source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39097 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
A certain article of clothing was discounted during a special sale to 4/5 of its original retail price. When the clothing didn't sell, it was discounted even further to 1/2 of its original retail price during a second sale. By what percent did the price of this article of clothing decrease from the first sale to the second sale? | [
"50%",
"33.33%",
"40%",
"16.66%"
] | C | Say the original retail price of the item was $200.
The price after the first sale = 4/5 * $200 = $160.
The price after the second sale = 1/2 * $200 = $100.
The percent change from the first sale to the second = (160 - 100) / 150 = 1/3 = 40%.
Answer: C. |
AQUA-RAT | AQUA-RAT-39098 | 10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?
11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.
12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?
13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.
14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.
15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.
16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.
17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.
18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.
19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.
The following is multiple choice question (with options) to answer.
The height of the wall is 6 times its width and length of the wall is 7 times its height .if the volume of the wall be 86436 cu.m.its width is | [
"4m",
"5m",
"6m",
"7m"
] | D | Explanation:
Let width = x
Then, height=6x and length=42x
42x × 6x × x = 86436
x = 7
Answer: D |
AQUA-RAT | AQUA-RAT-39099 | Now, $30N= 2^{p+1}(3^{q+1})(5^{r+1})$ and that must be divisible by $2^3(3^2)$ so p+ 1 must be larger than or equal to 3- p must be larger than or equal to 2- and q+ 1 must be larger than or equal to 2: q must be larger than or equal to 1.
$72N= 2^{p+3}3^{q+ 1}5^r$ and that must be divisible by 2(3)(5) so p+3 must be larger than or equal to 1, which is neccesarily true since 0+ 3> 1, q+ 1 must be larger than 1, which, again, is necessarily true, and r must be larger than or equal to 1.
So we have: $N= 2^p(3^q)(5^r)$ where p, q, and r are non-negative integers satisfying p larger than or equal to 2, q larger than or equal to 1, and r larger than or equal to 1.
We find the smallest possible value of k by taking p, q, and r to be the smallest possible.
5. May 17, 2010
### Fragment
Thank you very much for the help, it cleared up my confusion. I understand a little more now.
-F
The following is multiple choice question (with options) to answer.
If m = 3^n, what is the greatest value of n for which m is a factor of 31! | [
"8",
"10",
"12",
"14"
] | D | Solution-
Consider multiples of 25!=> 3,6,9,12,15,18,21,24,27,30
Count no. of 3 in each multiple.
3=3x1->1
6=3x2->1
9=3x3->2
12=3x4->1
15=3x5->1
18=3x3x2->2
21=3x7->1
24=3x8->1
27=3x3x3->3
30=3x10->1
---- count 3's =14 so answer is 14
Answer : D |
AQUA-RAT | AQUA-RAT-39100 | Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
Algebraic Solution: If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/B *100 = 100T/(NB)
That Algebraic solution was elegant if you saw it, but if not, here’s a full solution with picking numbers.
The following is multiple choice question (with options) to answer.
A watch was sold at a loss of 12%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price? | [
"875",
"2876",
"1977",
"2778"
] | A | 88%
104%
--------
16% ---- 140
100% ---- ? => Rs.875
Answer: A |
AQUA-RAT | AQUA-RAT-39101 | # 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.
On a sum of money, the simple interest for 2 years is Rs. 330, while the compound interest is Rs. 340, the rate of interest being the same in both the cases. The rate of interest is | [
"15%",
"14.25%",
"6.06%",
"10.5%"
] | C | Explanation:
The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum
= (R × SI)/(2 × 100)
Difference between the compound interest and simple interest = 340 - 330 = 10
(R × SI)/(2 × 100) = 10
(R × 330)/(2 × 100) = 10
R = 6.06%
Answer: Option C |
AQUA-RAT | AQUA-RAT-39102 | arithmetic, boolean-algebra, integers
Prove that for $A, B \in \mathbb{Z}$, $A + B$ $= (A\operatorname{\&}B) + (A \mid B)$ $= (A \oplus B) + 2(A\operatorname{\&}B)$ where $\operatorname{\&}$ is bitwise AND, $\mid$ is bitwise OR and $\oplus$ is bitwise XOR. (It is reasonable to assume that the bit representation of integers is two's complement.)
You have interpreted item 23 so that it is almost correct.
If we allow $A$ or $B$ to be negative, however, it is hard/impossible to assign/determine the sign of $A\operatorname{\&}B$, $A\mid B$ or $A\oplus B$ if we use the natural/standard representation of a number in base $2$. A different method that can represent signed integers without using the sign $\pm$ has to come into play. You selected two's complement. Then means you should also specify the length of a binary representation as well as restrict the magnitude of $A$ and $B$ to avoid overflow. That kind of specification and restriction is unlikely to be the intention of Rich Schroeppel, the author of this standalone item.
On the other hand, the equalities hold for all integers if integers are represented by two's complement with infinite length, as mentioned by hobbs and the Wikipedia article Two's complement and 2-adic numbers.
Let $A$ and $B$ be nonnegative integers
Here is a simpler way to interpret item 23. The only change is $A$ and $B$ are natural numbers (including $0$) instead.
For all $A, B \in \Bbb N$, $\ A + B$ $= (A\operatorname{\&}B) + (A\mid B)$ $= (A \oplus B) + 2(A\operatorname{\&}B)$ where $\operatorname{\&}$ is bitwise AND, $\mid $ is bitwise OR and $\oplus$ is bitwise XOR.
The following is multiple choice question (with options) to answer.
A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B. Which of the following can be the difference of A and B ? | [
"73",
"44",
"27",
"21"
] | C | A and B are positive two-digit integers such that the tens digit in A is the same as the units digit in B, and the units digit in A is the same as the tens digit in B:
A = 10x + y
B = 10y + x
A - B = (10x + y) - (10y + x) = 9(x - y). As you can see the difference must be a multiple of 9. Only option C is a multiple of 9.
Answer: C. |
AQUA-RAT | AQUA-RAT-39103 | ### Show Tags
04 Aug 2019, 03:54
generis wrote:
Ayush1692 wrote:
If y is a positive integer, and |x| < 5 − y, then what is the least possible value of x ?
A. 4
B. 1
C. 0
D. -1
E. -4
Don't remove brackets and solve. The question asks for the logic behind absolute value.
$$y$$ is a positive integer
$$|x| < 5 − y$$
LHS is nonnegative - it is positive or 0.
Least value for x is ZERO
Does RHS work?
For LHS to be less than RHS:
RHS cannot be negative
RHS cannot be 0
RHS must be positive
RHS = (5 - pos. integer)
y could = 4, 3, 2, or 1, and
RHS can = (5 - y) = 1, 2, 3, 4
That works.
The least possible value of |x| = 0
The absolute value of 0 is 0
Or: the distance of 0 from 0 is 0
Least possible value: x = 0
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Why can't x be a negative integer, in that case it should be -1 i.e. option D.
Intern
Joined: 21 Mar 2019
Posts: 22
Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
08 Aug 2019, 18:54
I don't understand why x can't be negative
Intern
Joined: 06 Nov 2014
Posts: 17
Location: Viet Nam
GMAT 1: 720 Q50 V36
GMAT 2: 740 Q50 V40
GPA: 3.67
Re: If y is a positive integer, and |x| < 5 − y, then what is the least po [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
y = 2x^2 + bx + 256 cuts the x axis at (h, 0) and (k, 0). If h and k are integers, what is the least value of b? | [
"-120.5",
"-124.5",
"-128.5",
"-132.5"
] | C | 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 of ax^2+bx+c=0,
The product of the roots =c/a= 256/2=128 and the sum of the roots =-b/a=-b
128 can be expressed as product of two numbers in the following ways:
1 * 256
2 * 128
4 * 64
8 * 32
16 * 16
The sum of the roots is maximum when the roots are 1/2 and 128 and the maximum sum is 1/2 + 128 = 128.5
The least value possible for b is therefore -128.5.
C |
AQUA-RAT | AQUA-RAT-39104 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
Alice leaves her home and walks toward Bob's house. One hour later, Bob leaves his home and walks toward Alice's house. The distance between their homes is 41 kilometers, Alice's walking speed is 5 km/h, and Bob's walking speed is 4 km/h. How many kilometers will Alice walk before she meets Bob? | [
"22",
"25",
"27",
"38"
] | B | Alice walks 5 km in the first hour so there are 36 km remaining.
When Bob starts walking, they complete a total of 9 km per hour.
They will meet four hours after Bob starts walking.
Since Alice walks for 5 hours, she walks 25 km.
The answer is B. |
AQUA-RAT | AQUA-RAT-39105 | ### Question 3
Multiple choice!
The decimal $0.23$ equals:
(A) $\dfrac {23}{10}$
(B) $\dfrac {23}{100}$
(C) $\dfrac {23}{1000}$
(D) $\dfrac {23}{10000}$
### Question 4
The decimal $0.0409$ equals:
(A) $\dfrac {409}{100}$
(B) $\dfrac {409}{1000}$
(C) $\dfrac {409}{10000}$
(D) $\dfrac {409}{100000}$
## SIMPLE FRACTIONS AS DECIMALS
Some decimals give fractions that simplify further.
For example,
$0.5 = \dfrac{5}{10} = \dfrac{1}{2}$
and
$0.04 = \dfrac{4}{100} = \dfrac{1}{25}$.
Conversely, if a fraction can be rewritten to have a denominator that is a power of ten, then we can easily write it as a decimal.
For example,
$\dfrac{3}{5} = \dfrac{6}{10}$ and so $\dfrac{3}{5} = 0.6$
and
$\dfrac{13}{20} = \dfrac{13 \times 5}{20 \times 5} = \dfrac{65}{100} = 0.65$.
What fractions (in simplest terms) do the following decimals represent?
$0.05$, $0.2$, $0.8$, $0.004$
### Question 6
Write each of the following fractions as a decimal.
$\dfrac {2} {5}$, $\dfrac {1} {25}$, $\dfrac {1} {20}$, $\dfrac {1} {200}$, $\dfrac {2} {2500}$
### Question 7
MULTIPLE CHOICE!
The decimal $0.050$ equals
The following is multiple choice question (with options) to answer.
Which of the following fractions, if written as a decimal, would have a 2 in the thousandths place ? | [
"3/11",
"7/9",
"1/8",
"4/7"
] | A | Here rewriting the fractions
A)0.2727272727272727...
B)0.777777
C)0.25
D)0.5714285...
E)0.1666666
Clearly A will be our pick as the thousandths digit is 2
Hence A |
AQUA-RAT | AQUA-RAT-39106 | What you've said is correct, but it's not the usual way that we deal with "false cases".
Consider the statement $P(n)$ that says $$3\ln(1 + n) < n.$$ You might try proving this is true for every $n$, but you'd fail: it's false for $n = 0$ and $n = 1$. This is fairly common: namely, it's often the case that for finitely many values of $n$, $P(n)$ is false. Clearly you therefore can't prove $P(n)$ true for all $n$. You can, however, often find some $n_0$ with the property that all the "bad" $n$s are less than $n_0$. In the example, $n_0 = 6$ works.
So you typically prove instead that the statement is true for all $n > n_0$.
Alas, the principle of induction doesn't apply directly.
The usual subterfuge is to consider a new $S(n) = P(n + n_0)$. (So $S(0)$ says that $P(n_0)$ is true, and $S(1)$ says that $P(n_0 + 1)$ is true, and so on.)
The principle of induction then applies to the statments $S(n)$, for $n = 0, 1, ...$ and from their truth, you can deduce the truth of $P(n)$ for $n = n_0, n_0 + 1, n_0 + 2, \ldots$.
• What about the possibility of a mathematical formula that is provable by induction for $n_0-1$ till $n =100000$, and then fails at one point $n+1 =100001$. Why it is not possible is the key. Dec 26 '17 at 5:10
The following is multiple choice question (with options) to answer.
A sheet of paper has statements numbered from 1 to 40. For all values of n from 1 to 40, statement n says: ‘Exactly n of the statements on this sheet are false.’ Which statements are true and which are false? | [
"The even numbered statements are true and the odd numbered statements are false.",
"The odd numbered statements are true and the even numbered statements are false.",
"All the statements are false.",
"The 39th statement is true and the rest are false"
] | B | Assume there is only one statement is there. The statement should read "Exactly 1 statement on this sheet is false" . If the truth value of the statement is true, then given statement should be false. This is contradiction. If the statement is false, Then the given statement is true. but there is not other true statement.
