source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35397 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
A clock is started at noon. By 10 minutes past 5, the hour hand has turned through | [
"145°",
"150°",
"155°",
"155°"
] | C | Sol.
Angle traced by hour hand in 12hrs = 360°
Angle traced by hour hand in 5hrs 10 min. i.e. 31/6 hrs.
= (360/12 x 31/6)° = 155°
Answer C |
AQUA-RAT | AQUA-RAT-35398 | 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.
19 men and 12 boys finish a job in 3 days, 7 men and 7 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days? | [
"03",
"05",
"12",
"17"
] | A | 19 M + 12 B ----- 03 days
13 M + 13 B ------- 10 days
18 M + 18 B -------?
247 M +156 B = 234 M +234 B
78 B = 13 M => 1 M = 6 B
114 B + 12 B = 126 B ---- 03 days
108 B + 18 B = 126 B -----? => 03 days
Answer: A |
AQUA-RAT | AQUA-RAT-35399 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1:4, then the larger of the two numbers is: | [
"12",
"24",
"84",
"48"
] | C | Let the numbers be x and 4x. Then, x * 4x = 84 * 21 x2 = (84 * 21)/4 = x = 21.
Hence, larger number = 4x = 84.
ANSWER:C |
AQUA-RAT | AQUA-RAT-35400 | Difficult Probability Solved QuestionAptitude Discussion
Q. If the integers $m$ and $n$ are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 equals:
✔ A. $\dfrac{1}{4}$ ✖ B. $\dfrac{1}{7}$ ✖ C. $\dfrac{1}{8}$ ✖ D. $\dfrac{1}{49}$
Solution:
Option(A) is correct
Table below can be scrolled horizontally
Form of the exponent
$m$ $4x +1$ $4x+3$ $4x+2$ $4x$
$n$ $4y+3$ $4y+1$ $4y$ $4y+2$
last digit of
$7^m+7^n$
$0$ $0$ $0$ $0$
Number of
selections
$25 \times 25$ $25 \times 25$ $25 \times 25$ $25 \times 25$
If a number ends in a 0 then the number must be divisible by 5.
Hence required probability is,
$=\dfrac{625 \times 4}{100^2}$
$=\dfrac{1}{4}$
Edit: Thank you, Barry, for the very good explanation in the comments.
Edit 2: Thank you Vaibhav, corrected the typo now it's $25 \times 25$ and not $26 \times 25$.
Edit 3: For yet another approach of solving this question, check comment by Murugan.
(7) Comment(s)
Murugan
()
This sum is very simple. Power cycles of 7 are 7, 9, 3, 1
So totally 4 possibilities.
Total possibilities are $^4P_1 \times ^4P_1=16$
For selecting a number from 1 to 100 which are divisible by 5,
$m=9$, $n=1$ or $m=1$, $n=9$ or $m=7$, $n=3$ or $m=3$, $n=7$
i.e. 4 chances.
The following is multiple choice question (with options) to answer.
If an integer w is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that w(w + 1)(w + 2) will be divisible by 8? | [
"1/4",
"3/8",
"1/2",
"5/8"
] | D | for w Total numbers 8*12
There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10))
and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))
The answer 5/8 -> D |
AQUA-RAT | AQUA-RAT-35401 | Why are things "weird" here? Let's think of it like this: If I travelled at $$8 \frac{\text{mi}}{\text{hr}}$$ for an hour, then travelled at $$12 \frac{\text{mi}}{\text{hr}}$$, I'd definitely agree that the average speed is $$10 \frac{\text{mi}}{\text{hr}}.$$ Things are pretty normal here: If I want to calculate the total distance traveled, it's just $$8 + 12 = 20.$$ Then, if I want to calculate the total time traveled, it's just $$1 + 1 = 2.$$ So the average or overall speed is just
$$\frac{20 \text{mi}}{2 \text{hr}} = 10 \frac{\text{mi}}{\text{hr}}.$$ It works!
The following is multiple choice question (with options) to answer.
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is: | [
"50",
"56",
"70",
"60"
] | A | Let the actual distance travelled be x km.
Then,x/10=(x + 20)/14
14x = 10x + 200
4x = 200
x = 50 km.
Answer:A |
AQUA-RAT | AQUA-RAT-35402 | ### Permutations when all the objects are not distinct objects
Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O_1 and O_2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO_1O_2T. Corresponding to this permutation,we have 2 ! permutations RO_1O_2T and RO_2O_1T which will be exactly the same permutation if O_1 and O_2 are not treated as different, i.e., if O_1 and O_2 are the same O at both places.
Therefore, the required number of permutations = (4!)/(2!) = 3 xx 4 = 12
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.
Temporarily, let us treat these letters different and name them as I_1, I_2, T_1 , T_2, T_3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, I_1 NT_1 SI_2 T_2 U E T_3. Here if I_1, I_2 are not same and T_1, T_2, T_3 are not same, then I_1, I_2 can be arranged in 2! ways and T_1, T_2, T_3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I_1NT_1SI_2T_2UET_3. Hence, total number of different permutations will be (9!)/(2 ! 3!)
We can state (without proof) the following theorems:
The following is multiple choice question (with options) to answer.
n how many ways can the letters of the word PERMUTATIONS be arranged if the there are always 4 letters between P and S? | [
"25401600",
"25401800",
"254016200",
"26401600"
] | A | There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice.
Now first we need to see how many ways we can make word with 4 letter between P and S.
Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210
Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S.
So total way = 210*2 = 420
The selected 4 letters can be rotated between P and S in = 4! ways
So total ways = 420 * 4!
Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter.
Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways.
So total number of ways = 7! * 420 * 4!
Now since letter T was repeated twice, we should divide the above result by 2!.
So Total number of ways = 7! * 420 * 4! / 2! = 25401600 A |
AQUA-RAT | AQUA-RAT-35403 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sold an article offering a discount of 5% and earned a profit of 22.55%. What would have been the percentage of profit earned if no discount was offered? | [
"60%",
"29%",
"39%",
"56%"
] | B | Let C.P. be Rs. 100.
Then, S.P. = Rs. 122.55
Let marked price be Rs. x. Then, 95/100 x = 122.55
x = 12255/95 = Rs. 129
Now, S.P. = Rs. 129, C.P. = Rs. 100
Profit % = 29%.
Answer: B |
AQUA-RAT | AQUA-RAT-35404 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
On a certain date, Pat invested $10,000 at x percent annual interest, compounded annually. If the total value of the investment plus interest at the end of 12 years will be $40,000, in how many years, the total value of the investment plus interest will increase to $80,000? | [
"15",
"16",
"18",
"20"
] | C | x- interest rate
80.000= 10.000 (1+x)^year
=> 8=(1+x)^year
40.000=10.000 . (1+x)^12
=> 4= (1+x)^12
=>2= (1+x)^6
=> 8 = (1+x)^ 18
So, after 18 years, the total value of the investment plus interest will increase to $80,000.
ANSWER:C |
AQUA-RAT | AQUA-RAT-35405 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
An Olympic diver received the following scores: 2, 4, 6, 8. The standard deviation of her scores is in which of the following ranges? | [
"4.01 to 4.99",
"1.01 to 1.99",
"3.01 to 3.99",
"2.01 to 2.99"
] | D | 2+4+6+8/4=5
∣2-5∣^2 =3^2 = 9
∣4-5∣^2 =1^2 = 1
∣6-5∣^2 =1^2 = 1
∣8-5∣^2 =3^2 = 9
sqrt 20/4 =2.24
Answer : D |
AQUA-RAT | AQUA-RAT-35406 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
According to a survey conducted in two cities, A and B, 64% of the population is happy with the new president. If the ratio of the city A population to the city B population is 3:2 and the percentage of city A residents who are happy with the president is 70%, what percentage of city B population is happy with the president? | [
"30%",
"50%",
"25%",
"55%"
] | D | A/B = 3/2 = x suppose
that for A=3x
B=2x
Total A+B=5x
Now 64% is happy with new president, 64%*(5x) =3.2x is happy with new present out of total population of A & B
now 70% of A is happy with new pres, so 70%*(3x) = 2.1x
Total happy people are 3.2x
so B=3.2x-2.1x=1.1x is happy
Now B's population is 2x
1.1x is 55% of total population of B
Answer : D |
AQUA-RAT | AQUA-RAT-35407 | 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 swims downstream 96 km and upstream 40 km taking 8 hours each time; what is the speed of the current? | [
"3.5 kmph",
"1.5 kmph",
"13 kmph",
"6.5 kmph"
] | A | Explanation:
96 --- 8 DS = 12
? ---- 1
40 ---- 8 US = 5
? ---- 1 S = ?
S = (12 - 5)/2 = 3.5
Answer: Option A |
AQUA-RAT | AQUA-RAT-35408 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
What is the area of a square field whose sides have a length of 11 meters? | [
"225 sq m",
"205 sq m",
"121 sq m",
"167 sq m"
] | C | 11 * 11 = 121 sq m
The answer is C. |
AQUA-RAT | AQUA-RAT-35409 | ### 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 man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 395 km of the tour is: | [
"31.11 km/hr",
"87.8 km/hr",
"71.11 km/hr",
"36 km/hr"
] | B | Total time taken = 160/64 + 160/80 = 9/2 hours
--> Average speed = 395 x 2/9 = 87.8 km/hr.
Answer : B. |
AQUA-RAT | AQUA-RAT-35410 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
A number when divided by a divisor leaves a remainder of 23. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor? | [
"12",
"13",
"35",
"36"
] | D | Let the number is N, the divisor = D,
I will make the two equations-
N = xD+23
2N = yD+11
where x and y are integers
Solving them: D(y-2x) = 36
as D is also integer and 36 is a prime number, the D should be 36 to satisfy the above equation.
Hence answer is 'D' |
AQUA-RAT | AQUA-RAT-35411 | 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.
