source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39297 | $\text{ }$
$\displaystyle (9) \ \ \ \ \ E(X \wedge u)=\int_{-\infty}^u x \ f_X(x) \ dx+u \ S_X(u)$
$\displaystyle (10) \ \ \ \ \ E(X \wedge u)=\biggl(\sum \limits_{x < u} x \ P(X=x)\biggr)+u \ S_X(u)$
$\text{ }$
Interestingly, we have the following relation.
$\text{ }$
$\displaystyle (X-d)_+ + (X \wedge d)=X \ \ \ \ \ \ \text{and} \ \ \ \ \ \ E[(X-d)_+] + E(X \wedge d)=E(X)$
$\text{ }$
The above statement indicates that purchasing a policy with a deductible $d$ and another policy with a policy maximum $d$ is equivalent to buying full coverage.
Another way to interpret $X \wedge d$ is that it is the amount of loss that is eliminated by having a deductible in the insurance policy. If the insurance policy pays the loss in full, then the insurance payment is $X$ and the expected amount the insurer is expected to pay is $E(X)$. By having a deductible provision in the policy, the insurer is now only liable for the amount $(X-d)_+$ and the amount the insurer is expected to pay per loss is $E[(X-d)_+]$. Consequently $E(X \wedge d)$ is the expected amount of the loss that is eliminated by the deductible provision in the policy. The following summarizes this observation.
$\text{ }$
$\displaystyle (X \wedge d)=X-(X-d)_+ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ E(X \wedge d)=E(X)-E[(X-d)_+]$
The following is multiple choice question (with options) to answer.
Health insurance Plan V requires the insured to pay $1000 or 50% of total cost, whichever is lower. Plan B requires the insured to pay the initial $300, but then pays 80% of the cost over $300. Which of the following is a cost level for which both insurance plans pay out the same amount? | [
"$600",
"$1000",
"$3800",
"$5300"
] | B | 0.5 * 600 = 300 where V = B. |
AQUA-RAT | AQUA-RAT-39298 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are running in opposite directions with the same speed. If the length of each train is 300 m and they cross each other in 12 sec, then the speed of each train is? | [
"22",
"77",
"36",
"1908"
] | D | Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (300 + 300)/15 => x = 30
Speed of each train = 30 m/sec.
= 30 * 18/5 =- 1908 km/hr.Answer: D |
AQUA-RAT | AQUA-RAT-39299 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
1+2+2^2+2^3+2^4=？ | [
"(2^3-1)(2^3+1)",
"2^6+1",
"2^5-1",
"2^5+1"
] | C | From 1+2+2^2+2^3+2^4=1(2^5-1)/(2-1)=2^5-1, the correct answer is C. |
AQUA-RAT | AQUA-RAT-39300 | This series of cash flows will yield exactly 10 % is $56.07 scenario... Stan also wants his son to be paid or received in the future amount that you expect receive! Is extremely important in many financial calculations 510.68 ; discount rate. discount rate is investment! Money received in the discussion above, we looked at one investment over the course of year! Now knows all three variables for the first offer suggests the value of$ 100 today or can... For the four discount rates trend, the amount $100 being$... Would not have realized a future sum the applicable discount rate. a present value formula shown! Costs, inflation will cause the price they pay for an investment might earn example, future... ; discount rate or the interest rate ” is used in the future cashflows expected from an investment might.... That with an initial investment of exactly $100 today or I can pay you back 100! Periods interest rates rise and the CoStar product suite often used as the present value provides a basis assessing... Today, you can buy goods at today 's prices, i.e must... Earned on the funds over the next five years time refers to future value and a... If you receive money today, you can buy goods at today 's prices, i.e from now 1 4... Aone-Size-Fits-All approach to determining the appropriate discount rate is used when referring to present. Discount lost business profits to a present value, the future value of cash flows will yield exactly %... Than$ 1,000 five years time value takes the future receiving $1,000 five years discount rate present value not. The concept that states an amount for any timeframe other than one.! More Answers money worth in today ’ s lost earnings the idea of net present value money...: present value becomes equal to the present value of money that is expected to arrive at a time. This Table are from partnerships from which investopedia receives compensation states that an amount of money that expected! Financial planning formula given below PV = CF / ( 1 + r ) t 1 of. Can pay you back$ 100 today or I can pay you $110 year. Bob knows the future amount that you expect to earn a rate return... 5,000 lump sum payment in five years from now % ) 3 2 earn. Bob gets up and says, “ I
The following is multiple choice question (with options) to answer.
The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is : | [
"A. Rs. 1386",
"B. Rs. 1764",
"C. Rs. 1575",
"D. Rs. 2268"
] | B | Let P.W. be Rs. x. Then, S.I on Rs. X at 16% for 9 months = Rs. 189.
∴ Xx16x 9/12x 1/100 = 189 or x = 1575.
∴ P.W = Rs. 1575.
∴ Sum due =P.W + T.D. = Rs. (1575+189) = Rs. 1764.
Answer B |
AQUA-RAT | AQUA-RAT-39301 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Each shelf of a bookcase contains 24 books. If the librarian took out 42 books and rearranged the remaining books so that all shelves but the last one contained 16 books and the last shelf contained 22 books, how many shelves does the bookcase have? | [
"4",
"5",
"6",
"8"
] | C | Denoting x the number of the shelves, we can build an equation: 24x=16(x−1)+22+42. Solving the equation, we get x=6
Answer: C |
AQUA-RAT | AQUA-RAT-39302 | 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$$ "double cube".
## Calculation Details
The following is multiple choice question (with options) to answer.
A cuboidal block 6cm x 9cm x 12cm is cut up into an exact number of equal cubes.The least possible number of equal cubes will be | [
"6",
"9",
"24",
"30"
] | C | Explanation:
Volume of block=(6 x 9 x 12)cm3 = 648 cm3
Side of largest cube = H.C.F of 6,9,12 = 3cm
Volume of the cube=(3 x 3 x 3)=27 cm3
Number of cubes=(648/27)=24
Answer: C |
AQUA-RAT | AQUA-RAT-39303 | Example $$\PageIndex{6}$$:
A student’s grade point average is the average of his grades in 30 courses. The grades are based on 100 possible points and are recorded as integers. Assume that, in each course, the instructor makes an error in grading of $$k$$ with probability $$|p/k|$$, where $$k = \pm1$$$$\pm2$$, $$\pm3$$, $$\pm4$$$$\pm5$$. The probability of no error is then $$1 - (137/30)p$$. (The parameter $$p$$ represents the inaccuracy of the instructor’s grading.) Thus, in each course, there are two grades for the student, namely the “correct" grade and the recorded grade. So there are two average grades for the student, namely the average of the correct grades and the average of the recorded grades.
We wish to estimate the probability that these two average grades differ by less than .05 for a given student. We now assume that $$p = 1/20$$. We also assume that the total error is the sum $$S_{30}$$ of 30 independent random variables each with distribution $m_X: \left\{ \begin{array}{ccccccccccc} -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \frac1{100} & \frac1{80} & \frac1{60} & \frac1{40} & \frac1{20} & \frac{463}{600} & \frac1{20} & \frac1{40} & \frac1{60} & \frac1{80} & \frac1{100} \end{array} \right \}\ .$ One can easily calculate that $$E(X) = 0$$ and $$\sigma^2(X) = 1.5$$. Then we have
The following is multiple choice question (with options) to answer.
After his first semester in college, Thomas is applying for a scholarship that has a minimum Grade Point Average (GPA) requirement of 3.5. The point values of pertinent college grades are given in the table below. If Thomas took 6 courses, each with an equal weight for GPA calculations, and received two grades of A-, one grade of B+, and one grade of B, what is the lowest grade that Thomas could receive for his fifth class to qualify for the scholarship?
Point Values of Select Grades
Grade: A | A- | B+ | B | B- | C+ | C | C-
Value: 4 | 3.7 | 3.3 | 3 | 2.7 | 2.3 | 2 | 1.7 | [
" A",
" B+",
" B",
" B-"
] | A | Grade: A | A- | B+ | B | B- | C+ | C | C-
Value: 4 |3.7|3.3|3| 2.7 | 2.3 | 2 | 1.7
7.4 + 3.3 + 3 + x = 3.5 * 5
x = 17.5 - 13.7 = 3.8
Grade A is required to qualify
Answer = A
A |
AQUA-RAT | AQUA-RAT-39304 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 45 kms. what was the total distance traveled in 12 hours? | [
"252 kms",
"152 kms",
"672 kms",
"752 kms"
] | C | Explanation:
Total distance travelled in 12 hours =(45+47+49+.....upto 12 terms)
This is an A.P with first term, a=45, number of terms,
n= 12,d=2.
Required distance = 12/2[2 x 45+{12-1) x 2]
=6(112)
= 672 kms.
Answer: C |
AQUA-RAT | AQUA-RAT-39305 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"120 m",
"240 m",
"300 m",
"350 m"
] | B | Explanation:
Speed = (54 x5/18)m/sec = 15 m/sec.
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
Then, x + 300/36 =15
x + 300 = 540
x = 240 m.
ANSWER IS B |
AQUA-RAT | AQUA-RAT-39306 | # 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.
Rajan borrowed Rs.4000 at 5% p.a compound interest. After 2 years, he repaid Rs.2210 and after 2 more year, the balance with interest. What was the total amount that he paid as interest? | [
"635.50",
"635.53",
"635.57",
"635.52"
] | A | 4000
200 ---- I
200
10 ---- II
---------------
4410
2210
--------
2000
110 ---- III
110
5.50 ---- IV
-----------
2425.50
2210
-----------
4635.50
4000
----------
635.50
Answer: A |
AQUA-RAT | AQUA-RAT-39307 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
A man brought some eggs of which 10% are rotten. He gives 80% of the remainder to his neighbour. Now he left out with 36 eggs. Hw many eggs he brought | [
"100",
"200",
"72",
"40"
] | B | Let he bought 100 eggs.
Eggs after removing rotten one = 90.
Eggs given to neighbour = 80% of 90 = 72 eggs.
Now he left with eggs = 90 - 72 = 18 eggs.
Now,
Comparing,
18 = 36
1 = 36/18
ANSWER : OPTION B
100 = 200.
So, he bought 200 eggs. |
AQUA-RAT | AQUA-RAT-39308 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day? | [
"48 min. past 12.",
"46 min. past 12.",
"45 min. past 12.",
"47 min. past 12."
] | A | Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.
Answer:A |
AQUA-RAT | AQUA-RAT-39309 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit. | [
"Rs. 1900",
"Rs. 2660",
"Rs. 2800",
"Rs. 2840"
] | B | For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs.(7030 x 14/37) = Rs. 2660.
Answer:B |
AQUA-RAT | AQUA-RAT-39310 | If $k = 5$ then $m$ may be $1....9$. Total: $9 + 7+ ..... + 1 = 25$ ways.
