source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-33797 | $a_{18}=$
07545265152740184887140788322673806569482388835389
5577110370797470603035554930
$a_{19}=$
02591621894664804222839429868664505564743756550515
2520842332602724614579447809
$a_{20}=$
04791002227899384351266879075743764807247161403811
8767378458621521760044966007
$a_{21}=$
03251071871924939761772100645669847224066002842238
6690935371046248267119967874
$a_{22}=$
07211128555514235391448579740428274673170438137060
9390617781010839144521896079
$a_{23}=$
02839820419745979344283855308465698534375525126267
1701870835230228506944995955
$a_{24}=$
06304631891686637702274634195264042846471748931602
4893381338158934204519928855
$a_{25}=$
06492095235781034422561843267711627481401158404402
2978856782776323231230432687
$a_{26}=$
11078868891712009912929762366314190797941038596568
The following is multiple choice question (with options) to answer.
If the price of a certain computer increased 30 percent from a dollars to 351 dollars, then 2a = | [
"540",
"570",
"619",
"649"
] | A | Before Price increase Price = a
After 30% Price increase Price = a+(30/100)*a= 1.3a = 351 (Given)
i.e. a= 351/1.3 = $270
i.e. 2a = 2*270 = 540
Answer: option A |
AQUA-RAT | AQUA-RAT-33798 | # In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty?
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.
My attempt:-
First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.
Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.
Total number of ways $= 60\cdot9 = 540$.
Where am I going wrong ?
• When you place the last two balls, you are over counting. Your answer better be less than $243$. Oct 1 '17 at 10:51
There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are $$\binom{3}{1}2^5$$ ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is $$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$ by the Inclusion-Exclusion Principle.
Where am I going wrong?
The following is multiple choice question (with options) to answer.
A man has a certain number of small boxes to pack into parcles. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcle, none is left over. What is the number of boxes, he may have to pack? | [
"106",
"301",
"309",
"400"
] | B | Explanation:
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.
Answer: B) 301 |
AQUA-RAT | AQUA-RAT-33799 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A completes a work in 12 days and B complete the same work in 36 days. If both of them work together, then the number of days required to complete the work will be? | [
"9 days",
"11 days",
"21 days",
"22 days"
] | A | If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 12 × 36/ 48 = 9 days.
Option 'A' |
AQUA-RAT | AQUA-RAT-33800 | # how many cubes have at least $1,2,3$ colors on them
I have a painted cube, which is cut into $n^3$ smaller cubes. I now want to find the number of cubes which have $1$,$2$,$3$ sides painted. I know the long way round of taking each cube and putting them into different categories... but is there a short way or formula to do it?
$1$ face painted - You have to consider all the faces. There are $6$ faces, each with $(n-2)^2$ cubes with $1$ face painted . That gives you the formula $$6*(n-2)^2$$
$2$ faces painted - You have to take the edges. There are $12$ edges, each edge having $n-2$ cubes with $2$ faces painted . That gives you the formula $$12*(n-2)$$
$3$ faces painted - You have to take the corners alone. That gives you the formula $$8$$
Extra - $0$ faces painted - You have to consider the interior alone which gives $$(n-2)^3$$
• I think that you want $8$ corners. ;-) – Sammy Black Feb 23 '16 at 7:20
• @SammyBlack Thank you for that. Edited. – Win Vineeth Feb 23 '16 at 7:20
• saying side is not corrrect it is face which is painted – Bhaskara-III Feb 23 '16 at 13:17
• @Bhaskara-III The question asked for side, hence, I used side. – Win Vineeth Feb 23 '16 at 13:20
• oops you should have correctly mentioned that it should be a painted face not a painted side in your answer because it's very confusing term – Bhaskara-III Feb 23 '16 at 13:22
The following is multiple choice question (with options) to answer.
All the faces of cubes are painted with red colour. It is then cut into 64 equal small cubes.Find How many small cubes are there whose two adjacent faces are coloured red ? | [
"0",
"8",
"16",
"24"
] | D | There are 64 small cubes,Hence one side side of the big cube is 3√64=4 cm
Number of small cubes having two adjacent faces coloured red = (x - 2) x No. of edges
= (4 - 2) x 12
= 24
Answer :D |
AQUA-RAT | AQUA-RAT-33801 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Hitesh is 40 years old and Ronnie is- 60 years old. How many years ago was the ratio of their ages 3:5 ? | [
"5 years",
"10 years",
"20 years",
"15 years"
] | B | Suppose, the ratio was 3 : 5, x years ago.
Then,
(40-x)/(60-x)=3/5
=5 (40 - x) = 3 (60 - x)
=200-5x =180-3x
=2x = 20
= x = 10.
Answer: B |
AQUA-RAT | AQUA-RAT-33802 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
Evaluate the expression
m - (m - n) : (-2) ⋅ (-5) when m = -4, n = -6 | [
"8",
"9",
"5",
"7"
] | B | Solution:
m - (m - n) : (-2) ⋅ (-5) = (-4) - [-4 - (-6)] : (-2) ⋅ (-5) = -4 - 2 : (-2) ⋅ (-5) = -4 - (-1) ⋅ (-5) = -4 - (+5) = -9
Answer B |
AQUA-RAT | AQUA-RAT-33803 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy? | [
"8",
"9",
"7",
"5"
] | B | 3x + 5x + 7x = 45
x =3
3x = 9
Answer:B |
AQUA-RAT | AQUA-RAT-33804 | # Math Help - ChessBoard
1. ## ChessBoard
Theres a chess board.. regular one with 64 squares.
If two squares are selected at random Find the probability that they have a common side..
2. Originally Posted by umangarora
Theres a chess board.. regular one with 64 squares.
If two squares are selected at random Find the probability that they have a common side..
Consider three cases: 1. first square choosen is a corner sq, 2. first square on a side but not a corner sq 3. first square an interior square.
CB
3. plz can u show working n ans.. im clueless.
4. Originally Posted by umangarora
plz can u show working n ans.. im clueless.
There are 4 corner squares, out of a total of 64, therefore the probability of the first sqyuare being a corner is 1/16. Then there are 8 squares adjacent to a corner square so the probability of choosing one of thes3es as the second square is 8/63. So the probability of choosing two adjacent squares the first of which is a corner is 1/126.
Now repeat for edges and interior first choices.
CB
5. Also, be careful that you do not double-count "pairs of squares" when dealing with the edge squares,
since the "order" in which squares are chosen does not matter.
6. Hello, umangarora!
There's a chessboard ... regular one with 64 squares.
If two squares are selected at random,
find the probability that they have a common side.
Choosing 2 squares from the available 64 squares,
. . there are: . $_{64}C_2 \:=\:{64\choose2} \:=\:2016$ choices.
How many of these are "dominos"? . $\square\!\square$
In a row, $\square\!\square\!\square\!\square\!\square\!\squa re\!\square\!\square$, there are 7 ways to place a domino.
. . With 8 rows, there are 56 possible "horizontal" dominos.
The same is true for the columns: .56 possible "vertical" dominos.
Hence, there are: . $56 + 56 \:=\:112$ possible dominos.
The following is multiple choice question (with options) to answer.
A certain board game has a row of squares numbered 1 to 100. If a game piece is placed on a random square and then moved 13 consecutive spaces in a random direction, what is the probability the piece ends no more than 13 spaces from the square numbered 49? | [
"28%",
"27%",
"30%",
"37%"
] | B | No more than 13 spaces from 49 means in the rangefrom 49-13=36 to 49+13=62, inclusive. Total numbers in this range 62-36+1=27, the probability favorable/total=27/100
Answer: B |
AQUA-RAT | AQUA-RAT-33805 | Now find the time Rick spends running.
$\displaystyle t_{r,R}=\frac{D}{2v_r}$
Now just add the two times up and you’re done.
$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$
#### PART B. Find Rick’s average speed for covering the distance D.
You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.
$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$
#### PART C. How long does it take Tim to cover the distance?
Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.
$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$
The time is just the distance divided by the average speed.
$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$
#### PART D. Who covers the distance D more quickly?
Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?
#### PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?
$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$
This is just simple subtraction between the two computed times.
The following is multiple choice question (with options) to answer.
Walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in. | [
"12 min",
"40 min",
"60 min",
"30 min"
] | B | dist = 4*22/4 = 22km
new speed = 16.5 kmph
=>D/S = 11/16.5 = 40 min
ANSWER B |
AQUA-RAT | AQUA-RAT-33806 | 1) What the probability that exactly two of the group of friends is chosen?
2 of the friends chose means $$P(Friend~draw1) = \frac{7}{35}$$ and then $$P(Friend~draw2) = \frac{6}{34}$$ and finally $$P(Not~ Friend~draw3) = \frac{28}{33}$$ This give a probability of $$P(2~friends~on~3~draws) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}$$ However you can draw {friend,friend,Not friend} or {friend,Not friend,friend} or {Not friend,friend,friend} there are $3 \choose 2$ ways of picking 2 friends out of 3. In general we can write the probability of picking from a group as $\frac{f}{f+n}$ and $\frac{n}{f+n}$ where $f$ and $n$ are the number of remaining people in each group respectively. $$~~P(friend,friend,Not friend) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}$$ $$~~P(friend,Not friend,friend) = \frac{7}{35}\times\frac{28}{34}\times\frac{6}{33}$$ $$+~P(Not friend,friend,friend) = \frac{28}{35}\times\frac{7}{34}\times\frac{6}{33}$$ Since multiplication is both associative and commutative these 3 probabilities are equivalent. Therefore we can write $$~~P(2 ~friends ~ on ~3 ~draws) = \frac{7}{35}\times\frac{6}{34}\times\frac{28}{33}\times 3$$
The following is multiple choice question (with options) to answer.
In a friendship gang Aravind has 2 gang, in how many ways can he invite one or more of the gang to his house? | [
"25",
"49",
"3",
"87"
] | C | Aravind can select one or more than one of his 8 gang.
=> Required number of ways = 2^2 – 1= 3.
C |
AQUA-RAT | AQUA-RAT-33807 | 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.
Shyam visited Ram during his brief vacation. In the mornings they both would go for yoga. In the
evenings they would play tennis. To have more fun, they indulge only in one activity per day, i.e.
either they went for yoga or played tennis each day. There were days when they were lazy and
stayed home all day long. There were 24 mornings when they did nothing, 14 evenings when they stayed at home, and a total of 22 days when they did yoga or played tennis. For how many days Shyam stayed with Ram? | [
"32",
"24",
"30",
"None of these"
] | C | Explanation :
Let ,
P days : they play tennis.
Y days : they went to yoga.
T days : total duration for which Ram and Shyam were together.
=> P + Y = 22.
Also, (T - Y ) = 24 & (T - P) = 14.
Adding all of them,
2T = 22 + 24 + 14
=> T = 30 days.
Answer : C |
AQUA-RAT | AQUA-RAT-33808 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.630 find the share of A? | [
"240",
"882",
"727",
"199"
] | A | (3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 630 = 240
Answer: A |
AQUA-RAT | AQUA-RAT-33809 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If 36 men do a work in 70 days, in how many days will 40 men do it? | [
"64",
"63",
"65",
"66"
] | B | 36 * 70 = 40 * x
x = 63 days
Answer: B |
AQUA-RAT | AQUA-RAT-33810 | $$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games
The following is multiple choice question (with options) to answer.
