source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36397 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
There are m cities. What is the number of airlines that connected 2 different cities (airline from city A to city B is different with airline from city B to city A)? | [
"n(n-1)",
"m(m-1)",
"n(n-1)/2",
"n(n+1)/2"
] | B | First of all, the possible number of airlines is m. Then, the possible number of airlines departing from one city to another becomes (m-1). Hence, the answer is m(m-1), which makes A an answer choice.
B |
AQUA-RAT | AQUA-RAT-36398 | # Given a number $n \in \Bbb{N}$. In how many ways can $n$ be written as $\prod_{i=1}^{k}n_i$ such that $n1|n2|\ldots|n_k|n$?
The following is multiple choice question (with options) to answer.
In how many ways can the integer 800 be expressed as a product of two different positive integers? | [
"10",
"7",
"5",
"4"
] | B | 800=(2^5)*(5^2)
Since 800 is not a perfect square, no of ways=7
Answer B |
AQUA-RAT | AQUA-RAT-36399 | The game is "fair".
Playing repeatedly, you can expect to break even.
4. thanks all for the answers.
but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?
The following is multiple choice question (with options) to answer.
One day, Connie plays a game with a fair 6-sided die. Connie rolls the die until she rolls a 6, at
which point the game ends. If she rolls a 6 on her first turn, Connie wins 6 dollars. For each
subsequent turn, Connie wins 1
6 of the amount she would have won the previous turn. What is
Connie's expected earnings from the game? | [
"32/31",
"33/31",
"34/31",
"36/31"
] | D | Connie has a 1
6 chance of winning 6 dollars her first turn. She has a 5/6
1/6 chance of
winning 1 dollar her second turn. Next, she has a 25
36
1/6 chance of winning 1/6 dollars her third turn.
Generalizing, Connie's expected earnings form a geometric series with initial term 1/6 *6 = 1 and
common ratio 5/6* 1/6 = 5/36 . Hence, Connie's expected earnings are
1/1- 5/36
=
36/31
correct answer D |
AQUA-RAT | AQUA-RAT-36400 | ### Show Tags
23 Oct 2009, 23:42
14
KUDOS
Expert's post
12
This post was
BOOKMARKED
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.
In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.
Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).
Consider:
$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:
$$ttttt|||$$
Means that first nephew will get all the tickets,
$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1
And so on.
How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.
The following is multiple choice question (with options) to answer.
Visitors to show were charged Rs.15 each on the first day. Rs.7.50 on the second day, Rs.2.50 on the third day and total attendance on the three days were in ratio 2:5:13 respectively. The average charge per person for the whole show is? | [
"Rs.6.00",
"Rs.5.50",
"Rs.5.00",
"Rs.7.00"
] | C | Answer: Option C
2: 5: 13
2x 5x 13x
15 7.5 2.5
30x + 37.5x + 32.5x = 100x/20x
average = 5 |
AQUA-RAT | AQUA-RAT-36401 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average weight of a group of boys is 30 kg. After a boy of weight 37 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally ? | [
"A)4",
"B)8",
"C)6",
"D)2"
] | C | Let the number off boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 37 kg joins the group, total weight of boys = 30x + 37
So 30x + 37 = 31(x + 1) = > x = 6.
Answer:C |
AQUA-RAT | AQUA-RAT-36402 | # System of three equations
$$3x+2y-z=4$$ $$4y-2z+6x=8$$ $$x-2y=5$$
I found that the first two equations were the same. Then, looking at the second and third equations, I thought the answer was No Solution, because I couldn't find a way to solve it, but instead the answer was Infinitely Many Solutions. Can someone explain this to me?
• "I couldn't find a way to solve it": there are zillion equations that have solutions though we don't know how to address them ! – Yves Daoust Sep 20 '16 at 12:49
• Usually I can solve such problems though – thunderbolt Sep 21 '16 at 2:59
## 4 Answers
As one of the equations is a duplicate, you can drop it and solve.
$$\begin{cases}3x+2y-z=4\\ x-2y=5\end{cases}$$
Let us move $z$ to the RHS and consider it as known to obtain a system of two equations in two unknowns,
$$\begin{cases}3x+2y=z+4\\ x-2y=5.\end{cases}$$
Then by elimination
$$\begin{cases}4x=z+9\\ 8y=z-11.\end{cases}$$
We can assign $z$ any value, yet get values of $x,y,z$ that satisfy all equations. For instance $(3,-1,3)$ or $(1,-2,-5)$. There are infinitely many others. In the solution space, $\mathbb R^3$, the solutions describe a straight line.
This is the same as$$\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\8\\5\end{bmatrix}$$
The following is multiple choice question (with options) to answer.
6x – 5y + 3z = 23
4x + 8y – 11z = 7
5x – 6y + 2z = 11
Given the equations above, x + y + z = ? | [
" 11",
" 12",
" 13",
" 14"
] | B | (6x – 5y + 3z) - (5x – 6y + 2z ) = 23-11
or, x+y+z = 12
Option B is the ans |
AQUA-RAT | AQUA-RAT-36403 | comes up heads. The student will count the coins and write their answer to the right of each problem. The probability of all three tosses is heads: P ( three heads) = 1 × 1 + 99 × 1 8 100. ' 'The coin is just as likely to land heads as tails. Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. problems as “if you tossed a coin 6 times, what is the probability of getting two heads?” Let p denote the probability of the outcome of interest, Hence, the probability of the other outcome must be (1 − p). Practice Problem. A coin has a 50% chance of landing on heads the each time it is thrown. Determine the probability of each event: a) an odd number appears in a toss of a fair die; b) one or more heads appear in the toss of four fair. If it isn’t a trick coin, the probability of each simple outcome is the same. An experiment could be rolling a fair 6-sided die, or. Subjective Probability. Coin Probability Problems Coin is a currency token which has two faces, one is head and other is tail. For a fair coin, what is the probability that the longest run of heads or tails in a sequence of 30 tosses is less than or equal to 5? (pg 107) Because the coin toss is the simplest random event you can imagine, many questions about coin tossing can be asked and answered in great depth. (15 – 20 min) Homework Students flip a coin. Describe the sample space. Think of it this way: What is the probability of tossing 2 heads in a row if you toss a fair coin 7 times? Multiplication would lead you to think the probability is 6*1/4=1. 125 Stacy and George are playing the heads or tails game with a fair coin. Well, that is unless you failed to spin the coin, there is probability involved there too. Assume that the probability a girl is born is the same as the probability a boy is born. Jack has coins C_1, C_2,. This page continues to illustrate probability facts using the flip-a-coin-4-times-and-count-the-number-of-heads problem. What is the probability that you’ll toss a coin and get heads? What about twice in a row? Three times? Probability questions ask you determine the likelihood that an
The following is multiple choice question (with options) to answer.
Three coins are tossed. Find the probability of at least 2 tails ? | [
"1/2",
"2/3",
"1/4",
"1/8"
] | A | n(s) = 2^3 = 8
let E is the event of getting at least 2 tails
n(E) = TTT, TTH , HTT, THT = 4
P(E) = n(E) / n(s)
= 4/8
= 1/2
Ans- A |
AQUA-RAT | AQUA-RAT-36404 | Quick way
Use Smart Numbers
Give 100 for the initial amount
Then you will have 50-0.25x = 30
x = 80
So % is 80/100 is 80%
Hope it helps
Cheers!
J
SVP
Joined: 06 Sep 2013
Posts: 1647
Concentration: Finance
Re: If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
29 May 2014, 11:41
Or one can use differentials to slve
Initially 50% alcohol
Then 25% alcohol
Resulting mixture 30% alcohol
Therefore, 20X - 5Y= 0
5X = Y
X/Y = 1/4
Now, mixture is 20% over total (1/5).
Therefore 80% has been replaced by water.
Hope this helps
Cheers
J
Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 463
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
If a portion of a half water/half alcohol mix is replaced [#permalink]
### Show Tags
02 Jul 2014, 17:12
Bunuel wrote:
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
The following is multiple choice question (with options) to answer.
The amount of water (in ml) that should be added to reduce 12 ml. Lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is? | [
"8 ml",
"5 ml",
"4 ml",
"6 ml"
] | A | 6 6
30% 70%
30% ----- 6
70% ------? => 14 - 6 = 8 ml
Answer:A |
AQUA-RAT | AQUA-RAT-36405 | ### 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.
Mary and Mike enter into a partnership by investing $550 and $450 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received $1000 more than Mike did, what was the profit made by their business in that year? | [
"$8000",
"$10000",
"$15000",
"$12000"
] | C | Explanatory Answer
Let the profit made during the year be $3x
Therefore, $x would have been shared equally and the remaining $2x would have been shared in the ratio 5.5 : 4.5
i.e., 55% of 2x would go to Mary and 45% of 2x would go to Mike.
Hence, Mary would get (55 - 45)% of 2x more than Mike
Or10% of 2x = $1000
i.e.,(60/100)*2x = 1000
or 2x = 10000.
Hence, the profit made by the company during the year $3x = $15000.
Answer : C |
AQUA-RAT | AQUA-RAT-36406 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election, candidate A got 79% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid vote polled in favour of candidate. | [
"357003",
"357090",
"37667",
"379888"
] | B | Total number of invalid votes = 15 % of 560000
= 15/100 × 560000
= 8400000/100
= 84000
Total number of valid votes 560000 – 84000 = 476000
Percentage of votes polled in favour of candidate A = 75 %
Therefore, the number of valid votes polled in favour of candidate A = 75 % of 476000
= 75/100 × 476000
= 35700000/100
= 357090
Answer:B |
AQUA-RAT | AQUA-RAT-36407 | $$T = \frac {60}{160 + w} + \frac {60}{160-w} = \frac {60(160-w) + 60(160 + w)} {(160 + w)(160 - w)} = \frac {9600 - 60w + 9600 + 60w}{(160 + w)(160 - w)}= \frac {19200}{(160 + w)(160 - w)} = \frac {19200}{25600 +160w-160w-w^2}=\frac {19200}{25600-w^2}$$
For this scenario of finding round trip time where you have 2 islands, your wind direction is always parallel to your flight path, and the wind is always in the same direction you could reduce it to this formula.
r=rate (in this instance 160 km/h) w=wind speed (assuming here is in units km/h) d=km between the 2 islands
$$T = \frac {d(r-w) + d(r +w)}{(r+w)(r-w)} = \frac {dr+dr} {r^2-w^2} = \frac {2dr}{r^2-w^2}$$
If we take the problem with no wind.
r= 160 km/h w=wind speed (assuming here is in units km/h) d=60km $$\frac {(2)(60)(160)}{160^2-0^2} = \frac {19200}{25600} = \frac {3}{4}$$ of an hour
Problem with 30 km/h wind speed r= 160 km/h w=30km/h d=60km $$\frac {(2)(60)(160)}{160^2-30^2} = \frac {19200}{25600-900}= \frac {19200}{24700} = \frac {192}{247}$$ of an hour = ~$$0.777$$ of an hour
The following is multiple choice question (with options) to answer.
