Fastest way to solve Average Questions with tricks for RRB PO 2022 | Average Quiz for RBI Assistant 2022 SBI Clerk 2022

1

In an exam of 100 marks, the average marks of a class of 40 students are 76. If the top 3 scorers of the class leave, the average score falls “down by 1. If the other two toppers except “the highest topper scored not more than 85. “then what is the minimum score the topper can score?

» Explain it

Correct Option: C

Total score of 40 students = (40 × 76) = 3040

Total score of top 3 scorers = 3040 – (37 × 75) = 265

To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.

So, the top scorer scored = 265 – 170 = 95 marks.

Hence, option C is correct.

1

100 अंकों की एक परीक्षा में, 40 छात्रों की एक कक्षा के औसत अंक 76 हैं। यदि कक्षा के शीर्ष 3 स्कोरर चले जाते हैं, तो औसत स्कोर "1 से नीचे गिर जाता है। यदि" उच्चतम टॉपर को छोड़कर अन्य दो टॉपर्स ने 85 से अधिक अंक प्राप्त नहीं किए। "तो टॉप स्कोर करने वाला न्यूनतम स्कोर कितना कर सकता है?

» Explain it

सही विकल्प : C

Total score of 40 students = (40 × 76) = 3040

Total score of top 3 scorers = 3040 – (37 × 75) = 265

To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.

So, the top scorer scored = 265 – 170 = 95 marks.

Hence, option C is correct.

2

Average weight of three friends Amar, Visera and Daman is 70 kg. Another person Vishal joins the group and now the average is 66 kg. If another person Tahir whose weight is 6 kg more than Vishal, joins the group replacing Amar, then average weight of Visera, Daman, Vishal and Tahir becomes 75 kg. What is the weight of Amar (in kg)?

» Explain it

Correct Option: D

Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ......... (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ................... (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg

Hence, option D is correct.

2

तीन दोस्तों अमर , विसरा और दमन का औसत वजन 70 किलो है। एक अन्य व्यक्ति विशाल समूह में शामिल हो जाता है और अब औसत 66 किग्रा है। यदि एक अन्य व्यक्ति ताहिर जिसका वजन विशाल से 6 किग्रा अधिक है, अमर के स्थान पर समूह में शामिल हो जाता है, तो विसरा, दमन, विशाल और ताहिर का औसत वजन 75 किग्रा हो जाता है। तो अमर का वजन (किलो में) क्या है?

» Explain it

सही विकल्प : D

Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ......... (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ................... (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg

Hence, option D is correct.

3

The average salary of a company increases by 100 when the salary of the manager, which is Rs. 9500, is included. If the number of employees excluding the manager is the smallest cube divisible by 16, what is the final average of the company?

» Explain it

Correct Option: D

The smallest cube divisible by 16 is 64.

Lets assume the average salary before the manager's salary is included is x

After addition of Manager's salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100

Hence, option D is correct.

3

एक कंपनी के औसत वेतन में 100 की वृद्धि होती है जब प्रबंधक का वेतन, जो कि 9500 रुपये है, को शामिल किया जाता है। यदि प्रबंधक को छोड़कर कर्मचारियों की संख्या 16 से विभाज्य सबसे छोटा घन है, तो कंपनी का अंतिम औसत क्या है?

» Explain it

सही विकल्प : D

The smallest cube divisible by 16 is 64.

Lets assume the average salary before the manager's salary is included is x

After addition of Manager's salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100

Hence, option D is correct.

4

An exam was conducted in a state over 222 centres. The average number of applicants per centre was found to be 1560. However, it was later realized that in one centre, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants per centre (upto two decimals)?

» Explain it

Correct Option: D

Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = 110/222 = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51

Hence, option D is correct.

4

एक राज्य में 222 केंद्रों पर एक परीक्षा आयोजित की गई थी। प्रति केंद्र आवेदकों की औसत संख्या 1560 पाई गई। हालांकि, बाद में पता चला कि एक केंद्र में आवेदकों की संख्या 1747 के बजाय 1857 के रूप में गिना गया था। प्रति केंद्र आवेदकों की सही औसत संख्या क्या थी (दो दशमलव तक) )?

» Explain it

सही विकल्प : D

Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = 110/222 = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51

Hence, option D is correct.

5

In Champions league, Rohit scored an average of 120 runs per match in the first 3 match and an average of 140 runs per match in the last four match. What is Rohit’s average runs for the first match and the last two match if his average runs per match for all the five match is 122 and total number of matches are 5?

» Explain it

Correct Option: A

Rohit’s average score in the first 3 exams = 120

Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5

M1 + M2 + M3 = 120 × 3 = 360 .......(i)

Average of last 4 match = 140

⇒	M2 + M3 + M4 + M5	= 140
	4

⇒ M2 + M3 + M4 + M5 = 560 ......(ii)

Average of all the exams

⇒	M1 + M2 + M3 + M4 + M5	= 122
	5

∴ M1 + M2 + M3 + M4 + M5 = 122 × 5 = 610 .......(iii)

From solving above equation, we get M1 + M4 + M5 = 300

Required average runs =	300	= 100
	3

Hence correct option (A) is correct.

5

चैंपियंस लीग में, रोहित ने पहले 3 मैचों में प्रति मैच औसतन 120 रन और पिछले चार मैचों में प्रति मैच 140 रन का औसत बनाया। पहले मैच और आखिरी दो मैचों के लिए रोहित के औसत रन क्या हैं यदि सभी पांच मैचों के लिए प्रति मैच उसका औसत रन 122 है और मैचों की कुल संख्या 5 है?

» Explain it

सही विकल्प : A

Rohit’s average score in the first 3 exams = 120

Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5

M1 + M2 + M3 = 120 × 3 = 360 .......(i)

Average of last 4 match = 140

⇒	M2 + M3 + M4 + M5	= 140
	4

⇒ M2 + M3 + M4 + M5 = 560 ......(ii)

Average of all the exams

⇒	M1 + M2 + M3 + M4 + M5	= 122
	5

∴ M1 + M2 + M3 + M4 + M5 = 122 × 5 = 610 .......(iii)

From solving above equation, we get M1 + M4 + M5 = 300

Required average runs =	300	= 100
	3

Hence correct option (A) is correct.

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