source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38097 | \$7,382.94 3.2% 01.09.08 \$19.6879
\$7,402.63 3.2% 01.10.08 \$19.7404
\$7,422.37 3.2% 01.11.08 \$19.7930
\$7,442.17 3.2% 01.12.08 \$19.8458
The following is multiple choice question (with options) to answer.
After decreasing 24% in the price of an article costs Rs.912. Find the actual cost of an article? | [
"1218",
"2777",
"1200",
"2688"
] | C | CP* (76/100) = 912
CP= 12 * 100 => CP = 1200
Answer: C |
AQUA-RAT | AQUA-RAT-38098 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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Posts: 1892
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
The average age 9 members of a committee are the same as it was 2 years ago, because an old number has been replaced by a younger number. Find how much younger is the new member than the old number? | [
"65 years",
"88 years",
"18 years",
"55 years"
] | C | 9 * 2 = 18 years
Answer: C |
AQUA-RAT | AQUA-RAT-38099 | So on.
=====
Another thing to note:
(Assume only positive values)
$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$
So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$
That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.
But it doesn't go the other way!
$$\frac 12 < a < 4 \not \implies 1 < a < 2$$
Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$
[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)
• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06
The following is multiple choice question (with options) to answer.
If 0 < a< 1 < b, which of the following must be true? | [
"1 < 1/a< 1/b",
"1/a< 1 < 1/b",
"1/a< 1/b< 1",
"1/b< 1 < 1/a"
] | D | 0<a<1<b. Let a=1/2 and b=2
Substituting values:
A. 1<2<1/2 ---Wrong
B. 2<1< 1/2 ---Wrong
C. 2<1/2<1 ---Wrong
D. 1/2<1<2 ---Correct
E. 1/2<2<1 ---Wrong
Hence answer is D |
AQUA-RAT | AQUA-RAT-38100 | # 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.
4000 was divided into two parts such a way that when first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%? | [
"2178",
"1087",
"2800",
"2689"
] | C | (x*3*1)/100 + [(4000 - x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 è x = 2800
Answer: C |
AQUA-RAT | AQUA-RAT-38101 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A man gains 20% by selling an article for a certain price. If the sells it at double the price, the percentage of profit will be? | [
"76%.",
"66%.",
"89%.",
"140%."
] | D | Let C.P. = Rs. x.
Then, S.P. = Rs. (12% of x) = Rs. 6x/5
New S.P. = 2 * 6x/5 = Rs. 12x/5
Profit = 12x/5 - x = Rs. 7x/5
Profit = 7x/5 * 1/x * 100 = 140%.
Answer: D |
AQUA-RAT | AQUA-RAT-38102 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
How many times in a day, the hands of a clock are straight? | [
"42",
"41",
"44",
"45"
] | C | Explanation:
In 12 hours, the hands coincide or are in opposite direction 22 times.
In 24 hours, the hands coincide or are in opposite direction 44 times a day.
Answer is C |
AQUA-RAT | AQUA-RAT-38103 | Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Intern Joined: 12 Mar 2015 Posts: 4 Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink] ### Show Tags 05 Oct 2015, 23:05 Barkatis wrote: It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages? A. 5 B. 6 C. 8 D. 10 E. 12 Hello, why can't it be time taken by tom= T and time taken by Jack= T+2 ? And then the eq becomes $$\frac{20}{T}$$ + $$\frac{20}{T+2}$$ = $$\frac{25}{3}$$ Please help? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8196 Location: Pune, India Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink] ### Show Tags 05 Oct 2015, 23:34 2 amanlalwani wrote: Barkatis wrote: It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages? A. 5 B. 6 C. 8 D. 10 E. 12 Hello, why can't it be time taken by tom= T and time taken by Jack= T+2 ? And then the eq becomes $$\frac{20}{T}$$ + $$\frac{20}{T+2}$$ = $$\frac{25}{3}$$ Please help? It can be. If you solve it, you will get T = 4. Then, to get time taken by Jack, you just add 2 to it to get 6 hrs. Note that you can take either variable but it is often better to assume what you have to find (time taken by Jack) since sometimes, we forget the last step. If you directly get the value of J, great. If you first
The following is multiple choice question (with options) to answer.
Two consultants can type up a report in 12.5 hours and edit it in 7.5 hours. If Mary needs 30 hours to type the report and Jim needs 12 hours to edit it alone, how many E hours will it take if Jim types the report and Mary edits it immediately after he is done? | [
"41.4",
"34.1",
"13.4",
"12.4"
] | A | Break down the problem into two pieces: typing and editing.
Mary needs 30 hours to type the report--> Mary's typing rate = 1/30 (rate reciprocal of time)(point 1 in theory below);
Mary and Jim can type up a report in 12.5and --> 1/30+1/x=1/12.5=2/25 (where x is the time needed for Jim to type the report alone)(point 23 in theory below)--> x=150/7;
Jim needs 12 hours to edit the report--> Jim's editing rate = 1/12;
Mary and Jim can edit a report in 7.5and --> 1/y+1/12=1/7.5=2/15 (where y is the time needed for Mary to edit the report alone) --> y=20;
How many E hours will it take if Jim types the report and Mary edits it immediately after he is done--> x+y=150/7+20=~41.4
Answer: A. |
AQUA-RAT | AQUA-RAT-38104 | (A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
let total qty be x liters
so milk 9x/100
9 liters is withdrawn so left with 9x/100- (9/100) * 9 which becomes equal to 6x/100
we have
9/100 * ( x-9) = 6x/100
3x-27 = 2x
x= 27 litres
OPTION B
Director
Joined: 25 Jul 2018
Posts: 731
A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 11:44
$$\frac{91}{100 }$$—the amount of water in 1 liter mixture.
—>$$( \frac{91}{100})x —(\frac{91}{100})*9 + 9 = (\frac{94}{100})x$$
$$\frac{( 94 —91)}{100} x = \frac{(100 —91)}{100}*9$$
$$(\frac{3}{100})x = \frac{81}{100}$$
—> $$x = 27$$
Posted from my mobile device
Stern School Moderator
Joined: 26 May 2020
Posts: 268
Concentration: General Management, Technology
WE: Analyst (Computer Software)
Re: A vessel is full of a mixture of milk and water, with 9% milk. Nine li [#permalink]
### Show Tags
01 Jun 2020, 14:47
Bunuel wrote:
A vessel is full of a mixture of milk and water, with 9% milk. Nine litres are withdrawn and then replaced with pure water. If the milk is now 6%, how much does the vessel hold?
(A) 18 litres
(B) 27 litres
(C) 36 litres
(D) 40 litres
(E) 47 litres
The following is multiple choice question (with options) to answer.
A vessel of capacity 50 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution? | [
"23.89",
"72.9",
"33.62",
"78.3"
] | C | Explanation:
Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T - y)/T]^n * T
For the given problem, T = 50, y = 9 and n = 2.
Hence, quantity of milk finally in the vessel
= [(50 - 9)/50]^2 (50) = 33.62 litres.
Answer: Option C |
AQUA-RAT | AQUA-RAT-38105 | ## Solution 4
Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$.
Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$. Dividing everything by $b^2$, we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$.
Applying the quadratic formula....and following the restriction that $a>b>0$....
$\frac{a}{b}=\frac{10}{7}$.
Hence, $7a=10b$.
Since they are relatively prime, $a=10$, $b=7$.
$10-7=\boxed{\textbf{(C)}\ 3}$.
## Solution 5
Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{\textbf{(C)}\ 3}$ ~mathboy282
The following is multiple choice question (with options) to answer.
If 9a - b = 10b + 70 = -12b - 2a, what is the value of 11a + 11b? | [
"-4",
"-2",
"0",
"2"
] | C | (i) 9a - 11b = 70
(ii) 2a + 22b = -70
Adding (i) and (ii):
11a + 11b = 0
The answer is C. |
AQUA-RAT | AQUA-RAT-38106 | # STRNO - Editorial
Setter: Kritagya Agarwal
Tester: Felipe Mota
Editorialist: Taranpreet Singh
Easy
Number theory
# PROBLEM:
Given two integers X and K, you need to determine whether there exists an integer A with exactly X factors and exactly K of them are prime numbers.
# QUICK EXPLANATION
A valid choice for A exists if the prime factorization of X has the number of terms at least K. So we just need to compute prime factorization of X.
# EXPLANATION
Let’s assume a valid A exists, with exactly K prime divisors.
The prime factorization of A would be like \prod_{i = 1}^K p_i^{a_i} where p_i are prime factors of A and a_i are the exponents. Then it is well known that the number of factors of A are \prod_{i = 1}^K (a_i+1). Hence we have X = \prod_{i = 1}^K (a_i+1) for a_i \geq 1, hence (a_i+1) \geq 2
So our problem is reduced to determining whether is it possible to write X as a product of K values greater than 1 or not.
For that, let us find the maximum number of values we can split X into such that each value is greater than 1. It is easy to prove that all values shall be prime (or we can further split that value). If the prime factorization of X is \prod r_i^{b_i}, then \sum b_i is the number of terms we can split X into.
For example, Consider X = 12 = 2^2*3 = 2*2*3. Hence we can decompose 12 into at most 3 values such that their product is X. However, if K < \sum b_i, we can merge the values till we get exactly K values. Suppose we have X = 12 and K = 2, so after writing 2, 2, 3, we can merge any two values, resulting in exactly two values.
So a valid A exists when the number of prime factors of X with repetition is at least K.
The following is multiple choice question (with options) to answer.
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT | [
"-4",
"-2",
"-1",
"2"
] | B | There is probably an easier way, but I just used the picking numbers option for this.
I chose x=2
2(1)(2-k) then just plugged in the answer choices for K until one wasn't evenly divisible by 3.
B gives you 8. 8/3 is not an integer.
B is the answer |
AQUA-RAT | AQUA-RAT-38107 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
Walking at 4/5 of her normal speed, a worker is 10 minutes later than usual in reaching her office. The usual time (in minutes) taken by her to cover the distance between her home and her office is | [
"30",
"35",
"40",
"45"
] | C | Let V be her normal speed and let T be her normal time.
D = (4/5)V * (T+10)
Since the distance is the same we can equate this to a regular day which is D = V*T
V*T = (4/5)V * (T+10)
T/5 = 8
T=40
The answer is C. |
AQUA-RAT | AQUA-RAT-38108 | 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.
What number comes next?
