source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-38697 | The following is an example of a method that you could use to figure this out. The method is called bisection. It is not the fastest, but it is clear how to extract digits from it because it has explicit error bounds.
You know $1^2<2$. You know $2^2>2$. So $1<\sqrt{2}<2$ so its first decimal digit is $1$.
You check $1.5^2>2$. You check $1.25^2<2$. You check $1.375^2<2$. You check $1.4375^2>2$. You check $1.40625^2<2$. Now you know $1.40625<\sqrt{2}<1.4375$ so you know the first two digits are $1.4$.
You continue: you check $1.421875^2>2$. You check $1.4140625^2<2$. You check $1.41796875^2>2$. So $1.4140625<\sqrt{2}<1.41796875$, so you know the first three decimal digits now.
You can keep going; at each time you know $\sqrt{2}$ is in between two numbers getting closer together, so as soon as those numbers have a new digit in common, you know that digit of $\sqrt{2}$. On average it takes $\log_2(10) \approx 3.3$ steps to get a new correct decimal digit.
At the cost of slightly more iterations, you can make calculations easier (if you're doing it by hand) by rounding the lower bound down and/or the upper bound up. For instance back in the second paragraph of iterations you could have said $1.4<\sqrt{2}<1.44$ and then continued, obtaining $1.41<\sqrt{2}<1.415$ at the end of the third paragraph.
The following is multiple choice question (with options) to answer.
Rounded to three decimal places, 1.001^4 = | [
"1.004",
"1.006",
"1.008",
"1.012"
] | A | As compared to 1, 0.001 is a very small quantity. Thus, we can write (1+0.001)^4 is nearly equal to (1+4*0.001) = 1.004.As the question asks for approximation to three decimal places, the further terms will anyways not come into picture.
A |
AQUA-RAT | AQUA-RAT-38698 | sum(res==0)/B
[1] 0.120614
So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).
The following is multiple choice question (with options) to answer.
If 40% of a certain number is 160, then what is 90% of that number? | [
"270",
"380",
"260",
"360"
] | D | Explanation:
40% = 40*4 = 160
90% = 90*4 = 360
Answer: Option D |
AQUA-RAT | AQUA-RAT-38699 | ## 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.
An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 12 hours to fill the tank. If the tank is full, how much time will the leak take to empty it? | [
"4 hours",
"12hours",
"8hours",
"5hours"
] | A | Work done by the leak in 1 hour = 1/3 - 1/12 = 1/4
The leak will empty the tank in 4hours
Answer is A |
AQUA-RAT | AQUA-RAT-38700 | 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.
How many seconds does Sandy take to cover a distance of 600 meters, if Sandy runs at a speed of 18 km/hr? | [
"100",
"120",
"140",
"160"
] | B | 18 km/hr = 18000 m / 3600 s = 5 m/s
time = 600 / 5 = 120 seconds
The answer is B. |
AQUA-RAT | AQUA-RAT-38701 | 2. When you add one component to a solution, the amount of other component does not change. In milk and water solution, if you add water, amount of milk is the same (not percentage but amount)
3.
Amount of A = Concentration of A * Volume of mixture
Amount = C*V
( e.g. In a 10 lt mixture of milk and water, if milk is 50%, Amount of milk = 50%*10 = 5 lt)
When you add water to this solution, the amount of milk does not change.
So Initial Conc * Initial Volume = Final Conc * Final Volume
$$C_i * V_i = C_f * V_f$$
$$C_f = C_i * (V_i/V_f)$$
In the question above, we find the final concentration of wine. Initial concentration $$C_i$$ = 1 (because it is pure wine)
When you remove 1 lt out of 3 lt, the volume becomes 2 lt which is your initial volume for the addition step. When you add 2 lts, final volume becomes 4 lt.
So $$C_f = 1 * 2/4$$
Since it is done 19 times, $$C_f = 1 * (\frac{2}{4}) * (\frac{3}{5}) * (\frac{4}{6}) * (\frac{5}{7}) * .......* (\frac{19}{21}) * (\frac{20}{22})$$
The concentration of wine is 1/77 and since the final volume is 22 lt (the last term has $$V_f$$ as 22, you get amount of wine = 1/77 * 22 = 2/7 lt
Kudos +1 Karishma
Is there an fast way to compute the result of the multiplacation series like we have for $$Cf$$? I actually did the long way .
Re: A container has 3L of pure wine. 1L from the container is taken out an [#permalink] 06 Feb 2013, 22:54
Go to page 1 2 Next [ 30 posts ]
# A container has 3L of pure wine. 1L from the container is taken out an
The following is multiple choice question (with options) to answer.
Four cups of milk are to be poured into a 3-cup bottle and a 6-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 6-cup bottle? | [
" 2/3",
" 7/9",
" 24/9",
" 8/3"
] | C | Let x be the # of cups going into the 6 cup bottle.
so.... x/6 = ((4-x)/3)
24 - 6x = 3x
24 = 9x
x = 24/9.
Answer : C |
AQUA-RAT | AQUA-RAT-38702 | c#, formatting
/// <summary>
/// Calculates and returns the general discount
/// based on the <see cref="AccountStatus"/>
/// </summary>
/// <param name="accountStatus"></param>
/// <returns></returns>
private static decimal GetDiscountPercentage(AccountStatus accountStatus)
{
switch(accountStatus)
{
case AccountStatus.SimpleCustomer:
return 0.10m;
case AccountStatus.ValuableCustomer:
return 0.30m;
case AccountStatus.MostValuableCustomer:
return 0.50m;
default:
return 0.00m;
}
}
/// <summary>
/// Applying the discounts (if any) on the price and returns the final price (after discounts).
/// </summary>
/// <param name="price"></param>
/// <param name="accountStatus"></param>
/// <param name="timeOfHavingAccountInYears"></param>
/// <returns></returns>
public static decimal ApplyDiscount(decimal price , AccountStatus accountStatus , int timeOfHavingAccountInYears)
{
decimal loyaltyDiscountPercentage = GetLoyaltyDiscountPercentage(timeOfHavingAccountInYears);
decimal discountPercentage = GetDiscountPercentage(accountStatus);
decimal priceAfterDiscount = price * (1.00m - discountPercentage);
decimal finalPrice = priceAfterDiscount - ( loyaltyDiscountPercentage * priceAfterDiscount );
return finalPrice;
}
}
since all methods don't need several instance, and the nature of the class is unchangeable, making the class static would be more appropriate.
You can then reuse it :
var finalPrice = DiscountManager.ApplyDiscount(price, accountStatus, timeOfHavingAccountInYears);
The following is multiple choice question (with options) to answer.
A single discount equivalent to the discount series of 20%, 10% and 5% is? | [
"31.7",
"31.1",
"31.6",
"31.4"
] | C | 100*(80/100)*(90/100)*(95/100)
= 68.4
100 - 68.4
= 31.6
Answer: C |
AQUA-RAT | AQUA-RAT-38703 | Define y = 2^{2x}. Then
y^2 - y - 2 = 0
(y - 2)(y + 1) = 0
Thus y = 2 or y = -1.
So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.
Thus x = 1/2.
-Dan
5. Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?
In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216
(2*3)^x = 216
6^x = 6^3
Thus x = 3.
So keep an eye out for ones you can do.
-Dan
The following is multiple choice question (with options) to answer.
Let y = 2m + x^2 and m = 5x + 4. If m^2 =81, then which of the following can be a value of 2y + 3m? | [
"65",
"130",
"141",
"142"
] | A | m^2 = 81 => m=9
9= 5x+4 => x=1
y= 2*9+1^2 = 19
2y+3m = 2*19+3*9 = 65
Answer : A |
AQUA-RAT | AQUA-RAT-38704 | 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.
The ages of Patrick and Michael are in the ratio of 3 : 5 and that of Michael and Monica are in the ratio of 3 : 5. If the sum of their ages is 146, what is the difference between the ages of Patrick and Monica? | [
"27",
"48",
"45",
"72"
] | B | Ages of P and Mi=3x+5x
Ages of Mi and Mo=3x:5x
Rationalizing their ages. ratio of their ages will be 9x:15x:25x
Sum=47x=146
x=3
Difference if ages of Pa and Mo=25x-9x=16x=16*3=48
Answer B |
AQUA-RAT | AQUA-RAT-38705 | 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.
At a certain conference, 150 students spoke one or more of the following languages: Spanish, English, and French. If 51 students spoke Spanish, 90 spoke English, 65 spoke French, and 32 spoke only 2 languages, how many students spoke three different languages? | [
"7",
"9",
"12",
"15"
] | C | 150=51+90+65-32-2x
x=12
C. 12 |
AQUA-RAT | AQUA-RAT-38706 | # How many numbers made out of 5 distinct digits contain 4 and are even
This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.
I distinguished some cases (not sure if I distinguished in a 'smart' way)
• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.
The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.
This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?
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.
The following is multiple choice question (with options) to answer.
From the digits 2, 3, 4, 5, 6 and 7, how many 5-digit numbers can be formed that have distinct digits and are multiples of 12? | [
"36",
"60",
"84",
"72"
] | B | DETAILED SOLUTION
Any multiple of 12 should be a multiple of 4 and 3. First, let us look at the constraint for a number being a multiple of 3. Sum of the digits should be a multiple of 3. Sum of all numbers from 2 to 7 is 27. So, if we have to drop a digit and still retain a multiple of 3, we should drop either 3 or 6.
So, the possible 5 digits are 2, 4, 5, 6, 7 or 2, 3, 4, 5, 7.
When the digits are 2, 4, 5, 6, 7. the last two digits possible for the number to be a multiple of 4 are 24, 64, 52, 72, 56, 76. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 36 different numbers.
When the digits are 2, 3, 4, 5, 7. The last two digits possible for the number to be a multiple of 4 are 32, 52, 72, 24. For each of these combinations, there are 6 different numbers possible. So, with this set of 5 digits we can have 24 different numbers.
Overall, there are 60 different 5-digit numbers possible.
Correct Answer: B |
AQUA-RAT | AQUA-RAT-38707 | ### 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.
David invested certain amount in three different schemes. A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B? | [
"5000",
"5007",
"5006",
"5003"
] | A | Let x, y and z be the amount invested in schemes A, B and C respectively. Then,
(x * 10 * 1)/100 + (y * 12 * 1)/100 + (z * 15 * 1)/100 = 3200
10x + 12y + 15z = 320000
Now, z = 240% of y = 12/5 y
And, z = 150% of x = 3/2 x
x = 2/3 z = ( 2/3 * 12/5) y = 8/5 y
16y + 12y + 36y = 320000
y = 5000
Sum invested in scheme B = Rs. 5000.
Answer:A |
AQUA-RAT | AQUA-RAT-38708 | # Sort of a challenge
#### topsquark
##### Well-known member
MHB Math Helper
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
-Dan
#### Bacterius
##### Well-known member
MHB Math Helper
Well the Earth's radius is $6371 ~ \text{km}$, so the original rope has a length of $2 \pi \times 6371 ~ \text{km}$. If we add $6 ~ \text{m} = 6 \times 10^{-3} ~ \text{km}$ to this rope, its new radius is:
$$\frac{2 \pi \times 6371 + 6 \times 10^{-3}}{2 \pi} \approx 6371.0009 ~ \text{km}$$
So the rope now "floats" about $0.0009 ~ \text{km} = 90 ~ \text{cm}$ above the ground. To be exact, $95.5 ~ \text{cm}$ (this is $\frac{6}{2 \pi} ~ \text{m}$).
Wait, what? My mind is blown
#### MarkFL
Staff member
This isn't so much a challenge problem as much as it has a startling (at least I think so) answer.
Say we tie a rope around the Earth. Now we are going to cut it and add another 6 meters to it. If we pull the new rope tight (in a circle) how high is the new rope above the Earth's surface?
It's reasonably easy so if you've seen it before let someone else answer. (I first saw this problem in a "Dr. Crypton" article in "Science Digest.")
The following is multiple choice question (with options) to answer.
A rope of which a calf is tied is increased from 12 m to 23 m, how much additional grassy ground shall it graze? | [
"1218",
"1212",
"1210",
"277"
] | C | π (232 – 122) = 1210
Answer: C |
AQUA-RAT | AQUA-RAT-38709 | Since these 6 possibilities are exhaustive, we have $$(1+2+3+4+5+6)q/6 = 1 \implies q = {6 \over 21} = {1 \over 3.5}$$ as I originally suspected; in particular, this implies that $$P(X_\tau = M + j) = (6-j)q/6 = (6-j)/21,$$ agreeing with Mike's answer.
