51. What is the average of 2, 4, 6 ……, 100?

2, 4, 6 ……, 100 का औसत क्या होगा?

Option “B” is correct.

Concept:

Average of any A.P = (First term + Last term)/2

Calculation:

= (2 + 100)/2 = 102/2 = 51

Alternate approach

Here, d = 4 – 2 = 2

And, N  = 50

Sn = N/2[2a + (N – 1)d] = 50/2[2 × 2 + (50 – 1) × 2] = 50/2 × [4 + 98] = 50 × 51

Hence, average = Sn/N = (50 × 51)/50  = 51

52. The average of 39 numbers is zero. Out of them, how many may be greater than zero, at the most?

39 संख्याओं का औसत शून्य है। उनमें से, अधिकतम कितनी संख्याएं शून्य से बड़ी हो सकती हैं?

Option “B” is correct.
The maximum numbers greater than zero will be 38 because at least one number should be negative to make the sum of the other numbers equal to zero.

53. The average marks of 20 students in a test is 75. Later it is found that three marks 53, 60 & 76 were wrongly entered as 93, 64& 86. The average marks after mistakes were rectified is

एक परीक्षा में 20 छात्रों के औसत अंक 75 है। बाद में यह पाया गया कि तीन अंक 53, 60 & 76 को गलत तरह से 93, 64& 86 के रूप में लिखा गया। गलतियों को सुधारने के बाद औसत अंक है

Option “A” is correct.

Average marks of 20 students = 75

∴ Total marks of 20 students = 20 × 75 = 1500

∴ Actual total marks of 20 students,

⇒ 1500 – (93 + 64 + 86) + (53 + 60 + 76)

⇒ 1500 – 243 + 189

⇒ 1446

∴ Average marks of 20 students = 1446/20 = 72.3

54. While tabulation of marks scored in an examination by the students of a class, by mistake the marks scored by one student got recorded as 93 in place of 63, and thereby the average marks increased by 0.5. What was the number of students in the class?

एक कक्षा के छात्रों द्वारा एक परीक्षा में प्राप्त अंकों की तालिका बनाते समय, एक छात्र के द्वारा प्राप्त अंक 63 की जगह 93 अंकित हो जाता है, और जिसके कारण औसत में 0.5 की वृद्धि होती है। कक्षा में छात्रों की संख्या क्या थी?

Option “C” is correct.

Short trick:

Required number of students = (93 – 63)/0.5 = 30/0.5 = 60

Detailed solution:

Let the total number of students be x and average marks of the students be y, then

According to the question

xy + 93 – 63 = x × (y + 0.5)

xy + 30 = xy + 0.5x

0.5x = 30

x = 30/0.5

x = 60

55. The average weight of some students in a class was 60.5 kg. When 8 students, whose average weight was 65 kg. Joined the class, then the average weight of all the students increased by 0.9kg. The number of students in the class, initially, was:

एक कक्षा में कुछ छात्रों का औसत वज़न 60.5 किग्रा था। जब 8 छात्र, जिनका औसत वज़न 65 किग्रा था, कक्षा में शामिल हुए, तो सभी छात्रों का औसत वज़न 0.9 किग्रा बढ़ गया। शुरुआत में, कक्षा में छात्रों की संख्या क्या थी?

Option “B” is correct.

Let total number of students be x

Average weight of x students = 60.5 kg

Sum of weight of x students = 60.5x

Average weight of 8 students = 65

Sum of weight of 8 students = 65 × 8 = 520

According to the question

60.5x + 520 = (x + 8)(60.5 + 0.9)

60.5x + 520 = (x + 8) × 61.4

60.5x + 520 = 61.4x + 491.2

61.4x – 60.5x = 520 – 491.2

0.9x = 28.8

x = 28.8/0.9

x = 32

Short Trick :

0.9 kg weight increase for = 1 students

(65 – 60.5) × 8 kg weight increase for = (1/0.9) × 4.5 × 8 = 40 students

Number of students before = 40 – 8 = 32

56. The average weight of some students in a class was 60.5 kg. When 8 students, whose average weight was 65 kg. Joined the class, then the average weight of all the students increased by 0.9kg. The number of students in the class, initially, was:

एक कक्षा में कुछ छात्रों का औसत वज़न 60.5 किग्रा था। जब 8 छात्र, जिनका औसत वज़न 65 किग्रा था, कक्षा में शामिल हुए, तो सभी छात्रों का औसत वज़न 0.9 किग्रा बढ़ गया। शुरुआत में, कक्षा में छात्रों की संख्या क्या थी?