428, 693, 714, 286, 937, ? | [
"142",
"231",
"245",
"148"
] | A | A
142
The numbers 42869371 are being repeated in the same sequence. |
AQUA-RAT | AQUA-RAT-38109 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of 13 articles is equal to the selling price of 11 articles. Find the profit percent? | [
"18 2/18%",
"18 2/11%",
"18 3/11%",
"18 2/17%"
] | B | 13 CP = 11 SP
11 --- 2 CP
100 --- ? =>18 2/11%
Answer: B |
AQUA-RAT | AQUA-RAT-38110 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^5, then what was the number in the population 90 minutes later? | [
"2(10^5)",
"9(10^5)",
"(2^9)(10^5)",
"(10^9)(10^5)"
] | C | Every 10 minutes, the population is double the previous population.
In 90 minutes, the population doubles 9 times.
The population then is 2^9*10^5.
The answer is C. |
AQUA-RAT | AQUA-RAT-38111 | # If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$
If $a\lt{b}$ and $c\le{d}$, prove that $a+c\lt b+d$.
This seems like a basic proof and I think this is how it goes: $$c \le d, \text{ Given }$$ $$a+c \le a+d$$ $$a+c \lt b+d, \text{ since } a \lt b$$
Is that all I need? I'm thinking this does it all quickly and concisely, but I have had trouble with proofs in my classes.
• Yes, or just go $a < b \implies a+c<b+c$, so since $c \leq d \implies b + c \leq b + d$, we have $a+c < b+c \leq b+d$, i.e. $a+c < b+d$. – Ryker Feb 17 '14 at 20:10
• Is $(b+d)-(a+c)>0$? – David Mitra Feb 17 '14 at 20:10
• I'd do it $(a-b)+(c-d)\lt 0$ but it's all the same in the end – Mark Bennet Feb 17 '14 at 20:18
• What is the reason you did it your way, @Ryker, versus my way? It is just semantical? I've noticed that the things posted have a middle argument between a less than and a less than or equal to sign. Is that preferred when proving to see the logical reason? – Faffi-doo Feb 17 '14 at 20:25
• @Faffi-doo , guys, wow! :) Perhaps I should have added a smiley in the end of the comment to avoid misunderstandings.. 1. You don't need to be so apologetic! :) Choose what is most correct for you. Always, and not just in Math. 2. I will not hold anything against you even if you actually did not like my answer. – user76568 Feb 17 '14 at 21:21
The following is multiple choice question (with options) to answer.
If (a – b) is 17 more than (c + d) and (a + b) is 3 less than (c – d), then (a – c) is: | [
"6",
"2",
"3",
"7"
] | D | (a – b) – (c + d) = 17 and (c – d) – (a + b) = 3
=> (a – c) – (b + d) = 17 and (c – a) – (b + d) = 3
=> (b + d) = (a – c) – 17 and (b + d) = (c – a) – 3
=> (a – c) – 17 = (c – a) – 3 => 2(a – c) = 14 => (a – c) = 7
ANSWER:D |
AQUA-RAT | AQUA-RAT-38112 | recognize that if the event $P(A|A^cB^c)$ is the event that they both miss, and so they must start over again meaning $P(A|A^cB^c)=P(A)$ and therefore
$$P(A)=1 \cdot p_B(1-p_A)+0\cdot (1-p_B)p_A+1\cdot p_Ap_B+P(A)(1-p_B)(1-p_A)$$
solving this equation for $P(A)$ gives
$$P(A)=\frac{p_B}{1-(1-p_A)(1-p_B)}$$
the probability that both duelists are hit
using similar logic as in part (A)
$$P(AB)=P(AB|AB^c)P(AB^c)+P(AB|A^cB)P(A^cB)+P(AB|AB)P(AB)+P(AB|A^cB^c)P(A^cB^c)$$
$$=0+0+1\cdot p_Ap_B+P(AB)(1-p_A)(1-p_B)$$
$$P(AB)=\frac{p_Ap_B}{1-(1-p_A)(1-p_B)}$$
the probability that the duel ends after the nth round of shots
let $G_i$ be the event that the game goes longer than the $i_th$ duel
the game ends with more than one duel on the condition that
$$P(G_1)=0+0+0+1\cdot (1-p_A)(1-p_B)=(1-p_A)(1-p_B)$$
the game ends with more than two duels on the condition that
The following is multiple choice question (with options) to answer.
In a game of 100 points,A can give B 20 points and C 28 points.Then, B can give C : | [
"8 points",
"10 points",
"14 points",
"40 points"
] | B | Solution
A : B = 100 : 80 and A : C = 100 : 72.
∴ B/C=(B/C x A/C) = (80/100 x 100/72)= 10/9 =100/90= 100 : 90.
∴ B can give C 10 points.
Answer B |
AQUA-RAT | AQUA-RAT-38113 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The price for a loaf of bread in April was $1.17, in May the price was $1.32, and in June the price was $1.08. If 2/3 as much bread was sold in April as in May, and twice as much was sold in June as in April, what was the average price for a loaf of bread sold during the three-month period? | [
"$1.16",
"$1.17",
"$1.18",
"$1.19"
] | C | Let x be the number of loaves sold in May.
Then 2x/3 loaves were sold in April and 4x/3 loaves were sold in June.
The average price was (2x/3)(117) + 132x + (4x/3)(108) / (2x/3 + x + 4x/3) =
(78 + 132 + 144) / (3) = 354/3 = $1.18
The answer is C. |
AQUA-RAT | AQUA-RAT-38114 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A vendor buys 10 t-shirts at an average price of $14 per t-shirt. He then buys 15 more t-shirts at an average price of $11 per t-shirt. What is the average price B per t-shirt that the vendor paid for these purchases? | [
"$12.20",
"$12.50",
"$12.55",
"$12.70"
] | A | Correct Answer: A
Explanation: The relevant formula for this problem is Average B= (Sum)/(Number of Terms). Another way to look at the formula is Sum = Average x Number of Terms. For the first purchase, the vendor's sum (total cost) was $140, since 14 x 10 = 140. For the second purchase, the vendor's cost was $165, since 11 x 15 = 165. The grand sum is then $140 + $165, which equals $305. The total number of shirts purchased was 25, so to get the average price per shirt, we divide 305 by 25, which equals $12.20. As a result, the correct answer is A.
NOTE: A relative understanding of weighted average offers a shortcut to this problem. Because the true average of 11 and 14 is 12.5, but the vendor sells more shirts at the lower price than at the higher price, the weighted average must be less than $12.50; only answer choice A is a possibility. |
AQUA-RAT | AQUA-RAT-38115 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of 13 numbers is 60. Average of the first 7 of them is 54 and that of the last 7 is 61. Find the 8th number? | [
"25",
"83",
"45",
"53"
] | A | Sum of all the 13 numbers = 13 * 60 = 780
Sum of the first 7 of them = 7 * 54 = 378
Sum of the last 7 of them = 7 * 61 = 427
So, the 8th number = 427 + 378 - 780 = 25.
Answer:A |
AQUA-RAT | AQUA-RAT-38116 | # probability question on a customer
A very frustrated customer is trying to find an electronic receipt on their phone so they can return an item. The trouble is that the customer has three email accounts and can't remember which, one, account it was sent to. The customer assumes that there is an equal probability for each account. To add complexity the phone only has enough battery power to search one of the three email accounts
Unfortunately, the store says that this is the last day it will accept the return, and it is only 3 minutes till close, so there will not be any chance to get to a charger. The customer randomly decides to search one of the emails, without any bias.
Suppose, due to the organization and the various spam filters of the accounts, the chance of finding the receipt even if they were to search in the correct account is not a guarantee. The probability of finding it in account 1, assuming it was sent there is 62%, 54% if in account 2 and 56% for account 3.
Part (a) What is the probability that if the receipt is in account 2 that the customer will find it?
I think this one is (1/3)*(0.54) The third is for choosing the account and then times 0.54 is the prob for finding from the question. but the question seems like a conditional probability so I'm not sure.
Part (b) Calculate the probability the receipt was in account 2, if the search in account 2 is unsuccessful.
I think this one is ((1/3)(0.16))/((1/3)(0.31)+(1/3)(0.16)+(1/3)(0.49))
Part(c) What is the probability this person finds the email?
(1/3)(0.69)+(1/3)(0.84)+(1/3)(0.51) because we can either find it in account 1 or 2 or 3
Any help would be appreciated!
The following is multiple choice question (with options) to answer.
A shop that sells phone accessories had the following sales: 10 thousand earbuds at $10 each, 5300 phone cases at $12 each and 450 power banks at $50 each. What % of the shop's revenue is from power banks? | [
"11.1%",
"12.1%",
"10.1%",
"9.1%"
] | B | Correct Answer: B
The company's total revenue from earbuds=10000*$10=$100000
The company's total revenue from phone cases=5300*$12=$63600
The company's total revenue from power banks=450*$50=$22500
The combined total of all sales=$186100
Therefore % of power banks sales=22500/186100*100= 12.1% |
AQUA-RAT | AQUA-RAT-38117 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. 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 H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is | [
"276",
"299",
"312",
"322"
] | D | Solution
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Answer D |
AQUA-RAT | AQUA-RAT-38118 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
If the population of a certain country increases at the rate of one person every 15 seconds, by how many persons does the population increase in 25 minutes? | [
"80",
"100",
"150",
"240"
] | B | Since the population increases at the rate of 1 person every 15 seconds, it increases by 4 people every 60 seconds, that is, by 4 people every minute. Thus, in 25 minutes the population increases by 25 x 4 = 100 people.
Answer. B. |
AQUA-RAT | AQUA-RAT-38119 | Maybe the examiner didn't think through your method.
From 10,000 to 99,999 the numbers divisible by 5 end in 0 or 5.
The first (most significant) digit can be any of nine from 1 to 9.
The 2nd digit can be any of 10.
The 3rd digit can be any of 10.
The 4th digit can be any of 10.
The 5th digit can be either of 2.
That's 9(10)(10)(10)2=18,000
The following is multiple choice question (with options) to answer.
The greatest number of 5 digits = 99999. | [
"137",
"143",
"77",
"88"
] | B | Answer:B |
AQUA-RAT | AQUA-RAT-38120 | So, the number of dealings = 52!/(13!)^4.
How many dealings give each of A,B,C,D exactly one Ace (assuming 4 aces in the deck)? Well, there are 4! ways to give each player one ace, then there are
$$\frac{(N-4)!}{(a-1)! (b-1)! (c-1)! (d-1)!}$$ dealings of (N-4) non-aces to the players, giving (a-1) non-aces to Mr. A, etc. For each of these dealings, we just give each player one of the aces, so the total number of such dealings is
$$\frac{4! (N-4)!}{(a-1)! (b-1)! (c-1)! (d-1)!}.$$ In the example we have N=52, a=b=c=d=13, so the number of one-ace-per-player dealings is 4!*48!/(12!)^4. The required probability is
$$\text{prob.} = \frac{(13!)^4}{52!} \cdot \frac{4! 48!}{(12!)^4} = \frac{2197}{20825} \doteq 0.1050,$$ You would get exacctly the same probability by implementing the approach in my previous post.
