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ECAT and MCAT Preparation: Post your doubts here!

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well 32 must be B ISNT it smz imran ....mayb book is wrong.....
and question 36 i dint. undrstnd.........molecular weight of of that trivalent elements oxide u are asking???
32 i also marked B so book must be wrong

36 i also did not understand the question! thats why i posted
 
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well increasing area must have an effect..........its so obvious.....how that teacher cn say no
Alright, here you go, the explanation.
mathematically:
Friction = (co-efficient of friction)(Normal reaction force)
Here the variable factor is normal force R on which friction depends and Normal force is independent of area of contact. That is why friction on a horizontal surface and an inclined surface is different because even though surface area is same, the normal force changes. 8ecause the reaction force becomes equal to the sin component of weight instead of the actual weight.
and this is why the width of tires does not affect the grip of a car on road. And that is why a person with big feet does not necessarily have better grip while walking :p
 
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How many different 3-digit number divisible by 5 can be formed using the
elements of the set A = {1,2,3,4,5,6}
(A) 36 (B) 24 (C) 40 (D) none of the above
is it D....cos my anser comes to be 20....___ ___ ____ for 3rd blank 5 is a must...........this leaves u with 5 and 4 numbers for the remaining blanks respectively
 
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The answer is A 36.
Because it will be 5 permutation 2 + 5 permutation 2 - 4 permutation 1. Ah kaisay smjhaun? :p
 
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Lol okay i'll try explaining.
you see first of all, we count all the 3 digit numbers with 5 as last digit. That would be 5 permutation 2 = 20
Now count all the 3 digit numbers with 0 as last digit because they are also divisible by 5. Again 5 permutation 2 = 20
20 + 20 = 40
Now we exclude those numbers which will have zero as first digit and 5 as last (because that would make them 2 digit numbers). So 0 _ 5. The middle space is the only space that can be occupied by the remaining 4 digits. So 4 permutation 1 = 4
So subtracting these 4 numbers from 40, we get 36. Hope that helped :)
 
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Lol okay i'll try explaining.
you see first of all, we count all the 3 digit numbers with 5 as last digit. That would be 5 permutation 2 = 20
Now count all the 3 digit numbers with 0 as last digit because they are also divisible by 5. Again 5 permutation 2 = 20
20 + 20 = 40
Now we exclude those numbers which will have zero as first digit and 5 as last (because that would make them 2 digit numbers). So 0 _ 5. The middle space is the only space that can be occupied by the remaining 4 digits. So 4 permutation 1 = 4
So subtracting these 4 numbers from 40, we get 36. Hope that helped :)
There is no zero in the set!
 
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Hahaha oh sorry i confused it with another question i was solving :p sorry. Yeah you guys are right then, the answer is 20. My bad. Sorry again :p
 
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ohk this cis stands fr what?????

in maths complex no.

a question was there stating 7cis(pi/6)= ?

so this cis??
 
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it is the polar form of complex numbers. it means r( cos theeta + iota sine theeta) where theeta is the argument of the complex no. r is the modulus of complex no.
 
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it is the polar form of complex numbers. it means r( cos theeta + iota sine theeta) where theeta is the argument of the complex no. r is the modulus of complex no.


u mean that ( cos + sin) = cis????
 
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acha ohk one more thing chaptrs namely limits and groups(mathematics 1s year) are too comming in aptitude tests???
 
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i don't think so because it comes in advanced mathmetics which is usually not included in the maths portion of aptitude test. Ask some experienced petson too.
 
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isnt hid question rong...anser comes 24.7 stth
 

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