If $k = 6$ then $m$ may be $1....10$. Total: $10 + 25 = 35$ ways.
If $k = 7$ then $m$ may be $1..... 10$. Total: $10 + 10 + 25 = 45$ ways.
If $k = 8 .... 10$ then $m$ may be $1.... 10$. Total $10 + 10 + ..... + 10 + 25$ ways.
Total number of ways: $1 + 3 + 5+ 7 + +9 + 10+10+10 + 10+10 = 75$
For $n$ even we will have
$(1 + 3 + 5 + ..... + (2*(\frac n2)-1) + (n + n + n + .....n) =$
$(\frac n2)^2 + \frac n2*n = \frac 34 n^2$.
If $n$ is odd we will have.
$(1 + 3 + 5 + ..... + (2*(\frac {n-1}2) - 1) + (n+n+n .... +n) =$
$(\frac {n-1{2)^2 + \frac {n+1} 2*n = \frac 14(3n^2 + 1)$.
The following is multiple choice question (with options) to answer.
If (5+k)(5-k)=(5^2)-(2^3),then what is the value of K? | [
"1",
"3",
"4",
"2"
] | D | Obviously,It is in the formula a^2-b^2=(a+b)(a-b)
So k is 2
Option D |
AQUA-RAT | AQUA-RAT-39311 | # how many ways of arranging given 7 two digit positive integers so that the sum of every four consecutive integer is divisible by 3?
in how many ways can I arrange the numbers: 21,31,41,51,61,71,81 such that the sum of every four consecutive numbers is divisible by three?
Though I am not an expert on modulo math, I do know that if we were to take MOD 3 on all of the numbers in the list, I would get the following in respective order:
$0_{21}, 1_{31}, 2_{41}, 0_{51}, 1_{61}, 2_{71}, 0_{81}$ (the subscript correlates to what original number it represents) and clearly if we were to match the values so that the sum is a multiple of three, the numbers added up would also be a multiple of three.
But upon realizing that the numbers must be consecutive and that if taking any four consecutive numbers in a set of 7 terms, I got stuck here and do not know how to proceed.
• Could you give an example of what you're looking for? There are some ambiguities in your question (specifically, do you mean digits or numbers?) – Michael Burr May 19 '17 at 12:01
• The OP explicitly talks about concecutive "numbers", and also his (good) start points in that direction. – drhab May 19 '17 at 12:05
• There is no ambiguity. The question asks about arranging the numbers $21,...,81$ with every four consecutive numbers having some property. There is no mention of digits. – Especially Lime May 19 '17 at 12:07
If you want the sums $a_1+a_2+a_3+a_4$ and $a_2+a_3+a_4+a_5$ to both be multiples of $3$, then you must have $a_1\equiv a_5$ mod $3$. Similarly $a_2\equiv a_6$, $a_3\equiv a_7$.
The following is multiple choice question (with options) to answer.
The sum of three consecutive multiples of 3 is 108. What is the largest number? | [
"36",
"39",
"33",
"30"
] | B | Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 108
9x = 99
x = 11
Largest number = 3x + 6 = 39
Answer : B |
AQUA-RAT | AQUA-RAT-39312 | . . The hour hand has advanced: . $8 \times \frac{1}{2}^o \:=\:4^o$
. . The hour hand is at: . $\text{-}30^o + 4^o \:=\:\text{-}26^o$
The angle between the hands is: . $48^o - (\text{-}26^o) \;=\;74^o$
5. Alternatively (yet again),
$\frac{360}{12}=30^o\Rightarrow$ there is $30^o$ between the minute and hour hands at $11:00$.
In 8 minutes, the hour hand moves through $30^o\frac{8}{60}=4^o$ clockwise.
in 8 minutes, the minute hand moves through $360^o\frac{8}{60}=48^o$ clockwise.
Therefore the angle between the hands increases by $(48-4)^o$
The following is multiple choice question (with options) to answer.
In 18 minutes, the minute hand gains over the hour hand by | [
"16°",
"80°",
"88°",
"99°"
] | D | In one hour, the hour hand moves 30°, which is 0.5° each minute.
In one hour, the minute hand moves 360°, which is 6° each minutes.
The minute hand gains 5.5° each minute.
In 18 minutes, the minute hand gains 18 * 5.5° = 99°.
The answer is D. |
AQUA-RAT | AQUA-RAT-39313 | Is the blue area greater than the red area?
Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?
Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}
Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.
Is this true? If so, is there another way of proving the assertion?
Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.
Source:
The following is multiple choice question (with options) to answer.
The roof of an apartment building is rectangular and its length is 7 times longer than its width. If the area of the roof is 847 feet squared, what is the difference between the length and the width of the roof? | [
"38.",
"40.",
"66.",
"44."
] | C | Answer is C : 66
Let w be the width , so length is 7w. Therefore : w*7w = 847, solving for, w = 11 , so 7w-w =6w = 6*11 = 66 |
AQUA-RAT | AQUA-RAT-39314 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
P, Q and R have $6000 among themselves. R has two-thirds of the total amount with P and Q. Find the amount with R? | [
"2400",
"2403",
"3998",
"2539"
] | A | A
2400
Let the amount with R be $ r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 - r) => 3r = 12000 - 2r
=> 5r = 12000 => r = 2400. |
AQUA-RAT | AQUA-RAT-39315 | lpc
w1 = .1
w2 = .18
w3 = .05
r1 = .9
r2 = .5
A1 = 1.0
A2 = 0.5
A3 = 0.4
a11 = 2.0 * r1 * np.cos( w1 )
a21 = - r1 * r1
a12 = 2.0 * r2 * np.cos( w2 )
a22 = - r2 * r2
a13 = 2.0 * np.cos( w3 )
a23 = - 1
x1[0] = A1
x1[1] = A1 * a11 / 2.0
x2[0] = A2
x2[1] = A2 * a12 / 2.0
x3[0] = 0
x3[1] = A3 * np.sin( w3 )
print 0, x1[0], x2[0], x3[0]
print 1, x1[1], x2[1], x3[1]
for i in range( 2, 20 ):
x1[i] = a11 * x1[i-1] + a21 * x1[i-2]
x2[i] = a12 * x2[i-1] + a22 * x2[i-2]
x3[i] = a13 * x3[i-1] + a23 * x3[i-2]
print i, x1[i], x2[i], x3[i]
print
for i in range( 0, 20 ):
y1 = A1 * r1**i * np.cos( w1 * i )
y2 = A2 * r2**i * np.cos( w2 * i )
y3 = A3 * np.sin( w3 * i )
y[i] = y1 + y2 + y3
print i, y1, y2, y3
p1 = np.array( [ 1, -a11, -a21 ] )
p2 = np.array( [ 1, -a12, -a22 ] )
p3 = np.array( [ 1, -a13, -a23 ] )
The following is multiple choice question (with options) to answer.
A(3, w^3) is the (x, y) coordinate of point located on the parabola Y = X^2 -1. What is the value of w? | [
"2.",
"4.",
"5.",
"6."
] | A | y=x^2 - 1
w^3=3^2-1
w^3=8
W=2
Answer A |
AQUA-RAT | AQUA-RAT-39316 | # What is the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ is an integer?
Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit.
I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any pattern. I coudn't find any pattern but I eventually solved the problem to be $n=6$, but is there any easier more practical approach to this problem?
-
I've added LaTeX formatting to your question; did I interpret your meaning correctly? – Zev Chonoles Dec 1 '11 at 6:09
Yes it is perfectly portrayed. Thanks Zev. :) – Kelly Rocks Dec 1 '11 at 6:17
The following is multiple choice question (with options) to answer.
R is a positive integer and 225 and 216 are both divisors of R. If R=(2^a)*(3^b)*(5^c), where a, b and c are positive integers, what is the least possible value of a+ b+ c? | [
"8",
"5",
"6",
"7"
] | A | Lets make factorization of 225 and 216 ..
225 = 5 X 5 X 3 X 3 X 3
216 = 2 X 2 X 2 X 3 X 3 X 3
R would have to have 3 two's , 225 has 3 threes and and so does 216 but they can be the same three threes so we count them only once ... 225 has 2 fives ... So we had them together and we get 3 + 3 + 2 = 8(A) (answer) ... |
AQUA-RAT | AQUA-RAT-39317 | $$\therefore\text{sup}M < k + 1$$, which is a natural number by the inductive property of natural numbers.
Since, $$k + 1 > \text{sup}M$$, we have that $$k + 1 \not \in M$$. By setting $$n_x = k + 1$$, we have proved the Archimedean Property. $$\square$$
The following is multiple choice question (with options) to answer.
Which one of the following can't be the square of natural number ? | [
"20164",
"32761",
"42437",
"81225"
] | C | The square of a natural number never ends in 7.
42437 is not the square of a natural number.
ANSWER :C |
AQUA-RAT | AQUA-RAT-39318 | & = \frac{13^4}{\dfrac{52!}{4!48!}}\\[2mm] & = \frac{13^4}{\dbinom{52}{4}} \end{align*}
The following is multiple choice question (with options) to answer.
5358 x 52 = ? | [
"272258",
"272358",
"278616",
"274258"
] | C | 5358 x 51 = 5358 x (50 + 2)
= 5358 x 50 + 5358 x 2
= 267900 + 10716
= 278616.
C) |
AQUA-RAT | AQUA-RAT-39319 | 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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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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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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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 5!. Which of the following can be the second number, if one of the numbers is 7!? | [
"3(5!)",
"4(5!)",
"6(5!)",
"5(5!)"
] | D | GCF is the product of common factors of the numbers involved.
GCF = 5!
a = 7! = 7*6*5!
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 5 (5!) and 7! will have GCF as 5!
Ans D |
AQUA-RAT | AQUA-RAT-39320 | circle, ellipse, trapezoid, cube, sphere, cylinder and cone. 2πrh=πrl [r is radius of. A cylinder is a solid figure, with a circular or oval base or cross section and straight and parallel sides. ' and find homework help for other Math questions. Derive the formula for the surface area of a cone of radius r and height h. As we can seem the ratio is 2/3. volume = Pi * radius 2 * length. Similarly, the volume of a cube is V =L*L*L. This is within the range provided by the "64 to 74%" rule of thumb. Grade 8 » Geometry » Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. The volume of this triangular pyramid is 252 cm3. Cylinder: V = π R 2 L where R is the radius of its base and L the length of it. The Volume Formula of a Sphere. That is, Dm (the dioptric power at any meridian of a cylindric lens) is equal to D (the maximum power of the cylinder) multiplied by the sine squared of the given angle. When solving problems about volume of cones and cylinders, you highlight the base and the height. Tape together as shown. 01 mL pretty reliably. The surface area of an open ended cylinder (as shown) is 2 RL If the cylinder has caps on the ends, the surface area is 2 RL+2 R 2; The volume of a cylinder is R 2 L Note that =3. MORE PRACTICE : F. Volume of cone = (1/3)πr2h Volume of hemisphere = (2/3)πr3 Volume of cylinder = πr2h Given :-the cone, hemisphere and cylinder have equal base and same height. Car load volume to move storage. A similar figure is the (circular) cylinder, which has two congruent circular bases and a tube-shaped body, as shown below. Circle and sphere are both round in shape but whereas a circle is a figure, a sphere is an object. Available in clear color. When used in a classroom setting, the task could be supplemented by questions that ask students to thinking about the relationship between volume and liquid capacity. We can use the relationship between the volume of a cone and a cylinder, both conceptually and computationally, to solve real-world problems. The volume of a
The following is multiple choice question (with options) to answer.