Assume there are two statements. By the above logic, 2nd statement should not be true. But 1st statement is true as it truthfully says the truthfulness. By this logic we know that If there are "n" statements, (n-1)th statement is the only true statement And all other are false
Answer:B |
AQUA-RAT | AQUA-RAT-39107 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A train covers a certain distance at a speed of 200kmph in 4 hours. To cover the same distance in 2 hours, it must travel at a speed of | [
"400 km/hr",
"420 km/hr",
"450 km/hr",
"470 km/hr"
] | A | Explanation:
Distance = 200×4 = 800km
Required speed = (800/2) = 400km/hr
Answer: Option A |
AQUA-RAT | AQUA-RAT-39108 | # math
How is the circumference of a circle with radius 9cm related to the circumference of a circle with diameter 9cm?
i don't understand this.Help?? Thanks:)
1. 👍
2. 👎
3. 👁
c = pi x d.
The diamter is 2 x radius. So, the diameter of the first cirlce is 18cm. You know the diameter of the second circle. So, how do the circumferences compare?
1. 👍
2. 👎
2. is the second circle twice as big? like is that how they compare?
1. 👍
2. 👎
3. the circumference of the larger circle is indeed twice as long as that of the smaller one.
but...
the area of the larger circle is 4 times that of the area of the smaller.
Proof:
area of smaller = pi(4.5)^2 = 20.25pi
area of larger = pi(9^2) = 81pi
20.25pi/(81pi) = 1/4
1. 👍
2. 👎
4. alright thx!
1. 👍
2. 👎
## Similar Questions
1. ### Math ~ Check Answers ~
Find the circumference of the circle (use 3.14 for pi). Show your work. Round to the nearest tenth. the radius is 23 yd My Answer: ???? •Circumference = (2) (3.14) (23) •3.14 x 2 x 23 = 144.44
2. ### Mathematics
1. Find the circumference of the given circle. Round to the nearest tenth. (Circle with radius of 3.5 cm) -22.0 cm -38.5 cm -11.0 cm -42.8 cm 2. Find the area of the given circle. Round to the nearest tenth. (Circle with a
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4. ### MATH
The following is multiple choice question (with options) to answer.
There are two circles of different radii. The are of a square is 784 sq cm and its side is twice the radius of the larger circle. The radius of the larger circle is seven - third that of the smaller circle. Find the circumference of the smaller circle. | [
"99",
"277",
"12",
"989"
] | C | Let the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.
a2 = 784 = (4)(196) = (22).(142)
a = (2)(14) = 28
a = 2l, l = a/2 = 14
l = (7/3)s
Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s = 12∏ cm.
Answer: C |
AQUA-RAT | AQUA-RAT-39109 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
There are 13 which, 15 black, 16 red hats on a table. In one step, you may chose two hats of different colors and replace each one by a hat of the third color. After how many steps this can be reached? | [
"2",
"7",
"3",
"9"
] | C | Let the initial number of hats are (a, b, c). In one step the hats color may change into one of the following (a+2, b-1, c-1), (a-1, b+2, c-1), (a-1, b-1, c+2). In each of these cases, difference between any two colored hats should be divisible by 3 to reach mono chromatic stage. ie., a + b + c should also be divisible by 3.
In this case sum is not divisible by 3 so the final stage cannot be reached.
Answer:C |
AQUA-RAT | AQUA-RAT-39110 | biology, dimensional-analysis, scaling
short example as Sonny asked for in comment:
ant with 10 mm length & 10 mg mass
$\Rightarrow$ lets scale up to human size (2m) $\Rightarrow$ means a factor of 200. So the mass scales with 200x200x200=8000000 (Volume $\propto$ $l^{3}$ ) $\Rightarrow$ human sized ant=80 kg. But muscle forces scales only by factor 200x200=40000. The small ant can carry 100x10mg of her own mass=1g, the human sized ant should be able to carry 1g x 40000=40 kg.
Conclusion: pretty comparable to a avg. 80 kg human man able to carry 40 kg!
The following is multiple choice question (with options) to answer.
A row of ants comprised of 102 ants. A man walked upon it and all except 42 were crushed beneath his foot.
How many ants are left alive? | [
"41",
"42",
"43",
"44"
] | B | Solution:
All except 42 were crushed means that all the ants except 42 were crushed beneath the man's foot. Hence, 42 ants are alive now
Answer B |
AQUA-RAT | AQUA-RAT-39111 | # 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.
John purchased a grinder and a mobile for Rs. 15000 &Rs. 8000 respectively. He sold the grinder at a loss of 4% and the mobile phone at a profit of 10%. Overall how much he make aprofit. | [
"Rs. 90",
"Rs. 120",
"Rs. 200",
"Rs. 250"
] | C | Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 - 4/100) = 15000 - 600
m = 8000(1 + 10/100) = 8000 + 800
Total SP - Total CP = r + m - (15000 + 8000) = -600 + 800 = Rs. 200
As this is positive, an overall profit of Rs. 200 was made.
C |
AQUA-RAT | AQUA-RAT-39112 | Hint: You may suppose, w.l.o.g. that $$|x-2|<\frac12$$, in which case $$|x-1|>\frac12$$, so that $$\left|\frac{x-2}{x - 1}\right| <2|x-2|.$$
$$\delta<\min(1,\varepsilon)/2$$ should do the job.
The following is multiple choice question (with options) to answer.
If x > 2000, then the value of (x)/(2x+1) is closest to? | [
"1/6",
"1/3",
"10/21",
"1/2"
] | D | assume x = 1999
(x)/(2x+1) = 1999 / (1999*2+1)
=1999 / 3999
= = 1/2
Ans - D |
AQUA-RAT | AQUA-RAT-39113 | # 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.
If the cost price is 91% of sp then what is the profit % | [
"9.89%",
"8.90%",
"9.00%",
"8.00%"
] | A | Sol. sp=Rs100 : then cp=Rs 91:profit =Rs 9.
Profit={(9/91)*100}%=9.89%
Answer is A. |
AQUA-RAT | AQUA-RAT-39114 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
In a race of 2.4 Kms. A beats B by 400 m or 20 sec. What is A’s time over the course? | [
"1 min 30 sec",
"1 min",
"1 min 10 sec",
"1min 40 sec"
] | D | 2400*20/400=120
120-20=100sec=1 min 40 sec.
ANSWER:D |
AQUA-RAT | AQUA-RAT-39115 | also have a clearance (play) vertical to tooth depth. 14 * 5 2 * ((4/3) * 5 +10) Volume (V) = 1309 m 3. The Tank Capacity Calculator below allows you to type in your desired tank diameter and height and provides an estimated volume by gallon amount. White & blue vertical tanks are translucent for convenient product level viewing. Calculations are performed for: - Vertical tank - Horizontal tank - Rectangular tank - Ecliptic tank - Tanks with conical bottom, flat bottom, Torispherical head, Elliptical head, Hemispherical head. The reference point of the water level in the tank used to calculate the water rising level was made to the bell mouth top side for the vertical overflow outlet type as shown in Figure 4, and made a pipe inside bottom for the horizontal overflow outlet type. • if k should be negative, the vertical stretch or shrink is. Hence, we can describe the concentration of salt in the tank by concentration of salt = S 100 kg/l. User-defined axis scale type using formulas. Alternatively, you can use this tank volume calculator as a water volume calculator if you need to calculate some specific water volume. If you have a complex tank, or need something special created please don't hesitate contact us for more info. The radius BO of the hemisphere (as well as of the cone) = ( ½) × 4 cm = 2 cm. , the Volumes of complicated shapes can be calculated with integral calculus if a formula exists for the In other systems the conversion is not trivial; the capacity of a vehicle's fuel tank is rarely stated in cubic. This Excel spreadsheet helps you calculate the liquid volume in partially filled horizontal tanks. 000233 D3 ( D = ID in inches ) ( DR = ID ). Nonsymmetrical Vertical Curves G1 & G2 Tangent Grades in percent A The absolute of the Algebraic difference in grades in percent BVC Beginning of Vertical Curve EVC End of Vertical Curve VPI Vertical Point of Intersection l1 Length of first section of vertical curve l2 Length of second section of vertical curve L Length of vertical curve. Volume formula of a rectangular prism Volume of a right circular cone is equal to one third of the product of the area of its base times the height. Adding air manually is not the best way to ensure proper. Internal lining material is natural rubber with thickness 4. There are two main methods to tank
The following is multiple choice question (with options) to answer.
Tanks C and B are each in the shape of a right circular cylinder. The interior of tank C has a height of 10 meters and a circumference of 8 meters, and the interior of tank B has a height of 8 meters and a circumference of 10 meters. The capacity of tank C is what percent of the capacity of tank B? | [
"75%",
"80%",
"100%",
"120%"
] | B | B.
For C, r=8/2pi. Its capacity = (4pi)^2 * 10= 160pi
For B, r=10/pi. Its capacity = (5pi)^2 *8 = 200pi
C/B = 160pi/200pi = 0.8 |
AQUA-RAT | AQUA-RAT-39116 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people, the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people be solicited? | [
"300",
"250",
"400",
"500"
] | A | Solution: Let the number of people be x who has been asked for the donations.
People already solicited = 60% of x = 0.6x
Remaining people = 40% of x = 0.4x
Amount collected from the people solicited,
= 600 *0.6x = 360x
360x = 75% of the amount collected.
Remaining amount = 25% = 120x
Thus,
Average donations from remaining people,
= 120x /0.4x = 300.
Answer: Option A |
AQUA-RAT | AQUA-RAT-39117 | # 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.
Train A left Centerville Station, heading toward Dale City Station, at 3: 00 p.m. Train B left Dale City Station, heading toward Centerville Station, at 3: 20 p.m. on the same day. The trains rode on straight tracks that were parallel to each other. If Train A traveled at a constant speed of 30 miles per hour and Train B traveled at a constant speed of 10 miles per hour, and the distance between the Centerville Station and Dale City Station is 90 miles, when did the trains pass each other? | [
"4: 45 p.m.",
"5: 00 p.m.",
"5: 20 p.m.",
"5: 35 p.m"
] | C | the distance travelled by the train A in first 20 minutes will be 10.
The distance which will be remaining is 80.
Now both trains are running in opposite direction.Their speed will be added so 40.
Time at which they will meet =80/40=2
time of train B will be 3:20 +2=5:20
ANSWER:C |
AQUA-RAT | AQUA-RAT-39118 | # Math Help - need help.
1. ## need help.
A long rope is pulled out between two opposite shores of a lake. It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
2. Hello, Bobby!
A long rope is pulled out between two opposite shores of a lake.
It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
Code:
C
* * *
* :x *
* : 35 *
A * - - - - -+- - - - - * B
\ D: /
\ :R-x /
R \ : / R
\ : /
\:/
*
O
The center of the earth is $O.$ . $OA = OB = OC = R$ (radius of the earth).
The 70-km rope is $AB.$ .We see that: $AD = DB = 35.$
Let $x = CD$ be the distance the rope is underwater at its center.
Then $DO = R - x.$
From right triangle $ODB:\;\;DO^2 + DB^2\:=\:OB^2$
So we have: . $(R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$
Quadratic Formula: . $x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$
Since $R = 6370$, we have: . $x \;= \;6370 \pm \sqrt{6370^2 - 1225}$
The following is multiple choice question (with options) to answer.
A rectangular field 30 m long and 20 m broad. How much deep it should be dug so that from the earth taken out, a platform can be formed which is 8 m long, 5.5 m broad and 1.5 m high where as the earth taken out is increase by 10/5? | [
"12 cm",
"10 cm",
"83 cm",
"20 cm"
] | B | 30 * 20 * x
= (8 * 5.5 * 1.5)/2=10
Answer:B |
AQUA-RAT | AQUA-RAT-39119 | # Physics kinematics SIN question
1. Dec 30, 2011
### ShearonR
1. The problem statement, all variables and given/known data
A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?
Given: v=30m/s
vi=0
Need: vf=?
2. Relevant equations
vf=vi+αΔt
vf2=vi2+2αΔd
v=Δd/Δt
3. The attempt at a solution
So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
2. Dec 30, 2011
### Vorde
This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.
edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
3. Dec 30, 2011
### ShearonR
Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
4. Dec 30, 2011
### Staff: Mentor
Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.
