source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38297 | Amir has identified why the book's answer and your's differ, Felix Brauer
An urn contains 3 white balls and 4 black balls. Second urn contains 6 white balls and 4 black balls. From the first urn are draws 2 balls and they dropped in the second urn. Then from the second urn is drawn one ball which is white. What is the probability that two balls transferred in second urn were black balls ?
The solution to this is, using Conditional Probability and the Law of Total Probability: \begin{align} \mathsf P(F_{BB}\mid S_W) & = \frac{ \mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB}) }{ \mathsf P(S_W\mid F_{BB})\,\mathsf P(F_{BB})+\mathsf P(S_W\mid F_{BW})\,\mathsf P(F_{BW})+\mathsf P(S_W\mid F_{WW})\,\mathsf P(F_{WW}) } \\ & = \frac{\frac 6 {12}{4\choose 2}/{7\choose 2}}{\left(\frac 6 {12}{4\choose 2}+\frac 7{12}{3\choose 1}{4\choose 1}+\frac 8{12}{3\choose 2}\right)/{7\choose 2}} \\ & = \frac{6\cdot 6}{6\cdot 6+7\cdot 12+8\cdot 3} \\ & = 1 / 4 \end{align}
As you, and Amir, obtained.
The following is multiple choice question (with options) to answer.
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other
without replacement. What is the probability that both drawn balls are black? | [
"1/2",
"1/5",
"1/5",
"3/7"
] | D | Let E and F denote respectively the events that first and second ball drawn
are black. We have to find P(E n F) or P (EF).
Now P(E) = P (black ball in first draw) = 10/15
Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
That is P(F|E) = 9/14
By multiplication rule of probability, we have
P (E n F) = P(E) P(F|E)
= 10/15 × 9/14 = 3/7
D |
AQUA-RAT | AQUA-RAT-38298 | c#, formatting
/// <summary>
/// Calculates and returns the general discount
/// based on the <see cref="AccountStatus"/>
/// </summary>
/// <param name="accountStatus"></param>
/// <returns></returns>
private static decimal GetDiscountPercentage(AccountStatus accountStatus)
{
switch(accountStatus)
{
case AccountStatus.SimpleCustomer:
return 0.10m;
case AccountStatus.ValuableCustomer:
return 0.30m;
case AccountStatus.MostValuableCustomer:
return 0.50m;
default:
return 0.00m;
}
}
/// <summary>
/// Applying the discounts (if any) on the price and returns the final price (after discounts).
/// </summary>
/// <param name="price"></param>
/// <param name="accountStatus"></param>
/// <param name="timeOfHavingAccountInYears"></param>
/// <returns></returns>
public static decimal ApplyDiscount(decimal price , AccountStatus accountStatus , int timeOfHavingAccountInYears)
{
decimal loyaltyDiscountPercentage = GetLoyaltyDiscountPercentage(timeOfHavingAccountInYears);
decimal discountPercentage = GetDiscountPercentage(accountStatus);
decimal priceAfterDiscount = price * (1.00m - discountPercentage);
decimal finalPrice = priceAfterDiscount - ( loyaltyDiscountPercentage * priceAfterDiscount );
return finalPrice;
}
}
since all methods don't need several instance, and the nature of the class is unchangeable, making the class static would be more appropriate.
You can then reuse it :
var finalPrice = DiscountManager.ApplyDiscount(price, accountStatus, timeOfHavingAccountInYears);
The following is multiple choice question (with options) to answer.
A single discount equivalent to the discount series of 20%, 10% and 5% is? | [
"31.7",
"31.4",
"31.6",
"31.1"
] | C | 100*(80/100)*(90/100)*(95/100)
= 68.4
100 - 68.4
= 31.6
Answer:C |
AQUA-RAT | AQUA-RAT-38299 | => 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.
Mary's income is 60 percent more than Tim's income, and Tim's income is 60 percent less than Juan's income. What percent of Juan's income is Mary's income? | [
" 64%",
" 120%",
" 96%",
" 80%"
] | A | Juan's income = 100 (assume);
Tim's income = 40 (60 percent less than Juan's income);
Mary's income = 64 (60 percent more than Tim's income).
Thus, Mary's income (64) is 64% of Juan's income (100).
Answer: A. |
AQUA-RAT | AQUA-RAT-38300 | 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 work crew of 6 Men takes 8 days to complete one-half of a job. If 9 men are then added to the crew and the men continue to work at the same rate, how many days will it take the enlarged crew to do the rest of the job? | [
"2",
"3",
"3 1/5",
"4"
] | C | Suppose 1 man can do work in X days..
so 4 men will do in ..
6/X=1/8*1/2 as half job is done
X=96
now 9 more are added then
15/96=1/2*1/d for remaining half job
d=3 1/5 Number of days
C |
AQUA-RAT | AQUA-RAT-38301 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A 2-year certificate of deposit is purchased for K dollars. If the certificate earns interest at a n annual rate of 6 percent compound quarterly, which of the following represents the value, in dollars, of the certificate at the end of 2 years? | [
"k(1.06)^2",
"k(1.06)^8",
"k(1.015)^2",
"k(1.015)^8"
] | D | term (n) = 2
value of k at the end of 2 yrs, if compounded annually=k(1+r)^n=k(1+0.06)^2
quarterly int. rate (r) = 1.5%
term (n) = 2x4 = 8
value of k at the end of 2 yrs, if compounded qtrly =k(1+r)^n=k(1+0.015)^8
ANSWER:D |
AQUA-RAT | AQUA-RAT-38302 | thermodynamics, energy, home-experiment
Note 3: Peak hours tend to be during the day (this may reverse if we get wide-spread solar energy), so keeping the house cool during the day would cost even more. You pretty much have it right. We have two scenarios:
1 - Leave air conditioning on all day. Say that outside temperature is 90 degrees and inside temperature is 70 degrees. Then, all day long, the air conditioning has to remove any heat that gets into the house continuously. Heat transfer depends on the difference in temperature between outside and inside the house. In this case, heat transfer per unit time is equal to a constant K times 20 degrees. Over a whole day of D units of time, the total energy removed by the air conditioner is $20KD$ This assumes that convection (ie air draft) can be neglected.
2 - Stop the air conditioning during the day. If you stop the air conditioning, the flow of heat in the house initially will be the same. However, as time goes by, the inside temperature will increase. This will gradually reduce heat flow as it is proportional to the difference in temperatures. If the house is small enough and badly insulated, you could even get to a point where the inside temperature is equal to the outside temperature, at which point no more heat enters the house. When you get back from work, or if you have a timer a bit before, air conditioning is restarted. You now have to remove all the accumulated heat. However, since less heat entered the house over the duration of the day than if you had kept the air conditioning on, you have less work to do to cool down the house. Yes, the heat will have accumulated in insulation, furniture, walls, etc, but it doesn't matter. There is less heat total to remove.
The following is multiple choice question (with options) to answer.
An air conditioner can coo the hall in 20 minutes while another takes 25 minutes to cool under similar conditions. if both air conditioners are switched on at same instance then how long will it take to cool the room? | [
"about 14 minutes",
"about 12 minutes",
"about 15 minutes",
"about 16 minutes"
] | B | Take LCM of 20 and 25 = 100
20*5=100/ 25*4=100.
So if both are switched on together , room cooled in 100/(5+4)=100/9=11.11 minutes
11.11 can be round to 12 minutes , so about 12 minutes
ANSWER: B |
AQUA-RAT | AQUA-RAT-38303 | The totient of $210$ - the number of values between $1$ and $210$ that are relatively prime to $210$ - is $(2-1)(3-1)(5-1)(7-1)=48$. Using this, we can say that there are $48\cdot5=240$ numbers not divisible by these four numbers up to $1050$. Some of these of course are out of range of the original question; we'll have to figure out what those are.
The totient of $30$ is $8$; from $991$ to $1050$ there are $16$ numbers relatively prime to $30$. We'll take these out for now, leaving $224$. But two numbers we removed - $1001$ and $1043$ - are divisible by $7$ and so weren't part of the original $240$ so we have to un-remove them, adding them back to the total. Further, $991$ and $997$ are below $1000$ so shouldn't have been removed either. This gives $224+2+2=228$ numbers relatively prime to $210$, so $1000-228=772$ numbers are divisible by $2$, $3$, $5$, or $7$.
• Note that this doesn't work when any of the numbers we're testing divisibility for are composite. – Dan Uznanski Jan 1 '18 at 14:34
• how would you change it to work for composite numbers? – oliver Apr 27 '20 at 13:51
Your solution is correct (or, at least, your result matches mine). Huzzah!
The following is multiple choice question (with options) to answer.
What is 30% of 210? | [
"61",
"63",
"65",
"67"
] | B | X/210 = 30/100
100X = 210X30
X =(210X30)/100
X =6300/100
= 63
ANSWER = B = 63 |
AQUA-RAT | AQUA-RAT-38304 | $\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
4. Originally Posted by chancey
Right, but thats not the answer in the back of the book...
I did it this way:
$\sum_{n=1}^{20} (2^n - 1) = \sum_{n=1}^{20} (2^n) - 20$
$\sum_{n=1}^{20} (2^n) = \frac{1-2^{21}}{1-2} - 1 = 2097150$
$\because \sum_{n=0}^{k} (ar^n) = \frac{a(1-r^{k+1})}{1-r}$
$\therefore \sum_{n=1}^{20} (2^n) - 20 = 2097150 - 20 = 2097130$
Which is the same answer as yours, however the answer in the book says $2097170$. They must of added the 20 instead of subtracted it. OK thanks
Then the book is inncorect.
The following is multiple choice question (with options) to answer.
The sum of 2 numbers is 2000. Their L.C.M. is 21879. Their difference is? | [
"1845",
"1978",
"1945",
"1862"
] | B | Let the numbers be x and 2000-x
L.C.M. = x(200-x) = 21879
x^2 - 2000x + 21879 = 0
(x-1989) (x-11) = 0
x = 1989 and x = 11
Difference = 1989-11 = 1978
Answer is B |
AQUA-RAT | AQUA-RAT-38305 | investment Banking Course, Download Valuation... As are bonds less than 365 days, most developed countries use simple interest is bonds! Is 1-Feb-2020, and investors collect returns when the bond “ days maturity... The market price quarterly, or the coupon Equivalent rate ( CER.. For calculating return on investment and an investment that is taxed bankruptcy and liquidates... Second part is used to calculate the annual yield of Second Govt basis that incorporates! 42 every year for calculating return on investment and the yield to calculate bond Equivalent yield for company... Thus 11 % multiplied by two, which comes out to 22 % have to understand this! ( 5 ) /95 } * { 1.520833 } ], bond Equivalent yield formula be calculated using the function! 365/180 } ], bond Equivalent yield formula on how to calculate the bond happens! Is Rs value ) to pay ( coupon rated ) is an calculation. From an annual-pay bond can not be directly compared to find out the bond Equivalent formula... That of bond or call premium of years until maturity or until call until. Calculate current yield calculation, as are bonds can be calculated using following! 5 ) /95 } * { 1.520833 } ], bond Equivalent formula. To yield to calculate EAY as follows: EAY bond equivalent yield formula 1.0253 ( 365/90 ) = %... Different price and tenure discount and do not pay interest at all NCD, the to. /.75 ) = 10.66 % * 100 1 as the current market price the... Until maturity or until put is exercised Contents ) or bonds ( fixed income securities, which sold! Allows the investor to calculate the annualized yield of Second Govt also some bonds, do not annual. Issue, we must know the bond Equivalent yield formula = ( Face value Rs!
The following is multiple choice question (with options) to answer.
What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum? | [
"277",
"278",
"120",
"276"
] | C | Let the present worth be Rs. x. Then,
S.I. = (132 - x)
(x * 5 * 2) / 100 = (132 - x)
10x = 13200 - 100x
110x = 13200 => x = 120.
Answer:C |
AQUA-RAT | AQUA-RAT-38306 | # 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.
