Thanks Angadbir , that helps .
Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4- 1.6
b) 4+ 8.4
c) 4+ 10.4
d) 2- 1.6
e) 2- 0.8
Solution :
The two cars travel around the circumference of a circle, the measure of which is 2*pi*r = 2*pi* 10 = 20*pi miles.
Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point).
At Car A's start time, the additional distance to cover before the cars meet is (20*pi - 20) miles. We are interested in the moment when the cars have passed each other and have traveled another 12 miles: (20pi - 20) + 12 = (20pi - 8 ) miles.
Let us first use an RTD chart to determine the time it takes for cars A and B to cover that distance of interest. We will employ a shortcut: since the cars are traveling in opposite directions, they are Working Together to cover the distance, so we can add their rates.
R
T
D
Car A 3 mph
t hours
Car B
2 mph
t hours
Total:
5 mph
t hours
20pi - 8 miles
Note that the rates add, but the times do not: both cars are traveling simultaneously for t hours.
We solve for t using the RT = D formula, or T = D/R = (20pi - 8 ) miles / 5 mph = (4pi - 1.6) hours.
So Car A traveled (4pi - 1.6) hours, and Car B traveled the same time plus the 10 hours it traveled alone before the simultaneous travel began.
The total time for Car B is therefore 4pi 1.6 + 10 = 4pi + 8.4 hours.
The correct answer is B.
---------------
I'd like to know how "At Car A's start time, the additional distance to cover before the cars meet is (20*pi - 20) miles."
Why is one distance measures as a Circular distance and the other one as linear one?
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?
A. 3
B. 16
C. 75
D. 24
E. 26
A. 3
B. 16
C. 75
D. 24
E. 26
Thanks Nairpraveenk...hope to see more from you. I have just started my GMAT preparations and this will be helping me a lot
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?
A. 3
B. 16
C. 75
D. 24
E. 26
Correct Answer - C. 75
Number of integers between 200 and 300 divisible by 2 = 51
Number of integers between 200 and 300 divisible by 3 = 34
Number of integers between 200 and 300 divisible by 5 = 21
Number of integers between 200 and 300 divisible by both 2 and 3 = 16
Number of integers between 200 and 300 divisible by both 2 and 5 = 11
Number of integers between 200 and 300 divisible by both 3 and 5 = 8
Number of integers between 200 and 300 divisible by 2, 3 and 5 = 4
Hence, number of integers between 200 and 300 divisible by 2, 3 or 5 = 51 + 34 + 21 - 16 - 11 - 8 + 4 = 75
Car B starts at point X and moves clockwise around a circular track at a constant rate of 2 mph. Ten hours later, Car A leaves from point X and travels counter-clockwise around the same circular track at a constant rate of 3 mph. If the radius of the track is 10 miles, for how many hours will Car B have been traveling when the cars have passed each other for the first time and put another 12 miles between them (measured around the curve of the track)?
a) 4- 1.6
b) 4+ 8.4
c) 4+ 10.4
d) 2- 1.6
e) 2- 0.8
Solution :
The two cars travel around the circumference of a circle, the measure of which is 2*pi*r = 2*pi* 10 = 20*pi miles.
Car B has already traveled (2 mph)(10 hours) = 20 miles in a clockwise direction by the time Car A starts moving 10 hours later in a counterclockwise direction (and from the same starting point).
At Car A's start time, the additional distance to cover before the cars meet is (20*pi - 20) miles. We are interested in the moment when the cars have passed each other and have traveled another 12 miles: (20pi - 20) + 12 = (20pi - 8 ) miles.
Let us first use an RTD chart to determine the time it takes for cars A and B to cover that distance of interest. We will employ a shortcut: since the cars are traveling in opposite directions, they are Working Together to cover the distance, so we can add their rates.
R
T
D
Car A 3 mph
t hours
Car B
2 mph
t hours
Total:
5 mph
t hours
20pi - 8 miles
Note that the rates add, but the times do not: both cars are traveling simultaneously for t hours.
