Let 's' be the number of small packets and 'b' the number of large packets sold on that day.
Therefore, s + b = 5000 ... eqn (1)
Each small packet was sold for Rs.150.
Therefore, 's' small packets would have fetched Rs.150s.
Each large packets was sold for Rs.250.
Therefore, 'b' large packets would have fetched Rs.250b.
Total value of sale = 150s + 250b = Rs. 10.5 Lakhs (Given)
Or 150s + 250b = 10,50,000 ... eqn (2)
Multiplying equation (1) by 150, we get 150s + 150b = 7,50,000 ... eqn (3)
Subtracting eqn (3) from eqn (2), we get 100b = 3,00,000
Or b = 3000
We know that s + b = 5000
So, s = 5000 - b = 5000 - 3000 = 2000.
2000 small packets were sold.
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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