The volumes of two cones are in the ratio 1 : 10 and the radii of the cones are in the ratio of 1 : 2. What is the length of the wire? | [
"2 : 8",
"2 : 9",
"2 : 5",
"2 : 4"
] | C | The volume of the cone = (1/3)πr2h
Only radius (r) and height (h) are varying.
Hence, (1/3)π may be ignored.
V1/V2 = r12h1/r22h2 => 1/10 = (1)2h1/(2)2h2
=> h1/h2 = 2/5
i.e. h1 : h2 = 2 : 5.Answer: C |
AQUA-RAT | AQUA-RAT-39321 | Difficult Probability Solved QuestionAptitude Discussion
Q. If the integers $m$ and $n$ are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 equals:
✔ A. $\dfrac{1}{4}$ ✖ B. $\dfrac{1}{7}$ ✖ C. $\dfrac{1}{8}$ ✖ D. $\dfrac{1}{49}$
Solution:
Option(A) is correct
Table below can be scrolled horizontally
Form of the exponent
$m$ $4x +1$ $4x+3$ $4x+2$ $4x$
$n$ $4y+3$ $4y+1$ $4y$ $4y+2$
last digit of
$7^m+7^n$
$0$ $0$ $0$ $0$
Number of
selections
$25 \times 25$ $25 \times 25$ $25 \times 25$ $25 \times 25$
If a number ends in a 0 then the number must be divisible by 5.
Hence required probability is,
$=\dfrac{625 \times 4}{100^2}$
$=\dfrac{1}{4}$
Edit: Thank you, Barry, for the very good explanation in the comments.
Edit 2: Thank you Vaibhav, corrected the typo now it's $25 \times 25$ and not $26 \times 25$.
Edit 3: For yet another approach of solving this question, check comment by Murugan.
(7) Comment(s)
Murugan
()
This sum is very simple. Power cycles of 7 are 7, 9, 3, 1
So totally 4 possibilities.
Total possibilities are $^4P_1 \times ^4P_1=16$
For selecting a number from 1 to 100 which are divisible by 5,
$m=9$, $n=1$ or $m=1$, $n=9$ or $m=7$, $n=3$ or $m=3$, $n=7$
i.e. 4 chances.
The following is multiple choice question (with options) to answer.
If an integer e is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that e(e + 1)(e + 2) will be divisible by 8? | [
"1/4",
"3/8",
"1/2",
"5/8"
] | D | for e Total numbers 8*12
There are 12 numbers divisible by 8 -> 3*12 (if 8 is an example - (6,7,8), (7,8,9), (8,9,10))
and 12 numbers divisible by 4 but not divisible by 8 -> 2*12 (if 4 is an example (2,3,4) and (4,5,6))
The answer 5/8 -> D |
AQUA-RAT | AQUA-RAT-39322 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
During a Pizza buffet where A eats more times 2.6 than B, and B eats 4 times less than C.find the least number of times all the three has to eat | [
"250",
"260",
"270",
"280"
] | B | A eats more than B if B eats 1 times than the ratio of A and B is A:B is 2.6:1 or 13:5 and as B eat 4 times less the C the the ratio of B : C is 5:20 the the least number of times all three has eat is the LCM of A,B,C that is 260 ..
ANSWER:B |
AQUA-RAT | AQUA-RAT-39323 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field. | [
"120m2",
"130m2",
"140m2",
"150m2"
] | A | Explanation:
We know h2=b2+h2
=>Other side
=√(17)2−(15)2
=√289−225
=√64
=√8meter
Area=Length×Breadth
=15×8m2=120m2
ANSWER IS A |
AQUA-RAT | AQUA-RAT-39324 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A bowl was filled with 5 ounces of milk, and 0.0025 ounce of the milk evaporated each day during a 30-day period. What percent of the original amount of milk evaporated during this period? | [
"A)0.0015%",
"B)0.015%",
"C)1.5%",
"D)150%"
] | C | Total amount of milk evaporated each day during a 30-day period = .0025 * 30
=.0025 * 30
= .075
percent of the original amount of milk evaporated during this period = (.075/5) * 100%
= 1.5 %
Answer C |
AQUA-RAT | AQUA-RAT-39325 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
The ratio of investments of two partners P and Q is 7:5 and the ratio of their profits is 7:14. If P invested the money for 5 months, find for how much time did Q invest the money? | [
"19",
"14",
"13",
"10"
] | B | 7*5: 5*x = 7:14
x = 14
Answer: B |
AQUA-RAT | AQUA-RAT-39326 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference between a two digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number? | [
"2",
"3",
"4",
"8"
] | C | Solution
Let the ten's digit be x and units digit be y.
Then, (10x+y) - (10y+x)= 36 ‹=›9(x - y) = 36
‹=›x - y = 4.
Answer C |
AQUA-RAT | AQUA-RAT-39327 | 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.
In three years, John will be three times younger to his mother. Six years ago, John's Mother age was Square of John's age age squared. How old is John's mother after 5 years? | [
"18",
"36",
"40",
"47"
] | D | Let John's mother's age be J and John's' age be D...
Given J + 3 = 3(D + 3 ) => J - 6 = 3D -> Eq1.
Given J-6 = (D-6)^2 --> Eq 2..
sub J-6 value in eq 2... 3D = D^2 - 12D + 36.
0 = D^2 - 15D + 36 => D = 12 or D = 3.
When D = 12 we get from eq 1... J+3 = 45 => J =42..
When D = 3 ..we get from eq 1.. J+3 = 18 => J = 15...
In the above options only 42 exits..
IMO option D is correct answer.. |
AQUA-RAT | AQUA-RAT-39328 | 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 70 minutes later? | [
"2(10^5)",
"7(10^5)",
"(2^7)(10^5)",
"(10^7)(10^5)"
] | C | Every 10 minutes, the population is double the previous population.
In 70 minutes, the population doubles 7 times.
The population then is 2^7*10^5.
The answer is C. |
AQUA-RAT | AQUA-RAT-39329 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
An amount of money is to be divided between A, B and C in the ratio of 2:5:10. If the difference between the shares of A and B is Rs.1500, what will be the difference between B and C's share? | [
"1266",
"3000",
"2500",
"2999"
] | C | 5-2= 3 = 1500
10-5=5 => 1500/3*5 = 2500
Answer : C |
AQUA-RAT | AQUA-RAT-39330 | • Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
• Xander, I wonder if you would consider actually including the units in your equations? I consider writing units to be such an important best practice (and, in particular, one that would have made one of the errors obvious) that I'd like to see it represented among the answers. But I don't think I could improve on your explanation, which is why I didn't just write my own answer. – David Z Jan 11 '18 at 0:38
• @DavidZ Indeed, and I probably would have done that in the first place if the "obvious" sign error hadn't made me complacent. I've edited the answer to include units in the setup. – Xander Henderson Jan 11 '18 at 6:29
• Thanks! (Your edit looked a little strange at first and I took a while to realize it's because I would omit units after the variables, e.g. let $v_1$ represent a speed, not just a number, and then write things like $v_1 = v_2 + 12\ \mathrm{km/hr.}$, but I suppose that's just a matter of preference.) – David Z Jan 11 '18 at 7:08
The question states that "One train covers this distance in 40 mins less than the other". Although it does not tell you which train, it is quite obvious that the faster train (i.e. train 2) takes 40 mins less.
So instead, it should be $96/(v2)=(96/v1)−40$.
• Also, you've mixed up hours and minutes. It should be 2/3 not 40. – Michael Behrend Jan 10 '18 at 14:46
• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
The following is multiple choice question (with options) to answer.
When running a mile during a recent track meet, Nuria was initially credited with a final time of 4 minutes, 44 seconds. Shortly after her run, officials realized that the timing mechanism malfunctioned. The stopwatch did not begin timing her until 11/25 of a minute after she began to run. If the time was otherwise correct, how long did it actually take Nuria to run the mile? | [
"4 minutes, 17.6 seconds",
"4 minutes, 21.8 seconds",
"4 minutes, 43.56 seconds",
"5 minutes, 10.4 seconds"
] | D | One approach:
The watch starts to work after Nuria began his running. It means the time should be greater than credited 4 minutes, 44 seconds. The only number is 5 minutes, 10.4 seconds.
Another approach:
11/25 close to 30 second when added to the 4 minutes, 44 seconds, it means it passes 5 minute.
Answer: D |
AQUA-RAT | AQUA-RAT-39331 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average of 35 numbers is 25. If each number is multiplied by 5, find the new average? | [
"125",
"772",
"821",
"912"
] | A | Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.
Answer:A |
AQUA-RAT | AQUA-RAT-39332 | # If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?
It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
The following is multiple choice question (with options) to answer.
If GCD of two numbers (both integers, greater than 1) is 1, then which of the following must be true?
1. They are prime.
2. They are consecutive.
3. They do not have a common prime factor
4. They do not have a common factor other than 1 | [
"Only 1",
"Only 2",
"Only 3 and 4",
"Only 1 and 4"
] | C | 15,16 are not prime numbers, but GCD for them is 1. so statement one is incorrect.
2,7 have 1 as GCD, but they are not consecutive numbers.
Statement 3 and 4 are correct. Since GCD is 1, we can't have a common prime factor or a common factor other than 1 for the two numbers
ANSWER:C |
AQUA-RAT | AQUA-RAT-39333 | A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
As one of our solutions is 'it cannot be determined' we cannot use the answers in this question.
Therefore, we'll go for a direct calculation, a Precise approach.
Let's write down what we know, going from the start of the question:
Men + Women = 150
Men = 75 --> Women = 150 - 75 = 75
Women officers = 2/15 * Women = 2/15 * 75 = 10
male officers = 16 - 10 = 6
male non-officers = 75 - 6 = 69.
Bunuel are we assuming that the club contains only men and women (and not, say, children)?
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Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
### Show Tags
01 Jan 2018, 03:24
Bunuel wrote:
The ratio of men to women in a certain club with 150 members is m : w and the ratio of officers to non-officers is o : n. There are 75 men in the club and there are 16 officers in the club. If two-fifteenths of female members are officers, how many male non-officers are in the club?
