member is 17-2=15.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36731 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
70 students are required to paint a picture. 52 use green color and some children use red, 38 students use both the colors. How many students use red color? | [
"53",
"54",
"55",
"56"
] | D | 70 = red +green -both
Red = 70-green+both = 70-52+38=56
ANSWER:D |
AQUA-RAT | AQUA-RAT-36732 | # Math Help - Prob. Question
1. ## Prob. Question
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
2. Originally Posted by I-Think
Question
5 boys (call them A,B,C,D,E) place their books in a bag. The 5 books are are drawn out in random order and given back to the 5 boys. What is the probability that exactly 3 boys will receive their original book?
Help wanted please. The only way I could think of to solve this problem is to list the number of ways 3 boys could receive their original book and sum the individual probabilities of each of them occurring. This is tedious and can't be sure if I have them all.
If someone could list an alternative method, it would be greatly appreciated.
If exactly three boys have their original books then the other two books are switched. Do you see how this helps?
3. Hello, I-Think!
Five boys place their books in a bag.
The books are are drawn out in random order and given back to the boys.
What is the probability that exactly 3 boys will receive their original book?
The only way I could think of to solve this problem is to
list the number of ways 3 boys could receive their original books.
You don't have to list them . . .
Select three of the five boys who will get their own books.
. . There are: . $_5C_3 \:=\:{5\choose3} \:=\:10$ ways.
The other two boys have simply switched books: . $1$ way.
Hence, there are: . $10\cdot1 \:=\:10$ ways for 3 boys to get their own books.
There are: . $5! \,=\,120$ ways to return the books.
The following is multiple choice question (with options) to answer.
A bookshop had science and art books in the ratio of 2:5. By the end of the week, 20% of both type of books were sold and 2240 books were unsold. How many science books were there at the starting? | [
"500",
"560",
"670",
"800"
] | D | science books sold = 2x × 0.2 = 0.4x
science books unsold = 2x – 0.4x = 1.6x
art books sold = 5x × 0.2 = x
art books unsold = 5x – x = 4x
total books unsold = 1.6x + 4x = 5.6x
5.6x = 2240
x = 400
2x science = 800
D |
AQUA-RAT | AQUA-RAT-36733 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. The number is | [
"22030",
"220030",
"23030",
"24030"
] | B | Explanation:
(555 + 445) * 2 * 110 + 30 = 220000 + 30 = 220030
Option B |
AQUA-RAT | AQUA-RAT-36734 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
The Racing magic takes 40 seconds to circle the racing track once. The Charging bull makes 40 rounds of the track in an hour. If they left the starting point together, how many minutes will it take for them to meet at the starting point for the second time? | [
"3",
"6",
"9",
"12"
] | D | Time taken by Racing magic to make one circle = 40 seconds
Time taken byCharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds
LCM of 90 and 40 seconds = 360 seconds = 6 mins
Time taken for them to meet at the starting point for the second time = 6 mins *2 = 12 mins
Answer D |
AQUA-RAT | AQUA-RAT-36735 | # 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.
Balls are bought at 11 for Rs.10/- and sold at 10 for Rs.11/-. Find the gain or loss percentage? | [
"21 % Profit",
"22 % Profit",
"24 % Profit",
"31 % Profit"
] | A | 11 balls -----> Rs.10/-
10 balls -----> Rs.11/-
Note: Number of articles bought is must equal to number of articles sold.
10(11 10) ==> 110 100 = Cost price
11(10 11) ==> 110 121 = Selling price
Then, 21/100 × 100 = 21 % Profit
A) |
AQUA-RAT | AQUA-RAT-36736 | # how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
The following is multiple choice question (with options) to answer.
if there is one larger cube with surface area 600 and no. of smaller cubes with surface area 24.then how many small cubes are required to make a larger cube with surface area of 600? | [
"5",
"125",
"4",
"100"
] | B | the ratio between larger and smaller is
6a12:6a22=600:24
a12:a22=100:4
a1:a2=10:2
then length of cube is 10 and 2
so make a cube with length 10, we have to 5 small cubes each in length, breadth and height. so total 125 cubes are required.
then the answer is B |
AQUA-RAT | AQUA-RAT-36737 | ### Show Tags
19 Aug 2015, 01:34
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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]
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19 Aug 2015, 01:55
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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.
In a village the average age of n people is 42 years. But after the verification it was found that the age of a person had been considered 20 years less than the actual age, so the new average, after the correction,increased by 1. The value of n is? | [
"11",
"20",
"28",
"17"
] | B | Explanation:
It is the same as a person with 20 years more age replaces an existing person of the group ( or village)
Since the total age of the village having n persons, is being increased by 20 years and the average age of village is being increased by 1 year, hence there are total 20 people in the village.
Alternatively : ( n x 42 ) + 20 = ( n x 43 )
=> n=20
Answer: B |
AQUA-RAT | AQUA-RAT-36738 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Harry started a 7-mile hike with a full 11-cup canteen of water and finished the hike in 3 hours with 2 cup of water remaining in the canteen. If the canteen leaked at the rate of 1 cup per hour and Harry drank 3 cups of water during the last mile, how many cups did he drink per mile during the first 6 miles of the hike? | [
"2/6",
"3/6",
"4/6",
"6/4"
] | B | No of cups leaked during the trip = 3 cups.
No of cups Harry drank = 6 cups.
No of cups harry drank during the first 6 miles = 3.
drink / mile = 3/6
Answer: B |
AQUA-RAT | AQUA-RAT-36739 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into a partnership. A invests 3 times as much as B invests and 2/3 of what C invests. At the end of the year, the profit earned is Rs. 55000. What is the share of C? | [
"Rs. 12250",
"Rs. 13375",
"Rs. 16750",
"Rs. 15000"
] | D | Explanation:
Let the investment of C be Rs. x.
The inverstment of B = Rs.(2x/3)
The inverstment of A = Rs. (3 × (2/3)x) = Rs. (2x)
Ratio of capitals of A, B and C = 2x : 2x/3 : x = 6 : 2 : 3
C's share = Rs. [(3/11) × 55000] = Rs. 15000
Answer: Option D |
AQUA-RAT | AQUA-RAT-36740 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
A man invested Rs. 14,400 in Rs. 100 shares of a company at 25% premium.If the company declares 5% dividend at the end of the year,then how much does he get ? | [
"Rs. 576",
"Rs. 600",
"Rs. 650",
"Rs. 720"
] | A | Solution
Number of shares = (14400/125) = 115.2.
Face value = Rs. (100x115.2 ) = Rs. 11520.
Annual income = Rs.(5/100x11520) = Rs. 576.
Answer A |
AQUA-RAT | AQUA-RAT-36741 | python, python-3.x, programming-challenge
Title: Sum square difference Problem description:
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.
Square difference:
l = sum of the squares of the first n natural numbers
k = sum of n naturals numbers
m = differences between l and k
My Solution
This is my solution for problem 6 of Project Euler using Python:
def square_difference(n):
l = (n * (n + 1) * (2 * n + 1)) / 6
k = (n * (n + 1)) / 2
k = k ** 2
m = abs(l - k)
return m
The following is multiple choice question (with options) to answer.
If the sum and difference of two numbers are 5 and 10 respectively, then the difference of their square is: | [
"50",
"28",
"160",
"180"
] | A | Let the numbers be x and y.
