We can check divisibility of 195 + 215 by 10 by adding the unit digit of 95 + 15 which is equal to 9 + 1 = 10.
So it must be divisible by 10.
Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20.
So 195 + 215 is divisible by both 10 and 20
(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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