You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?
The following is multiple choice question (with options) to answer.
A thief goes away with a SANTRO car at a speed of 40 kmph. The theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start? | [
"2 hours",
"7 hours",
"8 hours",
"3 hours"
] | A | |-----------20--------------------|
50 40
D = 20
RS = 50 – 40 = 10
T = 20/10 = 2 hours
Answer:A |
AQUA-RAT | AQUA-RAT-39120 | EZ as pi
Featured 5 months ago
$\text{males : females } = 6 : 5$
#### Explanation:
When working with averages (means), remember that we can add sums and numbers, but we cannot add averages.
(An exception would be if there were the same number of males and females - in this case we may add the averages and divide by 2)
Let the number of females be $x$.
Let the number of males be $y$
Let's work with the $\textcolor{red}{\text{whole group first:}}$
The total number of people at the party is $\textcolor{red}{x + y}$
The sum of all their ages is $\textcolor{red}{\left(x + y\right) \times 29}$
Now let's work with $\textcolor{b l u e}{\text{males and females separately.}}$
The sum of the ages of all the females = $23 \times x = \textcolor{b l u e}{23 x}$
The sum of the ages of all the males = $34 \times y = \textcolor{b l u e}{34 y}$
The sum of the ages of all the people = $\textcolor{b l u e}{23 x + 34 y}$
The sum of the ages of all the people = $\textcolor{red}{29 \left(x + y\right)}$
We now have 2 different expressions for the same information, so we can make an equation.
$\textcolor{red}{29 \left(x + y\right)} = \textcolor{b l u e}{23 x + 34 y}$
$29 x + 29 y = 23 x + 34 y$
$34 y - 29 y = 29 x - 23 x$
$5 y = 6 x \text{ we need to compare } y : x$
$y = \frac{6 x}{5}$
$\frac{y}{x} = \frac{6}{5}$
$y : x = 6 : 5$
Notice that although we do not know the actual number of people at the party, we are able to determine the ratio.
$\text{males : females } = 6 : 5$
The following is multiple choice question (with options) to answer.
Find the mean proportional of the following pairs of numbers 6 and 24. | [
"27",
"28",
"29",
"23"
] | B | Required number of trees
= 24/36 * 42
= 28.
Answer:B |
AQUA-RAT | AQUA-RAT-39121 | Why are things "weird" here? Let's think of it like this: If I travelled at $$8 \frac{\text{mi}}{\text{hr}}$$ for an hour, then travelled at $$12 \frac{\text{mi}}{\text{hr}}$$, I'd definitely agree that the average speed is $$10 \frac{\text{mi}}{\text{hr}}.$$ Things are pretty normal here: If I want to calculate the total distance traveled, it's just $$8 + 12 = 20.$$ Then, if I want to calculate the total time traveled, it's just $$1 + 1 = 2.$$ So the average or overall speed is just
$$\frac{20 \text{mi}}{2 \text{hr}} = 10 \frac{\text{mi}}{\text{hr}}.$$ It works!
The following is multiple choice question (with options) to answer.
A person goes to his office at 1/3rd of the speed at which he returns from his office. If the avg speed during the whole trip is 12m/h. what is the speedof the person while he was going to his office? | [
"8km/h",
"12km/h",
"14km/h",
"16km/h"
] | A | u = k , v= 3k
\inline \therefore \frac{2uv}{u+v}\: \: \Rightarrow \frac{2\times k\times 3k}{(k+3k)}=12
\inline \Rightarrow 1.5k = 12
\inline \Rightarrow k=8km/h
A |
AQUA-RAT | AQUA-RAT-39122 | 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.
5 men and 3 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio | [
"1:1",
"1:3",
"2:1",
"2:3"
] | A | Explanation:
Let 1 man 1 day work = x
1 boy 1 day work = y
then 5x + 3y = 4(x+y)
=> x = y
=> x/y = 1/1
=> x:y = 1:1
Option A |
AQUA-RAT | AQUA-RAT-39123 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
Carol and Jordan draw rectangles of equal area. If Carol's rectangle measures 5 inches by 24 inches and Jordan's rectangle is 3 inches long, how wide is Jordan's rectangle, in inches? | [
"25",
"43",
"42",
"40"
] | D | Area of Carol's rectangle = 24*5 = 120
Let width of Jordan's rectangle= w
Since , the areas are equal
3w = 120
=>w = 40
Answer D |
AQUA-RAT | AQUA-RAT-39124 | # 1985 AIME Problems/Problem 12
## Problem
Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac n{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.
## Solution 1
Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$, we have $P(n + 1) = \frac13 (1 - P(n))$. We also know $P(0) = 1$, so we can quickly compute $P(1)=0$, $P(2) = \frac 13$, $P(3) = \frac29$, $P(4) = \frac7{27}$, $P(5) = \frac{20}{81}$, $P(6) = \frac{61}{243}$ and $P(7) = \frac{182}{729}$, so the answer is $\boxed{182}$. One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$.
## Solution 2
We can find the number of different times the bug reaches vertex $A$ before the 7th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A$.
The following is multiple choice question (with options) to answer.
Three cubes of metal whose edges are 9, 12 and 15 cm respectively, are melted and one new cube is made. Find the edge of the new cube? | [
"99",
"88",
"71",
"18"
] | D | 93 + 123 + 153 = a3 => a = 18
Answer: D |
AQUA-RAT | AQUA-RAT-39125 | Math Help - Sum of 2 squares
1. Sum of 2 squares
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
2. Originally Posted by Mukilab
Sorry before I tell you the question here was the question that led to this question.
Show that (m^2+1)(n^2+1)=(m+n)^2+(mn-1)^2
Done that successfully.
Using this results, write 500050 as the sum of 2 square numbers.
I have no idea how to make a knowledgeable guess at the two squares so I've tried using random (well, not exactly, generic random numbers) to get an answer but to no avail
Try m=7.
50 is very close to 49 and stands out that way..
3. Hello, Mukilab!
Show that: . $(m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2" alt=" (m^2+1)(n^2+1)\:=\m+n)^2+(mn-1)^2" />
Done that successfully. . Good!
Using this results, write 500,050 as the sum of 2 square numbers.
Note that: . $500,050 \;=\;(50)(10,\!001) \;=\;(7^2+1)(100^2+1)$
$\text{Let }m = 7,\;n = 100 \text{ in the formula:}$
. . $(7^2 + 1)(100^2 + 1) \;=\;(7 + 100)^2 + (7\cdot100 - 1)^2 \;=\;107^2 + 699^2$
The following is multiple choice question (with options) to answer.
If the sum of a number and its square is 210, What is the number? | [
"16",
"14",
"25",
"87"
] | B | Explanation:
Let the integer be x.
Then, x + x2 = 210
x2 + x - 210 = 0
(x + 15) (x – 14) = 0
x = 14
ANSWER: B |
AQUA-RAT | AQUA-RAT-39126 | 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.
During a clearance sale, a retailer discounted the original price of its TVs by 30% for the first two weeks of the month, then for the remainder of the month further reduced the price by taking 20% off the sale price. For those who purchased TVs during the last week of the month, what percent of the original price did they have to pay? | [
"40%",
"45%",
"56%",
"60%"
] | C | VERITAS PREPOFFICIAL SOLUTION:
C With percent problems, the key is often to make sure that you take the percent of the correct value. In this case, the initial 30% off means that customers will pay 70% of the original price. Then for the second discount, keep in mind that the discount is taken off of the sale price, not of the original price. So that's 20% off of the 70% that they did pay, which can be made easier by looking at what the customer does pay: 80% of the 70% sale price. Using fractions, that means they pay: 4/5*(0.7) of the original price, which nets to 0.56 of the original price, or 56%. |
AQUA-RAT | AQUA-RAT-39127 | the number of committees that exist where the man and women serve together is given by,
$$\begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 7 \\ 2 \end{pmatrix} = 210$$
so the total number of committees in this case amounts to,
$$1120 - 210 = 910$$
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
The following is multiple choice question (with options) to answer.
In how many ways a committee consisting of 3 men and 2 women can be formed from 4 men and 9 women? | [
"A)144",
"B)15",
"C)20",
"D)18"
] | A | Required number of ways = 4C3*9C2 = 4*36 = 144
Answer is A |
AQUA-RAT | AQUA-RAT-39128 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
### Show Tags
12 Aug 2013, 23:15
5
KUDOS
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This post was
BOOKMARKED
zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
Average of first five multiples of 10 is | [
"9",
"11",
"30",
"15"
] | C | Explanation:
Average=10(1+2+3+4+5) / 5=150 / 5=30
Option C |
AQUA-RAT | AQUA-RAT-39129 | ### 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.
Of the land owned by a farmer, 90 percent was cleared for planting. Of the cleared land, 60 percent was planted with grapes and 30 percent of the cleared land was planted with potato. If the remaining 360 acres of cleared land was planted with tomato, how many acres did the farmer own? | [
"3250",
"3450",
"4000",
"3750"
] | C | 60% of 90% = 54%, 30% 0f 90% = 27% so the remaining 90-54-27=9%=360 acres or 10% of 90%=9% --> 360/9*100=4000 acres Answer (C) |
AQUA-RAT | AQUA-RAT-39130 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
There are 2 friends Peter and Paul. Peter age is twice as old as Paul when peter was as old as Paul is now. Sum of the present ages of Peter and Paul is 35.What is the present age of Peter? | [
"10",
"15",
"20",
"25"
] | C | let pauls age be 'a' and peters be 'e'
e+a=35 (given)
e=35-a ------eq1
when peter was as old as paul that is when peter's age was 'a'
pauls age was a-(e-a) =2a-e
now peters age is e=2(2a-e)
e=4a-2e
3e=4a
3(35-a)=4a
105-3a=4a
a=15
hence e=20
ANSWER:C |
AQUA-RAT | AQUA-RAT-39131 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by: | [
"1/7",
"1/8",
"1/9",
"7/8"
] | A | Original share of 1 person = 1/8
New share of 1 person = 1/7
Increase =(1/7-1/8) = 1/56
Required fraction = (1/56)/(1/8)
= (1/56)*8
= 1/7
ANSWER:A |
AQUA-RAT | AQUA-RAT-39132 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In a mixture of 60 litres the ratio of milk to water is 5:1. Additional 20 litres of water is added to the mixture. Find the ratio of milk to water in the resulting mixture. | [
"2:3",
"5:3",
"1:4",
"3:2"
] | B | Given that Milk/Water=5x/x and 5x+x=60 --> x=10.
Thus Milk=5x=50 liters and Water=x=10 liters.
New ratio = 50/(20+10) = 50/30 = 5/3.
Answer is B |
AQUA-RAT | AQUA-RAT-39133 | water, flow
Title: How is it that a cistern can collect rainfall, but a well cannot, even though both of them are holes? I imagined that a regular downpour or flood of rain would fall into the well directly, so I did a search on Google and was surprised to find this answer. But the article didn't make sense other than that the water would seep through the ground and in ten years, it would be at the water table, where it might, perhaps, refill the well. Still, I'm not sure what sets a cistern apart at being able to collect rainwater, since the gravity should be able to pull rain down.
If there is a better SE site for this question, feel free to move it. Without directing runoff rainwater into an open well it can only catch the rain that lands on its exposed surface area. This could only add a few inches at most in a single rainfall. A cistern directs runoff from a larger area into it, so it is fed by a much larger area than just the surface area of the tank.
The following is multiple choice question (with options) to answer.
A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled? | [
"2.9 hrs",
"8.9 hrs",
"2.9 hrs",
"7.2 hrs"
] | D | Net part filled in 1 hour = (1/4 - 1/9)
= 5/36
The cistern will be filled in 36/5 hrs i.e., 7.2 hrs.
Answer:D |
AQUA-RAT | AQUA-RAT-39134 | To be more precise, we should be clear to distinguish numbers from the sequences of digits we might use to represent them. Standard notation unfortunately does not make this very clear. When we write $1+1=2$ normally, what we really mean is "the sum of the number represented by $1$ in decimal notation and the number represented by $1$ in decimal notation is the number represented by $2$ in decimal notation". So what "$1+1=2$ is valid in any base" really means is "for any base $b>2$, the sum of the number represented by $1$ in base $b$ notation and the number represented by $1$ in base $b$ notation is the number represented by $2$ in base $b$ notation." This is because, as mentioned above, "the number represented by $1$ in base $b$ notation" is the exact same number as "the number represented by $1$ in decimal notation", and similarly for $2$.
