Consider the following sum: I + 1I + 2I + I3 + I1 = 21I In the above sum, I stands for

Correct Option (D)

The problem presents an arithmetic sum where 'I' represents a single digit. The notation '1I', '2I', 'I3', 'I1', and '21I' indicates numbers formed by positional notation. To solve this, let 'I' be represented by the variable 'x'.

The terms in the sum can be expanded as follows:

• I = x
• 1I = 10 + x (since 1 is in the tens place)
• 2I = 20 + x (since 2 is in the tens place)
• I3 = 10x + 3 (since I is in the tens place)
• I1 = 10x + 1 (since I is in the tens place)
• 21I = 210 + x (since 21 is in the tens and hundreds place, and I is in the units place)

Substitute these expanded forms into the given equation:

x + (10 + x) + (20 + x) + (10x + 3) + (10x + 1) = 210 + x

Combine the 'x' terms and constant terms on the left side of the equation:

(x + x + x + 10x + 10x) + (10 + 20 + 3 + 1) = 210 + x

23x + 34 = 210 + x

To solve for x, rearrange the equation by bringing all 'x' terms to one side and constant terms to the other:

23x - x = 210 - 34

22x = 176

Divide both sides by 22 to find the value of x:

x = $\frac{176}{22}$22176​

x = 8

Therefore, 'I' stands for 8.

Incorrect Options:

• If I = 4 (Option A), substituting into the equation: 4 + (10+4) + (20+4) + (10*4+3) + (10*4+1) = 4 + 14 + 24 + 43 + 41 = 126. The right side would be 210 + 4 = 214. Since 126 ≠ 214, this option is incorrect.
• If I = 5 (Option B), substituting into the equation: 5 + (10+5) + (20+5) + (10*5+3) + (10*5+1) = 5 + 15 + 25 + 53 + 51 = 149. The right side would be 210 + 5 = 215. Since 149 ≠ 215, this option is incorrect.
• If I = 6 (Option C), substituting into the equation: 6 + (10+6) + (20+6) + (10*6+3) + (10*6+1) = 6 + 16 + 26 + 63 + 61 = 172. The right side would be 210 + 6 = 216. Since 172 ≠ 216, this option is incorrect.