The following is multiple choice question (with options) to answer.
In ΔPQS above, if PQ =5 and PS = 6, then | [
" 9/4",
" 12/5",
" 16/5",
" 30/7"
] | D | there are two ways to calculate area of PQS. area remains same, so both are equal.
5*6/2 = PR*7/2
PR = 30/7
D |
AQUA-RAT | AQUA-RAT-38710 | 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
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
The following is multiple choice question (with options) to answer.
P and Q can do a work in 30 days. Q and R can do the same work in 24 days and R and P in 20 days. They started the work together, but Q and R left after 12 days. How many days more will P take to finish the work? | [
"12",
"15",
"18",
"20"
] | A | Let work done by P in 1 day = p,
Work done by Q in 1 day = q,
Work done by R in 1 day = r
p + q = 1/30
q + r = 1/24
r + p = 1/20
Adding all the above, 2p + 2q + 2r = 1/30 + 1/24+ 1/20 = 15/120 = 1/8
=> p + q + r = 1/16
=> Work done by P,Q and R in 1 day = 1/16
Work done by P, Q and R in 12 days = 12 × (1/16) = 12/16 = 3/4
Remaining work = 1 = 3/4 = 1/4
Work done by P in 1 day = Work done by P,Q and R in 1 day - Work done by Q and R in 1 day
= 1/16 – 1/24 = 1/48
Number of days P needs to work to complete the remaining work = (1/4) / (1/48) = 12
Option A |
AQUA-RAT | AQUA-RAT-38711 | Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
(A) 1/5
(B) 1/4
(C) 1/2 *correct answer
(D) 3/4
(E) 4/5
Original Acid : Total ratio = 50:100 . let us say you replaced x part of this. So you are left with (1-x) of the original. So volume of acid left is (1-x)50.
in the new solution.. you added x of 30% solution. i.e 30x acid
(1-x)50 + 30x is the volume of acid. The total volume is still 100. And this concentration is 40%
[(1-x)50 + 30x] / 100 = 40/100
solve to get x=1/2
SVP
Joined: 30 Apr 2008
Posts: 1835
Location: Oklahoma City
Schools: Hard Knocks
Re: Some part of a 50% solution of acid was replaced with an [#permalink]
### Show Tags
12 Aug 2008, 15:16
The reason that 1/2 is correct is this:
What would happen if you have 1 liter of 60% solution (NaCl with water) and add 1 liter of 50% solution? I'm not chemist but lets mix $$H_2O$$ (water) with Sodium Chloride ($$NaCl$$).
The result will be a solution of 55%. 60% will have .6 liters of NaCl and the 50% solution will have .5 liters of NaCl. Together that's 1.1 liters out of the total of 2 liters.
Here we see that 50% solution becomes 40% from adding 30%. 40% is just the average between the two, so we know that there must now be equal parts 50% solution and 30% solution. The only way to get equal parts is if you have 1/2 50% solution and 1/2 30% solution meaning 1/2 was replaced.
chrissy28 wrote:
Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution of acid was obtained. What part of the original solution was replaced?
The following is multiple choice question (with options) to answer.
Some of the 50% solution of acid was removed and this was replaced with an equal amount of 10% solution of acid. As a result, a 20% solution of acid was obtained. What fraction of the original solution was replaced? | [
"1/2",
"2/3",
"3/4",
"3/5"
] | C | Let x be the fraction of the original solution that was replaced.
0.5*(1-x)+0.1(x)=0.2
0.4x=0.3
x=3/4
The answer is C. |
AQUA-RAT | AQUA-RAT-38712 | # Formulation and computation of “the” unique median of an even-sized list
Consider an even-sized set of numbers $X = \{x_k\}$, such as $X = \{1, 2, 7, 10\}$.
The median $m$ is defined as:
$$m = \mathrm{arg \min_x} \sum_k \lvert x_k - x\rvert^1$$
Any $m \in [2, 7]$ is a minimizer of this function, and is therefore "a" median of this list.
Now, it is common practice to take the average of 2 and 7 and call it "the" median.
But that's lame, and I think I have invented (?) a more logical way to find a unique median $m^*$:
$$m^* = \lim_{\epsilon \to 0^+} \mathrm{arg \min_x} \sum_k \left\lvert x_k - x\right\rvert^{1+\epsilon}$$
Diferentiation to find the minimum only gets us so far:
$$\sum_k \mathrm{sgn}{\left(x_k - m^*\right)}\left\lvert x_k - m^*\right\rvert^{\epsilon} = 0$$ This expression can be solved numerically for smaller and smaller $\epsilon$ to give $m^* \approx 4.85$ in this example, and I suspect the "correct" median is in fact $m^* = 34/7$, but I don't know how to prove it.
I have 3 questions:
1. First of all, is this a well-known and/or useful approach? Does it have a name?
I came up with the new formulation myself, but I've never seen it used anywhere.
2. Is there some way to directly find the exact value of $m^*$, without numerical optimization?
If not, is there a better/faster approach than brute-force numerical optimization techniques?
The following is multiple choice question (with options) to answer.
Set C consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set C? | [
"5",
"6",
"7",
"1"
] | D | Whenever a question asks you to deal with the MEDIAN of a group of numbers, you MUST put the numbers in order from least to greatest (or at the very least,group offthe numbers so that you can restrict the potential value of the median).
Here, we have 8 values (one of the values is the variable N). We're told that the values are UNIQUE, so N CANNOT be any of the other 7 values listed. We're asked which of the following answers COULD be the median, so if we determine that one of the answer choices is a POSSIBLE median, then we can stop working....
Since the group includes 8 values, the median will be the AVERAGE of the 4th and 5th values...
Putting the 7 numbers in order, we have:
-27, -3, -2, 2, 3, 15, 17
This means that the2will either be the 4th term or the 5th term (depending on the value of N).
IF....
N is REALLY BIG, then the 4th and 5th terms will be 2 and 3, so the median will = 2.5
N is REALLY SMALL, then the 4th and 5th terms will be -2 and 2, so the median will = 0
Neither of those answers is among the choices though, so N must be something relativelycloseto 2.....
IF....
N = 0, then the 4th and 5th terms will be 0 and 2, so the median will = 1. That answer IS among the choices, so we're done.
D |
AQUA-RAT | AQUA-RAT-38713 | Time needed for both machine to produce 1500 units is given.
1/8 + 1/12
= 5/24
I hrs work done by both machine.
24/5 = 4.8.
Time needed to complete the entire work.
******* 1500 units must be considered as a single work. units are same for both machine. Therefore we don't need to do anything with units.
C is the correct answer.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10543
Location: United States (CA)
Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink]
### Show Tags
12 Aug 2019, 10:53
Bunuel wrote:
A new copy machine can run off 1,500 workbooks in 8 hours, while it takes on older copy machine 12 hours to do the same job. What is the total number of hours that it would take both copy machines working at the same time, but independently, to run off the 1,500 workbooks?
(A) 4.4
(B) 4.6
(C) 4.8
(D) 5
(E) 10
Let T be the number of hours to complete the job when both machines are running. We can create the following equation:
1500/8 * T + 1500/12 * T = 1500
T/8 + T/12 = 1
Let’s multiply each side by 24:
3T + 2T = 24
5T = 24
T = 4.8 hours
_________________
# Scott Woodbury-Stewart
Founder and CEO
Scott@TargetTestPrep.com
202 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.
Re: A new copy machine can run off 1,500 workbooks in 8 hours, while it ta [#permalink] 12 Aug 2019, 10:53
# A new copy machine can run off 1,500 workbooks in 8 hours, while it ta
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
The following is multiple choice question (with options) to answer.
Printer P can print one lakh books in 8 hours. Printer Q can print the same number of books in 10 hours while Printer R can print the same in 12 hours. All the Printers started printing at 9 A.M. Printer P is stopped at 11 A.M. and the remaining 2printers complete work. Approximately at what time will the printing of 1 lac books be completed? | [
"10 pm",
"1 pm",
"12 pm",
"2 pm"
] | B | Work done by P in 1 hour = 1/8
Work done by Q in 1 hour = 1/10
Work done by R in 1 hour = 1/12
Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120
Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60
From 9 am to 11 am, all the Printer were operating.
Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60
Pending work = 1- 37/60 = 23/60
Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11
which is approximately equal to 2
Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm
B |
AQUA-RAT | AQUA-RAT-38714 | java
Any help will be welcomed. Thanks
EDIT:
Keypad format:
The problem statement:
We are given two inputs:
A phone number in string format.
A list of strings.
The output is the list containing only those strings which correspond to any substring of the given phone number when converted to the corresponding letters on the phone keypad.
Example test case:
Input:
3662277
words = ["foo", "bar", "baz", "emo", "foobar", "cap", "car", "cat"]
Output:
outputList = ["foo", "bar", "foobar", "emo", "cap", "car"]
The following is multiple choice question (with options) to answer.
How many times the keys of a writter have to be pressed in order to write first 400 counting no.'s? | [
"998",
"1000",
"1092 times",
"1100"
] | C | 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
-----------
1092
C |
AQUA-RAT | AQUA-RAT-38715 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
Three types of tea the a,b,c costs Rs. 95/kg,100/kg and70/kg respectively.How many kgs of each should be blended to produce 100 kg of mixture worth Rs.90/kg, given that the quantities of band c are equal | [
"70,15,15",
"50,25,25",
"60,20,20",
"40,30,30"
] | B | given quantities of b and c are equal .
therefore instead of considering them as a different quantities take average of both and consider it as a single entity .
so the cost of the mixture "D" (equal quantities of "B" and "C") is 170/2=85rs/kg
now the tea contains only "A" of 95 rs/per and "D" of 85 rs/kg.
it is clearly seen that "A" is 5 more and "D" is 5 less than the required cost per kg .
therefore 50/50 will go well .
A=50kg and D=50kg. (D= B&C in equal ratio). therefore B&C=25kg each .
(A:B:C)=(50,25,25)
ANSWER:B |
AQUA-RAT | AQUA-RAT-38716 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
There are four members in a family. Their average income is Rs.10000 per month. Three members get Rs.8000, Rs.15000, Rs.6000 respectively. Therefore what is the income of the fourth member of the family? | [
"Rs. 10000",
"Rs. 11000",
"Rs. 12000",
"Rs. 9000"
] | B | EXPLANATION
Total income of 3 members = Rs. (8000 + 15000 + 6000) = Rs. 29000.
Required income = Rs. [ (10000 x 4) – 29000 ]
= Rs. (40000 – 29000)
= Rs. 11000.
Answer B |
AQUA-RAT | AQUA-RAT-38717 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The cost price of 15 articles is equal to the selling price of 11 articles. Find the profit percent? | [
"18 2/91%",
"36 4/11%",
"18 9/11%",
"18 2/91%"
] | B | 15 CP = 11 SP
11 --- 2 CP
100 --- ? =>36 4/11%
Answer:B |
AQUA-RAT | AQUA-RAT-38718 | java, performance
Title: The car plate problem: generate from AAA0001 to ZZZ9999 skipping the 0000 I was given the problem (called "car plate problem") to write all the values that go from AAA0001 to ZZZ9999 with the following premises:
Create all the possibilities from AAA0001 to ZZZ9999;
Skip all the 0000, ie: AAB0000, AAC0000... ZZZ0000.
No repetitions. Once a value is "created" or "found", whatever, the value cannot repeat. So, once your code generated the AAA0002, it shouldn't generate the AAA0002 again.
So... I found a solution (posted below). It was "short" and "oh boy" it was weak! But I was coding against time so I had to go with that solution. I recognize the solution lacks creativity, better use of O.O. and probably there is a design pattern to be use here. None I could identify but still, there must be a better way to make it work. Is there a solution with better code quality and (if possible) less processing?
public class PlateIncrement {
private String plate;
public PlateIncrement () {
}
public void setPlate(String thePlate) {
plate = thePlate;
}
public String getIncrementedPlate() {
String finalNumber = "";
String finalLetters = "";
String letters = plate.substring(0,3);
String numbers = plate.substring(3,7);
letters = letters.toUpperCase();
if (Integer.parseInt(numbers) != 9999) {
finalNumber = incrementNumber(numbers);
finalLetters = letters;
} else {
finalLetters = getIncrementedLetters(letters);
finalNumber = "0001";
}
return finalLetters.concat(finalNumber);
}
The following is multiple choice question (with options) to answer.
