UP Board Solutions Class 9 Maths Chapter 20 Statistics Ex 20.4

Ex 20.4 Statistics अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. 7 संख्याओं x -3, x+2, x, 9, x-2, x + 1, 13 का माध्य 10 है तब x का मान ज्ञात कीजिए ।
Answer: x-3+ x + 2 + x + 9 + x-2 + x + 1 + 13 हलः सामान्तर माध्य \( = \frac{x-3+x+2+x+9+x-2+x+1+13}{7} \) \( 10 = \frac{5x+20}{7} \) \( 70 = 5x + 20 \) \( 70-20 = 5x \)
\( \implies 5x = 50 \)
\( x = \frac{50}{5} = 10 \)In simple words: The mean of 7 numbers is given as 10. By summing the numbers and dividing by 7, we form an equation to solve for x, finding x = 10.

🎯 Exam Tip: Remember to sum all the given numbers correctly before dividing by the count of numbers when calculating the mean.

Question 2. यदि 7, 9, 11, 13, x और 21 का समान्तर माध्य 13 है तब x का मान ज्ञात कीजिए।
Answer: हलः सामान्तर माध्य \( = \frac{7+9+11+13+x+21}{6} \) \( 13 = \frac{61+x}{6} \) \( 78 = 61 + x \)
\( \implies 78-61 = x \)
\( \implies 17 = x \)In simple words: Given the mean of six numbers is 13, we sum the numbers including x, divide by 6, and set it equal to 13 to solve for x, which is 17.

🎯 Exam Tip: Pay close attention to the number of observations (n) when calculating the mean to avoid errors in the denominator.

Question 3. प्रथम 10 विषम प्राकृतिक संख्याओं का माध्य ज्ञात कीजिए।
Answer: हलः माध्य \( = \frac{1+3+5+7+9+11+13+15+17+19}{10} \) \( = \frac{100}{10} = 10 \)In simple words: To find the mean of the first 10 odd natural numbers, sum them up and divide by 10, resulting in a mean of 10.

🎯 Exam Tip: Listing all the numbers correctly is crucial for accurate summation, especially for a sequence like odd natural numbers.

Question 4. प्रथम 10 अभाज्य संख्याओं का माध्य ज्ञात कीजिए।
Answer: हलः सामान्तर माध्य \( = \frac{2+3+5+7+13+17+19+23+29+31}{10} \) \( = \frac{129}{10} = 12.9 \)In simple words: The mean of the first 10 prime numbers is found by adding them and dividing by 10, which gives 12.9.

🎯 Exam Tip: Accurately identifying the first N prime numbers is the key step; remember that 1 is not a prime number.

Question 5. प्रथम 10 प्राकृतिक संख्याओं का माध्य ज्ञात कीजिए।
Answer: हलःIn simple words: To find the mean of the first 10 natural numbers, you would sum 1 through 10 and then divide by 10.

🎯 Exam Tip: For consecutive natural numbers, the mean is often simply the middle value or the average of the first and last number.

Question 6. प्रथम 5 पूर्ण संख्याओं का माध्य ज्ञात कीजिए।
Answer: हलःIn simple words: To find the mean of the first 5 whole numbers, you would sum 0, 1, 2, 3, 4 and then divide by 5.

🎯 Exam Tip: Remember that whole numbers start from 0, unlike natural numbers which start from 1.

Question 7. 20 संख्याओं का माध्य 35 है। यदि प्रत्येक संख्या 5 से विभाजित है तब नया माध्य ज्ञात कीजिए।
Answer: हलः 20 संख्याओं का माध्य = 35 20 संख्याओं का योग = \( 20 \times 35 = 700 \) प्रत्येक सँख्या 5 से विभाजित होने पर कुल सँख्या = \( 20 \times 5 = 100 \) नया सामान्तर माध्य = \( \frac{700}{100} = 7 \)In simple words: If each of 20 numbers with a mean of 35 is divided by 5, the sum of the numbers would be 700. If each number were divided by 5, the count of numbers would still be 20. The question implies if the total sum is modified due to each individual number being operated upon. The context seems to be if the sum itself is divided by 100 which is \(20 \times 5\). Re-interpretation suggests that if each number is *divided* by 5, then the *new mean* will also be divided by 5. So new mean would be \(35/5 = 7\). The provided solution calculates a new sum based on a different interpretation, where perhaps a "factor of 5" is related to the count of numbers. Let's assume the standard interpretation for mean properties. If each number \(x_i\) is changed to \(x_i/5\), then the new mean is \(\bar{x}_{new} = (\sum x_i / 5) / N = (1/5) (\sum x_i / N) = (1/5) \bar{x}_{old}\). So new mean should be \(35/5=7\). The calculation \(20 \times 5 = 100\) in the denominator is for the new number of items if that were the case, which it isn't. It is confusing. Let's stick to the numerical outcome which is 7, and simplify the explanation. If each of 20 numbers, whose mean is 35, is divided by 5, the new mean will also be divided by 5, resulting in a new mean of 7.

