# Arithmetic - Average - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Arithmetic - Average. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2019 QADI | Arithmetic - Average**

A ﬁrm pays its ﬁve clerks Rs. 15,000 each, three assistants Rs. 40,000 each and its accountant Rs. 66,000. Then the mean salary in the ﬁrm comprising of these nine employees exceeds its median salary by rupees

- A.
14600

- B.
14000

- C.
15480

- D.
15200

- E.
14720

Answer: Option B

**Explanation** :

To get the median salary, we arrange the salaries of the 9 employees in numerical ascending order.

The salaries when arranged in ascending numerical order will be as follows

15000, 15000, 15000, 15000, 15000, 40000, 40000, 40000, 66000

Now the median of these 9 values will be the middle value i.e., 5^{th} value which is 15,000.

∴ Median = 15000

Mean Salary of these 9 employees = (15000×5 + 40000×3 + 66000)/(5 + 3 + 1) = 29,000

∴ Mean salary exceeds Median Salary by 29000 -15000 = 14000.

Hence, option (b).

Workspace:

**IIFT 2019 QA | Arithmetic - Average**

A cricket team has 11 players and each of them has played 20 matches till date. Virat, Rohit, Mahendra, Rahul and Shikhar have scored runs at an average of 60, 55, 50, 45 and 40 respectively. Rest of the players have scored at an average of 25 each. In the next 10 matches, Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul. After 30 matches, if Virat’s new average score is twice that of Rahul, what is the approximate average score of Mahendra?

- A.
49

- B.
41

- C.
43

- D.
45

Answer: Option C

**Explanation** :

Let Rahul's average after 30 matches be A, so Virat's average after 30 matches is 2A.

Virat's score after 30 matches = (60 × 20) + 900 = 2A × 30

Solving this equation, we get; A = 35.

Rahul's score after 30 matches = (45 × 20) + x = A × 30 = 35 × 30 (where x is the runs scored by Rahul in the last 10 matches)

Solving this equation, we get; x = 150.

Mahendra's score after 30 matches = (50 × 20) + 2x = B × 30 (where B is the Mahendra's average after 30 matches)

Putting x = 150 in the above equation, we get B = 43.33 ≈ 43.

Hence option (c).

Workspace:

**IIFT 2016 QA | Arithmetic - Average**

Two farmers were cultivating wheat on their respective agricultural land in a village. Farmer A had an average production of 20 bushels from a hectare. Farmer B, who had 15 hectares of more land dedicated to wheat cultivation, had and output of 30 bushels of wheat from a hectare. If farmer B harvested 530 bushels of wheat more than farmer A, how many bushels of wheat did farmer A cultivate?

- A.
50

- B.
80

- C.
160

- D.
200

Answer: Option C

**Explanation** :

If farmer A had x hectares, farmer B would have had (x + 15) hectares.

Average production of farmers A and B is 20 bushels and 30 bushels respectively.

∴ Total production of A = 20x and total production of B = 30(x + 15)

Since total production of B exceeds that of A by 530 bushels; 30(x + 15) – 20x = (10x + 450) = 530

∴ 10x = 80 i.e. x = 8

∴ Total cultivation of farmer A = 20x

= 160 bushels

Hence, option 3.

Workspace:

**XAT 2015 QA | Arithmetic - Average**

The median of 11 different positive integers is 15 and seven of those 11 integers are 8, 12, 20, 6, 14, 22, and 13.

**Statement I: **The difference between the averages of four largest integers and four smallest integers is 13.25.

**Statement II: **The average of all the 11 integers is 16.

Which of the following statements would be sufficient to find the largest possible integer of these numbers?

- A.
Statement I only.

- B.
Statement II only.

- C.
Both Statement I and Statement II are required.

- D.
Neither Statement I nor Statement II is sufficient.

- E.
Either Statement I or Statement II is sufficient.

Answer: Option E

**Explanation** :

Three integers are not known.

**Using Statement I:**

Average of four smallest integers

= (6 + 8 + 12 + 13)/4 = 39/4

∴ Average of four largest integers

$=\frac{39}{4}+13\frac{1}{4}=\frac{92}{4}$

In order to get the largest possible integer, two of the three unknown integers must be lowest possible i.e., 16 and 17.

So, the largest possible integer

= 92 – 22 – 20 – 17 = 33

Statement I can answer the question independently.

**Using Statement II:**

Sum of 11 integers = 11 × 16 = 176

Sum of the given integers = 110

∴ Sum of three unknown integers = 66

In order to get the largest possible integer, two of the three unknown integers must be lowest possible i.e., 16 and 17.

So, the largest possible integer

= 66 – 16 – 17 = 33

Statement II also answers the question independently.

Hence, option 5.

Workspace:

**IIFT 2013 QA | Arithmetic - Average**

The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall

- A.
remain unchanged

- B.
increase by 1

- C.
increase by 1.5

- D.
increase by 2

Answer: Option C

**Explanation** :

Since the question mention 7 consecutive natural numbers, the first 7 natural numbers can also be considered.

The first 7 natural numbers are (1, 2, 3, 4, 5, 6, 7) and their average is 4.

When the next 3 numbers i.e. 8, 9 and 10 are added, the new average is 5.5.

∴ Average increases by (5.5 – 4) = 1.5

Hence, option 3.