During a certain season, a team won 85 percent of its first 100 games and 50 percent of its remaining games. If the team won 70 percent of its games for the entire season, what was the total number of games that the team played? | [
"180",
"175",
"156",
"150"
] | B | We are first given that a team won 85 percent of its first 100 games. This means the team won 0.85 x 100 = 85 games out of its first 100 games.
We are next given that the team won 50 percent of its remaining games. If we use variable T to represent the total number of games in the season, then we can say T – 100 equals the number of remaining games in the season. Thus we can say:
0.5(T – 100) = number of wins for remaining games
0.5T – 50 = number of wins for remaining games
Lastly, we are given that team won 70 percent of all games played in the season. That is, they won 0.7T games in the entire season. With this we can set up the equation:
Number of first 100 games won + Number of games won for remaining games = Total Number of games won in the entire season
85 + 0.5T – 50 = 0.7T
35 = 0.2T
350 = 2T
175 = T
Answer is B. |
AQUA-RAT | AQUA-RAT-33811 | A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:
(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:
99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)
Therefore, the probability is: 24,000 / 90,000 ==4 / 15
Feb 4, 2023
The following is multiple choice question (with options) to answer.
The sum of all two digit numbers divisible by 5 is | [
"945",
"545",
"745",
"645"
] | A | Required numbers are 10,15,20 . . . . . 95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,
10+5n-5=95
5+5n=95
5n=95-5
n = 90/5
then n=18.
Required sum
=n/2(a+l)
=18/2(10+95)
=945
Answer is A. |
AQUA-RAT | AQUA-RAT-33812 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
Find the principle on a certain sum of money at 5% per annum for 3 1/5 years if the amount being Rs.1160? | [
"Rs.1000",
"Rs.1100",
"Rs.1010",
"Rs.10000"
] | A | Explanation:
1160 = P [1 + (5*16/5)/100]
P= 1000
Answer: Option A |
AQUA-RAT | AQUA-RAT-33813 | ## 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 pump will fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ? | [
"20 hrs",
"21 hrs",
"22 hrs",
"23 hrs"
] | B | work done for 1 hr without leak = 1/3
work done with leak = 31/2 ==>7/2
work and leak in 1 hr = 2/7
work leak in 1 hr = 1/3 - 2/7 = 1/21
so 21 hrs
ANSWER B |
AQUA-RAT | AQUA-RAT-33814 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
At present the ratio between the ages of Arun and Deepak is 4:3. After 6years, Arun's age will be 26 years. What is the age of Deepak at present? | [
"15",
"21",
"23",
"12"
] | A | Let the present ages of Arun and Deepak be 4x years and 3x years respectively
4x+6 = 26
4x = 20
x = 5
Deepak's age = 3x = 15years
Answer is A |
AQUA-RAT | AQUA-RAT-33815 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
A class consists of 100 students, 20 of them are girls and 80 boys; 20 of them are rich and remaining poor; 40 of them are fair complexioned. The probability of selecting a fair complexioned rich girl is | [
"2/125",
"1/59",
"1/56",
"1/54"
] | A | The probability of selecting girl is: 20/100 = 1/5
The probability of selecting rich is: 20/100 = 1/5
The probability of selecting fair complexioned is: 40/100 = 2/5
Three are independent;probability of rich and fair complexioned girl is:
(1/5) ×(1/5) × (2/5) = 2/125
Answer: A |
AQUA-RAT | AQUA-RAT-33816 | main: # @main
pushq %rbp
movq %rsp, %rbp
pushq %r15
pushq %r14
pushq %r13
pushq %r12
pushq %rbx
pushq %rax
movl $1, %r15d leaq -42(%rbp), %r14 .LBB0_1: # %._crit_edge cmpl$100, %r15d
movl $0, %edi movl$60, %eax
cmovgl %eax, %edi
movslq %r15d, %r15
cmpl $101, %r15d setl %al movzbl %al, %esi orl %esi, %edi imulq$1431655766, %r15, %rcx # imm = 0x55555556
imulq $1717986919, %r15, %rax # imm = 0x66666667 movq %rax, %r10 shrq$63, %r10
shrq $32, %rax movl %eax, %edx sarl$2, %edx
leal (%rdx,%r10), %ebx
movq %rcx, %r11
shrq $63, %r11 cmpl$9, %r15d
setg %r8b
imull $10, %ebx, %ebx negl %ebx leal 48(%rdx,%r10), %edx movb %dl, -42(%rbp) leal 48(%r15,%rbx), %r9d shrq$32, %rcx
leal (%rcx,%rcx,2), %ecx
movl %r15d, %edx
subl %ecx, %edx
sete %r12b
movzbl %r8b, %ecx
movb %r9b, -42(%rbp,%rcx)
setne %bl
sarl %eax
leal (%rax,%rax,4), %eax
The following is multiple choice question (with options) to answer.
Express 25 mps in kmph? | [
"22",
"88",
"90",
"21"
] | C | 25 * 18/5 = 90 kmph
Answer: C |
AQUA-RAT | AQUA-RAT-33817 | the number of committees that exist where the man and women serve together is given by,
$$\begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$
so the total number of committees in this case amounts to,
$$1120 - 210 = 910$$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
The following is multiple choice question (with options) to answer.
A committee of 6 is chosen from 5 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed? | [
"150",
"700",
"1404",
"2620"
] | A | The only cases possible are :
1. 2 men and 4 women : 5C2 * 5C4 = 50
2. 3 men and 3 women: 5C3*5C3 = 100
Rest of the cases will either have 1 or 0 men (not allowed) or will have 1 or 2 or 0 women (not allowed)
Total possible combinations = 50+100 = 150. Thus A is the correct answer. |
AQUA-RAT | AQUA-RAT-33818 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
How many even multiples of 45 are there between 449 and 901? | [
"5",
"6",
"9",
"10"
] | B | 450 = 10*45
900 = 20*45
The even multiples are 45 multiplied by 10, 12, 14, 16, 18, and 20 for a total of 6.
The answer is B. |
AQUA-RAT | AQUA-RAT-33819 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
How many ways are there to split a group of 8 girls into two groups of 4 girls each? (The order of the groups does not matter) | [
"30",
"40",
"35",
"45"
] | C | the combination is 8C4 /2
= 8!/4!*4! *2 = 35
C |
AQUA-RAT | AQUA-RAT-33820 | +0
# At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
0
2788
3
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Feb 25, 2015
#3
+99377
+5
At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Mmm
Let there be k people at the party.
The first person shook with k-1 people.
the second with a further k-2 people
the kth person did not shake with anyone new.
So the number of handshakes was
1+2+3+.....+(k-1)
this is the sum of an AP S=n/2(a+L) = $$\frac{k-1}{2}(1+(k-1))=\frac{k(k-1)}{2}$$
so
$$\\\frac{k(k-1)}{2}=66\\\\ k(k-1)=132\\\\ k^2-k-132=0\\\\$$
$${{\mathtt{k}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{k}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{k}} = {\mathtt{12}}\\ {\mathtt{k}} = -{\mathtt{11}}\\ \end{array} \right\}$$
Obviously there is not a neg number of people so there must be 12 people.
Feb 25, 2015
#1
0
can you even solve that
Feb 25, 2015
#2
+98196
+5
We can solve this by this "formula"
n(n-1)/ 2 = 66 multiply by 2 on each side
n(n-1) = 132 simplify and rearrange
n^2 - n - 132 = 0 factor
The following is multiple choice question (with options) to answer.
49 members attended the party. In that 22 are males, 27 are females. The shake hands are done between males, females, male and female. Total 12 people given shake hands. How many such kinds of such shake hands are possible? | [
"66",
"291",
"26",
"29"
] | A | If only 12 people shaked their hands, then total hand shakes are 12C212C2 = 66
Answer:A |
AQUA-RAT | AQUA-RAT-33821 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
Martin has to secure 80% marks to clear his exam of class 9th.
He got 200 marks and failed by 200 marks.
What is the maximum marks ? | [
"500",
"505",
"510",
"515"
] | A | A
500
To pass the exam ravish needs 200 + 200 = 400 marks.
=> (400/80) *100 = 500 |
AQUA-RAT | AQUA-RAT-33822 | Of course not! It is actually due to the fact that $102 = 6 \cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \cdot 17$, then $391 = 17 \cdot (17 + 6) = 17 \cdot 23$, and $493 = 17 \cdot (23 + 6) = 17 \cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.
This sequence of multiples of 17 starts from $17 \cdot 5$, and if we continue it further we see that it contains several more numbers from our exceptional set as well:
$\begin{array}{rcl}85 &=& 17 \cdot 5 \\ 187 &=& 17 \cdot 11 \\ \mathbf{289} &=& 17 \cdot 17 \\ \mathbf{391} &=& 17 \cdot 23 \\ \mathbf{493} &=& 17 \cdot 29 \\ 595 &=& 17 \cdot 35 \\ \mathbf{697} &=& 17 \cdot 41 \\ \mathbf{799} &=& 17 \cdot 47 \\ \mathbf{901} &=& 17 \cdot 53 \end{array}$
What about if we start with $17 \cdot 7$ and keep adding $102$?
The following is multiple choice question (with options) to answer.
96 can be expressed as a product of primes as | [
"2 × 2 × 3 × 3 × 3 × 7",
"2 × 3 × 3 × 3 × 7 × 7",
"2 × 2 × 2 × 2 × 2 × 3",
"2 × 3 × 3 × 3 × 3 × 7"
] | C | Explanation:
It is clear that 96 = 2 × 2 × 2 × 2 × 2 × 3
Answer: Option C |
AQUA-RAT | AQUA-RAT-33823 | $$ii 9, 18, 31, 48, 69, 94,...$$
$$iii 14, 11.5, 9, 6.5, 4, 1.5,...$$
Consider the following sequence. $$\displaystyle{s}_{{n}}={2}{n}−{1}$$ (a) Find the first three terms of the sequence whose nth term is given. $$\displaystyle{s}_{{1}}={N}{S}{K}{s}_{{2}}={N}{S}{K}{s}_{{3}}=$$ (b) Classify the sequence as arithmetic, geometric, both, or neither. arithmeticgeometric bothneither If arithmetic, give d, if geometric, give r, if both, give d followed by r. (If both, enter your answers as a comma-separated list. If neither, enter NONE.)
Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference, if it is geometric, find the common ratio. If the sequence is arithmetic or geometric,
find the sum of the first 50 terms.
$$\{9=\frac{10}{11}n\}$$
What type of sequence is $$\{9=\frac{10}{11}n\}? asked 2021-02-10 The first four terms of a sequence are given. Can these terms be the terms of an arithmetic sequence? \(38, -22, -6, 10,...$$
Yes, the sequence is arithmetic.
No, the sequence is not arithmetic.
If the sequence is arithmetic, find the common difference d. (If the sequence is not arithmetic, enter DNE.)
$$d=?$$
The arithmetic mean (average) of two numbers c and d is given by $$\overline{x} = \frac{c+d}{2}$$
The following is multiple choice question (with options) to answer.
