source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35097 | • $P_2$ will fly $\big[1-(d+r+y)\big]$ distance away from the airport in the counter-clockwise direction to meet up with $P_3$.
• At this point, $P_2$ will donate $z$ fuel to $P_3$.
• $P_2$ and $P_3$ will then both fly back $z$ distance, arriving at a distance of $1-d-r-y-z$ from the airport with no fuel.
• After refuelling at the airport, $P_1$ will fly the distance towards $P_2$ and $P_3$ and refund each of them for that much fuel. All three planes will then head back to the airport together.
From this, we must have
• $0 \leqslant s\leqslant d/3$: $P_1$ can fly $s$ distance forward and backwards, and refund $P_2$ for $s$ distance
• $z\geqslant 0$: cannot donate negative fuel
• $2x + 1-d-r-y \leqslant d+r+y$: $P_3$ must not run out of fuel before $P_2$ can reach it again
• $1-d-r-y - z \leqslant d/4$: $P_1$ can reach $P_2$ and $P_3$, refund them both, and the three of them will have enough fuel to head back to the airport
• $2x + 2s + 1-d-r-y - z\leqslant d+r+y + z$: $P_2$ and $P_3$ must not run out of fuel before $P_1$ can reach them again
Putting these together:
The following is multiple choice question (with options) to answer.
A boy wants to go abroad by boat and return by flight. He has a choice of 3 different boat to go and 5 flight to return. In how many ways, can the boy perform his journey? | [
"15",
"30",
"43",
"45"
] | A | Ans.(A)
Sol. Number of choices to go by ship = 3 Number of choices to return by airline = 5 From, fundamental principle, total number of ways of performing the journey = 3 x 5 = 15 ways |
AQUA-RAT | AQUA-RAT-35098 | What this means is $2^2$ (which is $4$) is $1$ greater than some multiple of $3$ (that multiple, in this case is obviously, $3$)
So if $2^2\equiv 1 ~~~(\text{mod } 3)~~\implies (2^2)^n\equiv (1)^n ~~~(\text{mod } 3)\implies 2^{2n}\equiv 1~~(\text{mod } 3)$,
Again, this means that any even power of $2$ is $1$ greater than some multiple of $3$.
So if $2^{2n}$ is $1$ greater than some multiple of $3$ (another way to say this is that $2^{2n}$ leaves a remainder of $1$ when divided by $3$), then what can you say about $2^{2n}+5$?.
To answer this, forget about $2^{2n}$ and just think about the remainder (i.e $1$). This is how modular arithmetic makes our live a lot easier. If $2^{2n}$ is $1$ greater than some multiple of $3$, then $2^{2n}+5$ should be $1+5=6$ greater than that multiple of $3$, right? But you know that $6$ is, by itself, a multiple of $3$. So, this should mean that $3|2^{2n}+5$.
A more formal (and a neat) argument could be:
$3|2^{2n}-1\implies 3|2^{2n}+5$, since $2^{2n}+5=(2^{2n}-1+6)$
Now since it has been shown that $2^{2n}+5$ is infact a multiple of $3$, it should be apparent that $2^{2n}+5$ is after all, composite.
I Hope it helps!
The following is multiple choice question (with options) to answer.
If m is 3 more than a multiple of 5 and 2 more than a multiple of 6, which of the following could be m? | [
"14",
"28",
"58",
"68"
] | D | m is 3 more than a multiple of 5 i.e. m = 5a+3
m is 2 more than a multiple of 6 i.e. m = 6b+2
Just check option
A. 14 when divided by 5 leaves remainder 4 instead of 3 hence incorrect option
B. 28 when divided by 5 leaves remainder 3 and when divided by 6 leaves remainder 4 instead of 2 hence Incorrect option
C. 58 when divided by 5 leaves remainder 3 and when divided by 6 leaves remainder 4 instead of 2 hence Incorrect option
D. 68 when divided by 5 leaves remainder 3 and when divided by 6 leaves remainder 2 hence Correct option
E. 74
Answer: Option D |
AQUA-RAT | AQUA-RAT-35099 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
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Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
Sakshi can do a piece of work in 25 days.Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work : | [
"15",
"20",
"18",
"25"
] | B | Solution
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4 .
Suppose Tanya taken x days to do the work.
5 : 4 : : 25 : x ⇒ x = (25x4/5) ⇒ x = 20 days.
Hence,Tanya takes 16 days is complete the work.
Answer B |
AQUA-RAT | AQUA-RAT-35100 | \begin{align} \dbinom{6}{2} & \longleftrightarrow \dbinom{6}{6-2} \\[8pt] AB & \longleftrightarrow CDEF \\ AC & \longleftrightarrow BDEF \\ AD & \longleftrightarrow BCEF \\ AE & \longleftrightarrow BCDF \\ AF & \longleftrightarrow BCDE \\ BC & \longleftrightarrow ADEF \\ BD & \longleftrightarrow ACEF \\ BE & \longleftrightarrow ACDF \\ BF & \longleftrightarrow ACDE \\ CD & \longleftrightarrow ABEF \\ CE & \longleftrightarrow ABDF \\ CF & \longleftrightarrow ABDE \\ DE & \longleftrightarrow ABCF \\ DF & \longleftrightarrow ABCE \\ EF & \longleftrightarrow ABCD \end{align} There are exactly as many ways to choose $2$ out of $6$ as to choose $6-2$ out of $6$ because each way of choosing $2$ out of $6$ has a corresponding way of choosing $6-2$ out of $6$ and vice-versa.
• NOTE TO FUTURE USERS: This combined with TheSparkThatThought's answer and Ned's comment at the original question will guide you to the way someone must think. Jun 27 '14 at 23:12
Consider a collection of $n$ objects. Choosing $k$ of them to place into a set is equivalent to choosing $n-k$ to leave out.
Edit: Consider a high school dodgeball game with a red team and a blue team. There are $n$ total students, and the blue team has $k$ students. Since every student plays, there are $n-k$ students on the red team.
Because the PE teacher is biased, he lets the blue team pick all of their players first. They have $\binom{n}{k}$ ways to do this. After that, the red team has no choices. They must pick all of the remaining $n-k$ students.
The following is multiple choice question (with options) to answer.
In a college, 71 percent of students Play Football, 76 percent of students Play Hockey, 86 percent Play Basket Ball and 81 percent Play Volleyball.
Can you find out the percentage of students who study all 4 Games? | [
"14%",
"10%",
"16%",
"18%"
] | A | A
14% of student Play all four Games. |
AQUA-RAT | AQUA-RAT-35101 | $\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$
There are 12 in which the sum is greater than 9 or there is at least one 6.
Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$
Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$
And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$
You can expect to win an average of $\frac{1}{3}$ of a point per throw.
. . . as Plato already pointed out.
.
5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)
RonL
The following is multiple choice question (with options) to answer.
At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90? | [
"30 points",
"20 points",
"10 points",
"12 points"
] | C | a:b=60:45
a:c=60:40
b/a *a/c=45/60 *60/45=45/40=90/80 so 10 points
ANSWER:C |
AQUA-RAT | AQUA-RAT-35102 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
12 is subtracted from 95% of a number, the result is 178. Find the number? | [
"200",
"240",
"198",
"190"
] | A | (95/100) * X – 12 = 178
9.5X = 1900
X = 200
Answer:A |
AQUA-RAT | AQUA-RAT-35103 | ### 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 invested $50 in a business and B invested $100 in a business. At the end of the year they got $150 as the profit. Find their share? | [
"$50,$100",
"$150,$250",
"$210,$130",
"$120,$150"
] | A | A:B = 50:100
A:B = 1:2
A's share = 150*1/3 = $50
B's share = 150*2/3 = $100
Answer is A |
AQUA-RAT | AQUA-RAT-35104 | java, optimization, programming-challenge, dynamic-programming
But there's something more to learn from this...
If \$n_1 + n_2 = N\$, and \$(n_1, n_2)\$ are consecutive, then \$N\$ is of the form \$2n_1 + 1\$.
If \$n_1 + n_2 + n_3 = N\$, and \$(n_1, n_2, n_3)\$ are consecutive, then \$N\$ is of the form \$3n_2\$.
Think of the sum as \$(n_2 - 1) + n_2 + (n_2 + 1)\$.
If \$n_1 + n_2 + n_3 + n_4 = N\$, and \$(n_1 \ldots n_4)\$ are consecutive, then \$N\$ is of the form \$4n_2 + 2\$.
If \$n_1 + n_2 + n_3 + n_4 + n_5 = N\$, and \$(n_1 \ldots n_5)\$ are consecutive, then \$N\$ is of the form \$5n_3\$.
Think of the sum as \$(n_3 \underbrace{- 2) + (n_3 \underbrace{- 1) + n_3 + (n_3 +} 1) + (n_3 +} 2)\$.
The following is multiple choice question (with options) to answer.
The sum of two consecutive number is 115. Which is the larger number? | [
"42",
"58",
"44",
"45"
] | B | Let consecutive number be x, x+1
Therefore sum of the consecutive number is x + x+1=115
2x+1=115
2x=114
x=57
Therefore larger number is x+1=58
ANSWER:B |
AQUA-RAT | AQUA-RAT-35105 | 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.
Of the three numbers, the first is twice as second and three times the third. The average of the three numbers is 77, and the three numbers in order are? | [
"116,58,36",
"98,49,33",
"126,63,42",
"108,54,36"
] | C | Explanation :
Solution: Let A =x,
B =x/2, C = x/3.
=x + x/2 + x/3 / 3 = 77
By solving we will get x = 126.
Hence A= 126
B= 126/2= 63
C= 126/3 = 42
Answer : C |
AQUA-RAT | AQUA-RAT-35106 | $3$-cube faces are made by starting with a $3$-cube in one of the two n-cubes plus translation of the squares in the new dimension. So $C(n+1)=2C(n)+S(n)$.
• An expository article about how to think about n-cubes from a combinatorial point of view can be found here: york.cuny.edu/~malk/tidbits/n-cube-tidbit.html – Joseph Malkevitch Jan 5 '11 at 1:23
• @Joseph: Thanks. It shows that the recurrence will be the same for all dimensions of boundary: points, lines, squares, etc. – Ross Millikan Jan 5 '11 at 3:36
First you need to be clear about what you mean by face, edge, vertex, etc. In my view if we have an $n$ dimensional cube, then $n-1$ dimensional cubes form its faces, and $n-2$ dimensional cubes form its edges. Vertices as you define above are 0-D points. We do not have names for $n-3$, $n-4$ dimensional boundaries so I will not attempt to name them here.
If you are looking for an intuitive way to build up the relations you can define inductive relationships using the method of extrusion. You can start with $n=1$ if you like. In this case we have one 1-D cube (i.e. a segment) with two 0-D faces (i.e. points). So
$$C_1 = 1$$ $$F_1 = 2$$ $$E_1 = 0$$ $$V_1 = 2$$
Where $C_1$ is the number of cubes in 1-D space, $F_1$ is the number of faces, $E_1$ is the number of edges (not defined here), and $V_1$ is the number of vertices.
We create a 2-D cube (i.e. a square) by extruding this 1-D cube in a direction orthogonal all current dimensions. By doing this two things happen:
The following is multiple choice question (with options) to answer.
