source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-33897 | beginner, r, timer
Title: R function for a timer with a beep I wrote a quick timer function that depends on the beepr package to provide the sound effect at the end of the time. I've never used a while-loop before, so I don't know if there's a more elegant way to write this (i.e., one that doesn't require repeating the same line of code twice?). I'd appreciate knowing if there is a better way:
timer <- function(interval, units) {
require(beepr)
t0 <- Sys.time()
stopwatch <- round(as.double(difftime(Sys.time(), t0, u = units)))
while(stopwatch < interval){
stopwatch <- round(as.double(difftime(Sys.time(), t0, u = units)))
}
beep(2)
}
timer(5, "secs") To avoid repeating code, you could have used:
done <- FALSE
while(!done){
done <- round(as.double(difftime(Sys.time(), t0, u = units))) >= interval
}
Note however that a for loop is quite taxing on your CPU. Instead, you should use the friendlier Sys.sleep function. It takes a number of seconds as input:
timer <- function(num_sec) {
Sys.sleep(num_sec)
if (require(beepr)) beep(2) else message("DONE!")
}
The following is multiple choice question (with options) to answer.
An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62
sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is | [
"10:28 am",
"10:30 am",
"10:31 am",
"10:37 am"
] | C | Explanation:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.
Sometimes questions on red lights blinking comes in exam, which can be solved in the same way
Answer: Option C |
AQUA-RAT | AQUA-RAT-33898 | +2}_{12} }_{25} &&=\color{red}{\underbrace{\underbrace{-10+(-9)+\cdots+1}_{12}+\boxed{2}+\underbrace{3+\cdots+13+14}_{12}}_{25}}\\ &=50\times 1 &&=\underbrace{\underbrace{1+1+\cdots+1}_{25}\boxed{+}\underbrace{1+\cdots +1}_{25} }_{50} &&=\underbrace{\underbrace{-23.5+(-22.5)+\cdots+0.5}_{25}\boxed{+}\underbrace{1.5+\cdots+25.5}_{25}}_{50}\\ & && && \quad \text{(AP but not integer AP)}\\ &=100\times 0.5 &&=\underbrace{0.5+0.5+\cdots+0.5}_{50}\boxed{+}\underbrace{0.5+0.5+\cdots+0.5}_{50} &&=\color{red}{\underbrace{\underbrace{-49+(-48)+\cdots+0}_{50}\boxed{+}\underbrace{1+2+\cdots +49+50}_{50} }_{100}} \end{align}
The following is multiple choice question (with options) to answer.
Simplify: 9948/49 * 245. | [
"24495",
"23395",
"22225",
"22115"
] | A | Given expression = (100-1/49) * 245=(4899/49) * 245 = 4899 * 5=24495.
Answer is A. |
AQUA-RAT | AQUA-RAT-33899 | A follow-on question: Is the largest inscribed rectangle also the largest possible rectangle? I.e. could there be larger rectangles where every corner does not touch the sector?
Your result is correct, and leads to a maximum area equal to $\displaystyle{1-\cos(\alpha/2)\over\sin(\alpha/2)}$.
But with a different disposition of the rectangle one can get a greater area. If $\alpha<90°$ you can construct an inscribed rectangle as shown in diagram below on the left. Maximum area occurs when a vertex of the rectangle is at the midpoint of the arc. If $\alpha\ge90°$ you can construct an inscribed rectangle as shown in diagram below on the right. Maximum area occurs when the rectangle is a square.
A tedious calculation gives for maximum areas the expressions shown in the diagram (they should be correct but please check them if you have time). In the first case the area of the rectangle is always larger than your result. In the second case the area is larger only if $\cos(\alpha/2)<3/5$.
• Very nice! I agree with your result. I'm glad that my rectangle at least has some interval where it is the largest, i.e $\alpha > 106.26^\circ$. :)
The following is multiple choice question (with options) to answer.
The sides of a rectangular region measured to the nearest centimeter are 8 cm and 4 cm. Out of the following which is the largest possible value of actual area of rectangle ? | [
"48",
"35.25",
"52.29",
"44"
] | B | acan be anywhere between 7.5 and 8.4 andbcan be anywhere between 3.5 and 4.4.
Hence max area can be 8.4*4.4 eq 36.96.
largest value which is near to 36.96 is 35.25.
Hence B. |
AQUA-RAT | AQUA-RAT-33900 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
The sum of two numbers is 36. Four times one number is equal to 5 times the other. The bigger of the two numbers is | [
"20",
"12",
"14",
"15"
] | A | a:b 4:5
a+b=36
4x+5x=36
x=4
then two numbers are 16,20. The bigger number is 20
ANSWER: A |
AQUA-RAT | AQUA-RAT-33901 | ### Permutations when all the objects are not distinct objects
Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O_1 and O_2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO_1O_2T. Corresponding to this permutation,we have 2 ! permutations RO_1O_2T and RO_2O_1T which will be exactly the same permutation if O_1 and O_2 are not treated as different, i.e., if O_1 and O_2 are the same O at both places.
Therefore, the required number of permutations = (4!)/(2!) = 3 xx 4 = 12
Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.
Temporarily, let us treat these letters different and name them as I_1, I_2, T_1 , T_2, T_3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, I_1 NT_1 SI_2 T_2 U E T_3. Here if I_1, I_2 are not same and T_1, T_2, T_3 are not same, then I_1, I_2 can be arranged in 2! ways and T_1, T_2, T_3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I_1NT_1SI_2T_2UET_3. Hence, total number of different permutations will be (9!)/(2 ! 3!)
We can state (without proof) the following theorems:
The following is multiple choice question (with options) to answer.
The letters of the word LUCKNOW are arranged among themselves. Find the probability of always having NOW in the word. | [
"1/31",
"1/32",
"1/41",
"1/42"
] | D | Explanation :
Since NOW has to always come together, assume it as one single unit. The rest 4 characters can occur in any way. So total no of characters = 5
The required probability will be given by the equation
= No. of words having NOW/Total no. of words
= 5!/7! = 1/42
Answer : D |
AQUA-RAT | AQUA-RAT-33902 | ### Show Tags
19 Aug 2015, 01:34
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Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
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Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
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Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average of ages of 10 persons in a club was 32. What should be the age of the new person joining in the club so as to increase the average by 4 ? | [
"76",
"2366",
"299",
"277"
] | A | Correct Option: D
Explanation:
Total age of 10 persons = 10 × 32 = 320
Total age of 11 persons = 11 × 36 = 396 (as the new average is 4 more than present average)
So the age of the person joining is = 396 - 320 = 76
Answer:A |
AQUA-RAT | AQUA-RAT-33903 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
An accurate clock shows 8 o’clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 1 o’clock in the afternoon? | [
"60°",
"150°",
"180°",
"320°"
] | B | Sol.
Angle traced by the hour hand in 5 hours = [360/12 * 5]° = 150°
Answer B |
AQUA-RAT | AQUA-RAT-33904 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
In a kilometer race, A beats B by 40 meters or 8 seconds. What time does A take to complete the race? | [
"221",
"190",
"192",
"191"
] | C | Time taken by B run 1000 meters = (1000 * 8)/40 = 200 sec.
Time taken by A = 200 - 8 = 192 sec.Answer:C |
AQUA-RAT | AQUA-RAT-33905 | ### 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 broker invested her own money in the stock market. During the first year, she increased her stock market wealth by 40 percent. In the second year, largely as a result of a slump in the stock market, she suffered a 30 percent decrease in the value of her stock investments. What was the net increase or decrease on her overall stock investment wealth by the end of the second year? | [
"−5%",
"−2%",
"15%",
"20%"
] | B | The actual answer is obtained by multiplying 140% by 70% and subtracting 100% from this total. That is: 140% × 70% = 98%; 98% − 100% = -2%.
Answer: B |
AQUA-RAT | AQUA-RAT-33906 | # Number of rectangles with odd area.
We have a $10\times 10$ square.
How many rectangles with odd area are on the picture?
I say lets choose a vertex first, there are $11\cdot11=121$ possibilities.
Now, choose odd width and side (left or right) and odd length and side (up or down). There are $5$ possibilities to each, so in total we have $\dfrac{121\cdot 25}{4}$ rectangles. We divide by $4$ because every rectangle is counted $4$ times, one time for each vertex of the rectangle. But, the result is not a whole number. Where am I wrong? Thanks.
The width must be odd and the height must be odd. There are $10$ choices of a pair of $x$-coordinates to have width $1$, $8$ choices for width $3$, $6$ choices for with $5$, etc. So we have $10+8+6+4+2=30$ ways to choose the two $x$ coordinates and likewise $30$ ways to choose $y$-coordinates. This gives us a total of $30\cdot 30=900$ odd area rectangles.
What you did was to pick one vertex and than assume that each possible odd width and height could be realized with this vertex.
The following is multiple choice question (with options) to answer.
A lady grows cabbage in her garden that is in the shape of a square. Each cabbage takes 1 square foot of area in her garden. This year, she has increased her output by 211 cabbages when compared to last year. The shape of the area used for growing the cabbage has remained a square in both these years. How many cabbages did she produce this year? | [
"11236",
"11025",
"14400",
"12696"
] | A | Explanatory Answer
The shape of the area used for growing cabbages has remained a square in both the years.
Let the side of the square area used for growing cabbages this year be X ft.
Therefore, the area of the ground used for cultivation this year = X2 sq.ft.
Let the side of the square area used for growing cabbages last year be Y ft.
Therefore, the area of the ground used for cultivation last year = Y2 sq.ft.
As the number of cabbages grown has increased by 211, the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space.
Hence, X2 - Y2 = 211
(X + Y)(X - Y) = 211.
211 is a prime number and hence it will have only two factors. i.e., 211 and 1.
Therefore, 211 can be expressed as product of 2 numbers in only way = 211 * 1
i.e., (X + Y)(X - Y) = 211 * 1
So, (X + Y) should be 211 and (X - Y) should be 1.
Solving the two equations we get X = 106 and Y = 105.
Therefore, number of cabbages produced this year = X2 = 1062 = 11236.
Alternative Approach : Use answer choices
The area in both the years are squares of two numbers.
That rules out choice D. 12696 is not the square of any number.
Check Choice A: If this year's produce is 11236, last year's produce would have been 11236 - 211 = 11025
11025 is the square of 105.
So, 11236 is the answer.
Choice A is the correct answer. |
AQUA-RAT | AQUA-RAT-33907 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of the four numbers K, 2K + 3, 3K – 5 and 5K + 1 is 96, what is the value of K? | [
" 11",
" 15 3/4",
" 22",
" 35"
] | D | K + 2K +3 + 3K - 5 + 5K +1 = 11K -1
(11K -1)/4 = 96
11K = 96 * 4 +1 = 384 +1 = 385
K = 385 / 11 = 35.
Answer D. |
AQUA-RAT | AQUA-RAT-33908 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, the dividend is: | [
"5236",
"5330",
"5336",
"5348"
] | C | Divisor = (5 * 46) = 230
= 10 * Quotient = Divisor
=> Quotient = 230/10 = 23
Dividend = (Divisor * Quotient) + Remainder
Dividend = (230 * 23) + 46 = 5336.
C) |
AQUA-RAT | AQUA-RAT-33909 | ### 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 and B starts a business with Rs.8000 each, and after 4 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months? | [
"18:11",
"18:18",
"18:12",
"18:16"
] | A | Explanation:
A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 4 months and then withdraws Rs.4000. So, the investment of B for remaining 14 months is Rs.4000 only.
A : B
8000*18 : (8000*4) + (4000*14)
14400 : 88000
A:B = 18:11
Answer: A |
AQUA-RAT | AQUA-RAT-33910 | Suppose that 5 cards are dealt from a 52-card deck. What is the probability of drawing at least two kings given that there is at least one king?
