source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36597 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
2 men and 4 boys can complete a work in 4 days. 5 men and 6 boys can complete the same work in 3 days. The work done by 2 boys is equal to the work of how many men? | [
"3",
"4",
"5",
"6"
] | C | (2m + 4b)’s one day’s work = 1/4
(5m + 6b)’s one day’s work = 1/3
=> (8m + 16b)’s one day’s work = (15m + 18b) ’s one day’s work
=> 7 men’s work = 2 boy’s work
So, we should be employ 5 more men to be completed the work.
ANSWER:C |
AQUA-RAT | AQUA-RAT-36598 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
Given that x = 2^b – (8^34 + 16^5), which of the following values for b yields the lowest value for |x|? | [
"101",
"102",
"103",
"104"
] | B | 8^34 + 16^5 = 2^102 + 2^20
Compared to 2^102, the value of 2^20 is negligible.
2^102 - (2^102 +2^20) will minimize the value of |x|.
The answer is B. |
AQUA-RAT | AQUA-RAT-36599 | 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
St Charles Borromeo Liverpool Newsletter, Soft Skills Questions And Answers Pdf, Calphalon Classic Nonstick Stainless Steel 2-piece Fry Pan Set, Legendary Dragon Decks Price Guide, Toril Moi Feminist, Female, Feminine, Carlton Postcode Nsw, How To Build A Powerful Electric Motor From Scratch Pdf, Moca Poe Filter, Wow Console Commands, Is It Hard To Build An Acoustic Guitar, Yamaha Spare Parts Price List, Khao Soi Restaurant, 4 Pack - Lysol Concentrate Disinfectant, Original Scent 12 Oz, Hindrances To Fulfilling God's Purpose, Redmi Y2 4/64 Price In Bangladesh, How To Pronounce Photogenic, How To Shape Orecchiette, Purple Graphic Tee, Thistle Plants For The Garden, Naphthalene Is Acid Or Base, Prayer Points On Divine Touch, Beautiful Quilt Patterns, " />
The following is multiple choice question (with options) to answer.
If the length of the sides of two cubes are in the ratio 6:1, what is the ratio of their total surface area? | [
"6:1",
"12:1",
"24:1",
"36:1"
] | D | Let x be the length of the small cube's side.
The total surface area of the small cube is 6x^2.
The total surface area of the large cube is 6(6x)^2=216x^2.
The ratio of surface areas is 36:1.
The answer is D. |
AQUA-RAT | AQUA-RAT-36600 | In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
The following is multiple choice question (with options) to answer.
A number exceeds by 25 from its 3/8 part. Then the number is? | [
"32",
"35",
"39",
"40"
] | D | Explanation:
x – 3/8 x = 25
x = 40
ANSWER IS D |
AQUA-RAT | AQUA-RAT-36601 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
The sum of three consecutive even numbers is 42. Find the middle number of the three? | [
"14",
"16",
"18",
"22"
] | A | Three consecutive even numbers (2P - 2), 2P, (2P + 2).
(2P - 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.
Option A |
AQUA-RAT | AQUA-RAT-36602 | dataset
References:
https://github.com/edenton/svg/blob/master/data/convert_bair.py
The following is multiple choice question (with options) to answer.
CONVERT 1.6 hectares in ares | [
"130 ares.",
"160 ares.",
"180 ares.",
"230 ares."
] | B | 1.6 hectares in ares
1 hectare = 100 ares
Therefore, 1.6 hectares = 1.6 × 100 ares
= 160 ares.
ANSWER- B |
AQUA-RAT | AQUA-RAT-36603 | zoology, ethology, learning
Title: How do beavers learn how to build dams? I was wondering whether all beavers, from all around the world, know how to build dams and lodges? Do they need to learn it from their parents? If you release a group of beavers in the wild that haven't been in contact with their parents, would they start to build stuff? or just hopelessly die/starve to death? Question summary: is dam building learned or instinctive in beavers?
A blog post from 2011 references an article in the Juneau Empire titled Running water is sound of spring for beavers. This article is no longer hosted on the Juneau Empire website, but archived versions are available.
Here's an excerpt (emphasis mine) --
Swedish biologist Lars Wilsson spent years studying captive and wild beavers, and he gained remarkable insights into their behavior. He raised beavers in an outdoor enclosure and in a large indoor terrarium ...
Wilsson initially captured four adult beavers and later he raised a number of beavers from infancy, some in small colonies with their parents and some completely isolated from adult beavers. He isolated the young beavers to see what beavers learn from their parents and what behaviors are instinctive.
He found that young beavers - who had never even seen a beaver dam - were able to build almost-perfect dams at the first opportunity.
The foundation of sticks and logs anchored to the stream bottom, the interwoven lattice of trimmed branches, the mud chinking, every aspect of dam building was hard-wired. Beavers do get more skilled at dam building as they gain experience, but the building behavior is instinctive.
Wilsson learned that the sound of running water is the cue for dam building and dam repair. In one experiment, he played a recording of running water, and the young beavers built a dam in a tank of still water in the terrarium. In another peculiar experiment, his captive beavers built a "dam" on a concrete floor against a loudspeaker that played the sound of running water.
The following is multiple choice question (with options) to answer.
20 beavers, working together in a constant pace, can build a dam in 6 hours. How many hours will it take 12 beavers that work at the same pace, to build the same dam? | [
"10.",
"4.",
"5.",
"6"
] | A | Total Work = 20*6= 120 Beaver hours
12 Beaver * x =120 Beaver hours
x=120/12=10
ANSWER:A |
AQUA-RAT | AQUA-RAT-36604 | # 2010 AMC 10B Problems/Problem 25
## Problem
Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that
$P(1) = P(3) = P(5) = P(7) = a$, and
$P(2) = P(4) = P(6) = P(8) = -a$.
What is the smallest possible value of $a$?
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
## Solution
We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$.
Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$.
Then, plugging in values of $2,4,6,8,$ we get
The following is multiple choice question (with options) to answer.
If both 2^2 and 3^4 are factors of the number a*4^3*6^2*13^11, then what is the smallest possible value of a? | [
"6",
"9",
"12",
"15"
] | B | The number a must include at least 3^2 = 9
The answer is B. |
AQUA-RAT | AQUA-RAT-36605 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly? | [
"1 litre",
"7 litre",
"31 litre",
"41 litre"
] | C | Solution
Required measurement = H.C.F of (496, 403, 713) litres
‹=›31 litres.
Answer C |
AQUA-RAT | AQUA-RAT-36606 | By adding the first and the third equation we get $5x_2=0$ which implies $x_2=0$. If we now plug $x_2=0$ into remaining equations, we see, that they are all multiples of $x_1-x_2=0$.
Hence all solutions are of the form $(x_1,0,x_1)$. I.e., the solutions form the subspace $\{(x_1,0,x_1); x_1\in\mathbb R\}$.
Every vector in this subspace is a multiple of $(1,0,1)$, which means that vector $(1,0,1)$ generates the subspace.
Hence this subspace is one-dimensional and basis consists of a single vector $(1,0,1)$.
Of course, we could take any non-zero multiple of $(1,0,1)$ instead. For example, vector $(-2,0,-2)$ generates the same subspace.
The following is multiple choice question (with options) to answer.
If both a and b belongs to the set {1,2,3,4 }, then the number of equations of the form ax2 + bx + 1 = 0 having real roots is | [
"10",
"7",
"6",
"12"
] | B | Explanation :
ax2 + bx + 1 = 0 .
For real roots
=> b2 -4ac ≥ 0 .
i.e b2 - 4a(1) ≥ 0 .
i.e b2 ≥ 4a.
For a = 1, 4a = 4, ∴ b = 2, 3, 4
a = 2, 4a = 8, ∴ b = 3, 4
a = 3, 4a = 12, ∴ b = 4
a = 4, 4a = 16, ∴ b = 4
∴ Number of equations possible = 7.
Harsh Mishra a year ago
0 upvotes
Answer : B |
AQUA-RAT | AQUA-RAT-36607 | } \left( \dfrac { 2y+b }{ a-2y } \right)$$$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right)$$Maths
The following is multiple choice question (with options) to answer.
For all numbers a and b, the operationis defined by ab = (a + 2)(b – 3).
If 3y = –30, then y = | [
" –15",
" –6",
" -3",
" 6"
] | C | (3+2)(y-3)=-30..
x-3=-6..
x=-3
C |
AQUA-RAT | AQUA-RAT-36608 | # 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.
The difference between the compound interest compounded annually and simple interest for 2 years at 20% per annum is Rs.144. Find the principal? | [
"Rs.3000",
"Rs.3300",
"Rs.3600",
"Rs.3900"
] | C | P = 144(100/5)2 => P = 3600
ANSWER:C |
AQUA-RAT | AQUA-RAT-36609 | # 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.
The difference between the simple interest received from two different sources on Rs. 1500 for 3 years is Rs. 13.50. The difference between their rates of interest is? | [
"0.7%",
"8.3%",
"0.3%",
"0.39%"
] | C | (1500 * R1 * 3)/100 - (1500 * R2 * 3)/100
= 13.50 4500(R1 - R2)
= 1350
R1 - R2 = 0.3%
Answer: C |
AQUA-RAT | AQUA-RAT-36610 | => 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.
In a colony of 70 residents, the ratio of the number of men and women is 4 : 3. Among the women, the ratio of the educated to the uneducated is 2 : 3. If the ratio of the number of educated to uneducated persons is 8 : 27, then find the ratio of the number of educated to uneducated men in the colony? | [
"A)1:6",
"B)1:1",
"C)1:8",
"D)1:3"
] | D | Number of men in the colony = 4/7 * 70 = 40.
Number of women in the colony = 3/7 * 70 = 40.
Number educated women in the colony = 2/5 * 30 = 12.
Number of uneducated women in the colony = 3/5 * 50 = 24.
Number of educated persons in the colony = 8 /35 * 70 = 16.
As 6 females are educated, remaining 10 educated persons must be men.
Number of uneducated men in the colony = 40 - 10 = 30.
Number of educated men and uneducated men are in the ratio 10 : 30 i.e., 1:3.
Answer:D |
AQUA-RAT | AQUA-RAT-36611 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
How many words can be formed from the letters of the word ‘DIRECTOR’ So that the vowels are always together? | [
"3251",
"2160",
"1203",
"2564"
] | B | In the given word, we treat the vowels IEO as one letter.
Thus, we have DRCTR (IEO).
This group has 6 letters of which R occurs 2 times and others are different.
Number of ways of arranging these letters = 6!/2! = 360.
Now 3 vowels can be arranged among themselves in 3! = 6 ways.
Required number of ways = (360x6) = 2160.
Ans: B |
AQUA-RAT | AQUA-RAT-36612 | ~ Nafer
## Solution 3 (using PIE)
Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion.
The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees.
The total number of configurations is given by $\frac{12!}{3! \cdot 4! \cdot 5!}$. To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE.
$\#$(configurations with at least one pair of adjacent Birch trees) $=$ $\#$(configurations with one pair) $-$ $\#$(configurations with two pairs) $+$ $\#$(configurations with three pairs) $-$ $\#$(configurations with four pairs).
To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\frac{11!}{3! \cdot 3! \cdot 4!}$ configurations.
For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\frac{10!}{2! \cdot 3! \cdot 4!}$ cases. So our second term is $\frac{2 \cdot 10!}{2! \cdot 3! \cdot 4!}$.
The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\frac{2 \cdot 9!}{3! \cdot 4!}$ arrangements.
The final term can happen in one way (BBBBB). This gives $\frac{8!}{3! \cdot 4!}$ arrangements.
The following is multiple choice question (with options) to answer.
In a garden, 26 trees are planted at equal distances along a yard 600 metres long, one tree being at each end of the yard. What is the distance between two consecutive trees? | [
"10",
"8",
"12",
"24"
] | D | 26 trees have 25 gaps between them.
Length of each gap = 600/25 = 24
i.e., distance between two consecutive trees = 24
Answer is D. |
AQUA-RAT | AQUA-RAT-36613 | 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.
A sum of money becomes 7/6 of itself in 2 years at a certain rate of simple interest. The rate per annum is? | [
"50/9",
"50/7",
"50/6",
"50/8"
] | C | Let sum = x. Then, amount = 7x/6
S.I. = 7x/6 - x = x/6; Time = 2 years.
