Arun Sharma Test: Averages- 3 - CAT MCQ

# Arun Sharma Test: Averages- 3 - CAT MCQ

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## 15 Questions MCQ Test Quantitative Aptitude (Quant) - Arun Sharma Test: Averages- 3

Arun Sharma Test: Averages- 3 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Arun Sharma Test: Averages- 3 questions and answers have been prepared according to the CAT exam syllabus.The Arun Sharma Test: Averages- 3 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Arun Sharma Test: Averages- 3 below.
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Arun Sharma Test: Averages- 3 - Question 1

### The average marks of the girls in a class is equal to the number of boys and the average marks of boys is equal to the number of girls. If the class average is 4 less than the average of both the boys’ and the girls’ average marks, what will be the number of students in the class?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 1

Arun Sharma Test: Averages- 3 - Question 2

### A, B, C, D and E are five consecutive integers and the average of these five numbers is less than 1/4th of A. Then A is

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 2

As a is less than a negative number, so 'a' is also a negative number. Hence option C is correct.

Arun Sharma Test: Averages- 3 - Question 3

### Ram was born 30 years after his father was born and Ram's sister was born 25 years after Ram’s mother was born. The average age of the Ram family is 26.25 years right now. Ram's sister will get married 4 years from now and will leave the family. Then the sum of age of the family will be 107 years. What is the age of Ram's father?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 3

Let the present age of father be x and mother be y

∴ Ram age = (x – 30)

And Ram’s sister = (y – 25)

We know that

Average = Sum of all/total number of terms

⇒ 26.25 = Sum of all/4

Sum of their ages will be = 105

∴ x + y + (x - 30) + (y – 25) = 105

⇒ x + y = 80     ----(i)

After four years their total ages will be (excluding daughter age)

∴ (x + 4) + (y + 4) + (x – 30) + 4 = 107

⇒ 2x + y = 125     ----(ii)

By solving equation (i) and (ii) we get

⇒ x = 45 and y = 35

∴ The age of Ram’s father will be 45 years

Arun Sharma Test: Averages- 3 - Question 4

India is supposed to send its boxing team for Olympics in each of the following 10 weight group divisions.

A (48 kg -52 kg)        B (52 kg - 56 kg)

C (56 k g - 6 0 kg)     D (60 k g - 6 4 kg)

E (64 kg -68 kg)        F (68 kg - 72 kg)

G (72 k g - 76 kg)    H (76 k g - 80 kg)

I (80 kg - 84 kg)       J (84 kg - 88 kg)

After selecting one player from each group, their average weight comes out to be 68 kg. if one of the players named X leaves

If X leaves the team, and two new players join the group, then their average weight increases to 68 kg. These players can be from group....

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 4

Total weight of 11 players ( 68 x 11) kg = 748 kg

Increase in weight = (748 - 598.5) kg = 149.5

Arun Sharma Test: Averages- 3 - Question 5

In continuation with previous question, the average of all the groups together, which group contributes most in overall average?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 5

Exact weight of players are not known hence (d)

Arun Sharma Test: Averages- 3 - Question 6

The average age of Mr and Mrs Sinha at the time of their marriage in 1972 was 23 years. On the occasion of their anniversary in 1976, they observed that the average age of their family had come down by 4 years compared to their average age at the time of their marriage. This was due to the fact that their son Vicky was born during that period. What was the age of Vicky in 1980?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 6

Sum of ages of Mr and Mrs Sinha in 1972 = 46 years Sum of age of their family in 1976 = 19 x 3 = 57 years

Sum of ages of Mr & Mrs Sinha in 1976 = (46 + 8) years = 54 years

their son’s age in 1976 = (57 - 54) years = 3 years

Arun Sharma Test: Averages- 3 - Question 7

Sachin Tendulkar has a certain batting average N (a whole number) in his career of 86 innings. In the 87th inning, he gets out after scoring 270 runs which increases his batting average by a whole number. The batting average is calculated by dividing the total number of runs scored by the total number of innings played by the player. How many values of his new average is/are possible?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 7

Total number of runs scored till 86th inning = 86 N Now, 86 N + 270 = 87 (N + S), where S is the increase in batting average.
Different values of S possible now S = 0, 1, 2 and 3.