The edge of three cubes of metal is 3 dm, 4 dm and 5 dm. They are melted and formed into a single cube. Find the edge of the new cube? | [
"3 dm",
"4 dm",
"5 dm",
"6 dm"
] | D | 33 + 43 + 53 = a3 => a = 6
ANSWER:D |
AQUA-RAT | AQUA-RAT-35107 | $$W = 1$$ is impossible, because that would require $$H,E = 7,3$$, in some order, and $$(37) \times (73) > 1999.$$
So, at this point, you know that $$W \in \{2,4,6,7\}.$$
$$W = 7$$ is impossible, because that would require $$H,E = 9,3$$, in some order, and $$(39) \times (93) < 7000.$$
So, at this point, you know that $$W \in \{2,4,6\}.$$
Suppose that $$(10 \times H) + E \equiv k \pmod{9} ~:~ k \in \{1,2,\cdots,9\}.$$
By Casting out 9's you know that $$(10 \times E) + H$$ and $$(E + H)$$ are each also $$\equiv k\pmod{9}$$.
This implies that $$(2W) + H + E \equiv k^2 \pmod{9} \implies$$
$$2W \equiv (k^2 - k) = k(k-1) \pmod{9}.$$
As $$k$$ ranges from $$1$$ through $$9$$, $$k(k-1)$$ is never equal to either $$8$$ or $$4 \pmod{9}.$$
From this you know that $$w$$ can't equal either $$2$$ or $$4$$, so $$w = 6$$
and $$H,E \in \{2,3,4,7,8,9\}$$.
The following is multiple choice question (with options) to answer.
IF
1= N
2= W
3= H
4= O
5= I
THEN 6=? | [
"I",
"J",
"K",
"L"
] | A | IF
1= N... Second letter of ONE.
2= W.... Second letter of TWO.
3= H
4= O
5= I
THEN 6=I ... Second letter of SIX.
ANSWER:A |
AQUA-RAT | AQUA-RAT-35108 | 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
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of first 10 even numbers is? | [
"8",
"92",
"83",
"11"
] | D | Explanation:
Sum of 10 even numbers = 10 * 11 = 110
Average = 110/10 = 11
Answer: Option D |
AQUA-RAT | AQUA-RAT-35109 | @Rajeshwar: Because when you divide $6k+1$ by $5$, the $5k$ part is already a multiple of $5$. Adding or subtracting multiples of five form a number does not change the remainder when you divide by $5$. E.g., $2$, $7=2+5$, $22=2+5\times 4$, $177 = 2 + 5\times35$, etc., all have the same remainder when divided by $5$. So the remainder you get when dividing $n=6k+1$ by $5$ is the same as the remainder you get when dividing $n-5k = k+1$ by $5$. – Arturo Magidin Jul 20 '12 at 2:31
The following is multiple choice question (with options) to answer.
I chose a number and divide it by 5. Then I subtracted 154 from the result and got 6. What was the number I chose? | [
"800",
"237",
"2765",
"288"
] | A | Let xx be the number I chose, then
x5−154=6x5−154=6
x5=160x5=160
x=800
Answer:A |
AQUA-RAT | AQUA-RAT-35110 | 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 a platform in 8 seconds and also crosses the platform which is 273 metres long in 20 seconds. The length of the train (in metres) is: | [
"182",
"176",
"175",
"96"
] | A | Explanation:
Let the length of train be L m.
Acc. to question
(273+L)/20 = L/8
2184+8L=20L
L= 2184/12 = 182 m
Answer A |
AQUA-RAT | AQUA-RAT-35111 | # 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.
Two trains of length 120 m and 300 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet? | [
"21 sec",
"32 sec",
"82 sec",
"20 sec"
] | A | Relative speed = (42 + 30) * 5/18 = 4 * 5 = 20 mps.
Distance covered in passing each other = 120 + 300 = 420m.
The time required = d/s = 420/20 = 21 sec.
Answer:A |
AQUA-RAT | AQUA-RAT-35112 | (A) 10
(B) 45
(C) 50
(D) 55
(E) 65
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
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15 Nov 2017, 16:22
9
3
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
We can create the following equation:
Total houses = number with air conditioning + number with sunporch + number with pool - number with only two of the three things - 2(number with all three things) + number with none of the three things
150 = 0.6(150) + 0.5(150) + 0.3(150) - D - 2(5) + 5
150 = 90 + 75 + 45 - D - 10 + 5
150 = 205 - D
D = 55
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Re: Of the 150 houses in a certain development, 60 percent have [#permalink]
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29 May 2017, 05:24
8
5
ganand wrote:
Of the 150 houses in a certain development, 60 percent have air-conditioning, 50 percent have a sunporch, and 30 percent have a swimming pool. If 5 of the houses have all three of these amenities and 5 have none of them, how many of the houses have exactly two of these amenities?
(A) 10
(B) 45
(C) 50
(D) 55
(E) 65
This can be solved using Venn Diagram (refer the attachment)
AC = 60% of 150 = 90
Sunporch = 50% of 150 = 75
SP = 30% of 150 = 45
The following is multiple choice question (with options) to answer.
At a certain health club, 30 percent of the members use both the pool and sauna, but 30 percent of the members who use the pool do not use the sauna. What percent of the members of the health club use the pool? | [
"33 1/3%",
"40%",
"53 11/13%",
"62 1/2%"
] | C | P = pool S = sauna
given P+S = 30 then let only S be x and only P will be 100 - (30+x) = 70 -x
30% of (70-x) = x => 21 - 0.3x = x => x = 16 2/13% so only P = 70 -x = 53 11/13%
Answer C |
AQUA-RAT | AQUA-RAT-35113 | thermodynamics, energy, home-experiment
Note 3: Peak hours tend to be during the day (this may reverse if we get wide-spread solar energy), so keeping the house cool during the day would cost even more. You pretty much have it right. We have two scenarios:
1 - Leave air conditioning on all day. Say that outside temperature is 90 degrees and inside temperature is 70 degrees. Then, all day long, the air conditioning has to remove any heat that gets into the house continuously. Heat transfer depends on the difference in temperature between outside and inside the house. In this case, heat transfer per unit time is equal to a constant K times 20 degrees. Over a whole day of D units of time, the total energy removed by the air conditioner is $20KD$ This assumes that convection (ie air draft) can be neglected.
2 - Stop the air conditioning during the day. If you stop the air conditioning, the flow of heat in the house initially will be the same. However, as time goes by, the inside temperature will increase. This will gradually reduce heat flow as it is proportional to the difference in temperatures. If the house is small enough and badly insulated, you could even get to a point where the inside temperature is equal to the outside temperature, at which point no more heat enters the house. When you get back from work, or if you have a timer a bit before, air conditioning is restarted. You now have to remove all the accumulated heat. However, since less heat entered the house over the duration of the day than if you had kept the air conditioning on, you have less work to do to cool down the house. Yes, the heat will have accumulated in insulation, furniture, walls, etc, but it doesn't matter. There is less heat total to remove.
The following is multiple choice question (with options) to answer.
An air-conditioning unit costs $470. On December there was a discount for Christmas of 16%. Six months later, the holiday season was over so the company raised the price of the air-conditioning by 15%. How much will an air-conditioning unit cost in November? | [
"$454",
"$470",
"$472",
"$484"
] | A | if its previous November (before discount) then price is $470.
but if its November of next year then
16% discount on $470=470(1-16/100)=$394.8
again a corrected raised price of 15% over $394.8=394.8(1+15/100)=454.02~$454
Ans A |
AQUA-RAT | AQUA-RAT-35114 | => 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.
The salaries of A and B together is Rs. 14,000. A spend 80% of his salary and B spends 85% of his salary. What is the salary of B if their savings are equal? | [
"Rs. 6,000",
"Rs. 8,000",
"Rs. 7,500",
"Rs. 6,500"
] | B | Let the salaries of A and B are X and Y respectively
X + Y = 14,000
Savings of A = 20X/100 = Savings of B = 15Y/100
X = ¾ Y
3/4Y + Y = 14,000, 7Y/4 =14,000, Y =8,000
ANSWER:B |
AQUA-RAT | AQUA-RAT-35115 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A sum of Rs. 2040 has been divided among A, B and C such that A gets of what B gets and B gets of what C gets. B’s share is: | [
"Rs. 120",
"Rs. 160",
"Rs. 240",
"Rs. 360"
] | D | EXPLANATION
Let C’s share = Rs. x
Then, B’s share = Rs. x/4 , A’s share = Rs. (2/3 x x/4 ) = Rs. x/6
=x/6 + x/4 + x = 2040
=> 17x/12 =2040
=> 2040 x 12/ 17 = Rs.1440
Hence, B’s share = Rs. (1440/4) = Rs. 360.
Answer D |
AQUA-RAT | AQUA-RAT-35116 | combinatorics, sets, matching
Put your students in a row: $a!$ possibilities. The first $b$ students form the first group, the next $b$ students the second group, etc. The order in the line-up between each group of students does not matter, so we have to divide by th1s number of possibilities which is $(c!)^b$. Again we divide by $(a-bc)!$ the number of permutations of the not chosen ones. Also, the order of the groups themselves does not matter, we additionally divide by $b!$.
Answer: $a!\;/\;{(c!)^b (a-bc)! \;b!}$
Another way of obtaining the same number. Choose a group of $c$ students from total $a$, then a group of $c$ students from the remaining $c-b$, etc. This can be done in ${a \choose c}{a-c\choose c}\dots{a-c(b-1)\choose c}$. Again I divide by the possible orderings of the groups which is $b!$.
Thus $\frac{a!}{c!(a-c)!}\frac{(a-c)!}{c!(a-2c)!}\dots \frac{(a-(b-1)c)!}{c!(a-bc)!} / b!$
Your first example $a=12$, $b=4$, $c=3$. You have 880, this number is 15400.
Second example $a=12$, $b=3$, $c=2$. 13680, we agree.
The following is multiple choice question (with options) to answer.
Each student at a certain business school is assigned a 8-digit student identification number. The first digit of the identification number cannot be zero, and the last digit of the identification number must be prime. How many different student identification numbers can the school create? | [
"9,000",
"3,600",
"36,000,000",
"2,592"
] | C | The identification number is of the form _ _ _ _
1. First digit cannot be 0
2. Middle digits can be anything
3. Last digit has to be prime - 2, 3, 5, 7
We can have the following number of possibilities for each space
__ __ __ __
9 10 10 10 10 10 10 4
Total cases = 36,000,000
Option C |
AQUA-RAT | AQUA-RAT-35117 | ### Show Tags
03 Oct 2019, 11:34
OFFICIAL EXPLANATION
Hi All,
We're told that each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. We're asked for the number of cars that are in the LARGEST lot. This is an example of a 'System' question; it can be solved Algebraically, but it can also be solved rather quickly by TESTing THE ANSWERS...
Based on the information that we're given, the three parking lots clearly each end up holding a different number of cars. We're asked for the LARGEST number of the three, so we should look to TEST one of the larger answers first. Let's TEST Answer D...
IF....the largest lot holds 28 cars....
then the middle lot holds 28 - 8 = 20 cars...
and the smallest lot holds 28 - 16 = 12 cars...
Total = 28 + 20 + 12 = 60 cars
This is an exact MATCH for what we were told, so this MUST be the answer!
GMAT assassins aren't born, they're made,
Rich
_________________
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Re: Each of 60 cars is parked in one of three empty parking lots [#permalink]
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03 Oct 2019, 12:39
Top Contributor
EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Block 1, Question 5
Each of 60 cars is parked in one of three empty parking lots. After all of the cars have been parked, the largest lot holds 8 more cars than the middle lot and 16 more cars than the smallest lot. How many cars are in the largest lot?
A. 12
B. 20
C. 22
D. 28
E. 30
Let x = number of cars in the LARGEST lot
The following is multiple choice question (with options) to answer.
