source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37697 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 172 votes. Find the total number of valid votes? | [
"430",
"570",
"480",
"520"
] | A | Let the total number of valid votes be x.
70% of x = 70/100 * x = 7x/10
Number of votes secured by the other candidate = x - 7x/100 = 3x/10
Given, 7x/10 - 3x/10 = 172 => 4x/10 = 172
=> 4x = 1720 => x = 430.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37698 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
If a * b * c =(√(a + 2)(b + 3)) / (c + 1), find the value of 6 * 15 * 1. | [
"8",
"5",
"2",
"3"
] | C | 6 * 15 * 1 = (√(6 + 2)(15 + 3) )/ (1 + 1)
= (√8 * 18) / 2
= (√144) / 2
= 12 / 6 = 2
Answer is C. |
AQUA-RAT | AQUA-RAT-37699 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A certain school implemented a reading program for its students, with the goal of getting each student to read 6 books per month year-round. If the school has c classes made up of s students in each class, how many books will the entire student body read in one year? | [
"20cs",
"cs/2",
"72cs",
"(2cs)/12"
] | C | Ans: C
Solution: simple multiplication s students , c classes , 6 books/ month= 72 books a year
total number of books = 72cs |
AQUA-RAT | AQUA-RAT-37700 | wheeled-robot, rocker-bogie
Title: Rotation ratio between left rocker and right rocker in rocker-bogie system Following, the previous question, I am trying to calculate how much one rocker would rotate when the other is being rotated. I attached my calculation here.
I am trying calculate the rotation of gear B that connects to right rocket. Given gear A rotates at 0.05 rad, what is the rotation of gear B in rad? Gear ratio A:D is 4:1, and D:B is 1:4.
At the end, I ended up with rotational gear A = gear B. This somewhat puzzles me. Is my calculation correct? Your calculation is correct in magnitude but incorrect in sign, because gear B rotates oppositely to A (when the axis of D is fixed and D is not locked).
If D is locked (ie, the gear is not free to rotate in its plane) then A and B are locked together and rotate identically.
If the body V to which the axis of D is fixed rotates during rotation of A, then the rotation rate of B will differ from that of A. Example: Let rotation directions for A, B, and V be stated relative to a view from the left, and for D relative to a view from above. With V fixed, suppose A rotates CW at 40 rpm. Then D rotates CCW at 10 rpm, driving B CCW at 40 rpm. If V now begins to rotate CW at 20 rpm, D's rotation rate drops to 5 rpm, so that B begins to rotate at -20 rpm relative to A, 0 rpm relative to V, and 20 rpm to the frame of reference.
The following is multiple choice question (with options) to answer.
Circular gears L and R start to rotate at the same time at the same rate. Gear L makes 20 complete revolutions per minute and gear R makes 40 revolutions per minute. How many seconds after the gears start to rotate will gear R have made exactly 6 more revolutions than gear L? | [
"a) 6",
"b) 8",
"c) 10",
"d) 12"
] | A | Gear L -- 20 rotations per 60 seconds -- 2 rotation per 6 seconds.
Gear R -- 40 rotations per 60 seconds -- 4 rotations per 6 seconds.
First 6 seconds -- Gear L makes 1 rotation. -- Gear R makes 4 rotations -- Net difference -- 2 rotations
Hence every 6 seconds the difference between the number of rotations of R and L gear is 2 units.
Required net difference should be 6 rotations => 3 (6 seconds later) ==> 18 seconds.
Answer: A |
AQUA-RAT | AQUA-RAT-37701 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
In one hour, a boat goes 11 km along the stream and 3 km against the stream. the sped of the boat in still water (in km/hr) is : | [
"3",
"5",
"7",
"9"
] | C | Solution
Speed in still water= 1/2(11+3) km/hr= 7kmph. Answer C |
AQUA-RAT | AQUA-RAT-37702 | 'Y': 0.01842499064721287, 'X': 0.0014029180695847362, 'Z': 0.0006546950991395436}
The following is multiple choice question (with options) to answer.
Given that X and Y are events such that
Z(X)= 0.02
Z(Y)=0.10
Z(XnY)=0.10
Z(X|Y)=
find Z (X|Y) and Z(X|Y) | [
"1/5",
"1/6",
"1/6",
"1/4"
] | A | Here, X and Y are events
Z(X|Y) = Z(XnY)/Z(Y) = 0.10/0.10= 1
Q(G|H) = Z(XnY)/Z(X) = 0.4/0.02 = 1/5
Answer is A |
AQUA-RAT | AQUA-RAT-37703 | +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.
Patrick purchased 80 pencils and sold them at a loss equal to the selling price of 16 pencils. The cost of 80 pencils is how many times the selling price of 80 pencils? | [
" 0.75",
" 1.2",
" 1.8",
" 2.4"
] | B | Say the cost price of 80 pencils was $80 ($1 per pencil) and the selling price of 1 pencil was p.
Selling at a loss: 80 - 80p = 16p --> p = 5/6.
(cost price)/(selling price) = 1/(5/6) = 6/5 = 1.2.
Answer: B. |
AQUA-RAT | AQUA-RAT-37704 | everyday-chemistry, environmental-chemistry
You can also think of the compound like this, based on component masses:
$$\text{Molar mass} = 0.28 \times M(\ce{K2O}) + 0.575 \times M(\ce{SiO2}) + 0.145 \times M(\ce{H2O}) = \pu{63.532 g/mol},$$ and the average molar formula is $\ce{K2O - 3.2SiO2 - 2.7H2O}$.
Finally, I would recall that dissolution of $\ce{K2O}$ in general produces a basic solution, so if cucumbers are fond of acid (I don't know), you may consider adding a little acid ($\ce{HNO3}$, perhaps) to the solution to avoid perturbing the greens. But, you're the farmer!
The following is multiple choice question (with options) to answer.
Each of the cucumbers in 100 pounds of cucumbers is composed of 99% water, by weight. After some of the water evaporates, the cucumbers are now 97% water by weight. What is the new weight of the cucumbers, in pounds? | [
"2",
"33",
"92",
"96"
] | B | Out of 100 pounds 99% or 99 pounds is water and 1 pound is non-water. After somewaterevaporates the cucumbers become 97% water and 3% of non-water, so now 1 pound of non-water composes 3% of cucucmbers, which means that the new weight of cucumbers is 1/0.03=34 pounds.
Answer: B. |
AQUA-RAT | AQUA-RAT-37705 | 3. 0725. This calculator shows the steps and work to convert a fraction to a decimal number. Write the repeating part as the numerator of the fraction. Okay, so you can use division to see if a fraction has a repeating decimal, but how do you convert a repeating decimal into a fraction? Changing a repeating decimal in which all numbers after the decimal repeat, such as zero point one repeating, zero point two repeating, etc. Ask students to describe each piece as a fraction (1/8), a decimal (0. $$\frac{decimal}{1}$$ Step 2: For Every number after the decimal point multiply by 10 for both top and bottom (i. 125 Arranging decimal numbers by size When comparing decimal numbers and arranging them in order it is usually easiest to line up the numbers vertically with the decimal points in a vertical line. Convert fractions to repeating decimals Grade 5 Decimals Worksheet Convert to decimals, round to 3 digits if necessary. 5 ), this percentage is simpler to convert than was the previous one. Here is a simple online Recurring Decimal to Fraction Calculator to convert from Repeating Decimal to Fraction. to Fractions Focus 4 - Learning Goal #1: Students will know that there are numbers that are not rational, and approximate them with rational numbers. Writing Terminating Decimals as Fractions. 3 2 1 0. Example: Convert 0. 7025 = 3 + . 444444…. Name: _____ Converting Fractions, Decimals, and Percents fraction decimal percent a. 5555555555 Step 2: After examination, the repeating digit is 5 Step 3: To place the repeating digit ( 5 ) to the left of the decimal point, you need to move the decimal point 1 repeating part of the number is immediately next to the decimal. Converting decimals to fractions To convert a decimal to a fraction we rst split it into the whole number part added to the decimal part Example: 3. Convert decimal to fraction changing 1. Fractions as decimals. Decide how many numbers are repeating. org and *. Math Worksheets Fractions as Decimals Fractions as Decimals. org This file derived from G7-M2-TE-1. Use repeating DECIMALS INTO. To convert fractions to decimals, look at the fraction as a division problem. Students should be able to convert fractions to decimals, decimals
The following is multiple choice question (with options) to answer.
Half of 3 percent written as decimal is | [
"5",
"0.5",
"0.015",
"0.005"
] | C | Explanation:
It will be 1/2(3%) = 1/2(3/100) = 3/200 = 0.015
Option C |
AQUA-RAT | AQUA-RAT-37706 | ### 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]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
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Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Out of 40 applicants to a law school, 15 majored in political science, 20 had a grade point average higher than 3.0, and 10 did not major in political science and had a GPA equal to or lower than 3.0. How many E applicants majored in political science and had a GPA higher than 3.0? | [
"5",
"10",
"15",
"25"
] | A | Total applicants = 40
Political science = 15 and Non political science = 40 - 15 = 25
GPA > 3.0 = 20 and GPA <= 3.0 = 20
10 Non political science students had GPA <= 3.0 --> 15 Non political science students had GPA > 3.0
GPA > 3.0 in political science = Total - (GPA > 3.0 in non political science) E= 20 - 15 = 5
Answer: A |
AQUA-RAT | AQUA-RAT-37707 | (A) 10
(B) 45
(C) 50
(D) 55
(E) 65
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
15 Nov 2017, 16:22
9
3
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
We can create the following equation:
Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things
150 = 0.6(150) + 0.5(150) + 0.3(150) - D - 2(5) + 5
150 = 90 + 75 + 45 - D - 10 + 5
150 = 205 - D
D = 55
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
### Show Tags
29 May 2017, 05:24
8
5
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
This can be solved using Venn Diagram (refer the attachment)
AC = 60% of 150 = 90
Sunporch = 50% of 150 = 75
SP = 30% of 150 = 45
The following is multiple choice question (with options) to answer.