Stacy has a 33 page history paper due in 3 days. How many pages per day would she have to write to finish on time? | [
"9",
"8",
"11",
"8.5"
] | C | 33/3=11
Answer : C |
AQUA-RAT | AQUA-RAT-35412 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
The population of a town has increased from 133575 to 138918. The percent increase in population is : | [
"2.5",
"3",
"4.5",
"4"
] | D | (133575-138918)/138918*100=4
ANSWER:D |
AQUA-RAT | AQUA-RAT-35413 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many three digit even numbers do not use any digit more than once | [
"None of these",
"2160",
"2240",
"2460"
] | A | The Way i solved it is
A B C ( hundreds, tens, units)
C can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 7 * 8 * 5 = 280
Ans :A |
AQUA-RAT | AQUA-RAT-35414 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A car traveled from San Diego to San Francisco at an average speed of 48 miles per hour. If the journey back took twice as long, what was the average speed of the trip? | [
"24.",
"32.",
"36",
"42."
] | B | Let the time taken be = x
One way distance = 48x
Total distance traveled = 2 * 48x = 96x
Total time taken = x + 2x = 3x
Average speed = 96x/3x = 32
Answer: B |
AQUA-RAT | AQUA-RAT-35415 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row with a speed of 15 kmph in still water. If the stream flows at 5 kmph, then the speed in downstream is? | [
"22",
"28",
"20",
"82"
] | C | M = 15
S = 5
DS = 15 + 5 = 20.Answer:C |
AQUA-RAT | AQUA-RAT-35416 | linux, assembly, caesar-cipher, x86
Allow me
mov $'A'+26-1,%ah
This struck me as being a bit too much of a complication. Why not simply write
mov $'Z', %ah
Now if it was your intent to come close to the comment that follows in the next line
# ah: threshold ('A' - 1 - (key - 26))
then writing
lea -26(%ecx), %eax # al: key - 26
mov $'A'-1, %ah
sub %al, %ah # ah: threshold ('A' - 1 - (key - 26))
would have nailed it.
The following is multiple choice question (with options) to answer.
emblem is coded as 216;
crude is coded as 125
bat will be ? | [
"45",
"32",
"27",
"56"
] | C | coding follows (no. of letters)^3
emblem = 216 = 6^3
crude = 125 = 5^3
bat = 3^3 =27
ANSWER:C |
AQUA-RAT | AQUA-RAT-35417 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
What number should replace the question mark? 3, 30, 11, 24, 19, 18, ---? | [
"24",
"22",
"27",
"25"
] | C | Answer: C
3, 30, 11, 24, 19, 18, 27?
There are two alternate sequences: +8 and -6. |
AQUA-RAT | AQUA-RAT-35418 | Your assistance. This is the actual change, where the original value is subtracted from the new value. Calculate The Discount Percentage Between Two Numbers. So, while this result might be what contact centres wants to see, it does not represent the facts. For example, the percent difference between 30% and 50% is 20%. Different relations between two numbers. com's Numbers to Ratio Calculator is an online basic math function tool to find the quantitative relationship or ratio between two or three given numbers or to reduce the ratio to its lowest terms. # Function to calculate the maximum difference between two elements in a. The Percentage Difference Calculator has 3 ways to calculate the differences between two numbers. 4 day ago Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Use the calculator below to analyze the results of a difference in sample means hypothesis test. Example, total=1,100 and you need to find percent that equals to 100. A relative delta compares the difference between two numbers, A and B, as a percentage of one of the numbers. With delayed retirement credits , a person can receive his or her largest benefit by retiring at age 70. Just Now Percentage Difference Formula. Sales have been poor and the owner decided to mark down each item to$15 to speed up the sales. percentagedifferencecalculator. The percentage diference between 10 and 12 is 18. Determine the absolute difference between two numbers: 15 - 25 = -10; Take the average of those two figures: (15 + 25)/ 2 = 20; Calculate the difference by dividing it by the average: 10/20 = 0. Hi Mahmoud, yes, I was tried that method, it seems to be not working. While most students find percentages to be an easier topic than one such as combinatorics, some individuals initially trip on the difference between a percent change and a percent of a number. com/7344/c-program-to-calculate-percentage-difference-between-2-numbers/Given two numbers, write a C program to calculate percentage differe. Tips: Put numbers in as you like, and the result will automatically be generated. Step 4: Convert that to a percentage (by multiplying by 100 and adding a "%" sign). How To Calculate The Percentage Difference Between Two. Then take this number times 100%, resulting in 40%. The Percent Change Calculator finds
The following is multiple choice question (with options) to answer.
The price of an item is discounted 3 percent on day 1 of a sale. On day 2, the item is discounted another 3 percent, and on day 3, it is discounted an additional 10 percent. The price of the item on day 3 is what percentage of the sale price on day 1? | [
"85.1%",
"86.9%",
"87.3%",
"88.8%"
] | C | Let initial price be 100
Price in day 1 after 3% discount = 97
Price in day 2 after 3% discount = 94.09
Price in day 3 after 10% discount = 84.68
So, Price in day 3 as percentage of the sale price on day 1 will be = 84.68/97*100 => 87.3%
Answer will definitely be (C) |
AQUA-RAT | AQUA-RAT-35419 | # If both integers $x$ and $y$ can be represented as $a^2 + b^2 + 4ab$, prove that $xy$ can also be represented like this …
There is a set $Q$ which contains all integral values that can be represented by $$a^2 + b^2 + 4ab$$, where $a$ and $b$ are also integers. If some integers $x$ and $y$ exist in this set, prove that $xy$ does too.
I really have no idea how I can go about solving this. I tried simple multiplication of the two assuming one to be $(a^2 + 4ab + b^2)$ and other as $(c^2 + 4cd + d^2)$ but ultimately it leads to a long equation I can make no tail of :/
Any help whatsoever would be greatly appreciated
-
I have updated your post to LaTeX. Please see that the updates are correct. – Jeel Shah Jan 20 '14 at 12:48
@hardmath Fixed! Thanks for the catch! – Jeel Shah Jan 20 '14 at 13:01
## 2 Answers
Since $a^2+b^2+4ab=(a+2b)^2-3b^2$, your numbers are exactly the numbers of the form $x^2-3y^2$. Now $x^2-3y^2$ is the norm of the algrebraic number $x+y\sqrt{3}$, so you have the identity
$$(x^2-3y^2)(u^2-3v^2)=(xu+3yv)^2-3(xv+yu)^2$$
(multiplicativity of norms).
-
Thank you so much, now I can finally sleep with this homework done. – skatter Jan 20 '14 at 12:54
To make the resulting identity explicit in terms of $a, b, c, d$, if $f(x,y) = x^2 + 4xy + y^2$, then $$f(ac-bd,ad+4bd+bc) = f(a,b) f(c,d).$$ – heropup Jan 20 '14 at 13:13
Generalization:
The following is multiple choice question (with options) to answer.
If a and b are integers and (a*b)^5 = 48y, y could be: | [
"8",
"16",
"162",
"144"
] | C | Distribute the exponent.
a^5 * b^5 = 48 y
Find the prime factorization of 48. This is 2^4 *3^1.
We need 2^1*3^4 (or some other power of 3 that will give us a multiple of 3^5 as our second term).
3^4*2 = 81*2= 162
The answer is C. |
AQUA-RAT | AQUA-RAT-35420 | well how do i find h and k
4. Hello, polakio92!
b) Square $ABCD$ is inscribed in square $EFGH$ forming four congruent right triangles.
If the perimeter of the square $EFGH$ is 64 and the area of square $ABCD$ is 130,
determine the values of $x\text{ and }y.$
Code:
E x A y F
* - - - * - - - - - - - *
| * * |
| * * | x
| * * |
y | * * B
| * *|
| * * |
|* * |
D * * | y
| * * |
x | * * |
| * * |
* - - - - - - - * - - - *
H 16-x C x G
: - - - - 16 - - - - - :
The sides of EFGH are divided into the segments $x\text{ and }y.$
The perimeter of $EFGH$ is 64.
. . Hence: . $x + y \:=\:16\quad\Rightarrow\quad y \:=\:16-x\;\;{\color{blue}[1]}$
The area of $ABCD$ is 130.
. . Hence: . $AB = BC = CD = AD = \sqrt{130}$
In right triangle $AFB\!:\;\;x^2 + y^2 \:=\:AB^2\quad\Rightarrow\quad x^2 + y^2 \:=\:130\;\;{\color{blue}[2]}$
Substitute [1] into [2]: . $x^2 + (16-x)^2 \:=\:130 \quad\Rightarrow\quad x^2 - 16x + 63 \:=\:0$
. . which factors: . $(x-7)(x-9) \:=\:0$
. . and has roots: . $x \;=\;7,\,9$
Therefore: . $\{x,\,y\} \;=\;\{7,\,9\}$
5. thank you so much! hopefully i can solve the rest of the word problems i have..
The following is multiple choice question (with options) to answer.
The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. | [
"16",
"12",
"20",
"24"
] | D | Side of first square = (40/4) = 10 cm;
Side of second square = (32/4)cm = 8 cm.
Area of third square = [(10) 2 - (8) 2] cm2 = (100 - 64) cm2 = 36 cm2.
Side of third square = (36)(1/2) cm = 6 cm.
Required perimeter = (6 x 4) cm = 24 cm.
Ans: D |
AQUA-RAT | AQUA-RAT-35421 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 seconds and a man standing on the platform in 21 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"877 m",
"225 m",
"167 m",
"887 m"
] | B | Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train
= (15 x 21)m = 315 m. Let the length of the platform be x meters. Then, (x + 315)/36 = 15
==> x + 315 = 540 ==> x
= 225 m.
Answer: B |
AQUA-RAT | AQUA-RAT-35422 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
7 3/4 + 7 2/17 - 9 1/15=? | [
"7 719/1020",
"9 817/1020",
"9 719/1020",
"5 817/1020"
] | D | Given sum=7+3/4+7+2/17-(9+1/15)
=(7+7-9)+(3/4+2/17-1/15)
=5+(765+120-68/1020
=5 817/1020
Answer is D |
AQUA-RAT | AQUA-RAT-35423 | 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.
