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Test: Averages (January 22) - CAT MCQ


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Test: Averages (January 22) - Question 1

The average of a batsman after 25 innings was 56 runs per innings. If after the 26th inning his average increased by 2 runs, then what was his score in the 26th inning? 

Detailed Solution for Test: Averages (January 22) - Question 1

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 

Test: Averages (January 22) - Question 2

There are 7 members in a family whose average age is 25 years. Ram who is 12 years old is the second youngest in the family. Find the average age of the family in years just before Ram was born?

Detailed Solution for Test: Averages (January 22) - Question 2

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.

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Test: Averages (January 22) - Question 3

The average weight of a class of 10 students is increased by 2 kg when one student of 30kg left and another student joined. After a few months, this new student left and another student joined whose weight was 10 less than the student who left now. What is the difference between the final and initial averages?

Detailed Solution for Test: Averages (January 22) - Question 3

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...
 

Test: Averages (January 22) - Question 4

The average of 15 numbers is 18. If each number is multiplied by 9, then the average of the new set of numbers is:

Detailed Solution for Test: Averages (January 22) - Question 4

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

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

Test: Averages (January 22) - Question 5

The average number of runs scored by Virat Kohli in four matches is 48. In the fifth match, Kohli scores some runs such that his average now becomes 60. In the 6th innings he scores 12 runs more than his fifth innings and now the average of his last five innings becomes 78. How many runs did he score in his first innings? (He does not remain not out in any of the innings)

Detailed Solution for Test: Averages (January 22) - Question 5

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.

Test: Averages (January 22) - Question 6

If a – b : b – c : c – d = 1 : 2 : 3, then what is the ratio of (a + d) : c?

Detailed Solution for Test: Averages (January 22) - Question 6

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

Test: Averages (January 22) - Question 7

The average marks of a group of 20 students on a test is reduced by 4 when the topper who scored 90 marks is replaced by a new student. How many marks did the new student have? 

Detailed Solution for Test: Averages (January 22) - Question 7

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

Test: Averages (January 22) - Question 8

The average of the first ten composite numbers is 

Detailed Solution for Test: Averages (January 22) - Question 8

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

Test: Averages (January 22) - Question 9

The average weight of 3 boys Ross, Joey and Chandler is 74 kg. Another boy David joins the group and the average now becomes 70 kg. If another boy Eric, whose weight is 3 kg more than that of David, replaces Ross then the average weight of Joey, Chandler, David and Eric becomes 75 kg. The weight of Ross is:

Detailed Solution for Test: Averages (January 22) - Question 9

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

Test: Averages (January 22) - Question 10

The mean temperature of Monday to Wednesday was 35 °C and of Tuesday to Thursday was 30 °C. If the temperature on Thursday was 1/2 that of Monday, the temperature on Thursday was ______ .

Detailed Solution for Test: Averages (January 22) - Question 10

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

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