• I would go a bit further and say that these equations still make sense (and are still true) if you allow numbers beyond the base (which is done from time to time, e.g. in discussions of carrying). Then $1+1=2$ is valid in all bases, even binary and even unary. Aug 18, 2016 at 19:03
• I feel poorly for unary which is left out of so many discussion of base just because it's so often useless. It should be included with binary as a base in which $1+1=2$ is invalid. $1+1=11$ Aug 18, 2016 at 21:14
• It gets fun when you look at how the sum of digits of multiples of n - 1 in base n are multiples of n - 1. Aug 19, 2016 at 3:46
• @EngineerToast In binary, $1+1=10$ Aug 19, 2016 at 6:44
• @rexkogitans He's speaking about unary. Aug 19, 2016 at 6:53
Integers have their own existence, separate from how we may choose to represent them.
The statement
"(the integer referred to by the decimal symbol 1) + (the integer referred to by the decimal symbol 1) = (the integer referred to by the decimal symbol 2)"
The following is multiple choice question (with options) to answer.
If (10^67) – 63 is written as a base 10 integer, which of the following options is the sum of the digits in that integer? | [
"598",
"495",
"590",
"595"
] | D | We know that (10^67) is ending 00, so (10^67) – 63=9....9937
total number of digits in (10^20) – 63 is 67, or 65 digits of 9 and two digits 3 and 7.
answer choice is 65*9+10=595
Answer is D 595 |
AQUA-RAT | AQUA-RAT-39135 | # Conditional Probability with Circuits
I am very sorry for the myriad of probability questions but I have a dam hard time trying to understand these questions:
This is the question:
Consider the following portion of an electric circuit with three relays. Current will flow from point a to point b if there is at least one closed path when the relays are activated. The relays may malfunction and not close when activated. Suppose that the relays act independently of one another and close properly when activated, with a probability of .9.
The question is: What is the probability that current will flow when the relays are activated?
Answer:
1 - (0.1)^3
I don't understand why you are multiplying 0.1 * 0.1 * 0.1 and not adding 0.1 + 0.1 + 0.1. Since if any one of the three are open then the current will flow so isn't this technically a union which means you add?
• It's very simple: the multiplication principle. The relays are independent of each other. – Parcly Taxel Oct 12 '16 at 4:26
• @ParclyTaxel Oh ok. Thanks for pointing that out. So if they are dependent of each other then you would add them? – CapturedTree Oct 12 '16 at 5:23
## 2 Answers
Yes , it is a union . The current will flow when at least one of the relays will pass it through .
However , you do not simply add because the events are not disjoint . This then requires using the Principle of Inclusion and Exclusion (PIE) to avoid over-counting common outcomes .
The following is multiple choice question (with options) to answer.
A relay has a series of 5 circuits in a line. The even-numbered circuits are control circuits; the odd are buffer circuits. If both a control circuit and the buffer circuit immediately following it both fail in that order, then the relay fails. The probability of circuit one failing is 1/8; circuit two, 3/8; circuit three, 3/10; circuit four, 3/4; and circuit five, 2/5 .What is the probability that the relay fails? | [
"9/80",
"3/10",
"35/80",
"303/800"
] | D | The first circuit doesn't matter.
Prob(relay fails) = 1 - Prob(relay succeeds)
Prob(2+3 work) = 1 - 9/80 = 71/80
Prob(4+5 work) = 1 - 3/10 = 7/10
Prob(relay fails) = 1 - Prob(2+3 work AND 4+5 work) = 1 - (71/80)(7/10) = 1 - 497/800 = 303/800
D |
AQUA-RAT | AQUA-RAT-39136 | ## Effective interest rate of 12 compounded quarterly
Rate (EAR) from a stated nominal or annual interest rate and compounding frequency. APY Calculator to Calculate Annual Percentage Yield from a Stated Nominal Interest Rate For example, if one saving institution offers an annual interest rate of 1% compounded annually, whereas APY = (1 + .04875/12 )12 – 1. This words out to a 12% interest rate. However, since interest is compounded monthly, the actual or effective interest rate is higher because interest in the current
8 Sep 2014 But loan interest is almost never compounded annually! To convert a nominal interest rate to an effective interest rate, we have to pay close daily and annual compounding is a lot bigger at 12%/yr interest than at 4%/yr. 5 Feb 2019 It is likely to be either monthly, quarterly, or annually. Locate the stated interest rate in the loan documents. Enter the compounding period and Where r is the interest rate per period in decimal form so R = r * 100 and, i is the effective interest rate in decimal form so I = i * 100. P is the rate per compounding period where P = R/m. The effective interest rate is the interest rate on a loan or financial product restated from the nominal interest rate as an interest rate with annual compound interest payable in arrears. It is used to compare the annual interest between loans with different compounding terms (daily, monthly, quarterly, semi-annually, annually, or other).
## (b) The annual interest rate is 2.4%, and the number of interest periods is 12. Table 3 shows the effects of interest rates (compounded quarterly) on the Effective Rate of Interest Formula If interest is compounded m times per year, then.
The following is multiple choice question (with options) to answer.
Effective annual rate of interest corresponding to nominal rate of 6% per annum compounded half yearly will be | [
"6.09%",
"6.10%",
"6.12%",
"6.14%"
] | A | Explanation:
Let the amount Rs 100 for 1 year when compounded half yearly, n = 2, Rate = 6/2 = 3%
Amount=100(1+3/100)2=106.09
Effective rate = (106.09 - 100)% = 6.09%
Option A |
AQUA-RAT | AQUA-RAT-39137 | Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.
Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.
Back to the problem:
From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.
From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.
D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]
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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?
(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1
1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient
And is D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]
### Show Tags
28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of x + 1, x + 4, and x + 7 is 0, then x = | [
"–4",
"–3",
"–2",
"–1"
] | A | (x+1 + x+4 + x+7)/3 = 0
=>3x + 12 = 0
=>x = -4
Answer A |
AQUA-RAT | AQUA-RAT-39138 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 20 natural numbers. | [
"A)11.5",
"B)20.5",
"C)10.5",
"D)15.5"
] | C | Sum of first n natural numbers = n(n+1)/2
sum of first 20 natural numbers = 20*21/2 =210
average = 210/20 = 10.5
Answer is C |
AQUA-RAT | AQUA-RAT-39139 | 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.
Difference between the length and breadth of a rectangle is 23 m. Ifits perimeter is 206 m, then its area is?? | [
"1220 m^2",
"1520 m^2",
"2520 m^2",
"2600 m^2"
] | C | We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) m2 = 2520 m^2
C |
AQUA-RAT | AQUA-RAT-39140 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row upstream at 12kmph and downstream at 14kmph. Find man's rate in still water ? | [
"5km/hr",
"15km/hr",
"13km/hr",
"20km/hr"
] | C | Rate in still water = 1/2 (12+14) = 13km/hr
Answer is C |
AQUA-RAT | AQUA-RAT-39141 | \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \color{red}{ \frac{p_{11}}{20}}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}& \frac{p_{11}}{20}\\12& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \frac{p_{12}}{20}& \color{red}{ \frac{p_{12}}{20}}& \frac{p_{12}}{20}& \frac{p_{12}}{20}&
The following is multiple choice question (with options) to answer.
The forth proportional to 5,10,20 is? | [
"20",
"40",
"45",
"80"
] | B | Let the fourth proportional to 5,10,20 be x
then 5:10::20:x
5x = 10*20
x = 40
Answer is B |
AQUA-RAT | AQUA-RAT-39142 | The game is "fair".
Playing repeatedly, you can expect to break even.
4. thanks all for the answers.
but the answer of the expected value was certainly $1 was it required to find the expected value of "what is finally getting back" then the result should be the sum of the stake ($1) and the expected value of winning), right?
The following is multiple choice question (with options) to answer.
6 private companies bought a share of 2 million dollars. If the all companies have to pay the same amount and none of them have bills smaller than $1, how much money would they have to pay?
Options: | [
"$2,000,000",
"$2,000,001",
"$2,000,004",
"$2,000,007"
] | C | In order to divide the sum in 6 shares, the amount must be divisible by 6
Divisibility rule of 6: The sum of the digits must be divisible by 6
Sum of digits of 2,000,000 = 2 and 6 is divisible by 6.
Hence, we need to add 4 to this number for it to be divisible by 6
Correct Option: c
Correct Option: $2,000,004 |
AQUA-RAT | AQUA-RAT-39143 | For $N = 5$, $2A$ looks like:
$$\begin{array}{ccccc} &0 &1 &0 &0 &1 \\ &1 &0 &1 &0 &0 \\ &0 &1 &0 &1 &0 \\ &0 &0 &1 &0 &1 \\ &1 &0 &0 &1 &0 \end{array}$$
Solving this equation system for $\mathbf{x}$ with $N = 10$ and $\mathbf{b} = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$ gives
$x_6 = 1$.
6. For her to announce an average of 6, the two numbers she receives must add up to 12. This happens to be the average of the two averages announced to either side of her. This pattern holds around the circle, which makes sense if the announced average is also the number picked.
• Wait, I forgot to take into account that this is a circle. :( will edit answer once I reconsider – Irishpanda Feb 18 '16 at 13:08
• haha, I had this same thought process initially – question_asker Feb 18 '16 at 16:03
1
Let $g_n$ be the number picked by the girl who said $n$. The average of $g_6$ and $g_4$ is two more than the average of $g_2$ and $g_4$, so $g_6$ must be four more than $g_2$. Symmetrically $g_6$ four less than $g_{10}$. $g_6$ is therefore the average of $g_2$ and $g_{10}$, but we already have already been told what this is.
• Very elegantly thought and stated. Best answer. – Vynce Aug 30 '17 at 2:20
I know there are plenty of correct answers, but here is a super-simple one.
Let's note $$g_n$$ the n'th girl's secret number and $$a_n$$ the average she gave aloud.
The following is multiple choice question (with options) to answer.
The average (arithmetic mean) of four numbers is 4x + 6. If one of the numbers is x, what is the average of the other three numbers? | [
"x + 1",
"3x + 3",
"5x + 8",
"5x + 4"
] | C | The sum of all four numbers is 4 * (4x + 6) = 16x + 24.
If one of the numbers is x, the sum of the other 3 numbers is 15x+24.
The average of the remaining three numbers is (15x+24)/3 = 5x+8.
The answer is C. |
AQUA-RAT | AQUA-RAT-39144 | 3 BIOGRAPHIES books: $\dfrac{4\cdot 3 \cdot 2}{32!} \cdot 6 =24$
4 BIOGRAPHIES books: 1
then the result is 115.
-
Suppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$. Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$. So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$. That made a contribution of $1$ to your $168$.
But among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your "two more" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$.
Both of these ways of choosing end us up with $A$, $B$, $C$, and $N$. So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next. Still another contribution of $1$ to your $168$.
So your product counts the set $\{A, B, C, N\}$ three times. This it does for every combination of three biographies and one novel. It also overcounts the set $\{A, B, C, D\}$.
One could adjust for the overcount. In some problems that is a useful strategy. Here it takes some care.
But a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.
-
The following is multiple choice question (with options) to answer.
Tavid has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d? | [
" 5/6*d",
" 7/3*d",
" 10/3*d",
" 7/2*d"
] | C | Tavid has d books;
Jeff has d/3 books;
Paula has 2d books;
Total = d+d/3+2d=10d/3.
Answer: C. |
AQUA-RAT | AQUA-RAT-39145 | # 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 man purchased 3 blankets @ Rs.100 each, 1 blankets @ Rs.150 each and two blankets at a certain rate which is now slipped off from his memory. But he remembers that the average price of the blankets was Rs.150. Find the unknown rate of two blankets? | [
"1050",
"2770",
"1550",
"5102"
] | A | 10 * 150 = 1500
3 * 100 + 1 * 150 = 450
1500 – 450= 1050
Answer:A |
AQUA-RAT | AQUA-RAT-39146 | Is the blue area greater than the red area?
Problem:
A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.
Which area is greater?
Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.
Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.
But how can it be proven?
I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.
Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$
According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that \begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}
Assertion:
To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.
Is this true? If so, is there another way of proving the assertion?
Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.
This question is very similar to this post.
Source:
The following is multiple choice question (with options) to answer.
The side of a square is increased by 30% then how much % does its area increases? | [
"52.65",
"69",
"50.75",
"42.75"
] | B | a = 100 a2 = 10000
a = 130 a2 = 16900
----------------
10000 --------- 6900
100 -------? => 69 %
ANSWER: B |
AQUA-RAT | AQUA-RAT-39147 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
Some pens are divided among A, B, C and D. A gets twice the number of pens that B gets. C gets the same number of pens as D gets. If A gets 25 pens more than D and the ratio of the number of pens that B and C get is 2 : 3, then find the number of pens that D gets ? | [
"56 pens",
"76 pens",
"75 pens",
"87 pens"
] | C | Let the number of pens that A, B, C and D get be a, b, c and d respectively.
a : b = 2 : 1
a = c + 25
b : c = 2 : 3
a : b : c : d = 4 : 2 : 3 : 3
a,d get 4p, 3p pens
=> 4p - 3p = 25 (given)
p = 25
=> D gets 3p = 3 * 25 = 75 pens.
Answer:C |
AQUA-RAT | AQUA-RAT-39148 | 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 160 metres long passes a standing man in 18 seconds. The speed of the train is | [
"35 kmph",
"45 kmph",
"32 kmph",
"42 kmph"
] | C | From formula Speed = Distance / Time
Thus, Speed = 160 / 18 => 8.88 sec
or, Speed = 8.88 x 18/5 kmph
or, Speed = 32 kmph
ANSWER:C |
AQUA-RAT | AQUA-RAT-39149 | # Difference between revisions of "2017 AMC 10A Problems/Problem 25"
## Problem
How many integers between $100$ and $999$, inclusive, have the property that some permutation of its digits is a multiple of $11$ between $100$ and $999?$ For example, both $121$ and $211$ have this property.
$\mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486$
## Solution 1
There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.
Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.
There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each.
There are 110, 220, 330 ... 990, yielding 9 extra permutations
Also, there are 209, 308, 407...902, yielding 8 more permutations.
Now, just subtract these 17 from the total (243), getting 226. $\boxed{\textbf{(A) } 226}$
• Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer.
## Solution 3
We note that we only have to consider multiples of $11$ and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of $11$ has:
$\textbf{Case 1:}$ All three digits are the same. By inspection, we find that there are no multiples of $11$ here.
$\textbf{Case 2:}$ Two of the digits are the same, and the third is different.
The following is multiple choice question (with options) to answer.
How many multiples of 2 are there between 1 and 54, exclusive? | [
"21",
"22",
"24",
"26"
] | D | 26
multiples of 2 between 1 and 54 exclusive.
from 2 * 1 upto 2*26, (1,2,3,4,...,26). Hence, 26 multiples !
correct option is D |
AQUA-RAT | AQUA-RAT-39150 | filters
a7 = 512 + (128 *w2 * w1 * w3) + (32 *w2 * w1*w1 * w3) + (256 *w2 * w1 * w4) + (64 *w2 * w1*w1 * w4) + (256 *w1 * w3 * w4) + (128 *w2 * w3 * w4) - (256 *w2 * w1) + (64 *w2 * w1*w1) - (128 *w1*w1) + (128 *w2 * w1 * w3 * w4) - (192 *w2 * w1*w1 * w3 * w4) - (256 *w1 * w3) - (128 *w2 * w3) + (64 *w1*w1 * w3) - (512 *w1 * w4) - (256 *w2 * w4) - (256 *w3 * w4) + (128 *w1*w1 * w4) + (64 *w1*w1 * w3 * w4) - (128 *w4*w4) + (128 *w4*w4 * w1) + (32 *w4*w4 * w1*w1) - (192 *w4*w4 * w1 * w2 * w3) + (120 *w4*w4 * w1*w1 * w2 * w3) + (64 *w4*w4 * w1 * w2) - (96 *w4*w4 * w1*w1 * w2) + (64 *w4*w4 * w2) + (64 *w4*w4 * w3) + (64 *w4*w4 * w1 * w3) - (96 *w4*w4 * w1*w1 * w3) + (32 *w4*w4 * w2 * w3)
The following is multiple choice question (with options) to answer.
find the terms which do not contain 7 between 1000 and 9999 | [
"3168",
"3169",
"3170",
"3172"
] | A | Total number of four digit numbers =9000 (i.e 1000 to 9999 )
We try to find the number of numbers not having digit 2 in them.
Now consider the units place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
Tens place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
Hundreds place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
Thousands place can be selected in 8 ways (i.e 1,2,3,4,5,6,,8,9) here '0' cannot be taken
Total number of numbers not having digit 7 in it =9 x 9 x 9 x 8 =5832
Also,Total number of numbers having digit 7 in it = 9000-5832 =3168
ANSWER:A |
AQUA-RAT | AQUA-RAT-39151 | Hint:
$$\frac{2\sqrt x-3x+x^2}{\sqrt x}=2-3\sqrt x+x^{3/2}$$
$\dfrac{-3x}{\sqrt{x}} = -3\sqrt{x}$, and $\dfrac{x^2}{\sqrt{x}} = x^{\frac{3}{2}}$
The following is multiple choice question (with options) to answer.
Find the value for x from below equation: x/3=-2? | [
"-6",
"7",
"-3",
"4"
] | A | 1. Multiply both sides by 3:
x*3/3= -2/3
2. Simplify both sides:
x = -6
A |
AQUA-RAT | AQUA-RAT-39152 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 500 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"20 sec",
"10 sec",
"5 sec",
"30 sec"
] | D | Explanation:
Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 500 * 3/50 = 30 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-39153 | 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.
The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number? | [
"240",
"270",
"295",
"360"
] | B | Let the smallest number be x.
Then larger number = (x + 1365)
x + 1365 = 6x + 15
= 5x = 1350
x = 270
Smaller number = 270.
ANSWER:B |
AQUA-RAT | AQUA-RAT-39154 | The total is $1024+175=1199$. Subtracting out the $44$ numbers between $2003$ and $2048$ gives $1155$. Thus the answer is $155$.
## Solution 3
We will count the number of it $< 2^{11}=2048$ instead of $2003$ (In other words, the length of the base-2 representation is at most $11$. If there are even digits, $2n$, then the leftmost digit is $1$, the rest, $2n-1$, has odd number of digits. In order for the base-2 representation to have more $1$'s, we will need more $1$ in the remaining $2n-1$ than $0$'s. Using symmetry, this is equal to $\frac{2^9+2^7+..+2^1}{2}$ Using similar argument where there are odd amount of digits. The remaining even amount of digit must contains the number of $1$'s at least as the number of $0$'s. So it's equal to $\frac{\binom{10}{5}+2^{10}+\binom{8}{4}+2^8+\binom{6}{3}+2^6+...+\binom{0}{0}+2^0}{2}$ Summing both cases, we have $\frac{2^0+2^1+..+2^{10}+\binom{0}{0}+..+\binom{10}{5}}{2} = 1199$. There are $44$ numbers between $2004$ and $2047$ inclusive that satisfy it. So the answer is $1199-44=\boxed{155}$
The following is multiple choice question (with options) to answer.
How many odd integers Q between 200 and 600 are there such that their tens digit is even? | [
"20",
"25",
"100",
"150"
] | C | The hundreds digit can take 4 values: 2, 3, 4 and 5;
The tens digit can take 5 values: 0, 2, 4, 6 and 8;
The units digit can take 5 values: 1, 3, 5, 7 and 9;
Total Q: 4*5*5=100.
Answer: C. |
AQUA-RAT | AQUA-RAT-39155 | Veritas Prep Reviews
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### Show Tags
05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
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Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
A can copy 50 papers in 10 hrs, while A & B can copy 70 papers in 10 hrs. How many hours are required for B to copy 26 papers? | [
"11 hours",
"12 hours",
"13 hours",
"14 hours"
] | C | A can copy 50 papers in 10 hrs, while A & B can copy 70 papers in 10 hrs.
It means B can copy 20 papers in 10 hrs.
Then time taken by B to copy 26 papers = 26*10/20 = 13 hours
ANSWER:C |
AQUA-RAT | AQUA-RAT-39156 | Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.
Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.
-
M.Scott you are genius.Thanks so much.This is the best way.Thank you again. – vikiiii Mar 30 '12 at 15:13
The following is multiple choice question (with options) to answer.
The three digits of a number add up to 23. The number is divisible by 5. The leftmost digit is double the middle digit. What is the product of the three digits? | [
"350",
"720",
"360",
"380"
] | C | Say the three-digit number is abc.
The leftmost digit is double the middle digit --> a=2b;
The number is divisible by 5 --> c is either 0 or 5;
The three digits of a number add up to 23 --> a+b+c=23
So, either a+b+c=2b+b+0=23 or a+b+c=2b+b+5=23. The first equation does not give an integer value of b, so it's not the case. From the second we get that b=6 --> a=12 and c=5 --> a*b*c=360.
Answer: C. |
AQUA-RAT | AQUA-RAT-39157 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
34 people attend a party. 4 men are single and the rest are there with their wives. There are no children in the party. In all 22 women are present. Then the number of married men at the party is ? | [
"5",
"6",
"7",
"8"
] | D | Total people = number of men + number of women
34 = 4 + number of married men + 22
number of married men = 34 - 22-4 = 8 men
ANSWER:D |
AQUA-RAT | AQUA-RAT-39158 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
A, B, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2460? | [
"750",
"888",
"261",
"787"
] | A | 1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500
Shyam = 4/13 * 3250 = 1000
Mohan = 3/13 * 3250 = 750
Answer: A |
AQUA-RAT | AQUA-RAT-39159 | ### 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.
John bought 2 shares and sold them for $96 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had | [
"a profit of $10",
"a profit of $8",
"a loss of $8",
"a loss of $10"
] | C | Loss% = (%age Profit or loss / 10)^2 = (20/10)^2 = 4% loss
Total Selling Price = 96*2 = $192
Total Cost Price = 196/(0.96) = $200
Loss = 200-192 = $8
Answer: Option C |
AQUA-RAT | AQUA-RAT-39160 | b) choose(2) choose(4) choose(7) choose(1,3).
But in actual they are same. Therefore, Your intended solution over calculates number of ways.
• @Zephyr : Consider the following : state 1 has student number 1, 2 and 3. state 2 has student number 4, 5 and 6. state 3 has student number 7, 8 and 9. Now your method will consider these two ways different : a) choose(1) choose(4) choose(7) choose(2,3) b) choose(2) choose(4) choose(7) choose(1,3), but in actual they are same. Therefore, your solution over calculates number of ways. – JVJ Sep 21 '17 at 9:27
The following is multiple choice question (with options) to answer.
A student chose a number, multiplied it by 2, then subtracted 180 from the result and got 104. What was the number he chose? | [
"90",
"100",
"120",
"142"
] | D | Solution:
Let x be the number he chose, then
2*x*180=104
2x=284
x=142
correct answer D |
AQUA-RAT | AQUA-RAT-39161 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Bhanu spends 30% of his income on petrol on scooter 20% of the remaining on house rent and the balance on food. If he spends Rs.300 on petrol then what is the expenditure on house rent? | [
"2287",
"140",
"128",
"797"
] | B | Given 30% (Income ) = 300 ⇒⇒ Income = 1000
After having spent Rs.300 on petrol, he left with Rs.700.
His spending on house rent = 20% (700) = Rs.140
Answer:B |
AQUA-RAT | AQUA-RAT-39162 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
The average of 5 quantities is 12. The average of 3 of them is 4. What is the average of remaining 2 numbers? | [
"24",
"10",
"8",
"9.5"
] | A | ANSWER: A
(5x12-3x4)/2=24 |
AQUA-RAT | AQUA-RAT-39163 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
For every positive integer n, the nth term of a sequence is the total sum of three consecutive integers starting at n. What is the total sum of terms 1 through 50 of this series? | [
"3975",
"4125",
"4375",
"4525"
] | A | Each term of the series has the form (n+n+1+n+2) = 3n+3
Since the series goes from 1 to 50, the sum of the series is:
3(1+2+...+50) + 50(3) =
3(50)(51)/2 + 50(3) =
75*51 + 150 = 3975
The answer is A. |
AQUA-RAT | AQUA-RAT-39164 | c#, algorithm, palindrome
In my tests this is 3 times faster.