One night a certain hotel rented 2/5 of its rooms, including 2/3 of their air conditioned rooms. If 3/5 of its rooms were air conditioned, what percent of the rooms that were not rented were air conditioned? | [
"20%",
"25%",
"33%",
"36%"
] | C | The rooms which were not rented is 3/5
The AC rooms which were not rented is (1/3)*(3/5) = 1/5
The percentage of unrented rooms which were AC rooms is (1/5) / (3/5) = 1/3 = 33%
The answer is C. |
AQUA-RAT | AQUA-RAT-37708 | 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 job in 30 days and B can do it in 30 days. A and B working together will finish twice the amount of work in ------- days? | [
"14 days",
"26 days",
"30 days",
"31 days"
] | C | C
1/30 + 1/30 = 1/15
15*2 = 30 days |
AQUA-RAT | AQUA-RAT-37709 | -64 cube root of 64 = \., so factor out ( x + 4 ) ( cube root of 64 ) and imaginary! Real ( -4 ) ( x^2 -4x + 16 ) = 0 -64 =! Following numbers by prime factorisation method to explain, we can understand why such myth carries on explain. Of 2 up there symbol is called the radical spans over the entire equation for which the root to... Here is arrange the 2 multiplication so it looks like 8 * 8 you get! New password via email is -4: -4 * -4 = -64 ( ). 8 * 8 times will give you 64. which is 4 cubed ( or 43 )?... Note that the only thing we 've done here is arrange the 2 multiplication so it looks like *! Use synthetic division for this part ): ( x + 4 ) + 16 ) = 0 the multiplication! Equation for which the root is to be a perfect cube is 64, which happens to a. ) and two imaginary by giving you a small division lesson which does n't give perfect! Similar primes by 3 indeed \ [ \sqrt { 64 } =\not \pm 8\ ],. Part ): ( x + 4 ) like 8 * 8 get thousands of step-by-step solutions your! And the nearest next perfect cube is 125 because 4 x 4 x 4 is equal to.! The beach container at the beach to the fifth decimal place then similar... Know one root is -4, so factor out ( x + 4 ) done is! To the fifth decimal place for which the root is -4, so factor out ( x + 4.! See, 0.8 is greater than 0.64 receive a link and will create a password. So it looks like 8 * 8 to 64 -4 = -64 from to... Roots, start with: x^3 = -64 receive a link and will create a new password via email *. 64 radical notation exponential cube root of 64 Calculate the cube root of 0.64 is...., we used a scientific calculator and typed in 64 and the nearest next perfect.. -64, one real ( -4 ) ( x^2 -4x + 16 ) = ( -4 ) -4. This from problems you have done one is 65.The nearest previous perfect cube is 125: ( x +
The following is multiple choice question (with options) to answer.
Find k if 64 ÷ k = 4. | [
"8",
"12",
"16",
"25"
] | C | Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then
k = 16
correct answer is C)16 |
AQUA-RAT | AQUA-RAT-37710 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If [x] is the greatest integer less than or equal to x, what is the value of [-5.6]+[3.4]+[12.7]? | [
" 3",
" 5",
" 7",
" 9"
] | A | You are asked what the closest lesser integer value to [x] is.
[-5.6] = -6.0
[3.4] = 3.0
[12.7] = 12.0
Therefore , answer is: -6.0 + 3.0 + 12.0 = 9.0
Option D. |
AQUA-RAT | AQUA-RAT-37711 | ## Extra 01
How many arrangements can be made using the letters from the word COURAGE? What if the arrangements must contain a vowel in the beginning?
• $$4 \times 6!$$
## Extra Problem 02
How many arrangements are possible using the words
• EYE
• CARAVAN?
## 3(a)
There are (p+q) items, of which p items are homogeneous and q items are heterogeneous. How many arrangements are possible?
## 2(j)
There are 10 letters, of which some are homogeneous while others are heterogeneous. The letters can be arranged in 30240 ways. How many homogeneous letters are there?
Let, $$m = \text{number of homogeneous items}$$
• n(arrangements) = 30240 = $$\frac {10!}{m!}$$
• $$m! = \frac{10!}{30240}=120$$
• m = 5
## 2(k)
A library has 8 copies of one book, 3 copies of another two books each, 5 copies of another two books each and single copy of 10 books. In how many ways can they be arranged?
Total books = $$1 \times 8+3 \times 2+5 \times 2 + 8 \times 1 + 10$$ = 42
• n(arrangements) = $$\frac{42}{8!(3!)^2(5!)^2}$$
## 2(l)
A man has one white, two red, and three green flags; how many different signals can he produce, each containing five flags and one above another?
Flags: W = 2, R = 2, G = 3, Total = 7
Answer
## 2 (m)
A man has one white, two red, and three green flags. How many different signals can he make, if he uses five flags, one above another?
## 3(a)
How many different arragnements can be made using the letters of the word ENGINEERING? In how many of them do the three E’s stand together? In how many do the E’s stand first?
i
ii
iii
## 3(b)
In how many ways can the letters of the word CHITTAGONG be arranged, so that all vowels are together?
The following is multiple choice question (with options) to answer.
In how many different number of ways the letters of the word 'ENGINEERING' can be arranged | [
"276800",
"277000",
"277200",
"277400"
] | C | 11!/(3! × 3! × 2! × 2! × 1!)
= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 3 × 2 × 2 × 2 × 1)
= 11 × 10 × 8 × 7 × 5 × 3
= 277200
C) |
AQUA-RAT | AQUA-RAT-37712 | So:
W_2(L,F) = 0, If 2F+1 > L
W_2(L,F) = L - 2F, If 2F+1 <= L
We can compute W_2(L), for all valid F, by summing W_2(L,F) for F ranging from 1 to (L-1)/2:
...skipping the algebra...
If L is odd, W_2(L) = (L^2 - 2L + 1) / 4
If L is even, W_2(L) = (L^2 - 2L) / 4
Now, we can use this result in calculating the three-number problem:
If we have a first number F, then our second number, S, and third number, T, must satisfy: F < S < T F + S + T <= L
Assuming we have chosen an F small enough that there are S and T that can satisfy F + S + T <= L, then
W_3(L,F) = W_2(L - 3F)
W_3(L,F) = ((L-3F)^2-2*(L-3F)+1)/4, if 3F-L is odd
W_3(L,F) = ((L-3F)^2-2*(L-3F))/4, if 3F-L is even
Our upper limit for F is F + (F + 1) + (F + 2) <= L or F <= (L - 3) / 3
Finally, we can compute W_3(L) for all valid F, by summing W_3(L,F) for F ranging from 1 to I = Int((L-3)/3), where INT() rounds down to the nearest integer.
...skipping even more algebra...
If (L-3)/3 is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I)/8
The following is multiple choice question (with options) to answer.
If W is an integer greater than 6, which of the following must be divisible by 3? | [
"W(W+1)(W-4)",
"N(N+2)(N-1)",
"N(N+3)(N-5)",
"N(N+4)(N-2)"
] | A | Now take W = 3k , W = 3k+1 W = 3k+2..put in all the choices. If by putting all the values of W we get it is divisible by 3, then it is correct answer choice.
A is correct. It will hardy take 10 sec per choice as we have to consider only 3k+1 and 3k+2. |
AQUA-RAT | AQUA-RAT-37713 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface areas? | [
"81:121",
"81:122",
"81:129",
"81:120"
] | A | Ratio of the sides = ³√729 : ³√1331
= 9 : 11
Ratio of surface areas = 92 : 112
= 81:121
Answer: A |
AQUA-RAT | AQUA-RAT-37714 | Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$
-
Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring: \begin{align} \left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2 &=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\ &=8+2\sqrt{16-12}\\[6pt] &=12 \end{align} Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
-
Write as $\sqrt{4+2\sqrt{3}} = a+b\sqrt{3}$. Now square both sides, equate real and radical part. This gives two equations in $a$ and $b$. Now eliminate $a$, solve for $b$. Goes perfect. Same for the other term.
-
The following is multiple choice question (with options) to answer.
4/x + 3x = 3(x+12) | [
"-9",
"-1/3",
"-1/9",
"1/9"
] | D | We can solve - expand the right side, multiply by x on both sides and then subtract away the 3x^2 terms:
(4/X) + 3x = 3(x+12)
(4/x) + 3x = 3x + 36
4 + 3x^2 = 3x^2 + 36x
4 = 36x
1/9 = x
and to confirm, you can plug that answer back into the original equation to see that it makes the left and right sides equal.
D |
AQUA-RAT | AQUA-RAT-37715 | ### Show Tags
16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25 [#permalink]
### Show Tags
25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails
(2.60 * 12)/0.13 = 240 nails (Ans)
_________________
"Please hit +1 Kudos if you like this post"
_________________
Manish
The following is multiple choice question (with options) to answer.
The price of a table is marked at $12,000. If successive discounts of 15%, 10% and 5% be allowed, then at what price does a customer buy it? | [
"8721",
"2304",
"7720",
"6221"
] | A | A
8721
Actual price = 95% of 90% of 85% of $12000
= 95/100 * 90/100 * 85/100 * 12000 = $8721. |
AQUA-RAT | AQUA-RAT-37716 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
Difference between two numbers is 4, six times of the smaller lacks by 6 from the four times of the greater. Find the numbers? | [
"5,9",
"7,9",
"9,8",
"11,12"
] | A | Explanation:
x – y = 4
4x – 6y = 6
x = 9 y = 5
A) |
AQUA-RAT | AQUA-RAT-37717 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
Tom and Linda stand at point A. Linda begins to walk in a straight line away from Tom at a constant rate of 2 miles per hour. One hour later, Tom begins to jog in a straight line in the exact opposite direction at a constant rate of 8 miles per hour. If both Tom and Linda travel indefinitely, what is the positive difference, in minutes, between the amount of time it takes Tom to cover half of the distance that Linda has covered and the amount of time it takes Tom to cover twice the distance that Linda has covered? | [
"50",
"72",
"84",
"90"
] | A | A is the answer....
D = TS where D=distance, T=Time and S=Speed
To travel half distance, (2+2T) = 8T ==> T = 1/6 ==> 10 minutes
To travel double distance, 2(2+2T) = 8T ==> 1 ==> 60 minutes
Difference, 50 minutes
A |
AQUA-RAT | AQUA-RAT-37718 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is? | [
"16 km/hr",
"17 km/hr",
"60 km/hr",
"15 km/hr"
] | C | Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(100 + 100)/8 = 3x => x = 25/3.
So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.
Answer: C |
AQUA-RAT | AQUA-RAT-37719 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
An investment of $3000 was made in a certain account and earned interest that was compounded annually. The annual interest rate was fixed for the duration of the investment, and after 12 years the $3000 increased to $12000 by earning interest. In how many years after the initial investment was made the $3000 have increased to $24000 by earning interest at that rate? | [
"16",
"22",
"20",
"18"
] | D | In 12 years the investment quadrupled (from $3,000 to $12,000).
Thus, at the same rate compounded annually, it would need additional 12/2=6 years to double (from $12,000 to $24,000).
Therefore, 12+6=18 years are needed $3,000 to increase to $24,000.
Answer: D. |
AQUA-RAT | AQUA-RAT-37720 | (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.