Arun Sharma Test: Averages- 3 - Question 8

Read the passage below and solve the questions based on it.
The average score of a batsman for a certain number of innings was 21 75 per inning. He played 3 innings more and scored 28. 34 and 37 runs respectively, thus increasing his average by 1.125.

Q. How many innings in all did he play?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 8

Let number of innings be x

average of the first (x-3) innings = 21.75
Total of the first (x-3) innings = 21.75(x-3)

Average of the x innings = (21.75+1.125) = 22.875
Total of the x innings = 22.875x

21.75(x-3) + 28+34+37= 22.875x
21.75(x-3) + 99 = 22.875x
21.75x - 65.25 + 99 = 22.875x
1.125x = 33.75
x = 30

Arun Sharma Test: Averages- 3 - Question 9

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 Arun Sharma Test: Averages- 3 - Question 9

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

Arun Sharma Test: Averages- 3 - Question 10

The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 10

Arun Sharma Test: Averages- 3 - Question 11

Mukul has earned as an average of 4,200 dollars for the first eleven months of the year. If he justifies his staying in the US on the basis of his ability to earn at least 5000 dollars per month for the entire year, then how much should he earn (in dollars) in the last month to achieve his required average for the whole year?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 11

The salary expectation for the whole year is 5000. So, the total earning will be 60000.
Earning for 11 months is 4200 so the total earnings are 46200.
Now for the last month he has to earn 60000 – 46200 = 13800.

Arun Sharma Test: Averages- 3 - Question 12

The average of the 5 consecutive even numbers A,B,C,D ,E is 52.what is the product of B & E

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 12

Let the five consecutive even numbers A,B,C,D ,E = (x−4),(x−2),(x),(x+2),(x+4) respectively.
Average = A+B+C+D+E/5 = 52
=> (x−4)+(x−2)+(x)+(x+2)+(x+4)=52×5
=> 5x=52×5
=> x=52×55=52
=> x=52–2=50 and x=52+4=56
∴ Product of B & E = 50 × 56 = 2800

Arun Sharma Test: Averages- 3 - Question 13

If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 9th, 11th, 15 th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Then

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 13

This question is one standard example of the definition of average in terms of surplus and deficit of the values. (Refer to the definition given) Now, what we were expected to mark here was the fact that whatever is the average (in this case, it is A), surplus generated by the marks of 3rd student will be same as deficit incurred due to 15th student. So, rusticating both of them is not going to create any difference on average marks of the class (remember marks are in AP). And similar will be the impact of rusticating 7th student and 11th student and then finally 9th student.
So, A = B

Arun Sharma Test: Averages- 3 - Question 14

The average salary of 30 officers in a firm is Rs.120 and the average salary of laborers is Rs. 40. Find the total number of laborers if the average salary of the firm is Rs. 50.

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 14

The sum of the salary of officers will be = 30×120 = 3600
Let the number of labourers = X.
ATQ, 3600 + 40X = 50(30+X)
2100 = 10X
X= 210

Arun Sharma Test: Averages- 3 - Question 15

We write down all the digits from 1-9 side by side. Now. we put "+" between as many digits as we wish to. so that the sum of numbers become 666.It is explained below: 1  2  3  4  5  6  7  8  9 = 666`

Now suppose we put plus signs al following places: 12 + 3 5 + 67 + 89 = 513` Since there are four numbers, so the average can be calculated by dividing the bum by 4. What is the average if the sum is 666?

In the above question, what is the average if the sum is 261?

Detailed Solution for Arun Sharma Test: Averages- 3 - Question 15

261 is possible only if we take 123 +45 + 6 + 78 + 9 So, the average will be 261/5 = 52.2

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