source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-36897 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, there by increasing his average by 7 runs. His new average is: | [
"38 runs",
"37 runs",
"39 runs",
"40 runs"
] | A | Let average for 10 innings be x. Then,
(10x + 108)/11 = x + 7
= 11x + 77 = 10x + 108
= x = 31.
New average = (x + 7) = 38 runs.
Answer:A |
AQUA-RAT | AQUA-RAT-36898 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
Two men starting from the same place walk at the rate of 5 kmph and 5.5 kmph respectively. What time will they take to be 8.0 km apart, if they walk in the same direction? | [
"12 hrs",
"14 hrs",
"15 hrs",
"16 hrs"
] | D | Sol.
to be 0.5 km apart, they take 1 hour.
To be 8.0 km apart, they take [1/0.5 * 8.0] hrs = 16 hrs.
Answer D |
AQUA-RAT | AQUA-RAT-36899 | 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 speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is? | [
"388",
"150",
"288",
"269"
] | B | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
Answer:B |
AQUA-RAT | AQUA-RAT-36900 | # Math Help - Finding the values of a and b
1. ## Finding the values of a and b
Hello everyone. This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
The keyword in this irksome question would be the word 'or'. So it denotes that that this involves quadratic formulas.
Can anyone give me a clue so that I may make a breakthrough in understanding this problem? Thank you so much!
2. The wording is a bit iffy, but they actually meant that the solutions for that equation is x = 3 AND x = -4.
3. Originally Posted by PythagorasNeophyte
This question is apparently unsolvable:
If $x = 3$ or $-4$ are the solutions of the equation $x^2+ax+b=0$, find the values of $a$ and $b$.
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
$x=\dfrac{-a + \sqrt{a^2 - 4 \times 1 \times b}}{2}$ and $x=\dfrac{-a - \sqrt{a^2 - 4 \times 1 \times b}}{2}$
We know that the minus squareroot usually gives us the smaller answer of x. So then we just substitute in our x answers to get:
$3=\dfrac{-a + \sqrt{a^2 - 4 \times b}}{2}$ and $-4=\dfrac{-a - \sqrt{a^2 - 4 \times b}}{2}$
Now solve for a and b using simultaneous equation.
4. Originally Posted by Educated
No this question IS solvable. And quite easily I might add.
We know the quadratic formula as having a $\pm$ which yields 2 answers.
Substituting in values from $x^2+ax+b=0$ into the quadratic formula we get:
The following is multiple choice question (with options) to answer.
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 9x +3 = 0? | [
"28.3",
"1.45",
"2.78",
"6.84"
] | A | a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -9/1 = -9
ab = 3/1 = 3
Hence a/b + b/a = [(-9)2 - 2(3)]/3 = 56/4 = 28.3.
A) |
AQUA-RAT | AQUA-RAT-36901 | 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.
If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream is | [
"5 km/hr",
"4 km/hr",
"2 km/hr",
"1 km/hr"
] | D | Explanation:
Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph
Option D |
AQUA-RAT | AQUA-RAT-36902 | # Math Help - Perfect square s, odd number of factors
1. ## Perfect square s, odd number of factors
If a number is a perfect square, it will have an odd number of factors (e.g., 4 has factors 1, 2, 4), whereas all other numbers have an even number of factors.
Is the converse true? Please explain why?
2. ## Re: Perfect square s, odd number of factors
By converse, do you mean the following statement: "If a number has an odd number of factors, then it is a perfect square, and if a number has an even number of factors, then is it not a perfect square"? The original statement says that being a perfect square is equivalent to having an odd number of factors, so of course the converse is true.
3. ## Re: Perfect square s, odd number of factors
Yes, that's what I meant. But I am not convinced of the equivalence. Can you explain a bit further with the proof.
4. ## Re: Perfect square s, odd number of factors
Suppose that the number of factors of n is odd. We need to show that n is a perfect square. Suppose the contrary; then by the statement in post #1, n has an even number of factors, a contradiction.
In general, if you showed A implies B and (not A) implies (not B), then you showed that A and B are equivalent, so either one implies the other. Also, (not A) and (not B) are equivalent in this case.
5. ## Re: Perfect square s, odd number of factors
No, I meant if it's just given that, "If a number is a perfect square, it will have an odd number of factors (e.g., 4 has factors 1, 2, 4)"
Now how would you prove the converse?
6. ## Re: Perfect square s, odd number of factors
So, I understand that the part in post #1 after "whereas" is not given.
The following is multiple choice question (with options) to answer.
How many factors of 2475 are odd numbers greater than 1? | [
"3",
"4",
"17",
"6"
] | C | prime factors of 330 are 2^1,3^2,5^2 and 11^1
total divisors = (power if a prime factor+1)
total no. of odd factors(3,5,11) = (2+1)(2+1)(1+1) =18
since we need odd divisors other than 1 =>18-1 = 17 odd divisors
C is the answer |
AQUA-RAT | AQUA-RAT-36903 | 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 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 700 m long? | [
"11",
"190",
"188",
"177"
] | B | L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 1900 m
T = D/S
T = 1900/10
T = 190 Sec ,Answer:B |
AQUA-RAT | AQUA-RAT-36904 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
By selling 100 pencil, a trader gains the cost of 20 Pencil. Find his gain percentage? | [
"26 1/3%",
"51 1/3%",
"20%",
"53 1/3%"
] | C | C
20%
Let the CP of each pencil be Rs. 1.
CP of 100 pens = Rs. 100
Profit = Cost of 20 pencil = Rs. 20
Profit% = 20/100 * 100 = 20% |
AQUA-RAT | AQUA-RAT-36905 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
If the personal income tax rate is lowered from 45% to 30%, what is the differential savings for a tax payer having an annual income before tax to the tune of $48000? | [
"$3500",
"$5000",
"$3150",
"$7200"
] | D | Saving = (45-30)% of 48000 = 7200.
Answer:D |
AQUA-RAT | AQUA-RAT-36906 | The correct answer is 4. The chart below shows the remainder calculated vs. the remainder that is the answer (4). The values that don't match up are once again in red. Please note that I removed 3 from the table because it was a root of the equation, so the results weren't useful for my purposes.
In summary, does the remainder theorem not work for smaller values or does the method of my textbook not work for smaller values. If the latter, what is a better method that is still feasible under a short time limit (SAT 2 Subject Test).
Edit:I realized the remainder theorem is working (I'm getting the correct remainder), so it isn't the remainder theorem that is the problem. What explains the difference between the actual remainder and the remainder predicted by the answer of the textbook?
• @Masacroso Your comment made me realize that the issue doesn't lie with the remainder theorem, but with the solution/method of the textbook. What is the reason for this difference? – Eric Wiener Dec 18 '16 at 4:15
• Don't know that I like that textbook's advice. In both cases it is far easier to actually calculate the remainder (in any of a number of ways) than try to "guess" it. In the latter case, the remainder of any polynomial $P(x)$ divided by $x-3$ is always $P(3)$. – dxiv Dec 18 '16 at 4:18
• Wow, the textbook's solution is really terrible. It happens that their method works as long as you choose $x$ large enough (and the polynomials have integer coefficients and the polynomial you're dividing by is monic), but they haven't given any justification that the $x$ they have chosen is large enough in this particular problem. – Eric Wofsey Dec 18 '16 at 4:22
The following is multiple choice question (with options) to answer.
When p is divided by W, the remainder is 4. All of the following are possible values of W, except for which of the following? | [
"13",
"31",
"49",
"57"
] | D | The number can be expressed in the form of 9x+4 where x can 0,1,2,3,4....
by evaluating the answer choices carefully we can clearly observe that 57 is the only number which can't be W expressed in the form of 9x+4
In other words we can also say that the (answer - 4) will not be divisible by 9.57 is the number which doesn't follow this condition
Correct Answer - D |
AQUA-RAT | AQUA-RAT-36907 | Of course not! It is actually due to the fact that $102 = 6 \cdot 17$. Each of these numbers is a product of $17$ with some other prime, and the second prime increases by $6$ every time. That is, $289 = 17 \cdot 17$, then $391 = 17 \cdot (17 + 6) = 17 \cdot 23$, and $493 = 17 \cdot (23 + 6) = 17 \cdot 29$. Of course adding 6 to a prime doesn’t always get us another prime—but it works surprisingly often for smaller primes. And every prime besides 2 and 3 is either one more than a multiple of 6, or one less. So if we start with 5 and 7 and keep adding 6, we will definitely hit all the primes.
This sequence of multiples of 17 starts from $17 \cdot 5$, and if we continue it further we see that it contains several more numbers from our exceptional set as well:
$\begin{array}{rcl}85 &=& 17 \cdot 5 \\ 187 &=& 17 \cdot 11 \\ \mathbf{289} &=& 17 \cdot 17 \\ \mathbf{391} &=& 17 \cdot 23 \\ \mathbf{493} &=& 17 \cdot 29 \\ 595 &=& 17 \cdot 35 \\ \mathbf{697} &=& 17 \cdot 41 \\ \mathbf{799} &=& 17 \cdot 47 \\ \mathbf{901} &=& 17 \cdot 53 \end{array}$
What about if we start with $17 \cdot 7$ and keep adding $102$?
The following is multiple choice question (with options) to answer.
There arr 4 prime numbers in ascending order. The multiplication of 1st 3 is 385 and that of last 3is 1001. what is the last number is | [
"8",
"10",
"12",
"13"
] | D | Explanation:
abcbcd=3851001=>ad=513
abcbcd=3851001=>ad=513
So d = 13
D |
AQUA-RAT | AQUA-RAT-36908 | # smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2|<0.001$$
How to proceed? Thanks in advance.
• You are almost done ! – Khosrotash Aug 11 '17 at 14:49
$$|2[1-(-\frac{1}2)^n]-2|<0.001\\ |-2(\frac{-1}2)^n|<0.001\\ +2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\ +2.\frac{1}{2^n}<\frac{1}{1000}\\ \frac{2^n}{2}>1000\\ 2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
The following is multiple choice question (with options) to answer.
If n is an integer, then the least possible value of |33 - 5n| is? | [
"0",
"1",
"2",
"3"
] | C | |33 - 5n| represents the distance between 33 and 5n on the number line. Now, the distance will be minimized when 5n, which is multiple of 5, is closest to 33. Multiple of 5 which is closest to 33 is 35 (for n = 7), so the least distance is 2: |33 - 35| = 2.
Answer: C. |
AQUA-RAT | AQUA-RAT-36909 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
An error 2% in excess is made while measuring the side ofa square. The % of error in the calculated area of the square is? | [
"4.14%",
"4.04%",
"4.23%",
"4.67%"
] | B | 100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error = 404 x 100 % = 4.04%
100 x 100
B |
AQUA-RAT | AQUA-RAT-36910 | 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 25%, 20% and 15% is? | [
"31.8",
"31.1",
"31.6",
"49"
] | D | Explanation:
100*(75/100)*(80/100)*(85/100) = 51
100 - 51 = 49
Answer: D |
AQUA-RAT | AQUA-RAT-36911 | Problem #2: A box contains 18 tennis balls, 8 new 10 old. 3 balls are picked randomly and played with (so if any of them were new, they become 'old'), and returned to the box. If we pick 3 balls for the second time (after this condition), what is P that they are all new? I broke this down into 4 pieces: P(3 new second round|3 new first round)P(3 new first round) + P(3 new second round|2 new 1 old first round)P(2 new 1 old first round) + P(3 new second round|1 new 2 old first round)P(1 new 2 old first round) + P(3 new second round|3 old first round)(3 old first round). However, I was supposed to used binomials to count this. Instead I had a feeling that I should just multiply probabilities this way: \begin{align*} \frac{5\times4\times3}{18\times17\times16} &\times \frac{8\times7\times 6}{18\times 17\times 16} + \frac{6\times5\times 4}{18\times17\times16} \times \frac{8\times7\times10}{18\times17\times16}\\ &\quad + \frac{7\times6\times5}{18\times17\times16} \times \frac{8\times10\times9}{18\times17\times16} + \frac{8\times7\times6}{18\times17\times16} \times \frac{10\times9\times8}{18\times17\times16}. \end{align*} I get the correct answer with binomials, but this equation that I constructed undercounts the possibilities. Could you tell me what I am missing? ty!
-
Let's reduce problem 1 to see where you are going wrong. Let's say that there are 7 fishes, 4 trout and 3 carp, and you want to count how many ways there are of catching 2 fishes, at least one of them a carp.
The following is multiple choice question (with options) to answer.
In a certain archery competition, points were awarded as follows: the first place winner receives 11 points, the second place winner receives 7 points, the third place winner receives 5 points and the fourth place winner receives 2 points. No other points are awarded. John participated several times in the competition and finished first, second, third, or fourth each time. The product of all the points he received was 15400. How many times did he participate in the competition? | [
"5",
"6",
"7",
"8"
] | C | 15400 = 2*2*2*5*5*7*11
John participated 7 times.
The answer is C. |
AQUA-RAT | AQUA-RAT-36912 | earth, sun, moon, distance
Title: Moon vs Sun size and distance 400 times I have seen below statement, and it doesn't sound right:
The Sun and Moon seem to have the same size because of this amazing coincidence:
the moon is 400 times smaller than the Sun and 400 closer than the Sun.
Checking 400 from this source:
The Moon's distance from the Earth: 384,000 km and diameter: 3,480 km
The Sun's distance from the Earth: 149,000,000 km and diameter of 1,392,000 km
Distance: 149,000,000 / 384,000 = 388.02, OK almost 400
Size: 1,392,000 / 3,480 = 400 spot on.
Does being X times further make it look X times smaller?
I have seen this post but not sure it answer my question:
Can the apparent equal size of sun and moon be explained or is this a coincidence?
Note: First time post, let me know if the post is suitable.
