The given 3b - 7 < 32 can be solved as

3b - 7 < 32

3b < 39

b < 13.

given

  48 ? + 32 ? = 320 

=>  6 ? + 4 ? = 40

Now squaring on both sides

=>  36 x ? + 16 x ? + 48 x ? = 1600

=> 100 x ? = 1600

=>  ? = 16

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 integer be x. Then,

x2 - 20x = 96

(x + 4)(x - 24) = 0

x = 24

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

3 x + 7 = 7 x + 5 = > 7 x - 3 x = 2 = > 4 x = 2 = > x = 2

Now,  3 x + 7 = x 2 + p = 3 2 + 7 = 1 4 + p = > p = 17 2

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.

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.