Four car rental agencies A, B, C and D rented a plot for parking their cars during the night. A parked 15 cars for 12 days, B parked 12 cars for 20 days, C parked 18 cars for 18 days and D parked 16 cars for 15 days. If A paid Rs. 1350 as rent for parking his cars, what is the total rent paid by all the four agencies? | [
"Rs. 4500",
"Rs. 4800",
"Rs. 5250",
"Rs. 7380"
] | D | The ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15
= 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60
Let us consider the four amounts to be 45k, 60k, 81k and 60k respectively.
The total rent paid by the four agencies = 45k + 60k + 81k + 60k= 246k
It is given that A paid Rs. 1350
45k = 1350 => k = 30
246k = 246(30) = Rs. 7380
Thus the total rent paid by all the four agencies is Rs. 7380.
ANSWER:D |
AQUA-RAT | AQUA-RAT-35118 | 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
5
KUDOS
3
This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
The average of 10 numbers is 23. If each number is increased by 2, what will the new average be? | [
"36",
"25",
"72",
"29"
] | B | Sum of the 10 numbers = 230
If each number is increased by 2, the total increase =
2 * 10 = 20
The new sum = 230 + 20 = 250 The new average = 250/10
= 25.
Answer:B |
AQUA-RAT | AQUA-RAT-35119 | 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 180 m in length crosses a telegraph post in 20 seconds. The speed of the train is? | [
"37 kmph",
"35 kmph",
"32 kmph",
"38 kmph"
] | C | S = 180/20 * 18/5
= 32 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-35120 | mechanical-engineering, gears, machine-design
Title: Velocity or Speed Ratios in Gears and Gear Trains I have seen the different definitions of the velocity ratio for Gears and Gear trains. I have referred a book called Theory of machines by S. S. Ratan. In that book the definitions are as following:
For gears, velocity ratio is ratio of angular velocity of follower to that of the driving gear.
For gear train, speed ratio is given as the ratio of speed of driving to that of the driven shaft.
Why are these two definitions different (inverse of each other) as I think Velocity ratio and speed ratio are the same thing. Also if there is a difference, it will be helpful if a bit explanation is provided. TL;DR IMHO, this is something that think has its roots to older times, where the strength design of gears required tables and charts
Like solarMike said, they are essentially the same thing. I don't think many people nowadays pay too much attention to this detail.
IMHO, the reason for the existence stems from the different need. Like you stated, (some authors) use (note that I'll start the opposite way):
For gear train, speed ratio is given as the ratio of speed of driving to that of the driven shaft.
In that case, the big picture is important. I.e the entire drivetrain, and more specifically its kinematics. In that case the important is to define the angular velocities and torque on the various stages of the drivetrain (or maybe just at the end). Therefore, it that case, the speed ratio focuses on that.
For gears, velocity ratio is ratio of angular velocity of follower to that of the driving gear.
The following is multiple choice question (with options) to answer.
The speeds of three motor bikes are in the ratio 6 : 5 : 4. The ratio between the time taken by them to travel the same distance is : | [
"10 : 12 : 15",
"12 : 10 : 8",
"15 : 12: 10",
"10 : 15 : 12"
] | A | Ratio of time taken :
1/6 :1/5 : 1/4 = 10 : 12 : 15
ANSWER:A |
AQUA-RAT | AQUA-RAT-35121 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
Micheal and Adam can do together a piece of work in 20 days. After they have worked together for 15 days Micheal stops and Adam completes the remaining work in 10 days. In how many days Micheal complete the work separately. | [
"40 days",
"100 days",
"120 days",
"110 days"
] | A | Rate of both = 1/20
Together they do = 1/20*15 = 3/4
Left work = 1 - 3/4 = 1/4
Adam completes 1/4 work in 10 day
so he took 10*4 = 40 days to complete the left work alone.
Thus the rate of adam is 40/1
Rate of Micheal = 1/20 - 1/40 = 1/40
Thus micheal takes 40 days to complete the whole work.
ans. A. |
AQUA-RAT | AQUA-RAT-35122 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
There is 60% increase in an amount in 6years at SI. What will be the CI of Rs. 13,000 after 3 years at the same rate? | [
"2372",
"2572",
"4303",
"2343"
] | C | Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = 100 x 60 = 10% p.a.
100 x 6
Now, P = Rs. 13000. T = 3 years and R = 10% p.a.
C.I.
= Rs. 13000 x 1 + 10 3 - 1
100
= Rs. 13000 x 331
1000
= 4303.
C |
AQUA-RAT | AQUA-RAT-35123 | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
The following is multiple choice question (with options) to answer.
In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons? | [
"170",
"180",
"200",
"210"
] | D | 7p3 = 7 × 6 × 5 = 210
Option 'D' |
AQUA-RAT | AQUA-RAT-35124 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
Ashley and Vinny work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinny will be chosen? | [
"1/10",
"3/28",
"2/9",
"1/4"
] | C | I did slightly different and hence obviously ended up with the answer 2/9.
I assumed that because Ashley and Vinny have to be chosen, the remaining selection is 3 people from 8.
8C3/10C5= 2/9. Option C. |
AQUA-RAT | AQUA-RAT-35125 | 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.
Nil and Ethan are brothers. They left their home
at the same time and drove to the same beach.
Nil drove at a speed of 30 miles per hour. Ethan
drove at a speed of 20 miles per hour. Nil
arrived at the beach 0.5 hour earlier than Ethan.
What is the distance between their home and the
beach? | [
"20 miles",
"30 miles",
"40 miles",
"50 miles"
] | B | Every hour, Nil gets ahead of Ethan 30 - 20 =
10 miles. When Nil arrived at the beach, Ethan
is only 20 × 0.5 = 10 miles behind. That tells us
they only drove 1 hour when Nil arrived at the
beach.
The distance between their home and the beach
is
Nil’s speed × Nil’s time
= 30 × 1 = 30 miles.
correct answer B |
AQUA-RAT | AQUA-RAT-35126 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half | [
"15 mins",
"20 mins",
"25 mins",
"30 mins"
] | D | Explanation:
Let the total time be x mins.
Part filled in first half means in x/2 = 1/40
Part filled in second half means in x/2 =
1/60+1/40=1/24
Total = x/2∗1/40+x/2∗1/24=1
=>x/2(1/40+1/24)=1
=>x/2∗1/15=1
=>x=30mins
Option D |
AQUA-RAT | AQUA-RAT-35127 | The approach is correct, but the answer is wrong. There are 145 zeros from 1 to 751: 9 zeros from 1 to 99, 120 zeros from 100 to 699 (20 x 6) and 16 zeros from 700 to 759
Total: 9+120+16 = 145.
-
You are right - here is a Mathematica code to check With[{x = 751}, Count[Flatten[IntegerDigits /@ Table[x - n, {n, 0, x - 1}]], 0]] – Norbert Oct 17 '12 at 5:50
There's a nice algorithm for doing this calculation which is explained here.
If $x$ is the number we're given, $f(x)$ is the number of zeros that appear in the range $1..x$. Using a simple program we can calculate $f(x)$ for some small values to spot a pattern.
public int CountZerosInRangeOneTo(int end)
{
return Enumerable.Range(1, end)
.Select(i => i.ToString())
.SelectMany(s => s.ToCharArray())
.Count(c => c == '0');
}
For example:
f(5 ) = 0
f(52 ) = 5
f(523 ) = 102
f(5237 ) = 1543
f(52378) = 20667
If $y$ is the new single digit number added to the end each time, it would appear the following is true:
$f(10x + y) = 10 \cdot f(x) + x$
For example, if $x = 523$, $y = 7$, and $f(x) = 102$, then:
$f(10 \cdot 523 + 7) = f(5237) = 10 \cdot f(x) + x = 10 \cdot 102 + 523 = 1543$
Fantastic. However, where this breaks down is when $x$ contains zeros itself. For example:
The following is multiple choice question (with options) to answer.
Find the number of zeros in the product 5 x 10 x 25 x 40 x 50 x 55 x 65 x 125 x 80 | [
"8",
"9",
"12",
"13"
] | B | Explanation :
From the question understand whether number of 2's are more or number of 5's are more...count the lesser one only...the number of zero will be equal to that as they will be responsible in creating zeros by being multiplied by 5
Answer : B |
AQUA-RAT | AQUA-RAT-35128 | Your best bet in a situation like this, where there are relatively few numbers between $50$ and $100$ which are divisible by $7$, is to just list them, and count the number of entries on the list. (Of course, you'd like a more efficient means if the interval over which multiples of a given number span is very very large!) Note: When computing the number of integers between $50$ and $100$ that are divisible by both $7$ and $11$, there is only one such number: $7 \times 11 = 77$.
If, however, the question asked how many such numbers are divisible by $7$ OR $11$, then you'd need to list/count
(1) those divisible by $7$,
(2) those divisible by $11$;
(3) those divisible by both $7$ and $11$: any multiple of $7 \times 11$ in the given range;
Then add the number of entries on list (1) to those on (2), and then subtract the number of entries on (3) from that sum. (Otherwise, without subtracting, you'd be counting, e.g. 77, twice, since it would appear on both (1) and (2).)
In general, if the lower bound of the interval under consideration is $L$, and the upper bound of the interval under consideration is $U$, then the smallest multiple of a positive integer $n$ that is larger than or equal to $L$ is $L/n$, and if the result is not an integer, round up to the next integer (say $k_{min}$). On the other hand, the largest multiple of $n$ that is less than or equal to $U$ is $U/n$, but if the result is not an integer, round down (drop off the remainder), say $k_{max}$. Then, count all multiples of $n$ such that $L \leq nk_{min} < nk_i < nk_{max} \leq U$, where $nk_i$ are all the multiples of $n$ between the least and greatest integers in the given interval [$L, U$].
-
The following is multiple choice question (with options) to answer.
How many 7 in between 1 to 110? | [
"18",
"21",
"22",
"23"
] | B | 7,17,27,37,47,57,67,
70,71,72,73,74,75,76,77(two 7's),78,79,87,97,107
21 7's between 1 to 110
ANSWER:B |
AQUA-RAT | AQUA-RAT-35129 | This does not change even if we change the extras $4'$ and $3'$ to $40'$ and $30'$! A $40'$ wide extra "flank" along the $12'$ side plus a $30'$ flank along the $10'$ side are bigger than vice versa.
• great work, it helps – Complex Guy Jun 4 '13 at 17:34
Another simple method particularly suited to a deliberately 'symmetric' problem like this: $35{,}043\times 25{,}430=(25{,}043\times25{,}430)+(10{,}000\times25{,}430)$ while $25{,}043\times35{,}430=(25{,}043\times25{,}430)+(10{,}000\times25{,}043)$, and written out this way it's clear that the former has to be larger.
A = 35,043×25,430 = 35,043 × 25,043 + 35,043 × 0,387
B = 35,430×25,043 = 35,043 × 25,043 + 25,043 x 0,387
so A is bigger
$\begin{eqnarray}{\bf Hint}\quad 35043 \times 25430 &-\,&\ \, 35430 \times 25043 \\ A\ (B\! +\! N)&-\,& (A\!+\!N)\ B\ =\ (A\!-\!B)\,N > 0\ \ \ {\rm by}\ \ \ A > B,\ N> 0\end{eqnarray}$
from the first equation we get:
(35,000 + 43) * (25,000 + 430) =
= 35,000 * 25,000 + 430*35,000 + 43*25,000 + 43*430 ( let it be a)
from the second second equation we get:
(35,000 + 430)*(25,000 + 43) =
= 35,000 * 25,000 + 43 * 35,000 + 430 * 25,000 + 43 * 430 (let it be b)
suppose that a > b (1)
The following is multiple choice question (with options) to answer.
If the sides of a square are multiplied by sqrt(20), the area of the original square is how many times as large as the area of the resultant square? | [
"2%",
"5%",
"20%",
"50%"
] | B | Let x be the original length of one side.