Also called the intercept theorem.
The following is multiple choice question (with options) to answer.
The mass of the sun is approximately 4 × 10^30 kg and the mass of the moon is approximately 8 × 10^12 kg. The mass of the sun is approximately how many times the mass of the moon? | [
"4.0 × 10^(−18)",
"2.5 × 10^17",
"5 × 10^17",
"2.5 × 10^19"
] | C | Mass of sun = x * Mass of moon
x = Mass of sun / Mass of moon = (4 × 10^30) / (8 × 10^12) = 2^2*2^-3*10^18 = 10^18/2= 5*10^17
Ans. C) 2.5 × 10^17 |
AQUA-RAT | AQUA-RAT-36913 | 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.
One train is traveling 45 kmph and other is at 10 meters a second. Ratio of the speed of the two trains is? | [
"5:4",
"9:4",
"5:6",
"5:0"
] | A | Explanation:
45 * 5/18 = 10
25:20 => 5:4
Answer: Option A |
AQUA-RAT | AQUA-RAT-36914 | 4,639 views
In a tournament with $7$ teams, each team plays one match with every other team. For each match, the team earns two points if it wins, one point if it ties, and no points if it loses. At the end of all matches, the teams are ordered in the descending order of their total points (the order among the teams with the same total are determined by a whimsical tournament referee). The first three teams in this ordering are then chosen to play in the next round. What is the minimum total number of points a team must earn in order to be guaranteed a place in the next round?
1. $13$
2. $12$
3. $11$
4. $10$
5. $9$
I think possible with 9 only.
no 9 will not guarantee!!!
Let the $7$ Teams be $A,B,C,D,E,F,G$ and so each team plays total $6$ matches.
Suppose, Team $A$ wins over $E,F,G$ and draws with $B,C,D$ hence total points scored by Team A $= 9$ points
Now, Team $B$ wins over $E,F,G$ and draws with $A,C,D$ hence total points scored by Team B $= 9$ points
Similarly, happens for next two teams $C$ and $D$ .
Hence, Finalized scores are $\Rightarrow$
A = 9
B = 9
C = 9
D = 9
E = ? (Less than or equal to 4)
F = ? ("...")
G = ? ("...")
Given that the order among the teams with the same total are determined by a whimsical tournament referee.
So, He/She can order the top $3$ teams like $ABC,ABD,BCD,ACD,\ldots$
But, Question says " team must earn in order to be guaranteed a place in the next round "
Hence, Not to depend on that whimsical referee, the minimum total number of points a team must earn in order to be guaranteed a place in the next round = $9+1 = 10$ points
Correct Answer: $D$
by
So only 2 teams have 10 point right ??
why u considered A has won with 3 teams and tie with 3 teams?
The following is multiple choice question (with options) to answer.
The scoring system in a certain football competition goes as follows: 3 points for victory, 1 point for a draw, and 0 points for defeat. Each team plays 20 matches. If a team scored 14 points after 5 games, what is the least number of the remaining matches it has to win to reach the 40-point mark by the end of the tournament? | [
"6",
"7",
"8",
"9"
] | A | To get 40 points as end of season we need another 26 points or more from remaining 15 matches:
Option A= 6*3+9*1=27
hence option A-6 |
AQUA-RAT | AQUA-RAT-36915 | ### Show Tags
03 Jan 2019, 02:15
cfc198 wrote:
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A -1
B 4
C 1
D -4
E 0
y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.
The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.
The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.
The second root is the second point where the curve cuts the x-axis, which is -1.
If you liked the question and explanation, please do hit the kudos button
_______________
Merging topics.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 14467
Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
### Show Tags
22 Feb 2020, 15:03
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
The following is multiple choice question (with options) to answer.
y = x^2 + bx +64 cuts the x axis at (h, 0) and (k, 0). If h and k are integers, what is the least value of b? | [
"-65",
"-64",
"-32",
"32"
] | B | As the curve cuts the x-axis at (h,0) and (k,0). Therefore h,k are the roots of the quadratic equation.
For the quadratic equation is in the form ofax^2+bx+c=0,
The product of the roots =c/a= 64/1=256 and the sum of the roots =-b/a=-b
64 can be expressed as product of two numbers in the following ways:
1 * 64
2 * 32
4 * 16
The sum of the roots is maximum when the roots are 1 and 64 and the maximum sum is 1 + 64 = 65.
The least value possible for b is therefore -65.
B |
AQUA-RAT | AQUA-RAT-36916 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a covering a certain distance, the speeds of A and B are in the ratio of 3:4. A takes 30 minutes more than B to reach the destination. The time taken by A to reach the destination is? | [
"32 hrs",
"8 hrs",
"2 hrs",
"6 hrs"
] | C | Ratio of speeds = 3:4
Ratio of times taken = 4:3
Suppose A takes 4x hrs and B takes 3x hrs to reach the destination.
Then, 4x - 3x = 30/60 => x = 1/2
Time taken by A = 4x hrs = 4 * 1/2 = 2 hrs.
Answer: C |
AQUA-RAT | AQUA-RAT-36917 | Case 4: One digit is used four times, while two other digits are used once each.
Subcase 1: The leading digit is repeated.
We have nine ways of choosing the leading digit and $\binom{5}{3}$ ways of choosing the other three positions in which it appears. We have nine choices for the leftmost open position and eight choices for the remaining position. $$9 \cdot \binom{5}{3} \cdot 9 \cdot 8 = 6480$$
Subcase 2: The leading digit is not repeated.
We have nine ways of choosing the leading digit. We have nine ways of choosing the repeated digit and $\binom{5}{4}$ ways of selecting four of the five open positions in which to place it. We have eight ways of filling the remaining open position. $$9 \cdot 9 \cdot \binom{5}{4} \cdot 8 = 3240$$
That gives a total of $$9 + 405 + 81 + 810 + 405 + 6480 + 3240 = 11,430$$ excluded cases.
Hence, there are $$900,000 - 11,430 = 888,570$$ six-digit positive integers in which no digit appears more than three times.
This is neatly handled using exponential generating functions. Assuming first that we are allowed to have 0 as the first digit (e.g. we're talking about license plates or lock combinations): each of $10$ digits can occur up to $3$ times, and the order of the symbols matters. The answer is $$\left[\frac{x^6}{6!}\right] \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^{10} = 987,300.$$
The following is multiple choice question (with options) to answer.
A license plate in the country Kerrania consists of four digits followed by two letters. The letters A, B, and C are used only by government vehicles while the letters D through Z are used by non-government vehicles. Kerrania's intelligence agency has recently captured a message from the country Gonzalia indicating that an electronic transmitter has been installed in a Kerrania government vehicle with a license plate starting with 79. If it takes the police 60 minutes to inspect each vehicle, what is the probability that the police will find the transmitter within three hours? | [
"18/79",
"1/6",
"1/300",
"1/50"
] | C | So there are 900 cars that they have to search.. Each takes 60mins, total of 54000 mins.. Have to find in 180 mins.. Prob180/54000 = 1/300 (ANSWER C) |
AQUA-RAT | AQUA-RAT-36918 | 7. Given the following grades on a test: 86, 92, 100, 93, 89, 95, 79, 98, 68, 62, 71, 75, 88, 86, 93, 81, 100, 86, 96, 52
1. A stem-and-leaf plot is a quick way to construct a histogram by hand, whereby one lumps together all values that have common leading digits up to some level of precision (i.e., the "stem") using the remaining digit(s) of each data value to form the "bars" (i.e., the "leaves"), as shown below for the data set $13,15,17,17,19,24,25,25,25,26,29,30,31,33,46,62$:
6 | 2
5 |
4 | 6
3 | 0 1 3
2 | 4 5 5 5 6 9
1 | 3 5 7 7 9
Make a stem-and-leaf plot that represents the test grade data.
10 | 0 0
9 | 2 3 3 5 6 8
8 | 1 6 6 6 8 9
7 | 1 5 9
6 | 8
5 | 2
2. Find the mode, median, mean, range, standard deviation, and interquartile range
mode = 86, median = 87, mean = 84.5, range = 52 to 100, standard deviation = 13.17, interquartile range (IQR) = 77 to 94
In R:
grades = c(86, 92, 100, 93, 89, 95, 79, 98, 68, 62, 71, 75, 88,
86, 93, 81, 100, 86, 96, 52)
median(data)
mean(data)
range(data)
sd(data)
IQR(data)
# Sadly, the mode() function in R does something else.
# So you can write your own function, or if the data
# set is small enough (as it is here), you can use
# sort(data) to make it easier to count duplicate entries
TI-83:
Enter data in L1 with [STAT] : EDIT : Edit...
Then [STAT] : CALC : 1-Var Stats
The following is multiple choice question (with options) to answer.
Shekar scored 76, 65, 82, 67 and 95 marks in Mathematics, Science, Social studies, English and Biology respectively. What are his average marks? | [
"65",
"77",
"75",
"85"
] | B | Explanation :
Average= (76+65+82+67+95)/5 = 385/5 =77
Hence average=77
Answer : B |
AQUA-RAT | AQUA-RAT-36919 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
In a house, there are dogs, cats and parrot in the ratio 3:7:5. If the number of cats was more than the number of dogs by a multiple of both 9 and 7. What is the minimum number of pets in the house? | [
"945",
"630",
"252",
"238"
] | A | Explanation :
Let, the number of Dogs, Cat and Parrots be 3k, 7k and 5k respectively.
Then, According to the question,
=>7k−3k=63p. (where p is any positive integer).
As the number is a multiple of both 9 and 7, it has to be multiple of 63.
=>k=63p/4.
Minimum value of p for which k is a natural number is 4.
Thus, k=63.
Hence, the number of pets are (63x5) + (63x3) + (63x7) i.e 315 +189+ 441= 945.
Answer : A |
AQUA-RAT | AQUA-RAT-36920 | take a 5 = 48 ⇒ a r 5-1 = 48 ⇒ a 2 4 = 48 ⇒ a = 3. . To obtain the median, let us arrange the salaries in ascending order: The question asks for a three digit number: call the digits x, y and z. Example 1:Find the arithmetic mean of first five prime numbers. . . Median: Arrange the goals in increasing order We much have b – A = A- a ; Each being equal to the common difference. If we remove one of the numbers, the mean of the remaining numbers is 15. Mean with solution. The following table shows the grouped data, in classes, for the heights of 50 people. From equation ( ii ) , ( iii ) & b = -1. a = 2 and c = -4. Solution: Using the formula: Sum = Mean × Number of numbers Sum of original 6 numbers = 20 × 6 = 120 Sum of remaining 5 numbers = 15 × 5 = 75 Question 20. So the sequence is 2, -1, -4. The last two numbers are 10. Solution:First five prime numbers are 2, 3, 5, 7 and 11. Selection of the terms in an Arithmetic Progression, If number of terms is 3 then assume them as ” a-d, a & a+d” and common difference is “d”, If number of terms is 4 then assume them as ” a-3d, a-d, a+d & a+3d ” and common difference is “2d”, If number of terms is 5 then assume them as ” a-2d, a-d, a, a+d & a+2d ” and common difference is “d”, If number of terms is 6 then assume them as ” a-5d, a-3d, a-d, a+d , a+3d & a+5d ” and common difference is “2d”. . Detailed explanations and solutions to these questions … This GMAT averages problem is a vey easy question. The “n” numbers r1, r2, r3, r4, . G n are said to be Geometric means in between ‘a’ and ‘b’. a + c = 2b . Give One Example Each On Designed Experiment, Observational Study And Retrospective Study, T Distribution And Sampling Distribution Between The Difference Of Means And Justify. Solution: From graph, median = 54. Example -11: Four geometric means are
The following is multiple choice question (with options) to answer.
A certain bakery has 6 employees. It pays annual salaries of $14,000 to each of 2 employees, $16,000 to 1 employee, and $17,000 to each of the remaining 3 employees. The average (arithmetic mean) annual salary of these employees is closest to which of the following? | [
"$15,200",
"$15,500",
"$15,800",
"$16,000"
] | C | Let the salary for each person be 14. Thus, the bakery ends up getting 2 + 3*3 = 11 as extra. This again has to be divided among 6 people. 12/6 = 2 which makes it 14+2 =16. Thus, as it is 11/6, it will be somewhere close to 16 and less than that. Also, 1/6 = 0.166. Thus 10/6 = 1.66 which gives the final salary after redistribution as 15.66. Correct answer is C.
_________________ |
AQUA-RAT | AQUA-RAT-36921 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
One water pump can fill half of a certain empty tank in 4 hours. Another pump can fill half of the same tank in 4.5 hours. Working together, how long will it take these two pumps to fill the entire tank? | [
"1 7/13",
"4 5/8",
"3 1/4",
"4 4/17"
] | D | One pump can fill a tank in 4 hours and another in 4.5 hours
so the rate at which both can half fill the tank is (1/4+1/4.5) => 17/36
Thus half of the tank can be filled in 36/17
so for filling the complete tank => 36/17*2 = 72/17 =4 4/17
ANSWER:D |
AQUA-RAT | AQUA-RAT-36922 | 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.
Two numbers are less than third number by 35% and 37% respectively. How much percent is the second number less than by the first | [
"8%",
"10%",
"33%",
"41%"
] | C | Let the third number is x.
then first number = (100-35)% of x
= 65% of x =65x/100
Second number is (63x/100)
Difference =65x/100 - 63x/100 = x/50
So required percentage is, difference is what percent of first number
(x/50 * 100/65x * 100 )% = 33% ANSWER :C |
AQUA-RAT | AQUA-RAT-36923 | # Is a negative number proper fractions?