S is the infinite sequence S1 = 2, S2 = 22, S3 = 222,...Sk = Sk–1 + 2(10k–1). If p is the sum of the first 30 terms of S, what is the tenth digit of p, counting right to left from the units digit? | [
"1",
"2",
"4",
"6"
] | D | C
Sum of unit digits of first 30 terms = 60
Sum of tens digits of first 30 terms = 58
Sum of thousands digits of first 30 terms = 56
and so on..
p1 = 0
p2 = (6+58) = 4
p3 = (6+56) = 2
p4 = (6+54) = 0
p5 = (6+52) = 8
p6 = (5+50) = 5
p7 = (5+48) = 3
p8 = (5+46) = 1
p9 = (5+44) = 9
p10= (4+42) = 6
ANSWER : D |
AQUA-RAT | AQUA-RAT-33824 | # 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.
Vijay bought 160 shirts at the rate of Rs. 225 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%? | [
"Rs.282",
"Rs.229",
"Rs.208",
"Rs.285"
] | D | Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 20)/100]
= Rs.285.
Answer:D |
AQUA-RAT | AQUA-RAT-33825 | # Probability of having $n$ dice equal to or greater than $x$ when $m$ dice are rolled
I am trying to figure out how to do the math for something like this. The scenario:
• $n = 4$
• $x = 7$ (using d10s, where 0 = 10)
• $m = 8$
In words, if I roll 8d10 (eight 10-sided dice) what is the probability of having four dice greater than or equal to 7 (where 0 is the greateest [10])
I have seen a lot of sites which will do the calculation where $n = 1$, but I want to make $n$ a variable.
I know the probability of a 'successful' role is 4/10.
I know that I could write out all the possible combinations and count the ones that meet the criteria, but I'm sure this can be done with math.
How would you calculate this?
## migrated from stackoverflow.comNov 26 '16 at 5:07
This question came from our site for professional and enthusiast programmers.
Probability (in this case) is the ratio of desired outcomes to all outcomes. Let's count desired outcomes in your particular case (and by the way we derive a general formula to compute it).
Suppose you roll 4 dice and you have a total success - on every die you get "7 to 10". How many quadruplets of numbers "1 to 10" may cause it? It is not so difficult: $$4 \cdot 4 \cdot 4 \cdot 4$$ or $$4 ^ 4$$.
Then you roll the 4 remaining dice - but now you need a total failure - numbers 1 to 6 on every die. How many quadruplets is able to cause it? The answer is similar: $$6 ^ 4$$.
So for the first group of four dice to be totally successful and in the same time the second group of four dice to be totally unsuccessful - e. g. $$(8, 8, 8, 8, 1, 1, 1, 1)$$ - there is $$4^4 \cdot 6^4$$ possibilities.
The following is multiple choice question (with options) to answer.
In a throw of dice what is the probability of geƫng number greater than 4 | [
"1/2",
"1/3",
"1/5",
"1/6"
] | B | Explanation:
Number greater than 4 is 5 & 6, so only 2 number
Total cases of dice = [1,2,3,4,5,6]
So probability = 2/6 = 1/3
Answer: B |
AQUA-RAT | AQUA-RAT-33826 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum? | [
"338",
"277",
"229",
"646"
] | D | 840 = P [1 + (10*3)/100]
P = 646
Answer:D |
AQUA-RAT | AQUA-RAT-33827 | # how many ways of arranging given 7 two digit positive integers so that the sum of every four consecutive integer is divisible by 3?
in how many ways can I arrange the numbers: 21,31,41,51,61,71,81 such that the sum of every four consecutive numbers is divisible by three?
Though I am not an expert on modulo math, I do know that if we were to take MOD 3 on all of the numbers in the list, I would get the following in respective order:
$0_{21}, 1_{31}, 2_{41}, 0_{51}, 1_{61}, 2_{71}, 0_{81}$ (the subscript correlates to what original number it represents) and clearly if we were to match the values so that the sum is a multiple of three, the numbers added up would also be a multiple of three.
But upon realizing that the numbers must be consecutive and that if taking any four consecutive numbers in a set of 7 terms, I got stuck here and do not know how to proceed.
• Could you give an example of what you're looking for? There are some ambiguities in your question (specifically, do you mean digits or numbers?) – Michael Burr May 19 '17 at 12:01
• The OP explicitly talks about concecutive "numbers", and also his (good) start points in that direction. – drhab May 19 '17 at 12:05
• There is no ambiguity. The question asks about arranging the numbers $21,...,81$ with every four consecutive numbers having some property. There is no mention of digits. – Especially Lime May 19 '17 at 12:07
If you want the sums $a_1+a_2+a_3+a_4$ and $a_2+a_3+a_4+a_5$ to both be multiples of $3$, then you must have $a_1\equiv a_5$ mod $3$. Similarly $a_2\equiv a_6$, $a_3\equiv a_7$.
The following is multiple choice question (with options) to answer.
The sum of three consecutive multiples of 3 is 72. What is the largest number? | [
"21",
"24",
"27",
"36"
] | C | Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27.
ANSWER:C |
AQUA-RAT | AQUA-RAT-33828 | 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.
3 boys and 7 girls can complete John work in 10 days . But 4 boys and 6 girls need 8 days to complete the same work . In how many days will 10 girls complete the same work? | [
"10 days",
"20 days",
"40 days",
"56 days"
] | C | Work done by 4 boys and 6 girls in 1 day = 1/8
Work done by 3 boys and 7 girls in 1 day = 1/10
Let 1 man does m work in 1 day and 1 woman does w work in 1 day. The above equations can be written as
4m + 6w = 1/8 ---(1)
3m + 7w = 1/10 ---(2)
Solving equation (1) and (2) , we get m=11/400 and w=1/400
Amount of work 10 girls can do in John day = 10 × (1/400) = 1/40
Ie, 10 girls can complete the work in 40 days
C |
AQUA-RAT | AQUA-RAT-33829 | Problem: Point on a sphere with minimum distance to the plane.
Normal vector to the plane, $n_p$:
$\nabla f = (\frac {\partial f}{\partial x},\frac {\partial f}{\partial y},\frac {\partial f}{\partial z})$.
We get $n_p = (1,2,2)$.
Normal vector to the circle, $n_c$:
$\nabla g = (\frac {\partial g}{\partial x},\frac {\partial g}{\partial y},\frac {\partial g}{\partial z})$.
We get $n_c = (2x,2y,2z)$.
At the desired point on the circle the two normals are parallel or anti parallel.
$(2x,2y,2z) = \alpha (1,2,2)$.
Hence:
$\, 2x = \alpha, 2y = 2\alpha , 2z = 2\alpha$.
Combining the above with the equation of the circle:
$\alpha ^2 + ( 2\alpha)^2 + (2\alpha)^2 = 4$,
$\alpha ^2 + 4 \alpha ^2 + 4 \alpha ^2 = 4$,
$9 \alpha ^2 = 4$,
$\alpha_1 = 2/3$ and $\alpha_2 = - 2/3$.
The closest point:
1) $P_1 (1/3,2/3,2/3)$ for $\alpha_1 = 2/3$,
the farthest:
2) $P_2 (-1/3,-2/3,-2/3)$ for $\alpha_2 = - 2/3.$
The following is multiple choice question (with options) to answer.
What is the greatest possible (straight line) distance, between any two points on a hemisphere of radius 4? | [
"n/8",
"0.8",
"8",
"18"
] | C | Maximum distance straight line is diameter
d = 2r = 8..
ANS option C. |
AQUA-RAT | AQUA-RAT-33830 | • $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.
The pilot of a small aircraft uses 60 gallons of fuel to fly 300 miles. At this rate, how many gallons of fuel are needed for a 700 mile flight? | [
"140",
"155",
"160",
"170"
] | A | Number of gallons of fuel per mile = 300 /60 = 5 gallons per mile
Number of gallons of fuel for a 700 mile flight = 700/5 = 140
ANSWER:A |
AQUA-RAT | AQUA-RAT-33831 | # If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?
It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
The following is multiple choice question (with options) to answer.
If the integer n has exactly four positive divisors, including 1 and n, how many positive divisors does n^3 have? | [
"16",
"11",
"10",
"20"
] | A | take the example of 6...
it has 4 positive divisors (1,2,3,4)
Now, take the example of 216 ...
it has 16 divisors..
so A is the ans |
AQUA-RAT | AQUA-RAT-33832 | That is, the elevator travels at a rate of 8 floors per 2 minutes.
How many floors does an elevator travel in 30 seconds?
Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds.
So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors
Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2
(A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT!
(B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE
(C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE
(D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE
(E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE
For more information on this question type and this approach, we have some free videos:
- Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933
- Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934
- Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
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Joined: 07 Dec 2014
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Re: At the rate of f floors per m minutes, how many floors does an elevato [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
An escalator moves towards the top level at the rate of 12 ft.sec and its length is 160 feet. If a person walks on the moving escalator at the rate of 8 feet per second towards the top level, how much time does he take to cover the entire length. | [
"10sec",
"8sec",
"9sec",
"7sec"
] | B | Explanation :
Time taken to cover the entire length = tot.dist/resultant speed = 160/ (12+8) = 8sec
Answer : B |
AQUA-RAT | AQUA-RAT-33833 | 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.
The value of 489.1375 x 0.0483 x 1.956/0.0873 x 92.581 x 99.749 is closest to: | [
"0.04",
"0.08",
"0.07",
"0.06"
] | D | = 489.1375 x 0.0483 x 1.956/0.0873 x 92.581 x 99.749
= 489 x 0.05 x 2/0.09 x 93 x 100
= 489/9 x 93 x 10
= 163/279 x 1/10
= 0.58/10
= 0.058 (or) 0.06.
Answer is D. |
AQUA-RAT | AQUA-RAT-33834 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The simple form of the ratio 8/5:2/3 is? | [
"12:5",
"12:7",
"7:12",
"5:7"
] | A | 8/5:2/3 = 12:5
Answer: A |
AQUA-RAT | AQUA-RAT-33835 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
The cost of 4 bags and 12 purses is Rs.1600, what is the cost of 10 bags and 30 purses? | [
"Rs.3600",
"Rs.4000",
"Rs.3800",
"Rs.3900"
] | B | Explanation:
Cost of 4 bags + 12 purses = Rs.1600
Multiply each term by 5/2, we get,
Cost of 10 bags + 30 purses = Rs.4000.
Answer: Option B |
AQUA-RAT | AQUA-RAT-33836 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When positive integer n is divided by 7, the remainder is 1. When n is divided by 11, the remainder is 5. What is the smallest positive integer k such that k+n is a multiple of 77? | [
"2",
"4",
"6",
"8"
] | C | n = 7p + 1 = 11q + 5
n+6 = 7p + 7 = 11q + 11
n+6 is a multiple of 7 and 11, so it is a multiple of 77.
The answer is C. |
AQUA-RAT | AQUA-RAT-33837 | LCM, HCF/GCD etc. Learn more. Types of Numbers (Natural, Whole, Integer, Rational, Irrational, Real, Imaginary, Complex Numbers) Following is the Classification of various types of numbers- Natural Numbers (N) If N is a set of natural numbers, then we can write the set of natural numbers as N={1,2,3,4,5,6...}. Integers is a subset of real numbers. SHORTINTEGER. • Examples: • The factors of 6 are 1, 2, 3 and 6. Integers have negative numbers. From these two sets, we can see that every whole number is an integer, but not every integer is a whole number. For example, 21, 4, 0, and −2048 are integers, while 9.75, 5 + 1 / 2 , and √ 2 are not. For example, look at the below code. Irrational number - a number is irrational if it has infinite, nonrecurrent decimal places. It allows for 19 digits; positive or negative whole numbers between -9,223,372,036,854,775,807 (-2^63+1) and 9,223,372,036,854,775,806 (2^63-2). The main difference that lies between the two is that integers contain the negatives of the natural numbers, whereas whole number contains only zero and positive integers. Type int(x) to convert x to a plain integer. Number Type Conversion. DECIMAL. Whole Number – Represents a 64 bit (eight-byte) integer value. Integers are like whole numbers, but they also include negative numbers ... but still no fractions allowed!So, integers can be negative {-1, -2,-3, -4, -5, ... }, positive {1, 2, 3, 4, 5, ... }, or zero {0}We can put that all together like this:Integers = { ..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ... } These are all integers (click to mark), and they continue left and right infinitely: Before integers, students’ life with numbers had been all about whole numbers and some friendly fractions and decimals. Here is the
The following is multiple choice question (with options) to answer.