A plane flies 660 km with the wind and 570 km against the wind in the same length of time. If the speed of the wind is 15 km/h, what is the speed of the plane in still air? | [
"185 km/h",
"195 km/h",
"205 km/h",
"215 km/h"
] | C | The speed of the plane in still air = x km/h
The speed of the wind is 15 km/h
Speed with the wind = (x + 15) km/h
Speed against the wind = (x – 15) km/h
Time = Distance/ Speed
660 / (x+15) = 570 / (x-15)
660(x-15) = 570(x+15)
66x - 990 = 57x + 855
9x = 1845
x = 205
Therefore, the speed of the plane in still air is 205 km/h.
The answer is C. |
AQUA-RAT | AQUA-RAT-36408 | => 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.
Quentin's income is 60% less than Rex's income, and Sam's income is 25% less than Quentin's income. If Rex gave 60% of his income to Sam and 40% of his income to Quentin, Quentin's new income would be what fraction of Sam's new income? | [
"8/9",
"11/12",
"8/13",
"11/13"
] | A | We can take some easy numbers and make calculations simpler.
Let R ( Rex's income) = 100
Q (Quentin's income) = 40%R = 40
S (Sam's income) = 75% Q = (3/4)*40 = 30
Now, if Rex gives 40% to Quentin --> Q = 40 + 40 = 80
60% given to Sam --> S = 30+60 = 90
The ratio is : Q/S = 80/90 = 8/9
ANSWER:A |
AQUA-RAT | AQUA-RAT-36409 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
What is the principal sum?
I. The sum amounts to Rs. 660 in 3 years at S.I.
II. The sum amounts to Rs. 700 at S.I.
III. The rate of interest is 6% p.a. | [
"I and III only",
"II and III only",
"I and II only",
"Any two of the three"
] | A | Explanation:
Clearly, I and III is needed to solve this question where as II is not needed
Answer: A |
AQUA-RAT | AQUA-RAT-36410 | rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron’s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 × 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The "base times height" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b×h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal ‘d1’ of a rhombus and a side ‘a’, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of […] where b is the
The following is multiple choice question (with options) to answer.
The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is? | [
"388",
"267",
"298",
"480"
] | D | 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48 = 480.Answer:D |
AQUA-RAT | AQUA-RAT-36411 | # 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 wishes to give 5% commission on the marked price of an article but also wants to earn a profit of 10%. If his cost price is Rs. 95, then marked price is: | [
"Rs. 100",
"Rs.110",
"Rs. 120",
"Rs. 130"
] | B | Solution: CP = Rs. 95.
Then SP = 95 + 10% of 95 = Rs. 104.5.
Let MP = X. He gives 5% commission on MP.
So,
SP = X - 5% of X
SP = 0.95X.
104.5 = 0.95X.
X = 104.5/0.95 = 110.
Thus, MP = Rs. 110.
Short-cut
95===10%(gain)===>104.5===5%(Commission)===>109.72( = 110)
Answer: Option B |
AQUA-RAT | AQUA-RAT-36412 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can do a work in 9 days. If A alone can do it in 18 days. In how many days can B alone do it? | [
"10",
"99",
"18",
"55"
] | C | 1/9 – 1/18 = 1/18
=> 18
Answer: C |
AQUA-RAT | AQUA-RAT-36413 | evolution, zoology, anatomy, species
Title: Examples of animals with 12-28 legs? Many commonly known animals' limbs usually number between 0 and 10. For example, a non-exhaustive list:
snakes have 0
Members of Bipedidae have 2 legs. Birds and humans have 2 legs (but 4 limbs)
Most mammals, reptiles, amphibians have 4 legs
Echinoderms (e.g., sea stars) typically have 5 legs.
Insects typically have 6 legs
Octopi and arachnids have 8 legs
decapods (e.g., crabs) have 10 legs
....But I can't really think of many examples of animals containing more legs until you reach 30+ legs in centipedes and millipedes. Some millipedes even have as many as 750 legs! The lone example I am aware of, the sunflower sea star, typically has 16-24 (though up to 40) limbs.
So my question is: what are some examples of animals with 12-28 legs? As a couple of counterexamples, species in the classes Symphyla (Pseudocentipedes) and Pauropoda within Myriapoda have 8-11 and 12 leg pairs respectively, so between 16 to 24 legs (sometimes with one or two leg pair stronlgy reduced in size).
(species in Symphyla, from wikipedia)
Another common and species-rich group with 14 walking legs (7 leg pairs) is Isopoda.
(Isopod, picture from wikipedia)
You also need to define 'legs' for the discussion to be meaningful. As you say, decapods have 10 legs on their thoracic segments (thoracic appendages), but they can also have appendages on their abdomens (Pleopods/swimming legs), which will place many decapods in the 10-20 leg range.
(Decapod abdominal appendages/legs in yellow, from wikipedia)
So overall, in Arthropoda, having 12-28 legs doesn't seem all that uncommon. There are probably other Arthropod groups besides those mentioned here that also have leg counts in this range.
However, for a general account, the most likely answer (if there is indeed a relative lack of 12-28 legged animals) is probably evolutionary contingencies and strongly conservative body plans within organism groups.
The following is multiple choice question (with options) to answer.
There are some pigeons and hares in a zoo. If heads are counted, there are 250. If legs are counted, there are 580. The number of hares in the zoo is? | [
"40",
"50",
"60",
"70"
] | A | 250*2 = 500
580
-----
80
1----2
?----80 = 40
ANSWER:A |
AQUA-RAT | AQUA-RAT-36414 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4342
Location: India
GPA: 3.5
A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Manager
Joined: 18 Aug 2013
Posts: 128
Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 640 Q48 V28
GPA: 3.92
WE: Operations (Transportation)
Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
The average age of 30 students in a class is 14 years. If the age of teacher is also included, the average becomes 15 years, find the age of the teacher. | [
"43",
"44",
"45",
"47"
] | C | Explanation:
If teacher's age is 14 years, there is no change in the average. But teacher has contributed 1 year to all the students along with maintaining his age at 15.
Age of teacher = Average age of all + Total increase in age
= 15 + (1 x 30 ) = 45 years
Answer:C |
AQUA-RAT | AQUA-RAT-36415 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row 7.2 kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream. | [
"2.4 km/hr",
"12.5 km/hr",
"1 2.6 km/hr",
"22.5 km/hr"
] | A | Given that, time taken to travel upstream = 2 × time taken to travel downstream
When distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream
Let speed upstream = x
Then speed downstream = 2x
we have, 1/2(x+2x) = speed in still water
⇒1/2(3x)=7.2
3x = 14.4
x = 4.8
i.e., speed upstream = 4.8 km/hr
Rate of stream = 1/2(2x−x)=x/2=4.8/2=2.4 km/hr
Answer is A |
AQUA-RAT | AQUA-RAT-36416 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a cistern in 75/2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after: | [
"5 min.",
"9 min.",
"10 min.",
"15 min"
] | B | If B is closed after x mins, then
30*2/75 + x/45 = 1
x/45 = 1 - 4/5 = 1/5
x= 45/5 = 9 mins
ANSWER:B |
AQUA-RAT | AQUA-RAT-36417 | And also...there is no specific weight for the satellite...(Thinking)
• Nov 23rd 2008, 12:25 PM
skeeter
I get $4.23 \times 10^7$ meters
what values are you using?
btw ... mass of the satellite doesn't matter
• Nov 23rd 2008, 01:21 PM
realintegerz
I used
G = 6.67 x 10^-11 N m^2/kg^2
M = 5.98 x 10^24 kg
T = 86400s/1 rev
Am I wrong with T?
• Nov 23rd 2008, 01:33 PM
o_O
Those numbers should work.
$r = \sqrt[3]{\frac{(6.67 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}) (5.98 \times 10^{24} \ \text{kg})(86400 s)^2}{4\pi^2}} \approx 4.23 \times 10^{7} \ \text{m}$
• Nov 23rd 2008, 07:47 PM
realintegerz
oh i see my problem now, i didnt square the period T
The following is multiple choice question (with options) to answer.
During the months May through October, Teddy the bear sextuples his weight, and during the months November through April, Teddy loses 100 pounds. If at the beginning of November 1973, Teddy's weight was m pounds, what was Teddy's weight in pounds by the end of April 1976? | [
"36m - 1300",
"36m - 1200",
"6m - 1300",
"6m - 1200"
] | A | Nov ' 73 = m
April ' 74 = m - 100
Oct ' 74 = 6m - 300
April' 75 = 6m - 400
Oct' 75 = 36m - 1200
April'76 = 36m - 1300
Hence answer will be (A) 36m - 1300 |
AQUA-RAT | AQUA-RAT-36418 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A person goes to his office at 1/3rd of the speed at which he returns from his office. If the avg speed during the whole trip is 24m/h. what is the speedof the person while he was going to his office? | [
"8km/h",
"10km/h",
"12km/h",
"16km/h"
] | D | u = k , v= 3k
\inline \therefore \frac{2uv}{u+v}\: \: \Rightarrow \frac{2\times k\times 3k}{(k+3k)}=24
\inline \Rightarrow 1.5k = 24
\inline \Rightarrow k=16km/h
D |
AQUA-RAT | AQUA-RAT-36419 | homework-and-exercises, forces, kinematics
Title: Homework Question Transformation Energy A $1400kg$ car is approaching the hill shown in the figure at $14.0m/s$ when it suddenly runs out of gas.
What is the car's speed after coasting down the other side?
I think I have to use this equation
$mg(\Delta h)=\frac 12 m(\Delta v)^2 $
$(1400)(9.81)(\Delta h)= \frac12 (\Delta v)^2$
I am not sure if this is correct. If it is, what do I plug in for $\Delta h$? You have to use $\Delta h$=$5m$ since it is the height by which the car descends along the course of the journey. And instead of $$mg\Delta h=\frac12 m(\Delta v)^2$$ you should rather use $$mg\Delta h=\Delta KE=\Delta (\frac12 mv^2)=\frac12 m (v_f^2-v_i^2)$$
where $v_i=14ms^{-1}$ and evaluate for $v_f$.
The following is multiple choice question (with options) to answer.
A car travels uphill at 30 km/hr and downhill at 80 km/hr. It goes 100 km uphill and 50 km downhill. Find the average speed of the car? | [
"38kmph",
"33kmph",
"34kmph",
"35kmph"
] | A | avg speed=total distance/total time.
total distance traveled=100+50=150km;
time taken for uphill journey=100/30=10/3;
time taken for down hill journey=50/80=5/8;
avg speed=150/(10/3+5/8)=38kmph
ANSWER:A |
AQUA-RAT | AQUA-RAT-36420 | # 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.
The number of arrangements that can be made with the letters of the word WORDS so that the vowels occupy the even places? | [
"14",
"32",
"777",
"24"
] | D | The word MEADOWS has 7 letters of which 3 are vowels.
-V-V-V-
As the vowels have to occupy even places, they can be arranged in the 1 even places in 1! i.e., 1 ways. While the consonants can be arranged among themselves in the remaining 4 places in 4! i.e., 24 ways.
Hence the total ways are 24 * 1 = 24.
Answer:D |
AQUA-RAT | AQUA-RAT-36421 | 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.