I hate to break this to you, but there is a bug in your code, which will probably mean rewriting your algorithm. You algorithm assumes that the palindrome will be an odd number of characters. However, a palindrome can be an even number of characters and your code won't find it if it is.
Here's some code that will find the longest palindrome regardless:
static string LargestPalindrome(string input)
{
string output = "";
int minimum = 2;
for(int i = 0; i < input.Length - minimum; i++)
{
for(int j = i + minimum; j < input.Length - minimum; j++)
{
string forstr = input.Substring(i, j - i);
string revstr = new string(forstr.Reverse().ToArray());
if(forstr == revstr && forstr.Length > minimum)
{
output = forstr;
minimum = forstr.Length;
}
}
}
return output;
}
EDIT
The above code has a bug. Here it is reworked:
static string LargestPalindrome(string input)
{
int longest = 0;
int limit = input.Length;
for (int i = 0; i < limit; i++)
{
for (int j = limit-1; j > i; j--)
{
string forStr = input.Substring(i, j - i);
string revStr = new string(forStr.Reverse().ToArray());
if (forStr == revStr && forStr.Length > longest)
{
return forStr;
}
}
}
return "";
}
The following is multiple choice question (with options) to answer.
Consider the word RMTMR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5 letter palindromes. | [
"16500",
"17568",
"17576",
"13454"
] | C | The first letter from the right can be chosen in 26 ways because there are 26 alphabets.
Having chosen this, the second letter can be chosen in 26 ways
The first two letters can chosen in 26 x 26 = 676 ways
Having chosen the first two letters, the third letter can be chosen in 26 ways.
All the three letters can be chosen in 676 x 26 =17576 ways.
It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.
C |
AQUA-RAT | AQUA-RAT-39165 | 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 running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train. | [
"150 meter",
"299 meter",
"278 meter",
"208 meter"
] | A | Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9
= 150 meter
Answer:A |
AQUA-RAT | AQUA-RAT-39166 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
If the sum of n consecutive positive integers is 33, what of the following could be the value of n?
I. 3
II. 6
III. 11 | [
"I only",
"II only",
"III only",
"I and II"
] | D | solved it the way bunuel explained, although took some time to arrive to the answer choice.
3 -> 10+11+12=33.
6 ->3+4+5+6+7+8=33.
11 -> since we are told that we have a set of n consecutive POSITIVE numbers, we can have the smallest possible numbers:
1+2+3+4+5+6+7+8+9+10+11 - the sum is way over 33, it is actually 11*12/2 = 11*6=66.
thus, only I and II works. the answer is D |
AQUA-RAT | AQUA-RAT-39167 | 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 trained covered x km at 40 kmph and another 2x km at 20 kmph. Find the average speed of the train in covering the entire 3x km. | [
"22",
"99",
"24",
"77"
] | C | Total time taken = x/40 + 2x/20 hours
= 5x/40 = x/8 hours
Average speed = 3x/(x/8) = 24 kmph.Answer: C |
AQUA-RAT | AQUA-RAT-39168 | • Thanks for the reminder.
1. 133
2. 10
3. 666
4. 25
For #4, instead of doing it all out, you should instead list the numbers out as if you were going to add them:
(2 + 4 + 6 + 8 + 10 + … + 50)
– (1 + 3 + 5 + 7 + 9 + … + 49)
—————————————-
1 + 1 + 1 + 1 + 1 + … + 1 = 25
Notice you get a bunch of 1’s when you line them up and subtract. Now how many 1’s are there? Well, there are 25 even numbers and 25 odd numbers from 1 to 50, so you will have 25 columns lined up. Therefore, you will get 25 1’s.
3. Maroosh says:
Thank you very much for answers. Effort very much appreciated!! And thank you for explaining how to do question 4 in a shorter method. I got all right except question 2 for which i got 6 instead of 10, namely; 4567, 4568, 4569, 4678, 4679, 4789
i really don’t see any other possibilities because the digits always have to increase from left to right. Please help on this one as well. i will be very grateful
4. Maroosh Qazi says:
Thank you very much for the hint however I don’t know what I am doing but I am still only getting 9. The 3 more numbers I have figured out now are: 4578 (as you hinted), 4579 and 4689.
So the complete list is: 4567, 4568, 4569, 4578, 4579, 4678, 4679, 4689, 4789.
5. For your example #2, above, the “trifactorable” problem, I think you overlooked a tenth solution,
0 x 1 x 2 = 2, another positive integer under 1000. Just a quibble! Thanks for the lessons, which I am using for my daughter’s benefit.
6. Zeina says:
The following is multiple choice question (with options) to answer.
Look at this series: 53, 53, 40, 40, 27, 27, ... What number should come next? | [
"12",
"15",
"14",
"18"
] | C | In this series, each number is repeated, then 13 is subtracted to arrive at the next number.
ANSWER C |
AQUA-RAT | AQUA-RAT-39169 | b) choose(2) choose(4) choose(7) choose(1,3).
But in actual they are same. Therefore, Your intended solution over calculates number of ways.
• @Zephyr : Consider the following : state 1 has student number 1, 2 and 3. state 2 has student number 4, 5 and 6. state 3 has student number 7, 8 and 9. Now your method will consider these two ways different : a) choose(1) choose(4) choose(7) choose(2,3) b) choose(2) choose(4) choose(7) choose(1,3), but in actual they are same. Therefore, your solution over calculates number of ways. – JVJ Sep 21 '17 at 9:27
The following is multiple choice question (with options) to answer.
A certain school principal must choose 5 students to attend a field trip out of a group of 10 students. In addition, out of the 5 chosen students, the principal must select a note-taker and a treasurer. How many different ways are there for the principal to select the 5 students and then select the treasurer and the note-taker? | [
"1,260",
"2,520",
"5,040",
"6,020"
] | C | Out of 10 students, 5 students can be chosen in 10C5 ways.
Out of the 5 chosen students, note taker role can be taken in 5 ways
Since the note taker role is already taken by 1 student, the treasure role can be taken in 4 ways
Total number of ways = 10C5 * 5 * 4 = 5040
Answer: C |
AQUA-RAT | AQUA-RAT-39170 | # Find the Remainder when $792379237923$…upto 400 digits is divided by $101$?
Find the Remainder when $792379237923\ldots$upto 400 digits is divided by $101$?
MyApproaach
when ($792379237923\ldots$400 digts)/$101$=
I learned this approach that I have to calculate(let say U)=Is the sum all of all the alternate groups starting with the rightmost
and (let say)Th=Is the the sum all of all the alternate groups starting with the second rightmost
Rem(U-Th)/$101$=?
But I am not following how to calculate U and Th
Can anyone guide me how to approach this problem?
Let $N$ be the number. Then we really want to find $N \pmod {101}$.
Note that $N=7923\cdot10^{396}+7923\cdot10^{392}+\cdots+7923$.
Next note that $7923\equiv 45 \pmod {101}$
Also, for example, $10^{100}=100^{50}\equiv(-1)^{50}\equiv 1 \pmod {101}$
We get the same result for each term, and there are $100$ of these terms so $N\equiv 45\cdot 100\equiv 56$.
Since $100\equiv-1$ mod $101$, your number, mod $101$ is:$$-79+23-79+23-\cdots$$ which is $100$ copies of $-56$. And mod $101$, that makes $56$.
The following is multiple choice question (with options) to answer.
What is the remainder when 3990990900032 is divided by 32 ? | [
"0",
"8",
"4",
"2"
] | A | Though i was unaware of the divisibility test for 32 but i guessed the pattern!!
divisibility rule for 4- last two digits must be divisible by 4
divisibility rule for 8- last three digits must be divisible by 8
similarly, divisibility rule for 32 - last five digits must be divisible by 32
Hence, Ans A |
AQUA-RAT | AQUA-RAT-39171 | # 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.
From the below series find the remainder?
1201 × 1203 ×1205 × 1207 is divided by 6 | [
"3",
"6",
"8",
"4"
] | A | If you don't know the above rule, this problem is really calculation intensive.
But by applying the above rule, when 1201, 1201, 1203, 1204 divided by 6, leaves remainders 1, 3, 5, 1. The product of these remainders = 15.
When 15 is divided by 6, Remainder is 3
A |
AQUA-RAT | AQUA-RAT-39172 | automata, finite-automata
Title: NFA for the language that accepts binary strings ending with 00 This is the question:
1.7. Give state diagrams of NFAs with the specified number of states recognizing each
of the following languages. In all parts, the alphabet is {0,1}.
a. The language {w| w ends with 00} with three states
This is how my professor answered it:
But this is the answer I found on web:
But I think the second one should be the correct answer. The first answer is incorrect. The automaton does not accept, for example, the input word $0100$. In fact, the first automaton misses at least all words in the language that are not of the form $1^*0^*$.
The second answer is correct. It is not hard to prove it. Can you try?
The following is multiple choice question (with options) to answer.
Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A? | [
"8",
"6",
"9",
"3"
] | B | There are five letters in the given word.
Consider 5 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6.
Answer:B |
AQUA-RAT | AQUA-RAT-39173 | Your assistance. This is the actual change, where the original value is subtracted from the new value. Calculate The Discount Percentage Between Two Numbers. So, while this result might be what contact centres wants to see, it does not represent the facts. For example, the percent difference between 30% and 50% is 20%. Different relations between two numbers. com's Numbers to Ratio Calculator is an online basic math function tool to find the quantitative relationship or ratio between two or three given numbers or to reduce the ratio to its lowest terms. # Function to calculate the maximum difference between two elements in a. The Percentage Difference Calculator has 3 ways to calculate the differences between two numbers. 4 day ago Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Use the calculator below to analyze the results of a difference in sample means hypothesis test. Example, total=1,100 and you need to find percent that equals to 100. A relative delta compares the difference between two numbers, A and B, as a percentage of one of the numbers. With delayed retirement credits , a person can receive his or her largest benefit by retiring at age 70. Just Now Percentage Difference Formula. Sales have been poor and the owner decided to mark down each item to$15 to speed up the sales. percentagedifferencecalculator. The percentage diference between 10 and 12 is 18. Determine the absolute difference between two numbers: 15 - 25 = -10; Take the average of those two figures: (15 + 25)/ 2 = 20; Calculate the difference by dividing it by the average: 10/20 = 0. Hi Mahmoud, yes, I was tried that method, it seems to be not working. While most students find percentages to be an easier topic than one such as combinatorics, some individuals initially trip on the difference between a percent change and a percent of a number. com/7344/c-program-to-calculate-percentage-difference-between-2-numbers/Given two numbers, write a C program to calculate percentage differe. Tips: Put numbers in as you like, and the result will automatically be generated. Step 4: Convert that to a percentage (by multiplying by 100 and adding a "%" sign). How To Calculate The Percentage Difference Between Two. Then take this number times 100%, resulting in 40%. The Percent Change Calculator finds
The following is multiple choice question (with options) to answer.
The price of a jacket is reduced by 25%. During a special sale the price of the jacket is reduced another 30%. By approximately what percent must the price of the jacket now be increased in order to restore it to its original amount? | [
"100",
"125",
"122.2",
"105.5"
] | C | 1) Let the price of jacket initially be $100.
2) Then it is decreased by 25% , therefore bringing down the price to $75.
3) Again it is further discounted by 30%, therefore bringing down the price to $45.
4) Now 67.5 has to be added byX %in order to equal the original price.
45 + ( X% ) 45 = 100.
Solving this eq for X, we get X = 122.2
Ans is C. |
AQUA-RAT | AQUA-RAT-39174 | # 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.