Then the original area is x^2.
The new square has sides of length sqrt(20)*x, so the area is 20x^2.
The area of the original square is 1/20 = 5% times the area of the new square.
The answer is B. |
AQUA-RAT | AQUA-RAT-35130 | 4,8),(3,5,8),(2,6,8),(1,7,8),(1,9,8),(9,0,9),(8,1,9),(7,2,9),(6,3,9),(5,4,9),(4,5,9),(3,6,9),(2,7,9),(1,8,9)]
The following is multiple choice question (with options) to answer.
The forth proportional to 4,6,8 is? | [
"12",
"15",
"16",
"18"
] | A | Let the fourth proportional to 4,6,8 be x
then 4:6::8:x
4x = 6*8
x = 12
Answer is A |
AQUA-RAT | AQUA-RAT-35131 | investment over a time period, with compounding factored in. These details are very useful for when you are searching for … Conversely, the effective interest rate can be … For the hypothetical mortgage provided ($300,000,$750 in fees, 6.25% interest… If you don’t want to do the math on your own, effective interest rates are usually 1.8x to 2.5x higher than flat interest rates, after accounting for fees. Effective annual interest rate calculation The effective interest rate is equal to 1 plus the nominal interest rate in percent divided by the number of compounding persiods per year n, to the power of n, minus 1. … (Ed.). By calculating your effective interest rate, this calculator gives you your monthly instalment, your total payment, and your effective interest rate. effective interest rate is in terms of yearly periods and stated such as the As you can see in the example above, a nominal interest rate of 8.0% with 12 compounding … Effective Interest Rate Calculator This calculator is used to compute the effective period interest rate and the effective annual interest rate. Cite this content, page or calculator as: Furey, Edward "Effective Interest Rate Calculator"; CalculatorSoup, Where r is the interest rate per period in decimal form so R = r * 100 and, i is the effective interest rate in decimal form so I = i * 100. m is the compounding times per period. The reverse calculation would be 1.0241^4 – 1 = 10% effective annual interest rate. Price Button - Press to calculate … The effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. Calculate Effective Annual Rate using the information. To demonstrate how this works, the table … The effective interest rate is calculated through a simple formula: r = (1 + i/n)^n - 1. Using the calculator, your periods are years, nominal rate is 7%, compounding is monthly, 12 times per yearly period, and your number of periods is 5. You can look at the \"cash discount\" price this way -- as a true discount for paying early -- or you can consider the cash price as the regular price and calculate an effective \"interest rate\" that you
The following is multiple choice question (with options) to answer.
A mobile phone is available for $39000 cash or $17000 as down payment followed by five equal monthly instalments of $4800 each. The simple rate of interest per annum under the instalment plan would be | [
"18%",
"19%",
"21.2%",
"21.81%"
] | D | For this 22,000 a total sum of 5*4800 = 24,000 was pain. (Time is five months, so T = 5/12, as T is in years.)
thus, SI = 2000 or
2000 = (P*R*T)/100
2000 = (22,000*R*5)/(100*12)
R = (2,000*12*100)/22,000*5
R = 21.81%
Therefore answer is D |
AQUA-RAT | AQUA-RAT-35132 | # 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.
How much more would Rs.20000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly? | [
"Rs.482",
"Rs.424",
"Rs.842",
"Rs.512"
] | A | 20000(11/10)4 - 20000(6/5)2 = 482
ANSWER:A |
AQUA-RAT | AQUA-RAT-35133 | 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.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 16 hours and 25 hours respectively. The ratio of their speeds is? | [
"4:9",
"4:3",
"5:4",
"4:1"
] | C | Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √25 : √26
= 5:4
Answer:C |
AQUA-RAT | AQUA-RAT-35134 | 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 110 m long is running with a speed of 60 km/hr. In what time will it pass a man who is running at 6 km/hr in the direction opposite to that in which the train is going? | [
"67 sec",
"6 sec",
"7 sec",
"5 sec"
] | B | Answer: Option B
Speed of train relative to man = 60 + 6 = 66 km/hr.
= 66 * 5/18 = 55/3 m/sec.
Time taken to pass the men = 110 * 3/55 = 6 sec. |
AQUA-RAT | AQUA-RAT-35135 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month? | [
"$1592.33",
"$2692.33",
"$2492.33",
"$2592.33"
] | D | let the girl gets x$ per month,then
for 12 months she would get 12x$,she spent 12x/2 $ on clothes,now she will have 12x/2 $=6x$,1/3 of this is spent on games i.e (12x/2)*(1/3)=2x.then remaining money is 6x-2x=4x$.1/4 of this is spent on toys i.e 4x/4=x $.now she is left with 4x-x=3x$ which is given as 7777.so x=7777/3=2592.333$.so, she would get monthly allowance of $2592.33
ANSWER:D |
AQUA-RAT | AQUA-RAT-35136 | cylinder is similar to a prism, but its two bases are circles, not polygons. What is the relationship between the volumes of the cylinder and the cone when they have the same radius and height measurements?. A cylinder is bounded by two parallel planes or bases and by a surface generated by revolving a rectangle about one of its sides. Last week I wrote about the maximum (volume) cylinder it’s possible to fit inside a sphere. Volume of a sphere=4/3 ×r³. C The volume of the cone is three times the volume of the cylinder. 5 cm 3 of ice cream. And they are in the ratio of 1:3. 5 Investigate and describe the density of solids, liquids, and gases. To do so, they examine the relationship between a hemisphere, cone, and cylinder, each with the same radius, and for the cone and cylinder, a height equal to the radius. A cylinder and a cone have the same diameter: 8 inches. Find the the remaining variables at that instance. Answer by jsmallt9(3757) ( Show Source ):. For example, a student might compare the areas in a given cross-section, reducing the problem to a comparison of the area under a line and under a quadratic-like curve. Then, the key is placed in the graduated cylinder. The drag coefficients (C) used in our calculation are from Blevins (2003). Exploring the Relationship Between Mass and Volume Purpose: For this activity you will be performing a few measurements to help describe the relationship between mass and volume. For example, after part (b), the teacher could ask the students for other ways to determine which vase holds the most water, with the expectation that students might respond with. EXAMPLE 1 Finding the Volume of a Cylinder Find the volume of the cylinder. Calculate volume of a cone if you know radius and height ( V ) What is the formula of the volume of a cone - Calculator Online Home List of all formulas of the site. Derive the formula of the surface area of a cylinder of radius r and height h. Liu Hui proves that the assumption is incorrect by showing that this relation in fact holds between the volume of a sphere and that of another object, smaller than the cylinder. r is the volume charge density in coulombs per cubic meter. Ex of units: L or mL Solid Volume: When an object has a definite shape (ex. What fraction of the cup
The following is multiple choice question (with options) to answer.
The bases of a hemisphere and a cone are equal. If their heights are also equal then ratio between their curved surface area is— | [
"1 : √2",
"√2 ∶1",
"1 : 2",
"2 : 1"
] | B | both having same radius and the heights are same. So height of cone is same as the radius.
2πr^2 / πrl
2r/√(r^2+h^2) , where πr gets cancelled on both sides and l^2=r^2+h^2
2r/√(2r^2) , r=h
2/√2
√2/1
ANSWER:B |
AQUA-RAT | AQUA-RAT-35137 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
The instructions state that Jansi needs 1/4 square yards of one type of material and 1/3 square yards of another type of material for a project. She buys exactly that amount. After finishing the project, however, she has 8/24 square yards left that she did not use. What is the total amount of square yards of material Jansi used? | [
"1/4",
"1/9",
"2/3",
"1 1/9"
] | A | Total bought = 1/4+1/3
left part 8/24--->1/3
so used part 1/4+1/3-1/3=1/4
ANSWER:A |
AQUA-RAT | AQUA-RAT-35138 | induction that the product of three consecutive integers is divisible by 6. Real numbers class 10. In mathematics, the least common multiple, also known as the lowest common multiple of two (or more) integers a and b, is the smallest positive integer that is divisible by both. Click 'show details' to verify your result. Therefore 8n(n+1)(n+2) is always divisible by 6*8 or 48. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. [3] b) Give an example to show that the sum of four consecutive integers is not always divisible by 4. Prove that n2-n is divisible by 2 for every positive integer n. Exercise: 2. For any positive integer n, prove that n3 – n is divisible by 6. The number is divisible by 6 means it must be divisible by 2 and 3. He brought a _action against the company, claiming that the accident had been caused by a manufacturing fault in the automobile. What are the two odd integers? 12. 111 is the smallest possible magic. Exercise 12. case (3) z is a multiple of three. By the three cases, we have proven that the square of any integer has the form 3k or 3k +1. Some other very important questions from real numbers chapter 1 class 10. Prove that the product of any three consecutive positive integers is divisible by 6. Since n is a perfect square, n is congruent to 0 or 1 modulo 4. Solution for Determine whether the statement is true or false. 1, which is divisible by 9. 2 Exercise 12) Provide a direct proof that n2 n+ 5 is odd, for all integers n. : Therefore: n = 3p or 3p+1 or 3p+2, where p is some integer If n = 3p, then n is divisible by 3 If n = 3p+1, then n+2 = 3p+1+2 = 3p+3 = 3(p+1) is divisible by 3. Consider three consecutive integers, n, n + 1, and n+ 2. 7: given non-empty nite sets X and Y with jXj= jYj, a function X !Y is an injection if and only if it
The following is multiple choice question (with options) to answer.
The least common multiple of positive integer w and 3-digit integer n is 690. If n is not divisible by 3 and w is not divisible by 2, what is the value of n? | [
"115",
"230",
"460",
"575"
] | B | The LCM of n and w is 690 = 2*3*5*23.
w is not divisible by 2, thus 2 goes to n
n is not divisible by 3, thus 3 goes to w.
From above:
n must be divisible by 2 and not divisible by 3: n = 2*... In order n to be a 3-digit number it must take all other primes too: n = 2*5*23 = 230.
Answer: B. |
AQUA-RAT | AQUA-RAT-35139 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
The number a yields a remainder p when divided by 11 and a remainder q when divided by 7. If p = q + 7, then which one of the following could be the value of a ? | [
"45",
"43",
"72",
"85"
] | B | I solved this question in the following way:
Q<7 so Q can be from 0 to 6 and P<11 so P can be from 0 to 10, BUT the constraint is P=Q +7 so this will mean that P can be in the range from 7 to 10.
a=11k + P or a= 11k + 7 to 10 and look at the answer as, place different values for k, B will give 43 which is 11*3 + 10, the other answers are out of the range |
AQUA-RAT | AQUA-RAT-35140 | Difference between revisions of "2014 AMC 10A Problems/Problem 17"
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$
Solution 1 (Clean Counting)
First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.
--happiface
Solution 2 (Bashy Casework)
Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.
The following is multiple choice question (with options) to answer.
Three 6 faced dice are thrown together. The probability that all the three show the same number on them is? | [
"1/38",
"1/36",
"1/35",
"1/389"
] | B | It all 3 numbers have to be same basically we want triplets. 111, 222, 333, 444, 555 and 666. Those are six in number. Further the three dice can fall in 6 * 6 * 6 = 216 ways.
Hence the probability is 6/216 = 1/36
Answer:B |
AQUA-RAT | AQUA-RAT-35141 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
How many 9 digit numbers are possible by using the digits 1, 2, 3, 4, 5 which are divisible by 4 if the repetition is allowed? | [
"77",
"26",
"88",
"27"
] | A | : If A number has to be divisible by 4, the last two digits must be divisible by 4. So possibilities are, 12, 24, 32, 44, 52. And the of the remaining 7 places, each place got filled by any of the five digits. So these 7 places got filled by 5 x 5 x .....(7 times) = 5757 ways. So total ways are 5 x 5757 = 5/8
Answer:A |
AQUA-RAT | AQUA-RAT-35142 | ### 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.
