Exams

r=I/PT

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years. 

Therefore, 

S.I. = PTR/100

S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63

Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %

Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.

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For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893

Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = P1 + r100n

(P+C.I) = P1 + 101002

(P+420) = P[11/10][11/10]

P-1.21P = -420

0.21P = 420

Hence, P = 420/0.21 = Rs. 2000

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The Coffee Board of India was established by an act of Parliament in 1942.

After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days

-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

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