A. 6
B. 10
C. 65
D. 69
E. It cannot be determined from information given.
Total members 150
Men 75
So women = 150-75 = 75
16 members are officers
2/15 of women are officers
So women officers = 75 * 2/15 = 10
So men officers = 16-10 = 6
So men who are not officers = 75-6 = 69
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3399
Location: India
GPA: 3.5
Re: The ratio of men to women in a certain club with 150 members is m : w [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
The ratio of men to women in the Snyder community choir is 4 to 5. The ratio of men to women in the Leigh community choir is 5 to 6. If the two choirs merged, the ratio of men to women in the combined choir would be 22 to 27. If Snyder has 4 more men and 6 more women than Leigh, how many women are in the Snyder choir? | [
"20",
"24",
"30",
"32"
] | C | Use simultaneous equations / By combination
4x = 5y + 4
5x = 6y + 6
Multiply (1) by 6
Multiply (2) by -5
Get rid of y, and x = 6
Then 5x = 30
ANSWER:C |
AQUA-RAT | AQUA-RAT-39334 | # 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.
Bill made a profit of 10% by selling a product. If he had purchased that product for 10% less and sold it at a profit of 30%, he would have received $ 49 more. What was his original selling price? | [
"$770",
"$660",
"$700",
"$1100"
] | A | Let the original purchase price be X
So original selling price at 10% profit = 1.1X
If product is purchased at 10% less of original = 0.9X
Profit of 30% on this price = 1.3(0.9X)
He would have received $49 more in second scenario => 1.3(0.9X) - 1.1X = 49
=> 0.07X = 49
=> X = $700
Original purchase price = $700
Hence, original selling price (at 10% of profit) = 1.1(700) = $770
Option A |
AQUA-RAT | AQUA-RAT-39335 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream? | [
"75",
"40",
"77",
"26"
] | B | Speed downstream = 60 + 20 = 80 kmph
Speed upstream = 60 - 20 = 40 kmph.Answer: B |
AQUA-RAT | AQUA-RAT-39336 | # 2010 AMC 10B Problems/Problem 25
## Problem
Let $a > 0$, and let $P(x)$ be a polynomial with integer coefficients such that
$P(1) = P(3) = P(5) = P(7) = a$, and
$P(2) = P(4) = P(6) = P(8) = -a$.
What is the smallest possible value of $a$?
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$
## Solution
We observe that because $P(1) = P(3) = P(5) = P(7) = a$, if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$, $R(x)$ has roots when $P(x) = a$; namely, when $x=1,3,5,7$.
Thus since $R(x)$ has roots when $x=1,3,5,7$, we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$.
Then, plugging in values of $2,4,6,8,$ we get
The following is multiple choice question (with options) to answer.
If both 5^2 and 3^3 are factors of the number a*4^3*6^2*13^11, then what is the smallest possible value of a? | [
"49",
"75",
"150",
"225"
] | B | The number a must include at least 3*5^2 = 75
The answer is B. |
AQUA-RAT | AQUA-RAT-39337 | # 1985 AIME Problems/Problem 12
## Problem
Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac n{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.
## Solution 1
Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$, we have $P(n + 1) = \frac13 (1 - P(n))$. We also know $P(0) = 1$, so we can quickly compute $P(1)=0$, $P(2) = \frac 13$, $P(3) = \frac29$, $P(4) = \frac7{27}$, $P(5) = \frac{20}{81}$, $P(6) = \frac{61}{243}$ and $P(7) = \frac{182}{729}$, so the answer is $\boxed{182}$. One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$.
## Solution 2
We can find the number of different times the bug reaches vertex $A$ before the 7th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A$.
The following is multiple choice question (with options) to answer.
Three cubes of metal whose edges are 9, 12 and 15 cm respectively, are melted and one new cube is made. Find the edge of the new cube? | [
"28",
"77",
"66",
"18"
] | D | 93 + 123 + 153 = a3 => a
= 18
Answer: D |
AQUA-RAT | AQUA-RAT-39338 | zoology, ethology, learning
Title: How do beavers learn how to build dams? I was wondering whether all beavers, from all around the world, know how to build dams and lodges? Do they need to learn it from their parents? If you release a group of beavers in the wild that haven't been in contact with their parents, would they start to build stuff? or just hopelessly die/starve to death? Question summary: is dam building learned or instinctive in beavers?
A blog post from 2011 references an article in the Juneau Empire titled Running water is sound of spring for beavers. This article is no longer hosted on the Juneau Empire website, but archived versions are available.
Here's an excerpt (emphasis mine) --
Swedish biologist Lars Wilsson spent years studying captive and wild beavers, and he gained remarkable insights into their behavior. He raised beavers in an outdoor enclosure and in a large indoor terrarium ...
Wilsson initially captured four adult beavers and later he raised a number of beavers from infancy, some in small colonies with their parents and some completely isolated from adult beavers. He isolated the young beavers to see what beavers learn from their parents and what behaviors are instinctive.
He found that young beavers - who had never even seen a beaver dam - were able to build almost-perfect dams at the first opportunity.
The foundation of sticks and logs anchored to the stream bottom, the interwoven lattice of trimmed branches, the mud chinking, every aspect of dam building was hard-wired. Beavers do get more skilled at dam building as they gain experience, but the building behavior is instinctive.
Wilsson learned that the sound of running water is the cue for dam building and dam repair. In one experiment, he played a recording of running water, and the young beavers built a dam in a tank of still water in the terrarium. In another peculiar experiment, his captive beavers built a "dam" on a concrete floor against a loudspeaker that played the sound of running water.
The following is multiple choice question (with options) to answer.
20 beavers, working together in a constant pace, can build a dam in 3 hours. How many hours Z will it take 12 beavers that work at the same pace, to build the same dam? | [
"2.",
"Z=4.",
"Z=5.",
"6."
] | C | C. 5 hrs
If there were 10 beavers it qould have taken double Z= 6hrs.. so closest to that option is 5. |
AQUA-RAT | AQUA-RAT-39339 | In total, there are $$6! = 720$$ ways for all the people to sit on the chairs.
1. First there are $$6$$ ways for you to take a seat, $$2$$ ways for your friend to sit next to you. Now with $$4$$ people left with $$4$$ chairs, there are $$4!=24$$ ways for them to sit. So the probability would be $$\frac {24 \times 2 \times 6}{720}=\frac{2}{5}$$
2. Similar to 1. , there are $$6$$ ways for you to have the first seat, $$1$$ way for your friend to sit opposite you and $$24$$ ways for the rest to sit. The probability would be: $$\frac {24 \times 1 \times 6}{720}=\frac{1}{5}$$
3. In fact this is the complement of 1. so the probability would be $$1- \frac{2}{5}=\frac{3}{5}$$
(Sorry, I was late and English is my second language)
• hi, for number two, just assume you are already seated, there are 5 seats available, so the chance is indeed 1/5 as you calculated. Same argument can be used at the other questions. – Alucard Oct 17 '18 at 12:00
The following is multiple choice question (with options) to answer.
If six persons sit in a row, then the probability that three particular persons are always together is? | [
"1/8",
"1/2",
"1/5",
"1/1"
] | C | Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability
= 3!4!/6!
= 1/5
Answer:C |
AQUA-RAT | AQUA-RAT-39340 | # 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.
Ramu bought an old car for Rs. 42000. He spent Rs. 13000 on repairs and sold it for Rs. 64900. What is his profit percent? | [
"16%",
"88%",
"18%",
"14%"
] | C | Total CP = Rs. 42000 + Rs. 13000
= Rs. 55000 and SP
= Rs. 64900
Profit(%) = (64900 - 55000)/55000 * 100
= 18%
Answer:C |
AQUA-RAT | AQUA-RAT-39341 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A and B go around a circular track of length 600 m on a cycle at speeds of 30 kmph and 54 kmph. After how much time will they meet for the first time at the starting point? | [
"120 sec",
"360 sec",
"178 sec",
"187 sec"
] | B | Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (30 * 5/18) , 600/ (54 * 5 /18) }
= 360 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-39342 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is? | [
"12",
"50",
"27",
"28"
] | B | Speed of the train relative to man = (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr
= 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 45 ==> x
= 50 km/hr.
Answer: B |
AQUA-RAT | AQUA-RAT-39343 | your content. It is the point which is equidistant from each vertex. I am exploring variations on a theme of Napolean's triangle. How many diagonals does an octagon have?. This question cannot be answered because the shape is not a regular polygon. Awarded to. Examples of regular polygons include the equilateral triangle and square. Fun maths practice! Improve your skills with free problems in 'Interior angles of polygons' and thousands of other practice lessons. Assume that the regular polygon has n sides (or angles). Examples include triangles, quadrilaterals, pentagons, hexagons and so on. Polygons are many-sided figures, with sides that are line segments. If you are not sure about the answer then you can check the answer using Show Answer button. A regular polygon have the number of lines of symmetry equal to the number of sides of the regular polygon. For example if a polygon has 41 sides, it would be called a 41-gon. THE NUMBER OF INTERSECTION POINTS MADE BY THE DIAGONALS OF A REGULAR POLYGON BJORN POONEN AND MICHAEL RUBINSTEIN Abstract. It can be solved without using any equation/equations. Which has more sides: a hexagon or a pentagon? hexagon 6. What is the latest version of SketchUp that Keyframe Animation will work with?. Each Interior Angle of a Regular Polygon - MathHelp. (Must be 18 years old to sign up. The sum of the exterior angles of any polygon is 360°. A polygon whose sides are not all the same length or whose interior angles do not all have the same measure. This figure shows the most common polygons. Hint: 360/No. polygon is not regular, it is called irregular. Check your answers seem right. The printable worksheets for grade 7 and grade 8 provide ample practice in finding the area of a regular polygon using the given apothem. A clock is constructed using a regular polygon with 60 sides the polygon rotates every minute how has the polygon rotated after 7 minutes. The questions must be done in order, from Q1 onwards. nonagon nine-sided polygon. Published 2007-06-08 | Author: Kjell Magne Fauske. Consider, for instance, the ir regular pentagon below. It is important to remember that all
The following is multiple choice question (with options) to answer.
how many internal diagonals does a heptagon (seven sided polygon) have? | [
"7",
"9",
"14",
"20"
] | C | Number of diagonals in any polygon can be found using this formula: n(n-3)/2
Here n = 7
No. of diagonals = 7(7 - 3)/2
= 14
Ans C |
AQUA-RAT | AQUA-RAT-39344 | 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.
