source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36497 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
Three candidates contested an election and received 1130, 7630 and 15620 votes respectively. What percentage of the total votes did the winning candidate get? | [
"57%",
"60%",
"64%",
"90%"
] | C | total votes are 24380 ( sum of votes obtained by 3 candidates)
percentage of the total votes recd by the winning candidate = 100*15620/24380 = 64%
ANSWER:C |
AQUA-RAT | AQUA-RAT-36498 | # 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 least value should be replaced by * in 2551112* so the number become divisible by 6 | [
"3",
"4",
"5",
"6"
] | B | Explanation:
Trick: Number is divisible by 6, if sum of all digits is divisible by 3 and 2, so (2+5+5+1+1+1+2+*) = 17+* should be divisible by ,
17+1 will be divisible by 3,
but we can't take this number because 1 is not dividable by 2(2 only dividable by those numbers who contain even number at last position)
so that least number is 4.
Answer: Option B |
AQUA-RAT | AQUA-RAT-36499 | # Calculate the number of ways to paint $5$ buildings with $4$ colours such that all $4$ colours must be used
A developer has recently completed a condominium project in a valley. There are blocks of buildings $A$, $B$, $C$, $D$ and $E$ as shown in the diagram below.
The developer has colours available to paint the buildings. Each block can only be painted using a single colour. Find the number of ways to paint all the blocks if all $4$ colours must be used.
My attempt: We have $5$ ways to choose $4$ buildings with $4$ different colours. Among the $4$ buildings, we have $4!$ ways to paint them using $4$ different colours.
So my answer is $120$. However, the answer given is $240$. What is my mistake?
• Um, what do paint the fifth building? Don't you have 4 chooses for that? I'd get that you method out to give 4*120 =480. Which is also wrong as we double counted. We'd choose a building to be a duplicate color. there are 5 choices for that. We'd paint the other four. 24 four that. We'd pick one of the four buildings to duplicate the fifth one. There are four choices of that. Which ever building we choose to duplicate is a double counting as we could have choosen that building to be a duplicate. so divide by two. – fleablood May 12 '17 at 15:24
You're on the right track. After choosing the four buildings with different colours, what about the fifth? It will be a repeated colour, and there are four colours to choose from. However, we will have overcounted by a factor of $2$, so the final answer will be $$120\cdot\frac42 = 240$$ To explain the overcounting, suppose that you first choose $A,B,C,D$ as the set of four buildings. Then $E$ is the same as one of them, say $A$. But then we will also later consider $E,B,C,D$ as the set of four, with $A$ the same as $E$.
The following is multiple choice question (with options) to answer.
A new condo development is offering buyers the option to choose from 4 different counter top styles, 5 floor types and 4 colors of paint. In order to keep costs to the minimum, the buyer would like to choose from the lower cost options. In how many ways can 3 styles of counter tops, 2 colors floor types, and 3 colors of paint be selected from the given group? | [
"112 ways",
"99 ways",
"80 ways",
"64 ways"
] | C | The number of ways of selecting three styles of counter tops, two floors and three colors of paint is:
= ⁴C₃ * ⁵C₂ * ⁴C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (5 * 4)/(2 * 1) * (4 * 3)/(3 * 2 * 1)
= 4 * 10 * 2
= 80 ways.
Answer: C |
AQUA-RAT | AQUA-RAT-36500 | => 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.
A certain sum of money is divided among A, B and C such that A gets one-third of what B and C together get and B gets two-seventh of what A and C together get. If the amount received by A is $30 more than that received by B, find the total amount shared by A, B and C. | [
"$1080",
"$1180",
"$1280",
"$1380"
] | A | A = 1/3 (B+C) => C = 3A - B ---(1)
B = 2/7 (A+C) => C = 3.5 B - A --(B)
A-B = $30
A = 30+B
(1)===> C = 90+3B - B = 2B+90 ==> 2B-C = -90 ---(3)
(2)===> C = 3.5 B - B-30 = 2.5B-30==>2.5B-C = 30---(4)
from (4) and (3) 0.5B = 120
B = $240
A= $270
C =810-240=$570 Total amount = 270+240+570 = $1080
Answer: A |
AQUA-RAT | AQUA-RAT-36501 | +0
# Melody
0
437
3
+644
How many of the integers in the collection {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} are relatively prime to 13?
waffles Aug 23, 2017
#1
+90023
+1
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} are relatively prime to 13?
Since 13 is prime, they are all relatively prime to 13 except for 1
CPhill Aug 23, 2017
#3
+1
13 isn't.
Guest Aug 23, 2017
#2
+20025
+1
How many of the integers in the collection {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} are relatively prime to 13?
Using the notation gcd(m,n) to denote the greatest common divisor, two integers and are relatively prime if gcd(m,n) = 1.
The following is multiple choice question (with options) to answer.
How many prime numbers are between 51/13 and 89/9? | [
"0",
"1",
"2",
"3"
] | C | 51/13= 4-
89/9= 10-
Prime numbers between 4 and 10 are 5 and 7
- sign signifies that the number is marginally less .
Answer C |
AQUA-RAT | AQUA-RAT-36502 | 4. ## Re: find length of rectangle given diagonal and area
Originally Posted by Bonganitedd
Rectangle has area=168 m^2 and diagonal of 25. Find length
This is how tried to attempt the problem
Area= L X W
168 = L x W ..........(1)
L^2 + W^2 =25^2 ............(2)
From (1) L = 168/W...........(3)
Substitute (3) into (2)
(168/W)^2 +W^2 = 625
28224/W^2 + W^2 = 625
The problem gets complicated as I proceed
Is this aproach correct if it is,
Is there a convinient method
Have a look at this webpage.
5. ## Re: find length of rectangle given diagonal and area
Hello, Bonganitedd!
Rectangle has area=168 m^2 and diagonal of 25. Find the length.
This is how tried to attempt the problem
$\text{Area} \:=\: L\cdot W \:=\:168 \quad\Rightarrow\quad L \,=\,\frac{168}{W}\;\;[1]$
$L^2 + W^2 \:=\:25^2\;\;[2]$
$\text{Substitute [1] into [2]: }\;\left(\frac{168}{W}\right)^2 +W^2 \:=\:625 \quad\Rightarrow\quad \frac{28,\!224}{W^2} + W^2 \:=\: 625$
Is this approach correct? . Yes
If it is, is there a convinient method?
We have: . $\frac{28,\!224}{W^2} + W^2 \:=\:625$
Multiply by $W^2\!:\;\;28,\!224 + W^4 \:=\:625W^2 \quad\Rightarrow\quad W^4 - 625W^2 + 28,\!224 \:=\:0$
The following is multiple choice question (with options) to answer.
If the area of a square with sides of length 4 centimeters is equal to the area of a rectangle with a width of 4 centimeters, what is the length of the rectangle, in centimeters? | [
"14",
"8",
"12",
"4"
] | D | Let length of rectangle = L
4^2 = L*4
=> L = 16/4 = 4
Answer D |
AQUA-RAT | AQUA-RAT-36503 | • It $2 = 100 - 7\times 14$ (which it does) then $$2 \times ANYTHING = (100 -7\times 14)ANYTHING = 100\times ANYTHING - 7\times 14\times ANYTHING$$. And if $$ANYTHING = 142857 = \frac {1000000 - 1}7$$ then $$2\times 142857 = 100\times 142857 - 7\times 14\times\frac {1000000 -1}7 = 100\times 142857 - 14\times 1000000 + 14$$. $$=\color{green}{14}\color{red}{2857}\color{purple}{00} - \color{green}{14}\color{red}{0000}\color{purple}{00} + \color{purple}{14}= \color{red}{2857}\color{purple}{00}$$That's all. Sep 22 at 20:11
• Yes I got confused, sorry about that
– Jim
Sep 23 at 20:50
The following is multiple choice question (with options) to answer.
If 144/0.144 = 14.4/x, then the value of x is: | [
"0.144",
"1.44",
"0.0144",
"0.014"
] | C | =144/0.144 = 14.4/x
=144 x 1000/144
=14.4/x
=x = 14.4/1000
= 0.0144
Answer is C. |
AQUA-RAT | AQUA-RAT-36504 | Thanks
• Your "40% overlap" would mean taht the surface overlaping is 40% of the surface of which rectangle ? – Evargalo Sep 28 '17 at 16:13
• I want to know if, say, 40% of the other rectangle is contained inside the self rectangle – John Lexus Sep 28 '17 at 16:14
For the equations, I will let the left of the first rectangle be $l_0$, the right be $r_0$, the top $t_0$ and the bottom $b_0$. The second rectangle is $l_1, r_1,$ etc. Their areas will be $A_0$ and $A_1$.
If the boxes don't overlap, obviously the percentage overlap is $0$.
If your boxes are found to be colliding, simply use this formula to calculate the area that is overlapping:
$$A_{overlap} = (\max(l_0, l_1)-\min(r_0, r_1))\cdot(\max(t_0, t_1)-\min(b_0, b_1)).$$
Now there are two ways to calculate a percentage error that could make sense for your explanation. If we're just checking the percentage within the first rectangle (or "self" in your program), the percent overlap is simple: $$P_{overlap} = \frac{A_{overlap}}{A_{self}}.$$
If you want the percentage to be equal whether it's calculated from either rectangle, the equation you're looking for is: $$\frac{A_{overlap}}{A_0+A_1-A_{overlap}}.$$
The following is multiple choice question (with options) to answer.
How much 45% of 60 is greater than 35% of 40? | [
"18",
"13",
"15",
"17"
] | B | (45/100) * 60 – (35/100) * 40
27 - 14 = 13
Answer: B |
AQUA-RAT | AQUA-RAT-36505 | Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:37
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:53
Bunuel wrote:
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!
this might clear my entire probability confusion i hope...
Kudos [?]: 16 [0], given: 17
Math Expert
Joined: 02 Sep 2009
Posts: 43322
Kudos [?]: 139450 [1], given: 12790
The following is multiple choice question (with options) to answer.
For a certain alarm system, each code is comprised of 5 digits, but no digit can be used more than twice. How many codes can be made? | [
"30,240",
"60,480",
"91,440",
"98,240"
] | C | Case 1: All digits distinct
Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240
Case 2: Two digits are same, other 3 are distinct e.g. 45722
Select a digit which is to be repeated in 10C1 ways, select other 3 digits in 9C3 ways and arrange them in 5!/2! ways to get 10C1 x 9C3 x 5!/2! = 50400 ways
Case 3: Two digits repeated and one other digit e.g. 33448
Select 2 digits to be repeated in 10C2 ways, select the single digit in 8C1 ways and arrange them all in 5!/2!.2! ways to get 10C2 x 8C1 x 5!/2!.2! = 10800
Sum of all = 30240 + 50400 + 10800 = 91440 ways
ANS:C |
AQUA-RAT | AQUA-RAT-36506 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
There are five newly appointed teachers in a college. The weight of first teacher is 200 kg and the weight of the 2nd teacher is 20% higher than the weight of the third teacher, whose weight is 25% higher than the first teacher's weight. The fourth teacher at 350 kgs is 30% lighter than the fifth teacher. Find the average weight of all the five teachers. | [
"310",
"320",
"240",
"260"
] | B | Explanation :
Weight of first teacher = 200 kg
Weight of 3rd teacher = 250kg
=> Weight of 2nd teacher = 300 kg
Weight of 4th teacher = 350 kg
=> Weight of 5th teacher = 350/0.7 = 500 kg
Average weight = 1600/5 = 320 kg.
