Test Level 2: Averages - 1 - CAT MCQ

Let the number of innings be x.
Total runs scored by the cricketer = 15x
New number of innings = x + 1
Also, new average = 13.5

15x = 13.5(x + 1)
15x = 13.5x + 13.5
15x - 13.5x = 13.5
1.5x = 13.5

Number of innings played, including the last one = 9 + 1 = 10

To get the greatest possible value for one number, the other two should be as small as possible sum of the three numbers = 33.

Since the numbers are positive distinct integers, the two numbers can be 1 and 2.

So, the third will be 30, which is the maximum possible value.

Let a, b and c be the three numbers.
a + b + c = 3x
b = 9 + a and c = x + 5

a + a + 9 + x + 5 = 3x
2a + 14 = 2x
∴ x - a = 7

Let the number of boys in the class be 'b' and the girls be 'g'.

Now, as the average score is 82, we have (86b + 80g)/(b + g) = 82
86b + 80g = 82b + 82g
4b = 2g
b/g = 1/2

Thus the required ratio is 1 : 2.
Hence, answer option (4) is correct.

Total employees = 600 + 400 = 1000
Total income at an average Rs. 2.55/day = 2.55 × 1000 = Rs. 2550 per day
Let the daily wage of a man be Rs. x.
Then, their total wages = Rs. 600x
A woman got 50 paise less than a man.
∴ Their total wages = 400(x - 0.5)
⇒ 600x + 400(x - 0.5) = 2550
⇒ 600x + 400x - 200 = 2550
⇒ 1000x = 2750
⇒ x = Rs. 2.75
∴ Man`s daily wage = Rs. 2.75
Woman`s daily wage = 2.75 - 0.50 = Rs. 2.25

Numbers that are divisible by 3, 6 and 9 = Multiple of L.C.M(3, 6, 9) = 18, 36, 54,......
The required numbers are 18, 36, 54, … and 90.

Let x, x + 1, x + 2, ...., (x + n) be n consecutive integers, then their sum is

as it is starting from 0 + 1 + 2 + 3 +  ..  . . .+  n so there are n-1 terms for addition if you look carefully

Assume after deleting any term, the new mean is , then their sum is 

Subtract equation (1) from equation (2) we get

So, the answer is a - 1/2

Average of 21 results = 54
Total of 21 results = 54 x 21 = 1134
Average of the first 11 results = 51
Total of the first 11 results = 51 x 11 = 561
Average of the last 11 results = 57
Total of the last 11 results = 57 x 11 = 627
Therefore, 11th result = 561 + 627 - 1134 = 1188 - 1134 = 54

Sum of 10 numbers = 200
Assume that x is the maximum.

If x is deleted, average

According to the question,
Total number of chocolate = 46n

2n² + 2n + n + 1 = 276

2n² + 3n - 275 = 0