Rate = (100 * x) / (x * 6 * 2) = 50/6 %.
Answer:C |
AQUA-RAT | AQUA-RAT-36614 | Goal: 25 KUDOZ and higher scores for everyone!
Senior Manager
Joined: 13 May 2013
Posts: 429
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
30 Jul 2013, 16:45
1
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip & then at a average speed of 60 miles/hr for the remaining 30 miles of the trip if she made no stops during the trip what was susan's avg speed in miles/hr for the entire trip
What we have here are equal distances for both segments.
First segment: 30 miles/hour and covered 30 miles, therefore it took one hour.
Second segment: 60 miles/hour and covered 30 miles, therefore it took 1/2 hour.
(Total distance / total time)
(60 / [1hr+ 1/2hr])
(60 / 1.5) = 40 miles avg. speed.
A. 35
B. 40
C. 45
D. 50
E. 55
(B)
When don't we simply add the distances/speeds together to get the average?
Intern
Joined: 23 Dec 2014
Posts: 48
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
03 Feb 2015, 16:58
Rate x Time = Distance
Going: 30 x 1 = 30
Returning: 30 x .5 = 30
Avg speed = Total distance/Total Time
=(30+30)/ (1+.5)
=40
Intern
Joined: 25 Jan 2016
Posts: 1
Re: Susan drove an average speed of 30 miles per hour for the [#permalink]
### Show Tags
10 Feb 2016, 21:17
Narenn wrote:
jsphcal wrote:
Susan drove an average speed of 30 miles per hour for the first 30 miles of a trip and then at an average speed of 60 miles per hour for the remaining 30 miles of the trip. If she made no stops during the trip, what was Susan's average speed, in miles per hour, for the entire trip?
a. 35
b. 40
c. 45
d. 50
e. 55
The following is multiple choice question (with options) to answer.
Gill drives 120 miles from Los Angeles to San Diego to fetch a package. On her way there she drives at 40 miles per hour. On her way back she drives 50% faster. What is Gill's average velocity W for the round trip? | [
"24 miles per hour",
"48 miles per hour",
"50 miles per hour",
"53 1/3 miles per hour"
] | B | Here's an Average Speed question in which the prompt gives you almost all of the immediate numbers needed to work with.
Driving from LA to SD, we have a distance of 120 miles and a speed of 40 miles/hour.
D = (R)(T)
120 = (40)(T)
120/40 = 3 = T
3 hours to drive to SD
On the way back, she drives 50% FASTER. Gill's return speed is (1.5)(40) = 60 miles/hour.
D = (R)(T)
120 = (60)(T)
120/60 = 2 = T
2 hours to drive to SD
Total Distance = 240 miles
Total Time = 3+2 = 5 hours
Average Speed = 240/5 = 48 miles/hour
Final Answer:
B |
AQUA-RAT | AQUA-RAT-36615 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
Andrew went to a shop and bought things worth Rs. 25, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items? | [
"18",
"19",
"19.7",
"21"
] | C | Total cost of the items he purchased = Rs.25
Given that out of this Rs.25, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (25−30/100−x)6/100 = 30/100
⇒ 6(25 −0.3 −x) = 30
⇒ (25 − 0.3 − x) = 5
⇒ x = 25 − 0.3 − 5 = 19.7
C |
AQUA-RAT | AQUA-RAT-36616 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
B invested $4000 in a business and C invested $3000 in a business. At the end of the year they got $14000 as the profit. Find their share? | [
"$8000,$6000",
"$7000,$5000",
"$6000,$4000",
"$6000,$8000"
] | A | B:C = 4000:3000
A:B = 4:3
A's share = 14000*4/7 = $8000
B's share = 14000*3/7 = $6000
Answer is A |
AQUA-RAT | AQUA-RAT-36617 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
40% of a number is more than 20% of 650 by 190. Find the number? | [
"997",
"127",
"800",
"128"
] | C | (40/100) * X – (20/100) * 650 = 190
2/5 X = 320
X = 800
Answer: C |
AQUA-RAT | AQUA-RAT-36618 | Each of your methods gives you $1.496 \times 10^9 ~\rm{m}$ - accurate to 4 significant digits. That's better than you should expect.
The accuracy of your result is only ever as good as the accuracy of the inputs. Some of the data you have is given to 5 significant digits - that will give you the "more accurate" results.
I recommend that you learn about error propagation. There are millions of resources online for this - a relatively basic introduction can be found here
• Sorry for the late reply! I have been really busy with the project and school. I actually totally forgot about error propagation and simply kept the numbers intact without thinking!
– user156610
May 30, 2017 at 9:01
$a = 149598023$
$e = 0.0167086$
$p = a(1-e^2)$
$p = a(0.99972082268604) = 149,556,258.62576513369892$
$b = \frac{p}{ \sqrt{1-e^2}}$
$= \frac{p} {0.99986040159916324190990908617052}$
$= 149,577,139.35522085944399314340085$
$b = \sqrt{p*a}$ = 149,577,139.35522085944399314340085
$pa = rmax * rmin$
$22,373,320,617,691,160.86368910923516 = rmax * rmin$
$rmax = \frac{p}{(1-e)} = \frac{p}{0.9832914} =152,097,596.5270978$
$rmin = \frac{p}{(1+e)} = \frac{p}{1.0167086} =147,098,449.4729022$
The following is multiple choice question (with options) to answer.
In expressing a length 81.475 km as nearly as possible with three significant digits, find the percentage error? | [
"0.075%",
"0.156%",
"0.031%",
"0.048%"
] | C | Error = 81.5-81.475 = 0.028
Required percentage = (0.028/81.475)*100 = 0.031%
Answer is C |
AQUA-RAT | AQUA-RAT-36619 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A person can row at 9 kmph and still water. He takes 7 1/2 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 1 kmph? | [
"33",
"28",
"29",
"54"
] | A | Let the distance between A and B be x km.
Total time = x/(9 + 1) + x/(9 - 1) = 7.5
=> x/10 + x/8 = 15/2 => (4x + 5x)/40 = 15/2 => x
= 33 km.
Answer:A |
AQUA-RAT | AQUA-RAT-36620 | Another answer that use almost the same idea: the sum or subtraction of two even or odd number is an even number. How many odd number we have?
Replacing 100 with $n$ and using Brian M. Scott's solution, we want a partition of $\{1, 2, ..., n+1\}$ into two sets with equal sums.
The sum is $\frac{(n+1)(n+2)}{2}$, and if $n=4k$, this is $(4k+1)(2k+1)$ which is odd and therefore impossible.
If $n = 4k+1$, this is $(2k+1)(4k+3)$ which is also odd, and therefore impossible.
If $n = 4k+2$, this is $(4k+3)(2k+2)$, so it is not ruled out, and each sum must be $(4k+3)(k+1)$.
if $n = 4k+3$, this is $(2k+2)(4k+5)$ which is also not ruled out, and each sum must be $(k+1)(4k+5)$.
Now I'll try to find a solution for the not impossible cases. (I am working these out as I enter them.)
The following is multiple choice question (with options) to answer.
Set A contains all the even numbers between 8 and 50 inclusive. Set B contains all the even numbers between 108 and 150 inclusive. What is the difference between the sum of elements of set B and the sum of the elements of set A? | [
"2200",
"2550",
"5050",
"6275"
] | A | Set A contains 8, 10, 12... 50
Set B contains 108 , 110 , 112 ... 150
Number of terms in each set = 22
Difference between corresponding terms in set A and B = 100
Difference between Sum of set B and set A = 100*22 = 2200
Answer A |
AQUA-RAT | AQUA-RAT-36621 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
_________________
Intern
Joined: 26 May 2010
Posts: 10
Followers: 0
Kudos [?]: 33 [5] , given: 4
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 10 numbers is 23. If each number is increased by 6, what will the new average be? | [
"36",
"29",
"72",
"29"
] | B | Sum of the 10 numbers = 230
If each number is increased by 6, the total increase =
6 * 10 = 60
The new sum = 230 + 60 = 290 The new average = 290/10
= 29.
Answer:B |
AQUA-RAT | AQUA-RAT-36622 | (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.
Little Praveen had $50.50. He spent $12.25 on sweets and gave same money to his two friends and the amount left with him is $31.25. Find how much he spent to two of his friends? | [
"$ 16.65",
"$ 18.65",
"$ 19.25",
"$ 16.55"
] | B | Praveen spent and gave to his two friends a total of
12.25 + 2x= $4.65, x is the amount spent to one friend.
Money left
50.50 -2x = $31.85. Then 2x=$18.65.
correct answer is B) $ 18.65 |
AQUA-RAT | AQUA-RAT-36623 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
There are six leopards and one sheep.
Leopard can eat sheep but since as the land is magical, the leopards who eats the sheep , turns into sheep and then can be eaten by the remaining leopard(s).
If we leave them for some time then how many sheep and leopard will be there , when we come back ? | [
"7 leopard and 3 sheep",
"4 leopard and 7 sheep",
"6 leopard and 4 sheep",
"5 leopard and 1 sheep"
] | D | D
5 leopard and 1 sheep
since all animal are rational , once 1st leopard eats the sheep all the rest of leopard would know about this. |
AQUA-RAT | AQUA-RAT-36624 | Finish solving the differential equation; then use the second sentence to find A and B. Note that when t=0, y=600; and when t=10, y=30,000.
We’ll be seeing how to finish in the following answers. But for this version, the equation will turn out to be $$y = \frac{p}{1 + ke^{-rt}}$$ where $$k = e^{-Bp}$$ and $$r = Ap$$ are new constants derived from the arbitrary constants A and B; it doesn’t matter what A and B actually were. Instead, let’s think about what r and k themselves mean.
Observe that as t increases, y approaches p, as expected. (The whole population will eventually be infected in this model, which has no isolation and no immunity.)
The initial number infected (t = 0) is $$\displaystyle p_0 = \frac{p}{1+k}$$, so we find that $$k = \frac{p – p_0}{p_0},$$ which in our problem is $$\displaystyle k = \frac{260,000-600}{600} = 432.333$$. This is the initial ratio of uninfected to infected.
Similarly, if we know that after T days, the number infected is $$p_T$$, then $$\displaystyle p_T = \frac{p}{1 + ke^{-rT}}$$; solving this for r, we get $$r = \frac{1}{T}\ln\frac{p-p_T}{kp_T} = \frac{1}{T}\ln\frac{p_T(p-p_0)}{p_0(p-p_T)}$$
In our problem, after 10 days, the number infected is 30,000; so $$\displaystyle r = \frac{1}{10}\ln\frac{30,000(260,000-600)}{600(260,000-30,000)} = 0.40323$$.
The following is multiple choice question (with options) to answer.
At 1:00 pm, there were 10.0 grams of bacteria. The bacteria increased to x grams at 4:00 pm, and 19.6 grams at 7:00 pm. If the amount of bacteria present increased by the same fraction during each of the 3-hour periods, how many grams of bacteria were present at 4:00 pm? | [
"13.7",
"14.0",
"14.3",
"14.6"
] | B | Let x be the factor by which the bacteria increases every three hours.
At 4:00 pm, the amount of bacteria was 10x and at 7:00 pm it was 10x^2.
10x^2 = 19.6
x^2=1.96
x=1.4
At 4:00 pm, the amount of bacteria was 10(1.4)=14 grams.
The answer is B. |
AQUA-RAT | AQUA-RAT-36625 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
Over the course of a year, a certain microbrewery increased its beer output by 70 percent. At the same time, it decreased its total working hours by 20 percent. By what percent did this factory increase its output per hour? | [
"50%",
"90%",
"112.5%",
"210%"
] | C | Let's take the original output to be 100 and the new output to be 170% or 170.
Let's take the original work hours to be 100 hours and the new work hours to be 80% or 80 hours.
The old output per work hour = 100/100 hours or 1.
The new output per work hour = 170/80 or 2.125. So the resulting increase is actually new minus old or 1.125 or 112.5%.