If in a $180 mln portfolio of stock, twice as much is invested in stock A as is invested in all the other stocks, how much is invested in all the other stocks? | [
"$60 mln",
"$80 mln",
"$100 mln",
"$120 mln"
] | A | Let x denote the amount invested in all the other stocks. Then the amount invested in stock A is 2x. As we have the $180 mln portfolio, x+2x=3x=180 mln from where x=60 mln.
Answer: A |
AQUA-RAT | AQUA-RAT-35143 | 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.
Two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 21 kmph respectively. When they meet, it is found that one train has traveled 60 km more than the other one. The distance between the two stations is? | [
"444 km",
"445 km",
"500 km",
"300 km"
] | A | 1h ----- 5
? ------ 60
12 h
RS = 16 + 21 = 37
T = 12
D = 37 * 12 = 444
ANSWER A |
AQUA-RAT | AQUA-RAT-35144 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
here's an easy question of averages, but let's try to see innovative ways of solving this.
A class has 12 boys and x girls. Average score of boys and girls is 83 and 92 respectively. the average of the whole class is 86, what is the value of x? | [
"a) 6",
"b) 7",
"c) 8",
"d) 10"
] | A | 12(83)+92x/12+x=86
996+92x/12+x=86
996+92x=86(12+x)
996+92x=1032+86x
X's one side, numbers one side we get,
92x - 86x=1032-996
6x=36
hence, x=6
Answer A |
AQUA-RAT | AQUA-RAT-35145 | => 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.
In May Mrs Lee's earnings were 45 percent of the Lee family's total income. In June Mrs Lee earned 20 percent more than in May. If the rest of the family's income was the same both months, then, in June, Mrs Lee's earnings were approximately what percent of the Lee family's total income ? | [
"50%",
"68%",
"72%",
"76%"
] | A | Let in May Lee family's total income =100
in May Mrs Lee's income =45
in May rest of the family's income = 55
In june Mrs Lees income = 45*120/100 = 54
In june Total income = 54+ 55= 109
% of Mrs Lee's income =54/109 =49.54
(A) |
AQUA-RAT | AQUA-RAT-35146 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed at which a man can row a boat in still water is 18 kmph. If he rows downstream, where the speed of current is 3 kmph, what time will he take to cover 60 metres? | [
"16 seconds",
"15.4 seconds",
"26 seconds",
"12 seconds"
] | B | Speed of the boat downstream = 18 + 3
= 21 kmph
= 21 * 5/18 = 35/9 m/s
Hence time taken to cover 60 m
= 60*9/35
= 15.4 seconds.
Answer: B |
AQUA-RAT | AQUA-RAT-35147 | Note that $-169 + ( 4 \times 52 ) = 39$, so $-169 \equiv 39 \mod 52$.
Try this yourself for the second question.
• how did you get the four? – Nicole Oct 13 '14 at 23:36
• @Nicole Four is the number of times you need to add 52 to -169 to get a positive number. Note that adding any lesser multiple of 52 will leave you with a negative number. – Nick Oct 13 '14 at 23:38
The following is multiple choice question (with options) to answer.
If 5395 / 4.15 = 1300, then 539.5 / 41.5 is equal to ? | [
"17",
"13",
"15",
"14"
] | B | Answer
Given expression 539.5 / 41.5 = 5395 / 415
= 5395 / (415 x 100)
= 1300 / 100
=13
Correct Option: B |
AQUA-RAT | AQUA-RAT-35148 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
Two pipes A and B can separately fill a cistern in 10 and 15 minutes respectively. A person opens both the pipes together when the cistern should have been was full he finds the waste pipe open. He then closes the waste pipe and in another 4 minutes the cistern was full. In what time can the waste pipe empty the cistern when fill? | [
"2 min",
"8 min",
"9 min",
"3 min"
] | B | 1/10 + 1/15 = 1/6 * 4 = 2/3
1 - 2/3 = 1/3
1/10 + 1/15 - 1/x = 1/3
x = 8
Answer: B |
AQUA-RAT | AQUA-RAT-35149 | equilibrium, precipitation
Now the OP has modified the problem statement to be:
Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.
The following is multiple choice question (with options) to answer.
An iron mining company depletes 1/11 of the iron ore in a pit it mines over a year. If the company opens a new iron ore mining pit in year 1 and if the initial reserves of the pit were estimated to be 9000 MT (million tons), what is the approximate total quantity of iron ore that would have been mined from that pit by the end of year 4? | [
"5728 MT",
"2853 MT",
"4514 MT",
"3152 MT"
] | B | X is the initial quantity of the iron ore in the pit.
After a year of mining the quantity of iron ore left would be X1=X-1/11X i.e. X1=10/11X
Same thing for X2, X3, X4, respectively representing the quantity left after 2, 3, 4 years of mining. (quantity at the end of the year)
So we have:
X1=10/11 X
X2=10/11 X1
X3=10/11 X2
X4=10/11 X3
By multiplying both ends of the equations we get:
X1*X2*X3*X4 = (10/11)^4 X*X1*X2*X3
Simplification leaves us with:
X4 = (10/11)^4 X
X4 is the quantity left at the pit by the end of year4. So the cumulative quantity (Q4) that has been mined from the pit after 4 years is: Q4= X-X4=X-(10/11)^4X i.e. Q4 = X (1-(10/11)^4)
In this case, given X was: X=9000 MT
So we have Q4 = 9,000 * (1-(10/11)^4) = 9000 * (1- 10,000/14,641) = 9,000 * (14,641-10,000)/14,641 = 9,000 * 4,641/14641 = 2852.87 rounded to 2,853 So Q4 = 2,853 MT.
Answer: B |
AQUA-RAT | AQUA-RAT-35150 | soft-question
Title: Strange equation on notebook It could sound like a silly question, but I found this equation written in my notebook and never knew what was the meaning and who wrote it. I think it could be related to some physics equation, and that's why I'm adding it here, but please tell my if it is off topic. Can you interpret it? The second term on the left has what looks like a square root symbol with nothing underneath.
The first term has an integral sign with no measure.
The second character in the second and third terms is not a symbol I've ever seen (though it is slightly similar to a capital sigma, $\Sigma$.)
I highly doubt that it is an actual mathematical statement, at least not with any symbol set I've seen. My first guess is that it's just someone writing down mathematical symbols that look cool. I have certainly done that before.
It might also be some sort of code or joke that uses math symbols instead of letters like leet-speak. But nothing stands out to me.
The following is multiple choice question (with options) to answer.
What is it answer?& it is a letter. | [
"22",
"77",
"397",
"27"
] | B | Explanation:
It is ASHOK as 01100101 10000011 01110010 01111001 01110101
65 83 72 79 75
A S H O K
Answer: B |
AQUA-RAT | AQUA-RAT-35151 | combinatorics, sets, matching
Put your students in a row: $a!$ possibilities. The first $b$ students form the first group, the next $b$ students the second group, etc. The order in the line-up between each group of students does not matter, so we have to divide by th1s number of possibilities which is $(c!)^b$. Again we divide by $(a-bc)!$ the number of permutations of the not chosen ones. Also, the order of the groups themselves does not matter, we additionally divide by $b!$.
Answer: $a!\;/\;{(c!)^b (a-bc)! \;b!}$
Another way of obtaining the same number. Choose a group of $c$ students from total $a$, then a group of $c$ students from the remaining $c-b$, etc. This can be done in ${a \choose c}{a-c\choose c}\dots{a-c(b-1)\choose c}$. Again I divide by the possible orderings of the groups which is $b!$.
Thus $\frac{a!}{c!(a-c)!}\frac{(a-c)!}{c!(a-2c)!}\dots \frac{(a-(b-1)c)!}{c!(a-bc)!} / b!$
Your first example $a=12$, $b=4$, $c=3$. You have 880, this number is 15400.
Second example $a=12$, $b=3$, $c=2$. 13680, we agree.
The following is multiple choice question (with options) to answer.
How many ways are there to split a group of 14 students into two groups of 7 students each? (The order of the groups does not matter) | [
"1016",
"1218",
"1412",
"1716"
] | D | 14C7 =3432
If we consider these groups, each group will be counted twice.
The number of ways to choose 2 groups of 7 is 3432/2 = 1716
The answer is D. |
AQUA-RAT | AQUA-RAT-35152 | # Math Help - Permutation Help!
1. ## Permutation Help!
Problem:
---------
There are 120 five-digit numbers that can be formed by permuting the digits 1,2,3,4 and 5 (for example, 12345, 13254, 52431). What is the sum of all of these numbers?
--------
399 960
I can't get this question...I don't know what to do...
All I think of doing is 120P5 = 2.29 x 10^10 .... but that's wrong
2. I hope that I am wrong. How about that?
But I think that this is really a programming question as opposed to a mathematical problem.
I really don’t see it as otherwise. But number theory is my weakness.
3. Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
Therefore, since we have sixty pairs of these permutations, the sum is 66666*60 = 399960.
4. Originally Posted by DivideBy0
Ah, this is a very nice problem, and there is an elegant solution.
For each permutation, there is another one that can be added to it so that the sum equals 66666.
Examples: For 12345, there exists exactly one other permutation that sums with it to 66666, and that is 54321.
For 13245 it is 53421, for 34251 it is 32415.
I don't have a formal proof for this, but after some consideration it does seem very intuitively correct.
The following is multiple choice question (with options) to answer.
How may 5 digit odd numbers can be formed from 1 2 3 4 5? | [
"36",
"48",
"72",
"68"
] | C | Since the number is odd last digit is 1,3,or 5
When last digit is 1 , remaining 4 digits can be arranged in 4!= 24 ways
Similarly when last digit is 3, number of ways = 24
and last digit 5, number of ways = 24
Total number of 5 digit odd numbers = 24*3 = 72
ANSWER:C |
AQUA-RAT | AQUA-RAT-35153 | ### 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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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.
City A to city B, Ameer drove for 1 hour at 60 mph and for 3 hours at 50 mph. What was the average speed for the whole trip? | [
"57",
"52.5",
"58.2",
"59"
] | B | The total distance is 1×60+3×50=210
And the total time is 4 hours. Hence,
Average Speed=(Total Distance/Total Time)=210/4=52.5
Answer : B |
AQUA-RAT | AQUA-RAT-35154 | 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.
An aeroplane covers a certain distance of 590 Kmph in 8 hours. to cover the same distance in 2 3/4 hours, it Must travel at a speed of | [
"1780",
"1716",
"1890",
"1980"
] | B | Speed of aeroplane = 590 Kmph
Distance travelled in 8 hours
= 590 * 8 = 4720 Km
Speed of aeroplane to acver 4720 Km in 11/4
= 4720*4/11 = 1716 Km
Answer B. |
AQUA-RAT | AQUA-RAT-35155 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed? | [
"16 days",
"19 days",
"11 days",
"13 days"
] | D | One day work of A, B and C = 1/24 + 1/30 + 1/40 = 1/10 Work done by A and B together in the last 4 days = 4*(1/24 + 1/30) = 3/10
Remaining work = 7/10
The number of days required for this initial work = 7 days.
The total number of days required = 4 + 7 = 11 days.