I was asked this question by a kid, is -$\frac{4}{7}$ a proper fraction or not? As per my knowledge $\frac{4}{7}$ is a proper fraction. If it has a -ve number does it make any difference? Definition says A number whose numerator is smaller than denominator is called a proper fraction. Can we consider -$\frac{4}{7}$as a proper fraction? If not why not please explain. This is my first question I don't have much idea about tags of mathematics if it is tagged wrongly please edit it.
Thank you
Dibya
• mathworld.wolfram.com/ProperFraction.html So it's neither proper or improper. – Lazar Ljubenović Mar 10 '13 at 12:23
• It is proper: en.wikipedia.org/wiki/… – Dennis Gulko Mar 10 '13 at 12:23
• If I had to define a convention that makes sense in the context of where questions like this have relevance, I would say that -4/7 isn't a fraction at all; it is a numeral consisting of a negative sign - and a (proper) fraction 4/7. – Hurkyl Mar 10 '13 at 12:26
From Wikipedia:
Common fractions can be classified as either proper or improper. When the numerator and the denominator are both positive, the fraction is called proper if the numerator is less than the denominator, and improper otherwise. In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is between $-1$ and $1$.* [Italics mine]
So $\;-1 < \left(-\dfrac 47\right) < 1$ is considered a proper fraction; (alternatively $\;\;0 < \Big|-\dfrac 47 \Big| = \dfrac 47 < 1$.
The following is multiple choice question (with options) to answer.
Which of the following is closest to the difference between sum of all proper fractions (fractions less than 1) in the form 1/x , where x is a positive digit, and the product of all proper fractions in the form y/(y+1), where y is a positive digit? | [
"2.82",
"2.72",
"1.82",
"1.72"
] | D | First term = 1/2 +1/3 +1/4 +1/5 +1/6 +1/7 +1/8 + 1/9
Second term = 1/2 * 2/3 * 3/4 * 4/5 * 5/6 * 6/7 * 7/8 * 8/9 * 9/10 = 1/10
First term - Second term = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 +1/8 + 1/9 - 1/10
Now rearrange the above equation for simplification:
= (1/2 + 1/5 - 1/10) +( 1/3 + 1/6 + 1/9) + (1/4 + +1/8) +1/7
= 6/10 + 11/18 + 3/8 +1/7
6/10 , 11/18, 3/8 are approximately equal to 0.5
=0.5+0.5+0.5+0.14 = 1.64
The closest answer choice is D |
AQUA-RAT | AQUA-RAT-36924 | #### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).
With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)
The following is multiple choice question (with options) to answer.
What annual instalment will discharge a debt of Rs 1092 due in 3 years at 12% simple interest? | [
"Rs.325",
"Rs.545",
"Rs.560",
"Rs.550"
] | A | Explanation:
Let each instalment be Rs.x .
Then, [x + (x × 12 × 1)/100] + [ x + (x × 12 × 2)/100] + x =1092
( 28x/25 ) + ( 31x/25 ) + x = 1092
(28x + 31x + 25x) = (1092 * 25)
x = (1092 × 25)/84= 325
∴ Each instalment = Rs. 325
Answer: A |
AQUA-RAT | AQUA-RAT-36925 | 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 (64)2 - (36)2 = 20 x a, then a = ? | [
"70",
"120",
"180",
"140"
] | D | 20 x a = (64 + 36)(64 - 36) = 100 x 28
a = (100 x 28)/20 = 140
ANSWER:D |
AQUA-RAT | AQUA-RAT-36926 | # 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.
The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use π = \\frac{22}{7}\\) in your calculations. | [
"1696 sq m",
"1694 sq m",
"1594 sq m",
"1756 sq.m"
] | B | Explanatory Answer
The cow can graze the area covered by the circle of radius 19m initially, as the length of the rope is 19m.
Area of a circle = π * (radius)2
Therefore, the initial area that the cow can graze = 22/7 * 19(2) sq m.
When the length of the rope is increased to 30m, grazing area becomes = 22/7 * 30(2) sq m.
The additional area it could graze when length is increased from 19m to 30m
= 22/7 * (30square - 19square) sq m.
22/7 * (30 + 19)(30 - 19) = 22/7 * 49 * 11 = 1694 sq m.
Choice B |
AQUA-RAT | AQUA-RAT-36927 | 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.
A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 20 per kg, then his gain percent is | [
"7%",
"13%",
"14%",
"15%"
] | A | Explanation:
Suppose he bought 5 kg and 3 kg of tea.
Cost Price = Rs. (5 x 18 + 3 x 20) = Rs. 150.
Selling price = Rs. (8 x 20) = Rs. 160.
Profit = 160 - 150 = 10
So, Profit % = (10/150) * 100 = 7%
Option A |
AQUA-RAT | AQUA-RAT-36928 | Can you finish the problem?
• November 15th 2007, 05:39 PM
poofighter
d=r(t)
distance= rate(km/hr) X time from noon to 2pm is 2 hrs
70km= 35km/hr x 2 hrs
the distance is the hypothnuse of the open triangle below. use pythrugrams therom to solve for it. a^2+b^2=c^2
. l
. l70km
. l
---70km--A--125km---B
195km
• November 15th 2007, 05:50 PM
singh1030
thanks...that helps a lot. but one question. does the part about the rate of change of distance at 2 PM have no relavence to the problem? does it matter whether they ask how fast is the distance changing at 2 PM, 3 PM, 4PM and so on??
The following is multiple choice question (with options) to answer.
If a man can cover 24 metres in one second, how many kilometres can he cover in 3 hours 45 minutes? | [
"118 km",
"162 km",
"324 km",
"1887 km"
] | C | 24 m/s = 24 * 18/5 kmph
3 hours 45 minutes = 3 3/4 hours = 15/4 hours
Distance = speed * time = 24 * 18/5 * 15/4 km = 324 km.
Answer: C |
AQUA-RAT | AQUA-RAT-36929 | 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.
If 7 cats can kill 7 rats in 7 minutes, how long will it take 100 cats to kill 100 rats? | [
"5 minutes",
"6 minutes",
"7 minutes",
"8 minutes"
] | C | It will take 7 minutes for 100 cats to kill 100 rats.
1 cat can kill 1 rat in 7 minutes, so 100 cats can kill 100 rats in 7 minutes
Answer C |
AQUA-RAT | AQUA-RAT-36930 | Question
# In measuring the length and breadth of a rectangle, errors of 5% and 3% in excess, respectively, are made. The error percent in the calculated area is
A
7.15%
B
6.25%
C
8.15%
D
8.35%
Solution
## The correct option is C 8.15% Let actual length and breadth of the rectangle be 'l' and 'b' respectively. The measured sides are (l+0.05l) and (b+0.03b) i.e. 1.05l and 1.03b ∴ Measured area =(1.05l)(1.03b) Actual area =lb Error in area=1.0815lb−lb =0.0815lb ∴% error=Error in areaActual area×100 ⇒% error=0.0815lblb×100%=8.15%Mathematics
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The following is multiple choice question (with options) to answer.
An error 2% in excess is made while measuring the side ofa square. The % of error in the calculated area of the square is? | [
"4%",
"4.04%",
"4.15%",
"4.34%"
] | B | 100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error = 404 x 100 % = 4.04%
100 x 100
B |
AQUA-RAT | AQUA-RAT-36931 | 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 covers a distance of 12 km in 10 min. If it takes 11 sec to pass a telegraph post, then the length of the train is? | [
"177 m",
"189 m",
"220 m",
"178 m"
] | C | Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec
= 20 m/sec.
Length of the train = 20 * 11
= 220 m.
Answer: C |
AQUA-RAT | AQUA-RAT-36932 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
The following is multiple choice question (with options) to answer.
A cricketer scored 152 runs which included 12 boundaries and 2 sixes. What percent of his total score did he make by running between the wickets. | [
"46.2%",
"54.54 %",
"60.52%",
"70%"
] | C | Explanation :
Number of runs made by running = 152 - (12 x 4 + 2 x 6)
= 152 - (60)
= 92
Now, we need to calculate 92 is what percent of 152.
=> 92/152 * 100 = 60.52 %
Answer : C |
AQUA-RAT | AQUA-RAT-36933 | # Placing m books on n shelves such that there is at least one book on each shelf
Given $m \ge n \ge 1$, how many ways are there to place m books on n shelves, such that there is at least one book on each shelf?
Placing the books on the shelves means that:
• we specify for each book the shelf on which this book is placed, and
• we specify for each shelf the order (left most, right most, or between other books) of the books that are placed on that shelf.
I solve this problem in the following way:
If $m=n$, there are $m!$ or $n!$ ways to do it
Else:
1. Place $n$ books on $n$ shelves: $n!$ ways to do it
2. Call the set of $m-n$ remaining books $T=\{t_1, t_2,..,t_{m-n}\}$
The procedure for placing books on shelves: choose a shelf, choose a position on the shelf
We know choosing a shelf then place the book on the far left has $n$ ways
For book $t_1$, there is a maximum of $1$ additional position (the far right). Thus there is $n+1$ ways to place book $t1$.
For book $t2$, there is a maximum of $2$ additional positions. Thus there is $n+2$ ways for book $t_2$
...
For book $t_i$, there is a maximum of $i$ additional positions. Thus there is $n+i$ ways for book $t_i$
In placing $m-n$ books, we have $(n+1)(n+2)...(n+m-n)$ or $(n+1)(n+2)..m$ ways
In total, we have $n!(n+1)(n+2)...m$ or $m!$ ways
Is there any better solution to this problem?
The following is multiple choice question (with options) to answer.
Michael is interested in dividing all of his boxed food equally among the shelves in his pantry. Unfortunately, the boxes don't divide evenly. It turns out that Michael will need to add three more shelves to the pantry and throw away four boxes of food. Which of the following can describe the initial number of boxes of food and the initial number of pantry shelves (in the order boxes of food; shelves)? | [
"71; 4.",
"28; 6.",
"65; 7.",
"53; 4."
] | D | Option D.) Initially 53 boxes of food then reduced to 49 boxes of food; and Initially 4 pantry shelves then increased to 7 pantry shelves.
49 / 7 = 7 boxes of food per pantry shelf. |
AQUA-RAT | AQUA-RAT-36934 | 5. Why we can't solve it by 100+15/2=125
6. Why we can't solve it by 100+15/2=125
7. nice
8. The speed of the train going from Nagpur to Allahabad is 100 km/h while when coming back from Allahabad to Nagpur, its speed is 150 km/h and again going to the same way Nagpur to Allahabad at a speed of 120km/hr.Find the average speed during whole journey...
9. Average Speed = 2ab/(a + b)
Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. On similar lines, you can modify this formula for one-third distance.
10. Yes
11. A+B=100
B+A=150
(100+150) /2= 120
12. Dear Sir,
Can I get pdf file of aptitude part to learn it withous using internet.
13. So the formula of average speed with same distance can be broken as such:
Let the distance be n
So the average speed = total distance covered/total time taken
= 2n/[(n/100)+(n/150)]
= 2n/[(n/x)+(n/y)]
= 2n[(nx+ny)/xy]
= 2n[n(x+y)/xy]
= 2(x+y)/xy
14. Can you please point me to some article which would help understanding this formulae for finding the average speed . Or some different formula for calculating train speed average
15. Let the distance between them is 300 km,so total distance covered is 600 km...from Nagpur to Allahabad speed is 100 kmph so time taken is 3 hrs. And while retiring it's speed is 150kmph so time taken is 2 hrs...so total distance is 600 km nd total time taken is 5 hrs and by applying the basic formula we will get 120 kmph
16. Its is the formula of the finding average speed of two variant of speed for the same distance. where 'x' and 'y' are the speed of the train receptively
17. Why we should multiply 2 before it?
The following is multiple choice question (with options) to answer.
X and Y are two towns. Ganesh covers the distance from X to Y at an average speed of 44 Km/hr. However, he covers the distance from Y to X at an average speed of 36 Km/hr. His average speed during the whole journey in km/hr. is : | [
"39.6",
"43",
"40",
"38"
] | A | Solution: Average speed = 2XY / X+Y
= 2*44*36 / 44+36
= 39.6
Answer : A |
AQUA-RAT | AQUA-RAT-36935 | The next one up is $$2 \times 3\times 4=$$ 24. If you keep going, you should get this list:
$$1 \times 2 \times 3= 6$$
$$2 \times 3 \times 4= 24$$
$$3 \times 4 \times 5= 60$$
$$4 \times 5 \times 6= 120$$
$$5 \times 6 \times 7= 210$$
$$6 \times 7 \times 8= 336$$
$$7 \times 8 \times 9= 504$$
$$8 \times 9 \times 10= 720$$
$$9 \times 10 \times 11= 990$$
Look how quickly we got there! Good things can happen if you just keep going.
If you count them up, you should see that the answer is 9.
Some of you might complain, "Ok it happened to work for this one question. What do you do when the list keeps on going and going?"
When that happens, making a list will still help, but another step or tactic will often be necessary.
Here's one example:
3. How many multiples of 3 are there from 100 to 500?
Again, let's start by listing some of them out:
$102,\, 105,\, 108,\, 111,\, \ldots$
At this point, it's not hard to see that this will take forever. Here's the trick—instead of listing out all the numbers in the middle, let's list out a few at the end:
$102,\, 105,\, 108,\, 111,\, \ldots 492,\, 495,\, 498$
Now let's divide each number in the list by 3:
$34,\, 35,\, 36,\, 37,\, \ldots 164,\, 165,\, 166$
See what happened? The numbers are now consecutive. Now the question becomes, "How many numbers are there from 34 to 166?"
If you're not sure, ask yourself how many numbers there are from 5 to 10.
There are 6.
Now how did you get that?
Perhaps you didn't realize it, but you subtracted 5 from 10 and then added 1:
$10 - 5 + 1 = 6$
The following is multiple choice question (with options) to answer.
If
1 = 6
2 = 12
3 = 18
4 = 24
5 = 30
6 = 36
Then 24 = ?