How many different number-plates for cars can be made if each number-plate contains four of the digits 0 to 9 followed by a letter A to Z, assuming that
(a) no repetition of digits is allowed? | [
"131040",
"116700",
"142369",
"147225"
] | A | we want 4 digit of cars and repetition is not allowed
1st digit can be fill with 0 to 9 i.e = 10
2nd digit can be fill with 0 to 9 except earlier selected i.e = 9
3rd digit can be fill with 0 to 9 except earlier selected i.e = 8
4th digit can be fill with 0 to 9 except earlier selected i.e = 7
No of ways = 10 * 9 * 8 * 7 * (total aplphabets)
= 5040 * 26
= 1,31,040
Ans - A |
AQUA-RAT | AQUA-RAT-38719 | homework-and-exercises, fluid-dynamics
Title: Time needed to empty a syringe A) a syringe is filled with 10mL water.
The diameter of the barrel of the syringe is 15.9 mm and the diameter of the tip is 1.2mm. With how much force do we need to push on the plunger to empty the syringe from water in 10 seconds? Neglect friction of the plunger and the waters viscosity.
B)Now we add a needle to the syringe, with a length of 25mm and diameter 0.210 mm. If you push on the plunger with a force 0.10 N, how much time will it take to empty the syringe? Now we should consider the viscosity of water: 0.0010Pas For the first case:
Neglect friction of the plunger and the waters viscosity.
This means the only force needed is that to overcome inertia, acc. $\text{N2L}$ $(^*)$. This force is usually very small and most often neglected in similar problems.
For the second case we also ignore the minute inertial forces and you rightly suggest to use the Hagen-Poiseuille equation for flow through pipes:
$$\Delta p=\frac{8 \mu LQ}{\pi R^4}\tag{1}$$
(for list and meaning of symbols, see link above)
The first thing to determine is $\Delta p$. The stated force of $0.1\mathrm{N}$ applied to the plunger and the plunger diameter allow to calculate the pressure in the barrel of the syringe:
$$p_1=\frac{F}{A}$$
where $F$ is the force and $A$ the cross-section of the plunger.
Now we make a very reasonable assumption, i.e. that the pressure loss due to flow through the barrel is very low compared to the pressure loss through the actual needle.
Because the pressure at the end tip of the needle is equal to atmospheric pressure $p_{atm}$, we can then write:
$$\Delta p \approx p_1-p_{atm}$$
We now have everything to apply $(1)$.
The following is multiple choice question (with options) to answer.
Two taps can separately fill a cistern 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 9 minutes. The waste pipe can empty the full cistern in? | [
"18",
"17",
"16",
"19"
] | A | 1/10 + 1/15 - 1/x = 1/9
x = 18
Answer:A |
AQUA-RAT | AQUA-RAT-38720 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
If john makes a contribution to a charity fund at school, the average contribution size will increase by 50% reaching $75 per person. If there were 3 other contributions made before john's, what is the size of his donation? | [
"$100",
"$150",
"$200",
"$250"
] | B | Cavg = average contribution before John
Cavg*1.5 = 75, therefore the average cont is $50 before John.
If he needs to increase the average contribution by $25, he must put in $25 for each of the 3 people.
so $75.
But, he also has to put in the average for himself (the fourth person), so add $75.
So $150 is your answer.
Answer B |
AQUA-RAT | AQUA-RAT-38721 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
The average weight of 5 students decreases by 12 kg when one of them weighing 72 kg is replaced by a new student. The weight of the student is | [
"12 kg",
"10 kg",
"20 kg",
"72 kg"
] | A | Explanation :
Let the weight of student be x kg.
Given, Difference in average weight = 12 kg
=> (72-x)/5 = 12
=> x = 12
Answer : A |
AQUA-RAT | AQUA-RAT-38722 | (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 = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A container has a capacity of 20 gallons and is full of spirit. 4 gallons of spirit is drawn out and the container is again filled with water. This process is repeated 4 times. Find how much spirit is left in the resulting mixture finally? | [
"8(257/525) gallons",
"8(346/625) gallons",
"8.192 gallons",
"8.25 gallons"
] | C | There is a direct formula to calculate such repetition:
Left QTY / Orig QTY = ( (Orig QTY - Rep QTY) / Orig QTY )^n
where n is number of times mixing done:
Left QTY / 20 = (20 -4)^4 / 20^4
Left QTY = 20 (16/20)^4
Left QTY = 20 ( 4/5 )^4
Left QTY = 8.192
Answer is C |
AQUA-RAT | AQUA-RAT-38723 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\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.
Find the cost of fencing around a circular field of diameter 28 m at the rate of Rs.1.50 a meter? | [
"289",
"132",
"279",
"277"
] | B | 2 * 22/7 * 14 = 88
88 * 1 1/2
= Rs.132
Answer: B |
AQUA-RAT | AQUA-RAT-38724 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
If the price of an article decreases by 111/9 % and the sale of the article increases by 12 ½ %, what is the net effect on revenue? | [
"1% loss",
"1% gain",
"No loss or no gain",
"Cannot be determined"
] | C | M. F 1 = 100-100/9/100 =800/900 = 8/9
M. F 2 = 100+25/2/100 =225/200 =9/8
Total M. F =8/9*9/8 = 1
Overall % change = ( M. F – 1 ) * 100 = ( 1 – 1 ) * 100 = 0
ANSWER:C |
AQUA-RAT | AQUA-RAT-38725 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 280 m long, running with a speed of 63 km / hr will pass a tree in | [
"12 sec",
"15 sec",
"16 sec",
"20 sec"
] | C | Sol.
Speed = (63 x 5/18) m/sec.
= 35 /2 m / sec.
Time taken = (280 x 2/35) sec = 16 sec
Answer C |
AQUA-RAT | AQUA-RAT-38726 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A salesman travels a distance of 50 km in 2 hours and 30 minutes. How much faster in kilometer per hour, on an average, must he travel to make such trip in 5/6 hour less time? | [
"10km/hr",
"15km/hr",
"20km/hr",
"30km/hr"
] | A | Time required = 2 hrs 30 min - 50 min = 1 hr 40 min = 1 2/3 hrs
Required speed = 50*3/5 = 30km/hr
Original speed = 50*2/5 = 20 km/hr
Difference = 30-20 = 10 km/hr
Answer is A |
AQUA-RAT | AQUA-RAT-38727 | doesn't rule out anything and thus it could be either coin with equal probability. to bet you even money that it is the two headed coin. One coin has been specially made and has a head. He selects a coin at random and flips it twice. One coin is chosen at random and flipped, coming up heads. Perform the following experiment. on each side. If the same coin is tossed twice, find the probability that it is the two-headed coin. There are three coins One is two headed coin, another is biased coin that comes up tails 25% of the times and the other is unbiased coin One of the three coins is chosen at random and tossed , it shows head What is the probability that - Math - Probability. I've read that if you flip a coin 10 times and it comes up heads every time then it's still a 50/50 chance to be a heads or tails on the 11th flip. I'm not a mathematician so please bear with me. The hypotheses areH1-the coin is two headed, and H2 the coin is fair. Condtion that the face observed is already heads. Find the probability that heads appears twice. A coin is chosen at random and tossed 2 times. Remember that P(A given B) = P(A and B)/P(B) So let's say that A is the event that he chose the 2-headed coin, and B is an event denoted by H(N), which indicates that the coin was tossed N times, and came up heads each time, so the answer in our first case is P(A given H(1)), and the answer our last case is P(A given H(3)). What is the conditional prob-ability that both are boys given that at least one of them is a boy? 2. There are three coins. This is the currently selected item. (The fair coin lands on heads 50% of the time). All k times the coin landed up heads. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of time. If you toss a coin, you cannot get both a head and a tail at the same time, so this has zero probability. What is the conditional probability that it is the fair coin. The rest are fair. onditional probability is a tool for
The following is multiple choice question (with options) to answer.
John tossed a fair coin 2 times. What is the probability that the coin landed heads up exactly twice? | [
"0.5",
"0.55",
"0.6",
"0.65"
] | A | The probability is exactly same as the cion landed tail up once. That is 2*(1/2)^2 = 2/4 =0.5.
The answer is, therefore, (A). |
AQUA-RAT | AQUA-RAT-38728 | computational-chemistry
( 5 0 | 4 0) = 0.000862792421254
( 5 2 | 1 0) = 0.000000000000000
( 5 3 | 1 0) = 0.049151047350638
( 5 4 | 1 0) = 0.041228155506758
( 5 1 | 2 0) = 0.000000000000000
( 5 1 | 3 0) = 0.005091750233326
( 5 1 | 4 0) = 0.004270986718219
( 5 0 | 2 1) = 0.000000000000000
( 5 0 | 3 1) = 0.001923027401919
( 5 0 | 4 1) = 0.001613045439387
( 5 2 | 1 1) = 0.000000000000000
( 5 3 | 1 1) = 0.184701361441018
( 5 4 | 1 1) = 0.154928467698285
( 5 1 | 2 1) = 0.000000000000000
( 5 1 | 3 1) = 0.038683534029509
( 5 1 | 4 1) = 0.032447950603007
( 5 2 | 2 0) = 0.009336058122476
( 5 2 | 3 0) = 0.000000000000000
( 5 2 | 4 0) = 0.000000000000000
( 5 3 | 2 0) = 0.000000000000000
( 5 3 | 3 0) = 0.011813922734513
( 5 3 | 4 0) = 0.002078445792233
( 5 4 | 2 0) = 0.000000000000000
( 5 4 | 3 0) = 0.002078445792233
( 5 4 | 4 0) = 0.011079469320272
( 5 0 | 2 2) = 0.054104219459936
( 5 0 | 3 2) = 0.000000000000000
The following is multiple choice question (with options) to answer.
0.002 x 0.5 = ? | [
"0.0001",
"0.001",
"0.01",
"0.1"
] | B | 2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
Answer: Option B |
AQUA-RAT | AQUA-RAT-38729 | 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.
Andy completes a piece of work in 10 days, Raju completes the same work in 40 days. If both of them work together, then the number of days required to complete the work is | [
"6 days",
"8 days",
"9 days",
"10 days"
] | B | 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.
That is, the required No. of days = 10 × 40/50 = 8 days.
B |
AQUA-RAT | AQUA-RAT-38730 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
The price of commodity X increases by 40 paise every year, while the price of commodityY increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ? | [
"2010",
"2011",
"2012",
"2013"
] | B | EXPLANATION
Suppose commodity X will cost 40 paise more than Y after z years.
Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40
0.25z = 0.40 + 2.10
x =2.50/ 0.25 =250/25 =10
X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.
Answer B |
AQUA-RAT | AQUA-RAT-38731 | Let P=[$(\sqrt{n}+0.7)^2$]
Second Hint
given $n \geq 1$, put n=1 gives P=2
n=2 gives P=4
n=3 gives P=5
n=4 gives P=7
n-5 gives P=8
n=6 gives P=9
n=7 gives P=11
Final Step
here missing number are
1,3,6,10,… which is following a certain pattern
1, 1+2, 3+3, 6+4, 10+5, 15+6, 21+7, 28+8, 36+9, 45+10, 55+11, 66+12.
so, $n_{12}$=78.
The following is multiple choice question (with options) to answer.
Let's find out how sharp your mind is. Let's see if you can find the missing numbers in the series:
192, 021, 222, 324, 252, 627, 2__, 9__? | [
"281 and 930.",
"282 and 930.",
"283 and 930.",
"None"
] | B | Solution:
If you look closely, you will find a sequence in every alternating number.
192, 222 and 252 form one series where 30 is added each time.
021, 324 and 627 form one series where 303 is added each time.
Thus the next two numbers will be 282 and 930.