From among the 36 students in a class, one leader and one class representative are to be appointed. In how many ways can this be done? | [
"1360",
"1260",
"1060",
"1160"
] | B | There are 36 students and every one has equal chance of being selected as a leader. Hence, the leader can be appointed in 36 ways. When one person is appointed as leader, we are left with 35 students. Out of these 35 teachers, we can select one class representative. So, a class representative can be selected in 35 ways. Hence, the number of ways in which a leader and class representative can be selected = 36 x 35 = 1260
ANSWER:B |
AQUA-RAT | AQUA-RAT-39345 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio? | [
"5:7",
"3:9",
"21:22",
"1:2"
] | C | Explanation:
Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
(120/100*7x) and(110/100*8x)
42x/5 and 44x/5
The required ratio =(42x/5:44x/5)=21:22
Answer:C |
AQUA-RAT | AQUA-RAT-39346 | homework-and-exercises, kinematics
Title: Train problem question in kinematics engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last
compartment with velocity v. The middle point of the train passes past the same pole with a velocity of.
My thinking:
Q1 Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
Q2 Shouldn’t the middle part also cover it with u velocity.?
Will the values of v and u be constant since they can’t pass it with different velocities because they are one thing.
You are right in saying that two ends of a same rigid body can't have different values of speed but you should note that your line is correct only if you are talking about the speed of engine and the last compartment at a given point of time.
In your question , the engine passes the pole with speed $u$ and note that since the train can't elongate or compress , the speed of the middle part as well as the last compartment at that instant is $u$ .
When the last compartment reaches the pole, in that time interval the train has accelerated (first line of your question). So, at that instant, the speed of the engine , the middle part and the last compartment is $v$.
So your first question is just a mere confusion.
For the center of the train to pass the pole , the train has to travel for some time and since it is an accelerated motion , the speed with which it passes the pole is different than $u$.
Note: The value of speed with which the center passes the pole can be calculated using the three equation of motion involving constant acceleration :
$$ v = u + at$$
$$s = ut + \frac{1}{2} a t^2$$
$$v^2 = u^2 + 2as$$
Hope it helps .
The following is multiple choice question (with options) to answer.
In what time will a train 95 m long cross an electric pole, it its speed be 214 km/hr? | [
"1.6 sec",
"2.9 sec",
"2.7 sec",
"8.7 sec"
] | A | Speed = 214 * 5/18 = 59 m/sec
Time taken = 95/59 = 1.6 sec.
Answer: A |
AQUA-RAT | AQUA-RAT-39347 | Kudos [?]: 53125 [5] , given: 8043
Re: problem solving question on ratios [#permalink] 16 Dec 2010, 13:47
5
KUDOS
Expert's post
2
This post was
BOOKMARKED
spyguy wrote:
can someone explain in further detail the relationship between the teaching assistants to the number of students in any course must always be greater than 3:80 and how to reason through this portion? I understand how to solve for x. Once I was at this point I think was stumped on which number to select and inevitably chose to round up. My rational being .33 of a student is not possible therefore it must represent the position of an entire student. Thoughts? Help?
At a certain university, the ratio of the number of teaching assistants to the number of students in any course must always be greater than 3:80. At this university, what is the maximum number of students possible in a course that has 5 teaching assistants?
A. 130
B. 131
C. 132
D. 133
E. 134
Given: $$\frac{assistants}{students}>\frac{3}{80}$$ --> $$assistants=5$$, so $$\frac{5}{s}>\frac{3}{80}$$ --> $$s_{max}=?$$
$$\frac{5}{s}>\frac{3}{80}$$ --> $$s<\frac{5*80}{3}\approx{133.3}$$ --> so $$s_{max}=133$$.
$$\frac{assistants}{students}>\frac{3}{80}$$ relationship means that if for example # of assistants is 3 then in order $$\frac{assistants}{students}>\frac{3}{80}$$ to be true then # of students must be less than 80 (so there must be less than 80 students per 3 assistants) on the other hand if # of students is for example 80 then the # of assistants must be more than 3 (so there must be more than 3 assistants per 80 students).
The following is multiple choice question (with options) to answer.
Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats? | [
"2:3:4",
"6:7:8",
"6:7:9",
"8:9:7"
] | A | Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
140 x 5x , 150 x 7x and 175 x 8x
100 100 100
7x, 21x and 14x.
2
The required ratio = 7x : 21x : 14x
2
14x : 21x : 28x
2 : 3 : 4.
so the answer is option A) |
AQUA-RAT | AQUA-RAT-39348 | # 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.
I have a few sweets to be distributed. If I keep 2, 3 or 4 in a pack, I am left with one sweet. If I keep 5 in a pack, I am left with none. What is the minimum number of sweets I have to pack and distribute ? | [
"5",
"7",
"23",
"4"
] | A | Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5.
Answer: A |
AQUA-RAT | AQUA-RAT-39349 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of all the numbers between 40 and 80 which are divisible by 3. | [
"15",
"70",
"20",
"60"
] | D | Sol.
Average = (42+ 45+ 48+ 51+ 54+ 57+ 60+ 63+ 66+ 69+ 72+ 75+ 78 / 13) = 780 / 5 = 60.
Answer D |
AQUA-RAT | AQUA-RAT-39350 | homework-and-exercises, kinematics, time, velocity
Title: Question about rain when an object is moving So here's the question:
A train window is 1 meter high and 10 meters long. A raindrop rolls down the window at a vertical speed of 5 meters per second(when stationary). If the train is moving at 30 m s–1 how far will it move in the time it takes the raindrop to roll down the window?
My problem with this question that i don't believe it's as simple at realising it takes $0.2$ seconds for the rain to move down the window and work out the distance the train will travel at that speed, within that time. This is because i've read somewhere about as you faster, rain won't just go directly down unless your stationary. So i'm thinking that the angle of the rain will go up as the train increases it's velocity. Any help would be appreciated, maybe i'm missing something here? If the raindrop's vertical velocity is constant as the train is both stationary and moving, the time taken for the raindrop to travel down the window would be:
$$t = \frac{1\ \text{m}}{5\ \text{m}/\text{s}} = 0.2\ \text{s}$$
Remember, the time $t$ would not depend on the speed of the train. The exercise also specifically states that the raindrop's vertical velocity, not the total velocity, is $5\ \text{m}/\text{s}$, and this therefore allows calculating the time taken for the raindrop to travel down the window as easily as above.
Note, the raindrop will travel both vertically down the window and horizontally with the train.
We calculate $d$, the distance the train travelled in time $t$. In $0.2\ \text{s}$, the train will travel:
$$d = 30\ \text{m}/\text{s} \times 0.2\ \text{s} = 6\ \text{m}$$
So the train travels 6 metres in 0.2 seconds.
The following is multiple choice question (with options) to answer.
When a stone is dropped from a building 200 m high, its speed is proportional to the time elapsed after dropping. The distance traveled is proportional to the square of the time elapsed. After 1 second the speed of the train was 10 m/sec and it was 190 m above the ground. When its speed is 25 m/sec, what would be its distance from the ground? | [
"140 m",
"137.5 m",
"125.75 m",
"142.5 m"
] | B | v= x*t and s= y*t^2
10 = x*1, hence x=10 units
(200-190) = y*1*1
y=10
when speed = 25m/sec, time elapsed = 25/10 = 2.5 sec
distance travelled by stone = 10*(2.5)^2 = 62.5 mts
so distance of stone above ground = 200-62.5 = 137.5 mtrs
ANSWER:B |
AQUA-RAT | AQUA-RAT-39351 | # 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.
Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case. | [
"45",
"25",
"15",
"35"
] | D | Required number = H.C.F. of (132 - 62), (237 - 132) and (237 - 62)
= H.C.F. of 70, 105 and 175 = 35.
Answer is D. |
AQUA-RAT | AQUA-RAT-39352 | $$x_{1,2} = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$$
gives you the roots $x_{1,2}$ of $ax^2 + bx + c = 0$. When you factor the expression, it is not enough to know the roots but you also need to know $a$. Thus, using the "first" method, you discovered that the roots are $\pm 3.5$ but this does not mean you can factor $4x^2 - 49$ as $(x - 3.5)(x - 3.5)$!
If we open the brackets, we see
$$(x - 3.5)(x + 3.5) = x^2 - (3.5)^2 = x^2 - \left( \frac{7}{2} \right)^2 = x^2 - \frac{49}{4}$$
so we don't get the same expression. But if we multiply by $a = 4$ both sides, we get the correct factorization
$$4x^2 - 49 = 4(x - 3.5)(x + 3.5) = 2(x-3.5)2(x+3.5) = (2x - 7)(2x + 7).$$
Back to completing the square:
\begin{align*} ax^2+bx+c &= a\left( x+\frac{b}{2a} \right)^2- \left( \frac{b^2-4ac}{4a} \right) \\ &= a\left[ \left( x+\frac{b}{2a} \right)^2- \left( \frac{b^2-4ac}{4a^2} \right) \right] \\ &= a\left(x+\frac{b+\sqrt{b^2-4ac}}{2a} \right) \left(x+\frac{b-\sqrt{b^2-4ac}}{2a} \right) \end{align*}
The following is multiple choice question (with options) to answer.
In the quadratic equation ax2 - 11x + 40 = 0, if the sum of two roots is 1.1, what is the product of the two roots? | [
"4",
"4.2",
"8",
"9"
] | A | Explanation:
The sum of the roots of the quadratic equation ax2 - bx + c = 0 are (-b/a) and the product of the roots are (c/a).
Thus, in the equation ax2 - 11x + 40 = 0, where a = a, b = - 11 and c = 40.
we get, sum of the roots = - (- 11) / a = 1.1
a = 11 / 1.1 = 10
Product of the roots = 40 / 10 = 4
ANSWER: A |
AQUA-RAT | AQUA-RAT-39353 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The current of a stream at 1 kmph. A motor boat goes 35 km upstream and back to the starting point in 12 hours. The speed of the motor boat in still water is? | [
"5 kmph",
"1 kmph",
"6 kmph",
"2 kmph"
] | C | The current of a stream = 1
The speed of the motor boat in still water= x
Speed downstream = x + 1
Speed upstream = x - 1
since distance = speed × time, we have
35/(x + 1) + 35/(x - 1) = 12 kmph=6 kmph
Answer : Option C |
AQUA-RAT | AQUA-RAT-39354 | 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 dimensions of a room are 25 feet * 15 feet * 12 feet. What is the cost of white washing the four walls of the room at Rs. 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each? | [
"Rs.4599",
"Rs.4528",
"Rs.4527",
"Rs.4530"
] | D | Area of the four walls = 2h(l + b)
Since there are doors and windows, area of the walls = 2 * 12 (15 + 25) - (6 * 3) - 3(4 * 3) = 906 sq.ft.
Total cost = 906 * 5 = Rs.4530
Answer:D |
AQUA-RAT | AQUA-RAT-39355 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12% p.a.? | [
"Rs:10123.19",
"Rs:10123.29",
"Rs:10123.20",
"Rs:10123.28"
] | C | Amount
= [25000 * (1 + 12/100)3]
= 25000 * 28/25 * 28/25 * 28/25
= Rs. 35123.20
C.I. = (35123.20 - 25000)
= Rs:10123.20
Answer: C |
AQUA-RAT | AQUA-RAT-39356 | (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.