The following is multiple choice question (with options) to answer.
The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip? | [
"1/336",
"1/120",
"1/56",
"1/720"
] | C | P=P(ace in the first draw)*P(ace in the second draw)*P(ace in the third draw)=3/8*2/7*1/6=1/56.
Answer: C. |
AQUA-RAT | AQUA-RAT-38121 | 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 goods train runs at the speed of 72 km/hr and crosses a 150 m long platform in 26 sec. What is the length of the goods train? | [
"299",
"277",
"276",
"370"
] | D | Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 150)/26 = 20
x = 370 m.
Answer: D |
AQUA-RAT | AQUA-RAT-38122 | 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.
If a certain number of people can dig earth 100m deep, 25m long, and 30m broad in 12 days, then find the number of days the same number of people will require to dig earth 75m deep, 20m long and 50m broad. | [
"12",
"18",
"6",
"1"
] | A | Explanation:
More number of days means – more length, more depth and more width. Hence, it’s a direct proportion.
(100*25*30):(75*20*50)::12:x
75000:75000::12:x
x = 12
ANSWER A |
AQUA-RAT | AQUA-RAT-38123 | At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
5. Hmmm....I'm not a maths expert but let's see what we get if we reverse the ratios. Perhaps there was a communication error.
$p:a=3:5$
$\frac {1}{10}$of a=wooden
$\frac{1}{10}\times\frac{5}{8}=\frac{5}{80}$
= number of wooden.
number of brick = 1-number of wooden
$=\frac{75}{80} = \frac{15}{16}$
Edit: Are you sure you wrote the ratio the correct way around? The way you wrote it suggests that the person above has the correct answer.
I don't know how it got 15/16. I actually can't figure out why that is.
At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
The following is multiple choice question (with options) to answer.
In a certain apartment building, there are one-bedroom and two-bedroom apartments. The rental prices of the apartment depend on a number of factors, but on average, two-bedroom apartments have higher rental prices than do one-bedroom apartments. Let M be the average rental price for all apartments in the building. If M is $700 higher than the average rental price for all one-bedroom apartments, and if the average rental price for all two-bedroom apartments is $2,100 higher that M, then what percentage of apartments in the building are two-bedroom apartments? | [
"25%",
"15%",
"20%",
"40%"
] | A | Ratio of 2 Bedroom Apartment: 1 Bedroom Apartment = 700 : 2100 -----> 1 : 3
Let total number of Apartments be X
No. of 2 Bedroom Apartment = (1 / 4) * X
percentage of apartments in the building are two-bedroom apartments ---->
(1/4) * 100 ---> 25%
Answer : A |
AQUA-RAT | AQUA-RAT-38124 | # 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 person purchased a TV set for Rs. 2700 and a DVD player for Rs. 200 . He sold both the items together for Rs. 3180. What percentage of profit did he make? | [
"20%",
"6%",
"40%",
"5%"
] | B | The total CP = Rs. 2700 + Rs. 300
= Rs. 3000 and SP = Rs. 3180
Profit(%)
= (3180- 3000)/3000 * 100
= 6%
Answer: B |
AQUA-RAT | AQUA-RAT-38125 | The $$26$$ blocks consist of three groups: six of them are face-to-face with $$V_1$$, twelve of them are edge-touching-edge-only with $$V_1$$, and the remaining eight are vertex-touching-vertex-only with $$V_1$$.
The value of each $$I_f$$, $$I_e$$, and $$I_v$$ will be obtained using $$E(a,b,c)$$, successively building up starting from $$I_f$$.
It will involve a $$2$$-$$1$$-$$1$$ long box, a $$2$$-$$2$$-$$1$$ flat box, then finally the $$2$$-$$2$$-$$2$$ "double cube".
## Calculation Details
The following is multiple choice question (with options) to answer.
Three cubes of edges 6 cms, 8 cms and 10 cms are meted without loss of metal into a single cube. The edge of the new cube will be: | [
"8 cms",
"12 cms",
"14 cms",
"16 cms"
] | B | Volume of new cube = Volume of cube 1 + cube 2 + cube 3 = 63 + 83 + 103, = 216 + 512 + 1000
a3 = 1728,
a = (1728)1/3 = 12
Answer: B |
AQUA-RAT | AQUA-RAT-38126 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
The ratio of the ages of Maala and Kala is 1 : 4. The total of their ages is 2.0 decades. The proportion of their ages after 0.5 decades will be
[1 Decade = 10 years] | [
"6:5",
"3:4",
"2:5",
"7:9"
] | C | Let, Maala’s age = A and Kala’s age =4A
Then A + 4A = 20
A =5
Maala’s age = 5 years
and Kala’s age = 20 years
Proportion of their ages after 5 is = (5 + 5) : (20 + 5)
= 10 : 25
= 2 : 5
Answer : C |
AQUA-RAT | AQUA-RAT-38127 | mass, weight
Title: Mass versus Weight What are the difference between mass and weight? I keep getting confused in my physics class, and I am in 8th grade. Thank you in advance. Mass is a measure of how much matter (atoms) make up an object. Weight is a force that results from that quantity of matter accelerating in a gravitational field. Weight = mass*acceleration due to gravity. Weight can change depending on what gravitational field you are in, mass cannot change.
The following is multiple choice question (with options) to answer.
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 13 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.) | [
"25 pounds",
"46 pounds",
"92 pounds",
"130 pounds"
] | D | Since only water evaporates, then the weight of pulp (non-water) in grapes and raisins is the same. Thus 0.08*{weight of grapes}=0.8*{weight of raisins} --> 0.08x = 0.8*13 --> x = 130.
Answer: D. |
AQUA-RAT | AQUA-RAT-38128 | # 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.
Simple interest on a certain sum at a certain annual rate of interest is 1/4 of the sum. If the numbers representing rate percent and time in years be in the ratio 1:4, then the rate of interest is: | [
"4 1/2 %.",
"3 1/2 %.",
"2 1/2 %.",
"2 3/4 %."
] | C | Explanation :
Let sum = x. Then, S.I. = x/4
Let rate = R% and time = R years.
[x * R * 4R / 100] = x / 4 ? R^2 = 25/4
R = 5/2 = 2 1/2.
Hence, R = 2 1/2 %.
Answer : C |
AQUA-RAT | AQUA-RAT-38129 | # 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 trader sells 85 meters of cloth for Rs. 8925 at the profit of Rs. 15 per metre of cloth. What is the cost price of one metre of cloth? | [
"26",
"88",
"90",
"42"
] | C | SP of 1m of cloth = 8925/85 = Rs. 105
CP of 1m of cloth = SP of 1m of cloth - profit on 1m of cloth
= Rs. 105 - Rs. 15 = Rs. 90
Answer:C |
AQUA-RAT | AQUA-RAT-38130 | # smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2|<0.001$$
How to proceed? Thanks in advance.
• You are almost done ! – Khosrotash Aug 11 '17 at 14:49
$$|2[1-(-\frac{1}2)^n]-2|<0.001\\ |-2(\frac{-1}2)^n|<0.001\\ +2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\ +2.\frac{1}{2^n}<\frac{1}{1000}\\ \frac{2^n}{2}>1000\\ 2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
The following is multiple choice question (with options) to answer.
Which of the following fraction is smallest? | [
"23/28",
"14/15",
"15/19",
"21/24"
] | C | 23/28 =0.821
14/15 = 0.933
15/19 = 0.7894
21/24 = 0.875
So, 15/19 = 0.7894 is smallest.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38131 | the acceptance volume is the volume of water that the bladder is designed to hold, which is smaller than the overall tank size. The volume flow, Q = ∫ u(2π r)dr, can be estimated by a numerical quadrature formula such as Simpson’s rule. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308. This process is carried out in a structure called sedimentation tank or settling tank. Indirect tank heating uses a heat transfer medium to apply the heat to the tank. Motor fuel 60 to 250 gallons (Primary or alternate fuel system tanks for autos, trucks, buses. It is defined as the ratio of pressure stress to volumetric strain. The barometric formula is used in both ISA and USSA to model how the pressure and density of air changes with altitude. When the substance in the tank is near the mid-fill level, each inch of depth corresponds to a much larger volume that when the substance is near the top or bottom of the tank. We manufacture our tanks to UL-142, UL-2085, API-650, and other industry-proven standards for Above Ground Storage of Flammable and Combustible Liquids. calculate the volume of ellipsoidal head or dish 2 1. Major Axis of an Ellipse. Stated as a mathematical formula, taking into consideration the total volume of a pressure tank, it looks like this: Drawdown = P1V / P2 – P1V / P3 where, P1 is the pre-charge pressure. Inspiratory Reserve Volume (IRV). The problem can be generalized to other cones and n-sided pyramids but for the moment consider the right. Many times, this formula will use the height of the prism, or depth (d), rather than the length (l), though you may see either abbreviation. We use cookies to improve your navigation experience. Volume (V) = π * R 2 * ((4/3) * R + H) Volume (V) = 3. This means that the normal line at this point is a vertical line. Conservation. por | set 15, 2020 | Sem categoria | 0 Comentários | set 15, 2020 | Sem categoria | 0 Comentários. When a sphere is falling through a fluid it is completely submerged, so there is only
The following is multiple choice question (with options) to answer.
A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.2 m and its walls are 5 cm thick. The thickness of the bottom is: | [
"90 cm",
"2 dm",
"1 m",
"1.1 cm"
] | B | Explanation:
Let the thickness of the bottom be x cm.
Then , [(330 - 10) × (260 - 10) × (120 - x)] = 8000 × 1000
=> 320 × 250 × (120 - x) = 8000 × 1000
=> (120 - x) = 8000×1000/320=
100
=> x = 20 cm = 2 dm.
Answer: B |
AQUA-RAT | AQUA-RAT-38132 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A classroom has equal number of boys and girls. Eight girls left to play kho-kho, leaving twice as many boys as girls in the classroom. What was the total number of girls and boys present initially? | [
"16",
"24",
"32",
"48"
] | C | after 8 girls left remaining 8 girls
now boys 16 are twice as many as remaining girls.
initially boys=16 and girls=16.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38133 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
The average runs scored by a batsman in 20 matches is 40. In the next 10 matches the batsman scored an average of 30 runs. Find his average in all the 30 matches? | [
"31",
"36.67",
"88",
"13"
] | B | Total score of the batsman in 20 matches = 800.
Total score of the batsman in the next 10 matches = 300.