If the selling price is quadrupled,the profit is pentadruple, what is the percentage of profit?? | [
"50%",
"100%",
"150%",
"200%"
] | B | we can make two equations from the given info..
let selling price be S and cost price be C..
so profit=P=S-C...
if S is doubled profit is tripled..
so 5P=4S-C..
or S=4P...
therefore C=P..
% of profit=P/P=100%
B |
AQUA-RAT | AQUA-RAT-38307 | 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
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
The batting average of a particular batsman is 60 runs in 46 innings. If the difference in his highest and lowest score is 180 runs and his average excluding these two innings is 58 runs, find his highest score. | [
"179",
"194",
"269",
"177"
] | B | Explanation:
Total runs scored by the batsman = 60*46 = 2760 runs
Now excluding the two innings the runs scored = 58*44 = 2552 runs
Hence the runs scored in the two innings = 2760 – 2552 = 208 runs.
Let the highest score be x, hence the lowest score = x – 180
x + (x - 180) = 208
2x = 388
x = 194 runs
ANSWER: B |
AQUA-RAT | AQUA-RAT-38308 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Find the simple interest on Rs.500 for 9 months at 6 paisa per month? | [
"867",
"270",
"267",
"278"
] | B | I = (500*9*6)/100
= 270
Answer: B |
AQUA-RAT | AQUA-RAT-38309 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
50% of a number is added to 120, the result is the same number. Find the number? | [
"300",
"277",
"240",
"99"
] | C | :
(50/100) * X + 120 = X
2X = 480
X = 240
Answer: C |
AQUA-RAT | AQUA-RAT-38310 | => 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 labourer was engaged for 25 days on the condition that for every day, he works, he will be paid Rs. 2 and for every day, he is absent he will be fined 50p. If he receives only Rs. 37•50, find the no. of days he was absent is_____ | [
"3 days",
"4 days",
"5 days",
"6 days"
] | C | if a labour worked for 25 days he should get 50 Rs.
but in this he is getting 13 Rs less. if he don't go for work for 1 day he will loose 2.5 Rs then
no of days(Absent)=13/2.5=5.2. Means he was absent for 5 days
ANSWER:C |
AQUA-RAT | AQUA-RAT-38311 | The midpoint of $\displaystyle DE$ is: $\displaystyle M(3\tfrac{1}{2},\,2).$
The median to side $\displaystyle DE$ starts at $\displaystyle \,M$, passes through $\displaystyle \,C,$
. . and extends to $\displaystyle \,F$, where: .$\displaystyle FC \,=\,2\!\cdot\!CM.$
Going from $\displaystyle \,M$ to $\displaystyle \,C$, we move up 2 and left $\displaystyle \frac{1}{2}$
Hence, going from $\displaystyle \,C$ to $\displaystyle \,F$, we move up 4 and left 1.
Therefore, we have: .$\displaystyle F(2,8).$
The following is multiple choice question (with options) to answer.
What is the median from the below series 90, 92, 93, 88, 95, 88, 97, 87, 108and 98 | [
"80",
"93.5",
"92",
"98"
] | B | Ordering the data from least to greatest, we get:
87, 88, 88, 90, 92, 93, 95, 96, 98,108
The median quiz score was 93.5 (Four quiz scores were higher than 93.5 and four were lower.)
B |
AQUA-RAT | AQUA-RAT-38312 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
How many numbers are there between 100 and 1000 in which all digits are distinct. | [
"900",
"648",
"720",
"252"
] | B | We know that numbers between 100 and 1000 will be 3 digit integers.
Let the arrangement be ABC where A,B,C represent distinct digits.
We have 9 options for A (as 0 can not start a 3 digit number), 9 options again for B and 8 for C, giving a total of 9*9*8 = 648 options.
ANSWER B |
AQUA-RAT | AQUA-RAT-38313 | # Ten-digit number that satisfy divisibilty rules for 2,3,4,5,6,7,8,9,10&11
Question: Arrange the digits 1,2,3,4,5,6,7,8,9,0 to make a ten-digit Number that satisfies all of the divisibility rules for 2,3,4,5,6,8,9,10,&11. BONUS: make the number also divisible by 7
• Should I add no-computer tag? – Omega Krypton Oct 27 '18 at 2:33
• Why did you have to take down the post? By posting on the Stack Exchange network, you've granted a non-revocable right for SE to distribute that content (under the CC BY-SA 3.0 license). By SE policy, deletions like this will be reverted. – Glorfindel Oct 28 '18 at 18:56
• @Glorfindel thanks for informing me. I am slowing learning how this all works – DeNel Oct 28 '18 at 20:17
• What does "satisfies all of the divisibility rules" mean? That the number needs to actually be divisible by all those divisors? Or just that it needs to follow some well-known necessary (but not necessarily sufficient) rules to be divisible by them? – R.. GitHub STOP HELPING ICE Oct 28 '18 at 21:27
Alright. First of all,
3 and 9 are automatically okay. The sum of the digits from 0 to 9 is 45, which is divisible by 3 and 9, so any number made out of these ten will be divisible by 3 and 9.
Now,
the last digit MUST be 0, to satisfy 2 and 5 and 10. This also satisfies 6 as well, because it's now divisible by 2 and 3.
Now, let's look at the second-last and third-last digits.
The second-last digit must be divisible by 4, because the number must be divisible by 4 and 8. The third to last digit must also be even, to make it divisible by 8.
Now
The following is multiple choice question (with options) to answer.
How many numbers from 10 to 43 are exactly divisible by 3? | [
"13",
"11",
"16",
"17"
] | B | 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42.
11 Numbers.
10/3 = 3 and 43/3 = 14 ==> 14 - 3 = 11. Therefore 11 digits
B) |
AQUA-RAT | AQUA-RAT-38314 | quantum-mechanics, soft-question, popular-science
The situation here is uncertain because of issues far displaced from chaos or quantum. The challenges arise because it is massively difficult to isolate the input factors and determine them.
School performance, for example, could esily be expected to be influenced by many different inputs. Family patterns, income, culture, etc. etc. It would be a chore to even list them all, never mind determine them and extract the effect. Thus, supposing there is an effect due to nutrition, it may be required to extract the effect from the effect of large numbers of other input factors. Each of which may be known to a different degree of accuracy.
This school uses this set of textbooks and that another set. It already begins to be difficult to extract results. This school runs for 7 hours per day and that one 9. This one starts students at age 6 and that at age 5. And so on and so on. This one is in the innner city and that one is far out in rural areas. All of these could easily be expected to have some effect, though in advance it would be difficult to know exactly what.
And it is also massively difficult to be confident in the output factors. It may be that different school systems do things very differently and so performance is a major challenge to extract and compare. This school uses numerical grades and that letter grades is just a trivial example. This school uses standardized state-wide tests, this one uses surprise quizzes, and that one uses interviews of the students. This one does a lot of written work with written tests, and that one does a lot of hands-on work with group presentations. It is a challenge to extract the results and present them in a manner that fairly compares them. Sometimes even for the same school from one year to the next.
And, even when two schools use the same method of evaluating performance, it is not instantly obvious that this method is a good predictor of the academic future of a given student. Written tests, for example, can be well correlated on average. But the individual student is subject to a huge collection of factors that may mean he does not align with the results his tests would suggest.
The following is multiple choice question (with options) to answer.
Statements: Prime school-going children in urban India have now become avid as well as more regular viewers of TV, even in households without a TV. As a result there has been an alarming decline inthe extent of readership of newspapers.
Conclusions:
1) Method of increasing the readership of newspapers should be devised.
2) A team of experts should be sent to other countries to study the impact of TV. on the readership of newspapers. | [
"Only conclusion I follows",
"Only conclusion II follows",
"Either I or II follows",
"Neither I nor II follows"
] | D | The statement concentrates on the increasing viewership of TV. and does not stress either on increasing the readership of newspapers or making studies regarding the same. So, neither I nor II follows.
D |
AQUA-RAT | AQUA-RAT-38315 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
If y > 0, (6y)/20 + (3y)/10 is what percent of y? | [
"40%",
"50%",
"60%",
"70%"
] | C | can be reduced to 3y/10+3y/10 =3y/5=60%
C |
AQUA-RAT | AQUA-RAT-38316 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
Of the three- digit integers greater than 700, how many have two digits Q that are equal to each other and the remaining digit different from the other two? | [
"90",
"82",
"80",
"45"
] | C | three-digit integers greater than 700: 701 to 999, inclusive.
possible values for hundreds-digit--> 7,8,9
possible values for tens-digit and ones-digit --> 0, 1,2,3,4,5,6,7,8,9
when hundreds-digit and tens-digit are the same: (3x1x10)-3=27 ---> we minus three to exclude 777, 888 and 999
when hundreds-digit and ones-digit are the same: (3x10x1)-3=27 ---> we minus three to exclude 777, 888 and 999
when tens-digit and hundreds-digit are the same:[(3x10x1)-3]-1=26 ---> we minus three to exclude 777, 888 and 999; we minus one to exclude 700
Q=27+27+26 = 80
Answer: C |
AQUA-RAT | AQUA-RAT-38317 | ### Show Tags
19 Aug 2015, 01:34
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Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
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Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average monthly salary of 15 employees in an organisation is Rs. 1800. If the manager's salary is added, then the average salary increases by Rs. 150. What is the manager's monthly salary? | [
"Rs.3601",
"Rs.3618",
"Rs.4200",
"Rs.3619"
] | C | Manager's monthly salary
= Rs. (1950 * 16 - 1800 * 15)
= Rs. 4200 Answer:C |
AQUA-RAT | AQUA-RAT-38318 | 7,7),(1,8,7),(2,9,7),(8,0,8),(7,1,8),(9,1,8),(6,2,8),(5,3,8),(4,4,8),(3,5,8),(2,6,8),(1,7,8),(0,8,8),(1,9,8),(9,0,9),(8,1,9)
The following is multiple choice question (with options) to answer.
0, 3, 8, 15, 24, ? | [
"28",
"35",
"48",
"25"
] | B | The sequence is a series of squares-1
0, 3, 8, 15, 24, 35
Answer : B. |
AQUA-RAT | AQUA-RAT-38319 | # Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
The following is multiple choice question (with options) to answer.
A rectangular tiled patio is composed of 90 square tiles. The rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles. After the change in layout, the patio will still have 90 tiles, and it will still be rectangular. How many rows are in the tile patio before the change in layout? | [
"5",
"6",
"10",
"11"
] | D | r*c=90 and (11+4)(c-2)=90 --> r=11 and c=8.
Answer:D. |
AQUA-RAT | AQUA-RAT-38320 | # 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.
John and Peter are among the seven players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter? | [
"5/11",
"8/15",
"9/20",
"10/21"
] | D | The total possible ways of selecting a 5-member team is 7C5 = 21
The possible ways which include John and Peter is 5C3 = 10
The probability of choosing both John and Peter is 10/21
The answer is D. |
AQUA-RAT | AQUA-RAT-38321 | # 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.
If money is invested at r percent interest, compounded annually, the amount of the investment will double in approximately 48/r years. If Pat's parents invested $5,000 in a long-term bond that pays 8 percent interest, compounded annually, what will be the approximate total amount of the investment 18 years later, when Pat is ready for college? | [
" $20000",
" $15000",
" $12000",
" $10000"
] | B | Since investment doubles in 48/r years, then for r=8 it'll double in 48/8=~6 years (we are not asked about the exact amount so such an approximation will do). Thus after 18 years investment will become $5,000*3=$15,000 .
Answer: B. |
AQUA-RAT | AQUA-RAT-38322 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
My son adores chocolates. He likes biscuits. But he hates apples. I told him that he can buy as many chocolates he wishes. But then he must have biscuits twice the number of chocolates and should have apples more than biscuits and chocolates together. Each chocolate cost Re 1. The cost of apple is twice the chocolate and four biscuits are worth one apple. Then which of the following can be the amount that I spent on that evening on my son if number of chocolates, biscuits and apples brought were all integers? | [
"Rs 34",
"Rs 33",
"Rs 8",
"None of the above"
] | A | Explanation :
Let the person buy c chocolates, 2c buiscuits. Therefore, apples is >3c.
