source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36097 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains 6 black and 7 white balls. One ball is drawn at random. What is the probability that the ball drawn is white? | [
"3/4",
"7/13",
"1/7",
"1/8"
] | B | Let number of balls = (6 + 7) = 13.
Number of white balls = 7.
P (drawing a white ball) =7/13
Option B. |
AQUA-RAT | AQUA-RAT-36098 | RonL
Quite right!
I have made a picture of how could these streets look like.
There is also second question.
How many there are areas which are edged with streets from every side?
I have counted from picture that there are 15, I must say that I can't think another way at the moment. :D
• Apr 24th 2006, 05:50 PM
ThePerfectHacker
If you have $n$ lines than at most you can have.
$\frac{n(n-1)}{2}$ intersection points.
Thus,
$\frac{n(n-1)}{2}=21$
Thus,
$n^2-n+42=0$
Thus, $n=-6,7$
Disregard the negative which yields $n=7$
Proof,
If you have $n$ lines than to maximize the number of intersetion points in a plane you need them to be non-parallel such as no two are parallel to each other* then they need to intersect in such a way that no two pass through the same point. Now for the first line you have $n-1$ intersection points because it intersects each line EXCEPT for itself thus there are a total of $n-1$. While the second line interesects at $n-2$ because it cannot intersect with itself and not with the first because two lines intersect in only one point and it already intersected with the first line, while with the other lines it intersects with other points other than the first. The third one intersects with $n-3$ and so on. Thus you need to find the sum:
$(n-1)+(n-2)+(n-3)+...+2+1$
Which is $\frac{n(n-1)}{2}$ by the fact of the sum of an arithmetic series.
*)I did not find a rigorous demonstartion of this argument but if you think about it it is obvious. By that I of course mean the construction is possible.
• Apr 24th 2006, 05:57 PM
ThePerfectHacker
I once tried to develope a math riddle based on what you said. Never got to solving it.
-----------
The first question of the riddle I answered, i.e. the max number of distinct points.
The following is multiple choice question (with options) to answer.
Apple Street begins at Peach Street and runs directly east for 3.2 kilometers until it ends when it meets Cherry Street. Apple Street is intersected every 200 meters by a perpendicular street, and each of those streets other than Peach Street and Cherry Street is given a number beginning at 1st Street (one block east of Peach Street) and continuing consecutively (2nd Street, 3rd Street, etc...) until the highest-numbered street one block west of Cherry Street. What is the highest-numbered street that intersects Apple Street? | [
"14th",
"15th",
"16th",
"17th"
] | B | 3.2 km / 200 m = 16.
However, the street at the 3.2-km mark is not 16th Street; it is Cherry Street.
Therefore, the highest numbered street is 15th Street.
The answer is B. |
AQUA-RAT | AQUA-RAT-36099 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When the integer n is divided by 17, the quotient is x and the remainder is 5. When n is divided by 23, the quotient is y and the remainder is 13. Which of the following is true? | [
" 23x + 17y =19",
" 17x –23y = 8",
" 17x +23y =19",
" 14x + 5y = 6"
] | B | From the problem it follows:
n=17x+5
n=23y+13
So, 17x+5=23y+13
17x-23y=8
The answer is B |
AQUA-RAT | AQUA-RAT-36100 | 8 Since g(X)>f(X) we may conclude that g is always greater than f.
3. Mar 13, 2010
### Werg22
Re: x=2^x
Much simpler is to write 2^x - x = 0, and take the derivative of LHS: ln(2)2^x - 1. On [0, ∞), this is clearly strictly positive, so the function 2^x - x is strictly increasing on that interval. At x = 0, 2^x - x = 1, so it never attains 0 on [0, ∞) (it only gets larger than 1). It also can never attain 0 on (-∞, 0), so there is no solution.
4. Mar 13, 2010
### arildno
Re: x=2^x
Most definitely not!
At x=0, for example, ln(2)-1 is negative.
5. Mar 13, 2010
### Werg22
Re: x=2^x
Ah, then I retract my solution.
6. Mar 14, 2010
### Dragonfall
Re: x=2^x
Ah, thanks for the replies!
7. Mar 15, 2010
### uart
Re: x=2^x
This question could be turned into a nice little exercise for introductory calculus.
Q. Find the largest value of "a" for which the equation, $a^x = x$ has a real solution? It has a fairly cute solution.
a = e^(1/e), with the solution occurring at x = e.
The following is multiple choice question (with options) to answer.
f(x) = x! g(x)=x^10 h(x)= 10^x for large values of x which is greater? | [
"f(x)",
"g(x)",
"h(x)",
"both"
] | C | let x=100
then g(x)=100^10=10^20
h(x)=10^100
so h(x) is greater
ANSWER:C |
AQUA-RAT | AQUA-RAT-36101 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
If (2^x)(3^y) = 108, where x and y are positive integers, then (2^x-1)(3^y-2) equals: | [
"6",
"24",
"48",
"96"
] | A | So I would start attacking this problem by quickly performing the prime factorization of 288. With that it is easy to count the 5 twos and the 2 threes that are the prime factors. So x=2, y=3. now quickly 2^1(3^1)=6. Than answer should be number 1.
A |
AQUA-RAT | AQUA-RAT-36102 | # How many distinct ways to climb stairs in 1 or 2 steps at a time?
I came across an interesting puzzle:
You are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?
Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?
Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.
The following is multiple choice question (with options) to answer.
Mr. Shah decided to walk down the escalator of a tube station. He found  that if he walks down 26 steps, he requires 30 seconds to reach the bottom. However, if he steps down 34 stairs he would only require 18 seconds to get to the bottom. If the time is measured from the moment the top step begins  to descend to the time he steps off the last step at the bottom, find out the height of the stair way in steps? | [
"44",
"45",
"46",
"47"
] | C | (s1*t2~s2*t1)/(t2~t1)=(26*18~34*30)/(18~30)=46
ANSWER:C |
AQUA-RAT | AQUA-RAT-36103 | # Probability interview question
Suppose we have three positive integers $A, B, C$. We randomly choose an integer $a$ in the range $[0,A]$ and an integer $b$ in the range $[0,B]$. Find the probability that $a + b\leq C$.
I am unable to proceed in this question. Any hints .
• Well, what's C? – Newb Oct 3 '14 at 9:14
• The source where I read the question nothing has been given about C. It's just a positive integer – abkds Oct 3 '14 at 9:15
• We will have to take all possible cases – abkds Oct 3 '14 at 9:15
Assuming that each integer in given range has the same probability to be choosen, you have to count the cases when the inequality holds and divide it by all cases, namely: $$\frac{\sum_{a=0}^A\sum_{b=0}^B\mathbf{1}(a+b\leq C)}{(A+1)\cdot (B+1)}$$ where $\mathbf{1}(P)=1$ if proposition $P$ is true and $\mathbf{1}(P)=0$ otherwise.
• Shouldn't denominator be $(A+1).(B+1)$ ? Because zero is also included in the interval – abkds Oct 3 '14 at 9:20
• @AbhishekKDas Yes, and $a$ and $b$ should both start at $0$. However, the idea is correct, which of course is the main thing. – drhab Oct 3 '14 at 9:28
• If A and B are large (say around a million) you would obviously not count the numbers, but calculate them. At least you would for each a, 0 ≤ a ≤ A, calculate how many values 0 ≤ b ≤ B there are such that a + b ≤ C; that number would be min (B + 1, max (C - A + 1, 0)). – gnasher729 Oct 3 '14 at 10:08
The following is multiple choice question (with options) to answer.
Three people each took 3 tests. If the ranges of their scores in the 3 practice tests were 17, 28 and 35, what is the minimum possible range in scores of the three test-takers? | [
"a 17",
"b) 28",
"c) 35",
"d) 45"
] | B | It is finding the minimum range between all their scores, if all test taker scores are between 0 and maximum range we will have:
A- 0 and 17
B- 0 and 28
therefore the minimum range is 35, it cant be any lower however you play with the numbers.
B |
AQUA-RAT | AQUA-RAT-36104 | ## 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) | [
"0.5",
"1.5",
"2.5",
"1.0"
] | 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-36105 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 90 kmph and 72 kmph. After how much time will the trains meet? | [
"27/7 sec",
"20/9 sec",
"33/7 sec",
"21/7 sec"
] | B | They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (90 + 72)*5/18 = 9*5 = 45 mps.
The time required = d/s = 100/45 = 20/9 sec.
Answer : B |
AQUA-RAT | AQUA-RAT-36106 | python, beginner, algorithm, strings, regex
Test 8: 1e is an invalid number.
--------------------------------------------------
Test 9: e3 is an invalid number.
--------------------------------------------------
Test 10: 6e-1 is a valid number.
--------------------------------------------------
Test 11: 99e2.5 is an invalid number.
--------------------------------------------------
Test 12: 53.5e93 is a valid number.
--------------------------------------------------
Test 13: --6 is an invalid number.
--------------------------------------------------
Test 14: -+3 is an invalid number.
--------------------------------------------------
Test 15: 95a54e53 is an invalid number.
The following is multiple choice question (with options) to answer.
Find the invalid no.from the following series 13, 17, 22, 26, 31, 38 | [
"13",
"38",
"40",
"37"
] | B | The differences between two successive terms from the beginning are 4, 5, 4, 5, 4, 5. So, 38 is wrong.
Answer : B |
AQUA-RAT | AQUA-RAT-36107 | 2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)
3.
Amount of A = Concentration of A * Volume of mixture
Amount = C*V
( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt)
When you add water to this solution, the amount of milk does not change.
So Initial Conc * Initial Volume = Final Conc * Final Volume
$$C_i * V_i = C_f * V_f$$
$$C_f = C_i * (V_i/V_f)$$
In the question above, we find the final concentration of wine. Initial concentration $$C_i$$ = 1 (because it is pure wine)
When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt.
So $$C_f = 1 * 2/4$$
Since it is done 19 times, $$C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})$$
The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has $$V_f$$ as 22, you get amount of wine = 1/77 * 22 = 2/7 lt
Kudos +1 Karishma
Is there an fast way to compute the result of the multiplacation series like we have for $$Cf$$? I actually did the long way .
Re: A container has 3L of pure wine. 1L from the container is taken out an [#permalink] 06 Feb 2013, 22:54
Go to page 1 2 Next [ 30 posts ]
# A container has 3L of pure wine. 1L from the container is taken out an
The following is multiple choice question (with options) to answer.
In some quantity of ghee, 60% is pure ghee and 40% is vanaspati. If 10kg of pure ghee is added, then the strength of vanaspati ghee becomes 20%. The original quantity was? | [
"10",
"15",
"20",
"18"
] | A | Let the original quantity be x
then, vanaspati ghee in xkg = 40x/100 kg= 2x/5kg
(2x/5)/(x+10) = 20/100
2x/(5x+50) = 1/5
x = 10
Answer is A |
AQUA-RAT | AQUA-RAT-36108 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
What will come in place of the x in the following Number series? 25, 100, x, 1600, 6400 | [
"2345",
"3579",
"6400",
"6799"
] | C | (C)
25 x 4 = 100, 100 x 4 = 400, 400 x 4 = 1600, 1600 x 4 = 6400. |
AQUA-RAT | AQUA-RAT-36109 | # Sum and Product of n-positive integers
If I have $n$-positive integers, and I compute their sum and product, is there any different group of $n$-positive integers that will have the same sum and product?