Total score of the batsman in the 30 matches = 1100.
Average score of the batsman = 1100/30 = 36.67.
Answer:B |
AQUA-RAT | AQUA-RAT-38134 | $11 + 6.5 = 17.5 \text{ or } 24 - 6.5 = 17.5$.
Therefore, 17.5 is the number in the middle of $11 \mathmr{and} 24.$
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The following is multiple choice question (with options) to answer.
Find the value of * in the following
(1 5/3) ÷ 3/11 × */11 = (2 2/3 × 7/5 × 6/7) | [
"3.1",
"3.4",
"3.6",
"4"
] | C | 3.6
Option 'C' |
AQUA-RAT | AQUA-RAT-38135 | Change the chapter
Question
Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.) (b) Compare this force with the weight of the jogger.
a) $16300 \textrm{ N}$
b) $22.2$
Solution Video
OpenStax College Physics Solution, Chapter 7, Problem 53 (Problems & Exercises) (3:42)
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The following is multiple choice question (with options) to answer.
I have to run a 100 meter dash.
How many steps would I need to take if with each step i cover 20 centimeters ? | [
"20",
"500",
"50",
"10"
] | B | 100centimeters=1meter
x centimeters=100meters
(20 centimeters per step) x (nos running steps)=10000 centimeters
nos running steps= 10000 centimeters /20 centimeters
500steps
The answer is B |
AQUA-RAT | AQUA-RAT-38136 | rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron’s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 × 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The "base times height" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b×h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal ‘d1’ of a rhombus and a side ‘a’, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of […] where b is the
The following is multiple choice question (with options) to answer.
The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is? | [
"218",
"265",
"268",
"480"
] | D | 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48
= 480
Answer: D |
AQUA-RAT | AQUA-RAT-38137 | 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.
If the price of sugar rises from Rs.10 per kg to Rs. 12 per kg, a person, to have no increase in the expenditure on sugar, will have to reduce his consumption of sugar by | [
"17%",
"20%",
"25%",
"30%"
] | A | Sol.
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 10 = x × 12 = x = 83 kg.
∴ Reduction in consumption = 17%.
Answer A |
AQUA-RAT | AQUA-RAT-38138 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of 20 number is zero of them at the most, how many may be greater than zero ? | [
"0",
"1",
"10",
"19"
] | D | Solution
average of 20 numbers = 0.
∴ Sum of 20 numbers = (0×20) = 0.
it is quite possible that 19 of these number may be positive and if their sum is a, then 20th number is (-a). Answer D |
AQUA-RAT | AQUA-RAT-38139 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The average of 35 students in a class is 16 years. The average age of 25 students is 14. What is the average age of remaining 10 students? | [
"21 years",
"22 years",
"23 years",
"24 years"
] | A | Sum of the ages of 14 students
= (16 * 35) - (14 * 25) = 560 - 350 = 210
Required average = (210/10) = 21 years.
Answer:A |
AQUA-RAT | AQUA-RAT-38140 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 7 days. With the help of C they finish the work in 5 days. C alone can do that piece of work in? | [
"11 days",
"66 days",
"30 days",
"77 days"
] | C | C = 1/5 – 1/6 = 1/30 => 30 days
Answer: C |
AQUA-RAT | AQUA-RAT-38141 | 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.
A man rides at the rate of 40km/hr. But stops 20minutes to change horses at the end of every 25th kilometer. How long will he take to go a distance of 200 kilometers? | [
"5hr",
"6hr 30min",
"8hr 10min",
"7hr 20min"
] | D | speed of man = 40km/hr
number of rests = (200/25)-1 = 7
time taken for the man = (200/40)+7*(20/60) = 7 hr 20 min
Answer is D |
AQUA-RAT | AQUA-RAT-38142 | # String probability (with conditional prob and combinations)
I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.
SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.
Q1. What is the probability that it contains exactly one letter that is a vowel?
A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)
Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?
A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))
SET2: How many distinct permutations of the letters in "letters"...
Q1. Begin with two vowels?
A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...
Q2. Begin with two e's or end with two t's?
A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.
Q3. Have the vowels together?
A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.
Thank you!
The following is multiple choice question (with options) to answer.
The letters U and S are rearranged to form the word 'US'. Find its probability. | [
"1/4",
"1/3",
"1/2",
"1/7"
] | C | Explanation :
There are total 2 letters. The probability that S gets the first position is 1/2.
The probability that S is in the second position is 1/1.
Hence, the required probability is:-
=> (1/2) x 1.
=> 1/2
Answer : C |
AQUA-RAT | AQUA-RAT-38143 | -
No, please see what I added for c) – André Nicolas Nov 18 '12 at 3:52
Your answer of $\binom{20}5$ for (a) is correct.
(b) To get the probability that $1$ and $2$ are not in the sample, you must count the $5$-card sets that don’t contain either $1$ or $2$. These are the $5$-card sets chosen from $\{3,4,5,\dots,20\}$; how many of them are there? This is really a problem just like (a). Once you have that, what fraction must you form to get the desired probability?
(c) Now observe that the events in (b) and (c) are complementary, so their probabilities add up to a known amount; what is that amount? Use it and your answer to (b) to get the answer to (c).
-
The following is multiple choice question (with options) to answer.
In a nationwide poll, N people were interviewed. If 1/6 of them answered yes to question 1, and of those, 1/2 answered yes to question 2, which of the following expressions represents the number of people interviewed who did NOT answer yes to both questions? | [
"N/8",
"7N/8",
"11N/12",
"17N/24"
] | C | The number of people who answered yes to both questions is (1/2)(1/6)N = N/12
The number of people who did not answer yes to both is 1 - N/12 = 11N/12
The answer is C. |
AQUA-RAT | AQUA-RAT-38144 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
If n is an integer, when (2n + 2)2 is divided by 4 the remainder is | [
"0",
"1",
"2",
"3"
] | A | We first expand (2n + 2)2
(2n + 2)2 = 4n 2 + 8 n + 4
Factor 4 out.
= 4(n 2 + 2n + 1)
(2n + 2)2 is divisible by 4 and the remainder is equal to 0. The answer is A. |
AQUA-RAT | AQUA-RAT-38145 | Also: explore what must happen if one of the variables, say x, is zero.
4. Aug 9, 2013
Thanks for the hints , but I used a different method to solve this equation. From eq. 1 I got y=(6x)/(5x-6) .
I substituted that in eq. 2 to get x in terms of z. ( x=(126z)/(126+45z) ) then I substituted that into eq. 3 to get a quadratic equation in z with solutions z=0,7. From z=7 , I got x= 2 . Then I got y=3 from the relationship between x and y obtained earlier from eq. 1 . But I felt that the arithmetic involved was quite daunting and there must be a better way to solve it. I tried using the method that pasmith recommended but am feeling a bit lost. Can anyone please explain a simpler method to solve this system ?
5. Aug 9, 2013
### Ray Vickson
Let $a = xy, \;b = yz, \; c = zx \:\text{ and }\: p = xyz.$ Multiply the first equation by z, the second by x and the third by y to get
$$6b+6c=5p\\ 21a+21c=10p \\ 14a+14b=9p$$
This is a simple 3x3 linear system which can be solved using grade-school methods, to get
$a = p/7,\: b = p/2, \: c = p/3.$ Assuming $x,y,z,\neq 0$ we have
$$xy = xyz/7 \Longrightarrow z = 7 \\ yz = xyz/2 \Longrightarrow x = 2\\ zx = xyz/3 \Longrightarrow y=3$$
6. Aug 10, 2013
The following is multiple choice question (with options) to answer.
Given 2x + y + z = 3 and 5x + 3y + z = 11, what is the value of x + y - z? | [
"3",
"4",
"5",
"6"
] | C | (1) 5x + 3y + z = 11
If 2x + y + z = 3, then (2) 4x + 2y + 2z = 6
Let's subtract equation (2) for equation (1).
x + y - z = 5
The answer is C. |
AQUA-RAT | AQUA-RAT-38146 | I am also in High School, and it is my understanding that if there is a number squared and then square rooted then it is (by definition) just the original number. Taking your example of -5, plugging it in you would get sqrt (-5)^2 = -5. (-5)^2 is 25 and square rooting it is 5 which is not equal to -5. To avoid this, you must specify that if X is negative then sqrt X^2 = -x and if X is positve then sqrt X^2 = X. This would be expressed mathematically using inequalites as seen in your question.
• "Square Rooting" does NOT by definition mean just take the original number. This is because for $\sqrt{}$ to behave like a function from $\Bbb{R} \to \Bbb{R}$ (which we want for a lot of reasons), it must take in a single number. This makes $\sqrt{x^2} = \sqrt{(-x)^2}$ because before you can take the root, you need to simplify the inside, deleting its "past". If we wanted to keep the "past", it would need to come from $\Bbb{R}^n, n > 1$. To do so, it takes only one branch, and mathematicians decided for a variety of reaasons that it would be the positive branch. Thus, $\sqrt{x^2} = |x|$. Sep 25 at 5:58
The following is multiple choice question (with options) to answer.
A positive number x is multiplied by 2, and this product is then divided by 5. If the positive square root of the result of these two operations equals x, what is the value of x ? | [
" 9/4",
" 3/2",
" 4/3",
" 2/3"
] | B | We need to produce an equation from the information given in the problem stem. We are first given that x is multiplied by 2 and then the product is divided by 3. This gives us:
2x/3
Next we are given that the positive square root of the result (which is 2x/3) is equal to x. This gives us
√(2x/3) = x
2x/3 = x^2
2x = 3x^2
3x^2 – 2x = 0
x(3x – 2) = 0
x = 0 or
3x – 2 = 0
3x = 2
x = 3/2
Because x is positive, x = 3/2. The answer is B. |
AQUA-RAT | AQUA-RAT-38147 | 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 teacher gave the same test to three history classes: A, B, and C. The average (arithmetic mean) scores for the three classes were 65, 85, and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined? | [
"74",
"75",
"76",
"77"
] | D | ratio is 4:6:5 , numbers are 4x, 6x, 5x
total scores of each class is (65*4x + 6x * 85 + 77*5x) = 260x+510x+385x = 1155x
total number of students = 15x
average = 1155x/15x = 77
D is the answer |
AQUA-RAT | AQUA-RAT-38148 | |x - a| = b
x is a distance 'b' away from 'a' on either side on the number line.
-4 < x < 8
Mid point of -4 and 8 is 2. Both -4 and 8 are 6 away from 2.