John bought a shirt on sale for 25% off the original price and another 25 % off the discounted price. If the final price was $15, what was the price before the first discount? | [
"$45.10",
"$34.31",
"$28.44",
"$26.66"
] | D | let x be the price before the first discount. The price after the first discount is
x - 25%x (price after first discount)
A second discount of 25% of the discounted price after which the final price is 15
(x - 25%x) - 25%(x - 25%x) = 15
Solve for x
x = $26.66
correct answer D |
AQUA-RAT | AQUA-RAT-37721 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 13 Jul 2018, 06:28.
Last edited by EgmatQuantExpert on 12 Aug 2018, 23:04, edited 2 times in total.
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Joined: 04 Aug 2010
Posts: 322
Schools: Dartmouth College
Re: Question of the Week- 7 ( Three pipes P, Q, and R are attached to a..) [#permalink]
### Show Tags
Updated on: 18 Jul 2018, 12:09
3
EgmatQuantExpert wrote:
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
Let the tank = 60 liters.
Since P takes 3 hours to fill the 60-liter tank, P's rate $$= \frac{60}{3} = 20$$ liters per hour.
Since Q takes 4 hours to fill the 60-liter tank, Q's rate $$= \frac{60}{4} = 15$$ liters per hour.
Since R takes 5 hours to empty the 60-liter tank, R's rate $$= \frac{60}{-5} = -12$$ liters per hour.
Since R works to EMPTY the tank, R's rate is negative.
The following is multiple choice question (with options) to answer.
Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full is? | [
"8 hrs",
"6 hrs",
"7 hrs",
"3 hrs"
] | C | (A + B)'s 1 hour work = (1/12 + 1/15) = 3/20
(A + C)'s 1 hour work = (1/12 + 1/20) = 2/15
Part filled in 2 hrs = (3/20 + 2/15) = 17/60
Part filled in 6 hrs = 3 * 17/60 = 17/20
Remaining part = 1 - 17/20 = 3/20
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
Total time taken to fill the tank = (6 + 1) = 7 hrs.
Answer: C |
AQUA-RAT | AQUA-RAT-37722 | • In other words, if you have $x$ going from $0$ to $1$ and then for any fixed value of $x$ you have $y$ going from $0$ to $3(1-x),$ and that's the same as saying $y$ is between $0$ and $3,$ and for each fixed value of $y$ between $0$ and $3$ you have $x$ going from $0$ to $(3-y)/3. \qquad$ – Michael Hardy Dec 25 '18 at 19:37
The following is multiple choice question (with options) to answer.
If x > 3 and y < -3, then which of the following must be true ? | [
"x/y > 1",
"x/y < -1",
"x/y < 0",
"x + y > 0"
] | D | Pick x=3, y = -3
A) x/y > 1 - Incorrect as x/y = -1
B) x/y < -1 - Incorrect as x/y = -1
C) x/y < 0 -Correct. It will hold for all values x > 2 and y < -2 as x/y = -1 < 0
D) x + y > 0 - Incorrect. x + y = 0
E) xy > 0 - Incorrect. XY = -9 which is less than zero.
D should be the answer. |
AQUA-RAT | AQUA-RAT-37723 | 2021-02-22
Step 1
Given $$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$Q(x) = (((3x - 5)x + 1)x-3)x+5$$
$$R(x) = x^{5} -2x^{4} + 3x^{3} - 2x^{2} + 3x + 4$$
Expand Q
$$Q(x) = (((3x - 5)x + 1)x-3)x + 5$$
$$=((3x^{2} - 5x + 1)x-3)x + 5$$
$$=(3x^{3} - 5x^{2} + x - 3)x + 5$$
$$= 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$\text{So}, P(x) = Q(x) = 3x^{4} -5x^{3} +x^{2} - 3x + 5$$
Hence proved
Step 2
Evaluate P(2) and Q(2)
$$P(x) = 3x^{4} - 5x^{3} + x^{2} - 3x + 5$$
$$P(2) = 3(2)^{4} - 5(2)^{3} + (2)^{2} - 3(2) + 5$$
$$= 48 - 40 + 4 - 6 + 5$$
$$=11$$
$$Q(2) = (((3(2) - 5)2+1)2- 3) 2 + 5$$
$$=((3(2) + 1)2 - 3)2 +5$$
$$=((3(2) - 3)2 + 5$$
$$= (3)2+5$$
$$= 11$$
Nested form of R(x)
The following is multiple choice question (with options) to answer.
(3X+2)(2X-5)=AX^2+ KX+N;
THEN WHAT IS THE VALUE OF A-N+K? | [
"4",
"5",
"6",
"7"
] | B | (3x+2)(2x-5)=6x^2-11x-10
So,A=6, K=-11, N=-10
Finally,
A-N+K=6-(-10)+(-11)
=6+10-11
=5
ANSWER:B |
AQUA-RAT | AQUA-RAT-37724 | So:
W_2(L,F) = 0, If 2F+1 > L
W_2(L,F) = L - 2F, If 2F+1 <= L
We can compute W_2(L), for all valid F, by summing W_2(L,F) for F ranging from 1 to (L-1)/2:
...skipping the algebra...
If L is odd, W_2(L) = (L^2 - 2L + 1) / 4
If L is even, W_2(L) = (L^2 - 2L) / 4
Now, we can use this result in calculating the three-number problem:
If we have a first number F, then our second number, S, and third number, T, must satisfy: F < S < T F + S + T <= L
Assuming we have chosen an F small enough that there are S and T that can satisfy F + S + T <= L, then
W_3(L,F) = W_2(L - 3F)
W_3(L,F) = ((L-3F)^2-2*(L-3F)+1)/4, if 3F-L is odd
W_3(L,F) = ((L-3F)^2-2*(L-3F))/4, if 3F-L is even
Our upper limit for F is F + (F + 1) + (F + 2) <= L or F <= (L - 3) / 3
Finally, we can compute W_3(L) for all valid F, by summing W_3(L,F) for F ranging from 1 to I = Int((L-3)/3), where INT() rounds down to the nearest integer.
...skipping even more algebra...
If (L-3)/3 is even, W_3(L) = (6I^3-3(2L-5)I^2+2*(L^2-5L+5)I)/8
The following is multiple choice question (with options) to answer.
when W is divided by 14, the reminder is 0. if W is three more than it value and when divided by 9 its remainder is 0. what is the value of W ? | [
"42",
"14",
"28",
"30"
] | A | W is divided by 14 so that is multiple of 14 as 14,28,42...
W+3 is divided by 9 the remainder is 0 so it is divisible by 9. Consider from option let us take the number is 42 it is divisible by 14 but 42+3 is divisible by 9 so ans is A |
AQUA-RAT | AQUA-RAT-37725 | We can argue that the are 21/7= 3 times as many men now so they will require three times as much water, 3(13)= 39 quarts just for the same two weeks. And they need the water for 4/2= 2 times as many weeks so need 2(39)= 78 quarts of water.
The following is multiple choice question (with options) to answer.
If 20 men can build a water fountain 56 metres long in 21 days, what length of a similar water fountain can be built by 35 men in 3 days? | [
"10 m",
"14 m",
"17 m",
"19 m"
] | B | Explanation :
Let the required length be x metres
More men, More length built (Direct Proportion)
Less days, Less length built (Direct Proportion)
Men 20: 35
Days 21: 3 : : 56 : x
Therefore (20 x 21 x x)=(35 x 3 x 56)
x=(35 x 3 x 56)/420=14
Hence, the required length is 14 m.
Answer : B |
AQUA-RAT | AQUA-RAT-37726 | 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, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together? | [
"3 days",
"4 days",
"5 days",
"6 days"
] | B | Explanation:
1/7 + 1/14 + 1/28 = 7/28 = 1/4 => 4 days
ANSWER IS B |
AQUA-RAT | AQUA-RAT-37727 | probability of guessing the correct password
Suppose I need to guess a password of one digit. Each time I'm wrong the password increases by another digit. What's the probability that I can correctly guess the password (assuming I have unlimited amount of time and the number is all uniform)?
My approach: since the number is generated random, it doesn't matter what strategy we adopt to guess the number. So consider the following sequence of guess $$1, 21, 221, 2221, 22221, ...$$. The probability that the password results in $$k$$-th guess is then $$\frac{1}{10^n}$$. Thus the total probability is $$\sum_n \frac{1}{10^n}$$. So the probability is $$0.11111111111... = \frac{1}{9}$$.
Is my approach correct, and if so, is there any better solution for this. The fraction $$\frac{1}{9}$$ seems to suggest an easier perspective to look at this puzzle.
The following is multiple choice question (with options) to answer.
A password for a computer uses five digits from 0 to 9, inclusive. What is the probability that the digits of the password solely consist of prime numbers or zero? | [
"1/32",
"1/16",
"1/8",
"2/5"
] | A | Favorable outcomes will be obtained by making Password using {0, 2, 3, 5, 7} i.e. 5 digits
Total Such Passwords = 5*5*5*5*5 = 5^5
Total Possible Password = 10*10*10*10*10 = 10^5
Probability = (5^5) / (10^5) = 1/2^5 = 1/32
Answer: option A |
AQUA-RAT | AQUA-RAT-37728 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
A truck driver starts with the speed of 70 kmph with the truck driver decreasing speed every two hours by 10 kmph. The truck driver's speed does not drop below 50 kmph. In how many hours will it take the truck driver to travel 500 kms? | [
"480 minutes",
"492 minutes",
"504 minutes",
"528 minutes"
] | D | Distance covered in first two hours = 70 × 2 = 140 km
Distance covered in next two hours = 60 × 2 = 120 km
Distance covered in next two hours = 50 × 2 = 100 km
Distance covered in next two hours = 50 × 2 = 100 km
Distance covered in first eight hours
140 + 120 + 100 + 100 = 460 km
Remaining distance = 500 – 460 = 40 km.
Now, this distance will be covered at the speed of 50 km/hr.
∴ Time taken = 40â„50 = 4â„5 hour.
Total time= 8 + 4â„5 = 8 4â„5 hour
Answer D |
AQUA-RAT | AQUA-RAT-37729 | Let's say P wins the first time they toss the coin. But since Q starts the game, Q must get a tails first. For both to happen, the probability is (0.8)(0.2).
Let's say P wins in the second chance they get to toss the coin. But in order for that to happen, Q should get a tails in both the chances they toss the coin and P should get a tails the first time and heads the second time. The probability of that happening is (0.8)(0.8)(0.8)(0.2).
This pattern progresses until such a time that P gets a head. Therefore, the sum of all these probabilities gives us the total probability of P winning the competition.
P(P) = (0.8)(0.2) + (0.8)(0.8)(0.8)(0.2) + ...
The following is multiple choice question (with options) to answer.