If x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year, the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $38. If $2,000 is the total amount invested, how much is invested at 8 percent? | [
"$700",
"$800",
"$900",
"$1100"
] | C | 0.1x = 0.08(2000-x)+38
0.18x = 198
x = 1100
Then the amount invested at 8% is $2000 - $1100 = $900
The answer is C. |
AQUA-RAT | AQUA-RAT-39175 | # remainder of $a^2+3a+4$ divided by 7
If the remainder of $$a$$ is divided by $$7$$ is $$6$$, find the remainder when $$a^2+3a+4$$ is divided by 7
(A)$$2$$ (B)$$3$$ (C)$$4$$ (D)$$5$$ (E)$$6$$
if $$a = 6$$, then $$6^2 + 3(6) + 4 = 58$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
if $$a = 13$$, then $$13^2 + 3(13) + 4 = 212$$, and $$a^2+3a+4 \equiv 2 \pmod 7$$
thus, we can say that any number, $$a$$ that divided by 7 has remainder of 6, the remainder of $$a^2 + 3a + 4$$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
• Did you mean to write "$a^2+3a+4 \equiv 2\pmod{7}$" instead of "$a \equiv 2\pmod{7}$"? Jul 28 '15 at 7:22
• @JimmyK4542 yeah, was a little bit rush, so mistakes happen. Alot.. Jul 28 '15 at 7:23
$a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$.
The following is multiple choice question (with options) to answer.
a is a positive integer and multiple of 2; p = 4^a, what is the remainder when p is divided by 10? | [
"10",
"6",
"4",
"0"
] | B | It is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer. To help see this, consider the following examples:
4/10 is 0 with a remainder of 4
14/10 is 1 with a remainder of 4
5/10 is 0 with a remainder of 5
105/10 is 10 with a remainder of 5
It is also essential to remember that the a is a positive integer and multiple of 2. Any integer that is a multiple of 2 is an even number. So, a must be a positive even integer.
With these two observations, the question can be simplified to:what is the units digit of 4 raised to an even positive integer?
The units digit of 4 raised to an integer follows a specific repeating pattern:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
4^(odd number) --> units digit of 4
4^(even number) --> units digit of 6
There is a clear pattern regarding the units digit. 4 raised to any odd integer has a units digit of 4 while 4 raised to any even integer has a units digit of 6.
Since a must be an even integer, the units digit of p=4^a will always be 6. Consequently, the remainder when p=4^a is divided by 10 will always be 6.
In case this is too theoretical, consider the following examples:
a=2 --> p=4^a=16 --> p/10 = 1 with a remainder of 6
a=4 --> p=4^a=256 --> p/10 = 25 with a remainder of 6
a=6 --> p=4^a=4096 --> p/10 = 409 with a remainder of 6
a=8 --> p=4^a=65536 --> p/10 = 6553 with a remainder of 6
Answer: B. |
AQUA-RAT | AQUA-RAT-39176 | ## 1 Answer
You also cannot succeed with $$8$$ tiles. Each of the tiles can only cover one of the squares marked with an $$\times$$: $$\begin{array}{|c|c|c|c|c|c|c|} \hline \times &\;\,&\;\,&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline \end{array}$$
Here is a solution with $$9$$ tiles:
• Wow that's so nice... and I am blind. – abc... Feb 23 at 3:53
• Very nice. For a slightly prettier solution with 9 tiles, you can start with a rotationally symmetric arrangement of 8 tiles covering all but the center square. Namely, put tiles on b1&c2, d2&e1, g2&f3, f4&g5, f7&e6, d6&c7, a6&b5, b4&a3; every square but d4 is covered. – bof Feb 23 at 5:40
The following is multiple choice question (with options) to answer.
A room 8 m 47 cm long and 7m 77 cm broad is to be paved with square tiles. Find the least number of square tiles required to cover the floor. | [
"13636",
"13440",
"13647",
"13431"
] | D | Explanation:
Area of the room = (847 x 777) cm2.
Size of largest square tile = H.C.F. of 847 cm and 777 cm = 7 cm.
Area of 1 tile = (7 x 7) cm2.
Number of tiles required =(847×777)/(7×7)=13431
Answer: Option D |
AQUA-RAT | AQUA-RAT-39177 | ### 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 the product of 3 consecutive integers, which of the following must be true about n?
I. n is a multiple of 2
II. n is a multiple of 3
III. n is a multiple of 4 | [
"I only",
"II only",
"III only",
"I and II"
] | D | n is the product of 3 consecutive integers.
So, according to the rule, n must be divisible by 3, 2 and 1
So, we already know that statements I and II must be true.
Do we need to check statement III? No.
Notice that NONE of the answer choices include all 3 statements. Since we've already concluded that statements I and II are true, the correct answer is D. |
AQUA-RAT | AQUA-RAT-39178 | 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.
36 men can complete a piece of work in 18 days. In how many days will 72 men complete the same work ? | [
"24",
"17",
"18",
"9"
] | D | Explanation:
Less Men, means more Days {Indirect Proportion}
Let the number of days be x
then,
72 : 36 :: 18 : x
x = 9
Answer: D) 9 days |
AQUA-RAT | AQUA-RAT-39179 | Since 2m+1 is greater than P1, x1 is positive. An even number minus an odd plus an odd equals an even, and half an even number is an integer. Hence, x1 is a solution to S.
In the second case, where 2m+1 is less than P1, let’s try N = 2m+1 as a solution. This yields: $x_1 = (\frac{2S}{2^{m+1}} – 2^{m+1} + 1)/2 \\ S = 2^m\cdot P_1 \cdot S_2 \implies \\ x_1 = (\frac{2 \cdot 2^m \cdot P_1 \cdot S_2}{2^{m+1}} – 2^{m+1} + 1)/2 \\ = (P_1 \cdot S_2 – 2^{m+1} + 1)/2$
Since P1 is greater than 2m+1, x1 is again positive. P1 and S2 are both odd, hence P1S2 is odd. An odd number minus an even plus an odd is an even number, and half of an even number is an integer. Hence, x1 is a solution to S.
This proves that S has a solution R for all values that are not powers of 2; it also illustrates why 23•16 requires 23 numbers (23 < 32) while 23•8 requires only 16 (16 < 23). As we would expect, 592 (37•16) requires 32 numbers, not 37 (3+4+...+34); 1184 (37•32) requires 37 (14+15+...+50). All of this proves the original hypothesis: A positive integer S is equal to the sum of consecutive positive integers iff it has odd prime factors.
## 1 Comment
1. paulhartzer
It occurred to me later that this provides a direct proof all odd numbers have N=2 solutions: All odd numbers are of the form 2^0•P_1•S_2. Since 2^(0+1) = 2 will always be lower than P_1, N=2^(0+1)=2 will always provide a solution.
The following is multiple choice question (with options) to answer.
If x is a positive integer, which of the following must be odd? | [
"x+1",
"x^2+x",
"x^2+x+7",
"x^2−1"
] | C | A. X+1 = can be odd or even. Since O + O =E or E + O = O
B. X^2 + X = X(X+1). Since from the above derivation we already know the term X+1 can be Odd or Even, directly substitute here. X(ODD) = Even (When X is Even) or X(EVEN) = EVEN [When X is ODD]
C. Here's the answer. Since we know the term x^2 + X can always take a EVEN number, EVEN + 7 = ODD
Hence C. |
AQUA-RAT | AQUA-RAT-39180 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
The length of the bridge, which a train 110 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"265",
"288",
"245",
"776"
] | A | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (110 + X)/30 = 25/2 ==> 2(110 + X) = 750 ==> X
= 265 m.
Answer:A |
AQUA-RAT | AQUA-RAT-39181 | # Given a bag of 3 red marbles, 5 black, and 8 green, what is the probability that?
A bag has:
$3$ red
$5$ black
$8$ green
marbles.
Total of 16 marbles.
You select a marble, and then another one right after. (without replacement).
What is the probability that $both$ are red?
Probability that first pick is red: $\frac{3}{16}$
Probability that second pick is red: $\frac{2}{15}$ (since one ball is removed)
Probability of both marbles being red is: $\frac{3}{16} \cdot \frac{2}{14} = \frac{1}{40}$
How do I do this using combinations only?
• How exactly did that $\frac{2}{15}$ become $\frac{2}{14}$??? Dec 11 '16 at 7:54
• @barakmanos, as $\frac 3{16}\cdot\frac 2{15}=\frac 1{40}$ it looks like a sinple typo. Dec 11 '16 at 7:59
Hint: $$\frac {\text { number of ways in which you can choose 2 red balls without replacement from 15 balls}}{ \text {number of ways in which you can choose 2 balls of any colour without replacement from 15 balls}}=\frac {\binom {3}{2}}{ \binom {16}{2}}=??$$
Hope this helps you.
The following is multiple choice question (with options) to answer.
A bag contains 8 red, 6 blue and 4 green balls. If 2 ballsare picked at random, what is the probability that both are red? | [
"1/13",
"2/23",
"5/26",
"7/30"
] | D | P(Both are red),
=8C216C2=8C216C2
=28/120=7/30
D |
AQUA-RAT | AQUA-RAT-39182 | 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 passes a km stone in 15 seconds and another train of the same length travelling in opposite direction in 8 seconds.The speed of the second train is | [
"60 kmph",
"66 kmph",
"72 kmph",
"99 kmph"
] | D | Given that two trains are of same length i.e..150 mtrs
first train passes the km stone in 15 seconds. here we have time and distance so speed=150/15=10 m/s
we need to find out the second train speed.
suppose the speed of the 2nd train is x m/s
Relative speed of two trains is (10+x)
==> (150+150)/(10+x)=8
==> (300)/(10+x)=8
==> 300=80+8x
==> 300-80=8x
==> 220=8x
:- x=55/2 m/s
convert m/s into km/ph
(55/2)*(18/5)=99kmph
ANSWER:D |
AQUA-RAT | AQUA-RAT-39183 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
A committee has 4 men and 6 women. What are the number of ways of selecting 2 men and 3 women from the given committee? | [
"120",
"160",
"200",
"240"
] | A | The number of ways to select two men and three women = 4C2 * 6C3 = 6 * 20 = 120
The answer is A. |
AQUA-RAT | AQUA-RAT-39184 | but x, y must satisfy the negative Pell equation x2 − 2y2 = −1.
nigiri sushi. and the thrust pressure, )
The original cube (1m sides) has a surface area to volume ratio of 6:1. {\displaystyle x^{3}+(-x)^{3}+n^{3}=n^{3}} The figure above shows a cube. (If I didn't remember, or if I hadn't been certain, I'd have grabbed my calculator and tried cubing stuff until I got the right value, or else I'd have taken the cube root of 64.). Yes, a2 – 2ab + b2 and a2+ 2ab + b2 factor, but that's because of the 2's on their middle terms.
( You know that L and h have to be the same because, by definition, in a cube, all sides are the same. (Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade. Face diagonals are line segments linking the opposite corners of a face. + cubes of numbers in arithmetic progression with common difference d and initial cube a3, is known for the special case of d = 1, or consecutive cubes, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[6]. 3
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The following is multiple choice question (with options) to answer.
The sides of a cube are in the ratio of 1:2 the ratio of their volumes is ? | [
"1:2",
"1:4",
"1:8",
"2:1"
] | C | 1:8
Answer : C |
AQUA-RAT | AQUA-RAT-39185 | 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 alone can do a work in 12 days. A and B together finished work and got rs.54 and rs.18 respectively. In how many days was the work finished? | [
"12",
"14",
"9",
"10"
] | C | A does the work in 12 days
and A and B together finished work and got rs.54 and rs.18 respectively
so B will do work in 12*3 = 36 days (because he got 1/3 money relative to other)
so one day work = 1/12 + 1/36 = 4/36 = 1/9
so total days required = 9
ANSWER:C |
AQUA-RAT | AQUA-RAT-39186 | Overcounting cricketer combinations
Following this, Navneet had a new problem:
I am stuck on another problem:
A team of 11 is to be selected out of 10 batsmen, 5 bowlers, and 2 keepers such that in the team at least 4 bowlers should be included. Find the number of possible ways of selection.
I tried to solve this question like this:
First select 4 bowlers out of 5 = 5C1
Then, remaining candidates = 10+2+(5-4) = 13
Hence, select the remaining 7 players out of 13 = 13C7
So, my final answer is 5C4*13C7
But, this is a wrong answer.
The correct answer given is (5C4*12C7)+(5C5*12C6)
Please explain me where I am doing the error?
Also, can you please tell me what should I check or do in order to avoid such errors in future?
Again, a well-asked question, showing his thinking along with the problem and the provided answer, so we have all we need. He got 8580, while they say 4884. Why?
Doctor Rick responded:
I thought first of the same approach you took. Then I considered it more carefully, looking to see if I had missed any possibilities or if I had counted any selection more than once.