There are 50 students in a class. If 14% are absent on a particular day, find the number of students present in the class. | [
"43",
"36",
"28",
"129"
] | A | Number of students absent on a particular day = 14 % of 50
i.e., 14/100 × 50 = 7
Therefore, the number of students present = 50 - 7 = 43 students.
Answer:A |
AQUA-RAT | AQUA-RAT-36422 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 27 km/hr crosses a post in 20 seconds. What is the length of the train? | [
"200",
"150",
"300",
"175"
] | B | Speed=(27x5/18)=7.5 m/sec.
Length of the train=(Speed x Time).
Length of the train=7.5 x20 m = 150 m.
Answer: B |
AQUA-RAT | AQUA-RAT-36423 | # Thread: What is the inverse of this matrix?
1. ## What is the inverse of this matrix?
What is the inverse of:
[3 7]
[1 3]
I've tried working it out, but I keep getting the wrong answer.
Here's my working out:
[3 7 1 0]
[1 3 0 1]
Divide row 1 by 3
[1 2.3 0.3 0]
[1 3 0 1]
Row 1 minus (1 x row 1)
[1 2.3 0.3 0]
[0 1 -0.43 1.43]
Row 2 divided by 0.7
[1 2.3 0.3 0]
[0 1 -0.43 1.43]
Row 1 minus (2.3 x row 2)
[1 0 1.289 -3.289]
[0 1 -0.43 1.43]
The answer is supposed to be:
[3/2 -7/2]
[-1/2 3/2]
But my answers don't match the ones I got above even when I try and change the decimals to fractions.
Help would be much appreciated. Sorry if the matrices look messy, I hope you get the idea.
2. Originally Posted by brumby_3
What is the inverse of:
[3 7]
[1 3]
I've tried working it out, but I keep getting the wrong answer.
Here's my working out:
[3 7 1 0]
[1 3 0 1]
Divide row 1 by 3
[1 2.3 0.3 0]
[1 3 0 1]
Row 1 minus (1 x row 1)
[1 2.3 0.3 0]
[0 1 -0.43 1.43]
Row 2 divided by 0.7
[1 2.3 0.3 0]
[0 1 -0.43 1.43]
Row 1 minus (2.3 x row 2)
[1 0 1.289 -3.289]
[0 1 -0.43 1.43]
The answer is supposed to be:
[3/2 -7/2]
[-1/2 3/2]
But my answers don't match the ones I got above even when I try and change the decimals to fractions.
Help would be much appreciated. Sorry if the matrices look messy, I hope you get the idea.
The following is multiple choice question (with options) to answer.
The inverse ratio of 3: 2: 1 is? | [
"2:3:7",
"2:3:9",
"2:3:2",
"2:3:6"
] | D | 1/3: 1/2: 1/1 = 2:3:6
Answer:D |
AQUA-RAT | AQUA-RAT-36424 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
Luke travelled 12 2/3 miles in 1 hours and 40 minutes. What was his average rate of speed in miles per hour? | [
"7",
"7 3/5",
"8 1/5",
"9"
] | B | D = 12(2/3) = 38/3
T = 1(2/3) = 5/3
S = D/T = 7 3/5
Answer = B |
AQUA-RAT | AQUA-RAT-36425 | #### Opalg
##### MHB Oldtimer
Staff member
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
First, the minimum value of $xy$ must be positive, because if $xy\leqslant 0$ then $(x+y)z\geqslant 4$. So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and $$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$$ Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. But that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}$ is $8$ (occurring when $z^2 = 4$).
So we may assume that $xy>0$. Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$ Therefore $xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.
The following is multiple choice question (with options) to answer.
If x = 7 and y = −2, what is the value of (x − 2y)^y? | [
"−100",
"0.001",
"0.25",
"4"
] | B | Quickly we can spot that answer is neither integer nor negative. Eliminate A, DE
by inversing and squaring 0.001
Answer: B |
AQUA-RAT | AQUA-RAT-36426 | javascript, datetime, converting
if(minutes < 10) minutes = '0' + minutes;
return hours + ':' + minutes + ' ' + suffix;
}
totalMinutes rolls over after rounding, so passing in 23.999 is equivalent to passing in 0, and 27 becomes 3, etc.
hours also roll over but start at 12 instead of zero, then 1, 2, 3 and so on.
minutes should be straight-forward, and
suffix basically looks at whether the totalMinutes are greater than or equal to noon-in-minutes.
Snippet with some test cases below (format borrowed from rolfl).
function original(info) {
var suffix='AM';
var hrs = parseInt(Number(info));
var min = Math.round((Number(info)-hrs) * 60);
if (hrs >12) { hrs = hrs - 12; suffix='PM'; }
if (min < 10) {min='0'+min;}
return hrs + ':' + min + ' ' + suffix;
}
function reviewed(fractionalHours) {
var totalMinutes = Math.round(fractionalHours * 60) % (24 * 60),
hours = Math.floor(totalMinutes / 60) % 12 || 12,
minutes = totalMinutes % 60,
suffix = totalMinutes >= 12 * 60 ? 'PM' : 'AM';
if(minutes < 10) minutes = '0' + minutes;
return hours + ':' + minutes + ' ' + suffix;
}
var testCases = [
["12:00 AM", 0], // midnight
["12:30 AM", 0.5],
["1:00 AM", 0.99999999],
["11:30 AM", 11.5],
["12:00 PM", 11.9999999], // noon
["12:00 PM", 12],
["11:30 PM", 23.5],
["12:00 AM", 23.9999999],
["12:00 AM", 24],
["1:00 AM", 25]
];
The following is multiple choice question (with options) to answer.
The sum of 4 hours 45 minutes and 5 hours 55 minutes is approximately what percent of a day? | [
"32%",
"36%",
"40%",
"44%"
] | D | Since the question is asking for an approximate percentage
4:45+5:55 ~ 11 hours
% of day = 11*100/24 ~ 11*100/25= 44%
Answer is D. |
AQUA-RAT | AQUA-RAT-36427 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is: | [
"8.5 km/hr",
"10 km/hr",
"12.5 km/hr",
"9 km/hr"
] | B | Explanation:
Man's speed with the current = 15 km/hr
=> speed of the man + speed of the current = 15 km/hr
speed of the current is 2.5 km/hr
Hence, speed of the man = 15 - 2.5 = 12.5 km/hr
man's speed against the current = speed of the man - speed of the current
= 12.5 - 2.5 = 10 km/hr
ANSWER IS B |
AQUA-RAT | AQUA-RAT-36428 | Smallest number of children such that, rounding percentages to integers, $51\%$ are boys and $49\%$ are girls [closed]
I faced a very confusing question during my preparation for Mathematics olympiad, here is the question:
The percentage of boys in a gathering, rounded to an integer, is 51 percent. and the percentage of girls in this gathering, rounded to an integer, is 49 percent. What is the minimum possible number of participants in this gathering?
Could anyone help me?
• Confusing, as in hard to understand or not sure what to do? If the latter, please at least show some observations that you have made. – player3236 Oct 27 at 11:57
• @Reza, did the question have any options ? – Spectre Oct 27 at 11:57
• How are you starting to think about this? Are there any bounds you can easily find on the solution? Is there anything you have tried at all? These Olympiad questions are about building problem-solving resilience as much as anything - and any observation which gets you anywhere can potentially be a way in. – Mark Bennet Oct 27 at 11:58
The following is multiple choice question (with options) to answer.
90 students represent x percent of the boys at Jones Elementary School. If the boys at Jones Elementary make up 20% of the total school population of x students, what is x? | [
"125",
"150",
"212",
"250"
] | C | 90=x/100*20/100*x=>x^2 = 9*10000/2 = > x =212
C |
AQUA-RAT | AQUA-RAT-36429 | • One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37
The following is multiple choice question (with options) to answer.
If z + | z | = 0, which of the following must be true? | [
"z > 0",
"z≥0",
"z< 0",
"z≤0"
] | D | Manipulate the equations:
z + | z | = 0
|z| = -y
-z > 0 OR -z = 0
This means z could be 0 or z is less than 0.
D. z≤0
E. z=0
Answer: D |
AQUA-RAT | AQUA-RAT-36430 | \mathit{skill}_{p,s} x_{p,t}\\ & f_{\mathit{preft}} = \sum_{p,t} \mathit{prefteam}_{p,t} x_{p,t} \\ & f_{\mathit{prefw}} = \sum_{p,p',t|p' >p} \mathit{prefworker}_{p,p'} x_{p,t} x_{p',t} \\ &\sum_t x_{p,t} \le 1 \\ &\sum_p x_{p,t} = \mathit{demand}_{t}\\ &x_{p,t} \in \{0,1\} \end{align}}
The following is multiple choice question (with options) to answer.
If 20% of s = t, then t% of 20 is the same as | [
"10% of s",
"4% of s",
"20% of s",
"None"
] | B | Explanation :
20% of s = t
=> t = 20/100s
Then,
t% of 20 = (t/100)*20=((20/100)*s)*20/100
=(2020s)/(100100) =4s/100 = 4% of s
Answer : B |
AQUA-RAT | AQUA-RAT-36431 | #### Opalg
##### MHB Oldtimer
Staff member
Find the minimum value of $xy$, given that $x^2+y^2+z^2=7$, $xy+xz+yz=4$, and $x, y$ and $z$ are real numbers.
First, the minimum value of $xy$ must be positive, because if $xy\leqslant 0$ then $(x+y)z\geqslant 4$. So $(x+y)^2\geqslant\dfrac{16}{z^2}$, and $$7-z^2 = x^2+y^2 = (x+y)^2 - 2xy \geqslant (x+y)^2 \geqslant \frac{16}{z^2}.$$ Thus $z^2 + \dfrac{16}{z^2} \leqslant 7$. But that cannot happen, because the minimum value of $z^2 + \dfrac{16}{z^2}$ is $8$ (occurring when $z^2 = 4$).
So we may assume that $xy>0$. Let $u = \sqrt{xy}$ and $v = x+y$. Then we can write the equations as $v^2-2u^2 + z^2 = 7$, $vz+u^2=4$. Therefore $v^2 - 2(4-vz) + z^2 = 7,$ so $(v+z)^2 = 15$. But $vz\leqslant\bigl(\frac12(v+z)\bigr)^2 = \frac{15}4.$ Therefore $xy = u^2 \geqslant 4-\frac{15}4 = \frac14$.
The following is multiple choice question (with options) to answer.
What is the minimum value of z for which z^2 + z - 3/6 > 0 is not true? | [
" -5/2",
" -3/2",
" -1",
" 1/4"
] | C | I think C -1 is the best ans... |
AQUA-RAT | AQUA-RAT-36432 | This does not change even if we change the extras $4'$ and $3'$ to $40'$ and $30'$! A $40'$ wide extra "flank" along the $12'$ side plus a $30'$ flank along the $10'$ side are bigger than vice versa.
• great work, it helps – Complex Guy Jun 4 '13 at 17:34
Another simple method particularly suited to a deliberately 'symmetric' problem like this: $35{,}043\times 25{,}430=(25{,}043\times25{,}430)+(10{,}000\times25{,}430)$ while $25{,}043\times35{,}430=(25{,}043\times25{,}430)+(10{,}000\times25{,}043)$, and written out this way it's clear that the former has to be larger.