Option “B” is correct.

Let total number of students be x

Average weight of x students = 60.5 kg

Sum of weight of x students = 60.5x

Average weight of 8 students = 65

Sum of weight of 8 students = 65 × 8 = 520

According to the question

60.5x + 520 = (x + 8)(60.5 + 0.9)

60.5x + 520 = (x + 8) × 61.4

60.5x + 520 = 61.4x + 491.2

61.4x – 60.5x = 520 – 491.2

0.9x = 28.8

x = 28.8/0.9

x = 32

Short Trick :

0.9 kg weight increase for = 1 students

(65 – 60.5) × 8 kg weight increase for = (1/0.9) × 4.5 × 8 = 40 students

Number of students before = 40 – 8 = 32

57.The average of 36, 28, 43, 56, 74, 65, 12 and x is 45. What is the value of x?

36, 28, 43, 56, 74, 65, 12 और x का औसत 45 है। x का मान क्या है?

Option “A” is correct.

Given:

The average of 36, 28, 43, 56, 74, 65, 12 and x is 45.

Concept used:

Total = Number of entities × Average

Calculation:

(36 + 28 + 43 + 56 + 74 + 65 + 12 + x)/8 = 45

⇒ 314 + x = 360

⇒ x = 360 – 314

∴ The value of x is 46.

58. The average of a series of five numbers, in which each term is greater than its previous term by 4, is 24. What is the value of the greatest term?

पाँच संख्याओं की एक श्रृंखला का औसत 24 है, जिसमें प्रत्येक पद पिछली संख्या से 4 अधिक है। सबसे बड़ी संख्या का मान क्या है? 

Option “C” is correct.

Let the number be x, x +4, x +8, x + 12, x +16

According to question

(x + x +4 + x +8 + x + 12 + x +16)/5 = 24

⇒ 5x + 40 = 120

⇒ 5x = 80

⇒ x = 16

∴ Greatest number = 16 + 16 = 32 

59. A batsman scores 91 runs in his 18th innings. Due to this his average increased by 4 runs. Find his current average.

एक बल्लेबाज ने अपनी 18वीं पारी में 91 रन बनाए। इसके कारण उनका औसत 4 रन बढ़ गया। उसका वर्तमान औसत ज्ञात कीजिए।

Option “B” is correct.

We know that [old average × old number + something] = sum = new average × new number

Thus, Let the old average of batsman be x, then current average = x + 4

⇒ x × 17 + 91 = (x + 4) × 18

⇒ 17x + 91 = 18x + 72 x = 91 – 72 = 19

∴ Current average = x + 4 = 19 + 4 = 23

60. In an examination, the average score of students in a class who passed is 69 and that of who failed is 61. If the average score of all the students, who appeared in examination is 66, then the percentage of students who passed is:

एक परीक्षा में, एक कक्षा में उत्तीर्ण होने वाले छात्रों का औसत अंक 69 है और अनुत्तीर्ण होने वाले छात्रों का औसत अंक 61 है। यदि परीक्षा में शामिल होने वाले सभी छात्रों का औसत अंक 66 है, तो उत्तीर्ण होने वाले छात्रों का प्रतिशत क्या है?

Option “D” is correct.

Using allegation method

69     61

    66

5   :   3

Ratio of students who passed to fail = 5 : 3

Let passed students = 5, Failed students = 3

Hence, total students = 8

Percentage of passed students = (5/8) × 100 = 62.5%

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2 thoughts on “Average Questions for Competitive Exams”

  1. Arvind kumar

    App is very good only need to option for zoom the graph and photo for clear view

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