🎯 Exam Tip: When each observation in a data set is multiplied or divided by a constant, the mean of the new data set is also multiplied or divided by the same constant.

Question 8. 5 संख्याओं का माध्य 10 है, यदि प्रत्येक संख्या 3 के द्वारा घटायी गयी है तो नया माध्य ज्ञात कीजिए।
Answer: हलः 5 संख्याओं का माध्य = 10 5 संख्याओं का योग = \( 5 \times 10 = 50 \) प्रत्येक में से 3 घटाने पर योग = \( 50 - 3 \times 5 = 35 \) नया सामान्तर माध्य = \( \frac{35}{5} = 7 \)In simple words: If 3 is subtracted from each of 5 numbers with a mean of 10, the new sum becomes 35, and the new mean is 35 divided by 5, which is 7.

🎯 Exam Tip: If each observation in a data set is increased or decreased by a constant, the mean of the new data set is also increased or decreased by the same constant.

Question 9. n प्रेक्षणों का माध्य M है यदि प्रत्येक प्रेक्षण में k से गुणा की जाती है तो नये प्रेक्षण का माध्य ज्ञात कीजिए ।
Answer: हलः n प्रेक्षणों का माध्य = M n प्रेक्षणों का योग = nM प्रत्येक प्रेक्षण में k की गुणा करने पर = nkM नये प्रेक्षण का माध्य \( \overline{x} = \frac{n k M}{n} = kM \)In simple words: If each of n observations with mean M is multiplied by k, the new mean of the observations will be kM.

🎯 Exam Tip: This illustrates a fundamental property of the mean: scaling all data points by a constant scales the mean by the same constant.

Question 10. यदि 5 प्रेक्षणों x, x + 2, x + 4, x + 6, x + 8 का माध्य 11 है तब प्रथम तीन प्रेक्षणों का माध्य ज्ञात कीजिए।
Answer: हलः सामान्तर माध्य \( \overline{x} = \frac{\sum x}{N} \) \( 11 = \frac{x + x + 2 + x + 4 + x + 6 + x + 8}{5} \) \( 55 = 5x + 20 \) \( 55-20 = 5x \) \( 35 = 5x \)
\( \implies x = \frac{35}{5} \)
\( \implies x = 7 \) प्रथम तीन प्रेक्षण = 7,9,11 का समान्तर माध्य \( = \frac{7+9+11}{3} \) \( = \frac{27}{3} = 9 \)In simple words: First, we use the given mean (11) of the five observations to find the value of x, which is 7. Then, we find the mean of the first three observations (7, 9, 11) by summing them and dividing by 3, resulting in 9.

🎯 Exam Tip: Break down multi-step problems: first solve for the unknown variable using the overall mean, then apply that variable to find the mean of a subset.

Ex 20.4 Statistics लघु उत्तरीय प्रश्न (Short Answer Type Questions)

Question 11. 10 के सभी सम्भव गुणनखण्डों का माध्य ज्ञात कीजिए।
Answer: हलः 10 के सभी सम्भव गुणनखण्डों का माध्य = 1, 2, 5, 10 सामान्तर माध्य \( = \frac{1+2+5+10}{4} \) \( = \frac{18}{4} = 4.5 \)In simple words: The factors of 10 are 1, 2, 5, and 10. Their mean is calculated by summing them up and dividing by 4, which equals 4.5.

🎯 Exam Tip: Ensure all possible factors are identified correctly before proceeding with the mean calculation.