A = 35,043×25,430 = 35,043 × 25,043 + 35,043 × 0,387
B = 35,430×25,043 = 35,043 × 25,043 + 25,043 x 0,387
so A is bigger
$\begin{eqnarray}{\bf Hint}\quad 35043 \times 25430 &-\,&\ \, 35430 \times 25043 \\ A\ (B\! +\! N)&-\,& (A\!+\!N)\ B\ =\ (A\!-\!B)\,N > 0\ \ \ {\rm by}\ \ \ A > B,\ N> 0\end{eqnarray}$
from the first equation we get:
(35,000 + 43) * (25,000 + 430) =
= 35,000 * 25,000 + 430*35,000 + 43*25,000 + 43*430 ( let it be a)
from the second second equation we get:
(35,000 + 430)*(25,000 + 43) =
= 35,000 * 25,000 + 43 * 35,000 + 430 * 25,000 + 43 * 430 (let it be b)
suppose that a > b (1)
The following is multiple choice question (with options) to answer.
One side if a square is doubled. If the parameter is now 12x. What was the length of one of the sides before it was doubled? | [
"12x/4",
"12x/2",
"6x/2",
"12x/6"
] | B | If the parameter is now 12x, the each side is 12x/4. To get the original each side is divided by 2. One side is represented by 1 since it's a square. 1/2 is 0.5, which when times by 4 is 2. Thus the answer is 12x/2. Answer option B. |
AQUA-RAT | AQUA-RAT-36433 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
A lady builds 10cm length, 13cm width, and 5cm height box using 5cubic cm cubes. What is the minimum number of cubes required to build the box? | [
"107",
"108",
"130",
"150"
] | C | Number of cubes required = Volume of box/Volume of cube
= 10*13*5/5
=130 cubes
ANSWER:C |
AQUA-RAT | AQUA-RAT-36434 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
Find the smallest number of 6 digits which is exactly divisible by 111? | [
"12000",
"15550",
"100011",
"158993"
] | B | Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 - 100) - 11.
Hence, required number = 100011
Answer B |
AQUA-RAT | AQUA-RAT-36435 | or, $x^{2}$+10x+100=$x^{2}$+6x+36+196
or, 4x=132
or, x=33.
Categories
## Algebra and Positive Integer | AIME I, 1987 | Question 8
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.
## Algebra and Positive Integer – AIME I, 1987
What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?
• is 107
• is 112
• is 840
• cannot be determined from the given information
### Key Concepts
Digits
Algebra
Numbers
AIME I, 1987, Question 8
Elementary Number Theory by David Burton
## Try with Hints
Simplifying the inequality gives, 104(n+k)<195n<105(n+k)
or, 0<91n-104k<n+k
for 91n-104k<n+k, K>$\frac{6n}{7}$
and 0<91n-104k gives k<$\frac{7n}{8}$
so, 48n<56k<49n for 96<k<98 and k=97
thus largest value of n=112.
Categories
## Positive Integer | PRMO-2017 | Question 1
Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.
## Positive Integer – PRMO 2017, Question 1
How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$
• $9$
• $7$
• $28$
### Key Concepts
Algebra
Equation
multiplication
Answer:$28$
PRMO-2017, Problem 1
Pre College Mathematics
## Try with Hints
Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$
Can you now finish the problem ……….
The following is multiple choice question (with options) to answer.
If m is a positive integer and m^2 is divisible by 24, then the largest positive integer that must divide m is? | [
" 3",
" 6",
" 8",
" 12"
] | C | M^2 is divisible by 48 so M^2 must be multiple of 48.
If the value of M is Multiples of 8 then it will satisfy the condition. If we If M is 12 or 24 or 36 then it ans is C but if M = 48 then answer should be 16.
Is the question right? Or am i missing some thing?
C |
AQUA-RAT | AQUA-RAT-36436 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
a person travels from his house to office from 11:10am to 11:50 am and covers a distence of 13.4km. find the average speed of the person. | [
"20.1km/h",
"20.5km/h",
"21.1km/h",
"22.1km/h"
] | A | time=11:50-11:10=40min=2/3hour
distance=13.4km
speed=13.4*3/2=20.1km/h
ANSWER:A |
AQUA-RAT | AQUA-RAT-36437 | 1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2)
3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$
So I would answer problem 1 as a is not equal to either 0 or 1.
In the second problem, it is clear that $x \ne 3.$
Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$
$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$
Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.
Does this help
5. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by JeffM
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
The following is multiple choice question (with options) to answer.
3/x + 3x = 3(x-9) | [
"-9",
"-1/6",
"-1/9",
"1/9"
] | B | We can solve - expand the right side, multiply by x on both sides and then subtract away the 3x^2 terms:
(3/X) + 3x = 3(x-9)
(3/x) + 3x = 3x - 18
3 + 3x^2 = 3x^2 - 18x
3 = -18x
-1/6 = x
and to confirm, you can plug that answer back into the original equation to see that it makes the left and right sides equal.
B |
AQUA-RAT | AQUA-RAT-36438 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
A bowl of fruit contains 14 apples and 21 oranges. How many oranges must be removed so that 70% of the pieces of fruit in the bowl will be apples? | [
" 3",
" 15",
" 14",
" 17"
] | B | Number of apples = 14
number of oranges = 23
let number of oranges that must be removed so that 70% of pieces of fruit in bowl will be apples = x
Total number of fruits after x oranges are removed = 14+(21-x) = 35-x
14/(35-x) = 7/10
=>20 = 35-x
=>x= 15
Answer B |
AQUA-RAT | AQUA-RAT-36439 | 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 210 meters long is running with a speed of 54 kmph. The time taken by it to cross a tunnel 120 meters long is? | [
"29",
"22",
"48",
"99"
] | B | D = 210 + 120 = 330
S = 54 * 5/18 = 15 mps
T = 330/15 = 22 sec
Answer: B |
AQUA-RAT | AQUA-RAT-36440 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
At the store opening, Larry's Grocery had 50 Lemons and 60 Oranges. By closing, the store at 20 Lemons and 40 Oranges left. By approximately what percent did the ratio of Lemons to Orange decrease from opening to closing. | [
"50.0%",
"40.0%",
"30.0%",
"20.0%"
] | B | Opening: Lemons/Oranges = 50/60 =100/120
Closing: Lemons/Oranges = 20/40 =60/120
ASIDE: It's useful to write both ratios with the same denominator. This allows us to IGNORE the denominator and focus solely on the numerators.
So, our ratio went from 100/120 to 60/120
Ignoring the denominators, we went from 100 to 60
The percent change = 100(difference in values)/(original value)
= (100)(100-60)/100
= (100)(40)/100)
= 40
Answer: B |
AQUA-RAT | AQUA-RAT-36441 | 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.
.
On dividing a number by 357, we get 40 as remainder. On dividing the same number 17, what will be the remainder ? | [
"6",
"3",
"5",
"8"
] | A | Let x be the number and y be the quotient. Then,
x = 357 x y + 40
= (17 x 21 x y) + (17 x 2) + 6
= 17 x (21y + 2) + 6)
Required remainder = 6.
Answer: Option A |
AQUA-RAT | AQUA-RAT-36442 | to solve for the other. Logic To Calculate Percentage Difference Between 2 Numbers. Sale Discount Calculator - Percent Off Mortgage Loan Calculator - Finance Fraction Calculator - Simplify Reduce Engine Motor Horsepower Calculator Earned Value Project Management Present Worth Calculator - Finance Constant Acceleration Motion Physics Statistics Equations Formulas Weight Loss Diet Calculator Body Mass Index BMI Calculator Light. The commonly used way is to go from right to left, which gives us a positive number. Posted: (1 week ago) Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Just subtract the past value from the current value. Click the calculate button 3. Calculate percentage difference between two columns I have a input text file in this format: ITEM1 10. The percentage difference between the two values is calculated by dividing the value of the difference between the two numbers by the average of the two numbers. First, you need Excel to subtract the first number from the second number to find the difference between them. Listing Results about Calculate Variance Between Two Numbers Real Estate. To write an increase or decrease as a percentage, use the formula actual increase or decrease original cost × 100%. The growth rate can be listed for real or nominal GDP. This difference needs to be divided between the first number (the one that doesn't change). We then append the percent sign. Follow 52 views (last 30 days) Show older comments. Enter an old number in cell A1 and a new number in cell B1. The percent difference formula or the percent difference equation of two numbers a and b is: ((a - b) / (a+b)/2) × 100, where a > b Calculate the percentage difference between the numbers 35 and 65. The first step to the equation is simple enough. Step 1: Calculate the difference (subtract one value from the other) ignore any negative sign. With a Difference From, Percent Difference From, or Percent From calculation, there are always two values to consider: the current value, and the value from which the difference should be calculated. A percentage variance, aka percent change, describes a proportional change between two numbers, an original value and a new value. It has more of an impact when you say, "There was a 50 percent increase in attendance at the concert compared to last year," versus when you say, "There were 100 more people at the concert this year than
The following is multiple choice question (with options) to answer.
A number x is 7 times another number y. The percentage that y is less than x is | [
"12.5%",
"85.7%",
"80%",
"11%"
] | B | Say y=1 and x=7.
Then y=1 is less than x=7 by (7-1)/7*100=6/7*100=85.7%.
Answer: B. |
AQUA-RAT | AQUA-RAT-36443 | 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.
There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is | [
"100",
"110",
"120",
"140"
] | A | Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 x - y = 20 .... (i)
and x + 20 = 2(y - 20) x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100. |
AQUA-RAT | AQUA-RAT-36444 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made? | [
"1.2",
"3.78",
"16.5",
"5.84"
] | C | Explanation:
Let the quantities of A and B mixed be 3x kg and 7x kg.
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
Correct Option : C |
AQUA-RAT | AQUA-RAT-36445 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
Legendary Member
Posts: 1892
Joined: 14 Oct 2017
Followed by:3 members
### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Legendary Member
Posts: 2663
Joined: 14 Jan 2015
Location: Boston, MA
Thanked: 1153 times
Followed by:127 members
GMAT Score:770
by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
From a group of 7 men and 6 women, 4 persons are to be selected to form a committee so that at most 2 men are there in the committee. In how many ways can it be done? | [
"480",
"520",
"760",
"470"
] | D | Total Men = 7
Total Women = 6
Persons in committee = 4
No. of ways to select at most 2 men = 6C4 + 7C1 * 6C3 + 7C2*6C2
= 470
Ans -D |
AQUA-RAT | AQUA-RAT-36446 | By brute force, here are the possible sequences.
(8,0,...),(7,1,0,...),(6,2,0,...),(6,1,1,0,...),(5,3,0,...),(5,2,1,0,...)(5,1,1,1,0,...),(4,4,0,...),(4,3,1,0...),(4,2,2,0,...),(4,2,1,1,0,...),(4,1,1,1,1,0,...),(3,3,2,0,...),(3,3,1,1,0,...),(3,2,2,1,0,...),(3,2,1,1,1,0,...),(3,1,1,1,1,1,0,...),(2,2,2,2,0,...),(2,2,2,1,1,0,...),(2,2,1,1,1,1,0,...),(2,1,1,1,1,1,1,0,...),(1,1,1,1,1,1,1,1,0,...)
