Solution SAMPLE 6

Q.1

Q. 2 Let the median mc be double and construct the parallelogram ACBP. For a paralleglogram, we know : The sum of the squares of the diagonals of a parallelogram in equal to the sum of the squares of all of its sides.

We get :

Q.3 We know

The nth term of series

Now

… … …

S =

Q.4 Let the age of woman = x From the question :

it can be solved to get x = 18

Q.5 Solution :

Q. 6 From the symmetry the quadrilateral EKMP is a square

So, the desired are to be calculated in square EKMP plus four equal segments

DWKZ is equilateral,

i.e. Ð KWZ = 600

\ Ð XWK = 330

ÐMWZ = 330

So ÐKWM = 330

We have, area of the segment

S =

q is measured in radians

So, area of the segment = …(1)

For D WKM, by cosines law :

So, the desired area = Area of square + 4 x area of segment

Q.7

= 0

So, a = b, b = c, and c = a

So, a+b+c=1, gives

So,

= 1+1+1

= 3

and

Q. 8 The solution is

Lets know the power and number of digits.

No of digit

21 = 2 1

22 = 4 1

23 = 8 1

24 = 16 2

25 = 32 2

26 =64 2

27=128 2

28=256 3

29=512 3

210=1024 4

211=2048 4

212=4096 4

213=8192 4

214=16384 5

215=32768 5

216=65536 5

217=131072 6

218=262144 6

219 =524288 6

220 =1048576 7

221 =2097152 7

222 =4194304 7

223 =838868 7

224 = 16777216 8

225 = 33554432 8

226 = 67108864 8

227 = 134xxxxxx 9

228 = 268 xx xx xx 9

229 = 536 xx xx xx 9

230 = 1073 xx xx xx 10

231 = 2147 xx xx xx 10

232 = 4294 xx xx xx 10

233 = 8589 xx xx xx 10

234 = 17179 xx xx xx 11

235 = 34315 xx xx xx 11

236 = 68631 xx xx xx 11

237 = 137263 xx xx xx 12

238 = 27 xx xx xx xx xx 12

239 = 54 xx xx xx xx xx 12

240 = 108 xx xx xx xx xx 13

241 = 216 xx xx xx xx xx 13

242 = 432 xx xx xx xx xx 13

243 = 864 xx xx xx xx xx 13

244 = 1728 xx xx xx xx xx 14

The pattern

Power / 1 / 4 / 8 / 12 / 16
No. of digit / 1 / 2 / 3 / 4 / 5
Power / 20 / 24 / 28 / 32 / 36
No. of digit / 7 / 8 / 9 / 10 / 11
Power / 40 / 44 / 48 / 52 / 56
No. of digit / 13 / 14 / 15 / 16 / 17
Power / 60 / 64 / 68 / 72 / 76
No. of digit / 19 / 20 / 21 / 22 / 23
Power / 1000 / 1004 / 1008 / 1012 / 1016
No. of digit / 301 / 302 / 303 / 304 / 305

So 21000 have 301 digits.

Q. 10 The solution using Pascals principle of triangles.

So numbers of routes – 110.