Choice C |
AQUA-RAT | AQUA-RAT-36626 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
If it is possible to from a word with the first, fourth, seventh and eleventh letters in the word
'SPHERVLVODS' write the second letter of thet word. Otherwise, X is the answer.? | [
"T",
"E",
"R",
"HG"
] | B | The first, fourth, seventh and eleventh letters of the word 'SPHERVLVODS'
The word formed is LESS
The second letter is E.
Answer: Option 'B' |
AQUA-RAT | AQUA-RAT-36627 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
In a camp, there is a meal for 100 men or 200 children. If 130 children have taken the meal, how many men will be catered to with remaining meal? | [
"20",
"30",
"35",
"10"
] | C | There is a meal for 200 children. 130 children have taken the meal.
Remaining meal is to be catered to 70 children.
Now, 200 children 100 men.
70 children (100/200) * 70 = 35 men
Answer is C. |
AQUA-RAT | AQUA-RAT-36628 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
15 men can complete a work in 10 days while 20 boys can complete the same work in 15 days .How many days will 10 men and 10 boys together take to complete the same work | [
"10 days",
"11 days",
"12 days",
"13 days"
] | A | Solution:
Work done by one man in one day=1/(15*10)
Work done by one boy in one day=1/(20*15)
Work done by 10 men and 10 boys in one day=10[ 1/(15*10) + 1/(20*15)]
Days taken=1/Work done by 10 men and 10 boys in one day
=10 days
Answer A |
AQUA-RAT | AQUA-RAT-36629 | = \sqrt{2}.$$ Obviously $$q \neq 0$$. We can assume for convenience that $$p$$ and $$q$$ are both positive. We can actually iterate $$q$$ through several positive integers, choosing a $$p$$ that will get the division arbitrarily close to $$\sqrt{2}$$, but we can't possibly iterate $$q$$ through enough possibilities to be absolutely certain. The negation of $$A$$ is therefore much easier: we assume that the $$p$$ and $$q$$ we're looking for do exist. If the pair does exist, we will be able to deduce a lot of facts about these numbers. But if the pair does not exist, we should arrive at a contradiction. • I don't think your post addresses the question at all. There are existence theorems that cannot be proven without showing that its negation leads to contradiction. Feb 1, 2021 at 16:01 easy to follow but not exacting or rigorous answer... Yes as long as the contrapositive is constructed in such a way that the answer space is complete and you disprove the contrapositive in the general case. so picture a venn diagram that where there is clear lint between true and false and all the space is filled with one or the other with no overlaps. Then find a way to contradict false in such generality that you can cross out all of false. Well all you have left is true so accept it. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true". I think it's worth considering several cases. 1. If there is a contradiction between two unproven statements, that just means that they can't be true at the same time. You can't use the contradiction to directly prove or disprove either one. 2. If one of them appears particularly plausible, then that would be a good fit for your “big problem” description. But then what you have to do is conjecture that the plausible thing holds true, and prefix every statement you derive from this with “if the … conjecture holds, then …” or “assuming the …
The following is multiple choice question (with options) to answer.
If pq represents a positive two digit number, where p and q are single digit integers, which of the following cannot be true? | [
"p+q =9",
"(p)(q)=9",
"p-q = 9",
"q-p= 9"
] | D | (A) pq = 45 or 63 or 27
(B) pq = 33 or 19
(C) 90
(D) impossible
(E) pq = 91
(D) is impossible because 09 is not a valid two digit number --- if there's a zero in the ten's place, that's just 9, a single digit number. Therefore, no valid two digit number would satisfy (D).
ans D |
AQUA-RAT | AQUA-RAT-36630 | kinetics, medicinal-chemistry, pharmacology
Title: What is the maximum amount of medicine that could accumulate in the body if the compound has a half-life of 24 hours?
If a patient is prescribed 25 mg per day of a compound that has a half-life of roughly 24 hours, what is the maximum accumulated amount of the medicine that would build in the patients body? 1st day: 25 mg
2nd day: (12.5 + 25) mg
3rd day: (6.25 + 12.5 + 25) mg
4th day: (3.125 + 6.25 + 12.5 + 25) mg
The amount get closer and closer to 50 mg but never exceeds it.
The following is multiple choice question (with options) to answer.
Roy was suffering from severe headaches. He went to see his doctor and the doctor gave him five tablets asking him to take one tablet every 15 minutes.
How much time will it take Roy to consume all the five tablets? | [
"60 minutes",
"65 minutes",
"50 minutes",
"45 minutes"
] | A | A
60 minutes
Roy will be able to consume all the five tablets in an hour.
Tablet 1 will be taken in 0 min.
Tablet 2 will be taken in 15 min.
Tablet 3 will be taken in 30 min.
Tablet 4 will be taken in 45 min.
Tablet 5 will be taken in 60 min. |
AQUA-RAT | AQUA-RAT-36631 | c#, formatting
/// <summary>
/// Calculates and returns the general discount
/// based on the <see cref="AccountStatus"/>
/// </summary>
/// <param name="accountStatus"></param>
/// <returns></returns>
private static decimal GetDiscountPercentage(AccountStatus accountStatus)
{
switch(accountStatus)
{
case AccountStatus.SimpleCustomer:
return 0.10m;
case AccountStatus.ValuableCustomer:
return 0.30m;
case AccountStatus.MostValuableCustomer:
return 0.50m;
default:
return 0.00m;
}
}
/// <summary>
/// Applying the discounts (if any) on the price and returns the final price (after discounts).
/// </summary>
/// <param name="price"></param>
/// <param name="accountStatus"></param>
/// <param name="timeOfHavingAccountInYears"></param>
/// <returns></returns>
public static decimal ApplyDiscount(decimal price , AccountStatus accountStatus , int timeOfHavingAccountInYears)
{
decimal loyaltyDiscountPercentage = GetLoyaltyDiscountPercentage(timeOfHavingAccountInYears);
decimal discountPercentage = GetDiscountPercentage(accountStatus);
decimal priceAfterDiscount = price * (1.00m - discountPercentage);
decimal finalPrice = priceAfterDiscount - ( loyaltyDiscountPercentage * priceAfterDiscount );
return finalPrice;
}
}
since all methods don't need several instance, and the nature of the class is unchangeable, making the class static would be more appropriate.
You can then reuse it :
var finalPrice = DiscountManager.ApplyDiscount(price, accountStatus, timeOfHavingAccountInYears);
The following is multiple choice question (with options) to answer.
A single discount equivalent to the discount series of 20%, 10% and 5% is? | [
"31.9",
"31.1",
"31.6",
"31.2"
] | C | :
100*(80/100)*(90/100)*(95/100) = 68.4
100 - 68.4 = 31.6
Answer: C |
AQUA-RAT | AQUA-RAT-36632 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
Eight mobsters have arrived at the theater for the premiere of the film “Goodbuddies.” One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied? | [
"20160",
"2400",
"12000",
"36002"
] | A | Arrangement of 8=8!. In half of the cases Frankie will be behind Joey and in half of the cases Joey will be behind Frankie (as probability doesn't favor any of them). So, the needed arrangement is 8!/2=20160.
Answer: A |
AQUA-RAT | AQUA-RAT-36633 | ### Show Tags
17 Oct 2010, 20:24
3
C.
I am not sure if this approach is correct. I used Elimination. There can be only 5 possible values of C if we fix A. So the number of triangles possible has to be multiple of 5. The only answer that satisfies the criterion is C.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 418
Concentration: Marketing, Finance
GPA: 3.23
Re: Right triangle ABC is to be drawn in the xy-plane so that [#permalink]
### Show Tags
25 Jan 2013, 07:08
4
2
hrish88 wrote:
Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?
A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824
First, get the integer points available for x-axis: 2 - (-6) + 1 = 9
Second, get the interger points available for y-axis: 9-4+1 = 6
How many ways to select the location of line AB in the x-axis? 9
How many ways to select the location of point C in the x-axis? 8 (Note: we cannot select the location of line AB)
How many ways to select the location of the base? 2 (Is it BC or AB?)
How many ways to position line AB parallel to y axis? 6!/2!4! = 15
Multiple all that:$$9*8*2*15 = 2,160$$
Answer: C
_________________
Impossible is nothing to God.
Intern
Joined: 01 Apr 2013
Posts: 16
Schools: Tepper '16 (S)
Re: tough problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Line Q has the equation 5y – 3x = 90. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.) | [
"58",
"61",
"64",
"67"
] | D | 5y - 3x = 90 and so y = 3x/5 + 18
When x = 0, then y = 18.
When y = 0, then x = -30
The slope is 3/5, so the slope of line S is -5/3.
Through the point (-30, 0), 0 = -5(-30)/3 + c
The y-intercept is c = -50.
Thus the perpendicular line S can have y-intercepts from -49 up to 17.
The number of possible lines is 49 + 17 + 1 = 67
The answer is D. |
AQUA-RAT | AQUA-RAT-36634 | 5. Originally Posted by Nath
"I have 2 children. One is female. What is the probability the other is also female?"
This puzzle appeared in New Scientist magazine, and a similar one in Scientific American several months back. The consensus among mathematicians (according to the articles in the magazines), including a friend of mine (who has PhD in Maths from Cambridge), is that the answer is 1/3.
I've seen several statements which claim to "mathematically prove" the answer to the question is 1/3, but I'm still not convinced. I think it's 1/2.
I'm interested to get some more opinions...
Here is an alternative "logical" way to consider this.
There are 4 possibilities in regard to the children being boy or girl.
The probability of having a boy and a girl is twice the probability of having 2 girls,
and is also twice the probability of having 2 boys.
(i) 1st and 2nd children are girls
(ii) 1st child is a girl and the 2nd is a boy
(iii) 1st child is a boy and the 2nd is a girl
(iv) 1st and 2nd children are boys
GG
GB
BG
BB
One is female.
This reduces to...
GG
GB
BG
3 cases of equal probability and in only 1 of these cases is the other child also a girl.
Hence the probability of the 2nd child also being a girl is 1/3.
Also note that in 2 of these 3 cases, the 2nd child is a boy,
therefore, if one of the children is a girl, the probability that the other is a boy is 2/3.
6. ## Solutions
This was my solution, but alas my friend disagrees:
I have 2 children, 1 is female. This gives four possible scenarios:
1. The female has a younger sister
2. The female has an older sister
3. The female has a younger brother
4. The female has an older brother
Each of these events has a 25% probability, so the probablity the female has a sister = 25% + 25% = 50%.
---
My friend, with the PhD in Maths thinks the following:
The following is multiple choice question (with options) to answer.
If there is an equal probability of a child being born a boy or a girl, what is the probability that a couple who have 5 children have two children of the same sex and one of the opposite sex? | [
"1/3",
"2/3",
"1/4",
"5/16"
] | D | No of ways of selecting a gender - 2C1
No of ways of selecting any 2 children out of 5 = 5C2
Total possible outcomes - 2^5 (each child can be either a girl or a boy)
Probability=2C1*5C2/2^5= 2*5/2*2*2*2*2=10/32=5/16
Ans=D |
AQUA-RAT | AQUA-RAT-36635 | # math
How is the circumference of a circle with radius 9cm related to the circumference of a circle with diameter 9cm?
i don't understand this.Help?? Thanks:)
1. 👍
2. 👎
3. 👁
c = pi x d.
The diamter is 2 x radius. So, the diameter of the first cirlce is 18cm. You know the diameter of the second circle. So, how do the circumferences compare?
1. 👍
2. 👎
2. is the second circle twice as big? like is that how they compare?
1. 👍
2. 👎
3. the circumference of the larger circle is indeed twice as long as that of the smaller one.
but...
the area of the larger circle is 4 times that of the area of the smaller.
Proof:
area of smaller = pi(4.5)^2 = 20.25pi
area of larger = pi(9^2) = 81pi
20.25pi/(81pi) = 1/4
1. 👍
2. 👎
4. alright thx!
1. 👍
2. 👎
## Similar Questions
1. ### Math ~ Check Answers ~
Find the circumference of the circle (use 3.14 for pi). Show your work. Round to the nearest tenth. the radius is 23 yd My Answer: ???? •Circumference = (2) (3.14) (23) •3.14 x 2 x 23 = 144.44
2. ### Mathematics
1. Find the circumference of the given circle. Round to the nearest tenth. (Circle with radius of 3.5 cm) -22.0 cm -38.5 cm -11.0 cm -42.8 cm 2. Find the area of the given circle. Round to the nearest tenth. (Circle with a
3. ### Math
The formula C = 2 x Pi x r is used to calculate the circumference of a circle when given the radius. What is the formula for calculating the radius when given the circumference? C/2Pi = R*** C/2R = Pi 2 x Pi x C = R 2 x Pi/ C = R
4. ### MATH
The following is multiple choice question (with options) to answer.
