Averages concepts for CAT

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Examinations like the CAT, XAT, IIFT, CMAT, GMAT, SSC CGL, and Bank PO include straight-forward questions from this chapter. According to past statistics, they consistently ask at least one question in the quantitative aptitude portion and at least four in the data interpretation area each year.

Properties of Average

The average of any two or more quantities (or data) necessarily lies between the lowest and the highest values of the given data. i.e., if x_land x_hbe the lowest and highest (or greatest) values of the given data (x₁,x₂,...x_l,...x_h,...,x_n) then

x_l < Average < x_h; x_l = x_hi.e. x_l< (x₁ + x₂ + x₃ + x_l…+ x_h ….+x_n)/n < x_h

If each quantity is increased by a certain value ‘K’ then the new average is increased by K.
If each quantity is decreased by a certain value K, then the new average is also decreased by K.
If each quantity is multiplied by a certain value K, then the new average is the product of old average with K.
If each quantity is divided by a certain quantity ‘K’ ,then the new average becomes 1 times of the initial average, K where K ≠
If ‘A’ be the average of x₁,x₂,x_m,... y₁, y₂,..., y_n. where x₁, x₂,..., x_m be the below A and y₁, y₂, y₃,..., y_n be the above A, then (A − x₁) + (A − x₂) +... (A − x_m) =(y₁ −A)+(y₂ −A)+...(y_n −A) i.e., the surplus above the average is always equal the net deficit below average.
The term "weighted average" refers to the calculation of the average of sets or groups with various numbers of elements as opposed to individuals. As the number of items in this instance varies for the various sets, their relative weights also vary. If K₁, K₂, K₃, K₄,..., Knis the number of components in n groups, and A₁ , A₂ , A₃ , A₄ ... An is the average of each group, then

Weighted Average = (K₁ A₁+ K₂ A₂ + K₃ A₃+ K₄ A₄+………+ K_n A_n)/(K₁+ K₂+ K₃ + K₄+……+K_n)

Problems Based on Age :

(i) If a family has "n" members and their average age is x years, then K years ago, then the family's average age is (x – K) years, assuming no one has died or been born during that time.

(ii) If there are 'n' family members and their average age is x years now, then K years from now, their average age will be (x + K) years.

Problems Based on Income/Salary: Income = Expenditure + Savings
Problems Based on Time, Speed and Distance :

Case 1. When the distance travelled in different time slots or parts is same i. e., if a person or vehicle moves x km at a speed of u km/hr and further he goes or comes back the same distance x km at a speed of v km/hr.

Then the average speed = $\frac{2uv}{u+ v}$

If there are 3 parts of distance x km travelled with 3 different speeds i.e., if a person goes first x km @ speed of u km/hr and next x km @ v km/hr and the last x km @ w km/hr.

Then the average speed = $\frac{3uvw}{uv+vw+wu}$

General Formula :

Average speed = $\frac{Total\ distance}{Total\ Time}$

The above two formulae are derived with the help of the general formula of average speed.

Proportion Method : If the half of the distance is covered at u km/hr and the rest half of the journey is covered at v km/hr then the average speed can be found as follows :

Step 1. Divide the difference of u and v in the ratio of u : v (where u < v)

Step 2. u + ((u ~ v) × u) or v − (u ~ v) × v (u + v) u + v

Case 2. When the distances travelled at different speeds are different then we calculate the average speed with the help of general formula of average speed. e. g., A person first goes x₁ km at the speed of u km/hr and x₂ km at the speed of v km/hr and x₃ km at the speed of w km/hr and so on, then the

Average Speed = Total Distance/Total time = (x₁ + x₂ + x₃+_…..+ x_n)/(t₁+ t₂ + t_3+…+ t_n)

Value of An Overlapping Element
(i) a + b = k and b + c = l and a + b + c = m

then [(a + b) + (b + c)] − (a + b + c) = (k + l) − m or b = k + l − m

(ii) a+b=k, d+e=l and a+b+c+d+e = m

Then c = (a+b+c+d+e)−[(a+b)+(d+e)]=m−[k +l]

Average of Some Important Series of Numbers :

(i) Average of first ' n' natural numbers (n+1)/ 2

(ii) Average of first ' n' even numbers (n+1)

(iii) Average of first ' n' odd numbers n

(iv) If there are ( p + q) elements in a set or group but the average of p elements is r and the average of q elements is s, then the average of all the elements of

the set (or group) is ( pr + sq)/( p + q)

Averages Questions for CAT

Ques: In 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. (CAT 1997)

55
62
60
Cannot be determined

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Ans: C

The total number of points for 10 papers is 800. 81 times 8 papers = 648

(800 – 648) = 152 If the highest number is 92, the lowest number is (152 – 92) = 60.