Answer B |
AQUA-RAT | AQUA-RAT-38732 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a loss of 21%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price? | [
"560",
"882",
"799",
"778"
] | A | 79%
104%
--------
25% ---- 140
100% ---- ? => Rs.560
Answer: A |
AQUA-RAT | AQUA-RAT-38733 | distance from a given point called centre. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. As you can imagine, as the discs become thinner, the volume of the sphere gets more accurate. (Assume ≈ 3. Diameter - a line going through the circle from edge to edge, dividing circle in half. The cubic volume of a cylinder is found by multiplying the radius times the radius times pi times the height. Cylinder, hollow Calculate the volume, height, inner or outer radius of hollow cylinder. Cylinder Volume Formula Calculator - How to Calculate the volume of a cylinder. Net of a Cone. Find the area of a circle when you know the diameter. When the piston has moved up to the top of its stroke inside the cylinder, and the remaining volume inside the head or combustion chamber has been reduced to 100 cc, then the compression ratio would be proportionally described as 1000:100, or with fractional reduction, a 10:1 compression ratio. Show that the rectangle of maximum area that can be inscribed in a circle is a square. The Volume of a Cylinder is. The volume of each cone is equal to ⅓Bh = ⅓(28. Volume calculator will determine the volume of the most common geometric solids. The base of the cylinder is large circle and the top portion is smaller circle. Volume of a cylinder : V = πr 2 h where r is the radius and h is the height of the cylinder. It is the same measurement for circles of any size. 25 × 6 Inches Height = 37. cm, the base ring area is 115. What is the value of pi, rounded to the nearest hundredth? 3. However (a) the statements are in the incorrect order; (b) the function calls are incorrect: (c) the logical expression in the while loop is incorrect; and (d) function definitions are incorrect. Their radius (r) is therefore 3 m. Usually the pipe line would be in the shape of cylinder. Now you see that the ratio of the volume of a sphere to the volume of a cylinder is 2/3. 14 x 9 2 x 7. Therefore, the volume of a cylinder = πr2h cubic units. Also, this is important to know that the radius of a circle is always the half of its. Calculate the volume of a
The following is multiple choice question (with options) to answer.
The radius of a cylinder is 8 m, height 13 m. The volume of the cylinder is: | [
"2200",
"2614.9",
"3300",
"1100"
] | B | Cylinder volume = πr(power2)h
= 22/7 × 8 × 8 × 13
= 2614.9 m(power3)
Answer is B. |
AQUA-RAT | AQUA-RAT-38734 | ### 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 started a business with an investment of Rs. 70000 and after 6 months B joined him investing Rs. 120000. If the profit at the end of a year is Rs. 26000, then the share of B is? | [
"A)14500",
"B)14000",
"C)12000",
"D)13007"
] | C | Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 26000
Share of B = 6/13 (26000)
= Rs. 12000
Answer:C |
AQUA-RAT | AQUA-RAT-38735 | # If $(a+b)(a+c)(b+c)=8abc$ prove $a=b=c$
For positive numbers $$a,b,c$$ we have $$(a+b)(a+c)(b+c)=8abc$$. Prove $$a=b=c$$
I tried expanding the expression. after simplifying we have,
$$a^2b+ab^2+b^2c+ca^2+ac^2+bc^2=6abc$$ But not sure how to continue.
I also noticed that we have,
$$(a+b)(a+c)(b+c)=(2a)(2b)(2c)$$ $$(a+b)+(a+c)+(b+c)=(2a)+(2b)+(2c)$$ But I don't know if it helps.
• Have you tried AM-GM inequality...? Dec 14, 2021 at 14:12
• Dec 14, 2021 at 14:27
By AM-GM, $$a+b \geqslant 2\sqrt{ab}$$ with equality if and only if $$a=b$$. Multiplying together the three similar inequalities we get $$(a+b)(b+c)(c+a) \geqslant 8abc$$ with equality if and only if $$a=b=c$$.
Continuing from where you left, you have: $$b(a-c)^2+a(b-c)^2+ b^2c+ca^2-2abc=0$$,which is same as $$b(a-c)^2+a(b-c)^2+c(b-a)^2=0$$ So you now have sum of three non-negative numbers equal to $$0$$. Can you take it from here?
The following is multiple choice question (with options) to answer.
If a+b+c=13, a2+b2+c2=69, then find ab+bc+ca | [
"50",
"60",
"69",
"75"
] | A | (a+b+c)2= a2+b2+c2+2(ab+bc+ca)
2(ab+bc+ca)= (a+b+c)2- a2+b2+c2
=169-69
=100
ab+bc+ca=50
ANSWER A 50 |
AQUA-RAT | AQUA-RAT-38736 | ### Show Tags
13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
_________________
Thanks,
Lucky
_______________________________________________________
Kindly press the to appreciate my post !!
The following is multiple choice question (with options) to answer.
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3, 4, 5}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative? | [
"5/8",
"1/3",
"2/5",
"1/2"
] | A | Answer A.
Total # of outcomes: 5*8= 40
# of outcomes where Product is -ve : (-6,1) ,(-6,2), (-6,3)... Hence, total: 25
Probability: 25/40 = 5/8 |
AQUA-RAT | AQUA-RAT-38737 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 108 as interest at the end of the loan period, what was the rate of interest? | [
"3.6",
"3",
"18",
"24"
] | B | Let rate = R% and time = R years.
Then, (1200 x R x R)/100 = 108
12R2 = 108
R2 =9
R = 3.
Answer: Option B |
AQUA-RAT | AQUA-RAT-38738 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A sum of money is to be distributed among A, B, C, D in the proportion of 5:2:4:3. If C gets Rs. 1000 more than D, what is B's share? | [
"2778",
"2679",
"2000",
"2699"
] | C | Let the shares of A, B, C and D be 5x, 2x, 4x and 3x Rs. respectively.
Then, 4x - 3x = 1000 => x = 1000.
B's share = Rs. 2x = 2 * 1000 = Rs.2000.
Answer:C |
AQUA-RAT | AQUA-RAT-38739 | java, performance
Title: The car plate problem: generate from AAA0001 to ZZZ9999 skipping the 0000 I was given the problem (called "car plate problem") to write all the values that go from AAA0001 to ZZZ9999 with the following premises:
Create all the possibilities from AAA0001 to ZZZ9999;
Skip all the 0000, ie: AAB0000, AAC0000... ZZZ0000.
No repetitions. Once a value is "created" or "found", whatever, the value cannot repeat. So, once your code generated the AAA0002, it shouldn't generate the AAA0002 again.
So... I found a solution (posted below). It was "short" and "oh boy" it was weak! But I was coding against time so I had to go with that solution. I recognize the solution lacks creativity, better use of O.O. and probably there is a design pattern to be use here. None I could identify but still, there must be a better way to make it work. Is there a solution with better code quality and (if possible) less processing?
public class PlateIncrement {
private String plate;
public PlateIncrement () {
}
public void setPlate(String thePlate) {
plate = thePlate;
}
public String getIncrementedPlate() {
String finalNumber = "";
String finalLetters = "";
String letters = plate.substring(0,3);
String numbers = plate.substring(3,7);
letters = letters.toUpperCase();
if (Integer.parseInt(numbers) != 9999) {
finalNumber = incrementNumber(numbers);
finalLetters = letters;
} else {
finalLetters = getIncrementedLetters(letters);
finalNumber = "0001";
}
return finalLetters.concat(finalNumber);
}
The following is multiple choice question (with options) to answer.
When I acquired my Mercedes-Benz car in Germany, the first thing I had to do was to get a license plate. The plate I got had a peculiar number on it. It consisted of 5 different numbers and by mistake when I fixed it upside down the number could be still read, but the value had increased by 78633. | [
"10966",
"10968",
"10967",
"10966"
] | B | 5 differ numbers can be read upside down same are 0 , 1 , 6 , 8 , 9
try with these no.s
number is 10968.
10968
78633
-------
89601
--------
ANSWER:B |
AQUA-RAT | AQUA-RAT-38740 | • The first remark makes me more confident that the first approach is the correct one. Thanks for that. However, I do not think I forgot the possibility to have a six. My probability is $\frac{4}{6}$ so I am considering the four cases in which the third number is greater than $2$. Maybe also look at my edit. – AndreasS Jun 4 '16 at 0:30
• @AndreasS It's not that you haven't included the possibility, it's that you over counted how often it happens. You've counted the ways to select two sixes, $(6,6,x), (6,x,6), (x,6,6)$ then said the x can be replaced with any one of four values, including $(6,6,6), (6,6,6), (6,6,6)$ . – Graham Kemp Jun 4 '16 at 0:34
• Ah, yes. Now I understand what you mean. However, the second approach does not count those three ways but rather give the probability that the third dice (so only one) has a number in $\{3,4,5,6\}. – AndreasS Jun 4 '16 at 0:45 • Which is the third die when two are six and the third is a six? – Graham Kemp Jun 4 '16 at 2:01 • You are given that two of the dice are 6. So assume you have already two 6. Then you look at the probability that the third one, which we can now consider to be only a single die, is either$3,4,5,6\$, i.e. 4/6. So the third die "when two are six and the third is a six" is the last one. – AndreasS Jun 4 '16 at 8:00
The following is multiple choice question (with options) to answer.
Ram has two dies which has a nos 2,3,4,6 when he throws the two dies together find the no of possibility so that the some of the two dies should be even...??? | [
"4/8",
"5/8",
"6/8",
"7/8"
] | B | total possible combinations=(2,2)(2,4)(2,6)(4,2)(4,4)(4,6)(3,3)(6,2)(6,4)(6,6)=10
so probability=10/16=5/8
ANSWER:B |
AQUA-RAT | AQUA-RAT-38741 | "Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Expert Reply GMAT Club Legend Joined: 11 Sep 2015 Posts: 4959 Location: Canada GMAT 1: 770 Q49 V46 When a positive integer n is divided by 5, the remainder is 2. What is [#permalink] ### Show Tags 13 Apr 2018, 06:48 1 Top Contributor 4 MathRevolution wrote: [GMAT math practice question] When a positive integer $$n$$ is divided by $$5$$, the remainder is $$2$$. What is the remainder when $$n$$ is divided by $$3$$? 1) $$n$$ is divisible by $$2$$ 2) When $$n$$ is divided by $$15$$, the remainder is $$2$$. Target question: What is the remainder when n is divided by 3? Given: When positive integer n is divided by 5, the remainder is 2 ----ASIDE---------------------- When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. ----------------------------------- So, from the given information, we can conclude that some possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, etc Statement 1: n is divisible by 2 When we examine our list of possible n-values (2, 7, 12, 17, 22, 27, 32, 37, ... ), we see that n could equal 2, 12,
The following is multiple choice question (with options) to answer.
When positive integer N is divided by positive integer J, the remainder is 16. If N/J = 127.05, what is value of J? | [
"300",
"320",
"370",
"400"
] | B | When a number is divided by another number, we can represent it as : Dividend = Quotient * Divisor + Remainder
So, Dividend/Divisor = Quotient + Remainder/Divisor
Given that N/J = 127.05
Here 127 is the quotient.
Given that Remainder = 16
So, 127.05 = 127 + 16/J
So, J = 320
Answer: B |
AQUA-RAT | AQUA-RAT-38742 | algorithms, graphs, optimization
There is one final component of your problem we have not addressed, i.e., there is a set of books $S_1$ that MUST go into box 1 and a set $S_2$ that MUST go into box 2.
Off the top of my head, I do not see a way to fix this without incorporating a source and sink $s$ and $t$ (for box 1 and 2 respectively). The formulation I've given above is almost sufficient. If $j\in S_1$ (book j MUST go into box 1), make $p_{sj}$ arbitrarily large (much more than $W$). This ensures an arbitrarily large cost will be charged if you try to but a book that must go into box 1 into box 2. Similarly, if $i\in S_2$, make $p_{jt}$ arbitrarily large.
Assuming $S_1$, and $S_2$ are non-empty (that books are already forced into each box), $W$ can be taken as 0 (since a proper partition is enforced by $S_1,S_2$) and $p_{sj}=p_{it}$ for $j\in S_1, i\in S_2$ can be as little as $\sum_{{i,j} \in N:\ i<j} p_{ij}$ (assuming each $p_{ij}\geq 0$).
The following is multiple choice question (with options) to answer.
When Greenville State University decided to move its fine arts collection to a new library, it had to package the collection in 20-inch by 20-inch by 15-inch boxes. If the university pays $1.20 for every box, and if the university needs 3.06 million cubic inches to package the collection, what is the minimum amount the university must spend on boxes? | [
"$255",
"$275",
"$510",
"$612"
] | D | Total no. Of boxes = 3060000/(20×20×15) = 510
Total cost = 510×$1.20=$612
Answer D |
AQUA-RAT | AQUA-RAT-38743 | Once you have $9\ge x^2$, I would recommend first solving the associated equation. That is, $9= x^2$ which does, in fact, have x= 3 and x= -3 as solutions. That divides the number line into 3 intervals, $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$ on each interval of which the inequality is either true for every point in the interval or false for every point in the interval. (That is true because for any continuous function, f(x), f(x)< 0 can only change to f(x)> 0, and vice-versa, where f(x)= 0.)
