31. The average of a series of five numbers, in which each term is greater than its previous term by 3, is 12. What is the value of the greatest term?

पाँच संख्याओं की एक श्रृंखला का औसत, जिसमें प्रत्येक पद अपने पिछले पद से 3 अधिक है, 12 है। सबसे बड़े पद का मान क्या है?

Option “D” is correct.

Let first number be x, then

Then second, third, fourth and fifth numbers will be (x + 3), (x + 6), (x + 9) and (x + 12) respectively.

Average of 5 numbers = 12

Hence, sum of 5 numbers = 12 × 5 = 60

According to the question

x + x + 3 + x + 6 + x + 9 + x + 12 = 60

⇒ 5x + 30 = 60

⇒ 5x = 60 – 30

⇒ 5x = 30

⇒ x = 30/5

⇒ x = 6

∴ The greatest number = x + 12 = 6 + 12 = 18

32. The average of 20 values is 27. If each value is multiplied by 3, then the new average is:

20 मानों का औसत 27 है। यदि प्रत्येक मान को 3 से गुणा किया जाता है, तो नया औसत है:

Option “D” is correct.

Average of 20 values is = 27

Sum of 20 values = 27 × 20 = 540

If each value is multiplied by 3, then

New sum of 20 values = 540 × 3 = 1620

New average of 20 values = 1620/20 = 81

Short trick:

Required average = 27 × 3 = 81

33. The average of 50 numbers is 75. If the average of the first set of 25 numbers is 65, then what is the average of the second set of 25 numbers?

50 संख्याओं का औसत 75 है। यदि 25 संख्याओं के पहले समूह का औसत 65 है, तो 25 संख्याओं के दूसरे समूह का औसत क्या है?

Option “C” is correct.

Average of 50 numbers = 75

Sum of 50 numbers = 75 × 50 = 3750

The average of fist 25 numbers = 65

Sum of 25 numbers = 65 × 25 = 1625

Sum of last 25 numbers = 3750 – 1625 = 2125

Average of last 25 numbers = 2125/25 = 85

Short Trick:

Let the average of 2 numbers = 75

First number is = 65

Let second number be x, then

⇒ x + 65 = 2 × 75

⇒ x = 150 – 65 = 85

34. An hour-long test has 60 problems. If a student completes 30 problems in 25 minutes, then the required seconds he has taken on average for computing each of the remaining problems is

एक घंटे की परीक्षा में 60 प्रश्न हैं। यदि एक छात्र 25 मिनट में 30 प्रश्नों को हल करता है, तो उसके द्वारा शेष सभी प्रश्नों को हल करने के लिए आवश्यक औसत सेकंडों की संख्या है

Option “A” is correct.

Total number of question = 60

If a student completes 30 questions in 25 minutes,

Remaining minutes = 60 – 25 = 35

Remaining questions = 60 – 30 = 30

30 question complete in = 35 min

1 question complete in = 35/30 × 60 = 70 sec

35. The average of the first five consecutive odd numbers is m. If the next three odd numbers are also included, then what is the increase in the average?

पहले पाँच लगातार विषम संख्याओं का औसत m है। यदि अगली तीन विषम संख्याओं को भी सम्मिलित कर लिया जाता है, तब औसत में कितनी वृद्धि होगी?

Option “B” is correct.

The five consecutive odd number are 1, 3, 5, 7, 9

Average of five consecutive odd number = (1 + 3 + 5 + 7 + 9)/5 = 25/5 = 5

So m = 5

If next three odd number are included, then the average of 8 consecutive odd number

(25 + 11 + 13 + 15)/8

⇒ 64/8

⇒ 8

We can write 8 = 5 + 3 or m + 3

36. The average marks obtained by a student in 9 subjects is 98. On subsequent verification, it was found that the marks obtained by him in a subject were wrongly copied as 86 instead of 68. The correct average of the marks obtained by him is:

9 विषयों में एक छात्र द्वारा प्राप्त औसत अंक 98 है। बाद में जाँच करने पर यह पाया गया कि किसी विषय में उसके द्वारा प्राप्त अंकों को गलत तरीके से 68 के स्थान पर 86 लिख दिया गया था। उसके द्वारा प्राप्त अंकों का सही औसत क्या है?


Option “C” is correct.

The average mark obtained by a student in 9 subjects is 98.

It was found that the marks obtained by him in a subject were wrongly copied as 86 instead of 68.

Formula used:

Average = The sum of all the values / Total number of values

Calculation:

Average marks of 9 subjects = 98

Sum of marks of 9 subjects = 98 × 9

⇒ 882

Correct sum of marks of 9 subjects = 882 – 86 + 68

⇒ 864

The correct sum of marks of 9 subjects = 864/9

⇒ 96

∴ The correct average of the marks obtained by him is 96.

37.The average age of A, B and C is 20 years, and that of B and C is 25 years. What is the age of A?

A, B और C की औसत आयु 20 वर्ष है, तथा B और C की औसत आयु 25 वर्ष है। A की आयु क्या है?

Option “D” is correct.

Average age of A, B and C = 20

Sum of age of A, B and C = 20 × 3 = 60

Average age of B and C = 25

Sum of age of B and C = 25 × 2 = 50

Age of A = 60 – 50 = 10 years

38. The average of 4 terms is 30 and the 1st term is 1/3 of the sum of the remaining terms. What is the first term?

4 पदों का औसत 30 है और पहला पद शेष पदों के योग का 1/3 है। पहला पद क्या है?

Option “B” is correct.

Let the sum of last three terms be 3x.

According to the Question, the first term will be:

= 3x × 1/3 = x

Average of 4 terms is given as 30, i.e.

(Sum of 4 terms)/4 = 30

Sum of 4 terms = 120

From the above concluded data, we can write:

x + 3x = 120

4x = 120

x = 120/4

x = 30

First term is 30.

39.The average of the first 20 multiples of 12 is:

12 के पहले 20 गुणज का औसत:

A. 124

B. 120

C. 126

D. 130

Option “C” is correct.

First multiple of 12 = 12 × 1 = 12

20th multiple of 12 = 12 × 20 = 240

∴ required average = (12 + 240)/2 = 252/2 = 126

IMPORTANT POINT

The average of the first 20 multiples of 12 = (First multiple of 12 + 20th multiple of 12)/2

40.The sum of the ages of 4 children born at the intervals of 4 years is 48. Find the age of the youngest child.

4 वर्ष के अंतराल में जन्मे 4 बच्चों की आयु का योगफल 48 है। सबसे छोटे बच्चे की आयु ज्ञात कीजिए।

A. 4 years

B. 5 years

C. 6 years

D. 7 years

Option “D” is correct.

Let age of younger child be x

According to the question

x + x + 4 + x + 8 + x + 12 = 48

⇒ 4x + 24 = 48

⇒ 4x = 48 – 24

⇒ x = 6

∴ Age of youngest child is 6 years.

Pages: 1 2 3 4 5 6 7 8

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