Starting with 0, a mathematician labels every non-negative integer as one of five types: alpha, beta, gamma, delta, or epsilon, in that repeating order as the integers increase. For instance, the integer 8 is labeled delta. What is the label on an integer that is the sum of an alpha raised to the seventh power and a delta raised to the seventh power? | [
"alpha",
"beta",
"gamma",
"delta"
] | C | Let the alpha be 5k and let the delta be 5j+3.
(5k)^7 + (5j+3)^7 which has the form 5k+2, which is a gamma.
The answer is C. |
AQUA-RAT | AQUA-RAT-33838 | 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$$
To access all the questions: Question of the Week: Consolidated List
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The following is multiple choice question (with options) to answer.
Two pipes P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first is turned off. How much longer will the cistern take to fill? | [
"11 1/9",
"11 1/4",
"19 1/9",
"11 1/2"
] | B | 3/12 + x/15 = 1
x= 11 1/4
Answer: B |
AQUA-RAT | AQUA-RAT-33839 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
To access all the questions: Question of the Week: Consolidated List
_________________
Number Properties | Algebra |Quant Workshop
Success Stories
Guillermo's Success Story | Carrie's Success Story
Ace GMAT quant
Articles and Question to reach Q51 | Question of the week
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities
Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
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The following is multiple choice question (with options) to answer.
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is? | [
"60 gallons",
"100 gallons",
"120 gallons",
"180 gallons"
] | C | Explanation:
Work done by the waste pipe in 1 minute = 1/15 - (1/20 + 1/24) = - 1/40
Volume of 1/40 part = 3 gallons\
Volume of whole = 3 * 40 = 120 gallons.
ANSWER IS C |
AQUA-RAT | AQUA-RAT-33840 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B undertake to do a piece of work for Rs. 1200. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. Find the share of B. | [
"250",
"450",
"750",
"800"
] | B | C's 1 day's work = 1/3-(1/6+1/8)=24
A : B : C = Ratio of their 1 day's work = 1/6:1/8:1/24= 4 : 3 : 1.
A’s share = Rs. (1200 *4/8) = Rs.600, B's share = Rs. (1200 *3/8) = Rs. 450
C's share = Rs. [1200 - (300 + 225») = Rs. 150.
Answer is B |
AQUA-RAT | AQUA-RAT-33841 | Difference between revisions of "2014 AMC 10A Problems/Problem 17"
Problem
Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?
$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$
Solution 1 (Clean Counting)
First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is $$\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.$$ Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.
--happiface
Solution 2 (Bashy Casework)
Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.
The following is multiple choice question (with options) to answer.
Given two fair dice, what is the probability Q that the sum of their numbers is 4 if exactly one die shows a 3? | [
"a) 2/11",
"b) 1/18",
"c) 3/11",
"d) 2/39"
] | A | If exactly one die shows 3, there are 10 such combinations (given that we have 2 fair dice).
1:3, 2:3, 4:3, 5:3, 6:3,
3:1, 3:2, 3:4, 3:5, 3:6
The sum of numbers would be 4 for two events: 1:3, 3:1.the probability Q that the sum of their numbers is 4 if exactly one die shows a 3
2/11 would be the answer if 'at least one of the die shows a 3'.A |
AQUA-RAT | AQUA-RAT-33842 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
60 + 5 * 12 / (180/3) = ? | [
"60",
"120",
"13",
"61"
] | D | Explanation:
60 + 5 * 12 / (180/3) = 60 + 5 * 12 / (60)
= 60 + (5 * 12)/60 = 60 + 1 = 61.
ANSWER IS D |
AQUA-RAT | AQUA-RAT-33843 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
A brick measures 20cm*10cm*7.5cm how many bricks will be required
for a wall 25m*2m*0.75m ? | [
"24000",
"23000",
"22000",
"25000"
] | D | 25*2*0.75=20/100*10/100*0.75/100*X
25=1/100*X
X=25000
Answer :D |
AQUA-RAT | AQUA-RAT-33844 | The error lies in case $C1:$
. . the people are divided into two pairs.
Call the people $A,B,C,D.$
How many pairings are possible?
The answer is three: . $\{AB|CD\},\;\{AC|BD\},\;\{AD|BC\}$
. . Note that $\{CD|AB\}$ is not a different pairing.
Make this correction and you will get 256.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I first ran into this snag many years ago.
There are four bridge players.
In how many ways can the players be paired?
I immediately said: . $_4C_2 \:=\:\dfrac{4!}{2!\,2!} \:=\:6$
I was surprised to learn that the answer was three.
(And no explanation was given.)
I finally reasoned it out.
When we select a pair of players to be partners,
. . we have automatically assigned the other two to be partners.
Call the players $A,B,C,D.$
With whom can $A$ be paired?
$A$ can be paired with $B,C,\text{ or }D$ . . . 3 choices.
And that's it!
. . Get it?
The following is multiple choice question (with options) to answer.
The New York Classical Group is designing the liner notes for an upcoming CD release. There are 10 soloists featured on the album, but the liner notes are only 5 pages long, and therefore only have room for 5 of the soloists. The soloists are fighting over which of them will appear in the liner notes, though not about which page they appear on. How many different combinations E of soloists can appear in the liner notes? | [
"5!",
"10!/(5!5!)",
"10!/5!",
"10!"
] | B | Total Soloists to choose from = 10
Soloist Required = 5
Question : How many different combinations E of soloists can appear in the liner notes?
The language of the question clearly mentions that only the combinations need to be calculated
i.e. Arrangement of the Soloists does NOT matter(Refer theHighlightedpart of the question)
Method-1:
Total Ways to Choose 5 out of 10 soloists = 10C5 = 10! / (5!*5!)
Method-2:
No. of ways to select and arrange 5 soloist on 5 Page = 10*9*8*7*6
Since the arrangement of 5 selected Notes (which can happen in 5! ways) doesn't NOT matter,
Therefore total ways to pick 5 out of 10 soloists = 10*9*8*7*6 / 5! = 10*9*8*7*6 *5! / (5!*5!) = 10! / (5!*5!)
Answer: Option B |
AQUA-RAT | AQUA-RAT-33845 | python, programming-challenge, time-limit-exceeded, primes, factors
Title: Work out largest prime factor of a number I am trying to get better at problem solving, so I am going through challenges on Project Euler.
Problem:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
My solution:
My approach to this was the following process:
Get all factors of the number
Identify prime factors by looping from 1 up to each factor. If the factor MOD the numbers going up is equal to 0 and the number is not 1 or the number itself, then it is not a prime. Then another for loop goes through each factor again and checks if it is not in the not primes list, hence making it a prime.
The last item in the primes list gives the biggest prime number
Code:
def get_factors(n):
factors = []
for i in range(1,n+1):
if n % i == 0:
factors.append(i)
return factors
def get_prime_factors(factors):
print(factors)
primes = []
not_primes = [1]
for i in range(len(factors)):
for k in range(1,factors[i]+1):
if factors[i] % k == 0 and factors[i] !=1 and k!=1 and k!=factors[i]:
not_primes.append(factors[i])
for i in factors:
if i not in not_primes:
primes.append(i)
return primes
def biggest_prime(primes):
biggest = primes[-1]
print(biggest)
factors = get_factors(600851475143)
primes = get_prime_factors(factors)
biggest_prime(primes)
The following is multiple choice question (with options) to answer.
The largest prime number is? | [
"11",
"13",
"5",
"7"
] | B | The largest prime number is 13.
B) |
AQUA-RAT | AQUA-RAT-33846 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A collection of books went on sale, and 2/3 of them were sold for $3 each. If none of the 36 remaining books were sold, what was the total amount received for the books that were sold? | [
"$216",
"$135",
"$90",
"$60"
] | A | Since 2/3 of the books in the collection were sold, 1/3 were not sold. The 36 unsold books represent 1/3 of the total number of books in the collection, and 2/3 of the total number of books equals 2(36) or 72. The total proceeds of the sale was 72($3) or $216. The best answer is therefore A.
Answer: A. |
AQUA-RAT | AQUA-RAT-33847 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A goods train runs at the speed of 72 km/hr and crosses a 250 m long platform in 26 sec. What is the length of the goods train? | [
"278",
"166",
"151",
"270"
] | D | Speed = 72 * 5/18 = 20 m/sec.
Time = 26 sec.
Let the length of the train be x meters.
Then, (x + 250)/26 = 20
x = 270 m.
Answer:D |
AQUA-RAT | AQUA-RAT-33848 | 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.
One of the following numbers can be marked out as the odd one out. can you find which one?
1) 680986879
2) 716089780
3) 820670987
4) 932967879 | [
"The odd number is 680986879",
"The odd number is 820670987",
"The odd number is 716089780",
"The odd number is 932967879"
] | B | Correct Answer : B
if you add up the first three digits of all the numbers, it adds up to 14 except for second option 820670987 |
AQUA-RAT | AQUA-RAT-33849 | Someone above told me not to give out the answer.. and btw your answer is correct.. and yes n = 10 not 9 . There are 10 terms of the GP.
11. Jun 5, 2008
### HallsofIvy
Staff Emeritus
spdeyunlimit, n= 9, not 10 because because the sum starts with k= 1. He is thinking of this as 6+ 6(2)+ 6(22)+ 6(2n) with n running from 0 to 9.
Another way to do the problem would be to use the same formula with n= 10, a= 3, but then subtract off 3- to allow for the missing 3(20) term.
12. Jun 5, 2008
### spideyunlimit
for the GP
a, ar, ar^2, ar^3 ... and so on
Sum of first n terms = a(1 - r^n) / (1-r)
where in the first term the power of r is 0.. whereas here it is one.. so yeah i got that.. it will be for n = 9
The following is multiple choice question (with options) to answer.
If each term in the sum R=a1+a2+a3+...+an either 2 or 22 and the sum equals 100, which of the following could be equal to n? | [
"38",
"39",
"40",
"41"
] | C | Min value of 'n' can be 10 i.e 4*22+6*12=R => 22+22+22+22+2+2+2+2+2+2 = 100
Since we don't have 10 in the options proceed further, (10-1)+22/2 => 20 digits, which is again not in the options
(20-1) + 22/2 = 30 digits ( not in options)
(30-1) + 22/2 = 40 digits
Hence C. |
AQUA-RAT | AQUA-RAT-33850 | entomology, ant
Title: In an ant (or bee) colony, what is the very approximate ratio of new breeders to workers? For example, out of every 1000 eggs laid, X mature into drones and/or virgin queens. That seems impossibly precise but it illustrates the kind of number I want well. I'll accept answers for any species and any number of species, even one, with any amount of precision or lack thereof, because right now I can't even feel confident saying that there are more workers or more breeders, though I obviously suspect more workers.