For example, if $a,...,z$ denote 26 positive integers, and we define:
\begin{align} a+b+c+d+....+z &= \text{Sum} \\ a \cdot b \cdot c \cdot d \cdot .... \cdot z &= \text{Product} \end{align}
Is there any way I can get the same Sum and Product from a different group of 26 (in this example) positive integers?
EDIT:
A friend of mine pointed out that knowing that we have a group of 3 that works, we can show that it works for all positive groups of $n$ integers.
For Example: $\{3,3,10 \}$ and $\{2,5,9 \}$ both yield Sum $=16$ and Product $=90$.
Now we can just continually add a number (let's say 1) as the next integer to get multiple solutions for $n =4,5,6,...$.
Explicitly, $\{ 3,3,10,1 \}$ and $\{ 2,5,9,1 \}$ both give Sum$=17$ and Product$=90$.
The following is multiple choice question (with options) to answer.
If 6x = 8y = 10z, then what is a possible sum of positive integers x, y, and z? | [
"52",
"58",
"84",
"94"
] | D | 6x = 8y = 10z
3x = 4y = 5z
3(4*5) = 4(3*5) = 5(3*4)
Addition = 20+15+12 = 47
Answer would be multiple of 47 which is 94
Answer :D |
AQUA-RAT | AQUA-RAT-36110 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
2^(24) + 2^(25) + 2^(26) + ... + 2^(43) + 2^(44) = | [
"2^24(2^(21)-1)",
"2^24(2^(20)-1)",
"2^24(2^(23)-1)",
"2^21(2^(20)-1)"
] | A | 2^(24) + 2^(25) + 2^(26) + ... + 2^(43) + 2^(44) =
2^(24)*(1+2+4+...+2^(20)) =
2^(24)*(2^(21) - 1)
The answer is A. |
AQUA-RAT | AQUA-RAT-36111 | They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
David pays the simple interest on an amount he borrowed from his boss, if interest rate is halved, then simple interest is multiplied by | [
"2",
"1",
"1/4",
"1/2"
] | D | Formula for SI = (P*T*R)/100
SI = Simple Interest
P = Principle Amount
T = Time
R = Interest Rate
Say r=2 => SI1 = 2*SI
when r=1; SI2 = SI
SI2 = 1/2*SI1.
Answer : D |
AQUA-RAT | AQUA-RAT-36112 | Back
## A Test Question
Today, Pearl’s $9$ grandchildren are coming to visit! She loves to spoil them, so she opens her purse and finds $13$ dollar bills.
In how many different ways can Pearl distribute those dollars amongst her grandchildren? Keep reading to find out, or skip to today’s challenge for a similar problem.
As we’ll see, there are a lot of ways for Pearl to distribute her dollars! So, let’s start with a smaller example. Last week, Pearl’s $3$ favorite grandchildren visited, and at that time, she had $4$ dollar bills to give them. To visualize how they could be distributed, she laid them out in a row, along with some pencils to divide them into $3$ groups.
We’ll represent the dollars with stars $\large \star$ and divisions between groups with bars $\large{|}.$ One arrangement that Pearl found was $\large \star \; | \, \star \star \; | \; \star$ which represents $1$ dollar for the first grandchild, $2$ dollars for the second, and $1$ dollar for the third. Another arrangement was $\large \star \; | \: | \, \star \star \, \star$ which represents $1$ dollar for the first grandchild, $0$ dollars for the second, and $3$ dollars for the third.
To create $3$ groups, we need $2$ bars to separate the stars. So, to count the total number of arrangements into groups, we can count where in the line of stars and bars we can place those bars to define the groups.
The following is multiple choice question (with options) to answer.
Three people have $30, $72, and $98, respectively. If they pool their money then redistribute it among them, what is the maximum value for the median amount of money? | [
"$72",
"$85",
"$98",
"$100"
] | D | Solution -
Total money distributed is $200. In order to maximize median, one person has to accept $0 and remaining two people share $100 each. Hence median is $100.
ANS D |
AQUA-RAT | AQUA-RAT-36113 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area? | [
"40%",
"30%",
"50%",
"70%"
] | C | C
50%
Length is halved.
i.e., length is decreased by 50%
Breadth is tripled
i.e., breadth is increased by 200%
Change in area
=(−50+200−(50×200/100))%=50%
i.e., area is increased by 50% |
AQUA-RAT | AQUA-RAT-36114 | &\equiv2^3.3^8.7^5.9^3\cr &\equiv8.1.7.9\cr &\equiv4\ .\cr}
The following is multiple choice question (with options) to answer.
What is value of ((2^5)*(9^2))/((8^2)*(3^5))=? | [
"1/6",
"1/3",
"2/3",
"1/4"
] | A | =>((2^5)*((3^2)^2)/((2^3)^2)*(3^5))
=>((2^5)*(3^4))/((2^6)*(3^5))
=>1/(2*3)
=>1/6
Option A is answer |
AQUA-RAT | AQUA-RAT-36115 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train? | [
"120 metres",
"180 metres",
"324 metres",
"150 metres"
] | D | Explanation:
Speed= 60 x 5/18 m/sec = 50/3 m/sec.
Length of the train = (Speed x Time).
Length of the train = 50/3 x 9 m = 150 m.
ANSWER IS D |
AQUA-RAT | AQUA-RAT-36116 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, which A and C together can do it in 2 hours. How long will B alone take to do it? | [
"1/19",
"1/11",
"1/12",
"1/17"
] | C | A's 1 hour work = 1/4;
(B + C)'s 1 hour work = 1/3;
(A + C)'s 1 hour work = 1/2
(A + B + C)'s 1 hour work = (1/4 + 1/3) = 7/12
B's 1 hour work = (7/12 + 1/2) = 1/12
B alone will take 12 hours to do the work.
Answer: C |
AQUA-RAT | AQUA-RAT-36117 | Difference between revisions of "2014 AMC 10A Problems/Problem 17"
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$
Solution 1 (Clean Counting)
First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.
--happiface
Solution 2 (Bashy Casework)
Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.
The following is multiple choice question (with options) to answer.
Three 6 faced dice are thrown together. The probability that all the three show the same number on them is ? | [
"1/37",
"1/36",
"1/31",
"1/33"
] | B | It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36.Answer: B |
AQUA-RAT | AQUA-RAT-36118 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man’s speed with the current is 14 km/hr and the speed of the current is 2.5 km/hr. the man’s speed against the current is : | [
"9 km/hr",
"8.5 km/hr",
"10 km/hr",
"12.5 km/hr"
] | A | Solution
Man's rate in still water = (14-2.5) km/hr = 11.5 km/hr.
Man's rate against the current = (11.5-2.5) km/hr = 9 km/hr Answer A |
AQUA-RAT | AQUA-RAT-36119 | 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 motor cycles A & B are started from one point at 4 Kmph & 6 Kmph; After 45 min B starts returning, at what time they will meet ? | [
"54 mins",
"55 mins",
"56 mins",
"57 mins"
] | A | Distance covered by A in 45 mins= 3 kms.
Distance covered by B in 45 mins= 4.5 kms.
Distance between and B after 45 mins= 1.5 kms.
Relative speed when B is returning = 4+6=10 kmph
Distance between A and B is covered in 1.5/10 = 0.15 hrs = 9 mins
so they will meet again after 54 mins from starting time.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36120 | astrophysics, stars, hydrogen, fusion, stellar-physics
How big sphere do you need? If we start with hydrogen gas ($\mu=3.32\times 10 ^{-27}$ kg) at room temperature $T=300$ K and one atmosphere $\rho=0.0899$ kg/m$^3$ I get $\lambda_J=498,000$ km. That is however just 0.0000234 solar masses (about 7.8 earth masses), so it will be a dud. It is not the Jeans length that sets the necessary radius of your container.
You need at least 0.079 solar masses to get fusion, so we can estimate using $M=(4\pi/3)R^3 \rho$ that $R=\left((3/4\pi)M/\rho\right )^{1/3}$, in this case $R=7.343$ million km. Compared to actual star formation this is an extremely small radius, but one atmosphere hydrogen is also extremely dense and 300 K rather hot, so it is an odd case.
As a Dyson sphere this is tiny, just 0.05 AU in radius. But the real problem is gathering so much gas in space, where it is far, far less dense. You will run into cooling issues long before you get to the Jeans collapse since the ideal gas law $PV=nRT$ points out that if you decrease the volume $V$ the pressure must increase proportionally or the temperature $T$ will start going up instead - as you do work to compress the gas, resisting the pressure, you will heat it. The funny thing is that nature already is supplying us with pre-compressed stars to build Dyson spheres around.
The following is multiple choice question (with options) to answer.
How many shots of 1cm radius can be prepared from a sphere of 6cm radius? | [
"33",
"88",
"216",
"288"
] | C | 4/3 π * 6 * 6 * 6 = 4/3 π * 1 * 1 * 1 * x
x = 216
Answer: C |
AQUA-RAT | AQUA-RAT-36121 | Yes, those are exactly the equations you give. Now you need to solve those equations for a, b, and c, by the usual method of solving systems of equations: removing one variable at a time. For example, if you subtract the second equation from the third, (4a- 2b+ c)- (4a+ 2b+ c)= 49+ 3, both b and c cancel and giving -4b= 52. Dividing both sides by -4, b= -13. If you subtract the first equation from the second, (4a+ 2b+ c)- (a+ b+ c)= -3- 4, the "c" terms cancel giving 3a+ b= -7. Since b= -13, 3a+ b= 3a- 13= -7. Adding 13 to both sides of the equation, 3a= 6 so a= 2. Finally put a= 2, b= -13, in any one of the equations, say a+ b+ c= 4, to get 2- 13+ c= 4, -11+ c= 4 or c= 15. a= 2, b= -13, c= 15 gives $f(x)= 2x^2- 13x+ 15$.
The following is multiple choice question (with options) to answer.
a=2, b=3, c=6 Find the value of c/(a+b)-(a+b)/c | [
"8/30",
"9/30",
"10/30",
"11/30"
] | D | c/(a+b)-(a+b)/c
= (c/(a+b))-((a+b)/c)
= (6/(2+3))-((2+3)/6)
= (6/5)-(5/6)
= 11/30
ANSWER:D |
AQUA-RAT | AQUA-RAT-36122 | # 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.
