QUESTION

CSAT

Easy

Reasoning

Prelims 2021

Consider the following addition problem : 3P+4P+PP+PP=RQ23P + 4P + PP + PP = RQ2; where P, Q and R are different digits. What is the arithmetic mean of all such possible sums?

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Explanation

3P+4P+PP+PP=RQ23P + 4P + PP + PP = RQ2

Or 30+P+40+P+10P+P+10P+P=100R+10Q+230 + P + 40 + P + 10P + P + 10P + P = 100R + 10Q+2

Or 24P+70=100R+10Q+224P + 70 = 100R + 10Q+ 2

Or 20P+70+4P=100R+10Q+220P + 70 + 4P = 100R + 10Q + 2

The unit digit of the resultant is 2. It will be obtained when 4 is multiplied by P. So, P must be 3, or 8.

If P = 3, then: 24P+70=24×3+70=72+70=14224P + 70 = 24 \times 3 + 70 = 72 + 70 = 142

If P = 8, then: 24P+70=24×8+70=192+70=26224P + 70 = 24 \times 8 + 70 = 192 + 70 = 262

Arithmetic sum of 142 and 262 = (142+262)/2=202(142 + 262)/2 = 202

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