IIFT questions on Arithmetic - Average

No of persons who drink coke = 2x + 57/x
As the no. of persons who drink coke will be an integer so 57/x must be an integer so x can be 1, 3, 19 or 57.
No. of persons who drink neither fruit juice nor coke = 57/3x = 19/x this also should be an integer, so x= 1 or 19 only
When x=1, no. of persons who drink coke = 2+57 = 59
When x=19, no. of persons who drink coke = 38 + 3 = 41

It is given that,
The mean of 29 observations is 337
Therefore we can say:
337 = $\frac{S}{29}$ where S is sum of observations
S = (337)(29) (1)
The first term is increased by 1
2nd by 2 .. and so on

New sum = S + $\frac{29\left(30\right)}{2}$

Therefore, mean will be $\frac{\left(S+\frac{29\left(30\right)}{1}\right)}{29}$

= 337 + 15 = 352

Answer is option C.

Let Sumit's age at the time of graduating MBA [in 2013] be years. Thus, his wife's age that year will be X - 3 years.
The average age of the family in 2013 = 41 years.
Thus, the total age of the family in 2013 = 7 x 41 = 287
Sumit got married in 2015. Thus, his wife's age in 2015 will added to the family's total.
The total age of family in 2015 = 287 + 14 + - 3 + 2 = X + 300
In 2017, Sumit had a daughter.
The total age of the family in 2017 = X + 300 + 8(2) = X + 316
The total age of the family in 2019 = X + 316 + 9(2) = X + 334
We are given that the average age of the family in 2019 is the same as that in 2013.
Thus, X + 334 = 41 x 9 = 369
X = 369 - 334 = 35 years
Thus, the age of Sumit at the time of graduating MBA is 35 years.
Hence, the answer is option B.

From the given data, we can create the following diagram:

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Hence, the required answer = 100 + 200 + 300 = 600

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).

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 (c). 

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 (c).

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 (a).

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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 (b).

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 (a).

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

Hence, option (c).