Answer : B |
AQUA-RAT | AQUA-RAT-36507 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
A professional athlete was offered a three-year contract to play with Team K that provided for an annual salary of $100,000 in the first year, an increase in annual salary of 20% over the previous year for the next two years, and a bonus of $50,000 on signing. Team L offered a three-year contract providing for an annual salary of $150,000 in the first year, an increase in annual salary of 20% over the previous year for the next two years, and no signing bonus. If he accepts the offer of Team L and fulfills the three-year contract terms, the athlete will receive how much more money by choosing Team L over Team K ? | [
"$132,000",
"$50,000",
"$82,500",
"$92,000"
] | A | Team K's contract = $100,000 + $100,000*1.2 + $100,000*1.2*1.2 + $50,000 = $414,000
Team L's contract = $150,000 + $150,000*1.2 + $150,000*1.2*1.2 = $546,000
The difference = $132,000.
Answer: A. |
AQUA-RAT | AQUA-RAT-36508 | What length of string will make 4 complete revolutions in a length of 12 inches around the cylinder of radius $\frac{2}{\pi}$?.
$t_{0}=8{\pi}$, because we have 4 rev.
a=radius of cylinder = $\frac{2}{\pi}$
$c=\frac{12}{8\pi}=\frac{3}{2\pi}$
$\boxed{8{\pi}\sqrt{\left(\frac{2}{\pi}\right)^{2}+ \left(\frac{3}{2\pi}\right)^{2}}=20}$
The following is multiple choice question (with options) to answer.
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high? | [
"10 cm",
"12 cm",
"15 cm",
"13 cm"
] | D | d2 = 122 + 42 + 32 = 13
ANSWER:D |
AQUA-RAT | AQUA-RAT-36509 | I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:
There are 7 ways in which Team 1 can play with another team.
Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent.
But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)
So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28
Hope this was useful!
Best Regards
Japinder
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Re: There are 8 teams in a certain league and each team plays each of the [#permalink]
### Show Tags
25 May 2016, 22:25
1
Attached is a visual that should help.
Attachments
Screen Shot 2016-05-25 at 9.41.57 PM.png [ 75.15 KiB | Viewed 78729 times ]
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The following is multiple choice question (with options) to answer.
The table below shows how many coaches work with each of the major sports teams at Kristensen School. Although no single coach works with all three teams, 3 coaches work with both the Track and Tennis teams, 2 coaches work with both the Track and Baseball teams, and 1 coach works with both the Tennis and Baseball teams. How many different coaches work with these three teams?
Sports No of coaches
Track 7
Tennis 5
Baseball 4 | [
"6",
"10",
"11",
"12"
] | B | X = 7+ 5 + 4 - (3 + 2 + 1) = 10
Answer is B |
AQUA-RAT | AQUA-RAT-36510 | # 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.
Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 5 times his original investment. What is the value of n? | [
"50 years",
"25 years",
"12 years 6 months",
"37 years 6 months"
] | A | Explanatory Answer
Let us say Braun invested $100.
Then, at the end of 'n' years he would have got back $500.
Therefore, the Simple Interest earned = 500 - 100 = $400.
We know that Simple Interest = (Principal * number of years * rate of interest) / 100
Substituting the values in the above equation we get 400 = (100 * n * 8) / 100
Or 8n = 400
Or n = 50 years.
correct choice is (A) |
AQUA-RAT | AQUA-RAT-36511 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
Of the 20 guests at a party, 10 percent are women. How many men must leave in order to bring the number of women up to 50 percent of the guests? | [
"18",
"14",
"16",
"4"
] | C | Now, there are 2 women (10% of 20) and 20-2=18 men at a party. We need there to be 2 women and 2 men, in this case there will be 50% women and 50% men.
Therefore 18-2=16 men must leave.
Answer: C. |
AQUA-RAT | AQUA-RAT-36512 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
The entire contents of a full sack of flour can be used to make 15 cupcakes and 10 pizzas. The same full sack of flour can be used to make 7 cupcakes and 14 pizzas. If a full sack of flour is used to make only pizzas, how many pizzas can be made? | [
"10",
"21 1/2",
"18 3/4",
"19"
] | B | 15x+10y=7x+14y
x=1/2y. Answer B |
AQUA-RAT | AQUA-RAT-36513 | # Find the least next N-digit number with the same sum of digits.
Given a number of N-digits A, I want to find the next least N-digit number B having the same sum of digits as A, if such a number exists. The original number A can start with a 0. For ex: A-> 111 then B-> 120, A->09999 B-> 18999, A->999 then B-> doesn't exist.
You get the required number by adding $9$, $90$, $900$ etc to $A$, depending on the digits of $A$.
First Case If $A$ does not end in a row of $9$s find the first (starting at the units end) non-zero digit. Write a $9$ under that digit and $0$s under all digits to the right of it. Add the two and you get $B$.
Example: $A=3450$. The first non-zero digit is the 5 so we write a $9$ under that and a $0$ to its right and add:
\begin{align} 3450\\ 90\\ \hline 3540 \end{align}
There is a problem if the digit to the left of the chosen non-zero digit is a $9$. In this case we write a $9$ under that $9$ and $0$s to its right. And if there are several $9$ we put our $9$ under the highest one.
Example: $A=3950$. The first non-zero digit is the 5 but there is a $9$ to its left. We write a $9$ under that $9$ instead and $0$s to its right and add:
\begin{align} 3950\\ 900\\ \hline 4850 \end{align}
Second case If $A$ does end in a row of $9$s write a $9$ under the highest of the row of $9$s and $0$s under all digits to the right of it. Add the two and you get $B$. As you say, if $A$ is entirely $9$s there is no solution.
The following is multiple choice question (with options) to answer.
A number consists of two digits. If 3/5 of 1/5 of the number is 9. Find the sum of its two digits? | [
"28",
"189",
"17",
"12"
] | D | Explanation:
x * 3/5 * 1/5 = 9
x = 75 => 7 + 5 = 12
Answer: D |
AQUA-RAT | AQUA-RAT-36514 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Set S is the prime integers between 0 and 35. If three numbers are chosen randomly from set S, what is the probability that the sum of these three numbers is odd? | [
"15/56",
"3/8",
"15/28",
"8/11"
] | D | Total number of combinations=11C3=165................(1)
Total number of combinations including 2 and two other numbers=10C2=45......................(2)
Therefore the difference ie. (1)-(2) is the set of 3 primes without 2=(165-45)=120
So, probability=120/165=8/11
D |
AQUA-RAT | AQUA-RAT-36515 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
In wardrobe Karan has 3 trousers. One of them is grey the 2nd blue and the 3rd brown. In his wardrobe he also has 4 shirts. One of them is grey and the other 3are white. He opens his wardrobein the dark and pick out one shirttrouser pair without examination the color. What is the likelihood that neither the shirt nor the trouser is grey? | [
"1/2",
"2/3",
"4/5",
"6/11"
] | A | Probability that trouser is not grey = 2/3
Probability that shirt is not grey = 3/4
∴ Required probability = 2/3 x 3/4 = 1/2
A |
AQUA-RAT | AQUA-RAT-36516 | inorganic-chemistry, extraction
I want to use household vinegar - 6% apple vinegar if possible because it is easily available. Also, I want to use 3% pharmacy grade $\ce{H2O2}$ for the same reasons.
I have done this, kind of successfully, but I want to know what is the chemistry behind all that, and how would you calculate the amounts of reagents needed? If you want to save the iodine $\ce{I_2}$ from its solution, you may add enough $\ce{KOH}$ or $\ce{NaOH}$ in the mixture to transform all $\ce{I_2}$ into a colorless mixture of $\ce{KI}$ and $\ce{KIO_3}$. $$\ce{3I_2 + 6OH^- -> IO_3^- + 5I^- + 3H_2O}$$The remaining ethanol can then be evaporated by heating the nearly colorless solution to a temperature higher than $80$°C. When the temperature reaches $100$°C, the ethanol is totally removed from the solution : the hot solution can be cooled down to room temperature, and some acid should be added in order to destroy the excess of $\ce{NaOH}$ ou $\ce{KOH}$, and then to recover the iodine $\ce{I_2}$ according to $$\ce{IO_3^- + 5 I^- + 6 H^+ -> 3 I_2 + 3 H_2O}$$ The totality of the iodine $\ce{I_2}$ dissolved in the original tincture is recovered, without any amount of ethanol, and can be saved by filtration.
The following is multiple choice question (with options) to answer.
If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a five-percent vinegar solution, what was the concentration of the original solution? | [
"19.3%",
"25.8%",
"16.67%",
"15.5%"
] | B | Let X be the quantity of non-vinegar in the strong vinegar solution
Thus vinegar quantity will be 12 - X
When 50 ounces of water were added the percentage of vinegar becomes 5%, thus (12 - X)/62 = 5/100
From this equation X = 8.9
Answer (12-8.9)/12 = 25.8%
ANSWER:B |
AQUA-RAT | AQUA-RAT-36517 | # 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.
Lucy deposited $62500 in an investment fund that provided 24 percent annual return compounded half yearly.If she made no other transactions with the fund, in how much time, in years, did her investment earn a total interest of $10400? | [
"0.5",
"1",
"3",
"6"
] | B | A = P + I = 62500 + 10400 = 72900
72900 = 62500(1 + 24/2*100)^(2t)
(676/625) = (112/100)^(2t)
(27/25)^2 = (27/25)^2t
t = 1yrs =
Answer: B |
AQUA-RAT | AQUA-RAT-36518 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work? | [
"90",
"125",
"145",
"225"
] | D | 1 man’s 1 day’s work = 1⁄100
(10 men + 15 women)’s 1 day’s work = 1⁄6
15 women’s 1 day’s work
= (1⁄6 - 10⁄100) = (1⁄6 - 1⁄10) = 1⁄15
∴ 1 woman’s 1 day’s work = 1⁄225
∴ 1 woman alone can complete the work in 225 days
Answer D |
AQUA-RAT | AQUA-RAT-36519 | # Making Friends around a Circular Table
I have $n$ people seated around a circular table, initially in arbitrary order. At each step, I choose two people and switch their seats. What is the minimum number of steps required such that every person has sat either to the right or to the left of everyone else?
To be specific, we consider two different cases:
1. You can only switch people who are sitting next to each other.
2. You can switch any two people, no matter where they are on the table.
The small cases are relatively simple: if we denote the answer in case 1 and 2 for a given value of $n$ as $f(n)$ and $g(n)$ respectively, then we have $f(x)=g(x)=0$ for $x=1, 2, 3$, $f(4)=g(4)=1$. I’m not sure how I would generalize to larger values, though.
(I initially claimed that $f(5)=g(5)=2$, but corrected it based on @Ryan’s comment).
If you’re interested, this question came up in a conversation with my friends when we were trying to figure out the best way for a large party of people during dinner to all get to know each other.
Edit: The table below compares the current best known value for case 2, $g(n)$, to the theoretical lower bound $\lceil{\frac{1}{8}n(n-3)}\rceil$ for a range of values of $n$. Solutions up to $n=14$ are known to be optimal, in large part due to the work of Andrew Szymczak and PeterKošinár.
The moves corresponding to the current best value are found below. Each ordered pair $(i, j)$ indicates that we switch the people in seats $(i, j)$ with each other, with the seats being labeled from $1 \ldots n$ consecutively around the table.
The following is multiple choice question (with options) to answer.
Find the no.of ways of arranging the boy and 8 guests at a circular table so that the boy always sits in a particular seat? | [
"5!",
"8!",
"9!",
"12!"
] | B | Ans.(B)
Sol. Total number of persons = 9 Host can sit in a particular seat in one way. Now, remaining positions are defined relative to the host. Hence, the remaining can sit in 8 places in 8P8 = 8! Ways ... The number of required arrangements = 8! x 1= 8! = 8! ways |
AQUA-RAT | AQUA-RAT-36520 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
If the tens digit of positive integer x is 2 and the tens digit of positive integer y is 2, how many possible values of the tens digit of 2(x+y) can there be? | [
"2",
"3",
"4",
"5"
] | C | We only need to consider the tens and ones digits of x and y, and the last two digits can be represented as 20 + p and 20 + q respectively.
