Simplify  

$$\begin{gathered} 1.53 + 4.73 + 3.83 - 3 x 1.5 x 4.7 x 3.81.52 + 4.72 + 3.82 - 1.5 x 4.7 - 4.7 x 3.8 - 1.5 x 3.8 \\ \end{gathered}$$

$$\begin{gathered} ?2 = 2×78-21+8+49-148 \\ = 57 - 21 \\ =36 \\ => ? = ±6 \end{gathered}$$

Given 92613 x 62742241 = ? + 89

Using unit digit method,

On LHS,

1 x 1 x 1 = 1

sqrt of last digit 1 => it must end with either 9 or 1

so now

1 x 1 = 1

1 x 9 = 9

On RHS,

? + 89

=> ? = ...1 - 89 = last digit is 11- 9 = 2

or

=> ? = ...9 - 89 = last digit is 9 - 9 = 0

Here in the options last digit 2 is option C.

1397 x 1397

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

 0.48z = 1.40

 z = 140/18 = 35/12 = 2.9 (approx.)

Let the Oranges be 'O', Bananas be 'B' and Apples be 'A'

From the given data,

2O + 3B + 4A = 15 ...... (I)

3O + 2B + A = 10 ..... (II)

On adding (I) and (II),

5O + 5B + 5A = 25

O + B + A = 5

4O + 4B + 4A = 20.

Given 373+353+283-3x37x35x28372+352+282-37x35-35x28-37x28

It is in the form of

 a3+b3+c3-3abca2+b2+c2-ab-bc-ca= a + b + c

Here a = 37, b = 35 & c = 28 

=> a + b + c = 37 + 35 + 28 = 100

Therefore,   = 100

Total strength of the class is given by

15 + 44 + 4 + 3 - 1 = 65

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.