I would also be ecstatic to have any live count just before the nuptial flight, i.e. for this colony in this study there were X workers, Y drones, and Z virgin queens just before the nuptial flight, or X workers and (Y+Z) breeders, or X% of the colony was breeders, or for this species on average X% are breeders just before the nuptial flight.
Anything. Any one thing and I can accept it as an answer.
I can find any number of studies that talk about the sex ratio between drones and virgin queens, so I know someone is counting. Maybe I'm not reading closely enough, but they always seem to slip away from giving all the numbers I need to figure this out for myself. So.
This answer is specific to the western honeybee, Apis mellifera, as there are massive amounts of data on them; more, possibly, than any other insect species. There has certainly been more data collected about them than any other hymenopteran.
At around the time of the nuptial flight, there may be as many as 60,000 workers in the hive, though likely number is more like 15,000 - 20,000. There will be either one (virgin) or two (one mated, one virgin) queens (the old queen will stay with the hive, if she is alive). There may be as many as 400 drones from the original colony (though usually the number is less, around 150 is typical; 10 - 50 of them will actually mate with the queen), and an equal number may join in the flight drawn from other colonies, especially in commercial beekeeping operations. Somewhere between 1000 - 6000 workers will take part in the nuptial flight with the virgin queen and the drones.
The following is multiple choice question (with options) to answer.
The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 32,000 locusts? | [
"2",
"4",
"6",
"8"
] | C | - 4 hours: 1,000
-2 hours: 2,000
Now: 4,000
+ 2 hours: 8,000
+ 4 hours: 16,000
+ 6 hours: 32,000
Answer : C |
AQUA-RAT | AQUA-RAT-33851 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
The price of a book is increased from $300 to $330. What is the % of increase in its price? | [
"10%",
"20%",
"30%",
"35%"
] | A | Explanation: Change in the price = Rs 330 – Rs 300
= Rs 30
Percentage of increase = Change in the price Initial Price
* 100.
Percentage increase in price =( 30 / 300
) *100 = 10%
A |
AQUA-RAT | AQUA-RAT-33852 | ### Show Tags
23 Oct 2009, 23:42
14
KUDOS
Expert's post
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BOOKMARKED
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute $$x-8$$ vouchers, so that each can get from zero to $$x-8$$ as at "least 2", or 2*4=8, we already booked. Let $$x-8$$ be $$k$$.
In how many ways we can distribute $$k$$ identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.
Let $$k=5$$. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).
Consider:
$$ttttt|||$$
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:
$$ttttt|||$$
Means that first nephew will get all the tickets,
$$|t|ttt|t$$
Means that first got 0, second 1, third 3, and fourth 1
And so on.
How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 $$t$$'s and 3 $$|$$'s are identical, so $$\frac{8!}{5!3!}=56$$. Basically it's the number of ways we can pick 3 separators out of 5+3=8: $$8C3$$.
The following is multiple choice question (with options) to answer.
The 44 parents participating in the Smithville PTA have been assigned to at least 1 of 3 committees: festival planning, classroom aid, and teacher relations. 21 parents are assigned to the festival planning committee, 18 parents are assigned to the classroom aid committee, and 19 parents are assigned to the teacher relations committee. If 5 parents are assigned to all 3 committees, how many parents are assigned to exactly 2 committees? | [
"4",
"6",
"8",
"9"
] | A | The formula is Total = A+B+C - sum of exactly two + 2*all three + neither
21+18+19-x-2*5=44
solving for x you get 4
Answer A |
AQUA-RAT | AQUA-RAT-33853 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B together can do a piece of work in 2 days. If A alone can do the same work in 22 days, then B alone can do the same work in? | [
"0.35 days",
"0.45 days",
"0.55 days",
"0.25 days"
] | B | B = 1/2 – 1/22 =0.45 days
ANSWER:B |
AQUA-RAT | AQUA-RAT-33854 | ## 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 farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden. | [
"20 m * 58 m",
"20 m * 7 m",
"20 m * 5 m",
"80 m * 5 m"
] | C | From the question, 2b+l = 30
=> l = 30-2b
Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0
Area=100m2=>l×b=100=>b(30−2b)=100b2−15b+50=0=>(b−10)(b−5)=0
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m
Answer: C |
AQUA-RAT | AQUA-RAT-33855 | 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.
The length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"27",
"82",
"245",
"288"
] | C | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (130 + X)/30 = 25/2 ==> 2(130 + X) = 750 ==> X = 245 m.
Answer:C |
AQUA-RAT | AQUA-RAT-33856 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 27,000, then what is B's share in the profit? | [
"Rs. 2000",
"Rs. 3000",
"Rs. 4000",
"Rs. 5000"
] | D | Explanation :
Ratio of the initial investment = 7/2 : 4/3 : 6/5
= 105 : 40 : 36
From this ratio, we can assume that actual initial investments of A, B and C
are 105x, 40x and 36x respectively
A increases his share 50% after 4 months. Hence the ratio of their investments =
(105x * 4) + (105x * 150/100 * 8) : 40x * 12 : 36x : 12
= 105 + (105 * 3/2 * 2) : 40*3 : 36 * 3
= 105 * 4 : 40 *3 : 36 * 3
= 35 * 4 : 40 : 36
= 35 : 10 : 9
B's share = total profit * (10/54) = 27,000 * 10/54 = 5000. Answer : Option D |
AQUA-RAT | AQUA-RAT-33857 | . . The hour hand has advanced: . $8 \times \frac{1}{2}^o \:=\:4^o$
. . The hour hand is at: . $\text{-}30^o + 4^o \:=\:\text{-}26^o$
The angle between the hands is: . $48^o - (\text{-}26^o) \;=\;74^o$
5. Alternatively (yet again),
$\frac{360}{12}=30^o\Rightarrow$ there is $30^o$ between the minute and hour hands at $11:00$.
In 8 minutes, the hour hand moves through $30^o\frac{8}{60}=4^o$ clockwise.
in 8 minutes, the minute hand moves through $360^o\frac{8}{60}=48^o$ clockwise.
Therefore the angle between the hands increases by $(48-4)^o$
The following is multiple choice question (with options) to answer.
In 16 minutes, the minute hand gains over the hour hand by | [
"16°",
"80°",
"88°",
"96°"
] | C | In 1 hour, the minute hand gains 330° over the hour hand.
i.e. in 60 minute, the minute hand gains 330° over the hour hand.
∴ In 16 minutes, the minute hand gains over the hour hand by 330°⁄60 × 16° = 88°
Answer C |
AQUA-RAT | AQUA-RAT-33858 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A, B and C play a cricket match. The ratio of the runs scored by them in the match is A:B = 2:3 and B:C = 2:5. If the total runs scored by all of them are 75, the runs scored by B are? | [
"11",
"18",
"19",
"12"
] | B | A:B = 2:3
B:C = 2:5
A:B:C = 4:6:15
6/25 * 75 = 18
Answer: B |
AQUA-RAT | AQUA-RAT-33859 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day? | [
"10 days",
"12 days",
"15 days",
"20 days"
] | C | A's 2 day's work = 1 x 2 = 1 .
20 10
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 .
20 30 60 60 10
Work done in 3 days = 1 + 1 = 1 .
10 10 5
Now, 1 work is done in 3 days.
5
Whole work will be done in (3 x 5) = 15 days.
Option C |
AQUA-RAT | AQUA-RAT-33860 | Now, let's handle potential factors of $$\3\$$ in $$\n\$$. We know that $$\10^{\varphi(n)}-1\$$ is a multiple of $$\n\$$, but $$\\frac{10^{\varphi(n)}-1}9\$$ might not be. But, $$\\frac{10^{9\varphi(n)}-1}9\$$ is a multiple of $$\9\$$ because it consists of $$\9\varphi(n)\$$ ones, so the sum of its digits a multiple of $$\9\$$. And we've noted that multiplying the exponent $$\k\$$ by a constant preserves the divisibility.
Now, if $$\n\$$ has factors of $$\2\$$'s and $$\5\$$'s, we need to add zeroes to end of the output. It way more than suffices to use $$\n\$$ zeroes (in fact $$\\log_2(n)\$$ would do). So, if our input $$\n\$$ is split as $$\n = 2^a \times 5^b \times m\$$, it suffices to have $$\9\varphi(m)\$$ ones to be a multiple of $$\n\$$, multiplied by $$\10^n\$$ to be a multiple of $$\2^a \times 5^b\$$. And, since $$\n\$$ is a multiple of $$\m\$$, it suffices to use $$\9\varphi(n)\$$ ones. So, it works to have $$\9\varphi(n)\$$ ones followed by $$\n\$$ zeroes.
The following is multiple choice question (with options) to answer.
What is the greatest positive integer E such that 3^E is a factor of 9^10? | [
"E=5",
"E=9",
"E=10",
"E=20"
] | D | What is the greatest positive integer E such that 3^E is a factor of 9^10?
9^10 = (3^2)^10 = 3^20
D. 20 |
AQUA-RAT | AQUA-RAT-33861 | The probabilities I got for parking between 4 and 22 cars are:
As it can be seen, there is a 'high' probability that the number of cars is going to move in the range $$(6,15)$$ (more or less). But as the question is for the most likely number, this number is $$10$$.
• Edited - replaced text $>=$ with TeX $\ge$ (\ge). – CiaPan Oct 9 '14 at 9:44
Joannes' answer shows that that if there are $N$ parking spaces, the probability that exactly $n$ vehicles get parked is $$p_N(n)=\frac{(N-n)!}{N!}\left[(N-n+1)^n-(N-n)^n\right].$$ I'm going to analyze where this is maximized.
The following is multiple choice question (with options) to answer.
A parking garage rents parking spaces for $10 per week or $35 per month. How much does a person save in a year by renting by the month rather than by the week? | [
" $100",
" $160",
" $220",
" $240"
] | A | 10$ per week!
An year has 52 weeks.
Annual charges per year = 52* 10 = 520$
35$ per month!
An year has 12 months.
Annual charges per year = 12 * 35 = 420$
520 - 420 = 100
Ans A |
AQUA-RAT | AQUA-RAT-33862 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
How many arrangements of the letters of the word CONTINUES can be made, without changing the place of the vowels in the word? | [
"130",
"140",
"150",
"120"
] | D | O,I,E,U fixed. Consonants can be arrangements in 5P5 = 5! = 120 ways
Option 'D' |
AQUA-RAT | AQUA-RAT-33863 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train passes a station platform in 48 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform? | [
"476 m",
"420 m",
"440 m",
"407 m"
] | B | Speed = [54 * 5/18] m/sec
= 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 48 = 15
x + 300 = 720
x =420 m.