The total circumference of two circles is 25. If the first circle has a circumference that is exactly twice the circumference of the second circle, then what is the approximate sum of their two radii? | [
"5.7",
"6.0",
"4.0",
"9.7"
] | C | Let r= Radius of smaller circle. Let R = Radius of larger circle
Therefore:
2πr + 2πR = 25
where 2r = R
Thus:
2πr + 4πr = 25
6πr = 25
r = approx 1.33
πR + 2Rπ = 25
3πR = 25
R = approx 2.65
r + R = approx 3.98 = 4.0
Answer : C |
AQUA-RAT | AQUA-RAT-36636 | # Problem regarding percentage increases.
#### Strikera
##### New member
Ran into a problem which is driving me nuts and would appreciate any assistance on the problem.
Problem:
A factory increases it's production by 10%. The factory then increases it's production by another 20%. To return to the original production (before the 10 and 20% increases) how much production would the factory need to reduce?
The answer is 24% but I have no idea on how they arrived at that number or the steps needed to properly approach the problem.
Thanks in advance for any help!
#### skeeter
##### Senior Member
let P = production level
increase P by 10% = (1.10)P
increase the new level by another 20% = (1.20)(1.10)P = (1.32)P
to reduce back to P ... P = (1/1.32)(1.32)P
1/1.32 = approx 0.76 of the last level of production ... a 24% decrease.
#### Denis
##### Senior Member
Striker, when you have "no idea", make up a simple example,
like let initial production = 1000:
1000 + 10% = 1000 + 100 = 1100
1100 + 20% = 1100 + 220 = 1320
Now you can "see" that 1320 needs to be reduced back to 1000,
so a reduction of 320: kapish?
#### pka
##### Elite Member
Here is a sure-fire method to do all these problems:
$$\displaystyle \frac{{New - Old}}{{Old}}$$
This works for % of increase or decrease.
#### tkhunny
##### Moderator
Staff member
...unless, of course, Old = 0.
#### stapel
##### Super Moderator
Staff member
tkhunny said:
...unless, of course, Old = 0.
But if you're starting from zero, then "percent change" has no meaning, so it's a moot point, isn't it...?
Eliz.
#### tkhunny
The following is multiple choice question (with options) to answer.
The salary of a person was reduced by 14%. By what percent should his reduced salary be raised so as to bring it at par with his original salary? | [
"10%",
"16.3%",
"13.7%",
"15.1%"
] | B | Let the original salary be $100
New salary = $86
Increase on 86=14
Increase on 100 = 14/86* 100 = 16.3%
Answer is B |
AQUA-RAT | AQUA-RAT-36637 | 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 and B can complete a work in 30days and 15day. They started doing the work together but after 5days B had to leave and A alone completed the remaining work. The whole work was completed in? | [
"10days",
"12days",
"20days",
"18days"
] | C | A+B 1day work = 1/30 + 1/15 = 1/10
Work done by A and B in 10days = 1/10 * 5 = 1/2
Remaining work = 1-1/2 = 1/2
Now 1/30 work is done by A in 1day
1/2 work will be done by A in 30*1/2 = 15days
Total time taken = 15+5 = 20days
Answer is C |
AQUA-RAT | AQUA-RAT-36638 | # find the number of possible positive integer solutions when the inequality is $a \times b \times c \lt 180$
As most of you know there is a classical question in elementary combinatorics such that if $$a \times b \times c = 180$$ , then how many possible positive integer solution are there for the equation $$?$$
The solution is easy such that $$180=2^2 \times 3^2 \times 5^1$$ and so , for $$a=2^{x_1} \times 3^{y_1} \times 5^{z_1}$$ , $$b=2^{x_2} \times 3^{y_2} \times 5^{z_2}$$ , $$c=2^{x_3} \times 3^{y_3} \times 5^{z_3}$$ .
Then: $$x_1+x_2+x_3=2$$ where $$x_i \geq0$$ , and $$y_1+y_2+y_3=2$$ where $$y_i \geq0$$ and $$z_1+z_2+z_3=1$$ where $$z_i \geq0$$.
So , $$C (4,2) \times C(4,2) \times C(3,1)=108$$.
Everything is clear up to now.However , i thought that how can i find that possible positive integer solutions when the equation is $$a \times b \times c \lt 180$$ instead of $$a \times b \times c = 180$$
The following is multiple choice question (with options) to answer.
Let a1, a2, . . . , ar be r not necessarily distinct positive integers such that
(x + a1)(x + a2) · · · (x + ar) = xr + 230xr−1 + · · · + 2007,
where each missing term on the right has degree between 1 and r − 2
inclusive. What is the value of r ? | [
"4",
"12",
"63",
"669"
] | A | The conditions imply a1 + a2 + · · · + ar = 230 and a1a2 · · · ar = 2007. One checks that
2007 = 32 · 223, where 223 is prime. We deduce that some aj is 223, two aj’s are 3, and the
remaining aj’s are 1. Since their sum is 230 = 223 + 3 + 3 + 1, we obtain r = 4.
correct answer A |
AQUA-RAT | AQUA-RAT-36639 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
Advertisement Remove all ads
#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
Advertisement Remove all ads
#### APPEARS IN
The following is multiple choice question (with options) to answer.
In a school with 652 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years 9 months, then the number of girls in the school is | [
"150",
"200",
"250",
"163"
] | D | Sol.
Let the number of grils be x.
Then, number of boys = (652 - x).
Then, (11 3/4 × 652)
⇔ 11x + 12(652 - x) ⇔ x = 7824 - 7661 ⇔ 163.
Answer D |
AQUA-RAT | AQUA-RAT-36640 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
Find the annual income derived by investing $ 6800 in 30% stock at 136. | [
"550",
"1500",
"250",
"1300"
] | B | By investing $ 136, income obtained = $ 30.
By investing $ 6800, income obtained = $ [(30/136)*6800] = $ 1500.
Answer B. |
AQUA-RAT | AQUA-RAT-36641 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A sum of money at simple interest amounts to $ 680 in 3 years and to $710 in 4 years. The sum is: | [
"$153",
"$698",
"$590",
"$549"
] | C | C
$590
S.I. for 1 year = $(710- 680) = $30.
S.I. for 3 years = $(30 x 3) = $90.
Principal = $(680 - 90) = $590. |
AQUA-RAT | AQUA-RAT-36642 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
A car covers a distance of 984 km in 6 ½ hours. Find its speed? | [
"104 kmph",
"187 kmph",
"164 kmph",
"175 kmph"
] | C | 984/6
= 164 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-36643 | # Division of a straight line in given ratio
The problem in the book states:
"Find the coordinates of the point that divides AB in the given ratio in each of the following cases: a) A(2,4), B(-3,9) 1:4 internally"
Given the point that we are looking for is $P(X,Y)$.
Since I am not familiar with the "line division" conventions, I went to find a point (-2,8) with the following formula: $$X=\frac{\lambda x_2+\mu x_1}{\lambda+\mu}$$, where $\lambda =1$ and $\mu=4$. And a similar formula is used to find coordinate of $Y$.
I was having a cartesian coordinate system in mind when I was dividing the line into ratios. So that the portion on the left is 1 unit and the portion on the right is 4 units. However, the book assumed inversely, so that:
Would you say that my answer is incorrect, or both answers are correct? I would imagine that it would be more accurate for the book to give the ratio 5:-1, if it wanted me to arrive at the answer they are giving.
• Probably the ratio 1:4 is to be taken not from left to right, but from the first point given (A) to the second one (B). However your answer is correct, because the book should have specified more clearly what was intended. Sep 13 '15 at 14:23
• MyPoint=$P_1$ is such that $BP_1/AP_1=1/4$. TheirPoint $P_2$ is such that $AP_2/BP_2 =1/4$. So, without specifing the order of the ratio we have two solutions. Sep 13 '15 at 14:32
The following is multiple choice question (with options) to answer.
A straight line is formed using two points A(3,3) and B(6,6). Another point P(x,y) lies between A and B such that AP/BP = 9. What are the coordinates of P? | [
"(57/10, 57/10)",
"(3/10, 3/10)",
"(63/10, 63/10)",
"(27/10, 27/10)"
] | A | The equation of the straight line is y = x, so the x- and y-coordinates of P are the same.
x = 3 + (6-3)*(9/10) = 3 + 27/10 = 57/10 = y
The answer is A. |
AQUA-RAT | AQUA-RAT-36644 | So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
VP
Joined: 07 Dec 2014
Posts: 1128
Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 48 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 45 cents? | [
"1",
"2",
"3",
"4"
] | B | Let number of Apples = A
number of oranges = B
A+B=10 --- 1
.48 =(.4A + .6 B)/10
=> 48 = 4A + 6B ----2
Solving 1 and 2, we get
A= 6
B= 4
Let the number of oranges put back = C
45*(10-c) = 40*6 + 60(4-C)
=> C= 2
Answer B |
AQUA-RAT | AQUA-RAT-36645 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
_________________
Intern
Joined: 26 May 2010
Posts: 10
Followers: 0
Kudos [?]: 33 [5] , given: 4
Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of five consecutive negative integers is 10k – 1, what is the difference between the greatest and least of the five integers? | [
"4K",
"4k^2 -4k",
"4-4k",
"4"
] | D | (n-2 + n - 1 + n + n + 1 + n + 2)/5 = 10k - 1
=> n = 50k - 5
Greatest = n + 2 = 50k - 5 + 2 = 50k - 3
Least = n - 2 = 50k - 5 - 2
=> Difference = 50k - 3 - (50k - 7)
= 4
Answer - D |
AQUA-RAT | AQUA-RAT-36646 | If $a + b = c \ge N; 0 \le a < N; 0 \le b < N$ then in base N $a + b = a*N^0 + b*N^0 = (a+b)*N^0 = c*N^0 = (N + (c - N))*N^0 = N^1 + (c-N)*N^0$. Note that $0 \le c-N < N$ so there is a digit $M = c-N$. So $N^1 + (c-N)*N^0 = 1*N^1 + M*N^0 = 1M$.
==== old answer ====
Um.... 1 and 2 are symbols and mean nothing by themselves.
When we do base arithmetic we share symbols for digits.
Base1 = has one digit-symbol {1}
Base2 = has two digit-symbols {0,1}
Base3 = has three {0,1,2}
Base 4 = has four {0,1,2,3}
...
And so on.
For all bases the symbol "1" means the "base single number.
For all bases that contain the symbol "2", "2" means the "number after 1".
For all bases that contain the symbol "3", "3" means the "number after 2".
And so on.
If "2" is a symbol in the base then, yes, "2" = "the number after the base number" = "the base number plus the base number" = "1+1". That will always be true.
But Base 1 and Base 2 do not have the symbol "2".
In base 2, 1+ 1 = 10.
In base 1, 1 + 1 = 11.
But for all bases > 2 which do have the symbol "2". 1 + 1 = 2.
The following is multiple choice question (with options) to answer.
In some code, letters, a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. However, we don’t know which letter represent which number.
Consider the following relationships:
i. a + c = e
ii. b – d = d
iii. e + a = b | [
"b = 4, d = 2",
"a = 4, e = 6",
"b = 6, e = 2",
"a = 4, c = 6"
] | B | Explanation :
From the given equations, we see that , b=10 and d=5.
Now 'a' and 'c' must be 4 and 2 respectively.
So, e = 6.
Answer : B |
AQUA-RAT | AQUA-RAT-36647 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of a radio is Rs.2400 and it was sold for Rs.2100, find the loss %? | [
"12.5%",
"11%",
"13%",
"15%"
] | A | 2400 ---- 300
100 ---- ? => 12.5%
Answer:A |
AQUA-RAT | AQUA-RAT-36648 | 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.
