Quant Questions for IIFT Set-2 PDF

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Quant Questions for IIFT Set-2 PDF
Quant Questions for IIFT Set-2 PDF

Quant Questions for IIFT Set-2 PDF

Download important IIFT Quant Set-2 Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Quant Set-2 Questions and answers for IIFT and other  exams.

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Question 1: Consider the set S = {2, 3, 4, …., 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 ?

a) 0

b) 1

c) (1/2)*n

d) (n+1)/2n

e) 2008

Question 2: Three classes X, Y and Z take an algebra test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average for all the three classes?

a) 81

b) 81.5

c) 82

d) 84.5

Question 3: Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student
can get an A grade in the course if the average of her scores is more than or equal to 90.Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he will score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?

a) 6

b) 7

c) 8

d) 9

e) None of these

Question 4: The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

a) 65 percent

b) 60 percent

c) 70 percent

d) None of the above

Question 5: The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘A’ is 3 :4. In alloy ‘B’ same metals are mixed in the ratio 5:8. If 26 kg of alloy ‘B’ and 14 kg of alloy ‘A’ are mixed then find out the ratio of ‘metal 1’ and ‘metal 2’ in the new alloy.

a) 3:2

b) 2:5

c) 2:3

d) None of the above

Question 6: A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]

a) 2 : 3

b) 1 : 2

c) 1 : 3

d) 3 : 4

Question 7: A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a) 2

b) 3

c) 4

d) 5

Question 8: The remainder, when $(15^{23} + 23^{23})$ is divided by 19, is

a) 4

b) 15

c) 0

d) 18

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Question 9: If R = $(30^{65}-29^{65})/(30^{64}+29^{64})$ ,then

a) $0<R\leq0.1$

b) $0.1<R\leq0.5$

c) $0.5<R\leq1.0$

d) $R>1.0$

Question 10: Let $n!=1*2*3* …*n$ for integer $n \geq 1$.
If $p = 1!+(2*2!)+(3*3!)+… +(10*10!)$, then $p+2$ when divided by 11! leaves a remainder of

a) 10

b) 0

c) 7

d) 1

Instructions

Directions for the following two questions:

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is $240 + bx + cx^2$ , where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.67%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

Question 11: What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

a) 620

b) 920

c) 840

d) 760

e) Cannot be determined

Instructions

Books and More sells books, music CDs and film DVDs. In December 2009, they earned 40% profit in music CDs and 25% profit in books. Music CDs contributed 35% towards their total sales in rupees. At the same time total sales in rupees from books is 50% more than that of music CDs.

Question 12: If Books and More have earned 20% profit overall, then in film DVDs they made

a) 15.2% profit

b) 10.0% profit

c) 10.0% loss

d) 16.3% loss

e) 23.4% loss

Question 13: I sold two watches for Rs. 300 each, one at the loss of 10% and the other at the profit of 10%. What is the percentage of loss(-) or profit(+) that resulted from the transaction?

a) (+)10

b) (-)1

c) (+)1

d) (-)10

Question 14: In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs. 10,000 and in the beginning of 2010, the accumulated interest becomes Rs. 25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is:

a) Rs. 16,000

b) Rs. 18,000

c) Rs. 20,000

d) Rs. 25,000

e) None of the above

Question 15: A man borrows 6000 at 5% interest, on reducing balance, at the start of the year. If he repays 1200 at the end of each year, find the amount of loan outstanding, in , at the beginning of the third year.

a) 3162.75

b) 4125.00

c) 4155.00

d) 5100.00

e) 5355.00

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Answers & Solutions:

1) Answer (B)

The odd numbers in the set are 3, 5, 7, …2n+1

Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$

Average of odd numbers = $n^2 + 2n$/n = n+2

Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

2) Answer (B)

Let x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $\frac{83x+76y+85z}{x+y+z}$

Substitute the relations in above equation we get,

$\frac{83x+76y+85z}{x+y+z}$  = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

3) Answer (D)

Grade A $\geq$ 90 and Grade B = 87 to 89

If Ramesh scores 70 instead of 97, => Change of marks = 97 – 70 = 27

It creates a change from grade A to B, this means an overall change in average by

= Minimum marks for grade A – Minimum marks for Grade B = 90 – 87 = 3

$\therefore$ Number of subjects = $\frac{27}{3} = 9$

4) Answer (A)

The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

Let ‘9x’ be the number of total journalists in the office. Then, we can say that the number of male and female journalists are ‘5x’ and ‘4x’ respectively.