Ques: Ram purchased a flat at Rs. 1 lakh and Prem purchased a plot of land worth = Rs. 1.1 lakh. The respective annual rates at which the prices of the flat and the plot increased were 10% and 5%. After two years, they exchanged their belongings and one paid the other the difference. Then, (CAT 1995)

Ram paid Rs. 275 to Prem
Ram paid Rs.475 to Prem
Ram paid Rs.375 to Prem
Prem paid Rs. 475 to Ram

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Ans: A

Value of flat after two years = 1 × 1.1 × 1.1 = 1.21 lakh

Value of plot after two years= 1.1 × 1.05 × 1.05 = 1.21275 lakh

Hence, difference in price after two years (1.21275 − 1.21) × 100000 = Rs. 275

Ques: The A change making machine contains 1 rupee, 2 rupee and 5 rupee coins. The total number of coins is 300. The amount is Rs 960.

If the number of 1 rupee coins and the number of 2 rupee coins are interchanged, the value comes down by Rs 40. The total number of 5 rupee coins is (CAT 2001)

100
140
60
150

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Ans: B

Let number of 1 rupee coins = x;

2 rupees coin = y and

5 rupees coin = z

x + y + z = 300 ...(i)

x + 2y + 5z = 960 ...(ii)

(x + 2y + 5z) (y + 2x + 5z) = 40 ...(iii)

yx = 40 or y = 40+x

Multiply (i) by 5 and subtract (ii),

540 = 4x + 3y put y=40 + x

549 = 4x + 3x + 120 or 420 = 7x

or x = 60, y = 100 & z = 140

Ques: If At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes, and one order of fries for Rs 120 exactly. At the same place it would cost Rs 164.5 for 4 burgers, 10 shakes, and one order of fries. How much would it cost for an ordinary meal of one burger, one shake, and one order of fries? (CAT 2001)

Rs 31
Rs 41
Rs 21
Cannot be determined

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Ans: A

Let the cost of burger, shake and one order of fries be x, y and z respectively.

According to question

3x + 7y + z = 120....(i)

and 4x +10y + z = 164.5....(ii)

Subtracting (i) from (ii),

x + 3y = 44.5

Multiplying (i) by 4, (ii) by 3 and subtracting (ii) from (i)

We get,

x + y + z = 44.5 – 13.5 = 31

Ques: Mayank, Mirza, Little and Jaspal bought a motorbike for $60.00. Mayank paid one half of the sum of the amounts paid by the other boys, Mirza paid one third of the sum of the amounts paid by the other boys; and Little paid one fourth of the sum of the amounts paid by the other boys. How much did Jaspal have to pay? (CAT 2002)

15
13
17
None of these

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Ans: B

Let Mayank paid $x, Little paid $ z, Jaspal paid $u and Mirza paid $z

Based on the question

x = (y + z + u)/2,

y = (x + z + u)/2

z = (x + y + u)/4

Provided in the question => x + y + z + u = 60

x = (60 – x )/2 =20

y = (60 – y)/3 =15

z = (60 – z)/4 = 12

x + y + z + u = 60

20 + 15 + 12 + u = 60

u= 60 – 47 = 13

Ques: If $\frac{a}{b} = \frac{1}{3}, \frac{b}{c} = 2, \frac{c}{d}= \frac{1}{2}, \frac{d}{e}= 3 \ and \ \frac{e}{f}= \frac{1}{4}$ then what is the value of $\frac{abc}{def}$? (CAT 2006)

3/8
27/8
3/4
27/4

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Ans: A

a/b=1/3 => a=3, b=3

b/c=2/1 => b=3, c=3/2

c/d=1/2 => c=3/2, d=3

d/e=3/1 => d=3, e=1

e/f=1/4 => e=1, f=4

abc/def= (1x3x3/2)/(3x1x4) = 3/8

Ques: In a class of 5 students, the average weight of the 4 lightest students is 40 kg, average weight of the 4 heaviest students is 45 kg. What is the difference between the maximum and minimum possible average weight overall? (CAT 2016)

2.8 kg
3.2 kg
3 kg
4 kg

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Ans: C

Let say that the students are named a, b, c, d and e, in increasing order of weights. The average of a, b, c and d is 40 kg, where as the average of b, c, d and e is

45 kg. The sum of a, b, c and d is 160 kg and the sum of b, c, d and e is 180 kg.