Hint: Its a logic Riddle not a mathematical riddle | [
"1",
"2",
"4",
"6"
] | C | C
24
As stated
4=24 => 24=4
Answer is C |
AQUA-RAT | AQUA-RAT-36936 | # 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.
What amount does Kiran get if he invests Rs.7680 at 8% p.a. compound interest for two years, compounding done annually? | [
"8957",
"8978",
"7890",
"3408"
] | A | A= P{1 + R/100}n
=> 7680{1 + 8/100}2
= Rs.8957
Answer: A |
AQUA-RAT | AQUA-RAT-36937 | ### Show Tags
22 Mar 2017, 01:52
If n is the product of 3 consecutive integers, which of the following must be true?
I. a multiple of 2 II. a multiple of 3 III. a multiple of 4
A. I only
B. II only
C. III only
D. I and II
E. II and III
_________________
The following is multiple choice question (with options) to answer.
If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 294, what is the least possible value of n? | [
"8",
"10",
"12",
"14"
] | D | 294 = 2*3*7*7, so n must be at least 14.
The answer is D. |
AQUA-RAT | AQUA-RAT-36938 | • Thank you for taking the time to help me!! How can you say that $x$ and $y$ are co-prime? – user342661 Aug 6 '16 at 2:46
• Suppose $x$ and $y$ are not co-prime. Then let $x = gw$ and $y=gz$, where $g \neq 1$. Then note that $c=xd=gwd$,and $m = yd = gzd$. Note that $gd$ divides both $x$ and $y$ and is greater than $d$ because $g>1$. This contradicts the fact that $d$ is the gcd of $m$ and $c$. Hence $x$ and $y$ are co-prime. – астон вілла олоф мэллбэрг Aug 6 '16 at 2:51
• You are welcome. – астон вілла олоф мэллбэрг Aug 6 '16 at 4:16
Assuming integers for all variables, and $c' = c/d$, $m' = m/d$ :
\begin{align} ac \equiv bc \pmod m &\iff \exists k ~~ ac - bc = km \\ % &\iff \exists k ~~ ac'd - bc'd \equiv km \\ % &\iff \exists k ~~ a - b \equiv (k/c')(m/d) \end{align}
So we have to establish that $k / c'$ is an integer. From $\gcd(c, m) = d$ we can infer $\gcd(c', m') = 1$, so
$$ac - bc = km$$ $$ac'd - bc'd \equiv k(m'd)$$ $$c'(a - b) = km'$$
So $c'$ divides $k$, so $k/c'$ is an integer.
The following is multiple choice question (with options) to answer.
If y = 30p, and p is prime, what is the greatest common factor of y and 20p, in terms of p? | [
"p",
"2p",
"5p",
"10p"
] | D | y = 30p = 2*3*5*p
20p = 2^2*5*p
The greatest common factor of 30p and 20p is the product of all the common prime factors, using the lower power of repeated factors. The greatest common factor is 2*5*p = 10p
The answer is D. |
AQUA-RAT | AQUA-RAT-36939 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ? | [
"15%",
"41%",
"37%",
"25%"
] | D | Required percentage = [(20*100)/(100-20)]%=25%.
Answer is D. |
AQUA-RAT | AQUA-RAT-36940 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group? | [
"76",
"28",
"23",
"98"
] | C | Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n - 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
Answer:C |
AQUA-RAT | AQUA-RAT-36941 | ### Show Tags
23 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?
Aman
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
24 Jun 2019, 03:25
pkloeti wrote:
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.
Does that approach hold up in general? Bunuel VeritasPrepKarishma
Thanks a lot for the feedback!
I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
_________________
Karishma
Veritas Prep GMAT Instructor
Intern
Joined: 12 Feb 2018
Posts: 9
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
04 Aug 2019, 10:54
i tried like this:
Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85
The following is multiple choice question (with options) to answer.
15 lts are taken of from a container full of liquid A and replaced with Liquid B. Again 15 more lts of the mixture is taken and replaced with liquid B. After this process, if the container contains Liquid A and B in the ratio 9:16,What is the capacity of the container M? | [
"A:45",
"B:25",
"C:37.5",
"D:36"
] | C | If you have a 37.5 liter capacity, you start with 37.5 L of A and 0 L of B.
1st Replacement
After the first replacement you have 37.5-15=22.5 L of A and 15 L of B. The key is figuring out how many liters of A and B, respectively, are contained in the next 15 liters of mixture to be removed.
The current ratio of A to total mixture is 22.5/37.5; expressed as a fraction this becomes (45/2) / (75/2), or 45/2 * 2/75. Canceling the 2s and factoring out a 5 leaves the ratio as 9/15. Note, no need to reduce further as we're trying to figure out the amount of A and B in 15 L of solution. 9/15 of A means there must be 6/15 of B.
Multiply each respective ratio by 15 to get 9 L of A and 6 L of B in the next 15L removal.
Final Replacement
The next 15L removal means 9 liters of A and 6 liters of B are removed and replaced with 15 liters of B. 22.5-9=13.5 liters of A. 15 liters of B - 6 liters + 15 more liters = 24 liters of B.
Test to the see if the final ratio = 9/16; 13.5/24 = (27/2) * (1/24) = 9/16. Choice C is correct. |
AQUA-RAT | AQUA-RAT-36942 | 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 150 m long crosses a platform 130 m long in 20 sec; find the speed of the train? | [
"50",
"60",
"55",
"45"
] | A | D = 150 + 130 = 280
T = 20
S = 280/20 * 18/5 = 50 kmph.
Answer: A |
AQUA-RAT | AQUA-RAT-36943 | +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.
If 5 - 5/X = 4 + 4/X, then X = | [
"1",
"9",
"-1",
"5/4"
] | B | We're given the equation 5 - 5/X = 4 + 4/X. We're asked for the value of X.
The common-denominator of these 4 numbers is X, so we need to multiply both sides of the equation by X, giving us...
5X - 5X/X = 4X + 4X/X
We can then eliminate that denominator, which gives us....
5X - 5 = 4X + 4
9 = X
B |
AQUA-RAT | AQUA-RAT-36944 | = 4 52. Probability (statistics) What is the probability of getting a sum of 8 in rolling two dice? Update Cancel. Two fair dice are rolled and the sum of the points is noted. For example: 1 roll: 5/6 (83. of ways are - 1 , 1 1 , 2 2 , 1 1 , 4 4 , 1 1 , 6. The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6). No, other sum is possible because three dice being rolled give maximum sum of (6+6+6) i. and only one way to roll a 12 (6-6, or boxcars). EXPERIMENTAL PROBABILITIES Simulate rolling two dice 120 times. The sum of two dice thrown can be 7 and 11 in the following cases : (6,1) (1,6) (3,4) (4,3) (5,6) (6,5) (2,5) (5,2) The total possible cases are = 36 Favorable cas. Two dice are tossed. Probability of Rolling Multiple of 6 with 2 dice - Duration: 4:19. Sample space S = {H,T} and n(s) = 2. That intuition is wrong. When you roll a pair of dice there are 36 possible outcomes. To find the probability we use the mutually exclusive probability formula P(A) + P(B). A sum less than or equal to 4. Isn’t that kind of cool?. Two dice are tossed. So the probability of getting a sum of 4 is 3/36 or 1/12. 10 5 13 ! Find the probability distribution. So 1/36 is part of the. Sum of Two Dice. hi Dakotah :) A number cube is rolled 20 times and lands on 1 two times and on 5 four times. Rolling two dice. Good morning Edward, I liked your dice probability work on the chances of getting one 6 when rolling different number of dice. The fundamental counting principle tells us there are 6*6=36 ways to roll two dice, all of them equally likely if the dice are fair. , in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and 1. Let B be the event - The sum
The following is multiple choice question (with options) to answer.
A dice is rolled twice. What is the probability of getting sum 9? | [
"1/9",
"1/8",
"1/7",
"1/5"
] | A | Explanation:
Possible event = 6*6 = 36
Favourable outcomes= {(3, 6), {4, 5}, {5, 4}, (6, 3)}
Thus probability; favourable/total
= 4/36
=> 1/9
ANSWER: A |
AQUA-RAT | AQUA-RAT-36945 | ### Show Tags
23 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?
Aman
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9558
Location: Pune, India
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
24 Jun 2019, 03:25
pkloeti wrote:
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.
Does that approach hold up in general? Bunuel VeritasPrepKarishma
Thanks a lot for the feedback!
I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
_________________
Karishma
Veritas Prep GMAT Instructor
Intern
Joined: 12 Feb 2018
Posts: 9
Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
04 Aug 2019, 10:54
i tried like this:
Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85
The following is multiple choice question (with options) to answer.
A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%. Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available? | [
"20%",
"25%",
"33.33%",
"40%"
] | A | 100 * (b + f) - 100 * b = (25/100) * 100 * b
b = 4 * f
Percentage of f = (f / (b + f)) * 100 = (f/5f) * 100 = 20%
ANSWER:A |
AQUA-RAT | AQUA-RAT-36946 | 1 = 1
4 = 2^2
6 = 2 * 3
9 = 3^2
13 = 13
16 = 2^4
22 = 2 * 11
24 = 2^3 * 3
25 = 5^2
33 = 3 * 11
36 = 2^2 * 3^2
37 = 37
46 = 2 * 23
49 = 7^2
52 = 2^2 * 13
54 = 2 * 3^3
61 = 61
64 = 2^6
69 = 3 * 23
73 = 73
78 = 2 * 3 * 13
81 = 3^4
88 = 2^3 * 11
94 = 2 * 47
96 = 2^5 * 3
97 = 97
100 = 2^2 * 5^2
109 = 109
117 = 3^2 * 13
118 = 2 * 59
121 = 11^2
132 = 2^2 * 3 * 11
141 = 3 * 47
142 = 2 * 71
144 = 2^4 * 3^2
148 = 2^2 * 37
150 = 2 * 3 * 5^2
157 = 157
166 = 2 * 83
169 = 13^2
177 = 3 * 59
181 = 181
184 = 2^3 * 23
193 = 193
196 = 2^2 * 7^2
198 = 2 * 3^2 * 11
208 = 2^4 * 13
213 = 3 * 71
214 = 2 * 107
216 = 2^3 * 3^3
222 = 2 * 3 * 37
225 = 3^2 * 5^2
229 = 229
241 = 241
244 = 2^2 * 61
249 = 3 * 83
253 = 11 * 23
256 = 2^8
262 = 2 * 131
276 = 2^2 * 3 * 23
277 = 277
286 = 2 * 11 * 13
289 = 17^2
292 = 2^2 * 73
294 = 2 * 3 * 7^2
297 = 3^3 * 11
312 = 2^3 * 3 * 13
313 = 313
321 = 3 * 107
324 = 2^2 * 3^4
325 = 5^2 * 13
333 = 3^2 * 37
334 = 2 * 167
337 = 337
349 = 349
352 = 2^5 * 11
358 = 2 * 179
361 = 19^2
366 = 2 * 3 * 61
373 = 373
376 = 2^3 * 47
382 = 2 * 191
384 = 2^7 * 3
388 = 2^2 * 97
393 = 3 * 131
The following is multiple choice question (with options) to answer.
What number is obtained by adding the units digits of 734^98 and 347^81? | [
"10",
"11",
"12",
"13"
] | D | The units digit of 734^98 is 6 because 4 raised to the power of an even integer ends in 6.
The units digit of 347^81 is 7 because powers of 7 end in 7, 9, 3 or 1 cyclically. Since 81 is in the form 4n+1, the units digit is 7.
Then 6+7 = 13.
The answer is D. |
AQUA-RAT | AQUA-RAT-36947 | P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2 P ( A | C ) = 0 . 4 P ( B | C ) = 0 . 6 P ( A | C c ) = 0 . 3 P ( B | C c ) = 0 . 2
(1)
Show whether or not the pair {A,B}{A,B} is independent.
### Solution
P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc)P(A)=P(A|C)P(C)+P(A|Cc)P(Cc),P(B)=P(B|C)P(C)+P(B|Cc)P(Cc), and P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc)P(AB)=P(A|C)P(B|C)P(C)+P(A|Cc)P(B|Cc)P(Cc).
PA = 0.4*0.7 + 0.3*0.3
PA = 0.3700
PB = 0.6*0.7 + 0.2*0.3
PB = 0.4800
PA*PB
ans = 0.1776
PAB = 0.4*0.6*0.7 + 0.3*0.2*0.3
PAB = 0.1860 % PAB not equal PA*PB; not independent
## Exercise 2
Suppose {A1,A2,A3} ci |C{A1,A2,A3} ci |C and ci |Cc ci |Cc, with P(C)=0.4P(C)=0.4, and
The following is multiple choice question (with options) to answer.
If c is 25% of a and 50% of b, what percent of a is b? | [
" 2.5%",
" 50%",
" 25%",
" 35%"
] | B | Answer = B
25a/100 = 50b/100
b = 25a/50 = 50a/100 =50% |
AQUA-RAT | AQUA-RAT-36948 | 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 is twice as fast as B. If B alone can do a piece of work in 10 days, in what time can A and B together complete the work? | [
"10/3",
"8",
"9",
"7/3"
] | A | A can do the work in10/2 i.e., 5 days.
A and B's one day's work = 1/5+ 1/10 = (2 + 1)/10 = 10/3
So A and B together can do the work in 10/3 days.
Answer: A |
AQUA-RAT | AQUA-RAT-36949 | 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 a carpenter completes 3/8th of his work in 6 days, then how many more days will he require completing his remaining work? | [
"10",
"99",
"77",
"66"
] | A | Explanation:
As 3/8th of the work is completed, the work remaining is 5/8th of the total.
(3/8):6::(5/8):x
x = 6*(5/8)/(3/8)
x = 10
ANSWER: A |
AQUA-RAT | AQUA-RAT-36950 | Dividing into cases where there is one chocolate ball and more than one.