Then, x + y = 5 and x - y = 10
x2 - y2 = (x + y)(x - y) = 5 * 10 = 50.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36742 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
47% of the students in a class of 38 students has glasses or contacts. How many students in the class have either glasses or contacts? | [
"15",
"16",
"17",
"18"
] | D | a=r⋅b
47%=0.47a
=0.47⋅34
a≈18
Option D is correct. |
AQUA-RAT | AQUA-RAT-36743 | Zz.
8. Jul 17, 2004
ohh, k , thanx
9. Jul 17, 2004
### BobG
Self Adjoint explained the math to it.
The reason behind it is this. Acceleration is fow fast your velocity is increasing. If you started out from rest and accelerated at 10 m/s^2, you would be travelling 10 m/s after one second. But you didn't travel 10 m/s for the entire second. You were travelling 0 m/s at time 0, 1 m/s after .1 seconds, 2 m/s after .2 seconds, etc. Your average velocity was 5 m/s (1/2 your acceleration).
This continues on even after the first second, except now you're starting at 10 m/s and reach 20 m/s by the end of the 2nd second. In other words, your average velocity for the 2nd second is 15 m/s. You total distance, so far, is 20 meters (5 meters the first second and 15 meters the 2nd second). You can keep going on second by second like this. The 1/2 a accounts for the non-constant velocity while the t^2 accounts for the accumulating distance from second to second.
10. Jul 17, 2004
### Entropy
Yeah back when I was a freshmen and just started physics I wondered what was up with that equation. Since I learned a little calculus I understand.
11. Jul 18, 2004
### JohnDubYa
I vote BobG has having the best answer.
12. Jul 19, 2004
### ArmoSkater87
its good, but i like gerben's better
13. Jul 19, 2004
### robphy
The following is multiple choice question (with options) to answer.
A person travels at a speed of 10 mph from one city to another. However while returning as there was not much traffic. He came with a speed of 15mph. what is his average speed? | [
"10Kmph",
"11Kmph",
"12Kmph",
"13Kmph"
] | C | Average speed = 2xy/(x+y) = (2*15*10)/25 = 12Kmph
ANSWER:C |
AQUA-RAT | AQUA-RAT-36744 | (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.
In a barrel of juice there is 20 liters; in a barrel of beer there are 80 liters. If the price ratio between barrels of juice to a barrel of beer is 3:4, what is the price ratio between one liter of juice and one liter of beer? | [
"3:2.",
"2:1.",
"3:1.",
"4:3."
] | C | Price of 20 L juice= 3x
1L= 3x/20
Price of 80 L beer= 4x
1L= 4x/80
Ratio of 1 L price = 3x/20/4x/80= 3:1
C is the answer |
AQUA-RAT | AQUA-RAT-36745 | ### Show Tags 23 Aug 2014, 11:31 kanusha wrote: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at$0.60 per glass on the first day, what was the price per glass on the second day?
The following is multiple choice question (with options) to answer.
A fruit seller had some oranges. He sells 10% oranges and still has 360 oranges. How many oranges he had originally? | [
"420",
"700",
"220",
"400"
] | D | Explanation :
He sells 10% of oranges and still there are 360 oranges remaining
=> 90% of oranges = 360
⇒ (90 × Total Oranges)/100 = 360
⇒ Total Oranges/100 = 4
⇒ Total Oranges = 4 × 100 = 400
Answer : Option D |
AQUA-RAT | AQUA-RAT-36746 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containg 19% alcohol and now the percentage of alcohol was found to be 26%. What quantity of whisky is replaced ? | [
"1/3",
"2by3",
"2/5",
"3/5"
] | B | Let us assume the total original amount of whiskey = 10 ml ---> 4 ml alcohol and 6 ml non-alcohol.
Let x ml be the amount removed ---> total alcohol left = 4-0.4x
New quantity of whiskey added = x ml out of which 0.19 is the alcohol.
Thus, the final quantity of alcohol = 4-0.4x+0.19x ----> (4-0.21x)/ 10 = 0.26 ---> x = 20/3 ml.
Per the question, you need to find the x ml removed as a ratio of the initial volume ---> (20/3)/10 = 2/3.
Hence, B is the correct answer. |
AQUA-RAT | AQUA-RAT-36747 | 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 sec and a man standing on the platform in 25 sec. If the speed of the train is 54 km/hr. What is the length of the platform? | [
"288",
"165",
"881",
"1277"
] | B | Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 25 = 375 m.
Let the length of the platform be x m . Then,
(x + 375)/36 = 15 => x = 165 m.
Answer: B |
AQUA-RAT | AQUA-RAT-36748 | (ii) 1/(√7 1/(√5 + √2) = 1/(√5 + √2) × (√5 − √2)/(√5 − √2) Solution: We rationalize the denominator of x: \begin{align} x &= \frac{{11}}{{4 - \sqrt 5 }} \times \frac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}\\ &= \frac{{11\left( {4 + \sqrt 5 } \right)}}{{16 - 5}}\\ &= 4 + \sqrt 5 \\ \Rightarrow x - 4 &= \sqrt 5 \end{align}. By using this website, you agree to our Cookie Policy. Oh No! That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. Ex 1.5, 5 Ask questions, doubts, problems and we will help you. &= 2 - \sqrt 3 \hfill \\ The denominator contains a radical expression, the square root of 2.Eliminate the radical at the bottom by multiplying by itself which is \sqrt 2 since \sqrt 2 \cdot \sqrt 2 = \sqrt 4 = 2.. the smallest positive integer which is divisible by each denominators of these numbers. $\begin{array}{l} 4\sqrt {12} = 4\sqrt {4 \times 3} = 8\sqrt 3 \\ 6\sqrt {32} = 6\sqrt {16 \times 2} = 24\sqrt 2 \\ 3\sqrt {48} = 3\sqrt {16 \times 3} =12\sqrt 3 \end{array}$, $\boxed{\begin{array}{*{20}{l}} To get the "right" answer, I must "rationalize" the denominator. &= \frac{{3 + 2\sqrt 3 }}{{5 - 2\sqrt 3 }} \times \frac{{5 + 2\sqrt 3 }}{{5 +
The following is multiple choice question (with options) to answer.
What is 2/5 of 5/9 of 1/2? | [
"1/4",
"1/9",
"9/16",
"5/8"
] | B | 2/5 * 5/9 * 1/2= 1/9
Answer: B |
AQUA-RAT | AQUA-RAT-36749 | Kudos [?]: 13 [0], given: 21
GMAT 1: 570 Q46 V24
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?
Then is 28*2 the correct approach?
Kudos [?]: 13 [0], given: 21
Math Expert
Joined: 02 Sep 2009
Posts: 42302
Kudos [?]: 133018 [1], given: 12402
Re: There are 8 teams in a certain league and each team plays [#permalink]
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27 Dec 2013, 03:08
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
Kudos [?]: 133018 [1], given: 12402
Manager
Joined: 07 Apr 2014
Posts: 138
Kudos [?]: 31 [0], given: 81
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
12 Sep 2014, 06:36
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
total 8 teams & each game by 2 pair then 8C2
Kudos [?]: 31 [0], given: 81
Intern
Joined: 20 Sep 2014
Posts: 9
Kudos [?]: 4 [1], given: 49
There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Nov 2014, 02:14
1
KUDOS
1
This post was
BOOKMARKED
SreeViji wrote:
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
Hey SreeViji,
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
The following is multiple choice question (with options) to answer.