2(20+p+20+q)= 80 + 2(p + q)
2(p+q) can range from 0 up to a maximum value of 36 when p=q=9.
Then the tens and ones digits of 2(x+y) can range from 80 up to 16.
That is, the tens digit can be 8, 9, 0, or 1.
There can be 4 possibilities for the tens digit.
The answer is C. |
AQUA-RAT | AQUA-RAT-36521 | so $$4c^2-4c(b+d)+(b+d)^2-(b+d)^2+4d^2+3b^2-4d +\frac85 \leq 0$$ so $$(2c-b-d)^2+3d^2+2b^2-2bd-4d+\frac85 \leq 0$$ so
$$3d^2+2b^2-2bd-4d+\frac85 \leq 0\;\;\;/\cdot 2$$ so $$4b^2-4bd+d^2+5d^2-8d+\frac{16}5 \leq 0 \;\;\;/\cdot 5$$ so $$5(2b-d)^2+25d^2-40d+16 \leq 0$$ so $$5(2b-d)^2+(5d-4)^2 \leq 0$$ which means that $d=4/5$ and $b=2/5$ and $c=(b+d)/2=...$ and $a=...$
The following is multiple choice question (with options) to answer.
If a + b - c = 4d, and if a - b + c = 4e, then a = | [
"2(d + e)",
"d – e",
"3(d + e)/2",
"d-2e"
] | A | Add both equations:
a + b - c = 4d
a - b + c = 4e
a+b-c+a-b+c=4d+4e
2a=4d+4e
a=2(d+e)
Ans:A |
AQUA-RAT | AQUA-RAT-36522 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Rose is two years older than Bruce who is twice as old as Chris. If the total of the age of Rose, B and Chris be 47 years, then how old is Bruce ? | [
"18 years",
"10 years",
"12 years",
"13 years"
] | A | Let Chris's age be x years. Then, Bruce's age = 2x years.Rose's age = (2x + 2) years.
(2x + 2) + 2x + x = 47
5x = 45
x = 9
Hence, Bruce's age = 2*9 = 18 years.
Answer : A |
AQUA-RAT | AQUA-RAT-36523 | 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.
First train 108 m long moving at a speed of 50 km/hr crosses a second train 112 m long coming from opposite direction in 6 seconds. Find the speed of the second train ?? | [
"82 km/hr",
"80 km/hr",
"90 km/hr",
"60 km/hr"
] | A | speed of the second train consider as x
relative speed = (x+50)*5/18 m/sec
(250+5x/18) m/sec
distance = (108 + 112) = 220 m
==> 220/(250+5x/18) = 6
250 + 5x = 660
=>x = 82 km/hr
ANSWER A |
AQUA-RAT | AQUA-RAT-36524 | # 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.
Bombay Express left Delhi for Bombay at 14:30hrs travelling at a speed of 60kmph and Rajdhani Express left Delhi for Bombay on the same day at 16:30hrs travelling at a speed of 80kmph. How far away from Delhi will the two trains meet | [
"120km",
"360km",
"480km",
"500km"
] | C | Explanation:
Suppose they meet x hours after 14:30hrs
Then, 60x = 80(x-2) => x=8
Required distance = 60×8 = 480km
Answer: Option C |
AQUA-RAT | AQUA-RAT-36525 | concentration
Title: Concentration of solutions I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%? The answer is that for a simple dilution the following formula applies:
$$c_1m_1 = c_2m_2$$ $$ m_2 = \frac{c_1m_1}{c_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$
Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$).
The following is multiple choice question (with options) to answer.
20 liters of a mixture is created by mixing Liquid P and Liquid Q in the ratio 3:2. How many liters of Liquid Q must be added to make the ratio 4:3? | [
"1",
"3",
"5",
"7"
] | A | Let x be the amount of liquid Q to be added.
(2/5)*20 + x = (3/7)*(20+x)
280 + 35x = 300 + 15x
20x = 20
x = 1
The answer is A. |
AQUA-RAT | AQUA-RAT-36526 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
An accurate clock shows 8 o’clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 3 o’clock in the afternoon? | [
"60°",
"90°",
"180°",
"210°"
] | D | Sol.
Angle traced by the hour hand in 7 hours = [360/12 * 7]° = 210°
Answer D |
AQUA-RAT | AQUA-RAT-36527 | 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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Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Dacid obtained 45, 35, 52, 47 and 55 marks (out of 100) in English, Mathematics, Physics, Chemistry and Biology. What are his average marks? | [
"79",
"99",
"47",
"89"
] | C | Average = (45+35+ 52+ 47+55)/5
= 234/5 = 47.
Answer:C |
AQUA-RAT | AQUA-RAT-36528 | (D)
Manager
Joined: 03 Aug 2017
Posts: 64
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink]
### Show Tags
05 Oct 2019, 22:16
Bunuel wrote:
If 20 typists can type 48 letters in 20 minutes, then how many letters will 30 typists working at the same rate complete in 1 hour?
A. 63
B. 72
C. 144
D. 216
E. 400
This i how i solved...
20 Typist can type 48 letters / 20 Min therofre in 1 Min 20 Typist can type = 48/20 = 2.4 Letters per Min
In 60 Min 20 Typist can type = 2.4 *60 =144 Words per min...
We are also told now there are 30 Typist so if 20 typist can type 144 words so 30 Typist can type X words per min.....
Solve for X
20/144 = 30 / x .....
x = 72*3 = 216 Words per min
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink] 05 Oct 2019, 22:16
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Rates for having a manuscript typed at a certain typing service are $5 per page for the first time a page is typed and $3 per page each time a page is revised. If a certain manuscript has 200 pages, of which 80 were revised only once, 20 were revised twice, and the rest required no revisions, what was the total cost of having the manuscript typed? | [
"$1360",
"$1620",
"$1650",
"$680"
] | A | For 200-80-20=100 pages only cost is 5$ per page for the first time page is typed - 100*5=500$;
For 80 pages the cost is: first time 5$ + 3$ of the first revision - 80*(5+3)=640$;
For 20 pages the cost is: first time 5$ + 3$ of the first revision + 3$ of the second revision - 20(5+3+3)=220$;
Total: 500+640+220=1360$.
Answer: A. |
AQUA-RAT | AQUA-RAT-36529 | # How many ways we can choose items from different boxes
I searched through the internet but couldn't find my answer, which can either be a very simple or a hard one.
Assume there are $3$ boxes, which carry, respectively, $1$, $4$, $2$ items. My question is how many ways we can select $3$ items from these boxes. I am looking for a formula rather than a solution for these specific values.
If I choose (take away) $3$ items by trying one by one. \begin{array}{c c c} 0 & 3 & 1\\ 0 & 2 & 2\\ 0 & 4 & 0\\ 1 & 1 & 2\\ 1 & 3 & 0\\ 1 & 2 & 1 \end{array}
Items remain each time, so the answer seems to be $6$ different ways. But I am not sure.
The following is multiple choice question (with options) to answer.
In Sam's hanger there are 23 boxes, 18 out of the boxes are filled with toys and the rest are filled with electrical appliances. 8 boxes are for sale, 5 of them are filled with toys. How many boxes with electrical appliances are in Sam's hanger that is not for sale? | [
"1.",
"2.",
"3.",
"4."
] | B | Total boxes = 23
Filled with toys = 18
Filled with appliance = 5
Total boxes for sale = 8
Toy boxes for sale = 5
Appliance boxes for sale = 3
Appliance boxes not for sale = 5 - 3 = 2
Correct Option: B |
AQUA-RAT | AQUA-RAT-36530 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into partnership. A invests 3 times as much as B invests and B invests two third of what C invests. At the end of the year, the profit earned is Rs. 8500. What is the share of B? | [
"1545.45",
"1545.48",
"1545.42",
"1545.46"
] | A | Explanation:
Let C's capital = Rs. y. Then, B's capital = Rs. (2/3)y
A's capital = Rs. (3 × (2/3)y) = Rs. 2y
Therefore, ratio of their capitals
= 2y : (2/3)y : y
= 6 : 2 : 3
Hence, B's share = Rs. (8500 x 2/11) = Rs. 1545.45
ANSWER: A |
AQUA-RAT | AQUA-RAT-36531 | Math Help - N positive integer problem with a + exponent
1. N positive integer problem with a + exponent
Here it is a problem that is driving me nuts...
Note: ^ means the exponent, in this case the exponent is "+".
If n is a positive integer, then n^+ denotes a number such that n < n^+ < n+1.
So decide which of the following options is greater (or if both ofthem are equal or if it is impossible to determine it).
Option A: 20^+ / 4^+
Option B: 5^+
What I did: 20^+ / 4^+ SO, (20/4)^+ , SO (5)^+ , so I concluded that both of them are equal, but the correct answer is " It can´t be determined which option is greater". Could anyone explain me why??
2. We have:
$20< 20^+ < 21$
$4< 4^+ < 5$
$5<5^+<6$
Clearly we cannot divide these inequalities.
$\frac{20< 20^+ < 21}{4< 4^+ < 5} = ???????$
But we can determine bounds.
How do we obtain the smallest possible value of $\frac{20^+}{4^+}?$ We minimize the numerator and maximize the denominator. The numerator's smallest value is close to $20,$ and the denominator's largest value is close to $5.$ So, the fractional value is $\frac{20^+}{4^+}> \frac{20}{5}=4$.
Now, How do we obtain the largest possible value of $\frac{20^+}{4^+}?$ We maximize the numerator and minimize the denominator. We would end up with $\frac{20^+}{4^+} < \frac{21}{4} = 5.25.$
So, $4<\frac{20^+}{4^+}<5.25,$ Which could be greater or less than $5 <5^+<6$ depending on which value $5^+$ were to take on.
The following is multiple choice question (with options) to answer.
If @ is a binary operation defined as the difference between an integer n and the product of n and 5, then what is the largest positive integer n such that the outcome of the binary operation of n is less than 16? | [
"1",
"2",
"3",
"4"
] | C | @(n) = 5n - n
We need to find the largest positive integer such that 5n - n < 16.
Then 4n < 16 and n < 4.
The largest possible integer is n = 3.
The answer is C. |
AQUA-RAT | AQUA-RAT-36532 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
Find avrg speed if a man travels at speed of 24km/hr up and 36km/hr dawn at an altitude of 200m. | [
"25.8",
"26.8",
"27.8",
"28.8"
] | D | avg speed=2*x*y/(x+y)
=2*24*36/(24+36)=28.8
ANSWER:D |
AQUA-RAT | AQUA-RAT-36533 | combustion, stoichiometry, isotope
wasn't that easier?
3.
35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948
This is slightly wrong.
0.00006, x, 0.99994-x
The following is multiple choice question (with options) to answer.
34.97 + 240.016 + 23.98 = ? | [
"298.966",
"298.694",
"289.496",
"289.469"
] | A | 34.97
240.016
+ 23.98
--------
298.966
Answer is A. |
AQUA-RAT | AQUA-RAT-36534 | ## eliassaab Group Title Let m and n be two positive integers. Show that (36 m+ n)(m+36 n) cannot be a power of 2. 2 years ago 2 years ago
1. sauravshakya
LET (36m+n)=2^x AND (m+36n)=2^y THEN, (36m+n)(m+36n)=2^(x+y) AlSO, x and y must me both integers
2. sauravshakya
Now, n=2^x - 36m THEN, m+36n=2^y m+36(2^x-36m)=2^y 36 * 2^x -1295m=2^y
3. sauravshakya
Now, I think I have to prove that y is never a integer
4. sauravshakya
2^y=36*2^x-1295m 2^y=2^x(36-1295m/2^x) Now, 2^y must be positive so, (36-1295m/2^x) must be positive... Now let 2^z=36-1295m/2^x HERE z also must be an positive integer..... So, 2^z can be 4,8,16,32 NOW,
5. sauravshakya
36-1295m/2^x= 4, 8, 16, 32 1295m/2^x=32, 28 ,16 ,4 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4
6. sauravshakya
Now, 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4 2^x=40.47 m , 46.25m , 80.94m ,323.75m Thus, for no positive integer value of m we will get x a integer value......