In any case, total cost will be
=>cost of chocolates + cost of biscuits + cost of apples.
=> (c*1) + [2c*0.5] + (2*something greater than 3c).
Thus, Total cost=>
=> 2c + (2* something greater than 3c).
Here, the total cost is an even number. Thus ans should be an even number.
Taking c=1, the total minimum cost we get is 10.
Thus the only other option that satisfies the constraints is 34.
Answer : A |
AQUA-RAT | AQUA-RAT-38323 | java, algorithm, knapsack-problem
Title: Stampcalculator - Given a set of stamps, what combinations are there to reach a certain amount? Background
My mother has a hobby of buying and reselling books via online trading sites. After a price is agreed on, the books have to be put into an envelope for mailing. On this envelope, stamps have to be applied to pay for postage (mailing fee). My mother buys stamps in bulk at various prices, varying from 75 to 90% of face value. As the postage prices change every year, new stamp combinations have to be calculated that will satisfy the postage fee with a minimum amount of hassle.
My mother is not all that good with calculations. And having to manually recalculate what stamp combinations are best to get to the postage required for a 250 - 500 gram package takes a lot of time. Therefore, she calculates a few options via trial and error - sometimes overshooting the postage amount by a few cents - and just uses those.
Some stamps are preferred to be used, whereas others aren't, due to the cost not always being the face value of a stamp. Asking my mother about the stamp prices individually would have taken too much time, so I just made a program that generates a list of options. It requires recompilation when you need to add different stamps or want to get to a different limit, but that's okay. The whole point of this was to not spend too much time on it (the goal is to save time!) - which is why I wrote it in the total time of 1 hour (from idea to solution).
Problem description
Given a target amount, a set of stamp denominations, and a maximum amount of usable stamps, print a list of stamp combinations that will get to the target amount.
Programmed in 1 hour. I have placed everything in one file to ensure that it can run in Ideone, so that she could theoretically make the changes herself, if needed.
How it works
Using a list of stamps, make partial combinations that are at or below the target amount. For each combination, make a new combinations via duplicating the current combination and adding a stamp with a value equal to or below the last added stamp. Combinations that go past the goal are discarded. Combinations that meet the goal are added to a solutions list.
The following is multiple choice question (with options) to answer.
A company hired a printer to produce a total of x + 2 envelopes. The job consisted of two types of envelopes, 3¢ envelopes and 7¢ envelopes. If the company requested 2 more 3¢ envelopes than 7¢ envelopes, which of the following expressions denotes the cost, in cents, of the total x + 2 envelopes ? | [
"(9x-5)/2",
"(7x+4)/2",
"5x+6",
"6x+5"
] | C | we can take
X+2 = 10
X= 8
So 3c envelopes will be 6
and 7c envelopes will be 4
So total cost will be 46
Then that is our target
Plug x =8 in answer choices
Only c works
Correct Answer : C |
AQUA-RAT | AQUA-RAT-38324 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
In a school of 400 boys, 44% of Muslims, 28% Hindus, 10% Sikhs and the remaining of other communities. How many belonged to the other communities? | [
"41",
"63",
"72",
"63"
] | C | 44 + 28 + 10 = 82%
100 – 82 = 18%
400 * 18/100 = 72
ANSWER:C |
AQUA-RAT | AQUA-RAT-38325 | # Lottery problem - dependent vs. independent events?
A lottery sells $10,000,000$ tickets with probability of winning at $\large\left(1\over 5000000\right)$
Let $t$ to represent the number of tickets sold. Using the formula $1-\left(4999999\over 5000000\right)\large^t$, the probability of there being a winner is approximately $86.5\%$ (or for no winner, $13.5\%$). Okay, I understand that.
Using the same numbers, what if the lottery guarantees there will be $\large2$ winners.
This is problematic because the probability of $0$ winners calculates as $13.5\%$, which is impossible since we know that there will be $2$ winners out of $\large t$ tickets sold. Ultimately my question and confusion is how would you calculate the probability of winner/no winner in such a scenario?
Thanks
Edit:
Let me clarify what I'm basically asking. I know the events are not independent because the lottery is guaranteeing exactly 2 winners out of 10 million. So, how do you calculate probabilities of $X$ winners at various points of tickets sold. For example, they have sold 9,000,000 out of the possible 10,000,000 tickets. What is the probability of only one winner so far? The winners are not selected after 10,000,000; they are picked randomly.
I assume for 1 winner it would be $9,000,000 \over 10,000,000$
But what about non-obvious figures, i.e. 2 winners?
Would it simply be $\left(9,000,000\over 10,000,000\right)\large^2$
-
(1) you have omitted something in calculating 0.1353; and (2) everything depends on precisely how the lottery proposes to satisfy the guarantee of two winners. – Pieter Geerkens Sep 1 '13 at 3:52
What did I miss? And the lottery issues the tickets a la raffle. Out of the 10 million sold, 2 are winners. This is determined before they are sold. – Peeping Tom Sep 1 '13 at 3:55
The following is multiple choice question (with options) to answer.
In a certain lottery drawing, two balls are selected at random from a container with 10 balls, numbered from 1 to 10, inclusive. If the winner of the lottery is awarded a cash prize in the amount of $1,000 times the product of the numbers on the two selected balls, which of the following is a possible amount of the cash prize given to the winner? | [
"$9 x 10^4",
"$9.90 x 10^6",
"$1.00 x 10^7",
"$1.05 x 10^7"
] | A | The max product of any two balls can be 9 * 10 = 9900 = 9 * 10
If we multiply it by $1000 the max amount will be 9 * 10^4.
A |
AQUA-RAT | AQUA-RAT-38326 | & \text{if b\equiv0\pmod{4}}\\ \\ \binom{n^2/4}{(b-1)/4} & \text{if b\equiv1\pmod{4}}\\ \\ 0 & \text{otherwise.}\end{cases}$$ Specializing to $b=n,$ we get $$C(n,n,C_4)=\begin{cases}\binom{n^2/4}{n/4} & \text{if n\equiv0\pmod{4}}\\ \\ \binom{(n^2-1)/4}{(n-1)/4} & \text{if n\equiv1\pmod{4}}\\ \\ 0 & \text{otherwise.}\end{cases}$$
The following is multiple choice question (with options) to answer.
If A : B = 4 : 3 and B : C = 5 : 9 then A : B : C is : | [
"20 : 35 : 63",
"20 : 15 : 27",
"30 : 35 : 65",
"25 :34: 68"
] | B | Expl : A : B = 4 : 3
B : C = 5 :9 = 5*3/5 : 9 *3/5 = 3 : 27/5
A : B : C = 4 : 3 : 27/5 = 20 : 15 : 27
Answer: B |
AQUA-RAT | AQUA-RAT-38327 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A third of Arun’s marks in mathematics exeeds a half of his marks in english by 80.if he got 240 marks In two subjects together how many marks did he got inh english? | [
"140",
"150",
"160",
"170"
] | C | M=mathematics marks
E=english marks
M+E=240-(1eqution)
M=E+E/2-(2 equation)
put 2nd equation in 1st equation
E+E/2+E=240
3E/2=240
3E=240*2
3E=180
E=180/3
E=160
ANSWER:C |
AQUA-RAT | AQUA-RAT-38328 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
There are 3 mixtures when mixed in the ratio of 3:4:5 price was 240.When mixed with 6:4:9 price was 340.what's the price when mixed in 5:6:8 ration price is_________ | [
"346",
"356",
"366",
"376"
] | D | 3x+4x+5x=240
x=20
6x+4x+9x=340
x=17.89
5*20+6*20+8*20=380-4=376
ANSWER:D |
AQUA-RAT | AQUA-RAT-38329 | The square of every integer is of the form either $3k+1$ or $3k$
How can I prove that square of every integer is of the form either $3k+1$ or $3k$, but not $3k+2$?
My approach
I considered first, the integer $n$ to be even and then $n= 2m$; and if $n$ is odd then $n=2m+1$ but this step takes me to nowhere. I get usually stuck in these kind of questions. and this time I seriously need help . So, please post answer in detail and also mention how and why you arrived at that particular step. Thanks in advance.
-
I did not see you take the square of the integers you made. – Raskolnikov Jul 24 '12 at 7:01
Take $n=3k$ or $n=3k\pm 1$ – Kns Jul 24 '12 at 7:05
Nearly the same: math.stackexchange.com/questions/172535/… – Zander Jul 24 '12 at 14:31
The following is multiple choice question (with options) to answer.
The number 81 can be written as the sum of squares of 3 integers. Which of the following could be the difference between the largest and smallest integers of the 3? | [
"2",
"5",
"8",
"7"
] | D | Notice that the question asks which of the followingcouldbe the difference between the largest and smallest integers, not must be. The 3 integers could be: +/-1, +/-4 and +/-8, so the difference could be 7,9, or 12. Since only one of them is among the choices, then it must be the correct answer.
Answer: D. |
AQUA-RAT | AQUA-RAT-38330 | 2. ### math
list all the 3 digit numbers that fit these clues. the hundreds digits is less than 3. the tens digit is less than 2. the ones digit is greater than 7
3. ### Math
Think of a five-digit number composed of odd numbers. The thousands digit is two less than the ten thousands digit but two more than the hundreds digit. The tens digit is two more than ones digit which is four less than the
4. ### Math (Confused)
How many international direct-dialing numbers are possible if each number consists of a four-digit area code (the first digit of which must be nonzero) and a five-digit telephone numbers (the first digit must be nonzero)? a.
1. ### math
how many positive 4-digit numbers are there with an even digit in the hundreds position and an odd digit in the tens position? a. 10,000 b. 5,040 c. 2,500 d. 2,250
2. ### math
Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3-digit integer to the original 4-digit
3. ### Math
7 digit number no repetition of numbers digit 5 in thousands place, greatest digit in the millions place, digit in the hundred thousands place is twice the digit in the hundreds place, digit in the hundreds place is twice the
4. ### math
the hundred thousands digit of a six-digit even numbers is 3 more than the thousand digit,which is twice the ones digit.give at least four numbers that satisfy the given condition.
The following is multiple choice question (with options) to answer.
E is an even integer greater than 300,000 and smaller than 1,000,000. How many numbers can E be? | [
"300,000",
"349,999",
"350,000",
"399,999"
] | B | E 1,000,000-300,000=700,000 integers
700,000/2= 350,000 even integers.
350,000+1 inclusive. But since 1,000,000 and 300,000 are not included.
350,001-2=349,999
B |
AQUA-RAT | AQUA-RAT-38331 | 05 Jan 2010, 14:01
IMO it's (A)
From Statement 1:
n = a * a * a * a * b * b * b
n has 13 factors: 12 different combinations of a & b + 1.
Statement 2 tells nothing: n could have only 2 prime factors but what about non-prime factors?! we should consider them too. f.e. 35, 25, 49 and so on.
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Posts: 170
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Re: How many different factors does the integer n have? [#permalink]
### Show Tags
05 Jan 2010, 14:37
ohhh I didnt read the question correctly...anyways thanks a lot...
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Posts: 287
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Re: How many different factors does the integer n have? [#permalink]
### Show Tags
06 Jan 2010, 04:57
Thanks for clarification, I've missed $$a^0$$ and $$b^0$$
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Re: How many different factors does the integer n have? [#permalink]
### Show Tags
14 Jan 2017, 04:54
1
Nice Question.
Here is what i did in this one.
We need to get the number of factors of positive integer n.
Statement 1=>
As a and b are "different" prime numbers => Number of factors of a must be 5*4=20
Hence sufficient .
Statement 2=>
There exist ∞ numbers with the same set of prime numbers.
E.g
5*7=> Four factors.
5^2*7^2=> Nine factors.
Etc.
Hence not sufficient.
Hence A.
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The following is multiple choice question (with options) to answer.