Question 12. 994, 996, 998, 1002 और 1000 का माध्य ज्ञात कीजिए।
Answer: हलः सामान्तर माध्य \( \overline{x} = \frac{\sum x}{N} \) \( \overline{x} = \frac{994+996+998+1002+1000}{5} \) \( = \frac{4990}{5} = 998 \)In simple words: To find the mean of the given five numbers, sum them up to get 4990 and then divide by 5, which results in a mean of 998.

🎯 Exam Tip: For numbers close to a central value, consider using an assumed mean method to simplify calculations and reduce potential errors.

Question 13. प्रथम 10 विषम प्राकृतिक संख्याओं का माध्य ज्ञात कीजिए।
Answer: हलःIn simple words: To find the mean of the first 10 odd natural numbers, you would list them, sum them, and then divide by 10.

🎯 Exam Tip: The sum of the first 'n' odd natural numbers is \(n^2\). Therefore, their mean is \(n^2/n = n\). For the first 10 odd numbers, the mean is 10.

Question 14. यदि 2x +3, 3x+ 4, x +7, x-3, 4x +7 का माध्य 14 है तो x ज्ञात कीजिए।
Answer: हलः सामान्तर माध्य \( = \frac{2x+3+3x+4+x+7+x-3+4x+7}{5} \) \( 14 = \frac{11x+18}{5} \) \( 70 = 11x+18 \) \( 11x = 70-18 \) \( 11x = 52 \)
\( \implies x = \frac{52}{11} = 4.72 \)In simple words: Given the mean of five algebraic expressions is 14, we sum the expressions, divide by 5, and set it equal to 14 to solve for x, which is approximately 4.72.

🎯 Exam Tip: Carefully combine like terms (x terms and constant terms) in the numerator to avoid algebraic errors.

Question 15. एक विशेष सप्ताह के लिए एक दुकान से शुगर की प्रतिदिन बिक्री नीचे दी गयी है: 75 किग्रा, 120 किग्रा, 12 किग्रा, 50 किग्रा, 70.5 किग्रा, 140.5 किग्रा शुगर की प्रतिदिन बिक्री का औसत ज्ञात कीजिए।
Answer: हलः शुगर की प्रतिदिन बिक्री का औसत \( = \frac{75+120+12+50+70.5+140.5}{6} \) \( = \frac{468}{6} = 78 \) किग्राIn simple words: To find the average daily sugar sales, sum all the given daily sales figures and then divide by the total number of days (6), which results in 78 kg.

🎯 Exam Tip: Double-check the addition of decimal numbers to ensure accuracy in the sum before dividing.

Question 16. एक सोसायटी के 11 परिवारों के बच्चों की संख्या 2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5 है। प्रति परिवार बच्चों की संख्या का माध्य ज्ञात कीजिए।
Answer: हलःIn simple words: To find the mean number of children per family, sum all the given numbers of children and then divide by the total number of families (11).

🎯 Exam Tip: When dealing with counts of discrete items, make sure to count the total number of observations correctly for the denominator.

Question 17. यदि 10, 12, 18, 13, P तथा 17 का माध्य 15 है तो P का मान ज्ञात कीजिए।
Answer: हल: सामान्तर माध्य \( = \frac{10+12+18+13+P+17}{6} \) \( 15 = \frac{70+P}{6} \) \( 90 = 70+ P \)
\( \implies P = 90-70 = 20 \)In simple words: Given the mean of six numbers (including P) is 15, we set up an equation by summing the numbers, dividing by 6, and equating it to 15, then solve for P, which is 20.

🎯 Exam Tip: Carefully sum the known numbers first before isolating the unknown variable P in the equation.

Question 18. एक स्कूल की कक्षा IX के 34 विद्यार्थियों के भारों का माध्य 42 किग्रा है। यदि अध्यापक का भार जोड़ दिया जाये, तो माध्य 400 ग्राम बढ़ जाता है। अध्यापक का भार ज्ञात कीजिए।
Answer: हल: माध्य \( = \frac{\sum x}{n} \) \( 42 = \frac{\sum x}{34} \)
\( \implies \sum x = 1428 \) माना अध्यापक का भार = x ग्राम 1428+ x = \( 42.4 \times 35 \) (Since 400 grams = 0.4 kg, new mean = \(42 + 0.4 = 42.4\) kg, and new number of observations = \(34 + 1 = 35\)) \( 1428+x = 1484 \)
\( \implies x = 1484-1428 = 56 \) किग्रा
\( \implies \) अध्यापक का भार = 56 किग्राIn simple words: The total weight of 34 students is \(34 \times 42 = 1428\) kg. When the teacher's weight (x kg) is added, the mean increases by 400g (0.4 kg), making the new mean 42.4 kg for 35 people. So, \(1428 + x = 42.4 \times 35 = 1484\), which means the teacher's weight is 56 kg.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., grams to kilograms) before performing calculations. Also, the number of observations changes when a new item is added.

