Correct option is A
Given: 50 × 3 + 12 × 5 – (66 ÷ 2) + 11 × 4 – 22 = 160
After interchanging the numbers in the given equation will be:
Option (a): 3 and 2
50 × 2 + 12 × 5 – (66 ÷ 3) + 11 × 4 – 22 = 160
50 × 2 + 12 × 5 – 22 + 11 × 4 – 22 = 160
100 + 60 - 22 + 44 -22 = 160
160 + 44 - 22 - 22 = 160
204 - 44 = 160
160 = 160
It is correct.
Option (b): 5 and 3
50 × 5 + 12 × 3 – (66 ÷ 2) + 11 × 4 – 22 = 160
50 × 5 + 12 × 3 – 33 + 11 × 4 – 22 = 160
250 + 36 - 33 + 44 - 22 = 160
250 + 36 + 44 - 33 - 22 = 160
330 - 33 - 22 = 160
330 - 55 = 160
275 ≠ 160
It is incorrect.
Option (c): 22 and 2
50 × 3 + 12 × 5 – (66 ÷ 22) + 11 × 4 – 2 = 160
50 × 3 + 12 × 5 – 3 + 11 × 4 – 2 = 160
150 + 60 - 3 + 44 - 2 = 160
210 + 44 - 3 - 2 = 160
254 - 5 = 160
249 ≠ 160
It is incorrect.
Option (d):12 and 11
50 × 3 + 11 × 5 – (66 ÷ 2) + 12 × 4 – 22 = 160
50 × 3 + 11 × 5 – 33 + 12 × 4 – 22 = 160
150 + 55 - 33 + 48 - 22 = 160
205 + 48 - 33 - 22 = 160
253 - 33 - 22 = 160
198 ≠ 160
It is incorrect.
Thus, the correct option is (a) 3 and 2.
English
80 Questions
160 Marks
60 Mins
