503 x 201=?

? = 6885.009 - 419.999 - 94.989
= 6885 - 420 - 95 ? 6370

? = 840.003 ÷ 23.999
Here one number is increased and other is decreased to their nearest whole number
= 840 ÷ 24 = 35

? = 16.003 x 27.998 - 209.010
? ? 16 x 28 - 210 = 448 - 210
? = 238 ? 240

125.009 + 69.999 + 104.989 = ?
Lets assume, each value is approximated to nearest whole number
? ? ? 125 + 70 + 105
? ? ? 300

Let number of students in the sections
A, B, C and D be a, b, c and d respectively.
Then, total weight of students of sections A = 45a
Total weight of students of section B = 50b
Total weight of students of section C = 72c
Total weight of students of section D = 80d

According to the question,
Average weight of students of sections of A and B = 48 kg
? (45a + 50b)/(a + b) = 48
? 45a +50b = 48a + 48b
? 3a = 2b
? 15a = 10b

And average weight of students of sections B and C = 60 kg
? 50b + 72c = 60(b + c)
? 10b = 12c

Now, average weight of students of A, B, C and D = 60 kg
? 45a + 50b + 72c + 80d = 60(a + b + c + d)
? 15a + 10b - 12c - 20d = 0
? 15a = 20d
? a : d = 4 : 3

(2/3) x (6/8) x (2/3) x (3/4) = 1/4 = 0.25

? = 127/8 x 116/5 + 19/5 x 41/8
= 14732/40 + 779/40 = 15511/40 = 387.7 ? 385

? = 2³/₅ x 15/26 x 283.75
? 13/5 x 15/16 x 284 = 426 ? 425

36.98276421 x 21.00002
Using nearest value in whole number for 36.9827421 is 37 and 21.0002 is 21
? 37 x 21 = 777 ? 775

? = 14.995 x 8.001 x 20.991
? 15 x 8 x 21 = 2520