Test Level 1: Averages - 1 - CAT MCQ

The consecutive even numbers are a, b, c, d, e.
Clearly, b = a + 2, c = a + 4, d = a + 6, e = a + 8

Note: Whenever difference between any two consecutive terms is constant (means the series is an arithmetic series), the average of all the numbers is

(i) the middle number if the number of terms is odd.

(ii) the average of the middle two numbers if the number of terms is even.

Let present age of father = F years
Let present age of son = S years
Let present age of mother = M years
Now, F + S = 29 × 2 = 58 years
Also, F + M + S + 15 = 3 × 37 = 111 years
On solving, we get
M = 38 years
Thus, present age of mother = 38 years

According to the question,
480 - 86 - x = 79  4 = 316
Or, lowest score = x = 480 - 86 - 316 = 78

The mean of 19 observations is 4.

New data when 24 is added:
76 + 24 = 100
New number of observations = 19 + 1 = 20

Let the number of officers be x.
Number of workers = 400 - x
10,000x + 2000(400 - x) = 3000(400)
10x + 800 - 2x = 1200
8x = 400
x = 50
Hence, the number of officers is 50.

Let the sum of all 10 numbers be S.
S = 10 × 27 = 270
Let the smallest number be s and the largest number be L.
So, S - s = 9x ... (i)
Also, S - L = 9y ... (ii)
Adding (i) and (ii), we get
2S - (s + L) = 9(x + y) = 9(x + y) = 9(10) = 90
Or 2S - 90 = (s + L)

Or 270 - 45 = 225

= 

Thus, answer option (1) is correct.

Average of 2 numbers is greater than 50 means that the sum should be greater than 100.
Different pairs are as follows:
(1, 100) … (1 pair)
(2, 99), (2, 100) … (2 pairs)
(3, 98), (3, 99), (3, 100) … (3 pairs)
... so on
(49, 52), (49, 53) ... (49, 100) ... (49 pairs)
(50, 51), (50, 52) … (50, 100) ... (50 pairs)
So, 1 + 2 + 3 + 4 + ... + 50 = 1275
Again
(51, 51), (51, 52), (51, 53), ... (51, 100) ... (50 pairs)
(52, 52), (52, 53), (52, 54), ... (52, 100) ... (49 pairs)
(53, 53), (53, 54), (53, 55), ... (53, 100) ... (48 pairs)
... so on
(99, 99), (99, 100) ... (2 pairs)
(100, 100) ... (1 pair)
So, 1 + 2 + 3 + 4 + ... + 50 = 1275
Hence, total = 1275 + 1275 = 2550 pairs

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 - 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

⇒ y = 500 and x = 5000

X + 100y = 5000 + 50000 = 55000

Average expense = 55000/100 = Rs. 550

Required average = (2 + 3 + 5 + 7 + 11 + 13 + 17
+ 19 + 23 + 29)/10 = 129/10 = 12.9.