Answer:D |
AQUA-RAT | AQUA-RAT-35156 | 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 travelling 80kmph to pass through a tunnel of 70km at 5:12am. the train leaves the tunnel at 5:18am. find the length of train? | [
"1",
"2",
"3",
"4"
] | A | Distance=speed*time
let x is length of train
5:18-5:12=6minutes=6/60 hr=1/10hr
now D=(x+7)=80kmph*1/10
x=1km
ANSWER:A |
AQUA-RAT | AQUA-RAT-35157 | 5. Thanks!
6. Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?
7. Originally Posted by ThePerfectHacker
Welcomed.
That happens to be a nice problem which I really like. Where did you get it from?
From high school book. Author has also wrote book with variety of problems which i find most interesting.
The following is multiple choice question (with options) to answer.
If Matthew and Fred are both offering Chemistry and Mathematics, which of the following could be true?
I. Matthew and Fred both offer Chemistry.
II. Fred does not love Chemistry but offers it.
III. Matthew loves Chemistry but Fred does not. | [
" I only",
" I and II only",
" I and III only",
" II and III only"
] | A | Let’s check each statement.
Statement I:Matthew and Fred both offer Chemistry. Fred offers Chemistry, Matthew offers Chemistry in which case Statement I is TRUE.
Statement II: Fred does not love Chemistry but offers it. There is no mention of Fred's love for Chemistry in which case Statement II can NEVER BE TRUE.
Statement III: Matthew loves Chemistry but fred does not. There is no mention of Fred's and Matthew's love for Chemistry, in which case Statement III can NEVER BE TRUE.
Only Statement I is true.
So,the correct answer is A |
AQUA-RAT | AQUA-RAT-35158 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern is filled by a tap in 4 1/2 hours. Due to leak in the bottom of the cistern, it takes half an hour longer to fill the cistern. If the cistern is full how many hours will it take the leak to empty it? | [
"30",
"36",
"42",
"45"
] | D | filling rate - leak rate = net rate
1/4.5 - leak rate = 1/5
leak rate = 2/9 - 1/5 = 1/45
The answer is D. |
AQUA-RAT | AQUA-RAT-35159 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
Find the odd man out. 5, 10, 40, 81, 320, 640, 2560 | [
"81",
"5",
"10",
"81"
] | A | Alternatively 2 and 4 are multiplied with the previous terms
5
5 × 2 = 10
10 × 4 = 40
40 × 2 = 80
80 × 4 = 320
320 × 2 = 640
640 × 4 = 2560
Hence, 81 is wrong. 80 should have come in place of 81.
Answer is A |
AQUA-RAT | AQUA-RAT-35160 | # What is the probability distribution of the points obtained by player A in this game?
The game is as follows:
• Two players A and B play consecutive rounds, and the winner of each round obtains one point.
• Each round is independent of the others, and player A has probability $p$ of winning, player B has probability $1-p$.
• The game ends when one of the players obtain N points.
If $n_A$ is the number of points obtained by player A, what is the value of the probability $P(n_A=i) ~,~i=0,1,...,N$? i.e. what is the probability mass function of $n_A$?
Background:
I'm currently learning probability on my own, using the textbook A First Course in Probability - Sheldon Ross, 8th edition". When doing the self-test exercise 4.11, I had the idea of generalizing the problem to the one I posted here.
Similar, but different question:
## 3 Answers
You just need to separate the winning state where the points will be N from the other states.
Define $f(N,k,a) = \binom {N+k-1}{k}a^k$ as an auxiliary function to simplify the following.
We can write the points distribution for player $A$ $$n_A(k) = \left\{ \begin{array}{ll} q^Nf(N,k,p) & k={0 \dots (N-1)} \\ p^N \sum_{k=0}^{N-1} f(N,k,q) & k=N \end{array} \right.$$
simply swap $p$ and $q$ for player $B$.
The following is multiple choice question (with options) to answer.
In a certain game, each player scores either 2 points or 5 points. If n players score 2 points and m players score 5 points, and the total number of points scored is 50, what is the least possible positive difference Q between n and m? | [
"1",
"3",
"5",
"7"
] | B | We have equation 2n + 5m = 50
We have factor 2 in first number and we have factor 5 in second number.
LCM(2, 5) = 10
So we can try some numbers and we should start from 5 because it will be less list than for 2
2 * 5 = 10 and n should be equal 20
4 * 5 = 20 and n should be equal 15
6 * 5 = 30 and n should be equal 10
8 * 5 = 40 and n should be equal 5
10 * 5 = 50 and n should be equal 0
third variant give us the mininal difference
n - m = 10 - 6 = 4
And there is some mistake in my way of thinking because we don't have such answer )
If we change the task and will seek for difference between m and n
than minimal result Q will be 8 - 5 = 3
And answer B |
AQUA-RAT | AQUA-RAT-35161 | javascript, datetime
// a simple helper function
function nextDay(date) {
return new Date(date.getFullYear(), date.getMonth(), date.getDate() + 1);
}
// as long we're in the same year, keep adding 1 day,
// and store the ones that match the weekday we're looking for
while(current.getFullYear() === year) {
if(current.getDay() === weekday) {
dates.push(current);
}
current = nextDay(current);
}
return dates;
}
No need to build ranges, map, and filter them. Just a while loop and push.
Incidentally, if you want to learn more about why programming calendar and time things are just hideously complex, check out this video on youtube
1) Happy Guy Fawkes day everyone
The following is multiple choice question (with options) to answer.
Today is Wednesday. After 29 days, it will be: | [
"Sunday",
"Saturday",
"Friday",
"Thursday"
] | D | Each day of the week is repeated after 7 days.
So, after 28 days, it will be Wednesday.
After 29 days, it will be Thursday.
Answer :D |
AQUA-RAT | AQUA-RAT-35162 | 0.104768, 0.0591745, 0.0475319, 0.0452583}, {0.0510719, 0.0599374,
0.0730602, 0.0975814, 0.101289, 0.0691997, 0.0498054, 0.044892,
0.043122}, {0.0460517, 0.0567025, 0.0574044, 0.0587778, 0.0537118,
0.0487221, 0.0474098, 0.0413977, 0.04477}}
The following is multiple choice question (with options) to answer.
4.036 divided by 0.04 gives : | [
"10.09",
"1.06",
"10.06",
"100.9"
] | D | = 4.036/0.04
= 403.6/4
= 100.9
Answer is D. |
AQUA-RAT | AQUA-RAT-35163 | 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.
Anne can do a piece of work in 2 days and Emma alone can do it in 5 days. How much time will both take to finish the work ? | [
"A)5",
"B)6.56",
"C)1.42",
"D)8.333"
] | C | This question can be solved by different methods. We need to conserve time in exams so solving this problem using equations is the good idea.
Time taken to finish the job = XY / (X + Y)
= 2 x 5 / (2 + 5)
= 10 / 7
= 1.42 days
Answer: C |
AQUA-RAT | AQUA-RAT-35164 | Tools » Screen Aspect Ratio & Dimension Calculator Screen Aspect Ratio & Dimension ... with a diagonal screen size of and ... ) takes up ??? Then drag the corners to create an arbitrary rectangle.Calculate the length of the diagonals.Click 'show details' to verify your answer. Mathematics - Mathematical rules and laws - numbers, areas, volumes, exponents, trigonometric functions and more ; Related Documents . Height. A = a × b A = a × b. where a a and b b are the length and width of the rectangle, respectively. Keep in mind that this is a decimal fraction; for example .5 is 1/2 inch and .75 is 3/4 inch. Length of Diagonal of Rectangle Formula: The diagonal of a rectangle is determined by the following formula. Width = 10 in Simply enter the length of a side of the square and the diagonal will be calculated quickly. Use our online diagonal of a rectangle calculator to find diagonal of rectangle by entering the width and height. Diagonal of a Square Calculator - calculate the diagonal of a square. Use this square calculator to find the side length, diagonal length, perimeter or area of a geometric square. Enter the measurement that you know (diagonal, width or height) and the other two will be calculated. Area = length x width It is necessary to follow the next steps: Enter the length and width of a rectangle in the box. Related Topics . Right triangle calculator to compute side length, angle, height, area, and perimeter of a right triangle given any 2 values. Diagonal Matrix Calculator is a free online tool that displays the result whether the given matrix is a diagonal or not for the given matrix. % of the device surface area. iForce Systems LLC, Through two sides and the angle between them, Through the radius of the inscribed circle, Through the radius of the circumscribed circle, By the diagonals and the angle between them, Through the diagonals and the angle between them, Through the sides and the angle between them, Total surface area of the regular pyramid across the height, Lateral surface area of the regular pyramid through the height, Lateral surface area of the regular pyramid through the apothem, Isosceles triangle, through side and height, Isosceles triangle, through side and angle. Calculate screen dimensions (height/width/area, in inches or
The following is multiple choice question (with options) to answer.
What would be the length of the diagonal of a square plot whose area is equal to the area of a rectangular plot of 45 m length and 40 m width? | [
"42.5 m",
"60 m",
"4800 m",
"Data inadequate"
] | B | a2 = 45 × 40 = 1800
∴ a = √1800=30√2
∴ Diagonal of the square = √2 a = √2 × 30√2
= 30 × 2 = 60 m
Answer B |
AQUA-RAT | AQUA-RAT-35165 | To get the required result in the book you first have to calculate the amount of the equivalent annual payment. The formula for payments at the beginning of every month is
$C_1=12\cdot r+\frac{\color{blue}{13}\cdot r\cdot i}{2}$
In case of payments at the end of every month $\color{blue}{13}$ has to be replaced by $11$.
$C_1=12\cdot 200+\frac{\color{blue}{13}\cdot 200\cdot 0.045}{2}=2458.5$
To get the Future value after 10 years we use the formula for annual payments.
$C_{10}=2458.5\cdot \frac{1-1.045^{10}}{1-1.045}=30210.56$
But in general I wouldn´t say that your method is worse then the method above. Your result differs from my result about $0.015\%$ only.
The following is multiple choice question (with options) to answer.
What annual payment dischargea debit of Rs.12900, due in 4yrs.At 5% rate? | [
"2000",
"2300",
"3000",
"3300"
] | C | A.P.=(200X12900)/[4(200+5X3)]=3000 Ans
Alternative
100+105+110+115=12900
430=12900
100=12900/(430)X100=3000
C |
AQUA-RAT | AQUA-RAT-35166 | • Thanks for the reminder.
1. 133
2. 10
3. 666
4. 25
For #4, instead of doing it all out, you should instead list the numbers out as if you were going to add them:
(2 + 4 + 6 + 8 + 10 + … + 50)
– (1 + 3 + 5 + 7 + 9 + … + 49)
—————————————-
1 + 1 + 1 + 1 + 1 + … + 1 = 25
Notice you get a bunch of 1’s when you line them up and subtract. Now how many 1’s are there? Well, there are 25 even numbers and 25 odd numbers from 1 to 50, so you will have 25 columns lined up. Therefore, you will get 25 1’s.
3. Maroosh says:
Thank you very much for answers. Effort very much appreciated!! And thank you for explaining how to do question 4 in a shorter method. I got all right except question 2 for which i got 6 instead of 10, namely; 4567, 4568, 4569, 4678, 4679, 4789
i really don’t see any other possibilities because the digits always have to increase from left to right. Please help on this one as well. i will be very grateful
4. Maroosh Qazi says:
Thank you very much for the hint however I don’t know what I am doing but I am still only getting 9. The 3 more numbers I have figured out now are: 4578 (as you hinted), 4579 and 4689.
So the complete list is: 4567, 4568, 4569, 4578, 4579, 4678, 4679, 4689, 4789.
5. For your example #2, above, the “trifactorable” problem, I think you overlooked a tenth solution,
0 x 1 x 2 = 2, another positive integer under 1000. Just a quibble! Thanks for the lessons, which I am using for my daughter’s benefit.