There are 19 teams in the hockey league, and each team faces all the other teams 10 times each. How many games are played in the season? | [
"1710",
"1920",
"2440",
"2860"
] | A | The number of ways to choose two teams is 19C2 = 19*18/2 = 171
The total number of games in the season is 10*171 = 1710.
The answer is A. |
AQUA-RAT | AQUA-RAT-36750 | 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.
The food in a camp lasts for 36 men for 45 days. If twelve more men join, how many days will the food last? | [
"80 days",
"30 days",
"65 days",
"34 days"
] | D | one man can consume the same food in 36*45 = 1620 days.
12 more men join, the total number of men = 48
The number of days the food will last = 1620/48
= 34 days.
Answer:D |
AQUA-RAT | AQUA-RAT-36751 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
The total of the ages of Amar, Akbar and Anthony is 80 years. What was the total of their ages three years ago ? | [
"71 years",
"72 years",
"74 years",
"77 years"
] | A | Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.
Answer: Option A |
AQUA-RAT | AQUA-RAT-36752 | It is under the interpretation of Option B that the "official" answer is $3/8$. But what about Option A? Let's do the computation. Let $X = 1$ if the true value of the die is six, and $X = 0$ otherwise. Let $Y = 1$ if the reported value of the die is six, and $Y = 0$ otherwise. Let $L = 1$ if the man lies, and $L = 0$ otherwise. Then we know $\Pr[X = 1] = \frac{1}{6}$, $\Pr[L = 1] = \frac{1}{4}$. We also know $$\Pr[Y = 1 \mid X = 1] = \Pr[L = 0] = \frac{3}{4},$$ and $$\Pr[Y = 1 \mid X = 0] = \Pr[Y = 1 \mid (X = 0 \cap L = 1)]\Pr[L = 1] = \frac{1}{5} \cdot \frac{1}{4} = \frac{1}{20},$$ because two things must happen for the man to report a six if no six was rolled: he has to flip two heads (and thus be allowed to lie) with probability $\frac{1}{4}$, and he must randomly and uniformly choose to report six with probability $\frac{1}{5}$. From this, we can now compute \begin{align*} \Pr[X = 1 \mid Y = 1] &= \frac{\Pr[Y = 1 \mid X = 1] \Pr[X = 1]}{\Pr[Y = 1]} \\ &= \frac{\Pr[Y = 1 \mid X = 1]\Pr[X = 1]}{\Pr[Y = 1 \mid X = 1]\Pr[X = 1] + \Pr[Y = 1 \mid X = 0]\Pr[X = 0]} \\ &= \frac{\frac{3}{4} \cdot \frac{1}{6}}{\frac{3}{4}
The following is multiple choice question (with options) to answer.
The “c-number” of a number x is defined as the ones digit of 2^x. Antony rolls a die with 6 sides labeled with the integers from 1 to 6, each of which has an equal probability of landing face-up. He then takes 3^c, where c is the c-number of the result of his die roll, and plots 3^c on a number line as the point A. Finally, he repeats this entire process, this time plotting the result as the point B. What is the probability that the distance between A and B is greater than the value of B? | [
" 3/8",
" 13/36",
" 17/36",
" 19/36"
] | B | If you calculate 3^c for 1st roll, all 6 results will be 9, 81, 6561, 729, 9, 81. This result is the same for 2nd roll. 9, 81, 6561, 729, 9, 81.
About distance: If the first result is 9 and the second is also 9, the distance is 9-9=0 which is smaller than 9. If the first result is 9 and the second is 81, the distance is 81-9=72 which is also smaller than B which has the value of 81. If the first result is 81 and the second is 9, the distance will be greater than B. Distance 81-9=72>9.
On the first roll, the probability of getting result 9 is 2/6. In this case no other alternative values for second roll which would make the distance greater than B. So probability is 0. So next estimations are:
probability of getting 81 on the first roll (2/6) * probability of getting 9 on the second roll (2/6) = 1/9
probability of getting 729 on the first roll (1/6) * probability of getting 9, 81 on the second roll (4/6) = 1/9
probability of getting 6561 on the first roll (1/6) * probability of getting 9, 81, 729 on the first roll (5/6) = 5/36
All together: 1/9 + 1/9 + 5/36 = 13/36=B |
AQUA-RAT | AQUA-RAT-36753 | algorithms
Title: How to calculate least common multiple (LCM) for two numbers with constants (shifts)? What's the most effective algorithm to calculate LCM for 2 numbers where each of them has their own "shift"?
Example:
We want to find LCM for 25 + 5 as c and 30 + 10 as c
Then for the first case, the sequence will look like this:
25*1+5, 25*2+5, 25*3+5, 25*4+5...
And the second one:
30*1+10, 30*2+10, 30*3+10, 30*4+10...
In this case, the result should be 130 - 5th member of the first sequence and 4th of the second one. Given integers $a,b \geq 1$ and $c,d \geq 0$, you are looking for the set of solutions of $ax + c = by + d$ over integers $x,y \geq 0$. We can assume without loss of generality that $d = 0$ (replace $c,d$ with $c-d,0$ or $0,d-c$) and that $(a,b,c) = 1$ (otherwise, divide everything by the GCD).
I claim that if there is any solution then necessarily $(a,b) = 1$. Indeed, suppose that $(a,b) = g > 1$. If $ax + c = by$ then $g \mid ax,by$ implies $g \mid c$, and so $(a,b,c) = g > 1$, contrary to assumption. Therefore $(a,b) = 1$.
If $ax + c = by$ then $ax + c \equiv 0 \pmod{b}$ and so $x \equiv -c/a \pmod{b}$ (note that we can divide by $a$ since $(a,b)=1$). Similarly, $y \equiv c/b \pmod{a}$. These equations essentially give us both the minimal solution and all solutions.
The following is multiple choice question (with options) to answer.
If the LCM of 9, 12, 5, and x is 180, which of the following could be x? | [
"10",
"12",
"13",
"9"
] | A | Using elimination method, eliminate the option to arrive at 10
if x is replaced by 10 and the LCM is calculated you arrive at 180
A |
AQUA-RAT | AQUA-RAT-36754 | meteorology, snow, radar
Also note that winter precipitation adds an extra complication because the particles are lighter in weight and can thus be blown about more by vertical and horizontal winds. Raindrops (and hail) are quite likely to fall unless extreme updrafts exist because they are heavy. But drizzle, snow, and sleet may be blown around quite a bit. Without a time-intensive dual-Doppler analysis, you cannot know the wind motion in the storm thoroughly, and therefore will have varying results at times.
And finally, the big wrench is unfortunate inherent to how radars work. They measure the percentage of their sent energy that is reflected back to them. That's great because that's directly connected to the diameter of the item falling (to the 6th power). But unfortunately the grand problem is that in a storm, there is a huge variety of drop/flake sizes mixed together at once... such that we can't extract which combination of particle sizes created it (and thus can't calculate volume to actually know the rain/snow amount that falls). It could be like 6 medium size flakes causing the 10 dBZ echo... or 2 large flakes and 10 small flakes... and each combination is a different volume/snow total. (to see the nitty-gritty math details on this, read more here.) So we can never know for sure the exact rain/snow falling using just radar. The good news is we've at least done lots of experiments and come up with some fairly useful best-practice formulas for using the Z-R ratio in different scenarios. Good, but not perfect.