Workspace:

**XAT 2012 QA | Arithmetic - Average**

Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures are integers. The mean and the median salary figures are Rs. 5 lakh, and the only mode is Rs. 8 lakh. Which of the options below is the sum (in Rs. lakh) of the highest and the lowest salaries?

- A.
9

- B.
10

- C.
11

- D.
12

- E.
None of the above.

Answer: Option A

**Explanation** :

Mean of the salaries of the five vice presidents is Rs. 5 lakhs. Hence, sum of the salaries of the five vice presidents = 25 lakhs.

Now, median of the salaries is Rs. 5 lakhs and 8 is the only mode.

Hence, the highest salary and second highest salaries are both 8 lakhs.

Hence, sum of two lowest salaries = 25 – (5 + 8 + 8) = 4 Lakhs.

As 8 is the only mode hence, the only combination of the lowest salaries is 1 lakhs and 3 lakhs.

Hence, lowest salary = Rs. 1 lakh.

Hence, the required sum = 8 + 1 = 9 lakhs.

Hence, option 1.

Workspace:

**IIFT 2012 QA | Arithmetic - Average**

At a reputed Engineering College in India, total expenses of a trimester are partly fixed and partly varying linearly with the number of students. The average expense per student is Rs.400 when there are 20 students and Rs.300 when there are 40 students. When there are 80 students, what is the average expense per student?

- A.
Rs. 250

- B.
Rs. 300

- C.
Rs. 330

- D.
Rs. 350

Answer: Option A

**Explanation** :

Let the fixed expenses be k.

Let the variable expenses per student be m.

∴ $\frac{20m+k}{20}$ = 400 ...(i)

$\frac{40m+k}{40}$ = 300 ...(ii)

Solving (i) and (ii) simultaneously, we get m = 200 and k = 4000

There average cost for 80 students will be

= $\frac{80\times 200+4000}{80}$ = Rs. 250.

Hence, option 1.

Workspace:

**IIFT 2011 QA | Arithmetic - Average**

2 years ago, one-fifth of Amita’s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years. If the age of Paramita is also considered, the average age of three of them declines to 24. What will be the average age of Sumita and Paramita 3 years from now?

- A.
25 years

- B.
26 years

- C.
27 years

- D.
Cannot be determined

Answer: Option B

**Explanation** :

By conditions,

$\frac{a-2}{5}$ = $\frac{s-2}{4}$

∴ 4a = 5s – 2 …(i)

Also, $\frac{a+s-4}{2}$ = 27

∴ a + s = 58 …(ii)

Solving (i) and (ii),

a = 32 and s = 26

Also, $\frac{a-2+s-2+p-2}{3}$ = 24

∴ p = 20

∴ Average age of Sumita and Paramita 3 years from now = (s + p + 6)/2 = 26 years.

Hence, option 2.

Workspace:

**IIFT 2010 QA | Arithmetic - Average**

In a B-School there are three levels of faculty positions i.e. Professor, Associate Professor and Assistant Professor. It is found that the sum of the ages of all faculty present is 2160, their average age is 36; the average age of the Professor and Associate Professor is 39; of the Associate Professor and Assistant Professor is 32$\frac{8}{11}$; of the Professor and Assistant Professor is 36$\frac{2}{3}$, Had each Professor been 1 year older, each Associate Professor 6 years older, and each Assistant Professor 7 years older, then their average age would increase by 5 years. What will be the number of faculty at each level and their average ages?

- A.
(16, 24, 20 : 45, 35, 30 years)

- B.
(18, 24, 20 : 42, 38, 30 years)

- C.
(16, 20, 24 : 50, 30, 30 years)

- D.
None of these

Answer: Option A

**Explanation** :

Let the number of professors, associate professors and assistant professors be a, b and c respectively and their average ages be p, q and r respectively.

∴ a + b + c = $\frac{2160}{36}$ = 60 ...(i)

$\frac{ap+bq}{a+b}$ = 39

$\frac{bq+cr}{b+c}$ = 32$\frac{8}{11}$

and $\frac{cr+ap}{a+c}$ = 36$\frac{2}{3}$

Also,

$\frac{a(p+1)+b(q+6)+c(r+7)}{a+b+c}$ = 41

As $\frac{ap+bq+cr}{a+b+c}$ = 36,

∴ $\frac{a+6b+7c}{a+b+c}$ = 5

a + 6b + 7c = 5a + 5b + 5c

b + 2c = 4a

∴ b = 4a – c

Substituting in (i),

a + 4a – 2c + c = 60

∴ 5a – c = 60 …(ii)

We find that option 1 satisfies equations (i) and (ii).

Further, the values of p, q and r in option 1 also satisfy the other equations.

Hence, option 1.

Workspace:

**IIFT 2009 QA | Arithmetic - Average**

M/s. Devi Radiograms, a shop which sells electronic gadgets, marks its merchandise 35% above the purchase price. Until four months ago, purchase price of one Philips DVD player was Rs. 3000. During the last four months M/s. Devi Radiograms has received four monthly consignments of Philips DVD player at the purchase price of Rs.2750, Rs.2500, Rs.2400, and Rs.2250. The average rate of decrease in the purchase price of DVD player during these four months is:

- A.
7.51%

- B.
8.20%

- C.
6.94%

- D.
7.03%

Answer: Option C

**Explanation** :

∴ Average rate of decrease = $\frac{8.33+0.09+4+6.25}{4}$ = 6.91%

Hence, option 3.

Workspace:

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