7. sauravshakya
Hence, (36 m+ n)(m+36 n) cannot be a power of 2.
8. sauravshakya
I have jumped some step......... I hope I made it clear
9. sauravshakya
The following is multiple choice question (with options) to answer.
If 42.36 = k(14 + m/50), where k and m are positive integers and m < 50, then what is the value of k + m ? | [
"6",
"7",
"8",
"9"
] | D | 42.36 = 14K + Km/50...we can rewrite the number as follows:
42+ 0.36= 14K + Km/50........Since K is integer, then 42=14K..........K=3
0.36=Km/50......36/100=3m/50......m=6
k+m=3+6=9
Answer: D |
AQUA-RAT | AQUA-RAT-36535 | 5%------------------20%
so ratio is 1:4 in final mixture
Earlier type 1 alcohol was 1
Now it is 1/5 ----> so loss of 4/5 = 80%...
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 56303
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 08:52
11
14
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
Question can be solved algebraically or using allegation method.
Algebraic approach:
Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.
Let the portion replaced be $$x$$ and the volume of initial solution be 1 unit.
Then the amount of alcohol after removal of a portion will be $$0.5(1-x)$$ and the amount of alcohol added will be $$0.25x$$, so total amount of alcohol will be $$0.5(1-x)+0.25x$$. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was $$0.3*1$$.
So $$0.5(1-x)+0.25x=0.3$$ --> $$x=0.8$$, or 80%.
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Joined: 06 Jul 2010
Posts: 6
Re: Mixture problem-Can someone explain this [#permalink]
### Show Tags
02 Sep 2010, 10:11
zest4mba wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
a 3%
b 20%
c 66%
d 75%
e 80%
The following is multiple choice question (with options) to answer.
A jar is filled of liquid which is 3 parts water and 5 parts alcohol. How much of this mixture should be drawn out and replaced such that this mixture may contain half
water and half alcohol? | [
"25%",
"20%",
"14.29%",
"33.33%"
] | B | Explanation:
Let the jar initially contain 8 litres of mixture:3 litres water and 5 litres alcohol
Let x litres of this mixture is drawn out and is replaced by x litres of water.
Amount of water now in the solution: 3-(3/8)x+x.
Amount of alcohol now in the solution: 5-(5/8)x
Desired ratio: 1/1
=>3-(3/8)x+x=5-(5/8)x
x=8/5 litres
%x=((8/5)/8)*100=20%
ANSWER B |
AQUA-RAT | AQUA-RAT-36536 | # 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 merchant purchased a jacket for $60 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 25 percent of the selling price. During a sale, the merchant discounted the selling price by 20 percent and sold the jacket. What was the merchant’s gross profit on this sale? | [
"$0",
"$3",
"$4",
"$12"
] | C | Markup = x
x = 0.25(60 + x)
=> 3x/4 = 15
=> x = 20
So SP = 80
But Actual SP = 0.8 * 80 = 64
So profit = $4
Answer - C |
AQUA-RAT | AQUA-RAT-36537 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A car traveled 75% of the way from town A to town B at an average speed of 50 miles per hour. The car travels at an average speed of S miles per hour for the remaining part of the trip. The average speed for the entire trip was 60 miles per hour. What is S ? | [
"150",
"200",
"250",
"300"
] | A | Total distance = 100 miles (easier to work with %)
75% of the distance = 75 miles
25% of the distance = 25 miles
1st part of the trip → 75/50 = 1.5
2nd part of the trip → 25/S = t
Total trip → (75+25)/60 = 1.5 + t » 100/60 = 1.5 + t » 2.5 = 1.5 + t » t = 0.1667
Back to 2nd part of the trip formula: 25/S = 0.1667 » S = 150
Ans A |
AQUA-RAT | AQUA-RAT-36538 | java, datetime, sorting, collections, lambda
Initially I had implemented it using Comparators but the Lambda made the code much cleaner.
Desired output:
MM - DD - YYYY
7 - 20 - 2023
7 - 18 - 2027
8 - 30 - 2010
8 - 12 - 2020
9 - 21 - 2006
10 - 23 - 2008
11 - 19 - 2000
11 - 16 - 2033
11 - 13 - 1989
12 - 13 - 2019
1 - 10 - 1985
1 - 15 - 2016
4 - 10 - 2021
4 - 13 - 2022
4 - 14 - 2004
5 - 1 - 2025
5 - 4 - 2014
5 - 20 - 2023
5 - 21 - 2022
6 - 29 - 1990 A Comparator is the correct approach, you just need to mod the month by the current month before comparing them. The easiest way to do this is to introduce a static variable currentMonth to hold the current month for all instances of DateObj.
Add DateObj.setCurrentMonth(CURRENTMONTH); to your Main, and use the code below:
//Custom Date Object
class DateObj implements Comparable<DateObj>{
private int year;
private int month;
private int date;
private static int currentMonth;
DateObj(int year, int month ,int date){
this.year = year;
this.month = month;
this.date = date;
}
public Integer getYear() {return year;}
public Integer getMonth() {return month;}
public Integer getDate() {return date;}
public Integer getCurrentMonth() { return DateObj.currentMonth;}
public static void setCurrentMonth(int currentMonth){
DateObj.currentMonth = currentMonth;
}
@Override
public int compareTo(DateObj o) {
int months = (month+(12-currentMonth))%12 - (o.month+(12-currentMonth))%12;
if(months == 0)
return this.date-o.date;
else{
return months;
}
}
}
The following is multiple choice question (with options) to answer.
What is the year next to 1990 which will have the same calendar as that of the year 1990? | [
"1992",
"2001",
"1995",
"1996"
] | B | Explanation:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.
Take the year 1992 from the given choices.
Total odd days in the period 1990-1991= 2 normal years
≡ 2 x 1 = 2 odd days
Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days
Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995= 5 normal years + 1 leap year
≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.
Take the year 2001 from the given choices.
Number of odd days in the period 1990-2000= 8 normal years + 3 leap years
≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001
Answer: Option B |
AQUA-RAT | AQUA-RAT-36539 | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
The following is multiple choice question (with options) to answer.
A meeting has to be conducted with 5 managers. Find the number of ways in which the managers be selected from among 9 managers, if 2 managers will not attend the meeting together? | [
"35",
"91",
"120",
"126"
] | B | we can either choose all 5 people from 7 manager who have no problems or choose 4 from the 7 and 1 from the 2 managers who have a problem sitting together
so 7C5 + (7C4 * 2C1)
this is 21 + 70
91 ANS:B |
AQUA-RAT | AQUA-RAT-36540 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
If 4 (P's Capital ) = 6 ( Q's Capital ) = 10 ( R's Capital ) , then out of the total profit of Rs 4030 , R will receive | [
"600",
"780",
"800",
"900"
] | B | Explanation :
Let P's capital = p, Q's capital = q and R's capital = r
Then
4p = 6q = 10r
=> 2p = 3q = 5r
=>q = 2p/3
r = 2p/5
P : Q : R = p : 2p/3 : 2p/5
= 15 : 10 : 6
R's share = 4030 * (6/31) = 130*6 = 780. Answer : Option B |
AQUA-RAT | AQUA-RAT-36541 | # One-zero dividend
## Challenge description
For every positive integer n there exists a number having the form of 111...10...000 that is divisible by n i.e. a decimal number that starts with all 1's and ends with all 0's. This is very easy to prove: if we take a set of n+1 different numbers in the form of 111...111 (all 1's), then at least two of them will give the same remainder after division by n (as per pigeonhole principle). The difference of these two numbers will be divisible by n and will have the desired form. Your aim is to write a program that finds this number.
## Input description
A positive integer.
## Output description
A number p in the form of 111...10...000, such that p ≡ 0 (mod n). If you find more than one - display any of them (doesn't need to be the smallest one).
## Notes
Your program has to give the answer in a reasonable amount of time. Which means brute-forcing is not permited:
p = 0
while (p != 11..10.00 and p % n != 0)
p++
Neither is this:
do
p = random_int()
while (p != 11..10.00 and p % n != 0)
Iterating through the numbers in the form of 11..10..00 is allowed.
Your program doesn't need to handle an arbitrarily large input - the upper bound is whatever your language's upper bound is.
## Sample outputs
2: 10
3: 1110
12: 11100
49: 1111111111111111111111111111111111111111110
102: 1111111111111111111111111111111111111111111111110
The following is multiple choice question (with options) to answer.
How many positive integers will divide evenly into 370? | [
"4",
"6",
"8",
"12"
] | C | The question is asking how many factors 370 has.
370 = 2*5*37
The number of factors is 2^3 = 8
The answer is C. |
AQUA-RAT | AQUA-RAT-36542 | ## Circular Combination
1. Arrange 8 dancers in circular fashion.
2. Use 8 pearls in a band to make a necklace.
3. Arrange 8 science and 7 arts students circularly so no two arts students are together.
• 7!
• $$\frac{(8-1)!}{2}$$
• $$^8P_7 \times \space 7!$$
# Combination
## Concept
A, B, C
In how many ways can we select 2?
• AB, AC, BC
• When making teams $$AB \equiv BA$$
## Formulae And Notation
Formulae (Don’t Memorize!)
• $$^nC_r = {n! \over r!(n-r)!}$$
• $$^nC_r \times ^rP_r = ^nP_r = ^nP_r \times r!$$ (Permutation-combination relationship)
• $$^{n+1}C_r = ^nC_r + ^nC_{r-1}$$
• $$^nC_r=?$$ (from above, expanding twice up to n-2)
• $$^nC_r = ^nC_{n-r}$$
Notation
• $$^nC_r \equiv$$ $$n \choose r$$
• Select 5 people from 6 $$\rightarrow ^6C_5 =$$ $$6 \choose 5$$
## Expaniosn of $$^nC_r$$
The following is multiple choice question (with options) to answer.
A composer’s guild is planning its spring concert, and eleven pieces have been submitted for consideration. The director of the guild knows that they will only have time to present three of them. If the pieces can be played in any order, how many combinations of pieces are possible? | [
"402",
"165",
"154",
"512"
] | B | Combination question..
11C3 = 165..
IMO option B is correct.. |
AQUA-RAT | AQUA-RAT-36543 | # 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 cash difference between the selling prices of an article at a profit of 8% and 6% is Rs 3. The ratio of two selling prices is | [
"51:52",
"52:53",
"53:54",
"54:53"
] | D | Explanation:
Let the Cost price of article is Rs. x
Required ratio = (108% of x) / (106% of x)
=108/106
=54/53 = 54:53.
Answer: D |
AQUA-RAT | AQUA-RAT-36544 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
A car travels from point A to point B. The average speed of the car is 60 km/hr and it travels the first half of the trip at a speed of 48 km/hr. What is the speed of the car in the second half of the trip? | [
"72",
"75",
"80",
"84"
] | C | Let D be the distance and let V be the speed in the second half.
The total time = T1 + T2
D/60 = D/96 + (D/2) / V
8D/480 - 5D/480 = (D/2) / V
D/160 = D/2V and so V = 80 km/hr
The answer is C. |
AQUA-RAT | AQUA-RAT-36545 | 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.