Some of the workers at a hedge fund are analysts; the others are not analysts. There are a total of 2x/3 analysts at the hedge fund. A total of x hedge fund workers have an MBA. Furthermore, x/3 analysts have an MBA, y/4 of the hedge fund workers are neither analysts nor have an MBA, and 80 workers are not analysts. If there are y total workers at the hedge fund, x is equal to which of the following? | [
"45",
"50",
"60",
"72"
] | D | AnalystsNot AnalystsTotal
MBAx/32x/3x
No MBAy/4
Total2x/350y
We are asked for the value of x.
From the table: 2x/3 = y - 80
y - 80 + (y/4) = 80
5y/4 =160
y = 128
2x/3 = 128 - 80 = 48
x = 144/2 = 72
Answer: D |
AQUA-RAT | AQUA-RAT-38332 | Suppose an urn contains 8 red,5 white and 7 blue marbles.a.If 3 marbles are drawn at random from the urn with replacement,what is the probability that three marbles are red?
8. ### Math
there are 9 blue marbles, 4 black marbles, 5 white marbles and 6 red marbles. If the probability of drawing a blue marble is now 1/3, how many of the 6 marbles removed were blue?
9. ### Finite Math
An urn contains 8 blue marbles and 7 red marbles. A sample of 6 marbles is chosen from the urn without replacement. What is the probability that the sample contains at least one blue marble?
10. ### math
You randomly draw marbles from a bag containing both blue and green marbles, without replacing the marbles between draws. If B=drawing a blue marble and G=drawing a green marble, which represents the probability of drawing a blue marble …
More Similar Questions
The following is multiple choice question (with options) to answer.
An urn contains g green marbles and b blue marbles. If 2 green marbles and 5 blue marbles are added, and then one marble is selected at random from the urn, what is the probability that a blue marble will be drawn? | [
"b/(b+g)",
"(b+5)/(g+3)",
"(b+5)/(b+g+7)",
"(b+5)/(b+g+3)"
] | C | no of fav outcomes i.e. blue MArbles = b+5
Total no of outcomes i.e. total Marbles = (b+5)+(g+2)= b+g+7
probability = (b+5) /(b+g+7)
Answer C |
AQUA-RAT | AQUA-RAT-38333 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
P and Q are both positive integers. When P is divided by Q, the remainder is some positive integer D, and when P is divided by (Q + 7), the remainder is also D. If P/Q = 1020.75 and P/(Q + 7) = 816.6, then which of the following gives the correct set of {D, Q}? | [
"{6, 12}",
"{6, 15}",
"{21,28}",
"{9, 15}"
] | C | p/q= d=p/q+7
d= .75 q
d= .6(q+7)
we get d = 21 and q = 28. Answer = C. |
AQUA-RAT | AQUA-RAT-38334 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
Of the 50 electronics components that a factory must
manufacture, 70 percent would be most e¢ ciently manufactured
by Machine A and the remaining 30 percent would be most
e¢ ciently manufactured by Machine B, though either machine
could manufacture any of the 50 components. If 36 percent of
the components were manufactured by Machine A and the
remainder were manufactured by Machine B, what is the
highest possible number of components that were
manufactured by the machine that would manufacture them
the most efficiently? | [
"30",
"31",
"32",
"33"
] | D | # components A and B could manufacture efficiently:
A: 50*70% = 35
B: 50*30% = 15
# components A and B effectively manufacture:
A: 50*36% = 18
B: 50*64% = 32
Max. # components efficiently manufactured:
18 (from A) and 15 (from B) = 33
ANSWER:D |
AQUA-RAT | AQUA-RAT-38335 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
It is known that no more than 5 children will be attending a party. What is the smallest number of cookies that must be brought to the party so that each child receives the same number of cookies? | [
"35",
"105",
"60",
"210"
] | C | The question states that there are no more than 5 children in the party. Thus, the possibility is that there could be 1, 2, 3, 4 or 5 children.
If you assume answer as 35 and there are 3 children, you may not distribute be able to distribute cookies equally.
similarly if there were 105 cookies, and 2 children, cookies cannot be distributed equally.
or if there were 210 cookies, and 4 children, cookies cannot be distributed equally.
Thus, the question asks for a number of cookies which can be distributed to any number of children (from 1 to 5).
And therefore the smallest number of cookies would be lcm of (1, 2, 3, 4, 5), i.e., 60.
Answer (C) |
AQUA-RAT | AQUA-RAT-38336 | # Expected number of hours that all employees take a break at the same time?
This is derived from a real question my company has, so the wording might sound a bit weird.
There are 4 employees, each working 8 hours a day. Breaks are in 1 hour blocks and employees may only take breaks at the beginning of an hour (so there are 8 chunks to choose from). Each employee is entitled 2 breaks per day and are not required to use them. (EDIT) Each employee will choose uniformly choose among 8 break time slots available.
What is the expected number of hours that all 4 employees take a break at the same time?
Here's what I got so far.
The probability of everybody using 0 hours to break is _I_DONT_CARE_ because it doesn't affect the expected hours.
The probability of everybody using the same break, GIVEN THAT everybody used 1 hour break is $$x = (\frac1{8 \choose 1})^4$$
The probability of everybody using the same breaks, GIVEN THAT everybody used 2 hour breaks is $$y = (\frac1{8 \choose 2})^4$$
The expected number of hours that everybody takes a break at the same time is
$$1x + 2y$$
Is my calculation right? Is there an easier way to look at the problem?
The following is multiple choice question (with options) to answer.
Company A has 13 employees, 6 of whom belong to the union. If 5 people work any one shift, and the union contract specifies that at least 4 union members work each shift, then how many different combinations of employees might work any given shift? | [
"81",
"231",
"336",
"350"
] | A | Looks easy: -- 6C4*5C1 + 6C5 -- 81 (A). |
AQUA-RAT | AQUA-RAT-38337 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A is 30% more efficient than B. How much time will they working together take to complete a job which A alone could have done in 23 days? | [
"11",
"13",
"20 3/17",
"14"
] | B | Explanation:
Ratio of times taken by A and B= 100:130 =10:13
suppose B takes x days to work the 10:13::23:x =>x=23 x 13/10 = 299/10
A’s one day work = 1/23 and B’s is 10/299
Then A+B ‘s work = 1/23 + 10/299 = 23/299 =1/13 => 13 days
Answer: Option B |
AQUA-RAT | AQUA-RAT-38338 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
How many bricks each measuring 21cm x 10cm x 8cm, will be needed to build a wall 9m x 5m x 18.5m | [
"4955",
"4899",
"4650",
"7200"
] | A | Explanation:
no. of bricks= volume of the wall/volume of 1 brick = (900 x 500 x 18.5) / (21 x 10 x 8) = 4955
Answer: A |
AQUA-RAT | AQUA-RAT-38339 | 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
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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.
John has taken four (4) tests that have an average of 82. In order to bring his course grade up to a ‘B’, he will need to have a final average of 88. What will he need to average on his final two tests to achieve this grade? | [
"100",
"90",
"92",
"94"
] | A | 88-82=6*4=24 points
2*88=176
176+24=200 points to be scored in the 2 tests.
Avg of 2 tests = 200/2=100
Answer: A
Traditional Method:
Total scored till now 82*4=328
Total score to avg 88 in 6 tests = 88*6=528
Total to be scored on 2 tests = 528-328=200
Avg on 2 tests = 200/2 = 100
Answer A |
AQUA-RAT | AQUA-RAT-38340 | 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.
If shares of two persons in profits are Rs.300 and Rs.500 then ratio of their capitals is | [
"3:4",
"3:5",
"4:3",
"1:3"
] | B | total profit =1000
ratio=300/500
=3:5
ANSWER:B |
AQUA-RAT | AQUA-RAT-38341 | speed of the car in miles per hour? what dimensions should he use? Custom Solutions. Find the fraction. Find the fraction. Given : This year, the chickens laid 30% less eggs than they did last year and they laid 3500 eggs this year. This was$14 less than twice what she spent for a blouse. So, the chickens laid 5000 eggs last year. Since in one year the man will be six time as old as the daughter is now, the man's present age is. Time and work word problems. … So, the three angles of a triangle are 60°, 72° and 48°. Find the number. Get help with your Math Word Problems homework. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Given : Difference between x and âx = 12. x = 9 does not satisfy the condition given in the question. Stay Home , Stay Safe and keep learning!!! Let x be a …, Applications: Number problems and consecutive integers Sum of 3 consecutive odd integers is -3, what are the integers? Basic-mathematics.com. Then the length of the rod 2 meter shorter is (x - 2) and the total cost of both the rods is $60 (Because cost would remain unchanged). If 18 be subtracted from the number, the digits are reversed. Let x, then, be how much she spent for the blouse. The fourth part of a number exceeds the sixth part by 4. Math Word Problems with Answers - Grade 8. In this problem, it is the price of the blouse. A number consists of three digits of which the middle one is zero and the sum of the other digits is 9. Elsa used her card only once to make a long distance call. MathHelp.com. These word problems worksheets are a good resource for students in the 5th Grade through the 8th Grade. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Sign up today! To solve word problems in algebra, start by reading the problem carefully and determining what you’re being asked to find. Step 1:: A problem involving work can be solved using the formula , where T = time working together, A = the time for person A working alone, and B = the time for person B working alone. Given : Sum of the reciprocals of the parts is 1/6. Let
The following is multiple choice question (with options) to answer.
A tailor had a number of shirt pieces to cut from a roll of fabric. He cut each roll of equal length into 10 pieces. He cut at the rate of 45 cuts a minute. How many rolls would be cut in 24 minutes ? | [
"32 rolls",
"54 rolls",
"108 rolls",
"120 rolls"
] | D | Explanation:
Number of cuts made to cut a roll into 10 pieces = 9.
Therefore Required number of rolls = (45 x 24)/9 = 120.
Answer: D |
AQUA-RAT | AQUA-RAT-38342 | # 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.
Rama bought goods at a discount of 20% of list price.He marks them at a price as search he can give discount of 20% on a marked price and still makes the profit of 20% on selling price.What percent of list price is the marked price | [
"100%",
"175%",
"125%",
"180%"
] | C | If marked price is Rs 100/-
Selling price = Rs 80/-
Profit = 80*0.2= Rs 16
CP= Rs 64
List price = 64*100/80 = Rs 80
so marked price is 100*100/80 = 125%
ANSWER:C |
AQUA-RAT | AQUA-RAT-38343 | time, sun
d represents the day of the year for 2000. For example d=7
would be January 7th, 2000.
If you're willing to lose some precision in exchange for
convenience, you can compute the day of the current year (instead
of counting all the way back to 2000), since the sun's declination
repeats yearly (roughly speaking).
To calculate the day of the year, remember that December 31st
(noon) of the previous year is day 0. This means day 1 is January
1st, and day 31 is January 31st. Day 31 is also "February 0", so if
you need a date in February, just add. February 19th, for example,
would be 31+19 or 50. Since February has 29 days this year, February
29th would be day 31+29 or day 60, which is also "March 0". However,
at our level of precision, it doesn't matter whether you could the
leap day or not: the results will be approximately the same.
The formula above is accurate to about 0.1 degrees for this
century. Since the sun's declination can change by as much as 0.4
degrees in a day, this amount of precision should suffice.
Technically, the formula above computes the sun's declination at
Greenwich noon for a given day, which is the time when most of the
world is observing the same day. Again, the inaccuracies from using
Greenwich noon (instead of the actual, as yet unknown, time) are
small enough to ignore for our purposes.
The following is multiple choice question (with options) to answer.
if 3rd december 1999 is sunday what day is 3rd january 2000 | [
"wednesday",
"thursday",
"friday",
"saturday"
] | A | There is difference of 31 days from 3rd Dec 1999 to 3rd Jan 2000..
31/7 left with 3 odd days..
that is wednesday
ANSWER:A |
AQUA-RAT | AQUA-RAT-38344 | We have 8 members and a 3 slots. So total number of ways is 8*7*6, divided by 3! for double counts. 8*7*6/3!=56 .
Now what are the number of ways can we have two teams members on the committee?
We can pick any of the 8 for the first slot. The second slot is reserved for the team member of the first slot, so 1 ways. And the last slot can be any of the remaining 6.