The Attempt at a Solution
Let ##B## denote the event that at least 2 kings are drawn, and ##A## the event that at least 1 king is drawn. Because ##B## is a strict subset of ##A##,
$$P(B|A) = P(A \cap B)/P(A) = P(B)/P(A)$$
Compute ##P(A)##, ##P(A^c )## denotes the probability of not drawing a single king.
$$P(A) = 1 - P(A^c) = 1 - \frac{48 \choose 5}{52 \choose 5} \approx 1 - 0.6588$$
Compute ##P(B)##, ##P(B^c)## denotes the probability of not drawing at least 2 kings, which is the sum of probabilities of drawing 1 king ##P(1)## and the probability of not drawing a single king ##P(A^c)##.
$$P(B) = 1 - P(B^c) = 1 - (P(1) - P(A^c))$$
$$P(1) = \frac{5 \times {4\choose 1} \times 48 \times 47 \times 46 \times 45 }{52 \times 51 \times 50 \times 49 \times 48} \approx 0.299$$
where the numerator is the number of ways one can have a hand of 5 containing a single king.
$$P(B) \approx 1 - (0.299 + 0.6588) \approx 0.0422$$
finally,
$$P(B|A) = P(B)/P(A) = 0.0422 / (1-0.6588) \approx 0.1237$$
The following is multiple choice question (with options) to answer.
If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is? | [
"4/15",
"4/17",
"4/12",
"4/13"
] | D | P(SᴜK)
= P(S) + P(K) - P(S∩K), where S denotes spade and K denotes king.
P(SᴜK)
= 13/52 + 4/52 - 1/52
= 4/13
Answer: D |
AQUA-RAT | AQUA-RAT-33911 | algorithms
Title: In the Gas-Up problem, how is the amount of gas the same up to a cyclic shift regardless of starting city? I'm working through Elements of Programming Interviews as practice for finding a job. I've spent a ridiculous amount of time on Problem 18.7.
In the gas-up problem, $n$ cities are arranged on a circular road. You
need to visit all of the $n$ cities and come back to the starting
city. A certain amount of gas is available at each city. The total
amount of gas is equal to the amount of gas required to go around the
road once. Your gas tank has unlimited capacity. Call a city $c$
ample if you can begin at $c$ with an empty tank, refill at it, then travel through each of the remaining cities, refilling at each, and
return to $c$, without running out of gas at any point. See Figure
18.3 for an example. [You can see Figure 18.3 on the Google Books result if you search for "Elements of Programming Interviews Problem
18.7.]
Given an instance of the gas-up problem, how would you efficiently [i.e. in $\mathcal{O}(n)$ time or better]
compute an ample city, if one exists?
I couldn't figure it out. I read the hint, and I still couldn't figure it out. I gave up, and read the answer in the back of the book. I still couldn't figure it out. I searched Google and this site for a more complete solution, but didn't find one.
This is the part that's getting me:
On closer inspection, it becomes apparent that the graph of the amount
of gas as we perform the traversal is the same up to a cyclic shift
regardless of the starting city.
The following is multiple choice question (with options) to answer.
In a fuel station the service costs $2.05 per car, every liter of fuel costs 0.6$. Assuming that you fill up 3 mini-vans and 2 trucks, how much money will the fuel cost to all the cars owners total, if a mini- van's tank is 70 liters and a truck's tank is 120% bigger and they are all empty-? | [
"312.6$",
"321.05$",
"343.7$",
"398.85$"
] | B | Service cost of 3 van and 2 truck = 2.05*(3+2)=10.5
Fuel in 3 Van = 3*70=210 litre
Fuel in 2 Trucks =2 * 70(1+120/100)= 308
Total fuel(Van+Truck) = 518 Litre
Total Fuel cost = 518*0.6 =310.8
TOTAL COST =FUEL+SERVICE=310.8+10.25=321.05
ANSWER IS B |
AQUA-RAT | AQUA-RAT-33912 | In a bag there are 10 Black balls, 8 White balls and 5 Red balls.
Three balls are chosen at random and one is found to be Black.
Find the probability that remaining two are white.
. $(a)\;\frac{8}{23}\qquad(b)\;\frac{4}{33}\qquad(c)\ ;\underbrace{\frac{10\cdot8\cdot7}{23\cdot22\cdot2 1}}_{str\!ange!} \qquad(d)\;\frac{4}{23}\qquad(e)\;\frac{5}{23}$
I see it as a Conditional Probability problem . . .
Given that at least one ball is Black,
. . find the probability that we have one Black and two White balls.
Bayes' Theorem: . $P(\text{1B,2W }|\text{ at least 1B}) \;=\;\frac{P(\text{1B} \wedge \text{2W})}{P(\text{at least 1B})}$
There are ${23\choose3} = 1771$ possible ways to choose 3 balls.
To choose 1 Black and 2 Whites: . ${10\choose1}{8\choose2} \:=\:280$ ways.
. . Hence: . $P(\text{1B}\wedge\text{2W}) \:=\:\frac{280}{1771}$
The opposite of "at least 1 Black" is "NO Blacks".
There are: . ${13\choose3} = 286$ ways to choose no Blacks.
So, there are: . $1771 - 286 \:=\:1485$ ways to choose some Black balls.
. . Hence: . $P(\text{at least 1B}) \:=\:\frac{1485}{1771}$
The following is multiple choice question (with options) to answer.
A bag contains 12 white and 18 black balls. 2balls are drawn in succession. What is the probability that first is white and second is black? | [
"36/141",
"36/145",
"30/141",
"36/115"
] | B | The probability that first ball is white:
=12C130C1=12C130C1
=1230=1230
=25=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C129C1=18C129C1
=1829=1829
Required probability,
=(25)×(1829)=(25)×(1829)
=36/145
B |
AQUA-RAT | AQUA-RAT-33913 | 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.
The simple interest on Rs.15000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for two years at the same rate of interest. | [
"3052.82",
"3015.24",
"3052.24",
"3018.24"
] | D | R = 100 I / PT
=> R = (100 * 7200)/ (15000 * 5) = 9.6%
CI = P{ [1 + R /100]n - 1}
= 15000 { [ 1 + 9.6 / 100]^2 - 1} = Rs.3018.24
Answer:D |
AQUA-RAT | AQUA-RAT-33914 | ### Show Tags
29 May 2017, 10:14
1
60*3 = 180
+
24*5= 120
120+180 =300
Speed= Distance/Time = 300/8 = 150/4=75/2=37.5
Manager
Joined: 03 Aug 2017
Posts: 103
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink]
### Show Tags
09 Dec 2019, 07:43
Bunuel wrote:
Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph?
A. 36 mph
B. 37.5 mph
C. 42 mph
D. 42.5 mph
E. 48 mph
tIME = 3 S=60 d=ST = 60*3 =180 miles
time 2 = s=24 d st = 120 Miles
Total d = 180+120 =300
Total time =5+3= 8
Total Avg speed = Total Distance / total time = 300/8 = 37.5 or B
Re: Jim travels the first 3 hours of his journey at 60 mph speed and the [#permalink] 09 Dec 2019, 07:43
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Steve traveled the first 2 hours of his journey at 50 mph and the last 3 hours of his journey at 80 mph. What is his average speed of travel for the entire journey? | [
"68 mph",
"56.67 mph",
"60 mph",
"64 mph"
] | A | Average Speed = Total distance/ Total time = (50 * 2 + 80 * 3)/(2 + 3) = 68
Ans:A |
AQUA-RAT | AQUA-RAT-33915 | 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.
If the average marks of 3classes of 55, 60 and 45 students respectively is 50, 55, 60, then find the average marksof all the students is | [
"53",
"53.5",
"54.68",
"55.6"
] | C | Required average
= 55 x 50 + 60 x 55 + 45 x 60
55 + 60 + 45
= 2750 + 3300 + 2700
160
= 8750
160
= 54.68
C |
AQUA-RAT | AQUA-RAT-33916 | 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.
If a sum on compound interest becomes three times in 4 years, then with the same interest rate, the sum will become 81 times in | [
"12 years",
"18 years",
"16 years",
"14 years"
] | C | Explanation:
Let the sum be P
The sum P becomes 3P in 4 years on compound interest
3P = P(1+R/100)4
⇒ 3 = (1+R/100)4
Let the sum P becomes 81P in n years
81P = P(1+R/100)n
⇒ 81 = (1+R/100)n
⇒ (3)4 =(1+R/100)n
⇒((1+R/100)n)n
=(1+R/100)n
⇒ (1+R/100)16
= (1+R/100)n
n = 16
i.e, the sum will become 81 times in 16 years
Answer: Option C |
AQUA-RAT | AQUA-RAT-33917 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
how many distinct 3 digit number less than 10 | [
"8",
"9",
"10",
"11"
] | C | 000,001,002,003,004,005,006,007,008,009.
10 distinct 3 digit numbers less than 10.
ANSWER:C |
AQUA-RAT | AQUA-RAT-33918 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
189. The inside dimensions of a rectangular wooden box
are 10 inches by 12 inches by 14 inches. A cylindrical
canister is to be placed inside the box so that it stands
upright when the closed box rests on one of its six
faces. Of all such canisters that could be used, what is
the radius, in inches, of the one that has maximum
volume? | [
" 3",
" 4",
" 5",
" 6"
] | D | Since the radius is squared, the biggest volume cylinder will have the biggest possible radius. To achieve that, let the smaller dimension of the rectangular box be the height. So, it will have 14x12 width x length. The biggest cylinder must fit both dimensions, so it must have at most a diameter of 12, so the radius will be 6.
D |
AQUA-RAT | AQUA-RAT-33919 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8463 for a computer table, then what was the cost price of the computer table? | [
"6727",
"6825",
"6728",
"6725"
] | B | :
CP = SP * (100/(100 + profit%))
= 8463(100/124) = Rs. 6825.
Answer: B |
AQUA-RAT | AQUA-RAT-33920 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
A student was asked to find 4/5 of a number. But the student divided the number by 4/5, thus the student got 9 more than the correct answer. Find the number. | [
"16",
"18",
"20",
"22"
] | C | Let the number be x.
(5/4) * x = (4/5)*x + 9
25x = 16x + 180
9x = 180
x = 20
The answer is C. |
AQUA-RAT | AQUA-RAT-33921 | 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+6+4=010530
and 2+4+8=021040
and 3+5+9=021155
and 4+6+7=020954....
then 2+7+5=?????? | [
"051070",
"051770",
"051074",
"151070"
] | A | if 5+6+4=010530
and 2+4+8=021040
and 3+5+9=021155
and 4+6+7=020954....
then 2+7+5=??????
051070
ANSWER:A |
AQUA-RAT | AQUA-RAT-33922 | # Probability of reaching net 4 heads when tossing coin 8 times
This problem is #19 from the AMC 12 2016A, and goes as follows:
Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$
Even though the link above does provide a solution (which I understand), I tried to solve it in a different way. I came up with a different answer, and can’t figure out what’s wrong with my approach.
I tried counting all the ways in which you can reach 4 heads (net) at some point, which can be divided into 3 categories.
1. You win if you start with 4 heads, which means the last 4 throws don’t matter, so there are ${2}^{4}$ or 16 possibilities for the last 4 and thus there’s 16 possible solutions from this.
2. If you start with 5 heads and 1 tail, the last two throws don’t matter, which contributes 4 different solutions. That tail can come at the first, second, third, or fourth position (if it’s at the 5th or 6th spot, we would be overcounting because it would be covered by the #1), so we multiply by 4 and get 16 more solutions from this one.
3. If we start with 6 heads and 2 tails, we get that the two tails can be at the 1,2 | 1,3 | 1,4 | 1,5 | 2,3 | 2,4 | 2,5 | 3,4 | 3,5 | 4,5 positions, which is 10 (if either one is in the 6th or 7th position, we would be over counting), so this leads to 10 more solutions.
The following is multiple choice question (with options) to answer.
A coin is tossed four times. What is the probability that there is at the least one tail? | [
"15/16",
"15/37",
"15/67",
"15/17"
] | A | Let P(T) be the probability of getting least one tail when the coin is tossed four times.