I then realized that I was overcounting, and here’s why: You’re selecting four bowlers to include first, and then maybe the fifth bowler will be among the remaining 7 players you choose. But if you chose a different set of four bowlers to start, and then the fifth bowler, you’d end up with the same set of 11 players — you just picked all five bowlers in a different order.
More specifically:
The following is multiple choice question (with options) to answer.
A batsman scored 130 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? | [
"45(4/11) %",
"45 %",
"53(11/13) %",
"44(5/11) %"
] | C | Explanation :
Total runs scored = 130
Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60
Total runs scored by running between the wickets = 130 - 60 = 70
Required % = (70/130) × 100 = 700/13 = 53(11/13)%
Answer : Option C |
AQUA-RAT | AQUA-RAT-39187 | $38k - (2*38 - 7)l = 34$
$38(k-2l) + 7l = 34$. Let $m=k-2l$
$38m + 7l = 34$.
$(3+ 5*7)m + 7l = 34$
$3m + 7(5m + l) = 34$. Let $n = 5m+l$.
$3m + 7n = 34$
$3m + 2*3n + n = 34$
$3(m + 2n) + n = 34$. Let $a = m+2n$.
$3a + n = 34$.
Let $a=11; n = 1$. So $a=m+2n; 11=m + 2; m = 9$. and $n= 5m + l; 1=5*9 +l; l = -44$. and $m = k - 2l; 9=k +88; k = -79$.
So $25 - 79*38 = -2977$ and $59 - 44*69=-2977$.
So $-2977\equiv 25 \mod 38$ and $-2977 \equiv 59 \mod 69$.
That's of course not positive but.
$25 - 79*38 = 59- 44*69 \iff$
$25 - 79*38 + 69*38 = 59 - 44*69 + 69*38 \iff$
$25 - 10*38 = 59 - 6*69 \iff$
$25 - 10*38 + 69*38 = 59 - 6*69 + 69*38 \iff$
$25 + 59*38 = 59 + 32*69$
And $25+59*38 = 59 + 32 * 69 =2267$.
$2267\equiv 25\mod 38$ and $2267\equiv 59\mod 69$ and as the lowest common multiple of $38$ and $69$ is $2622$ this is the smallest positive such number.
The following is multiple choice question (with options) to answer.
The L.C.M of two numbers is 48. The numbers are in the ratio 4:3. The sum of numbers is: | [
"28",
"30",
"40",
"50"
] | A | Let the numbers be 4x and 3x.
Then, their L.C.M = 12x. So, 12x = 48 or x = 4.
The numbers are 16 and 12.
Hence, required sum = (16 + 12) = 28.
ANSWER:A |
AQUA-RAT | AQUA-RAT-39188 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
A salesman sold twice as much pears in the afternoon than in the morning. If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? | [
"150",
"190",
"100",
"240"
] | D | Let x be the number of kilograms he sold in the morning.Then in the afternoon he sold 2x kilograms. So, the total is x+2x=3x This must be equal to 360.
3x=360
x=360/3
x=120
Therefore, the salesman sold 120 kg in the morning and 2⋅120=240 kg in the afternoon. |
AQUA-RAT | AQUA-RAT-39189 | # 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 motorcycle importer is planning on increasing the price of a certain model by $1000. At this new price 8 fewer motorcycles will be sold per month, but the total revenues will increase by $26,000 to $594,000. What is the number of motorcycles the manufacturer will sell at this new price? | [
"61",
"62",
"65",
"63"
] | D | Responding to a pm: To solve this question, I will make an equation in x and then make educated guesses. Here's how:
Assuming x motorcycles were sold every month initially.
(568000/x + 1000)(x-8) = 594000
(568/x + 1)(x-8) = 594
Now 568 = 8*71
Assuming x = 71 (We have all integers so it is obvious that 568/x should be an integer.
We get 9*66 = 594 (matches)
So he will sell 71 - 8 = 63 bikes this month
Answer (D) |
AQUA-RAT | AQUA-RAT-39190 | ros-kinetic
[ INFO] [1550419387.914285308, 142.078000000]: HectorSM p_odom_frame_: odom_combined
[ INFO] [1550419387.914301138, 142.078000000]: HectorSM p_scan_topic_: lidar
[ INFO] [1550419387.914315614, 142.078000000]: HectorSM p_use_tf_scan_transformation_: true
[ INFO] [1550419387.914329710, 142.078000000]: HectorSM p_pub_map_odom_transform_: true
[ INFO] [1550419387.914342660, 142.078000000]: HectorSM p_scan_subscriber_queue_size_: 5
[ INFO] [1550419387.914357817, 142.078000000]: HectorSM p_map_pub_period_: 0.250000
[ INFO] [1550419387.914371499, 142.078000000]: HectorSM p_update_factor_free_: 0.450000
[ INFO] [1550419387.914386448, 142.078000000]: HectorSM p_update_factor_occupied_: 0.900000
[ INFO] [1550419387.914398388, 142.078000000]: HectorSM p_map_update_distance_threshold_: 0.020000
[ INFO] [1550419387.914434465, 142.078000000]: HectorSM p_map_update_angle_threshold_: 0.100000
[ INFO] [1550419387.914461370, 142.078000000]: HectorSM p_laser_z_min_value_: -1.000000
[ INFO] [1550419387.914487060, 142.078000000]: HectorSM p_laser_z_max_value_: 1.000000
The following is multiple choice question (with options) to answer.
Convert the following unit:
2.5 hectares in ares | [
"200 ares.",
"210 ares.",
"220 ares.",
"250 ares."
] | D | 2.5 hectares in ares
1 hectare = 100 ares
Therefore, 2.5 hectares = 2.5 × 100 ares
= 250 ares.
ANSWER : OPTION D |
AQUA-RAT | AQUA-RAT-39191 | One of you all sent a fairly interesting problem, so I thought I would work it out. The problem is I have a group of 30 people, so 30 people in a room. They're randomly selected 30 people. And the question is what is the probability that at least 2 people have the same birthday? This is kind of a fun question because that's the size of a lot of classrooms. What's the probability that at least someone in the classroom shares a birthday with someone else in the classroom? That's a good way to phrase as well. This is the same thing as saying, what is the probability that someone shares with at least someone else. They could share it with 2 other people or 4 other people in the birthday. And at first this problem seems really hard because there's a lot of circumstances that makes this true. I could have exactly 2 people have the same birthday. I could have exactly 3 people have the same birthday. I could have exactly 29 people have the same birthday and all of these make this true, so do I add the probability of each of those circumstances? And then add them up and then that becomes really hard. And then I would have to say, OK, whose birthdays and I comparing? And I would have to do combinations. It becomes a really difficult problem unless you make kind of one very simplifying take on the problem. This is the opposite of-- well let me draw the probability space. Let's say that this is all of the outcomes. Let me draw it with a thicker line. So let's say that's all of the outcomes of my probability space. So that's 100% of the outcomes. We want to know-- let me draw it in a color that won't be offensive to you. That doesn't look that great, but anyway. Let's say that this is the probability, this area right here-- and I don't know how big it really is, we'll figure it out. Let's say that this is the probability that someone shares a birthday with at least someone else. What's this area over here? What's this green area? Well, that means if these are all the cases where someone shares a birthday with someone else, these are all the area where no one shares a birthday with anyone. Or you could say, all 30 people have different birthdays. This is what
The following is multiple choice question (with options) to answer.
A teacher will pick a group of 4 students from a group of 9 students that includes Bart Lisa and john. If one of all the possible four-student groups is picked at random, what is the probability of picking a group that includes john Bart and Lisa? | [
"1/7",
"3/14",
"1/4",
"1/21"
] | D | Probability = Favorable Outcomes / Total Outcomes
Total Outcomes= Total No. of ways of Picking Group of 4 out of 9 = 9C4 = 9! / (5! * 4!) = 126
Favorable Outcomes= Total No. of ways of Picking Group of 4 out of 9 such that J, B and L are always in the group (i.e. we only have to pick remaining one out of remaining 6 as J, B and L must be there is group) = 6C1 = 6
Hence,Probability=6/126=1/21
Answer: Option D |
AQUA-RAT | AQUA-RAT-39192 | # 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.
At what price must an article costing Rs.47.50 be marked in order that after deducting 5% from the list price. It may be sold at a profit of 25% on the cost price? | [
"62.5",
"62.7",
"62.9",
"62.1"
] | A | CP = 47.50
SP = 47.50*(125/100) = 59.375
MP*(95/100) = 59.375
MP = 62.5
Answer: A |
AQUA-RAT | AQUA-RAT-39193 | # Is “divisible by 15” the same as “divisible by 5 and divisible by 3”?
Is stating that a number $x$ is divisible by 15 the same as stating that $x$ is divisible by 5 and $x$ is divisible by 3?
-
The two assertions are equivalent. – André Nicolas Feb 1 '12 at 23:38
may be worth noting for somebody who might be new to this, that it happens to work in both directions here, because 3 and 5 are co-prime. i.e. if you ask the same question for a different example you may not get the converse implication. – Beltrame Feb 1 '12 at 23:49
If you want to figure out the general facts at work here, you could read this Keith Conrad handout. – Dylan Moreland Feb 1 '12 at 23:55
Yes, if a number $n$ is divisible by $15$, this means $n=15k$ for some integer $k$. So $n=5(3k)=3(5k)$, so it is also divisible by $3$ and $5$.
Conversely, if $n$ is divisible by $3$ and $5$, it is a simple lemma that it is divisible by the least common multiple of $3$ and $5$. Since $3$ and $5$ are coprime, their lcm is just $15$.
-
I feel that this answer is incomplete. It does not prove that x|n and y|n implies that lcm(x,y)|n; it just claims it's "a simple lemma". A better answer would show why this is true. – user22805 Feb 2 '12 at 8:32
Suppose there's a number n such that x|n and y|n, but lcm(x,y) does not divide n. Then write m = lcm(x,y) and n = pm+q, where 0 < q < m and p is an integer. Then x and y must both divide q, so m is not the lcm of x and y - contradiction. – user22805 Feb 2 '12 at 8:35
The following is multiple choice question (with options) to answer.
Which of the following must be an integer if the integer x is divisible by both 15 and 21? | [
"x/252",
"x/189",
"x/126",
"x/105"
] | D | the integer has to be multiple of LCM of 15 and 21, which is 105..
so x/105 will be an integer
ans D |
AQUA-RAT | AQUA-RAT-39194 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
If you cut a 20ft piece of wood into two pieces making one piece 6ft longer than the other. What size is the smaller piece? | [
"4ft",
"7ft",
"8ft",
"10ft"
] | B | Total length is 20ft, one piece is 6ft longer (x+6), leaving the other piece to figure out (x).
(x)+ (x+4)=20
x+x+6-6=20-6
2x=14
2x/2=14/2
x=7
The piece is B) 7ft. |
AQUA-RAT | AQUA-RAT-39195 | ## Solution 4
Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$.
Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$. Dividing everything by $b^2$, we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$.
Applying the quadratic formula....and following the restriction that $a>b>0$....
$\frac{a}{b}=\frac{10}{7}$.
Hence, $7a=10b$.
Since they are relatively prime, $a=10$, $b=7$.
$10-7=\boxed{\textbf{(C)}\ 3}$.
## Solution 5
Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{\textbf{(C)}\ 3}$ ~mathboy282
The following is multiple choice question (with options) to answer.
If 2a = 3b and ab ≠0, what is the ratio of a/3 to b/2? | [
"27/8",
"8/27",
"1",
"9/4"
] | C | A nice fast approach is the first find a pair of numbers that satisfy the given equation: 2a = 3b
Here's one pair: a =3 and b =2
What is the ratio of a/3 to b/3?
In other words, what is the value of (a/3)/(b/2)?
Plug in values to get: (a/3)/(b/2) = (3/3)/(2/2)
= 1/1
= 1
C |
AQUA-RAT | AQUA-RAT-39196 | 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.
If 12 men and 16 boys can do a piece of work in 5 days; 13 men and 24 boys can do it in 4 days, then the ratio of the daily work done by a man to that of a boy is? | [
"2:1",
"2:7",
"2:2",
"2:6"
] | A | Let 1 man's 1 day work = x and 1 boy's 1 day work = y.
Then, 12x + 16y = 1/5 and 13x + 24y = 1/4
Solving these two equations, we get:
x = 1/100 and y = 1/200
Required ratio = x:y = 1/100 : 1/200 = 2:1.
Answer:A |
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