Answer: B |
AQUA-RAT | AQUA-RAT-33864 | java, performance, json, memory-optimization
"2020-06-06",
"2020-06-08",
"2020-06-09"
]
},
{
"firstName": "Dan",
"lastName": "Scavona",
"email": "dscavona@xyz.com",
"country": "Wales",
"availableDates": [
"2020-05-24",
"2020-05-25",
"2020-05-29",
"2020-06-04",
"2020-06-05",
"2020-06-07",
"2020-06-09"
]
},
{
"firstName": "Mark",
"lastName": "Kuehler",
"email": "mkuehler@xyz.com",
"country": "Wales",
"availableDates": [
"2020-05-24",
"2020-05-25",
"2020-06-01",
"2020-06-04",
"2020-06-05",
"2020-06-06",
"2020-06-09",
"2020-06-10"
]
},
{
"firstName": "Dang",
"lastName": "Come",
"email": "dcome@xyz.com",
"country": "Wales",
"availableDates": [
"2020-05-24",
"2020-05-25",
"2020-05-31",
"2020-06-04",
"2020-06-05",
"2020-06-10",
"2020-06-15"
]
},
{
"firstName": "Neha",
"lastName": "Torchia",
"email": "ntorchia@xyz.com",
"country": "Wales",
"availableDates": [
"2020-06-03",
"2020-06-06"
]
},
{
"firstName": "Anita",
"lastName": "Pujals",
"email": "apujals@xyz.com",
"country": "Wales",
"availableDates": [
"2020-05-26",
"2020-05-30",
The following is multiple choice question (with options) to answer.
What was the day of the week on 17th June 1998? | [
"Monday",
"Sunday",
"Wednesday",
"Friday"
] | C | Explanation:
17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17-Jun-1998)
We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)
Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)
Number of odd days in the period 1901-1997
= 73 normal years + 24 leap year
= 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days
Number of days from 1-Jan-1998 to 17-Jun-1998
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun)
= 168 = 0 odd day
Total number of odd days = (0 + 1 + 2 + 0) = 3
3 odd days = Wednesday
Hence 17th June 1998 is Wednesday.
Answer: Option C |
AQUA-RAT | AQUA-RAT-33865 | 4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN
Target question: Is integer n EVEN?
Statement 1: n² - 1 is an odd integer
n² - 1 = (n + 1)(n - 1)
So, statement 1 is telling us that (n + 1)(n - 1) = ODD
From rule #4 (above), we can conclude that BOTH (n + 1) and (n - 1) are ODD
If (n + 1) is ODD, then n must be EVEN (since 1 is ODD, we can apply rule #2 to conclude that n is EVEN)
If (n - 1) is ODD, then n must be EVEN (by rule #2 )
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: 3n + 4 is an even integer
In other words, (3n + EVEN) is EVEN
From rule #3, we can conclude that 3n is EVEN
Since 3 is odd, we can write: (ODD)(n) = EVEN
From rule #5, we can conclude that n is EVEN
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Cheers,
Brent
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Re: If n is an integer, is n even? [#permalink]
### Show Tags
20 Jun 2019, 03:27
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
The following is multiple choice question (with options) to answer.
If p is an odd integer and q is an even integer, which of the following must be an even integer? | [
"2q-p",
"3p+q",
"2p+3q",
"4(p+q)"
] | D | p is odd; for this assume p=1
q is even; for this assume q = 2
(2*2)-1 = 3 Odd
(3*1)+2 = 5 Odd
(2*1)+(3*2) = 9 Odd
4(1+2) = 12 Even
1+(2^2) = 5 Odd
Answer:D |
AQUA-RAT | AQUA-RAT-33866 | Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Which of the following CANNOT be the sum of two prime numbers? [#permalink]
### Show Tags
11 Feb 2015, 22:09
6
KUDOS
Expert's post
6
This post was
BOOKMARKED
banksy wrote:
187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88
We know that other than 2, all prime numbers are odd. When you add two odd numbers, you get an even sum. To get an odd sum, one number must be even and then other odd. So to get 19, 45 and 79, one prime must be 2.
Now we just need to subtract 2 out of each of these three options to see whether we get another prime. 79 - 2 = 77 which is not prime. So 79 CANNOT be the sum of two prime numbers.
Note that there is a conjecture that every even number greater than 2 can be written as sum of two prime numbers. So we don't even need to check for the even sum options.
For more on this, check: http://www.veritasprep.com/blog/2014/08 ... t-part-iv/
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Re: Which of the following CANNOT be the sum of two prime numbers? [#permalink]
### Show Tags
04 Sep 2015, 17:56
Bunuel wrote:
banksy wrote:
187. Which of the following CANNOT be the sum of two prime numbers?
(A) 19
(B) 45
(C) 68
(D) 79
(E) 88
The following is multiple choice question (with options) to answer.
Which of the following cannot be the sum of 2 different prime numbers? | [
"10",
"13",
"14",
"15"
] | C | Option A: 10 = 7 + 3. Sum of 2 different prime numbers.
Option B: 13 = 11 + 2. sum of 2 different prime numbers
Option C: 4 = This is not a sum of 2 different prime numbers.
Option D: 15 = 13 + 2. Sum of 2 different prime numbers
Option E: 19 = 17 + 2. Sum of 2 different prime numbers
Correct Option: C |
AQUA-RAT | AQUA-RAT-33867 | ### 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 man buys $50 shares in a company which pays 10% dividend. If the man gerts 12.5% on his investment, at what price did he buy the shares? | [
"37.5",
"40",
"50",
"48"
] | B | Dividend on 1 share = $(10/100*50)
=$5
$12.5 is an income on an investment of $100
$5 is an income on an investment of $(100*2/25*5)=40
Thus, cost of 1 share =$40
ANSWER B |
AQUA-RAT | AQUA-RAT-33868 | 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.
There are 418 doctors and nurses in a hospital. If the ratio of the doctors to the nurses is 8 : 11, then how many nurses are there in the hospital? | [
"152",
"209",
"242",
"171"
] | C | Given,
The ratio of the doctors to the nurses is 8 : 11
Number of nurses = 11/19 x 418 = 242
ANSWER:C |
AQUA-RAT | AQUA-RAT-33869 | ### Show Tags
18 May 2010, 15:36
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.
This seems wrong and no one's explanation makes any sense.
Math Expert
Joined: 02 Sep 2009
Posts: 58335
Re: If x<0, then (-x*|x|)^(1/2) is: [#permalink]
### Show Tags
19 May 2010, 01:21
1
5
shammokando wrote:
Hey I'm sorry guys, this still does not make sense. Everyone's argument here is that the square root of 4 is 2, that is just not true! The square root of 4 is 2 OR -2. We're just accustomed to thinking that 2 is the "standard root" but -2 is just as correct. Therefore the square of -2 (which is x in this case) is 4, and the squareroot of that is 2 OR -2! So it could be x or -x.
This seems wrong and no one's explanation makes any sense.
Red part is not correct.
THEORY:
GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers.
When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root.
That is, $$\sqrt{25}=5$$, NOT +5 or -5. In contrast, the equation $$x^2=25$$ has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
The following is multiple choice question (with options) to answer.
Two-third of two-fifth of three-fourth of a number is 36. What is the square root of four-fifth of that number? | [
"9",
"12",
"14",
"16"
] | B | Explanation:
We have, 2/3 * 2/5 * 3/4 * X = 36
X = 180
Now, 4/5 * X = 4/5 * 180 = 144
√144 = 12
ANSWER: B |
AQUA-RAT | AQUA-RAT-33870 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A grocer has a sale of Rs. 5921, Rs. 5468, Rs. 5568, Rs. 6088 and Rs. 6433 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 5900? | [
"4902",
"4922",
"5922",
"5924"
] | C | Total sale for 5 months = Rs. (5921 + 5468 + 5568 + 6088 + 6433) = Rs. 29478.
Required sale = Rs. [ (5900 x 6) - 29478 ]
= Rs. (35400 - 29478)
= Rs. 5922.
ANSWER:C |
AQUA-RAT | AQUA-RAT-33871 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
Four packages have an average weight of 14.5 pounds. What is the minimum possible weight of the heaviest package in pounds if the median is 12 pounds? | [
"25",
"24",
"23",
"22"
] | B | Let us denote the weights of the packages in pounds by a, b, c, d naming from the lightest one to the heaviest one. The median is 12 pounds. Therefore (b + c) / 2 = 12.
b + c = 24
The average is 14.5 pounds. Therefore (a + b + c + d) / 4 = 14.5.
a + (b + c) + d = 58
a + 24 + d = 58
a + d = 34
The weight a must be no greater than 12, since 12 is the median. Therefore the minimum possible weight of the heaviest package is 34 – 12 = 24 pounds (all the other packages would weigh 12 pounds in this case).
Answer: B |
AQUA-RAT | AQUA-RAT-33872 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The average price of 10 books is Rs.12 while the average price of 8 of these books is Rs.11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books? | [
"Rs. 5, Rs.7.50",
"Rs. 8, Rs. 12",
"Rs. 16, Rs. 10",
"Rs. 12, Rs. 14"
] | C | Solution:
Total cost of 10 books = Rs. 120
Total cost of 8 books = Rs. 94
=> The cost of 2 books = Rs. 26
Let the price of each book be x and y.
=> x + y = 26 ---------------- (1)
(160/100)y+y =26
On Solving for y, we get
y = 10.
Now, Substituting y = 10 in (1) we get,
x + 10 = 26
x = 16.
So the price of each book is Rs. 16 and Rs. 10 respectively.
Answer: Option C |
AQUA-RAT | AQUA-RAT-33873 | $$f(2 ± δ) = 2 ( (2 ± δ) + 2) \qquad\qquad\text{ eq. 2.1x-5}$$ Now remembering that $δ$ is always greater than zero (and more importantly it never is zero), we can see that in the above, we still never have to evaluate $f(x)$ at the forbidden value. So we just multiply out the above expression: $$f(2 ± δ) = 8 ± 2 δ \qquad\qquad\text{ eq. 2.1x-6}$$ The requirement is that we have to be able to choose $δ$ so that the value above is no farther from the limit (which in this case is $8$) than the $ε$ that you might give me, no matter how small an $ε$ you do give me. So by giving me an $ε$, you are telling me to make it so that: $$|f(2 ± δ) - 8| ≤ ε \qquad\qquad\text{eq. 2.1x-7}$$ But we can get an expression for what's inside the absolute value brackets from stuff we have already done. Just take equation 2.1x-6 and subtract $8$ from both sides. If you substitute that in for $2 ± δ$, you get: $$|± 2 δ| ≤ ε \qquad\qquad\text{eq. 2.1x-8}$$ and since the absolute value brackets make the ± sign moot, we have simply: $$2 δ ≤ ε \qquad\qquad\text{ eq. 2.1x-9}$$ or
The following is multiple choice question (with options) to answer.
If the operation ∆ is defined by a ∆ b = (b - a)^2/a^2 for all numbers a and b, and a ≠ 0, then what is the result when we evaluate this expression: (-4) ∆ (4 ∆ −4) ? | [
"2",
"4",
"6",
"8"
] | B | (-4) ∆ (4 ∆ −4) =
(-4) ∆ (8^2/4^2) =
(-4) ∆ (4) =
(-8^2) / (-4)^2 = 4
The answer is B. |
AQUA-RAT | AQUA-RAT-33874 | Alice speaks the truth with probability 3/4 and Bob speaks the truth with probability 2/3. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is 3 while Bob says (to Carl) the number is 1. Find the probability that the number is actually 1.
UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ({1,2,⋯,6} - {The number that actually showed up}). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.
My attempt:
Case 1: Number 1 showed up.
Chance of all this happening = $\frac{1}{6}\cdot \frac{2}{3}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)=\frac{1}{180}$
Case 2: Number 3 showed up
Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{120}$
Case 3: Other number showed up
Chance of all this happening = $\frac{4}{6}\cdot \left(\frac{1}{4}\cdot \frac{1}{5}\right)\cdot \left(\frac{1}{3}\cdot \frac{1}{5}\right)=\frac{1}{450}$
So, total = $\overline{)\frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}$
Is my attempt correct? If not, how to do this problem?
