Let circumference = 100 cm .
Then, ? 2?r = 100
? r = 100/2?
=50/?
? New circumference
= 105 cm
Then, 2?R = 105
? R = 105 / (2?)
&rArr Original area = [ ? x (50/?) x (50/?) ]
= 2500/? cm2
? New Area = [? x (105/2?) x (105/2?)]
= 11025 / (4?) cm2
? Increase in area = [11025/(4?)] - 2500/? cm2
= 1025 / 4? cm2
Required increase percent [1025/(4?)] x 2500/? x 100 = 41/4%
= 10.25%
(A + B)'s 2 day's work = 2 x (1/3) = 2/3
Remaining work = 1 - (2/3) = 1/3
A will complete 1/3 work in 2
A will complete 1 work in 6
A's 1 days work = 1/6
B's 1 day's work = (1/3) - (1/6) = 1/6
? B will take 6 days to complete the work alone.
Distance covered in one revolution = total distance travelled / total number of revolution.
= ( 88 x 1000) / 1000 m
= 88 m
We know that the distance covered in one revolution = circumference of the wheel.
? ?d = 88
? 22d / 7 = 88
? d = 28 m
So, 36 is wrong.
Given expression = (.896 x .752 +.896 x .248) / (.7 x .034 + .7 x.966)
=.[896 x (.752+.248)] / [.7 x (.034+.966)]
= .(896 x 1) / ( .700 x 1)
=896/700
= 1.28
Average speed =2 x 40 x 60 / ( 40+ 60)
= 4800 / 100
= 48 km/hr
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
After 61 days, it will be Saturday.
9548 16862 = 8362 + x + 7314 x = 16862 - 8362 ----- = 8500 16862 -----
We know that
BG = SI on TD = (240 x 30 x 1 x 1)/100
= ? 72
BG = BD - TD
? BD = BG + TD
= 72 + 240
= ? 312
If previous year is leap year then calendar of May is similar to July
In tossing a coin 2 times the sample space is 4 i,e (H, H), (H, T), (T, H), (T, T)
(1) If A1 denotes exactly one head
then A1 = {(H, T) (T, H) }
So, P(A1 ) = 2/4 = 1/2
(2) If A denotes at least one head
then A = {(H, T) (T, H) (H, H)}
? A = {(H, T) (T, H) (H, H )}
? P(A) = 3/4
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