So the absolute value form will be |x - 2| < 6
For more, check: http://www.veritasprep.com/blog/2011/01 ... edore-did/
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Veritas Prep GMAT Instructor
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Re: Which of the following inequalities is equivalent to −4 < x < 8? [#permalink]
### Show Tags
22 Aug 2016, 02:45
2
We know that |x| < a means -a < x < a, where Sum of lower limit of x (i.e -a) and the upper limit of x (i.e a), is 0
Given is, -4 < x < 8, let's say by adding y to this inequality we will get into the above format
-4+y < x+y < 8+y
Now, to move this into the mod format, we need to have (-4+y) + (8+y) = 0 => y = -2
Thus, -6< x-2 < 6 => |x-2| < 6.
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Re: Which of the following inequalities is equivalent to −4 < x < 8? [#permalink]
### Show Tags
22 Aug 2016, 19:28
2
2
Bunuel wrote:
Which of the following inequalities is equivalent to −4 < x < 8?
A. |x - 1| < 7
B. |x + 2| < 6
C. |x + 3| < 5
D. |x - 2| < 6
E. None of the above
Given −4 < x < 8,
Now try to get into the mod form,
subtract two from both sides of the equation, (graphically the line remains the same)
The following is multiple choice question (with options) to answer.
Which of the following is equivalent to the pair of inequalities x + 8 > 10 and x - 4 <= 5 ? | [
" 2 < x < 16",
" 2 <= x < 4",
" 2 < x <= 9",
" 4 < x <= 8"
] | C | Solution:
Let’s isolate x in both inequalities, starting with x + 8 > 10.
x + 8 > 10
x > 2
Next we isolate x in the inequality x – 4 5.
x – 4 <= 5
x <= 9
Bringing these two inequalities together we know:
2 < x <= 9
The answer is C. |
AQUA-RAT | AQUA-RAT-38149 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? | [
"0",
"1",
"10",
"19"
] | A | EXPLANATION
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
Answer A |
AQUA-RAT | AQUA-RAT-38150 | 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.
Ratio between Rahul and Deepak is 4:3, After 10 Years Rahul age will be 26 years. What is Deepak present age | [
"12",
"15",
"20",
"22"
] | A | Explanation:
Present age is 4x and 3x,
=> 4x + 10 = 26 => x = 4
So Deepak age is = 3(4) = 12
Answer: Option A |
AQUA-RAT | AQUA-RAT-38151 | (ones digit is $$0$$, because it is the difference of two numbers whose ones digit is $$1$$).
Now if $$3n$$ is divisible by $$10$$, for an integer $$n$$, then $$n$$ is divisible by $$10$$. QED.
The following is multiple choice question (with options) to answer.
The number N is 5,3H7, where H represents the ten's digit. If N is divisible by 9, what is the value of H? | [
"4",
"3",
"5",
"8"
] | B | If the number is divisible by 9, it must also be divisible by 3. Only 3 yields such a number.
Answer: B |
AQUA-RAT | AQUA-RAT-38152 | Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
### Show Tags
16 Apr 2017, 07:45
1
KUDOS
Top Contributor
1
This post was
BOOKMARKED
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Another approach.
Take the task of arranging the 5 letters and break it into stages.
Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways
IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.
Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)
[Reveal] Spoiler:
D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.
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The following is multiple choice question (with options) to answer.
Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word MANGO could be arranged to create a five-letter code? | [
"5!",
"5! − (3! + 2!)",
"5! − (3! × 2!)",
"5!/(3! + 2!)"
] | A | MANGO - five letters can be arranged in 5! ways
since there is no repetition in MANGO so
5!
Ans. A) 5! |
AQUA-RAT | AQUA-RAT-38153 | You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
09 Mar 2018, 03:18
HarveyKlaus wrote:
Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
Question:
The following is multiple choice question (with options) to answer.
If x < y < 0, which of the following must be true? | [
"(-x)+ (-y) is positive integer",
"2x+y is a negative integer",
"x-y is a positive integer",
"2xy/3x is a negative integer"
] | A | Given x, y are less than 0. So both will be negative integers
negative + negative = negative
negative * negative = positive
negative / negative = positive
x-y is false since xSo, option A is true , because if we multiple negative number with negative it becomes positive(x, y are negative numbers)
Answer : A |
AQUA-RAT | AQUA-RAT-38154 | java
public class TestStudents
{
public static void main(String[] args) throws IOException
{
System.out.println("============" + "=================");
System.out.println("Students " + "Personal Details");
System.out.println("============" + "=================");
String name;
int age;
int year;
String studentNum;
List<Student> studentsList = new ArrayList<Student>();
for (int i = 0; i < 2; i++)
{
int studentNumber = (i + 1);
System.out.println("");
System.out.println("Please enter " + "data for student " + studentNumber);
InputStreamReader converter = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(converter);
System.out.println("Enter Student "+ studentNumber + " Name:");
name = in.readLine();
System.out.println("Enter Student " + studentNumber + " Age (Integer):");
age = Integer.valueOf(in.readLine());
System.out.println("Enter Student " + studentNumber + " Year (Integer):");
year = Integer.valueOf(in.readLine());
System.out.println("Enter Student " + studentNumber + " Student Number:");
studentNum = in.readLine();
Student student = new Student(name, age, year, studentNum);
studentsList.add(student); // add student
}
for (int j = 0; j < studentsList.size(); j++)
{
Student st = studentsList.get(j);
The following is multiple choice question (with options) to answer.
The average age of a class of 22 students is 10 yrs. if the teacher's age is also included, the average increases by one year. Find the age of the teacher | [
"30 Years",
"31 Years",
"33 Years",
"34 Years"
] | C | Total age of students is 22X10 = 220 Years
Total age inclusive of teacher = 23X (10+1) = 253
So, Teacher's age is 253-220 = 33 Yrs
There is a shortcut for these type of problems
Teacher's age is 10+(23X1) = 33 Years
C |
AQUA-RAT | AQUA-RAT-38155 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
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BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
Given that a is the average (arithmetic mean) of the first seven positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b? | [
"3:4",
"8:13",
"5:6",
"13:10"
] | B | The first nine positive multiples of six are {6, 12, 18, 24,30, 36, 42}
The first twelve positive multiples of six are {6, 12, 18, 24, 30,36,42, 48, 54, 60, 66, 72}
Both sets are evenly spaced, thus their median=mean:
a=24 and b=(36+42)/2=39 --> a/b=24/39=8/13.
Answer: B. |
AQUA-RAT | AQUA-RAT-38156 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
If the sum of 9 consecutive integers is 90, what is the sum of the following 9 consecutive integers? | [
"99",
"135",
"145",
"171"
] | D | 1) greatest of each will have a difference of 9 and next greatest too will have a difference of 9 and so on..
Therefore all nine terms will have a term MORE by 9..
Sum =90+(9*9)=171
2) if sum is 90, the average= median=90/9=10..The median of next 9 will be 10+9=19..
And sum would be 19*9=171
ANSWER:D |
AQUA-RAT | AQUA-RAT-38157 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Question: A sporting good store sells one type of baseball bat and one type of baseball. The cost for 2 bats and 4 balls is $180. The cost for 1 bat and 6 balls is $190, as well. If someone were to buy an equal number of bats and balls, at most how many bats can he purchase if he has a budget of $195 for the purchase?
Options: | [
"1",
"2",
"3",
"4"
] | C | IMO it should be C that is C
reason:
formed an equation... bat = b ball = c
2b+4c=180
1b+6c=190
solving both we get b that is bat = 40 and c that is ball = 25
new equation 195 to be divided in equal
3b+3c=195
3*40 + 3*25 = 195
120+75= 195 |
AQUA-RAT | AQUA-RAT-38158 | At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
5. Hmmm....I'm not a maths expert but let's see what we get if we reverse the ratios. Perhaps there was a communication error.
$p:a=3:5$
$\frac {1}{10}$of a=wooden
$\frac{1}{10}\times\frac{5}{8}=\frac{5}{80}$
= number of wooden.
number of brick = 1-number of wooden
$=\frac{75}{80} = \frac{15}{16}$
Edit: Are you sure you wrote the ratio the correct way around? The way you wrote it suggests that the person above has the correct answer.
I don't know how it got 15/16. I actually can't figure out why that is.
At most, 90% of the private homes are brick. So that's 9/10 x 3/8 = 27/80.
100% of the apartments are brick. So that's 5/8 = 50/80.
Add those two together and you get 77/80 not 15/16 (=75/80 as the answer suggests)...
Look at it this way.
Say there were 80 houses in total.
The ration tells us that 30 would be private homes and 50 would be apartments.
10% of the private homes are wooden which makes 3.
All 50 apartments are brick and the remaining 27 private houses are unknown which could mean that all 27 are brick.
Hence a maximum of 77 brick houses out of 80.
The following is multiple choice question (with options) to answer.
In a neighborhood having 90 households, 11 did not have either a car or a bike. If 18 households had a both a car and a bike and 44 had a car, how many had bike only? | [
"30",
"35",
"20",
"18"
] | B | {Total}={Car}+{Bike}-{Both}+{Neither} --> 90=44+{Bike}-18+11 --> {Bike}=53 --> # those who have bike only is {Bike}-{Both}=53-18=35.
Answer: B. |
AQUA-RAT | AQUA-RAT-38159 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
In what time will Rs.4200 lent at 3% per annum on simple interest earn as much interest as Rs.5000 will earn in 5 years at 4% per annum on simple interest? | [
"8 1/3",
"8 1/6",
"8",
"8 1/2"
] | C | (4200*3*R)/100 = (5000*5*4)/100
R = 8
Answer: C |
AQUA-RAT | AQUA-RAT-38160 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief steels a car at 2.30p.m. and drives it at 60kmph. The theft is discovered at 3p.m. and the owner sets off in another car at 75 kmph. When will he overtake the thief? | [
"5p.m.",
"6p.m.",
"4p.m.",
"7p.m."
] | A | Suppose the thief overtakes x hours after 2.30p.m.
Distance covered by the thief in x hrs = distance covered by the owner in (x-1/2 hours)
60x = 75(x-1/2)
15x = 75/2
x = 5/2 hrs
Thief is overtaken at 5p.m.
Answer is A |
AQUA-RAT | AQUA-RAT-38161 | 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.
B completes a work in 3 days. A alone can do it in 10 days. If both work together, the work can be completed in how many days? | [
"2.31 days",
"4.31 days",
"5.31 days",
"6.31 days"
] | A | 1/3 + 1/10 = 13/30
30/13 = 2.31 days
ANSWER:A |
AQUA-RAT | AQUA-RAT-38162 | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
The following is multiple choice question (with options) to answer.