At the recent downhill mountain bike race 4 entrants entered a challenging slalom event. Prem came first. The entrant wearing no 2 wore red. whereas kian did not wear yellow. the loser wore blue and rahul wore no 1.Sanjay beat rahul and the person who came second wore no 3.te entrant in yellow beat the entrant in green. Only one of the entrant s wore the same number and the final position. Determine who finished 3rd, the no and the color they wore? | [
"Rahul 1 green",
"rahul 2 red",
"Kiran 4 blue",
"Kiran 7 blue"
] | A | the names are prem, sanjay, rahul, kian
colors are red,yellow,green,blue
since prem came 1st and sanjay beat rahul so, given that sanjay came 2nd, wore no 3, yellow color
and rahul wore no 1, green color, came 3rd
hence,
name position wore t-shirt no color
prem 1 2 red
sanjay 2 3 yellow
rahul 3 1 green
kian 4 (1,2,3) blue
ANSWER:A |
AQUA-RAT | AQUA-RAT-36123 | is a free online tool that assists students to calculate the area of a parallelogram easily. b vector = 3i vector − 2j vector + k vector. Area Ar of a parallelogram may be calculated using different formulas. Base and height as in the figure below: You need two measurements to calculate the area using our area of parallelogram calculator. The area of a polygon is the number of square units inside the polygon. If diagonals are 59 cm, 50 cm and the side length is 12 cm. Please fill in the details below: INNER … - CLICK HERE ADVERTISE GET FREE CHATBOT . Multiply the obtained result with a sine function of angle to check the area. Here are some samples of Area of a Parallelogram calculations(Base and Height), Here are some samples of Area of a Parallelogram calculations(Sides and Angle), Here are some samples of Area of a Parallelogram calculations(Diagonals and Angle). Provide base, height or parallel sides length, angle or diagonals lengths and angle in the specified input fields and finally press on the calculate button to get the output i.e area in a short span of time. C = A Formulas, explanations, and graphs for each calculation. You can find all the details without any hassle by simply providing the side length or any other parameters metrics in the input fields of the calculator. Get the product and multiply it with the sine of the corner angle. Multiply those values to get the accurate answer. There are lots of options below. The leaning rectangular box is a perfect example of the parallelogram. 2. These calculators will be useful for everyone and save time with the complex procedure involved to obtain the calculation results. Also, the interior opposite angles of a parallelogram are equal in measure. Area of an Equilateral Triangle Calculator, Perimeter of an Equilateral Triangle Calculator, Area of an Irregular Quadrilateral Calculator, Perimeter of a Regular Polygon Calculator, Area of a Parallelogram when Base and Height is Given, Area of a Parallelogram when Sides Angle Between Them is Given, Area of a Parallelogram when Diagonals Angle Between Them is Given, Area of a Parallelogram base 66 cm height 75 cm, Area of a Parallelogram base 18 cm height 98 cm, Area of a Parallelogram base 97 ft
The following is multiple choice question (with options) to answer.
Find the area of a parallelogram with base 32 cm and height 18 cm? | [
"576 cm2",
"384 cm2",
"672 cm2",
"267 cm2"
] | A | Area of a parallelogram
= base * height
= 32 * 18
= 576 cm2
Answer: A |
AQUA-RAT | AQUA-RAT-36124 | # 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 the first of the year, James invested x dollars at Proudstar bank in an account that yields 0.5% in interest every quarter year. At the end of the year, during which he made no additional deposits or withdrawals, he had y dollars in the account. If James had invested the same amount in an account which pays interest on a yearly basis, what must the interest rate be for James to have y dollars at the end of the year? | [
"2.04%",
"4.12%",
"5%",
"8.25%"
] | A | Per quarter, interest = 0.5% So for a year, interest = 2% Due to quarter cumulation, effective yield (YTM) would be slight higher than 2% Answer = 2.04% =ANSWER:A |
AQUA-RAT | AQUA-RAT-36125 | combinatorics, sets, python
Note that we've used all $10$ elements from $S$. All sets $A,..,E$ have $6$ elements, so we just select one of them to be $S_1$, and arrange elements in the sets so that $S_1 = \{0,..,5\}$
The following is multiple choice question (with options) to answer.
A set consists of 15 numbers, all are even or multiple of 5. If 8 numbers are even and 10 numbers are multiple of 5, how many numbers is multiple of 10? | [
" 0",
" 1",
" 2",
" 3"
] | D | {Total} = {Even} + {Multiple of 5} - {Both} + {Nether}.
Since{Neither}=0(allare even or multiple of 5) then:
15 = 8 + 10 - {Both} + 0;
{Both}=3(so 1 number is both even AND multiple of 5, so it must be a multiple of 10).
Answer: D. |
AQUA-RAT | AQUA-RAT-36126 | # 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 purchased 70 kg of potatoes for Rs. 420 and sold the whole lot at the rate of Rs. 6.20 per kg. What will be his gain percent? | [
"8 1/7 %",
"2 1/3 %",
"3 1/3 %",
"8 1/8 %"
] | C | C.P. of 1 kg = 420/70 = Rs. 6
S.P. of 1 kg = Rs. 6.20
Gain % = 0.20/6 * 100 = 10/3
= 3 1/3 %
Answer:C |
AQUA-RAT | AQUA-RAT-36127 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
What should be the least number to be added to the 51234 number to make it divisible by 9? | [
"12",
"17",
"18",
"77"
] | C | Explanation:
The least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits.
Sum of digits = 15.
Nearest multiple of 9 greater than sum of digits = 18.
Hence 3 has to be added.
Answer: C |
AQUA-RAT | AQUA-RAT-36128 | Case 1: $$t\ge \frac{n}{2}$$:
For each $$i \in S$$, we must have a set containing the pair $$(t+1,i)$$, and as we can't have $$2$$ elements from $$S$$ together in another set, we must have $$t$$ new sets.
As $$t+2$$ can be in at most $$1$$ of the above sets, we need at least $$t-1$$ new sets for $$t+2$$ (to have the edges $$(t+2,i): i\in S$$). We can continue similarly until $$n=t+(n-t)$$ intoduces $$t-(n-t-1)$$ new sets.
We got a total of $$1+t(n-t)-(0+1+...+n-t-1)=1+t(n-t)-\frac{(n-t)(n-t-1)}{2}=$$ $$1+\frac{n-t}{2}(3t-n+1)$$ sets.
Treating $$t$$ as a variable and $$n$$ fixed, we can see that the function has a global maxima at $$t \sim \frac{2}{3}n$$ so it is enough to check for $$t=n-1, \frac{n}{2}$$.
For $$t=n-1$$ we get $$n$$ subsets, which is exactly enough.
For $$t=\frac{n}{2}$$ we get $$\ge \frac{n}{4}(\frac{n}{2}+1)$$. We want it to be $$\ge n$$ so we need $$\frac{n}{2} + 1 \ge 4$$ or $$n\ge 6$$.
The following is multiple choice question (with options) to answer.
A set of numbers has the property that for any number t in the set,t+2 is in the set. If -2 is in the set, which of the following must also be in the set? I. -3 II. 1 III. 4 | [
"II only",
"I,III only",
"III only",
"II,III only"
] | C | question is asking for must be there elements.
According to the question if t is there t+2 must be there.
if -2 is the starting element the sequence is as follows.
S ={-2,0,2,4,6....}
if -2 is not the starting element the sequence is as follows
S = {...-4,-2,0,2,4,6...}
By observing the above two sequences we can say that 4 must be there in set S.
Answer : C |
AQUA-RAT | AQUA-RAT-36129 | # 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.
After allowing a discount of 15% on the marked price, the selling price is Rs. 6800 for an article. If it was sold at marked price, there would have been a profit of 60%. The cost price of the article is? | [
"Rs.2989",
"Rs.5029",
"Rs.5000",
"Rs.5197"
] | C | Given SP = Rs. 6800
Marked price = [SP(100)]/(100 - d%) = (6800 * 100)/(100 - 15)
= Rs. 8000
If SP = Rs. 8000, profit
60%
CP = [SP(100)]/(100 + 60)
= (8000 * 100)/160
= Rs.5000
Answer: C |
AQUA-RAT | AQUA-RAT-36130 | # How can I find the minimum number of tiles?
Consider an $$n\times n$$ chessboard whose top-left corner is colored white. But Alice likes darkness, so she wants you to cover those white cells for her. The only tool you have are black L-shaped tiles each of which covers $$3$$ unit cells.
Formally, each tile covers unit cells satisfying the following:
1. Two of the cells are adjacent to the third (shares a side).
2. All three of the cells do not lie on the same row or same column.
3. No two tiles should overlap (cover the same cell) or go outside the board.
Since these tiles cost a lot, you have to cover all the white cells using the minimum number tiles.
### Example: $$1\times 1$$
Answer: Impossible, there's a single cell which is white. Since one tile needs $$3$$ empty cells, there's no way to cover this cell.
### Example: $$4\times 4$$
Answer: $$4$$ ($$4$$ tiles can be placed as shown)
### Example: $$7 \times 7$$
If each tile can be represented by a number, and each uncovered piece of board can be represented by 'zero', then the answer for a $$7 \times 7$$ board is $$16$$:
$$\begin{bmatrix} 16& 16& 15& 15& 14& 14& 13 \\ 16& 12& 15& 11& 14& 13& 13 \\ 12& 12& 11& 11& 10& 10& 9 \\ 8& 8& 7& 6& 10& 9& 9 \\ 8& 7& 7& 6& 6& 2& 2 \\ 5& 5& 4& 3& 3& 1& 2 \\ 5& 0& 4& 4& 3& 1& 1\\ \end{bmatrix}$$
### Question
For any given $$n$$, what will be the minimum number of tiles?
(Note: Answer exists for odd value of $$n \geq 7$$)
An image of the 7×7 solution as mentioned by OP:
And one possible way to extend this to a 9×9 solution (and further):
The following is multiple choice question (with options) to answer.
What is the least number of squares tiles required to pave the floor of a room 2 m 42 cm long and 1 m 76 cm broad? | [
"88",
"82",
"84",
"44"
] | A | Length of largest tile = H.C.F. of 242 cm and 176 cm = 22 cm.
Area of each tile = (22 x 22) cm2.
Required number of tiles =242 x 176/(22^2)= 88.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36131 | int <- .1/100 # annual interest rate of 0.1%
inf <- 2/100 # annual inflation rate 2%
n <- 10 # number of years
The following is multiple choice question (with options) to answer.
A money lender finds that due to a fall in the annual rate of interest from 8% to 7 ¾ % his yearly income diminishes by Rs.61.50. his capital is | [
"22378",
"3778",
"24600",
"27888"
] | C | Explanation:
Capital = Rs.x , then
x = 24600
Answer: C) Rs.24600 |
AQUA-RAT | AQUA-RAT-36132 | # 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 man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and black. How far is the place? | [
"2.89",
"2.88",
"2.82",
"2.84"
] | B | M = 6
S = 1.2
DS = 6 + 1.2 = 7.2
US = 6 - 1.2 = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
Answer:B |
AQUA-RAT | AQUA-RAT-36133 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is? | [
"272",
"111",
"822",
"150"
] | D | Let the length of the train be x meters and its speed be y m/sec.
They, x / y = 15 => y = x/15
x + 100 / 25 = x / 15
x = 150 m.
Answer:D |
AQUA-RAT | AQUA-RAT-36134 | # 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.
Indu gave Bindu Rs.2500 on compound interest for 2 years at 4% per annum. How much loss would Indu has suffered had she given it to Bindu for 2 years at 4% per annum simple interest? | [
"1",
"2",
"7",
"4"
] | D | 2500 = D(100/4)2
D = 4
Answer:D |
AQUA-RAT | AQUA-RAT-36135 | $y = x$ if $x \geq -1$.
It also means
$y = -(x + 1) - 1$ if $x + 1 < 0$, i.e. $x < -1$
$y = -x - 2$ if $x < -1$.
So all the points which satisfy these two linear equations will be your solution.
