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Practice Test: Averages - 1 - CAT MCQ


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15 Questions MCQ Test - Practice Test: Averages - 1

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Practice Test: Averages - 1 - Question 1

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 Practice Test: Averages - 1 - Question 1

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.

Practice Test: Averages - 1 - Question 2

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 Practice Test: Averages - 1 - Question 2

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...
 

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Practice Test: Averages - 1 - Question 3

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 Practice Test: Averages - 1 - Question 3

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

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

Practice Test: Averages - 1 - Question 4

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 Practice Test: Averages - 1 - Question 4

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.

Practice Test: Averages - 1 - Question 5

Dev and Om are among 22 students who write an examination. Dev scores 82.5. The average score of the 21 students other than Om is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Dev. The score of Om is. 

Detailed Solution for Practice Test: Averages - 1 - Question 5

Let the score of Om =x
Total marks of all the students = 21*62 +x
As per the question, The average score of all the 22 students is one more than the average score of the 21 students other than Dev,
Or (21*62 +x)/22 -1 = (21*62 +x – 82.5)/21
(21*62 + x -22)/22 = (21*62 +x – 82.5)/21. Therefore, x =51

Practice Test: Averages - 1 - Question 6

There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is

Detailed Solution for Practice Test: Averages - 1 - Question 6

Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.
1st condition

2nd condition

Adding both the equations, we get:

Given that 12x = 12 ⇒ x = 1.

Practice Test: Averages - 1 - Question 7

The average of the first ten composite numbers is 

Detailed Solution for Practice Test: Averages - 1 - Question 7

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

Practice Test: Averages - 1 - Question 8

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 Practice Test: Averages - 1 - Question 8

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

Practice Test: Averages - 1 - Question 9

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 Practice Test: Averages - 1 - Question 9

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

Practice Test: Averages - 1 - Question 10

The average age of a family of 5 members is 20 years. If the age of the youngest member is 10 years, what was the average age of the family at the birth of the youngest member?

Detailed Solution for Practice Test: Averages - 1 - Question 10

At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Therefore, average age of the family at the time of birth of the youngest member,
= 50/4 =12.5

Practice Test: Averages - 1 - Question 11

The average weight of 10 men is decreased by 2 kg when one of them weighing 140 kg is replaced by another person. Find the weight of the new person.

Detailed Solution for Practice Test: Averages - 1 - Question 11

Shortcut:
The decrease in weight would be 20kgs (10people’s average weight drops by 2 kgs). Hence, the new person’s weight = 140 - 20 = 120.

Detailed Solution:

Let weight of 9 men =x.
Weight of new men =y

According to the question:

((x+140)/10) ​− 2 = (x+y​)/10
y = 120

Practice Test: Averages - 1 - Question 12

Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be

Detailed Solution for Practice Test: Averages - 1 - Question 12

Savings target in a year = 550*12 = Rs 6600

Saving in first 9 months = 9(4000-3500) = Rs 4500

Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100

Savings for each month in last 3 months = 2100/3 = Rs 700

It is given, monthly expenses in last 3 months = Rs 3700

This implies, his monthly earnings from 10th month should be 3700 + 700, i.e. Rs 4400

The answer is option A.

Practice Test: Averages - 1 - Question 13

The average weight of a class is 54 kg. A student, whose weight is 145 kg, joined the class and the average weight of the class now becomes a prime number less than 72. Find the total number of students in the class now.

Detailed Solution for Practice Test: Averages - 1 - Question 13

Let the original number of students in the class be N.
Total weight of the class = 54N
New total weight of the class = 54N + 145
New average weight of the class = (54N + 145)/(N+1) = (54N + 54)/(N+1) + 91/(N+1) = 54 + 91/(N+1).
Since the new average is an integer, (N+1) should be a factor of 91.
If N+1 = 7, the new average becomes 54 + 91/7 = 54 + 13 = 67
and if N+1 = 13, then the new average becomes 54 + 91/13 = 54 + 7 = 61
Both 67 and 61 are prime numbers less than 72. So, we cannot uniquely determine the number of students in the class.

Practice Test: Averages - 1 - Question 14

Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum possible average weight of the class now?

Detailed Solution for Practice Test: Averages - 1 - Question 14

To determine the maximum possible average weight of the class after new students join, let's break down the problem step by step.

  1. Initial Class Details:

    • Number of Students: 40
    • Average Weight: 40 kg
    • Total Weight: 40 students * 40 kg/student = 1600 kg
  2. New Students Joining:

    • Number of New Students: m
    • Average Weight of New Students: n kg
    • Given Constraint: m + n = 50
  3. Total Class After New Students Join:

    • Total Number of Students: 40 + m
    • Total Weight: 1600 kg + (m * n) kg
  4. Combined Average Weight Formula:

  • Given the constraint m + n = 50, we can express n as (50 - m).

  • Substituting n in the Formula:

  • Finding the Maximum Average: We aim to find the value of m (and consequently n) that maximizes the combined average. We'll evaluate this for integer values of m from 1 to 49 (since n has to be at least 1).

    Calculating a few values:

    • When m = 5:

      • n = 50 - 5 = 45
      • Total Weight = 1600 + (5 * 45) = 1600 + 225 = 1825 kg
      • Total Students = 40 + 5 = 45
      • Combined Average = 1825 kg / 45 students ≈ 40.5556 kg
    • When m = 4:

      • Combined Average ≈ 40.5455 kg
    • When m = 6:

      • Combined Average ≈ 40.5217 kg

    Observing these calculations, the maximum average occurs when m = 5.

  • Conclusion:

    • Number of New Students: 5
    • Average Weight of New Students: 45 kg
    • Combined Average Weight: Approximately 40.5556 kg
Practice Test: Averages - 1 - Question 15

The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

Detailed Solution for Practice Test: Averages - 1 - Question 15

C)

It is given that the average of the 30 integers = 5
Sum of the 30 integers = 30*5=150
There are exactly 20 integers whose value is less than 5.
To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers
So the sum of 10 integers = 10*6=60
The sum of the 20 integers = 150-60= 90
Average of 20 integers = 90/20  = 4.5

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