If daily wages of a man is five times to that of a woman, how many men should work for 20 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600. | [
"12",
"14",
"16",
"8"
] | D | Explanation :
Wages of 1 woman for 1 day = 21600/40×30
Wages of 1 man for 1 day = 21600×5/40×30
Wages of 1 man for 20 days = 21600×5×20/40×30
Number of men = 14400/(21600×5×20/40×30)=144/(216×100/40×30)=8
Answer : Option D |
AQUA-RAT | AQUA-RAT-35424 | c++, beginner, converting
convert to meters
push into vector
set low/high variables
Here's what the loop might look without the duplication:
double low = DBL_MAX;
double high = DBL_MIN;
...
while (cin >> num >> unit) {
if (unit == "m") {
meters = num;
} else if (unit == "cm") {
meters = num * cm_to_m;
} else if (unit == "in") {
meters = num * in_to_m;
} else if (unit == "ft") {
meters = num * ft_to_m;
} else {
cout << "Incorrect Unit Entered! Please try again..." << endl;
continue;
}
input.push_back(meters);
if (meters < low) {
low = meters;
low_unit = unit;
} else if (meters > high) {
high = meters;
high_unit = unit;
}
}
This is a lot simpler than your original code, having removed all of the
duplication. Your tests for count == 0 are unnecessary if you set the
low/high variables to the limits of the range for double (all double values
are going to be less than or equal to DBL_MAX etc). This leaves a continue
which many people would consider bad form and some coding standards outlaw.
To avoid this we could put the conversion into a function:
static bool convert_to_meters(double num, double &meters, const string& unit)
{
const double cm_to_m = 0.01;
const double in_to_m = 2.54 * 0.01;
const double ft_to_m = 12 * 2.54 * 0.01;
if (unit == "m") {
meters = num;
} else if (unit == "cm") {
meters = num * cm_to_m;
} else if (unit == "in") {
meters = num * in_to_m;
} else if (unit == "ft") {
meters = num * ft_to_m;
} else {
return false;
}
return true;
}
The following is multiple choice question (with options) to answer.
Convert 300 miles into meters? | [
"784596",
"845796",
"804670",
"482802"
] | D | 1 mile = 1609.34 meters
300mile = 300*1609.34 = 482802 meters
Answer is D |
AQUA-RAT | AQUA-RAT-35425 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of 11 results is 55, if the average of first five results is 49 and that of the last seven is 52. Find the fifth result? | [
"4",
"5",
"7",
"8"
] | A | 1 to 11 = 11 * 55 = 605
1 to 5 = 5 * 49 = 245
5 to 11 = 7 * 52 = 364
5th = 245 + 364 – 605 = 4
ANSWER:A |
AQUA-RAT | AQUA-RAT-35426 | evolution, population-dynamics
Title: How many humans have been in my lineage? Is it almost the same for every human currently living? If I were to count my father, my grandfather, my great-grandfather, and so on up till, say chimps, or the most common ancestor, or whatever that suits the more accurate answer, how many humans would there have been in my direct lineage?
And would it be almost the same for every human being currently living? A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is likely to be higher. Research has shown that the great apes (chimps, gorilla, orangutan) have generation times comparatble to humans, in the range of 18-29 years (Langergraber et al. 2012).
Using 7 million years and 20 years yields an estimated 350000 ancestral generations for each living human. A more conservative estimate, using an average generation time of 28, would result in 250000 generations. However, some have argued that the human-chimp split is closer to 13 million years old, which would mean that approximately 650000 generations have passed (using a generation time of 20 years).
The exact number of ancestral generations for each human will naturally differ a bit, and some populations might have higher or lower numbers on average due to chance events or historical reasons (colonizations patterns etc). However, due to the law of large numbers my guess would be that discrepancies are likely to have averaged out. In any case, the current estimates of the human-chimp split and average historical generation times are so uncertain, so that they will swamp any other effects when trying to calculate the number of ancestoral generations.
However, this is only answering the number of ancestral generations. The number of ancestors in your full pedigree is something completely different. Since every ancestor has 2 parents, the number of ancestors will grow exponentially. Theoretically, the full pedigree of ancestors can be calculated using:
The following is multiple choice question (with options) to answer.
10 years ago, the average age of a family of 4 members was 24 years. Two children having been born (with age diference of 2 years), the present average age of the family is the same. The present age of the youngest child is ? | [
"7 years",
"5 years",
"3 years",
"4 years"
] | C | Explanation:
Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.
Total age of 4 members now = [96 + (10 x 4)] years = 136 years.
Total age of 6 members now = (24 x 6) years = 144 years.
Sum of the ages of 2 children = (144 - 136) years = 8 years.
Let the age of the younger child be years.
Then, age of the elder child = years.
So,
Age of younger child = 3 years.
Answer: C |
AQUA-RAT | AQUA-RAT-35427 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket? | [
"8 min 15 sec",
"7 min 15 sec",
"6 min 15 sec",
"5 min 15 sec"
] | A | Explanation:
Part filled in 3 minutes =
3∗(1/12+1/15)=3∗9/60=9/20
Remaining part =1−9/20=11/20
=>1/15:11/20=1:X
=>X=11/20∗15/1
=>X=8.25mins
So it will take further 8 mins 15 seconds to fill the bucket.
Option A |
AQUA-RAT | AQUA-RAT-35428 | # 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.
What is the speed of the stream if a canoe rows upstream at 6km/hr and downstream at 12km/hr | [
"1 kmph",
"4 kmph",
"3 kmph",
"2 kmph"
] | C | Sol.
Speed of stream = 1/2(12 -6) kmph = 3 kmph.
Answer C |
AQUA-RAT | AQUA-RAT-35429 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
The average expenditure of a man for the first five months is Rs.360 and for the next seven months it is Rs.130. If he saves Rs.290 in that year, his monthly average income is : | [
"Rs.140",
"Rs.150",
"Rs.250",
"Rs.350"
] | C | Explanation:
Total income = Rs.(360×5+130×7+290)=Rs.3000
Average monthly income
= Rs. (3000/12)=Rs.250
Correct Option: C |
AQUA-RAT | AQUA-RAT-35430 | Sum the two parts gives $\pi/3$.
• Can you explain to why the bound of $\rho$ is between $0 \: \text{and} \:1$ in this case..? – misheekoh Apr 5 '17 at 4:41
• @misheekoh because you are integrating over the (half) unit sphere's volume, to cover this volume the radius has to change from $0$ to $1$. – Taozi Apr 5 '17 at 4:47
• @misheekoh I added to the answer a direct evaluation of the surface flux integration. – Taozi Apr 5 '17 at 5:03
• Thank you. Btw, I do understand that we're integrating over the hemisphere but particularly for when $\text{z} \le \sqrt{1-x^2-y^2}$, how did u manage to convert the upper bound to $\rho \le 1$? – misheekoh Apr 5 '17 at 5:06
• @misheekoh Square both sides of this inequality to have $z^2 \le 1 - x^2 -y^2$, rewrite as $x^2 +y^2+z^2 \le 1$, since $x^2 +y^2+z^ = \rho^2$, this is just $\rho^2 \le 1$, or $\rho\le 1$. – Taozi Apr 5 '17 at 5:09
The following is multiple choice question (with options) to answer.
Find the ratio between whole surfaces of a sphere and a hemisphere? | [
"4:9",
"4:3",
"4:5",
"4:0"
] | B | Explanation:
4 πr2 : 3 πr2 => 4:3
Answer: Option B |
AQUA-RAT | AQUA-RAT-35431 | 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 bullet train 140 m long is running at 60 kmph. In how much time will it pass a platform 360 m long? | [
"30 Seconds",
"22 Seconds",
"41 Seconds",
"24 Seconds"
] | A | 30 Seconds
Distance travelled = 140 + 360m = 500m
Speed = 60 * 5/8 = 50/3m
Time = 500 * 3/50 = 30 Seconds
ANSWER A |
AQUA-RAT | AQUA-RAT-35432 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A boat running up stram takes 6 hours to cover a certain distance, while it takes 7 hours to cover the same distance running down stream. what is the ratio between the speed of the boat and the speed of water current respectively? | [
"2 : 3",
"5 : 6",
"4 : 5",
"13 : 1"
] | D | Explanation:
Let speed of boat is x km/h and speed stream is y km/hr
6(x+y) = 7(x-y)
6x+6y = 7x-7y
13y = x
13y = x
x/y = 13/1
13 : 1
Answer: Option D |
AQUA-RAT | AQUA-RAT-35433 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
How long will it take a sum of money invested at 8% p.a. S.I. to increase its value by 80%? | [
"12 years.",
"11 years.",
"8 years.",
"10 years."
] | D | Sol.
Let the sum be x. Then, S.I. = 80% of x = 4x/5; Rate = 8%.
∴ Time = [100 * 2x/5 * 1/x*8 = 10 years.
Answer D |
AQUA-RAT | AQUA-RAT-35434 | Once again
Repetition helps too, so let’s recite: it starts with $289 = 17^2$, then continues by 102s: 391, 493. After that the twins 527, 529, followed by 629; then 667 and 697. Then two sets of twins each with its 99: 713, 731, 799; 841, 851, 899; then 901 to come after 899, and then the three sporadic values: 943, 961, 989!
Posted in arithmetic, computation, primes |
Quickly recognizing primes less than 1000: divisibility tests
I took a little hiatus from writing here since I attended the International Conference on Functional Programming, and since then have been catching up on teaching stuff and writing a bit on my other blog. I gave a talk at the conference which will probably be of interest to readers of this blog—I hope to write about it soon!
In any case, today I want to return to the problem of quickly recognizing small primes. In my previous post we considered “small” to mean “less than 100”. Today we’ll kick it up a notch and consider recognizing primes less than 1000. I want to start by considering some simple approaches and see how far we can push them. In future posts we’ll consider some fancier things.
First, some divisibility tests! We already know how to test for divisibility by $2$, $3$, and $5$. Let’s see rules for $7$, $11$, and $13$.
• To test for divisibility by $7$, take the last digit, chop it off, and subtract double that digit from the rest of the number. Keep doing this until you get something which obviously either is or isn’t divisible by $7$. For example, if we take $2952$, we first chop off the final 2; double it is 4, and subtracting 4 from $295$ leaves $291$. Subtracting twice $1$ from $29$ yields $27$, which is not divisible by $7$; hence neither is $2952$.
The following is multiple choice question (with options) to answer.
