Test: Averages (January 22) - CAT MCQ

Normal process:
Runs in 26th inning = Runs total after 26 innings – Runs total after 25 innings
= 26 X 58 – 25 X 56

For Easy calculation use:

= (56 + 2) X 26 – 56 X 25 )
= 2 X 26 + (56 X 26 – 56 X 25)
= 52 + 56 = 108

Since the average increases by 2 runs per innings, it is equivalent to 2 runs being added to each score in the first 25 innings. Now, since these runs can only be added by the runs scored in the 26th inning, the score in the 26th inning must be 25 X 2 = 50 runs higher than the average after 26 innings (i.e. new average = 58).

Hence, runs scored in 26th inning:
= New Average + Old innings X Change in average

= 58 + 25 X 2
= 108 

In order to find the average age of the family before Ram was born, we need to know the age of the youngest member of the family. 
Since, we do not know the age of the youngest member, we can not calculate the total age of the family before Ram was born.
Hence, we can not calculate the answer with the given conditions.

Thus, D is the right choice.

Change in total weight of 10 students = difference in weight of the student who joined and the student

=> weigth of first student who left = 30 + (10×2) = 50

weight of the student who joined last = 50 – 10 = 40...
Thus change in average weight = (40 – 30)/10 = 1...
 

When we multiply each number by 9, the average would also get multiplied by 9.

Hence, the new average = 18 X 9 = 162.

Runs scored by Kohli in first 4 innings = 48*4 = 192
Average of 5 innings is 60, so total runs scored after 5 innings = 60*5 = 300
Hence runs scored by Kohli in fifth inning = 300 – 192 = 108
It is given that in 6th innings he scores 12 runs more than this, so he must score 120 in the sixth inning. Hence total runs scored in 6 innings = 300+120 = 420
Now average of last five innings is 78, so runs scored in last innings = 390
Hence runs scored in first inning = 420 – 390 = 30.

Let a – b = x, b – c = 2x and c – d = 3x
Thus,
c = 3x + d
b = 2x + c = 5x + d
a = x + b = 6x + d
Hence,
(a + d )/ c =  (6x + d + d) / (3x + d) =  2/1

Let initial average be x.
Then the initial total is 20x and the New average will be (x – 4),

The new total will be:
20(x – 4) = 20x – 80.

The reduction of 80 is created by the replacement. Hence, the new student has 80 marks less than the student he replaces. Hence, he must have scored 10 marks.

Short Cut:
The replacement has the effect of reducing the average marks for each of the 20 students by 4. Hence, the replacement must be 20 X 4 = 80 marks below the original.

Hence, answer = 10 marks. 

The first ten composite numbers are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. 

Required average:

= (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18) / 10
= 112 / 10
= 11.2

David's Weight = (4 x 70) – (3 x 74) = 280 – 222 = 58
Eric’s weight = 58 + 3 = 61

Now, we know that:
Ross + Joey + Chandler + David = 4 x 70 = 280
Joey + Chandler + David + Eric = 75 x 4 = 300.

Hence, Ross’s weight is 20 kg less than Eric’s weight. Ross = 61 - 20 = 41 kg. 

Mon + Tue + Wed = 35*3 = 105  ---------(1)
Tue + Wed + Thu = 30*3 = 90  -------------(2)
Thu = (1/2) Mon  ------------(3)

Eqn (1)-(2):
Mon-Thu = 15 ------------(4)

⇒ Mon - (1/2) Mon = 15
⇒ (1/2) Mon = 15
⇒ Mon =30
⇒ Thu = 30/2=15