What is the total weight of all the students?

There are two ways of looking at this

(a) 160 + e (b) 180 + a Or e is 20 more than a.

The total weight is 160 + e. So, the highest value of e will correspond to the highest possible average.

The highest possible value of e occurs when it is 20 higher than the highest possible value for a, which is 40 (all the first 4 scores are equal to 40).

So, the highest possible average is (160 + 60)/5 = 44.

This will be the case when the weights are 40 kg,

40 kg, 40 kg, 40 kg and 60 kg.

Conversely, the least possible value for the average occurs when a is the least. This happens when e is the least too (since, a is 20 less than e).

The least possible for e is 45= 180/4

So, the least possible value for a would be 25. The least possible average = (180 + 25)/5 = 41

This will be the case when the weights are 25 kg, 45 kg, 45 kg, 45 kg and 45 kg.

So, the difference between maximum possible and minimum possible average = 3 kg.

Ques: There are nine three-digits numbers with distinct unit's digits. Each number is reversed and the reversed number is subtracted from the original number. The results were found to have an average of 0. If for each number, the hundred's digit is not less than its unit's digit, then find the average of the hundred's digits of the greatest and the least numbers. (CAT 2014)

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Ans: B

Let the numbers be a₁b₁c₁, a₂b₂c_2,a₃b₃c_3…… a₉b₉c₉

Average of the numbers= 1/9 ( a₁b₁c₁– c₁b₁a₁+ a₂b₂c₂– c₂b₂a₂+……… a₉b₉c₉– c₉b₉a₉)

=1/9[99(a₁– c₁+ a₂– c_{2 +} a₃– c_3+…….. a₉– c₉)

For i=1 to 9, a_i>= c_i

As the average of the results is 0, it follows that a₁= c₁.

As the unit’s digits of the numbers are distinct, the unit’s digits must be from 1 to 9. The greatest and the least hundred’s digits are 9 and 1, respectively.

∴Required average = (1 + 9)/2 = 5

Ques: There are two classes A and B. The average weight of the students in class A is 40 kg. The average weight of the students in class B is 60 kg. A student, whose weight is x kg left A and joined B. As a result, the average weight of A, as well as that of B decreased. Which of the following must be true? (CAT 2014)

35< x≤ 40
40< x< 60
60< x< 65
30< x≤ 35

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Ans: B

As the average weight of A decreased after the student left, his weight must be more than the average weight of A. As the average weight of B decreased after the student joined, his weight must be less than the average weight of B.

So, his weight must be between 40 kg and 60 kg.

Ques: Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is ` 700 when there are 25 boarders and ` 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders? (CAT 1999)

550
580
540
570

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Ans: A

Let X be the fixed cost and Y be the variable cost, then according to question.

X + 25Y = 17500 ...(i)

X + 50Y = 30000 ...(ii) Solving the Eqs. (i) and (ii), we get

X =5000,Y = 500

Now, if average expenses of 100 boarders be k.

Then, 100 × k =5000 + 500 ×100⇒ A =550

Ques: A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight of the heaviestBox? (CAT 2000)

60 kg
64 kg
62 kg
Cannot be determined

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Ans: C

If each box is to be weighed with every other box (in pairs), there would be ten different weight combinations. option is the best way to answer the question.

The maximum weight can't be 60 kg because the combined weight of two boxes should be 121 kg, which means that the other should be 61 kg. Again, if the maximum weight is 64 kg, the next box weight should be 57 kg for a total weight of 121 kg and 63 kg for a total weight of 120 kg. Now, if we add the weights of the two boxes, which are 63 kg and 64 kg, we get 127, which is not in the question. But if we say the maximum weight is 62 kg, then the other boxes weigh 59 kg, 54 kg, 58 kg, and 5 kg. So, 62 kg would be the most you could weigh.

Ques: In Rajdhani Express, there are 10 boggies which carries on an average of 20 passengers per boggie. If at least 12 passengers were sitting in each boggie and no any boggie has equal number of passenger then maximum, how many passengers can be accommodated in a boggie? (CAT 2013)

45
64
56
None of these

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Ans: C

The Rajdhani Express has a total of 200 people on board, which is 10 times 20.

So, the least number of people who could fit in the nine boggies is 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 144 So, the least number of people who can fit in one boggie is (200 144) = 56.