6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in - days. | [
"4 days",
"6 days",
"2 days",
"8 days"
] | A | Explanation :
Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b
Work done by 6 men and 8 women in 1 day = 1/10
=> 6m + 8b = 1/10
=> 60m + 80b = 1 --- (1)
Work done by 26 men and 48 women in 1 day = 1/2
=> 26m + 48b = ½
=> 52m + 96b = 1--- (2)
Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200
Work done by 15 men and 20 women in 1 day
= 15/100 + 20/200 =1/4
=> Time taken by 15 men and 20 women in doing the work = 4 days
Answer IS A |
AQUA-RAT | AQUA-RAT-36649 | ### Show Tags 23 Aug 2014, 11:31 kanusha wrote: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at$0.60 per glass on the first day, what was the price per glass on the second day?
The following is multiple choice question (with options) to answer.
A fruit shop on its opening in a day, had 400 mangoes and 600 oranges. If 57 percent of the mangoes and 42 percent of the oranges got sold during the day, what percent of the total number of planted fruits were sold ? | [
" 45.5%",
" 46.5%",
" 48.0%",
" 49.5%"
] | C | 57% of 400 mangoes sold, hence 0.47*200 = 228 sold.
42% of 600 oranges sold, hence 0.42*600 = 252 sold.
Thus (germinated)/(total) = (228 + 252)/(400 + 600) = 480/1000 = 48% fruits sold.
Answer: C. |
AQUA-RAT | AQUA-RAT-36650 | # 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.
Walking 7/6 of his usual rate, a boy reaches his school 6 min early. Find his usual time to reach the school? | [
"22",
"28",
"99",
"42"
] | D | Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 -------- 7
6 --------- ?
=42 m
Answer:D |
AQUA-RAT | AQUA-RAT-36651 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article for Rs 2524.36. Approximately what was his profit percent if the cost price of the article was Rs 2400 | [
"4%",
"5%",
"6%",
"7%"
] | B | Explanation:
Gain % = (125.36*100/2400) = 5.2 % = 5% approx
Option B |
AQUA-RAT | AQUA-RAT-36652 | # 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.
Sheela deposits Rs. 3800 in bank savings account. If this is 32% of her monthly income. What is her monthly income in? | [
"22000",
"20000",
"11875",
"12340"
] | C | Explanation :
32% of income = Rs. 3800
100% of income = 3800 X 100/32 = Rs. 11875
Answer : C |
AQUA-RAT | AQUA-RAT-36653 | The totient of $210$ - the number of values between $1$ and $210$ that are relatively prime to $210$ - is $(2-1)(3-1)(5-1)(7-1)=48$. Using this, we can say that there are $48\cdot5=240$ numbers not divisible by these four numbers up to $1050$. Some of these of course are out of range of the original question; we'll have to figure out what those are.
The totient of $30$ is $8$; from $991$ to $1050$ there are $16$ numbers relatively prime to $30$. We'll take these out for now, leaving $224$. But two numbers we removed - $1001$ and $1043$ - are divisible by $7$ and so weren't part of the original $240$ so we have to un-remove them, adding them back to the total. Further, $991$ and $997$ are below $1000$ so shouldn't have been removed either. This gives $224+2+2=228$ numbers relatively prime to $210$, so $1000-228=772$ numbers are divisible by $2$, $3$, $5$, or $7$.
• Note that this doesn't work when any of the numbers we're testing divisibility for are composite. – Dan Uznanski Jan 1 '18 at 14:34
• how would you change it to work for composite numbers? – oliver Apr 27 '20 at 13:51
Your solution is correct (or, at least, your result matches mine). Huzzah!
The following is multiple choice question (with options) to answer.
How many prime numbers between 1 and 100 are factors of 210? | [
"5",
"4",
"3",
"2"
] | B | factor of 210= 2*3*5*7--- 4 prime numbers
B |
AQUA-RAT | AQUA-RAT-36654 | Formula The slope m of a line passing through points (Xl, y;) and (x2, Y2) is the ratio of the difference in the y coordinates to the corresponding difference in the x coordinates. It does not matter which points you pick for and nor does it matter which formula you use. Students use the formula slope = rise over run to solve the problems in this lesson. The slope m determines if the graph is increasing or decreasing, and how quickly. The numbers will update as you interact with the graph. , using different sets of axes) where both graphs have the same scale. y = mx + c is the equation of Straight Line. 6 Represent linear relationships graphically, algebraically (including the slope-intercept form) and verbally and relate a change in the slope or the y-intercept to its effect on the various representations; Slope Definition. Given two points on a line, you are free to label either one of them as (x 1, y 1) and the other as (x 2, y 2). If you enter this in the formula you get. 99 POINTS QUESTION 1 Use the slope formula to find the slope of the line passing through the given points. We have a very nice graph, but we have not determined if the experiment actually verified the formula C = 2pr! ©T q280w1 D17 YK2u itIaT 1S lo wfTt QwUaPrUea cL lL wCj. Finding the Equation of a Line Given a Point and a Slope. Find Slope Using a Graph 2 Find the slope of the line. A zero-slope = belongs to a horizontal line. Throughout this page we will teach you to calculate the slope of a line when you are given: The graph of the line . The SLOPE function’s syntax is: SLOPE(data_y, data_x). This does not necessarily mean that the slope of the tangent line is undefined, but it does mean that the slope cannot be determined by examining the behavior of the graph at the point P. A , B , and C are three real numbers. The larger the value is, the steeper the line. S s 5M da0d je2 Vw0i XtBhu yI2nUfOiin yi mtYeP UAyl jg NeFb 6r4aF D1R. Locate the y-intercept on the graph and plot the point. Parallel Lines: The line
The following is multiple choice question (with options) to answer.
Line m lies in the xy-plane. The y-intercept of line m is -2, and line m passes through the midpoint of the line segment whose endpoints are (2, 8) and (8,-2). What is the slope of line m? | [
"-2",
"-1",
"0",
"1"
] | D | The midpoint of (2,8) and (8,-2) is (5,3).
The slope of a line through (0,-2) and (5,3) is (3-(-2))/(5-0) = 5/5 = 1
The answer is D. |
AQUA-RAT | AQUA-RAT-36655 | (ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED.
The following is multiple choice question (with options) to answer.
If n is a 36-digit positive integer, all of whose digits are the same, which of the following must be true?
I. n is divisible by 3
II. n is divisible by 9
III. n is divisible by 36 | [
"I only",
"I and II only",
"I ,II and III only",
"II and III only"
] | C | Rule for divisibility by 3: The sum of the digits of the number should be multiple of 3
Rule for divisibility by 9: The sum of the nos digits of the number should be divisible by 9 or the number should be divisible by 3 two times.
Rules for divisibility by 36: The sum of the digits should a multiple of 36
Consider no 11111111...36 times = The sum 36*1=36----> divisbible by 3,9 and 36
consider number to be 222....36 times, then sum of the no. 36*2=72 divisibly by 3,9 and 36
So why so because when you sum the numbers either you can add the digits 36 times or multiply the digit *36..
Note that since 36 is divisble by 36,9 and 3 and thus the sum of the nos will be divisible by all the nos.
Ans is C |
AQUA-RAT | AQUA-RAT-36656 | python, programming-challenge, primes, factors
Title: Calculate the first largest factor of 600851475143 ٍProblem description:
The prime factors of 13195 are 5, 7, 13 and 29. Largest prime factor 13195 is 29.
What is the largest prime factor of the number 600851475143 ?
Prime Factor:
any of the prime numbers that can be multiplied to give the original number.
Example:
Find the prime factors of 100:
100 ÷ 2 = 50; save 2
50 ÷ 2 = 25; save 2
25 ÷ 2 = 12.5, not evenly so divide by next highest number, 3
25 ÷ 3 = 8.333, not evenly so divide by next highest number, 4
But, 4 is a multiple of 2 so it has already been checked, so divide by next highest number, 5
25 ÷ 5 = 5; save 5
5 ÷ 5 = 1; save 5
List the resulting prime factors as a sequence of multiples, 2 x 2 x 5 x 5 or as factors with exponents, 2^2 x 5^2.
My Solution
This is my solution for problem 3 of Project Euler using Python:
def FLPF(n):
'''Find Largest Prime Factor
'''
PrimeFactor = 1
i = 2
while i <= n / i:
if n % i == 0:
PrimeFactor = i
n /= i
else:
i += 1
if PrimeFactor < n: PrimeFactor = int(n)
return PrimeFactor Couple of points:
The following is multiple choice question (with options) to answer.
The second largest prime number is? | [
"1",
"2",
"3",
"4"
] | C | The second largest prime number is 3.
C) |
AQUA-RAT | AQUA-RAT-36657 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
If there are 4 peanuts in a box and Mary puts 2 more peanuts inside, how many peanuts are in the box? | [
"8",
"9",
"6",
"11"
] | C | 2+4=6
correct answer is C)6 |
AQUA-RAT | AQUA-RAT-36658 | # Evaluate the sum of the reciprocals
#### anemone
##### MHB POTW Director
Staff member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
#### mente oscura
##### Well-known member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
Hello.
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$
$$=qrs+prs+pqs+pqr=$$
$$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$
$$=-rrs-srs+pqs+pqr$$, (*)
$$(p+q)^3=-(r+s)^3$$
$$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$
$$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$
$$661+p^2q+pq^2=-r^2s-rs^2$$, (**)
For (*) and (**):
The following is multiple choice question (with options) to answer.
Two-third of a positive number and 16/216 of its reciprocal are equal. The number is: | [
"5/12",
"12/5",
"1/3",
"144/25"
] | C | Let the number be x. Then,
2/3 x = 16/216 * 1/x
x2 = 16/216 * 3/2 = 16/144 = 1/9
x = 1/3
ANSWER:C |
AQUA-RAT | AQUA-RAT-36659 | ### 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.
A certain business produced x rakes each month form November through February and shipped x/2 rakes at the beginning of each month from March through October. The business paid no storage costs for the rakes from November through February, but it paid storage costs of $0.50 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form November through October? | [
"0.40x",
"1.20x",
"1.40x",
"1.60x"
] | D | because we have a total of 4X
Also from Mar- Oct the rakes will be deducted by 1/8 X
so
In Apr they pay for the storage 0.5 * 4X * 7/8
In May they pay for the storage 0.5 * 4X * 6/8
In Jun they pay for the storage 0.5 * 4X * 5/8
In Jul they pay for the storage 0.5 * 4X * 4/8
In Aug they pay for the storage 0.5 * 4X * 3/8
In Sep they pay for the storage 0.5 * 4X * 2/8
In Oct they pay for the storage 0.5 * 4X * 1/8
total = 0.5 * 4X * 1/8 * [ 1+2+3+4+5+6+7]
= 0.5 * X/2 * (28)
= 7.0X
D |
AQUA-RAT | AQUA-RAT-36660 | classical-mechanics, soft-question
Title: Physics of a fixed wheel moving on a flat surface held at distance from a center point Ok, so my boss is trying to make a car turntable. In essence, he has a two boards that sit atop a rotating ring. He wants to put two wheels at the end of each board (8 wheels total). He thinks that you can angle wheels properly (2d) such that they can be flat (axis is parallel to the floor) and the whole contraption move smoothly.
Take a look at this picture (Edit: copied below - Mark)
The following is multiple choice question (with options) to answer.
There are 27 bicycles and 10 tricycles in the storage area at Stanley's flat building. Each bicycle has 2 wheels and each tricycle has 3 wheels. How many wheels are there in all? | [
"84",
"67",
"39",
"21"
] | A | Step 1: Find the number of bicycle wheels.
2 × 27 = 54
Step 2: Find the number of tricycle wheels.
3 × 10 = 30
Step 3: Find the total number of wheels.
54 + 30 = 84
There are 84 wheels in all.
Answer is A. |
AQUA-RAT | AQUA-RAT-36661 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A grocer has a sale of Rs. 5400, Rs. 9000, Rs. 6300, Rs. 7200 and Rs.4500 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 5600? | [
"s. 1200",
"s. 5400",
"s. 5400",
"s. 4999"
] | A | Total sale for 5 months = Rs. (5400 + 9000 + 6300+ 7200 + 4500) = Rs. 32400.
Required sale = Rs. [ (5600 x 6) - 32400 ]
= Rs. (33600 - 32400)
= Rs.1200.
Option A |
AQUA-RAT | AQUA-RAT-36662 | ### Show Tags
23 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?