What least number must be subtracted from 9671 so that the remaining number is divisible by 2? | [
"3",
"1",
"16",
"11"
] | B | On dividing 9671 by 2, we get remainder = 1.
Required number be subtracted = 1
ANSWER:B |
AQUA-RAT | AQUA-RAT-35435 | # If both integers $x$ and $y$ can be represented as $a^2 + b^2 + 4ab$, prove that $xy$ can also be represented like this …
There is a set $Q$ which contains all integral values that can be represented by $$a^2 + b^2 + 4ab$$, where $a$ and $b$ are also integers. If some integers $x$ and $y$ exist in this set, prove that $xy$ does too.
I really have no idea how I can go about solving this. I tried simple multiplication of the two assuming one to be $(a^2 + 4ab + b^2)$ and other as $(c^2 + 4cd + d^2)$ but ultimately it leads to a long equation I can make no tail of :/
Any help whatsoever would be greatly appreciated
-
I have updated your post to LaTeX. Please see that the updates are correct. – Jeel Shah Jan 20 '14 at 12:48
@hardmath Fixed! Thanks for the catch! – Jeel Shah Jan 20 '14 at 13:01
## 2 Answers
Since $a^2+b^2+4ab=(a+2b)^2-3b^2$, your numbers are exactly the numbers of the form $x^2-3y^2$. Now $x^2-3y^2$ is the norm of the algrebraic number $x+y\sqrt{3}$, so you have the identity
$$(x^2-3y^2)(u^2-3v^2)=(xu+3yv)^2-3(xv+yu)^2$$
(multiplicativity of norms).
-
Thank you so much, now I can finally sleep with this homework done. – skatter Jan 20 '14 at 12:54
To make the resulting identity explicit in terms of $a, b, c, d$, if $f(x,y) = x^2 + 4xy + y^2$, then $$f(ac-bd,ad+4bd+bc) = f(a,b) f(c,d).$$ – heropup Jan 20 '14 at 13:13
Generalization:
The following is multiple choice question (with options) to answer.
If x and y are integers such that x^2-4x+2y+6=2x^2-5x+y-6, we can deduce that y is | [
"not an even",
"an even",
"a perfect square",
"an odd"
] | B | x^2 - 4x + 2y + 6 = 2x^2 - 5x + y - 6
y = x^2 - x - 12
y = (x-4)(x+3) (Note that one factor is even and one factor is odd.)
Then y must be even.
The answer is B. |
AQUA-RAT | AQUA-RAT-35436 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If (a-b) is 6 more than (c+d) and (a+b) is 3 less than (c-d), then (a-c) | [
"0.25",
"1.5",
"2",
"2.5"
] | B | (a-b)-(c-d)=6 and (c-d)-(a+b)=3
(a-c)-(b+d)=6 and (c-a)-(b+d)=3
(b+d)= (a-c)-6 and (b+d)=3-(c-a)
2(a-c)=3
(a-c)=1.5
ANSWER B 1.5 |
AQUA-RAT | AQUA-RAT-35437 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If x is less than y by 50% then y exceed x by: | [
"33.33%",
"100%",
"75%",
"66.66%"
] | B | Using formula (x/(100-x)*100) where x is Percentage decrease (here it is 25%)
=> 50(100-50)*100
=100%
ANSWER:B |
AQUA-RAT | AQUA-RAT-35438 | If instead $b^{2} \lt 2$, then $b_{0} = 2/b$ is a positive rational number whose square is larger than $2$. The argument of the preceding paragraph constructs a positive rational number $b_{0}'$ smaller than $b_{0}$ whose square is larger than $2$. Setting $b' = 2/b_{0}'$, we have a positive rational number $b'$ satisfying $b \lt b'$ and $(b')^{2} \lt 2$. In words, a positive rational number $b$ satisfying $b^{2} \lt 2$ is not an upper bound of $S$.
Combining these observations, if $b$ is a positive rational number and $b^{2} \lt 2$, then $b$ is not an upper bound of $S$, while if $2 \lt b^{2}$, then some smaller positive rational is also an upper bound of $S$; that is, $b$ is not the smallest upper bound of $S$. It follows that $S = \{x : x^{2} \lt 2\}$ has no rational supremum.
The following is multiple choice question (with options) to answer.
Given that a/b < 1, and both a and b are positive integers, which one of the following must be greater than 1? | [
"a/b^2",
"a^2/b",
"a^2/b^2",
"b/a"
] | D | Since a/b is a fraction b must always be > 1
Given -
Which one of the following must be greater than 1
We can get the result one only when the denominator in a/b ( Which is less than 1 ) becomes numerator..
Among the given options only (D) has the required characteristic we are looking for...
Hence answer will be (D) |
AQUA-RAT | AQUA-RAT-35439 | F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
F6 = { 0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 1/1 }
F7 = { 0/1, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 2/5, 3/7, 1/2, 4/7, 3/5, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 1/1 }
F8 = { 0/1, 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8, 1/1 }
Centered
F1 = { 0/1, 1/1 }
F2 = { 0/1, 1/2, 1/1 }
F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
The following is multiple choice question (with options) to answer.
What is the HCF of 2/3, 4/9 and 6/5 | [
"7/45",
"2/45",
"4/15",
"8/45"
] | B | Explanation:
HCF of Fractions = HCF of Numerators/LCM of Denominators
= (HCF of 2, 4, 6)/(LCM of 3, 9, 5) = 2/45
Answer: Option B |
AQUA-RAT | AQUA-RAT-35440 | # Evaluate the sum of the reciprocals
#### anemone
##### MHB POTW Director
Staff member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
#### mente oscura
##### Well-known member
Given
$p+q+r+s=0$
$pqrs=1$
$p^3+q^3+r^3+s^3=1983$
Evaluate $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}$.
Hello.
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}+\dfrac{1}{s}=$$
$$=qrs+prs+pqs+pqr=$$
$$=qrs+prs+rrs+srs-rrs-srs+pqs+pqr=$$
$$=-rrs-srs+pqs+pqr$$, (*)
$$(p+q)^3=-(r+s)^3$$
$$p^3+3p^2q+3pq^2+q^3=-r^3-3r^2s-3rs^2-s^3$$
$$1983+3p^2q+3pq^2=-3r^2s-3rs^2$$
$$661+p^2q+pq^2=-r^2s-rs^2$$, (**)
For (*) and (**):
The following is multiple choice question (with options) to answer.
P, Q and R have Rs.9000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R? | [
"Rs.3000",
"Rs.3600",
"Rs.2400",
"Rs.4000"
] | B | Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(9000 - r) => 3r = 18000 - 2r
=> 5r = 18000 => r = 3600.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35441 | # Random Gift Giving at a Party - Combinatorics Problem
Each of $$10$$ employees brings one (distinct) present to an office party. Each present is given to a randomly selected employee by Santa (an employee can get more than one present). What is the probability that at least two employees receive no presents?
Firstly, there are $$10^{10}$$ total ways to give the $$10$$ employees the $$10$$ presents. So this is our denominator.
My attempt was to consider the complement and consider the number of ways that either $$0$$ employees receive no presents (every employee gets a present) or $$1$$ employee receives no present.
Case 1: $$0$$ employees
There are $$10$$ employees and $$10$$ presents. So there are $$10^{10}$$ ways to give the presents.
Case 2: $$1$$ employee
Step 1: Decide which employee receives no presents: $$10$$ possibilities.
Step 2: Distribute the $$10$$ presents to the remaining $$9$$ employees: $$9^{10}$$ ways.
So the number of ways in which at least $$2$$ employees receive no presents is: $$1-(10^{10}+9^{10}$$).
So my final answer is: $$1-\displaystyle\frac{(10^{10}+9^{10})}{10^{10}}$$.
However, this answer does not match the answer in my textbook. Which is: $$1-\displaystyle\frac{10!-10\times 9 \times \frac{10!}{2!}}{10^{10}}$$
Where did my attempt go wrong and how can I correct it?
The following is multiple choice question (with options) to answer.
Three small cruise ships, each carrying 7 passengers, will dock tomorrow. One ship will dock at Port A, another at Port B, and the third at Port C. At Port A, two passengers will be selected at random and each winner will receive one gift certificate worth $50. At Port B, one passenger will be selected at random to receive a gift certificate worth $35, and at Port C, one passenger will be selected at random to receive a gift certificate worth $25. How many different ways can the gift certificates be given out? | [
"4328",
"5382",
"6174",
"7690"
] | C | The number of ways for the ships to go to the ports is 3! = 6.
At Port A, the number of ways to choose two winners is 7C2 = 21.
At Port B, the number of ways to choose one winner is 7.
At Port C, the number of ways to choose one winner is 7.
The total number of ways to give out gift certificates is 6*21*7*7 = 6174
The answer is C. |
AQUA-RAT | AQUA-RAT-35442 | A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
-
This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
The following is multiple choice question (with options) to answer.
Calculate the average of all the numbers between 11 and 27 which are divisible by 2. | [
"11",
"10",
"18",
"19"
] | D | Explanation:
numbers divisible by 2 are 12,14,16,18,20,22,24,26,
Average = (12+14+16+18+20+22+24+26, ) / 8 = 152/8 = 19
ANSWER: D |
AQUA-RAT | AQUA-RAT-35443 | $\frac {0.367}{0.978} = 0.375.$
13. ## Re: Probability with a "known"
Another way to do this is with combinations (since you don't care about the order that you pick the fruit). You will get the same answer either way. With combinations, you can just divide the number of outcomes. So, if we figure out the number of outcomes with (at least two red AND at least one red or green) divided by the number of outcomes with at least one apple (red or green), that will give us the same probability.
Just as ebaines did, we want the number of outcomes with exactly 0 red and exactly 1 red, then take the compliment (total number of outcomes minus outcomes that give us only 0 or 1 red). For exactly 0 red, that is $\binom{24}{5}$ outcomes.
For exactly 1 red, that is $\binom{24}{4}\binom{8}{1}$ outcomes (note: since we are doing combinations, we don't care which of the apples is red). So, there are $\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}$ outcomes with at least 2 red.
For at least one apple (red or green), we want the number of outcomes with 0 apples and take the compliment: $\binom{32}{5} - \binom{16}{5}$.
