किसी परीक्षा में किसी विद्यार्थी के अंक में संशोधन किए जाने पर कक्षा के औसत अंक x में 4 की वृद्धि होती है; इस विद्यार्थी को 0 के स्थान पर संशोधित अंक 100 मिलते हैं। बाद में, यह पता चलता है कि दूसरे विद्यार्थी के अंक 56 के स्थान पर 81 दर्ज हुए हैं। यदि और कोई संशोधन नहीं होता है तथा अंतिम संशोधित औसत y है, तो y - x क्या है?
सही उत्तर: (b) 3
मान लीजिए कक्षा में विद्यार्थियों की संख्या N है और प्रारंभिक औसत x है। एक विद्यार्थी का अंक 0 के स्थान पर 100 दर्ज करने पर औसत में 4 की वृद्धि होती है। कुल योग में वृद्धि = 100 - 0 = 100। विद्यार्थियों की संख्या N = 100 / 4 = 25 है। बाद में दूसरे विद्यार्थी के अंक 56 के स्थान पर 81 दर्ज पाए गए (अर्थात 25 की अधिकता)। इस दूसरी त्रुटि सुधार से औसत में कमी = 25 / 25 = 1 की होती है। अतः अंतिम संशोधित औसत y और प्रारंभिक औसत x का अंतर y - x = 4 - 1 = 3 होगा।
In English (Question & Model Answer)
The class average x in a test increases by 4 when the score of a student is rectified, whose corrected score is 100 instead of 0. Later, the score of another student was found to have been recorded as 81 in place of 56. If there are no other corrections and the final corrected average is y, then y - x is
Correct Option: (b) 3
Let's analyze the changes in the class average. Let there be N students in the class, and the initial recorded average be x. A student's score was incorrectly recorded as 0 instead of 100. The increase in the total sum of scores is 100 - 0 = 100. This increase causes the class average to increase by 4. Therefore, Change in average = (Increase in sum) / N => 4 = 100 / N => N = 100 / 4 = 25 students. So there are 25 students in the class. Later, another correction was found: the score of another student was recorded as 81 instead of 56. The change in the sum of scores for this second correction = Corrected score - Recorded score = 56 - 81 = -25. This second correction changes the average from x (after first correction) to y. The net change in the average due to the second correction is (Change in sum) / N = -25 / 25 = -1. Wait! Let's re-read carefully: 'If there are no other corrections and the final corrected average is y, then y - x is...' Let's check: x is the average *after* the first correction (where score 0 was corrected to 100). The second correction changes the score from 81 to 56, which decreases the sum by 25. Thus, the final average y is 1 less than x, meaning y = x - 1. Therefore, y - x = -1? Wait, let's check the options: (a) 2, (b) 3, (c) 5, (d) 6. Let's re-calculate carefully: 'class average x in a test increases by 4 when the score of a student is rectified, whose corrected score is 100 instead of 0.' This means x is the *new* average after the first correction! Wait, let's re-read: 'The class average x ... increases by 4 when ...' If x is the new average, then increase = 4, so N = 100 / 4 = 25. Then the second correction changes score from 81 to 56 (error was 81, actual was 56? Wait: 'recorded as 81 in place of 56' means recorded = 81, actual = 56). So the sum decreases by 81 - 56 = 25. The change in average is -25 / 25 = -1. Wait, why would y - x be positive in the options? Let's check if the first correction increased average by 4, meaning x was the *original* average and the new average is x + 4. If x is original average, then y - x = +3? Let's re-read: 'The class average x in a test increases by 4 when...'. If x denotes the initial average, then new average is x + 4. Then the second correction adds (56 - 81) = -25, changing the average by -1, so final average y = (x + 4) - 1 = x + 3. Thus y - x = 3. This matches option (b) 3.