So 8*1*6. But we have to account for double counts. Do we divide by 3! or 2!? This is were I am stuck. If we divide by 2!, we get 8*1*6/2=24
56-24=32, our answer. But I don't understand why we divide by 2!??
I hope I was a little more clearer. Thank you again for responding Bunuel!
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Re: Committee Combination Tough Problem [#permalink]
### Show Tags
02 Oct 2012, 04:36
Bunuel wrote:
alphabeta1234 wrote:
A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A)20
B)22
C)26
D)30
E)32
I'm not sure what are you exactly dong in your last approach.
Anyway, if you want to solve with slot method the simpler solution would be: 8*6*4/3!=32, 8 ways to choose for the first slot, 6 ways to choose for the second slot, 4 ways to choose for the third slot and dividing by 3! to get rid of duplication.
The following is multiple choice question (with options) to answer.
A committee has 5 men and 6 women. What are the number of ways of selecting a group of eight persons? | [
"165",
"187",
"277",
"266"
] | A | Total number of persons in the committee = 5 + 6 = 11
Number of ways of selecting group of eight persons = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2)
= 165 ways
Answer: A |
AQUA-RAT | AQUA-RAT-38345 | Hint: if $$8$$ is the largest single digit factor than doesn't it make sense that the largest such three digit number would be in the $$8$$ hundreds.
Now, that was the hard way. Were there any handy mathematical observations you might have used to make this easier? Would knowing that $$24 = 2^3*3$$ is the unique prime factorization of $$24$$ have helped you find the ways to find three term factorizations? Would knowing there are $$k!$$ ways to arrange $$k$$ objects have helped? What if some of the objects were indistinguishable?
$$xyz= 2^3 .3$$
powers of $$2$$ can be divided among $$x,y,z$$ in
$$^{3+(3-1)} \ C _{3-1}= 10$$ ways
and power of $$3$$ can be divided into $$^{1+(3-1)} \ C _{3-1}= 3$$ ways
thus giving total $$10\times 3 =30$$ways
but here we also counted arrangements like $$2 \times 1\times 12$$ and $$1\times 1\times 24$$ which violates 3 digit number policy so, by subtracting these i.e, $$\left(3! + \dfrac{3!}{2!}\right)=9$$ ways
we get
total number of $$3$$ digit numbers having product $$4 != (30-9)=21$$
and out of all such $$3$$ digit numbers largest is $$8 3 1$$
The following is multiple choice question (with options) to answer.
Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 1200 | [
"30",
"120",
"210",
"300"
] | B | 27 = 3*3*3, and its GCF with x is 3, which implies that x has ONE factor of 3, but not TWO factors of 3.
100 and x have a GCF of 10, which implies that x has ONE factor of 10, but not TWO factors of 10.
Then we want to know what is the largest possible GCF of x and 2100.
Well 1200 = 3 * 4 * 10 * 10
We want x to include as many factors in common with 2100 as possible, to make the GCF with 2100 as big as possible.
We know x has one factor of 3, but not two factors --- that takes the 3.
We know x has one factor of 10, but not two factors --- we can take one of those 10's, but we have to leave the other
No other restrictions, so we can also grab that factor of 7 --- nothing saying that we can't, and it's there for the taking.
3*4*10 = 120
If we allow x to include as many factors as possible, within the constraints given, that is the most x could have in common with 1200.
B |
AQUA-RAT | AQUA-RAT-38346 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A person lent a certain sum of money at 4% per annum at simple interest and in 8 years the interest amounted to Rs.238 less than the sum lent. What was the sum lent? | [
"228",
"350",
"289",
"500"
] | B | P - 238 = (P*4*8)/100
P = 350
Answer: B |
AQUA-RAT | AQUA-RAT-38347 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains, 250 metres and 150 metres long respectively, are running on parallel lines. If they are running in the same directions, the faster train crosses the slower train in 40 seconds. If they are moving in the opposite direction they pass each other in eight seconds. What is the speed of the slower train? | [
"108 kmph",
"82 kmph",
"92 kmph",
"72 kmph"
] | D | Let the speed of faster train be f and slower train be y.
Time taken to cross each other traveling in the same direction = 250 +150/ (x-y) = 40………(1)
Time taken to cross each other traveling in the opposite direction = 250+150 /(x+y) = 8 …….(2)
From (1) and (2) f= 30 m/s and s = 20 m/s
Speed of slower train = 20 * 18/5 = 72 kmph
ANSWER:D |
AQUA-RAT | AQUA-RAT-38348 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
### Show Tags
02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
If a sum of money borrowed at 5 percent/annum SI amounts to Rs.1020 in 4 yrs, then the sum of money borrowed is? | [
"Rs. 850",
"Rs. 900",
"Rs. 980",
"Rs. 990"
] | A | Sol We have, A : Rs.1U2D
Let the pr xnpal be R5 x
Then mterest u — 12 : wzo — x
Therefore, by u xglurmula.
_umxx
’ Rx’: (
, _ma>< 1nzn—xv
\4Ve|\a\e,Jc— 5“
:> ~(:51DD—S*(or6x:51UU
or, J: : w 350
. The sum of money burrowed : Rs. 850
A) |
AQUA-RAT | AQUA-RAT-38349 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
In the third grade of Windblown School there are 102 students, one third of them failed the math test and 1/6 failed that literature test. At least how many students failed both tests? | [
"0.",
"6.",
"8.",
"10."
] | A | Total = 102
failed in math = 102/3=34
failed in literature=108/6=17
the least failed in both can be 0
while max can be 17
Answer A |
AQUA-RAT | AQUA-RAT-38350 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
If a number is formed by writing integers 1 to 150 one after another, like 12345...150. How many R digits does this integer consists of? | [
"150",
"339",
"342",
"359"
] | C | Number of single digit integers = 9
Number of 2 digit integers = 99 - 10 + 1 = 90
Number of 3 digit integers = 150 - 100 + 1 = 51
Number of digits in the integer R= 1*9 + 2*90 + 3*51 = 9 + 180 + 153 = 342
Answer: C |
AQUA-RAT | AQUA-RAT-38351 | Since we are not interested on the numerical value of this equation, but that the result of the equation to be an integer, not a fraction, the n outside of the parenthesis is irrelevant.
So the equation could be written as, (3n ± 1)/2, and the result of this equation is always an integer for any odd or even value of n.
-
We know that every prime $> 3$ is of the form $6n+1$ or $6n-1$. Now find two numbers $a$ and $b$ such that $a+b = p$ and $a-b =1$, where $p$ is either $6n+1$ or $6n-1$. Suppose $p = 6n+1$ Solving $a+b=6n+1$ and $a-b = 1$, we get $a= 3n+1$ and $b=3n$ Thus $ab$ is a multiple of $6$. Therefore $4ab$ is a multiple of $24$. But $4ab = (a+b)^2 – (a-b)^2$. Thus we see that $p^2-1$ is a multiple of $24$.
-
Note that any prime $p$ which is $>3$ we have $p-1$ or $p+1$ should be divisible by $3.$ $$(2m+1)^2=4m(m+1)+1\equiv1\mod(8)$$ Hence $p^2-1=(p-1)(p+1)$ is divisible by $3\times 8=24.$
-
The great clue, I believe, is the perennially useful fact that p^2-1 is (p+1)(p-1). Since p>3, p must be odd and cannot be divisible by three; since there is one number divisible by three in any set of three consecutive numbers either p+1 or p-1 must be divisible by three; and since every other even number is divisible by four, one of these factors must be divisible by two and the other by four. Simple arithmetic follows: 4x3x2=24.
The following is multiple choice question (with options) to answer.
Integers 3n+4 and 8n+9 are divisible by an integer p. If p is not equal to 1, then p equals to? | [
"2",
"3",
"5",
"7"
] | C | If p divides 3n+4, then p divides 24n+32.
If p divides 8n+9, then p divides 24n+27.
Then p must equal 5.
The answer is C. |
AQUA-RAT | AQUA-RAT-38352 | The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.
Therefore, $$\frac{x+y}{2}= 18.5 => x+y = 37$$ where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
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27 Dec 2017, 11:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
Let the numbers be x-2, x-1, x, x+1, x+2 & x+3
Sum of six numbers $$= 6x+3= \frac{37}{2}*6$$
$$=>x=18$$
if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, $$x=Average =18$$
Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]
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27 Dec 2017, 11:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?
(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
$$n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111$$
The following is multiple choice question (with options) to answer.
Calculate the average of first 18 even numbers is? | [
"21",
"29",
"19",
"14"
] | C | Explanation:
Sum of 10 even numbers = 18 * 19 = 342
Average = 342/18 = 19
Answer: Option C |
AQUA-RAT | AQUA-RAT-38353 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
find next in series 31,29,31,30,31,? | [
"30",
"31",
"32",
"33"
] | B | leave the number 31 in the series,then note that 29,30,31
hence 31 is the result
ANSWER:B |
AQUA-RAT | AQUA-RAT-38354 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?
I. 15.15
II.18.18
III. 17.17 | [
" I only",
" II only",
" III only",
" I and II only"
] | A | Problem statement was difficult to parse. but some how I managed to understand what we can do.
1) x/y = 15 + 0.15
2) x/y = 18 + 0.18
3) x/y = 17 + 0.17
w.k.t x = yn (i.e Q) + yk=60 (i.e remainder)
therefore :
1) y*0.15=60 -- we can get y an integer.
2) y*0.18=60 -- we can't get y an integer.
3) y*0.17=60 -- we can't get y an integer.
Therefore 1 are possible answers.
Ans A |
AQUA-RAT | AQUA-RAT-38355 | For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB.
The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126.
The following is multiple choice question (with options) to answer.
The number of new words that can be formed by rearranging the letters of the word 'NUMBERS' is | [
"2880",
"7670",
"1192",
"5039"
] | D | Explanation:
Number of words which can be formed = 7! - 1 = 5040- 1 = 5039.
Answer: D |
AQUA-RAT | AQUA-RAT-38356 | # 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.
Vindy paddles her kayak upstream at w kilometers per hour, and then returns downstream the same distance at n kilometers per hour. How many kilometers upstream did she travel if she spent a total of p hours for the round trip? | [
"wnp",
"wn/p",
"(w + n)/p",
"wnp/(w + n)"
] | D | let t1 and t2 be the time taken to row upstream and downstream respectively
now,
t1=distance/speed=d/w
similarly,
t2=d/n (as same distance has to be rowed)
also,
t1+t2=p
therefore,
p=(d/w)+(d/n)
=d(w+n)/wn
d=wmn/(w+n)=D |
AQUA-RAT | AQUA-RAT-38357 | python, python-3.x, programming-challenge
Title: Sum square difference Problem description:
The sum of the squares of the first ten natural numbers is,
1² + 2² + … + 10² = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + … + 10)² = 55² = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is:
3025 – 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Square difference:
l = sum of the squares of the first n natural numbers
k = sum of n naturals numbers
m = differences between l and k
My Solution
This is my solution for problem 6 of Project Euler using Python:
def square_difference(n):
l = (n * (n + 1) * (2 * n + 1)) / 6
k = (n * (n + 1)) / 2
k = k ** 2
m = abs(l - k)
return m
The following is multiple choice question (with options) to answer.
The sum of the squares of two numbers is 3341 and the diference of their squares is 891. The numbers are : | [
"35 and 46",
"35 and 49",
"35 and 42",
"35 and 41"
] | A | Explanation:
Let the numbers be x and y. Then,
Adding (i) and (ii), we get :
Subtracting (ii) from (i), we get :
So, the numbers are 35 and 46
Answer: A |
AQUA-RAT | AQUA-RAT-38358 | # 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 shipment of 1500 heads of cabbage, each of which was approximately the same size was purchased for $600.The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head.The following day the rest were sold at a price per head equal to 10% less than the price each head sold for the day before.what was the gross profit on this shipment? | [
"$100",
"$115",
"$125",
"$130"
] | C | Sol:
1500 heads -> $600
1 head -> $600/1500
1 head -> $(2/5)
25% more of (2/5) -> 125/100 * 2/5 = $(1/2)
He sold 2/3*1500 = 1000 heads for $(1/2) per head
Total revenue by selling 1000 heads = 1000 * 1/2 = $500
Heads left: 500
Cost per head: 90% of the previous price: 90/100 * 1/2 = $(9/20)
Total revenue by selling 500 heads = 9/20 * 500 = 225
Total revenue after selling 1500 cabbage heads - 225+500 = $725
Money spent on the purchase: $600
Profit = 725-600 = $125
Ans: "C |
AQUA-RAT | AQUA-RAT-38359 | # Write 100 as the sum of two positive integers
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.
Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then $100 = 44 + 56 = 4 \times 11 + 8 \times 7$.
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.
• I think you want to reword this 'Scalable algorithm to write N as the sum of two positive integers, for large N'
– smci
May 4, 2015 at 7:13
• This seems like a rather badly designed exercise since, looking at the answers, it's clear that just checking multiples of 7 and 11 is by far the simplest way of solving it. I once saw an exam question that made things much clearer by saying the equivalent of, "Proceed as if 100 is a very large number and you do not know that 100=44+56." May 4, 2015 at 21:07
From Bezout's Lemma, note that since $\gcd(7,11) = 1$, which divides $100$, there exists $x,y \in \mathbb{Z}$ such that $7x+11y=100$.
A candidate solution is $(x,y) = (8,4)$.
The following is multiple choice question (with options) to answer.
The number 90 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers? | [
"17",
"16",
"15",
"14"
] | B | 7^2+5^2+4^2=49+25+16 = 90
7+5+4 =16
Hence answer is B |
AQUA-RAT | AQUA-RAT-38360 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A milk man has 10 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk.If the cost of pure milk is Rs.18 per liter,then the profit of the milkman, when he sells all the mixture at cost price is: | [
"20%",
"15%",
"33.33%",
"18%"
] | B | Explanation:
when the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 10 liters of milk.
Therefore, profit percentage = 15 %. Answer: B |
AQUA-RAT | AQUA-RAT-38361 | They're compounding this much every day, so if I were to write this as a decimal ... Let me just write that as a decimal. 0.06274%. As a decimal this is the same thing as 0.0006274. These are the same thing, right? 1% is .01, so .06% is .0006 as a decimal. This is how much they're charging every day. If you watch the compounding interest video, you know that if you wanted to figure out how much total interest you would be paying over a total year, you would take this number, add it to 1, so we have 1., this thing over here, .0006274. Instead of just taking this and multiplying it by 365, you take this number and you take it to the 365th power. You multiply it by itself 365 times. That's because if I have$1 in my balance, on day 2, I'm going to have to pay this much x $1. 1.0006274 x$1. On day 2, I'm going to have to pay this much x this number again x $1. Let me write that down. On day 1, maybe I have$1 that I owe them. On day 2, it'll be $1 x this thing, 1.0006274. On day 3, I'm going to have to pay 1.00 - Actually I forgot a 0. 06274 x this whole thing. On day 3, it'll be$1, which is the initial amount I borrowed, x 1.000, this number, 6274, that's just that there and then I'm going to have to pay that much interest on this whole thing again. I'm compounding 1.0006274. As you can see, we've kept the balance for two days. I'm raising this to the second power, by multiplying it by itself. I'm squaring it. If I keep that balance for 365 days, I have to raise it to the 365th power and this is counting any kind of extra penalties or fees, so let's figure out - This right here, this number, whatever it is, this is - Once I get this and I subtract 1 from it, that is the mathematically
The following is multiple choice question (with options) to answer.
A man borrowed Rs 24000 from two money lenders. For one loan, he paid 15% per annum and for the other 18% per annum. At the end of one year, he paid Rs 4050. How much did he borrow at each rate ? | [
"Rs.16000",
"Rs.12000",
"Rs.15000",
"Rs.13000"
] | C | Let the sum at 15% be Rs x and that at 18% be Rs (24000 - x).
{(x * 15 * 1)/100 } + { [(24000 – x) * 18 * 1]/100 } = 4050
or 15 x + 432000 - 18x = 405000 or x = 9000.
Money borrowed at 15% = Rs 9000 .
Money borrowed at 18% = Rs 15000.
answer :C |
AQUA-RAT | AQUA-RAT-38362 | I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?
6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
The following is multiple choice question (with options) to answer.
How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? | [
"03",
"4",
"5",
"1"
] | A | You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.
AB , AC
BA , BC
CA , CB
There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.
The lines are: AB, BC and AC ; 3 lines only.
So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.
Answer:A |
AQUA-RAT | AQUA-RAT-38363 | This seems way too high, and there's probably something I'm missing here. Any thoughts on this calculation, or what it should be?
• Your locker's combination is now the target. You are hoping either the one on your left matches it OR the one to your right matches it. It seems to me is a simple '+' instead of '*'. – Mick Dec 21 '13 at 5:00
• Is your anecdote really true!? – Sammy Black Dec 21 '13 at 5:00
• Mick's right. The odds are therefore about 1 in 30,000. That's small, but not insane. Then again, it suggests that your estimate of the number of combinations might be wrong. I seem to (vaguely) recall that the first number on a Master combination lock was always congruent to 1 mod 4 (or something like that), and a similar rule held for the second and third, i.e., there are about 64 times FEWER possibilities than you might imagine. If so, then yours was about a 1-in-500 chance, which is the sort of thing that does really happen. – John Hughes Dec 21 '13 at 5:05
• Sammy: It is true! I confronted the guy at the end of the semester about it, and asked to buy his lock. He sold it to me and said the combination, asking if I needed to write it down. I told him this anecdote about how I accidentally opened his locker once and thought someone had stolen everything of mine and replaced it. I think he was too shocked that someone had been in his locker to be sufficiently shocked by the chances of it actually happening. – AdamFortuna Dec 23 '13 at 4:31
As you correctly calculated, there is a $1$ in $40 \times 39 \times 39 = 60\,840$ of another lock having the same permutation (they're not combination locks, as order does matter). So what is the probability that either of your neighbors has the same permutation?
The following is multiple choice question (with options) to answer.
a locker opened by combination of 4 keys.a man have four key and he know that 3rd no. key will be placed at first key hole but don't know about rest.so tell me maximum no. of attempt he will make to open locker. | [
"6",
"8",
"10",
"5"
] | A | as no. 3rd key position get fixed.so we have to calculate total no. of configuration of rest three key.
which is permutation of three item,3P3=3!=6
answer A |
AQUA-RAT | AQUA-RAT-38364 | # Calculating probability of 8 letter String [closed]
Computer generate 8 string password with letters a,b,c,d,e. Each letter is chosen independently and with equal liklihood of being selected.
What is the probability that the string has 8 of the same letter?
What is the probability that there is at least one a and one b?
Can someone help me with these? I calculated the total outcomes, but I' stuck there.
## closed as off-topic by Saad, farruhota, max_zorn, Cesareo, JaviApr 18 at 10:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, farruhota, max_zorn, Cesareo, Javi
If this question can be reworded to fit the rules in the help center, please edit the question.
The total number of the possible passwords are $$5^8=390625$$. All of these possible passwords have the same probability.
What is the probability that the string has 8 of the same letter?
It is only 5 possibility of the 390625, so the probability is $$\frac{1}{5^7}=1.28\cdot 10^{-5}$$.
What is the probability that there is at least one a and one b?
It is more complex:
1. If there is no "a" and no "b": then we have 8 character long passwords, but only from c, d and e. Thus, it is $$3^8=6561$$.
2. If there is no "a": so we have 8 character long passwords, but only from b, c, d, and e. Thus, they are $$4^8=65536$$.
3. If there is no "b": the calculation is the same.
The following is multiple choice question (with options) to answer.
A computer system uses alphanumeric case sensitive characters for its passwords. When the system was created it required users to create passwords having 5 characters in length. This year, it added the option of creating passwords having 6 characters in length. Which of the following gives the expression for the total number of passwords the new computer system can accept?
Assume there are 72 unique alphanumeric case sensitive characters. | [
"63^4",
"62^5",
"72^4 * 73",
"73(62^4)"
] | C | Total number of passwords = number of 5 character password +number of 6 character password
= 72^4 + 72^5 (Since there is no limitation on repetition , each character can be chosen in 72 ways )
=72^4(1+ 72)
=72^4 * 73
Answer C |
AQUA-RAT | AQUA-RAT-38365 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
A can run a kilometer race in 4 1/2 min while B can run same race in 5 min. How many meters start can A give B in a kilometer race, so that the race mat end in a dead heat? | [
"228",
"2197",
"666",
"100"
] | D | A can give B (5 min - 4 1/2 min) = 30 sec start.
The distance covered by B in 5 min = 1000 m.
Distance covered in 30 sec = (1000 * 30)/300 = 100 m.
A can give B 100m start.Answer: D |
AQUA-RAT | AQUA-RAT-38366 | 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 man is 25 years older than his daughter. In two years, his age will be twice the age of his daughter. The present age of his daughter is: | [
"20",
"21",
"22",
"23"
] | D | Let the daughter's present age be x years. Then, man's present age = (x + 25) years.
(x + 25) + 2 = 2(x + 2)
x + 27 = 2x + 4
x = 23.
Answer:D |
AQUA-RAT | AQUA-RAT-38367 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a race of 1000 meters, A can beat B by 100 meters, in a race of 800 meters, B can beat C by 100 meters. By how many meters will A beat C in a race of 600 meters? | [
"125.5 meters",
"126.5 meters",
"127.5 meters",
"128.5 meters"
] | C | Explanation:
When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.
Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.
So, in a race of 1000 meters, A beats C by (1000 - 787.5) = 212.5 meters to C.
So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.
Answer: C |
AQUA-RAT | AQUA-RAT-38368 | 4. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
Psec/1bolt * 5bolts/1rev * 1rev/78.4in *...
The OP specified that the circumference of the wheel is 78.4in. Why do we need to include the radius in the conversion?
Last edited: Sep 8, 2015
5. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
Because I missed that little detail?
6. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
But the circumference of the wheel is irrelevant -- it's the circumference of the tire that matters. Perhaps the TS means the tire and not the wheel, but I'm not positive about that.
7. ### djsfantasi AAC Fanatic!
Apr 11, 2010
2,805
833
A circumference of 78.4" is a diameter of 10". Much more likely to be the diameter of a wheel.
Update: Helps to use the right equation...
Last edited: Sep 8, 2015
8. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
Or it's one of the new micro-cars that are becoming the snob-appeal rage.
9. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
As sensors go I like it. It would also be good for odometers too.
10. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
I think driveshaft sensors are probably better since you typically get several rotations of the shaft per rotation of the tire (and you can mount multiple magnets on the shaft to further improve resolution). It also tends to average out the distance traveled by the tires as you drive due to the natural behavior of the differential.
Jan 26, 2014
5
0
12. ### rjwheaton Thread Starter New Member
Jan 26, 2014
5
0
Thank you!
Not only does that answer my question, but it helps me understand how you did the calculations.
The following is multiple choice question (with options) to answer.
The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 500 resolutions? | [
"724 m",
"704 m",
"287 m",
"278 m"
] | B | In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 500 resolutions.
= 500 * 2 * 22/7 * 22.4 = 70400 cm
= 704 m
Answer:B |
AQUA-RAT | AQUA-RAT-38369 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
P is three times as old as Q. In 11 years, P will be twice as old as Q. How old was P three years ago? | [
"21",
"24",
"27",
"30"
] | D | P = 3Q so Q = P/3
P+11 = 2(Q+11) = 2(P/3+11)
P/3 = 11
P = 33
Three years ago, P was 30.
The answer is D. |
AQUA-RAT | AQUA-RAT-38370 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Two pipes A and B can fill a tank in 10hour and 15 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank? | [
"2hrs",
"6hrs",
"5hrs",
"4hrs"
] | B | Part filled by A in 1 hour = 1/10
part filled by B in 1 hour = 1/15
Part filled by A+B in 1 hour = 1/10 + 1/15 = 1/6
Both the pipes will together will fill the tank in 6 hours.
