Q.1. The average marks of 13 papers is 40. The average marks of the first 7 papers are 42 and that of the last seven papers is 35. Find the marks obtained in the 7th paper.
A. 23
B. 38
C. 19
D. 57
Sol : Option C
The average of 13 papers is 40, so the sum = 13 × 40 = 520
The average of first 7 papers is 42, so the sum will be = 7 × 42 = 294
The average of last 7 papers is 35, so the sum will be = 7×35 = 245
So, the marks obtained in the 7th paper will be = 539 – 520 = 19
Q.2. Mukul has earned as an average of 4,200 dollars for the first eleven months of the year. If he justifies his staying in the US on the basis of his ability to earn at least 5000 dollars per month for the entire year, then how much should he earn (in dollars) in the last month to achieve his required average for the whole year?
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Sol : Option B
The salary expectation for the whole year is 5000. So, the total earning will be 60000.
Earning for 11 months is 4200 so the total earnings are 46200.
Now for the last month he has to earn 60000 – 46200 = 13800.
Q.3. The average of x, y and z is 45. x is as much more than the average as y is less than the average. Find the value of z.
A. 45 years
B. 35 years
C. 60 years
D. 15 years
Sol : Option A
Sum of x, y and z will be 135. As the average is 45 and Also given that ,
X-45= 45-Y, X+Y= 90, therefore, 90+Z=135;Z=45
Q.4. The average for some number of terms (say ‘n’ ) is zero. How many maximum numbers of terms can be negative?
A. 0
B. n
C. n+1
D. n-1
Sol : Option D
Here, as sum of n terms will 0, so we can have at the most n-1 terms negative and the last term whose magnitude will be equal to the sum of all the negative terms.
Q.5. The average salary of 30 officers in a firm is Rs.120 and the average salary of laborers is Rs. 40. Find the total number of laborers if the average salary of the firm is Rs. 50.
A. Rs. 180
B. Rs. 420
C. Rs. 240
D. Rs. 210
Sol : Option D
The sum of salary of officers will be = 30×120 = 3600
Let the number of labourers = X.
ATQ, 3600 + 40X = 50(30+X)
2100 = 10X
X= 210
Q.6. In a class, the average marks of 40 students was calculated to be 52.15. It was later discovered that the marks of a student were taken to be 49, instead of 85. Find the real average of the class.
A. 53.05
B. 53.15
C. 52.85
D. 52.95
Sol : Option A
Average marks for 40 students is equal to 52.15 , the marks were taken as 49 instead of 85 so there will be an increase of 36 which is now to be distributed equally amongst 40 students , so 36/40 = 0.9 which is to be distributed amongst all.
So, new average stands out to be 52.15 + 0.9 = 53.05
Q.7. For 9 innings, Aman has a certain run rate. In the tenth inning, he scores 100 runs, thereby increasing his run rate by 8 runs. What is his new run rate?
A. Rs. 22
B. Rs. 26
C. Rs. 28
D. Rs. 32
Sol : Option C
Let initial run rate be x so, the final run rate would have be X+8
ATQ, 9×X + 100 = 10× (X+8)
Solving, we get X = 20
So, the new average will be 28.
Q8. The average of 5 consecutive numbers is n. What will be the average if the next two numbers are included?
A. n+2
B. n-1
C. n-2
D. n+1
Sol : Option D
The average of 5 consecutive terms is n, implies that the 3rd term is n. Now as the next 2 terms are included implies that the new average for 7 terms would be the 4th term. So, the 4th term would be n+1.
Q9. In 40 overs game, in first 20 overs of a game of cricket, the run rate was only 5. What should be the run rate for the remaining overs so that the total score reaches 300?
A. 15
B. 10
C. 28
D. 20
Sol : Option B
Runs scored for 1st 20 overs = 100
Total required = 300
So, in the next 20 overs the team has to score 200 runs.
So, run rate required = 200/20 = 10
Q10. The average weight of 39 men travelling to Ladakh is 30. If an obese man with weight 130 kg join them. What will be the average weight of the people travelling to Ladakh?
A. 52
B. 30
C. 32.5
D. 130
Sol : Option C
If the weight of the man would have been 30, then the average weight would have been the same. So, the extra 100 kg that the obese man brings with him would be distributed equally amongst all of them, i.e. 100/40 = 2.5
So, the average becomes 30 + 2.5 = 32.5