Try this
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
a 5
b 5(x-y)
c 20x
d 20y
e 35x
Answer is 32.
No. of ways in which Drivers seat can be occupied: 2
In each of the case, consider two other case
Case 1. One of the daughter sits on the front seat --> 2 ways
People on the backseat can sit in 6 ways
Therefore a total of 12 ways.
Case 2: Both daughters sit in the back.
No. of arrangements: 2 (for daughters) * 2 (for other people) = 4 ways.
Therefore total = 2* (12 + 4) = 32.
Try this
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
a 5
b 5(x-y)
c 20x
d 20y
e 35x
answer shd b option C.. dont know ne direct method 2 solve this question.. tried using substituting values for x and y.. was left wid option C alone..
still trying 2 figure out a better shorter method.. will put the solution again if i can figure out something shorter..
I think A is the answer
answer shd b option C.. dont know ne direct method 2 solve this question.. tried using substituting values for x and y.. was left wid option C alone..
still trying 2 figure out a better shorter method.. will put the solution again if i can figure out something shorter..
answer is c
Try this
If x and y are positive integers, which of the following CANNOT be the greatest common divisor of 35x and 20y?
a 5
b 5(x-y)
c 20x
d 20y
e 35x
CANNOT = Not possible
35X = 5*7*X
20Y = 2*2*5*Y
a) obviously is - 5 can divide both.
b) 35X/5(x-y)=7x/(x-y) - can be an integer = K implies y=((K-7)/k)x. K is an integer. Put K as 8. y = ( 1/8 )x, which is a possibility. similar logic applies for 20y
c) 35x/20x = 7/4 (definitely not an integer) - So this is the correct choice.
d) 35X/20Y = 7x/4Y can be an integer. 20Y/20Y =1 (Integer).
e) 35X/35x=1(Integer), 20Y/35x = 4y/7x = can be an integer.
566. A right-angled triangle XYZ is in the xy-plane so that the right angle is at X and XZ is parallel to the x-axis... x, y coordinates of X, Y and Z are integers that satisfy the inequalities -4
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
566. A right-angled triangle XYZ is in the xy-plane so that the right angle is at X and XZ is parallel to the x-axis... x, y coordinates of X, Y and Z are integers that satisfy the inequalities -4
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
Am I missing something or do we need to have a range for Z? Otherwise it looks like we can have infinite right triangles for any number of values of Z.(Or my ignorance knows no depths)
phemanth SaysAm I missing something or do we need to have a range for Z? Otherwise it looks like we can have infinite right triangles for any number of values of Z.(Or my ignorance knows no depths)
y do u want the range 4 Z separately?? the given x-y coordinates r for all the three vertices X, Y and Z..
i think i got wat u r confused wid..
capital X and Y are vertices..
small-case x and y are co-ordinates..
566. A right-angled triangle XYZ is in the xy-plane so that the right angle is at X and XZ is parallel to the x-axis... x, y coordinates of X, Y and Z are integers that satisfy the inequalities -4
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
Point (x,y) is such that x belongs to {-4,...,5} total 10; y belongs to {6,...,16) total 11.
No. of ways of choosing the co-ordinates X,Y,Z are as follows:-
X = No of ways X1 * No of ways Y1 = 10C1 * 11C1
Y relative to X = 10C1 since x is fixed coz of X and Y has only 10 remaining choices.
Z relative to X = 9C1 since Y is fixed and X has 9 remaining choices.
Hence total no. of ways are 10C1 * 11C1* 10C1 * 9 C1 = 9900.
Option (D).
566. A right-angled triangle XYZ is in the xy-plane so that the right angle is at X and XZ is parallel to the x-axis... x, y coordinates of X, Y and Z are integers that satisfy the inequalities -4
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
y do u want the range 4 Z separately?? the given x-y coordinates r for all the three vertices X, Y and Z..
i think i got wat u r confused wid..
capital X and Y are vertices..
small-case x and y are co-ordinates..
According to me the answer is D (9900).
For a given y coordinate of X you can have 10 integer y coordinates for the vertex Y ranging from 6 to 16 except that integer y co-ordinate that X is already holding. So if X = (-4,6) then Y can be (-4,7.....16). Now for vertex Z the y co-ordinate is already fixed at 6 but it's x co-ordinate can have any value except that integer coordinate that X already has, i.e. 9 values.
So for a particular y coordinate of Vertex X we have: 10 (possible x coordinates of Vertex X) * 10 (possible y co-ordinates for vertex Y) * 9 (possible x coordinates of Vertex Z) = 10*10*9 = 900 triangles.
but the y co-ordinate of vertex X itself can take 11 values from 6 to 16. therefore we have 900 * 11 = 9900 possible triangles.