So, the probability that you get at least 2 red apples given that you pick at least one apple (red or green):
$\dfrac{\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}}{\binom{32}{5} - \binom{16}{5}} = \dfrac{73,864}{197,008} = \dfrac{1319}{3518} \approx 0.375$
14. ## Re: Probability with a "known"
The following is multiple choice question (with options) to answer.
A box contains 12 mangoes out of which 4 are spoilt. If four mangoes are chosen at random, find the probability that Exactly three are not spoiled. | [
"116/495",
"224/495",
"129/495",
"187/495"
] | B | equired probability = 8C3 . 4C1/12C4 = 56 x 4/495
= 224/495.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35444 | 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 120 m long crosses a platform 220 m long in 20 sec; find the speed of the train? | [
"87 kmph",
"65 kmph",
"54 kmph",
"61 kmph"
] | D | D = 120 + 220 = 340
T = 20
S = 340/20 * 18/5 = 61 kmph
Answer:D |
AQUA-RAT | AQUA-RAT-35445 | # 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.
Kamal started a business investing Rs. 9000. After five months, Sameer joined with a capital of Rs. 8000. If at the end of the year, they earn a profit of Rs. 6970, then what will be the share of Sameer in the profit? | [
"2318",
"2380",
"1277",
"2662"
] | B | Kamal : Sameer = (9000 * 12) : (8000 * 7)
= 108:56 = 27:14
Sameer's share = 6970 * 14/41 = Rs. 2380.
Answer: B |
AQUA-RAT | AQUA-RAT-35446 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man buys a cycle for Rs. 1800 and sells it at a loss of 25%. What is the selling price of the cycle? | [
"s. 1090",
"s. 1160",
"s. 1190",
"s. 1350"
] | D | S.P. = 75% of Rs. 1800 = Rs.75/100x 1800 = Rs. 1350
ANSWER :D |
AQUA-RAT | AQUA-RAT-35447 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
Milk contains 5% water. What content of pure milk should be added to 10 liters of milk to reduce this to 2%? | [
"10liters",
"15liters",
"20liters",
"18liters"
] | B | Quantity of water in 10 liters = 5% of 10 liters = 0.5 liters
Let x liters of pure milk be added.
Then, 0.5/(10+x) = 2/100
2x = 30
x=15 liters
Answer is B |
AQUA-RAT | AQUA-RAT-35448 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a tank is 8 minutes and 14 minutes respectively. If both the pipes are opened simultaneously, and the pipe A is closed after 3 minutes, then how much more time will it take to fill the tank by pipe B? | [
"5 minute 25 second",
"5 minute 45 second",
"5 minute 35 second",
"6 minute 45 second"
] | B | Part filled by pipe A in 1 minute =1/8
Part filled by pipe B in 1 minute =1/14
Part filled by pipe A and pipe B in 1 minute =1/8+1/14=11/56
Pipe A and B were open for 3 minutes.
Part filled by pipe A and B in 3 minute=3 × 11/56=33/56
Remaining part =1-33/56=23/56
Time taken by pipe B to fill this remaining part=(23/56)/(1/14)=23/4
=5 3/4 minutes
=5 minute 45 second
Answer is B |
AQUA-RAT | AQUA-RAT-35449 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If x > 3000, then the value of (x)/(2x-42) is closest to? | [
"1/6",
"1/3",
"10/21",
"1/2"
] | D | assume x = 3002
(x)/(2x-42) = 3002 / (3002*2-42)
=3002 / 5962
= = 1/2
Ans - D |
AQUA-RAT | AQUA-RAT-35450 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In a mixture of milk and water, the proportion of milk by weight was 90%. If, in a 120-gram mixture, 30 grams of pure milk is added, what would be the percentage of milk in the resulting mixture? | [
"91.4%",
"91.6%",
"91.8%",
"92.0%"
] | D | The amount of milk is 0.9(120) + 30 = 138 grams.
The proportion is 138/150 = 0.92 which is 92%
The answer is D. |
AQUA-RAT | AQUA-RAT-35451 | If the numbers can be negative, then this conclusion no longer holds. For example, if $a_1=1$, $a_2=2$ and $b_1=b_2=-1$, then $\min(a_1,a_2)\min(b_1,b_2)=1\cdot(-1)=-1>-2=\min(-1,-2)=\min(a_1b_1,a_2b_2)$,
I am extremely grateful
Sarrah
The following is multiple choice question (with options) to answer.
If (a-7)(b-2) =0, which of the following must be true ?
I. a=7 and b=2
II. if a is not 7, then b=2
III. if a=7, then b is not=2 | [
"I only",
"II only",
"III only",
"I and II"
] | B | (a-7)(b-2)=0
So either a=7 or b=2 or both. It is 'or' not 'and', so it could be any of the three possibilities. Thus I and III are not necessarily true.
The answer is B. |
AQUA-RAT | AQUA-RAT-35452 | java, beginner, homework, palindrome
int Palindrome = reverse(input);
if (Palindrome == input){
return true;
}
else
return false;
}
public static void main(String[] args) {
int integer = 0;
Scanner input = new Scanner(System.in);
System.out.print("Enter a positive, multi-digit integer: ");
integer = input.nextInt();
while (integer <= 9 && integer > 0)
{
System.out.println(integer + " is a single digit. Please re-enter another integer: ");
integer = input.nextInt();
if (isPalindrome(integer) && (integer > 0 && integer > 9))
{
System.out.println(integer + " is a palindrome");
return;
}
else if (!isPalindrome(integer) && (integer > 0 && integer > 9))
{
System.out.println(integer + " is not a palindrome");
return;
}
The following is multiple choice question (with options) to answer.
A palindrome is a number that reads the same front-to-back as it does back-to-front (e.g. 202, 575, 1991, etc.) p is the smallest integer greater than 100 that is both a prime and a palindrome. What is the sum of the digits of p? | [
"3",
"4",
"5",
"6"
] | C | Given that p is smallest integer greater than 200 - assume there is a 3-digit that satisfies the above conditions. Let the number be xyx ; question asks us the values of 2x+y
We can straight away cross out options A) and D) - sum of digits 3 or 6 implies it is divisible by 3 ---> we know that p is a prime number
Coming to option B) 2x + y = 4 --> only x = 2 and y = 0 satisfy this equation ( x> 2 will never give sum of digits = 4) ; but 202 is divisible by 2 ; we know that p is a prime number
Similarly option C) 2x+y = 5 --> only x = 2 and y = 1 satisfy this equation ( x> 2 will never give sum of digits = 5) ; but 212 is divisible by 2 ; we know that p is a prime number
Therefore answer option should be E ---> can be verified by taking 2x+y = 7 ---> x = 3 and y = 1 ; gives 313
C |
AQUA-RAT | AQUA-RAT-35453 | It is easy to see from this website, problem/solution 23, http://books.google.com.au/books?id=8NtJLfGf94QC&pg=PA431&lpg=PA431&dq=monkey+climbing+ladder+on+pulley&source=bl&ots=2tWFRhdKME&sig=9ocddC7lk53A1_XdrVAwPm7MBw0&hl=en&sa=X&ei=Jv9SUevUEoTziAfSs4HIBg&ved=0CDAQ6AEwAA#v=onepage&q=monkey%20climbing%20ladder%20on%20pulley&f=false, that as the man moves up a distance $l'$, the centre of gravity moves up by a distance $l = \frac{ml'}{2M}$. Where $m$ is the mass of the man, and $M$ is the mass of the counter-weight.
This means that if the man is moving at a velocity of $V$ meters per second, then the centre of gravity will move up at a rate of $\frac{mV}{2M}$ meters per second as well.
I finally think something works. I don't need to imagine anything either. So, I assume the block's lower part is at height $H$ from the ground, ladder's last step at height $h$, the height of ladder being $l$ and the distance of man from the bottom being $x$ (where $x<l$ ).
The following is multiple choice question (with options) to answer.
A monkey start climbing up a tree 19 ft tall.Each hour it hops 3 ft and slips back 2 ft. How much time would it take the monkey to reach the top. | [
"15 hrs",
"18 hrs",
"19 hrs",
"17 hrs"
] | D | if monkey hops 3ft and slips back 2ft in a hour, it means the monkey hops (3ft-2ft)= 1ft/hr.
similarly in 16hrs it wil be 16ft.
Bt since the height of the tree is 19ft, so if the monkey hops up the tree in the next hr i.e 17th hr then it reaches at the top of the tree.
hence it takes 17 hrs for monkey to reach at the top
ANSWER:D |
AQUA-RAT | AQUA-RAT-35454 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
In a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man. What is weight of new comer? | [
"70",
"66",
"68",
"69"
] | C | Solution: Member in group * age increased = difference of replacement
Or, 8*1 = new comer - man going out
Or, new comer = 8+60;
Or, new comer = 68 years.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35455 | 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.
If there are 10 apples and 20 oranges, what is the ratio of apples to oranges? | [
"10:20",
"2:4",
"1:2",
"4:8"
] | C | A ratio should be reduced to the lowest common denominator.
10:20 means 10x:20x which can be written as 1(10):2(10)...
ANSWER:C |
AQUA-RAT | AQUA-RAT-35456 | 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.
The speed at which a man can row a boat in still water is 15 kmph. If he rows downstream, where the speed of current is 5 kmph, what time will he take to cover 60 metres? | [
"16 seconds",
"76 seconds",
"10.8 seconds",
"12 seconds"
] | C | Speed of the boat downstream = 15 + 5
= 20 kmph
= 20 * 5/18 = 50/9 m/s
Hence time taken to cover 60 m
= 60*9/50
= 10.8 seconds.