Knowing that the three intervals are $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$. We need only look at a single point in each interval. x= -4 is in $(-\infty, -3)$ and $(-4)^2= 16> 9$ so $x^2> 9$ for all x in $(-\infty, -3)$. x= 0 is in $(-3, 3)$ and $(0)^2= 0< 9$ so $x^2< 9$ for all x in $(-3, 3)$. Finally, $x= 4$ is in (3, \infty) and $4^2= 16> 9$ so $x^2> 9$ for all x in $(3, \infty)$. Add to this the fact that $3^2= (-3)^2= 9$ and we have that $x^2\le 9$ if and only if x is in [-3, 3] or $-3\le x\le 3$.
The following is multiple choice question (with options) to answer.
Which of the following describes all values of x for which 9–x^2 >= 0? | [
" x >= 1",
" x <= –1",
" -3 <= x <= 3",
" x <= –1 or x >= 1"
] | C | 9 - x^2 >=0
means x^2 - 9 <=0
=> (x-3)(x+3) <= 0
=> -3 <= x <= 3
Answer - C |
AQUA-RAT | AQUA-RAT-38744 | 2. ### math
list all the 3 digit numbers that fit these clues. the hundreds digits is less than 3. the tens digit is less than 2. the ones digit is greater than 7
3. ### Math
Think of a five-digit number composed of odd numbers. The thousands digit is two less than the ten thousands digit but two more than the hundreds digit. The tens digit is two more than ones digit which is four less than the
4. ### Math (Confused)
How many international direct-dialing numbers are possible if each number consists of a four-digit area code (the first digit of which must be nonzero) and a five-digit telephone numbers (the first digit must be nonzero)? a.
1. ### math
how many positive 4-digit numbers are there with an even digit in the hundreds position and an odd digit in the tens position? a. 10,000 b. 5,040 c. 2,500 d. 2,250
2. ### math
Rich chooses a 4-digit positive integer. He erases one of the digits of this integer. The remaining digits, in their original order, form a 3-digit positive integer. When Rich adds this 3-digit integer to the original 4-digit
3. ### Math
7 digit number no repetition of numbers digit 5 in thousands place, greatest digit in the millions place, digit in the hundred thousands place is twice the digit in the hundreds place, digit in the hundreds place is twice the
4. ### math
the hundred thousands digit of a six-digit even numbers is 3 more than the thousand digit,which is twice the ones digit.give at least four numbers that satisfy the given condition.
The following is multiple choice question (with options) to answer.
How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit? | [
"1339",
"2300",
"4200",
"1340"
] | A | There is one number in hundred with 2 in th tens digit and 1 in the units digit: 21, 121, 221, 321, ...
The difference between 324,700 and 458,600 is 458,600-324,700=133,900 - one number per each hundred gives 133,900/100=1,339 numbers.
Answer: 1,339 A |
AQUA-RAT | AQUA-RAT-38745 | ### 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.
In business, A and C invested amounts in the ratio 4:2, whereas the ratio between amounts invested by A and B was 4:1, If Rs 20,000 was their profit, how much amount did A receive. | [
"13333",
"26758",
"45896",
"14639"
] | A | Explanation:
A:B = 4:1 = 4:1
=> A:C = 4:2 = 4:2
=> A:B:C = 4:1:2
A share = (4/6)*20000
= 13333
Option A |
AQUA-RAT | AQUA-RAT-38746 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
The arithmetic mean and standard deviation of a certain normal distribution are 14.5 and 1.7, respectively. What value is exactly 2 standard deviations less than the mean? | [
"10.5",
"11",
"11.1",
"12"
] | C | Mean = 14.5
Two standard deviations is 1.7 + 1.7 = 3.4
There could be two calues for this.
Mean + two standard deviations = 17.9
Mean - Two standard deviations = 11.1
Answer choice has 11.1 and so C is the answer. |
AQUA-RAT | AQUA-RAT-38747 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
What number comes next?
482, 693, 754, 826, 937, ? | [
"113",
"231",
"245",
"548"
] | D | D
548
The numbers 48269375 are being repeated in the same sequence. |
AQUA-RAT | AQUA-RAT-38748 | logic, first-order-logic, propositional-logic
Title: Relations between statements involving universal quantifier, conditional and biconditional If we consider two predicates:
$b(x)$: x is a boy
$c(x)$: x is clever
Then, there are four statements involving $∀, b(x), c(x), →$ and $↔$ . These are below along with my interpretation of their meaning. Correct me if I am wrong in any of these interpretations.
(a) $∀x(b(x)→c(x))$ - All boys are clever
(b) $∀x(b(x))→∀x(c(x))$ - If all are boys, then all are clever
(c) $∀x(b(x)↔c(x))$ - All boys are clever & all those who are clever are boys
(d) $∀x(b(x))↔∀x(c(x))$ - If all are boys, then all are clever & if all are clever then all are boys
I am trying to determine all conditional $(→,↛)$ / bi-conditional $(↔)$ relations between each two of above four. Between any two of these, if there is $↔$, then there is only one relation. However, if there is conditional relation in one direction, then conditional relation in other direction will be invalid. That is there can be $(→,↛)$. So at max there will be $4C_2\times 2=12$ such relations.
Now to understand these relations, I prepared below table. I considered that there are only two person in the universe $x_1,x_2$. $\{(b,c),(¬b,¬c)\}$ means $x_1$ is boy and is clever and $x_2$ is not a boy and is not clever.
The truth values (in table) in boldface (all other than vi, viii and xiv) are one which I feel are correct, though I am not sure. The non bold-faced truth values (vi, viii and xiv) are ones which I am unsure. So, I am more likely make mistake with them.
Based on the truth values in above table, I prepared following relations:
The following is multiple choice question (with options) to answer.
There is a certain relation between two given words on one side of : : and one word is given on another side of : : while another word is to be found from the given alternatives, having the same relation with this word as the given pair has. | [
"Chauffeur",
"Steering",
"Mechanic",
"Brake"
] | A | Explanation:
Horse is driven by a jockey .
Similarly,car is driven by a chauffeur
Answer: A) |
AQUA-RAT | AQUA-RAT-38749 | ### 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.
Anil spends 40% of his income on rent, 30% of the remaining on medicines and 20% of the remaining on education. If he saves Rs. 840 every month, then find his monthly salary? | [
"2882",
"2672",
"2288",
"2500"
] | D | Let's Anil's salary be Rs. 100.
Money spent on Rent = 40% of 100 = Rs. 40.
Money spent on medical grounds = 30% of (100 - 40) = 3/10 * 60 = Rs. 18.
Money spent on education = 20% of (60 - 18) = 1/5 * 42 = Rs. 8.40
Anil saves 100 - (40 + 18 + 8.40) i.e., Rs. 33.60
for 33.6 ---> 100 ; 840 ---> ?
Required salary = 840/33.6 * 100 = Rs. 2500
Answer: D |
AQUA-RAT | AQUA-RAT-38750 | Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
Hm, i got stuck cuz I got something a little different:
YOURS: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
MINE: 3 men would do the same work 5 days sooner than 9 women --> $$\frac{3}{m}=\frac{9}{w}+5$$
In the above equation you also have for 2 men: $$\frac{2}{m}$$ - so why do u suddenly use the reciprocal? And why don't we add the 5 to women, because they take longer, hence their side is smaller...
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
15 men take 21 days of 8 hrs. each to do a piece of work. How many days of 3 hrs. each would it take for 21 women if 3 women do as much work as 2 men? | [
"60",
"20",
"19",
"29"
] | A | Let 1Man does 1unit/hr of work
15M in 21 days of 8hrs will do (15*21*8) units
3W = 2M
1W=(2/3)units/hr
21W with 3hrs a day will take (15*21*8)/(21*3*(2/3)) days
=> 60 Days
ANSWER:A |
AQUA-RAT | AQUA-RAT-38751 | # 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.
Shopkeeper rise price by 31% and gives successive discount of 10% and 15%. What is overall % gain or loss? | [
"0.215%",
"1.215%",
"3.215%",
"6.215%"
] | A | Let d initial price be 100
31 % rise
now price = 131/100*100 = 131
10% discount
Then price = 131 * 90/100 = 117.9
15 % discount
Then price = 117.9 * 85/100 = 100.215
So Gain = 100.215 - 100 = 0.215
Gain % = Gain * 100 /CP
==> 0.215 * 100 /100 = 0.215%
ANSWER:A |
AQUA-RAT | AQUA-RAT-38752 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
Jane covered a distance of 330 miles between city A and city B taking a total of 5 hours. If part of the distance was covered at 60 miles per hour speed and the balance at 80 miles per hour speed, how many hours did she travel at 60 miles per hour? | [
"2 hours 30 minutes",
"3 hours",
"2 hours",
"1 hour 45 minutes"
] | A | Answer
Let Jane travel 'x' at 60 miles per hour.
As the total time taken to cover 340 miles is 5 hours, Jane would have traveled (5 - x) hours at 80 miles per hour.
Distance covered at 60 miles per hour = Speed * time = 60 * x = 60x miles.
Distance covered at 80 miles per hour = Speed * time = 80 (5 - x) = 400 - 80x miles.
Total distance covered = Distance covered at 60 miles per hour + Distance covered at 80 miles per hour.
Therefore, total distance = 60x + 400 - 80x.
Total distance travelled = 340 miles.
Therefore, 330 = 60x + 400 - 80x
20x = 50 or x = 2 hours 30 minutes.
Choice A |
AQUA-RAT | AQUA-RAT-38753 | 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$
The following is multiple choice question (with options) to answer.
1^2 – 2^2 + 3^2 – 4^2 + 5^2 – 6^2 + 7^2 – 8^2 + 9^2 – 10^2=? | [
"-21",
"-53",
"-35",
"-55"
] | D | 1-4+9-16+25-36+49-64+81-100= -55
ANSWER:D |
AQUA-RAT | AQUA-RAT-38754 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Find 4 + 7 + 10 + 13 + 16 + . . . up to 20 terms | [
"650",
"640",
"630",
"620"
] | A | a = 4
d = 7 – 4 = 3
Sum of first 20 terms, S20
ANSWER IS A |
AQUA-RAT | AQUA-RAT-38755 | ### 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 invested Rs. 26000 in 5% stock at 104. He sold the stock when the price rose to Rs. 120 and invested the sale proceeds in 6% stock. By doing this his income increased by Rs. 2500. At what price did he purchase the second stock? | [
"125",
"40",
"48",
"35"
] | C | Assuming that face value of the first stock = Rs.100 as it is not given in the question
Since it is a 5% stock, we can take the dividend per stock = Rs.5
Market Value of the first stock = Rs.104
Investment on the first stock = Rs.26000
Number of stocks purchases = 26000/104 = 250
His total income from all these stocks = Rs.250 × 5 = Rs.1250
He sells each of this stock at Rs.120
ie, amount he earns = Rs.120 × 250 = Rs.30000
He invest this Rs.30000 in 6% stock (here also face value is not given and hence take it as Rs.100)
His new income = Rs.(1250 + 2500) = Rs.3750
ie, By Rs.30000 of investment , he earns an income of Rs.3750
To get an income of Rs.6, investment needed = 30000×6/3750=Rs.48
This is the market value of the second stock
Answer is C. |
AQUA-RAT | AQUA-RAT-38756 | binary, numeral-representations
= \sum_{i=0}^{N-1} 2^i - \sum_{i=0}^{N-1} b_i 2^i + 1 = 2^N - b$$
since the two's complement is defined as the complement of the representation $b_{N-1} \cdots b_0$, plus one, and $\sum_{i=0}^{N-1}2^i = 2^N - 1$. Since $a$ is negative, $a_{N-1} = 1$ and then:
$$a = -b
= -2^N + \sum_{i=0}^{N-1} a_i 2^i
= -2^N + 2^{N-1} + \sum_{i=0}^{N-2} a_i 2^i
= -2^{N-1} + \sum_{i=0}^{N-2} a_i 2^i
$$
The following is multiple choice question (with options) to answer.
