Weighted Averages Questions for CAT

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Weighted Averages Questions for CAT
Weighted Averages Questions for CAT

Weighted Averages Questions for CAT:

Download important CAT Weighted Averages Questions PDF based on previous asked questions in CAT and other MBA exams. Top 25 Weighted Averages questions for CAT quantitative aptitude.

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Question 1: A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?

a) Rs. 300

b) Rs. 250

c) Rs. 400

d) 500

Question 2: The average marks in English of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 in lieu of the actual marks 48, 59 and 67 respectively, then what would be the correct average?

a) 56.5

b) 59

c) 57.5

d) 58

e) None of these

Question 3: In an enterance examination, Ritu scored 56 percent marks, Smita scored 92 percent marks and Rina scored 634 marks. The maximum marks of the examination is 875. What is the average marks scored by all the three girls together?

a) 1929

b) 815

c) 690

d) 643

e) None of these

Question 4: Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is:[CAT 2000]

a) n

b) n+1

c) kn, where k is a function of n

d) n+(2/7)

Question 5: The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

a) 55

b) 60

c) 62

d) Cannot be determined


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Question 6: Total expenses of a boarding house are partly fixed and partly varying linearly with tile number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

a) 550

b) 580

c) 540

d) 560

Question 7: A group of 10 students have an average weight of 55.The heaviest person in the group is replaced by 2 other persons.As a result the average weight of the group decreases by 8.63636.What can be the weight of the person who was removed if one of new person weighs 20 kg and the weight of the other person is a multiple of 7?

a) 75

b) 74

c) 72

d) 68

Question 8: The average weight of twenty one students is 70 kilograms. When the teacher’s weight of 100 kilograms is added to the total, what’s the new average?

a) 71.36 kg

b) 78.56 kg

c) 80.96 kg

d) 85.56 kg

e) 90.46 kg

Question 9: Find the average of the following scores: 111, 121, 145, 134, 0, 154, 198

a) 111.29

b) 123.29

c) 133.29

d) 143.29

e) None of the above

Question 10: Ranjit got the following scores in five subjects in school: 135 in science, 115 in maths, 120 in english, 130 in hindi and 105 in computing. The maximum score in each subject was 140. Calculate the average score in all the five subjects expressed as a percentage?

a) 121

b) 131

c) 90.43

d) 86.43

e) None of the above

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Question 11: The average of five numbers is 101. The average of two of them is 104. The average of two more of those numbers is 110. How many numbers can be accurately identified with this information?

a) none

b) one

c) two

d) three

e) all

Question 12: The average of n integers is 50 . When 76 is added to this set, the average of the numbers increases by 2. Find the maximum value of any integer in this set, given that every integer in this set is a positive integer.

a) 590

b) 566

c) 589

d) 534

Question 13: Raghu bought 13 apples at Rs 20 each and 17 bananas at Rs 14 each. Find the average price he paid per fruit.

a) Rs 15.60

b) Rs 17.10

c) Rs 16.60

d) Rs 18.20

Question 14: 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 ?

a) 15.167

b) 18.2

c) 13

d) cannot be determined

Question 15: 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.

a) 7

b) 13

c) 15

d) Cannot be determined

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Question 16: There are some bags numbered 1, 2, 3 and so on. Each bag contains exactly as many balls as the number on the bag. Out of these bags one is emptied and the average number of balls per bag becomes 14. What was the average number of balls per bag that were there initially?
Given that,
X = 14
Y = 14.5
Z = 15

a) Both X and Y

b) Both Y and Z

c) Both X and Z

d) All X, Y and Z

Question 17: In an exam taken by 10 students the average marks out of 50 was 30. If the 5 toppers, who scored distinct marks, are not included, the average decreases by 5. The passing score was 30 and all the 5 toppers passed. What is the maximum possible score of the topper?

a) 46

b) 48

c) 50

d) None of these

Question 18: The average weight of a class of 40 students is 40 kg. After some students, whose average weight is 47 kg, joined the average weight of the class increased by 5%. How many students joined the class?

Question 19: What is the average of the first 79 terms of the sequence
3, 7, 13, 21, 31 . . . . . ?

a) 2973

b) 2161

c) 2371

d) 2767

Question 20: In a class of 60 students the average weight of all the students is 68 kg. The ratio of men and women in the class is 8:7 and the ratio of average weight of women and men is 4:5. What is the average weight of all the men in the class (in kg)?