In a game, a player can win by getting five consecutive tails in six throws of a coin. What is the probability of winning if the player gets a tail on the first throw? | [
"1/32",
"1/16",
"3/16",
"1/8"
] | B | The only way to win is to get a tail on each of the next four throws.
P(four tails)=(1/2)^4=1/16
The answer is B. |
AQUA-RAT | AQUA-RAT-37730 | 3,5,-5... The first term in the sequence of #'s shown above is 3. Each even # term is 2 more than the previous term and each odd # term, after the first, is -1 times the previous term. For example, the second term is 3+2, and the
8. ### Math *URGENT
Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646? 3. Determine
9. ### maths
The fifth term of an arithmetic sequence is 23 and the 12th term is 72. What is the value of the 10th term. Which term has a value of 268.
10. ### Sj
Mathematics......Under number patterns-Geometric series If a question goes.determine the expression for the nth term of the following sequence if the a) 4th term is 24 and the 7th term is 192 in a geometric sequence.what formula
More Similar Questions
The following is multiple choice question (with options) to answer.
Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It continues in such a way that by adding 5 to the nth term, one obtains the (n + 2)th term. What is the sum of the first 17 members of this sequence? | [
"878",
"900",
"743",
"928"
] | C | I would split them up like this.
23 28 33 and so on (9 terms)....the 9th term = 23+8*5 = 63
27 32 37 and so on (8 terms)......the 8th term = 27+7*5 = 62
Since the distance between any two numbers is the same we can use arithmetics
first+last/2 times no of numbers = the sum
(23+63)/2 * 9 = 387
(27+62)/2 * 8 = 356
= 743
C |
AQUA-RAT | AQUA-RAT-37731 | ### Show Tags
23 Oct 2009, 23:42
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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
Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.
In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.
Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).
Consider:
$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:
$$ttttt|||$$
Means that first nephew will get all the tickets,
$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1
And so on.
How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.
The following is multiple choice question (with options) to answer.
Visitors to show were charged Rs.15 each on the first day. Rs.7.50 on the second day, Rs.2.50 on the third day and total attendance on the three days were in ratio 2:5:13 respectively. The average charge per person for the whole show is? | [
"2",
"3",
"5",
"6"
] | C | 2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
Average = 5.Answer: C |
AQUA-RAT | AQUA-RAT-37732 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
If A : B : C = 3 : 4 : 7, then what is the ratio of (A / B) : (B / C) : (C / A)? | [
"63 : 48 : 196",
"63 : 48 : 18",
"63 : 48 : 1928",
"63 : 48 : 1918"
] | A | Hint: If a = kb for some constant k, then we can say that a is directly proportional to b.
A : B : C = 3 : 4 : 7
Assume, A = 3 k, B = 4 k, C = 7 k
Therefore,
A = (3k) , B = (4k) , C = (7k)
B (4k) C (7k) A (3k)
A = (3) , B = (4) , C = (7)
B (4) C (7) A (3)
L.C.M of 3, 4, 7 is 84
(3 x 84) / 4 = 63
(4 x 84) / 7 = 48
(7 x 84) / 3 = 196\
Correct option :A |
AQUA-RAT | AQUA-RAT-37733 | Case 1: $$n = m + w$$
Since the women must be separated by the men, there must be at least as many men as women, so we require that $$m \geq w$$.
Hand each person a chair. Suppose Andrew is one of the $$m$$ men. Seat him first. The other men can be seated around the table in $$(m - 1)!$$ ways as we proceed clockwise around the table from Andrew. Seating the $$m$$ men creates $$m$$ spaces in which we can place a woman. To separate the women, we must choose $$w$$ of those $$m$$ spaces, which can be done in $$\binom{m}{w}$$ ways. The women can be arranged in those $$w$$ spaces in $$w!$$ ways as we proceed clockwise around the table from Andrew. Hence, there are $$(m - 1)!\binom{m}{w}w!$$ seating arrangements in which no two of the women are adjacent.
Case 2: $$n \geq m + w$$
Since the women must be separated by the men or by an empty seat, we require that $$w \leq \left\lfloor \dfrac{n}{2} \right\rfloor$$.
The following is multiple choice question (with options) to answer.
Along a road lie an odd number of stones placed at intervals of 10m. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of 4.8 km. Then the number of stones is | [
"35",
"15",
"29",
"31"
] | D | Explanation :
If he travels 10 m for the first stone he has to travel that dist back as well...so he travels 2 * 10 m dist for the first stone and so on.
=> 2400 = 2 * (10 + 20 + 30 .... 10n) using formula for AP
=> 2400 = 2 * n/2*(2*10 + (n-1)*10) .
=> 2400 = 10n2 + 10n .
On solving this, we get n = 15 as the positive root, so total numer of stones on both sides= 30.
and including the one which he was at = 31.
Answer : D |
AQUA-RAT | AQUA-RAT-37734 | Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Another approach.
Take the task of arranging the 5 letters and break it into stages.
Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways
IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.
Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)
[Reveal] Spoiler:
D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.
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The following is multiple choice question (with options) to answer.
How many words can be formed by using all letters of TIHAR | [
"100",
"120",
"140",
"160"
] | B | Explanation:
First thing to understand in this question is that it is a permutation question.
Total number of words = 5
Required number =
5P5=5!=5∗4∗3∗2∗1=120
Option B |
AQUA-RAT | AQUA-RAT-37735 | PRIME NUMBERS:
1. 1 is not a prime, since it only has one divisor, namely 1.
2. Only positive numbers can be primes.
3. There are infinitely many prime numbers.
4. the only even prime number is 2. Also 2 is the smallest prime.
5. All prime numbers except 2 and 5 end in 1, 3, 7 or 9.
PERFECT SQUARES
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;
2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;
3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);
4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.
IRRATIONAL NUMBERS
1. An irrational number is any real number that cannot be expressed as a ratio of integers.
2. The square root of any positive integer is either an integer or an irrational number. So, $$\sqrt{x}=\sqrt{integer}$$ cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).
This week's PS question
This week's DS Question
The following is multiple choice question (with options) to answer.
How many odd prime numbers are there less than 100? | [
"78",
"5",
"24",
"12"
] | C | Odd prime number less than 100:3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
There is 24 the odd prime number
Answer is C |
AQUA-RAT | AQUA-RAT-37736 | # Seating couples around 2 tables
Here's my question and possible answer.
How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:
1) each couple must sit at the same table, and,
2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).
My solution:
Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)
$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)
$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$
$=2 * 8!$
Can someone verify this solution or provide the correct one?
First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$
Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4}$ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.
Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.
Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.
Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$
And again, if you do not differentiate between the tables, then divide this by $2$
The following is multiple choice question (with options) to answer.
A newly-wed couple is using a website to design an eBook Wedding Album to distribute to their friends and families. The template they have chosen has places for 3 large photos and 19 smaller photos. The couple has 6 large photos they could use for those three slots, and 20 smaller photos they could use for those 19 slots. Given these choices, how many different possible albums could they create? | [
"3,150",
"400",
"5,040",
"20,520"
] | B | MAGOOSHOFFICIAL SOLUTION:
For the large photos, we need 6C3, which we calculated in the article:
6C3 = 20
For the smaller photos, we need 20C19, which by symmetry must equal 20C1, and we have a formula for that. In fact, in the article above, we already calculated that 21C2 = 20.
Now, by the FCP, we just multiply these: total number of possible albums = 20*20 = 400.
Answer = B |
AQUA-RAT | AQUA-RAT-37737 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of a train and that of a platform are equal. If with a speed of 90 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is? | [
"202",
"278",
"377",
"750"
] | D | Speed = [90 * 5/18] m/sec = 25 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 25 è x = 25 * 60 / 2
= 750
Answer: D |
AQUA-RAT | AQUA-RAT-37738 | You've got what it takes, but it will take everything you've got
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Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
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15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
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The following is multiple choice question (with options) to answer.
Marge has 3 pumps for filling her swimming pool. When all 3 pumps work at their maximum rates, the swimming pool is filled in 90 minutes. Pump 1's maximum rate is twice the maximum rate of pump 2 and four times the maximum rate of pump 3. How long would it take Marge to fill the pool if she used only pump 2 at its maximum rate ? | [
"2hrs, 48mins",
"5hrs, 15mins",
"7hrs, 12mins",
"13hrs, 4mins"
] | B | The rate of pump 1 = 4x job/minute.
The rate of pump 2 = 2x job/minute.
The rate of pump 3 = x job/minute.
Given that x+2x+4x=1/90 --> x=1/630 --> (time) = (reciprocal of rate) = 630 minutes x/2 = 5 hours and 15 minutes.
Answer: B. |
AQUA-RAT | AQUA-RAT-37739 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
Out of a group of 10 contestants, 2 are to be selected at random. What is the maximum number of male contestants possible if the probability that both selected contestants are male is less than 60% ? | [
" 0",
" 1.66",
" 2.55",
" 1.22"
] | D | Let x be the number of males.
Total contestants = 10
Probability that the first contestant selected is a male = x/10
Probability that the second contestant selected is also a male = x-1/9
Total probability = x(x-1)/ 90 < 1/60 (this probability is less than 60%)
Solving for x we get, x(x-1)< 3/2. This implies that the maximum value of x can be 1.22.
Answer D! |
AQUA-RAT | AQUA-RAT-37740 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 6 2/3 days and 5 days respectively. They work together for 2 days and then A leaves. In how many days after that B will complete the work alone. | [
"5 days",
"18 days",
"1 ½ days",
"25 days"
] | C | Explanation:
3/20 * 2 + (2 + x)/5 = 1
x = 1 ½ days
Answer C |
AQUA-RAT | AQUA-RAT-37741 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
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Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities
Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
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The following is multiple choice question (with options) to answer.
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 2 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is? | [
"60 gallons",
"100 gallons",
"120 gallons",
"80 gallons"
] | D | Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24) = - 1/40
Volume of 1/40 part = 2 gallons\
Volume of whole = 2 * 40 =80 gallons.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37742 | They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
You are so right, soroban. I actually solved for time instead of amount. I've edited my post to indicate that. Your conclusion using 10.5 years is good. I forgot what it was I was solving for. It's Christmas. I'm still full of eggnog.
• December 28th 2008, 07:21 AM
magentarita
yes...
Quote:
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
The following is multiple choice question (with options) to answer.