The S.I. on a certain sum of money for 3 years at 8% per annum is half the C.I. on Rs. 4000 for 2 years at 10% per annum. The sum placed on S.I. is? | [
"2197",
"1267",
"1750",
"2267"
] | C | Explanation:
C.I. = [4000 * (1 + 10/100)2 - 4000]
= (4000 * 11/10 * 11/10 - 4000) = Rs. 840.
Sum = (420 * 100)/(3 * 8) = Rs. 1750
Answer:C |
AQUA-RAT | AQUA-RAT-38163 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio of two numbers is 3:4 and their sum is 28. The greater of the two numbers is? | [
"16",
"18",
"20",
"24"
] | A | 3:4
Total parts = 7
= 7 parts --> 28 (7 × 4 = 28)
= 1 part ---->4 (1 × 4 = 4)
= The greater of the two number is = 4
= 4 parts ----> 16 (4 × 4 = 16)
A |
AQUA-RAT | AQUA-RAT-38164 | ### Factoring 2e^2+4e-15 Solution
The variable we want to find is e
We will solve for e using quadratic formula -b +/- sqrt(b^2-4ac)/(2a), graphical method and completion of squares.
${x}_{}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Where a= 2, b=4, and c=-15
Applying values to the variables of quadratic equation -b, a and c we have
${e}_{}=\frac{-4±\sqrt{{4}^{2}-4x\mathrm{2x}\mathrm{-15}}}{2x2}$
This gives
${e}_{}=\frac{-4±\sqrt{{4}^{2}-\mathrm{-120}}}{4}$
${e}_{}=\frac{-4±\sqrt{136}}{4}$
${e}_{}=\frac{-4±11.6619037897}{4}$
${\mathrm{e1}}_{}=\frac{-4+11.6619037897}{4}$
${\mathrm{e2}}_{}=\frac{-4-11.6619037897}{4}$
${e}_{1}=\frac{7.66190378969}{4}$
${e}_{1}=\frac{-15.6619037897}{4}$
The e values are
e1 = 1.91547594742 and e2 = -3.91547594742
### Factoring Quadratic equation 2e^2+4e-15 using Completion of Squares
2e^2+4e-15 =0
Step1: Divide all terms by the coefficient of e2 which is 2.
The following is multiple choice question (with options) to answer.
If the function Q is defined by the formula Q = 5e/(4x(z^2)), by what factor will Q be multiplied if e is quadrupled, x is doubled, and z is tripled? | [
"1/9",
"2/9",
"4/9",
"3/9"
] | B | We just need to find the factor thats all,
e -> quadrupled -> 4e
x-> doubled -> 2x
z-> tripled -> 3Z
Hence, Z^2 = 9Z^2
e is in numerator, and x*z in denominator.
Hence,
Additional factor being introduced = 4/2*9
=4/18 = 2/9 = B |
AQUA-RAT | AQUA-RAT-38165 | Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
Originally posted by EgmatQuantExpert on 08 May 2015, 00:28.
Last edited by EgmatQuantExpert on 12 Jun 2015, 03:42, edited 1 time in total.
##### General Discussion
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Joined: 12 Nov 2014
Posts: 62
Re: How many positive factors do 180 and 96 have in common? [#permalink]
### Show Tags
07 May 2015, 21:28
2
The number of common factors will be same as number of factors of the Highest Common Factor(HCF)
HCF of 180 and 96 is 12
Number of factors of 12 = 6
Ambarish
_________________
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Ambarish
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Joined: 02 Sep 2009
Posts: 47981
Re: How many positive factors do 180 and 96 have in common? [#permalink]
### Show Tags
08 May 2015, 03:40
1
1
Baten80 wrote:
How many positive factors do 180 and 96 have in common?
A. 6
B. 12
C. 16
D. 18
E. 24
To find the number of common factors of two integers:
1. Find the greatest common divisor (GCD) of the two integers;
2. Find the number of factors of that GCD.
GCD of 180 and 96 is 12 = 2^2*3. The number of factors of 12 is (2 + 1)(1 + 1) = 6.
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Joined: 09 Sep 2013
Posts: 7757
Re: How many positive factors do 180 and 96 have in common? [#permalink]
### Show Tags
11 Feb 2018, 07:22
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The greatest common factor of two numbers is 6!. Which of the following can be the second number, if one of the numbers is 8!? | [
"3(5!)",
"4(5!)",
"5(5!)",
"6(6!)"
] | D | GCF is the product of common factors of the numbers involved.
GCF = 6!
a = 8! = 8*7*6!
b will certainly have 5! and cannot have any more common factors with a (as this will increase the GCF)
Looking at the answers only 6 (6!) and 8! will have GCF as 6!
Ans D |
AQUA-RAT | AQUA-RAT-38166 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
if the ratio of simple interest on certain sum with same rate is 4:3.what is the ratio of time. | [
"3:5",
"6:5",
"5:4",
"4:3"
] | D | let ratio constant be x.
4x/3x=(p*r*t1/100)/(p*r*t2/100)
t1/t2=4:3.
answer D |
AQUA-RAT | AQUA-RAT-38167 | \text{19 is prime} & \color{gray}{\enclose{circle}{2}}&\color{gray}{\enclose{circle}{3}}&\color{gray}{\cancel{4}}&\color{gray}{\enclose{circle}{5}}&\color{gray}{\cancel{6}}&\color{gray}{\enclose{circle}{7}}&\color{gray}{\cancel{8}}&\color{gray}{\cancel{9}}&\color{gray}{\cancel{10}}&\color{gray}{\enclose{circle}{11}}&\color{gray}{\cancel{12}}&\color{gray}{\enclose{circle}{13}}&\color{gray}{\cancel{14}}&\color{gray}{\cancel{15}}&\color{gray}{\cancel{16}}&\color{gray}{\enclose{circle}{17}}&\color{gray}{\cancel{18}}&\enclose{circle}{19}&\color{gray}{\cancel{20}}&\color{gray}{\cancel{21}}&\color{gray}{\cancel{22}}&23&\color{gray}{\cancel{24}}&\color{gray}{\cancel{25}}\\ \text{23 is prime} &
The following is multiple choice question (with options) to answer.
The average of the first five prime numbers greater than 20 is: | [
"32.2",
"32.22",
"31.22",
"32.27"
] | A | Explanation:
Required prime numbers are 23, 29, 31, 37, 4.
Average will be (23 + 29 + 31 + 37 + 41)/5 = 32.20
Answer: A |
AQUA-RAT | AQUA-RAT-38168 | # Difference between revisions of "1984 AIME Problems/Problem 4"
## Problem
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$?
## Solution 1 (Two Variables)
Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$
We are given that \begin{align*} \frac{s+68}{n+1}&=56, \\ \frac{s}{n}&=55. \end{align*} Clearing denominators, we have \begin{align*} s+68&=56n+56, \\ s&=55n. \end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$
The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\cdot1=\boxed{649}.$
~JBL (Solution)
~MRENTHUSIASM (Reconstruction)
## Solution 2 (One Variable)
The following is multiple choice question (with options) to answer.
Four numbers are in the ratio 2:3:4:5 add up to give a sum of 1344. Find the biggest number. | [
"480",
"239",
"270",
"282"
] | A | ANSWER : A |
AQUA-RAT | AQUA-RAT-38169 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A gardener increased the length of a rectangle-shaped garden by 20 percent and decreased its width by 10 percent. The area of the new garden | [
"has increased by 20 percent",
"has increased by 12 percent",
"has increased by 8 percent",
"is exactly the same as the old area"
] | C | Let L be the original length and let W be the original width.
The original area was L*W.
The new area is 0.9W*1.2L = (0.9)(1.2)L*W = 1.08*L*W
The answer is C. |
AQUA-RAT | AQUA-RAT-38170 | 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.
Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ? | [
"2",
"2.5",
"3",
"3.5"
] | C | Solution
part filled by (A+B+C) in 1 hour = (1/5 + 1/6 + 1/30)
‹=› 1/3.
All the three pipes together will fill the tank in 3 hours.
Answer C |
AQUA-RAT | AQUA-RAT-38171 | homework-and-exercises, kinematics
Title: Average Velocity A car travels 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg to average 100mph over the total journey.
My thoughts on this are that it is impossible as if the total average was 100mph then the total time would be 2 hours but that can't be if the first leg took 2 hours.
Please tell me if I am missing something Are you missing something?
You probably are if this question was asked during a course on relativity. Anyway, this is a physics site and I'm going to make the question a bit more precise on the reference frames in which the measurements might have taken place:
We observe a car travel 100 miles in 2 hours, it then completes the return leg of the journey. How fast must it travel on the return leg for the driver to have done the full 200 miles in 2 hours?
The answer starts from the observation that during the first leg the driver will have aged $2\sqrt{1-\frac{v^2}{c^2}}$ hours, with $v/c \approx 50/670616629 \approx 7.5 \ 10^{-8}$. That is a fraction $5.6 \ 10^{-15}$ short of 2 hours.
So, the second leg the car should travel at a speed $v'$ such that the driver ages $\sqrt{1-\frac{v'^2}{c^2}} \frac{100 mi}{c}= 11 \ 10^{-15}$ hr. It follows that $v'$ needs to be a fraction $3 \ 10^{-15}$ short of the speed of light.
The following is multiple choice question (with options) to answer.
On a trip, a cyclist averaged 9 miles per hour for the first 18 miles and 10 miles per hour for the remaining 12 miles. If the cyclist returned immediately via the same route and took a total of 7.2 hours for the round trip, what was the average speed (in miles per hour) for the return trip? | [
"6.9",
"7.2",
"7.5",
"7.8"
] | C | The time to go 30 miles was 18/9+12/10=2+1.2=3.2 hours.
The average speed for the return trip was 30 miles/4 hours= 7.5 mph.
The answer is C. |
AQUA-RAT | AQUA-RAT-38172 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
The cricket team of 11members is 28yrs old & the wicket keeper is 3 yrs older. If the ages ofthese 2are excluded, the average age of theremaining players is 1 year less than the average age of the whole team. What is the average age of the team ? | [
"21",
"22",
"23",
"25"
] | D | Let the average age of the whole team be x years.
11x - (28 + 31) = 9 (x - 1)
=> 11x - 9x = 50
=> 2x = 50
=> x = 25.
So, average age of the team is 25 years.
D |
AQUA-RAT | AQUA-RAT-38173 | 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.
A group of students was interviewed for that if it was asked whether or not they speak French and / or English. Among those who speak French, 30 speak English well, while 40 of them do not speak English. If 86% of students do not speak French, how many students were surveyed? | [
"500",
"450",
"250",
"350"
] | A | Number of students who speak French are 30 + 40 = 70
Of total students, the percentage of students who do not speak French was 86% --> percentage of who do is 14%
70-------14%
x ------- 100%
x = 70*100/14 = 500 = number of all students
Answer is A |
AQUA-RAT | AQUA-RAT-38174 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. 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 product of two numbers is 4107. If the H.C.F of these numbers is 37, then the greater number is: | [
"101",
"107",
"111",
"112"
] | C | Let the numbers be 37a and 37b.