6. Zeina says:
The following is multiple choice question (with options) to answer.
Insert the missing number.
3, 6, 6, 12, 9, 18, 12, 24, 15, (....) | [
"30",
"36",
"42",
"40"
] | A | Explanation:
There are two series, beginning respectively with 3 and 6. In one 3 is added and in another 6 is added.
The next number is 24+ 6 = 30.
Answer: A) 30 |
AQUA-RAT | AQUA-RAT-35167 | # If $28a + 30b + 31c = 365$, then what is the value of $a +b +c$?
Question: For 3 non negative integers $a, b, c$; if $28a + 30b + 31c = 365$ what is the value of $a +b +c$ ?
How I approached it : I started immediately breaking it onto this form on seeing it :
$28(a +b +c) +2b +3c = 365 .......(1)$ $30(a +b +c) -2a +c = 365 .......(2)$ $31(a +b +c) -b -3a = 365 .......(3)$
And then I find out that
$365 = 28*13 + 1......(1')$; $365 = 30*12 + 5......(2')$ $365 = 31*11 + 24......(3')$
Now as we see (1) and (1') as well as (3) and (3') or even equations $2$ and $2'$ do not combine quiet congruently, so I meet with a dead end here.
my issue : how should I approach such problems where we are given no other equations or data? Basically I am asking what are a few ways to get a solution for this problem.
• $28\times 2 + 30\times 7 + 31\times 3 = 359\ne 365$ – Joey Zou Aug 11 '16 at 21:25
• This is a trick question, you are supposed to observe that these are the lengths of months and 365 is the length of a normal year. – f'' Aug 11 '16 at 21:31
More direct path to $a+b+c = 12$:
The following is multiple choice question (with options) to answer.
If 3a – 2b – 2c = 30 and √3a-√(2b+2c)=4, what is the value of a + b + c? | [
"3",
"9",
"10",
"12"
] | C | when we look at the two equations, we can relize some similarity, so lets work on it..
3a – 2b – 2c = 32 can be written as √3a^2-√(2b+2c)^2=32
{√3a-√(2b+2c)}{√3a+√(2b+2c)}=32..
or 4*√3a+√(2b+2c)=32..
or √3a+√(2b+2c)=8..
now lets work on these two equations
1)√3a-√(2b+2c)=4..
2)√3a+√(2b+2c)=8..
A) add the two eq..
√3a+√(2b+2c)+√3a-√(2b+2c)=12..
2√3a=12..
or √3a=6..
3a=36..
a=12.
B) subtract 1 from 2..
√3a+√(2b+2c)-√3a+√(2b+2c)=4..
2√(2b+2c)=4..
√(2b+2c)=2..
2b+2c=4..
or b+c=2..
from A and B a+b+c=12+2=14..
C |
AQUA-RAT | AQUA-RAT-35168 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The average age of 15 students of a class is 15 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years. Tee age of the 15th student is: | [
"11",
"14",
"15",
"13"
] | A | Age of the 15th student=
[15 * 15 - (14 * 5 + 16 * 9)] = (225 - 214) = 11 years.
ANSWER A |
AQUA-RAT | AQUA-RAT-35169 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed? | [
"56m",
"52m",
"27m",
"28m"
] | A | Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed
along the perimeter of the rectangular plot, not in a single straight line which is
very important.
Hence number of poles required = 280 / 5 = 56
Answer: A |
AQUA-RAT | AQUA-RAT-35170 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
How long will a boy take to run round a square field of side 35 meters, if he runs at the rate of 9km/hr? | [
"56sec",
"45sec",
"1min",
"32sec"
] | A | Speed = 9 km/hr = 9*5/18 = 5/2 m/sec
Distance = 35*4 = 140m
Time taken = 140*2/5 = 56sec
Answer is A |
AQUA-RAT | AQUA-RAT-35171 | 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.
Find the compound interest on $ 46,000 at 20% per annum for 9 months, compounded quarterly | [
"2532.05",
"2552.32",
"2524.34",
"7250.75"
] | D | Principal = $ 46000; Time = 9 months =3 quarters;
Rate = 20% per annum = 5% per quarter.
Amount = $ [46000 x (1+(5/100))^3] = $ 53250.75
CI. = $ (53250.75 - 46000) = $ 7250.75
Answer D. |
AQUA-RAT | AQUA-RAT-35172 | 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.
If a particular is never chosen, in how many ways can a cricketbe chosen out of 15 players? | [
"345",
"350",
"364",
"370"
] | C | A particular players is never chosen, it means that 11 players are selected out of 14 players.
=> Required number of ways = 14C11
= 14!/11!x3! = 364
C |
AQUA-RAT | AQUA-RAT-35173 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
A rectangle having length 150 cm and width 40 cm. If the length of the rectangle is increased by fifteen percent then how much percent the breadth should be decreased so as to maintain the same area. | [
"25%",
"9.09%",
"40%",
"75%"
] | B | Explanation :
Solution: (15/(150+15) * 100)% = 9.09%
Answer : B |
AQUA-RAT | AQUA-RAT-35174 | => 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.
Johnny makes $7.35 per hour at his work. If he works 6 hours, how much money will he earn? | [
"$30",
"$54",
"$48.32",
"$44.10"
] | C | 4.75*6=28.50. Answer is C. |
AQUA-RAT | AQUA-RAT-35175 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains 3 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green. | [
"19/20",
"17/20",
"21/40",
"7/10"
] | D | Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
P(A) = (1/2) x (6/10) = 3/10.
P(B) = (1/2) x (4/10) = 1/5.
Hence, the required probability is P(A) + P(B) which is nothing but 7/10
ANSWER:D |
AQUA-RAT | AQUA-RAT-35176 | # 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 mean daily profit made by a shopkeeper in a month of 30 days was Rs. 350. If the mean profit for the first fifteen days was Rs. 275, then the mean profit for the last 15 days would be | [
"Rs. 200",
"Rs. 350",
"Rs. 275",
"Rs. 425"
] | D | Average would be : 350 = (275 + x)/2
On solving, x = 425.
Answer: D |
AQUA-RAT | AQUA-RAT-35177 | 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 person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? | [
"5.6",
"8.9",
"7.2",
"2.5"
] | C | Explanation:
Speed = Distance/Time
Speed =(600/(5*60))m/sec
Converting m/sec to km/hr
=(2*18/5)km/hr=7.2 km/hr
Answer C |
AQUA-RAT | AQUA-RAT-35178 | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
The following is multiple choice question (with options) to answer.
In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons? | [
"200",
"210",
"230",
"250"
] | B | 7p3 = 7 × 6 × 5 = 210
B) |
AQUA-RAT | AQUA-RAT-35179 | Hi, Oscar.
We can certainly say that this is another ambiguity of the original question (which I am presuming you quoted exactly). But just as I read it as saying that two people are drivers and two OTHER people are twins, I definitely read it as asking about ways to choose the five people, without distinguishing who is driving. (Depending on distance, there's good reason to suppose that if there are two drivers in the car, they might take turns, anyway.)
This is similar to the sort of question where we ask, on one hand, how many ways there are to choose a committee of two from a club of 20, and, on the other hand, how many ways there are to choose a president and a treasurer from a club of 20. The members of the committee may well play different roles, but the first question does not distinguish the two committee positions, and so is a combination problem; whereas the second explicitly distinguishes them, making it a permutation problem. In the same way, choosing five people is distinct from choosing what seats some or all sit in.
If we don’t care who sits in the back seat, we don’t care who sits in the driver’s seat either, as long as there is someone who can do so. If the question were stated differently, though, we might care.
## Repairing the problem
Now the discussion changed direction, as I continued:
Rather than disagree about how to interpret the problem you stated, perhaps a more useful thing to do is to work on how to state the problem clearly, given any of the various interpretations we have been considering. I would certainly want to do this before putting the problem on a test.
Here is my version, edited from yours:
Twenty students must return from a school camp.
There are two ways of getting back home. There
is a 5-seat car and a bus. Two of the students
have a license to drive and two other students
are twins who have to go home together.
How many ways can 5 people be chosen to go home
in the car, given that there must be at least
one licensed driver in the car, but it is not
necessary to choose who will actually drive?
Would you like to write the problem in a way that makes your interpretation explicit? You might also try to write a version that explicitly allows one or both twins to be drivers, though that would be considerably harder.
The following is multiple choice question (with options) to answer.
11 speakers gathered to deliver speeches on two topics. Even though 11 speeches were initially planned to be delivered, due to lack of time only 2 speeches were finally decided to be delivered. Topics are to be spoken in a specific order and the speeches differ with respect to speaker or topic. How many such two speeches are possible? | [
"2",
"121",
"50",
"100"
] | B | 1)there are 11 speeches in ONE topic and 11 in OTHER..
2)there is a specific order for topics, so
a) both the topics are to be spoken on- so ONE from each
b) the order does not matter since there is ONLY one specific order..
so 11C1∗11C1=11∗11=121
ANSWER:B |
AQUA-RAT | AQUA-RAT-35180 | $$L(100) \pmod{20} = \color{red}7$$
The remainder when the 100th term of the sequence is divided by 20 is 7
May 23, 2021
edited by heureka May 23, 2021
edited by heureka May 23, 2021
The following is multiple choice question (with options) to answer.
1!+2!+3!+4!+5!+..100! is divided by 24. Find the remainder? | [
"2",
"5",
"9",
"10"
] | C | By applying rule 2, we divide the terms of the above expression individually, and add them to get the final remainder. But from 4! onwards all the terms leave a remainder 0 when divided by 24.
So the remainder = 1 + 2 + 6 + 0 + 0....... =
C |
AQUA-RAT | AQUA-RAT-35181 | ## 5 Feb 2019 Enter the compounding period and stated interest rate into the effective interest rate formula, which is: r = (1 + i/n)^n-1. Where: r = The effective
1 Apr 2019 Based on the method of calculation, interest rates are classified as nominal interest rate, effective interest rate and annual percentage yield 10 Jan 2018 The simple interest rate is the interest rate that the bank charges you for taking the loan. It is also commonly known as the flat rate, nominal rate or 10 Apr 2019 The advertised rate (also known as nominal rate) is the interest the bank charges you on the sum you borrow. Note that there are different ways to 10 Feb 2019 TaxTips.ca - The effective rate of interest depends on the frequency of compounding. (e.g. 6% compounded monthly), the stated rate is the nominal rate. Interest earned on chequing and savings accounts is usually 3 Oct 2017 In this situation, with an effective interest rate of 17.2737 percent, there is very little margin for missing out on making an amortization payment. 1) If I'm given a 7% semi-annual nominal rate, does that mean the annual nominal rate is simply 14%?. No. 7% semi-annual is 3.5% every six months. So annual 27 Nov 2016 Going further, since a nominal APR of 12% corresponds to a daily interest rate of about 0.0328%, we can calculate the effective APR if this
### 19 Apr 2013 The interest rate per annum is only the nominal interest rate. This nominal rate is equal to the effective rate when a loan is on annual-rest basis
The following is multiple choice question (with options) to answer.
The simple interest on Rs. 20 for 6 months at the rate of 5 paise per rupeeper month is | [
"1.2",
"1.4",
"6",
"7"
] | C | Sol.
S.I. = Rs. [20 * 5/100 * 6] = Rs. 6
Answer C |
AQUA-RAT | AQUA-RAT-35182 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
C and D started a business investing Rs. 85,000 and Rs. 15,000 respectively. In what ratio the profit earned after 2 years be divided between P and Q respectively? | [
"10 : 3",
"12 : 3",
"17 : 3",
"20 : 15"
] | C | Sol.
