71. There are 40 students in a class. The average age of the first 15 students is 14 years and the average age of remaining 25 students is 15.4 years. What is the average age of the whole class?

एक कक्षा में 40 छात्र हैं, पहले 15 छात्रों की औसत आयु 14 वर्ष है और शेष 25 छात्रों की औसत आयु 15.4 वर्ष है। पूरी कक्षा की औसत आयु क्या है?

Option “B” is correct.

Average age of first 15 students = 14 years

∴ Total age of first 15 students = 15 × 14 = 210 years

Average age of remaining 25 students = 15.4 years

∴ Total age of remaining 25 students = 25 × 15.4 = 385 years

∴ Total age of 40 students = (210 + 385) = 595 years

∴ Average age of the whole class = 595/40 = 14.875 years

72. The first and the last term of an A.P. is x and y, respectively. What is the average of that A.P.?

एक समान्तर श्रेणी का प्रथम और अंतिम पद क्रमशः x और y है। उस समान्तर श्रेणी का औसत क्या है?

Option “D” is correct.

As we know, average of an A.P = (First term + Last term)/2

∴ Average of A.P = (x + y)/2

73. The average of 48 numbers is zero. Out of them, how many numbers can be greater than zero, at the most?

48 संख्याओं का औसत शून्य है। तो, अधिकतम कितनी संख्याएँ शून्य से अधिक हो सकती हैं?

Option “B” is correct.

Let there are 47 numbers value of each number is 1 then we can say sum of all 47 number is 47 and another 48th number whose value is (-47), then

Average of all 48 numbers = (47 – 47)/48 = 0

Now, we can say there are 47 numbers may be greater than zero.

74. The average of n numbers is 36. If each of 75% of the numbers is increased by 6 and each of the remaining numbers is decreased by 9, then the new average of the numbers is:

n संख्याओं का औसत 36 है। यदि 75% संख्याओं में से प्रत्येक में 6 की वृद्धि की जाती है और शेष संख्याओं में 9 की कमी की जाती है, तो संख्याओं का नया औसत क्या होगा?

Option “A” is correct.

Let the total number = 100

Average = 36 (given)

Total = 36 × 100 = 3600

75% of 100 = 75

Total increase in 75 numbers = 75 × 6 = 450

Total decrease in 25 numbers = 25 × -9 = 225

∴ Total increase in 100 numbers = 450 – 225 = 225

Now

Total = 3600 + 225 = 3825

Average = 3825/100 = 38.25%

75. The average of n numbers is 42. If 75% of the numbers are increased by 4 each and the remaining numbers are decreased by 8 each, then what is the average of the numbers, so obtained?

n संख्याओं का औसत 42 है। यदि 75% संख्याओं में प्रत्येक 4 की वृद्धि होती है और शेष संख्याओं में प्रत्येक 8 की कमी आती है, तो इस प्रकार प्राप्त संख्याओं का औसत क्या है?

Option “C” is correct.

Given:

The average of n numbers is 42.

75% of the numbers are increased by 4 each and the remaining numbers are decreased by 8 each.

Formula used:

Average = sum of observations/Number of observations

Calculation:

Average of n numbers = 42

Sum of n numbers = 42n

New sum of n number = 42n + {n × (75/100) × 4} – {n × (25/100) × 8}

⇒ 42n + 3n – 2n

⇒ 43n

New average of n numbers = 43n/n

⇒ 43

∴ The average of the numbers, so obtained is 43.

76. What is the average of all natural numbers from 101 to 200?

101 से 200 तक की सभी प्राकृतिक संख्याओं का औसत क्या है?

Option “A” is correct.

Short Trick:

Remember the result that the average of any consecutive series is the median.

So average from 101 to 200 = 150.5

Detailed Solution:

As we know,

Sum of first n natural number = [n (n + 1)] /2

Sum of first 200 natural number = (200 × 201) /2 = 20100

Sum of first 100 natural number = (100 × 101) /2 = 5050

Sum of the numbers from 101 to 200 = 20100 – 5050 = 15050

Average of the numbers from the 101 to 200 = 15050/100 = 150.5

77.The average of 13 numbers is 42. If a 14th number is included, then the average becomes 44. What is the 14th number?

13 संख्याओं का औसत 42 है। यदि 14 वें नंबर को शामिल किया जाता है, तो औसत 44 हो जाता है। 14 वीं संख्या क्या है?

Option “A” is correct.

Calculation:

⇒ Sum of 13 numbers = 13 × 42 = 546

⇒ Sum of 14 numbers = 14 × 44 = 616

⇒ 14th number = 616 – 546 = 70

78. The average marks of 5 students are increased by 3 marks when a new candidate comes in place of one of them, who had scored 65 marks. What might be the marks of the new candidate?

5 छात्रों के औसत अंकों में 3 अंकों की वृद्धि होती है, जब उनमें से एक के स्थान पर एक नया उम्मीदवार आता है, जिसने 65 अंक प्राप्त किए थे। नए उम्मीदवार के अंक क्या हो सकते हैं?

Option “C” is correct.

Given:

Increased marks = 3 marks 

New candidate score =  65 marks

Calculations:

The total score increased = (5 × 3) = 15 marks. 

Among the 5, one has scored 65 marks and it’s replaced by the score of the new candidate. 

Then the score of the new candidate = (65 + 15) = 80 marks. 

∴ The marks of the new candidate is 80 marks. 

Alternate Method

Let the average marks of 5 candidates be x marks and marks of new candidate be y

Total marks of 5 candidates = x × 5 = 5x marks. 

⇒ New average = (x + 3)  

⇒ Total marks = 5x – 65 + y 

According to question,

⇒5x – 65 + y = 5 (x + 3)  

⇒ – 65 + y = 15

⇒ y = 65 + 15

⇒ y = 80

∴ The marks of the new candidate is 80 marks. 

MISTAKE POINT

​in place of one of them means interchange not addition of new candidate. 

79. There are 179 consecutive even numbers and their average is 179.2, then find the difference between smallest and largest number.

179 क्रमागत सम संख्याएं हैं और उनका औसत 179.2 है, तो सबसे छोटी और सबसे बड़ी संख्या के बीच का अंतर ज्ञात कीजिए।

Option “B” is correct.

Sum of 179 numbers = 179 × average = 179 × 179.2 = 32076.8

Now, Sn = n/2[2a + (n – 1)d]

⇒ 32076.8 = 179/2[2a + (179 – 1)2]

⇒ 32076.8 = 179/2[2a + 178 × 2]

⇒ 32076.8 = 179/2[2a + 356]

⇒ 2a + 356 = 353.6

⇒ 2a = 2.4

⇒ a = 1.2

Now, last number = a + (n -1)d = 1.2 + 178 × 2 = 1.2 + 356 = 357.2

∴ Required difference = 357.2 – 1.2 = 356

80. The average of 3 consecutive integers is ‘k’. If the next two numbers are added, then the new average is:

3 क्रमागत पूर्णांकों का औसत ‘k’ है। यदि अगली दो संख्याएं जोड़ी जाती हैं, तो नया औसत क्या है?

Option “A” is correct.

Let the three consecutive integers be 1, 2 and 3

Average of first three number is = (1 + 2 + 3)/3 = 2

⇒ k = 2

The average of five consecutive integers = (1 + 2 + 3 + 4 + 5)/5 = 3

⇒ 3

⇒ 2 + 1

⇒ k + 1


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