Ques: Mohan is a carpenter who specializes in making chairs. For every assignment he undertakes, he charges his commission and cost. His commission is fixed and equals Rs. 560 per assignment while the cost equals Rs. 2n² , where n is the total number of chairs he makes. If for a certain assignment the average cost per chair is not more than Rs.68, then the minimum and maximum possible numbers of chairs in the assignment are, respectively. (CAT 2011)

13 and 19
13 and 20
14 and 19
14 and 20

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Ans: D

Fixed cost (commission) = Rs.560/assignment

Variable cost = 2n²

Let the assignment be of x chairs,

then total cost = 560 + 2x².

Average cost = (560 + 2x²)/ x = 68

⇒ 2x² − 68x + 560 ≤ 0

⇒ (x−14)(x−20) ≤ 0

⇒14 ≤ x ≤ 20

Hence, minimum number of chair is 14 and maximum number of chair is 20.

Ques: What is the present worth of a house which would be worth Rs. 50000 after 3 years, if it depreciates at the rate of 10%? (CAT 2009)

Rs.35765.74
Rs. 67655.74
Rs. 67560.74
Rs. 68587.10

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Ans: D

Value of the house after 3 years = Rs. 50000

∴Present worth = 50000/(1-(10/100))³

= 50000/(0.9)³ ≈ Rs. 68587

Ques: Consider the set S = {2, 3, 4, K, 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of (X-Y)? (CAT 2007)

1
(1/2)n
(n+1)/2n
(n+1)
0

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Ans: A

Sum of odd integers in the set

S=(n/2) x {2×3+(n−1)×2} = (n(2n+4))/2=n×(n+2)

Therefore, the average of the odd integers in the set S = n+2

Sum of even integers in the set S = (n/2)x{2×2+(n−1)×2} = (n/2)x(2n+2) = n(n+1)

Therefore, the average of the even integers in the set S = n + 1.

Therefore, X − Y = (n + 2) − (n + 1) = 1

Averages concepts for CAT

Properties of Average

Averages Questions for CAT

CAT Related Questions

1.
Let $x$, $y$, and $z$ be real numbers satisfying
\(4(x^2 + y^2 + z^2) = a,\)
\(4(x - y - z) = 3 + a.\)
Then $a$ equals ?

2.
If the equations $x^2 + mx + 9 = 0$, $x^2 + nx + 17 = 0$, and $x^2 + (m+n)x + 35 = 0$ have a common negative root, then the value of $(2m + 3n)$ is ?

3.
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is ?

4.
Let $a_n$ be the largest integer not exceeding $\sqrt{n}$. Then the value of $a_1 + a_2 + \dots + a_{50}$ is

5.
In the $XY$-plane, the area, in sq. units, of the region defined by the inequalities $y \ge x + 4$ and $-4 \le x^2 + y^2 + 4(x - y) \le 0$ is

6.
ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of \(\triangle ADE\) is

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Averages concepts for CAT

Properties of Average

Averages Questions for CAT

CAT Related Questions

1. Let $x$, $y$, and $z$ be real numbers satisfying\(4(x^2 + y^2 + z^2) = a,\)\(4(x - y - z) = 3 + a.\)Then $a$ equals ?

2. If the equations $x^2 + mx + 9 = 0$, $x^2 + nx + 17 = 0$, and $x^2 + (m+n)x + 35 = 0$ have a common negative root, then the value of $(2m + 3n)$ is ?

3. The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is ?

4. Let $a_n$ be the largest integer not exceeding $\sqrt{n}$. Then the value of $a_1 + a_2 + \dots + a_{50}$ is

5. In the $XY$-plane, the area, in sq. units, of the region defined by the inequalities $y \ge x + 4$ and $-4 \le x^2 + y^2 + 4(x - y) \le 0$ is

6. ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of \(\triangle ADE\) is

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1.
Let $x$, $y$, and $z$ be real numbers satisfying
\(4(x^2 + y^2 + z^2) = a,\)
\(4(x - y - z) = 3 + a.\)
Then $a$ equals ?

2.
If the equations $x^2 + mx + 9 = 0$, $x^2 + nx + 17 = 0$, and $x^2 + (m+n)x + 35 = 0$ have a common negative root, then the value of $(2m + 3n)$ is ?

3.
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is ?

4.
Let $a_n$ be the largest integer not exceeding $\sqrt{n}$. Then the value of $a_1 + a_2 + \dots + a_{50}$ is

5.
In the $XY$-plane, the area, in sq. units, of the region defined by the inequalities $y \ge x + 4$ and $-4 \le x^2 + y^2 + 4(x - y) \le 0$ is

6.
ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of \(\triangle ADE\) is