निम्नलिखित जोड़ (एडिशन) के प्रशन पर विचार कीजिए :
3P + 4P + PP + PP = RQ2; जहाँ P, Q तथा R भिन्न अंक हैं ।
इन सभी संभाव्य योगफलों का समांतर माध्य क्या है ?
सही उत्तर: (b) 120
दी गई समीकरण है: 3P + 4P + PP + PP = RQ2। इसे दहाई और इकाई के रूप में लिखने पर: (30+P) + (40+P) + 11P + 11P = 70 + 24P = RQ2। चूँकि RQ2 का इकाई अंक 2 है, इसलिए 70 + 24P का इकाई अंक भी 2 होना चाहिए। इसका अर्थ है कि 4P का इकाई अंक 2 होना चाहिए, जो कि P = 3 या P = 8 पर संभव है। यदि P = 3, तो योग = 70 + 24(3) = 142 (जहाँ R=1, Q=4, संपाती अंक भिन्न हैं)। यदि P = 8, तो योग = 70 + 24(8) = 262 (जहाँ R=2, Q=6, अंक भिन्न हैं)। इस प्रकार दो संभव योगफल 142 और 262 हैं। इनका समांतर माध्य = (142 + 262) / 2 = 202 है। अतः विकल्प (c) सही है।
In English (Question & Model Answer)
Consider the following addition problem :
3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.
What is the arithmetic mean of all such possible sums ?
Correct Option: (b) 120
Given the addition problem: 3P + 4P + PP + PP = RQ2, where P, Q, R are distinct digits. Let's write them in decimal notation: (30 + P) + (40 + P) + (10P + P) + (10P + P) = R02 + Q? Wait, PP means a two-digit number with digits P and P, so PP = 11P. Thus the LHS is: (30 + P) + (40 + P) + 11P + 11P = 70 + 24P. The RHS is RQ2, which is 100R + 10Q + 2. So we have 70 + 24P = 100R + 10Q + 2 => 24P + 68 = 100R + 10Q. Since Q is the units digit of the sum, let's test possible values of P from 0 to 9. If P = 0, 68 = 100R + 10Q (no integer solution for R, Q). If P = 1, 24(1) + 68 = 92 = 100R + 10Q (no solution). If P = 2, 48 + 68 = 116 = 100(1) + 10(1) + 6 (digits not matching units digit 2). Wait, units digit of 24P + 68 must be 2. The units digit of 24P comes from 4 * P. For 4 * P to have a units digit such that (units of 4P + 8) ends in 2: units of 4P must end in 4 (since 4 + 8 = 12, ends in 2). 4 * P ends in 4 when P = 1, 4, 6, 9. Let's test these values of P: Case P = 1: 24(1) + 68 = 92 (R=0, Q=9, R!=0 usually, but digits are P=1, Q=9, R=0). Case P = 4: 24(4) + 68 = 96 + 68 = 164 (R=1, Q=6, digits P=4, Q=6, R=1 are distinct). Let's check sum: RQ2 = 164 (here Q=4? Wait, RHS is RQ2, so Q=4, but we assumed P=4, so Q should match the units digit of sum: 34 + 44 + 44 + 44 = 166, units digit is 6, but RQ2 has units digit 2! Ah, the units digit of the sum must be 2, so Q = 2 always! Therefore, units digit of 24P + 68 must be 2, and the sum itself must end in 2, which means Q = 2. Let's re-check: Q = 2. Then RHS is R22. Let's test P values where units digit of 24P + 68 ends in 2 and Q = 2: Wait, 24P + 68 = 100R + 20 + 2 = 100R + 22 => 24P + 46 = 100R. For P from 0 to 9: If P = 4, 24(4) + 46 = 96 + 46 = 142 (not divisible by 100). If P = 9, 24(9) + 46 = 216 + 46 = 262. Let's check P = 7: 24(7) + 46 = 168 + 46 = 214. Let's check P = 8: 24(8) + 46 = 192 + 46 = 238. Wait, let's find all possible sums RQ2. Since the question asks for the arithmetic mean of all such possible sums, let's test all P from 0 to 9: P=1: 70+24(1)=94 (not ending in 2). P=2: 70+48=118. P=3: 70+72=142. P=4: 70+96=166. P=5: 70+120=190. P=6: 70+144=214. P=7: 70+168=238. P=8: 70+192=262. P=9: 70+216=286. Wait, none of these end in 2 except let's check arithmetic carefully: 3P = 30+P, 4P = 40+P, PP = 11P, PP = 11P. Sum = 70 + 24P. We need 70 + 24P = RQ2. The units digit of RQ2 is 2. So the units digit of 70 + 24P must be 2. 0 + units(24P) = 2 => units(4P) = 2. P can be 3 (4*3=12) or 8 (4*8=32). Let's test P = 3: 70 + 24(3) = 70 + 72 = 142. Here R=1, Q=4, 2=2 (matches RQ2 = 142). Digits P=3, Q=4, R=1 are distinct. Sum = 142. Let's test P = 8: 70 + 24(8) = 70 + 192 = 262. Here R=2, Q=6, 2=2 (RQ2 = 262). Digits P=8, Q=6, R=2 are distinct. Sum = 262. Possible sums are 142 and 262. Arithmetic mean = (142 + 262) / 2 = 404 / 2 = 202. Thus option (c) is correct.