The following is multiple choice question (with options) to answer.
Last week we got 3.5 inches of snow. Six-tenths of an inch melted before another storm added
8.3 inches. Since then we have lost 4.2 inches to melting or evaporation. How may inches of snow
are left on the ground? | [
"7\" remaining",
"5\" remaining",
"8\" remaining",
"6\" remaining"
] | A | 3.5 - .6 = 2.9"
2.9 + 8.3 = 11.2"
11.2 - 4.2 = 7" remaining
correct answer A |
AQUA-RAT | AQUA-RAT-36755 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man buy a book in Rs50 & sale it Rs100. What is the rate of profit ??? | [
"10%",
"100%",
"30%",
"25%"
] | B | cp=50
sp=100
profit=100-50=50
%=50/50*100=100%
ANSWER:B |
AQUA-RAT | AQUA-RAT-36756 | F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
F6 = { 0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 1/1 }
F7 = { 0/1, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1/1 }
F8 = { 0/1, 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8, 1/1 }
Centered
F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 1:2:3 and their H.C.F is 12. The numbers are? | [
"12, 24, 38",
"12, 24, 56",
"12, 24, 31",
"12, 24, 36"
] | D | Let the required numbers be x, 2x and 3x. Then, their H.C.F = x. So, x = 12.
The numbers are 12, 24, 36.
Answer: D |
AQUA-RAT | AQUA-RAT-36757 | 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.
When a train travels at a speed of 90kmph,it reaches the destination on time.when the same train travels at a speed of 50kmph,it reaches its destination 15min late.what is the length of journey? | [
"18.75km",
"50km",
"60km",
"85km"
] | A | Let x be the time reached with the speed 90km/h
50km/h ----> x+15
Distance is equal so
90(km/h)× xhr = 50(km/h) × (x+15) hr
So
90 x = 50x + 750
So the would be in km
And x = 18.75
ANSWER:A |
AQUA-RAT | AQUA-RAT-36758 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
A man has Rs.680 in the denominations of two-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? | [
"160",
"40",
"680",
"120"
] | D | Let number of notes of each denomination be x.
Then 2x + 5x + 10x = 680
17x = 680
x = 40
Hence, total number of notes = 3x = 120.
Answer is D |
AQUA-RAT | AQUA-RAT-36759 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two? | [
"90",
"82",
"80",
"45"
] | C | Three digit number can have only following 3 patterns:
A. all digits are distinct;
B. two digits are alike and third is different;
C. all three digits are alike.
We need to calculate B. B=Total - A - C
Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);
A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);
C. all three are alike = 3 (777, 888, 999).
So, 299-216-3=80.
Answer: C. |
AQUA-RAT | AQUA-RAT-36760 | ### Solution 1
Notice that the difference between consecutive squares are consecutively increasing odd numbers. Thus, there are $3$ numbers between $1$ (inclusive) and $4$ (exclusive), $5$ numbers between $4$ and $9$, and so on. The number of numbers from $n^2$ to $(n + 1)^2$ is $(n + 1 - n)(n + 1 + n) = 2n + 1$. Whenever the lowest square beneath a number is odd, the parity will be odd, and the same for even. Thus, $a = [2(1) + 1] + [2(3) + 1] \ldots [2(43) + 1] = 3 + 7 + 11 \ldots 87$. $b = [2(2) + 1] + [2(4) + 1] \ldots [2(42) + 1] + 70 = 5 + 9 \ldots 85 + 70$, the $70$ accounting for the difference between $2005$ and $44^2 = 1936$, inclusive. Notice that if we align the two and subtract, we get that each difference is equal to $2$. Thus, the solution is $|a - b| = |b - a| = |2 \cdot 21 + 70 - 87| = \boxed{025}$.
### Solution 2
Similarly, $b = (3^2 - 2^2) + (5^2 - 4^2) + \ldots + (45^2 - 44^2) - 19$, where the $-19$ accounts for those numbers between $2005$ and $2024$.
Thus $a = (2^2 - 1^2) + (4^2 - 3^2) + \ldots + (44^2 - 43^2)$.
The following is multiple choice question (with options) to answer.
The difference between the squares of two consecutive numbers is 35. The numbers are | [
"14,15",
"15,16",
"17,18",
"18,19"
] | C | Explanation:
Let the numbers be a and (a+1)
(a+1)2−a2=35
⇒a2+2a+1−a2=35
⇒2a=34⇒2a=34 or a = 17
The numbers are 17 & 18.
Correct Option: C |
AQUA-RAT | AQUA-RAT-36761 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B started a business with investments of Rs. 42000, and Rs. 63000 respectively. After 4 months B withdraws from the business. At the end of a year they got Rs. 9600 as total profit. Find the share of B. | [
"Rs. 5600",
"Rs. 2800",
"Rs. 3200",
"Rs. 6400"
] | C | Explanation:
Ratio of capitals of A and B = (42000 × 12) : (63000 × 4) = 2 : 1
B's share = Rs. [(1/3) × 9600] = Rs. 3200
Answer: Option C |
AQUA-RAT | AQUA-RAT-36762 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a profit of 15%. If its cost had been 5% less and it had been sold for Rs. 21 less, than the profit would have been 10%. Find the cost price of the watch | [
"100",
"200",
"120",
"140"
] | B | let the price of the watch is x
it was sold at a profit of 15%
then the price of the watch is 1.15x
again cost had been decreased by 5% i.e 0.95x
according to the question
1.15x=0.95x*(110/100)+21
then x=200
ANSWER:B |
AQUA-RAT | AQUA-RAT-36763 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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Posts: 1892
Joined: 14 Oct 2017
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Legendary Member
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
The average age 9 members of a committee are the same as it was 2 years ago, because an old number has been replaced by a younger number. Find how much younger is the new member than the old number? | [
"27 years",
"87 years",
"18 years",
"19 years"
] | C | 9 * 2 = 18 years
Answer: C |
AQUA-RAT | AQUA-RAT-36764 | 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 128sq m and its altitude is twice the corresponding base. Then the length of the base is? | [
"8",
"9",
"5",
"2"
] | A | 2x * x = 128 => x= 8
Answer: A |
AQUA-RAT | AQUA-RAT-36765 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Calculate the amount that an investor needs to be invest to earn $1005 in interest in 12 months if the investor plans to invest x dollars in a savings account that pays interest at an annual rate of 11% compounded semi-annually? | [
"8000",
"8100",
"8900",
"8600"
] | C | the approach is substitution,
our interest requirement is $1005 after 12 months, 2 compounding period.
calculate the compound interest on each option and find out the one that yields 460 in 12 months
8900 yielded $1005
using the formula
A = P(1 + r/n)nt
hence answer is C |
AQUA-RAT | AQUA-RAT-36766 | # How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
How many numbers between 1 and 1000 are divisible by 2, 3, 5 or 7?