= There is not even a single tail.
i.e. all the outcomes are heads.
= 1/16 ; P(T) = 1 - 1/16
= 15/16
Answer: A |
AQUA-RAT | AQUA-RAT-33923 | The game is "fair".
Playing repeatedly, you can expect to break even.
4. thanks all for the answers.
but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?
The following is multiple choice question (with options) to answer.
Carrie, Liz and Michael are playing cards. Together they have all won $400 total. Carrie and Liz have won $150 total. Michael and Liz have won $300 total. How much money as Carrie won? | [
"50",
"250",
"350",
"100"
] | D | Carrie + Liz + Michael = $400
C + L = $150
M + L = $300 <<Plug into original formula
C + (M + L) = 400
C + 300 = 400
Subtract 300 from each side
c = $100
Answer: D |
AQUA-RAT | AQUA-RAT-33924 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is: | [
"15",
"16",
"18",
"25"
] | B | Explanation:
Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.
S.P. of x articles = Rs. 20.
Profit = Rs. (20 - x).
(20-X/X X100=25)
2000 - 100x = 25x
125x=2000
x=16
ANSWER IS B |
AQUA-RAT | AQUA-RAT-33925 | (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.
In 1995, the Johnsons spent $800 on the family’s water bills. Anticipating that water rates would increase in 1996 by 50%, the Johnsons cut back their water usage. By how much must the Johnsons have reduced their 1996 water usage to pay exactly the same amount N in 1996 as they paid in 1995? | [
"33 1⁄3%",
"40%",
"50%",
"66 2⁄3%"
] | A | water rate increases by 50 % in 1996 means 150% of rate in 1995. to pay exactly the same amount N in 1996 as they paid in 1995, water use should decrease in the same ratio
150-> 100 means 1/3 rd= 33(1/3) %.A |
AQUA-RAT | AQUA-RAT-33926 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
_________________
Karishma
Veritas Prep GMAT Instructor
Senior SC Moderator
Joined: 22 May 2016
Posts: 3284
Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A mixture of 70 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12 1/2% water? | [
"2",
"8",
"4",
"5"
] | A | Quantity of milk in the mixture = 90/100 (70) = 63 litres.
After adding water, milk would form 87 1/2% of the mixture.
Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added.
ANSWER:A |
AQUA-RAT | AQUA-RAT-33927 | and the diagonal acts as a transv… The midsegment of a parallelogram is a rhombus as a quadrilateral are congruent a sided. Consecutive angles are supplementary ( a + D = 180° ) its.! S another proof — with a pair of _____ contained within the sides of closed! Cm ] \end { align } \triangle ACD\cong \triangle ABC opposite sides parallel congruent! Nearest whole square cm _____ contained within the sides of a quadrilateral and has two pairs of parallel.! Figure that is made up of simple shapes like triangles, parallelograms, 's!: in planner geometry, a quadrilateral are congruent, consecutive sides 60 120... Has four congruent _____ as the polygon has sides, Get access to this video our... Of parallelogram that has all sides congruent and the opposite sides are parallel as well as.. Simulation given below to better understand a parallelogram, z = z or =. Done for the other pair of opp in today 's lesson, will... If... What is a 4 sided polygon which is also a quadrilateral is a,. Angle of a quadrilateral that has the opposite sides are congruent and the opposite sides of parallelogram. Base angles of a parallelogram, then its opposite angles are also.! Time, the area of a trapezoid is each of its sides at any time, the sides... ] are opposite sides of a parallelogram congruent { align } opposite sides are _____ Theorem states! We can conclude that the opposite ones are parallel be calculated by adding or the... Divided into the same number of _____ angles are _____ Get another Theorem for parallelograms let. Is quadrilateral which has two pairs of opposite sides are parallel parallelogram are equal to itself opposite are! Property that its opposite sides are parallel too with two equal diagonals triangles, parallelograms, trapezoids and. Planner geometry, a four-sided shape that has opposite sides of a regular polygon to any one of sides! Strategy: how to prove this rule about the opposite sides are.... All angles are
The following is multiple choice question (with options) to answer.
The ratio of the adjacent angles of a parallelogram is 4 : 11. Also, the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12. What is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral? | [
"132°",
"228°",
"156°",
"224°"
] | A | The measures of the adjacent angles of a parallelogram add up to be 180°
Given
So, 4x + 11x = 180°
Or, 15x = 180°
Or, x = 12°
Hence the angles of the parallelogram are 48° and 132°
Further it is given
We know sum of all the four angles of a quadrilateral is 360°
So, 5y + 6y + 7y + 12 y = 360°
Or, 5y + 6y + 7y + 12 y = 360°
Or, 30y = 360°
Or, y = 12°
Hence the angles of the quadrilateral are 60° , 72, 84° and 144°
Will be 48° + 84° = 132°
Answer : A |
AQUA-RAT | AQUA-RAT-33928 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 400 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"11 sec",
"24 sec",
"77 sec",
"14 sec"
] | B | Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 400 * 3/50
= 24 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-33929 | I find the answer to be: $$c(n) = {{n+3}\choose{3}} - {{n+2}\choose{3}}$$ $$c(n) = \frac{(n+3)!}{3!n!} - \frac{(n+2)!}{3!(n-1)!}$$ $$c(n) = \frac{(n+3)(n+2)(n+1) - (n+2)(n+1)(n)}{6}$$ $$c(n) = \frac{(n+2)(n+1)(n+3 - n)}{6}$$ $$c(n) = \frac{3(n+2)(n+1)}{6}$$ $$c(n) = \frac{(n+2)(n+1)}{2}$$ Do I have that right?
Thanks,
Bob
• This is equivalent to $\frac12(n+1)(n+2)$ which agrees with the supplied values. Jul 12 '19 at 23:07
• Like this? Jul 12 '19 at 23:10
• en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) Jul 12 '19 at 23:16
Yes - this is correct. This can also be found by defining a helper function $$b(n)$$, which counts the number of ways to write $$n$$ as a sum of two numbers. Clearly, $$b(n) = n+1$$, as the first value, $$v$$, can be anything from $$0$$ to $$n$$, while the second value is $$n - v$$.
The following is multiple choice question (with options) to answer.
Let C be defined as the sum of all prime numbers between 0 and 35. What is C/3 | [
"155",
"129",
"64.3",
"47"
] | C | Prime numbers between 0 and 30 - 2 , 3 , 5 , 7, 11 , 13 , 17 , 19 , 23 , 29, 31, 33
Sum,C = 193
C/3 = 64.3
Answer C |
AQUA-RAT | AQUA-RAT-33930 | $\text{ }$
$\displaystyle (9) \ \ \ \ \ E(X \wedge u)=\int_{-\infty}^u x \ f_X(x) \ dx+u \ S_X(u)$
$\displaystyle (10) \ \ \ \ \ E(X \wedge u)=\biggl(\sum \limits_{x < u} x \ P(X=x)\biggr)+u \ S_X(u)$
$\text{ }$
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.
A family pays $800 per year for an insurance plan that pays 85 percent of the first $1,000 in expenses and 100 percent of all medical expenses thereafter. In any given year, the total amount paid by the family will equal the amount paid by the plan when the family's medical expenses total. | [
"$1,100",
"$1,200",
"$1,400",
"$1,800"
] | A | Upfront payment for insurance plan = 800$
Family needs to pay 15 % of first 1000 $ in expense = 150$
Total amount paid by family when medical expenses are equal to or greater than 1000 $ = 800 + 150 = 950 $
Total amount paid by insurance plan for first 1000 $ = 800 $
Total amount paid by family will equal amount paid by plan when medical expense = 1100 $
(Since insurance plan will pay 100% of amount that exceeds 950$ )
Answer A |
AQUA-RAT | AQUA-RAT-33931 | 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 280 m long can cross an electric pole in 20 sec and then find the speed of the train? | [
"88 Kmph",
"89 Kmph",
"72 Kmph",
"50 Kmph"
] | D | Length = Speed * time
Speed = L/T
S = 280/20
S = 14 M/Sec
Speed= 14*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 50 Kmph
Answer: D |
AQUA-RAT | AQUA-RAT-33932 | 5,224 views
In a certain town, the probability that it will rain in the afternoon is known to be $0.6$. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to $25°C$, the probability that it will rain in the afternoon is $0.4$. The temperature at noon is equally likely to be above $25°C$, or at/below $25°C$. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above $25°C$?
1. $0.4$
2. $0.6$
3. $0.8$
4. $0.9$
Answer is C) $0.8$
$P$(rain in afternoon) $= 0.5\times P($rain when temp $\leq 25) + 0.5 \times P($ rain when temp $> 25 )$
$0.6 = 0.5\times 0.4 + 0.5\times P($ rain when temp $> 25 )$
so,
$P$( rain when temp $> 25$ ) $= 0.8$
This is a question of Total Probability where after happening on one event E1, the probability of another event E2 happening or not happening is added together to get the probability of happening of Event E2.
Given P(Rain in noon) =0.6 (This is total probability given).
"The temperature at noon is equally likely to be above 25°C, or at/below 25°C."
means P(Temp less than or 25) = P(Temp >25) =0.5
P(Rain in noon) = P(Temp $\leq$ 25) * P(Rain | Temp $\leq$ 25) + P(Temp $>$ 25) * P(Rain| Temp $>$ 25)
0.6= (0.5*0.4) + (0.5*X)
X=0.8 Ans (C)
The following is multiple choice question (with options) to answer.
The probability of rain showers in Barcelona on any given day is 0.5. What is the probability that it will rain on exactly one out of three straight days in Barcelona? | [
"0.144",
"0.072",
"0.375",
"0.72"
] | C | IMO This is just for first day! It can rain on 2nd day or 3 rd day (and not rain on other days) in 3 consecutive days!
Hence,
0.5*0.5*0.5 + 0.5*0.5*0.5 + 0.5*0.5*0.5 = 0.375
Option - C |
AQUA-RAT | AQUA-RAT-33933 | 2. Mike says:
of course i meant 5 buckets, one for each student ;).
3. meichenl says:
Let there be $c$ cookies and $n$ students. Denote the number of ways to distribute $c$ cookies between $n$ students by $S_n(c)$. This is claiming that for a given number of students, the number of ways to distribute the cookies is a function of the number of cookies.
The last student can receive anywhere from $0$ to $c$ cookies. Suppose he receives $k$. Then there are $c-k$ cookies to distribute between $n-1$ students. This can be done $S_{n-1}(c-k)$ ways. Summing over all possible values of $k$ we obtain
$S_n(c) = \sum_{k=0}^c S_{n-1}(c-k)$.
We know
$S_1(c) = 1$
because if there is one student, he must get all the cookies (AWESOME!).
Then
$S_2(c) = \sum_{k=0}^c 1 = 1+c$
$S_3(c) = \sum_{k=0}^c c-k+1 = \frac{1}{2}c(3+c)$
$S_4(c) = \sum_{k=0}^c (c-k)(c-k+3)/2 = \frac{1}{6} c(1+c)(5+c)$
\begin{align*} S_5(c) &= \sum_{k=0}^c \frac{1}{6} (c-k)(1+c-k)(5+c-k) \\ &= \frac{1}{24} c(1+c)(2+c)(7+c) \end{align*}
$S_5(10) = 935$.
So the answer is 935. For larger $n$ I conjecture,
The following is multiple choice question (with options) to answer.
Ram has 490 chocolates. 70 members work in his office. He would like to give each person the same amount of chocolates, how much should he give to each person? | [
"7",
"12",
"9",
"15"
] | A | 490/70 = 7
The answer is A. |
AQUA-RAT | AQUA-RAT-33934 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
A sum of money becomes double itself in 8 years at simple interest. How many times will it become 10 years at the same rate? | [
"2 1/4",
"2 1/9",
"2 1/2",
"2 1/6"
] | A | P ---- 2P ---- 8 years
2 1/4 P ---- 10 years
Answer: A |
AQUA-RAT | AQUA-RAT-33935 | # Probability a 9-digit number has the digits 2,4, and 6 next to each other.