One chocolate ball of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=4$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor and $1$ of another - cases where there are $4$ of one flavor):
${{11+4-1=14}\choose 4}-{11\choose 1}{10\choose 1}-{11\choose 1}=1001-110-11=880$
Two chocolate balls of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=3$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor):
${{11+3-1=13}\choose 3}-{11\choose 1}=286-11=275$
For a grand total of $1155$.
The following is multiple choice question (with options) to answer.
A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $141 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist? | [
"5",
"7",
"9",
"11"
] | C | $141 = 27*$5 + 2*$3
We can find other combinations by reducing the $5 chocolates by 3 (which is -$15)and increasing the $3 chocolates by 5 (which is +$15).
The number of $5 chocolates can be 27, 24, 21, 18, 15,...,3 for a total of 9 combinations.
The answer is C. |
AQUA-RAT | AQUA-RAT-36951 | $$3 = (321)(-18+41k) + (123)(47-107k) \text{ for any } k$$
so for instance we also get ($k=1$):
$$3 = (321)(23) + (123)(-60)$$
-
The following is multiple choice question (with options) to answer.
Evaluate:59- 12*3*2 =? | [
"42",
"51",
"62",
"72"
] | B | According to order of operations, 12?3?2 (division and multiplication) is done first from left to right
12**2 = 4* 2 = 8
Hence
59 - 12*3*2 = 59- 8 = 51
correct answer B |
AQUA-RAT | AQUA-RAT-36952 | # Kinematics Problem
1. Feb 24, 2008
### undefinable
1. The problem statement, all variables and given/known data
Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially travelling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.
Who wins the race? when and where did ophelia catch up? (both metres and time)
2. Relevant equations
d=vit+1/2(a)(t)2+di
3. The attempt at a solution
Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)
I got stuck trying to find out WHEN they caught up. I tried setting the equation to
vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di
and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
2. Feb 24, 2008
### naele
Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or $\triangle D_P = \triangle D_A + 13$
Last edited: Feb 24, 2008
3. Feb 24, 2008
### Mentz114
If you know where they crossed, plug that x value into Alvin's EOM to get t.
4. Feb 24, 2008
### cepheid
Staff Emeritus
Start by listing the information you have:
df = 100 m
Alvin
di = 13 m
v(t) = vi = 7 m/s
a = 0
==> d(t) = di + vit = 13 + 7t
Ophelia
di = 0 m
v(t) = vi = 1.2 m/s
a = 1.5 m/s2
==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2
The following is multiple choice question (with options) to answer.
Meg and Bob are among the 3 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob? | [
"24",
"30",
"3",
"90"
] | C | Total # of ways the race can be finished is 3!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 3!/2=3.
Answer: C. |
AQUA-RAT | AQUA-RAT-36953 | metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of the sector is. - 1133971 2:15 75.0k LIKES. Dec 15,2020 - If the perimeter of a sector of a circle of radius 5.2 cm. Let the angle subtended by the arc of the sector at the centre of the circle (radius assumed to be $r$) be $\theta$. What multiple of the radius is the area of the sector?a)5thb)3rdc)4thd)2ndCorrect answer is option 'B'. (a) Write down, in terms of r and θ, expressions for P and A. A sector of a circle is like a slice of pizza or pie. Units Metric units. In the Given Figure, Radius of Circle is 3.4 Cm and Perimeter of Sector P-abc is 12.8 Cm . Thus the perimeter of the sector is r+θr+r. Problem 6 : Find the length of arc, if the perimeter of sector is 45 cm and radius is 10 cm. Or P = d + L ( arc) => P = 2r + {(β*2 pi r)/360°} , where β is sector angle. = π* 25 * 42 /180+2* 25. thus the perimeter of the sector is L+2r units. Sector of a circle ↺ Theta is an angle that can be defined as the figure formed by two rays meeting at a common endpoint. The most common sector of a circle is a semi-circle which represents half of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. Perimeter of a Circle Sector A region bounded by two radii and an arc is called a sector of a circle. 15.8k SHARES. Click hereto get an answer to your question ️ The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Working with ratios (one question) Angles in quadrilateral (one question) We are learning about: The area and perimeter of a sector. If and the area of the
The following is multiple choice question (with options) to answer.
The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector so formed has a central angle of 80°, what fraction of the wheel’s area is represented by the smallest sector? | [
"1/72",
"1/36",
"1/18",
"1/12"
] | B | Largest angle = 80
Let, Difference between any two angles in A.P. = d
i.e. Sum of all angles will be
80 + (80-d) + (80-2d) + (80-3d) + (80-4d) + (80-5d) + (80-6d) + (80-7d) = 640 - 28d
But sum of all central angles in a circle = 360
i.e. 640 - 28d = 360
i.e. d = 280/28 = 10
Smallest Sector = (80-7d) = 80-7*10 = 10
Smallest sector as Fraction of entire circle = 10/360 = 1/36
Answer: option B |
AQUA-RAT | AQUA-RAT-36954 | Denote $K=\left\{x\in {\mathbb{R}}^{n}\mid \sum _{k=1}^{n}{x}_{k}=0,\sum _{k=1}^{n}{x}_{k}^{2}={a}^{2}\right\}$ and let $f\left(x\right)=\sum _{k=1}^{n}\left(a+{x}_{k}{\right)}^{-1}$ attain its minimum at $x=\overline{x}\in K$ (it does so, as a continuous function on $\left\{x\in K\mid f\left(x\right)⩽f\left(0\right)\right\}$ which is compact). Then, by Lagrange multiplier theorem, there are ${\lambda }_{1},{\lambda }_{2}\in \mathbb{R}$ such that for each k we have $\left(a+{\overline{x}}_{k}{\right)}^{-2}={\lambda }_{1}+{\lambda }_{2}{\overline{x}}_{k}$. Then the positive numbers ${y}_{k}=a+{\overline{x}}_{k}$ are solutions of
${y}^{2}\left({\lambda }_{1}-{\lambda }_{2}a+{\lambda }_{2}y\right)=1.$
But this equation has at most two positive solutions. Thus, at most two values among ${\overline{x}}_{k}$ are distinct, and in fact exactly two. So, let $m$ values of ${\overline{x}}_{k}$ equal $b>0$, where $0, and the remaining $n-m$ values equal $c<0$. We get a system for $b$ and $c$, solve it, and finally obtain
The following is multiple choice question (with options) to answer.
If the equation x^2−6x=-k has at least one solution, which of the following must be true? | [
"k > 9",
"k < −9",
"k = 9",
"k ≤ 9"
] | D | x^2−6x=-k
=> x^2−6x + k = 0
For a quadratic equation to have at least one solution, its discriminant (b^2-4ac) should be non-negative.
b^2-4ac ≥0
=> 36 - 4*1*(-k) ≥0
=> 36 - 4k ≥0
=> k ≤9
Answer D |
AQUA-RAT | AQUA-RAT-36955 | 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500
501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
The following is multiple choice question (with options) to answer.
Look at this series: 544, 509, 474, 439, 404 ... What number should come next? | [
"206",
"408",
"369",
"507"
] | C | 369
This is a simple subtraction series. Each number is 35 less than the previous number.
C |
AQUA-RAT | AQUA-RAT-36956 | 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 112 metres long in 6 days, what length of a similar wall can be built by 40 men in 3 days? | [
"65mtr.",
"52mtr",
"70mtr.",
"112mtr."
] | D | 20 men is 6 days can build 112 metres
25 men in 3 days can build = 112*(40/20)x(3/6)
= 112 meters
Answer: D. |
AQUA-RAT | AQUA-RAT-36957 | 1. ## Combinatorics
hi
i don't know where else to post this but i need some help on a textbook question i was given about combinatorics.
there is a game that involves a murderer, a mansion, and the weapon used to commit the murder. there are 5 characters in all, 7 rooms in the mansion, and 3 weapons.
the question is.. how many kinds of guesses can i make from choosing one character, one mansion room, and one weapon?
what kind of formula do i use for this, nPr or nCr? or can i use the fundamental counting principle?
any help?
2. What you want is $n_1!*n_2!*n_3!$, where $n_1$ is the Character, $n_2$ is the room in the mansion, and $n_3$ is the weapon.
For example: Lets say Professor Plum we know is the murderer. He could have done it in any of the 7 rooms correct? So there are (at the moment) 7 possible scenarios for Plum to have committed the murder. But we also know he can use 3 weapons in each of the 7 scenarios: so thats three weapons in Room 1, three weapons in Room 2, three weapons in Room 3, etc. to three weapons in room 7 - for a total of 21 possible scenarios that Professor Plum (Character), could have gone somewhere (Room) and used a weapon (Weapon) to murder someone. But Plum isn't the only character. We have 5.
Thus you would use the basic methods of counting to find out the total possible scenarios that can play out here.
As an aside, remember what permutations and combinations are: Permutations are just an arrangement of $n$ objects, and combinations are the same arragement with uniqueness factored in.
3. Originally Posted by ANDS!
What you want is $n_1!*n_2!*n_3!$, where $n_1$ is the Character, $n_2$ is the room in the mansion, and $n_3$ is the weapon.
The following is multiple choice question (with options) to answer.
Sherlock Homes and Dr. Watson have to travel from Rajeev Gandhi chowk to Indira
Gandhi International Airport via the metro. They have enough coins of 1, 5, 10 and 25
paise. Sherlock Homes agree to pay for Dr. Watson only if he tells all the possible
combinations of coins that can be used to pay for the ticket.
How many combinations are possible if the fare is 50 paise ? | [
"52",
"49",
"45",
"44"
] | B | Let the number of 1, 5, 10 and 50 paise coins required are x, y, z and w respectively.
So when paid total amount is: x + 5y + 10z + 25w, which is equal to the fare paid.
So, x + 5y + 10z + 25w = 50
The number of non negative integral solution of this equation will be equal to the
number of ways the fare can be paid.
x + 5y + 10z + 25w = 50
In this equation w can be either 0, 1 or 2
Case 1: When w is 2
No other coins are required to pay
So there is only one way to pay: (0, 0, 0, 2)
Total no of ways : 1
Case 2: When w is 1
x + 5y + 10z = 25
In this case z can have only three values possible: 0, 1 or 2
Case I: When z is 0
x + 5y = 25
possible values are: (25, 0, 0, 1), (20, 1, 0, 1), (15, 2, 0, 1), (10, 3, 0, 1),
(5, 4, 0, 1), (0, 5, 0, 1)
Case II: When z is 1
x + 5y = 15
possible values are: (15, 0, 1, 1), (10, 1, 1, 1), (5, 2, 1, 1), (0, 3, 1, 1)
Case III: When z is 2
x + 5y = 5
possible values are: (5, 0, 2, 1), (0, 1, 2, 1)
Total no of ways : 12
Case 3: When w is 0
x + 5y + 10z = 50
In this case z can have only three values possible: 0, 1, 2, 3, 4 or 5
Case I: When z is 0
x + 5y = 50
possible values are: (50, 0, 0, 0), (45, 1, 0, 0), (40, 2, 0, 0), (35, 3, 0, 0),
(30, 4, 0, 0), (25, 5, 0, 0), (20, 6, 0, 0), (15, 7, 0, 0), (10, 8, 0, 0), (5, 9, 0, 0),
(0, 10, 0, 0)
Case II: When z is 1
x + 5y = 40
possible values are: (40, 0, 1, 0), (35, 1, 1, 0), (30, 2, 1, 0), (25, 3, 1, 0),
(20, 4, 1, 0), (15, 5, 1, 0), (10, 6, 1, 0), (5, 7, 1, 0), (0, 8, 1, 0)
Case III: When z is 2
x + 5y = 30
possible values are: (30, 0, 2, 0), (25, 1, 2, 0), (20, 2, 2, 0), (15, 3, 2, 0),
(10, 4, 2, 0), (5, 5, 2, 0), (0, 6, 2, 0)
Case IV: When z is 3
x + 5y = 20
possible values are: (20, 0, 3, 0), (15, 1, 3, 0), (10, 2, 3, 0), (5, 3, 3, 0), (0,
4, 3, 0)
Case V: When z is 4
x + 5y = 10
possible values are: (10, 0, 4, 0), (5, 1, 4, 0), (0, 2, 4, 0)
Case VI: When z is 5
1 2 3 4 5 6 7 8 9
x + 5y = 0
possible values are: (0, 0, 5, 0)
Total no of ways : 36
So total no of ways = 1 + 12 + 36 = 49
ANSWER:B |
AQUA-RAT | AQUA-RAT-36958 | ### 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.
Nirmal and Kapil started a business investing Rs. 9000 and Rs. 12000 respectively. After 6 months, Kapil withdrew half of his investment. If after a year, the total profit was Rs. 4600, what was Kapil’s share initially ? | [
"Rs 2300",
"Rs 2400",
"Rs 2500",
"None of above"
] | A | Explanation:
Nirmal:Kapil = 9000*12:(12000*6+6000*6) = 1:1
Kapils share = Rs. [4600 *(1/2)) = Rs. 2300
Option A |
AQUA-RAT | AQUA-RAT-36959 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
There are t members in a certain department, including Michael. Two representatives are to be selected to attend a company conference. If there are 55 possible combinations in which Michael is not selected, what is the value of t? | [
"11",
"12",
"15",
"18"
] | B | Combinations of two persons, in which Michael was not selected = 55
number of ways two persons can be selected from m people = m*(m-1)/2
Let m be the number of people excluding Michael, then m*(m-1) = 110 => m=11
Thus, t = m + 1(Michael) = 12
option B |
AQUA-RAT | AQUA-RAT-36960 | Method 2: We subtract the probability that she is not poisoned from $1$.