A certain family has 3 sons: Richard is 6 years older than David, and David is 8 years older than Scott. If in 8 years, Richard will be twice as old as Scott, then how old was David 6 years ago? | [
" 8",
" 10",
" 12",
" 14"
] | A | Let's say
Age of Richard isR
Age of David isD
Age of Scott isS
Now
Richard is 6 years older than David,
i.e. R = D +6
David is 8 years older than Scott
i.e. D = S +8
If in 8 years, Richard will be twice as old as Scott
i.e. R+8 = 2x(S+8)
i.e. R+8 = 2S + 16
i.e. R = 2S+8
But R = D+6 = (S+8)+6 = S+14
therefore, 2S + 8 = S +14
i.e. S = 6
i.e. R = 20
i.e. D = 14
Now,
how old was David 6 years ago?
i.e. D-6 = 14-6 = 8 years
Answer: Option
A |
AQUA-RAT | AQUA-RAT-36546 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of a train and that of a platform are equal. If with a speed of 126 k/hr, the train crosses the platform in one minute, then the length of the train (in meters) is? | [
"227",
"299",
"1050",
"750"
] | C | Speed = [126 * 5/18] m/sec = 35 m/sec; Time = 1 min. = 60 sec.
Let the length of the train and that of the platform be x meters.
Then, 2x/60 = 35 è x = 35 * 60 / 2 = 1050
Answer: C |
AQUA-RAT | AQUA-RAT-36547 | # Identifying Prime Numbers from 507 to 10647
#### ErikHall
##### New member
So lets say you have the Numbers (2677; 6191 and 1091), how would you go about the check that 2677 and 1091 are Prime ?
The conditions are:
- You dont have a calculator
- You just have a Pen
Is there a clever way to do it except looking at the Last Digit ?
Thanks for the Help !
Also small fun fact, if you use the Google Calculator and divide by Zero, it says "Infinity". Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
#### lookagain
##### Senior Member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
#### lev888
##### Full Member
Is there a clever way to do it except looking at the Last Digit ?
Could you explain the last digit method?
#### ErikHall
##### New member
Could you explain the last digit method?
So if you see a Number like 546342 you know its not Prime since you can divide it by 2. Same with 4 and 8. Those numbers are, as far as i know, never Prime. So this is a way to see non Primes. But it dosnt help you do see if a Number is Prime. It only shows the ones who are not. And not even that many of them
#### ErikHall
##### New member
Look at 1,091, for instance. Look at the prime numbers whose squares are
less than or equal to 1,091. If none of those divide 1,091, then 1,091 is prime.
Interessting way, but that would mean you have to know the Primes up to like 10.000 right ? Or at least the ones, that are near to 10.000 Squared.
#### Subhotosh Khan
##### Super Moderator
Staff member
Meaning in Google´s World 2:0 = 173279384:0. Same thing as you can see.
Incorrect.
The following is multiple choice question (with options) to answer.
which of the following is a prime number ? | [
"17",
"18",
"15",
"20"
] | A | Clearly, 17 is a prime number.
Option A |
AQUA-RAT | AQUA-RAT-36548 | # 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 takes 3 hours 45 minutes to row a boat 25 km downstream of a river and 2 hours 30 minutes to cover a distance of 15 km upstream. Find the speed of the current. | [
"1 km/hr",
"2/3 km/hr",
"1/3 km/hr",
"4 km/hr"
] | C | Explanation:
First of all, we know that
speed of current = 1/2(speed downstream - speed upstream) [important]
So we need to calculate speed downstream and speed upstream first.
Speed = Distance / Time [important]
Speed upstream =(25/3 3/4)km/hr
=20/3km/hr
Speed Downstream = (15/2 1/2)km/hr
=6km/hr
So speed of current = 1/2(20/3−6)
=1/3 km/hr
Option C |
AQUA-RAT | AQUA-RAT-36549 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon? | [
"144°",
"150°",
"168°",
"180°"
] | D | Solution
Angle traced by hour hand in
5 hrs 10 min. = (360/12 x 6)°
= 180°.
Answer D |
AQUA-RAT | AQUA-RAT-36550 | Why are things "weird" here? Let's think of it like this: If I travelled at $$8 \frac{\text{mi}}{\text{hr}}$$ for an hour, then travelled at $$12 \frac{\text{mi}}{\text{hr}}$$, I'd definitely agree that the average speed is $$10 \frac{\text{mi}}{\text{hr}}.$$ Things are pretty normal here: If I want to calculate the total distance traveled, it's just $$8 + 12 = 20.$$ Then, if I want to calculate the total time traveled, it's just $$1 + 1 = 2.$$ So the average or overall speed is just
$$\frac{20 \text{mi}}{2 \text{hr}} = 10 \frac{\text{mi}}{\text{hr}}.$$ It works!
The following is multiple choice question (with options) to answer.
A person goes to his office at 1/3rd of the speed at which he returns from his office. If the avg speed during the whole trip is 12m/h. what is the speedof the person while he was going to his office? | [
"8km/h",
"10km/h",
"12km/h",
"13km/h"
] | A | u = k , v= 3k
\inline \therefore \frac{2uv}{u+v}\: \: \Rightarrow \frac{2\times k\times 3k}{(k+3k)}=12
\inline \Rightarrow 1.5k = 12
\inline \Rightarrow k=8km/h
A |
AQUA-RAT | AQUA-RAT-36551 | # Need Help With Simple Mathematical Sets Difference (Intersection) Problem
I'm having trouble with a seemingly simple problem in Math and I need some help with it. The problem states:
A hospital is doing a treatment research on 50 volunteers. Of those, some have had a few reactions: 12 have had headaches, 8 felt nausea and 4 had headaches and nausea. How many volunteers had headaches and not nausea? How many volunteers didn't feel neither (headache and nausea simultaneously)?
It's not a homework as I'm not in school anymore but studying by myself to try my country's equivalent of the SAT to get into college.
So, I know that my Universe (U) in this case is 50. I know the sets (let's say A and B) are A={12}, B={8} and A∩B={4}, so the number of volunteers that had only headaches exclusively is 8 (12 - 4 people that also had nausea) but i can't figure out the number of volunteers that haven't had neither.
I've tried putting together a simple equation to find the number: 12 volunteers that felt headache + 8 that felt nausea + x people that felt neither = 50 that will amount to x=30 volunteers but the book (without explaining why) says the correct answer is 34.
I've tried breaking the problem down by its inquiries, writing them one by one (which I won't do here to not extend the question) and it seems to be a simple thing, really, and that what I've done is correct.
Is the book wrong (unlikely)? Is my solution correct? How should I go about when solving this problem?
The following is multiple choice question (with options) to answer.
A marketing firm determined that, of 200 households surveyed, 80 used neither Brand R nor Brand B soap, 60 used only Brand R soap, and for every household that used both brands of soap, 3 used only Brand B soap. How many of the 200 households surveyed used both brands of soap? | [
" 15",
" 20",
" 30",
" 40"
] | D | SOLUTION FOR SOAP R AND SOAP B
(D) 40 |
AQUA-RAT | AQUA-RAT-36552 | Decide the seating order of the people, starting from one of the brothers, say Ivan. Then position the other brother, Alexei, in one of the two slots (fourth and fifth) that fulfill the "separated by two others" condition - $2$ options. Then with Ivan and Alexei resolved, order the remaining five people in one of $5!=120$ ways. Finally add the empty chair to the right of someone, $7$ options, giving $2\cdot 120\cdot 7 = 1680$ options.
$\underline{Get\;the\;bothersome\;empty\;chair\;out\;of\;the\;way\;\;as\;a\;marker\;at\;the\;12\;o'clock\;position}$
• Brother $A$ has $7$ choices of seats
• Brother $B$ now has only $2$ choices (one clockwise and one anticlockwise of $A$ )
• the rest can be permuted in $5!$ ways
• Thus $7\cdot2\cdot5!\;$ways
The following is multiple choice question (with options) to answer.
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 14 family members in a row for the picture? | [
"12!",
"13×12!",
"12×13!",
"14!/2!"
] | C | The number of ways to arrange 14 people is 14!
We need to subtract the arrangement when these two people are together.
Let's think of these two people as a single unit so there are 13 units.
The number of ways to arrange 13 units is 13!
We need to multiply this by 2 since these two people could switch places.
The total number of valid arrangements is 14! - 2*13! = 13!*(14-2) = 12x13!
The answer is C. |
AQUA-RAT | AQUA-RAT-36553 | 28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
_________________
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
### Show Tags
28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
The following is multiple choice question (with options) to answer.
What is the units digit of 33^2 * 17^3 * 39^2? | [
"1",
"3",
"5",
"7"
] | D | The units digit of 33^2 is the units digit of 3*3 = 9 which is 9.
The units digit of 17^3 is the units digit of 7*7*7 = 343 which is 3.
The units digit of 39^2 is the units digit of 9*9 = 81 which is 1.
The units digit of 9*3*1 = 27 is 7.
The answer is D. |
AQUA-RAT | AQUA-RAT-36554 | materials, energy
With these 3 we can find $f$ from (4): 0.25
Remainder of $n$ is trivial as we have column $1$ - multiply the $1$ column value by 0.25 for column $2$ and 0.75 for column $3$, e.g. $n_{2,b} = 10\cdot0.25 = 2.5$
For totals in kg/h, simply take the units given to find the factor by which to multiply:
$$total_{c} = \sum_i^{a,b,c}{MW_i \cdot n_{c,i}} $$
(remembering the 1000 factor of $kmol$ vs $mol$ - multiply $n$ by 1000 first): For column $1$
(4*10 + 10*20 + 6*30)*1,000 = 40+200+180 = 420,000 kg/h.
The following is multiple choice question (with options) to answer.
A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.800, what is the reduced price for kg? | [
"22",
"99",
"40",
"28"
] | C | 800*(25/100) = 200 ---- 5
? ---- 1 => Rs.40
Answer: C |
AQUA-RAT | AQUA-RAT-36555 | # 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.
Saskia needs to catch the 3.15pm train in order to get to her Mixed Martial Arts fight on time. It is 2.55 pm and Saskia is 2 miles from the train station. Saskia can run at an average speed of 10 miles per hour. However Saskia is carrying a sack of bricks which slows her down by 20%. If she runs to the station how many minutes early or late will she be for the train? | [
"10 minutes early",
"5 minutes early",
"0 minutes early",
"5 minutes late"
] | B | Saskia can run at an average speed of 10 miles per hour however she is slowed down by 20% by the sack of bricks and so can only run at 8 miles per hour. (0.2 x 10 = 2; 10 - 2 = 8)
At a speed of 8 miles per hour, Saskia can run 2 miles in 15 minutes. (8 miles / 2 miles = 4; 60 minutes/4 = 15 minutes)
Saskia has 20 minutes until her train but can run the distance in 15 minutes therefore she will be 5 minutes early (20 - 15 = 5 minutes).
ANSWER: B |
AQUA-RAT | AQUA-RAT-36556 | Therefore: . $p \:\leq \:0.1591$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
You can solve the problem "head on" if you dare.
But there are fifteen cases you must consider . . .
. . $C \:=\:\text{Closed, }\;O \:=\:\text{O{p}en}$
. . $\begin{array}{|c|c|c|c| } 1&2&3&4 \\ \hline \hline C&C&C&C \\ C&C&C&O \\ C&C&O&C \\ C&C&O&O \\ C&O&C&C \\ C&O&C&O \\ C&O&O&C \\ C&O&O&O \\ \end{array}$. . . $\begin{array}{|c|c|c|c|} 1&2&3&4 \\ \hline O&C&C&C \\ O&C&C&O \\ O&C&O&C \\ O&C&O&O \\ O&O&C&C \\ O&O&C&O \\ O&O&O&C \end{array}$
Hi Soroban
Thanks for the help.
The following is multiple choice question (with options) to answer.
The H.C.F of two numbers is 15 and their L.C.M is 3300. If one of the numbers is 225, then the other is? | [
"200",
"210",
"220",
"230"
] | C | Other number = (15 * 3300)/225
= 220.