The following is multiple choice question (with options) to answer.
x = y - (1/y), where x and y are both > 0. If the value of y is doubled in the equation above, the value of x will | [
"decrease",
"stay the same",
"increase four fold",
"just more than double"
] | D | Alternate approach
Plug in some values of y > 0
Let y = 10
So, x = y - (1/y)
Or, x = 10 - (1/10) = 9.9
Let y = 20
So, x = y - (1/y)
Or, x = 20 - (1/20) =>19.95
So, x increases from 9.9 to 19.95
Hence the number increases 2 times, thus the correct answer will be (D) |
AQUA-RAT | AQUA-RAT-36136 | $\text{(ii) }\;1 - P(A\cap B\cap C) \;= \;0.88$ . . . . Right!
(b) (i) only the New York flight is full. (ii) exactly one of the three flights is full.
$\text{(i) }\;P(A \cap \overline{B} \cap \overline{C}) \;=\;(0.6)(0.5)(0.6) \;=\;0.18$
$\text{(ii)}\;\begin{array}{ccccc}P(A \cap \overline{B} \cap \overline{C}) & = & (0.6)(0.5)(0.6) & = & 0.18 \\
P(\overline{A} \cap B \cap \overline{C}) &=& (0.4)(0.5)(0.6) &=& 0.12 \\
P(\overline{A} \cap \overline{B} \cap C) &=& (0.4)(0.5)(0.4) &=& 0.08\end{array}$
$P(\text{exactly one full}) \;=\;0.18 + 0.12 + 0.08 \;=\;0.38$
The following is multiple choice question (with options) to answer.
A tourist purchased a total of $3,500 worth of traveler’s checks in $10 and $50 denominations, During the trip the tourist cashed 7 checks and then lost all of the rest. If the number of $10 checks cashed was one more or one less than the number of $50 checks cashed, what is the minimum possible value of the checks that were lost? | [
"$1,430",
"$2,310",
"$2,290",
"$3,270"
] | D | Let the number of $10 checks cashed beT
Let the number of $50 checks cashed beF
7 checks cashed;
T+F=7
Now; T can be F+1 OR T can be F-1
Let's check both conditions;
T=F+1
T+F=7
F+1+F=7
2F=6
F=3
T=4
Value cashed = 3*50+4*10=150+40=$190
Let's check the other condition as well;
T=F-1
T+F=7
F-1+F=7
2F=8
F=4
T=3
Value cashed = 4*50+3*10=200+30=$230
The more money he cashes, the less loss he incurs. Thus, we must consider the latter case.
Value cashed = $230
Value lost = 3000-230 = $3270
Ans:D |
AQUA-RAT | AQUA-RAT-36137 | First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.
Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$
Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$
And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$
So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
The following is multiple choice question (with options) to answer.
Dany bought a total of 20 game cards some of which cost $0.25 each and some of which cost $0.15 each. If Dany spent $4.2 to buy these cards, how many cards of each type did he buy? | [
"2",
"3",
"5",
"8"
] | D | Let X be the number of cards that cost $0.25 each and Y the number of cards that cost $0.15 each. The total number of cards is 20. Hence
X + Y = 20
If X is the number of cards at $0.25, then the X cards cost
0.25 X
If Y is the number of cards at $0.15, then the Y cards cost
0.15 Y
The total cost of the X cards and the the Y cards is known to be $4.2 and also given by
0.25 X + 0.15 Y = 4.2
We now need to solve the system of equations
X + Y = 20
0.25 X + 0.15 Y = 4.2
The first equation gives Y = 20 - X. Substitute y by 20 - x in the second equation and solve
0.25 X + 0.15 (20 - X) = 4.2
X(0.25 - 0.15) + 3 = 4.2
0.1 X = 1.2
X = 12 and Y = 20 - 12 = 8
correct answer D |
AQUA-RAT | AQUA-RAT-36138 | Difference in work = $$\frac{11}{18}$$ - $$\frac{7}{18}$$ = $$\frac{4}{18}$$
HTH
'
Aah, I understand! Thank you for you explanation
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Re: Elana was working to code protocols for computer processing. She did [#permalink]
### Show Tags
24 Oct 2016, 08:11
SamsterZ wrote:
Where do you get the number 4?
Its from here -
Attachment:
Capture.PNG [ 12.36 KiB | Viewed 1149 times ]
11 - 7 = 4
gauravk wrote:
I guess there is a typo you used the correct value in calculation but in the highlighted part it should be 18 units instead of 22.
Indeed it was a TYPO, I have edited my post...
I sincerely apologize for the same...
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Re: Elana was working to code protocols for computer processing. She did [#permalink]
### Show Tags
02 Nov 2016, 10:26
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? a)$252.00
b) $269.50 c)$369.00
d) $423.50 e)$693.00
The following is multiple choice question (with options) to answer.
What is vasu salary if salary of vasu is more than rajan salary working in same company i)vasu salary is 100 more than rajan salary. ii)rajan found 2000 allowns which is 50 less than vasu. (iii)basic salry of rajan is 1000. | [
"only i is required",
"i & ii is required",
"i& iii is required",
"i&ii&iii is required"
] | B | i and iii is enough
ANSWER:B |
AQUA-RAT | AQUA-RAT-36139 | astrophotography
if it is 13 billion light years away wouldn't it take 26 billion light years to take those pictures?
as if light years are a measure of time. A light year is a measure of distance, the distance light travels in a year in a vacuum.
The following is multiple choice question (with options) to answer.
The distance light travels in one year is approximately 5,850,000,000,000 miles. The distance light travels in 100 years is | [
"A) 587 × 108 miles",
"B) 587 × 1010 miles",
"C) 587 × 10-10 miles",
"D) 585 × 1012 miles"
] | D | The distance of the light travels in 100 years is:
5,850,000,000,000 × 100 miles.
= 585,000,000,000,000 miles.
= 585 × 1012 miles.
The answer is (D) |
AQUA-RAT | AQUA-RAT-36140 | face observed is already heads. find probability that a:heads appear on the second toss. There are three coins in a box. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. He selects one of the coins at random; when he flips it, it shows heads. 01, what is his posterior belief ? Note. Depending on which coin you have, there is a 50% chance that the other side is tails (regular coin) and a 50% chance that the other side is heads (two-headed coin). And if you roll a standard die, there’s a 1/6 probability that you’ll roll a six. This is, however, wrong, because given that heads came up, it is more likely that the two-headed coin was chosen. The probability that the coin is two-headed, given that it shows heads, is given by P (E1|A). If you pick the two headed coin, you have a 100% probability of getting three heads. A coin is randomly selected and ipped. (a) He selects one of the coins at random, and when he °ips it, it shows heads. In the case of the coins, we understand that there's a $$\frac{1}{3}$$ chance we have a normal coin, and a $$\frac{2}{3}$$ chance it's a two-headed coin. Two coins are available, one fair and the other two-headed. Given that a tail appears on the third toss, then the probability that it is the two-headed coin is 0, so the probability that it is the fair coin is 1 in this case. At the root (level 0) we choose randomly the first coin. a coin is selected at random and tossed. Figure:Probability of mother being carrier free, given n sons are disease free for n = 1(black), 2(orange),3(red),4(magenta), and5(blue), The vertical dashed line at p = 1=2is the case for the boxes, one with a fair coin and one with a
The following is multiple choice question (with options) to answer.
A box contains four nickel coins, of which two coins have heads on both their faces, one coin has tail on both its faces and the fourth coin is a normal one. A coin is picked at random and then tossed. If head is the outcome of the toss, then find the probability that the other face (hidden face) of the coin tossed is also a head. | [
"3/5",
"2/5",
"4/5",
"1/5"
] | C | There are four ways (2 heads in one two-headed coin and 2 heads in the other two-headed coin)
How many ways can you get heads from this entire set of coins? There are five ways (four as identified above and one from the normal coin)
Therefore probability = (number of ways to get heads from a two-headed coin)/(number of ways to get heads from any coin)
= 4/5
C |
AQUA-RAT | AQUA-RAT-36141 | # 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 60% more than the cost price. If a customer paid Rs. 1000 for a computer table, then what was the cost price of the computer table? | [
"750",
"650",
"625",
"115"
] | B | CP = SP * (100/(100 + profit%))
= 1000(100/160) = Rs. 625.
Answer: B |
AQUA-RAT | AQUA-RAT-36142 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
For each hour of production, a certain factory requires 1 assembly line worker for every 25 units to be produced in that hour. The factory also requires 12 managers regardless of the number of units to be produced. Which of the following expressions represents the total number of assembly line workers and managers that this factory requires to produce 75N in one hour, where N is a positive integer? | [
"12 + 50N",
"12 + 3N",
"62N",
"37N"
] | B | 25 units = 1 worker;
75N units = 75N/25 = 3N workers.
So, the answer is 3N workers plus 12 managers.
Answer: B. |
AQUA-RAT | AQUA-RAT-36143 | 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.
calculate the time required to become sum 2 times of itself at rate of 10%. | [
"5year",
"15year",
"10year",
"8year"
] | C | if sum become 2 time then simple interest equal to x.
x=x*10*t/100
t=100/10=10year
answer C |
AQUA-RAT | AQUA-RAT-36144 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
find 3 consecutive odd integers, all less than 100, such that the sum of their squares forms a Unidigit (that is, all its digits are the same). Thus, for integer abc, a^2 + b^2 +c^2 equals a Unidigital. | [
"39,43,45",
"43,45,47",
"45,47,49",
"41,43,45"
] | D | If numbers are x,x-2,x+2, then
x^2+(x+2)^2 =(x-2)^2 = x^2+2x^2+8 = 3x^2+8
Now 3x^2+8 should be unidigital.
checking different combinations , we get
x=43.
so numbers are 41,43,45
ANSWER:D |
AQUA-RAT | AQUA-RAT-36145 | 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 bridge of length 1500 m in 70 seconds and a lamp post on the bridge in 20 seconds. What is the length of the train in metres? | [
"375 m",
"750 m",
"250 m",
"600 m"
] | D | Let Length of Train = L
Case-1: Distance = 1500+L (While crossing the bridge)
Time = 70 Seconds
i.e. Speed = Distance / Time = (1500+L)/70
Case-2: Distance = L (While passing the lamp post)
Time = 20 Seconds
i.e. Speed = Distance / Time = (L)/20
But Since speed has to be same in both cases so
(1500+L)/70 = (L)/20
i.e. 1500 + L = 3.5 L
i.e. 2.5L = 1500
i.e. L = 600
Answer: option D |
AQUA-RAT | AQUA-RAT-36146 | 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 55 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"180 m",
"525 m",
"240 m",
"320 m"
] | B | Sol.
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x metres.
Then, x + 300 / 55 = 15 ⇔ x + 300 = 825 ⇔ x = 525 m.
Answer B |
AQUA-RAT | AQUA-RAT-36147 | => 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 salary of a typist was at first raised by 10% and then the same was reduced by 5%. If he presently draws Rs.1045.What was his original salary? | [
"28788",
"2788",
"1000",
"2877"
] | C | X * (110/100) * (95/100) = 1045
X * (11/10) * (1/100) = 11
X = 1000
Answer: C |
AQUA-RAT | AQUA-RAT-36148 | $$\frac{\theta-1}{\theta+1}$$
So when $$\theta = 3$$, this equals $$(3-1)/(3+1) = 1/2$$
We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.
The following is multiple choice question (with options) to answer.