Answer is B |
AQUA-RAT | AQUA-RAT-38371 | 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.
How many seconds will a 800 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"26 sec",
"32 sec",
"13 sec",
"48 sec"
] | D | Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 800 * 3/50
= 48 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-38372 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is: | [
"18",
"24",
"42",
"81"
] | B | Let the two digit number is 10x+y
Product of the digits x*y=8
10x+y+18=10y+x
on solving this we get X-y=-2
x+y=6
by solving the above two equations we get x=2 and y=4
therefore the digit is 10*2+4=24
ANSWER:B |
AQUA-RAT | AQUA-RAT-38373 | # 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.
Mohit sold an article for Rs. 18000. Had he offered a discount of 10% on the selling price, he would have earned a profit of 8%. What is the cost price of the article? | [
"15000",
"15029",
"26882",
"82976"
] | A | Let the CP be Rs. x.
Had he offered 10% discount, profit = 8%
Profit = 8/100 x and hence his SP = x + 8/100 x = Rs. 1.08x
= 18000 - 10/100(18000)
= 18000 - 1800
= Rs. 16200
=> 1.08x = 16200
=> x = 15000
Answer:A |
AQUA-RAT | AQUA-RAT-38374 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average weight of 16 children in a class is 50.25 kg and that of the remaining 8 children is 45.15 kg. Find the average weights of all the children in the class. | [
"A)48.55",
"B)49",
"C)51",
"D)61"
] | A | Required average
= (50.25 x 16 + 45.15 x 8)/(16 + 8)
= (804 + 361.20)/24
= 1165.20/24
= 48.55kg
Answer is A |
AQUA-RAT | AQUA-RAT-38375 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The price of a microchip declines by 67 percent every 6 months. At this rate, approximately how many years will it take for the price of an $81 microchip to reach $1?
81- 67% * 81 = 27 approx
27 - 0.67*27 = 9 approx
9- 0.67*9 = 3 approx
3-0.67*3= $1 approx | [
"1.5 years",
"2 years",
"3 years",
"13 years"
] | B | not sure whether you like this:
81 - (81 x 67%)/3 = 81 - (81 x 2) / 3 = 27 = 81 / 31 = 81 / (3x3x3x3) = 81 / 3^4
so the time = 4 x 6 months = 24 months or 2 years. answer is B |
AQUA-RAT | AQUA-RAT-38376 | c#, winforms, fixed-point
Title: Divide decimal by range of fractions, and then convert the fraction that is closest to a whole number to a mixed fraction My program will divide the decimal (\$30.0575\$) by a range of fractions (\$\frac{1}{1}, \frac{1}{2}, \frac{1}{4}, \dots, \frac{1}{48})\$ leaving me with list of improper fractions that look like this:
\$\frac{1}{1} * 30.0575\$
\$\frac{1}{2} * 60.115\$
\$\frac{1}{3} * 90.1725\$
\$\dots\$
\$\frac{1}{48} * 1442.76\$
My program will then "pick out" the improper fraction that contains the decimal that is closest to a whole number. It will finally convert the improper fraction that is closest to a whole number into a mixed fraction.
How can I make my program more readable?
Are there any "shortcuts" I could implement that would reduce the lines of code used?
How can I overall optimize my code so it completes the same task in a more efficient way, while still keeping readability?
private void FractionsForm_Load(object sender, EventArgs e)
{
var myDecimal = 30.0575;
var results = new List<double>();
for (double i = 1; i < 49; i++)
{
var fraction = 1 / i;
var result = myDecimal / fraction;
results.Add(result);
Console.WriteLine("1/{0} * {1}", i, result);
}
WriteLineSeparator();
var remainders = new List<double>();
foreach (var result in results)
{
var roundedResult = Math.Round(result);
var remainder = roundedResult - result;
remainders.Add(Math.Abs(remainder));
}
The following is multiple choice question (with options) to answer.
Convert 0.32 in to a vulgar fraction ? | [
"8/25",
"16/50",
"17/50",
"19/50"
] | A | Answer
0.32 = 32/100 = 8/25
Correct Option: A |
AQUA-RAT | AQUA-RAT-38377 | 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",
"99",
"278",
"277"
] | A | (3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A |
AQUA-RAT | AQUA-RAT-38378 | # Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$
Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an equation using trigonometry and pythagoras. half the height height of the intersection is $$\sqrt{r^2-\left(r-\frac{s}{2}\right)^2}$$ where $$r-\frac{s}{2}$$ is the distance between a circle radius and the centre of the height of the intersection. From there I could work out the full height, then the area of the sector formed from the height as a chord and from that the area of the intersection, of which I know is $$\frac{\pi r^2}. {2}$$ due to the fact that all areas are equal. After working out the area of the triangle formed by the height and two radii, I found the angle of the sector with trig ($$2\cos^{-1}\left(\frac{r-\frac{s}{2}}{r}\right)$$). In conclusion the resultant equation is (with $$r=x$$ and $$s=y$$):
$$\frac{\pi x^2}{2}=2\left(\frac{2\cos^{-1}\left(\frac{x-\frac{y}{2}}{x}\right)}{2\pi}\pi x^2-\frac{2\sqrt{x^2-\left(x-\frac{y}{2}\right)^2}\left(x-\frac{y}{2}\right)}{2}\right)$$
The following is multiple choice question (with options) to answer.
The ratio of the radius of two circles is 2: 5, and then the ratio of their areas is? | [
"4:25",
"2:9",
"1:9",
"3:7"
] | A | r1: r2 = 2: 5
Î r1^2: Î r2^2
r1^2: r2^2 = 4:25
Answer: A |
AQUA-RAT | AQUA-RAT-38379 | ### Show Tags
04 Aug 2019, 03:54
generis wrote:
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?
A. 4
B. 1
C. 0
D. -1
E. -4
Don't remove brackets and solve. The question asks for the logic behind absolute value.
$$y$$ is a positive integer
$$|x| < 5 − y$$
LHS is nonnegative - it is positive or 0.
Least value for x is ZERO
Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.
The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0
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Why can't x be a negative integer, in that case it should be -1 i.e. option D.
Intern
Joined: 21 Mar 2019
Posts: 22
Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
08 Aug 2019, 18:54
I don't understand why x can't be negative
Intern
Joined: 06 Nov 2014
Posts: 17
Location: Viet Nam
GMAT 1: 720 Q50 V36
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GPA: 3.67
Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
For any number y, y* is defined as the greatest positive even integer less than or equal to y. What is the value of 7.2 – 7.2*? | [
"1.2",
"0.2",
"1.8",
"2.2"
] | A | Since y* is defined as the greatest positive even integer less than or equal to y, then 7.2* = 6 (the greatest positive even integer less than or equal to 7.2 is 6).
Hence, 7.2 –7.2* = 7.2 - 6 = 1.2
Answer: A. |
AQUA-RAT | AQUA-RAT-38380 | 0\cr −1 \cr −1\cr 1 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 1\cr 1 \cr 3\cr 0 \cr 3 } \right ]\right \}\right \rangle & & & & \cr ℒ\kern -1.95872pt \left (G\right ) = ℛ\kern -1.95872pt \left (L\right ) & = \left \langle \left \{\left [\array{ 1\cr 0 \cr 0\cr 0 \cr −{1\over 3} } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0\cr 1 \cr −{1\over 3} } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1\cr 1 \cr −1 } \right ]\right \}\right \rangle & &\text{@(a href="fcla-jsmath-latestli34.html#theorem.BRS")Theorem BRS@(/a)} & & & & \cr & = \left \langle \left \{\left [\array{ 3\cr 0 \cr 0\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 3 \cr 0\cr 3\ \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1\cr 1 \cr −1 } \right ]\right \}\right \rangle & & & & }
The following is multiple choice question (with options) to answer.
B @ K I 9 D © 5 2 T $ M 4 E J # U R 1 A δ K 5 W 6 % F Q 7 H P 8 Z
Which of the following is the ninth to the right of the twentieth element from the right end of the above arrangement? | [
"9",
"7",
"$",
"5"
] | D | Explanation:
5 is the ninth to the right of the twentieth element from the right end of the above arrangement
ANSWER IS D |
AQUA-RAT | AQUA-RAT-38381 | 2nd largest value = 231+510.222
so largest difference between them = 222
C . $2^{22}$
111111111 can be written as $2^{10}-1$
So Largest No. 1.111111111 x $2^{31}$ equivalent to ($2^{10}-1$) x $2^{-9}$ x $2^{31}$
Second largest No. 1.111111110 x $2^{31}$ equivalent to ($2^{10}-2$) x $2^{-9}$ x $2^{31}$
difference of above equation:
= [($2^{10}-1$) x $2^{-9}$ x $2^{31}$ ] - [($2^{10}-2$) x $2^{-9}$ x $2^{31}$]
= $2^{22}$
1
The following is multiple choice question (with options) to answer.
10^222 ÷ 10^220 = ? | [
"10",
"1000",
"100",
"10000"
] | C | 10^222 ÷ 10^220 = 10^(222 - 220) = 10^2 = 100
Answer is C |
AQUA-RAT | AQUA-RAT-38382 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a man standing on the platform. If the train is 180 meters long and its speed is 36 kmph, how much time it took in doing so? | [
"15 sec",
"18 sec",
"19 sec",
"28 sec"
] | B | D = 180
S = 36 * 5/18 = 10 mps
T = 180/10 = 18 sec
Answer:B |
AQUA-RAT | AQUA-RAT-38383 | To get the required result in the book you first have to calculate the amount of the equivalent annual payment. The formula for payments at the beginning of every month is
$C_1=12\cdot r+\frac{\color{blue}{13}\cdot r\cdot i}{2}$
In case of payments at the end of every month $\color{blue}{13}$ has to be replaced by $11$.
$C_1=12\cdot 200+\frac{\color{blue}{13}\cdot 200\cdot 0.045}{2}=2458.5$
To get the Future value after 10 years we use the formula for annual payments.
$C_{10}=2458.5\cdot \frac{1-1.045^{10}}{1-1.045}=30210.56$
But in general I wouldn´t say that your method is worse then the method above. Your result differs from my result about $0.015\%$ only.
The following is multiple choice question (with options) to answer.
A startup company which received a government loan is expected to make a monthly payment till it repays a total of $1093500. Every month it is expected to triple the amount it paid the previous month. The payment for the first month is $500. | [
"6",
"7",
"9",
"10"
] | B | First payment--$500
Total amount to be repaid--$1093500
First month payment--$500
Second month payment--$500*3=$4500
Third month payment--$4500*3=$13500
Fourth month payment--$13500*3=$40500
Fifth month payment--$40500*3=$121500
Sixth month payment--$121500*3=$364500
Seventh month payment--$364500*3=$1093500
Answer is B |
AQUA-RAT | AQUA-RAT-38384 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
Frances can complete a job in 12 hours, and Joan can complete the same job in 9 hours. Frances starts the job at 9 a.m., and stops working at 5 p.m. If Joan starts working at 5 p.m. to complete the job, at what time is the job finished? | [
"6 p.m.",
"7 p.m.",
"8 p.m.",
"10 p.m."
] | C | Let the total work be 24 units.
Frances can complete 2 units/hr
Joan can complete 2.67 units/hr
Work done by Frances in 8 hours is 16 units ; work left for Joan is 8 units ( 24 - 16)
Time required by Jones to complete the remaining work will be 8/2.67 => 3 hours
Given Jones starts at 5PM , so she would complete the work at 8 PM
Answer will be (C) |
AQUA-RAT | AQUA-RAT-38385 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100? | [
"11",
"88",
"25",
"88"
] | C | A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = (100 * 30)/40 = 25 * 3 = 75.
