There is a 10% chance that it won^t snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter? (A) 55% (B) 60% (C) 70% (D) 72% (E) 80%
There is a 10% chance that it won^t snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter? (A) 55% (B) 60% (C) 70% (D) 72% (E) 80%
hi, probability that it won't snow all winter long = 10% so, probability that it will snow all winter long = 90%
probability that schools will not be closed all winter long = 20% so, probability that schools will be closed all winter long = 80%
probability that it will snow and schools will be closed during the winter = 90% * 80% / 100% = 72%
Yes even i agree with the same explaination... I want two things i.e. 1) will snow and 2) schools not closed Therefore i need to multiply the probabilities...Ans 0.72 or 72%
Friends, Two days ago I gave Kaplan free test. To my surprise the quant was pretty easy and I could easily have managed 100% marks in it if I would have not come across about a weird fact that 0 is a multiple of 5 Can someone ascertain this fact.. I am confused!!!
Yes even i agree with the same explaination... I want two things i.e. 1) will snow and 2) schools not closed Therefore i need to multiply the probabilities...Ans 0.72 or 72%
-Ankita
Ankita & Raghav, Even I did the same mistake, We can multiply the probability of two events only when they are independent of each other, In this case both the events are dependent on each other so we cannot multiply their probabilities. Since the question has asked us to find out the greatest possibility so the answer is 80% not 72%.. OA is E.
I think 0 is not the factor of any number. 1 is a factor of all the nos.
Deepak, Even I feel that zero is not a multiple of any no. as a result I have marked a different answer. It means Kaplan may be giving wrong explanation by assuming 0 as multiple of 5. Cheers Aman
One more problem which I have come across.. If a and b are consecutive positive integers, and ab = 30x is x a non-integer? (1) a^2 is divisible by 21 (2) 35 is a factor of b^2 Cheers Aman
One more problem which I have come across.. If a and b are consecutive positive integers, and ab = 30x is x a non-integer? (1) a^2 is divisible by 21 (2) 35 is a factor of b^2 Cheers Aman
statement 1: ============= a = 21 and b = 22 and x= 15.4 a = 630 and b = 631 and x = 13251
Nt suff
Statement 2: ============ a = 36 b = 35 x= 42 a = 34 b = 35 x = 39.67
Nt suff
combining both we can't get values for a and b.So my take would be option E
One more problem which I have come across.. If a and b are consecutive positive integers, and ab = 30x is x a non-integer? (1) a^2 is divisible by 21 (2) 35 is a factor of b^2 Cheers Aman
a will be a multiple of 21 and b will be a multiple of 35, as both these numbers have co prime factors. 1.a=21k=>21K*b=30X=>x=21b/30, Not suffecient as b is not known 2.Same case as 1 3.Both statement together-21K*35P/30=x x=3*7*5*7*K*P/30=49PK/2 since a,b are consecutive numbers so one of them will be even, means one of K,P will be even hence C is the answer.
a will be a multiple of 21 and b will be a multiple of 35, as both these numbers have co prime factors. 1.a=21k=>21K*b=30X=>x=21b/30, Not suffecient as b is not known 2.Same case as 1 3.Both statement together-21K*35P/30=x x=3*7*5*7*K*P/30=49PK/2 since a,b are consecutive numbers so one of them will be even, means one of K,P will be even hence C is the answer.
we have to find consective nos from 7*3*k series and 7*5*m series.Is it possible to get consective nos from the above series?.Need to think.
we have to find consective nos from 7*3*k series and 7*5*m series.Is it possible to get consective nos from the above series?.Need to think.
No I am not saying that the numbers will be from 21k or 35m series,but if the numbers are consecutine one of it will be even. the numbers may be configuration of any other choices. whats about numbers- 84 and 85?
No I am not saying that the numbers will be from 21k or 35m series,but if the numbers are consecutine one of it will be even. the numbers may be configuration of any other choices. whats about numbers- 84 and 85?
Hi Guy, I agree tht if the nos are consecutive one of them has to be even and other one has to be odd.I'm jus thinking is it possible to get consecutive nos satisyfing both the conditions stated in the 2 statements.Then we can tell if X will be an int or not.For e.g: 84,85
84*84 is perfectly divisible by 21 but 35 doesn't divide 85*85.So 84,85 will not suffice our conditions.
There is a 10% chance that it won^t snow all winter long. There is a 20% chance that schools will not be closed all winter long. What is the greatest possible probability that it will snow and schools will be closed during the winter? (A) 55% (B) 60% (C) 70% (D) 72% (E) 80%
Ankita & Raghav, Even I did the same mistake, We can multiply the probability of two events only when they are independent of each other, In this case both the events are dependent on each other so we cannot multiply their probabilities. Since the question has asked us to find out the greatest possibility so the answer is 80% not 72%.. OA is E.
@ash16,
i have a doubt...how can we conclude that the 2 events are dependent, coz the ques doesn't mention so....?
One more problem which I have come across.. If a and b are consecutive positive integers, and ab = 30x is x a non-integer? (1) a^2 is divisible by 21 (2) 35 is a factor of b^2 Cheers Aman
1. Which of the following is the smallest? a. 5^1/2 b. 6^1/3 c. 8^1/4 d. 12^1/5
i picked up the option a, as the answer. However option d is the official answer.
Option a is close to 2.1... as it is just above 4 and a sq rt of 4 is 2.
I guess using same logic we can eliminate the rest of the options.
If i am right, then how option d is the answer???????
your approach is correct... by that we get a) lil more than 2 b) lil less than 2 c) lets take root(root(8 )) ~ root(2.8 ) ~ lil less than root(3) ~ d) Now lets look at 12^1/5 as
root ( cube root (12)) Now cuberoot(12) ~ bet 2 & 3 ~ 2.2** Root(2.2) ~ a lil more than root(2) ~ >1.4
** i tried 2.5^3 and it's arnd 15.. IMO D 12^1/5
Boy!!! this took a lil while.. But thrs got to be a simpler way to do this... Puys?