Aman
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
24 Jun 2019, 03:25
pkloeti wrote:
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.
Does that approach hold up in general? Bunuel VeritasPrepKarishma
Thanks a lot for the feedback!
I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
_________________
Karishma
Veritas Prep GMAT Instructor
Intern
Joined: 12 Feb 2018
Posts: 9
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
04 Aug 2019, 10:54
i tried like this:
Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85
The following is multiple choice question (with options) to answer.
A dishonest person wants to make a profit on the selling of milk. He would like to mix water (costing nothing) with milk costing 33 $ per litre so as to make a profit of 20% on cost when he sells the resulting milk and water mixture for 36$. In what ratio should he mix the water and milk? | [
"1:20",
"1:10",
"1:8",
"1:4"
] | B | He needs to make a profit of 20%, which means he needs to sell the milk +water combination for 39.6 (20% of 33 +33)
Now, he plans to sell 1 liter of this solution of milk and water for 36. which means he needs to sell 1.1 liters of this solution to earn 39.6. (36 for 1 liter and so 39.6 for 1.1 liters)
Hence the water added would be .1 liter.
Therefore the ratio of water to milk is .1/1 = 1/10
ANSWER:B |
AQUA-RAT | AQUA-RAT-36663 | +\frac1{271} +\frac1{273} +\frac1{275} -\frac1{66} +\frac1{277} +\frac1{279} +\frac1{281} +\frac1{283} -\frac1{68} +\frac1{285} +\frac1{287} +\frac1{289} +\frac1{291} -\frac1{70} +\frac1{293} +\frac1{295} +\frac1{297} +\frac1{299} +\frac1{301} -\frac1{72} +\frac1{303} +\frac1{305} +\frac1{307} +\frac1{309} -\frac1{74} +\frac1{311} +\frac1{313} +\frac1{315} +\frac1{317} -\frac1{76} +\frac1{319} +\frac1{321} +\frac1{323} +\frac1{325} -\frac1{78} +\frac1{327} +\frac1{329} +\frac1{331} +\frac1{333} +\frac1{335} -\frac1{80} +\frac1{337} +\frac1{339} +\frac1{341} +\frac1{343} -\frac1{82} +\frac1{345} +\frac1{347} +\frac1{349} +\frac1{351} -\frac1{84} +\frac1{353} +\frac1{355} +\frac1{357} +\frac1{359} -\frac1{86} +\frac1{361} +\frac1{363} +\frac1{365} +\frac1{367} -\frac1{88}
The following is multiple choice question (with options) to answer.
(212 + 222 + 232 + ... + 302) = ? | [
"385",
"2485",
"2570",
"3255"
] | C | Solution
(212 + 222 + 232 + ... + 302) = a+(n-1)d=last number .....so 212+(n-1)10=302.....so n=10. so sum= n/2(2a+(n-1)d) = 10/2(2*212+9*10) = 2570.
Answer C |
AQUA-RAT | AQUA-RAT-36664 | # A positive integer (in decimal notation) is divisible by 11 $\iff$ …
(I am aware there are similar questions on the forum)
## What is the Question?
A positive integer (in decimal notation) is divisible by $11$ if and only if the difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by $11$.
For example consider the integer 7096276.
The sum of the even positioned digits is $0+7+6=13.$ The sum of the odd positioned digits is $7+9+2+6=24.$ The difference is $24-13=11$, which is divisible by 11.
Hence 7096276 is divisible by 11.
## (a)
Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11.
## (b)
Show that divisibility statement is true for three-digit integers $abc$. Hint: $100 = 99+1$.
## What I've Done?
I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But...the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing.
I need some help on the modular arithmetic on these.
For example... Is 77 divisible by 11? $$7+7 = 14 \equiv ...$$
I don't know what to do next.
Thanks very much, and I need help on both (a) and (b).
In order to apply the divisibility rule, you have to distinguish between odd and even positioned digits. (It doesn't matter how you count.)
Example: In 77 the first position is odd, the second even, so you would have to calculate $7-7=0$, which is divisible by 11.
The following is multiple choice question (with options) to answer.
If k is a positive integer, which of the following must be divisible by 22? | [
" (k – 4)(k)(k + 3)(k + 7)",
" (k – 4)(k – 2)(k + 3)(k + 5)",
" (k – 2)(k + 3)(k + 5)(k + 6)",
" (k + 1)(k + 3)(k + 5)(k + 7)"
] | D | 22=11*2.
Note that the product of two consecutive even integers is always divisible by 8 (since one of them is divisible by 4 and another by 2). Only option B offers two consecutive even numbers for any integer value of k: k-4 and k-2, if k=even or k+3 and k+5 if k=odd.
Also from the following 3 consecutive integers:(k-4), (k-3),(k-2)one must be divisible by 3, if it's not k-4 or k-2 then it must be k-3 (if it's k-4 or k-2 option B is divisible by 3 right away). But if it's k-3 then (k-3)+6=k+3must also be divisible by 3.
So, option B: (k – 4)(k – 2)(k + 3)(k + 5) is divisible by 8 and 3 in any case.
Answer: D. |
AQUA-RAT | AQUA-RAT-36665 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
98 students represent x percent of the boys at a school. If the boys at the school make up 50% of the total school population of x students, what is x? | [
"110",
"140",
"220",
"250"
] | B | Let B be the number of boys in the school.
98 = xB/100
B = 0.5x
9800=0.5x^2
x^2 = 19600
x = 140
The answer is B. |
AQUA-RAT | AQUA-RAT-36666 | Alternatively, the lcm of 54 and 72 can be found using the prime factorization of 54 and 72:
• The prime factorization of 54 is: 2 x 3 x 3 x 3
• The prime factorization of 72 is: 2 x 2 x 2 x 3 x 3
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(54,54) = 216
In any case, the easiest way to compute the lcm of two numbers like 54 and 72 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 54,72. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 54 and 72
What is the least common multiple of 54 and 72 used for? Answer: It is helpful for adding and subtracting fractions like 1/54 and 1/72. Just multiply the dividends and divisors by 4 and 3, respectively, such that the divisors have the value of 216, the lcm of 54 and 72.
$\frac{1}{54} + \frac{1}{72} = \frac{4}{216} + \frac{3}{216} = \frac{7}{216}$. $\hspace{30px}\frac{1}{54} – \frac{1}{72} = \frac{4}{216} – \frac{3}{216} = \frac{1}{216}$.
## Properties of LCM of 54 and 72
The most important properties of the lcm(54,72) are:
• Commutative property: lcm(54,72) = lcm(72,54)
• Associative property: lcm(54,72,n) = lcm(lcm(72,54),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The following is multiple choice question (with options) to answer.
What is the sum of the greatest common factor and the lowest common multiple of 72 and 36? | [
"192",
"220",
"224",
"108"
] | D | Prime factorization of 36 = 2 x 2 x 3 x 3
Prime factorization of 72 =2 x 2 x 2 x 3 x 3
GCF = 36
LCM = 72
Sum = 108.
Answer D. |
AQUA-RAT | AQUA-RAT-36667 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold 30articles at the cost price of 25 articles. Then find the profit% or lost% | [
"15%",
"18%",
"20% loss",
"25%"
] | C | here 30 articles selling price = 25 articles cost price
so the difference = 25-30 =-5
% of loss = 5*100/25 =20%
correct option is C |
AQUA-RAT | AQUA-RAT-36668 | 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 30 and that of another class of 50 students is 60. Find the average marks of all the students? | [
"52.5",
"52.9",
"52.1",
"48.75"
] | D | Sum of the marks for the class of 30 students = 30 * 30 = 900
Sum of the marks for the class of 50 students = 50 * 60 = 3000
Sum of the marks for the class of 80 students =
900 + 3000 = 3900
Average marks of all the students = 3900/80
= 48.75
Answer:D |
AQUA-RAT | AQUA-RAT-36669 | For an arbitrary natural number $N$, we get the inequality $$\sum_{i > k} \frac{N}{p_i} < \frac{N}{2}$$
Now, we call all the primes $p_1, p_2, \dots , p_k$ the small primes, and all the other primes the big primes.
Let $N_b$ denote the number of positive integers $n \leq N$ that have at least one divisor that is a big prime. And $N_s$ denote the number of positive integers less than $N$ that have only small prime divisors.
We will show that for a suitable $N$, $N_b + N_s < N$, which is a contradiction because their sum should equal $N$.
First, we estimate $N_b$ $$N_b \leq \sum_{i > k}\big\lfloor \frac{N}{p_i} \big\rfloor < \frac{N}{2}$$
Now, we look at $N_s$. We write every $n \leq N$ as $a_nb_n^2$ where $a_n$ is the square free part. $a_n$ is a product of different small primes. There are only $k$ small primes, and each prime may either be chosen or not chosen. So, there are only $2^k$ different values of $a_n$. Furthermore, \begin{align} b_n \leq \sqrt n \leq \sqrt N \\ \text{There are at most $\sqrt N$ different values of $b_n$} \\ \implies N_s \leq 2^k\sqrt N \end{align}
The following is multiple choice question (with options) to answer.
A number is said to be prime saturated if the product of all the different positive prime factors of w is less than the square root of w. What is the greatest two digit prime saturated integer ? | [
"99",
"98",
"97",
"96"
] | D | w 96 = 3 * 32 = 3 * 2^5
Answer is D. |
AQUA-RAT | AQUA-RAT-36670 | 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 running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . If they cross each other in 24 seconds, what is the ratio of their speeds? | [
"1:3",
"3:1",
"7:3",
"3:2"
] | C | Let the speed of the trains be x and y respectively
length of train1 = 27x
length of train2 = 17y
Relative speed= x+ y
Time taken to cross each other = 24 s
= (27x + 17 y)/(x+y) = 24
= (27x + 17 y)/ = 24(x+y)
= 3x = 7y
=x/y = 7/3
Answer :C |
AQUA-RAT | AQUA-RAT-36671 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A committee has 5 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee? | [
"150",
"200",
"250",
"300"
] | B | The number of ways to select two men and three women = ⁵C₂ * ⁶C₃
= (5 *4 )/(2 * 1) * (6 * 5 * 4)/(3 * 2)
= 200
ANSWER:B |
AQUA-RAT | AQUA-RAT-36672 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum? | [
"Rs. 4462.50",
"Rs. 8032.50",
"Rs. 8900",
"Rs. 8925"
] | D | Principal
= (100 x 4016.25) /(9 x 5)
= 401625/45
= Rs. 8925.
Answer: Option D |
AQUA-RAT | AQUA-RAT-36673 | You might want to memorize the following logarithms, which can then allow you to approximate most others quite well:
(In base $10$)
$\log2=0.30$
$\log3=0.48$
$\log5=0.70$
$\log7=0.85$
Now, just remember the following logarithm properties: $\log{ab}=\log{a}+\log{b},$ $\log{\frac{a}{b}}=\log{a}-\log{b}$, and $\log_ab=\frac{\log_{10}{a}}{\log_{10}{b}}$. Using these identities, you can approximate, relatively, precisely, most logarithms.
Example 1: $\log_2{36}=\log_2{3}+\log_2{3}+\log_2{2}+\log_2{2}=\frac{\log3}{\log2}+\frac{\log3}{\log2}+1+1=\frac{0.48}{0.30}+\frac{0.48}{0.30}+1+1=1.60+1.60+1+1=5.20$
The actual answer is around $5.17$, but this is just an approximation.
Example 2: $\log_5{100}=\log_5{5}+\log_5{5}+\log_2{5}+\log_2{5}=1+1+\frac{\log2}{\log5}+\frac{\log2}{\log5}=1+1+\frac{0.30}{0.70}+\frac{0.30}{0.70}=1+1+0.43+0.43=2.86$
The actual answer is, when rounded to $2$ decimal places, $2.86.$
The following is multiple choice question (with options) to answer.
If log10 2=0.30103,find the value of log10 50 | [
"1.69897",
"2.69897",
"1.99897",
"0.69897"
] | A | log10 50=log10(100/2)=log 100-log 2
=2log10-log 2=2(1)-0.30103=1.69897
ANSWER:A |
AQUA-RAT | AQUA-RAT-36674 | 28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
_________________
Senior Manager
Joined: 23 Oct 2010
Posts: 318
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
### Show Tags
28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
The following is multiple choice question (with options) to answer.