We solve for t using the RT = D formula, or T = D/R = (20pi - 8 ) miles / 5 mph = (4pi - 1.6) hours.
So Car A traveled (4pi - 1.6) hours, and Car B traveled the same time plus the 10 hours it traveled alone before the simultaneous travel began.
The total time for Car B is therefore 4pi 1.6 + 10 = 4pi + 8.4 hours.
The correct answer is B.
---------------
I'd like to know how "At Car A's start time, the additional distance to cover before the cars meet is (20*pi - 20) miles."
Why is one distance measures as a Circular distance and the other one as linear one?
Circumference of the circle = 2*pi*r = 2*pi*10 = 20pi
When Car A starts, Car B has already covered 10*2 = 20 miles. Assume this as a start point of Car B. This should be circumference - 20 miles = 20pi - 20. This means that the distance between point P to point Q is 20pi - 20. However, the actual distance in question is 20pi - 20 + 12 = 20pi -8.
Its a linear distance. Only the path is circular. i.e. the length of the path is in the form a circle's circumference.
Puys,
Are questions on Surds and Logarithm common in GMAT or are these topics something which have never been asked?
How many positive integers between 200 and 300 (both inclusive) are not divisible by 2, 3 or 5?
A. 3
B. 16
C. 75
D. 24
E. 26
From 210 to 300, there are (300 - 210)*(1-1/2)(1-1/3)(1-1/5) = 24 such no.s
From 200 - 210 there is 203 and 209 as well.
So total 26 such no.s
Puys,
Are questions on Surds and Logarithm common in GMAT or are these topics something which have never been asked?
Certainly common.
Thanks Angadbir,
How difficult/lenghty the questions can be? Also in terms of difficulty where does it stands when compared with CAT?
Angadbir SaysCertainly common.
Thanks Angadbir,
How difficult/lenghty the questions can be? Also in terms of difficulty where does it stands when compared with CAT?
Soad, i'm not very well versed with the vastness of GMAT, but from what i've seen, Questions from Surds and Logs usually come up as the first initial problems in the Quant section. With that, u can assume that they are of basic level.
But as you progress through the exam and start getting tougher questions, questions on Surds and Logs are rare to come by. They are rather from DS,probability,functions,graphs,percentages. DS being the most tricky ones.
From 210 to 300, there are (300 - 210)*(1-1/2)(1-1/3)(1-1/5) = 24 such no.s
From 200 - 210 there is 203 and 209 as well.
So total 26 such no.s
Is there any generalization that defines such problems?
For example : what would be the number of positive integers between 300 and 400 such that they are not divisible by 2, 3 or 5?
Got it....
The mistake that I did was taking the order of the teams into account, which is not required for this question.
What this means is, Say we have 8 people numbered from 1-8.
One set of the four teams: (1,2), (3,4), (5,6), (7,8 )
Now even if we change the order of the teams to (5,6), (3,4), (1,2), (7,8 ),
it does not change the number of TEAMS, that, remains exactly same, that is why we need to divide the whole answer by 4!, taking into account the 4 Teams.
I am confused about this.
How is the above problem different from the below?:w00t:
What is the number of ways of dividing 6 boys into 2 equal groups?
The method here would be 6!/(3!^2) = 720/36 = 20
(Since boys will be distinct)
For the problem given, in case the groups are not distinct(for example, If they are some groups of identical boxes), it would be
= (8!)/(4! * 2!^4)
= 40320/24 * 16)
=105.
But which is not the case.The groups formed will be distinct from each other and hence it is 2520..
Correct me if I am wrong..
Regards,
Sabby
For the triangle shown above, where A, B and C are all points on a circle, and line segment AB has length 18, what is the area of triangle ABC?(1) Angle ABC measures 30.
(2) The circumference of the circle is 18
.Happy Solving..