Answer: C |
AQUA-RAT | AQUA-RAT-36557 | base and triangular sides which rise to meet at the same point. Volume of Hollow Cylinder Equation and Calculator. Volume of a Cone Quiz – Solution Find the formula for the volume of a cone of radius r and height h using volumes of rotation. The following sketch shows the. 18) A cone with diameter 16 m and a height of 16 m. More information about applet. Volume of a cone = 1/3πr 2 h The slant of a right circle cone can be figured out using the Pythagorean Theorem if you have the height and the radius. Surface Integrals Surface integrals are a natural generalization of line integrals: instead of integrating over a curve, we integrate over a surface in 3-space. Mar 12, 2017 · If top radius of a cone R1 , bottom radius R2 (where R1>R2), total height h and another height p (where p<=h) Then how can I calculate the volume of lower part with height of p ?. Several Web pages derive the formula for the surface area of a cone using calculus. Since the surface area of a sphere of radius r is 4πr 2, the volume of a spherical shell of radius r and thickness dr must be. Now let's find the volume V. Jan 21, 2015 · Calculate the volume of frusta using similar triangles to calculate the missing height of the smaller cone first. In order to determine how many seconds it will take for the tank to fill, we must divide the volume by the rate of flow of the water. In this case, we will only concern ourselves with the surface area. Collimating the primary x-ray beam limits x radiation exposure to the ROI. The volume of a cylinder of height h having circular base of radius r, is Vcylinder = πr2h. You can easily find out the volume of a cone if you have the measurements of its height and radius. Mar 29, 2010 · Homework Statement Evaluate the volume under z^2 = x^2 + y^2 and the disc x^2 + y^2 < 4. After plugging in all the values into this equation, you will be able to find the volume of any cone when given the require values. The volume of irregular shape is calculated by number of blocks present in it. If we revolve line OB around the x-axis it creates the cone we see in the figure. Figure 2 shows
The following is multiple choice question (with options) to answer.
The slant height of a cone is 12 cm and radius of the base is 4cm, find the curved surface of the cone? | [
"26",
"28",
"48",
"26"
] | C | π * 12 * 4
= 48
Answer:C |
AQUA-RAT | AQUA-RAT-36558 | 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 sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 4 years hence? | [
"50",
"60",
"68",
"80"
] | C | A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 44 and 24.
Sum of their ages = 44 + 24 = 68.
ANSWER:C |
AQUA-RAT | AQUA-RAT-36559 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
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. 58? | [
"3944",
"2287",
"2977",
"2668"
] | A | 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 * 58 = Rs. 3944.
Answer:A |
AQUA-RAT | AQUA-RAT-36560 | 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.
B and C walk around a circular track. They start at 9 a.m from the same point in the opposite directions. B and C walk at a speed of 5 rounds per hour and 4 rounds per hour respectively. How many times shall they cross each other before 1 p.m | [
"34",
"35",
"36",
"37"
] | C | Explanation:
Relative speed = (5+4) =9 rounds per hour
So, they cross each other 9 times in an hour hence, they cross 36 times before 1 p.m
Answer: Option C |
AQUA-RAT | AQUA-RAT-36561 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Sixteen men complete a work in 24 days while 48 children can do it in 16 days. Twelve men started the work, after 14 days 12 children joined them. In how Many days will all of them together complete the remaining work? | [
"12",
"62",
"27",
"29"
] | A | Explanation:
Let man capacity = 2 units/day. Then total work = 16 × 2 × 24 = 768
Let the children capacity is k units/ days. So total work = 48 × k × 16
Equating above two equations we get k = 1. So children capacity = 1 unit / day.
Twelve men did 14 days of job. So they completed 12 × 2 ×14 = 336.
Remaining work = 768 - 336 = 432.
Now 12 children joined them. So per day capacity of entire team = 12 × 2 + 12 × 1 = 36.
So they complete the remaining work in 432/36 = 12 days.
Answer:A |
AQUA-RAT | AQUA-RAT-36562 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
If a man walks to his office at ¾ of his usual rate, he reaches office 1/3 of an hour late than usual. What is his usual time to reach office? | [
"1",
"4",
"5",
"6"
] | A | Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 -------- 1/3
3 --------- ? è 1 hour
Answer: A |
AQUA-RAT | AQUA-RAT-36563 | # 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.
Rs.500 amounts to Rs.620 in 2 years at simple interest. If the interest is increased by 2%, it would amount to how much? | [
"120",
"25",
"614",
"530"
] | D | (500*3*2)/100 = 30
500+ 30 = 530
Answer: D |
AQUA-RAT | AQUA-RAT-36564 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
Yearly subscription to professional magazines cost a company $840.00. To Make a 30% cut in the magazine budget, how much less must be spent? | [
"654",
"655",
"656",
"588"
] | D | total cost 840
840*30/100=252
so the cut in amount is 252
the less amount to be spend is 840-252=588
ANSWER:D |
AQUA-RAT | AQUA-RAT-36565 | yields b -> 27.52553345330246, so we will use b = 28.
To check,
s = Sum[(-1)^n/(3 n + 1), {n, 0, \[Infinity]}];
s-(2Sum[(-1)^n/(3n+1),{n,0,b}]+(-1)^(b+1)/(3(b+1)+1))/2/.b->28.//Chop
(*-0.00009679302515119836 *)
We see we do indeed have an absolute value error of less than .0001 and have used just 29 terms to get it.
• You can use Chop to get rid ofthe spurious imaginary part of the final result. – corey979 Oct 12 '16 at 7:33
• @corey979 I have added that, thanks! – bobbym Oct 12 '16 at 10:17
The following is multiple choice question (with options) to answer.
What is the sum of the integers from -190 to 191 inclusive? | [
"0",
"191",
"375",
"875"
] | B | sum/n= average.
sum=(average)(n)
average=a+b/2=190+191/2=0.5
number of items(n)=B-A+1=191-(-190)+1=195+191=382.
sum=average*n=0.5*382=191.
answer is B |
AQUA-RAT | AQUA-RAT-36566 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
The number of ways in which six boys and six girls can be seated in a row for a photograph so that no two girls sit together is ? | [
"6! * 7P6",
"2(6!)",
"6!)2",
"6! * 7"
] | A | We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways
= 6! * 7P6
Answer:A: |
AQUA-RAT | AQUA-RAT-36567 | distance from a given point called centre. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. As you can imagine, as the discs become thinner, the volume of the sphere gets more accurate. (Assume ≈ 3. Diameter - a line going through the circle from edge to edge, dividing circle in half. The cubic volume of a cylinder is found by multiplying the radius times the radius times pi times the height. Cylinder, hollow Calculate the volume, height, inner or outer radius of hollow cylinder. Cylinder Volume Formula Calculator - How to Calculate the volume of a cylinder. Net of a Cone. Find the area of a circle when you know the diameter. When the piston has moved up to the top of its stroke inside the cylinder, and the remaining volume inside the head or combustion chamber has been reduced to 100 cc, then the compression ratio would be proportionally described as 1000:100, or with fractional reduction, a 10:1 compression ratio. Show that the rectangle of maximum area that can be inscribed in a circle is a square. The Volume of a Cylinder is. The volume of each cone is equal to ⅓Bh = ⅓(28. Volume calculator will determine the volume of the most common geometric solids. The base of the cylinder is large circle and the top portion is smaller circle. Volume of a cylinder : V = πr 2 h where r is the radius and h is the height of the cylinder. It is the same measurement for circles of any size. 25 × 6 Inches Height = 37. cm, the base ring area is 115. What is the value of pi, rounded to the nearest hundredth? 3. However (a) the statements are in the incorrect order; (b) the function calls are incorrect: (c) the logical expression in the while loop is incorrect; and (d) function definitions are incorrect. Their radius (r) is therefore 3 m. Usually the pipe line would be in the shape of cylinder. Now you see that the ratio of the volume of a sphere to the volume of a cylinder is 2/3. 14 x 9 2 x 7. Therefore, the volume of a cylinder = πr2h cubic units. Also, this is important to know that the radius of a circle is always the half of its. Calculate the volume of a
The following is multiple choice question (with options) to answer.
V is the volume of a cylinder; the radius of the cylinder is 3.5. The height of the cylinder is 650% more than the radius. Which of the following is true? | [
"100 < V < 300",
"300 < V < 500",
"500 < V < 700",
"700 < V < 900"
] | D | As we see the answers are in form of range we can use approximation
Volume of cylinder is πr^2h
Given π= 22/7 and r = 3.5 so r^2 ~ 12 and h = 6.5 * 3.5 ~ 23
So 22/7 * 12 * 23 ~ 868
so answer should be
D. 700 < V < 900
Answer : D |
AQUA-RAT | AQUA-RAT-36568 | This online aptitude test on Compound Interest is useful for candidates preparing for banking exams - Bank PO, IBPS PO, SBI PO, RRB PO, RBI Assistant, LIC,SSC, MBA - MAT, XAT, CAT, NMAT, UPSC, NET etc. COMPOUND INTEREST TABLES 561 1 / 2 %Compound Interest Factors 1 2 Single Payment Uniform Payment Series Arithmetic Gradient Compound Present Sinking Capital Compound Present Gradient Gradient. Compound Interest Invest €500 that earns 10% interest each year for 3 years, where each interest payment is reinvested at the same rate: End of interest earned amount at end of period Year 1 50 550 = 500(1. In compound interest calculations interest earned, or due, for each period, is added to the principal. Fill in the table below by calculating the interest, new balance and total payment for each month for the two payment scenarios,$25. 1) Year 3 60. So we are sharing It here with you so that It can be helpful for the aspirants who are going to appear in upcoming exams like SSC CPO 2019, SSC CGL 2019, SSC CHSL 2019, SSC MTS 2020, RRB NTPC 2019, RRC Group D 2019 and other state exams. Feb 1, 2017 - Compound Interest worksheet with answer key (pdf). The stochastic calculus part of these notes is from my own book: Probabilistic Techniques in Analysis, Springer, New York, 1995. Compound interest is incredibly powerful. Multiply the principal amount by one plus the annual interest rate to the power of the number of compound periods to get a combined figure for principal and compound interest. The chart below from JP Morgan shows how one saver (Susan) who invests for only 10 years early in her career, ends up with more wealth than another saver. MONTHLY COMPOUNDING ANNUAL COMPOUNDING STRATEGY 3 VS. In the formula, A represents the final amount in the account after t years compounded 'n' times at interest rate 'r' with starting. As you can see, at the end of 10 years, you receive more than 50 percent more money in interest payments with compound interest ($15,939) than you do with simple interest ($10,000). Now, let's say you deposited the same amount of money on a bank for 2 years at 3% annual interest compounded annually. 2 : Nov 20,
The following is multiple choice question (with options) to answer.
Rs 1000 is being charged at 50% per annum. what is the interest for 3rd year at compound interest? | [
"3277",
"2667",
"1125",
"2867"
] | C | Explanation:
Total amount for two years = \inline 1000(1.5)^2=2250
Now , interest for third year = \inline \frac{2250\times 50\times 1}{100}=1125
Answer: C) 1125 |
AQUA-RAT | AQUA-RAT-36569 | Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.
Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.
Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.
So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.
We can generalize our results from this simplified example to your 100 bag example.
Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99$.
So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$
But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$
Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.
The following is multiple choice question (with options) to answer.
The total weight of six nougat nuggets is 3 oz. What is the average (arithmetic mean) weight of one such nougat nugget? | [
"0.18 oz.",
"0.5 oz.",
"1.8 oz.",
"5 oz."
] | B | Average = Total/Number of items
= 3/6 = 0.5.
Hence, B |
AQUA-RAT | AQUA-RAT-36570 | Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?
The Attempt at a Solution
Let ##B## denote the event that at least 2 kings are drawn, and ##A## the event that at least 1 king is drawn. Because ##B## is a strict subset of ##A##,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute ##P(A)##, ##P(A^c )## denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute ##P(B)##, ##P(B^c)## denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king ##P(1)## and the probability of not drawing a single king ##P(A^c)##.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$
The following is multiple choice question (with options) to answer.