C : D = 85000 : 15000 = 85 : 15 = 17 : 3.
Answer C |
AQUA-RAT | AQUA-RAT-35183 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
Braun invested a certain sum of money at 8% p.a. simple interest for 'n' years. At the end of 'n' years, Braun got back 4 times his original investment. What is the value of n | [
"55/2",
"65/2",
"75/2",
"85/2"
] | C | PA=p
t=n
r=8%
According to question
p+(pnr/100)=4p
p[1+n8/100]=4p
n8/100=3
n=300/8
n=75/2
ANSWER:C |
AQUA-RAT | AQUA-RAT-35184 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 5 p.c? | [
"38008",
"38000",
"38029",
"380219"
] | B | CP + CP/8 = 45000
CP = 40000
SP = 40000*(95/100) = 38000
Answer:B |
AQUA-RAT | AQUA-RAT-35185 | # 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.
I bought two books; for Rs.480. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss? | [
"188",
"280",
"166",
"199"
] | B | x*(85/100) = (480 - x)119/100
x = 280
Answer:B |
AQUA-RAT | AQUA-RAT-35186 | 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.
Otto and Han are driving at constant speeds in opposite directions on a straight highway. At a certain time they are driving toward each other and are 20 miles apart. One and a half hours later, they are again 20 miles apart, driving away from each other. If Otto drives at a speed of x miles per hour, then, in terms of x, Han drives at a speed of how many miles per hour? | [
"a) 80-x",
"b) 40-x",
"c) 26.67-x",
"d) 120-x"
] | C | Let's say the two cars have speeds V1 and V2. The fact that they are moving in opposite direction means that their relative speed is (V1 + V2). In other words, any gap between them will be changing in size at a rate of (V1 + V2). It doesn't matter whether they are moving toward each other or away from each other. If they are approaching each other, the gap between them is decreasing at a rate of (V1 + V2). If they are moving away from each other, the gap between them is increasing at a rate of (V1 + V2). Either way, the number for the rate of change remains the same.
Here, the two cars approach a distance 20 mi, then move away from each other another distance of 20 miles. That's a total distance of 40 miles in 1.5 hr, which gives a rate of:
R = (40 mi)/(1.5) = 26.67 mph
That's the rate of change of the gap, so it must equal the sum of the speeds of the two cars.
One of the speeds is x, and let's call the other y. We want y.
x + y = 26.67
y = 26.67 - x
Answer =(C) |
AQUA-RAT | AQUA-RAT-35187 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Ann invested a certain sum of money in a bank that paid simple interest. The amount grew to $240 at the end of 2 years. She waited for another 3 years and got a final amount of $300. What was the principal amount that she invested at the beginning? | [
"$150",
"$175",
"$200",
"$210"
] | C | Amount at end of 2 years = 240$
Amount at end of 5 years = 300$
Difference in amount for time period of 3 years = 60$
Annual Interest earned = 60/3 = 20$
Principal amount invested at the beginning = 240 - 2*20 = 200$
Answer C |
AQUA-RAT | AQUA-RAT-35188 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A shopkeeper sells 400 metres of cloth for Rs. 18000 at a loss of Rs.5 per metre. Find his cost price for one metre of cloth? | [
"Rs.55",
"Rs.57",
"Rs.54",
"Rs.50"
] | D | SP per metre = 18000/400
= Rs. 45 Loss per metre
= Rs. 5 CP per metre = 45 + 5
= Rs.50
Answer: D |
AQUA-RAT | AQUA-RAT-35189 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
Find large number from below question The difference of two numbers is 1335. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder | [
"1254",
"1376",
"1456",
"1599"
] | D | Let the smaller number be x. Then larger number = (x + 1335).
x + 1335 = 6x + 15
5x = 1320
x = 264
Large number = 264+1335 = 1599
D |
AQUA-RAT | AQUA-RAT-35190 | &\equiv2^3.3^8.7^5.9^3\cr &\equiv8.1.7.9\cr &\equiv4\ .\cr}
The following is multiple choice question (with options) to answer.
((7.3)*(10^2))-((7)*(10^2))=? | [
"300",
"30",
"730",
"7300"
] | B | 10^2 is nothing but a 100.
7.3*(10^2)=7.3*100=730.
7*(10^2)=7*100=700
=730-700=30
The answer is option B |
AQUA-RAT | AQUA-RAT-35191 | 5. ## yes...
Originally Posted by Soroban
Hello, magentarita!
I got a different result . . .
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
So we have: . $6(1.11^n) \:=\:8(1.08^n) \quad\Rightarrow\quad \frac{1.11^n}{1.08^n} \:=\:\frac{8}{6} \quad\Rightarrow\quad \left(\frac{1.11}{1.08}\right)^n \:=\:\frac{4}{3}$
Take logs: . $\ln\left(\frac{1.11}{1.08}\right)^n \:=\:\ln\left(\frac{4}{3}\right) \quad\Rightarrow\quad n\cdot\ln\left(\frac{1.11}{1.08}\right) \:=\:\ln\left(\frac{4}{3}\right)$
. . Hence: . $n \;=\;\frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\ frac{1.11}{1.08}\right)} \;=\;10.4997388$
They will have the same amount of money in about ${\color{blue}10\tfrac{1}{2}}$ years.
At that time, they will have: . $6(1.11^{10.5}) \:=\:8(1.08^{10.5}) \:\approx\:{\color{blue}\17.95}\;\;{\color{red}(d )}$
Yes, the answer is (d) and I want to thank you.
The following is multiple choice question (with options) to answer.
Jadeja has x dollars more than Sharma has, and together they have a total of y dollars. Which of the following represents the number of dollars that Sharma has? | [
"3(y – x)/2",
"5(y – x)/2",
"(y – x)/5",
"(y – x)/2"
] | D | To solve, we will set up two equations. Let's start by defining two variables.
A = number of dollars Sharma has
B= number of dollars Jadeja has
We are given that Jadeja has x dollars more than Sharma. We set up an equation:
B= x + A
We are next given that together they have a total of y dollars. We can set up our second equation:
A + B = y
Since we know that B = x + A, we can substitute x + A for B into the second equation A+ B = y.
Notice that, after the substitution, we will only have variables of A, x, and y. Thus, we have:
A + x + A = y
2A + x = y
2A = y – x
A = (y – x)/2
Answer : D |
AQUA-RAT | AQUA-RAT-35192 | You can see why, in the case of five students, your method gives twice the correct count: each configuration has two descriptions, according to which of the two students who get the same book is included in the four who all get different books. So dividing your answer by $$2$$ gives the correct answer.
With more than five students, the overcounting factor in your method is not so predictable. In the case of six students, you can have three students all getting the same book and the other three each getting one of the remaining books, or two students both getting one book, two other students both getting some other book, and the other two each getting one of the two remaining books. The overcounting factor is different in these two cases: you're a factor of $$3$$ too high in the first case, but a factor of $$4$$ too high in the second. So the correct count is $$1560$$, of which $$480$$ are of the first type and $$1080$$ are of the second type. Your method gives an incorrect answer of $$5760$$, which breaks down as $$3\cdot480+4\cdot1080=5760.$$ You can see that it's going to become increasingly difficult to correct your method as the number of students grows.
• you mean that my solution counts : (1 ,2, 3, 4) = (H,R,L,E) and then (5) = (R) , as different from (1,3,4,5) = (H,L,E,R) and then (2)=R , if I am getting this right? amazing how wrong I was.. Jun 18, 2020 at 10:52
• Yes, that was what the issue was. Jun 18, 2020 at 12:56
The following is multiple choice question (with options) to answer.
The number of people who purchased book A is twice the number of people who purchased book B. The number of people who purchased both books A and B is 500, which is twice the number of people who purchased only book B. What is the number of people Z who purchased only book A? | [
"250",
"500",
"750",
"1000"
] | D | This is best solved using overlapping sets or a Venn Diagram. We know that A = 2B, and that 500 people purchased both A and B. Further, those purchasing both was double those purchasing B only. This gives us 250 people purchasing B only. With the 500 that pruchased both, we have a total of 750 that purchased B and this is 1/2 of those that purchased A. So, 1500 purchased A. Less the 500 that purchased both, Z=1000 purchased A only. (This is much simpler to solve using the Venn diagram).
Correct answer is D. 1000 |
AQUA-RAT | AQUA-RAT-35193 | which means you can rewrite the right hand side as $\frac 12 \left( \cos\left( \frac14 \pi \right) - \cos \left( \frac 12 \pi \right)\right) = \frac 12 \left( \frac 1 {\sqrt{2}} - 0 \right)$, so our ration is: $\frac 12 : \frac 1 {2\sqrt{2}}$, or, more neatly, $\sqrt{2}:1$. What do you know? It was a nice number after all!
The following is multiple choice question (with options) to answer.
What number has a 150:1 ratio to the number 2? | [
"130",
"300",
"200",
"30"
] | B | 150:1 = x: 2
x = 150*2
x=300
ANSWER:B |
AQUA-RAT | AQUA-RAT-35194 | $I-1$ is divisible by $4$
$I=69$ is the largest number that meet the two constraints we have.
A brute force approach abusing symmetries. Notice that if $(x,y)$ is a solution then so is $(y,x)$ and if $(x,y)$ is a solution then so is $(\pm x,\pm y)$.
Solutions including zero are $(1,0),(2,0),(3,0),(4,0),(5,0)$ and each of the pairs has $4$ symmetries hence $5\cdot 4=20$ and plus $(0,0)$ so $21$ solutions with $0$.
Now solutions where $x=y\neq 0$ also have $4$ symmetries $(1,1),(2,2),(3,3)$ so $3\cdot 4=12$ solutions and the rest solutions have $8$ symmetries and thoose are $(4,3),(4,2),(4,1),(3,2),(3,1),(2,1)$ so $6\cdot 8=48$ so in total there are $48+12+21=81$ solutions.
This is not an answer but a comment or a hint (so please don't down-vote it).
Consider this figure:
The following is multiple choice question (with options) to answer.
What is the greatest number of four digits which is divisible by 8, 16, 32 and 40 ? | [
"9930",
"9940",
"9920",
"9950"
] | C | Greatest number of four digits = 9999
LCM of 8, 16, 32 and 40 = 160
9999 ÷ 160 = 62, remainder = 79
Hence, greatest number of four digits which is divisible by 8, 16, 32 and 40
= 9999 - 79 = 9920
answer : C |
AQUA-RAT | AQUA-RAT-35195 | mechanical-engineering, fluid-mechanics, fluid
I am doubting my calculations because 3500 litres going past a hole of radius 0.157304491 millimetres every hour seems absurd.
This is area. You need diameter. Multiply it by 4 then divide it by pi then take root of the result. That will be diameter of hole
The following is multiple choice question (with options) to answer.
Find the expenditure on digging a well 14m deep and of 3m diameter at Rs.15 per cubic meter? | [
"Rs.1185",
"Rs.1285",
"Rs.1385",
"Rs.1485"
] | D | 22/7 * 14 * 3/2 * 3/2 = 99 m2
99 * 15 = 1485
ANSWER:D |
AQUA-RAT | AQUA-RAT-35196 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
5 men and 11 boys finish a job in 13 days, 11 men and 11 boys finish it in 10 days. 6 men and 6 boys shall finish it in how many days? | [
"12",
"17",
"10",
"13"
] | D | 5 M + 11 B ----- 13 days
11 M + 11 B ------- 10 days
06 M + 06 B -------?
55 M + 121 B = 66 M +66 B
55 B = 11 M => 5 B = 1 M
25 B + 11 B = 36 B ---- 13 days
30 B + 6 B = 36 B -----? => 13 days
Answer: D |
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