Regards,
Sabby
Correct Answer - C. 75
Number of integers between 200 and 300 divisible by 2 = 51
Number of integers between 200 and 300 divisible by 3 = 34
Number of integers between 200 and 300 divisible by 5 = 21
Number of integers between 200 and 300 divisible by both 2 and 3 = 16
Number of integers between 200 and 300 divisible by both 2 and 5 = 11
Number of integers between 200 and 300 divisible by both 3 and 5 = 8
Number of integers between 200 and 300 divisible by 2, 3 and 5 = 4
Hence, number of integers between 200 and 300 divisible by 2, 3 or 5 = 51 + 34 + 21 - 16 - 11 - 8 + 4 = 75
Almost dude.. the question is how many are not divisible..
so answer is 101 - 75 = 26
(1) Angle ABC measures 30.
(2) The circumference of the circle is 18
Happy Solving..
Regards,
Sabby
Is this a problem solving question, or a data sufficiency one?
It appears to be a DS type ques, so I'll go woth C ie. both statements are needed.
The second statement confirms that AB is the diameter, and so angle ACB = 90.
Now from statement I and basic trigo, we can find the area (but no need to solve in the context of DS).
Puys,
Can anyone please help me with the below questions
1) The probability that a given problem will be solved by A, B and C are 2/3, 5/7, 4/5 resp.
Find the probability of the problem being solved.
(a) 40/105 (b) 101/105 (C) 103/105 (d) None of the above
2) In how many ways can 5 boys and 3 girls sit around a table in such a way that no two girls sit together?
(a) 480 (b) 960 (c) 320 (d) 1500 (e) 1440
3) There are 5 balls of different colors and 5 boxes of colors the same as those of the balls. The no of ways in which the balls, one in each box can be placed such that a ball does not go to a box of its own color is
(a) 40 (b) 44 (c) 42 (d) 36 (e) 34
Puys,
Can anyone please help me with the below questions
1) The probability that a given problem will be solved by A, B and C are 2/3, 5/7, 4/5 resp.
Find the probability of the problem being solved.
(a) 40/105 (b) 101/105 (C) 103/105 (d) None of the above
2) In how many ways can 5 boys and 3 girls sit around a table in such a way that no two girls sit together?
(a) 480 (b) 960 (c) 320 (d) 1500 (e) 1440
3) There are 5 balls of different colors and 5 boxes of colors the same as those of the balls. The no of ways in which the balls, one in each box can be placed such that a ball does not go to a box of its own color is
(a) 40 (b) 44 (c) 42 (d) 36 (e) 34
Puys,
Can anyone please help me with the below questions
1) The probability that a given problem will be solved by A, B and C are 2/3, 5/7, 4/5 resp.
Find the probability of the problem being solved.
(a) 40/105 (b) 101/105 (C) 103/105 (d) None of the above
Probability that no one solves the problem = 1/3*2/7*1/5=2/105
Probability that the problem will be solved = 1-(2/105)= 103/105
Puys,
Can anyone please help me with the below questions
3) There are 5 balls of different colors and 5 boxes of colors the same as those of the balls. The no of ways in which the balls, one in each box can be placed such that a ball does not go to a box of its own color is
(a) 40 (b) 44 (c) 42 (d) 36 (e) 34
This is a problem based on the Inclusion-Exclusion principle, sometimes also known as De-arrangement. (Read more here : Math Forum - Ask Dr. Math)
The formula is : 1/2!-1/3!+1/4!-1/5!.......+*(1/n!), but this gives the probability.
For n=5, P = 1/2-1/6+1/24-1/120 = 44/120,
since we know the total ways is 5!, or 120, the possible ways are 44.
Puys,
Can anyone please help me with the below questions
2) In how many ways can 5 boys and 3 girls sit around a table in such a way that no two girls sit together?
(a) 480 (b) 960 (c) 320 (d) 1500 (e) 1440
I am not quite sure how to go about this one, will need some help....
total people = 8, no of ways to arrange them = 7!
no of arrangements where 2 girls sit together = 3P2(select 2 girls out of the three and arrange them)*6! (treat them as one and total no of people becomes 7) = 6*6!
Thus no of ways when no 2 girls sit together = 7!-(6*6!) = 720....oops........
can anyone point out the mistake in this solution