The probability that the first cup she selects is not poisoned is $3/4$ since three of the four cups do not contain poison. If the first cup she selects is not poisoned, the probability that the second cup she selects is not poisoned is $2/3$ since two of the three remaining cups do not contain poison. If both of the first two cups she selects are not poisoned, the probability that the third cup she selects is also not poisoned is $1/2$ since one of the two remaining cups is not poisoned. Hence, the probability that she is not poisoned if she drinks three of the four cups is $$\Pr(\text{not poisoned}) = \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$$ Hence, the probability that she is poisoned is $$\Pr(\text{poisoned}) = 1 - \Pr(\text{not poisoned}) = 1 - \frac{1}{4} = \frac{3}{4}$$
Addendum: We can relate this method to the first method by using the hypergeometric distribution.
She is not poisoned if she selects all three cups which do not contain poison when selecting three of the four cups. Hence, the probability that she is not poisoned is $$\Pr(\text{not poisoned}) = \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = \frac{1}{4}$$ so the probability she is poisoned is $$\Pr(\text{poisoned}) = 1 - \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = 1 - \frac{1}{4} = \frac{3}{4}$$
Method 3: We calculate the probability that the person is poisoned by adding the probabilities that she is poisoned with the first cup, the second cup, and the third cup.
Let $P_k$ denote the event that she is poisoned with the $k$th cup.
The following is multiple choice question (with options) to answer.
There are 7 cups placed on a table; 4 are juice cups and the other 3 are tea cups. If 2 cups are to be selected at random from the 7 cups, what is the probability that at least one of the juice cup will be selected? | [
"6/7",
"2/7",
"3/7",
"4/7"
] | A | Total probability=7C2=21
4C2 +4C1*3C1=6+12=18
therefore the probability that at least one of the fashion magazines will be selected= 18/21=6/7
answer is A |
AQUA-RAT | AQUA-RAT-36961 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
The positive integers a and b leave remainders of 4 and 7, respectively, when divided by 9. What is the remainder when a − 2b is divided by 9? | [
"5",
"6",
"7",
"8"
] | D | a=9x+4
so a could be 4, 13,22, 31,40,49,58,67,76,85
b=9y + 7
so b could be 7,16,25,34,43
Selecting any random number for a and b such that a-2b is positive number,
say a=22, b=7 => 2b=14
Hence a-2b = 8 Hence remainder when divided by 9 =8
say a=67, b=25 => 2b=50
Hence a-2b = 17 Hence remainder when divided by 9 is again 8
say a=49, b=7 => 2b=14
Hence a-2b = 35 Hence remainder when divided by 9 is again 8
Answer is 8.
ANSWER:D |
AQUA-RAT | AQUA-RAT-36962 | • Wooow, nice answear man, I really appreciate that and it actually opened my eyes to this economics field. But, your billion dollars example just take into account 1 iteration, I too would choose $10^9$, because is guaranteed, and because, as you said, $10^9$ to $10^{10}$ is nothing in this order of magnitude. In my example, I want to take into account multiple operations, not just one; and favorable odds, not a coin flip. Nov 27, 2021 at 7:03
• As I said, I made a Python simulation, and using x = 100 (100 USD), in 1 iteration, 10 interations, 1,000,000 iterations (that took a while to compute, almost burned my computer hahaha), the income for the first option was getting closer to 148.02 USD and the second option to 115.19 USD. Nov 27, 2021 at 7:03
• @rookie Well, I believe your mathematics is correct, so I'm guessing your code is the one that is wrong. Could you provide the code? Nov 27, 2021 at 7:29
• Okay, I see the problem. I thought that when you "fail", you'll lose all the interest not just for that month, but all the interest for all the months. This is what your mathematics do. Nov 27, 2021 at 7:52
The following is multiple choice question (with options) to answer.
A certain sum becomes four times itself at simple interest in nine years. In how many years does it become ten times itself? | [
"27 years.",
"25 years.",
"29 years.",
"17 years."
] | A | Let the sum be Rs. x, then it becomes Rs. 4x in eight years Rs. 3x is the interest on x for eight years.
R = (100 * 3x)/(x * 9) = 100/3 %
If the sum becomes ten times itself, then interest is 9x.
The required time period = (100 * 9x)/(x * 100/3) = (100 * 9x * 3)/(x * 300) = 27 years.
ANSWER:A |
AQUA-RAT | AQUA-RAT-36963 | ### 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.
Three business people who wish to invest in a new company. Each person is willing to pay one third of the total investment. . After careful calculation, they realize that each of them would pay $ 7400 less if they could find two more equal investors. How much is the total investment in the new business. | [
"a)\t$ 64,000",
"b)\t$ 54,000",
"c)\t$ 55,500",
"d)\t$ 5,400"
] | C | Initially each invest in x. Hence total investment is 3x.
Total investment is also 5(x-7400).
3x = 5(x-7400)
x= 5*7400/2 = 18500
3x = 55500 and the answer is C. |
AQUA-RAT | AQUA-RAT-36964 | # In 30 boxes are 15 balls. Chance all balls in 10 or less boxes?
Question1: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. Afer i collected all 15 balls i put them randomly inside the boxes.
How much is the chance that all balls are in only 10 boxes or less?
Question2: I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. In two of the boxes i could find 3 balls. (So in one box has to be 2 balls and in the other seven boxes have to be 1 ball.) Afer i collected all 15 balls i put them randomly inside the boxes.
How much is the chance that i find in only 2 boxes 6 balls or more?
I wrote a c# programm and tried it 1 million times. My solution was: With a chance of 12,4694% all balls are in 10 boxes or less.
Random trials/Monte Carlo simulations are notoriously slow to converge, with an expected error inversely proportional to the square root of the number of trials.
In this case it is not hard (given a programming language that provides big integers) to do an exact count of cases. Effectively the outcomes are partitions of the 15 balls into some number of boxes (we have thirty boxes to work with, so at least half will be empty).
I wrote a Prolog program to do this (Amzi! Prolog has arbitrary precision integers built in), and got the following results:
$$Pr(\text{10 or fewer boxes occupied}) = \frac{59486359170743424000}{30^{14}} \approx 0.124371$$
$$Pr(\text{2 boxes hold 6 or more balls}) = \frac{30415369655816064000}{30^{14}} \approx 0.063591$$
The following is multiple choice question (with options) to answer.
A box contain the number of balls which is as much greater than 20 is less than 30. The number of balls is? | [
"40",
"25",
"30",
"50"
] | B | Answer
Let the number be x.
Then, x - 20 = 30- x
⇒ 2x = 30+ 20= 50
⇒ 2x = 50
∴ x = 25
Correct Option: B |
AQUA-RAT | AQUA-RAT-36965 | Fortunately that extra condition is automatically satisfied, so here you can use stars and bars without any annoying constraints.
• Indeed $(-x_1) + (-x_2) + (-x_3) = -4$ so adding $6 = 2+ 2+ 2$ to both sides gives $y_1 + y_2 + y_3 = 2$. Then the stars and bars give $\binom{2+2}{2} = 6$ again. Nice and short. Also $x_i \le 2 \leftrightarrow y_i \ge 0$ and $x_i \ge 0 \leftrightarrow y_i \le 2$. But this is indeed automatic. – Henno Brandsma Sep 9 '17 at 9:52
• Sorry, how did you get " =2 " ? – user478905 Sep 9 '17 at 9:52
• @user478905 I expanded it in the comment. – Henno Brandsma Sep 9 '17 at 9:55
The following is multiple choice question (with options) to answer.
If x is even and y is odd, then x-y | [
"Even whenever xy is odd",
"Odd",
"Odd whenever xy is even",
"Even"
] | B | x is even and y is odd, then x + y is
Let's take x=4 and y=7.
x+y=4+7=11.
x+y=11=odd.
Option: B |
AQUA-RAT | AQUA-RAT-36966 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The price of 2 saris and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ? | [
"Rs. 2200",
"Rs. 2300",
"Rs. 2400",
"Rs. 2480"
] | C | Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 .... (i)
and x + 6y = 1600 .... (ii)
Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. --- (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
----------------
4y = 800
----------------
Therefore, y = 200.
Now apply value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
C |
AQUA-RAT | AQUA-RAT-36967 | 9. Men's weights follow a normal distribution with a mean of 172 pounds and a standard deviation of 29 pounds.
1. What is the probability that a randomly selected man carrying a 20 lb bag collectively weighs more than 195 lbs.
2. If an airplane is full of 213 men (and no women or children), each with a 20 lb bag, what is the probability that the total weight is greater than 41535 lbs (the weight limit for the airplane)?
1. With the bag the mean weight is $\mu = 192$. The standard deviation remains the same. $z_{195} \doteq 0.1034$. So $P(z \gt 0.1034) \doteq 0.4588$
2. If the total weight is 41535 lbs, the average weight of the 213 men is 195 lbs. Central limit theorem applies. $\mu = 192$, $\displaystyle{\sigma = \frac{29}{\sqrt{213}} \doteq 1.9870}$. Thus $z_{195} = 2.1282$. So the probability of exceeding the weight limit is $P(z \gt 2.1282) \doteq 0.0167$.
The following is multiple choice question (with options) to answer.
The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : | [
"70 kg",
"72 kg",
"75 kg",
"80 kg"
] | C | Explanation:
A+B+C = 3×84=252
A+B+C+D= (4×80)=320
D = (320-252)=68 and E = (68+3)=71
Now, B+C+D+E = (4×79)=316
B+C+D=(316-71)=245 kg
So, A = (320-245)=75 kg
Correct Option: C |
AQUA-RAT | AQUA-RAT-36968 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is : | [
"124 m2",
"120 m2",
"108 m2",
"58 m2"
] | A | Area of the outer rectangle = 19 × 16 = 304 m2
Area of the inner rectangle = 15 × 12 = 180 m2
Required area = (304 – 180) = 124 m2
Answer A |
AQUA-RAT | AQUA-RAT-36969 | 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.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be? | [
"4",
"5",
"6",
"7"
] | A | Let 1 men's 1 day work = x and 1 boy's 1 day work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2
Solving these two equations, we get:
x = 1/100 and y = 1/200
(15 men + 20 boys)'s 1 day work = (15/100 + 20/200) = 1/4
15 men and 20 boys can do the work in 4 days.
Answer:A |
AQUA-RAT | AQUA-RAT-36970 | The answer is: 133,320.
List the $4! = 24$ permutations of the four digits
,. . and consider their sum.
. . $\begin{array}{cc}
&2468 \\ &2486 \\ &2648 \\ &2684 \\ &2846 \\ &2864 \\ &\vdots \\
&8246 \\ &8264 \\ &8426 \\ &8462 \\ &8624 \\ + & 8642 \\ \hline
\end{array}$
We find that each column has: six 2's, six 4's, six 6's, six 8's.
The total of each column is: . $6\!\cdot\!2 + 6\!\cdot\!4 + 6\!\cdot\!6 + 6\!\cdot\!8 \:=\:120$
Hence, the addition has the form:
. . $\begin{array}{cccccc}
&&& 1 & 2 & 0 \\ && 1 & 2 & 0 \\ & 1 & 2 & 0 \\ 1 & 2 & 0 \\ \hline 1 & 3 & 3 & 3 & 2 & 0 \end{array}$
Therefore, the sum is: . $133,\!320$
Edit: Plato beat me to it . . . *sigh*
.
The following is multiple choice question (with options) to answer.
How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits S equal to 28? | [
"14",
"24",
"28",
"48"
] | D | First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.
Start from the largest number possible 9874.
Then the next largest number possible is 9865.
After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.
So you have two solutions. Each can be rearranged in 4!=24 ways. So S=24+24=48.D |
AQUA-RAT | AQUA-RAT-36971 | # We roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by 6?
So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?
Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six?
• Almost a duplicate of this question. Using the same logic, the answer is easily seen to be $1/6$. – TonyK Nov 11 '15 at 15:33
Hint.
Roll $9$ times and let $x$ be the total.
For exactly one number $n\in\{1,2,3,4,5,6\}$ we will have $6 \mid (x+n)$ (i.e. $x+n$ is divisible by $6$).
The following is multiple choice question (with options) to answer.
Exactly three sides of a certain 10-sided die are red. What is the probability that Kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll? | [
"0.184",
"0.16",
"0.18",
"0.147"
] | D | Total no. of sides = 10
Sides that are red =3
Probability that the die lands with red side up = 3/10
Therefore, Probability that the die does not land with red side up = 1 - 3/10 = 7/10
Probability that Kumar rolls the die 3 times and the die lands with a red side up for the first time on the third roll
= (1st roll - non red face) X (2nd roll - non red face) X (3rd roll - red face)
= (7/10) X (7/10) X (3/10)
= 128/1000
= 0.147
D |
AQUA-RAT | AQUA-RAT-36972 | • Hmm. Okay, I'll show you everything I did, maybe you can point out my mistake: $$V=2\int_0^1 \pi ((y^{1/3})^2 - y^2)dy$$ Next, $$= 2\pi \int_0^1 y^{2/3}-y^2dy$$ And, $$2\pi({{y^{5\over3}}\over{5\over3}}-{{y^3}\over3}) ]_0^1$$ So, $$2\pi({3\over5}-{1\over3})-2\pi(0-0)$$ Equals, $${6\pi \over 5} - {2\pi \over 3}$$ – JustHeavy Oct 1 '17 at 19:35
The following is multiple choice question (with options) to answer.
