1) let total capacity of 1st jar=100l
" " 2nd jar=200l
jar1 wine:water =50:50
jar2 wine:water =50:150
jar3 =jar1 + jar2 => wine:water =100:200 =1:2

2) the situation can be visualized as a 11*11 matrix
we know that the total number of elements in such a matrix =
11*11=121 trees.
3) let x be the number of cuffs given and y be the number of collars.
so eq. 1 => x+y=30 ................(1)
let a be the cost of 1 cuff and b be the cost of 1 collar
money given by A--- ax/2 + by/3 =27
3ax +2by =162
from (1) y=30-x
3ax+2b(30-x)=162........(2)
given 4a=5b
possible solutions are
(2.5,2) or multiples of them
from (2) we see only (2.5,2) satisfies the condition as others give negetive answers.
so cost of 1 cuff 'a' =2.5cents
cost of 1 collar 'b' =2cents
substituting in (2), we get
totla number of cuffs 'x' =12
total number of collars 'y' =18
so Bs things are 6 cuffs and 12 collars and the amount which she owes to the laundry is 6*2.5+12*2=39cents.
4) let the distance travelled on the first day be 'a'k.m
day 2 =a+20*1
day 3 =a+20*2
.
.
.
.
day 9=a+20*8
total distance covered in 9 days =9/2 *(a+(a+20*8))=1080
or a=40 k.m
so distance covered on the 4th day =a+20*3=40+60=100k.m
distance covered on the last day =a+20*8=40+160=200km
5) answer is 1/1000*997/1000 +1/1000*996/1000 ................................................1/1000*1/1000
=1/1000000(997+996+............+1)=.497503
6) on the second day the watch is 10s fast
3rd day 20s fast
so on the 28the day the watch is watch is 270s or 4.5min fast and that night it gains .5 min and overall it becomes 5mins fast in the night of 28th may.

7) A might have got everything correct, i.e 30 correct answers.
this is the only no. which one can be confident upon. if one thinks he got 29 correct answers, he cant say whether he got the other one wrong or he left it, for that matter every other numbers.
so if he gets 30 correct, then this is unique, no other combinations are possible.
9) ithink he's in X road now.X road is actually a twisted road with 5k.m facing down and 5km to the right so totally he walked 5+5=10km along X road though it's not straight.then he reached the junction and turned left to the Y road ,walked straight for 10km.reached another junction and turned left and walked straight along Y road for 10km.reached another junction, turned left and walked straight along Z road for 10km.reached another junction and turned left and walked straight for 15km, now note that after walking 10km thro this road, he'll be retracing his initial path through the X road(X road has a laterally inverted L bend with arms measuring 5km each.

10) i think the answer is 66 =12*11/2.if the person clinging is so friendly that he doesnt hear his cling as he was completely engaged in his activity,then we can subtract 11 from 66 to get 55.

8) let cost per yard of drape 1 =no.of yard =x
let cost per yard of drape 2 =no. of yard =y
total cost for yard 1 =x^2
total cost for yard 2 =Y^2
total cost =X^2 + y^2<10
also given Y^2=x^2+$2.15

solving we get:
first drape 1.98yard
second drape 2.464yard