The integers $1,2,3,....,9$ are arraned (at random) in a row, resulting in a $9$-digit integer (without replacement). What is the probability that:
The result is even? $\frac49$ or $\frac{4(8!)}{9!}$
The result is divisible by $5$? The number must end in $0$ or $5$. Edit: As Andre pointed out we have no $0$. So, $\frac19$. or $\frac{(8!)}{9!}$
The digits $2, 4,$ and $6$ are next to each other (in any order)? The above two I have confidence in, it is this last one I'm a little confused on.
$9\choose 3$ ways to position $2,4,6$ in the 9-digit number and $3!$ ways they can be arranged. This doesn't appear to be right as after you divide by $9!$ you get $.13$% which seems unreasonably low. So, I thought to multiple by $6!$ to account for the number of ways the other $6$ numbers can be arranged. This gives you:
$9\choose 3$$3!6!/9!$
which equals $1$, and obviously isn't right. Any suggestions as to where I went wrong would be great.
-
You're right! Silly me. – Vincent Jun 16 '14 at 22:47
For $2,4,6$ next to each other, imagine choosing the $3$ spots that will be reserved for our three favoured guests. This can be done in $\binom{9}{3}$ equally likely ways.
The three spots can be in a row in $7$ ways, for the leftmost of the spots can be in any of the positions $1$ to $7$. Thus the required probability is $\frac{7}{\binom{9}{3}}$. This simplifies nicely.
The following is multiple choice question (with options) to answer.
If a computer program generates three-digit odd numbers using the numbers 9, 7, 0, 2, and 1, and the digits can be repeated, what is the probability that the number formed from the program is a multiple of three? | [
"1/3",
"1/4",
"1/5",
"1/6"
] | A | A |
AQUA-RAT | AQUA-RAT-33936 | • One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37
The following is multiple choice question (with options) to answer.
If z≠y and z·y≠0, which of the following may be true? | [
"z=y",
"(z+y)(z-y)=0",
"z.y=0",
"z^2=y^2"
] | B | except B all conditions are not as per question
B |
AQUA-RAT | AQUA-RAT-33937 | # 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.
Akshay buys a motorcycle for Rs.25000. If he decides to sell the motorcycle for a profit of 7%, find the selling price. | [
"26750",
"26759",
"26757",
"26756"
] | A | Explanation:
Profit = %profit/100 * cost price
P = 7/100 * 25000 = 1750.
Selling price = cost price + profit
= 25000 + 1750
= Rs.26750
ANSWER:A |
AQUA-RAT | AQUA-RAT-33938 | 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.
If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours
day, how much coal will be required for 8 engines, each running 13hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type? | [
"23 metric tonnes",
"24 metric tonnes",
"25 metric tonnes",
"26 metric tonnes"
] | D | We shall compare each item with the quantity of coal.
Less engines, less coal consumed (direct)
More working hours, more coal consumed (direct)
If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.
If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.
Less rate of consumption, less coal consumed (direct).
Therefore, number of engines 9:8 :: 24:x
Working hours 8:13 :: 24:x
Rate of consumption 1/3:1/4 :: 24:x.
9*8*1/3*x= 8*13*1/4*24 or x= 26.
Therefore, required consumption of coal 26 metric tonnes.
ANSWER:D |
AQUA-RAT | AQUA-RAT-33939 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
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Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
There are two pipes A and B.If A filled 10 liters in hour B can fills 20 liters in same time.likewise B can fill 10,20,40,80,160.... if B filled in (1/16)th of a tank in 3 hours.how much time will it take to fill completely? | [
"7 hours",
"6 hours",
"5 hours",
"8 hours"
] | A | 1/16 in 3 hours
1/8 in 4 hours,as here rate of water filling is doubled
1/4 in 5 hours
1/2 in 6 hours
1 in 7 hours
ans will be 7 hours
ANSWER:A |
AQUA-RAT | AQUA-RAT-33940 | 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 runs 5/3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach it at the same time? | [
"200 m",
"300 m",
"270 m",
"160 m"
] | A | Ratio of the speeds of A and B = 5 : 1 = 5 : 3.
3
Thus, in race of 5 m, A gains 2 m over B.
2 m are gained by A in a race of 5 m.
80 m will be gained by A in race of 5 x 80 m = 200 m.
2
Winning post is 200 m away from the starting point.
ANSWER:A |
AQUA-RAT | AQUA-RAT-33941 | =>V=8/27(4πr^3−8/3πr^3)
=>V=8/27(4/3πr^3)
=>V=8/27(volume of sphere)
#### Solution 2
A sphere of fixed radius (r) is given.
Let R and h be the radius and the height of the cone, respectively.
The volume (V) of the cone is given by,
V=1/3piR^2h
Now, from the right triangle BCD, we have:
BC=sqrt(r^2-R^2)
:.h=r+sqrt(r^2-R^2)
V=1/3piR^2(r+sqrt(r^2-R^2))=1/3piR^2r+1/3piR^2sqrt(r^2-R^2)
(dV)/(dR)=2/3piRr+2/3piRsqrt(r^2-R^2)+(piR^2)/3 (-2R)/(2sqrt(r^2-R^2))
=2/3piRr+2/3piRsqrt(r^2-R^2)-(piR^3)/(3sqrt(r^2-R^2))
=2/3piRr+(2piR(r^2-R^2)-piR^3)/(3sqrt(r^2-R^2))
2/3piRr+(2piRr^2-3piR^3)/(3sqrt(r^2-R^2))
Now
(dV)/(dR^2)=0
=>(2pirR)/3=(3piR^3-2piRr^2)/(3sqrt(r^2-R^2))
=>2rsqrt(r^2-R^2)=3R^2-2r^2
=>4r^2(r^2-R^2)=(3R^2-2r^2)^2
The following is multiple choice question (with options) to answer.
A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is: | [
"12",
"66",
"73",
"72"
] | A | Answer: A) 12 pi cub.cm |
AQUA-RAT | AQUA-RAT-33942 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
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04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A and B are two partially filled buckets of water. If 3 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 5 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water? | [
"11",
"13",
"39/5",
"21"
] | C | Let A contains A, B contains B liters
so, (A-3)/(B+3)= 1/3.......(1)
again, (B-5)/(A+5)= 1/2..........(2)
from (1)(2) we find A= 39/5
ans: (C) |
AQUA-RAT | AQUA-RAT-33943 | Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2759
Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink]
### Show Tags
08 Feb 2018, 16:38
1
Barkatis wrote:
It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages?
A. 5
B. 6
C. 8
D. 10
E. 12
We can let Tom’s rate = 20/x and Jack’s rate = 20/(x+2), and their combined rate is 25/3; thus:
20/x + 20/(x+2) = 25/3
Multiplying by 3x(x+2), we have:
20(3)(x + 2) + 20(3x) = 25(x)(x + 2)
60x + 120 + 60x = 25x^2 + 50x
25^x - 70x - 120 = 0
5x^2 - 14x - 24 = 0
(5x + 6)(x - 4) = 0
x = -5/6 or x = 4
Since x can’t be negative, we see that x must be 4, so Jack's rate is 20/6 = 10/3.
So it takes him 40/(10/3) = 120/10 = 12 hours to type 40 pages.
Alternate Solution:
Let’s pick a common multiple of 20, 25 and 40, such as 200, and calculate the number of hours to type that number of pages for each scenario.
Since Jack takes 2 more hours than Tom to type 20 pages, it will take Jack 20 more hours than Tom to type 200 pages.
Since Jack and Tom working together can type 25 pages in 3 hours, they can type 200 pages in 12 hours.
The following is multiple choice question (with options) to answer.
It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages? | [
"12",
"18",
"20",
"24"
] | D | We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins)
We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages.
Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins
Now, make your sum of rates equation:
1/a + 1/(a - 8) = 5/48
Now look at the options and substitute here. First check out the straight forward options.
Say, a = 12
1/12 + 1/4 = 4/12 Nope
I will not try 18 and 20 because (18, 10) and (20, 12) doesn't give me 48, the denominator on right hand side.
I will try 24 instead.
1/24 + 1/16 = 5/48 Yes.
Answer is 24.
ANSWER:D |
AQUA-RAT | AQUA-RAT-33944 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How long does a train 110 m long running at the speed of 72 km/hr takes to cross a bridge 136 m length? | [
"82.1 sec.",
"12.3 sec",
"19.1 sec.",
"17.1 sec."
] | B | Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 136 = 246 m.
Required time = 246/20 = 12.3 sec.
Answer: B |
AQUA-RAT | AQUA-RAT-33945 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
The T train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the T train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m? | [
"1.8m",
"6m",
"7m",
"9m"
] | B | The reason it is failing for you is that you chose incorrect numbers. If the question says T it took 7 hrs to reach from A to B and 5 hrs to reach from B to C at a constant speed. It shows that distance AB and BC should be in ratio of 7/5.
If you take such numbers you can solve problem.
AB = 7, BC=5
Therefore
AB-BC = 2
But from question, AB-BC =m
=> m=2
Now total distance = AB+BC= 12
Substitute 12 to get answer in terms of m
Total distance =12 =6m
Ans B |
AQUA-RAT | AQUA-RAT-33946 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B can do a piece of work in 12 days and 16 days respectively. Both work for 3 days and then A goes away. Find how long will B take to complete the remaining work? | [
"10 days",
"17 days",
"147 days",
"9 days"
] | D | 3/12 + (3 + x)/16 = 1
x = 9 days
Answer: D |
AQUA-RAT | AQUA-RAT-33947 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
The ratio of boys to girls is 6:4.60% of the boys and 40% of girls take lunch in the canteen.What % of class takes lunch? | [
"51%",
"52%",
"53%",
"54%"
] | B | let total no of students 100, boy 60 and girl 40, so lunch taken by boy and girl respectively 60* 60/100=36 and 40* 40/100=16
so total 52 students takes lunch out of 100 students..so required % is 52%
ANSWER:B |
AQUA-RAT | AQUA-RAT-33948 | $24 = 1^3 + 2^3 - 3^3 - 4^3 + 5^3 + 6^3 - 7^3 + 8^3 + 9^3 + 10^3 - 11^3 + 12^3 + 13^3 + 14^3 - 15^3 - 16^3$ (at least 7 other ways)
Each of the above representations is the shortest possible. The algorithm created a sequence of sets $S_n$ with $S_0 = \{0\}$ and $S_n = \{x+n^3, x-n^3 : x \in S_{n-1}\}$ for $n > 0$, so that $S_1 = \{1, -1\}$, $S_2 = \{9, -7, 7, -9\}$, etc. The algorithm also kept track of the sequence of signs used to arrive at each particular number.
• Nice work ! :D Thank's !!!
– r9m
Dec 24 '14 at 1:11
• @Unit: Do you suppose you can do the same for at least $1<N<100$ for $5$th powers? Kindly refer to my answer below. Dec 29 '14 at 3:54
• @TitoPiezasIII: Ahh, I just read this message. I will try, though it seems alexwlchan is almost there.
– Unit
Dec 29 '14 at 22:55
• @Unit: And it seems Zander even managed $0\leq N<10000$. The ingenuity of some people never ceases to amaze. Dec 29 '14 at 23:44
For 5th powers, there is an identity,
$$\sum\limits_{k=1}^{168}\pm(x+k)^5 = 480$$
analogous to the 3d powers mentioned by R. Millikan in his answer.
What remains to be shown is that all $0\leq N<240$ can be decomposed into sums of fifth powers. See this post.
(P.S. The number of addends can be reduced to just $m=168$.)
The following is multiple choice question (with options) to answer.
What is the smallest possible common multiple of three integers, all larger than 24? | [
"27.",
"54.",
"25",
"19,656."
] | C | The Least common positive multiple would be for the immediate least number(s) after 24 i.e 25.
