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

[siddooba]

Let x1,x2,...x20 be the 20 numbers placed around the clock.

We know x1+x2+...x20 = 1+2+...20 (in some order)
= 20(21)/2
= 210

Taking the sum of 3 consecutive numbers , you will have 20C3 such pairs = 1140

Average of the sums of 3 consectutive numbers
= ((x1+x2+x3) +(x2+x3+x4) + ..... (x20+x1+x2) )/ 1140

= 19C2*(x1+x2+x3+....xn)/(1140)
= 171*(x1+x2+x3+....xn)/(1140)
= 171*210/1140
= 31.5

Now the sum of 3 integers cannot be equal to 31.5 so there has to be at least one pair of 3 consecutive numbers above 31.5 i.e the sum of some 3 consecutive numbers must be at least 32

[tarunad]

6! /(6+6)=120

[smartie]

2)Inverting 6 becomes 9
so (6/6)+9=10

[srikar2097]

Answer to knight - knave puzzle

knights - truth
knaves - lie
Pavan said "At least one amongst us is a knave"

4 cases arise-
1) If pavan is knight
=> pavan tells truth
=>slam is knave
possible case

2) if pavan is knave
=>pavan lies - no knave amongst them
contradiction case

3) if slam is knight
=>slam tells truth
=>pavan can be knight or knave

if pavan is knight - truth - but both are knights - contradiction.
if pavan is knave - lies - so there should be no knave - again contradiction.
contradiction case

4) if slam is knave
=>slam lies (actually this step is not relevant here)
=>pavan can be knight or knave

if pavan is knave - lies - so there should be no knave -contradiction.
if pavan is knight - truth - possible case

So the only possible case is
Paven->knight; slam->knave


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Problem 2


1) 6+6 - floor[(6!)^1/9] = 12 - floor(2.07) = 12-2 = 10

Some other ways of doing the same-

2) ceiling(ln6 + ln6) + 6 = ceiling(1.79+1.79)+6 = 10

3) floor ([[e^6]^0.5]0.5 + 6) = floor ([20.08]^0.5 + 6) = floor(4.48 + 6)=10

4) (6)^1.2 + floor(ln(6)) = 8.58 + floor(10.5 = 10

The last method uses only two 6's.

---------
Problem 3


At last found time to answer this!

Sum of all nos. from 1 to 20 = 210
average value of a number in this sequence = 210/20 = 10.5

so the sum of any 3 nos. on average is 3(10.5) = 31.5

But sum of 3 integers cannot be a decimal. So there must be some set (of 3 nos.) which evens this out. which means adding to this value. So, the sum of any 3 nos. has to be at least 32.

[smartie]

question 4.

pawan cannot be a knave as the statement "Atleast one of us is a knave" would be false then and he himself being a knave does not satisfy that.

When pawan is a knight the statement "atleast one of us is a knave" would be true -->so Slam would be a knave.

Clearly both pawan and slam cannot be knights.

So Pawan is a knight and Slam is a knave.

[tarunad]

small typo in my solution about 6!/(6+6)

a1,a2,a3..a20 are round the table, n1,n2...n20 corespond to sum of 3 consec nos,
n1+n2+...n20=630, all are less than or equal to 32 and also unequal =>n1+..n20 <<630 ,
thus disproving the hypothesis

[nikijpr]

Originally Posted by siddooba

a + 1/b = b + 1/c = c + 1/d = d + 1/a = t

a+b=bt

b+c=ct

c+d=dt

d+a=at
c/(t-1) + a = at
b/(t-1)^2 + a = at
a/(t-1)^3 + a = at
a/(t-1)^3 = a(t-1)

(t-1)^4 = 1

(c) t=0 is the solution

Might be a weird doubt but I did not understand the part in blue..

Thanks

[nikijpr]

Problem 4:

In an island, knights always tell the truth while knaves always lie. An islander Pavan, in the presance of another islander Slam said: "At least one of us is a knave". Is Pavan a knight or a knave. What about Slam?


If we assume Pavan to be a knave, he says one of them is a knave which would be a truth, contradictory to the given condition, as a knave can never speak truth. Therefore Pavan is a knight and slam is a knave.

[pavanpadekal]

Problem 1:

Let a, b, c, d be distinct reals such that a + 1/b = b + 1/c = c + 1/d = d + 1/a = t then t equals

(a) Less than -1 (b) -1 (c) 0 (d) 1 (e) greater than 1

Proving for non negative reals
ab+1+bc+1+cd+1+ad+1 = t(a+b+c+d)
=>t=[4+(ab+bc+cd+ad)]/(a+b+c+d)
We have for non negative reals
ab<= (a+b)/2
=>t>(4+a+b+c+d)/(a+b+c+d)(equality doesn't hold as we need distinct reals)
=>t>1+4(a+b+c+d)
Hence As for non negative reals t >1 ,
we can choose (e) greater than 1 from the answers

Problem 4:

In an island, knights always tell the truth while knaves always lie. An islander Pavan, in the presance of another islander Slam said: "At least one of us is a knave". Is Pavan a knight or a knave. What about Slam?

Pavan -> knight => Atleast One of the two is a knave and hence Slam is a knave.
Pavan -> knave => None of the two are knaves and hence contradiction
Only possibility is Pavan -> Knight and Slam -> Knave

[smartie]

Let a, b, c, d be distinct reals such that a + 1/b = b + 1/c = c + 1/d = d + 1/a = t then t equals

(a) Less than -1 (b) -1 (c) 0 (d) 1 (e) greater than 1

I did this by substituting the given solns in the problem

(1)when t=0 -->a=-1/b=c (does not satisfy the distinct criteria)

(2)when t=1 , we get an equation a^2-a+1=0 which has complex solutions

(3) when t=-1 ,a becomes equal to d-->again does not satisfy the distinct criteria

(4) when t>1 (say 2), we get a=1-->b=c=d=1 (distinct criteria not satisfied)

(5)when t<-1 (say-2), we get a=-1+sqrt(3)and -1-sqrt(3) which satifies the criteia.

Hence answer=option(1) t<-1