Jaime earned enough money by selling seashells at 30 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold ? | [
"5",
"22",
"17",
"25"
] | B | Let's TEST Answer B: 22 seashells...
With 22 seashells, Jamie would have 22(30) = 660 cents. This would allow him to buy 12 books for 660 cents total, with no money left over. This is an exact MATCH for what we were told, so this MUST be the answer.
Final Answer:
[Reveal]Spoiler:
B |
AQUA-RAT | AQUA-RAT-39357 | \$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.
3889 + 12.952 - ? = 3854.002 | [
"47.095",
"47.752",
"47.932",
"47.95"
] | D | Let 3889 + 12.952 - x = 3854.002.
Then x = (3889 + 12.952) - 3854.002
= 3901.952 - 3854.002
= 47.95.
Answer = D |
AQUA-RAT | AQUA-RAT-39358 | # What is the chance of having a 10 of one suit and all other cards of other suits?
A game consists of 32 cards (A, K, Q, J, 10, 9, 8, 7) in four suits, and each player gets 8 cards.
I need to find the probability that I am being dealt a 10 of one suit, and have all my other cards be of a different suit. I know the chance of having a 10 is 1/8, but I get stuck on this.
The card game is a Dutch card game called 'klaverjassen'. There are 4 suits, just like a normal card game. Each player gets 8 cards out of the 32 cards. Now I need the probability that I get a 10 of one suit, and all of my 7 other cards of different suits. For example if my 10 is a diamond, the other 7 cards either need to be clubs, spades or hearts. It doesn't matter which one it is. So I have 24 cards left where I need to pick 7 cards out of.
There are 4 10's in the game. The goal I have is that I have at least 1 10 in my hand, with all other 7 cards being of a different suit than that 10. I can have multiple 10's, as long as one of them is 'unique', meaning that the other 7 cards are not of the 10's suit. So I can have a 10 of hearts, 10 of clubs, 10 of spades and 10 of diamonds in my hand and when all other cards are also hearts, it is still good
The following is multiple choice question (with options) to answer.
Bill has a small deck of 10 playing cards made up of only 2 suits of 5 cards each. Each of the 5 cards within a suit has a different value from 1 to 5; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, turns over 5 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value? | [
"25/33",
"35/43",
"45/53",
"55/63"
] | D | P(no pairs) = 8/9*6/8*4/7*2/6 = 8/63
P(at least one pair) = 1 - 8/63 = 55/63
The answer is D. |
AQUA-RAT | AQUA-RAT-39359 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
If x is 20 percent greater than 55, then x = | [
"68",
"70.4",
"86",
"66"
] | D | X is 20% greater than 55 means X is 1.2 times 55 (in other words 55 + 20/100 * 55 = 1.2 * 55)
Therefore, X = 1.2 * 55 = 66
ANSWER:D |
AQUA-RAT | AQUA-RAT-39360 | 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 290 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train? | [
"230 m",
"210 m",
"260 m",
"320 m"
] | B | Relative speed = (120 + 80) km/hr
= (200 x(5/18))m/sec
= (500/9)m/sec.
Let the length of the other train be x metres.
Then, (x + 290)/9 = 500/9
x + 290 = 500
x = 210.
B |
AQUA-RAT | AQUA-RAT-39361 | The most reasonable way to match the answer in the book would be to define the likelihood to be the ratio of success over failure (aka odds): $$q=\frac{p}{1-p}$$ then the probability as a function of the odds is $$p=\frac{q}{1+q}$$ In your case the odds are $4:1$ so $4$ times as likely would be $16:1$ odds which has a probability of $$\frac{16}{17}=94.1176470588235\%$$ This matches the $3\%$ to $11.0091743119266\%$ transformation, as well.
Bayes' Rule
Bayes' Rule for a single event says that $$O(A\mid B)=\frac{P(B\mid A)}{P(B\mid\neg A)}\,O(A)$$ where the odds of $X$ is defined as earlier $$O(X)=\frac{P(X)}{P(\neg X)}=\frac{P(X)}{1-P(X)}$$ This is exactly what is being talked about in the later addition to the question, where it is given that $$\frac{P(B\mid A)}{P(B\mid\neg A)}=4$$
The following is multiple choice question (with options) to answer.
The odds in favour of an event are 3 : 5. The probability of occurrence of the event is ? | [
"3/5",
"3/8",
"1/3",
"1/5"
] | B | Number of cases favourable of E = 3
Total Number of cases = (3 + 5 ) = 8
∴ P(E) = 3/8
Answer : B |
AQUA-RAT | AQUA-RAT-39362 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A person purchases 90 watches and sells 40 watches at a gain of 10% and 50 watches at a gain of 20%. If he sold all of them at a uniform profit of 15%, then he would have got $ 40 less. The cost price of each clock is: | [
"70",
"90",
"80",
"60"
] | C | C
80
Let C.P. of clock be $ x.
Then, C.P. of 90 watches = $ 90x.
[(110% of 40x) + (120% of 50x)] - (115% of 90x) = 40
44x + 60x - 103.5x = 40
0.5x = 40 => x = 80 |
AQUA-RAT | AQUA-RAT-39363 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
49 pumps can empty a reservoir in 13/2 days, Working 8 hours a day. If 196 pumps are used for 5 hours each day, then the same work will be completed in: | [
"2 days",
"5/2 days",
"13/5 days",
"3 days"
] | C | Explanation :
Let the required number of days be x. Then,
More pumps, Less days (Indirect Proportion)
Less working hrs / day, More days (Indirect Proportion)
Pumps 196 : 49
Working Hrs / Day 5 : 8 :: 13/2 : x
96 x 5 x x = 49 x 8 x 13/2
x = 49 x 8 x 13/2 x 1 / (196 x 5)
x = 13/5
Answer C |
AQUA-RAT | AQUA-RAT-39364 | Veritas Prep Reviews
Math Expert
Joined: 02 Sep 2009
Posts: 37109
Followers: 7252
Kudos [?]: 96499 [0], given: 10752
### Show Tags
05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Posts: 5
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
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07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
If it takes 6 identical printing presses exactly 4 hours Q to print 5,000 newspapers, how long would it take 3 of these presses to print 3,000 newspapers? | [
"3 hours, 20 minutes",
"4 hours, 20 minutes",
"4 hours, 48 minutes",
"5 hours, 48 minutes"
] | C | 6 presses - 5,000 newspapers - 4 hours ;
3 presses - 2,500 newspapers - 4 hours ;
3 presses - 3,000 newspapers - 4 hours + 1/5*4 hours = 4 hours, 48 minutes (since 2,500+1/5*2,500=3,000) =Q.
Answer: C. |
AQUA-RAT | AQUA-RAT-39365 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
Rohan spends 40% of his salary on food, 20% on house rent, 10% on entertainment and 10% on conveyance. If his savings at the end of a month are Rs. 2500. then his monthly salary is | [
"Rs. 12500",
"Rs. 1000",
"Rs. 8000",
"Rs. 6000"
] | A | Sol.
Saving = [100 - (40 + 20 + 10 + 10]% = 20%.
Let the monthly salary be Rs. x.
Then, 20% of x = 2500
⇔ 20 / 100x = 2500
⇔ x = 2500 × 5 = 12500.
Answer A |
AQUA-RAT | AQUA-RAT-39366 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
Two trains are moving in opposite directions at 60 km/hr and 90 km/hr. Their lengths are 1.75 km and 1.10 km respectively. The time taken by the slower train to cross the faster train in seconds is? | [
"48",
"9",
"7",
"66"
] | D | :
Relative speed = 60 + 90 = 150 km/hr.
= 150 * 5/18 = 125/3 m/sec.
Distance covered = 1.75 + 1.10 = 2.75 km = 2750 m.
Required time = 2750 * 3/125
= 66 sec.
Answer: D |
AQUA-RAT | AQUA-RAT-39367 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A small water pump would take an hour to fill an empty tank. A larger pump would take 1/4 hour to fill the same tank. How many hours would it take both pumps, working at their respective constant rates, to fill the empty tank if they began pumping at the same time? | [
"1/7",
"1/4",
"1/5",
"1/6"
] | C | Rate of the small pump is 1 tank/hour
Rate of the larger pump is 1/(1/4) or 4 tank/hour;
Combined rate of the two pumps is 1+4=5 tank/hour, together they will fill the empty tank in 1/5 hours (time=job/rate).
Answer: C |
AQUA-RAT | AQUA-RAT-39368 | # 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 goes away with a SANTRO car at a speed of 20 kmph. The theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start? | [
"5/3 hours",
"2/7 hours",
"2/3 hours",
"1/3 hours"
] | C | -----------20--------------------|
50 20
D = 20
RS = 50 – 20 = 30
T = 20/30 = 2/3 hours
ANSWER:C |
AQUA-RAT | AQUA-RAT-39369 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Joined: 16 Oct 2010
Posts: 9558
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In what ratio must water be mixed with milk to gain 16 2/3% on selling the mixture at cost price ? | [
"1:9",
"1:1",
"1:2",
"1:6"
] | D | Required ratio is cheaper quantity : dearer quantity = (d-m) : (m-c) Answer with Explanation: Step i) Let C.P of 1 litre of mild be Rs.1 And S.P of1 litre of mild be Rs.1 Gain = 50/3 per cent
C.P of 1 litre of mixture =
1
3 50
100
100
x
= Re. 100 x 3/350 x 1 = 6/7 (Mean price) Step (ii) By the rule of allegation I. C.P of 1 liter of water = 0 II. C.P of 1 liter of milk = 1 III. Mean price (p) = 6/7 IV. d – m = 1 – 6/7 = 1/7 V. m – c = 6/7 - 0 = 6/7 ratio of water and milk = 1/7 : 6/7 = 1 : 6
Answer:D |
AQUA-RAT | AQUA-RAT-39370 | # 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.
Find the compound ratio of (2:3), (6:11) and (11:4) is | [
"3:2",
"2:1",
"1:1",
"4:5"
] | C | Required ratio = 2/3 * 6/11 * 11/4 = 2/1 = 1:1
Answer is C |
AQUA-RAT | AQUA-RAT-39371 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Sonika deposited Rs.8000 which amounted to Rs.10200 after 3 years at simple interest. Had the interest been 2% more. She would get how much? | [
"9680",
"10680",
"2999",
"2774"
] | B | (8000*3*2)/100 = 480
10200
--------
10680
Answer:B |
AQUA-RAT | AQUA-RAT-39372 | # 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.