Answer: C |
AQUA-RAT | AQUA-RAT-35457 | it. Given the first term and the common ratio of a geometric sequence find the first five terms and the explicit formula. The Sum of Geometric Sequence. We will just need to decide which form is the correct form. As a check, $$\frac{u_5}{u_3}=4=r^2$$ so this does work. This Site Might Help You. Access this finite geometric series worksheets tenaciously prepared for high school students. Determine the common ratio and. Any term = constant = r. This ratio is called _____. The sum of an infinite geometric series is 24, and the sum of the first 200 terms of the series is also 24. Geometric Series Questions (b) Find, to 2 decimal places, the di erence between the 5th and 6th terms. A geometric series is the sum of the terms of a geometric sequence. Work out the missing term in this geometric sequence:. Determine the common ratio and. How do we find the nth term of a geometric sequence? 1. For example, if I know that the 10 th term of a geometric sequence is 24, and the 9 th term of the sequence is 6, I can find the common ratio by dividing the 10 th term by the 9 th term: 24 / 6 = 4. Common Ratio. If there are 160 ants in the initial population, find the number of ants. Given u 5 =1280 and u 8 =81920 , find the geometric sequence. This sequence starts at 10 and has a common ratio of 0. Click here 👆 to get an answer to your question ️ 2. r = 4 2 = 2. Find (1) the common ratio, (2) the ninth term, (3) a recursive rule for the nth term, and (4) an Log On Algebra: Sequences of numbers, series and how to sum them Section Solvers Solvers. Find the common ratio of the geometric sequence. Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. And each time I'm multiplying it by a common number, and that number is often called the common ratio. Finally, use the rule to find the tenth term in the sequence. Also, a geometric sequence has p as its common ratio. If you’re good at finding patterns, then you’ll
The following is multiple choice question (with options) to answer.
Find the first term and the common ratio of a G.P whose fourth term is 250 and seventh term is 31,250 | [
"2/5,25",
"4,5/2",
"1,16",
"2,5"
] | D | a*r ^3 = 250, a *r ^6 =31,250, r ^3 =31,250/250 = 125
r = 5, a* 125 = 250, a = 2
ANSWER:D |
AQUA-RAT | AQUA-RAT-35458 | 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 600 meter long train crosses a signal post in 40 seconds. How long will it take to cross a 3 kilometer long bridge, at the same speed? | [
"4 min",
"5 min",
"3 min 4 sec",
"4 min 5 sec"
] | A | Explanation:
S = 600/40 = 15 mps
S = 3600/15 = 240 sec = 4 min
ANSWER IS A |
AQUA-RAT | AQUA-RAT-35459 | vcf, shell
1 11790870 rs2274976 C T 777.01 PASS "BaseCounts=0,18,0,24;BaseQRankSum=-0.042;DB;Dels=0;FS=4.677;GC=56.11;HaplotypeScore=0.6651;MQ=60;MQ0=0;MQRankSum=1.204;QD=15.05;ReadPosRankSum=0.836;DP=355;AF=0.5;MLEAC=1;MLEAF=0.5;AN=20;AC=10"
1 11792243 rs1476413 C T 1751.01 PASS "BaseCounts=1,18,0,21;BaseQRankSum=1.028;DB;Dels=0;FS=0;GC=56.11;HaplotypeScore=1.7287;MQ=60;MQ0=0;MQRankSum=-0.408;QD=16.65;ReadPosRankSum=-0.099;DP=954;AF=0.5;MLEAC=1;MLEAF=0.5;AN=50;AC=29"
1 11794400 rs4846051 G A 2616.01 PASS "BaseCounts=46,0,0,0;DB;Dels=0;FS=0;GC=59.35;HaplotypeScore=6.2573;MQ=60;MQ0=0;QD=31.75;DP=1654;AF=1;MLEAC=2;MLEAF=1;AN=78;AC=78"
The following is multiple choice question (with options) to answer.
Calculate the HCF of 64, 592, 960 | [
"15",
"16",
"13",
"14"
] | B | Explanation:
Lets solve this question by factorization method.
The factors of 64 are: 1, 2, 4, 8, 16, 32, 64
The factors of 592 are: 1, 2, 4, 8, 16, 37, 74, 148, 296, 592
The factors of 960 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 160, 192, 240, 320, 480, 960
Then the greatest common factor is 16.
Answer: Option B |
AQUA-RAT | AQUA-RAT-35460 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
What is the next number in the series?
1,5,9,13... | [
"19",
"17",
"24",
"25"
] | B | Adding (0,2,4,6,8,10) to the odd numbers
1+0=1
3+2=5
5+4=9
7+6=13
9+8=17
answer B |
AQUA-RAT | AQUA-RAT-35461 | 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 takes 9 hrs to cover distance with the speed 85kmph. Find distance the train travelled. | [
"750",
"850",
"725",
"765"
] | D | Length = Speed * time
Length =85kmph * 9 hrs
Length =765km
Ans : (D) |
AQUA-RAT | AQUA-RAT-35462 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is? | [
"16%",
"15%",
"12%",
"22%"
] | C | S.I. for 3 years = (12005 - 9800) = Rs. 2205
S.I. for 5 years = Rs. 2205/3 * 5 = Rs. 3675.
Principal = (9800 - 3675) = Rs. 6125
Hence, rate = (100 * 3675) / (6125 * 5)
= 12%
Answer: C |
AQUA-RAT | AQUA-RAT-35463 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
There is enough provisions for 600 men in an army camp for 25 days. If there were 100 men less, how long will the provision last? | [
"10days",
"50days",
"100days",
"150days"
] | D | Exp: We have, M1D1 = M2D2
600*25= 100* D2
D2 = 600*25/100 = 150 days.
Answer: D |
AQUA-RAT | AQUA-RAT-35464 | 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 cheese factory sells its cheese in rectangular blocks. A normal block has a volume of 5 cubic feet. If a large block has twice the width, twice the depth, and the same length of a normal block, what is the volume of cheese in a large block in cubic feet? | [
"15",
"30",
"20",
"18"
] | C | Volume of cube=lbh=5
New cube l ,b, h are increases of l, 2b,2h
New volume of cube =l*2b*2h=4*lbh
=4*5
=20
Answer: C |
AQUA-RAT | AQUA-RAT-35465 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
If the sides of a triangle are 65 cm, 60 cm and 25 cm, what is its area? | [
"120 cm2",
"750 cm2",
"216 cm2",
"197 cm2"
] | B | The triangle with sides 65 cm, 60 cm and 25 cm is right angled, where the hypotenuse is 65 cm.
Area of the triangle
= 1/2 * 60 * 25
= 750 cm2
Answer:B |
AQUA-RAT | AQUA-RAT-35466 | ## Comments
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All regular polygons will have the same continuation. but with different G.P. ratios.. Suppose you have an n-sided regular polygon with vertices $$A_1,A_2,A_3,...,A_n$$. Say $$B_1$$ is the midpoint of line segment $$A_1A_2$$ and $$B_2$$ of $$A_2A_3$$, then if we join $$A_1$$ to $$A_3$$, then midpoint theorem gives us $$B_1B_2$$=$$1/2$$*$$A_1A_3$$. Also, $$A_1A_3$$=$$2n$$ $$\sin \theta/2$$, where n is the side of the polygon.
So $$B_1B_2$$= $$n$$ $$\sin \theta/2$$
also, in a polygon of n-sides(regular), $$\theta$$= $$\frac {180(n-2)}{n}$$, so $$B_1B_2$$=$$n$$ $$\sin \frac {90(n-2)}{n}$$
and the ratio of the sides of the new polygon to the old one, is thus $$\sin \frac {90(n-2)}{n}$$ and ther ratio of the areas is thus the required G.P. ratio, which is nothing but $$(\sin \frac {90(n-2)}{n})^2$$. It can be easily verified in all cases. For n=4, i.e., for a square, this becomes 1/2, for a triangle it becomes 1/4, etc.
- 4 years, 7 months ago
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for regular shapes may be existed ratio according to the number of sides
- 4 years, 7 months ago
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The following is multiple choice question (with options) to answer.
Let S1 be a square of side a. Another square S2 is formed by joining the mid-points of the sides of S1. The same process is applied to S2 to form yet another square S3, and so on. If A1, A2, A3 .. be the areas and P1, P2, P3,…….. be the perimeters of S1, S2, S3,..., respectively, then the ratio (P1 + P2 +P3+ ......) / (A1 + A2 + A3 +.......) equals :- | [
"2(1+√2)/a",
"2(2+√2)/a",
"2(2+√2)/a",
"2(1+2√2)/a"
] | B | Explanation :
side of S1= a, perimeter of S1= 4a, area of S1 = a2
side of S2= a/√2 , perimeter of S2= 2√2a , area of S2 = a2/2
side of S3= a/2 , perimeter of S3= 2a , area of S3= a2/4
So, for P1 + P2 + P3 .... we will have a GP as below
4a + 2√2a + 2a + .....
First term =4a, common ratio = 1/√2, Sum = 4a/( 1-1/√2) = 4√2a/(√2 - 1)
For A1 + A2 + A3....we will have another GP
a2 + a2/2 + a2/4 .....
First term = a2 , common ratio = 1/2 , Sum = 2a2.
Hence, the required ratio :-
i.e (P1 + P2 + P3....) / (A1+ A2+A3....).
=> 4√2a / (√2 - 1) / 2a2.
=> 2√2(√2 + 1)/a.
=> 2(2 +√2)/a .
Answer : B |
AQUA-RAT | AQUA-RAT-35467 | A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 =$54, and thus the average revenue per unit is thus 54/3 = $18. Answer: E _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Retired Moderator Joined: 21 Aug 2013 Posts: 1386 Location: India Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 02 Jul 2017, 20:49 1 Ratio of quantity of P and Q sold, P:Q = 1:2. Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 =$ 18
Intern
Joined: 28 Aug 2016
Posts: 11
Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
### Show Tags
18 Apr 2019, 14:30
1
1
AbdurRakib wrote:
A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
The following is multiple choice question (with options) to answer.
A survey reveals that the average income of a company’s customers is $45,000 per year. If 40 customers respond to the survey and the average income of the wealthiest 10 of those customers is $75,000, what is the average income of the other 30 customers?
Is there a way to solve this using weighted average concept instead of doing tedious calculations? | [
" $27,500",
" $35,000",
" $37,500",
" $42,500"
] | B | let x be the average of 30 customers
30*x + 10* 75000 = 40*45000
solving this we have x= 35000
Answer is B. |
AQUA-RAT | AQUA-RAT-35468 | 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 crosses a platform of 120 m in 15 sec, same train crosses another platform of length 180 m in 18 sec. then find the length of the train? | [
"780m",
"180m",
"170m",
"140m"
] | B | Length of the train be ‘X’
X + 120/15 = X + 180/18
6X + 720 = 5X + 900
X = 180m
Answer:B |
AQUA-RAT | AQUA-RAT-35469 | # Probability of a certain ball drawn from one box given that other balls were drawn
Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
The following is multiple choice question (with options) to answer.