As you can see, there are 22 of them. I'm not sure how to generalize this result at the moment, perhaps someone who is more familiar with the problem will come around with a counting method. Perhaps a recurrence relation.
Looking for a pattern at the sequence of numbers for this problem with 1,2,3,... number of objects: I get the amounts: 1,2,3,5,7,11,15,22.
By google searching I find http://mathworld.wolfram.com/PartitionFunctionP.html which has a great deal of history about the problem and methods of calculating the sequence.
The following is multiple choice question (with options) to answer.
14, 35, 56, 91, 126
In the sequence above, each term is related with the others. Which of the following could not be a term in the sequence? | [
"407",
"259",
"322",
"686"
] | A | Simplifying the question into easier words we just need to find which of the numbers is not divisible by 7
clearly the answer is 407 because if we divide 407 by 4 we get a remainder of 1
Correct answer - A |
AQUA-RAT | AQUA-RAT-36447 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. Find out his new time compared with the old time. | [
"twice as much",
"three times as much",
"the same",
"half as much"
] | A | Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.
We know that
Speed = Distance/Time.
Or, Time = Distance/Speed.
So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.
And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.
= 100/x hr.
So we can clearly see that his new time compared with the old time was: twice as much.
Answer is A. |
AQUA-RAT | AQUA-RAT-36448 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
The age of Ritu's mother is 6 times that of Ritu's age. After 6 years it will be 3 times that of Ritu's age. what is Ritu's mother present age? | [
"24 years",
"26 years",
"28 years",
"10 years"
] | A | Let present age of Ritu=x
then, present age of ritu's mother=6x
A/c
after 6 years
=>6x+6=3(6+x)
x=4
so, Ritu mother present age is 24 years
ANSWER:A |
AQUA-RAT | AQUA-RAT-36449 | Let's consider a word of length $m$ from the binary alphabet $\{0,X\}$, having a total of $s$ zero's.
$$\begin{array}{*{20}c} X &| & {0,} & {0,} & {0,} &| & X &| & 0 &| & {X,} & X &| & {0,} & 0 &| & X \\ 0 &| & {1,} & {2,} & {3,} &| & 0 &| & 1 &| & {0,} & {0,} &| & {1,} & 2 &| & 0 \\ \end{array}$$
Imagine to sequentially scan the word and count the number of consecutive zeros, resetting the counter when the character is different from $0$, as in the exaple above.
Then we can bi-ject each word with an hystogram which has :
- $j$ bars;
- sum of the bars equal $s$;
- $m-s$ Xs.
The following is multiple choice question (with options) to answer.
Every letter in the alphabet has a number value that is equal to its place in the alphabet. Thus, the letter A has a value of 1, the letter B has a value of 2, the letter C has a value of 3, etc... The number value of a word is obtained by adding up the value of the letters in the word and then multiplying that sum by the length of the word. What is the number value of the word "RAT"? | [
"108",
"111",
"114",
"117"
] | D | "RAT" = (18+1+20)*3=117.
The answer is D. |
AQUA-RAT | AQUA-RAT-36450 | 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.
For any non-zero a and b that satisfy |ab| = -ab and |a| = a, a^2+b^2+2|ab| =? | [
"ab-4",
"(a+b)^2",
"(a-b)^2",
"a^2-|ab+ab^2|"
] | C | Given: |ab| = -ab and |a| = a
Question: a^2+b^2+2|ab| =?
**** Looking at |ab| = -ab tells us that a and b are either positive or negative
**** Looking at |a| = a tells us that a must be positive
**** Combine two observations: a is positive, b is negative
so a^2+b^2+2|ab| = a^2+b^2-2ab.
as we know : (a-b)^2= a^2+b^2-2ab.
hence option c is correct. |
AQUA-RAT | AQUA-RAT-36451 | 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 speeds past a pole in 15 seconds and a platform 100 meter long in 25 seconds. What is length of the train ? | [
"140 meter",
"145 meter",
"150 meter",
"155 meter"
] | C | Explanation:
Let the length of the train is x meter and Speed of the train is y meter/second
Then x/y = 15 [because distance/speed = time]
=> y = 15/x
=>x+100/25=x/15
x=150 meters
So length of the train is 150 meters
Option C |
AQUA-RAT | AQUA-RAT-36452 | Let x and y be the dimensions of the rectangle. Then area is A=xy
But $(\frac{x}{2})^{2}+y^{2}=16$
This means $y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}$
Sub into A and we have:
$\frac{x}{2}\sqrt{64-x^{2}}$
$\frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}$
$32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}$
Max area is then $(4\sqrt{2})(2\sqrt{2})=16$
Now, try the same problem with an inscribed trapezoid
The following is multiple choice question (with options) to answer.
The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle? | [
"1: 96.",
"1: 99",
"1: 94",
"1: 93"
] | A | Explanation:
Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 24
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x = 1: 96.
Answer: Option A |
AQUA-RAT | AQUA-RAT-36453 | # 2008 AMC 12A Problems/Problem 17
## Problem
Let $a_1,a_2,\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$?
$\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004$
## Solution
All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.
• If $a_1=4n$, then $a_2=\frac{4n}{2}=2n.
• If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1.
• If $a_1=4n+2$, then $a_2=2n+1.
• If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.
The following is multiple choice question (with options) to answer.
A sequence of positive and negative numbers contain q numbers. If the sequence begins with a negative number and q is an odd number how many positive numbers are in it. | [
"q/(2+1)",
"(q+1)/2",
"(q-1)/2",
"q/(2-1)"
] | C | Lets put values for x, say q=7 since it's odd
and first number is neg so sequence will be {neg,pos,neg,pos,neg,pos,pos} we can see that it has 3 positive terms
so putting q=7 in the options,
option C = (q-1)/2 => 7-1/2 = 3 , the number of positive integers
So, C is the answer |
AQUA-RAT | AQUA-RAT-36454 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
There are 8 executives, including the CEO and CFO, that are asked to form a small team of 5 members. However, the CEO and CFO may not both be assigned to the team. Given this constraint, how many ways are there to form the team? | [
"34",
"35",
"36",
"37"
] | C | The total number of ways to form a team of 5 is 8C5=56.
We need to subtract the number of teams that have both the CEO and the CFO.
The number of teams with both the CEO and CFO is 6C3=20.
The number of ways to form an acceptable team is 56-20=36.
The answer is C. |
AQUA-RAT | AQUA-RAT-36455 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A car covers a distance of 720 km in 6 ½ hours. Find its speed? | [
"104 kmph",
"187 kmph",
"120 kmph",
"175 kmph"
] | C | 720/6
= 120 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-36456 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can swim in still water at 6 km/h, but takes twice as long to swim upstream than downstream. The speed of the stream is? | [
"2",
"1.4",
"1.1",
"1.5"
] | A | M = 6
S = x
6 + x = (4.5 - x)2
6 + x = 12 -2x
3x = 6
x = 2
Answer:A |
AQUA-RAT | AQUA-RAT-36457 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round? | [
"4",
"8",
"16",
"32"
] | C | Explanation :
There is more than 1 way to approach the solution; however, I will detail the easiest way to go about it here.
We want to find the ratio of time taken for nth round : time taken for (n-1)th round
It will be same as finding the ratio of time taken for 2nd round : Time taken for 1st round.
1 round = Circumference of the circle = 2πr
1st round :
Speed = πr for 30 seconds. So, total distance travelled = πr/2.
Speed = πr/2 for 1 minute. So, total distance travelled = πr/2.
Speed = πr/4 for 2 minutes. So, total distance travelled = πr/2.
Speed = πr/8 for 4 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 7.5 minutes.
2nd round:
Speed = πr/16 for 8 minutes. So, total distance travelled = πr/2.
Speed = πr/32 for 16 minutes. So, total distance travelled = πr/2.
Speed = πr/64 for 32 minutes. So, total distance travelled = πr/2.
Speed = πr/128 for 64 minutes. So, total distance travelled = πr/2.
So, for a distance of 2πr, time taken is 120 minutes.
Ratio is 120:7.5 = 16:1.
Answer : C |
AQUA-RAT | AQUA-RAT-36458 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
Anand and Deepak started a business investing Rs. 4,500 and Rs. 6,000 respectively. Out of a total profit of Rs. 420. Deepak's share is? | [
"s. 280",
"s. 310",
"s. 240",
"s. 840"
] | C | Ratio of their shares = 4500 : 6000 = 3 : 4
Deepak's share = Rs. (420 * 4/7) = Rs. 240 ANSWER "C |
AQUA-RAT | AQUA-RAT-36459 | >
Order each of the following pairs of numbers, using $<\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}>\text{:}\phantom{\rule{0.2em}{0ex}}0.18___0.1.$
>
Order $0.83___0.803$ using $<$ or $>.$
## Solution
The following is multiple choice question (with options) to answer.
Evaluate : (2.39)(power 2) - (1.61) (power 2) / 2.39 - 1.61 | [
"4",
"6",
"8",
"2"
] | A | Given Expression = a(power 2) - b(power 2) / a-b
= (a + b)(a - b) / (a-b)
= (a + b) = (2.39 + 1.61) = 4.
Answer is A. |
AQUA-RAT | AQUA-RAT-36460 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of first 16 even numbers is? | [
"10",
"11",
"17",
"13"
] | C | Sum of 16 even numbers = 16 * 17 = 272
Average = 272/16 = 17
ANSWER:C |
AQUA-RAT | AQUA-RAT-36461 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
A basket contains 9 marbles. 4 are red and 5 are blue marbles If two marbles are drawn at random what is the probability that at least one ball is red. | [
"19/18",
"17/18",
"11/18",
"13/18"
] | D | Total probability=9C2=36
4C2 +4C1*5C1=6+20=26
therefore the probability that at least one of the red ball will be selected= 26/36=13/18
answer is D |
AQUA-RAT | AQUA-RAT-36462 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
Kudos [?]: 24 [2], given: 0
Manager
Joined: 21 Jan 2015
Posts: 149
Kudos [?]: 121 [0], given: 24
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Other)
Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of 38 students in a group is 14 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years? | [
"31",
"36",
"53",
"58"
] | C | Sol.
Age of the teacher = ( 39 × 15 – 38 × 14 ) years = 53 years.
Answer C |
AQUA-RAT | AQUA-RAT-36463 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
How many three digit numbers N are divisible by 78 or 91? | [
"17",
"19",
"20",
"21"
] | B | The answer will be 19.
Explanation:
78= 2*3*13 now multiples of 78, 156 ....780, now 1000-780 = 220 only two more muktiples of 78 can exists. So total number of 3 digit Multiples of 78 are 9+2=11
91= 13*7--Total number of three digit multiples --9
No remember we have a common multiples as well N-- 13*7*6=91*6=546
So total number of multiples--11+9 -1 =19.
Hence Answer is 19.B |
AQUA-RAT | AQUA-RAT-36464 | The operations above are perfectly reasonable. Let me show you the kind of situation that I think you were worried about, and you'll see why it's different from the situations above:
I take 4 exams and have an average score of 80. Then I take 2 more exams and on those 2, my average is 100. What's my overall average for the course?