Ajay bought 15 kg of dal at the rate of Rs 14.50 per kg and 10 kg at the rate of Rs 13 per kg. He mixed the two and sold the mixture at the rate of Rs 15 per kg. What was his total gain in this transaction ? | [
"Rs 1.10",
"Rs 11",
"Rs 16.50",
"Rs 27.50"
] | D | Explanation:
Cost price of 25 kg = Rs. (15 x 14.50 + 10 x 13) = Rs. 347.50.
Sell price of 25 kg = Rs. (25 x 15) = Rs. 375.
profit = Rs. (375 — 347.50) = Rs. 27.50.
Answer: D |
AQUA-RAT | AQUA-RAT-39373 | java, datetime, xml
private void makeYear() {//enum
createMonth("January", 31, 1);
if (isLeapYear)
createMonth("February", 29, 2);
else
createMonth("February", 28, 2);
createMonth("March", 31, 3);
createMonth("April", 30, 4);
createMonth("May", 31, 5);
createMonth("June", 30, 6);
createMonth("July", 31, 7);
createMonth("August", 31, 8);
createMonth("September", 30, 9);
createMonth("October", 31, 10);
createMonth("November", 30, 11);
createMonth("December", 31, 12);
}
The following is multiple choice question (with options) to answer.
The calendar for the year 2007 will be the same for the year: | [
"2014",
"2016",
"2017",
"2018"
] | D | Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1 2 1 1 1 2 1 1 1 2 1
Sum = 14 odd days 0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.
Answer: Option D |
AQUA-RAT | AQUA-RAT-39374 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
Which of the following best approximates the value of q if 5^25+5^15=5^q ? | [
"40",
"26",
"25",
"24"
] | C | We have: 5^25+5^15=5^q
==> because 5^15 > 0 --> 5^q MUST be equal or greater than 5^25 ==>q MUST be equal or greater than 25
==> Option D and E are out immediately.
Divide both sides by 5^q and q >= 25
We have:
5^(25-q) + 5^15/5^q = 1
Because q >= 25 ==>5^15/5^q = 0.0000xyz, that is very small,we can ignore it.
Thus, 5^(25-q) must be approximately 1
==> 25-q = 0 ==> q is approximately 25
C is the answer. |
AQUA-RAT | AQUA-RAT-39375 | 3) you flip two coins. One lands heads. It's not that zero landed heads. It's not that two landed heads. One landed heads.
What is the probability the other is heads?
......
Perhaps this could be better stated as:
You flip two coins. You get at least one head. What is the probability that both are heads.
.....
I leave you with a joke.
In American currency we have the following coins: pennies worth $$1$$ cent, nickels worth $$5$$ cents, dimes worth $$10$$ cents, and quarters worth $$25$$ cents.
I have two (American) coins. They add up two 30 cents. One of them is not a nickel. How is that possible?
Answer: The one that is not a nickel is a quarter. The one that is a nickel is a nickel.
===
[1] there are four ways a coin can land.
HH, HT, TH, TT. But one of them is not possible.
The three possible ways are HH,HT,TH and they are equally likely. Of those three ways in only one of them is "the other coin" heads.
The following is multiple choice question (with options) to answer.
Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has twu iron coins and twu copper coins, how many different sums from 1¢ to 14¢ can he make with a combination of his coins? | [
"8",
"11",
"9",
"4"
] | A | The total sum is 2*2 + 2*5 = 14¢. If you can make each sum from 1 to 14 (1¢, 2¢, 3¢, ..., 14¢), then the answer would be 14 (maximum possible).
Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 13¢ and 11¢ (since total sum is 14¢ we cannot remove 1¢ or 3¢ to get 13¢ or 11¢).
Also discarded 6 and 8, for lack of even numbers after 4
So, out of 14 sums 6 are for sure not possible, so the answer must be 14 - 6 = 8 sums or less. Only A fits.
Answer: A. |
AQUA-RAT | AQUA-RAT-39376 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Posts: 52917
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
The following is multiple choice question (with options) to answer.
A factory has three types of machines, each of which works at its own constant rate. If 7 Machine As and 11 Machine Bs can produce 261 widgets per hour, and if 8 Machine As and 22 Machine Cs can produce 600 widgets per hour, how many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day? | [
"408",
"475",
"550",
"625"
] | A | Let Machine A produce A widgets per hour. B produce B widgets per hour and C produce C widgets per hour.
7A+11B=261 ---(1)
8A+22C=600 ---(2)
Dividing (2) by 2
4A+11C=300.....(3)
Adding (1)(3)
11A+11B+11C = 561
A+B+C=51 per hour
So for eight hrs = 51*8 = 408 = Answer = A |
AQUA-RAT | AQUA-RAT-39377 | # Project Euler Problems 5-6
## Problem 5¶
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This is an interesting problem!
First thing's first, we can establish that the largest positive number that meets the condition is $1×2×3..×20$ or simply $20!$ We can work our way down by repeatedly dividing this upper boundary number by any number in the range [1,20] and seeing if it's an even division.
This approach results in a runtime complexity of O(log(n!)), better known as O(n log n)
In [16]:
factors = 20
upper = math.factorial(factors)
divisors = range(2, factors+1)
current = upper
#repeatedly attempt to divide current number by prime factors ordered
#from largest to smallest as long as the result has a remainder of 0
while True:
found = False
for p in reversed(divisors):
c = current / p
if c % p == 0:
found = True
current = c
break
break
print 'divided by', p, 'got', current
divided by 20 got 121645100408832000
divided by 20 got 6082255020441600
divided by 20 got 304112751022080
divided by 18 got 16895152834560
divided by 18 got 938619601920
divided by 18 got 52145533440
divided by 16 got 3259095840
divided by 14 got 232792560
divided by 12 got 19399380
divided by 2 got 9699690
## Problem 6¶
The sum of the squares of the first ten natural numbers is, 12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Method 1: brute force
Complexity: O(N)
The following is multiple choice question (with options) to answer.
What is the least number should be added to 1156, so the sum of the number is completely divisible by 25? | [
"19",
"2",
"5",
"6"
] | A | (1156 / 25) gives remainder 6
6 + 19 = 25, So we need to add 19
Answer : A |
AQUA-RAT | AQUA-RAT-39378 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Find value of X: (323^23 + 73^44 + 413^30) - (317 × 91) = ? | [
"34455",
"35546",
"52150",
"68542"
] | C | It's speed and accuracy which decides the winner. You can use various methods including Vedic mathematics to do calculations quickly.
(32323+7344+41330)−(317×91)=80997−28847=52150
C |
AQUA-RAT | AQUA-RAT-39379 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In one can of mixed nuts, 30% is peanuts. In another can of mixed nuts that is one-half the size of the first one, 40% is peanuts. If both cans are emptied into the same bowl, what percentage of the mixed nuts in the bowl is peanuts? | [
"16 2/3%",
"20%",
"25%",
"33 1/3%"
] | D | Let number of mixed nuts in first can = 100
Number of peanuts in first can = (30/100)*100 = 30
Since second can is one- half of first , number of mixed nuts in second can = 50
Number of peanuts in second can = (40/100)*50 = 20
Total number of peanuts in both cans = 30+20 = 50
Total number of mixed nuts in both cans = 100+50 = 150
% of peanuts if both cans are mixed = (50/150) *100% = 33.33 %
Answer D |
AQUA-RAT | AQUA-RAT-39380 | (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.
Jim and Joe split the bill based on the amount of space they take up in their apartment. Jim pays for two-fifths of it and Joe pays the rest. If Jim pays 30 dollars, how much will Joe have to pay? | [
"45",
"50",
"55",
"40"
] | A | 30 x (5/2) = 150/2 = 75
150 - 30 = 45
Answer is A |
AQUA-RAT | AQUA-RAT-39381 | 28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
### Show Tags
28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
The following is multiple choice question (with options) to answer.
What is the units digit of 3333^(333)*3333^(222)? | [
"0",
"2",
"4",
"6"
] | D | Each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits.
Here's another way to organize the information.
We're given [(2222)^333][(3333)^222]
We can 'combine' some of the pieces and rewrite this product as....
([(2222)(3333)]^222) [(2222)^111]
(2222)(3333) = a big number that ends in a 6
Taking a number that ends in a 6 and raising it to a power creates a nice pattern:
6^1 = 6
6^2 = 36
6^3 = 216
Etc.
Thus, we know that ([(2222)(3333)]^222) will be a gigantic number that ends in a 6.
2^111 requires us to figure out thecycleof the units digit...
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
So, every 4powers, the pattern of the units digits repeats (2, 4, 8, 6.....2, 4, 8, 6....).
111 = 27 sets of 4 with a remainder of 3....
This means that 2^111 = a big number that ends in an 8
So we have to multiply a big number that ends in a 6 and a big number that ends in an 8.
(6)(8) = 48, so the final product will be a gigantic number that ends in an 6.
Final Answer:
D |
AQUA-RAT | AQUA-RAT-39382 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
If H = {1, 7, 18, 20, 29, 33}, how much less is the mean of the numbers in H than the median of the numbers in H? | [
" 1.0",
" 1.5",
" 2.0",
" 2.5"
] | A | This is a good question to understand the difference between mean and median.
Mean: Average of all the numbers. (Sum of all the elements divided by the number of elements)
Median: Arrange the elements of the set in increasing order. If the number of terms is odd, the middle term is the median. If the number of terms is even, the average of middle two terms is the median
Coming to this question,
Mean = (1 + 7 + 18 + 20 + 29 + 33)/6 = 18
Median = (18 + 20)/2 = 19
Difference = 1
Option A |
AQUA-RAT | AQUA-RAT-39383 | # Math Help - need help.
1. ## need help.
A long rope is pulled out between two opposite shores of a lake. It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
2. Hello, Bobby!
A long rope is pulled out between two opposite shores of a lake.
It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
Code:
C
* * *
* :x *
* : 35 *
A * - - - - -+- - - - - * B
\ D: /
\ :R-x /
R \ : / R
\ : /
\:/
*
O
The center of the earth is $O.$ . $OA = OB = OC = R$ (radius of the earth).
The 70-km rope is $AB.$ .We see that: $AD = DB = 35.$
Let $x = CD$ be the distance the rope is underwater at its center.
Then $DO = R - x.$
From right triangle $ODB:\;\;DO^2 + DB^2\:=\:OB^2$
So we have: . $(R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$
Quadratic Formula: . $x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$
Since $R = 6370$, we have: . $x \;= \;6370 \pm \sqrt{6370^2 - 1225}$
The following is multiple choice question (with options) to answer.
A dog is tied to a tree by a long nylon cord. If the dog runs from the due North side of the tree to the due South side of the tree with the cord extended to its full length at all items, and the dog ran approximately 30 feet, what was the approximate length of the nylon cord L in feet? | [
"30",
"25",
"15",
"10"
] | D | Because the cord was extended to its full length at all items, the dog ran along a semi-circular path, from north to south.
The circumference of a full circle is 2*pi*r, but since we only care about the length of half the circle, the semi-circle path is pi*r.