In a box, there are 7 red, 3 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green? | [
"2/3",
"8/21",
"3/7",
"7/15"
] | D | Explanation:
Total number of balls = (7 + 3 + 5) = 15.
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 7.
P(E) = 7/15.
Answer: Option D |
AQUA-RAT | AQUA-RAT-35470 | => 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.
The average salary of all the workers in a workshop is Rs. 8000. The average salary of 10 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is : | [
"22",
"21",
"88",
"30"
] | D | Explanation:
Lot the total number of workers be v Then,
8OOOv = (12000 * 10) + 6000 (v - 10) <=> 2000v = 60000 <=> v = 30
Answer: D) 30 |
AQUA-RAT | AQUA-RAT-35471 | Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:37
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
_________________
Kudos [?]: 139450 [0], given: 12790
Manager
Joined: 26 Mar 2010
Posts: 116
Kudos [?]: 16 [0], given: 17
Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:53
Bunuel wrote:
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!
this might clear my entire probability confusion i hope...
Kudos [?]: 16 [0], given: 17
Math Expert
Joined: 02 Sep 2009
Posts: 43322
Kudos [?]: 139450 [1], given: 12790
The following is multiple choice question (with options) to answer.
A 4-letter code word consists of letters D, E, and G. If the code includes all the three letters, how many such codes are possible? | [
"72",
"48",
"36",
"24"
] | C | Note that the correct answer to this question is 36, not 72.
D-DEG can be arranged in 4!/2!=12 ways;
E-DEG can be arranged in 4!/2!=12 ways;
G-DEG can be arranged in 4!/2!=12 ways;
Total: 12+12+12=36.
Answer: C. |
AQUA-RAT | AQUA-RAT-35472 | I think the method you used is the best way to go.
Still, if you want to do it via the Chinese Remainder theorem....
Note that $$5000=2^3\times 5^4$$ so solve the problem mod $$2^3$$ and mod $$5^4$$ separately. Clearly the answer is $$0\pmod {5^4}$$ so that just leaves $$2^3$$. But $$15\equiv -1\pmod {2^3}$$ so the answer is $$1\pmod {2^3}$$. Now apply the CRT to $$n\equiv 0 \pmod {625}\quad \&\quad n\equiv 1\pmod {8}$$
Since $$625\equiv 1 \pmod {8}$$ the answer is $$625$$.
• Worth emphasis: we can eliminate the CRT calculation by essentially factoring $\,5^4 = 625\,$ out of the mod computation - as explained in my answer and its link. This often (greatly) simplifies modular computations of this sort. Jan 25, 2020 at 18:37
• @BillDubuque Good point, and the discussion you link to is well worth studying.
– lulu
Jan 25, 2020 at 18:39
The following is multiple choice question (with options) to answer.
3/4 of 1/2 of 2/5 of 5000 = ? | [
"392",
"229",
"753",
"750"
] | D | D
750
? = 5000 * (2/5) * (1/2) * (3/4) = 750 |
AQUA-RAT | AQUA-RAT-35473 | This implies:
$\displaystyle y+3=x+2\:\therefore\:x-y=1\:\therefore\:x+y=2x-1$
Now solve:
$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$
To find x.
#### MarkFL
Staff member
Here's another method:
$\displaystyle\frac{x^2-2x-3}{x+2}=\frac{2}{3}$
implies:
(1) $\displaystyle 3x^2-8x-13=0$
$\displaystyle\frac{y^2-4}{y+2}=\frac{2}{3}$
implies:
(2) $\displaystyle 3x^2-2y-18=0$
Adding (1) and (2) we get:
(3) $\displaystyle 3x^2+3y^2-8x-2y-31=0$
If we multiply (1) by 2 we have:
$\displaystyle 6x^2-16x-26=0$
which we may write as:
$\displaystyle -2x-8(x-1)+6x(x-1)-34=0$
Using $\displaystyle y=x-1$ this becomes:
(4) $\displaystyle-2x-8y+6xy-34=0$
Adding (3) and (4) we obtain:
$\displaystyle 3x^2+6xy+3y^2-10x-10y-65=0$
(5) $\displaystyle 3(x+y)^2-10(x+y)-65=0$
Now we have a quadratic in $\displaystyle x+y$.
The following is multiple choice question (with options) to answer.
If x+y=3 and x2y3 + y2x3=27, what is the value of xy? | [
"1",
"2",
"3",
"4"
] | C | xy=3
As x+y=3
x2y3+y2x3=27
x2y2(y+x)=27
Substituting x+y
x2y2=9
xy=3
ANSWER:C |
AQUA-RAT | AQUA-RAT-35474 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
Without stoppages, a train travels certain distance with an average speed of 250 km/h, and with stoppages, it covers the same distance with an average speed of 125 km/h. How many minutes per hour the train stops ? | [
"25",
"30",
"35",
"40"
] | B | Due to stoppages, it covers 125 km less .
Time taken to cover 125 km = 125â„250h = 1â„2h
= 1â„2 × 60 min = 30 min
Answer B |
AQUA-RAT | AQUA-RAT-35475 | # 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 shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 14% while buying and by 20% while selling. What is his percentage profit? | [
"10.22%",
"20.22%",
"21.22%",
"42.5%"
] | D | The owner buys 100 kg but actually gets 114kg;
The owner sells 100 kg but actually gives 80kg;
Profit: (114-80)/80*100=~42.5%
Answer: D. |
AQUA-RAT | AQUA-RAT-35476 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 3:4:5 and their L.C.M. is 600. Their H.C.F is? | [
"10",
"30",
"40",
"50"
] | A | Let the numbers be 3x,4x and 5x
their L.C.M. = 60x
60x =600
x = 10
The numbers are 3*10 , 4*10 , 5*10
Hence required H.C.F. = 10
Answer is A |
AQUA-RAT | AQUA-RAT-35477 | 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.
A certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint. If 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans, what is the least amount of paint E, in gallons, that must be purchased in order to measure out the portions needed for the mixture? | [
"2",
"2 1/2",
"3",
"3 1/2"
] | B | Given W:B = 3:5
That means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture.
But we want least amount of whiteblack paints for minimum of 2 gallons of mixture, so lets reduce keeping same ratio,
1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture, but we want only 2 gallons, lets further reduce
0.75: 1.25 gives 1+1.5 =2.5 gallons of mixture. This looks ok, but lets reduce further just to be sure
0.375: 0.625 gives 0.5 + 1 = 1.5 gallons of mixture, thats less than 2 gallons of mixture, so not acceptable.
So correct ans is 2.5 gallons. B |
AQUA-RAT | AQUA-RAT-35478 | of }(2)\text{ and }(23) \\ \quad (25) & p \land \lnot q & \land\text{-Introduction of }(3)\text{ and }(24) \\ \end{array}\\ }\\ \begin{array} {rlr} (26) & p \implies (p \land \lnot q) & \implies\text{-Elimination of }(3)\text{ to }(16) \\ (27) & \lnot p \implies (p \land \lnot q) & \implies\text{-Elimination of }(17)\text{ to }(25) \\ (28) & p \lor \lnot p & \text{Law of the Excluded Middle} \\ (29) & p \land \lnot q & \lor\text{-Elimination of }(28),(26),(27) \\ \end{array}\\ \end{array}$$
The following is multiple choice question (with options) to answer.
Statement: The authorities in Society X are cracking down on street hawkers, blaming them for traffic jams near their society.
Actions:
I. Street hawkers should not be allowed during peak hours
II. Street hawkers should be warned and asked not to create chaos. | [
"Only I follows",
"Only II follows",
"Either I or II follows",
"Neither I nor II follows"
] | B | Explanation: Here a situation of traffic jam is depicted and the actions mentioned, give different ways for improving the situation.
ANSWER IS B |
AQUA-RAT | AQUA-RAT-35479 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B start a business, with A investing the total capital of Rs.50000, on the condition that B pays A interest @ 10% per annum on his half of the capital. A is a working partner and receives Rs.1500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year? | [
"33388",
"27889",
"27889",
"59000"
] | D | Interest received by A from B = 10% of half of Rs.50000 = 10% * 25000 = 2500.
Amount received by A per annum for being a working partner = 1500 * 12 = Rs.18000.
Let 'P' be the part of the remaining profit that A receives as his share. Total income of A = (2500 + 18000 + P)
Total income of B = only his share from the remaining profit = 'P', as A and B share the remaining profit equally.
Income of A = Twice the income of B
(2500 + 18000 + P) = 2(P)
P = 20500
Total profit = 2P + 18000
= 2*20500 + 18000 = 59000
Answer:D |
AQUA-RAT | AQUA-RAT-35480 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
After working for 4 days, Ashok finds that only 1/3 rd of the work has been done. He employs Ravi who is 60% as efficient as Ashok. How many days more would Ravi take to complete the work? | [
"13 1/3 days",
"14 1/3 days",
"15 1/3 days",
"3 1/3 days"
] | A | 1/3 ---- 4
1 -------? A = 12
R = 1/12 * 60/100 = 1/20
1 ----- 1/20
2/3 ----? => 13 1/3 days
ANSWER:A |
AQUA-RAT | AQUA-RAT-35481 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is: | [
"2:3",
"4:3",
"6:7",
"9:16"
] | B | Let us name the trains as A and B. Then,
(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
Answer :B |
AQUA-RAT | AQUA-RAT-35482 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man buys an article and sells it at a profit of 20%. If he had bought it at 20% less and sold it for Rs.95 less, he could have gained 25%. What is the cost price? | [
"372",
"375",
"278",
"475"
] | D | CP1 = 100 SP1 = 120
CP2 = 80 SP2 = 80 * (125/100) = 100
20 ----- 100
95 ----- ? => 475
Answer:D |
AQUA-RAT | AQUA-RAT-35483 | => 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.
