How many distinct rectangles of perimeter not more than 258 cm have both length and breadth as positive integers(in cm)???
Ente approach sheri aavunilla... Onnu paranju tharuo???
@Kevin88 For max % increase, there are instances when mostly, few would be around 30% and 40% and 1 would be greater than 50%. When you look at the question, check the options. Then look back at the bar charts. Mostly one/two would have more than halved or doubled. :)
L B
1 1-128
2 1-127
3 1-126
. .
. .
. .
128 1
So, 1+2+.....+128 alle???
@Ash27Winz said:How many distinct rectangles of perimeter not more than 258 cm have both length and breadth as positive integers(in cm)???Ente approach sheri aavunilla... Onnu paranju tharuo???
l + b
When l=1, b=1,2,3......128
l=2, b=1,2,3......127
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l=128, b=1
So, for l=1, b has 128 values, for l=2 b has 127 values etc. So add 1+2+3...128.
128*129/2 = 8256
Then eliminate double cases.
@seetharam7 said:l + bWhen l=1, b=1,2,3......127l=2, b=1,2,3......126.....l=127, b=1So, for l=1, b has 127 values, for l=2 b has 126 values etc. So add 1+2+3...127.127*128/2 = 8128Then eliminate double cases.l+bSo, 8256...Which are the double cases???Options are 4160, 129, 4158 and 65
@Kevin88 @Ash27Winz I have edited my post. Its "not more than". :)
And there will be double cases as when
l=40 b=60
l=60 b=40
They are same rectangles. So, we need to eliminate half of them. Ans should be somewhere around 4000. :)
i am getting 4160 as the answer
@seetharam7 said:@Kevin88 @Ash27Winz I have edited my post. Its "not more than".And there will be double cases as whenl=40 b=60l=60 b=40They are same rectangles. So, we need to eliminate half of them. Ans should be somewhere around 4000.
Arent l=1, b=2 and l=2 b=1 different rectangles???
i did this way.
L+B
Take L+B = 129. So the number of ordered pairs (L,B) can take
(128,1) .......(65,64) = 64 values.
For L+B = 128 also we have ordered pairs (127,1) .....(64,64) ie 64 values.
same way for L+B = 127 and 126 we can have 63 values each.
Proceeding further for L+B = 3 and L+B = 2 we have 1 value each.
Total sum. = 2( 1+2+........64) = 4160
@Ash27Winz said:Arent l=1, b=2 and l=2 b=1 different rectangles???
Breadth cannot be more than length. Among two values whichever is more we call it as length and the smaller side as breadth
@raku1989 said:Proceeding further for L+B = 3 and L+B = 2 we have 2 values.
For L+B = 2, there is just one value right ?
@Kevin88 Just saw your mail. And NOO !! 
I am just reading DI questions and answers. Not solving, just looking at the methods. 😃
I am just reading DI questions and answers. Not solving, just looking at the methods. 😃
@raku1989 said:Breadth cannot be more than length. Among two values whichever is more we call it as length and the smaller side as breadth
Ohhhh.... I didnt know that one!!
Thanks..
Trignometric formulae :-
1) sin (-x) = - sinx
2) Cos (-x) = cos x
3) Sin (x+y) = Sin x cos y + Cos x sin y
4) Sin (x-y) = Sin x cos y - cos x sin y
5) Cos (x+y) = Cos x cos y - Sinx siny
6) Cos (x-y) = Cos x Cos y + Sin x Sin y
7) tan (x+y) = (tan x + tan y)/ (1- tan x tan y )
8) tan (x-y) = (tan x - tan y)/ (1+tan x tan y)
9) cot (x+y) = (cot x cot y - 1)/(cot x + cot y)
10) cot (x-y) = (cot x cot y + 1)/(cot y - cot x)
11) sin 2x = 2 sinx cosx = 2 tan x / (1+tan^2 x)
12) Cos 2x = cos^2x - sin^2x = 2 cos^2x - 1 = 1 - 2sin^2x =(1-tan^2 x)/(1+tan^2 x)
13) tan 2x = 2 tan x/ (1-tan^2 x)
14) Sin x + sin y = 2 sin (x+y/2) cos (x-y/2)
15) Sin x - sin y = 2 cos (x+y/2) sin (x-y/2)
16) Cos x + cos y = 2 cos (x+y/2) cos (x-y/2)
17) cos x - cos y = -2 sin (x+y/2) sin (x-y/2)
18) 2 cos x cos y = cos (x+y) + cos (x-y)
19) - 2 sinx siny = cos (x+y) - Cos (x-y)
20) 2 sin x cos y = sin (x+y) + sin (x-y)
21) 2 cos x sin y = sin (x+y) - sin (x-y)
P.S. Today's paper had questions directly from trignometry.