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Question 21: In a class of 50 students the average weight of the students is 60 kg. If the average weight of girls and boys of the class is 45 kg and 70 kg respectively, then what is the ratio of girls and boys in the class?

a) 1:2

b) 2:1

c) 2:3

d) 3:2

e) 1:1

Question 22: The average of 9 numbers is found to be 24. However, it was later realized that 47 was counted as 74. If the error is rectified and the average is calculated again, then what will be the new average?

a) 23

b) 22

c) 21

d) 20

Question 23: There are n boxes in which each box is numbered 1, 2,3…till n. Each box contains oranges equal to the number written on the box. When the orange in one of the box is emptied the average number of oranges per box becomes 20. What could be the total number of oranges initially?

a) 741

b) 850

c) 861

d) 903

e) 990

Question 24: The average weight of the class increases by 2 kg when a student of weight 50kg joins the class. However, if a student of weight 40 kg leaves the class, the average weight of the class goes down by 1 kg. What is the current average weight of the class?

a) 28

b) 34

c) 17

d) Cannot be determined

Question 25: ‘x’ years ago the ages of a husband and wife were ‘4x’ and ‘3x’ respectively. How many times of the sum of their present ages would the average of their ages be ‘2x’ years later?

a) $\frac{11}{18}$

b) $\frac{13}{18}$

c) $\frac{15}{18}$

d) None of these

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Answers & Solutions:

1) Answer (A)

Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000/40 = 300

2) Answer (E)

Total marks = 24 x 56 = 1344
Sum of actual marks = 1344 – (44 + 45 + 61) + (48 + 59 + 67) = 1368
Actual Average = $\frac{1368}{24}$ = 57

3) Answer (D)

Marks obtained by Ritu = 56% of 875 = $\frac{56}{100}$ x 875 = 0.56 x 875 = 490

Marks obtained by Smita = 92% of 875 = $\frac{92}{100}$ x 875 = 0.92 x 875 = 805

Marks obtained by Rina = 634

Average = $\frac{490 + 805 + 634}{3}$ = $\frac{1929}{3}$ = 643

4) Answer (B)

The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be (5n+2n+7)/7 = n+1

5) Answer (B)

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 – 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

6) Answer (A)

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $\frac{55000}{100}$ = Rs.550.

7) Answer (B)

a+b+c+…..+j=55*10=550
Suppose j is removed from the group and 2 other persons x and y are added.
a+b+c…+i+x+y=$11\times(55-\frac{95}{11})$=510
550-j+x+y=510
j-x-y=40
j=40+20+7k=60+7k
Only 74 satisfy the given condition.

8) Answer (A)

Total weight of the students = 70 * 21 = 1470 kgs. Total weight including the teacher = 1570. New Average = 1570/22 = 71.36 kgs.

9) Answer (B)

The average = Total / number of values = 863/7 = 123.29

10) Answer (D)

Total score = 605. Average score = 605/5 = 121.
Average score as a percentage = 121/140* 100 = 86.43

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11) Answer (B)

Only one of the five numbers can be accurately identified with the given information.

Number = sum of the five numbers – sum of two numbers – sum of the other two numbers.
= 5*101 – 2*104 – 2*110 = 77

12) Answer (C)

Since the average of n integers is 50,
Sum of n integers = 50n
50n+76 = 52n+52
2n = 24
n = 12
Sum of n integers = 50*12 = 600
Now we have to find the maximum value of the element of this set
This happens when all 11 elements are equal to 1
11+x = 600
x = 589

13) Answer (C)

Average price = $\frac{(13*20)+(17*14)}{30}$ = $\frac{498}{30}$ = Rs 16.60

14) Answer (D)

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. Thus, D is the right choice.

15) Answer (D)

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.

16) Answer (B)

Let the number of bags be ‘x’ each containing exactly as many balls as the number on the bag.
The total number of balls = $ \frac{x(x+1)}{2}$
The average number of balls per bag = $ \frac{x+1}{2}$
Let the bag numbered ‘y’ be emptied.
New average number of balls per bag = $ \frac{\frac{x(x+1)}{2}-y}{x}$
Equating this to 14 we get,
$ \frac{\frac{x(x+1)}{2}-y}{x}=14$
Simplifying we get,
$ x(x-27)=2y$
x > 27 as y can’t be negative.
If x = 28, y = 14.
If x = 29, y = 29.
If x = 30, y = 45 but y can’t be greater than x.
So only two values of x are possible i.e. 28 or 29.