At a local appliance manufacturing facility, the workers received a 25% hourly pay raise due to extraordinary performance. If one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged, by approximately what percent would he reduce the number of hours that he worked? | [
"83%",
"80%",
"20%",
"17%"
] | C | Let's say he works usually 10 hours and earns 100 per hour.
10 * 100 = 1000
10 * 125 = 1250 (this are the new earnings after the raise)
To figure out how much he needs to work with the new salary in order to earn the original 1000:
1000/125 = 8
So he can reduce his work by 2 hours. Which is 20%.
Answer C |
AQUA-RAT | AQUA-RAT-35484 | 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}$$
The following is multiple choice question (with options) to answer.
50% of Ram's marks is equal to 20% of Rahim's marks which percent is equal to 30% of Robert's marks. If Robert's marks is 80, then find the average marks of Ram and Rahim? | [
"70",
"97",
"84",
"90"
] | C | Given, 50% of Ram's marks = 20% of Rahim's marks = 30% of Robert's marks.
Given, marks of Robert = 80
30% of 80 = 30/100 * 8 = 24
Given, 50% of Ram's marks = 24.
=> Ram's marks = (24 * 100)/50 =48
Also, 20% of Rahim's marks = 24
=> Rahim's marks = (24 * 100)/20 = 120
Average marks of Ram and Rahim = (48 + 120)/2 = 84.
Answer:C |
AQUA-RAT | AQUA-RAT-35485 | So on.
=====
Another thing to note:
(Assume only positive values)
$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$
So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$
That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.
But it doesn't go the other way!
$$\frac 12 < a < 4 \not \implies 1 < a < 2$$
Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$
[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)
• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06
The following is multiple choice question (with options) to answer.
If −y ≥ x, and −x < −7, then which of the following must be true? | [
"y = −7",
"y > −7",
"−y > 7",
"y ≤ −7"
] | C | −y ≥ x, and −x < −7
y <= -x < -7
As '-x' is less than '-7' and Y is less than or equal to '-x'
we have y < -7 which is similar to -y > 7.
C) -y > 7 |
AQUA-RAT | AQUA-RAT-35486 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 11220, how much money did he borrow? | [
"11000",
"80288",
"2668",
"2600"
] | A | Let the sum be Rs. x. Then,
[(x * 6 * 3)/100] + [(x * 9 * 5)/100] + [(x * 13 * 3)/100] = 11220
18x + 45x + 39x = (11220 * 100)
102x = 1122000 => x = 11000.
Answer:A |
AQUA-RAT | AQUA-RAT-35487 | # 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 owner of a furniture shop charges his customer 42% more than the cost price. If a customer paid Rs. 8300 for a computer table, then what was the cost price of the computer table? | [
"Rs. 5725",
"Rs. 5845",
"Rs. 6275",
"Rs. 6725"
] | B | CP = SP * (100/(100 + profit%))
= 8300(100/142) = Rs. 5845.
ANSWER:B |
AQUA-RAT | AQUA-RAT-35488 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
When Jack picks olives for two hours at three times his regular speed, he picks 20 pounds of olives more than Mac working for five hours at 80% of his regular speed. Therefore, if Mac picks olives for one hour at double his regular speeds, and Jack picks olives for four hours at 75% of his regular speed, then | [
"Jack picks double the amount of olives Mac picks",
"Mac picks 10 pounds more than Jack",
"Jack picks 10 pounds more than Mac",
"Mac picks 5 more pounds than Jack"
] | D | Let's say Jack's regular speed is J olives/hr and Mac's regular speed is M olives/hr
Given:
2*3J = 10 + 5*(4/5)M
3J = 5 + 2M
Question:if Mac picks olives for one hour at double his regular speeds, and Jack picks olives for four hours at 75% of his regular speed
Mac picks 2M and Jack picks 4*(3/4)J = 3J
They are asking you for the relation between 3J and 2M. You already know 3J = 5 + 2M
So Jack picks 5 pounds more olives than Mac.
D |
AQUA-RAT | AQUA-RAT-35489 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 3 seconds. What is the length of the fast train ? | [
"27 7/9",
"16 2/3",
"29 7/9",
"30 7/9"
] | B | Explanation:
As Trains are moving in same direction,
So, Relative Speed = 40-20 = 20 kmph
= 20*(5/18) = 50/9 m/sec
Length of Train= Speed * Time
Length=50/9∗3
=150/9
=16 2/3
Option B |
AQUA-RAT | AQUA-RAT-35490 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
If Teena is driving at 55 miles per hour and is currently 7.5 miles behind Poe, who is driving at 40 miles per hour in the same direction then in how many minutes will Teena be 15 miles ahead of Poe? | [
"15",
"60",
"75",
"90"
] | D | This type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30-40 seconds for a difficult one.
Teena covers 55 miles in 60 mins.
Poe covers 40 miles in 60 mins
So teena gains 15 miles every 60 mins
Teena need to cover 7.5 +15 miles.
Teena can cover 7.5 miles in 30 mins
Teena will cover 15 miles in 60 mins
So answer 30+60= 90 mins=D |
AQUA-RAT | AQUA-RAT-35491 | sampling
0, 0.25, 0.5, 0.75
1, 1.25, 1.5, 1.75
...
or four samples per period, so four samples "cover" or "represent" one period of one second.
The following is multiple choice question (with options) to answer.
What is the average of four tenths and five thousandths? | [
"0.2022",
"0.2012",
"0.2025",
"0.2021"
] | C | Four tenths = 0.4
Five thousandths = 0.005
The average is (0.4 + 0.005)/2 = 0.2025
Answer: C |
AQUA-RAT | AQUA-RAT-35492 | length, and more with flashcards, games, and distinguishes! Diagonal lengths d1 and d2 are the diagonals of a rhombus? to the longer diagonal of the rhombus a... Here d₁ and d₂ are diagonals x if BCA = 3x -2 and ACD 12... Same, making the shape a rhombus what a quadrilateral is a quadrilateral is to... 7 m and the area of the properties of a rhombus whose sides are congruent and parallel and! Opposite sides are of equal length MNOP, a rhombus are congruent parallel! Four sides of the rules that a parallelogram with four congruent sides every. Squares are rhombuses, but not all equal if one of the properties of rhombus parallelogram.... Angles is equal to 6 cm and 12 cm, respectively is.... Its area is 121 cm2 and Lets say d1 = 22 cm distance between each base the! Both diagonals are not all parallelograms equal in measure diagonal is 22 cm special kind of quadrilateral which. Printable charts vividly state the properties of rhombus = 121 cm2 and Lets d1. Theorem 6-16 if the side of a rhombus rhombus, diagonals bisect each other two... Is symmetric across each of these diagonals ( IK × HJ ) = 54 us know what quadrilateral! The following statements the properties of rhombuses, rectangles, and other tools..., MNOP, a = ( 1/2 ) x ( d₁ x ). Almost all the properties of both equilateral and quadrilateral Quiz Flash cards solid is equal to one of rhombus. Square are perpendicular, then it is a parallelogram are perpendicular and divides each other at 90°,,... Makes a parallel combination and the measure of diagonals SA is 24 and the diagonals of rhombus: has! Degrees, then the parallelogram is a type of parallelogram whose diagonals each! Example problems for rhombus properties since a rhombus = ( 1/2 ) x ( d₁ d₂... Perpendicular angles is equal to 6 cm 6th-grade students differentiate a rhombus a! The playing cards two adjacent angles
The following is multiple choice question (with options) to answer.
Find the area of the quadrilateral of one of its diagonals is 15 cm and its off sets 6 cm and 4 cm? | [
"50 cm2",
"60 cm2",
"75 cm2",
"90 cm2"
] | C | 1/2 * 15(6 + 4)
= 75 cm2
Answer: C |
AQUA-RAT | AQUA-RAT-35493 | # 2012 AMC 10B Problems/Problem 22
## Problem
Let ($a_1$, $a_2$, ... $a_{10}$) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$
## Solution 1
If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$.
If we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$, with the exception of the $1$. For example, $(2,3,1,4,5,6,7,8,9,10)$, or $(2,3,4,1,5,6,7,8,9,10)$. There are $\dbinom{9}{1}$ ways to do so.
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are $\dbinom{9}{2}$ ways to do this.
In the same way, the total number of lists is: $\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}$
By the binomial theorem, this is $2^{9}$ = $512$, or $\boxed{\textbf{(B)}}$
The following is multiple choice question (with options) to answer.
The number of ways in which ten candidates A1, A2, ...., A10 can be ranked so that A1 is always above A2 is | [
"10!⁄2",
"10!",
"9!",
"8!⁄2"
] | A | Ten candidates can be ranked in 10! ways. In half of these ways A1 is above A2 and in another half A2 is above A1.
So, required number of ways is 10!⁄2
Answer A |
AQUA-RAT | AQUA-RAT-35494 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
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BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
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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 avg. age of a group of 16 students is 20 years.If 4 more students join the group,the avg age increases by 1 year.The avg age of the new student is? | [
"22 years",
"23 years",
"24 years",
"25 years"
] | D | Total age of 16 students=16*20=320
If total age of 4 students=x
Then, (320+x)/(16+4) =(20+1), x=100
So average age of new students=100/4= 25 years
ANSWER:D |
AQUA-RAT | AQUA-RAT-35495 | ### Show Tags
13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
_________________
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Lucky
_______________________________________________________
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The following is multiple choice question (with options) to answer.
M = {-6, -5, -4, -3, -2, -1}
T = {-5, -4, -3, -2, -1, 0, 1, 2}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative? | [
"0",
"1/4",
"2/5",
"1/2"
] | B | We will have a negative product only if 1 or 2 are selected from set T.
P(negative product) = 2/8 = 1/4
The answer is B. |
AQUA-RAT | AQUA-RAT-35496 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern which could be filled in 9 hours takes 1 hour more to be filled owing to a leak in its bottom.If the cistern is full, in what time will the leak empty it? | [
"50hours",
"62hours",
"90hours",
"75hours"
] | C | Let the leak empty the full cistern in x hours
9x/x-9 = 9+1
x = 90 hours
Answer is C |
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