Sara and Alan start a company with Rs.100000 and Rs.50000 respectively. how should they share their profits at the end of one year? | [
"2:1",
"3:4",
"3:8",
"3:1"
] | A | The ratio of the investments made by A and B =
100000: 50000=> 2:1.Answer: A |
AQUA-RAT | AQUA-RAT-37743 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
In a question on division with zero remainder, a candidate took 15 as divisor instead of 5. The quotient obtained by him was 35. The correct quotient is? | [
"A)105",
"B)24",
"C)28",
"D)29"
] | A | Number = (35 * 15) = 525
Correct quotient = 525/5 = 105
A) |
AQUA-RAT | AQUA-RAT-37744 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
If John covers a certain distance in 1 hr. 24 min. by covering two third of the distance at 4 kmph and the rest at 5 kmph, then find the total distance | [
"4 km",
"6 km",
"6.8 km",
"7.2 km"
] | B | Explanation:
Let the total distance be y km. Then,
(2/3)y/4 +(1/3)y/5 =7/5
y/6 + y/15 = 7/5
7y =42
y = 6 km
ANSWER B |
AQUA-RAT | AQUA-RAT-37745 | I think the method you used is the best way to go.
Still, if you want to do it via the Chinese Remainder theorem....
Note that $$5000=2^3\times 5^4$$ so solve the problem mod $$2^3$$ and mod $$5^4$$ separately. Clearly the answer is $$0\pmod {5^4}$$ so that just leaves $$2^3$$. But $$15\equiv -1\pmod {2^3}$$ so the answer is $$1\pmod {2^3}$$. Now apply the CRT to $$n\equiv 0 \pmod {625}\quad \&\quad n\equiv 1\pmod {8}$$
Since $$625\equiv 1 \pmod {8}$$ the answer is $$625$$.
• Worth emphasis: we can eliminate the CRT calculation by essentially factoring $\,5^4 = 625\,$ out of the mod computation - as explained in my answer and its link. This often (greatly) simplifies modular computations of this sort. Jan 25, 2020 at 18:37
• @BillDubuque Good point, and the discussion you link to is well worth studying.
– lulu
Jan 25, 2020 at 18:39
The following is multiple choice question (with options) to answer.
Which of the following options is equal to 3.75 x 10(power 5)? | [
"175000",
"372000",
"375000",
"365000"
] | C | = 3.75 x 10(power 5)
= 3.75 x 100000
= 375000.
Answer is C. |
AQUA-RAT | AQUA-RAT-37746 | Probability of choosing all the numbers in a set
Suppose you have $m=5$ people and $n=4$ numbers. Each person chooses, independently and at random, one of these numbers.
What's the probability that all the numbers will be chosen?
I've been able to experimentally find a result, which is ~0.234146. I've also been able to enumerate the favorable outcomes and the possible outcomes, which gives me a probability of:
$P(\text{all numbers are chosen}) = \dfrac{240}{1024} = 0.234375$
which conforms to the experimental result.
What I'm not able to do is to get a combinatorial meaning for this number, even in the form of a general formula with parameters $m$ and $n$.
My (wrong) reasoning:
• Notice that, for all the numbers to be chosen, one of them must be chosen twice.
• Suppose that the duplicate number is chosen by person 1, then the other 4 people have $4!$ ways to choose the 4 numbers.
• Repeat the reasoning in case the duplicate number is chosen by person 2, 3, 4 or 5.
• Get a result of $\dfrac{n! \cdot m}{n^m} = \dfrac{4! \cdot 5}{4^5} = \dfrac{120}{1024}$, which is wrong.
So, where's the flaw in my reasoning, and how do you obtain the result in terms of $m$ and $n$?
The following is multiple choice question (with options) to answer.
A number is selected at random from the first 30 natural numbers. What is the probability that the number is a multiple of either 4 or 15? | [
"17/30",
"2/5",
"3/10",
"4/15"
] | C | Number of Multiples of 4 from 1 through 30 = 30/4 = 7
Number of Multiples of 15 from 1 through 30 = 30/15 = 2
Number of Multiples of 4 and 15 both from 1 through 30 = Number of Multiples of 15*4(=60) = 0
Total favourable cases = 7 + 2 - 0 = 9
Probability = 9 / 30 = 3/10
Answer: option C |
AQUA-RAT | AQUA-RAT-37747 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Before a salary hike, the weekly salary of a worker for 40 hours in a week was as much as he is paid now for 35 hours of work in a week. What is the percent increase in his salary for an hour? | [
"11 1/3",
"12 1/4",
"13 1/5",
"14 2/7"
] | D | 280 is multiple of both 40 and 35
so I would rather take this value as the salary.
40 hrs 280$
1 hr=7$
Similarly
35 hrs 280$
1 hr=8$
Now 8-7=1$ increase in salary per hour now to calculate %
1/7*100=14 2/7
ANSWER:D |
AQUA-RAT | AQUA-RAT-37748 | ## 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.
Length of a rectangular plot is 32 mtr more than its breadth. If the cost of fencin gthe plot at 26.50 per meter is Rs. 5300, what is the length of the plot in mtr? | [
"56 m",
"66 m",
"76 m",
"86 m"
] | B | Let breadth = x metres.
Then, length = (x + 32) metres.
Perimeter = 5300 m = 200 m.
26.50
2[(x + 32) + x] = 200
2x + 32 = 100
2x = 68
x = 34.
Hence, length = x + 32 = 66 m
B |
AQUA-RAT | AQUA-RAT-37749 | # 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.
Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart? | [
"17hr",
"14hr",
"12hr",
"19hr"
] | A | Explanation:
relative speed =5.5−5=0.5 kmph (because they walk in the same direction)
distance =8.5 km
time=distance/speed=8.5/0.5=17 hr
ANSWER IS A |
AQUA-RAT | AQUA-RAT-37750 | # 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.
In how many years will a sum of money doubles itself at 30% per annum on simple interest? | [
"7%",
"3%",
"5%",
"8%"
] | B | P = (P*30*R)/100
R = 3%
Answer:B |
AQUA-RAT | AQUA-RAT-37751 | 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.
A and B walk around a circular track. They start at 9 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m.? | [
"8",
"7",
"6",
"5"
] | B | Sol.
Relative speed = (2 + 3) = 5 rounds per hour.
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9.30 a.m.
Answer B |
AQUA-RAT | AQUA-RAT-37752 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 160 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"388",
"267",
"215",
"288"
] | C | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (160 + X)/30 = 25/2 ==> 2(160 + X) = 750 ==> X = 215 m.
Answer: C |
AQUA-RAT | AQUA-RAT-37753 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
If the price has fallen by 15% what percent of its consumption be: increased so that the expenditure may be the same as before? | [
"A)11%",
"B)10%",
"1/9 %",
"3/13 %"
] | D | 100 – 15 = 85
85------15
100------? => 3/13%
ANSWER:D |
AQUA-RAT | AQUA-RAT-37754 | moveit, roslaunch
00;36:*.oga=00;36:*.opus=00;36:*.spx=00;36:*.xspf=00;36:',
The following is multiple choice question (with options) to answer.
If 7:13::301:x then the value of ‘x’ is: | [
"493",
"537",
"559",
"587"
] | C | Given the question;
7:13::301:x
7/13 = 301/x
7x = 301*13
x = 301*13/7
x = 559
ANSWER:C |
AQUA-RAT | AQUA-RAT-37755 | $62 = 95% of$P or
$62 = 95/100$P so
$P =$6200/95 = $65.263 This is what your calculator is doing. Penny Hi Steve. The reason for this is that 65.263 - 5% = 62. It is a matter of perspective. If the regular price on a dress shirt is$50 and I put it on sale, selling it for 20% off, then you would rightly expect the selling price to be $40. But if I subsequently raise the price by 20%, then the new price is$48, not the original \$50.
This is because 50 - 50(0.20) = 40 + 40(0.25).
It becomes even more noticeable with higher markups like 80% (try it).
So when you mark up things manually, you are taking a 5% of the original cost and adding it to the original cost to make a new price. So your markup is 5% of the original cost. When you use the "markup" button on the calculator, you are using the manufacturer's definition of markup, which is to do it differently. The "markup" button says that 5% of the final price should be markup.
Stephen La Rocque >
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
The following is multiple choice question (with options) to answer.
If the S.P of Rs. 45 results in a 50% discount on the list price, What S.P would result in a 40% discount on the list price? | [
"s. 54",
"s. 56",
"s. 58",
"s. 60"
] | A | Let the list price be Rs. X,
50/100*x = 45, x = 45*100/50 = 90
Required S.P = 60% of Rs. 90
= 60*90/100
=54
ANSWER:A |
AQUA-RAT | AQUA-RAT-37756 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
The monthly incomes of A and B are in the ratio 5 : 2. B's monthly income is 12% more than C's monthly income. If C's monthly income is Rs. 12000, then find the annual income of A? | [
"Rs. 420000",
"Rs. 403200",
"Rs. 201600",
"Rs. 504000"
] | B | B's monthly income = 12000 * 112/100 = Rs. 13440
B's monthly income = 2 parts ----> Rs. 13440
A's monthly income = 5 parts = 5/2 * 13440 = Rs. 33600
A's annual income = Rs. 33600 * 12 = Rs. 403200
ANSWER:B |
AQUA-RAT | AQUA-RAT-37757 | P (not A). Describe the sample space. The maximum probability is 1=2 since the rst coin will land tails with probability at least 1=2. This probability is achievable if all three coins are equal, i. Let us see some examples of problems related to this. There are three boxes: a box containing two gold coins, a box containing two silver coins, a box containing one gold coin and one silver coin. 000015390771693 0. Probability definition is - the quality or state of being probable. Coin Probability Problem. Tossing a fair coin. If you tossed a coin twice and the first time it came up tails and the second time came up heads, you could assume that the probability the coin would land on head is 1/2. Theoretical probability is the ratio between the number of ways an event can occur and the total number of possible outcomes in the sample space. If a coin is tossed, there are two possible outcomes − Heads $(H)$ or Tails $(T)$ So, Total number of outcomes = 2. When a coin is tossed, there are two possible outcomes: heads (H) or ; tails (T) We say that the probability of the coin landing H is ½. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7500. In coin tossing example, the simple outcomes would be: heads or tails. In a desk drawer in the house of Mr Jay Parrino of Kansas City there is a coin, 1913 Liberty Head nickel. ) This problem has been solved!. 11) We have two coins, A and B. Number of heads. 5 – which means likely. The first coin is a fair coin painted blue on the head side and white on the tail side. We choose a coin at random and flip it once. Probability Tools. (a) Find P {X = 1}; (b) Determine E [X]. The branches emanating from any point must have probabilities that sum to 1. The probability of tossing a coin and getting 'heads' is 1 in 2. Tajdar Alam. If a coin is tossed, there are two possible outcomes − Heads $(H)$ or Tails $(T)$ So, Total number of outcomes = 2. If we use Bayes' Theorem from above, we can calculate. Describe the sample space. You can
The following is multiple choice question (with options) to answer.