In a game of 100 points, B gives (100 - 75) = 25 points to C.Answer: C |
AQUA-RAT | AQUA-RAT-38386 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
At present the ratio between the ages of Arun and Deepak is 4:3. After 6 years Arun's age will be 26 years. What is the age of the Deepak? | [
"15",
"20",
"18",
"22"
] | A | Let the present age of Arun and Deepak be 4x and 3x
4x+6 = 26
4x = 20
x = 5
Deepak's age = 3x = 15 years
Answer is A |
AQUA-RAT | AQUA-RAT-38387 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
In what time will a train 180 meters long cross an electric pole, if its speed is 154 km/hr | [
"5 seconds",
"4.3 seconds",
"3 seconds",
"2.5 seconds"
] | B | Explanation:
First convert speed into m/sec
Speed = 154*(5/18) = 42 m/sec
Time = Distance/speed
= 180/42 = 4.3 seconds
Option B |
AQUA-RAT | AQUA-RAT-38388 | Finally, if $$D=2$$ (i.e., $$ABCD=1002$$), we can only have $$c+f=3+9$$ or $$c+f=5+7$$. In either case we'll need $$b+e=9$$, which is possible only as $$2+7$$ or $$4+5$$. If $$c+f=3+9$$, either of these is possible, while neither is possible if $$c+f=5+7$$. So again we get just two solutions, namely $$243+759=1002$$ and $$423+579=1002$$.
Altogether we get six solutions with $$a\lt d$$, $$b\lt e$$ and $$c\lt f$$:
\begin{align} 423+579&=1020\\ 243+957&=1200\\ 342+759&=1101\\ 432+579&=1011\\ 243+759&=1002\\ 423+579&=1002 \end{align}
The total count, without the restrictions $$a\lt d$$ and $$b\lt e$$, is thus $$24$$; removing the restriction $$c\lt f$$ brings the total number of solutions to $$48$$.
• Thank you for the comprehensive and very well explained answer. Insightful. Much appreciate the time you took! – Parzifal Oct 4 '18 at 11:04
Hint1:
$$\sum_{k=0}^n 10^ka_k \equiv \sum_{k=0}^n a_k \mod{9}$$
Hint2:
Note, that the digital root for your numbers is always the same, regardless of your choice.
The following is multiple choice question (with options) to answer.
If (A-B-C+D = 13) and (A+B-C-D = 9), what is the value of (B-D)^2? | [
"4.",
"8.",
"12.",
"16."
] | A | EQ1: A - B - C + D = 13
EQ2: A + B - C - D = 9
(1) Subtract EQ1 from EQ2
A - B - C + D = 13
- A +B - C - D = 9
------------------------
-2B + 2D = 4
(2) Simplify
-B + D = 2
B - D = -2
(B-D)^2 = (-2)^2 = 4
My answer: A |
AQUA-RAT | AQUA-RAT-38389 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
The least number which when divided by 3, 4, 5 and 6 leaves a remainder 6, but when divided by 9 leaves no remainder, is | [
"117",
"126",
"252",
"336"
] | B | Explanation:
L.C.M of 3, 4, 5, 6 = 60
therefore Required Number is of the form 60k+6.
Least value of k for which (60k+6) is divisible by 9 is k = 2
therefore Required Number = (60 x 2+6)=126.
Answer: B |
AQUA-RAT | AQUA-RAT-38390 | # Probability a 9-digit number has the digits 2,4, and 6 next to each other.
The integers $1,2,3,....,9$ are arraned (at random) in a row, resulting in a $9$-digit integer (without replacement). What is the probability that:
The result is even? $\frac49$ or $\frac{4(8!)}{9!}$
The result is divisible by $5$? The number must end in $0$ or $5$. Edit: As Andre pointed out we have no $0$. So, $\frac19$. or $\frac{(8!)}{9!}$
The digits $2, 4,$ and $6$ are next to each other (in any order)? The above two I have confidence in, it is this last one I'm a little confused on.
$9\choose 3$ ways to position $2,4,6$ in the 9-digit number and $3!$ ways they can be arranged. This doesn't appear to be right as after you divide by $9!$ you get $.13$% which seems unreasonably low. So, I thought to multiple by $6!$ to account for the number of ways the other $6$ numbers can be arranged. This gives you:
$9\choose 3$$3!6!/9!$
which equals $1$, and obviously isn't right. Any suggestions as to where I went wrong would be great.
-
You're right! Silly me. – Vincent Jun 16 '14 at 22:47
For $2,4,6$ next to each other, imagine choosing the $3$ spots that will be reserved for our three favoured guests. This can be done in $\binom{9}{3}$ equally likely ways.
The three spots can be in a row in $7$ ways, for the leftmost of the spots can be in any of the positions $1$ to $7$. Thus the required probability is $\frac{7}{\binom{9}{3}}$. This simplifies nicely.
The following is multiple choice question (with options) to answer.
Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 12 phone numbers by using the readable digits followed by 12 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly? | [
"1/9",
"10/243",
"2/45",
"10/271"
] | C | If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).
If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).
P = 12/(9*9*3+9*3) = 2/45.
Answer: C. |
AQUA-RAT | AQUA-RAT-38391 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 56 years now, the son's age 4 years back was: | [
"12 years.",
"24 years.",
"17 years.",
"16 years."
] | B | Let the son's present age be x years. Then, (56 - x) = x
2x = 56.
x = 28
Son's age 4 years back (28- 4) = 24 years.
ANSWER :B |
AQUA-RAT | AQUA-RAT-38392 | # computing probability that at least one letter matches with envelope
Four letters to different insurers are prepared along with accompanying envelopes. The letters are put into the envelopes randomly. Calculate the probability that at least one letter ends up in its accompanying envelope.
### Attempt
Since this is tedioues, we can do
$$P(at \; least \; one ) = 1 - P( no \; match )$$
We notice that sample space is $4!$ since for letter 1 it has 4 choices but letter 2 has 3 choices and so on. Now, we wanna count in how many ways we get no match.
Let start with first one, we only have $3$ choices for this since it can go to either 2,3,4 envelope.
Now, as for the second one, we have to possibilities. If the first letter went to the second envelope, then the second letter now will have 3 different choices, but if the first letter didnt go to second letter, then the second letter will have 2 choices. Assume the former. Then we have 3 choices for this stage.
Now, for the third one (envelope 2 is taken already and assume letter 2 went to letter 1) then it would have 1 choice only and
last one must go to envelope 3.
Thus, we have $3 \times 3 \times 1 \times 1 = 9$ choices in total
Thus,
$$P(at \; least \; 1 \; letter) = 1 - \frac{9}{24}$$
IS this correct? I still feel as is something wrong because I assumed the letter 2 went to 1 and letter 1 went to envelope 2. Can we do that?
The following is multiple choice question (with options) to answer.
After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope? | [
"1/2",
"6/7",
"0",
"1/12"
] | C | Explanation :
Answer is Zero. Question is asking whether there is any chances of one letter getting inserted in an improper envelope, but there can-not be onr case as if one gets improper there is another letter who will be inserted in the improper envelope too...
Hence C is the answer.
Answer : C |
AQUA-RAT | AQUA-RAT-38393 | # Linear Algebra Money Question
bleedblue1234
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # $10 bills A+B+C = 32 1A+5B+10C = 100 ## The Attempt at a Solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. ## Answers and Replies E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. azizlwl You can narrow the selection.$1 can only be in a group of 5.
Homework Helper
This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.
Last edited by a moderator:
Homework Helper
Dearly Missed
## Homework Statement
I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?
## Homework Equations
A= # $1 bills B= #$5 bills
C= # \$10 bills
A+B+C = 32
1A+5B+10C = 100
## The Attempt at a Solution
The following is multiple choice question (with options) to answer.
SID has 13 ten-dollar bills, 11 five-dollar bills, and 17 one-dollar bills. If SID needs to pay exactly $128, what is the least number of bills he will need to use? | [
"16",
"121",
"45",
"456"
] | A | 128 is the total sum of money.
as we have 13 10 dollar bills so the closest we can get to 128 using the 10 dollar bills is by using 12 * 10 dollar bils
128-120 = 8
so now we need to get 8 from either 5 dollar bills or 1 dollar bills
8 - 5 (use only 1 5 dollar bill) = 3
so we can get 3 from using 3 $1 bills.
Hence 16 ( 12+1+3 )
ANSWER IS A |
AQUA-RAT | AQUA-RAT-38394 | I'm assuming that the four fingers are given, e.g., the fingers of one hand without the thumb. An arrangement of five distinguishable rings on the four labeled fingers amounts to a linear arrangement of $1$, $2$, $3$, $4$, $5$, and three indistinguishable zeros as separators. There are ${8!\over3!}=6720$ such arrangements. Note that a green ring and a blue ring on the index finger can be worn in $2$ ways.
• If we let $0$ denote the lack of a ring and if given a linear arrangement, the first pair go on the first finger (in the specified order), etc then the sequence $10234500$ and $01234500$ should be considered equivalent but I don't see how your solution accounts for that? – benguin Apr 9 '17 at 4:07
• @benguin : The zeros serve as separators. – Christian Blatter Apr 9 '17 at 7:05
• Thanks for the clarification, that makes more sense :) I was considering an argument that first orders the rings ($5!$ ways) then uses stars-and-bars to place the rings onto the fingers ($C(5+4-1,4-1)$ ways) for a total of $5!C(8,3)$ which gives the same result. – benguin Apr 9 '17 at 8:43
Suppose you had only one ring. It then would be $4^1=4$ different arrangements/placements. Now if we add one more ring, it will be $4\cdot4$ because we combine the four possible arrangements with yet another four arrangements which makes it $16$. This means you can have two rings on one finger. By the same token we would have $4^5=1024$ arrangements for five rings. And the person can have five rings on one finger or no ring at all on the finger.
If you had $5^4$ arrangements, it would mean $5^1=5$ different arrangements for one ring on four fingers. And that doesn’t make sense.
And finger $a$ can have six possible arrangements: either of five rings or no ring at all.
Hope my explanation was not confusing.
The following is multiple choice question (with options) to answer.
A young girl counted in the following way on the fingers of her left hand. She started collang the thumb 1, the index finger 2, the middle finger 3, the ring finger 4, the little finger 5, then reversed direction calling the ring finger 6, the the middle finger 7, the index finger 8, the thumb 9 then back to the index finger for 10, the middel finger for 11, and so on. She counted up to 1997. She ended on her | [
"a) thumb",
"b) the index finger",
"c) the middle finger",
"d) the ring finger"
] | D | Notice the pattern. The thumb lansd up at numbers 1,9,17,25,33,... so every 8th number
By this patter the thumb is also at number 1993 (multiple of 8 + 1)
Hence 1997 has to be the ring finger
Answer is (D) |
AQUA-RAT | AQUA-RAT-38395 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
How many positive integers between 1 and 400 are there such that they are multiples of 25? | [
"12",
"13",
"16",
"15"
] | C | Multiples of 25 = 25,50,75,-----400
Number of multiples of 25 => 25*16= 400
Answer is C |
AQUA-RAT | AQUA-RAT-38396 | # If $28a + 30b + 31c = 365$, then what is the value of $a +b +c$?
Question: For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?
How I approached it : I started immediately breaking it onto this form on seeing it :
$28(a +b +c) +2b +3c = 365 .......(1)$ $30(a +b +c) -2a +c = 365 .......(2)$ $31(a +b +c) -b -3a = 365 .......(3)$
And then I find out that
$365 = 28*13 + 1......(1')$; $365 = 30*12 + 5......(2')$ $365 = 31*11 + 24......(3')$
Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine quiet congruently, so I meet with a dead end here.
my issue : how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.
• $28\times 2 + 30\times 7 + 31\times 3 = 359\ne 365$ – Joey Zou Aug 11 '16 at 21:25
• This is a trick question, you are supposed to observe that these are the lengths of months and 365 is the length of a normal year. – f'' Aug 11 '16 at 21:31
More direct path to $a+b+c = 12$:
The following is multiple choice question (with options) to answer.
If A:B is 7:5, B:C is 3:4 and C is 30 then what is the value of A? | [
"31.5",
"25.5",
"32.5",
"33.8"
] | A | A/B=7/5,
B/C=3/4,C=30
B=90/4
4A/90=7/5
A=630/20
A=31.5
ANSWER:A |
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