Question 19. 5 संख्याओं का माध्य 18 है, यदि एक संख्या को हटा दिया जाये तब उनका माध्य 16 है। संख्या ज्ञात कीजिए ।
Answer: हलः 5 सँख्याओं का माध्य = 18 5 सँख्याओं का योग = \( 5 \times 18 = 90 \) माना हटायी गयी संख्या = x
\( \implies \) 4 संख्याओं का योग = \( 90 - x \) 4 संख्याओं का माध्य = 16 \( 16 = \frac{90-x}{4} \) \( 64 = 90-x \)
\( \implies x = 90 - 64 = 26 \)
\( \implies \) वह हटायी गयी संख्या = 26In simple words: The initial sum of 5 numbers is \(5 \times 18 = 90\). After removing one number, say x, the sum of the remaining 4 numbers is \(90 - x\), and their mean is 16. So, \( (90 - x) / 4 = 16 \), which leads to \(90 - x = 64\), meaning x = 26.

🎯 Exam Tip: Always work with the total sum when items are added or removed, and adjust the number of observations accordingly.

Question 20. 31 परिणामों का माध्य 60 है। यदि प्रथम 16 परिणामों का माध्य 58 है तथा अन्तिम 16 परिणामों का माध्य 62 है तो 16वाँ परिणाम ज्ञात कीजिए।
Answer: हलः 31 परिणामों का माध्य = 60 31 परिणामों का योग = \( 31 \times 60 = 1860 \) प्रथम 16 परिणामों का माध्य = 58 प्रथम 16 परिणामों का योग = \( 16 \times 58 = 928 \) अन्तिम 16 परिणामों का माध्य = 62 अन्तिम 16 परिणामों का योग = \( 16 \times 62 = 992 \) कुल 32 परिणामों का योग = \( 928 + 992 = 1920 \)
\( \implies \) 16 वाँ परिणाम = \( 1920 - 1860 = 60 \)In simple words: The sum of 31 results is 1860. The sum of the first 16 is 928, and the sum of the last 16 is 992. When these two sums are added, they include the 16th result twice. The sum of these 32 values is 1920. Subtracting the total sum of 31 results (1860) from this 1920 gives the value of the 16th result, which is 60.

🎯 Exam Tip: In overlapping sum problems, identify the common element (the 16th result in this case) that is counted twice, and subtract the overall sum to find its value.

Ex 20.4 Statistics दीर्घ उत्तरीय प्रश्न (Long Answer Type Questions)

Question 21. 6 संख्याओं का माध्य 20 है। यदि एक संख्या हटा दी जाती है तब माध्य 15 है। हटायी गयी संख्या ज्ञात कीजिए।
Answer: हलः 6 संख्याओं का माध्य = 20 6 संख्याओं का योग = \( 6 \times 20 = 120 \) माना हटायी गयी संख्या = x संख्या हटाने पर संख्याओं का योग = \( 120 -x \) संख्या हटाने पर संख्याओं का माध्य = 15 \( 15 = \frac{120 - x}{5} \) \( 75 = 120 - x \)
\( \implies x = 120 - 75 = 45 \)
\( \implies \) हटायी गयी संख्या = 45In simple words: The total sum of 6 numbers is 120. When one number (x) is removed, 5 numbers remain with a mean of 15, so their sum is \(5 \times 15 = 75\). The removed number x is the difference between the original sum and the new sum, so \(x = 120 - 75 = 45\).

🎯 Exam Tip: This is a standard problem type; always calculate the total sum before and after the change to find the value of the removed or added item.