1 + 1 ÷ {1 + 1 ÷ (1 + 1/3)} = ? | [
"1 1/3",
"1 4/7",
"1 1/8",
"1 2/3"
] | B | Answer
Given expression = 1 + 1 ÷ {1 + 1 ÷ (1 + 1/3)}
= 1 + 1 ÷ { 1 + 1 ÷ 4/3}
= 1 + 1 ÷ { 1 + 3/4}
= 1 + 1 ÷ 7/4
= 1 + 1 x 4/7
= 1 + 4/7
= 1 4/7
Correct Option: B |
AQUA-RAT | AQUA-RAT-36149 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
_________________
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Joined: 25 Feb 2013
Posts: 1059
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GPA: 3.82
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3400
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
### Show Tags
27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
If the sum of three consecutive even numbers is 48 more than the average of these numbers, then the largest of these numbers is? | [
"20",
"24",
"22",
"26"
] | D | Explanation:
Let the smallest of these number be x. The other two numbers are (x + 2) and (x + 4).
x + (x + 2) + (x + 4) = (X + (X+2) + (x+4)) / 3 + 48
3x + 3*(x + 2) + 3*(x + 4) = x + (x + 2) + (x + 4) + 144
9x + 18 = 3x + 150
6x = 132
x = 22
Therefore, the largest number is 26.
ANSWER D |
AQUA-RAT | AQUA-RAT-36150 | Player $C$ makes free throw shots with probability $P(A_j|C) = 0.7$, independently, so we have
\begin{align} P(A_1|C) &= P(A_2|C) = 0.7 \\ P(A_1 \cap A_2|C) &= P(A_1|C) P(A_2|C) = 0.49 \\ P(A_1 \cup A_2|C) &= P(A_1|C) + P(A_2|C) - P(A_1 \cap A_2|C) \\ &= 2 \times 0.7 - 0.49 \\ &= 0.91 \end{align}
And so we have our case where
\begin{align} P(A_1|B) = 0.8 &\gt P(A_1|C) = 0.7 \\ P(A_2|B) = 0.8 &\gt P(A_2|C) = 0.7 \\ \\ \text{ ... and yet... } \\ \\ P(A_1 \cup A_2|B) &\lt P(A_1 \cup A_2|C) ~~~~ \blacksquare \end{align}
The following is multiple choice question (with options) to answer.
A basketball player succeeds with 60% of her free throws. If she has a set of three free throws, what is the probability that she succeeds at least once? | [
"0.856",
"0.872",
"0.936",
"0.952"
] | C | P(missing all 3 free throws) = 0.4^3 = 0.064
P(succeeding at least once) = 1 - 0.064 = 0.936
The answer is C. |
AQUA-RAT | AQUA-RAT-36151 | (1\times3\times7\times9)^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon\equiv 9^{\left\lfloor\frac{n}{10}\right\rfloor}\times\epsilon$, where $\epsilon$ is either $1$, $3$ or $9$ depending on $n$ the last digit of $n$. These two reduction rules together with the modular reduction of $n\underset{5}{!}$ cut the size of $n$ in half, and repeated application will bring the size of $n$ down fairly fast. We have:
The following is multiple choice question (with options) to answer.
What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? | [
"814",
"820",
"842",
"844"
] | A | Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
Required number of tiles = (1517×902)/ (41×41) = 814
answer :A |
AQUA-RAT | AQUA-RAT-36152 | meters (m), centimeters (cm) & millimeters (mm). This packet covers all you need to know about sectors of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. A worksheet where you need to find the perimeter of sectors given the radius and angle of the arc. The perimeter of a rectangle is 40 cm. The portion of the circle's circumference bounded by the radii, the arc , is part of the sector. The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. Find the area of the sector. Information that I knew just from looking at the diagram that could prove useful. Find A(P-abc). If the angle is θ, then this is θ/2π the fraction of the full angle for a circle. Sector-A sector is a portion of a circle which is enclosed between its two radii and the arc adjoining them. (2) Given that r = 2√2 and that P = A, (b) show that θ = 2 1 The formula for the area of a sector is (angle / 360) x π x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. Videos, worksheets, 5-a-day and much more Formula: s=r x θ l=s+(2*r) Where, r = Radius of Circle θ = Central Angle s = Sector of Circle Arc Length l = Sector of Circle Perimeter Related Calculator: Perimeter of sectors. 15.8k VIEWS. Answer is: units. Simplifying expressions. A FULL LESSON on calculating the area and perimeter of a sector. Practice Question: Calculate perimeter of sector of circle for the following problems: N.B. Perimeter is the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of
The following is multiple choice question (with options) to answer.
The sector of a circle has radius of 21 cm and central angle 135o. Find its perimeter? | [
"91.5 cm",
"11.5 cm",
"91.8 cm",
"92.5 cm"
] | A | Perimeter of the sector = length of the arc + 2(radius)
= (135/360 * 2 * 22/7 * 21) + 2(21)
= 49.5 + 42
= 91.5 cm
Answer:A |
AQUA-RAT | AQUA-RAT-36153 | 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 jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. how much time does it take for the train to pass the jogger? | [
"46 seconds",
"36 seconds",
"38 seconds",
"26 seconds"
] | B | Distance = 240+ 120 = 360 m
speed = 36 km/hr = 36×10/36 = 10 m/s
Time = distance/speed = 360/10 = 36 seconds
answer : B |
AQUA-RAT | AQUA-RAT-36154 | 0.00698 acre1 acre = 43560 square feet304 sq ft * 1 acre/43560 sq ft = 0.00698 acre
### Number of sq ft per acre?
1 sq mi=5280ft x 5280ft = 27878400 sq ft/sq mi/ 27,878,400 sq ft/sq mi divided by 640acres/sq mi = 43,560 sq ft /acre 1 acre = 43,560 sq ft
### How many square feet are in 1.8 acre?
1 acre = 43,560 sq ft → 1.8 acre = 1.8 × 43,560 sq ft = 78,408 sq ft
### What does 0.2 acres equals sq feet?
1 acre = 43,560 sq ft .2 acre * 43560 sq ft = 8712 sq ft
### What percentage of acres in 160 feet x 180 feet?
160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.160 ft * 180 ft = 28800 sq ft = 66.1% of an acre.
### How many ft are in a acre?
There are 43559.66 sq ft in an acre.
### What is a fifth of an acre?
an acre is 43,560 sq ft 1/5th of that is 8,712 sq ft
### How many acres is 11026 sq ft?
43,560 sq ft = 1 acre11,026 sq ft = 0.2531 acre (rounded)
### Figure how many acres in 60000 sq ft?
(60,000 sq ft) / (43,560 sq ft / acre) = 1.3774 acre (rounded)
### What is the convertion of 435 acre to square feet?
1 acre = 43,560 sq ft → 435 acre = 435 x 43,560 sq ft = 18,948,600 sq ft
### How many sq ft for one acre 14 guntas in India?
The following is multiple choice question (with options) to answer.
What will be the cost of building a fence around a square plot with area equal to 289 sq ft, if the price per foot of building the fence is Rs. 56? | [
"3944",
"920",
"3808",
"6928"
] | C | Explanation:
Let the side of the square plot be a ft.
a2 = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 * 56 = Rs. 3808.
Answer: Option C |
AQUA-RAT | AQUA-RAT-36155 | \$7,382.94 3.2% 01.09.08 \$19.6879
\$7,402.63 3.2% 01.10.08 \$19.7404
\$7,422.37 3.2% 01.11.08 \$19.7930
\$7,442.17 3.2% 01.12.08 \$19.8458
The following is multiple choice question (with options) to answer.
After decreasing 24% in the price of an article costs Rs.684. Find the actual cost of an article? | [
"218",
"777",
"900",
"2688"
] | C | CP* (76/100) = 684
CP= 9 * 100 => CP = 900
Answer: C |
AQUA-RAT | AQUA-RAT-36156 | 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$
The following is multiple choice question (with options) to answer.
(3 x 10^5) + (2 x 10^3) + (4 x 10^2) = | [
"302400",
"32400",
"30240",
"3240"
] | A | When you multiply a number by 10 you just add a zero
=> 3 x 10^5 => add 5 zeros => 3 x 10^5 = 300000
=>2 x 10^3 = 2000
=> 4 x 10^2 = 400
=>3 x 10^5 + 2 x 10^3 + 4 x 10^2 = 300000 + 2000 + 400 = 302400
=> answer is A ( 302400) |
AQUA-RAT | AQUA-RAT-36157 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
There are C girls in a class, and their average height is M inches. In the same class, there are D boys with an average height of N inches. What is the average height of all the students in the class? | [
"(X + Y)/(M + N)",
"(M + N)/(X + Y)",
"(XM + YN)/(M + N)",
"(CM + DN)/(C+ D)"
] | D | Total height = CM+DN
No of students = C+D
Avg = (CM+DN)/(C+D)
Answer is D |
AQUA-RAT | AQUA-RAT-36158 | • In other words, if you have $x$ going from $0$ to $1$ and then for any fixed value of $x$ you have $y$ going from $0$ to $3(1-x),$ and that's the same as saying $y$ is between $0$ and $3,$ and for each fixed value of $y$ between $0$ and $3$ you have $x$ going from $0$ to $(3-y)/3. \qquad$ – Michael Hardy Dec 25 '18 at 19:37
The following is multiple choice question (with options) to answer.
When x is multiplied by 3, the result is 10 more than the result of subtracting x from 26. What is the value of x? | [
"-4",
"9",
"11",
"13"
] | B | The equation that can be formed is:
3x-10=26-x
or, 4x=36
or, x=9.
B answer. |
AQUA-RAT | AQUA-RAT-36159 | newtonian-mechanics, drag, relative-motion
Title: Relative motion-Acceleration My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.
Say a boat travels at 10 m/s in a water channel.
the water speed relative to ground is 0.
so the boat travels at 10 m/s relative to the ground.
now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
As after a while the boat speed relative to ground has become 0,
then from the ground-based observer's point of view, the boat has undergone a deceleration.
My question is;
Is this deceleration always necessarily equal to minus the water acceleration?
In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?
PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?
My question is; Is this deceleration always necessarily equal to minus the water acceleration?
The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:
$$f=kv\qquad \text{ for low speed}\\
f=kv^2\qquad \text{ for high speed}$$
where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.
The following is multiple choice question (with options) to answer.
A boat can move upstream at 50 kmph and downstream at 100 kmph, then the speed of the current is? | [
"10",
"45",
"25",
"40"
] | C | US = 50
DS = 100
M = (100- 50)/2
= 25
Answer:C |
AQUA-RAT | AQUA-RAT-36160 | orbit
Title: International dateline In the story of Magellan in which they circled the globe ,returning to Portugal to discover that they "lost" a day. As they were traveling westward. I don't get it. They were going so slowly in those times. Took 3 years. I get that they, were traveling west so the day was slightly shorter for them. But to end up exactly 1 day different? Puzzles me. The speed of travel does not matter. The amount of rotations around Earth's axis matters when it comes to counting the amount of sunrises or sunsets on a travel around the Earth.
Assume that you are travelling once around the world. You depart westward, and it takes you 24 hours to once travel around the world. You depart 1 hour before Sunrise, and you stay the whole time of the trip 1 hour before sunrise. When you are back on your original location, the people who stayed there will tell you that a whole day has passed with sunrise and sunset and the next sunrise is about to happen in an hour. So you "lost a day". You didn't really lose it - you also lived 24 hours. But you did experience one sunrise less than people staying on a fixed position on the globe due to the fact that you did one complete rotation against the direction of Earth's rotation. Would you have travelled East, you would have experienced two sunrises in that timeframe. You lose or gain a day only, if you count day as "sunrise to sunset" as that depends on the amount of rotations around Earth's axis you perform.