A coin is tossed live times. What is the probability that there is at the least one tail? | [
"31/32",
"31/38",
"31/34",
"31/31"
] | A | Let P(T) be the probability of getting least one tail when the coin is tossed five times.
= There is not even a single tail.
i.e. all the outcomes are heads.
= 1/32 ; P(T) = 1 - 1/32 = 31/32
Answer: A |
AQUA-RAT | AQUA-RAT-37758 | => $$\frac{74}{100} \times x$$ = 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.
If 4 (P's Capital ) = 6 ( Q's Capital ) = 10 ( R's Capital ) , then out of the total profit of Rs 4650 , R will receive | [
"Rs. 600",
"Rs. 700",
"Rs. 800",
"Rs. 900"
] | D | Solution: Let
P's capital = p,
Q's capital = q and
R's capital = r.
Then
4p = 6q = 10r
=> 2p = 3q = 5r
=>q = 2p/3
r = 2p/5
P : Q : R = p : 2p/3 : 2p/5
= 15 : 10 : 6
R's share = 4650 * (6/31) = 150*6 = Rs. 900.
Answer: Option D |
AQUA-RAT | AQUA-RAT-37759 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many different number of ways 8 men and 2 women can sit on a shopa which can accommodate persons? | [
"151252",
"151225",
"151211",
"151200"
] | C | 10p6 = 10 x 9 x 8 x 7 x 6 x 5 = 151200
C) |
AQUA-RAT | AQUA-RAT-37760 | III. All faces same There are 2 colors, so two ways for all faces to be the same.
Adding them up, we have a total of 20 ways to have four vertical faces the same color. The are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$
The following is multiple choice question (with options) to answer.
The faces of Cuboid are to be painted by 6 different colors. In how many ways can be this be done? | [
"A.720",
"B.256",
"C.1",
"D.12"
] | A | If I have to paint 6 sides with 6 different colour...
First face can have 6c1 options,
2nd would have 5c1, and subsequent ones would have 4c1, 3c1, 2c1 and 1 options respectively.
Total options = 6c1 X 5c1 x 4c1 x 3c1 x 2c1 x 1 = 720 distinct ways
Correct answer - A |
AQUA-RAT | AQUA-RAT-37761 | 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 goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 36 seconds. What is the length of the goods train? | [
"470 m",
"240 m",
"260 m",
"270 m"
] | A | Explanation:
Speed =[ 72 x (5/18) ]m/sec= 20 m/sec.
Time = 36 sec.
Let the length of the train be x metres.
Then,[ (x+250)/36 ]= 20
=> x + 250 = 720
=> x = 470. Answer: A |
AQUA-RAT | AQUA-RAT-37762 | Of course not! It is actually due to the fact that $102 = 6 \cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \cdot 17$, then $391 = 17 \cdot (17 + 6) = 17 \cdot 23$, and $493 = 17 \cdot (23 + 6) = 17 \cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.
This sequence of multiples of 17 starts from $17 \cdot 5$, and if we continue it further we see that it contains several more numbers from our exceptional set as well:
$\begin{array}{rcl}85 &=& 17 \cdot 5 \\ 187 &=& 17 \cdot 11 \\ \mathbf{289} &=& 17 \cdot 17 \\ \mathbf{391} &=& 17 \cdot 23 \\ \mathbf{493} &=& 17 \cdot 29 \\ 595 &=& 17 \cdot 35 \\ \mathbf{697} &=& 17 \cdot 41 \\ \mathbf{799} &=& 17 \cdot 47 \\ \mathbf{901} &=& 17 \cdot 53 \end{array}$
What about if we start with $17 \cdot 7$ and keep adding $102$?
The following is multiple choice question (with options) to answer.
If the sum of a set of 5 different positive prime numbers is 94, which of the following prime numbers must be in the set? | [
"2,7,43",
"3,11,31",
"5,71,31",
"7,43,53"
] | A | All prime numbers apart from 2 are odd.
Even + Even = Even
Odd + Even = Odd
Odd + Odd = Even
We are given 5 different prime numbers, whose sum is 94 i.e even
If we include 2, we will have 4 odd prime numbers and one even.
This sum would be odd
If we exclude 2, we will have 5 odd numbers.
This sum would be even
Hence 2 is included.
Option A |
AQUA-RAT | AQUA-RAT-37763 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
The average marks of a class of 30 students is 40 and that of another class of 50 students is 90. Find the average marks of all the students? | [
"71.25",
"52.9",
"52.1",
"52.3"
] | A | Sum of the marks for the class of 30 students = 30 * 40 = 1200
Sum of the marks for the class of 50 students = 50 * 90 = 4500
Sum of the marks for the class of 80 students =
1200 + 4500 = 5700
Average marks of all the students = 5700/80
= 71.25
Answer:A |
AQUA-RAT | AQUA-RAT-37764 | # 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 distributor sells a product through an on-line store, which take a commission of 20% of the price set by the distributor. The distributor obtains the product from a producer at the price of $15 per item. What is the price that the buyer observers on-line if the distributor wants to maintain a 10% profit on the cost of the item? | [
"18",
"21.6",
"20",
"22.5"
] | C | Producer price = $15;
The distributor wants to maintain a 20% profit on the cost of the item, thus he must get $15*1.2 = $18 after the store takes a commission of 10% of the final price --> (final price)*0.9 = $18 --> (final price) = $20.
Answer: C. |
AQUA-RAT | AQUA-RAT-37765 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
A man wants to sell his scooter .There are two offers one at Rs12000 cash and other at a credit of Rs12880 to be paid after 8 months ,money being at 18% per annum which is better offer? | [
"24277",
"12000",
"28372",
"36788"
] | B | Explanation:
PW of Rs.12,880 due 8 months hence
= Rs\inline \fn_jvn \left [ \left ( \frac{12880\times 100}{100+(18\times \frac{8}{12})} \right ) \right ] =Rs.11500
Clearly 12000 in cash is a better offer.
Answer: B) Rs.12000 |
AQUA-RAT | AQUA-RAT-37766 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
(6!-4!)/5!=? | [
"29/5",
"28/3",
"27/6",
"25/4"
] | A | (6!-4!)/5!
(4!*5*6-4!)/5!
4!(5*6-1)/4!*5
29/5
answer A |
AQUA-RAT | AQUA-RAT-37767 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
A scooter costs 20,000 when it is brand new. At the end of each year, its value is only 80% of what it was at the beginning of the year. What is the value of the scooter at the end of 3 years? | [
"10,240",
"12,500",
"12,800",
"12,000"
] | A | After first year, the value of the scooter = 16,000
After second year, the value of scooter = 12,800
After third year, the value of scooter = 10,240
Answer A |
AQUA-RAT | AQUA-RAT-37768 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What is the least number should be added to 1056, so the sum of the number is completely divisible by 26? | [
"10",
"20",
"30",
"40"
] | A | (1056 / 26) gives remainder 16
10 + 16 = 26, So we need to add 10
A |
AQUA-RAT | AQUA-RAT-37769 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is | [
"35",
"45",
"55",
"65"
] | A | Explanation:
Number is (5*27) - (4*25) = 135-100 = 35
Option A |
AQUA-RAT | AQUA-RAT-37770 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
In how many ways can an answer key for a quiz be written if the quiz contains 5 true-false questions followed by 3 multiple-choice questions with 4 answer choices each, if the correct answers to all true-false questions cannot be the same? | [
"1220",
"1650",
"1920",
"2440"
] | C | There are 2^5 = 32 possibilities for the true-false answers.
However we need to remove two cases for TTTTT and FFFFF.
There are 4*4*4 = 64 possibilities for the multiple choice questions.
The total number of possibilities is 30*64 = 1920.
The answer is C. |
AQUA-RAT | AQUA-RAT-37771 | # 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.
1. How much more would Rs.16000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly? | [
"A)Rs.4,225.60",
"B)Rs.4226.60",
"C)Rs.4,227.60",
"D)Rs.4,228.60"
] | A | 16000(11/10)4 - 16000(6/5)2 = 4,225.60
ANSWER:A |
AQUA-RAT | AQUA-RAT-37772 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average weight of a group of boys is 30 kg. After a boy of weight 39 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ? | [
"A)4",
"B)8",
"C)6",
"D)2"
] | B | Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 39 kg joins the group, total weight of boys = 30x + 39
So 30x + 39 = 31(x + 1) = > x = 8.
Answer:B |
AQUA-RAT | AQUA-RAT-37773 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
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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]
### Show Tags
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.
The total of the ages of Arun, Ajith, Ajay is 90 years. Ten years ago, the ratio of their ages was 1:2:3. What is the present age of Ajay? | [
"40years",
"35years",
"38years",
"30years"
] | A | Let ages of Arun, Ajith, Ajay 10years ago be x,2x,3x
x+10 + 2x+10 +3x+10 = 90
x = 10
Ramesh present age = 3*10+10 = 40 years
Answer is A |
AQUA-RAT | AQUA-RAT-37774 | @Rajeshwar: Because when you divide $6k+1$ by $5$, the $5k$ part is already a multiple of $5$. Adding or subtracting multiples of five form a number does not change the remainder when you divide by $5$. E.g., $2$, $7=2+5$, $22=2+5\times 4$, $177 = 2 + 5\times35$, etc., all have the same remainder when divided by $5$. So the remainder you get when dividing $n=6k+1$ by $5$ is the same as the remainder you get when dividing $n-5k = k+1$ by $5$. – Arturo Magidin Jul 20 '12 at 2:31
The following is multiple choice question (with options) to answer.
If the number is decreased by 5 and divided by 7 the result is 7. What would be the result if 4 is subtracted and divided by 10? | [
"4",
"7",
"8",
"5"
] | D | Explanation:
Let the number be x. Then,
(x - 5)/7 = 7 => x - 5 = 49
x = 54
.: (x - 4)/10 = (54 - 4)/10 = 5
Answer: Option D |
AQUA-RAT | AQUA-RAT-37775 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A certain sum of money at simple interest amounted Rs.934 in 8 years at 2% per annum, find the sum? | [
"338",
"805",
"890",
"870"
] | B | 934 = P [1 + (8*2)/100]
P = 805
Answer:B |
AQUA-RAT | AQUA-RAT-37776 | $\text{(ii) }\;1 - P(A\cap B\cap C) \;= \;0.88$ . . . . Right!
(b) (i) only the New York flight is full. (ii) exactly one of the three flights is full.
$\text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18$
$\text{(ii)}\;\begin{array}{ccccc}P(A \cap \overline{B} \cap \overline{C}) & = & (0.6)(0.5)(0.6) & = & 0.18 \\
P(\overline{A} \cap B \cap \overline{C}) &=& (0.4)(0.5)(0.6) &=& 0.12 \\
P(\overline{A} \cap \overline{B} \cap C) &=& (0.4)(0.5)(0.4) &=& 0.08\end{array}$
$P(\text{exactly one full}) \;=\;0.18 + 0.12 + 0.08 \;=\;0.38$
The following is multiple choice question (with options) to answer.