My try:
Let $A_2, A_3, A_5, A_7$ be the set of numbers between 1 and 1,000 that are divisible by 2, 3, 5, and 7 respectively. I used the inclusion-exclusion formula for $|A_2\cup A_3\cup A_5\cup A_7|= |A_2|+|A_3|+|A_5|+|A_7|-|A_2\cap A_3|-|A_2\cap A_5|-|A_2\cap A_7|-|A_3\cap A_5|-|A_3\cap A_7|-|A_5\cap A_7|+|A_2\cap A_3\cap A_5|+|A_2\cap A_3\cap A_7|+|A_2\cap A_5\cap A_7|+|A_3\cap A_5\cap A_7|-|A_2\cap A_3\cap A_5\cap A_7| = 500+333+200+142-166-100-71-66-47-28+33+23+14+9-4 = 772$
And the result I received was - 772.
I would appreciate if you could confirm my method and result, and I'd be happy to see a different, more elegant approach.
• It should be “2 , 3, 5 or 7”. – user371838 Jan 1 '18 at 13:01
• I am afraid there may be an elegant solution if you insist with and, and not so, if it is or. – user371838 Jan 1 '18 at 13:02
• It should be or. My apologies – Moshe King Jan 1 '18 at 13:03
• I haven't checked the numbers, but your inclusion-exclusion argument should work. – Paul Aljabar Jan 1 '18 at 13:10
The following is multiple choice question (with options) to answer.
What is the number of integers from 1 to 1000 (inclusive) that are divisible by neither 10 nor by 35? | [
"884",
"890",
"892",
"910"
] | D | Normally, I would use the method used by Bunuel. It's the most accurate. But if you are looking for a speedy solution, you can use another method which will sometimes give you an estimate. Looking at the options (most of them are spread out), I wont mind trying it. (Mind you, the method is accurate here since the numbers start from 1.)
In 1000 consecutive numbers, number of multiples of 11 = 1000/11 = 90 (Ignore decimals)
In 1000 consecutive numbers, number of multiples of 35 = 1000/35 = 28
Number of multiples of 11*35 i.e. 385 = 1000/385 = 2
Number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - (90 + 28 - 2) {Using the concept of sets here) = 910
Think: Why did I say the method is approximate in some cases?
Think what happens if the given range is 11 to 1010 both inclusive (again 1000 numbers)
What is the number of multiples in this case?
D |
AQUA-RAT | AQUA-RAT-36767 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
In a company with 48 employees, some part-time and some full-time, exactly (1/3) of the part-time employees and (1/4) of the full-time employees take the subway to work. What is the greatest possible number of employees who take the subway to work? | [
"12",
"13",
"14",
"15"
] | D | Let part time emp=x
Let full time emp=y
then,
48=x+y.........(1)
(1/3)x+(1/4)y=No. of ppl taking the subway
4x+3y/12=No. of ppl taking the subway
using 1
x/12+3*48/12=No. of ppl taking the subway
so, the minimum value of x has to be 12.
Hence to maximize the no. put 36 for x
36/12+12/4=15
ANSWER:D |
AQUA-RAT | AQUA-RAT-36768 | the area and volume of a sphere, using simple geometric arguments. The surface area is the surface part of a three-dimensional object. Find out what's the height of the cylinder, for us it's 9 cm. Volume of a cylinder is (to find half, divide by 2): Vcylinder = Abase * height In a cylinder, the base is a circle and the area of a circle can be found: Acircle = pi * radius2 After you find the. The area of some circle is known and equals 15. 809625 Cubic Inches The volume of a cylinder 5 inches in diameter and 6 inches high is 117. What is the formula for finding the volume of a cylinder? V = π r 2 h 3. The volume of the cylinder is the amount of space inside the cylinder. Visually Understanding Area of a Circle and Volume of a Cylinder. Then click Calculate. Don't forget the two end bits: Total Surface Area. 0021m^3 Please Help or try and Solve this for me, Thankyou Very Much. For the rectangular solid, the area of the base, B, is the area of the rectangular base, length × width. Volume of a Right Circular Cylinder. Volume of a cylinder. Pi is the ratio of a circle's circumference to its diameter: 3. Im doing Maths about 'Volume' one of the questions are finding the volume of a semi-cylinder. The volume of three cones is equal to the volume of one cylinder with the same base and height. The cylinder has two bases, the base has radius r. The length of each cylinder (h) is 15,000 m. Surface Area of a Cylinder= 2πr² + 2πrh. Surface area and volume of a cylinder. Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. Area of a Trapezoid. To find the volume of a cylinder we multiply the base area (which is a circle) and the height h. In this example, r and h are identical, so the volumes are πr 3 and 1 ⁄ 3 π r 3. Browse by Radius in Meters. For engine's, this would be: Swept
The following is multiple choice question (with options) to answer.
The radius of a cylindrical vessel is 7cm and height is 2cm. Find the whole surface of the cylinder? | [
"308 sq cm",
"220 sq cm",
"396 sq cm",
"132 sq cm"
] | C | r = 7 h = 2
2πr(h + r) = 2 * 22/7 * 7(9) = 396
ANSWER:C |
AQUA-RAT | AQUA-RAT-36769 | astrophotography
if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
as if light years are a measure of time. A light year is a measure of distance, the distance light travels in a year in a vacuum.
The following is multiple choice question (with options) to answer.
The distance that light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 1000 years is: | [
"587 x 10^13 miles",
"999× 1238 miles",
"346 × 1012 miles",
"247 × 1012 miles"
] | A | The distance of the light travels in one years is:
5,870,000,000,000 = 587 * 10 ^10
The distance of the light travels in 1000 years is:
= 587 * 10^10 x 10 ^3 = 587 x 10^13 miles
Answer :A |
AQUA-RAT | AQUA-RAT-36770 | # 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.
The number 149 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers? | [
" 19",
" 16",
" 15",
" 14"
] | A | 2^2 + 8^2 + 9^2 = 149 --> 2 + 9 + 8 = 19.
A |
AQUA-RAT | AQUA-RAT-36771 | ### 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.
A milkman mixed 10 liters of water to 50 litres of milk of Rs.16 per liter, then cost price of mixture per liter is | [
"Rs.15.16",
"Rs.13.33",
"Rs.15",
"Rs.14.23"
] | B | Total mixture = 10w + 50m = 60
Total S.P= 16*60= Rs.960 @ Rs.16/lit
Now one can easily see in the statement that the milkman will get profit of 10 litres.
So, Gain = 10*60= Rs.160
Also, Gain = SP - CP
160=960-CP
CP = 800 for 60litres
so CP for 1 litre = 800/60 = 13.33
ANSWER:B |
AQUA-RAT | AQUA-RAT-36772 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
5
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This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of the even integers from 0 to 100 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 50 inclusive? | [
"24",
"25",
"26",
"48"
] | B | So, according to A mean of a set of even numbers from 0 to 100 =(0+100)/2=50 and mean of a set of even numbers from 0 to 50 =(0+50)/2=25
Difference=50-25=25
Answer: B. |
AQUA-RAT | AQUA-RAT-36773 | => $$\frac{74}{100} \times x$$ = 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.
B and C start a business with Rs.4000 and Rs.12000 respectively. How should they share their profits at the end of one year? | [
"3:1",
"1:5",
"1:4",
"1:3"
] | D | They should share the profits in the ratio of their investments.
The ratio of the investments made by B and C =
4000 : 12000 => 1:3 .