Then, 37a * 37 b = 4107 => ab = 3
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 * 1, 37 * 3) i.e., (1, 111).
Greater number = 111.
ANSWER:C |
AQUA-RAT | AQUA-RAT-38175 | # Proving Congruence Modulo
If $x$ is positive integer, prove that for all integers $a$, $(a+1)(a+2)\cdots(a+x)$ is congruent to $0\!\!\!\mod x$.
Any hints? What are the useful concepts that may help me solve this problem?
-
hint: you multiply $x$ consecutive integers so that one... – Raymond Manzoni Sep 1 '12 at 18:35
There are no "concepts", just original thought. Try to prove that one of the brackets is divisible by $x$ by considering what $a$ is mod $x$. – fretty Sep 1 '12 at 18:37
Hint: can you think of any reason why $x$ must divide one of the numbers $a,\ldots,a+x$? – shoda Sep 1 '12 at 18:39
So I can use factorial notation? – primemiss Sep 1 '12 at 18:43
so that one of them must be a multiple of $x$ (every $x$-th integer is a multiple of $x$). – Raymond Manzoni Sep 1 '12 at 18:55
Note that there is some $k$ such that $0\leq k< x$ and $a\equiv k\mod x$. Then $$a+(x-k)\equiv k+x-k\equiv x\equiv 0\mod x$$ and what do you get when you multiply this by other stuff?
-
One of the numbers $a+1, a+2,\ldots, a+x$ is congruent to $0$ mod $x$. Multiplying by $0$ yields $0$.
The following is multiple choice question (with options) to answer.
When positive integer x is divided by 6, the remainder is 1. Which of the following must be true?
I. x is a prime number
II. x is odd
III. x is divisible by 7 | [
"I only",
"II only",
"III only",
"I and II only"
] | B | X=6q+1
Examine II) put q=0,1,2,3....... we have x=1,7,13,...25 so x must be odd
Examine III) 14/6 has reminder 2 & 49/7 has reminder 1 so not always true
Examine I) X= 7 true but while 2 is prime, it does not give reminder 1
Answer: B |
AQUA-RAT | AQUA-RAT-38176 | java, integer
Title: Return reversed integer - Java The question is from Leetcode.
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note: Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
My solution:
class Solution {
public int reverseInteger(int num) {
if(num >= 0) { // positive numbers
char[] arr = String.valueOf(num).toCharArray();
reverse(arr);
return Integer.parseInt(new String(arr));
} else { // negative numbers
num *= -1;
char[] arr = String.valueOf(num).toCharArray();
reverse(arr);
return -Integer.parseInt(new String(arr));
}
}
void reverse(char[] arr) {
int start = 0;
int end = arr.length - 1;
while (start < end) {
char temp = arr[start];
arr[start++] = arr[end];
arr[end--] = temp;
}
}
}
For one of the test cases, I am getting the following error:
java.lang.NumberFormatException: For input string: "9646324351"
at line 68, java.base/java.lang.NumberFormatException.forInputString
at line 658, java.base/java.lang.Integer.parseInt
at line 776, java.base/java.lang.Integer.parseInt
at line 6, Solution.reverse
at line 54, __DriverSolution__.__helper__
at line 84, __Driver__.main
The following is multiple choice question (with options) to answer.
If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 45. By how much do the two digits differ? | [
"3",
"4",
"5",
"6"
] | C | we are given that if the integer N has its digits reversed the resulting integer differs from the original by 45. First let’s express the reversed number in a similar fashion to the way in which we expressed the original integer.
10B + A = reversed integer
Since we know the resulting integer differs from the original by 45 we can say:
10B + A – (10A + B) = 45
10B + A – 10A – B = 45
9B – 9A = 45
B – A = 5
Since B is the tens digit and A is the units digit, we can say that the digits differ by 5.
The answer is C. |
AQUA-RAT | AQUA-RAT-38177 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Veena ranks 73rd from the top in a class of 198. What is her rank from the bottom if 22 students have failed the examination? | [
"88",
"104",
"110",
"90"
] | B | total student=198
failed=22
paasd student=198-22=176
from bottom her rank is=176-73+1=104
ANSWER:B |
AQUA-RAT | AQUA-RAT-38178 | Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
The following is multiple choice question (with options) to answer.
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is: | [
"20",
"30",
"40",
"50"
] | A | Let the numbers be a, b and c.
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138+2x131= 400.
(a + b + c) = square root of 400 = 20
answer :A |
AQUA-RAT | AQUA-RAT-38179 | Kudos [?]: 96083 [0], given: 10706
Re: Find the power of 80 in 40!??? [#permalink]
### Show Tags
19 Oct 2010, 12:42
Expert's post
1
This post was
BOOKMARKED
AtifS wrote:
This is regarding the topic "Factorial" written by Bunuel for GMAT Club's Math Book.
When finding the power of non-prime number in $$n!$$ we first do prime-factorization of the non-prime number and then find the powers of each prime number in $$n!$$ one by one using the following formula
$$\frac{n}{p}+\frac{{n}}{{p^2}}+\frac{{n}}{{p^3}}+....+\frac{{n}}{{p^x}}$$ such that $$p^x <n$$, where $$p$$ is the prime number.
Let's suppose, we want to find the powers of $$80$$ in $$40!$$.
Prime factorization of $$80=2^4 * 5^1$$.
Now first find the power of $$2$$ in $$40!$$;
$$\frac{{40}}{{2}}+\frac{{40}}{{2^2}}+\frac{{40}}{{2^3}}+\frac{{40}}{{2^4}}+\frac{{40}}{{2^5}}=20+10+5+2+1=38$$powers of $$2$$ in $$40!$$ --> $$2^{38}$$
Now find the powers of $$5$$ in $$40!$$;
$$\frac{{40}}{{5}}+\frac{{40}}{{5^2}}=8+1=9 --> 5^9$$
The following is multiple choice question (with options) to answer.
Which of the following has 60^80 as a factor? | [
"15^60",
"30^40",
"120^80",
"60^60"
] | C | Here 60^80 = 30^80 * 20^80 Only Option C has both .
Smash C |
AQUA-RAT | AQUA-RAT-38180 | Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:37
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
_________________
Kudos [?]: 139450 [0], given: 12790
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Joined: 26 Mar 2010
Posts: 116
Kudos [?]: 16 [0], given: 17
Re: How many 4 digit codes can be made, if each code can only contain [#permalink]
### Show Tags
05 Oct 2010, 12:53
Bunuel wrote:
utin wrote:
Hi Bunuel,
why can't i write TOO,OTO,OOT AS
(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???
Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be $$4^3*\frac{3!}{2!}=4^3*3$$.
Hope it's clear.
I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!
this might clear my entire probability confusion i hope...
Kudos [?]: 16 [0], given: 17
Math Expert
Joined: 02 Sep 2009
Posts: 43322
Kudos [?]: 139450 [1], given: 12790
The following is multiple choice question (with options) to answer.
Which of the following leads to the correct mathematical solution for the number of ways that the letters of the word TIE could be arranged to create a three-letter code? | [
"3!",
"3! − (3! + 2!)",
"3! − (2!)",
"3!/(3! + 2!)"
] | A | TIE - three letters can be arranged in 3! ways
since 'there is no repetition'
3!
Ans. A) 3! |
AQUA-RAT | AQUA-RAT-38181 | we need to know definitions... They survived choices in the wrong spot: you still haven ’ t answered the question... Out how to restore/save my reputation the process of trying to answer this I... © 2021 Stack Exchange is a typical Bayes Theorem problem, it ’ s the probability of a student the! Kountz, Ashwini Miryala, Kyle Scarlett, Zachary Zell conditional probability for multiple random variables 4 0! Will finish the test to measure what the student both knows the answer being correct and knowing the answer 0.5. 3 or 4 choices for each question without the table helps me see more clearly we! Because if you know the answer but are not correct Stuck in wrong. Fida uses K ’ to mean “ not K ” of choices in the of! ) / ( 1/4 ) = P ( K∩C ) = 0 — you work., M = multiple choice examination has 5 questions and move 1/12 to right! Level and filesystem for a large storage server knew the answer, you will guess one of subject! That where you put 1/12, and conditional probability is a bit more confusing for.. The subject obtain a particular one are free and with help function teaching... Properties of conditional probabilities and solve problems pen if it is entirely forgivable not conditional probability multiple choice questions seen. A question and answer site for people studying math at any level and filesystem for a large server. Is the probability they knew the answer terms of service, privacy policy and policy... Has 5 questions on some of the subject options randomly authors use “ ~K or. Way I thought about P ( C ’ ∩ K ), which know. Saw when I read this problem was a reversal of information how many choices there are options... Into your RSS reader will get the question in order to work any! A recent question about probability has ties to Venn diagrams, tables, and Miriam symbol some! Completed a multiple-choice test with four options for each question has only one true answer multiple-choice questions are ;! Result of your quiz after you finish the novel that I am reading no! Design / logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa codes and datasets: them... Can ’ t be true professionals in related fields to buy them of choices! ( bleeding '', he who fears will
The following is multiple choice question (with options) to answer.
----------------YES---------NO----UNSURE Subject M----500--------200-----100 Subject R----400--------100-----300 A total of 800 students were asked whether they found two subjects, M and R, interesting. Each answer was either yes or no or unsure, and the numbers of students who gave these answers are listed in the table above. If 150 students answered yes only for subject M, how many of the students did not answer yes for either subject? | [
"100",
"250",
"300",
"400"
] | B | Since 150 students answered yes only for subject M, then the remaining 350 students who answered yes for subject M, also answered yes for subject R. So, 350 students answered yes for both subjects.
If 350 students answered yes for both subjects, then 400-350=50 students answered yes only for subject R.
So, we have that:
200 students answered yes only for subject M;
50 students answered yes only for subject R;
300 students answered yes for both subjects;
Therefore 800-(200+50+300)=250 students did not answer yes for either subject.
Answer: B. |
AQUA-RAT | AQUA-RAT-38182 | # 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.
In 1982 and 1983, Company B’s operating expenses were $10.0 million and $12.0 million, respectively, and its revenues were $15.6 million and $18.8 million, respectively. What was the percent increase in Company B’s profit (revenues minus operating expenses) from 1982 to 1983 ? | [
"3%",
"16 2/3%",
"25%",
"21 3/7%"
] | D | Profit in 1982 = 15.6 - 10 = 5.6 million $
Profit in 1983 = 18.8 - 12 = 6.8 million $
Percentage increase in profit = (6.8-5.6)/5.6 * 100 %
= 21 3/7%
Answer D |
AQUA-RAT | AQUA-RAT-38183 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
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The following is multiple choice question (with options) to answer.