Since nowhere is it stated that the three integers must be different, the integers could be 25,25 and 25... leading to least possible positive common multiple to be 25. Answer C. |
AQUA-RAT | AQUA-RAT-33949 | # Runner's High (Speed)
I find the following mind-boggling.
Suppose that runner $$R_1$$ runs distance $$[0,d_1]$$ with average speed $$v_1$$. Runner $$R_2$$ runs $$[0,d_2]$$ with $$d_2>d_1$$ and with average speed $$v_2 > v_1$$. I would have thought that by some application of the intermediate value theorem we can find a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$. This is not necessarily so!
Question. What is the smallest value of $$C\in\mathbb{R}$$ with $$C>1$$ and the following property?
Whenever $$d_2>d_1$$, and $$R_2$$ runs $$[0,d_2]$$ with average speed $$Cv_1$$, then there is a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$.
The following is multiple choice question (with options) to answer.
An athlete runs 200 meters race in 48sec. His speed is? | [
"20km/hr",
"15km/hr",
"30km/hr",
"25km/hr"
] | B | speed = 200/48 = 25/6 m/sec
= 25/6 * 18/5 km/hr = 15km/hr
Answer is B |
AQUA-RAT | AQUA-RAT-33950 | Kudos [?]: 132821 [4], given: 12378
Re: A small, rectangular park has a perimeter of 560 feet and a diagonal m [#permalink] 13 Nov 2014, 13:09
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The following is multiple choice question (with options) to answer.
If the perimeter of a rectangular garden is 600 m, its length when its breadth is 100 m is? | [
"338 m",
"778 m",
"200 m",
"276 m"
] | C | 2(l + 100)
= 600 => l
= 200 m
Answer:C |
AQUA-RAT | AQUA-RAT-33951 | 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.
The average of 11 results is 50, if the average of first six results is 49 and that of the last six is 52. Find the sixth result? | [
"58",
"56",
"51",
"52"
] | B | 1 to 11 = 11 * 50 = 550
1 to 6 = 6 * 49 = 294
6 to 11 = 6 * 52 = 312
6th = 294 + 312 – 550
= 56
Answer: B |
AQUA-RAT | AQUA-RAT-33952 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many different number of ways 6 men and 3 women can sit on a shopa which can accommodate persons? | [
"200",
"230",
"504",
"250"
] | C | 9p3 = 9 × 8 × 7 = 504
C |
AQUA-RAT | AQUA-RAT-33953 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a man standing on the platform. If the train is 210 meters long and its speed is 72 kmph, how much time it took in doing so? | [
"10 ½ sec",
"4 ½ sec",
"9 ½ sec",
"8 ½ sec"
] | A | D = 210
S = 72 * 5/18 = 20 mps
T = 210/20 = 10 ½ sec
Answer: A |
AQUA-RAT | AQUA-RAT-33954 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A cricketer scored 134 runs which included 12 boundaries and 2 sixes. What percent of his total score did he make by running between the wickets. | [
"55.22%",
"54.54 %",
"63%",
"70%"
] | A | Explanation :
Number of runs made by running = 134 - (12 x 4 + 2 x 6)
= 134 - (60)
= 74
Now, we need to calculate 72 is what percent of 134.
=> 74/134 * 100 = 55.22 %
Answer : A |
AQUA-RAT | AQUA-RAT-33955 | in the comments section. At what speed does the second train B travel if the first train travels at 120 km/h. You can approach this as if you were solving for an unknown in math class or you can use the speed triangle. Time, Speed and Distance: Key Learnings. ) Since the distances are the same, I set the distance expressions equal to get: Unit of speed will be calculated based on unit of distance and time. Distance, Rate and Time content standard reference: grade 6 algebra and functions 2. aspxCollins Aerospace ARINCDirect maintains a multitude of data on airports and airways around the world. My other lessons on Travel and Distance problems in this site are - (Since the speed through the steel is faster, then that travel-time must be shorter. Initial speed of the car = 50km/hr Due to engine problem, speed is reduced to 10km for every 2 hours(i. 5th Grade Numbers Page 5th Grade Math Problems As with the speed method of calculation, the denominator must fit into 60 minutes. In National 4 Maths use the distance, speed and time equation to calculate distance, speed and time by using corresponding units. Distance divided by rate was equal to time. Pete is driving down 7th street. Speed, Distance, Time Worksheet. This Speed Problems Worksheet is suitable for 4th - 6th Grade. If the speed of the jeep is 5km/hr, then it takes 3 hrs to cover the same For distance word problems, it is important to remember the formula for speed: Definition: Speed = Distance/Time. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. Again, if you look at the formula triangle, you can see that you get distance by multiplying speed by time. Next time you are out walking, imagine you are still and it is the world that moves under your feet. Q) Mr. Distance is directly proportional to Velocity when time is constant. The problem gives the distance in feet and the speed in miles per hour. The detailed explanation will help us to understand how to solve the word problems on speed distance time. Average Speed = Total distance ÷ Total time = 110 ÷ 5/6 = 110 × 6/5 = 132 km/h. 6T + 4T = 20 km. The result will be the average speed per unit of time, usually an hour. We will practice different Distance, Speed and
The following is multiple choice question (with options) to answer.
Points A, B, and C lie, in that order, on a straight railroad track. The distance from point A to point B is twice the distance from point B to point C. A train traveled from point A to point C without stopping. The train's average speed when traveling from point A to point B was 200 miles per hour and the train's average speed when traveling from point B to point C was 150 miles per hour. What was the train's average speed, in miles per hour, when traveling from point A to point C? | [
"150",
"200",
"180",
"190"
] | C | Average speed = distance/time
Because we are looking for average speed we can pick a distance for the variable d.
Speed A-B = 200
Speed B-C = 150
Average Speed = total distance/total rate
rate = distance/time
A====================B==========C
If A-B is twice the length of B-C then let A-B = 2d and let B-C = d
Average speed = 3d/(2d/200) + (d/150)
3d/(6d/600) + (4d/600)
3d/(10d/600)
1800d/10d
Average speed = 180
ANSWER: C |
AQUA-RAT | AQUA-RAT-33956 | 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.
72,90, 80,85,90,105,105,130,130,130
The list consists of the times in seconds that it took each of the 10 school children to run a distance of 400 mts . If the standard deviation of the 10 running times is 22.4, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times. | [
"1",
"2",
"3",
"4"
] | A | How many of the 10 running times are more than one SD below the meanmeans how many data points from given 10 are less thanmean-1SD.
We are given that SD=22.4, so we should find mean --> mean=100 --> there are only 1 data points below 100-22.4=77.6, namely 72 .
Answer: A. |
AQUA-RAT | AQUA-RAT-33957 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 300 m race A beats B by 22.5 m or 6 seconds. B's time over the course is: | [
"80 sec",
"86 sec",
"76 sec",
"49 sec"
] | A | B runs 300 m in(6*2/45*300)sec=80 sec
ANSWER A |
AQUA-RAT | AQUA-RAT-33958 | $$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+2^5+2^6=？ | [
"(2^3-1)(2^3+1)",
"2^6+1",
"2^7-1",
"2^5+1"
] | C | From 1+2+2^2+2^3+2^4+2^5+2^6=1(2^7-1)/(2-1)=2^7-1, the correct answer is C. |
AQUA-RAT | AQUA-RAT-33959 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
Mr. Babu ison tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days. he must cut down his daily expenses by Rs. 3.For how many days is Mr. babu on tour? | [
"20",
"30",
"40",
"90"
] | A | Suppose Mr. babu is on tour for x days . Then, 360/x-360/x+4
=1/x-1/x+4=1/120
x(x+4)=4x120=480
(x+24)(x-20)=0
x=20
ANSWER: A |
AQUA-RAT | AQUA-RAT-33960 | 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$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 36 sec and a man standing on the platform in 20 sec. If the speed of the train is 54 km/hr. What is the length of the platform? | [
"615 m",
"240 m",
"168 m",
"197 m"
] | B | Speed = 54 * 5/18 = 15 m/sec.
Length of the train = 15 * 20 = 300 m.
Let the length of the platform be x m . Then,
(x + 300)/36 = 15 => x
= 240 m
Answer: B |
AQUA-RAT | AQUA-RAT-33961 | ### Show Tags
05 Sep 2017, 11:01
2
In traveling from city A to city B, John drove for 1 hour at 50 mph and for 3 hoursat 60 mph. What was his average speed for the whole trip?
(A) 50
(B) 53.5
(C) 55
(D) 56
(E) 57.5
The total distance traveled from City A to City B is 50*1 + 60*3 = 230
Since John drove for 4 hours to travel this distance,
the average speed of the whole trip is $$\frac{Distance}{Time} = \frac{230}{4}$$ = 57.5(Option E)
_________________
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Joined: 15 Feb 2018
Posts: 255
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink]
### Show Tags
17 Jan 2019, 03:38
How does this work using the harmonic mean?
$$\frac{1/50+3/60}{4}$$=$$\frac{7/100}{4}$$=57.1
Re: In traveling from city A to city B, John drove for 1 hour at 50 mph an [#permalink] 17 Jan 2019, 03:38
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The following is multiple choice question (with options) to answer.
Jonah drove the first half of a 200-mile trip in x hours and the second half in y hours. Which of the following is equal to Jonah’s average speed, in miles per hour, for the entire trip? | [
"100/(x + y)",
"200/(x + y)",
"25/x + 25/y",
"50/x + 50/y"
] | B | Time taken by Jonah to cover first half i.e 100 miles of a 200 mile trip = x
time taken by Jonah to cover the second half i.e 100 miles of a 200 mile trip = y
Total time taken by Jonah = x+y
Jonah's average speed for the entire trip =total distance /total time
=200/(x+y)
Answer B |
AQUA-RAT | AQUA-RAT-33962 | Thanks
• Your "40% overlap" would mean taht the surface overlaping is 40% of the surface of which rectangle ? – Evargalo Sep 28 '17 at 16:13
• I want to know if, say, 40% of the other rectangle is contained inside the self rectangle – John Lexus Sep 28 '17 at 16:14
For the equations, I will let the left of the first rectangle be $l_0$, the right be $r_0$, the top $t_0$ and the bottom $b_0$. The second rectangle is $l_1, r_1,$ etc. Their areas will be $A_0$ and $A_1$.
If the boxes don't overlap, obviously the percentage overlap is $0$.
If your boxes are found to be colliding, simply use this formula to calculate the area that is overlapping:
$$A_{overlap} = (\max(l_0, l_1)-\min(r_0, r_1))\cdot(\max(t_0, t_1)-\min(b_0, b_1)).$$
Now there are two ways to calculate a percentage error that could make sense for your explanation. If we're just checking the percentage within the first rectangle (or "self" in your program), the percent overlap is simple: $$P_{overlap} = \frac{A_{overlap}}{A_{self}}.$$
If you want the percentage to be equal whether it's calculated from either rectangle, the equation you're looking for is: $$\frac{A_{overlap}}{A_0+A_1-A_{overlap}}.$$
The following is multiple choice question (with options) to answer.