55
a6
+ 2b
____
96
If a and b represent positive single digits in the correctly worked computation above, what is the value of a + 2b ? | [
"2",
"7",
"9",
"11"
] | D | Adding the digits in unit's place , 5+ 6 + b = 16 (Since a and b are positive single digits )
=> b = 5
Now adding the digits in ten's place , 5+a + 2 + 1= 9 (1 has been carried over from unit's place addition)
=> a= 1
a+2b = 1+2*5
=11
Answer D |
AQUA-RAT | AQUA-RAT-38757 | 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 can do a piece of work in 4 hours; B and C together can do it in 3 hours, which A and C together can do it in 2 hours. How long will B alone take to do it? | [
"11",
"77",
"12",
"99"
] | C | A's 1 hour work = 1/4;
(B + C)'s 1 hour work = 1/3;
(A + C)'s 1 hour work = 1/2
(A + B + C)'s 1 hour work = (1/4 + 1/3) = 7/12
B's 1 hour work = (7/12 + 1/2) = 1/12
B alone will take 12 hours to do the work.
Answer: C |
AQUA-RAT | AQUA-RAT-38758 | ### 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.
Company KW is being sold, and both Company A and Company B were considering the purchase. The price of Company KW is 50% more than Company A has in assets, and this same price is also 100% more than Company B has in assets. If Companies A and B were to merge and combine their assets, the price of Company KW would be approximately what percent of these combined assets? | [
"66%",
"75%",
"86%",
"116%"
] | C | Let the price of Company A's assets be 100
Price of assets of KW is 50% more than company A's assets which is 150
Price of assets of KW is 100% more than company B's assets which means price of Company B's assets is half the price of KW = 75
A+B = 175
KW = 150
KW/(A+B)*100 = 150/175*100 = 85.7% or 86%
ANSWER:C |
AQUA-RAT | AQUA-RAT-38759 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
A certain boxer has agreed to pay his opponent a fee of 4% of his total purse for every pound over the specified weight limit he weighs in. If the boxer pays his opponent a fee of $26,250 after weighing in seven pounds over the specified limit, what was the boxer's purse? | [
"$65,625",
"$105,000",
"$150,000",
"$93,750"
] | D | .04*7 = .28
26250/.28 = $93,750
Answer: D |
AQUA-RAT | AQUA-RAT-38760 | \begin{align} \dbinom{6}{2} & \longleftrightarrow \dbinom{6}{6-2} \\[8pt] AB & \longleftrightarrow CDEF \\ AC & \longleftrightarrow BDEF \\ AD & \longleftrightarrow BCEF \\ AE & \longleftrightarrow BCDF \\ AF & \longleftrightarrow BCDE \\ BC & \longleftrightarrow ADEF \\ BD & \longleftrightarrow ACEF \\ BE & \longleftrightarrow ACDF \\ BF & \longleftrightarrow ACDE \\ CD & \longleftrightarrow ABEF \\ CE & \longleftrightarrow ABDF \\ CF & \longleftrightarrow ABDE \\ DE & \longleftrightarrow ABCF \\ DF & \longleftrightarrow ABCE \\ EF & \longleftrightarrow ABCD \end{align} There are exactly as many ways to choose $2$ out of $6$ as to choose $6-2$ out of $6$ because each way of choosing $2$ out of $6$ has a corresponding way of choosing $6-2$ out of $6$ and vice-versa.
• NOTE TO FUTURE USERS: This combined with TheSparkThatThought's answer and Ned's comment at the original question will guide you to the way someone must think. Jun 27 '14 at 23:12
Consider a collection of $n$ objects. Choosing $k$ of them to place into a set is equivalent to choosing $n-k$ to leave out.
Edit: Consider a high school dodgeball game with a red team and a blue team. There are $n$ total students, and the blue team has $k$ students. Since every student plays, there are $n-k$ students on the red team.
Because the PE teacher is biased, he lets the blue team pick all of their players first. They have $\binom{n}{k}$ ways to do this. After that, the red team has no choices. They must pick all of the remaining $n-k$ students.
The following is multiple choice question (with options) to answer.
In a class of 38 students 26 play football and play 20 long tennis, if 17 play above, many play neither? | [
"6",
"8",
"9",
"12"
] | C | 26 + 20 - 17 = 29
38 - 29 = 9 play neither
Answer is C |
AQUA-RAT | AQUA-RAT-38761 | ### Show Tags
26 May 2017, 05:36
1
Which of the following equals the ratio of 3 $$\frac{1}{3}$$to 1 $$\frac{1}{3}$$?
3$$\frac{1}{3}$$ = $$\frac{10}{3}$$
1 $$\frac{1}{3}$$ = $$\frac{4}{3}$$
Required ratio = (10/3) / (4/3) = $$\frac{10}{4}$$ = $$\frac{5}{2}$$
Director
Joined: 13 Mar 2017
Posts: 703
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
26 May 2017, 05:40
banksy wrote:
Which of the following equals the ratio of 3 1/3 to 1 1/3?
(A)1 : 3
(B)2 : 5
(C)5 : 2
(D)3 : 1
(E)40 : 9
Its a very simple question....
[m]3\frac{1}{3} = 10/3
1\frac{1}{3} = 4/3
Ratio = (10/3)/(4/3) = 5/2
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)
What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Intern
Joined: 03 Dec 2017
Posts: 18
Re: Which of the following equals the ratio of 3 1/3 to 1 1/3? [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
The ratio of money with Ram and Gopal is 7 : 17 and that with Gopal and Krishan is 7: 17. If Ram has Rs. 686, Krishan has ? | [
"s. 2890",
"s. 4046",
"s. 1190",
"s. 1620"
] | B | Ram : Gopal = 7 : 17 = 49 : 119
Gopal : Krishan = 7 : 17 = 119 : 289
Ram : Gopal : Krishan = 49 : 119 : 289
Ram : Krishan = 49 : 289
Thus, 49 : 289 = 686 : N
&there N = 289 x 686 / 49
= Rs. 4046
Answer :B |
AQUA-RAT | AQUA-RAT-38762 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is? | [
"2%",
"4%",
"6%",
"9%"
] | B | 65*(90/100)*((100-x)/100) = 56.16
x = 4%
Answer:B |
AQUA-RAT | AQUA-RAT-38763 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
A salesperson received a commission of 3 percent of the sale price for each of the first 100 machines that she sold and 4 percent of the sale price for each machine that she sold after the first 100. If the sale price of each machine was $10,000 and the salesperson received a $32,000 commission, how many machines did she sell? | [
" 90",
" 110",
" 105",
" 115"
] | B | First 100 machines = 3% commission = 0.03*100*10000 = 30000
Commission from sale of next machines = 34000 - 30000 = 4000
So 10 more machines..
Total = 110 machines
IMO B.. |
AQUA-RAT | AQUA-RAT-38764 | physical-chemistry, density
Title: Find the number of copper spheres in a box
A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?
The following is multiple choice question (with options) to answer.
An alloy weighing 60 ounces is 20 percent copper. How many ounces of copper must be added to create an alloy that is 60 percent copper? | [
"40",
"50",
"60",
"70"
] | C | An alloy of 60oz which is 20% copper means there is 12 oz of gold. To get to an alloy that is 60% copper, let's use this expression: (12 + x)/(60 + x) = 0.60 with x representing the amount of pure copper that must be added to get to 60%. The expression we are using represents the new total weight of pure copper over the new total weight of the alloy and this fraction should represent 60% or 0.6. you will see that 60 is the correct answer, as 72/120=0.6
Choose C |
AQUA-RAT | AQUA-RAT-38765 | # Probability: If I have a friend that likes half of the food he tries, what is the probability that he likes three of five foods that he's given?
I was thinking 1*1*1*2*2 = 4 out of 32, with LLLDD, LLLLL, LLLDL, LLLLD, with L as like and D as dislike. But if I can do LLLLD and LLLDL, why couldn't I do LDLLL or DLLLD? Any explanation would be appreciated.
EDIT: At least three (Sorry, forgot to mention)
-
Do you want the probability that he likes exactly three of the five, or at least three? – Brian M. Scott Jan 29 '13 at 0:04
He sounds too picky, I doubt he will like any of them. – Anon Jan 29 '13 at 0:04
Yes, we have to take into account $DLLL$, $DLDLL$, $DLLDL$, and so on. (There are $10$ of these like $3$, dislike the others.) And they are used in calculating the probability. – André Nicolas Jan 29 '13 at 0:09
Your confusion comes from the following: You are calculating the event that he will like the first, second, and the third food, and then you say, "I don't care about the last two foods," and you put $2$ and $2$. Here (in your question), the order is not important. – Anon Jan 29 '13 at 0:19
Because of this reason, your current solution does not take into account the case e.g. LDLLL, as you have mentioned. – Anon Jan 29 '13 at 0:21
This to me looks like a Bernoulli trial with $p=1/2$.
Probability that your friend like $k=3$ of $n=5$ foods he tries is
The following is multiple choice question (with options) to answer.
Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 45% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above? | [
"5%",
"10%",
"15%",
"35%"
] | D | The formula is
n(AUBUC) = n(A)+n(B)+n(C)-B+A-N
where B is both (it will sum of (AB), (BC), (CA)).
and A is All
and N is neither
so plugging in we get
100 = 60+50+45+10-B-0
B = 65.
exactly two = 65 - 3(10) = 35
subtracting 10% three times as this value is including in all the (AB), (BC), (CA).
Answer D |
AQUA-RAT | AQUA-RAT-38766 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
A number when divided by 296 leaves 75 as remainder. When the same number is divided by 37, the remainder will be: | [
"1",
"2",
"8",
"11"
] | A | Let x = 296q + 75
= (37 x 8q + 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.
Answer :A |
AQUA-RAT | AQUA-RAT-38767 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
Two cars start from the opposite places of a main road, 105 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point? | [
"65",
"38",
"20",
"28"
] | C | Answer: C) 20 km |
AQUA-RAT | AQUA-RAT-38768 | 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.
If N = 2^0.25 and N^b = 8, b must equal | [
"3/80",
"3/5",
"12",
"5/3"
] | C | 25/100 = 1/4
N = 2 ^ 1/4
N^b = 2^3
(2^1/4)^b = 2^3
b =12
Answer: C |
AQUA-RAT | AQUA-RAT-38769 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
If $400 is deposited into a savings account at an annual interest rate of 3% that compounds monthly, what is the value of the investment after 8 years? | [
"574.65",
"514.23",
"508.35",
"536.87"
] | C | Using CI Formula:
A=P(1+r/n)^n*t
A=400(1+3/2)^2*8
A=508.35.
Option: C |
AQUA-RAT | AQUA-RAT-38770 | mechanical-engineering, torque
Title: How to calculate if a motor is capable for a load? I am trying to figure out if a motor I have is capable of spinning a wheel. Here is all of the information I have:
MOTOR: 190 g.cm min.
WHEEL: 26 cm diameter, weighs about 59 grams
I imagine that I would have to calculate the linear speed of the wheel in order to convert it to the same units that the motor torque is listed in. However, I don't know if I am doing it correctly or what the next step would be to solve this. Check out my answer to a similar question at the Robotics SE. You are asking only about a motor and a wheel, and you do not give the weight of the vehicle (assuming there is one), so I cannot comment regarding vehicular motion.
Basically, if you assume friction is negligible, the torque a motor outputs is used to accelerate a load. Then you have to ask, "What is my desired acceleration?" A [Honda Civic] can accelerate 0-60mph in about 9 seconds, while a [1968 VW bus] went the same 0-60mph in 37 seconds. Both vehicles eventually get to the same top speed, but higher torque values give you a higher acceleration rate.
You can calculate your own desired acceleration rate by giving a top speed you would like to reach and a time limit in which you would like to achieve that speed. So, for example, if you wanted to get to 10rpm in 4 seconds, you would need:
$$
a = (\mbox{Top speed} - \mbox{Starting speed})/(\mbox{Time limit}) \\
a = (10\mbox{rpm} - 0\mbox{rpm})/(4\mbox{s}) \\
a = 2.5 \frac{\mbox{rpm}}{\mbox{s}} \\
$$
To be used in an equation, you typically have to convert to radians per second, which is easy knowing that $2\pi \mbox{radians} = 1\mbox{revolution}$ and that $1\mbox{min} = 60\mbox{s}$:
$$
The following is multiple choice question (with options) to answer.
The wheel of a motorcycle, 70 cm in diameter makes 40 revolutions in every 10 seconds. What is the speed of the motorcycle in km/hr | [
"30.68 km/hr",
"31.68 km/hr",
"32.68 km/hr",
"33.68 km/hr"
] | B | Explanation:
In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference
So we will be having distance and time, from which we can calculate the speed. So let solve.
Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be
40 * Circumference =40 * 2*\pi*r =40∗2∗22/7∗35=8800cm
=8800/100m=88m
Distance covered in 1 sec =88/10=8.8m
Speed=8.8m/s=8.8∗18/5=31.68km/hr
Option B |
AQUA-RAT | AQUA-RAT-38771 | 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.
_________________
Re: A certain stock exchange designates each stock with a one-, [#permalink] 11 Aug 2015, 23:27
Similar topics Replies Last post
Similar
Topics:
Closing prices for a certain stock were recorded each day for a week 2 16 Aug 2016, 05:10
At the opening of a trading day at a certain stock exchange, the price 4 25 May 2016, 00:56
39 A certain stock echange designates each stock with a 27 10 Nov 2009, 14:50
23 A certain stock exchange designates each stock with a 1, 2 11 26 Oct 2009, 12:25
12 A certain stock exchange designates each stock with a one-, 3 21 Sep 2009, 17:02
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Yesterday's closing prices of 2,420 different stocks listed on a certain stock exchange were all different from today's closing prices. The number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price. How many of the stocks closed at a higher price today than yesterday? | [
"484",
"726",
"1,100",
"1,320"
] | D | Lets consider the below -
The number of stocks that closed at a higher price = H
The number of stocks that closed at a lower price = L
We understand from first statement -> H+L = 2420 ----(1)
We understand from second statement -> H = (120/100)L => H = 1.2L ----(2)
Solve eq (1) & (2) to get H = 1320.
D is answer. |
AQUA-RAT | AQUA-RAT-38772 | Substitute $$1/(b,d)=[b,d]/bd$$ and $$1/(a,c)=[a,c]/ac$$: $$=\left(a,d(a,c)\frac{[b,d]}{bd}\right)\left(c(b,d)\frac{[a,c]}{ac},b\right)$$ Cancel $$d$$'s from first term and $$c$$'s from second term and rewrite: $$=\left(a,\frac{(a,c)[b,d]}{b}\right)\left(\frac{(b,d)[a,c]}{a},b\right)$$ Treat the whole first term like $$m$$ in $$m(a,b)=(ma,mb)$$ and multiply it into the second term: $$=\left(\left(a,\frac{(a,c)[b,d]}{b}\right)\frac{(b,d)[a,c]}{a},\left(a,\frac{(a,c)[b,d]}{b}\right)b\right)$$ Multiply: $$=\left(\left(\frac{a(b,d)[a,c]}{a},\frac{(a,c)[b,d](b,d)[a,c]}{ba}\right),\left(ab,\frac{b(a,c)[b,d]}{b}\right)\right)$$ Simplify fractions, and note that $$(a,c)[a,c]=ac$$ and $$[b,d](b,d)=bd$$:
The following is multiple choice question (with options) to answer.
10 is 6% of a, and 6 is 10% of b. c equals b/a. What is the value of c? | [
"3/10",
"6/17",
"9/25",
"10/33"
] | C | 6a/100 = 10
a = 500/3
b/10 = 6
b = 60
c = b/a = 60*3/ 500 = 9/25
The answer is C. |
AQUA-RAT | AQUA-RAT-38773 | ## Wednesday, August 12, 2015
### The angle between the hour and minute hands of a clock.
Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?
A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.
Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.
But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.
Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.
## Monday, August 20, 2012
### Statistics Problem Set Aug-21-2012
1. Which of the following formulas measure symmetry of a sample data distribution?
(a)$(1/n) \sum (x-\overline{x})^2$ (b) $(1/n) \sum (x-\overline{x})^3$ (c)$(1/n) \sum (x-\overline{x})^4$ (d.) Not listed
2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since
The following is multiple choice question (with options) to answer.
At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but,not together? | [
"(5 + 5/11 )min past 5",
"(5 + 5/11 )min past 7",
"(5 + 5/11 )min past -0",
"(5 + 7/11 )min past 7"
] | B | Explanation:
When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min spaces are gained in min = min
so, Required time = min past 7
Answer: B) |
AQUA-RAT | AQUA-RAT-38774 | equinox
Title: Understanding date of astronomical events I have a masters in chemistry but pretty much, a layman in astronomy.
So, can you please explain to a novice like me, about this paragraph taken from the Wikipedia article on Makar Sankranti:
There are two different systems to calculate the Makara Sankranti date: nirayana (without adjusting for precession of equinoxes, tropical) and sayana (with adjustment, sidereal). The January 14 date is based on the nirayana system, while the sayana system typically computes to about December 23, per most Siddhanta texts for Hindu calendars.
Am I correct to understand that when the date was assigned in Hinduism some thousand (?) years back, the winter solistice used to be on 14th/15th January? Precession of equinoxes means that the North pole of the Earth is changing its direction. This also means, that the dates of the seasons (seen from outside the Solar System (= sidereal)) are changing. Nirayana calculates the date of Makar Sankranti with adding years to the first Makar Sankranti, but sayana calculates the date of Makar Sankranti relative to stars (sidereal).
In a nutshell: nirayana is on the same date as the first event (first Makar Sankranti), but sayana is on the same event as the first Makar Sankranti (winter solstice). Yes, winter solstice was once on the 14 January.
The following is multiple choice question (with options) to answer.
Valentine day 14 Feb. 2005 was celebrated by Anand and shilpa on Tuesday. They were very happy. What day of the week would come on 14 Feb. 2010? | [
"monday",
"tuesday",
"sunday",
"wednesday"
] | A | feb 14 2005 - tuesday
feb 14 2006 - wednesday
feb 14 2007 - thursday
feb 14 2008 - friday
feb 14 2009 - sunday[since 2008 is leap year , 2 days ll go forward]
feb 14 2010 - monday
ANSWER:A |
AQUA-RAT | AQUA-RAT-38775 | ### Remark 2
68% of all observations fall within 1 standard deviation from the mean.
### Example 5
Find the probability $$P(|Z| > 1)$$.
Key: We wish to find the area of the unshaded region in Figure 5. We may rewrite this probability as
$P( |Z| > 1) = 1 - P(|Z| < 1 )= 1 - .682 = .318$
### Exercise 1
Compute the following probabilities:
(a) $$P(|Z|<2)$$
(b) $$P(|Z|<3)$$
\begin{align} P(|Z|<2) &= .954 \\
P(|Z|<3) &= .997 \end{align}
### 68-95-99.7 Rule
68-95-99.7 Rule: 68% of all observations fall within 1 standard deviation, 95% of all observations fall within 2 standard deviations, and 99.7% of all observations fall within 3 standard deviations.
### Example 6
(a) Find the height of an individual in the $$75^{th}$$ percentile.
(b) Find the height of a student whose height is 16% from the tallest.
Inverse Normal
1. Press 2nd VARS to enter DISTR
2. Select 3:invNorm(
3. Enter percentile as decimal
4. Enter closing )
5. Press ENTER
**Remark**: Enter percentile as a decimal in **Area**, and select **Paste** and **ENTER**.
### Example 6 (a) Solution
We find that the $$75^{th}$$ percentile corresponds to the Z-score $$Z=.674$$. To find the height, we work backwards using $$(\ref{Z-score})$$:
\begin{align} .674 &= \frac{x - \mu}{\sigma}= \frac{x - 70.9}{2.75} \end{align}
We solve for $$x$$:
The following is multiple choice question (with options) to answer.
A certain characteristic in a large population has a distribution that is symmetric about the mean m. 84 percent of the distribution lies within one standard deviation d of the mean. If the shelf’s average life is 9.2 years and the standard deviation is 3.5 years, what percent of the distribution has more than 12.7 years as a shelf’s average life? | [
"18%",
"19%",
"20%",
"21%"
] | D | Average = 9.2
SD = 3.5
9.2 - 3.5 < 84% of distribution < 9.2 + 3.5
5.7 < 84% of distribution < 12.7
42% is outside this range.
Given: Distribution is symmetric. So 21% of distribution is less than 5.7 and the other 21% of distribution is greater than 5.7.
Answer: D |
AQUA-RAT | AQUA-RAT-38776 | 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 20 men can build a wall 66 meters long in 6 days, what length of a similar can be built by 86 men in 8 days? | [
"49",
"39",
"29",
"19"
] | A | Let the required length be x meters
More men, More length built (Direct Proportion)
Less days, Less length built (Direct Proportion)
Men 20: 35
Days 6: 3 : : 56 : x
Therefore (20 x 6 x x)=(35 x 3 x 56)= x=(35 x 3 x 56)/120=49
Hence, the required length is 49 m.
Answer is A. |
AQUA-RAT | AQUA-RAT-38777 | The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.
We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A"
So we end up with \frac{8x7}{2}= 28
If you still have trouble with combination and permutation check out this website it's well done,
http://www.mathsisfun.com/combinatorics ... tions.html
hope it helps.
Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.
Kudos [?]: 4 [1], given: 49
Intern
Joined: 15 Sep 2014
Posts: 8
Kudos [?]: 3 [0], given: 0
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Nov 2014, 02:57
If there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2
So 8*7/2 =28.
Posted from my mobile device
Kudos [?]: 3 [0], given: 0
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 437
Kudos [?]: 141 [0], given: 169
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
12 Jan 2015, 12:48
1
This post was
BOOKMARKED
I also used a table to do this, like that:
The following is multiple choice question (with options) to answer.
In a tournament, there are n teams T1 , T2 ....., Tn with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common: T1 & T2 , T2 & T3 ,......, Tn − 1 & Tn , and Tn & T1. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together? | [
"n(k-1)",
"k(n-1)",
"n(k-2)",
"k(n-2)"
] | A | Explanation :
Let T1 - T6 be the number of teams, and let 1,2,3,4,5,6….be the players in the teams.
Total players if there are no repetition is 6*4 here. Generally, it is n * k. (n = number of teams, k = player/team). However, we see that, in each team, 1 player is repeated. i.e.
T1 & T2 have 4 in common (So 1 can be subtracted)
T2 & T3 have 7 in common (So 1 can be subtracted)
T3 & T4 have 10 in common (So 1 can be subtracted)
T4 & T5 have 13 in common (So 1 can be subtracted)
T5 & T6 have 16 in common (So 1 can be subtracted)
T6 & T1 have 1 in common (So 1 can be subtracted)
So, totally from each column 1 can be subtracted.
There are “n” columns totally, so n can be subtracted ( Here 6 needs to be subtracted)
So, here we have 6 * 4 – 6 players. Generally we have n * k – n i.e n(k-1) players.
Answer : A |
AQUA-RAT | AQUA-RAT-38778 | ### 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.
From the sale of sleeping bags, a retailer made a gross profit of 13% of the wholesale cost. If each sleeping bag was sold for $28, what was the wholesale cost per bag? | [
"3.0",
"3.36",
"24.78",
"25.0"
] | C | Cost price * 1.13 = Selling price --> Cost price * 1.13 = $28 --> Cost price = $24.78.
Answer: C. |
AQUA-RAT | AQUA-RAT-38779 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
A furniture store owner decided to drop the price of her recliners by 20% to spur business. By the end of the week she had sold 70% more recliners. What is the percentage increase of the gross? | [
"10%",
"15%",
"20%",
"36%"
] | D | Say a recliner is actually worth $100
If she sells 100 recliners then she earns $10000
After the discount of 20%, she will earn $80 per recliner and she sells 70% more ie.,170 recliners
Hence her sales tields 170*80=$13600
Increase in sales=13600-10000=$3600
So% increase = 3600*100/10000 =36%
D is the answer |
AQUA-RAT | AQUA-RAT-38780 | Just need to verify if this one needs to be subtracted or no.
jaytheseer
New member
Mr. Gates owns 3/8 of Macrohard. After selling 1/3 of his share, how much more of Macrohard does Mr. Gates still own?
MarkFL
Staff member
Yes, I would view the subtraction in the form:
If Mr. Gates sold 1/3 of his share, how much of his share does he have left?
What portion of Macrohard is Mr. Gates' remaining share?
jaytheseer
New member
My solution so far:
3/8 = 9/24 and 1/3 = 8/24
9/24 - 8/24 = 1/24
But my book says a totally different thing which confuses me:
3/8 x 1/3 = 1/8.3/8 - 1/8 = 2/8 =1/4
Deveno
Well-known member
MHB Math Scholar
Mr. Gates owns 3/8 of Macrohard. This means that for every 8 shares of Macrohard out there, he owns 3 of them.
1/3 of 3, is of course, 1.
So if he sells 1/3 of his shares, he now only owns 2 shares out of every 8, which is 2/8 = 1/4.