The wrong way to do it is (80+100)/2 = 90. This is wrong because the averages were of different-sized groups. To get the correct answer, I know that on the first 4 exams I got 320 total points because when I divide 320 by 4, I get 80.
Similarly, for the last two exams, I must have gotten 200 points total. So for the six exams, I got 200+320 = 520 points and 520/6 = 86.66666 = my real grade average.
We saw this scenario last week: If we just average the two averages, the result can only be described as the average of the averages, not the average for the course. The latter has to be weighted, because for the course, each exam counts the same, not each of these two sets of exams.
As we saw last week, a weighted average can often be understood by breaking it down:
For your problem, the averages you are averaging are for the same period, so it works out. To convince you that it's true, let's just look at a situation where the averages came from 10 months of data.
Then in the first line of business, 500 people must have been hired, since 500/10 = 50. Similarly, 800 were hired in the other, since 800/10 = 80. Altogether, 1300 people were hired in the ten months, or 1300/10 = 130 per month, company-wide. Or if you're trying to get the average per line of work, 65 is right, since if each group had hired 65 people each month for 10 months, there would be 65*20 = 1300 total hires, so it works out.
Bottom line: always think about what you want, and what an average means, rather than use an average (or not!) unthinkingly.
## Using a weighted average
Averaging Averages
The following is multiple choice question (with options) to answer.
In three annual examinations, of which the aggregate marks of each was 500, a student secured
average marks 45% and 55% in the first and the second yearly examinations respectively. To secure 50% average total marks, it is necessary for him in third yearly examination to secure marks : | [
"300",
"250",
"400",
"450"
] | B | total marks:1500 for three exams
50% of 1500=750
first exam marks=45% of 500=225
second exam marks=55% of 500=275
let X be the third exam marks
225 + 275 + X =750
X=250
ANSWER:B |
AQUA-RAT | AQUA-RAT-36465 | # 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.1500 and it was sold for Rs.1230, find the loss %? | [
"18%",
"16%",
"17%",
"78%"
] | A | 1500 ---- 270
100 ---- ? => 18%
Answer:A |
AQUA-RAT | AQUA-RAT-36466 | Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240.
The following is multiple choice question (with options) to answer.
Two circular signs are to be painted. If the diameter of the larger sign is six times that of the smaller sign, how many times more paint is needed to paint the larger sign? (We can assume that a given amount of paint covers the same area on both signs.) | [
"6",
"12",
"24",
"36"
] | D | Let R be the radius of the smaller sign.
Then the diameter of the smaller sign is 2R, the diameter of the larger sign is 12R, and the radius of the larger sign is 6R.
The area A of the smaller sign is A = piR^2.
The area of the larger sign is pi(6R)^2=36piR^2=36A.
Since the area is 36 times larger, we need 36 times more paint for the larger sign.
The answer is D. |
AQUA-RAT | AQUA-RAT-36467 | in fabricating stock, custom & engineered-to-order vessel designs. The eye coordinates are now multiplied with GL_PROJECTION matrix, and become the clip coordinates. We’d be happy to help! (800) 922-1932. No, water will not rise about the height of the surface of the tote. The general method is to use the fact that the volume at any level h is the integral of the surface area at h times h. Home » Solutions » Mechanics: Statics » Determine the horizontal and vertical components. 4 metre height of water = 0. The steel calculated in step 3 is for 1m height from the bottom of the water tank. It may be calculated using the formula: Bulk Modulus (K) = Volumetric stress / Volumetric strain. Because of the symmetry of the circle and therefore the torus with respect to the y axis, we integrate from x = 0 to x = r then double the answer to find the total volume. Experiment confirms the Planck relationship. A f is the cross-sectional area of the fluid in a horizontal tank's cylindrical section. It is given by the formula: where r is the radius of the base and h is the perpendicular height of the cone. Any interaction of the fluid jet with air is ignored. Standard baffle configuration utilises 3 or 4 equally spaced vertical baffles T/12 where T is the internal tank diameter. Volume (V) = π * R 2 * ((4/3) * R + H) Volume (V) = 3. Volume of 2 1 elliptical head formula. x 81"H: A-VT0300-35: $367. Domestic 124 to 2,000 gallons. If this strip is revolved about the x-axis, we obtain a. (Back up fuel for remotely located generators, i. Factor K varies depending on the shape of the ellipse. The graph shows lines for tank diameters from 4 to 10 ft, and tank lengths from 1 to 50 ft. Volume in cubic inches. With a one-litre volume, it allows you to do short or long sessions. A obelisk is the polyhedron formed by two parallel rectangles whose side faces are trapezoids. A vertical cylinder with a movable piston contains 1. Engineering designs, specifications and schematics We measure a volume flow rate in the SI units of meters cubed per second (m3/s) but you
The following is multiple choice question (with options) to answer.
A cube of edge 5cm is immersed completely in a rectangular vessel containing water. If the dimensions of the base of vessel are 10cm * 5cm , find the rise in water level? | [
"2.5cm",
"3.6cm",
"5cm",
"6.43cm"
] | A | Increase in volume = volume of the cube = 5*5*5 cm^3
Rise in water level = volume / area = 5*5*5/10*5 = 2.5cm
Answer is A |
AQUA-RAT | AQUA-RAT-36468 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
A baker filled with a measuring cup with 1/4 cup water. He poured 1/2 of the water into the batter, and then spilled 1/8 of the water on the floor. How much water will the baker needed to add what is left in the cup to have 50% more than what he started with? | [
"1/8 cup",
"3/8 cup",
"1/4 cup",
"1/2 cup"
] | B | B
1/4 is the original water in cup .half in batter.So left is 1/8 out which 1/8 is spilled out.So again left with 0.
so 50% more than what he started was = 1/4+1/2*(1/4)=3/8
Amount of water needed to add = 3/8 - 0=3/8
B |
AQUA-RAT | AQUA-RAT-36469 | "How does this help me?" you say
I'm very glad you asked
Let's think about $8$ as our 100% and we want to find 2 parts of 8, so using what we just learned we say that "2 parts of 8 is the same as $\frac{2}{8}$." To turn $\frac{2}{8}$ into a percentage we just need the denominator to equal 100 since earlier we said that 8 is our 100%
We could use an equation for this, so let's use an equation for this :D
We know that $8 \cdot \left(\text{some number}\right) = 100$ so;
$8 x = 100$
$\frac{1}{8} \cdot 8 x = 100 \cdot \frac{1}{8} = \frac{100}{8} = \frac{25}{2} = 12.5$
Now we know that if we multiply the denominator by 12.5 we will get 100 and if we multiply the numerator by 12.5 we will get a ratio with 100 as the denominator, so our percentage must be the numerator!
2/8*12.5/12.5 = 25/100 = 25%
Was this helpful? Let the contributor know!
##### Just asked! See more
• 9 minutes ago
• 9 minutes ago
• 11 minutes ago
• 11 minutes ago
• 55 seconds ago
• 2 minutes ago
• 2 minutes ago
• 4 minutes ago
• 7 minutes ago
• 8 minutes ago
• 9 minutes ago
• 9 minutes ago
• 11 minutes ago
• 11 minutes ago
The following is multiple choice question (with options) to answer.
If two positive numbers are in the ratio 1/8 : 1/5, then by what percent is the second number more than the first? | [
"67%.",
"70%.",
"60%.",
"68%."
] | C | Given ratio = 1/8 : 1/5 = 5 : 8
Let first number be 5x and the second number be 8x.
The second number is more than first number by 3x.
Required percentage = 3x/5x * 100 = 60%.
Answer: C |
AQUA-RAT | AQUA-RAT-36470 | # 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 8articles at the cost price of 10 articles. Then find the profit% or lost% | [
"10%",
"25%",
"20%",
"30%"
] | B | here 8 articles selling price = 10 articles cost price
so the difference = 10-8 = 2
% of profit = 2*100/8 = 25%
correct option is B |
AQUA-RAT | AQUA-RAT-36471 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
Compound interest earned on a sum for the second and the third years are Rs.1200 and Rs.1440 respectively. Find the rate of interest? | [
"65% p.a",
"88% p.a",
"20% p.a",
"66% p.a"
] | C | Rs.1440 - 1200 = Rs.240 is the interest on Rs.1200 for one year.
Rate of interest = (100 * 240) / (100 * 1) = 20% p.a
Answer: C |
AQUA-RAT | AQUA-RAT-36472 | of the half of the and! = 18π: PM as the perimeter of a circle ) ] or\ c.: 6169ec2ffad00696 • Your IP: 185.2.4.37 • Performance & security by cloudflare, Please complete the security check access. ) r units that area of the window is 20 ft, find the perimeter and area 7!, y }, { rx, ry } ] gives an ellipse! Two congruent triangles d or πr, and area 1470 perimeter ( circumference ) of a semicircle has a is! You temporary access to the circumference of a semicircle including its diameter arc depends on. Gives a circle ) of that circle and the 2r because those diameters are part of semicircle.
The following is multiple choice question (with options) to answer.
A certain rectangular window is (1/3) times as long as it is wide. If its perimeter is 28 feet, what are its dimensions in terms of length by width? | [
"12 by 2",
"11 by 3",
"10.5 by 3.5",
"10 by 4"
] | C | 2x+2y=28
x+y=14
x+(1/3)x=14
4x=14*3
x=10.5
Answer C |
AQUA-RAT | AQUA-RAT-36473 | 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.
In how many years Rs 100 will produce the same interest at 5% as Rs. 200 produce in 2 years at 10% | [
"8",
"10",
"9",
"12"
] | A | Explanation:
Clue:
Firstly we need to calculate the SI with prinical 200,Time 2 years and Rate 10%, it will be Rs. 40
Then we can get the Time as
Time = (100*40)/(100*5) = 8
Option A |
AQUA-RAT | AQUA-RAT-36474 | Since we have $21$ terms taking $20$ possible values, there are some $0 \le i < j \le 20$ such that $S_i = S_j$. It follows that the total number of hours of study between days $i+1$ and $j$ (inclusive) is a multiple of $20$.
If it is not exactly equal to $20$ hours, then it must be at least $40$ hours. However, this is over a span of at most $20$ days. Dividing this into three periods of at most $7$ days (say the first week, second week, and third week), by averaging we find that she must have worked at least $14$ hours during one of the weeks, which is not allowed.
Thus she must actually have studied exactly $20$ hours between days $i+1$ and $j$.
• Yes, that's a little cleaner as route to Wiley's lemma.Thanks. – Joffan Jul 17 '16 at 23:05
• @Joffan, agreed. It's indeed cleaner to use 20 "holes" with 21 "pigeons" rather than separating $S_l=S_i+20$ as an individual case in my proof. Thank you, Shagnik. – Wiley Jul 18 '16 at 3:59
The proof consists of two parts.
• Part I: Prove that a period of $20$ days is enough such that there must exist some period of consecutive days during which totally $20$ hours are spent on studying.
• Part II: A counterexample which shows that $19$ days are not enough is presented.
Proof of Part I
The following is multiple choice question (with options) to answer.