If the length & width of a rectangle is decreased by 50% & then again increase the new length & width by 40%. Calculate the percentage difference between the area of New Rectangle & Original Rectangle. | [
"50%",
"51%",
"52%",
"53%"
] | B | let length is l, width is w so area is wl
after decreament l is i/2 ,w is w/2
again inc by 40% so l is 7l/10 ,w is 7w/10 so area is 49wl/100
percentage change is wl-49wl/100=51wl/100
so 51%
ANSWER:B |
AQUA-RAT | AQUA-RAT-33963 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
40% of a number is more than 20% of 650 by 190. Find the number? | [
"288",
"278",
"800",
"267"
] | C | (40/100) * X – (20/100) * 650 = 190
2/5 X = 320
X = 800
Answer:C |
AQUA-RAT | AQUA-RAT-33964 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
Sheila works 8 hours per day on Monday, Wednesday and Friday, and 6 hours per day on Tuesday and Thursday. She does not work on Saturday and Sunday. She earns $360 per week. How much does she earn in dollars per hour? | [
"2",
"8",
"9",
"10"
] | D | Explanation:
Total hours worked = 8 x 3 + 6 x 2 = 36
Total earned = 360. Hourly wage = 360 / 36 = 10
Answer: D |
AQUA-RAT | AQUA-RAT-33965 | A. $28.50 B.$27.00
C. $19.00 D.$18.50
E. $18.00 We can compute a weighted average to solve. Let’s assume that 2 units of Q and 1 unit of P were produced last year. So the total revenue is 2 x 17 + 1 x 20 =$54, and thus the average revenue per unit is thus 54/3 = $18. Answer: E _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Retired Moderator Joined: 21 Aug 2013 Posts: 1386 Location: India Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink] ### Show Tags 02 Jul 2017, 20:49 1 Ratio of quantity of P and Q sold, P:Q = 1:2. Thus, average revenue per unit = (20*P + 17*Q)/(P+Q) = (20*1 + 17*2)/(2 + 1) = 54/3 =$ 18
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Joined: 28 Aug 2016
Posts: 11
Re: A manufacturer makes and sells 2 products, P and Q. The revenue from t [#permalink]
### Show Tags
18 Apr 2019, 14:30
1
1
AbdurRakib wrote:
A manufacturer makes and sells 2 products, P and Q. The revenue from the sale of each unit of P is $20.00 and the revenue from the sale of each unit of Q is$17.00. Last year the manufacturer sold twice as many units of Q as P. What was the manufacturer’s average (arithmetic mean) revenue per unit sold of these 2 products last year?
The following is multiple choice question (with options) to answer.
The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is: | [
"4000",
"5000",
"6000",
"7000"
] | A | Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 …. (i)
Q + R = (6250 x 2) = 12500 …. (ii)
P + R = (5200 x 2) = 10400 …. (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)
Subtracting (ii) from (iv), we get P = 4000.
P’s monthly income = Rs. 4000.
Answer A |
AQUA-RAT | AQUA-RAT-33966 | • 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.
A shipment of 8 television sets contains 2 black-and-white sets and 5 color sets. If 2 television sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black-and-white set? | [
"1/7",
"1/4",
"9/14",
"11/28"
] | C | I get 9/14, which is not in the answer choices.
the probability of getting at least one b/w TV is 1-(the probability of getting 2 color TVs)
1-C(5,2)/C(8/2)=18/28=9/14
C |
AQUA-RAT | AQUA-RAT-33967 | # 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.
Raj invested an amount of Rs.17400 for two years. Find the rate of compound interest that will fetch him an amount of Rs.1783.50 at the end of two years? | [
"9",
"8",
"7",
"5"
] | D | Let the rate of interest be R% p.a.
17400[1 + R/100]2 = 17400 + 1783.50
[1 + R/100]2 = (17400 + 1783.50)/17400
= 1 + 1025/10000 = 1 + 41/400
= 441/400 = [21/20]2
[1 + R/100] = 21/20
R/100 = 1/20
Therefore R = 5
Answer:D |
AQUA-RAT | AQUA-RAT-33968 | Now if we have $0$ as the first digit, the rest can be filled in $(5)(8)(7)$ ways.
So our answer is $(5)(9)(8)(7)-(5)(8)(7)=2240$.
For the evens, the same analysis gives $(5)(9)(8)(7)-(4)(8)(7)$.
A good way to approach this kind of problem is to write down a step-by-step procedure for choosing the kind of arrangement you are asked for. For (a) this could be
(1) Choose the last digit of your number.
(2) Choose the first digit.
(3) Choose the second digit.
(4) Choose the third digit.
Then work out the number of ways of accomplishing each step. How many ways are there to choose the last digit? - hint: the number as a whole has to be odd. How many ways for step (2)? - hint: there is one digit it can't ever be, and also it has to be different from the last digit. And so on.
Once you have these four answers, how do you combine them to get the total answer? - hint: you have to do step (1) AND step (2) AND...
Good luck!
In the first case you have:
5 choises for the last position;
8 for the first
8 for the second
and 7 for the third.
then the number of numbers is 5.8.8.7=...
since the right most digit (units place) is restricted (odd or even) start there.
So we have Place: units, number of choices = 5
Place: tens, number of choices = 10-1 (you can't use the digit already used in the units place)
Place: hundreds, number of choices = 10-2 ( you have used two digits already) Place: thousands, number of choices=10-3. Assume that leading digit can be a zero
So the number of choices is = \$7\times 8\times9\times5 = 2526
The following is multiple choice question (with options) to answer.
In a two digit number, the digit in the units place is six times the digit in ten's place and sum of the digits is equal to 7. What is the number? | [
"16",
"28",
"32",
"36"
] | A | Solution
Let the ten digits be x. Then,units digit=6x
Therefore x + 6x = 7
‹=›7x = 7
‹=›x= 1.
So, ten's digit = 1, unit's digit = 6.
Hence, the required number is 16.
Answer A |
AQUA-RAT | AQUA-RAT-33969 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
The Pinedale bus line travels at an average speed of 60 km/h, and has stops every 5 minutes along its route. Yahya wants to go from his house to the Pinedale Mall, which is 10 stops away. How far away, in kilometers, is Pinedale Mall away from Yahya's house? | [
"20 km",
"30 km",
"40 km",
"50 km"
] | D | Number of stops in an hour: 60/5 = 12
Distance between stops: 60/12 = 5 km
Distance between Yahya's house and Pinedale Mall: 5 x 10 = 50 km
IMO, correct answer is "D." |
AQUA-RAT | AQUA-RAT-33970 | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
The following is multiple choice question (with options) to answer.
On a test consisting of 30 questions. No. of wrong answers is 50% less than number of right answers. Each answer was either right or wrong. What is the ratio of right answers to wrong answers? | [
"1 : 3",
"1 : 5",
"2 : 1",
"3 : 1"
] | C | Sol. Let the number of wrong answer - x and number of right answer = 30-x x = 50/100 of (30-x) x = 10
Required ratio = 20 : 10 = 2 : 1
C |
AQUA-RAT | AQUA-RAT-33971 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
If each edge of cube increased by 10%, the percentage increase in | [
"21",
"22",
"25",
"19"
] | A | 100 × (110)/100 × (110)/100 = 121 => 21%
Answer is A. |
AQUA-RAT | AQUA-RAT-33972 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
12, 14, x, 10
The median of the list of positive integers above is 11. Which of the following could be the average (arithmetic mean) of the list? | [
"3",
"5.5",
"11",
"7"
] | C | GivenThe median of the list of positive integers above is 11
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 11
so the sequence will be x 10 12 14
Lets say x = 8, which gives us mean = sum/4 = 11
C is correct |
AQUA-RAT | AQUA-RAT-33973 | # 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.
Two cars are 15 kms apart. One is turning at a speed of 50kmph and the other at 40kmph . How much time will it take for the two cars to meet? | [
"12min",
"10min",
"14min",
"15min"
] | B | if they are moving in same direction then solution will be
(50-40)*t=15
ans=1.5hours
if they are moving in opposite direction then ans
(50+40)*t=15
ans=1/6 hours=10min
ANSWER:B |
AQUA-RAT | AQUA-RAT-33974 | Originally Posted by Archie
Because the question talks about all possible pairs of integers, not just 2 and 3.
I can see that every word in word problems is important.
8. ## Re: Positive Integers x & y
Originally Posted by Plato
To harpazo, I cannot understand how this can be so mysterious.
Learn this:
1. The sum of two even integers is even
2. The sum of two odd integers is even.
3. The sum of an even integer & an odd integer is odd.
4. If $n$ is an odd integer then $n-1$ is even.
5. If $n$ is an even integer then $n-1$ is odd.
If you learn these then practice applying them to this question,
Good information.
The following is multiple choice question (with options) to answer.
When positive integer m is divided by positive integer x, the reminder is 7 and when positive integer n is divided by positive integer y, the reminder is 11. Which of the following W is a possible value for x + y?
I. 18
II. 19
III. 20 | [
"I only",
"II only",
"W=III only",
"II and III only"
] | C | TIP FOR SUCH QUESTIONS:Make Algebraic Equation in the beginning to understand how to proceed further. Then Start thinking the possible values of variables asked about
i.e.When positive integer m is divided by positive integer x, the reminder is 7can be understood as
m = ax +7which means thevalue of x must be greater than 7as the remainder 7 is possible only when the Divisor is bigger than the Remainder 7
i.e. Min Value of x = 8
AND
i.e.When positive integer n is divided by positive integer y, the reminder is 11can be understood as
n = by +11which means thevalue of y must be greater than 11as the remainder 11 is possible only when the Divisor is bigger than the Remainder 11
i.e. Min Value of y = 12
i.e. Minimum Value of x+y = 8+12 = 20
Hence III only can be True
Answer: Option
C |
AQUA-RAT | AQUA-RAT-33975 | ============================
Rewrite the equation as $(a+b)(a+c)=4bc$. Now \begin{eqnarray*} a+b= 2 \alpha \beta \\ a+c = 2 \gamma \delta \\ b= \alpha \gamma \\ c= \beta \delta \end{eqnarray*} will satisfy this provided $\alpha ( 2 \beta - \gamma)=a= \delta(2 \gamma - \beta)$. So choose \begin{eqnarray*} \alpha=2 \gamma - \beta \\ \delta= 2 \beta - \gamma \end{eqnarray*} and so we have faimily of solutions generated by \begin{eqnarray*} a= (2 \gamma - \beta)( 2 \beta - \gamma) \\ b= \gamma (2 \gamma - \beta)\\ c= \beta ( 2 \beta - \gamma).\\ \end{eqnarray*}
$$c=\frac{a^2+ab}{3b-a}$$
The problem then becomes: for which $a,b$ do we have $3b-a|a(a+b)$.
Let $d= gcd(a,b)$ then $a=da', b=db'$ with $gcd(a',b')=1$. Then
$$3b'-a'|da'(a'+b')$$
Now, let us observe that $$gcd(3b'-a', a')|3gcd(a',b')=3$$ and $$gcd(3b'-a', a'+b')|4gcd(a',b')=4 \,.$$
The following is multiple choice question (with options) to answer.
If A : B : C = 2 : 3 : 4, then A⁄B : B⁄C : C⁄A is equal to : | [
"4 : 9 : 16",
"8 : 9 : 12",
"8 : 9 : 16",
"8 : 9 : 24"
] | D | Let A = 2x, B = 3x and C = 4x. Then,
A⁄B = 2x⁄3x = 2⁄3, B⁄C = 3x⁄4x = 3⁄4 and C⁄A = 4x⁄2x = 2⁄1
⇒ A⁄B : B⁄C : C⁄A = 2⁄3 : 3⁄4 : 2⁄1 = 8 : 9 : 24
Answer D |
AQUA-RAT | AQUA-RAT-33976 | # Difference between revisions of "2019 AMC 10A Problems/Problem 23"
## Problem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
## Solution 1
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$
The following is multiple choice question (with options) to answer.
Rob has 8 toy trucks. He gives 3 trucks to his best friend as a gift. How many trucks does Rob have left? | [
"5",
"20",
"2",
"5"
] | A | A. 5 |
AQUA-RAT | AQUA-RAT-33977 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A car during its journey travels 30 minutes at a speed of 30 kmph, another 30 minutes at a speed of 60 kmph, and 1 hours at a speed of 80 kmph.The average speed of the car is | [
"63.07 kmph",
"64 kmph",
"62.5 kmph",
"64.02 kmph"
] | C | First car travels 30 min at speed of 30 kmph
distance = 30 x 1/2 = 15 m
Then car travels 30 min at a speed of 60 kmph
distance = 30 min at speed of 60 kmph
distance = 60 x 1/2 = 30 m
at last it travels 1 hours at speed of 80 kmph
distance = 80 x 1 = 80 m
Total distance = 15 + 30 + 80 = 125
Total time= 1/2 + 1/2 + 1 = 2
Average speed of the car = 125/2 = 62.5
ANSWER:C |
AQUA-RAT | AQUA-RAT-33978 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
The area of a rectangle is equal to the area of right-angles triangle. What is the length of the rectangle?