When we take a fraction OF something, it means: "multiply".
So 1/3 OF 3/8 means:
MULTIPLY (1/3)*(3/8), from which we get: (1/3)*(3/8) = 1/8 <---how much he sold.
If we want to know how much he has LEFT, then we SUBTRACT, so:
3/8 - 1/8 = ...?
MarkFL
Staff member
The way I look at it he has 2/3 of his shares left after selling 1/3. So the portion of Macrohard he still owns is:
$$\displaystyle \frac{2}{3}\cdot\frac{3}{8}=\frac{1}{4}$$
Prove It
The following is multiple choice question (with options) to answer.
A sum of money is divided among A, B, C & D in the ratio 3:5:8:9 respectively. If the share of D Rs.1872 more than the share of A, then what is the total amount of money of B & C together? | [
"3456",
"4056",
"4156",
"4567"
] | B | Sol. Share of B + C = -3 x (5 + 8) = Rs. 4056
B |
AQUA-RAT | AQUA-RAT-38781 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 4 hours and 16 hours respectively. The ratio of their speeds is | [
"4:1",
"4:2",
"4:5",
"4:3"
] | B | Ratio of their speeds = Speed of first train : Speed of second train
= √16−−√4
= 4:2
Answer is B. |
AQUA-RAT | AQUA-RAT-38782 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 40 minutes earlier than usual. How much time did the man spend walking? | [
"45 minutes",
"50 minutes",
"40 minutes",
"55 minutes"
] | C | As they arrived 40 minutes earlier than usual, they saved 40 minutes on round trip from home to station (home-station-home) --> 20 minutes in each direction (home-station) --> wife meets husband 20 minutes earlier the usual meeting time --> husband arrived an hour earlier the usual meeting time, so he must have spent waking the rest of the time before their meeting, which is hour-20 minutes=40minutes.
Answer: C |
AQUA-RAT | AQUA-RAT-38783 | # Difference between revisions of "2019 AMC 10A Problems/Problem 23"
## Problem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
## Solution 1
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$
The following is multiple choice question (with options) to answer.
Difficulty:
There are five sales agents in a certain real estate office. One month Andy sold five as many properties as Ellen, Bob sold 3 more than Ellen, Cary sold five times as many as Bob, and Dora sold as many as Bob and Ellen together. Who sold the most properties that month? | [
"Andy",
"Bob",
"Cary",
"Dora"
] | C | a= 5e
b =3 +e
c = 5b = 15+ 5e (highest of all)
d = b + e = 3+ e +e = 3 + 2e
Ans. C |
AQUA-RAT | AQUA-RAT-38784 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row his boat with the stream at 18 km/h and against the stream in 8 km/h. The man's rate is? | [
"5 kmph",
"2 kmph",
"7 kmph",
"8 kmph"
] | A | DS = 18
US = 8
S = ?
S = (18 - 8)/2 = 5 kmph
Answer:A |
AQUA-RAT | AQUA-RAT-38785 | # 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.
In how many ways can be 4 boys and 4 girls sit around circular table so that no two boys sit next to each other? | [
"(5!)^2",
"(6!)^2",
"3!4!",
"11!"
] | C | first fix one boy and place other 3 in alt seats so total ways is 3!
now place each girl between a pair of boys... total ways of seating arrangement of girls 4!
total is 4!*3!
ANS C |
AQUA-RAT | AQUA-RAT-38786 | 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.
In a bag, there are three 6-sided dice (numbered 1 to 6), three 12-sided dice (numbered 1 to 12), and two 20-sided dice (numbered 1 to 20). If four of these dice are selected at random from the bag, and then the four are rolled and we find the sum of numbers showing on the four dice, how many different possible totals are there for this sum? | [
"60",
"61",
"64",
"84"
] | B | As 'complex-looking' as this question might appear, it's actually rather simple. Pay careful attention to what the specific questions asks for - the number of DIFFERENT possible SUMS from 4 dice. Since we're dealing with some 'special' dice (some 12-sided and 20-sided dice), we have to adjust out math accordingly, but the possibilities are rather limited:
1) The minimum number on any given die is 1
2) The maximum possible sum would only occur if we took the 4 biggest possible dice and rolled the highest possible number on each.
With 4 dice, we could end up with any SUM between:
4 (if we rolled 1s on all 4 dice)
to
64 (if we rolled two 20s on the 20-sided dice and two 12s on the 12-sided dice).
Thus, there are only 61 possible sums.
B |
AQUA-RAT | AQUA-RAT-38787 | # Math Help - Calculus Help Please
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
2. Originally Posted by Luke007
Initially a population contains 1000 individuals. After t minutes the number N of individuals in the populataion is such that dN/dt = -0.25N.
a) write down the expression for N in terms of t
b) Find the rate at which the population is decreasing when it is half its orginal size.
This question is driving me nuts Any help would be much appreciated thanks
Reguards, Luke
a)
$\frac {dN}{dt} = -0.25N$
$\Rightarrow \frac {dN}{N} = -0.25 dt$
$\Rightarrow \ln N = -0.25t + C$
$\Rightarrow N = e^{-0.25t + C}$
$\Rightarrow N = e^C e^{-0.25t}$
$\Rightarrow N = Ae^{-0.25t}$ ......we could have jumped straight to here, but I wanted to show you how we got here. This formula should be in your book
when $t = 0$, $N = 1000$
$\Rightarrow N(0) = Ae^0 = 1000$
$\Rightarrow A = 1000$
$\Rightarrow N(t) = 1000e^{-0.25t}$
b)
$\frac {dN}{dt} = -0.25N$
when $N$ is half it's size:
$\frac {dN}{dt} = -0.25 (0.5N)$
$\frac {dN}{dt} = -0.125N$
So the population is decreasing at a rate of -0.125
The following is multiple choice question (with options) to answer.
If the population of a certain country increases at the rate of one person every 15 seconds, by how many persons does the population increase in 40 minutes? | [
"80",
"100",
"160",
"240"
] | C | Since the population increases at the rate of 1 person every 15 seconds, it increases by 4 people every 60 seconds, that is, by 4 people every minute. Thus, in 40 minutes the population increases by 40 x 4 = 160 people.
Answer. C. |
AQUA-RAT | AQUA-RAT-38788 | 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.
In what time will Rs.4000 lent at 3% per annum on simple interest earn as much interest as Rs.5000 will earn in 5 years at 4% per annum on simple interest? | [
"5 years",
"6 years",
"7 years",
"9 years"
] | A | (4000*T*5)/100 = (5000*5*4)/100
T = 5 years
ANSWER:A |
AQUA-RAT | AQUA-RAT-38789 | +0
# SOS!!!
0
230
7
+223
I cannot get the right answer for this problem for the life of me. If somebody would solve it I'd give them 5 stars.
THE PROBLEM:
(20+(1/4))x +(5+(1/2)) = (7+(1/16))
(solve for x)
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#1
+10
(20+(1/4))x + (5+(1/2)) = (7+(1/16))
(20(1/4))x = (7+(1/16)) - (5+(1/2))
x = ( (7+(1/16)) - (5+(1/2)) ) / (20+(1/4))
x = ((113/16) - (11/2)) / (81/4)
x = 25/324
Guest Mar 13, 2017
#6
+223
+5
This one seems to work! Thanks!
#5
+7153
+6
The following is multiple choice question (with options) to answer.
Replace X with the appropriate number in
1 , 2 , 3 , 4 , 7 , 10 , 11 , X | [
"12",
"13",
"15",
"16"
] | A | A
list of consecutive numbers that does not contain a 'i' in the spelling |
AQUA-RAT | AQUA-RAT-38790 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election only two candidates contested. A candidate secured 70% of the valid votes and won by a majority of 182 votes. Find the total number of valid votes? | [
"455",
"570",
"480",
"520"
] | A | Let the total number of valid votes be x.
70% of x = 70/100 * x = 7x/10
Number of votes secured by the other candidate = x - 7x/100 = 3x/10
Given, 7x/10 - 3x/10 = 182 => 4x/10 = 182
=> 4x = 1820 => x = 455.
ANSWER:A |
AQUA-RAT | AQUA-RAT-38791 | ### 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.
Rahim bought 50 books for Rs.1000 from one shop and 40 books for Rs.800 from another. What is the average price he paid per book ? | [
"28",
"20",
"27",
"29"
] | B | Average price per book
= (1000+ 800) / (50 + 40)
= 1800 / 90
= Rs.20
Answer:B |
AQUA-RAT | AQUA-RAT-38792 | 29. kc_kennylau
bye
30. anonymous
@aum I thought so. I see how to get the 4 but not the -1 could you help me solve this?
31. aum
My method is what @myininaya suggested. (2x - 3) / (x + 1) < 1 (2x - 3) / (x + 1) - 1 < 0 (2x-3 - x - 1) / (x + 1) < 0 (x - 4) / (x + 1) < 0 The x values that are of interest to us here are the ones that will make f(x) = (x-4)/(x+1) zero or undefined. f(x) = 0 when x = 4 f(x) is undefined when x = -1. So the number line is split into three intervals: (-infinity, -1); (-1, 4), (4, infinity). Pick a convenient number in each interval and see if the inequality is valid. f(x) = (x - 4) / (x + 1) < 0 when x = -2, f(x) = (-6) / (-1) = 6 which is greater than 0. So not a solution. when x = 0, f(x) = -4/1 = -4 which is less than 0. This is a solution. when x = 5, f(x) = 1 / 6 which is greater than 0. Not a solution. So the solution is x in the interval (-1, 4).
32. anonymous
thank you so much
33. aum
you are welcome.
34. aum
The following is multiple choice question (with options) to answer.
If [x] is the greatest integer less than or equal to x, what is the value of [-11.6]+[13.4]+[12.7]? | [
" 3",
" 10",
" 13",
" 16"
] | C | You are asked what the closest lesser integer value to [x] is.
[-11.6] = -12.0
[13.4] = 13.0
[12.7] = 12.0
Therefore , answer is: -12.0 + 13.0 + 12.0 = 13.0
Option A. |
AQUA-RAT | AQUA-RAT-38793 | 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}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A building contractor employs 20 male, 15 female and 5 child workers. To a male worker he pays Rs.25 per day, to a female worker Rs.20 per day and a child worker Rs.8 per day. The average wage per day paid by the contractor is? | [
"28",
"21",
"29",
"27"
] | B | 20 15 5
25 20 8
500 + 300 + 40 = 840/40 = 21
Answer:B |
AQUA-RAT | AQUA-RAT-38794 | Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$ ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in $$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$ ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are $$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$ ways to select five questions so that at least one is drawn from each of the three sections.
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $$A_1, A_2, A_3, B_1, C_1$$ are selected. You count this selection three times.
The following is multiple choice question (with options) to answer.
A question paper has 2 parts, A & B, each containing 5 questions. If a student has to choose 3 from part A & 4 from part B, in how many ways can he choose the questions? | [
"50",
"100",
"152",
"150"
] | A | There 3 questions in part A out of which 4 question can be chosen as = 5C3.
Similarly, 5 questions can be chosen from 10 questions of Part B as = 5C4.
Hence, total number of ways,
=5C3*5C4
=[5!/(2!3!)]*[5!/(4!*1)]
= {10}*{5*4!/(4!)}
= 10*5=50.
A |
AQUA-RAT | AQUA-RAT-38795 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4342
Location: India
GPA: 3.5
A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Manager
Joined: 18 Aug 2013
Posts: 128
Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 640 Q48 V28
GPA: 3.92
WE: Operations (Transportation)
Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
The average age of 20 men in the class is 15.6 years. Five new men join and the new average becomes 14.56 years. What was the average age of five new men? | [
"15.5",
"15.4",
"15.25",
"10.4"
] | D | Total age of 20 men = 15.6 x 20 = 312
Now, total age of 25 men = 364.
Total age of five men added later = 364 - 312 = 52.
Hence, the total average of five men = 52/5 = 10.4
ANSWER:D |
AQUA-RAT | AQUA-RAT-38796 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
In What proportion must flour at $0.8 per pound be mixed with flour at $0.9 per pound so that the mixture costs $0.865 per pound? | [
"1:3",
"1:2",
"1:1",
"2:1"
] | B | Using weighted average method:
Let x be the proportion in which the $0.8 per pound flour is mixed with $0.9 per pound flour.
Thus 0.8*x + 0.9*(1-x) = 0.865
0.9 – 0.1x = 0.865
x = 0.35
Thus ratio of both flours is 1:2
B |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.