The argument stays true regardless of the speed of travel: if you travel West, you will have one sunrise less than a person remaining on the same place; and if you travel East you will experience one more.
The following is multiple choice question (with options) to answer.
A person takes 5hrs 45 min in walking to a place and ridingback. He would have gained 2hrs by riding both ways. The time he would take to walk both ways is? | [
"5 hours 45 min",
"5 hours 55 min",
"6 hours 45 min",
"7 hours 45 min"
] | D | ----------------------------------------------------------------
Solution 1
----------------------------------------------------------------
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
D |
AQUA-RAT | AQUA-RAT-36161 | However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.
Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.
And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
8. Originally Posted by mr fantastic
My mistake. Yes there was confusion.
However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.
Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.
And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
It would depend on the marginal costs of providing a greater range (including consideration of the profit margin on the product). It is often supprising how large the impact of a 12.5% increase in some parameter in a product design can sometimes be (often because of their impact on other aspects of design).
I know of at least one product with some design parameters that could not be increased by even 5% without an additional development cost in the order of 10's if not 100's of millions of pounds!
CB
The following is multiple choice question (with options) to answer.
The price of the sugar rise by 25%. If a family wants to keep their expenses on sugar the same as earlier, the family will have to decrease its consumption of sugar by | [
"25%",
"20%",
"80%",
"75%"
] | B | Solution: Let the initial expenses on Sugar was Rs. 100.
Now, Price of Sugar rises 25%. So, to buy same amount of Sugar, they need to expense,
= (100 + 25% of 100) = Rs. 125.
But, They want to keep expenses on Sugar, so they have to cut Rs. 25 in the expenses to keep it to Rs. 100.
Now, % decrease in Consumption,
(25/125)*100 = 20%.
Mind Calculation Method;
100-----25%↑---→125------X%↓---→100.
Here, X = (25/125)*100 = 20%.
Answer: Option B |
AQUA-RAT | AQUA-RAT-36162 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
How many integers between 300 and 1100 are there such that their unit digit is odd? | [
"200",
"300",
"400",
"500"
] | C | There are 800 numbers from 301 to 1100 (inclusive).
Half of the numbers are odd, so there are 400 odd numbers.
The answer is C. |
AQUA-RAT | AQUA-RAT-36163 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:50 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train X’s speed is r miles per hour, and Train Y’s speed is s miles per hour, how many hours after 1:00 PM, in terms of p, r, and s, do the two trains pass each other? | [
"0.5 + (p - 0.5s)/(r + s)",
"(p - 0.5s)/(r + s)",
"0.5 + (p - 0.5r)/r",
"(p - 0.5r)/(r + s)"
] | A | The distance A is going to cover between 1:00 and 1:30
= .5r
now the distance between the two trains = (p-.5r)
the relative velocity = (r-(-s)) = r+s
From 1:30, time is going to take when they meet = (p-.5r)/(r+s)
số thẻ ans is .5+((p-.5r)/(r+s)) [.5 is added for the time from 1:00 tờ 1:30]
ans is A |
AQUA-RAT | AQUA-RAT-36164 | # Remainder on division with $22$
What is the remainder obtained when $$14^{16}$$ is divided with $$22$$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $$22$$, except the remainder?
How should I proceed?
• $14^{16} = (22 - 8)^{16}$ – ab123 Jul 3 '18 at 10:37
• $8^{16} = 64^8 = (66 - 2)^8$ – ab123 Jul 3 '18 at 10:41
• @ab123 Why not make an answer of that? – Arthur Jul 3 '18 at 10:41
You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,
$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$
Since $$14^2 \equiv -2$$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$
or
$$14^2 = 22k -2$$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$
A method that uses FLT.
Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.
The following is multiple choice question (with options) to answer.
On dividing number by 56, we get 29 as remainder. On dividing the same number by 8, what will be the remainder? | [
"4",
"5",
"6",
"7"
] | B | Let x be the number and y be the quotient.
Then, x = 357 * y + 39
= (17 * 21 * y) + (17 * 2) + 5
= 17 * (21y + 2) + 5.
Required remainder = 5.
ANSWER:B |
AQUA-RAT | AQUA-RAT-36165 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
In 10 years, A will be twice as old as B was 10 years ago. If A is now 7 years older than B the present age of B is | [
"37",
"38",
"39",
"40"
] | A | let present age of A be a and B be b
a+10 = 2*(b-10)
=> 2b-a = 30 ......(i)
a = b+7
=>2b-b-7 = 30
b=37
so the present age of b is 37 years
ANSWER:A |
AQUA-RAT | AQUA-RAT-36166 | # Venn Diagram Problem
• October 16th 2009, 07:34 AM
fifthrapiers
Venn Diagram Problem
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?
My solution:
I let x = number of students in both the club and fencing team and A = the number of students on the fencing team
Then, the number of students in the club is 70 - x
And then we know:
A + (70-x) + 20 - x = 100
This means A = 10
So now I used the same formula (Total # = A + B - (A^B) + neither)
100 = 10 + (70-x) - x + 20
But this didn't work... why isn't this formula working?
For 3 sets, Total # = A + B + C - A^B - A^C - B^C + (AUBUC) + "others"
So I think my formula for 2 sets is right..
The answer is meant to be 61 but plugging into that formula doesn't work:
10 + 9 -61 + 20 =! 100
Help?
• October 16th 2009, 08:08 AM
Plato
Quote:
Originally Posted by fifthrapiers
In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?
Think of it this way. There are 80 students in French or fencing.
Of those 70 are in French. So there must be 10 in fencing that are not in French.
But the number in fencing must be more than 70.
So what in the minimum in both?
• October 16th 2009, 09:49 AM
Soroban
Hello, fifthrapiers!
Quote:
In a group of 100 students, more students are in Fencing than in French.
If 70 are in French and 20 are neither in Fencing nor French,
what is the minimum number of students who could in both actitvities?
I placed the information into a chart . . .
The following is multiple choice question (with options) to answer.
In a class of 60 students 41 are taking French, 22 are taking German. Of the students taking French or German, 9 are taking both courses. How many students are not enrolled in either course? | [
"6",
"15",
"24",
"33"
] | A | Formula for calculating two overlapping sets:
A + B - both + NOT(A or B) = Total
so in our task we have equation:
41 (french) + 22 (german) - 9 (both) + NOT = 60
54 + NOT = 60
NOT = 60 - 54 = 6
So answer is A |
AQUA-RAT | AQUA-RAT-36167 | Continue with what we have left off, we could divide through the equation [MATH]u\left(\frac{2}{u^2-9}+\frac{2}{u^2-4}\right)=u[/MATH] by $u$ when $u\ne 0$ to look for the rest of the four solutions of the system.
[MATH]\frac{2}{u^2-9}+\frac{2}{u^2-4}=1[/MATH]
[MATH]\frac{2(u^2-4)}{u^2-9}+\frac{2(u^2-9)}{u^2-4}=1[/MATH]
[MATH]2(u^2-4)+2(u^2-9)=(u^2-4)(u^2-9)[/MATH]
[MATH]4u^2-26=u^4-13u^2+36[/MATH]
[MATH]u^4-17u^2+62=0[/MATH]
Solving the equation for $u^2$ by using the quadratic formula, we see that:
[MATH]u^2=\frac{17\pm\sqrt{(-17)^2-4(1)(62)}}{2}=\frac{17\pm\sqrt{41}}{2}[/MATH]
[MATH]u=\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]
Back substituting $u=x-4$ into the above equation we therefore get the rest of the solutions for the system:
[MATH]x-4=\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]
[MATH]x=4\pm \sqrt{\frac{17\pm\sqrt{41}}{2}}[/MATH]
Thus,
The following is multiple choice question (with options) to answer.
Which of the following equations is NOT equivalent to 4u^2 = y^2 - 9 ? | [
" 4u^2 + 9 = y^2",
" 4u^2 - y^2 = -9",
" 4u^2= (y + 3)(y - 3)",
" 2u = y - 3"
] | D | 4X^2 = Y^2 - 9
The basic rule when dealing with equations is that you can do ANYTHING to both sides of an equation as long as you do it EQUALLY to BOTH sides. Everything gets a bit more complex when you're dealing with variables in the denominator of a fraction and/or inequalities, but neither of those subjects is a factor in this prompt.
Looking at Answer D, we have....
2u = Y - 3
While you COULD take the square root of both sides of the original equation, it's important to note that the square root of Y^2 - 9 is NOT (Y-3).
Y^2 - 9 can be factored into (Y-3)(Y+3), but neither of these parentheses is the square root of Y^2 - 9.
You can see the proof when you square either of the two parentheses:
(Y-3)^2 = Y^2 - 6u + 9
(Y+3)^2 = Y^2 + 6u + 9
Thus, D is NOT equivalent to the prompt. |
AQUA-RAT | AQUA-RAT-36168 | So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$.
To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39
The following is multiple choice question (with options) to answer.
A used cars salesman receives an annual bonus if he meets a certain quota. This year, the salesman has so far sold 1/5 of last year's quota. If this year's quota is reduced by 25 percent from last year, the quantity that he still needs to sell this year in order to receive the annual bonus is what fraction of the preceding year's car quota? | [
"4/15",
"4/20",
"3/4",
"11/20"
] | D | Let Last year Quota be = 100x
This year Quota = 100x - 25%of100x = 75x
Salesman has so far sold = 1/5(100x) = 20x
Quantity still need to be sold = 75 x - 20x = 55x
Question :: What fraction of 100x is 55x = 55x/100x = 11/20.
Ans:: D |
AQUA-RAT | AQUA-RAT-36169 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
If the average marks of three batches of 56, 60 and 45 students respectively is 50, 55, 60, then the
average marks of all the students is | [
"54.48",
"54.68",
"54.65",
"54.58"
] | C | Explanation:
(56×50)+(60×55)+(45×60)/(56+60+46)=54.65
Answer: Option C |
AQUA-RAT | AQUA-RAT-36170 | 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
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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 toy store's revenue in November was 3/5 of its revenue in December and its revenue in January was 3/4 of its revenue in November, then the store's revenue in December was how many times the average (arithmetic mean) of its revenues in November and January? | [
" 1/4",
" 1/2",
" 2/3",
" 2"
] | D | Let Dec rev =100
Then Nov rev is 3/5 (100) => 60
Therefore Jan rev = 3/4(Nov rev) = 3/4(60) => 45
Hence Dec rev = x*( Nov rev+Jan rev)/2
100 = x* (60+45)/2
x = 100/52.5 => 1.90=2
Ans) D |
AQUA-RAT | AQUA-RAT-36171 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
The average age of three boys is 20 years and their ages are in the proportion 2 : 3 : 5. The age of the youngest boy is: | [
"21 years",
"18 years",
"15 years",
"9 years"
] | B | Total age of 3 boys = (20 × 3) years = 60 years. Ratio of their ages = 2 : 3 : 3.
Age of the youngest = (60 × 3â„10) years = 18 years.