A certain airline's fleet consisted of 90 type A planes at the beginning of 1980. At the end of each year, starting with 1980, the airline retired 3 of the TYPE A planes and acquired 4 new type B plans. How many years did it take before the number of type A planes left in the airline's fleet was less than 50 percent of the fleet? | [
"11",
"12",
"13",
"14"
] | C | Let x be the number of years.
4x > 90 - 3x
7x > 90
x > 12 + 6/7
The answer is C. |
AQUA-RAT | AQUA-RAT-37777 | You can reduce the labour a little by writing
$\dfrac{1}{x(x-1)^3(x-2)^2} = \dfrac{(1-x+x^2)+x(x-1)}{x(x-1)^3(x-2)^2}$
$=\dfrac{x(x-2)^2-(x-1)^3}{x(x-1)^3(x-2)^2} + \dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3} - \dfrac{1}{x(x-2)^2}+\dfrac{1}{(x-1)^2(x-2)^2}$
$=\dfrac{1}{(x-1)^3}- \dfrac{1}{x(x-2)^2}+ \dfrac{1}{(x-1)^2}+\dfrac{1}{(x-2)^2}+2 \left[\dfrac{1}{x-1} - \dfrac{1}{x-2} \right]$
If you wish you can further decompose $\dfrac{1}{x(x-2)^2}= \dfrac{1}{(x-2)^2}+\dfrac{1}{2} \left[\dfrac{1}{x} - \dfrac{1}{x-2} \right]$
This expression can be readily integrated.
The following is multiple choice question (with options) to answer.
Reduce
803/876
to the lowest terms. | [
"11/12",
"23/24",
"26/27",
"4/7"
] | A | Explanation:
HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12
Option A |
AQUA-RAT | AQUA-RAT-37778 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Rs.705 is divided amongst A, B, C so that 3 times A's share, 5 times B's share and 4 times C's share are all equal. Find B's share? | [
"177",
"150",
"180",
"716"
] | C | A+B+C = 705
3A = 5B = 4C = x
A:B:C = 1/3:1/5:1/4
= 20:12:15
12/47 * 705
= Rs.180
Answer: C |
AQUA-RAT | AQUA-RAT-37779 | Note that $-169 + ( 4 \times 52 ) = 39$, so $-169 \equiv 39 \mod 52$.
Try this yourself for the second question.
• how did you get the four? – Nicole Oct 13 '14 at 23:36
• @Nicole Four is the number of times you need to add 52 to -169 to get a positive number. Note that adding any lesser multiple of 52 will leave you with a negative number. – Nick Oct 13 '14 at 23:38
The following is multiple choice question (with options) to answer.
When 2 + 3 = 65, 3 + 4 = 125, 4 + 5 = 205, then 5 + 6 = ? | [
"285",
"305",
"315",
"350"
] | B | 2 + 3 => 2 × 3 = 6 => 6 × 10 + 5 = 65
3 + 4 => 3 × 4 = 12 => 12 × 10 + 5 = 125
4 + 5 => 4 × 5 = 20 => 20 × 10 + 5 = 205
Then 5 + 6 => 5 × 6 = 30 => 30 × 10 + 5 = 305
Answer : B |
AQUA-RAT | AQUA-RAT-37780 | algorithms, dynamic-programming, np-hard
if left_is_greater(greater, precincts, n):
greater.append(precincts.popleft())
continue
p = precincts.pop()
# if either of the districts is "full", we don't have a choice
# and must select the opposite district
if len(right) == n / 2:
left.append(p)
elif len(left) == n / 2:
right.append(p)
# we always prefer the lesser summed district. if gerrymandering
# is feasible, we won't want to put all of our party's votes
# into one district, but prefer to have each district barely win.
elif sum(left) < sum(right):
left.append(p)
elif sum(right) < sum(left):
right.append(p)
# if the districts sum to the same amount, we prefer the district
# with less precincts, as this maximizes future decision branches
elif len(right) < len(left):
right.append(p)
elif len(left) < len(right):
left.append(p)
# we choose left as the default seed case, all else equal
else:
left.append(p)
if sum(left) > 0 and sum(right) > 0:
return True, left, right
else:
return False, left, right
def solve(precincts):
maximized = deque(sorted(precincts))
minimized = deque(sorted(x * -1 for x in precincts))
for polarized_precincts in [maximized, minimized]:
is_gerrymanderable, left, right = maximize(polarized_precincts)
if is_gerrymanderable:
print("It's gerrymanderable! Take a look: ", left, right)
return
print("Sorry, doesn't seem to be gerrymanderable.")
The following is multiple choice question (with options) to answer.
A city with a population of 173,460 is to be divided into 9 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have? | [
"17,400",
"17,500",
"17,600",
"17,700"
] | D | The minimum possible population occurs when all the other districts have a population that is 10% greater than the least populated district.
Let p be the population of the least populated district.
Then 173,460 = p + 8(1.1)p
9.8p=173,460
p = 17,700
The answer is D. |
AQUA-RAT | AQUA-RAT-37781 | &= 2 \times {\frac{22}{7}} \times 35 \\\\ &= 220 \text{ feet}\end{align}. Equations of a circle with given centre and radius in different forms. How to find the area of a circle? Note: With this tool, you can know the radius of a circle anywhere on Google Maps by simply clicking on a single point and extending or moving the circle to change the radius on the Map. The fixed point is called the origin or center of the circle. Let's take a fixed point $$P$$ and try to draw lines from it. Here are the three basic circle formulas that are used in calculating the various dimensions of a circle. Circles are one of the most commonly found shapes in the world. The calculator will generate a step by step explanations and circle graph. The radius – the distance from the circle's center or origin to the edge, one half the diameter The circumference – the length of the outside boundaries of the circle Starting from the diameter, you can easily find the other two. And consult the table below our everyday lives enjoyed learning about circle with given and. Then one of those slices makes a quarter circle boundary or the length of a circle =.! Team of math experts is dedicated to making learning fun for our favorite readers the. More about pi, or radius from point \ ( P\ ) in directions... Places, then one of its area at solving a few interesting practice questions at the end of the is! 22 } { 7 } } \end { align } \pi = { \frac { 22 {! A carnival and in general form given the radius of the circle and \ ( P\ in... Floated away boundary of the circle ( pool ) fun for our favorite readers the! How the shape of a circle = π r 2 not display plot -- browser is out of date area. To compute are done live '': how to calculate the and... Compute something or you have and skip unknown values in, find its area = ø. Find area identify the parts of a circle can be calculated if the radius or.... Health, and the arc in it math experts is dedicated to making learning fun for favorite... Button to see the result about circle with radius 1 unit Google map labeled as distance map and directions... Circular chocolate cake from his favorite bakery squared
The following is multiple choice question (with options) to answer.
Find the area of circle whose radius is 7m? | [
"138",
"154",
"288",
"280"
] | B | 22/7 * 7 * 7
= 154
Answer:B |
AQUA-RAT | AQUA-RAT-37782 | to calculate area, perimeter or volume of a figure. Find the area of the face we are looking directly at. A (rectangular) cuboid is a closed box which comprises of 3 pairs of rectangular faces that are parallel to each other and joined at right angles. The Surface Area of a cuboid is the total of all the areas on each face added together. Then , Total surface Area of Cuboid = 2( lxb + bxh +hxl) square units. The injury is in the area of a small tarsal bone in the foot, the cuboid bone. Go to Surface Area or Volume. The cuboid bone is within the area of the mid-foot. Total surface area of cuboid is 1216 sq cm. The volume is found using the formula: Which is usually shortened to: It doesn't really matter which one is length, width or height, so long as you multiply all three together. For More Videos - Subscribe In this video, we will solve first two problem statements from class 9 NCERT, Exercise 13. Sol: Total area that can be painted = 9. Irregular Prism Volume Calculator - A prism has the same cross section all along its length. Write a Python program to calculate surface volume and area of a cylinder. 8 cm cubic and length of the two edges are 2 cm and 6 cm. You can change the Name, Class, Course, Date, Duration, etc. Some of the worksheets for this concept are Common 2 d and 3 d shapes, Geometric nets pack, Volume and surface area work, Surface area of solids, Surface area, Surface areas of prisms, Surface area of 3d shapes, Module mathematical reasoning. 6 x 3² = 54. So, total surface area of cuboid = 2(lb + bh + lb) = 2(5x × 4x + 4x × 2x + 5x × 2x) = 2(20x2 + 8x2 + 10x2) = 2(38x2) = 76x2 Total surface area of cuboid = 1216 cm2 76x2 = 1216cm2 x2 = 16 x = 4 Dimensions of cuboid are. Students learn how to find the surface area of cubes and cuboids using nets and 3D drawings. Submitted by IncludeHelp, on May 08, 2020. Answers included. It is also known as a right rectangular prism. A patient with cuboid
The following is multiple choice question (with options) to answer.
A cube of edge 7 cm is cut into cubes each of edge 1 cm. The ratio of the total surface area of one of the small cubes to that of the large cube is equal to: | [
"1:25",
"1:49",
"1:52",
"1:522"
] | B | Sol.
Required ratio = 6 * 1 * 1 / 6 * 7 * 7 = 1/49 = 1:49.
Answer B |
AQUA-RAT | AQUA-RAT-37783 | 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.
An art gallery has only paintings and sculptures. Currently, 1/3 of the pieces of art are displayed, and 1/6 of the pieces on display are sculptures. If 1/3 of the pieces not on display are paintings, and 800 sculptures are not on display, how many pieces of art does the gallery have? | [
"360",
"1800",
"540",
"640"
] | B | Too many words and redundant info there.
(i) 1/3 of the pieces of art are displayed, hence2/3 of the pieces of art are NOT displayed.
(ii) 1/6 of the pieces on display are sculptures, hence 5/6 of the pieces on display are paintings.
(iii) 1/3 of the pieces NOT on display are paintings, hence2/3 of the pieces NOT on display are sculptures.
800 sculptures are not on display, so according to (iii) 2/3*{not on display} = 800 --> {not on display} = 1200.
According to (i) 2/3*{total} = 1200 --> {total} = 1800.
Answer: B. |
AQUA-RAT | AQUA-RAT-37784 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
(8^16)+(16^13)+(4^20) = ? | [
"(4)*(2^29+1)",
"(6)*(2^48)",
"(9)*(2^49)",
"(28)*(2^53)"
] | B | I think you are assuming that (X^Y) + (X^Z) is = (X^Y+Z) which is not true, as it only applies to multiplication. Rather it would have to be (X^Y)(X^Z) = (X^YZ)
So when you simplify down to 2^48 + 2^52 + 2^48 you cannot just add exponents.