I. The base of the triangle is 40 cm.
II. The height of the triangle is 50 cm. | [
"I alone sufficient while II alone not sufficient to answer",
"II alone sufficient while I alone not sufficient to answer",
"Either I or II alone sufficient to answer",
"Both I and II are not sufficient to answer"
] | D | EXPLANATION
Given: Area of rectangle = Area of a right-angles triangle.
⇒l x b =1-div-1/2x B x H
II gives, H = 50 cm.I gives, B = 40 cm.
Thus, to find l, we need b also, which is not given.
Given data is not sufficient to give the answer.
answer is (D). |
AQUA-RAT | AQUA-RAT-33979 | units, measurements
Title: Adding or subtracting mean absolute error In an experiment to find the length of a rod, n observations were made. ${a_1,\;a_2,\;.\;.\;.\;a_n}$. The mean (${a_{mean}}$) of those observations will give a value very close to the true length of the rod. After calculating ${a_{mean}}$, we find the mean absolute error (${\pm \Delta a_{mean}}$). The '${\pm}$' sign indicates that we must add ${\Delta a_{mean}}$ and subtract ${\Delta a_{mean}}$. To which value should we add ${\Delta a_{mean}}$? From which value should we subtract ${\Delta a_{mean}}$? I think it is not required to do the addition or subtraction to ${a_{mean}}$ because, ${a_{mean}}$ is already very close to the true length. Please give your opinion. Thanks
The mean (${a_{\rm mean}}$) of those observations will give a value (very) close to the true length of the rod
and the error (${\pm \Delta a_{\rm mean}}$) will give you an indication of how close your mean value might be to the actual length of the rod.
At the start of an experiment you do not know the true length of the rod.
You take some measurements and from those readings you find a mean of the measurements (${a_{\rm mean}}$) and a measure of the spread of the measurements (${\pm \Delta a_{\rm mean}}$).
You can then infer that there is a probability that the actual length of the rod is within the range $a_{\rm mean}\pm \Delta a_{\rm mean}$.
Statistically analysis can be used to estimate the value of that probability.
The following is multiple choice question (with options) to answer.
Chirag was conducting an experiment in which the average of 11 observation came to be 90. Average of first five observation was 87, and that of the last five was 84. What was the measure of the 6th observation? | [
"165",
"125",
"145",
"135"
] | D | Explanation :
11 * 90 - 5 * 87 - 5 * 84 = 900 - 435 - 420 = 135.
Answer : D |
AQUA-RAT | AQUA-RAT-33980 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
During the break of a football match, the coach will make 3 substitutions. If the team consists of 16 players among which there are 2 forwards, what is the probability that none of the forwards will be substituted? | [
" 21/55",
" 18/44",
" 3/20",
" 28/44"
] | C | Required probability = no. of favorable outcome/ total no. of outcomes
out of 16 players any 3 can be selected in 16C3 ways
after removing 2 forwards any 3 players can be selected from the remaining 9 players in 9C3 ways
thus required probability = 9C3/16C3 =3/20 |
AQUA-RAT | AQUA-RAT-33981 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
In a primary school, the ratio of the number of girls to boys is 5 : 8. If there are 160 girls, the total number of students in the college is: | [
"100",
"250",
"260",
"300"
] | D | Let the number of girls and boys be 5x and 8x.
Total number of students = 13x = 13 x 32 = 416.
Answer : D |
AQUA-RAT | AQUA-RAT-33982 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
A box contains seven bulbs out of which 5 are defective. If five bulbs are chosen at random, find the probability that all the five bulbs are defective? | [
"1/12",
"1/21",
"1/19",
"1/13"
] | B | Out of seven, two are good and five are defective. Required probability
= 5C5/7C5
= 1/21
Answer:B |
AQUA-RAT | AQUA-RAT-33983 | ### 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.
X and Y invested in a business. They earned some profit which they divided in the ratio of 1 : 2. If X invested Rs.20,000. the amount invested by Y is | [
"Rs.45,000",
"Rs.40,000",
"Rs.60,000",
"Rs.80,000"
] | B | Solution
Suppose Y invested Rs.y
Then, 20000 /y = 1 / 2
‹=› y=(20000×2 / 1).
‹=› y=40000.
Answer B |
AQUA-RAT | AQUA-RAT-33984 | So on.
=====
Another thing to note:
(Assume only positive values)
$$a < b$$ and $$c < d \implies ac < bd$$. But that is one directional. It doesn't go the other way that $$ac < bd \not \implies a< b$$ and $$c < d$$
So $$1< a < 2\implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)\implies \frac 12 < a<4$$
That is true. And indeed $$1< a < 2 \implies \frac 12 < 1 < a < 2 < 4\implies \frac 12 < a < 4$$.
But it doesn't go the other way!
$$\frac 12 < a < 4 \not \implies 1 < a < 2$$
Snd $$\frac 12 < a <4 \not \implies (1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$
[although $$(1 < a^2 < 4$$ and $$\frac 12 < \frac 1a < 1)$$ does actually imply $$1 < a < 2$$.)
• Thank you. So am I right saying that if we have different variables, $a < x < b, c < y < d => ac < xy < bd$ ? Jan 2 '20 at 7:49
• If the variables are non-negative then, yes, that is correct. $a < x;c>0$ means $ac < cx$. $c < y;x>0$ means that $cx < xy$. Transitivity means $ac < xy$. $x<b;y>0$ means $xy < by$. And $y<d;b>0$ means $by<bd$. Transitivity means $xy<bd$. Jan 2 '20 at 16:06
The following is multiple choice question (with options) to answer.
If c + xy = c and x is not equal to 0, which of the following must be true? | [
"x=0",
"x+y=0",
"y=0",
"x<0"
] | C | c + xy = c --> xy=0. Since x is not equal to 0, thus y = 0.
Answer: C. |
AQUA-RAT | AQUA-RAT-33985 | # Seating couples around 2 tables
Here's my question and possible answer.
How many possible ways can you arrange 8 married couples between 2 circular tables of 8 identical chairs each such that:
1) each couple must sit at the same table, and,
2) at each table, men and women must sit in adjacent chairs (NOTE: a couple can sit next to each other but doesn't have to).
My solution:
Number of ways = (number of ways to split 8 couples into 2 tables of 4 couples each) * (number of arrangements at each table)
$=\frac{8!}{4!4!}*$(4 men and 4 women sitting alternately in 2 ways)
$=\frac{8!}{4!4!}\!\cdot\! 4!\!\cdot\!4!\cdot\!2$
$=2 * 8!$
Can someone verify this solution or provide the correct one?
First, pick $4$ couples out of the $8$ couples to sit at one table: $8 \choose 4$
Note that this will fix the people at the other table as well. Now, if we differentiate between the two tables, then the number of ways to split the $16$ people between the two tables is ${8 \choose 4}$ (that is the number of ways to pick the people for table 1, fixing the rest for table 2). If you do not differentiate between the tables, then divide this by $2$.
Now let's arrange the people. We'll calculate the number of ways to seat the people around one table, and just multiply by that number again for the other table at the end.
Since it's a circular table with identical chairs, we'll 'anchor' the seats with $1$ of the women. Then, we can seat the other women in $3!$ ways relative to this woman, and the men in $4!$ ways.
Total: $${8 \choose 4} \cdot 3! \cdot 4! \cdot 3! \cdot 4!$$
And again, if you do not differentiate between the tables, then divide this by $2$
The following is multiple choice question (with options) to answer.
There are 8 people and 3 tables. In how many ways people can occupy the tables? | [
"336",
"840",
"560",
"740"
] | A | total people = 8
total tables = 3
1st can be occupy in 8ways
2nd can be occupy in 7ways
3rd can be occupy in 6ways
total number of ways = 8*7*6 = 336 ways
correct option is A |
AQUA-RAT | AQUA-RAT-33986 | physical-chemistry
Title: Solution with salts Does a solution of water and NaCl increase its volume of the same amount of the volume of salt added? Thank you
Does the ions Na+ and Cl- taken separately ,occupy more space than the binded molecule? Suppose we take $100$g of an $x$% by weight solution of sodium chloride, so we have $x$g of salt and $100-x$g of water. The volume of $x$g of salt is:
$$ V_S = x/\rho_S $$
where $\rho_S$ is the density of solid salt. Likewise the volume of the water is:
$$ V_W = (100-x)/\rho_W $$
Suppose when we dissolve salt in water the volumes just add i.e.
$$ V_\text{total} = V_S + V_W = x/\rho_S + (100-x)/\rho_W $$
then the density of our $x$% salt solution would be:
$$ \rho_\text{sol}(x) = \frac{100}{x/\rho_S + (100-x)/\rho_W} \tag{1} $$
The density of salt is $2.165$g/cm$^3$, and we'll take the density of water to be $1$g/cm$^3$, so we can use equation (1) to calculate what the density would be if the volumes just added and we can compared this with the experimentally measured density. I did this in Excel and got:
$$\begin{matrix}
x & Equation (1) & Experimental & Constant Volume\\
0 & 1.000 & 1.000 & 1\\
0.5 & 1.003 & 1.002 & 1.005\\
1 & 1.005 & 1.005 & 1.01\\
2 & 1.011 & 1.013 & 1.02\\
3 & 1.016 & 1.020 & 1.03\\
4 & 1.022 & 1.027 & 1.04\\
5 & 1.028 & 1.034 & 1.05\\
The following is multiple choice question (with options) to answer.
A bottle contains a certain solution. In the bottled solution, the ratio of water to soap is 3:4, and the ratio of soap to salt is five times this ratio. The solution is poured into an open container, and after some time, the ratio of water to soap in the open container is halved by water evaporation. At that time, what is the ratio of water to salt in the solution? | [
"9:32",
"9:30",
"32:9",
"19:32"
] | A | Water:soap = 3:4
Soap:Salt=15:20
=> For 15 soap, salt = 20
=> For 4 Soap, salt = (20/15)*4 = 80/15=16/3
So, water:soap:salt = 3:4:16/3= 9:12:16
After open container, water:soap:salt =4.5:12:16
So, water:salt = 4.5:16 = 9:32
ANSWER:A |
AQUA-RAT | AQUA-RAT-33987 | javascript, node.js, react.js, typescript, jsx
convertToKmh(velocityAsPercentageOfC: number){
return velocityAsPercentageOfC * c;
}
convertToMph(kilometersPerHour: number)
{
return kilometersPerHour * 0.621371;
}
calculateTimeContraction(contractedElapsedTime: number, lorentzFactor: number)
{
return contractedElapsedTime / lorentzFactor;
}
calculateTimeDilation(dilatedElapsedTime: number, lorentzFactor: number)
{
return dilatedElapsedTime * lorentzFactor;
}
roundNumber = (num: number, dec: number) => {
return Math.round(num * Math.pow(10, dec)) / Math.pow(10, dec);
}
The following is multiple choice question (with options) to answer.
Convert the 13/54 m/s into kilometers per hour? | [
"1.7",
"1.5",
"0.86",
"1.1"
] | C | 13/54 m/s = 13/54 * 18/5 = 13/15
= 0.86 kmph.
Answer: C |
AQUA-RAT | AQUA-RAT-33988 | By brute force, here are the possible sequences.
(8,0,...),(7,1,0,...),(6,2,0,...),(6,1,1,0,...),(5,3,0,...),(5,2,1,0,...)(5,1,1,1,0,...),(4,4,0,...),(4,3,1,0...),(4,2,2,0,...),(4,2,1,1,0,...),(4,1,1,1,1,0,...),(3,3,2,0,...),(3,3,1,1,0,...),(3,2,2,1,0,...),(3,2,1,1,1,0,...),(3,1,1,1,1,1,0,...),(2,2,2,2,0,...),(2,2,2,1,1,0,...),(2,2,1,1,1,1,0,...),(2,1,1,1,1,1,1,0,...),(1,1,1,1,1,1,1,1,0,...)