Answer B |
AQUA-RAT | AQUA-RAT-36172 | combinatorics, sets, matching
Put your students in a row: $a!$ possibilities. The first $b$ students form the first group, the next $b$ students the second group, etc. The order in the line-up between each group of students does not matter, so we have to divide by th1s number of possibilities which is $(c!)^b$. Again we divide by $(a-bc)!$ the number of permutations of the not chosen ones. Also, the order of the groups themselves does not matter, we additionally divide by $b!$.
Answer: $a!\;/\;{(c!)^b (a-bc)! \;b!}$
Another way of obtaining the same number. Choose a group of $c$ students from total $a$, then a group of $c$ students from the remaining $c-b$, etc. This can be done in ${a \choose c}{a-c\choose c}\dots{a-c(b-1)\choose c}$. Again I divide by the possible orderings of the groups which is $b!$.
Thus $\frac{a!}{c!(a-c)!}\frac{(a-c)!}{c!(a-2c)!}\dots \frac{(a-(b-1)c)!}{c!(a-bc)!} / b!$
Your first example $a=12$, $b=4$, $c=3$. You have 880, this number is 15400.
Second example $a=12$, $b=3$, $c=2$. 13680, we agree.
The following is multiple choice question (with options) to answer.
Each student at a certain business school is assigned a 3-digit student identification number. The first digit of the identification number cannot be zero, and the last digit of the identification number must be prime. How many different student identification numbers can the school create? | [
"9,000",
"3,600",
"2,700",
"360"
] | D | The identification number is of the form _ _ _ _
1. First digit cannot be 0
2. Middle digits can be anything
3. Last digit has to be prime - 2, 3, 5, 7
We can have the following number of possibilities for each space
__ __ __ __
9 10 4
Total cases = 360
Answer : D |
AQUA-RAT | AQUA-RAT-36173 | # Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$
Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $$\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$$ where $$r-\frac{s}{2}$$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $$\frac{\pi r^2}. {2}$$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($$2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$$). In conclusion the resultant equation is (with $$r=x$$ and $$s=y$$):
$$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$$
The following is multiple choice question (with options) to answer.
The radius of the two circular fields is in the ratio 3: 5 the area of the first field is what percent less than the area of the second? | [
"84%",
"24%",
"54%",
"64%"
] | D | r = 3 πr2 = 9
r = 5 πr2 = 25
25 π – 16 π
100 ---- ? => 64%
Answer: D |
AQUA-RAT | AQUA-RAT-36174 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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$$.Physics
The following is multiple choice question (with options) to answer.
A train 125 m long passes a man, running at 15 km/hr in the same direction in which the train is going, in 10 sec. The speed of the train is? | [
"40",
"50",
"60",
"79"
] | C | Speed of the train relative to man = 125/10 = 25/2 m/sec.
= 25/2 * 18/5 = 45 km/hr
Let the speed of the train be x km/hr. Then, relative speed = (x - 15) km/hr.
x - 15 = 45 => x = 60 km/hr.
Answer: Option C |
AQUA-RAT | AQUA-RAT-36175 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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 moves at average speed of 40kmph reaches its destination on time.what its average speed becomes 35kmph , then it reaches its destination 15 late. find the length of the journey? | [
"70km",
"85km",
"38km",
"14km"
] | A | Difference between timings=15 mis=1/4hr
let the length of the journey be x km.
then x/35-x/40=1/4
8x-7x=70
x=70km
Answer (A) |
AQUA-RAT | AQUA-RAT-36176 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a sports club with 30 members, 16 play badminton and 19 play tennis and 2 do not play either. How many members play both badminton and tennis? | [
"7",
"8",
"9",
"10"
] | A | Let x play both badminton and tennis so 16-x play only badminton and 19-x play only tennis. 2 play none and there are total 30 students. hence,
(16-x)+(19-x)+x+2=30
37-2x+x=30
37-x=30
x=7
So 7 members play both badminton and tennis.
A |
AQUA-RAT | AQUA-RAT-36177 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
If (x/y)=(8/5), find the value (x^2+y^2)/(x^2-y^2) | [
"89/39",
"59/11",
"51/77",
"41/11"
] | A | = (x^2+y^2)/(x^2-y^2) = ( x^2 /y^2+ 1)/ ( x^2 /y^2-1) = [(8/5)^2+1] / [(8/5)^2-1]
= [(64/25)+1] / [(64/25)-1] = 89/39
Answer is A. |
AQUA-RAT | AQUA-RAT-36178 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A person can row at 10 kmph and still water. He takes 5 hours to row from A to B and back. What is the distance between A and B if the speed of the stream is 2 kmph? | [
"32 km",
"25 km",
"28 km",
"24 km"
] | D | Let the distance between A and B be x km.
Total time = x/(10 + 2) + x/(10 - 2) = 5
=> x/12 + x/8 = 5 => (2x + 3x)/24 = 5 => x = 24 km.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36179 | # Is “divisible by 15” the same as “divisible by 5 and divisible by 3”?
Is stating that a number $x$ is divisible by 15 the same as stating that $x$ is divisible by 5 and $x$ is divisible by 3?
-
The two assertions are equivalent. – André Nicolas Feb 1 '12 at 23:38
may be worth noting for somebody who might be new to this, that it happens to work in both directions here, because 3 and 5 are co-prime. i.e. if you ask the same question for a different example you may not get the converse implication. – Beltrame Feb 1 '12 at 23:49
If you want to figure out the general facts at work here, you could read this Keith Conrad handout. – Dylan Moreland Feb 1 '12 at 23:55
Yes, if a number $n$ is divisible by $15$, this means $n=15k$ for some integer $k$. So $n=5(3k)=3(5k)$, so it is also divisible by $3$ and $5$.
Conversely, if $n$ is divisible by $3$ and $5$, it is a simple lemma that it is divisible by the least common multiple of $3$ and $5$. Since $3$ and $5$ are coprime, their lcm is just $15$.
-
I feel that this answer is incomplete. It does not prove that x|n and y|n implies that lcm(x,y)|n; it just claims it's "a simple lemma". A better answer would show why this is true. – user22805 Feb 2 '12 at 8:32
Suppose there's a number n such that x|n and y|n, but lcm(x,y) does not divide n. Then write m = lcm(x,y) and n = pm+q, where 0 < q < m and p is an integer. Then x and y must both divide q, so m is not the lcm of x and y - contradiction. – user22805 Feb 2 '12 at 8:35
The following is multiple choice question (with options) to answer.
The integer x is divisible by both 15 and 25. Which of the following must be an integer? | [
"x/11",
"x/75",
"x/84",
"x/36"
] | B | Prime factorization of 15 = 5*3
Prime factorization of 25= 5*5
LCM of 15 and 25 = 5^2 * 3 = 75
Therefore x/75 must be an integer
Answer B |
AQUA-RAT | AQUA-RAT-36180 | &\equiv2^3.3^8.7^5.9^3\cr &\equiv8.1.7.9\cr &\equiv4\ .\cr}
The following is multiple choice question (with options) to answer.
The H.C.F of 2^4*3^2*5^3*7, 2^3*3^3*5^2*7^2 and 3*5*7*11 is? | [
"105",
"1155",
"2310",
"27720"
] | A | H.C.F= Product of lowest power of factors=3*5*7=105
Correct Option: A |
AQUA-RAT | AQUA-RAT-36181 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Four brothers Adam, Bill, Charles and David together contributed certain amount of money and purchased a car. The contribution made by Adam is half that of the total contribution made by the other three brothers, contribution made by Bill is one-fourth of the total contribution made by the other three brothers and the contribution made by Charles is two-third of the total contribution made by the other three brothers. If the total price of the car is $9000, find the contribution made by David. | [
"$540",
"$580",
"$600",
"$660"
] | C | Given that A:(B+C+D)= 1:2.
Hence A contributed 1/3rd of the total price.
Given B:(A+C+D)=1:4
Hence B contributed 1/5th of the total price.
Given C:(A+B+D)=2:3
Hence C contributed 2/5th of the total price.
Thus the contribution made by A,B and C= 1/3+1/5+2/5=14/15
So, the remaining 1/15th of the price is contributed by David.
Hence David's contribution= 1/15×9000=$600.
The answer is C. |
AQUA-RAT | AQUA-RAT-36182 | r
#DECILE 5
t_prag_top_total_income_50<-decili_total_income_neto\[5,1\]
t_prag_top_total_income_filter_50<-filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50)
t_prag_top_total_income_filter_50_tax<-sum(t_prag_top_total_income_filter_50$TAX)
t_tax_share_50<-((t_prag_top_total_income_filter_50_tax)/ZBIR_TOTAL_TAX)*100
t_prag_top_total_income_filter_50<-sum(t_prag_top_total_income_filter_50$NET_INCOME)
t_prag_top_total_income_filter_50a<-nrow(filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50))
t_prag_top_total_income_50b<-((t_prag_top_total_income_filter_50)/ZBIR_TOTAL_NET_INCOME)*100
FINAL_CENTILE_TABLE50<-data.frame(cbind(t_prag_top_total_income_50,t_prag_top_total_income_filter_50,t_prag_top_total_income_filter_50a,t_prag_top_total_income_50b,t_prag_top_total_income_filter_50_tax , t_tax_share_50))
colnames(FINAL_CENTILE_TABLE50)<-c("Decile threshold","Total income in the decile","Number of persons in the centile","Share of the decile in total income (%)","Tax","Share tax(%)")
FINAL_DECILE_TABLE <- rbind(FINAL_DECILE_TABLE, FINAL_CENTILE_TABLE50)
The following is multiple choice question (with options) to answer.
The salary of A, B, C, D, E is Rs. 8000, Rs. 5000, Rs. 16000, Rs. 7000, Rs. 9000 per month respectively, then the average salary of A, B, C, D, and E per month is | [
"Rs. 7000",
"Rs. 8000",
"Rs. 8500",
"Rs. 9000"
] | D | Answer
Average salary
= 8000 + 5000 + 16000 + 7000 + 9000 / 5
= Rs. 9000
Correct Option: D |
AQUA-RAT | AQUA-RAT-36183 | # Given an Alphabet, how many words can you make with these restrictions.
I'm trying to understand from a combinatoric point of view why a particular answer is wrong. I'm given the alphabet $\Sigma = \{ 0,1,2 \}$ and the set of 8 letter words made from that alphabet, $\Sigma_8$ . There are $3^8 =6561$ such 8 letter words.
How many words have exactly three 1's?
How many words have at least one each of 0,1 and 2?
In the first question I reasoned that first I choose $\binom{8}{3}$ places for the three 1's. Then I have 5 place left where I can put 0's and 2's which is $2^5$. Since I can combine each choice of 1 positions with every one of the $2^5$ arrangements of 0's and 2's then I get $\binom{8}{3}\cdot 2^5 = 1792$ which is correct.
I tried applying the same reasoning to the second question and got $\binom{8}{3}\cdot 3^5 = 13608$ which is obviously wrong.
Was my reasoning sound in the first question or did I just happen to get the correct answer by chance? If it is sound, why doesn't it work with the second question?
-
Why was this question marked down, especially more than two years after it was asked? – Robert S. Barnes Mar 15 '14 at 17:43
The following is multiple choice question (with options) to answer.
How many words can be formed by using all letters of the word ‘COMPUTER’ ? | [
"32240",
"36180",
"40320",
"45860"
] | C | The word contains 8 different letters.
