Let the radius of the circle be r cm. Then, 

120360πr2=667 

r2=667*722*3 

r=3.

Let the inner and outer radii be r and R metres.

Then  2πR=3527=>R=3527*722*12=8m 

 2πR=5287=>R=5287*722*12=12m

=> Width of the ring = (R - r) = (12 - 8) m = 4 m.

The toys are different; The boxes are identical
If none of the boxes is to remain empty, then we can pack the toys in one of the following ways
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2.

Therefore, total number of ways of achieving the 2−2–1 option is:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3−1–1 option is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Let inner radius be r metres. Then, 2πr = 440 ; r = 440×744= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

Distance covered in one revolution = 88×10001000= 88m. 

2πR2=88 ⇒2×227×R=88 ⇒R=88×744

Circumference = No.of revolutions * Distance covered

Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.

Area = (13.86 x 10000) sq.m = 138600 sq.m

πR2=138600⇒R2=138600×722⇒R=210m 

Circumference = 2πR2=2×227×210=1320m 

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32