A library has an average of 600 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is: | [
"250",
"260",
"300",
"285"
] | C | Since the month begins with sunday,to there will be five sundays in the month
Average required = (600x5 + 240x25) /30) = 300
Answer: Option C |
AQUA-RAT | AQUA-RAT-36475 | What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$
The following is multiple choice question (with options) to answer.
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high? | [
"7",
"9",
"11",
"13"
] | D | Explanation:
d2 = 122 + 42 + 32 = 13
D) |
AQUA-RAT | AQUA-RAT-36476 | # 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 buys two articles for Rs.1000 each and then sells them, making 15% profit on the first article and 15% loss on second article. Find the net profit or loss percent? | [
"200",
"768",
"150",
"280"
] | C | Profit on first article = 150% of 1000 = 150.
This is equal to the loss he makes on the second article. That, is he makes neither profit nor loss.
Answer: C |
AQUA-RAT | AQUA-RAT-36477 | python, beginner, algorithm, strings, regex
Test 8: 1e is an invalid number.
--------------------------------------------------
Test 9: e3 is an invalid number.
--------------------------------------------------
Test 10: 6e-1 is a valid number.
--------------------------------------------------
Test 11: 99e2.5 is an invalid number.
--------------------------------------------------
Test 12: 53.5e93 is a valid number.
--------------------------------------------------
Test 13: --6 is an invalid number.
--------------------------------------------------
Test 14: -+3 is an invalid number.
--------------------------------------------------
Test 15: 95a54e53 is an invalid number.
The following is multiple choice question (with options) to answer.
Find the invalid no.from the following series 2,3,12,37,86,166,288 | [
"2",
"3",
"166",
"12"
] | C | 3-2=1
12-3=9
37-12=25
86-37=49
166-86=80(this must be 81=167-86=81)
so wrong number in the sequence is 166
difference between two sucessive numbers sequence would be
1,9,25,49,81,121(square of 1,3,5,7,9,11)
C |
AQUA-RAT | AQUA-RAT-36478 | # Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$
Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $$\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$$ where $$r-\frac{s}{2}$$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $$\frac{\pi r^2}. {2}$$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($$2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$$). In conclusion the resultant equation is (with $$r=x$$ and $$s=y$$):
$$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$$
The following is multiple choice question (with options) to answer.
The volumes of two cones are in the ratio 1 : 12 and the radii of the cones are in the ratio of 1 : 3. What is the ratio of their heights? | [
"A)3:4",
"B)2:7",
"C)2:2",
"D)2:1"
] | A | The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r12h1/r22h2 => 1/12 = (1)2h1/(3)2h2
=> h1/h2 = 3/4
i.e. h1 : h2 = 3: 4
Answer: A |
AQUA-RAT | AQUA-RAT-36479 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
In a room there are 8 people. Each person shook hands with every other person. How many hand shakes were there? | [
"26",
"28",
"30",
"32"
] | B | 8C2 = 28
The answer is B. |
AQUA-RAT | AQUA-RAT-36480 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
Stay hungry, Stay foolish
PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1059
Location: India
GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3400
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
The average (arithmetic man) of three integers a, b, and c is exactly four times the median. If a < b < c and a = 0, what is the value of c/b? | [
"2",
"3",
"4",
"11"
] | D | The average of three integers a, b, and c is exactly twice the median --> (a+b+c)/3=4b --> since a=0, then (0+b+c)/3=4b --> c=11b --> c/b=11.
Answer: D. |
AQUA-RAT | AQUA-RAT-36481 | 321
a) f(x)=abs(x), g(x)=-x
b) f(x)=cos(x), g(x)=x+2pi
6. ### csprof2000
287
Good point.
I guess then that my argument only works for functions which don't have an inverse where Domain(inv f) = Range(f). This, naturally, precludes functions such as abs(x) and cos(x)...
So I guess more though will have to be put into functions which are not bijections.
7. ### mnb96
638
Thanks a lot. You all made very good observations that helped me a lot.
BTW, it seems that if the function f admits an inverse there are not many choices, while if the function f is not invertible, many solutions exist but the problem is non-trivial, and it is difficult to say what kind of functions f and g would have to be, in order to satisfy f(x)=f(g(x)).
At the moment I am trying to solve the following (similar) problem:
$$f = (f \circ g) g'$$
where g is invertible, and g' denotes its derivative
If you find it interesting, suggestions are always welcome.
Thanks!
8. ### csprof2000
287
Well, similar suggestions - cases - are possible.
Assume f(x) = c, c constant. Then c = cg', g' = 1, and g = x + k, k constant.
Assume f(x) = x. Then x = gg', gdg = xdx, and (g^2)/2 = (x^2)/2 + k, k constant.
Assume f(x) = x^n. Then x^n = g^n g', g^n dg = x^n dx, [1/(n+1)]g^(n+1) = [1/(n+1)]x^(n+1) + k, k constant
The following is multiple choice question (with options) to answer.
If f(x)= 5x^2- 2x+6 and g(y)= 3y-4, then g(f(x)) | [
"82x^2-9x+38",
"15x^2-6x+14",
"58x^2-4x+58",
"87x^2-5x+96"
] | B | g(f(x))=3(5x^2-2x+6)-4
=15x^2-6x+18-4
=15x^2-6x+14
the answer is B |
AQUA-RAT | AQUA-RAT-36482 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
In a village there are 150 men and 90 women in present time.if in next year population will be P=(a^2 +b^2)^1/2 , and in every year men are reduces 8%.what is population of after 2 year. | [
"140",
"141",
"142",
"143"
] | D | next year total population=[150^2+90^2]^.5=174.92=175
man decreased by 8% so total man =150*.92=138
women will be= 175-138=37
so population after two years= [135^2+37^2]^.5=142.87=143
so population after two year= 143
ANSWER:D |
AQUA-RAT | AQUA-RAT-36483 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Ramesh spends 50% of his monthly salary on food, and saves 90% of the remaining amount. If his monthly salary is Rs.10,000, how much money does he save every month ? | [
"Rs.2500",
"Rs. 4500",
"Rs.8000",
"Rs.3000"
] | B | Explanation:
Ramesh's monthly income = Rs.10,000
He spends 50% on food.
The total money spent on food = 50/100 * 10000 = Rs. 5000
Now, his monthly remaining income = Rs.10000 – Rs.500 = Rs. 5000
Out of Rs. 5000, he saves 90%.
Amount saved = 90/100 * 5000 = Rs. 4500
ANSWER:B |
AQUA-RAT | AQUA-RAT-36484 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A rectangular room 14 m long, 12 m broad is surrounded by a varandah, 3 m wide. Find the area of the varandah? | [
"170",
"180",
"190",
"210"
] | B | Area of varandah = (l+b+2p)2p
= (14 + 12 + 6 ) 6
= 180 m(power)2
Answer is B. |
AQUA-RAT | AQUA-RAT-36485 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
What is the probability that a 3 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains exactly 1 woman? | [
"77/204",
"77/832",
"11/77",
"385/816"
] | D | Total possible selections = 3 out of 18 group members = 18C3
Favorable selections = 1 out of 7 women and 2 out of 11 (= 6 men + 5 children) = 7C1 * 11C2
Thus the required probability = 7C1*11C2 / 18C3 = 385/816. Thus D is the correct answer. |
AQUA-RAT | AQUA-RAT-36486 | # Moderate Number System Solved QuestionAptitude Discussion
Q. A boy writes all the numbers from 100 to 999. The number of zeroes that he uses is '$a$', the number of 5's that he uses is '$b$' and the number of 8's he uses is '$c$'. What is the value of $b+c-a$?
✖ A. 280 ✔ B. 380 ✖ C. 180 ✖ D. 80
Solution:
Option(B) is correct
We can see by symmetry $b=c$ and hence all we need to calculate $b$ and $a$
$b= 280\text{ and }a= 180$
⇒ $2b-a =$ 380
Edit: For an alterntive solution, check comment by Abdulkhader.
Edit 2: For yet another alternative solution, check comment by Sravan Reddy.
## (7) Comment(s)
Sravan Reddy
()
I think it better to memorize few things for quick calculations
From 1 to 99, each number will be used $20$ times (Exception: 0 - only 10 times as we do not count zeros for 00,01,02,03.....09. Else it would also be 20)
If we remember the value 20, then coming to the problem. There are total 9 (00,99) sets in 100 to 999.
number of zero's $= 20*9 = 180$
Number of 5's/8's $= 20*9 + 100 = 280$ (100 is for the 100's digit)
$b+c-a = 280+280-180 = 380$
()
if we think , then the numbers like 100, 200 300 are 9 in total then the zeroes in them are 9*2=18
now if we see numbers like 101 ,102, 103,201 are 9 in each set of 100 numbers. then in 9 sets the total zeroes are 9*9=81
then we come to numbers like 110 120 920 etc . so these numbers are 9 in each set of hundred then we have total zeroes are 9*9=81,
then in total number of zeroes are 81+81+18=180
now to count 5
if we conlude a number in ones place, are 105 205 125etc 10 per set so total are 9*10=90
The following is multiple choice question (with options) to answer.
If you multiply all the numbers on your mobile phone except 0, what is the answer? | [
"256485",
"362880",
"125425",
"0"
] | B | We have to multiply 1 to 9 to find the answer.
Therefore 1*2*3*4*5*6*7*8*9=362880
Answer is B |
AQUA-RAT | AQUA-RAT-36487 | # Geometry
Two cylinders have the same volume. If the radius of cylinder I is 3 times the radius of cylinder II, then the height of cylinder II is how many times the height of cylinder I?
A. 12
B. 9
C. 6
D. 3
1. 👍 0
2. 👎 0
3. 👁 96
1. Cylinder 1 has nine times the base area of cylinder 2.
Volume is (base area) x (height)
For the volumes to be equal, cylinder 2 must have 9 times the height of cylinder 2.
1. 👍 0
2. 👎 0
posted by drwls
2. pi*r^2*h2 = pi*(3r)^2*h1.
pi*r^2*h2 = pi*9r^2*h1
Divide both sides by pi*r^2:
h2 = 9*h1.
1. 👍 0
2. 👎 0
posted by Henry
3. For the volumes to be equal, cylinder 2 must have 9 times the height of cylinder 1
1. 👍 0
2. 👎 0
posted by drwls
4. msh fahma 7aga
1. 👍 0
2. 👎 0
posted by osamya
## Similar Questions
1. ### physics
Consider two solid uniform cylinders that have the same mass and length, but different radii: the radius of cylinder A is much smaller than the radius of cylinder B. Rolling down the same incline, which one of the two cylinders
asked by rosa on November 18, 2013
2 cylinders are proportional the smaller cylinder has a radius of 4 centimeters, which is half as large as the radius of the larger cylinder. the volume of smaller cylinder =250 cubic cemtimeters. what is the volume of the larger
asked by brianna on December 14, 2008
3. ### math
2 cylinders are proportional. The smaller cylinder has a radius of 4 cm. which is half as large as the radius of the larger cylinder.The volume of the smaller cylinder is 250 cubic cm. What is the approximate volume of the larger
asked by brianna on December 14, 2008
4. ### College Chemistry
If the density of cylinder 1 is 3.55 , what is the density of cylinder 2? (The masses and volumes of two cylinders are measured. The mass of cylinder 1 is 1.35 times the mass of cylinder 2. The volume of cylinder 1 is 0.792 times
The following is multiple choice question (with options) to answer.
