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.
Now,
Total strength of the class is given by
15 + 44 + 4 + 3 - 1 = 65
Let the integer be x. Then,
x2 - 20x = 96
(x + 4)(x - 24) = 0
x = 24
If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.
Let the required two numbers be a, b
From the given data,
2a + 3b = 36 .....(1)
3a + 2b = 39 .....(2)
Now, by solving 1 and 2 eqnts, we get
5a = 45
a = 9 ....(3)
Put (3) in (1), we get
2(9) + 3b = 36
=> 18 + 3b = 36
3b = 36 - 18
=> b = 18/3 = 6
Required, difference between the numbers = a + b = 9 + 6 = 15.
2ab=
29-9=20
=>ab=10
x = y <=> 1 - q = 2q + 1 <=> 3q = 0 <=> q = 0.
Given Expression = = = 2
50.003% of 99.8 ÷ 49.988 = ?
? ~= (50 x 100/100)/50
? ~= (50 x 100)/(100 x 50)
? ~= 1
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