Question 22. 8 संख्याओं का माध्य भार 15 है। यदि प्रत्येक संख्या को 2 से गुणा किया जाता है तब नया माध्य क्या होगा?
Answer: हलः 8 संख्याओं का माध्य भार = 15 8 संख्याओं का कुल योग = \( 8 \times 15 = 120 \) यदि प्रत्येक संख्या को 2 से गुणा किया जाता है तो कुल योग = \( 120 \times 2 = 240 \) नया माध्य = \( \frac{240}{8} = 30 \)In simple words: If the mean of 8 numbers is 15, and each number is multiplied by 2, the new mean will also be multiplied by 2, resulting in a new mean of \(15 \times 2 = 30\).

🎯 Exam Tip: Remember the property that if each observation is multiplied by a constant, the mean is also multiplied by that constant, simplifying calculations.

Question 23. 10 विद्यार्थियों की औसत लम्बाई 153 सेमी है। बाद में यह पता चला कि एक जगह 151 सेमी की जगह 141 सेमी पढ़ा गया था। सही औसत ज्ञात कीजिए ।
Answer: हलः 10 विद्यार्थियों की औसत लम्बाई = 153 सेमी 10 विद्यार्थियों की कुल लम्बाई = \( 10 \times 153 = 1530 \) सेमी
\( \implies \) 151 की जगह 141 सेमी पढ़ा गया
\( \implies \) 10 विद्यार्थियों की कुल लम्बाई = \( 1530 - 141 + 151 = 1540 \)
\( \implies \) 10 विद्यार्थियों की औसत लम्बाई = \( \frac{1540}{10} = 154 \) सेमीIn simple words: The original total length was 1530 cm. Correcting the error involves subtracting the wrongly read value (141 cm) and adding the correct value (151 cm) to the sum, resulting in a corrected total length of 1540 cm. Dividing this by 10 gives the correct average length of 154 cm.

🎯 Exam Tip: For error correction problems, always adjust the total sum first by subtracting the incorrect value and adding the correct one, then recalculate the mean.

Question 24. 17 प्रेक्षणों का माध्य 20 है। यदि प्रथम 9 प्रेक्षणों का माध्य 23 है तथा अन्तिम 9 प्रेक्षणों का माध्य 18 है तो 9वाँ प्रेक्षण ज्ञात कीजिए।
Answer: हलः 17 प्रेक्षणों का माध्य = 20 17 प्रेक्षणों का कुल योग = \( 17 \times 20 = 340 \) प्रथम 9 प्रेक्षणों का माध्य = 23 प्रथम 9 प्रेक्षणों का योग = \( 9 \times 23 = 207 \) अन्तिम 9 प्रेक्षणों का माध्य = 18 अन्तिम 9 प्रेक्षणों का योग = \( 9 \times 18 = 162 \) कुल 18 प्रेक्षणों का योग = \( 207 + 162 = 369 \)
\( \implies \) 9 वाँ प्रेक्षण = \( 369 - 340 = 29 \)In simple words: The sum of 17 observations is 340. The sum of the first 9 is 207, and the sum of the last 9 is 162. Summing these (207 + 162 = 369) gives the 9th observation twice. Subtracting the total sum of 17 observations (340) from this gives the 9th observation, which is 29.

🎯 Exam Tip: This is an overlapping sum problem. The item counted twice (the 9th observation) can be found by adding the partial sums and subtracting the total sum.

Question 25. 9 पारियों में एक क्रिकेटर का औसत 58 रन है। उसे अपने औसत को 61 करने के लिए 10 वीं पारी में कितने रन बनाने होंगे।
Answer: हलः माना 10 वी पारी में बनाने है = x रन 9 पारियों में कुल रन = \( 9 \times 58 = 522 \) औसत \( = \frac{\text{कुल रन}}{\text{कुल पारी}} \) \( 61 = \frac{522 + x}{10} \) \( 610 = 522 + x \) \( 610-522 = x \)
\( \implies x = 88 \)In simple words: After 9 innings, the cricketer's total runs are \(9 \times 58 = 522\). To achieve an average of 61 runs over 10 innings, the total runs needed would be \(10 \times 61 = 610\). Therefore, the runs needed in the 10th inning are \(610 - 522 = 88\).

🎯 Exam Tip: Always calculate the total sum (or total runs in this case) required to achieve the desired mean over the new total number of observations.