hello everyone
Please try problem no 10 of Arun sharma,page no 202,geometry level I.
It has not been mentioned that
That is surely circumcenter. Otherwise it can't be solved and it shouldn't be in LOD I.
That is surely circumcenter. Otherwise it can't be solved and it shouldn't be in LOD I.
it can be solved by assumin x to be at circumcentre only
LOD III Number System Q.22
The Q seems inconsistent...what say guys....31...shud be 41...den the ans shud be option(d).
regards
mejogi
The Q seems inconsistent...what say guys....31...shud be 41...den the ans shud be option(d).
regards
mejogi
LOD II Mensuration Q.42
The diagram seems inconsistent have anybody corrected the figure,if so then...write or attach the correct diagram....Better write..no need to waste time.
The diagram seems inconsistent have anybody corrected the figure,if so then...write or attach the correct diagram....Better write..no need to waste time.
NUMBER THEORY-LOD 2
When 3^52 is divided by 50, reminder occurs,
the fractional part is of the form bx..
the quest was wat is b or x..nt very sure of it though..
thanks in advance
LOD II Mensuration Q.42
The diagram seems inconsistent have anybody corrected the figure,if so then...write or attach the correct diagram....Better write..no need to waste time.
can u post the question here ???
Q.42 Find the sum of the areas of the shaded sectors given that ABCDEFE is any hexagon and all the cirlces are of same radius r with different vertices of the hexagon as their centres as shown in the figure.
a) pi*r^2 b)2*pi*r^2 c) (5*pi*r^2)/4 (d) 3*pi*r^2
The Q doen't seems tough.But,The diagram is very obscure in the book itself,though i m attaching it.
I m thinking..that let's correct the diagram.
To get the desired ans we have to think that (1/3)rd of each circle is in the Hexagon and it is a regular one(which is not given in Q).
What u say?
NUMBER THEORY-LOD 2
When 3^52 is divided by 50, reminder occurs,
the fractional part is of the form bx..
the quest was wat is b or x..nt very sure of it though..
thanks in advance
It is sure that last digit of 3^32 is 1
divide any number of this form 331/50=6.62 or 341/50=6.62every time we are getting 2 as last digit in decimal part,though 331 or 341 is not comes frm the form of 3^n.
So any number with 1 as unit digit divided by 50 gives this 0.b2 form as fraction part.
chk out this 221/50=4.42.
41/50=0.82
1/50=0.02
Hi,g_sowmya
See this when 1 is in unit place of any number and if we divide it by 5 we will surely get 0.2 as fraction part. So in case of 50 we must expect to get 0.b2 as fraction. Doesn't seems simple !!
Never hesitate to ask if have still doubt ?
Though this explanation is better and easier than earlier one.
#Note: It is also true for any number if it's unit digit is 6
reagrds
mejogi
yeah..this s clear..thanks:-)
Q.42 Find the sum of the areas of the shaded sectors given that ABCDEFE is any hexagon and all the cirlces are of same radius r with different vertices of the hexagon as their centres as shown in the figure.
a) pi*r^2 b)2*pi*r^2 c) (5*pi*r^2)/4 (d) 3*pi*r^2
The Q doen't seems tough.But,The diagram is very obscure in the book itself,though i m attaching it.
I m thinking..that let's correct the diagram.
To get the desired ans we have to think that (1/3)rd of each circle is in the Hexagon and it is a regular one(which is not given in Q).
What u say?
Hi
i don't think we should bother abt the diagram.
since all the circles are of same radius, total area of circles is 6* pi* r^2
now sum of angles in hexagon is 720
fraction of shaded area is (6*pi*r^2) * 720/(6*360)=2*pi*r^2
hope i have clearly expressed .
n thanks 4 the problem. i haven't solved arun sharma
It seems that nobody is interested in this thread or Is it due to the CAT 2007 knocking on the door?.
Otherwise it seems we have to post queries and doubt in Q's of Arun Sharma book in the other threads? 😞
Hi
i don't think we should bother abt the diagram.
since all the circles are of same radius, total area of circles is 6* pi* r^2
now sum of angles in hexagon is 720
fraction of shaded area is (6*pi*r^2) * 720/(6*360)=2*pi*r^2
hope i have clearly expressed .
n thanks 4 the problem. i haven't solved arun sharma
Thanx dude nicely solved...It didn't came to my mind
360 = pi*r^2
hence 720= 2*pi*r^2. Hence the ans option(b)
regards
Mejogi
wrong post!!!
Q.42 Find the sum of the areas of the shaded sectors given that ABCDEFE is any hexagon and all the cirlces are of same radius r with different vertices of the hexagon as their centres as shown in the figure.
a) pi*r^2 b)2*pi*r^2 c) (5*pi*r^2)/4 (d) 3*pi*r^2
The Q doen't seems tough.But,The diagram is very obscure in the book itself,though i m attaching it.
I m thinking..that let's correct the diagram.
To get the desired ans we have to think that (1/3)rd of each circle is in the Hexagon and it is a regular one(which is not given in Q).
What u say?
yup sir absolutely if the answer has to -> 2*pi*r^2 ........then it has to be a regular hexagon , i think we can take tat quite confirmly and proceed. 😃
cloudsonfire Saysyup sir absolutely if the answer has to -> 2*pi*r^2 ........then it has to be a regular hexagon , i think we can take tat quite confirmly and proceed. :)
No..coludsonfire bhai...For any hexagon ..regular or non regular ,covex or concave the some of internal angles is 720 degrees.Since it will surely have 4 triangles inside it.
So the Q is correct and the figure.There is no need to take a regular hexagon. The important thing here is that center of each circle must lie at vertices and have equal radius and doesn't intersect each other.
So if 360 degress give rise to pi*r^2 so 720 will give rise to 2*pi*r^2.
This has been also sated by frigid sapphire
yup got tat 😃 nice sol. 😃
@mejogi
well i'l be d one flodding this thread with ques....just a few days...
thnx
regards
Gautam
@mejogi
well i'l be d one flodding this thread with ques....just a few days...
thnx
regards
Gautam
Not only solution. We will discuss the required Q's in detail...U will get my full cooperation. In many Q's I have found ..by changing options the ans wud be diff. and certain Q's r quite Interesting to solve without option...and through diff. approaches. Thanx for ur response !!
regards
Jogi
hi have a few dbts..
1. find the number of zeros in the prdt (LOD 2 number theory)
1^1*2^2*3^3*.....*100^100
2. find the last digit in ((3)^3)^4n