Practice Test: Averages - 1 - CAT MCQ

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 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

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.

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

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

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

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.

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.

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

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