You can still ask an expert for help
## Want to know more about Probability?
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
The following is multiple choice question (with options) to answer.
The probability that A speaks truth is 3/5 and that of B speaking truth is 4/7. What is the probability that they agree in stating the same fact? | [
"18/35",
"18/35",
"18/39",
"18/31"
] | A | If both agree stating the same fact, either both of them speak truth of both speak false.
Probability = 3/5 * 4/7 + 2/5 * 3/7
= 12/35 + 6/35 = 18/35
Answer: A |
AQUA-RAT | AQUA-RAT-33875 | It is easy to see from this website, problem/solution 23, http://books.google.com.au/books?id=8NtJLfGf94QC&pg=PA431&lpg=PA431&dq=monkey+climbing+ladder+on+pulley&source=bl&ots=2tWFRhdKME&sig=9ocddC7lk53A1_XdrVAwPm7MBw0&hl=en&sa=X&ei=Jv9SUevUEoTziAfSs4HIBg&ved=0CDAQ6AEwAA#v=onepage&q=monkey%20climbing%20ladder%20on%20pulley&f=false, that as the man moves up a distance $l'$, the centre of gravity moves up by a distance $l = \frac{ml'}{2M}$. Where $m$ is the mass of the man, and $M$ is the mass of the counter-weight.
This means that if the man is moving at a velocity of $V$ meters per second, then the centre of gravity will move up at a rate of $\frac{mV}{2M}$ meters per second as well.
I finally think something works. I don't need to imagine anything either. So, I assume the block's lower part is at height $H$ from the ground, ladder's last step at height $h$, the height of ladder being $l$ and the distance of man from the bottom being $x$ (where $x<l$ ).
The following is multiple choice question (with options) to answer.
A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour. If he begins his ascent at 8.00 a.m., at what time will he first touch a flag at 120 feet from the ground? | [
"4 p.m.",
"5 p.m.",
"6 p.m.",
"7 p.m."
] | C | A monkey climbs 30 feet at the beginning of each hour and rests for a while when he slips back 20 feet before he again starts climbing in the beginning of the next hour.
It means monkey climbs effectively 10 ft in 1 hr.
so it climbs 90 ft in 9 hrs .
balance 30 ft is covered in 10th hr.
So total it will take 10 hrs.
Monkey will touch 120 ft flag at 8+10 = 1800 hrs or 6 PM.
ANSWER:C |
AQUA-RAT | AQUA-RAT-33876 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
An article costing rs.160 is sold at 15% discount on a mark-up price. What is the selling price after
discount? | [
"106",
"116",
"126",
"136"
] | D | Rs. 136
160-(160*15/100)=160*85/100=136
ANSWER:D |
AQUA-RAT | AQUA-RAT-33877 | # How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
The following is multiple choice question (with options) to answer.
For dinner at a restaurant, there are x + 1 choices of appetizers, y + 1 main courses, and z choices of dessert. How many total possible choices are there if you choose 1 appetizer, 1 main course, and 1 dessert for your meal? | [
"x + y + z + 1",
"xyz + xz",
"xyz+xz+yz+z",
"xyz + 1"
] | C | x +1 appetizers,
y + 1 main courses,
z dessert
Required: 1 appetizer, 1 main course, and 1 dessert
Number of ways possible = (x+1)C1*(y+1)C1*zC1 {NC1 = N! / (N-1)!*1! = N}
Hence, number of ways = (x+1)(y+1)z = (yz+z)(x+1)=xyz+xz+yz+z
Correct Option: C |
AQUA-RAT | AQUA-RAT-33878 | # Runner's High (Speed)
I find the following mind-boggling.
Suppose that runner $$R_1$$ runs distance $$[0,d_1]$$ with average speed $$v_1$$. Runner $$R_2$$ runs $$[0,d_2]$$ with $$d_2>d_1$$ and with average speed $$v_2 > v_1$$. I would have thought that by some application of the intermediate value theorem we can find a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$. This is not necessarily so!
Question. What is the smallest value of $$C\in\mathbb{R}$$ with $$C>1$$ and the following property?
Whenever $$d_2>d_1$$, and $$R_2$$ runs $$[0,d_2]$$ with average speed $$Cv_1$$, then there is a subinterval $$I\subseteq [0,d_2]$$ having length $$d_1$$ such that $$R_2$$ had average speed at least $$v_1$$ on $$I$$.
The following is multiple choice question (with options) to answer.
In a race of length L metres, Johnson beats Lewis by X metres and Greene by Y metres, By how many metres does Lewis beat Greene in the same race ? (X<Y) | [
"L(L-Y) / L-X",
"L(Y-X) / L-X",
"L-Y",
"X-Y"
] | B | As the time is constant, the ratio of distances will be the same as the ratio of speeds.
If Sj,Sl,Sg are the speeds of Johnson, Lewis, and Greene respectively, then
Sj/Sl = L/(L-X)
and Sj/Sg = L/(L-Y)
=> Sg/Sl = (L-Y)/(L-X)
Therefore the speeds of Lewis and Greene are in the ratio (L-X)/(L-Y)
When Lewis finishes the race, the time run by him and Greene are same
=> The ratio of the speeds of Lewis and Greene will be the same as the ratio of distances run by them.
=> Distance run by Greene when Lewis finishes the race = (L-Y)/(L-X) * L
=> Lewis beats Greene by L - L*(L-Y)/(L-X) = L [ 1 - (L-Y)/(L-X)] = L (Y-X) / (L-X)
Option (B) is therefore correct. |
AQUA-RAT | AQUA-RAT-33879 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A green grocer received a boxful of tomatoes and on opening the box found that several had gone bad. He then counted them up so that he could make a formal complaint and found that 68 were mouldy, which was 16 per cent of the total contents of the box. How many tomatoes were in the box? | [
"336",
"425",
"275",
"235"
] | B | B
425
(68 ÷ 16) × 100 |
AQUA-RAT | AQUA-RAT-33880 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
Find the perimeter and area of the rectangle of length 15 cm and breadth 13 cm. | [
"71 cm2",
"121 cm2",
"141 cm2",
"195 cm2"
] | D | length = 15 cm, breadth = 13 cm
Perimeter of rectangle = 2 (length + breadth)
= 2 (15 + 13) cm
= 2 × 28 cm
= 56 cm
We know that the area of rectangle = length × breadth
= (15 × 13) cm22
= 195 cm2
ANSWER : D |
AQUA-RAT | AQUA-RAT-33881 | # Triangles in 20 non-straight-line "dots"
• August 22nd 2010, 07:07 AM
grottvald
Triangles in 20 non-straight-line "dots"
Twenty points/dots are given so that the three of them are never in a straight line. How many triangles can be formed with corners(as in vertex i think?) in the dots?
• August 22nd 2010, 07:21 AM
Plato
Calculate the number of ways to choose three points from twenty.
• August 22nd 2010, 10:36 AM
grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)
• August 22nd 2010, 10:57 AM
Plato
Quote:
Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct
It is not a large number at all: $\displaystyle \binom{20}{3}=\frac{20!}{3!\cdot 17!}=1140$
• August 22nd 2010, 11:58 AM
Quote:
Originally Posted by grottvald
C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)
You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.
The number of ways to do this is $\binom{19}{2}=171$
Therefore, there are 171 triangles that can be drawn which include the leftmost point.
If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.
These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.
Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.
The following is multiple choice question (with options) to answer.
6 points are marked on a straight line and another 7 points are marked on a second straight line with no points in common. How many triangles can be constructed with vertices from among the above points? | [
"21",
"281",
"168",
"196"
] | B | Solution: select 2 from 6 and one from 7 + select 2 from 7 and 1 from 6
6C2*7C1 + 7C2*6C1
= 281
Answer : B |
AQUA-RAT | AQUA-RAT-33882 | 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 Contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by three-fourth of the scheduled time, so he at once doubled the no of workers and thus he managed to finish the road on the scheduled time. How much work he had been completed, before increasing the number of workers? | [
"14 9/7 %",
"14 2/7 %",
"14 2/3 %",
"14 2/5 %"
] | B | Explanation:
Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.
D * x +(100- D) * 2x= 175x
=> D= 25 days
Now , the work done in 25 days = 25x
Total work = 175x
therefore, workdone before increasing the no of workers = \frac{25x}{175x}\times 100=14\frac{2}{7} %
Answer: B) 14 2/7 % |
AQUA-RAT | AQUA-RAT-33883 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
In an examination, 25% of total students failed in Hindi, 40% failed in English and 35% in both. The percentage of these who passed in both the subjects is : | [
"10%",
"20%",
"30%",
"40%"
] | C | pass percentage = 100 - (25+40-35) = 100-70 = 30
ANSWER:C |
AQUA-RAT | AQUA-RAT-33884 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
3
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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Joined: 09 Sep 2013
Posts: 12145
Followers: 538
Kudos [?]: 151 [0], given: 0
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.
A set of 11 positive integers has an average of 20. Which of the following is the greatest possible value for the median of this set? | [
"25",
"30",
"35",
"45"
] | C | for the median to bé max , we should take all the values below meadian to be the lowest and the remaining values to be the same..
here 1 is the lowest value(positive int) số thẻ lowest 5 int will equal 5..
remaining 6= 20*11-5=215..
therefore each valueof these 6 int =215/6=35
ans C |
AQUA-RAT | AQUA-RAT-33885 | 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.
here are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period? | [
"1234",
"4512",
"1800",
"1203"
] | C | 5 subjects can be arranged in 6 periods in 6P5 ways.
Any of the 5 subjects can be organized in the remaining period (5C1 ways).
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements
= 6P5× 5C1 /2!=1800
Ans: C |
AQUA-RAT | AQUA-RAT-33886 | # (GR. 10) 10 people are to be seated in a row. What is the total number of ways if…
Please help me! I understand what the question is asking for, but I can’t seem to get the right answer. The correct no. of ways should be $$645,120$$, though that may be incorrect. If anyone is kind enough to show me the solution, I would be very grateful.
$$10$$ people are to be seated in a row. What is the total number of ways in which this can be done if Eric and Carlos always have exactly one of the other people sitting between them?”
EDIT: Oh wow that was fast! Thank you for your kind hints! I was finally able to get the answer.
• Please show us your calculation. – saulspatz Feb 23 at 14:54
• I think it should be $8!*8*2$.I think your answer is correct. Cheers :) – Abhinav Feb 23 at 15:00
The possible positions of the two people are $$1-3,2-4,\cdots ,8-10$$ that is $$8$$ possibilities. We can swap the places, so multiply with $$2$$. Then, multiply with $$8!$$ because the other people can have $$8!$$ possible orders.
Here's a hint to get started. Suppose Alice is seated between Eric and Carlos. Then we can treat Eric-Alice-Carlos as a block to be arranged with the other $$7$$ students.
There are a total of $$10$$ people so there are $$8$$ people who could be seated between Eric and Carlos. There are $$2$$ ways of seating "Eric, other person, Carlos" or "Carlos, other person, Eric". Now treat those $$3$$ people as a single "person"- there are $$8!$$ ways to seat those $$8$$ "people". There are, then, $$8!(2)(8)= 645120$$ ways to do this. That is the same as Peter's answer.
The following is multiple choice question (with options) to answer.