A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour? | [
"71.11",
"71.12",
"71.1",
"71.17"
] | A | Car travels first 160 km at 64 km/hr
Time taken to travel first 160 km = distancespeed=16064
Car travels next160 km at 80 km/hr
Time taken to travel next 160 km = distancespeed=16080Total distance traveled = 160+160=2×160Total time taken = 16064+16080Average speed = Total distance traveledTotal time taken=2×16016064+16080=2164+180=2×64×8080+64=2×64×80144=2×8×8018=6409=71.11 km/hr
Answer:A |
AQUA-RAT | AQUA-RAT-38184 | # If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term. - Mathematics
Sum
If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also find that term.
#### Solution
Let the first term, common difference
And number of terms of the AP: 9, 7, 5,…. are
a1, d1 and n1, respectively
i.e., first term (a1) = 9
And common difference (d1) = 7 – 9 = –2
⇒ T"'"_(n_1) = a_1 + (n_1 - 1)d_1
⇒ T"'"_(n_1) =9 + (n_1 - 1)(-2)
⇒ T"'"_(n_1) = 9 - 2n_1 + 2
⇒ T"'"_(n_1) = 11 - 2n_1 ......(i)
∵ nth term of an AP, Tn = a + (n – 1)d
Let the first term, common difference and the number of terms of the AP: 24, 21, 18, … are a2, d2 and n2, respectively
i.e., first term, (a2) = 24
And common difference (d2) = 21 – 24 = – 3
∴ Its nth term, T"'"_(n_2) = a_2 + (n_2 - 1)d_2
⇒ T"'"_(n_2) = 24 + (n_2 - 1)(-3)
⇒ T"'"_(n_2) = 24 - 3n_2 + 3
⇒ T"'"_(n_2) = 27 - 3n_2 .....(ii)
Now, by given condition,
nth terms of the both APs are same
i.e., 11 - 2n_1 = 27 - 3n_2 ......[From equations (i) and (ii)]
The following is multiple choice question (with options) to answer.
Two terms are in 3 : 4, if 10 is added to each of the terms, they will be 4 : 5. Find second term? | [
"10",
"30",
"20",
"40"
] | D | Solution:
Let the two numbers be 3x and 4x. Then, (3x+10)/(4x+10) = 4/5.
=> 15x+50 = 16x+40 => x = 10.
.'. Second term = 4*10 =40.
ANSWER IS D |
AQUA-RAT | AQUA-RAT-38185 | # Question involving inequalities
Question: Sam has put sweets in five jars in such a way that no jar is empty and no two jars contain the same number of sweets. Also, any three jars contain more sweets in total than the total of the remaining two jars.
What is the smallest possible number of sweets altogether in the five jars?
My solution:
Let the number of sweets in jars 1, 2, 3, 4 & 5 be a, b, c, d & e respectively.
We have a>0, b>0, c>0, d>0 & e>0
We also have a ≠ b ≠ c ≠ d ≠ e
Let a>b>c>d>e
What do I do next to solve this question. Is my approach good?
• Well, you haven't done much yet but it's good to start by naming the variables, as you have done. What is the least value $e$ could be? Given that, what's the least value $d$ could be? Continue in this spirit. – lulu Sep 2 '19 at 18:28
• e=4 d=5 c=6 b=7 a=8 is the best combination I got – Sina Babaei Zadeh Sep 2 '19 at 18:30
Yes, it does make sense to order the numbers as $$a>b>c>d>e$$, because, if we make only the following true: $$e+d+c>b+a$$ (the smallest three vs. the biggest two), then automatically we will have any other sum of three jars bigger than any other sum of two jars.
Now, the inequalities above mean that $$b\ge c+1$$ and $$a\ge b+1\ge c+2$$ and $$d\le c-1$$ and $$e\le d-1\le c-2$$, and finally $$e+d+c\ge b+a+1$$, so by substituting we also conclude that:
$$3c-3\ge 2c+4$$
The following is multiple choice question (with options) to answer.
We have 36kg flour and 60kg sugar in stock and we would like to use the minimum number of packs that would have the same weight for both flour and sugar. How much more sugar is required if we would like to full 10 packs either of flour or sugar? | [
"12",
"18",
"20",
"24"
] | D | k is the integer that represents kg of pack
ak = 36 --> k = 36/a
bk = 60 --> k = 60/b
36/a = 60/b --> 5a = 3b and then k=12 for min(a+b)
a=36/12=3 , b=60/12=5
Total number of packs = 3+5 = 8
10-8=2 packs more required
2 *12 = 24 kg more sugar required to have 10 packs.
ANSWER: D |
AQUA-RAT | AQUA-RAT-38186 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
You are visiting your friend. When you walk in her room, you find that there are seven girls. Each of the seven girls have seven bags which consist of 7 adult cats each. Now, each of the adult cat has seven little cats as well along with them.
We know that each cat has four legs, can you find out the total number of legs in the room? | [
"26197",
"23578",
"10992",
"14728"
] | C | C
10992
Each girl has seven bags with seven adult cats each.
Also, each adult cat has seven little cats.
Thus each girl has 49 adult cats and 343 little cats.
Hence, seven girls will have
343 adult cats and 2410 little cats.
If we calculate the number of legs now.
Girls = 7 * 2 = 14
Cats + Little cats = 2744 * 4 = 10976
Total legs = 10976 + 14 = 10990
Since I also walked in the room, there will be two more legs.
Now total legs = 10990 + 2 = 10992. |
AQUA-RAT | AQUA-RAT-38187 | => 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 and B together have Sterling 1260. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have? | [
"Sterling 460",
"Sterling 484",
"Sterling 550",
"Sterling 504"
] | D | Explanation: 4/15A = 2/5B
A = (2/5 x 15/4)B
A = 3/2 B
A/B = 3/2
A : B = 3 : 2.
B's share = Sterling (1260 x 2/5) = Sterling 504.
Answer: Option D |
AQUA-RAT | AQUA-RAT-38188 | # 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.
In a shop 80% of the articles are sold at a profit of 10% and the remaining at a loss of 40%.what is the overall profit/loss? | [
"10% profit",
"10% loss",
"15% profit",
"no profit, no loss"
] | D | 80*1.1+20*0.6/100=1
ANSWER:D |
AQUA-RAT | AQUA-RAT-38189 | 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.
Some ladies can do a piece of work in 12 days. Two times the number of such ladies will do half of that work in : | [
"6 days",
"4 days",
"12 days",
"3 days"
] | D | Let x ladies can do the work in 12 days. More ladies, less days (Indirect)
Less work, less days (direct)
Ladies 2x : x Work 1 : ½
2x : x , 1 : ½ : : 12 : y
:. 2x * 1*y = x* ½ *12 or y = 3
Hence the required number of days = 3
ANSWER:D |
AQUA-RAT | AQUA-RAT-38190 | The $i$-th row has $18+2i$ seats, for $i=0,...,20$. The number of seats is thus $$\sum_{i=0}^{20}18+2i=18\times21+\sum_{i=0}^{20}2i=378+2\frac{20\times 21}{2}=378+420=798$$
-
You have a 0-th row... – Thomas Apr 6 '12 at 19:28
I know, and the answer is correct and all, but with the way the question is worded, we have 1st row to 21st row. (just a tiny thing that would improve the answer IMO) – Thomas Apr 6 '12 at 19:41
The following is multiple choice question (with options) to answer.
A bus can hold 108 passengers. If there are 12 rows of seats on the bus, how many seats are in each row? | [
"1",
"2",
"9",
"25"
] | C | Total number of passengers = 108
There are 12 rows of seats on the bus.
To find how many seats are there in each row, divide the total number of passengers by the number of rows of seats on the bus.
We get, divide 108 by 12
108 ÷ 12 = 9
Therefore, there are 9 seats in each row
ANSWER IS C |
AQUA-RAT | AQUA-RAT-38191 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
The average age of M boys is ‘b’ years and of ‘n’ girls ‘c’ years. The average age of all together is? | [
"(mb + nc)/ (m + b) years",
"(mb + bc)/ (m + n) years",
"(mb + nc)/ (m + n) years",
"(nb + nc)/ (m + n) years"
] | C | (mb + nc) / (m+ n).nswer: C |
AQUA-RAT | AQUA-RAT-38192 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
An automobile financier claims to be lending money at S.I., but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes? | [
"10.28%",
"10.25%",
"10.85%",
"30.25%"
] | B | Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 - 100) = 10.25%.
Answer: B |
AQUA-RAT | AQUA-RAT-38193 | 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 3000 m long can cross an electric pole in 120 sec and then find the speed of the train? | [
"90",
"95",
"100",
"105"
] | A | Length = Speed * time
Speed = L/T
S = 3000/120
S = 25 M/Sec
Speed= 25*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 90 Kmph
Answer:A |
AQUA-RAT | AQUA-RAT-38194 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square? | [
"876 cm",
"800 cm",
"167 cm",
"765 cm"
] | B | Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200
= 800 cm.
Answer:B |
AQUA-RAT | AQUA-RAT-38195 | # 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.
By selling 200 apples, a fruit-seller gains the selling price of 50 apples. Find the gain percent? | [
"33.33%",
"35%",
"27.75%",
"35.75%"
] | A | SP = CP + g
200 SP = 200 CP + 50 SP
150 SP = 200 CP
150 --- 50 CP
100 --- ? => 33.33%
Answer: A |
AQUA-RAT | AQUA-RAT-38196 | temperature, unit-conversion
Title: Values of coefficient of expansion differ,when temperatures are measured on centigrade scale and on Fahrenheit scale separately In solving above question, it was given that
Change of $5 ℃ = $ change of $9 ℉$
The following is multiple choice question (with options) to answer.
If C is the temperature in degrees Celsins and F is the temperature in degrees Fahrenheit, then the relationship between temperatures on the two scales is expressed by the equation 9C=5(F-32). In a weekly range of temperature recorded at a certain weather station differed by 25 degrees on the Fahrenheit scale, by how many degrees did the temperature extremes differ on the Celsius scale? | [
"a) 65/9",
"b) 13",
"c) 25",
"d) 45"
] | B | The relationship is 9/5 - 32, so approximately half of difference in Celsius (the -32 part won't be used in this case). Just looking at the answer choices you can pick B which is the only number near half of 25. The other answer choices are too far apart so no need to calculate. |
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