Here's how my brain works with this one,
Step 1: Recognize a common base.
(8^16) + (16^13) + (4^24) = ((2^2)^16) + ((2^4)^13) + ((2^2)^24) = (2^48) + (2^52) + (2^48)
Step 2: Recognize the factor and pull out of the equation.
= (2^48)(1 + (2^4) + 1)
= (2^48)(1 + 16 + 1)
= (2^48)(18)
Step 3: Recognize this is not an answer and adapt.
(2^48)(18) = (2^48)(2)(9) = (2^48)(6)
Answer is B. |
AQUA-RAT | AQUA-RAT-37785 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
### Show Tags
02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
If A lends Rs.1500 to B at 10% per annum and B lends the same sum to C at 11.5% per annum then the gain of B in a period of 3 years is? | [
"112.5",
"122.5",
"132.5",
"67.5"
] | D | (1500*1.5*3)/100 => 67.5
ANSWER:D |
AQUA-RAT | AQUA-RAT-37786 | c++, beginner
if (pieces < 0)
{
std::cout << "Cannot have negative pieces!\n";
return -1;
}
std::cout << "Price per Piece?\n";
std::cin >> unit_price;
if (unit_price < 0)
{
std::cout << "Cannot have negative unit price!\n";
return -1;
}
// Calculate total price
gesamt = pieces * unit_price;
// Apply discount, if applicable
if (pieces > 10)
{
if (pieces > 50)
//10% discount
{
gesamt *= 0.9;
}
else //5% discount
{
gesamt *= 0.95;
}
}
std::cout << gesamt << '\n';
return 0;
}
The following is multiple choice question (with options) to answer.
If the price of a computer were reduced by 16 percent, which of the following could not be the final price of the computer? (Assume initial price to be integer in cents) | [
"$844.10",
"$896.70",
"$1,056.30",
"$1,136.10"
] | A | Let X be the initial price of computer without discount
then price (final) after discount should be=X(1-16/100)------->X(21/25)=A(say)
means X=A*(25/21).....
so initial price to be integer(As per stem) final price must be multiple of 21(3 or 7)
if we check options all were divisible by 3 except option (A)
Answer: A |
AQUA-RAT | AQUA-RAT-37787 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
the average of four consecutive even numbers is 27. find the largest of these numbers? | [
"10",
"20",
"30",
"40"
] | C | Sol: let the numbers be x,x+2,x+4 andx+6. then,
(x+(x+2)+(x+4)+(x+6))/4) = 27
(4x+12)/4 = 27
x+3=27 x=24.
Therefore the largest number=(x+6)=24+6=30.
ANSWER: C |
AQUA-RAT | AQUA-RAT-37788 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 1000 m race, A beats B by 250 meters or 25 seconds. Find the speed of B? | [
"10",
"9",
"7",
"5"
] | A | Since A beats B by 250 m or 25 seconds, i
t implies that B covers 250 m in 25 seconds.
Hence speed of B = 250/25 = 10 m/s.
Answer: A |
AQUA-RAT | AQUA-RAT-37789 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A certain number of men can finish a piece of work in 100 days. If there were 10 men less, it would take 10 days more for the work to be finished. How many men were there originally? | [
"120",
"105",
"115",
"110"
] | D | Originally let there be x men.
Less men, More days (Indirect Proportion)
therefore (x-10) : x :: 100 :110
=(x-10)x 110-x(100)
= 10x= 1100
x= 110
Answer is D. |
AQUA-RAT | AQUA-RAT-37790 | ## 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 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required ? | [
"34",
"40",
"68",
"88"
] | D | Solution
We have : l = 20 ft and lb = 680 sq. ft. So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft. Answer D |
AQUA-RAT | AQUA-RAT-37791 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
An investment yields an interest payment of $234 each month. If the simple annual interest rate is 9%, what is the amount of the investment? | [
"$28,300",
"$30,400",
"$31,300",
"$31,200"
] | D | Let the principal amount = P
Simple annual interest = 9%
Simple monthly interest = (9/12) = (3/4)%
(3/4)*(P/100) = 234
=>P = (234 *4 * 10^2 )/3
= 78*4*10^2
= 312 * 10^2=31200
Answer D |
AQUA-RAT | AQUA-RAT-37792 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
The organizers of a week-long fair have hired exactly eight security guards to patrol the fairgrounds at night for the duration of the event. Exactly three guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different groups of 3 guards will be available to patrol the fairgrounds on the following Saturday night? | [
"6",
"8",
"10",
"12"
] | C | For any given day, only the guards patrolling on the previous day won't be available. So, 3 guards who patrolled on Friday won't be available. We are thus left with 5 guards.
To choose 3 out of 5, we will have 5C3 = 10 different groups.
The answer is C. |
AQUA-RAT | AQUA-RAT-37793 | Are there any others? Suppose that $y\equiv x\pmod{m_1}$. Then $m_1$ divides $y-x$. So the possible remainders when $y-x$ is divided by $m$ are $0, m_1,2m_1,\dots, (d-1)m_1$. It follows that $y-x=im_1$ for some $i$ where $0\le i\le d-1$. This just says that $y\equiv x+im_1 \pmod{m}$, that is, $y\equiv x+\frac{mi}{d}\pmod{m}$, for some $i$ ranging from $0$ to $d-1$.
Remark: The important part was the uniqueness modulo $m_1$. The rest is straightforward, and you know it well. To give a simple numerical example, suppose that we know that $x\equiv 3\pmod{16}$. What can we say about $x$ modulo $80$? We can say that $x$ is congruent to one of $3$, $19$, $35$, $51$, $67$.
The following is multiple choice question (with options) to answer.
When positive integer x is divided by positive integer y, the remainder is 9. If x/y = 96.15, what is the value of y? | [
"96",
"60",
"48",
"25"
] | B | Guys, one more simple funda.
5/2= 2.5
now .5 x2 =1 is the remainder
25/4 = 6.25
now .25x4=1 is the remainder
32/5=6.4
now.4x5 = 2 is the remainder
given x/y = 96.15 and remainder is 9
So .15 X y = 9
hence y= 60
Ans B |
AQUA-RAT | AQUA-RAT-37794 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
If $64 amount to $ 83.20 in 2 years. what will $ 86 amount to in 4 years at the same rate percent per annum? | [
"$127.40",
"$124.70",
"$114.80",
"$137.60"
] | D | First, Simple interest formula is
S.I=PNR/100 Here P=$64, T=2 years and S.I=$19.20 (i.e 83.20-64)
R=((100x19.20)/(64x2))%=15%
Now P=$86, T=4 years, R=15%
so S.I=86x4x15/100=$137.60
Answer is D |
AQUA-RAT | AQUA-RAT-37795 | Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
Hence answer will be (E) $693 The wage difference is not with respect to time , correct approach will be with respect to work done.. ( Errorr is highlighted in red ) Further explanation as provide to you via PM Wages/Salary can be of 2 types ( Something You will certainly admit ) - 1. Hourly Wage payment System ( Most White collar Jobs ) 2. Unit of Work Done ( Most Blue collar Jobs ) Here it is given the part of job completed by A and B , both have diff efficiency.. Suppose total Work is 18 units Efficiency of A = 11 and Efficiency of B = 7 Now, tell me , whom will you give more money A or B ( According to the amount of work completed ) Certainly A !!! Continuing with the same example - Time Taken by A = 18/11 = 1.63 Hours Time Taken by B = 18/7 = 2.57 Hours Now, tell me , whom will you give more money A or B ??? B is inefficient , straightay, he takes more time to complete less amount of work than A... Hence Hourly wage approach is not suited here, hope this helps.. _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 110 Location: United States GPA: 3.37 Re: Elana was working to code protocols for computer processing. She did [#permalink] ### Show Tags 02 Nov 2016, 11:44 Elena did 11/18 which means that Andrew did 7/18. They both earn the same hourly rate. The difference between the work they did is 4/18. We know that 4/18 equals$154 because that is the difference in pay. Divide $154 by 4 so we can see how much 1/11th of the job is worth.$154/4=\$38.5
The following is multiple choice question (with options) to answer.
A certain bus driver is paid a regular rate of $15 per hour for any number of hours that does not exceed 40 hours per week. For any overtime hours worked in excess of 40 hours per week, the bus driver is paid a rate that is 75% higher than his regular rate. If last week the bus driver earned $982 in total compensation, how many total hours did he work that week? | [
"56",
"40",
"44",
"48"
] | A | For 40 hrs = 40*15=600
Excess = 982-600=382
For extra hours =.75(15)=11.25+16=27.25
Number of extra hrs =382/27.25=14
Total hrs =40+14=56
Answer A 56 |
AQUA-RAT | AQUA-RAT-37796 | u, v& was adjacent edges. Substituting this back into our formula for the volume of a parallelepiped we get that: We note that this formula gives up the absolute value of the scalar triple product between the vectors. Notice that we Notify administrators if there is objectionable content in this page. rev 2021.1.20.38359, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Theorem 1: If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$, then the volume of the parallelepiped formed between these three vectors can be calculated with the following formula: $\mathrm{Volume} = \mathrm{abs} ( \vec{u} \cdot (\vec{v} \times \vec{w}) ) = \mathrm{abs} \begin{vmatrix}u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3 \end{vmatrix}$. One such shape that we can calculate the volume of with vectors are parallelepipeds. How to get the least number of flips to a plastic chips to get a certain figure? c 1 c2 c3 In each case, choose the sign which makes the left side non-negative. How can I cut 4x4 posts that are already mounted? Suppose three vectors and in three dimensional space are given so that they do not lie in the same plane. Click here to toggle editing of individual sections of the page (if possible). It follows that is the volume of the parallelepiped defined by vectors , , and (see Fig. Tetrahedron in Parallelepiped. Change the name (also URL address, possibly the category) of the page. By the theorem of scalar product, , where the quantity equals the area of the parallelogram, and the product equals the height of the parallelepiped. View and manage file
The following is multiple choice question (with options) to answer.
The volume of a solid is given by the formula V = ab^2.
If a is doubled and b tripled, by what factor does the volume (V) increase? | [
"3",
"6",
"9",
"18"
] | D | We have,
V = ab^2
After doubling a and tripling b, we have:
V = (2a)*(3b)^2
V = 2a*9b^2
V = 18ab^2
Initially V = ab^2. Now, V = 18ab^2.
Therefore, V increases by a factor of 18.
Answer = D =18 |
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