How many E ways can Jason sit with his five friends in a row of six seats with an aisle on either side of the row, if Jason insists on sitting next to one of the aisles? | [
"120",
"240",
"360",
"540"
] | B | Jason can select his seat in 2 Ways (two aisles)
His 1st of 4 friends have 5 seats to select =>
His 2nd of remaining 3 friends will have 4 seat to chose from...and so on
Total ways E=> 2*5*4*3*2*1 = 240.B |
AQUA-RAT | AQUA-RAT-33887 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
At the end of three years what will be the compound interest at the rate of 10% p.a. on an amount of Rs.20000? | [
"Rs.6620",
"3378",
"2768",
"2999"
] | A | A = 20000(11/10) 3
= 26620
= 20000
----------
6620
Answer: A |
AQUA-RAT | AQUA-RAT-33888 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream? | [
"87 kmph",
"40 kmph",
"16 kmph",
"15 kmph"
] | B | Speed downstream = 60 + 20
= 80 kmph
Speed upstream = 60 - 20
= 40 kmph
Answer:B |
AQUA-RAT | AQUA-RAT-33889 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\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}$$ job/day.
The following is multiple choice question (with options) to answer.
Anita, Indu and Geeta can do a piece of work in 18 days, 27 days and 36 days respectively. They start working together. After working for 4 days. Anita goes away and Indu leaves 7 days before the work is finished. Only Geeta remains at work from beginning to end. In how many days was the whole work done? | [
"16 days",
"14 days",
"18 days",
"19 days"
] | A | 4/18 + (x -7)/27 + x/36 = 1
x = 16 days
Answer: A |
AQUA-RAT | AQUA-RAT-33890 | MHF Helper
Problem : A bike manufacturer has a plant in Minneapolis and another in Philly. The Minneapolis plant produces 70% of the bikes, of which 1% are defective. The Philly plant produces the other 30%, of which 0.5% are defective.
1: What percentage of bikes made by this company are defective? a) 1% b) 0.85% c) 0.5% d) 1.5%
I figured that 1% of 70% of production is 0.7% of total production. 0.5% of 30 is 30/100 = 0.3/2 = 0.15. 0.15+0.7= 0.85 so B
2. A bike made by this company is found to be defective. What is the probability that it was produced by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
3. non defective. What is the probability that is it made by the Minneapolis plant? a) .824 b) .176 c) .699 d) .301
4. A bike made by this company is found to be defective. Probability it was produced by the philly plant? a) .824 b) .176 c) .699 d) .301
5. Non defective. Probability made by Philly plant? a) .824 b) .176 c) .699 d) .301
Frankly I cannot follow what you posted.
Lets do #2. If a bike is found to be defective, then what is the probability that the bike was produced by the Minneapolis plant?
The bike was produced at one of two plants: $$\displaystyle \mathcal{P}(D)=\mathcal{P}(D\cap M)+\mathcal{P}(D\cap P)$$
Now let us work on the question, if a bike is defective what is the probability it came from Minneapolis?
The following is multiple choice question (with options) to answer.
In a certain shop , 70 percent of customers buy mobiles made by company (A) and the rest buy mobiles made by company (B) . A new mobile is made by company (A) . If 90 percent of customers who prefer company (A) and 50 percent of customers who prefer company (B) are expected to buy the new mobile , whats percent of customers are expected to buy the new mobile ? | [
"70",
"78",
"80",
"82"
] | B | Say there are total of 100 customers in the shop . Thus 70 customers prefer company (A) and 30 customers prefer company(B).
70*0.9=63 customers who prefer company (A) are expected to buy for the new mobile ;
30*0.5=15 customers who prefer company (B) are expected to buy for the new mobile .
Thus total of 63+15=78 members are expected to buy the new mobile, which is 78% of the total number of customers .
Answer: B. |
AQUA-RAT | AQUA-RAT-33891 | # 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 dishonest dealer professes to sell goods at the cost price but uses a weight of 750 grams per kg, what is his percent? | [
"11",
"33",
"77",
"88"
] | B | 750 --- 250
100 --- ? => 33%
Answer:B |
AQUA-RAT | AQUA-RAT-33892 | # Probability of having $n$ dice equal to or greater than $x$ when $m$ dice are rolled
I am trying to figure out how to do the math for something like this. The scenario:
• $n = 4$
• $x = 7$ (using d10s, where 0 = 10)
• $m = 8$
In words, if I roll 8d10 (eight 10-sided dice) what is the probability of having four dice greater than or equal to 7 (where 0 is the greateest [10])
I have seen a lot of sites which will do the calculation where $n = 1$, but I want to make $n$ a variable.
I know the probability of a 'successful' role is 4/10.
I know that I could write out all the possible combinations and count the ones that meet the criteria, but I'm sure this can be done with math.
How would you calculate this?
## migrated from stackoverflow.comNov 26 '16 at 5:07
This question came from our site for professional and enthusiast programmers.
Probability (in this case) is the ratio of desired outcomes to all outcomes. Let's count desired outcomes in your particular case (and by the way we derive a general formula to compute it).
Suppose you roll 4 dice and you have a total success - on every die you get "7 to 10". How many quadruplets of numbers "1 to 10" may cause it? It is not so difficult: $$4 \cdot 4 \cdot 4 \cdot 4$$ or $$4 ^ 4$$.
Then you roll the 4 remaining dice - but now you need a total failure - numbers 1 to 6 on every die. How many quadruplets is able to cause it? The answer is similar: $$6 ^ 4$$.
So for the first group of four dice to be totally successful and in the same time the second group of four dice to be totally unsuccessful - e. g. $$(8, 8, 8, 8, 1, 1, 1, 1)$$ - there is $$4^4 \cdot 6^4$$ possibilities.
The following is multiple choice question (with options) to answer.
In a single throw of a die, what is the probability of getting a number greater than 4? | [
"1/2",
"1/3",
"2/3",
"1/4"
] | B | When a die is thrown we have S = {1,2,3,4,5,6}
Let E = event of getting a number greater than 4 = {5,6}
P(E) = n(E)/n(S) = 2/6 = 1/3.
Answer B. |
AQUA-RAT | AQUA-RAT-33893 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B and C enter into partnership. A invests some money at the beginning, B invests double the amount after 6 months, and C invests thrice the amount after 8 months. If the annual gain be Rs.21000. A's share is? | [
"s.7000",
"s.7200",
"s.6000",
"s.5750"
] | A | x* 12 : 2x* 6: 3x* 4
1:1:1
1/3 * 21000 = 7000
ANSWER:A |
AQUA-RAT | AQUA-RAT-33894 | I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.
• If $$4$$ is in the first position, then the last position has $$4$$ possibilities to keep the number even ($$0$$, $$2$$, $$6$$, or $$8$$). In this case, you get the number of possibilities is: $$1\cdot 4\cdot 8\cdot 7\cdot 6=1344.$$ Here, the product order is "first, last, second, third fourth."
• If $$4$$ is in the last position, then the count above works. In this case, you get the number of possibilities is: $$1\cdot 8\cdot 8\cdot 7\cdot 6=2688.$$ Here, the product order is "last, first, second, third, fourth."
• If $$4$$ is in one of the middle positions, then the last position has $$4$$ possibilities. There are two possibilities here, either $$0$$ is the last position or it isn't.
• If $$0$$ is in the last position, then the number of possibilities is: $$3\cdot 1\cdot 8\cdot 7\cdot 6=1008.$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."
• If $$0$$ is not in the last position, then the number of possibilities is: $$3\cdot 3\cdot 7\cdot 7\cdot 6=2646$$ Here, the product order is "position of $$4$$, last, first, two remaining positions."
Adding all of these up gives $$7686$$, as desired.
• I like this answer more than mine, it is much clearer. Thank you! – Student Nov 5 '18 at 17:47
The following is multiple choice question (with options) to answer.
Three numbers are such that the second is as much
lesser than the third as the first is lesser than
the second. If the product of the two smaller numbers
is 85 and the product of two larger numbers is 115.
Find the middle number? | [
"10",
"12",
"14",
"16"
] | A | Let the three numbers be x,y,z
2y = x+z.....................1
Given that the product of two smaller numbers is 85
x y = 85................2
Given that the product of two larger numbers is 115
y z = 115...............3
Dividing 2 and 3 x y /y z = 85/115
x / z = 17 / 23
From 1
2y = x+z
2y = 85/y + 115/y
2y2 = 200
y2 = 100
y = 10
ANSWER A |
AQUA-RAT | AQUA-RAT-33895 | x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
$$(20\frac{1}{4})x + 5\frac{1}{2} = 7\frac{1}{16} \\~\\ (\frac{81}{4})x + \frac{11}{2} = \frac{113}{16} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11}{2} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{11(8)}{2(8)} \\~\\ (\frac{81}{4})x = \frac{113}{16} - \frac{88}{16} \\~\\ (\frac{81}{4})x = \frac{113-88}{16} \\~\\ (\frac{81}{4})x = \frac{25}{16} \\~\\ x = \frac{25}{16} / \frac{81}{4} \\~\\ x = \frac{25}{16} * \frac{4}{81} \\~\\ x = \frac{25*4}{16*81} \\~\\ x = \frac{100}{1296} = \frac{25}{324}$$
hectictar Mar 13, 2017
#7
+223
+5
Since this one's laid out so nicely I'll give it 5 stars also! Thank you for your help, too!
The following is multiple choice question (with options) to answer.
65% of x = 20% of 617.50. Find the value of x? | [
"100",
"190",
"150",
"180"
] | B | 65% of x = 20% of 615.50
Then, 65/100 * x = 20/100 * 6175/10
x = 190
Answer is B |
AQUA-RAT | AQUA-RAT-33896 | There are five types of candies: cherry, strawberry, orange, lemon, and pineapple. How many ways are there to choose $32$ candies?
Let $c$, $s$, $o$, $l$, and $p$ denote, respectively, the number of cherry, strawberry, orange, lemon, and pineapple candies selected. Then $$c + s + o + l + p = 32 \tag{1}$$ is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of four addition signs in a row of $32$ ones. For instance, $$1 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 +$$ corresponds to $c = 7$, $s = 5$, $o = 12$, $l = 8$, and $p = 0$. Therefore, the number of solutions of equation 1 is equal to the number of ways we can choose which four of the $36$ positions (for $32$ ones and four addition signs) will be filled with addition signs, which is $$\binom{32 + 4}{4} = \binom{36}{4}$$
More generally, the number of solutions of the equation $$x_1 + x_2 + x_3 + \ldots + x_k = n$$ in the nonnegative integers is $$\binom{n + k - 1}{k - 1}$$ since we must choose which $k - 1$ of the $n + k - 1$ symbols (for $n$ ones and $k - 1$ addition signs) will be filled with addition signs.
I think I understand how to go from $5$ to $4$. This is because, there are $5$ types of candies, so we need $4$ partitions.
Yes.
Why then do we add $4$ to $32$?
The following is multiple choice question (with options) to answer.
How many subset of the Fruit set of Banana, Apple, Orange, Grape and Lemon does not contain Apple? | [
"17",
"16",
"15",
"14"
] | B | This is a combination solution
Since we have 4 different kinds of fruits other than Apple we can take any fruit from the set of 4 to make a subset.
4C1 +4C2 + 4C3 + 4C4 = 15
Plus the one set that is null that is the Set having no element in it which is 4C0 =1
= 15 + 1 = 16
B = 16 |
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