From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is? | [
"1/26",
"1/24",
"1/20",
"1/21"
] | A | Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13 = 1/26
Answer:A |
AQUA-RAT | AQUA-RAT-36571 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A shop produces sarongs. The daily average production is given by 5n + 20, where n is the number of workers aside from the owner. In the first k days, 500 units are produced, and then 5 workers are added to the team. After another k days, the cumulative total is 800. How many workers were part of the latter production run? | [
"A)6",
"B)10",
"C)11",
"D)21"
] | D | The daily average production is given by 5n + 20- given
In the first k days, 500 units are produced
= (5n+20)K =500
k = 500/5n+20...................................1
5 workers were added = 5(n+5)+20 = 5n +45
cumulative is 1250 .. thus for the current period = 800 -500 = 300
(5n+45)K= 300
k = 300/5n+45........................................2
equate 1 and 2
500/5n+20 = 300 /5n+45
500(5n+45) = 300(5n+20)
2500n + 22500 = 1500n + 6000
1000n = -16500
n = 16
thus n+5 = 21
hence D |
AQUA-RAT | AQUA-RAT-36572 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If 0 < a < b and k = (2a + 4b)/b , which of the following must be true? | [
"k < 2",
"k < 6",
"k < 9",
"k > 9"
] | B | Here's another approach:
k = (2a + 4b)/b
= 2a/b + 4b/b
= 2(a/b) + 4
Since 0 < a < b, we know that a/b is less than 1, which means that 2(a/b) is some number less than 2.
So, we get k = (some number less than 2) + 4
From here, we can see that k must be less than 6
Answer: B |
AQUA-RAT | AQUA-RAT-36573 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
find four consecutive numbers such that the sum of the first three numbers is twelve more than the fourth number ? | [
"7,8,9,10",
"3,4,5,6",
"9,10,11,12",
"6,7,8,9"
] | D | Let
1st Integer = x
2nd Integer = x+1
3rd Integer = x+2
4th integer = x+3
=======================================
Given
x+x+1+x+2=(x+3)+12
3x+3=x+3+12
3x+3=x+15
3x=x+15-3
3x=x+12
3x-x=12
2x=12
x=12/2
x=6
========================================
1st Integer = x = 6
2nd Integer = x+1 = 6+1 = 7
3rd Integer = x+2 =6+2= 8
4th integer = x+3=6+3=9
Answer D |
AQUA-RAT | AQUA-RAT-36574 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
By selling 16 pencils for a rupee a man loses 28%. How many for a rupee should he sell in order to gain 28%? | [
"8",
"9",
"7",
"6"
] | B | 72% --- 16
128% --- ?
72/128 * 16 = 9
Answer: B |
AQUA-RAT | AQUA-RAT-36575 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A person spends 1/3rd of the money with him on clothes, 1/5th of the remaining on food and 1/4th of the remaining on travel. Now, he is left with Rs 200. How much did he have with him in the beginning? | [
"s 200",
"s 500",
"s 300",
"s 450"
] | B | Suppose the amount in the beginning was Rs ’x’
Money spent on clothes = Rs 1x/3 Balance = Rs 2x/3
Money spent on food = 1/5 of 2x/3 = Rs 2x/15
Balance = 2x/3 - 2x/15 = Rs 8x/15
Money spent on travel = 1/4 of 8x/15 = Rs 2x/15 = 8x/15 - 2x/15 = 6x/15 = Rs2x/5
Therefore 2x/5 = 200 = 500
ANSWER:B |
AQUA-RAT | AQUA-RAT-36576 | This one is different: there are 60 siblings in junior class and 60 their pair siblings in the senior class. We want to determine probability of choosing one sibling from junior class and its pair from senior.
What is the probability of choosing ANY sibling from junior class? 60/1000 (as there are 60 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only one pair of chosen sibling it would be 1/800 (as there is only one sibling pair of chosen one).
So the probability of that the 2 students selected will be a sibling pair is: 60/1000*1/800=3/4000
I see no argument of multiplying this by two.
This problem can be solved in another way and maybe this way shows that no need of multiplication:
In how many ways we can choose 1 person from 1000=1C1000=1000
In how many ways we can choose 1 person from 800=1C800=800
So total # of ways of choosing 1 from 1000 and 1 from 800=1C1000*1C800=1000*800 --> this is our total # of outcomes.
Let’s count favorable outcomes: 1 from 60=60C1=60
The pair of the one chosen=1C1=1
So total favorable outcomes=60C1*1C1=60
Probability=Favorable outcomes/Total # of outcomes=60/(1000*800)=3/4000
Let’s consider another example:
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings of four children, each consisting of 2 junior and 2 senior (I’m not sure whether it’s clear, I mean there are 60 brother and sister groups, total 60*4=240, two of each group is in the junior class and two in the senior). If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
The same way here:
What is the probability of choosing ANY sibling from junior class? 120/1000 (as there are 120 of them).
What is the probability of choosing PAIR OF CHOSEN SIBLING in senior class? As in senior class there is only two pair of chosen sibling it would be 2/800 (as there is only one sibling pair of chosen one).
The following is multiple choice question (with options) to answer.
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 40 siblings pairs each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair? | [
"3/40000",
"1/3600",
"1/20000",
"1/60"
] | C | Let's see
Pick 40/1000 first
Then we can only pick 1 other pair from the 800
So total will be 40 / 800 *1000
Simplify and you get 1/20000
Answer is C |
AQUA-RAT | AQUA-RAT-36577 | # 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 began a business with Rs. 85,000. He was joined afterwards by B with Ks. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 8 : 1 ? | [
"6 months",
"3 months",
"2 months",
"8 months"
] | B | Suppose B joined for x months . Then, ( 85000 * 12 )/(42500 * x) = 8. or x = (85000 * 12) / (42500 * 8) = 3.
So, B joined for 3 months.
Answer: B |
AQUA-RAT | AQUA-RAT-36578 | mass, weight
Title: How do you measure mass? I feel like I am missing something fundamental with regards to mass vs. weight.
The weight of an object on the Moon is approximately 16.5% what you would experience on Earth.
In other words, if an object weighs 100kg on Earth, it would weigh only 16.5kg on the Moon.
As I'm led to believe by this statement, weight is variable, and depends on the gravitational force exerted upon it, but it's the same object, with the same constant mass.
I am even more confused by the statement made in the following site: https://www.mathsisfun.com/measure/weight-mass.html
An object has mass (say 100kg). This makes it heavy enough to show a weight of 100kg.
What I am confused about here, is the statement "An object has mass (say 100kg)", surely that is based on its weight on Earth?
The following is multiple choice question (with options) to answer.
Dhoni weighs twice as much as Nameetha. Nameetha's weight is 55% of Bima's weight. Dravid weighs 35% of Leela's weight. Leela weighs 30% of Dhoni's weight. Who among these 5 persons weighs the least? | [
"Dravid",
"Dhoni",
"Bima",
"Nameetha"
] | A | if
Bima weight = x
we get
Nameetha's weight = 0.55x
Dhoni's weight = 1.1x
Leela's weight = 0.33x
and Dravid's weight = 0.1155x
Answer : A |
AQUA-RAT | AQUA-RAT-36579 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
A money lender lent Rs. 1000 at 3% per year and Rs. 1400 at 5% per year. The amount should be returned to him when the total interest comes to Rs. 350. Find the number of years. | [
"3.5",
"3.75",
"4",
"4.25"
] | A | (1000xtx3/100) + (1400xtx5/100) = 350 → t =3.5 answer A |
AQUA-RAT | AQUA-RAT-36580 | The $\sqrt{18}$ length is only possible for one of the long diagonals, so our path must start with one of the $4$ corners of the grid. Without loss of generality (since the grid is rotationally symmetric), we let the vertex be $(0,0)$ and the endpoint be $(3,3)$.
$[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3),s); draw((0,0)--(3,3)); [/asy]$
The $\sqrt{13}$ length can now only go to $2$ points; due to reflectional symmetry about the main diagonal, we may WLOG let the next endpoint be $(1,0)$.
$[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); dotfactor = 4; pen s = linewidth(4); pen c = rgb(0.5,0.5,0.5); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); dot((0,0)^^(3,3)^^(1,0),s); draw((0,0)--(3,3),c); draw((3,3)--(1,0)); [/asy]$
The following is multiple choice question (with options) to answer.
Find the length of the longest pole that can be placed in an indoor stadium 24 m long, 18 m wide and 16 m high | [
"31",
"32",
"33",
"34"
] | D | sqrt(1156)=34
becoz max length is cuboid diagonal d=sqrt(L^2+B^2+H^2)
where L lenght
B breadth & H height
ANSWER:D |
AQUA-RAT | AQUA-RAT-36581 | c, strings
I know the next bit is a bit opinionated, but does my code give an idea of where I'm at with logic and programming as a whole
I think so. The code matches where you said you're at, which is that you are "relearning C".
Working a 40-hour week makes about 2000 hours per year.
So an adult professional programmer will have spent many 1000s of hours of practice and studying.
The following is multiple choice question (with options) to answer.
During a given week A programmer spends 1/4 of his time preparing flow chart, 3/8 of his time coding and the rest of the time in debugging the programs. If he works 48 hours during the week , how many hours did he spend debugging the program. | [
"15 hrs",
"16 hrs",
"18 hrs",
"19 hrs"
] | C | programmer takes 1/4*48=12 hrs a week to make flowchrt
same way.....18 hrs a week for codin...
so....for debugging=48-(12+18)=18 hrs
ANSWER:C |
AQUA-RAT | AQUA-RAT-36582 | # Finding total after percentage has been used?
Tried my best with the title. Ok, earlier while I was on break from work (I have low level math, and want to be more fond of mathematics)
I work retail, and 20% of taxes are taken out, and I am wanting to find out how much I made before 20% is taken out.
So I did some scribbling, and this is without googling so please be gentle. but lets say I make 2000 dollars per paycheck, so to see how much money I get after taxes I do 2000 * .8 which gives me 1600. I can also do 2000 * .2 which gives me 400 and then do 2000 – 400 (imo I don’t like this way since I have a extra unnecessary step)
Anyways, I was pondering how to reverse that to have the answer be 2000, and what I did was 1600 * 1.20 (120%) which gave me 1920, and I thought that is odd so I did 1600 * 1.25 and answer was 2000 exact.
My question is why did I have to add extra 5% (25%) to get my answer? I am sure I did something wrong, and I fairly confident the formula I am using is a big no no.
edit; wow thank you all for your detailed answers. I am starting to like mathematics more and more.
#### Solutions Collecting From Web of "Finding total after percentage has been used?"
If $20\%$ of your pay is removed for taxes, then you retain $80\%$ of it. Thus, the amount you receive after taxes is
$$\text{net} = 80\%~\text{of gross} = \frac{80}{100}~\text{gross} = \frac{4}{5}~\text{gross}$$
To determine your gross pay from your net pay, you must multiply your net pay by $5/4$ since
$$\text{gross} = \frac{5}{4} \cdot \frac{4}{5}~\text{gross} = \frac{5}{4}~\text{net}$$
The following is multiple choice question (with options) to answer.
Priya went to the stationers and bought things worth Rs. 200, out of which 50 paise went on sales tax on taxable purchases. If the tax rate was 2%. then what was the cost of the tax free items? | [
"Rs. 175.50",
"Rs. 174.50",
"Rs. 174.00",
"Rs. 170.00"
] | B | Solution
Let the amount of taxable purchases be Rs.x.
Then, 2% of x = 50/100
x ‹=› (50/100×100/2)
= 25
Cost of tax free items = Rs.[200 - (25 + 0.50)]
= Rs. 174.50
Answer B |
AQUA-RAT | AQUA-RAT-36583 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
### Show Tags
03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
8 friends are living in 8 different flats of an apartment. Each of them was allotted a parking spot to park their cars in the ground floor. But they used to park their cars randomly in any of the 8 parking spots. In how many ways can they park their cars so that exactly 6 people park their cars in the spots allotted to them? | [
"20",
"28",
"36",
"44"
] | B | 8C6 = 28
The remaining 2 people in each case will be parked in each other's parking spots.