Answer:D |
AQUA-RAT | AQUA-RAT-36774 | computer-architecture
So, if you were to try to subtract $01110000−11101011$, let's get an answer first so that we can check ourselves. $64+32+16 = 112$, and $-128+64+32+8+2+1 = -21$. You are performing the subtraction $112 - -21$. This means that our final answer should be $133$.
So, just as you can subtract by negating before you add (ie. $112 + 21$ instead of $112 - -21$), we can perform a similar operation here. If we negate the second number, we flip the bits and add one: $00010100 + 1 = 00010101$. But now that we've negated the number, our subtraction has become an addition problem. So, our original $01110000−11101011$ has become $01110000 + 00010101$
01110000
+ 00010101
--------
10000101
The following is multiple choice question (with options) to answer.
Find the number which when added to itself 13 times, gives 112. | [
"7",
"8",
"9",
"11"
] | B | Explanation:
Let the number be x. Then, x + 13x = 112 14x = 112 x = 8.
Answer: B |
AQUA-RAT | AQUA-RAT-36775 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
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26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
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The following is multiple choice question (with options) to answer.
The marks obtained by Vijay and Amith are in the ratio 4:5 and those obtained by Amith and Abhishek in the ratio of 3:2. The marks obtained by Vijay and Abhishek are in the ratio of? | [
"6:5",
"5:3",
"8:6",
"7:6"
] | A | 4:5
3:2
------
12:15:10
12:10
6:5
ANSWER A |
AQUA-RAT | AQUA-RAT-36776 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
In a sports contest there were m medals awarded on n successive days (n > 1). On the first day 1 medal and 1/7 of the remaining m " 1 medals were awarded. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on. On the nth and last day, the remaining n medals were awarded. how many medals were awarded altogether? | [
"36 medals",
"37 medals",
"38 medals",
"35 medals"
] | A | 1st day=1+35/7=6 remaining 30 medals
2nd day=2+28/7=6 remaining 24 medals
3rd day=3+21/7=6 remaining 18 medals
...
6th day 6 medals
6*6=36 medals
6 days
ANSWER:A |
AQUA-RAT | AQUA-RAT-36777 | # How can I calculate the remainder of $3^{2012}$ modulo 17?
So far this is what I can do:
Using Fermat's Little Theorem I know that $3^{16}\equiv 1 \pmod {17}$
Also: $3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}$
So I am left with $3^{12}\pmod{17}$.
Again I'm going to use fermat's theorem so: $3^{12} = \frac{3^{16}}{3^{4}} \pmod{17}$
Here I am stuck because I get $3^{-4} \pmod{17}$ and I don't know how to calculate this because I don't know what $\frac{1}{81} \pmod{17}$ is.
I know $81 = 13 \pmod{17}$
But I know the answer is 4. What did I do wrong?
The following is multiple choice question (with options) to answer.
What will be the remainder when 17^200 is divided by 18 | [
"1",
"2",
"3",
"4"
] | A | 17^200 % 18
(17-18)^200 % 18
(-1)^200 % 18
1 % 18 = 1
ANSWER:A |
AQUA-RAT | AQUA-RAT-36778 | 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.
39 persons can repair a road in 12 days,working 5 hours a day.In how many days will 30 persons,working 6 hours a day,complete the work ? | [
"10",
"13",
"14",
"15"
] | B | Solution
Let the required number of hours per day be x.
Less persons, More days (Indirect Proportion)
More working hrs per days, Less days (Indirect Proportion)
∴ 30 × 6 × x = 39 × 5 × 12 ⇔ x = 39x5x12/30x6 =x =13. Answer B |
AQUA-RAT | AQUA-RAT-36779 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
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cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. Find his average after 17th inning? | [
"19",
"29",
"39",
"49"
] | C | Solution
Let the average after 17th inning = x. Then, average after 16th inning = (x - 3)
Average
=16 (x-3)+87
= 17x or x=(87-48)
= 39.
Answer C |
AQUA-RAT | AQUA-RAT-36780 | 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 row with a speed of 5 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is? | [
"A)15",
"B)18",
"C)10",
"D)11"
] | C | M = 5
S = 5
DS = 5 + 5 = 10
Answer:C |
AQUA-RAT | AQUA-RAT-36781 | 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
all we have is 11 (and 7 for extra credit). To satisfy this, we have to make it so that when placing alternating signs between the digits so the result is a multiple of 11. After a bit of trial and error, I came up with 4123975680, which works for everything. It's not divisible by 7, though. I'll keep looking.
As Martin Schulz pointed out in the comments (now deleted),
9165784320 is divisible by 7
• I have 1385679240 & 7165932840 neither divisible by 7 – DeNel Oct 27 '18 at 0:10
• My theory is that the number has to end in 240 or 640 or 840 but your answer totally disproved that. – DeNel Oct 27 '18 at 0:11
• "The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8." This is not true. For a counterexample, 120 is divisible by 4 and 8, but its second-last digit is not divisible by 4. – elias Oct 27 '18 at 2:08
• the correct wording would be the last 2 digits must be divisible by 4 and the last 3 digits must be divisible by 8. In fact we just need to check for divisibility by 8, since that implies divisibility by 4 as well – phuclv Oct 27 '18 at 6:24
This is a good problem to attack by computer:
It turns out that there are 7344 solutions, including 1056 to the bonus.
So I thought, how far can we go?
And it just happens that 2438195760, 3785942160, 4753869120 and 4876391520 are exactly those permutations of 0123456789 divisible by each of 1 through 18, and none of them are divisible by 19.
Code used (Python 3 IDLE):
The following is multiple choice question (with options) to answer.
A four digit number divisible by 7 becomes divisible by 3 when 19 is added to it. The largest such number is : | [
"4461",
"4473",
"4479",
"4487"
] | D | Out of all the 5 options, only 4487 is NOT divisible by 3. All others are divisible
So Answer = D (No further calculation required)
Addition of any two non-divisible numbers by 3 gives the resultant divisible by 3
19 is non-divisible by 3; we are adding a number to that so that the resultant becomes divisible by 3
Applying the above rule,It means that the number which we are going to add should be non-divisible by 3
So comes the answer = 4487
Answer : D |
AQUA-RAT | AQUA-RAT-36782 | That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
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The following is multiple choice question (with options) to answer.
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 41st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross? | [
"19",
"25",
"30",
"32"
] | B | Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 41st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?
30 floors/120 floors per minute=1/4 minutes
11+57/4=25
41-63/4=25
Answer : B |
AQUA-RAT | AQUA-RAT-36783 | &= 2 \times {\frac{22}{7}} \times 35 \\\\ &= 220 \text{ feet}\end{align}. Equations of a circle with given centre and radius in different forms. How to find the area of a circle? Note: With this tool, you can know the radius of a circle anywhere on Google Maps by simply clicking on a single point and extending or moving the circle to change the radius on the Map. The fixed point is called the origin or center of the circle. Let's take a fixed point $$P$$ and try to draw lines from it. Here are the three basic circle formulas that are used in calculating the various dimensions of a circle. Circles are one of the most commonly found shapes in the world. The calculator will generate a step by step explanations and circle graph. The radius – the distance from the circle's center or origin to the edge, one half the diameter The circumference – the length of the outside boundaries of the circle Starting from the diameter, you can easily find the other two. And consult the table below our everyday lives enjoyed learning about circle with given and. Then one of those slices makes a quarter circle boundary or the length of a circle =.! Team of math experts is dedicated to making learning fun for our favorite readers the. More about pi, or radius from point \ ( P\ ) in directions... Places, then one of its area at solving a few interesting practice questions at the end of the is! 22 } { 7 } } \end { align } \pi = { \frac { 22 {! A carnival and in general form given the radius of the circle and \ ( P\ in... Floated away boundary of the circle ( pool ) fun for our favorite readers the! How the shape of a circle = π r 2 not display plot -- browser is out of date area. To compute are done live '': how to calculate the and... Compute something or you have and skip unknown values in, find its area = ø. Find area identify the parts of a circle can be calculated if the radius or.... Health, and the arc in it math experts is dedicated to making learning fun for favorite... Button to see the result about circle with radius 1 unit Google map labeled as distance map and directions... Circular chocolate cake from his favorite bakery squared
The following is multiple choice question (with options) to answer.