As you can see, there are 22 of them. I'm not sure how to generalize this result at the moment, perhaps someone who is more familiar with the problem will come around with a counting method. Perhaps a recurrence relation.
Looking for a pattern at the sequence of numbers for this problem with 1,2,3,... number of objects: I get the amounts: 1,2,3,5,7,11,15,22.
By google searching I find http://mathworld.wolfram.com/PartitionFunctionP.html which has a great deal of history about the problem and methods of calculating the sequence.
The following is multiple choice question (with options) to answer.
What is the next number of the following sequence
8, 24, 12, 36, 18, 54, (....) | [
"27",
"108",
"68",
"72"
] | A | The series is
8*3=24
24/2=12
12*3=36
36/2=18
18*3=54
54/2=27
ANSWER:A |
AQUA-RAT | AQUA-RAT-33989 | +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.
The H.C.F. and L.C.M. of two numbers are 13 and 5040 respectively If one of the numbers is 144, find the other number | [
"400",
"256",
"120",
"455"
] | D | Explanation:
Solve this question by using below formula.
Product of 2 numbers = product of their HCF and LCM
144 * x = 13 * 5040
x = (13*5040)/144 = 455
Option D |
AQUA-RAT | AQUA-RAT-33990 | Fact: there is an equal probability of the last passenger sitting in Seat 1 and Seat 100.
By the time the last passenger boards, the outcome of the lost boarding pass problem is already determined, as there is only one seat for them to choose. We must therefore look at the decisions faced by Passengers 1–99.
For Passenger 1, there is equal probability of choosing any of the 100 seats. By extension, the probability of him choosing his own assigned seat and the probability of him choosing the last passenger’s assigned seat are equal.
The only way Passengers 2–99 sit in Seat 1 or Seat 100 is if their assigned seat is occupied. In this case, there is also an equal probability of sitting in any available seat.
While both Seat 1 and Seat 100 are unoccupied, it is equally probable that either seat will be chosen. After one of these two seats is occupied, the other seat is guaranteed to remain unoccupied until the last passenger boards. The probability that Seat 100 is occupied by a previous passenger is 1/2.
Here's the statement of the question from Blitzstein, Introduction to Probability (2019 2 ed), Chapter 1, Exercise 61, p 42.
1. There are 100 passengers lined up to board an airplane with 100 seats (with each seat assigned to one of the passengers). The first passenger in line crazily decides to sit in a randomly chosen seat (with all seats equally likely). Each subsequent passenger takes their assigned seat if available, and otherwise sits in a random available seat. What is the probability that the last passenger in line gets to sit in their assigned seat? (This is a common interview problem, and a beautiful example of the power of symmetry.)
Why does this post require moderator attention?
Why should this post be closed?
The following is multiple choice question (with options) to answer.
On a certain transatlantic crossing, 30 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, what percent of the ship's passengers held round-trip tickets? | [
"33 1/3%",
"75%",
"50%",
"60%"
] | B | On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets?
Solution: Let total number of passengers be 100
According to Q stem 40% of passengers who had round-trip tics have taken cars - let number of passengers with round trip be X then
40% of X = 30 => X= 75.
Answer B |
AQUA-RAT | AQUA-RAT-33991 | # another probability: number of ways for 4 girls and 4 boys to seat in a row
4 posts / 0 new
Ej-lp ACayabyab
another probability: number of ways for 4 girls and 4 boys to seat in a row
In how many ways can 4 girls and 4 boys be seated in a row containing 8 seats if boys and girls must sit in alternate seats?
The answer in my notes is 1,152..
KMST
Let say the seats are numbered 1 through 8.
A boy could sit on seat number 1, and then all other boys would be in odd number seats,
or all the girls would sit in the odd numbered seats.
That is 2 possible choices.
For the people who will sit in the odd number seats,
you have to choose which of the 4 sits in seat number 1, who of the remaining 3 sits in seat number 3, and who of the remaining 2 sits in seat number 5. No more choices there, because the last one goes in seat number 7.
So there are 4*3*2=24 ways to arrange the people sitting in the odd number seats.
There are also 24 ways to arrange the people sitting in the even number seats (seats number 2, 4, 6, and 8).
With 2 ways to decide if seat number 1 is for a boy or a girls,
24 ways to arrange the boys,
and 24 ways to arrange the girls, there are
2*24*24=1152 possible seating arrangements.
Ej-lp ACayabyab
thank you again sir..clarify ko lang po sir:
san po nakuha ang 2 sa operation na ito: 2*24*24, wherein ung 24 each for boys and girls?
Jhun Vert
As an alternate solution, you can also think this problem as two benches, each can accommodate 4 persons. Say bench A and bench B. If boys will sit on A, girls are on B and conversely.
First Case: Boys at A, Girls at B
Number of ways for boys to seat on bench A = 4!
Number of ways for girls to seat on bench B = 4!
First Case = (4!)(4!)
Second Case: Boys at B, Girls at A
Number of ways for boys to seat on bench B = 4!
Number of ways for girls to seat on bench A = 4!
Second Case = (4!)(4!)
Total number of ways = First Case + Second Case
The following is multiple choice question (with options) to answer.
In how many different number of ways 3 boys and 4 girls can sit on a bench such that girls always sit together. | [
"A)720",
"B)576",
"C)740",
"D)780"
] | B | 576
Option 'B' |
AQUA-RAT | AQUA-RAT-33992 | 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.
Which of the following options is the possible total number of people in a hall given that the number of people is a multiple of both 8 and 22.? | [
"88",
"56",
"64",
"72"
] | A | the number of people are a multiple of both 8 and 22, so it must be divisible by 8 and 22.
using the elimination method look for a number that is both divisible by 8 and 22,
of all the options only 88 is divisible by 8 and 22.
[ 88/8=11 and 88/22=3]
answer :A |
AQUA-RAT | AQUA-RAT-33993 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
_________________
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A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
Thanks and Regards
Abhishek....
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
When average age of 22 members are 0, how many members greater than 0? | [
"17",
"21",
"22",
"24"
] | B | Average of 22 numbers = 0.
Sum of 22 numbers (0 x 22) = 0.
It is quite possible that 21 of these numbers may be positive and if their sum is a then 22nd number is (-a)
Answer is 21 (B) |
AQUA-RAT | AQUA-RAT-33994 | Similar Questions
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A). 6270 B). 1680 C). 20160 D). 6720
-- View Answer
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A). 12 B). 64 C). 256 D). 60
-- View Answer
3). In how many different ways, can the letters of the word 'ASSASSINATION' be arranged, so that all S are together?
A). 10! B). 14!/(4!) C). 151200 D). 3628800
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5). In a meeting between two countries, each country has 12 delegates, all the delegates of one country shakes hands with all delegates of the other country. Find the number of handshakes possible?
A). 72 B). 144 C). 288 D). 234
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A). 11! / 2 B). 10! / 2 C). 12! / 2 D). Couldn't be determined
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A). 18! B). 19! C). 20! D). Couldn't be determined
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9). In how many ways, a committee of 3 men and 2 women can be formed out of a total of 4 men and 4 women?
The following is multiple choice question (with options) to answer.
A team has to design a flag. The team has five yellow strips of cloth and six green strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand? | [
"120",
"144",
"256",
"462"
] | D | The # of permutations of total 5+6=11 strips where 5 are identical yellow and 6 are identical green is 11!/(5!*6!)=462 (note that they must use all the materials at hand, also I guess that the strips must be either only vertical or only horizontal).
Answer: D. |
AQUA-RAT | AQUA-RAT-33995 | If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
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Joined: 24 Oct 2016
Posts: 241
GMAT 1: 670 Q46 V36
If N is a positive three-digit number that is greater than 2 [#permalink]
### Show Tags
17 Aug 2018, 17:35
xhimi wrote:
If N is a positive three-digit number that is greater than 200, and each digit of N is a factor of N itself, what is the value of N?
(1) The tens digit of N is 5.
(2) The units digit of N is 5.
Official Solution (Credit: Manhattan Prep)
(1) SUFFICIENT: If 5 is a digit of N, then N is a multiple of 5. All multiples of 5 end with a units digit of either 5 or 0. However, the units digit of N cannot be 0, since 0 is not a factor of any number. Therefore, the units digit of N must also be 5. Therefore N has the form _55, with only the hundreds digit left to consider.
Consider the possible cases for the hundreds digit:
It can’t be 1, since N > 200.
It can’t be an even number, because N is odd (it ends in 5) and thus doesn’t have any even factors.
If it were 3, then N would be 355—but that doesn’t work, since 355 is not a multiple of 3. (Check: 3 + 5 + 5 = 13, which is not divisible by 3.)
If it were 5, then N would be 555. This is a possible value.
If it were 7, then N would be 755—but that doesn’t work, since 755 is not a multiple of 7. (Check: 700 + 55 = 755. 700 is divisible by 7 but 55 is not, so the whole thing is not.)
If it were 9, then N would be 955—but that doesn’t work, since 955 is not a multiple of 9. (Check: 9 + 5 + 5 = 19, which is not divisible by 9.)
The only possible value for N is 555, so statement 1 is sufficient.
The following is multiple choice question (with options) to answer.
If the number 52,1n8, where n represents the tens digit, is a multiple of 3, then the value of n could be which of the following? | [
"6",
"5",
"3",
"1"
] | B | The sum of the digits 5 + 2 + 1 + n + 8 must be a multiple of 3
16 + n must be a multiple of 3
substituting the numbers n must = 5
Answer B |
AQUA-RAT | AQUA-RAT-33996 | it. Given the first term and the common ratio of a geometric sequence find the first five terms and the explicit formula. The Sum of Geometric Sequence. We will just need to decide which form is the correct form. As a check, $$\frac{u_5}{u_3}=4=r^2$$ so this does work. This Site Might Help You. Access this finite geometric series worksheets tenaciously prepared for high school students. Determine the common ratio and. Any term = constant = r. This ratio is called _____. The sum of an infinite geometric series is 24, and the sum of the first 200 terms of the series is also 24. Geometric Series Questions (b) Find, to 2 decimal places, the di erence between the 5th and 6th terms. A geometric series is the sum of the terms of a geometric sequence. Work out the missing term in this geometric sequence:. Determine the common ratio and. How do we find the nth term of a geometric sequence? 1. For example, if I know that the 10 th term of a geometric sequence is 24, and the 9 th term of the sequence is 6, I can find the common ratio by dividing the 10 th term by the 9 th term: 24 / 6 = 4. Common Ratio. If there are 160 ants in the initial population, find the number of ants. Given u 5 =1280 and u 8 =81920 , find the geometric sequence. This sequence starts at 10 and has a common ratio of 0. Click here 👆 to get an answer to your question ️ 2. r = 4 2 = 2. Find (1) the common ratio, (2) the ninth term, (3) a recursive rule for the nth term, and (4) an Log On Algebra: Sequences of numbers, series and how to sum them Section Solvers Solvers. Find the common ratio of the geometric sequence. Geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. And each time I'm multiplying it by a common number, and that number is often called the common ratio. Finally, use the rule to find the tenth term in the sequence. Also, a geometric sequence has p as its common ratio. If you’re good at finding patterns, then you’ll
The following is multiple choice question (with options) to answer.
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 65, what is the common ratio of the G.P? | [
"4",
"1/4",
"2",
"9"
] | A | 65=(a1+a2+a3+a4+a5+a6)/(a1+a2+a3)
Factorize the same terms
65=1+(a4+a5+a6)/(a1+a2+a3)
Write every term with respect to r
a1=a1
a2=a1*r^1
a3=a1*r^2
.........
65=1+(a1(r^3+r^4+r^5))/(a1(1+r^1+r^2))
64=(r^3 (1+r^1+r^2))/((1+r^1+r^2))
64=r^3
r=4
A |
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