help !

if f(x) is a polynomial in x. it leaves remainder of 2 when divided by x-1
and remainder 1 when divided by x-2. what is the remainder when the same no. is divided by (x-1)(x-2) ?

options are :-
1) 2
2) 3
3) 3-x
4) 0

ah m afraid but the answer given is
3) 3-x.
but a nice try,....... y isn't ur correct will answer that tom if it is.
thanks.

(3-x) is the right ans.
use the basic theorem

"If a polynomial f(x) be divided by (x - a), the remainder is f(a)."

so we hav f(1)=2 and f(2)=1

now,f(x)= p(x)*(x-1)(x-2) + r(x) ;(r(x) is remainder,also degree of r(x)< degree of p(x) )
put x=1; f(1)= r(1)=2
x=2; f(2)=r(2)=1
so r(x) =3-x

i want to edit my location but i can't find it here
PaGaLGuY.com - The Everything of MBA

and yes,'last edit by' is also missing(sometimes it appears,sometimes not).and many a times it automatically goes into advance mode when editing the post.

Solve this

Points D, E and F are the midpoints of the sides BC, CA and AB of a triangle ABC with
centroid at point G. Angle AGB is 90, side AC is 8 cm and side BC is 6 cm. What is AB2
(in cm2)?

1) 64
2) 100
3) Less than 20
4) Cannot be determined
5) None of these

ans (5)

AB^2= 20cm2

hi puys...

QQAD is a brilliant product. Even if the team is taking off for a year, can we be provided with the past few year's of questions to keep us going this year.

And puys!! do any one u have an archive of all ht e QQAD question for last year?? thanx in advance

for the benefit of all puys i hav attached the year 2007 and 2008 QQAD set(with ans) & QQAD 2008 practice set .

Here are few more ...

i)Find all the values of a , so that 6 lies in b/w the roots of the equation x^2 +2(a-3)x+9=0.
1) a <-3/4 2) a>3/4 3)a<0 or a>6 4 ) a >6

ii) If both the roots of equation x^2 -6ax +2 -2a +9a^2 =0 exceed 3 then
1) a >9/11 2) a>=9/11 3)a>11/9 4 ) a <11/9

1) ans (a)
D= 4a(a-6)
and to have distinct roots D>0 so we get a>6 or a<0
so option 2 can be ruled out.
now,x= -(a-3)+-sqrt (a*(a-6))
put x> 6 and this doesn't satisfie the condition so we r left with only 1 option and that's a<-3/4
so ans a.

2) D=8a-8
for root to be real a>=1
so only option which can fit into ans is a >11/9
so ans (3)

Three solvers Toxic, Parx and Abhay, work together to complete the above two questions. The time it takes them to do the work together is 2 minutes less than Toxic would have done alone, 10 minutes less than Parx to do the work alone and one-third of the time that Abhay would have taken working alone.
How long did it take them to do the above two questions, if Toxic, Parx and Abhay work together?



Is the ans 2 ?

yup..u got it.
ans is 2 min

Questions given are in the increasing order of their difficulty level.

1)
Shreyan and Shivani together can complete a piece of work in 4 days while Shivani and Stalwart together can complete the same work in 6 days. Shreyan and Stalwart , in fact complete this work in 10 days. How long will it take to complete the work if Shreyan, Shivani and Stalwart work together?
2)
Each of the three inlet taps fill up a tank independently in 2 hours while an outlet tap empties the completely fileed tank in 3 hours. If all the four taps are opened, then when will the tank start overflowing?
3)
Three solvers Toxic, Parx and Abhay, work together to complete the above two questions. The time it takes them to do the work together is 2 minutes less than Toxic would have done alone, 10 minutes less than Parx to do the work alone and one-third of the time that Abhay would have taken working alone.
How long did it take them to do the above two questions, if Toxic, Parx and Abhay work together?


Sayonara

1) work done by trio in 1 day =1/2(1/4 +1/6 +1/10) =31/120
so total days to complete the work = 120/31

2) tank filled in 1 min = 3*1/120 -1/180 = 7/360
so after 360/7 min tank wil start overflowing.

3) still working on......

It surely loads faster and now i don't need to reload every page(as was case of earlier).it's a great relief.

and yes like many puys i m also not feeling comfy wid colour combination .
particularly Thank! and Groan! ka color

rajat_b Says

it is given in the question that angles are not equal

it's given that "not necessarily equal angles" .that doesn't mean that not equal angles.
but i think in case of not equal angle we can't decide the side length.

Take any random point inside.
Join the point to all the vertices.....5 triangles are formed.

area of each triangle = 0.5*b*h where b is common for all traingles and h is perpendicular distance from the point to the side

sum of areas = 0.5*b*(h1+h2+h3+h4+h5) = 40
== b = 80/16
= 5.......

suppose pentagon has unequal angles.in this case pentagon wil be a concave pentagon(as it has eual sides so convex pentagon isn't possible).and then ans wil change......

take case of conclave pentagon(like "Draw an equilateral triangle.....draw a square with the base of triangle as its side...remove the base frm picture now.... ").here ans wil not be same as in case of pentagon having equal sides and equal angles.
what u say..