Find the area of circle whose radius is 35m? | [
"2883",
"1543",
"7673",
"3850"
] | D | 22/7 * 35 * 35
= 3850
Answer: D |
AQUA-RAT | AQUA-RAT-36784 | So, to create the desired rectangle, we need only choose 2 different x-coordinates and 2 different y-coordinates
So, let's take the task of creating rectangles and break it into STAGES
STAGE 1: Select the 2 x-coordinates
We can choose 2 values from the set {3, 4, 5, 6, 7, 8, 9, 10, and 11}
In other words, we must choose 2 of the 9 values in the set
Since the order in which we choose the numbers does not matter, we can use COMBINATIONS
We can select 2 number from 9 numbers in 9C2 ways (= 36 ways)
STAGE 2: Select the 2 y-coordinates
We can choose 2 values from the set {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
In other words, we must choose 2 of the 11 values in the set
We can select 2 number from 11 numbers in 11C2 ways (= 55 ways)
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a rectangle) in (36)(55) ways (= 1980 ways)
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### Re: Rectangle ABCD is constructed in the coordinate plane parallel
by swerve » Thu Oct 21, 2021 2:25 pm
BTGModeratorVI wrote:
Sun Jul 19, 2020 1:38 pm
Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 3 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD?
396
1260
1980
7920
15840
Source: Grockit
As the rectangle is parallel to coordinate axes, the coordinates of the points of the rectangle would be
$$(X_1, Y_1), (X_2, Y_1), (X_2, Y_2), (X_1,Y_2)$$
Given that $$X_1, X_2$$ lie between $$3$$ and $$11$$ ie., $$9$$ possible numbers
The following is multiple choice question (with options) to answer.
Rectangle ABCD is constructed in the coordinate plane parallel to the x- and y-axes. If the x- and y-coordinates of each of the points are integers which satisfy 7 ≤ x ≤ 11 and -5 ≤ y ≤ 5, how many possible ways are there to construct rectangle ABCD? | [
"396",
"1260",
"550",
"7920"
] | C | As the rectangle is parallel to coordinate axes, the coordinates of the points of the rectangle would be
(X1, Y1), (X2, Y1), (X2, Y2), (X1,Y2)
given that X1, X2 lie between 7 and 11..ie., 5 possible numbers
Possible combinations for X1,X2 would be 5C2 = 10
Similarly, Possible combinations for Y1, Y2 would be 11C2 = 55
Possible ways of constructing rectangle is by selecting any of the combination of X1,X2 and Y1,Y2
= 10 * 55 = 550
Ans. C |
AQUA-RAT | AQUA-RAT-36785 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
The cost to rent a small bus for a trip is x dollars, which is to be shared equally among the people taking the trip. If 10 people take the trip rather than 17, how many more dollars, in terms of x, will it cost per person? | [
" x/6",
" x/16",
" x/40",
" 3x/40"
] | D | Choose x as a multiple of 17, I chose 64:
So for 10 people, that's 6.4 each and for 16 people it's 4 USD each... Pick one of the options that gives you 6.4 - 4 = 2.4... The answer is D. |
AQUA-RAT | AQUA-RAT-36786 | Related Rate Prob - Ships sailing in different directions
• November 5th 2010, 09:15 PM
dbakeg00
Related Rate Prob - Ships sailing in different directions
Ship A is 15 miles east of point O and moving west at 20mi/h. Ship B is 60 miles south of O and moving north at 15mi/h. a)Are they approaching or seperating after 1 hour and at what rate? b)after 3hrs?
Let D=distance between the ships at time t
$D^2=(60-15t)^2+(15-20t)^2$
$2D*\frac{dx}{dt}=(2)(-15)(60-15t)+(2)(-20)(15-20t)$
$2D*\frac{dx}{dt}=(-30)(60-15t)+(-40)(15-20t)$
$2D*\frac{dx}{dt}= 1250t-2400$
$\frac{dx}{dt}=\frac{1250t-2400}{2D}$
$D=\sqrt{5^2+45^2}=5\sqrt{82}$
$\frac{dx}{dt}=\frac{1250t-2400}{10\sqrt{82}}$
a) $\frac{dx}{dt}=\frac{1250*1-2400}{10\sqrt{82}}=approaching\ at \frac{-115}{\sqrt{82}}$ mi/h
b) $\frac{dx}{dt}=\frac{1250*3-2400}{10\sqrt{82}}=seperating\ at \frac{135}{\sqrt{82}}$ mi/h
The book agrees with me on part A, but on part B it says the answer should be $seperating\ at\ \frac{9\sqrt{10}}{2}$ mi/h. Am I missing something obvious or is the book wrong here?
• November 6th 2010, 02:24 AM
Ackbeet
The following is multiple choice question (with options) to answer.
On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets? | [
"33 1/3 %",
"40%",
"50%",
"60%"
] | C | Let the total # of passengers be 100.
Now, 20 passengers held round-trip tickets AND cars.
As 60% of the passengers with round-trip tickets did not take their cars then 40% of the passengers with round-trip tickets did take their cars, so # of passengers with round-trip tickets AND cars is 40% of the passengers with round-trip tickets.
If we take # of the passengers with round trip tickets to be xx then we'll have 0.4x=200.4x=20 --> x=50x=50.
Answer: C. |
AQUA-RAT | AQUA-RAT-36787 | If $6\nmid k$, then $7^1$ is the exact power of $7$ dividing $n$; i.e., $7^1\mid\mid n$. This contradicts Fermat's result. Hence we can conclude that $6\mid k$. It implies $3\mid k$, so that $9\mid k$, because otherwise $3^1\mid\mid n$, which is impossible. Also $2\mid k$, so that $11\mid n$. This implies $198=11\cdot2\cdot 3^2\mid k$, because otherwise $11^1\mid \mid n$, which is impossible. This is what Martin already had. We may continue now that in particular $33\mid k$, i.e., $67\mid n$. Since if not $3\cdot 11\cdot 67\mid k$ then $67^1\mid \mid n$, we arrive at $13266=2\cdot 3^2\cdot 11\cdot 67\mid k$, etc. I have no idea whether this will continue, or whether we will finally find a possible $k$.
• Yes, that was what I was thinking. I am sure that this will continue in that way, but of course that is no proof – Martin Apr 28 '14 at 12:54
• Your non-answer (?) will sound much more impressive if you'd change Fermat with Jupiter: See end of first row. – Lucian Apr 28 '14 at 13:53
The following is multiple choice question (with options) to answer.