9 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 12 cm height. What is the diameter of each sphere? | [
"4 cm",
"6 cm",
"8 cm",
"10 cm"
] | C | Volume of cylinder = pi*r^2*h
Volume of a sphere = 4*pi*R^3 / 3
9*4*pi*R^3 / 3 = pi*r^2*h
R^3 = r^2*h / 12 = 64 cm^3
R = 4 cm
D = 8 cm
The answer is C. |
AQUA-RAT | AQUA-RAT-36488 | Similar Questions
1). How many different signals , can be made by 5 flags from 8 flags of different colors?
A). 6270 B). 1680 C). 20160 D). 6720
-- View Answer
2). A child has four pockets and three marbles. In how many ways, the child can put the marbles in the pockets?
A). 12 B). 64 C). 256 D). 60
-- View Answer
3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together?
A). 10! B). 14!/(4!) C). 151200 D). 3628800
-- View Answer
4). There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?
A). 120 B). 100000 C). 6720 D). 30240
-- View Answer
5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible?
A). 72 B). 144 C). 288 D). 234
-- View Answer
6). Find the number of ways, in which 12 different beads can be arranged to form a necklace.
A). 11! / 2 B). 10! / 2 C). 12! / 2 D). Couldn't be determined
-- View Answer
7). 20 persons were invited to a party. In how many ways, they and the host can be seated at a circular table?
A). 18! B). 19! C). 20! D). Couldn't be determined
-- View Answer
8). A committee of 5 members is going to be formed from 3 trainees, 4 professors and 6 research associates. How many ways can they be selected, if in a committee, there are 2 trainees and 3 research associates?
A). 15 B). 45 C). 60 D). 9
-- View Answer
9). In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
The following is multiple choice question (with options) to answer.
In how many ways Chief Minister and Minister be elected froma team of 12 members? | [
"126",
"128",
"132",
"146"
] | C | To do this, if captain is elected first, then we have 12 ways of doing this.
For election of vice-captain, we have only 11 ways left, coz 1 way is already consumed. (Situations like this is called dependent situation. One selection depends upon other selection.)
So, the ans is 12*11 = 132 ways.
C |
AQUA-RAT | AQUA-RAT-36489 | ## Exercise 21
An individual is to select from among n alternatives in an attempt to obtain a particular one. This might be selection from answers on a multiple choice question, when only one is correct. Let A be the event he makes a correct selection, and B be the event he knows which is correct before making the selection. We suppose P(B)=pP(B)=p and P(A|Bc)=1/nP(A|Bc)=1/n. Determine P(B|A)P(B|A); show that P(B|A)P(B)P(B|A)P(B) and P(B|A)P(B|A) increases with n for fixed p.
### Solution
P(A|B)=1P(A|B)=1, P(A|Bc)=1/nP(A|Bc)=1/n, P(B)=pP(B)=p
P ( B | A ) = P ( A | B ) P ( B ) P A | B ) P ( B ) + P ( A | B c ) P ( B c ) = p p + 1 n ( 1 - p ) = n p ( n - 1 ) p + 1 P ( B | A ) = P ( A | B ) P ( B ) P A | B ) P ( B ) + P ( A | B c ) P ( B c ) = p p + 1 n ( 1 - p ) = n p ( n - 1 ) p + 1
(22)
P ( B | A ) P ( B ) = n n p + 1 - p increases from 1 to 1 / p as n P ( B | A ) P ( B ) = n n p + 1 - p increases from 1 to 1 / p as n
(23)
## Exercise 22
The following is multiple choice question (with options) to answer.
If 20% of a = b, then b% of 20 is the same as : | [
"4% of a",
"6% of a",
"8% of a",
"10% of a"
] | A | Explanation:
20% of a = b =>20/100 a=b
b% of 20 =b/100 x 20 =20 a/100 x 1/100 x 20 =4a/100 =4% of a.
ANSWER IS A |
AQUA-RAT | AQUA-RAT-36490 | speed-of-light, measurements, si-units, metrology, length
Title: How is the Length of a Meter Physically Measured? I have two parts to this question.
First, I understand that the meter is defined as the distance light travels in 1/299,792,458 seconds. But how is this distance actually measured? The second is obviously from an atomic clock, but Wikipedia makes it appear that the distance is calculated by the counting of wavelengths. For the count of wavelength to be useful you must calculate the physical length of it, which is dependent on the speed of light, which is defined with meters and to increase the accuracy of the measurement, NIST recommends the use of a specific wavelength of laser (which is in meters). Does this not make the actual measurement of the length of the meter circular? Or does the fact that the speed of light is a constant defined in meter/second over come the appearance of the circular logic? The way I see the circular logic would be if the speed of light ever changed, the length of the meter would also change, which make it impossible for us to know the speed of light changed without referencing it back to an older physical object.
Further, the measurement is done in a vacuum but we can not actually create a perfect vacuum, so I assume there would be a pressure range allowed on the vacuum, and pressure measurements are also based on the definition of a meter (I would think this would further add circular logic to laboratory measurements performed in air and adjusted for refraction).
Second, how is this measurement for the meter actually used to calibrate physical objects? I.e. If I buy a meter stick that has been calibrated against the national standard, how do they actually compare the length of the stick vs the wavelength measurements? The length of a meter bar can be measured using a HeNe laser. The laser used is chosen because we are very good at stabilizing the frequencies it outputs. This means that if we can measure the frequencies before the testing, they will remain steady during the testing. A cesium clock can be used to determine the frequencies of light used with great precision, as the second is defined from said clocks.
Once you have a good measurement of frequency, you have a good measurement of wavelength (assuming a reasonable medium... a vacuum is best, as the speed of light is defined in a vacuum, thus no uncertainty). With a wavelength in hand, you can do interferometry.
The following is multiple choice question (with options) to answer.
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m
95 cm is | [
"16cm",
"25cm",
"15cm",
"35cm"
] | D | Explanation:
So by now, you must be knowing this is a question of HCF, right.
H.C.F. of (700 cm, 385 cm, 1295 cm) = 35 cm.
Answer: Option D |
AQUA-RAT | AQUA-RAT-36491 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The boat moves downside 100km and upside 20km in 2 hours each. Then find the boat speed in still water? | [
"15km/hr",
"25km/hr",
"30km/hr",
"20km/hr"
] | C | Down stream = 100/2 = 50km/hr
Up stream = 20/2 = 10km/hr
speed of boat in still water = 60/2 = 30km/hr
Answer is C |
AQUA-RAT | AQUA-RAT-36492 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
Find the odd man out
253, 136, 352, 326, 631, 244 | [
"326",
"136",
"352",
"631"
] | A | The sum all three digits comes to 10 except 326
ANSWER:A |
AQUA-RAT | AQUA-RAT-36493 | When two dice are thrown simultaneously, the possible outcomes are: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) Number of all possible outcomes = 36 (i) Let E1 be the event in which 5 will not come up on either of them. Then, the favourable outcomes are: (1,1), (1,2), (1,3), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,6), (6,1), (6,2), (6,3), (6,4) and (6,6). Number of favourable outcomes = 25 ∴ P (5 will not come up on either of them) = P (E1) = $\frac{25}{36}$ (ii) Let E2 be the event in which 5 will come up on at least one. Then the favourable outcomes are:
The following is multiple choice question (with options) to answer.
When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3? | [
"238",
"671",
"297",
"200"
] | B | When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6
= 1296 outcomes.
The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5
= 625 outcomes.
Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625
= 671
Answer: B |
AQUA-RAT | AQUA-RAT-36494 | $\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$
There are 12 in which the sum is greater than 9 or there is at least one 6.
Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$
Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$
And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$
You can expect to win an average of $\frac{1}{3}$ of a point per throw.
. . . as Plato already pointed out.
.
5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)
RonL
The following is multiple choice question (with options) to answer.
Eight persons participated in a shooting competition. The top score in the competition is 85 points. Had the top score been 92 points instead of 85 points, the average score would have been 84. Find the number of points actually scored in the competition. | [
"645",
"655",
"665",
"636"
] | C | Let the actual number of points scored be x,
Then [ x + (92 –85 )] /8 = 84, ( x + 7) /8 = 84 , x = ( 84*8 )- 7,
= 672 - 7 = 665
ANSWER:C |
AQUA-RAT | AQUA-RAT-36495 | Probability in getting two numbers exactly the same after rolling three dices?
What is the probability of getting two numbers the same having three dices? I had this on my exam, it sounded super easy but isn't so.
Solution: I know that there are $$6^3$$ combinations for all three dices. Then I just wrote the following:
And the combinations that I want are: $$first-second:(1,1,x),(2,2,x),(3,3,x),(4,4,x),(5,5,x),(6,6,x)$$
$$second-third:(x,1,1),(x,2,2),(x,3,3),(x,4,4),(x,5,5),(x,6,6)$$ $$first-third:(1,x,1),(2,x,2),(3,x,3),(4,x,4),(5,x,5),(6,x,6)$$ wich gives me the combination of $$6+6+6=18$$ and final result $$18/216$$, which is incorrect. What am I doing wrong?
• Exactly two the same, or at least two the same? – tilper Sep 7 '16 at 14:27
• Exactly the same – eugene_sunic Sep 7 '16 at 14:28
• Refer to Lovsovs' answer-hint. What you've done so far doesn't exclude, for example, (1,1,1). – tilper Sep 7 '16 at 14:29
• Still unclear. Do you mean exactly 2 the same? – drhab Sep 7 '16 at 14:29
• Yes exactly to the same... – eugene_sunic Sep 7 '16 at 14:29
Hint: For each of your eighteen cases, what can $x$ be?
The following is multiple choice question (with options) to answer.
Three 6 faced dice are thrown together. The probability that exactly two dice show the same number on them is | [
"5/18",
"5/12",
"5/15",
"5/14"
] | B | Using question number 11 and 12, we get the probability as
1 - (1/36 + 5/9) = 5/12.Answer:B |
AQUA-RAT | AQUA-RAT-36496 | ### 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.
In May, Xiang sold 15 used cars. For these 15 cars, the range of the selling price was $15,000 and the lowest selling price was $4,500. In June, Xiang sold 10 used cars. For these 10 cars, the range of the selling prices was $16,500 and the lowest selling price was $6,500. What was the range of the selling prices of the 25 used cars sold by Xiang in May and June? | [
"1. $15,600",
"2. $15,750",
"3. $16,820",
"4. $18,500"
] | D | Edited the question as there was a typo.
Range = Highest Value - Lowest Value.
Range in May was 15,000 and the lowest price in May was 4,500, thus the highest price in May was 15,000+4,500=19,500.
Range in June was 16,500 and the lowest price in June was 6,500, thus the highest price in June was 16,500+6,500=23,000.
Lowest price of a car in May-June was 4,500 (May) and the highest price of a car in May-June was 23,000, thus the range for May-June is 23,000-4,500=18,500.
Answer: D. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.