The value of Underrot 32pi/4 + Underrot 25 is most nearly equal to which of the following integers? (Please refer to the picture below for the exact version of the question) | [
"6",
"35",
"41",
"30"
] | D | Under root (32pi/4) ~ 25
Under root (25) = 5
Hence, 25+5 = 30 is the approx answer. D is the correct answer. |
AQUA-RAT | AQUA-RAT-36973 | 5 miles per hour. g. It shows how three runners ran a 100-meter race. Distance, Time and Speed Challenge. Calculate the average speed (in meters/sec) if a golf cart runs 140 meters in 10 seconds What is the average speed (in miles per hour) of the car that traveled a …Distance, Time and Speed Challenge. Speed Distance Time questions are a classic topic in the GCSE Maths syllabus. An airplane travels 4362 km against the wind in 6 hours and 5322 km with the wind in the same amount of time. Distance is inter-linked to a number of other important mathematical concepts. The rate is the speed at which an object or person travels. Home; Welcome; Videos and Worksheets; Primary; 5-a-day. A car takes 4 hours to cover a distance, if it travels at a speed of 40 mph. Speed Distance Time GCSE Revision and Worksheets. Explanation: We are having time and speed given, so first we will calculate the distance. You can choose the types of word problems, the Speed distance time problems rely on three key formulas. Let's do some Time and Distance problems, in case you face any difficulty then post a comment in the comments section. Speed = Distance/time = 15/2 = 7. Speed, Velocity and Acceleration Calculations Worksheet Calculate the speed for a car that went a distance of 125 miles in 2 hours time. To find Speed when Distance and Time are given. Interpret this relationship. 5 miles per hour. 0 meters across. For the first part of the trip, the average speed was 105 mph. 4MQuiz & Worksheet - Distance, Time & Speed | Study. Average speed for the entire trip is 10. ) I multiply the rate by the time to get the values for the distance column. The problem will have something to do with objects moving at a constant rate of speed or an average rate of speed. Solution to Problem 3: After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by D1 = 72 t and D2 = 78 t After t hours total distance D traveled by the two trains is given by D = D1 + D2 = 72 t …Solve for speed, distance, time and rate with formulas s=d/t, d=st, d=rt, t=d/s. He would have gained 2
The following is multiple choice question (with options) to answer.
A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour? | [
"3.6",
"7.2",
"8.4",
"10"
] | B | Speed =(600/5*60)m/sec.
= 2 m/sec.
Converting m/sec to km/hr
=(2*18)/5 km/hr
=7.2 km/hr.
Answer:B |
AQUA-RAT | AQUA-RAT-36974 | # Simple and Compound Interest Problem
• January 14th 2011, 01:41 AM
dumluck
Simple and Compound Interest Problem
Hi All,
Q:Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received$605 as interest. What was the value of his total savings before investing in these two bonds?
1. $5500 2.$ 11000
3. $22000 4.$ 2750
5. $44000 Answer Explanation... 1. Interest for the first year of the simple compound bond is 275/2 -$275.
2. So we need to determine the rate of interest based on this so...
605 - 550 = 55. That's the difference between the interest earned on the simple vs compound interest bonds.
55/275 * 100/1 = 11/55 * 100/1 = 20% Interest
3. 275 represents 20% interest of a number
275/20 * 100/1 = 55/4 * 100/1 = $1375. 4. This represents half the money so 1375*2 =$2750. (D).
My questions is: Why are we using 55. I.E. The difference between the two interest to determine the interest in 2. What does this 55 represent (besides the difference between the two?)
• January 14th 2011, 09:20 AM
Soroban
Hello, dumluck!
I'm not impressed with their explanation.
Quote:
Q: Shawn invested one half of his savings in a bond
that paid simple interest for 2 years and received $550 as interest. He invested the remaining in a bond that paid compound interest, compounded annually, for the same 2 years at the same rate of interest and received$605 as interest.
What was the value of his total savings before investing in these two bonds?
. . $1.\;\5500 \qquad 2.\;\11000 \qquad 3.\;\22000 \qquad 4.\;\2750 \qquad 5.\;\44000$
Let $\,r$ be the annual interest rate for both accounts.
Let $\,P$ be the amount invested in each account.
The following is multiple choice question (with options) to answer.
At what rate percent on simple interest will a sum of money double itself in 30 years? | [
"3 1/3%",
"8 1/3%",
"3 5/3%",
"5 1/3%"
] | A | P = (P*30*R)/100
R = 3 1/3%
Answer: A |
AQUA-RAT | AQUA-RAT-36975 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
A small company employs 3 men and 5 women. If a team of 3 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women? | [
" 1/14",
" 1/7",
" 2/7",
" 5/7"
] | D | Total ways to choose 4 employees from 8 employees = 8C3
Two women = 5C2
Two Men = 3C1
P = (5C2 * 3C1)/8C2 = 5/7 hence D. |
AQUA-RAT | AQUA-RAT-36976 | 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 speeds past a pole in 15 sec and a platform 120 m long in 25 sec, its length is? | [
"238",
"180",
"988",
"177"
] | B | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 120)/25 = x/15 => x = 180 m.
Answer: B |
AQUA-RAT | AQUA-RAT-36977 | ## 1 Answer
Case 2:
3 < x < 6
|x-3|= (x-3)
|x-6|= -(x-6)
(x-3)-(x-6)<5
3<5
3<5 is true, so 3 < x < 6 is also solution for |x − 3| + |x − 6| < 5
Finally we have |x − 3| + |x − 6| < 5 $\Leftrightarrow$ 2 < x < 7
The following is multiple choice question (with options) to answer.
If 6x^2 + x - 12 = (ax + b)(cx + d), then |a| + |b| + |c| + |d| | [
"10",
"12",
"15",
"18"
] | B | This is the hard one, definitely a 700+ level question. We need numbers a, b, c, and d such that
6x^2 + x - 12 = (ax + b)(cx + d)
This means that ac = 6, bd = –12, and ad + bc = 1. The a & c pair could be (1, 6) or (2, 3), in some order. The absolute values of the b & d pair could be (1, 12) or (2, 6) or (3, 4), and of course, in each case, one of the two would have to be negative. After some trial and error, we find:
6x^2 + x - 12 = (2x + 3)(3x - 4)
Thus, we see:
|a| + |b| + |c| + |d| = 2 + 3 + 3 + 4 = 12
Answer = B |
AQUA-RAT | AQUA-RAT-36978 | # Algebra
posted by Angela
If 3 is subtracted from the numerator of a fraction, the value of the fraction is 1/2. If 6 is added to the denominator of the ORIGINAL fraction, the value of the fraction is 1/2. What is the original fraction?
1. Damon
(x-3)/y = .5
x/(y+6) = .5
-----------------
.5 y = x - 3
.5 y + 3 = x
these two equations are the same, choose any old x (infinite number of solutions)
if x = 13
.5 y = 10 so y = 20
x/y = 13/20
check that
(13 -3)/20 = 1/2 sure enough
(13)/(26) = 1/2 sure enough
2. Angela
OK, THANK YOU. But I still do not understand why x=13. that is exactly what my text uses, as well. How did you know to use 13 for x?
3. Angela
I just reworked it using your method and subsituted as you did. Then I used x=15 and got 15/24 for my final answer. However text says the correct answer is 13/20, does not say infinite number. I am thinking this is a book error! Anyway, thank you!
4. Damon
I just used 13 because I wanted easy numbers
trying x = 15 now
.5 y = 12
y = 24
so
15/24 yes
15-3/24 = 1/2 yes
15/30 = 1/2 yes
so 15/24 is just as good as 13/20
## Similar Questions
1. ### algebra
The denominator of a fraction is 12 more than the numerator. If 16 is added to the numerator and 16 is subtracted from the denominator, the value of the resulting fraction is equal to 2/1. Find the original fraction
2. ### Maths
When the numerator and denominator of a fraction are each increased by 5, the value of the fraction becomes 3/5 When the numerator and denominator of that same fraction are each decreased by 5, the fraction is then 1/5 Find the original …
3. ### Algebra
The following is multiple choice question (with options) to answer.
The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction? | [
"3/15",
"3/10",
"3/11",
"3/12"
] | C | The denominator be P, the numerator will be (P - 8).
The fraction will be (P - 8)/P.
Adding 3 to the numerator and subtracting 3 from the denominator, (P - 8 + 3)/(P - 3) = 3/4.
(P - 5)/(P - 3) = 3/4
P = 20 - 9 => P = 11.
The fraction is: 3/11.Answer: C |
AQUA-RAT | AQUA-RAT-36979 | A and B are called independent if P(A|B)=P(A), i. Event B: The sum of the two numbers being odd. If the events A and B are independent, then P(A ∩ B) = P(A)P(B) and not necessarily 0. 𝑃 = P(A) and P(B|A)=P(B), if A and B are two independent events. Event #A# is "the sum of two dice is #7#" and Event #B# is "at least one die is #6#". Events A and B are independent events if the probability of Event B occurring is the same whether or not Event A occurs. For two independent events A and B, the probability that both A & B occur is 1 / 8 and the probability that neither of them occur is 3 / 8. The probability of occurrence of A may be 1 Verified Answer. Tossing two dice is a compound event. New York City, New York. (a) The probability that the next two babies are girls is (b) The probability that at least one baby is a boy is. P (A ∩ B c) is a. 08) which is 0. In other words, knowing that E occurred does not give any additional information about whether F will or will not occur; knowing that F occurred does not give any additional information about the occurance of E. 1/2 times 1/2 is 1/4. Answer to (10pts) Suppose A and B are two independent random events (means or , if and , find and. We can see here that two different events…will win this contest for you. In other words, the product of probability of events A and B equals to probability of intersection of events A and B. Prove that if events A and B are independent, then the complement events of A and B are also independent. Are these events independent? A fair coin is tossed two times. Problem 98SE from Chapter 3: Two events, A and B, are independent, with P(A) =. Two events are said to be independent if the occurrence of one has no effect on the probability of the occurrence of the other. Independent Documentary Films. More specifically, if A and B are independent events, then P(A and B) = P(A)P(B). Independent events---Events A and B are said to be independent events
The following is multiple choice question (with options) to answer.
The events A and B are independent, the probability that event A occurs is greater than 0, and the probability that event A occurs is twice the probability that event B occurs. The probability that at least one of events A and B occurs is 3 times the probability that both events A and B occur. What is the probability that event A occurs? | [
"44/7",
"9/8",
"25/8",
"3/4"
] | D | Let us say probability of A occuring is a.
Let us say probability of B occuring is b.
a = 2b
Probability (either A or B or both) = 3 times Probability (A and B)
a*(1-b) + b*(1-a) + ab = 3*ab
Substituting a=2b in the second equation:
2b*(1-b) + b*(1-2b) + 2b*b = 3*2b*b
3b-2b^2 = 6b^2
3b = 8b^2
b = 3/8 = 3/8
So, a = 2b = 3/4
THE CORRECT ANSWER IS D. |
AQUA-RAT | AQUA-RAT-36980 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?
I. 15.19
II.18.16
III. 17.17 | [
" I only",
" II only",
" III only",
" I and II only"
] | B | Problem statement was difficult to parse. but some how I managed to understand what we can do.
1) x/y = 15 + 0.19
2) x/y = 18 + 0.16
3) x/y = 17 + 0.17
w.k.t x = yn (i.e Q) + yk=60 (i.e remainder)
therefore :
1) y*0.19=60 -- we can't get y an integer.
2) y*0.16=60 -- we can get y an integer.
3) y*0.17=60 -- we can't get y an integer.
Therefore 2 are possible answers.
Ans B. |
AQUA-RAT | AQUA-RAT-36981 | # 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 111 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 40 km and the other at the rate of 50 kmph. In what time will they be completely clear of each other from the moment they meet? | [
"11.04",
"10.0",
"6.85",
"5.85"
] | A | T = (111 + 165)/ (40 + 50) * 18/5
T = 11.04
ANSWER:A |
AQUA-RAT | AQUA-RAT-36982 | gears, mechanisms, homework
Title: Conical pulley system and transmission ratio (closed) I tried asking this on other SE forum but i think this is the correct one:
Basically our teacher has sent us quite a lot of exercises but I can't get past this one which says: a pulley cone consists of 3 pulleys of 150, 250 and 350 mm of diameter that link an identical but inverted cone. Determine the three possible transmission ratios. Our teacher doesn't explain very well and I don't know where to search. If someone knows something about this, anything would be helpful.
PD: if you think I didn't tried to solve it, in all the book there's no information about this type of mechanism and we don't see our teacher until the day we need to bring the homework. There's no more info. in the exercise sorry.
(ignore, i messed up the question here)Edit: by the way what I'm trying to look for here is for a formula about the transmission ratio, but I can't understand how to do the formula with this specific mechanism because I don't see how to get the speed of the pulleys, here's an example given by the book.
Edit 2: i finally came up with this answer:
This question defines a system of adjustable speed pulleys, of this type of mechanisms what we can see as a ratio is that the bigger the gear we choose the pulley will go slower, this means that if we choose the gears that are parallel to each other the speed will always be the same because we will always have gears that are proportional to each other.
350mm -> 150mm
250mm -> 250mm
150mm -> 350mm Your question is describing an adjustable speed pulley system. These are commonly found in bench drill presses. (See the link below.)
Figure 1. A 5-speed adjustable speed pulley system. Image source: Toolbox Buzz.
Since the centres of the shafts are fixed and the length of the belt is fixed this arrangement requires that the path of the belt on each "level" of the system must be a constant (because the belt isn't elastic). Therefore, as you increase the pulley circumference on one shaft you must decrease it on the other to keep the path length constant.
Question to get you thinking:
The following is multiple choice question (with options) to answer.
An automobile manufacturer offers a station wagon with either a 14-cylinder engine or a 12-cylinder engine and with either a manual transmission or an automatic transmission. A trailer hitch is also offered, but only on a station wagon with a 14-cylinder engine. How many combinations of the five options listed does the manufacturer offer for its station wagon? | [
" 93",
" 94",
" 95",
" 96"
] | D | 14-cylinder engine wagons = 2*2*2*2*2*2= 64 (manual or automatic, with or without trailer);
12-cylinder engine wagons = 2*2*2*2*2 = 32 (manual or automatic).