In both the cases Average = $ \frac{x+1}{2}$ = 14.5 or 15 depending on the value of x.

17) Answer (D)

Overall sum of marks of the class = 10*30 = 300.
When top 5 students are not included the average becomes 25. So sum = 125.
Sum of marks of 5 toppers = 300 – 125 = 175.
Also all of the 5 toppers passed.
The scores of 4 of them have to be minimized to find the max possible score.
The possible values of marks are 30, 31, 32, and 33.
The sum = 126.
Hence the score of the topper = 175 – 126 = 49.

18) Answer: 16

The total weight of all the students = 40*40 = 1600 kg
Let the number of students who joined be ‘x’
The average weight of the class after the new students joined is 42 kg
The average weight of the new students is 47 kg
So, 42(40+x) – 1600 = 47x
=> x = 16.
Thus 16 new students joined.

19) Answer (B)

The second level difference of the terms is constant. Hence the general term of the equation will be $an^2 + bn + c = 0$
=> Putting the value of n = 1, we get a + b + c = 3
Similarly, Putting n = 2, we get 4a + 2b + c = 7 and 9a + 3b + c = 13.
Solving these three equations, we get a = 1, b = 1 and c = 1. So the general term is
$n^{2} + n + 1$
So the sum of first n terms of the sequence can be obtained by
Σ($n^{2} + n + 1$)
=> We know that average of n terms is given by
$\frac{\textrm{sum of all terms}}{n}$
=> ${\frac{n*(n+1)*(2n+1)}{6n} + \frac{n*(n+1)}{2n} + \frac{n}{n}}$
=> Average = (n+1)*(2n+1)/6 + (n+1)/2 + 1
= $\frac{n+1}{2}(\frac{2n+1}{3}+ 1) + 1$
= Putting n = 79 gives
=> $\frac{80}{2}(\frac{162}{3}) + 1$
=> 40*54 + 1 = 2161

20) Answer: 75

The ratio of men and women in the class is 8:7.
The number of men is 32 and the number of women is 28.
The ratio of average weight of women and men is 4:5.
Let the average weight of men and women in the class be 5x and 4x respectively.
$\frac{5x*32+4x*28}{60}=68$
=> $x=15$
The average weight of men = $5x=5*15=75$ kg

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21) Answer (C)

Let the number of girls be ‘x’
The number of boys will be ’50-x’
The total weight of all the students combined = 50*60 = 3000 kg
The total weight of all the girls combined = x*45 = 45x
The total weight of all the boys combined = (50-x)*70 = 3500-70x
Equating both we get,
45x + 3500 – 70x = 3000
=> 25x = 500
=> x = 20.
So the number of girls and boys are 20 and 30 respectively.
Hence the ratio is 2:3.

22) Answer (C)

Average of 9 numbers is 24. So the sum of all possible number would be 24*9 = 216.
Now it is given that 47 was counted as 74
So the actual sum should have been 216 – 74 + 47 = 189
Thus the correct average should be 189/9 = 21

23) Answer (C)

The total number of oranges in the box = n(n + 1)/2
Let the apples be emptied out from the mth box. Then the number of oranges after emptying one box = n(n + 1)/2 – m
Thus, the final averages of oranges in each box = $\frac{n(n + 1)/2 – m}{n}$ = 20
=> $n^2 + n -2m = 40n$
$n^2 – 39n = 2m$
$n(n – 39) = 2m$
n > 39 and m$\leq$n
Thus, the possible values are n = 40 and n = 41. For n greater than 41, m would be greater than n.
Thus, the total number of apples initially = n(n + 1)/2 = 40(41)/2 or 41(42)/2 = 820 or 861

24) Answer (B)

Let the initial average weight be A and the number or children in the class be k.
=> $\frac{Ak+50}{k+1}=A+2$ and $\frac{Ak-40}{k-1}=A-1$
Solving the first eqn we get, 50=A+2k+2 or A+2k=48
Solving the second eqn we get, A+k-1=40 or A+k=41
Subtracting 2nd from 1st we get k=7. Thus, A=34

25) Answer (B)

It is given that, ‘x’ years ago the ages of husband and wife were ‘4x’ and ‘3x’.
So their present ages will be ‘5x’ and ‘4x’ respectively.
The sum of their ages is ‘9x’
Their ages after ‘2x’ years will be ‘7x’ and ‘6x’ respectively.
Required ratio = $\frac{13x/2}{9x}=\frac{13}{18}$

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