If k is a non-negative integer and 18^k is a divisor of 624,938 then 6^k - k^6 = | [
"0",
"1",
"36",
"118"
] | B | 6+2+4+9+3+8 = 32, so this number is not divisible by 3 and thus not divisible by 18.
Therefore, k=0
6^k - k^6 =1-0=1
The answer is B. |
AQUA-RAT | AQUA-RAT-36788 | 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.
20 men do a work in 20days. How many men are needed to finish the work in 4days? | [
"50",
"20",
"30",
"100"
] | D | men required to finish the work in 4days = 20*20/4 = 100
Answer is D |
AQUA-RAT | AQUA-RAT-36789 | &\equiv2^3.3^8.7^5.9^3\cr &\equiv8.1.7.9\cr &\equiv4\ .\cr}
The following is multiple choice question (with options) to answer.
What is the value of: 5^10 - 5^9 + 5^8 ? | [
" 5^9",
" 15^9",
" 21(5^8)",
" 29(5^8)"
] | C | 5^10 - 5^9 +5^8 = 5^8 (5^2 - 5 + 1) = 5^8 (25 - 5 + 1) = 21 (5^8), Answer C. |
AQUA-RAT | AQUA-RAT-36790 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is | [
"8/17",
"7/15",
"3/15",
"8/15"
] | D | Person ( A ) ( B ) ( A+B )
Time - ( 15 ) ( 20 ) (-)
Rate - ( 20 ) ( 15 ) ( 35 )
Work -(300) (300) (300)
therefore A+B requires (300/35) days to complete entire work
For 1st 4 days they work 35 * 4 = 140
Remaining work is 300 - 140 = 160
Remaining fraction of work is = 160/300 = 8/15
Answer D |
AQUA-RAT | AQUA-RAT-36791 | 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.
When a=4+(2/4) and b=4-(2/4), (2^a^2)/(2^b^2)=? | [
"260",
"256",
"240",
"210"
] | B | --> (2^a^2)/(2^b^2) ={(2)^(a^2-b^2)}=2^(a-b)(a+b). Since a-b=4/4 and a+b=8, 2^(a-b)(a+b)=2^(4/4)(8)=2^8=256
Therefore, the answer is B |
AQUA-RAT | AQUA-RAT-36792 | Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2759
Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink]
### Show Tags
08 Feb 2018, 16:38
1
Barkatis wrote:
It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages?
A. 5
B. 6
C. 8
D. 10
E. 12
We can let Tom’s rate = 20/x and Jack’s rate = 20/(x+2), and their combined rate is 25/3; thus:
20/x + 20/(x+2) = 25/3
Multiplying by 3x(x+2), we have:
20(3)(x + 2) + 20(3x) = 25(x)(x + 2)
60x + 120 + 60x = 25x^2 + 50x
25^x - 70x - 120 = 0
5x^2 - 14x - 24 = 0
(5x + 6)(x - 4) = 0
x = -5/6 or x = 4
Since x can’t be negative, we see that x must be 4, so Jack's rate is 20/6 = 10/3.
So it takes him 40/(10/3) = 120/10 = 12 hours to type 40 pages.
Alternate Solution:
Let’s pick a common multiple of 20, 25 and 40, such as 200, and calculate the number of hours to type that number of pages for each scenario.
Since Jack takes 2 more hours than Tom to type 20 pages, it will take Jack 20 more hours than Tom to type 200 pages.
Since Jack and Tom working together can type 25 pages in 3 hours, they can type 200 pages in 12 hours.
The following is multiple choice question (with options) to answer.
Tom reads at an average rate of 30 pages per hour, while Jan reads at an average rate of 41 pages per hour. If Tom starts reading a novel at 3:20, and Jan begins reading an identical copy of the same book at 4:26, at what time will they be reading the same page? | [
"6:26",
"6:56",
"7:26",
"7:56"
] | C | Since Tom reads an average of 1 page every 2 minutes, Tom will read 33 pages in the first 66 minutes. Jan can catch Tom at a rate of 11 pages per hour, so it will take 3 hours to catch Tom.
The answer is C. |
AQUA-RAT | AQUA-RAT-36793 | # 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.
Jason can row his boat with the stream at 50 km/h and against the stream in 2 km/h. Jason's rate is? | [
"18 kmph",
"24 kmph",
"35 kmph",
"15 kmph"
] | B | DS = 50
US = 2
S = ?
S = (50 - 2)/2 = 24 kmph
Answer:B |
AQUA-RAT | AQUA-RAT-36794 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular field is to be fenced on three sides leaving a side of 8 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required? | [
"244ft",
"88ft",
"122ft",
"178ft"
] | D | Given that length and area, so we can find the breadth.
Length x Breadth = Area
8 x Breadth = 680
Breadth = 85 feet
Area to be fenced = 2B + L = 2 (85) + 8 = 178 feet
Answer: D) 178ft |
AQUA-RAT | AQUA-RAT-36795 | • Possible duplicate of Chess Master Problem – Aqua Dec 21 '17 at 20:01
• – Aqua Dec 21 '17 at 20:04
• – Aqua Dec 21 '17 at 20:05
We define a sequence $a_1,a_2,\ldots$, by letting $a_i$ be the amount of games played on day $i$. It is given that for all $n\in\mathbb{N}$, we have: $$a_{7n+1}+a_{7n+2}+a_{7n+3}+a_{7n+4}+a_{7n+5}+a_{7n+6}+a_{7n+7}\le 12$$ Now, define $$b_k=\sum_{i=1}^{k}a_i$$ for all $k=1,2\ldots$. By the pigeonhole principle, we can find some $l>m$ such that: $$b_l\equiv b_m\pmod {20}$$ and $l-m\le 20$. This means that: $$\sum_{i=m+1}^{l}a_i\equiv 0\pmod {20}$$ and $$1\le\sum_{i=m+1}^{l}a_i<3\cdot 12=36$$ Since the chess player plays at least one game per day and $l>m$. We conclude that: $$\sum_{i=m+1}^{l}a_i=20$$ and we are done.
The following is multiple choice question (with options) to answer.
20 chess players take part in a tournament. Every player plays twice with each of his opponents. How many games are to be played? | [
"380",
"470",
"560",
"650"
] | A | 2*20C2 = 2*190 = 380
The answer is A. |
AQUA-RAT | AQUA-RAT-36796 | computer-architecture
So, if you were to try to subtract $01110000−11101011$, let's get an answer first so that we can check ourselves. $64+32+16 = 112$, and $-128+64+32+8+2+1 = -21$. You are performing the subtraction $112 - -21$. This means that our final answer should be $133$.
So, just as you can subtract by negating before you add (ie. $112 + 21$ instead of $112 - -21$), we can perform a similar operation here. If we negate the second number, we flip the bits and add one: $00010100 + 1 = 00010101$. But now that we've negated the number, our subtraction has become an addition problem. So, our original $01110000−11101011$ has become $01110000 + 00010101$
01110000
+ 00010101
--------
10000101
The following is multiple choice question (with options) to answer.
Find the number which when added to itself 13 times, gives 112. | [
"7",
"8",
"9",
"10"
] | B | Let the number be x. Then, x + 13x = 112 14x = 112 x = 8.
ANSWER:B |
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