Let me know if where I went wrong. This is a nice problem but it I think it is pretty difficult to solve it in 2 mins if you haven't done such things before.
Thank you,
Hemanth
Well, Gail beat me to this one. Good work Gail.
566. A right-angled triangle XYZ is in the xy-plane so that the right angle is at X and XZ is parallel to the x-axis... x, y coordinates of X, Y and Z are integers that satisfy the inequalities -4
(A) 7,200
(B) 8,100
(C) 9,000
(D) 9,900
(E) 12,100
phemanth SaysWell, Gail beat me to this one. Good work Gail.
LOL...thanks Bud. Lets hope we both got it correct or else the question beats us both....
According to me the answer is D (9900).
For a given y coordinate of X you can have 10 integer y coordinates for the vertex Y ranging from 6 to 16 except that integer y co-ordinate that X is already holding. So if X = (-4,6) then Y can be (-4,7.....16). Now for vertex Z the y co-ordinate is already fixed at 6 but it's x co-ordinate can have any value except that integer coordinate that X already has, i.e. 9 values.
So for a particular y coordinate of Vertex X we have: 10 (possible x coordinates of Vertex X) * 10 (possible y co-ordinates for vertex Y) * 9 (possible x coordinates of Vertex Z) = 10*10*9 = 900 triangles.
but the y co-ordinate of vertex X itself can take 11 values from 6 to 16. therefore we have 900 * 11 = 9900 possible triangles.
Let me know if where I went wrong. This is a nice problem but it I think it is pretty difficult to solve it in 2 mins if you haven't done such things before.
Thank you,
Hemanth
both of u r rite.. answer is 9900..
this is not a difficult problem.. it is rather a very direct question wid some visualization of the fugure n do a bit of reasoning.. n m sure u took more time in writing this explanation than the time u took in solving it.. 😃
Somebody please explain me this
Either the question and its explaination are wrong or may be I am not able to understand..............
Q>If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?
A>There are two ways you can satisfy this condition: (a) You can get the Ace of Diamonds first and then a black card or (b) you can get a black card first and then the ace of diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case B: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case 2 is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case 1. Recall that the probability of A or B is P(A) + P(B) - P(A and B). In this problem, P(A and B) = 0 since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.
both of u r rite.. answer is 9900..
this is not a difficult problem.. it is rather a very direct question wid some visualization of the fugure n do a bit of reasoning.. n m sure u took more time in writing this explanation than the time u took in solving it.. :)
yeah...i spent a while trying to explain this. So I think it would be better if we can categorize a problem that we post into:
a) "hey this is a great problem, I have solved it and want to share with you"
b) "plz help me out with this"
That way others know how to respond. What say? Don't mean any offence but just a suggestion from a newbie pagalguy.
Somebody please explain me this
Either the question and its explaination are wrong or may be I am not able to understand..............
Q>If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card?
A>There are two ways you can satisfy this condition: (a) You can get the Ace of Diamonds first and then a black card or (b) you can get a black card first and then the ace of diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case B: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case 2 is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case 1. Recall that the probability of A or B is P(A) + P(B) - P(A and B). In this problem, P(A and B) = 0 since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.
Is the answer correct? I would have done it the same way (but wouldn't worry about the intersection so much as you would naturally take care of it) . I think the explanation is straightforward. Which part are you not able to understand? OR how do you think this needs to be done? Or does the question needs to be solved considering replacing the drawn card in the deck?
The probablily of Ace of Diamonds and a black card should be
1/52 * 26/51 = 1/102
tell me where I am wrong???
The probablily of Ace of Diamonds and a black card should be
1/52 * 26/51 = 1/102
tell me where I am wrong???
You are correct IF the order is important. i.e. if the question asks you the probability of drawing a ACE Diamond FIRST and then a black card.
But if the question is asking you the probability of the 2 cards drawn to be ACE Diamond and a black card then 2 events can occur - (1) Ace D + Black or (2) Black + Ace D. and hence 1/51.
Let me try and explain:
Consider a bag of 3 red and 4 white balls. What is the probability that the 2 drawn balls are of different color?
Approach 1: 1 - (P(both red or both white)) = 1 - (4/7 * 3/6 + 3/7*2/6) = 4/7
Approach 2: you can draw Red & then white or white and then red = 3/7*4/6 + 4/7*3/6 = 24/42 = 4/7.
Both approaches give the same result meaning that if the order is irrelevant there are 2 ways you can draw Ace diamond and a black card.
Hope this helps. Thanks.
Oh man,
U are right.....i got it.
thanks a ton for the wonderfull explaination dude...