The answer is B. |
AQUA-RAT | AQUA-RAT-36584 | # 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.
An article is bought for Rs.600 and sold for Rs.500, find the loss percent? | [
"16 2/6%",
"16 2/8%",
"16 3/3%",
"16 2/3%"
] | D | 600 ---- 100
100 ---- ? => 16 2/3%
Answer: D |
AQUA-RAT | AQUA-RAT-36585 | • $\frac{999,999}{7}=142857$. You used $1,000,000-1$ instead. But I don't understand how from $2\cdot 999,999 = 2\times \frac {1,000,000 -1}7 = 1,999,998$ we go to $285714$ via replacing $2$ with the formula for $100$.
– Jim
Sep 22 at 19:55
• If $100 = 7\times 14 + 2$ then $2 = 100 - 7\times 14$. So $2\times \frac {999,999}7 = (100-7\times 14)\times \frac {999999}7 = (100 - 7\times 14)\times \frac {10^6 -1}7$ Now just expand $(100-7\times 14)\times \frac{10^6 - 1}7 = 100\times \frac {10^6-1}7 - 7\times 14\times \frac {10^6-1}7= 100\times \frac {10^6-1}7 +-14\times (10^6 -1) = 100\times \frac {10^6-1}7 - 14\times 10^6 + 14$. ... Now $100\times {10^6-1}7$ will add two zeros to $142857$ to get $14285700$. And $-14\times 10^6=-14000000$ removes the leading $14$ to get $285700$ and $+14$ adds it back to the other side: $285714$. Sep 22 at 20:06
• $1999998 = 2\cdot 999,999 \ne 2\cdot \frac {999999}7=285714$. Why did you do anything with $2\times 999999$ that was not part of the question and never part of my solution? Sep 22 at 20:08
The following is multiple choice question (with options) to answer.
Find the value of 72519 x 9999 = m? | [
"709115678",
"723417481",
"710017481",
"725117481"
] | D | 72519 x 9999 = 72519 x (10000 - 1)
= 72519 x 10000 - 72519 x 1
= 725190000 - 72519
= 725117481
D |
AQUA-RAT | AQUA-RAT-36586 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Pipes A and B can fill a cistern in 8 and 24 minutes respectively. They are opened an alternate minutes. Find how many minutes, the cistern shall be full? | [
"10",
"26",
"12",
"24"
] | C | 1/8 + 1/24 = 1/6
6 * 2 = 12
Answer: C |
AQUA-RAT | AQUA-RAT-36587 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
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Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Jerry travels 8 miles at an average speed of 40 miles per hour, stops for 12 minutes, and then travels another 20 miles at an average speed of 60 miles per hour. What is Jerry’s average speed, in miles per hour, for this trip? | [
"42",
"42.5",
"44",
"50"
] | A | Total Time taken by jerry = (8/40)*60 Minutes + 12 Minutes + (20/60)*60 Minutes = 44 Minutes
Average Speed = Total Distance / Total Time = (8+20) miles / (44/60)Hours = 28*60 / 44 = 42 Miles per hour
Answer: option A |
AQUA-RAT | AQUA-RAT-36588 | As others have said, one can ask that the girls' chairs, and the block of chairs for the boys, be chosen together (in this way, we treat the group of boys as "another girl", for a total of 5 "girls" choosing among 5 "seats"); thus, we can have the boys choose one of the 5 "seats" (in reality, it is their block of chairs), the first girl chooses one of the 4 remaining chairs, the second girl chooses one of the 3 remaining chairs, etc. making for $$5!=5\times 4\times 3\times 2\times 1=120$$ ways of seating the girls and choosing the block of chairs for the boys. Then, all that remains is for the boys to arrange themselves within the block, which there are $$3!=3\times 2\times 1=6$$ ways to do, making (again) a total of $$(5!)\times (3!)=120\times 6=720$$ ways of arranging them.
-
+1 for the graphics! :-) – joriki Jul 8 '12 at 19:53
Thanks!$\text{}$ – Zev Chonoles Jul 8 '12 at 20:00
Treat the group of three boys as an additional girl. That makes five girls, which can be arranged in $5!$ ways. Then you can replace the additional girl by any permutation of the three boys, of which there are $3!$.
-
thanks joriki, but not quiet clear about what you say – user1419170 Jul 8 '12 at 19:49
@user1419170: Could you be more specific? What part isn't clear to you? – joriki Jul 8 '12 at 19:49
this part Treat the group of three boys as an additional girl – user1419170 Jul 8 '12 at 19:51
@user1419170: See Brian's and Saurabh's answers, which use the same idea but phrase it differently. – joriki Jul 8 '12 at 19:53
The following is multiple choice question (with options) to answer.
In a class composed of x girls and y boys, What part of the class is composed of girls | [
"y/(x + y)",
"x/xy",
"x/(x + y)",
"y/xy"
] | C | X/(X+Y).... ANSWER:C |
AQUA-RAT | AQUA-RAT-36589 | # Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have $$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$ There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B The answer for this question should be $0.35$, I am not sure where I did wrong.
I just figured maybe I should do this instead
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?
• Possible duplicate of Expected values of a dice game with a 30-sided die and a 20-sided die. – samjoe Mar 29 '17 at 6:00
• this is a different question – szd116 Mar 29 '17 at 6:05
• No, look properly. The first part of the question is same as yours, just it asks the probability of A winning. – samjoe Mar 29 '17 at 6:11
• I read it, but still couldn't figure out this problem – szd116 Mar 29 '17 at 6:13
• Look the answer to first part is 7/20 right? Now look at answer to that question. – samjoe Mar 29 '17 at 6:23
The following is multiple choice question (with options) to answer.
While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win? | [
"1/2",
"17/36",
"91/216",
"11/36"
] | C | Total outcomes possible: 36
Total outcomes possible with sum 7: 6
Probability to win when rolled once = 6/36
Probability not to win when rolled once = 30/36 = 5/6
Probability to win in three attempts= 1- Probability will not to win in all three attempts
= 1- (5/6* 5/6*5/6)
= 91/216
Answer is C. |
AQUA-RAT | AQUA-RAT-36590 | If stock price is $$X = (0,30)$$; we use the 7 puts (A) for profit = 7(40 - X); 9 (B) puts are used giving a loss = 9(30 - X). Total earning = 7(40-X) -9(30-X) + 2$= 2X + 12$ > 0
- 3 years, 2 months ago
Assume that the $40 put is still priced at$10.
What would be the price of the $30 put option, where there will be no arbitrage opportunity? Staff - 3 years, 2 months ago Log in to reply Let the price at which the 30$ put options is priced be $$M$$.
Now suppose that an arbitrage opportunity does exist. It is easily proved that the arbitrage opportunity(AO) must consist of buying 40$(A) put options and selling 30$ (B) put options. Let x (A) puts be bought and y (B) puts be sold for the AO.
So we spend $$10x$$ for the (A) puts and gain $$My$$ for the (B) puts. Total gain =$$My - 10x$$
If the stock price is above 40; both puts remain unused. Therefore net earnings = $$My - 10x.$$ Since earning is greater than 0 in an AO, $$My - 10x \geq 0 \implies My > 10x \implies (30-M)My > 10(30-M)x$$. Also, since $$M < 10$$, we have $$y > x$$
If the stock price is $$= P = (0,30)$$; both puts are used. Earning on (A) puts = $$x(40 - P)$$. Loss on B puts$$= y(30 - P).$$ Total earning = $$x(40 - P) - y(30-P) + My - 10x = 30x - (30 - M)y + (y-x)P$$
Since $$y - x > 0$$. Total earning is minimum when $$P = 0$$.
The following is multiple choice question (with options) to answer.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A. | [
"240",
"388",
"379",
"277"
] | A | Explanation:
(3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A |
AQUA-RAT | AQUA-RAT-36591 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled? | [
"2",
"2.2",
"3",
"3.5"
] | C | Part filled by A + B + C in 1 hour = 1/5 + 1/10 + 1/30 = 1/3
All the three pipes together will fill the tank in 3 hours.
ANSWER:C |
AQUA-RAT | AQUA-RAT-36592 | ### Bonus Question
As for the Bonus question, it is also an interesting exercise. If we start with $$n=1600$$ fair coins and at each turn we eliminate those landing tails, we wonder, how many rounds on average we will have before running out of coins.
A first approximation would be to say that, on average, we would expect to lose half of the coins each time. After the first toss we will end up with around $$800$$, then $$400$$, and after $$6$$ round we would expect to have only around $$25$$ coins. Now we are expecting to have an odd number of coins but we could keep halving until we have close to a single coin. That is, after $$10$$ rounds, $$\dfrac{1600}{2^{10}}\approx 1.5625$$ and so that we may expect to have at least an eleventh round before running out of coins. But, is the answer then $$11$$, $$12$$, $$13$$,…?
Just like with the main puzzle, by rephrasing the problem in terms of flipping all the $$n$$ coins and just considering a different notion of success we can come up with a better solution. Let us focus on a single coin; we would keep flipping it as long as it keeps turning heads. Well, the number of times we will flip said coin can be modeled as a Geometric random variable where now the success is landing tails. We see then that we will run out of coins the last time where a given coin lands tails for the first time. This is just to say that we are interested in the maximum of a series of $$n$$ independent Geometric random variables with parameter $$p=0.5$$ or, more precisely, in its expected value.
The following is multiple choice question (with options) to answer.
At a certain bowling alley, it costs $0.50 to rent bowling shoes for the day and $2 to bowl 1 game. If a person has $12.80 and must rent shoes, what is the greatest number of complete games that person can bowl in one day? | [
" 7",
" 8",
" 6",
" 10"
] | C | After renting bowling shoes the person is left with $12.80-$0.5=$12.30, which is enough for 12.3/2<7=~6.
Answer: C. |
AQUA-RAT | AQUA-RAT-36593 | 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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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A batsman has scored an average of 46 runs for a certain number of innings played in England. When he came back to India, he played another two test matches of two innings each and scored at an average of 55 runs. For the innings in England and in India taken together, he has improved his average by 2 runs over the matches played in England.Find the number of innings played in England. | [
"12",
"13",
"14",
"15"
] | C | Let the number of innings played in England be x.
∴ Total runs scored in England = 46x
Total runs scored for innings played in India
= 55 × 4 = 220
(∵ the number of innings played in India = 4)
Also, 46x+220 /x+4 = 48
⇒ 46x + 220 = 48 x + 192
⇒ 2x = 28
⇒ x = 14
Answer C |
AQUA-RAT | AQUA-RAT-36594 | 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? | [
"228",
"276",
"199",
"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-36595 | ## 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.
What values of x satisfy the equation x2/3 + x1/3 – 2 ≤ 0 ? | [
"– 8 ≤ x ≤ 1",
"– 1 ≤ x ≤ 8",
"1 ≤ x ≤ 8",
"– 8 ≤ x ≤ 8"
] | A | Explanation :
Let, x1/3 = y. The equation becomes y2 + y – 2 ≤ 0, or y2 + y ≤ 2.
Now, look at the upper limit. We know that y2 + y ≤ 2.
Among the positive numbers, 1 is the biggest number that satisfies the equation. No number greater than 1 can have y2 + y ≤ 2. So the upper limit is 1.
Maximum value of y is 1, hence maximum value of x1/3 is 1, and hence maximum value of x is 1.
This answer is there only for option 1. Hence that is the answer.
Answer : A |
AQUA-RAT | AQUA-RAT-36596 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 123, what is the least possible value of n? | [
"37",
"123",
"41",
"52"
] | C | 123 is 3 * 41, so n needs to be at least 41, so the answer is C. |
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