The number of permutations is 8! = 40320
The answer is C. |
AQUA-RAT | AQUA-RAT-36184 | Last edited: Mar 7, 2016
2. Mar 7, 2016
### SteamKing
Staff Emeritus
Is this a 25 gram burger or a 250 gram burger?
A 25 gram burger is about 1 bite.
Also, is the consumption rate $100 - (\frac{1}{900}m^2)$ or $(100 - \frac{1}{900m^2})$
3. Mar 7, 2016
### chwala
consumption rate is $100-(1/900)m^2$
4. Mar 7, 2016
### SteamKing
Staff Emeritus
What about the initial size of the burger - 25 grams or 250 grams?
5. Mar 7, 2016
### chwala
the initial burger is 250grams. it is eaten (all of it) until it becomes 0 grams.
6. Mar 7, 2016
### SteamKing
Staff Emeritus
After this point, you have some errors in your integration for the second partial fraction term.
7. Mar 8, 2016
### chwala
I have seen the error:
$t=1.5 ln u - 1.5 ln (600-u) + k$
eventually
$t = 1.5 ln 300 - 1.5 ln 300 - 1.5 ln 11$
$t = -3.6min$
why do we have a negative? is it taking care of the fact that the burger was initially whole and with every bite it kept on decreasing in mass value until the point
$m=0$ ?
8. Mar 8, 2016
### Samy_A
$m$ is the mass eaten:
9. Mar 8, 2016
### chwala
am not getting you $dt/dm=1.5/(300+m)-1.5/(300-m)$
and which is equal to
$1.5/u+1.5/(300-(u-300)$ =
$1.5/u-1.5/(600-u)$ =
$900/(u(600-u))$ =
$900/(300+m)(300-m)$ =
$900/(90000-m^2)$ as indicated in my original post. my question still is why do we have $-3.6$minutes?
10. Mar 8, 2016
### Samy_A
The following is multiple choice question (with options) to answer.
A side of beef lost 35 percent of its weight in processing. If the side of beef weighed 560 pounds after processing, how many pounds did it weigh before processing? | [
" 191",
" 355",
" 737",
" 861"
] | D | Let weight of side of beef before processing = x
(65/100)*x = 560
=> x = (560 * 100)/65 = 861
Answer D |
AQUA-RAT | AQUA-RAT-36185 | $\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$
There are 12 in which the sum is greater than 9 or there is at least one 6.
Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$
Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$
And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$
You can expect to win an average of $\frac{1}{3}$ of a point per throw.
. . . as Plato already pointed out.
.
5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)
RonL
The following is multiple choice question (with options) to answer.
In a game of 90 points, A can give B 20 points and C 25 points. then how many points can B give C in a game of 80? | [
"1.71",
"6.71",
"5.71",
"7.71"
] | C | In a game of 90 points,A give B 20 points and c 25 points means B=70,C=65
In 70 points B give 5 since 70-65=5
In 80 points B gives=80*5/70=5.71
ANSWER:C |
AQUA-RAT | AQUA-RAT-36186 | 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.
In how much time will a train of length 100 m, moving at 36 kmph cross an electric pole? | [
"17 sec",
"18 sec",
"19 sec",
"10 sec"
] | D | Convert kmph to mps. 36 kmph = 36 * 5/18 = 10 mps.
The distance to be covered is equal to the length of the train.
Required time t = d/s
= 100/10
= 10 sec.
Answer: D |
AQUA-RAT | AQUA-RAT-36187 | The pattern emerges and the final answer could be written as
x = (1/39)[4(10^d) - 16]
where is an integer and d > 0, which represents the number of digits in x.
For d = 5 we obtain our answer. You still have to try d for 1 through 4, but it's still easier to do it this way.
It literally took me 30 seconds to derive this expression and had the answer in no time.
What if the question was asking for the next smallest number that satisfies this condition? Yeah... try doing it the other way. For that situation d = 11, so you'd have to do it many time. That number is 10,256,410,256.
See a pattern?
The following is multiple choice question (with options) to answer.
If (10^4 * 3.456789)^9 is written as a single term, how many digits would be to the right of the decimal place? | [
"6",
"12",
"18",
"32"
] | C | 3.456789^9 has 6*9 = 54 decimal places.
10^36 moves the decimal place to the right 36 places.
(10^4 * 3.456789)^9 has 54-36 = 18 digits after the decimal point.
The answer is C. |
AQUA-RAT | AQUA-RAT-36188 | 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 560 m in length crosses a telegraph post in 16 seconds. The speed of the train is? | [
"126 kmph",
"77 kmph",
"54 kmph",
"71 kmph"
] | A | S = 560/16 * 18/5 = 126 kmph
Answer:A |
AQUA-RAT | AQUA-RAT-36189 | # 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.
A certain car dealership sells economy cars, luxury cars, and sport utility vehicles. The ratio of economy to luxury cars is 6:2. The ratio of economy cars to sport utility vehicles is 5:3. What is the ratio of luxury cars to sport utility vehicles? | [
"9:8",
"8:9",
"6:2",
"2:3"
] | B | The ratio of economy to luxury cars is 6:2 --> E:L = 6:2 = 30:10.
The ratio of economy cars to sport utility vehicles is 5:3 --> E:S = 5:3 = 30:18.
Thus, L:S = 10:18 = 5:6.
Answer: B. |
AQUA-RAT | AQUA-RAT-36190 | As an example, the encoding of $$42$$ consists of $$k=\lfloor \log_2 (44) \rfloor = 5$$ digits (since $$2^5=32$$ and $$2^6=64$$). These digits are $$01100$$ (notice the leading $$0$$) since $$(42-32+2)_{10} = (12)_{10} = (1100)_2$$.
To decode a number, you can do the reverse: count the number $$k$$ of digits in the encoded version, then add together $$2^{k}-2$$ and the number whose binary representation corresponds to the given digits. This also requires $$O(k)=O(\log n)$$ time.
For example, the encoded number $$01100$$ has $$5$$ digits, so the represented integer is $$(2^5 -2)_{10} + (01100)_2 = (30 + 12 )_{10} =(42)_{10}$$.
From shortlex to natural:
• add $$(10_b)^l-10_b$$ to the binary value, where $$l$$ stands for the length (number of bits).
E.g. $$011_s\to l=3\to11_b+1000_b-10_b=1001_b=9$$.
From natural to shortlex:
• the length is $$l=\lfloor\log_2(n+2)\rfloor$$,
• subtract $$2^l-2$$,
• fill left with zeroes.
E.g. $$9\to l=\lfloor\log_2(11_d)\rfloor=3\to9-2^3+2=3=11_b\to011_s$$
The following is multiple choice question (with options) to answer.
if log 5 = 0.69897, the number of digits in 5128 is | [
"90",
"39",
"88",
"28"
] | A | Explanation:
log(5128) = 128log(5) = 128 × 0.69897 ≈ 89.46
ie, its characteristic = 89
Hence, number of digits in 2128 = 89+1 = 90
Answer: Option A |
AQUA-RAT | AQUA-RAT-36191 | # 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.
Suganya and Suriya are partners in a business. Suganya invests Rs. 35,000 for 8 months and Suriya invests Rs.42,000 for 10 months. Out of a profit of Rs.31,570. Suganya's share is | [
"Rs.9471",
"Rs.12,628",
"Rs.18,040",
"Rs.18,942"
] | C | Solution
Ratio of their shares =(35000×8):(42000×10)
= 2 : 3.
Suganya's share = Rs.(31570 ×2/5)
= Rs.12628.
Answer C |
AQUA-RAT | AQUA-RAT-36192 | homework-and-exercises, kinematics
Title: rectlinear motion with constant acceleration Friends, this is a numerical homework problem. I tried my best to solve it but my answer is not matching with the one given at the back of the text book. Please help me out:
A motor car moving at a speed of 72 km/h can come to a stop in 3 seconds, while a truck can come to a stop in 5 seconds. On a highway, the car is positioned behind the truck, both moving at 72 km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it doesn't collide with the truck. The typical human response time is 0.5 sec.
My logic and answer: since car can decelerate to a stop much faster than the truck, it only need to worry about human response time which is 0.5sec. car would cover 10m in 0.5seconds at a speed of 72 km/h. so it just need to be 10m behind the truck minimum.
but the answer in the book is 1.25 m
How is this possible? You are missing the fact that the truck is still moving forwards during its decelleration interval.
The following is multiple choice question (with options) to answer.
Due to construction, the speed limit along an 5-mile section of highway is reduced from 52 miles per hour to 36 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | [
"A) 3.12",
"B) 8",
"C) 10",
"D) 2.57"
] | D | Old time in minutes to cross 5 miles stretch = 5*60/52 = 5*15/13 = 5.76
New time in minutes to cross 5 miles stretch = 5*60/36 = 5*5/3 = 8.33
Time difference = 2.57
Ans:D |
AQUA-RAT | AQUA-RAT-36193 | 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 is running with a speed of 60 kmph and its length is 110 metres. Calculate the time by which it will pass a man running opposite with speed of 6 kmph | [
"2 second",
"4 second",
"6 second",
"8 second"
] | C | Explanation:
From the given question,
we will first calculate the speed of train relative to man,
=> ( 60 + 6 ) = 66 km/hr ( we added 6 because man is running opposite )
Convert it in metre/second
=66×5/18
=55/3m/sec
Time it will take to pass man
=110×3/55
=6 seconds
ANSWER IS C |
AQUA-RAT | AQUA-RAT-36194 | 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 driver covers a certain distance by car driving at 50 km/hr and returns back to the starting point riding on a scooter at 10 km/hr. What was the average speed for the whole journey? | [
"12.5 km/h",
"14.8 km/h",
"16.7 km/h",
"17.5 km/h"
] | C | time 1 = d / 50
time 2 = d / 10
total time = d/50 + d/10 = 3d/25
average speed = total distance / total time = 2d / (3d/25) = 16.7 km/h
The answer is C. |
AQUA-RAT | AQUA-RAT-36195 | now subtract digit from it
$75800000-758=75799242$
this is the required ans.
sunitha naidu
2016-02-17 05:21:37
$758(100000-1)=75800000-758=75799242$
0 0
hari
2013-09-18 09:12:48
$1.~23458×999=\\2.~786×99=$
SANJAY PAUL
2015-09-15 12:19:30
Just quick trick.
$786×99$
First take last $2$ digit: $86$
$100-86=14$
Write down at right: $14$
Then add $1$ to first digit: $1+7=8$
Then $786-8=778$
Right answer is $77814$
0 0
123Next
The following is multiple choice question (with options) to answer.
When 20 is subtracted from a number, it reduces to seven-twelve of the number. What is the sum of the digit of the number | [
"40",
"44",
"46",
"48"
] | D | Explanation:
Let the number be x. Then,
x - 20 = 7x/12 => x - 7x/12 = 20
5x/12 = 20 => x = 48
Answer: Option D |
AQUA-RAT | AQUA-RAT-36196 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
A student earned a grade of 80% on a math test that had 20 problems. How many problems on this test did the student answer correctly? (round to the nearest whole number) | [
"36 problems",
"53 problems",
"27 problems",
"16 problems"
] | D | Multiply the opposites:
80/100=x/20
80 x 20 = 1600
Divide by the remaining number:
16/1600
*100
16 problems
correct answer D |
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