Wiley Edge Averages Quiz-4

Explanation:
Price of StringEnergy Drink in first year = Rs. 7.50 per litre.
∴ Quantity of StringEnergy Drink in first year,
= \frac{4080}{7.50}= 544Litres

Price of StringEnergy Drink in second year = Rs. 8.00 per litre.
∴ Quantity of StringEnergy Drink in second year,
= \frac{4080}{8.00}= 510Litres

Price of StringEnergy Drink in third year = Rs. 8.50 per litre.
∴ Quantity of StringEnergy Drink in third year,
= \frac{4080}{8.50}= 480Litres

Required Average = \frac{3\times 4080}{544+510+480}
= = \frac{12240}{1534}=7.98
=Rs 7.98

 

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Explanation:
Total age of 13 applicant = 36 × 13 = 468

Given that if the age of one more applicant is added the average decreases by half a year
∴ Total age of 14 applicants = 35.5 × 14 = 497

Now, Age of the new applicant = 497 – 468 = 29 years

Solution:

Sum of age of a,b,c and d = 20

We can write

 

A+B+8+ D = 20

A+ B + D = 12

 

Sum of age of A,B,D,E = 24

We can write

A+B+D+E = 24

12 + E = 24

E = 12

Solution:
Let his average x runs

total runs in 10 innings = 10x

Total runs present = 11x
We can write

=> 10x + 100/11 = x+5
=> 10x + 100 = 11x + 55
Therefore x = 45

Solution:
Sum of all runs in 10 innings= 500

Sum of runs in 11 innings= 11*52 =572

Score he need to score in 11th innings to maintain his average 2 more = 572-500
= 72 runs

Solution:
Sum of all 50 numbers= 50*38 = 1900

New average= (1900 - 100)/48
= 37.5

Solution:

Sum of 20 items= 20* 55 = 1100
Mean = 55

New mean = 1100 - 75 /18
=1025/18
= 56.94

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Explanation:

Let's say the first three numbers are a, b and c and the fourth number is d.

 

According to the question,

(a + b + c)/3 = 3d.

d = (a + b + c)/9.

 

The average of all four numbers is (a + b + c + d)/4 = 5.

 

Substituting the value of d from the previous equation, we get (a + b + c + (a + b + c)/9)/4 = 5. Solving for a + b + c, we get a + b + c = 45/2.

 

Substituting this value into the equation for d, we get d = (45/2)/9 =  5/2 . So the fourth number is  5/2 which is not one of the given options.