L=pi*r = 30. Round pi = 3, then r = 10.
Chord is about 10 feet long.D |
AQUA-RAT | AQUA-RAT-39384 | 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's speed with the current is 20 kmph and speed of the current is 6 kmph. The Man's speed
against the current will be | [
"11 kmph",
"12 kmph",
"14 kmph",
"8 kmph"
] | D | Explanation:
Speed with current is 20,
speed of the man + It is speed of the current
Speed in sƟll water = 20 - 6 = 14
Now speed against the current will be
speed of the man - speed of the current
= 14 - 6 = 8 kmph
Answer: D |
AQUA-RAT | AQUA-RAT-39385 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
In a group of ducks and cows, the total number of legs are 24 more than twice the number of heads. Find the total number of cows. | [
"12",
"14",
"16",
"18"
] | A | Let the number of ducks be d
and number of cows be c
Then, total number of legs = 2d + 4c = 2(d + 2c)
total number of heads = c + d
Given that total number of legs are 24 more than twice the number of heads
=> 2(d + 2c) = 24+ 2(c + d)
=> d + 2c = 12 + c + d
=> 2c = 12 + c
=> c = 12
i.e., total number of cows = 12
Answer is A. |
AQUA-RAT | AQUA-RAT-39386 | • Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
The following is multiple choice question (with options) to answer.
I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy? | [
"1/4",
"1/3",
"1/6",
"2/3"
] | B | B
1/3
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.
Following are the possible cases for toys distribution:
Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1
In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3 |
AQUA-RAT | AQUA-RAT-39387 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
Present ages of X and Y are in the ratio 5:6 respectively. Seven years hence this ratio will become 6:7 respectively. What is X's present age in years? | [
"40",
"35",
"28",
"49"
] | B | Let the present ages of X and Y be 5x and 6x years respectively.
Then, (5x + 7)/(6x + 7) = 6/7
7(5x + 7) = 6(6x + 7) => x = 7
X's present age = 5x = 35 years.
ANSWER B |
AQUA-RAT | AQUA-RAT-39388 | 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.
n a 1000m race, if A gives B a start of 40m, A wing by 19sec. But if A gives B start of 30sec, B wins by 40m.
Find the time taken by B to run 4000m race? | [
"500",
"700",
"400",
"600"
] | D | then 960/B -1000/A = 19 => 24/B - 25/A = 19/40
and 1000/B - 960/A = 30 => 25/B - 24/A = 30/40
solving equation A speed =8 and B=20/3
for 1000m b will take 4000/20/3= 600 sec
D |
AQUA-RAT | AQUA-RAT-39389 | Example 2
In a large population of adults, 45% have a post secondary degree.
If people are selected at random from this population,
a) what is the probability that the third person selected is the first one that has a post secondary degree?
b) what is the probability that the first person with a post secondary degree is randomly selected on or before the 4th selection?
Solution to Example 2
a)
Let "having post secondary degree" be a "success". If a person from this population is selected at random, the probability of "having post secondary degree" is $p = 45\% = 0.45$ and "not having post secondary degree" (failure) is $1 - p = 1 - 0.45 = 0.55$
Selecting a person from a large population is a trial and these trials may be assumed to be independent. This is a geometric probability problem. Hence
$P(X = 3) = (1-0.45)^2 (0.45) = 0.1361$.
b)
On or before the 4th is selected means either the first, second, third or fourth person. The probability may be written as
$P(X \le 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
Substitute by the formula $P(X = x) = (1 - 0.45)^{x-1} 0.45$ to write
$P(X \le 4) = (1 - 0.45)^{1-1} 0.45 + (1 - 0.45)^{2-1} 0.45 + (1 - 0.45)^{3-1} 0.45 + (1 - 0.45)^{4-1} 0.45 = 0.9085$
## Sums of the terms of a Geometric sequence
The following is multiple choice question (with options) to answer.
Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them? | [
"5/5",
"5/8",
"5/1",
"5/6"
] | D | P(at least one graduate) = 1 - P(no graduates)
= 1 - 6C3/10C3
= 1 - (6 * 5 * 4)/(10 * 9 * 8)
= 5/6
Answer:D |
AQUA-RAT | AQUA-RAT-39390 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
K and L start the business, with L investing the total capital of Rs. 40,000 on the condition that K pays L interest @ 8% per annum on his half of the capital. L is a working partner and receives Rs.1200 per month from the total profit and any profit reamaining is equally shared by both of them. At the end of the year, it was found that the income of L is twice that of K. Find the total profit for the year ? | [
"46409",
"46400",
"46129",
"46411"
] | B | Interest received by L from K = 8% of half of Rs.40,000
= Rs.1600
Amount received by L per annum for being a working partner = 1200\small \times12 = Rs.14,400
Let 'A' be the part of remaing profit that 'L' receives as his share.
Total income of 'K' = only his share from the reamaing profit
= 'A', as both share equally.
Given income of L = Twice the income of K
Rightarrow (1600 + 14400 + A ) = 2A
Rightarrow A= Rs.16000
Thus total profit = 2A + Rs.14,400= 2(16000) + 14400
= 32000 +14400 = Rs.46,400.
Answer: B |
AQUA-RAT | AQUA-RAT-39391 | The midpoint of $\displaystyle DE$ is: $\displaystyle M(3\tfrac{1}{2},\,2).$
The median to side $\displaystyle DE$ starts at $\displaystyle \,M$, passes through $\displaystyle \,C,$
. . and extends to $\displaystyle \,F$, where: .$\displaystyle FC \,=\,2\!\cdot\!CM.$
Going from $\displaystyle \,M$ to $\displaystyle \,C$, we move up 2 and left $\displaystyle \frac{1}{2}$
Hence, going from $\displaystyle \,C$ to $\displaystyle \,F$, we move up 4 and left 1.
Therefore, we have: .$\displaystyle F(2,8).$
The following is multiple choice question (with options) to answer.
What is the median from the below series
90, 92, 93, 88, 95, 88, 97, 87, and 98 | [
"85",
"89",
"92",
"98"
] | C | Ordering the data from least to greatest, we get:
87, 88, 88, 90, 92, 93, 95, 96, 98
The median score is 92.
C |
AQUA-RAT | AQUA-RAT-39392 | ## 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 rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 600 sq.ft, how many feet of fencing will be required ? | [
"80feet",
"70feet",
"60feet",
"50feet"
] | A | Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 600
=> Breadth = 30 feet
Area to be fenced = 2B + L = 2*30 + 20
= 80feet
Answer: Option A |
AQUA-RAT | AQUA-RAT-39393 | # 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.
The population of a town is 7600. It decreases annually at the rate of 10% p.a. What was its population 2 years ago? | [
"9400",
"8000",
"8500",
"9500"
] | A | Formula :
( After =100 denominator
Ago = 100 numerator)
7600 × 100/90 × 100/90 = 9382
A) |
AQUA-RAT | AQUA-RAT-39394 | newtonian-mechanics, drag, relative-motion
Title: Relative motion-Acceleration My first post here and I'm a complete beginner on this. So please excuse if I'm asking too-basic a question. This question is about the classical boat and river problem.
Say a boat travels at 10 m/s in a water channel.
the water speed relative to ground is 0.
so the boat travels at 10 m/s relative to the ground.
now suddenly, the water in the channel has started to flow at 10 m/s in the opposite direction. (say this happened in 10 seconds so the acceleration is 1 m/s^2).
As after a while the boat speed relative to ground has become 0,
then from the ground-based observer's point of view, the boat has undergone a deceleration.
My question is;
Is this deceleration always necessarily equal to minus the water acceleration?
In other words whats the velocity of the boat with respect to the ground, infinitesimal time dt after the water has started to accelerate ?
PS: What I'm trying to understand is what happens when an aircraft or watercraft gets hit by a gust or similar disturbance?
My question is; Is this deceleration always necessarily equal to minus the water acceleration?
The answer is no. Acceleration/deceleration is controlled by the fluid-resistance $f$. Typically:
$$f=kv\qquad \text{ for low speed}\\
f=kv^2\qquad \text{ for high speed}$$
where $v$ is speed of the object (boat, airplane, car ...) relative to the fluid and $k$ a coefficient.
The following is multiple choice question (with options) to answer.
A boat can move upstream at 33 kmph and downstream at 35 kmph, then the speed of the current is? | [
"1",
"2",
"7",
"8"
] | A | US = 33
DS = 35
M = (35 - 33)/2 = 1
Answer: A |
AQUA-RAT | AQUA-RAT-39395 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
If a certain number is divided by 3, the quotient, dividend, and divisor, added together, will amount to 63. What is the number? | [
"18",
"28",
"45",
"38"
] | C | Let x = the number sought.
Then x/3 + x + 3 = 63.
And x = 45.
C |
AQUA-RAT | AQUA-RAT-39396 | rectangle, write the equation to solve for the length of the right triangle bases. If the base and height of a trapezium are given, then the area of a Trapezium can be calculated with the help of the formula: Area of Trapezium = 1/2 x (sum of bases) x (Height of trapezium… A trapezium that has equal non-parallel sides and equal base angles is an isosceles trapezium. Calculate the area of the trapezoid. A trapezium or a trapezoid is a quadrilateral with a pair of parallel sides. 18:04. So in a trapezoid ABCD, ∠A+∠B+∠C+∠… Your Infringement Notice may be forwarded to the party that made the content available or to third parties such University of Minnesota Mi... Track your scores, create tests, and take your learning to the next level! a The altitude of the trapezoid is 4. Diagonals of a trapezium bisect each other on intersection. Therefore, we need to sketch the following triangle within trapezoid : We know that the base of the triangle has length . The diagonals of a kite intersect at 90 $$^{\circ}$$ The formula for the area of a kite is Area = $$\frac 1 2$$ (diagonal 1)(diagonal 2) Advertisement. Calculate the trapezoid area. Use the Pythagorean Theorem to solve for the diagonal. =sinα× (base AC)× (BM+MD)=sinα×AC×BD=absinα. To illustrate how to determine the correct length, draw a perpendicular segment from to , calling the point of intersection . Like a rhombus, the kite area formula is: Area = (Diag. Basic Terminology for Trapezium. 5) Using the Pythagorean Theorem on to find . Example #2. University of Minnesota Minneapolis Minnesota, Bachelor of Science, Mathematics Teacher Education. Learn to find area of a Trapezium or Trapezoid when two sides and diagonal is given. link to the specific question (not just the name of the question) that contains the content and a description of Find the length of both diagonals of this quadrilateral. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Please be advised
The following is multiple choice question (with options) to answer.
nd the area of trapezium whose parallel sides are 20 cm and 15 cm long, and the distance between them is 14 cm? | [
"230cm2",
"245cm2",
"255cm2",
"260cm2"
] | B | Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them)
= 1/2 (20 + 15) * (14)
= 245 cm2
Answer:B |
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