Let the present age of father be x years and the sum of present ages of 2 sons be y years.
? x = 3y ...(i)
? (x + 20) = (y + 20 + 20) ...(ii) [20 will be added twice as for 2 children]
Solving (i) and (ii), we get
x = 30 years
n(S) = 6, n(E) = (4, 6) = 2
? P(E) = 2/6 = 1/3
Here, A = ? 8800, T =2 yr, R = 5%
We know
SI = ART/(100 + RT) = (8800 x 5 x 2) / (100 + 5 x 2)
= (8800 x 10) / 110
= ? 800
3rd term = (2nd term) x 3 - 4 = 26 x 3 - 4 = 74.
4th term = (3th term) x 3 - 4 = 74 x 3 - 4 = 218.
5th term = (4th term) x 3 - 4 = 218 x 3 - 4 = 650.
∴ 5th term must be 650 instead of 654.
Other side = ?5 2 - 42
= ? 9
=3 m
So The area of the rectangular field = 4 x 3
= 12 m2
0.27 = (27 - 2)/90 = 25/90 = 5/18
For 50 students, food is sufficient for 45 days
? For 1 student, food is sufficient for 45 x 50 days
and for 75 students, food is sufficient for (45 x 50)/75 days. i,e., for 30 days.
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