What is the units digit of 2222^(333)*3333^(444)? | [
"0",
"2",
"4",
"6"
] | C | Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(3333)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out thecycleof the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4powers, the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 4.
Final Answer:
C |
AQUA-RAT | AQUA-RAT-36675 | # Linear Algebra Money Question
bleedblue1234
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # $10 bills A+B+C = 32 1A+5B+10C = 100 ## The Attempt at a Solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. ## Answers and Replies E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. azizlwl You can narrow the selection.$1 can only be in a group of 5.
Homework Helper
This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.
Last edited by a moderator:
Homework Helper
Dearly Missed
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # \$10 bills
A+B+C = 32
1A+5B+10C = 100
## The Attempt at a Solution
The following is multiple choice question (with options) to answer.
A man has Rs. 528 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? | [
"80",
"90",
"95",
"99"
] | D | Let number of notes of each denomination be x.
Then x + 5x + 10x = 528
16x = 528
x = 33.
Hence, total number of notes = 3x = 99.
D |
AQUA-RAT | AQUA-RAT-36676 | # What is the chance of having a 10 of one suit and all other cards of other suits?
A game consists of 32 cards (A, K, Q, J, 10, 9, 8, 7) in four suits, and each player gets 8 cards.
I need to find the probability that I am being dealt a 10 of one suit, and have all my other cards be of a different suit. I know the chance of having a 10 is 1/8, but I get stuck on this.
The card game is a Dutch card game called 'klaverjassen'. There are 4 suits, just like a normal card game. Each player gets 8 cards out of the 32 cards. Now I need the probability that I get a 10 of one suit, and all of my 7 other cards of different suits. For example if my 10 is a diamond, the other 7 cards either need to be clubs, spades or hearts. It doesn't matter which one it is. So I have 24 cards left where I need to pick 7 cards out of.
There are 4 10's in the game. The goal I have is that I have at least 1 10 in my hand, with all other 7 cards being of a different suit than that 10. I can have multiple 10's, as long as one of them is 'unique', meaning that the other 7 cards are not of the 10's suit. So I can have a 10 of hearts, 10 of clubs, 10 of spades and 10 of diamonds in my hand and when all other cards are also hearts, it is still good
The following is multiple choice question (with options) to answer.
A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 20 percent of the pack would be outfielder cards? | [
"10",
"5",
"6",
"7"
] | A | let the number of O card to be removed=X
then remaning cards=12-X
now this 12-X cards should count less than 20% of toatal Ocards
12-X<.2*12
X>12-2.4
X>9.6
X=10
Ans A |
AQUA-RAT | AQUA-RAT-36677 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 200m long passes a standing man in 24 seconds. What is the speed of the train? | [
"30km/hr",
"36km/hr",
"42km/hr",
"50km/hr"
] | A | Speed of the train = 200/24* 18/5 = 30km/hr
Answer is A |
AQUA-RAT | AQUA-RAT-36678 | 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 class consists of 3 boys and 6 girls. Given that one kid can only hold one title, in how many ways can you pick 2 boys to be the class clown and the teacher's pet or 2 girls to be the most beautiful girl in class and the smartest kid on the block? | [
"9",
"18",
"36",
"60"
] | C | Easiest way is to treat it like an arrangements question in the following manner:
From the boys we need to select 2 to be clown and pet: This can be done in 3*2 ways
Similarly for the girls, we have 6*5 ways.
Thus total = 30+6 = 36 ways. Thus C is the correct answer. |
AQUA-RAT | AQUA-RAT-36679 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
How much water must be added to 60 litres of milk at 11⁄2 litres for 20 so as to have a mixture worth 10 2⁄3 a litre? | [
"10 litres",
"12 litres",
"15 litres",
"18 litres"
] | C | C.P. of 1 litre of milk = (20 × 2⁄3) = 40⁄3
∴ Ratio of water and milk = 8⁄3 : 32⁄3 = 8 : 32 = 1 : 4
∴ Quantity of water to be added to 60 litres of milk
= (1⁄4 × 60) litres = 15 litres.
Answer C |
AQUA-RAT | AQUA-RAT-36680 | We give a quite formal, and unpleasantly lengthy, argument. Then in a remark we say what's really going on. Let $n$ be an integer. First note that there are integers $x$ such that $n-xd\ge 0$. This is obvious if $n\ge 0$. And if $n \lt 0$, we can for example use $x=-n+1$.
Let $S$ be the set of all non-negative integers of the shape $n-xd$. Then $S$ is, as we observed, non-empty. So there is a smallest non-negative integer in $S$. Call this number $r$. (The fact that any non-empty set of non-negative integers has a smallest element is a hugely important fact equivalent to the principle of mathematical induction. It is often called the Least Number Principle.)
Since $r\in S$, we have $r\ge 0$. Moreover, by the definition of $S$ there is an integer $y$ such that $r=n-yd$, or equivalently $n=yd+r$.
Note that $r\lt d$. For suppose to the contrary that $r \ge d$. Then $r-d\ge 0$. But $r-d=r-(y+1)d$, and therefore $r-d$ is an element of $S$, contradicting the fact that $r$ is the smallest element of $r$.
To sum up, we have shown that there is an $r$ such that $0\le r\lt d$ and such that there exists a $y$ such that $r=n-yd$, or equivalently $n=yd+r$.
Case (i): Suppose that $r\lt \dfrac{d}{2}$. Then let $k=y$ and $\ell=r$. We have then $n=kd+\ell$ and $0\le \ell\lt \dfrac{d}{2}$.
The following is multiple choice question (with options) to answer.
If X represents the product of the first 15 positive integers, then X is not a
multiple of: | [
"a) 99",
"b) 84",
"c) 72",
"d) 57"
] | D | a) 99 = 9*11
b) 84 = 4*7*3
c) 72 =9*11
d) 65 = 5*13
e) 57 =19*3
since 19 is not there in first 15 positive numbers it is the only possibility
D) |
AQUA-RAT | AQUA-RAT-36681 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
_________________
Intern
Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containg 19% alcohol and now the percentage of alcohol was found to be 22%. What quantity of whisky is replaced ? | [
"1/3",
"2/3",
"2/5",
"3/5"
] | D | Let us assume the total original amount of whiskey = 10 ml ---> 4 ml alcohol and 6 ml non-alcohol.
Let x ml be the amount removed ---> total alcohol left = 4-0.4x
New quantity of whiskey added = x ml out of which 0.19 is the alcohol.
Thus, the final quantity of alcohol = 4-0.4x+0.19x ----> (4-0.21x)/ 10 = 0.26 ---> x = 20/3 ml.
Per the question, you need to find the x ml removed as a ratio of the initial volume ---> (20/3)/10 = 3/5.
Hence, D is the correct answer. |
AQUA-RAT | AQUA-RAT-36682 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A pupil's marks were wrongly entered as 83 instead of 63. Due to the average marks for the class got increased by half. The number of pupils in the class is? | [
"33",
"38",
"40",
"21"
] | C | Let there be x pupils in the class.
Total increase in marks = (x * 1/2)
= x/2
x/2 = (83 - 63) => x/2 = 20 => x
= 40.
Answer:C |
AQUA-RAT | AQUA-RAT-36683 | 1. ## reivew for final
Three marbles are drawn at random from an urn containing 8 black, 7 white, and 5 red marbles.
What is the probability that none of them are white. Answer: $\frac{_{13}C_3}{_{20}C_3}$.
What is the probability that at least one of them is red? Answer: $1-\frac{_{15}C_3}{_{20}C_3}$
A drawer contains 8 blue socks, 4 green socks, and 6 brown socks. If you choose 2 socks at random, what is the probability that they are both brown? Answer: $\frac{_{6}C_2}{_{18}C_2}$
On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$
Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?
2. Originally Posted by dori1123
On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$
Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?
Hi!
For the 2nd last question, I think it should be:
Let X be the no. of correct answers:
$P(X \geq 8)= {10 \choose 8}(0.5)^8(0.5)^2+ {10 \choose 9}(0.5)^9(0.5)+(0.5)^{10}$
The following is multiple choice question (with options) to answer.
In a box of 10 pens, a total of 3 are defective. If a customer buys 2 pens selected at random from the box, what is the probability that neither pen will be defective? | [
"4/9",
"5/11",
"7/15",
"9/22"
] | C | # defective pens = 3
# good pens = 7
Probability of the 1st pen being good = 7/10
Probability of the 2nd pen being good = 6/9
Total probability = 7/10 * 6/9 = 7/15
The answer is C. |
AQUA-RAT | AQUA-RAT-36684 | Any particular sequence made up of $3$ A's, $2$ B's, and $1$ C has probability $a^3b^2c^1$. There are $\dbinom{6}{3}\dbinom{3}{2}$ such sequences, so our probability is $$\binom{6}{3}\binom{3}{2}a^3b^2c^1.$$ In the case of your question, we have $a=b=c=\dfrac{1}{3}$.
The same basic strategy can be used for any similar problem. For example, if we modify your problem slightly to have $4$ equally likely outcomes A, B, C, D, then the probability of $4$ A's, $2$ B's, $1$ C and $3$ D's in $10$ trials is $\dbinom{10}{4}\dbinom{6}{2}\dbinom{4}{1}$ divided by $4^{10}$.
-
The following is multiple choice question (with options) to answer.
The probabilities of solving a question by 3students A, B & C are 1/2,1/3 &1/4, respectively. The probability that the problem will be solved is? | [
"1/2",
"3/4",
"3/5",
"4/7"
] | B | Sol. First, we find the probability of not solving the problem x PTO x = (1 — D x - D x - D
1 2 3 1 =iXiX71=4;
1 3 Required probability 3/4
B |
AQUA-RAT | AQUA-RAT-36685 | Part (b)
A dot plot is shown below
In the figure, “×” represents company A and “◦” represents company B. The steel rods made by company B show more flexibility.
The following is multiple choice question (with options) to answer.
How many steel rods, each of length 14 m and diameter 4 cm can be made out of 1.76 cm3 of steel? | [
"80",
"90",
"100",
"110"
] | C | Volume of 1 rod = (22/7)x(2/100) x (2/100) x 14 ) m3 =11/625 m3
Volume of steel = 1.76 m3
Number of rods =(1.76 x 625/11) =100
answer :C |
AQUA-RAT | AQUA-RAT-36686 | # 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.
The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is? | [
"21",
"23",
"31",
"33"
] | B | L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23
B) |
AQUA-RAT | AQUA-RAT-36687 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
Sides of a rectangular park are in the ratio 3: 2 and its area is 3750 sq m, the cost of fencing it at 50 ps per meter is? | [
"287",
"1287",
"125",
"988"
] | C | 3x * 2x = 3750 => x = 25
2(75 + 50) = 250 m
250 * 1/2 = Rs.125
Answer: C |
AQUA-RAT | AQUA-RAT-36688 | ### What is the volume of the rectangular prism with the base area of 42 m2 and a height of 8 cm?
The volume of a rectangular prism is the length on the side times the width times the height. 42m2 is given as the width times the height, the area of the base. 8cm = .08m .08m * 42m2 = 3.36m3
### How do you find the length of a rectangular prism?
You can compute this only if you know the volume and height, or volume and cross-sectional area. The volume of a rectangular prism is Length X Width X Height. The volume is therefore Length X Area (cross-section). L = V/A L = V/(WH)
### How do you find the volume of a rectangular prism if you have the area?
All you need is the area of the base and then multiply that by the height and you have the volume of any 3-D shape.
### How do you find the base area of a rectangular prism?
If you know the Volume and the Height of the prism, then you can divide the volume by the height to get the base area. B=V/H If you know the side lengths of the base then you could simply multiply them together to get the base area.
### What is the formula for determining volume?
Volume of a right prism: Area of Base times Height. Volume of a cube: Vertex cubed. Volume of a rectangular prism: Length times Width times Depth.
### Formula for area of a rectangular prism?
The formula for the area of a rectangular prism is A= 2(wL+hL+hw). A refers to area, w is the width of the prism, h refers to height, and L is the length of the prism.