Total = 64 + 32 = 96.
Answer: D. |
AQUA-RAT | AQUA-RAT-36983 | 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.
On increasing the price of T.V. sets by 50%, their sale decreases by 20%. What is the effect on the revenue receipts of the shop ? | [
"4",
"5",
"56",
"20"
] | D | Explanation:
Let the price be = Rs.100, and number of units sold = 100
Then, sale value = Rs.(100 × 100) = Rs.10000
New sale value = Rs.(150 × 80) = Rs.12000
Increase% = 2000/10000 × 100 = 20%
Answer: D |
AQUA-RAT | AQUA-RAT-36984 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A trader has 200 kg of sugar, out of which he sells some with 10% profit and the rest with 25% profit. He gains 15% on the whole. Find the quantity of sugar sold at 25% profit? | [
"120 kg",
"96.5 kg",
"84 kg",
"67 kg"
] | D | Explanation:
Let CP of each kg. of sugar = Rs.1.
Cost Price of 200 kg of sugar = Rs.200.
Let quantity of sugar sold with 25% profit be X.
Therefore, (200 – X) kg of sugar sold at a profit of 10%
Then,
(125/100 * X) + 110/100 (200 – X) = 115/100 * 200
125X + 110(200 – X) = 115 * 200
125X + 22000 – 110X = 23000
125X – 110X = 23000 – 22000
15X = 1000
X = 1000/15
X = 66.66 kg
Quantity of sugar that was sold at 25% gain is 66.66 kg
Therefore, option nearby for 66.66 is 67 (d)
ANSWER: D |
AQUA-RAT | AQUA-RAT-36985 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
By selling 15 pencils for a rupee a man loses 15%. How many for a rupee should he sell in order to gain 15%? | [
"8",
"9",
"11",
"89"
] | C | 85% --- 15
115% --- ?
15/115 * 15 = 11
Answer: C |
AQUA-RAT | AQUA-RAT-36986 | As an example, let's see how many trailing zeros $$16!$$ (expressed in the usual base $$10$$) has in base $$b = 12$$. We have that $$b = 2^2\cdot 3$$, so $$p_1 = 2$$, $$p_2 = 3$$, and $$e_1 = 2$$, $$e_2 = 1$$. Now, we can compute $$m_1 = \left\lfloor \frac{1}{2}\sum_{s=1}^\infty \left\lfloor\frac{16}{2^s}\right\rfloor\right\rfloor = \left\lfloor \frac{8+4+2+1}{2}\right\rfloor = 7$$ $$m_2 = \left\lfloor \frac{1}{1}\sum_{s=1}^\infty \left\lfloor\frac{16}{3^s}\right\rfloor\right\rfloor = 5 + 1 = 6$$ so $$16!$$ (again, in decimal) has $$m = \min\{7,6\} = 6$$ trailing zeros when expressed in base $$12$$. We can check this in WolframAlpha, which tells us that $$(16!)_{10} = 241ab88000000_{12}$$ indeed has $$6$$ trailing zeros.
The following is multiple choice question (with options) to answer.
How many terminating zeroes T does 200! have? | [
" 40",
" 48",
" 49",
" 55"
] | C | You have 40 multiples of 5, 8 of 25 and 1 of 125. This will give 49 zeros.C |
AQUA-RAT | AQUA-RAT-36987 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them? | [
"44, 36",
"42, 33",
"43, 36",
"42, 36"
] | B | Let the marks secured by them be x and (x + 9)
Then sum of their marks = x + (x + 9) = 2x + 9
Given that (x + 9) was 56% of the sum of their marks
=>(x+9) = 56/100(2x+9)
=>(x+9) = 14/25(2x+9)
=> 25x + 225 = 28x + 126
=> 3x = 99
=> x = 33
Then (x + 9) = 33 + 9 = 42
Hence their marks are 33 and 42
B |
AQUA-RAT | AQUA-RAT-36988 | This has been offset by payments, whose value at the time of the $$(k)$$-th payment is
$$\displaystyle P[(1+j)^{k-1} + (1+j)^{k-2} + \cdots + 1] = P\left[\frac{(1+j)^k - 1}{(1 + j) - 1}\right] = P\left[\frac{(1+j)^k - 1}{j}\right].$$
This means that the loan balance, immediately after your $$(k)$$-th payment is
$$\displaystyle L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right].$$
During the period between the payments $$(k)$$ and $$(k+1)$$, the interest on this loan balance is
$$\displaystyle \left\{L(1 + j)^k - P\left[\frac{(1+j)^k - 1}{j}\right]\right\} \times j$$
$$\displaystyle = \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}.$$
Therefore, the principal reduction for payment $$(k+1)$$ is
$$\displaystyle P - \left\{L(1 + j)^k j - P\left[(1+j)^k - 1\right]\right\}$$
$$\displaystyle = P - L(1 + j)^k j + P\left[(1+j)^k - 1\right]$$
Using equation (1) above, this equals
$$\displaystyle = P - P \left[\frac{(1 + j)^{100} - 1}{j(1 + j)^{100}} \right](1 + j)^k j + P\left[(1+j)^k - 1\right]$$
The following is multiple choice question (with options) to answer.
A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was: | [
"Rs. 2000",
"Rs. 10,000",
"Rs. 15,000",
"Rs. 20,000"
] | C | Principal = Rs.(100 x 5400)/(12x3) = Rs. 15000.
Answer: Option C |
AQUA-RAT | AQUA-RAT-36989 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipes A and B can fill a tank in 4 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in? | [
"3 hrs",
"3 9/77 hrs",
"3 9/17 hrs",
"3 9/27 hrs"
] | A | Net part filled in 1 hour = 1/4 + 1/6 - 1/12
= 4/12 = 1/3
The tank will be full in 3 hrs
Answer:A |
AQUA-RAT | AQUA-RAT-36990 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
P is three times as fast as Q and working together, they can complete a work in 20 days. In how many days can Q alone complete the work? | [
"26 1/2 days",
"27 1/2days",
"29 3/4days",
"14 1/4 days"
] | A | P = 3Q
P + Q = 3Q + Q = 4Q
These 4Q people can do the work in 20 days, which means Q can do the work in 80 days.
Hence, P can do the work in 26 1/2 days.
Answer:A |
AQUA-RAT | AQUA-RAT-36991 | ### Which sequences will contain the number 1000 Charlie's way
If you look it at Charlie's way instead you can see that carrying on from where he started A2's sequence is the four times table just skipping out 1 number each time because it goes up in 8's so if you do 1000 divided by 4 =250 or divided by 8 just to make sure it's not a number that is skipped out it would =125 so A2=YES. A3's sequence is the 8 times table but skips out a numbers each time eg: 8-24(skips 16) because the sequence adds 16 each time so what you would do is 1000 divided by 8 = 125 = YES A4 is the 16 times table missing out a number each time so you would do 1000 divided by 16 which equals 62.5= NO A5 is the 32 times table missing out 1 each time so 1000 divided by 32 =31.25= NO A6 is the 64 times table missing out a number each time so do 1000 divided by 64 which equals 15.625=NO so out of the sequences shown only A2 and A3 have the number 1000
### Which sequences will contain the number 1000 Alison's way
If you do it Alison's way there is a bit of counting so if you think about what Alison said, each number is double the number in the row above.
To get to 1000 each sequence has to have the sequence above containing 500, which is less counting, and to make it even easier the row before that one has to have 250, and the one before that 125.
This way you can use the doubles as an advantage and do a lot less counting.
### A2 sequence
I was wrong on all three ways for sequence A2 as it does not have 1000 in its sequence as with Alison's way a1 doesn't reach 500 but 498 or 502 and with Bernard's way and Charlie's way it would equal 996 or 1004
### Persevering
Hi Sofia, well done for sticking at it with this problem even though you made a mistake initially, it's great to see you keeping going until you get the right answer!
### Math
The following is multiple choice question (with options) to answer.
How many 3's are there in the following sequence which are neither preceded by 6 nor immediately followed by 9 ?
9 3 6 6 3 9 5 9 3 7 8 9 1 6 3 9 6 3 9 | [
"1",
"2",
"3",
"4"
] | D | D
9 3 6 6 3 95 9 3 78 9 16 3 9 6 3 9 |
AQUA-RAT | AQUA-RAT-36992 | # Find the remainder when $10^{400}$ is divided by 199?
I am trying to solve a problem
Find the remainder when the $10^{400}$ is divided by 199?
I tried it by breaking $10^{400}$ to $1000^{133}*10$ .
And when 1000 is divided by 199 remainder is 5.
So finally we have to find a remainder of :
$5^{133}*10$
But from here I could not find anything so that it can be reduced to smaller numbers.
How can I achieve this?
Is there is any special defined way to solve this type of problem where denominator is a big prime number?
• $10^{400}=1000^{133}\times10$, not $1000^{333}\times10$. – Gerry Myerson Oct 9 '12 at 4:53
• A standard beginning (for prime moduli) is to use the fact that if $p$ does not divide $a$, then $a^{p-1}\equiv 1\pmod{p}$. Thus $10^{198}\equiv 1\pmod{199}$. It follows that $10^{396}\equiv 1\pmod{199}$ and therefore $10^{400}\equiv 10^4\pmod{199}$. Now we have to calculate. In this case, there is a further shortcut, since $1000=(5)(199)+5\equiv 5\pmod{199}$. – André Nicolas Oct 9 '12 at 5:24
• – Martin Sleziak Jun 17 '16 at 8:22
You can use Fermat's little theorem. It states that if $n$ is prime then $a^n$ has the same remainder as $a$ when divided by $n$.
So, $10^{400} = 10^2 (10^{199})^2$. Since $10^{199}$ has remainder $10$ when divided by $199$, the remainder is therefore the same as the remainder of $10^4$ when divided by $199$. $10^4 = 10000 = 50*199 + 50$, so the remainder is $50$.
The following is multiple choice question (with options) to answer.
Which of the following is closest to (-9/10)^199？ | [
"-1",
"-1/2",
"0",
"1"
] | C | (-9/10)^4 = 6561/10,000 which is already less than 2/3.
For larger exponents, the expression will get closer and closer to zero.
The answer is C. |
AQUA-RAT | AQUA-RAT-36993 | • Nice approach. +1. – N.S.JOHN Sep 24 '16 at 3:26
• And are you really just 12? – N.S.JOHN Sep 24 '16 at 3:26
• @N.S.JOHN Yes, I am twelve. Thank you for your appreciation. – астон вілла олоф мэллбэрг Sep 24 '16 at 3:28
The following is multiple choice question (with options) to answer.
"How old are you, Alchemerion?" asked one of the wizards appearances the wizards answer with a riddle, "I am still Very young as wizards go.I am only three times my son's age. My father is 40 year more than twice of my age. Together the three of us are a 1240 year old". How old is Alchemerion | [
"300",
"320",
"340",
"360"
] | D | let Alchemerion be A
let son be S
let father be F
A= 3* S
F= 40 +(2*A)
therefore 1240=(S+F+A)
1240=(A/3)+ (40+(2*A) )+ A
A=360
ANSWER:D |
AQUA-RAT | AQUA-RAT-36994 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is | [
"35",
"36 2/3",
"37 1/2",
"40"
] | D | Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get: x = 40.
Answer : D. |
AQUA-RAT | AQUA-RAT-36995 | • In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52
The following is multiple choice question (with options) to answer.
What is the height of the tower ? A man standing at a distance of 1km from the bottom of the tower makes an angle 300 degrees with the top of the tower .An insect starts from the bottom of the tower and reaches the top in 25sec. | [
"0.577km",
"0.477km",
"0.677km",
"0.577k7"
] | A | tan = opp side/adj side
tan 30= height/1
ht=0.577km
ANSWER:A |
AQUA-RAT | AQUA-RAT-36996 | . Sum of the First n Terms of an Arithmetic Sequence Suppose a sequence of numbers is arithmetic (that is, it increases or decreases by a constant amount each term), and you want to find the sum of the first n terms. Sum of First 50 Odd Numbers; Sum of First 50 Even Numbers; How to Find Sum of First 50 Natural Numbers? Example: Sum 1 to 100 = 100 * 101 / 2 = 50 * 101 = 5050. Are all integers even? Though both programs are technically correct, it is better to use for loop in this case. 1729 = 1 3 + 12 3 = 10 3 + 9 3. Problem 44. . . The sum of the integers from 1 to 25 is as follows: 325 To get the answer above, you could add up all the digits like 1+2+3... +25, but there is a much easier way to do it! Consecutive integers are numbers that follow each other in order. . In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100. . 200 xx (1+200)/2 = 20100 The sum of a finite arithmetic sequence is equal to the count of the number of terms multiplied by the average of the first and last terms. Put them in pairs: 1 & 500 2 & 499 3 & 498 etc There are 250 pairs, each pair = 501 250*501 = 125250-----How about only the odds: 1+3+5+7+...493+495+497+499? Therefore, 62500 is the sum of first 250 odd numbers. . Let there be n terms in this A.P. The sum of the odd numbers (from 1) up to to 500 is 62500. What is the sum of first 500 odd numbers? This is an A.P. The sum of 1+2+3+4+...496+497+498+499+500? Find the number and sum of all integer between 100 and 200, divisible by 9: ----- Numbers between 100 and 200, divisible by 9: 108 117 126 135 144 153 162 171 180 189 198 The sum : 1683 Flowchart: C++ Code Editor: Contribute your code and comments through Disqus. step 2 apply the input parameter values in the formulaSum =
The following is multiple choice question (with options) to answer.
Find the sum of first 35 natural numbers | [
"470",
"468",
"630",
"463"
] | C | Explanation:
Sum of n natural numbers
=n(n+1)/2
=35(35+1)/2=35(36)/2=630
Answer: Option C |
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