Question Detail

Average of 10 numbers is zero. At most how many numbers may be greater than zero

Answer: Option D

1. Marks of a student were wrongly entered in computer as 83, actual marks of that student were 63. Due to this mistake average marks of whole class got increased by half (1/2). Find the total number of students in that class.

Answer: Option D

Explanation:

Suppose total number of students are = X
$\begin{aligned} Total increase = x*\frac{1}{2} = \frac{x}{2} \\ => \frac{x}{2} = 83-63 = 20 \\ => x = 40 \end{aligned}$

2. Average of 10 numbers is zero. At most how many numbers may be greater than zero

Answer: Option D

3. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is

54.48
54.68
54.60
54.58

Answer: Option B

Explanation:

$\begin{aligned} \frac{(55 \times 50) +(60 \times 55) +(45 \times 60) }{55 + 60 + 45} \end{aligned}$

$\begin{aligned} \frac{8750}{160} = 54.68 \end{aligned}$

4. Average of all prime numbers between 30 to 50

37
37.8
39
39.8

Answer: Option D

Explanation:

Prime numbers between 30 and 50 are:
31, 37, 41, 43, 47

Average of prime numbers between 30 to 50 will be
$\begin{aligned} (\frac{31+37+41+43+47}{5}) = \frac{199}{5} = 39.8 \end{aligned}$

5. Find the average of all numbers between 6 and 34 which are divisible by 5

Answer: Option B

Explanation:

$\begin{aligned} Average = (\frac{10+15+20+25+30}{5}) = \frac{100}{5} =20 \end{aligned}$

PK2D 9 years ago

Here average of 10 numbers= 0.
So we have avg = sum/10.
That implies sum = 0.
Now for the sum of 10 numbers to be equal to zero, there may be all 9 numbers of them >0 and only one number which is -ve of all the 9 numbers sum.
So the answer is 9.