### Volume of a rectangular prism area of a trapezoid area of a rectangle?
Volume of a rectangular prism= LengthXWidthXHeight Area of a Rectangle= LengthXWidth Area of a Trapezoid= (Bottom+Top)/2)XHeight
### What is the base area of a rectangular prism with the height of 10 m and the volume of 320 m3?
The following is multiple choice question (with options) to answer.
A hall is 6 meters long and 6 meters wide. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, what is the volume of the hall (in cubic meters)? | [
"96",
"100",
"104",
"108"
] | D | 2HL + 2HW = 2LW
H = LW / (L+W)
Volume = LWH = (LW)^2 / (L+W) = 108
The answer is D. |
AQUA-RAT | AQUA-RAT-36689 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
Rakesh credits 15% of his salary in his fixed deposit account and spends 30% of the remaining amount on groceries. If the cash in hand is Rs.2380, what is his salary? | [
"Rs.3500",
"Rs.4000",
"Rs.4500",
"Rs.5000"
] | B | Explanation:
Let salary be Rs.x. Then,
x-15% of x -30% of 85% of x = 2380
or x- 15x/100−30×85×x/100×100=2380
or 200x-30x-51x=2380×2002380×200
or 119x = 2380×2002380×200 or x 2380×200/119=4000
Correct Option: B |
AQUA-RAT | AQUA-RAT-36690 | # What is the chance of having a 10 of one suit and all other cards of other suits?
A game consists of 32 cards (A, K, Q, J, 10, 9, 8, 7) in four suits, and each player gets 8 cards.
I need to find the probability that I am being dealt a 10 of one suit, and have all my other cards be of a different suit. I know the chance of having a 10 is 1/8, but I get stuck on this.
The card game is a Dutch card game called 'klaverjassen'. There are 4 suits, just like a normal card game. Each player gets 8 cards out of the 32 cards. Now I need the probability that I get a 10 of one suit, and all of my 7 other cards of different suits. For example if my 10 is a diamond, the other 7 cards either need to be clubs, spades or hearts. It doesn't matter which one it is. So I have 24 cards left where I need to pick 7 cards out of.
There are 4 10's in the game. The goal I have is that I have at least 1 10 in my hand, with all other 7 cards being of a different suit than that 10. I can have multiple 10's, as long as one of them is 'unique', meaning that the other 7 cards are not of the 10's suit. So I can have a 10 of hearts, 10 of clubs, 10 of spades and 10 of diamonds in my hand and when all other cards are also hearts, it is still good
The following is multiple choice question (with options) to answer.
Merry and Michelle play a card game. In the beginning of the game they have an equal number of cards. Each player, at her turn, gives the other a third of her cards. Michelle plays first, giving Merry a third of her cards. Merry plays next, and Michelle follows. Then the game ends. Merry ended up with 28 more cards than Michelle. How many cards did each player have originally? | [
"45",
"27",
"54",
"36"
] | C | GameMichelleMerry
Initially 54 54 assume
After game 1 36 72
After game 2 60 48
After game 3 40 68
Now Merry has 28 cards more than Michelle. This option gives us exactly what number of cards they had initially.
So the answer is C |
AQUA-RAT | AQUA-RAT-36691 | # STRNO - Editorial
Setter: Kritagya Agarwal
Tester: Felipe Mota
Editorialist: Taranpreet Singh
Easy
Number theory
# PROBLEM:
Given two integers X and K, you need to determine whether there exists an integer A with exactly X factors and exactly K of them are prime numbers.
# QUICK EXPLANATION
A valid choice for A exists if the prime factorization of X has the number of terms at least K. So we just need to compute prime factorization of X.
# EXPLANATION
Let’s assume a valid A exists, with exactly K prime divisors.
The prime factorization of A would be like \prod_{i = 1}^K p_i^{a_i} where p_i are prime factors of A and a_i are the exponents. Then it is well known that the number of factors of A are \prod_{i = 1}^K (a_i+1). Hence we have X = \prod_{i = 1}^K (a_i+1) for a_i \geq 1, hence (a_i+1) \geq 2
So our problem is reduced to determining whether is it possible to write X as a product of K values greater than 1 or not.
For that, let us find the maximum number of values we can split X into such that each value is greater than 1. It is easy to prove that all values shall be prime (or we can further split that value). If the prime factorization of X is \prod r_i^{b_i}, then \sum b_i is the number of terms we can split X into.
For example, Consider X = 12 = 2^2*3 = 2*2*3. Hence we can decompose 12 into at most 3 values such that their product is X. However, if K < \sum b_i, we can merge the values till we get exactly K values. Suppose we have X = 12 and K = 2, so after writing 2, 2, 3, we can merge any two values, resulting in exactly two values.
So a valid A exists when the number of prime factors of X with repetition is at least K.
The following is multiple choice question (with options) to answer.
For all positive integers f, f◎ equals the distinct pairs of positive integer factors. For example, 16◎ =3, since there are three positive integer factor pairs in 16: 1 x 16, 2 x 8, and 4 x 4.
What is the greatest possible value for f◎ if f is less than 100? | [
"6",
"7",
"8",
"9"
] | A | I took numbers in reverse order from 99 and got 6 distinct pairs in 96.
I thought it may be because 96=(2^5*3) has (5+1)*(1+1)=12(perhaps maximum) factors.
1*96,2*48*4*24,8*12,16*6,32*3
However, I was not sure while answering this. Please let us know if there is a better way.
Ans: "A" |
AQUA-RAT | AQUA-RAT-36692 | number the vertices in order, going either clockwise or counter-clockwise, starting at any vertex. Find the area of triangle whose vertices are (3, 1) (2, 2) and (2, 0). Select a coordinate system, (x,y), to measure the centroid location with. Now we have to take anticlockwise direction. In all cases, the (x, y) coordinates use the top-left corner as the origin (0, 0).. Coordinates for Rectangles Calculate area of any GPS tracklog, or trace boundaries on the map, and ExpertGPS automatically calculates the area in acreage, hectares, or square feet. Calculate area out of geo-coordinates on non-convex polygons. Let’s take a look at an example. Example: (0, 0), (5, 3), (5, 7), (0, 4). Click Run.This calculates the selected properties for each feature and adds them to the selected fields. Area of the triangle is a measure of the space covered by the triangle in the two-dimensional plane. Circumcenter calculator is used to calculate the circumcenter of a triangle by taking coordinate values for each line. If a formula exists, use it to find the perimeter and the area. The area is then given by the formula Where x n is the x coordinate of vertex n, y n is the y coordinate of the nth vertex etc. Arguably, the most useful feature of GPS Calculator is the calculation of Area. By (date), when given the coordinates of the vertices of a polygon (e.g., rectangle, triangle), (name) will find the length of the...dimensions (e.g., side lengths, altitude) using the distance formula and then calculate the perimeter and area for (4 out of 5) polygons. If you're seeing this message, it means we're having trouble loading external resources on our website. You can apply equations and algebra (that is, use analytic methods) to circles that are positioned in the x-y coordinate system. By (date), when given the coordinates of the vertices of a polygon (e.g., rectangle, triangle), (name) will find the length of the...dimensions (e.g., side lengths, altitude)
The following is multiple choice question (with options) to answer.
In a rectangular coordinate system, what is the area of a quadrilateral whose vertices have the coordinates (3,-1), (3, 8), (14, 2), (14,-5)? | [
"76",
"88",
"100",
"112"
] | B | By graphing the points, we can see that this figure is a trapezoid. A trapezoid is any quadrilateral that has one set of parallel sides, and the formula for the area of a trapezoid is:
Area = (1/2) × (Base 1 + Base 2) × (Height), where the bases are the parallel sides.
We can now determine the area of the quadrilateral:
Area = 1/2 × (9 + 7) × 11 = 88.
The answer is B. |
AQUA-RAT | AQUA-RAT-36693 | 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.
It was calculated that 75 men could complete a piece of work in 20 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work? | [
"10",
"30",
"20",
"15"
] | B | Before:
One day work = 1 / 20
One man’s one day work = 1 / ( 20 * 75)
Now:
No. Of workers = 50
One day work = 50 * 1 / ( 20 * 75)
The total no. of days required to complete the work = (75 * 20) / 50 = 30
ANSWER:B |
AQUA-RAT | AQUA-RAT-36694 | "Thursdays with Ron - Consolidated Verbal Master List - Updated"
Math Expert
Joined: 02 Sep 2009
Posts: 39609
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Dec 2016, 23:09
Nightfury14 wrote:
Bunuel wrote:
gdk800 wrote:
1) A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)
a. 48
b. 100
c. 120
d. 288
e. 600
[Reveal] Spoiler:
OA is B.
Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$;
So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.
Hi Bunuel
The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:
Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp
Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp
Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp
As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.
The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.
_________________
Intern
Joined: 18 May 2016
Posts: 20
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]
### Show Tags
18 Jan 2017, 17:21
Hello,
The following is multiple choice question (with options) to answer.
A company consists of 5 senior and 3 junior staff officers. If a committee is created with 3 senior and 1 junior staff officers, in how many ways can the committee be formed? | [
"12",
"30",
"45",
"80"
] | B | Choose 3 senior from 5 senior and choose 1 junior from 3 junior:
3C5 * 1C3 = 10*3 = 30
Ans: B |
AQUA-RAT | AQUA-RAT-36695 | kinetics, medicinal-chemistry, pharmacology
Title: What is the maximum amount of medicine that could accumulate in the body if the compound has a half-life of 24 hours?
If a patient is prescribed 25 mg per day of a compound that has a half-life of roughly 24 hours, what is the maximum accumulated amount of the medicine that would build in the patients body? 1st day: 25 mg
2nd day: (12.5 + 25) mg
3rd day: (6.25 + 12.5 + 25) mg
4th day: (3.125 + 6.25 + 12.5 + 25) mg
The amount get closer and closer to 50 mg but never exceeds it.
The following is multiple choice question (with options) to answer.
A patient was given a bottle of tablets by the doctor and he was asked to take five tablets in a gap of 20 minutes. In how much time will he be able to take all the five tablets? | [
"1 hour 10 min",
"1 hour",
"1 hour 20 min",
"None"
] | C | Suppose he takes the first tablet at 8:00 pm. Then the second will be consumed by him at 8:20, third at 8:40, fourth at 9:00 and fifth at 9:20.
Time = 1 hour 20 min
Answer C |
AQUA-RAT | AQUA-RAT-36696 | Cost of building the base in is
\begin{align}70\times ~\left( lb \right) & =70\left( 4 \right) \\ & =280 \end{align}
Cost of building the walls in is
\begin{align} 2h\left( l~+~b \right)~\times ~45&=2\times 2\left( 2+2 \right)\times 45 \\ & =720 \end{align}
Required total cost is is
$280 + 720 = 1000$
Thus, the total cost of the tank will be ₹ $$1000.$$
## Chapter 6 Ex.6.ME Question 10
The sum of the perimeter of a circle and square is $$k$$, where $$k$$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
### Solution
Let $$r$$ be the radius of the circle and $$a$$ be the side of the square.
Then, we have:
\begin{align}&2\pi r + 4a = k\\ &\Rightarrow a = \frac{{k - 2\pi r}}{4}\end{align}
The sum of the areas of the circle and the square $$\left( A \right)$$ is given by,
\begin{align}A &= \pi {r^2} + {a^2}\\ &= \pi {r^2} + \frac{{{{\left( {k - 2\pi r} \right)}^2}}}{{16}}\end{align}
Hence,
\begin{align}\frac{{dA}}{{dr}}& = 2\pi r + \frac{{2\left( {k - 2\pi r} \right)\left( { - 2\pi } \right)}}{{16}}\\& = 2\pi r - \frac{{\pi \left( {k - 2\pi r} \right)}}{4}\end{align}
Now,
The following is multiple choice question (with options) to answer.
The cost of cultivating a square field at the rate of Rs.135 per hectare is Rs.1215. The cost of putting a fence around it at the rate of 75 paise per meter would be : | [
"Rs.360",
"Rs.810",
"Rs.900",
"Rs.1800"
] | C | Area= Total cost/ Rate= (1215/135) hectares= (9*10000) sq.m.
Therefore, side of the square= Ö90000=300m.
Perimeter of the field= (300*4)m= 1200m
Cost of fencing= Rs.(1200*3/4)= Rs.900
ANSWER:C |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.