CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[dewan_iitr]

Originally Posted by implex

The RHS has an irrational number, so if x and y are integers they need to cancel out the irrational factor, that is sqrt(5)
so x and y are of the form 5k^2

my answer and method were same but this looks a better explanation.

[Aarav]

Ok -> this is the warm-up problem since I want Implex also to have a go at this

This is last year's QQAD problem that took Vineet, me and a friend of mine in Intel research team to come to any conclusion in 4 hours. Treat this as a warm up and real contest (easy + difficult problems) with points will be held tomorrow.

Allwin likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example, 89.34 is 89 and 99.54 is 100, and 115.5 is 116. Allwin always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true?

(a) The minimum possible number of courses is less than 15
(b) The maximum possible number of courses is not more than 50
(c) The minimum possible current rounded average is 95
(d) Either of 126 or 127 can be the current rounded average
(e) none of the foregoing

[nbangalorekar]

Originally Posted by Aarav

You are right about the fact stated above -> but that will require some explaining waise, the answer is OK.

Aarav for ne equation of the form 1/x+1/y=1/k, if we take the mean value (i.e. x=y) n then proceed from there to get the smaller values of 'x' (n thus 'y') wudnt it be a faster method for arriving at the answer?
for exmaple in this question 1/x+1/y=1/(2sqrt5)
so first we take x=y=4*(sqrt5).
then the next value 'x' can assume is 3*(sqrt5).


PS: i made a mistake in my previous answer.
'x' cannot assume the value of 2(sqrt5) since '1/y' will then be infinite
so no. of ordered pairs is 3.

Aarav please do comment on this method...

[vikas82]

------------------------------------------------------
Quantitative Question # 034
------------------------------------------------------

As there are 7 letters in the word JUPITER, they can be arranged amongst themselves in 7! ways. These arangements also include the arrangements for the three vowels (u,i,e) that can be arranged amongst themselves in 3! ways. However the condition given in the question requires vowels to be in alphabetical order implying there is only one way for the vowels to be arranged amongst themselves.
Hence the solution is 7!/3! = 840

(4) 840 is the correct answer.

[srikar2097]

Originally Posted by Aarav

------------------------------------------------------
Quantitative Question # 035
------------------------------------------------------

The number of ordered (x, y) such that 1/√x + 1/√y = 1/√20 is

(1) 1 (2) 3 (3) 5 (4) 7 (5) none of these

Given 1/√x + 1/√y = 1/√20 --->1

(√x + √y)/2 =√xy/2√20

Since AM >= GM
We get, √(xy)^1/2 <= √xy/4√5

Case 1) When x=y
this would yield x=y=80
So 1 pair here.

Case 2) when x & y not equal.
=> √xy < (xy)/80
=> xy > 80
but for the (1) we get value of y in terms of x. Stick it in the above equation.
We get a quadratic equation where one of the 2 roots is -ve and the other +ve.
By taking the +ve value of x. we get 1 value y. So 2 pairs here.

So there would be 3 ordered pairs. Hence choice (2)

[zion1729]

The number of ordered (x, y) such that 1/√x + 1/√y = 1/√20 is

(1) 1 (2) 3 (3) 5 (4) 7 (5) none of these


Sol:

the rhs should be 1/(2sqrt(5))

Hence if x and y are to be integers, we need to have x = a^2*5 , y = b^2*5, such that

1/a + 1/b = 1/2 --- this is have three ordered pairs (4,4), (3,6), (6,3) (since square root is always positive)

Hence answer is (3) for integers x and y

For real x and y, there are infinitely many solutions. The question should have been clear about the domain of x and y.

[sudeepdeb]

Originally Posted by implex

Will you give chocolates to the winner? I will miss it most probably, heading off to nepal today unless my friends decide not to wake up, and that looks a possibility!

Anyways, keep the good work going. You are the MAN

CFA ?? Katmandu ?
but then there would not have been possibility of waking up.....

[sudeepdeb]

My pick for Today's question :
3 sets .
(180,45),(80,80),(45,180).

rt(y) = a.rt(5),
on soving, rt(x) = {2a/(a-2) } .rt(y)
2a/(a-2) -> integer for a = 3,4,6.

[thebornattitude]

Solution to QQAD 35....

as x, y both have to be multiple of 5, to get integral solutions....
so let x= 5a,,, b=5b,,
so 1/a +1/b= 1/2
so (a-2)(b-2)=4

solving we get a= 3,4,6...and hence b=6,4,3

so 3 solutions

[Aarav]

Allwin likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example, 89.34 is 89 and 99.54 is 100, and 115.5 is 116. Allwin always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true?

(a) The minimum possible number of courses is less than 15
(b) The maximum possible number of courses is not more than 50
(c) The minimum possible current rounded average is 95
(d) Either of 126 or 127 can be the current rounded average
(e) none of the foregoing

This wouldn't be that straight -> someone who can solve this on his/her alone has to be a genius. Please treat this as a recreational problem than anything else.
But, I will urge each one of you to have a crack at this -> if not alone then you can collectively come out with a solution to this. Use of calculators, excel sheets is all allowed