It is given that 30 percent of the male journalists and 40 percent of the female journalists are covering political news. Hence, total number of journalists who are covering political news = 0.3*5x + 0.4*4x = 3.1x

Therefore, the total number journalists who are covering sports news = 9x – 3.1x = 5.9x.

Hence, the percentage of the journalists in the newspaper is currently involved in sports reporting = $\dfrac{5.9x}{9x}\times 100$ $\approx$ 65 percent. Therefore, option A is the correct answer.

5) Answer (C)

The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘A’ is 3 :4.Therefore, we can say that 14 kg of alloy ‘A’ will contain $\dfrac{3}{7} 14$ = 6 kg of ‘metal 1’ and $\dfrac{4}{7} 14$ = 8 kg of ‘metal 2’.

The ratio of ‘metal 1’ and ‘metal 2’ in alloy ‘B’ is 5 :8.Therefore, we can say that 26 kg of alloy ‘B’ will contain $\dfrac{5}{13} 26$ = 10 kg of ‘metal 1’ and $\dfrac{8}{13} 26$ = 16 kg of ‘metal 2’.

Hence, total weight of ‘metal 1’ in the new alloy = 6 + 10 = 16 kg
Total weight of ‘metal 2’ in the new alloy = 8 + 16 = 24 kg

Therefore, the ratio of ‘metal 1’ and ‘metal 2’ in the new alloy. = 16 : 24 = 2 :3. Hence, option C is the correct answer.

6) Answer (A)

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

7) Answer (C)

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

8) Answer (C)

The remainder when $15^{23}$ is divided by 19 equals $(-4)^{23}$
The remainder when $23^{23}$ is divided by 19 equals $4^{23}$
So, the sum of the two equals$(-4)^{23}+(4)^{23}=0$

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9) Answer (D)

$\frac{(30^{65}-29^{65})}{(30^{64}+29^{64})} = ((30-29)*\frac{(30^{64}+30^{63}*29+….+29^{64})}{(30^{64}+29^{64})}$ , which is greater than 1 . Hence option D.

10) Answer (D)

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!. So equation becomes p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +… + 11! – 10!.  So p = 11! – 1! = 11! – 1.  p + 2 = 11! + 1 .So when it is  divided by 11! gives a remainder of 1. Hence, option 4.

11) Answer (D)

Cost of 20 units = 240+20b+400c
Cost of 40 units = 240+40b+1600c = 5/3 * (240+20b+400c) => 720+120b+4800c = 1200+100b+2000c
=> 480 = 20b + 2800c => 120 = 5b + 700c
Cost of 60 units = 240+60b+3600c = 3/2 (240+40b+1600c) => 480 + 120b + 7200c = 720 + 120b + 4800c
=> 240 = 2400c => c = 1/10 and b = 10
Let the number of items needed for max profit be k
CP = $240+10k+k^2/10$
SP = 30k
Profit = SP – CP = $30k – 240 – 10k – k^2/10$ = $20k – 240 – k^2/10$
Maximum when 20 – k/5 = 0 or k = 100
Profit = 2000 – 240 – 1000 = 760

12) Answer (E)

Let total cost price of Books and More = $RS. 100x$

It is given that Books and More have earned 20% profit overall

=> Total S.P. = $\frac{120}{100} \times 100x = 120x$

Thus, S.P. of music CDs = $\frac{35}{100} \times 120x = 42x$

S.P. of books = $42x + \frac{50}{100} \times 42x = 63x$

=> S.P. of DVDs = $120x – 42x – 63x = 15x$

40% profit is earned in music CDs and 25% profit in books.

=> C.P. of music CDs = $\frac{100}{140} \times 42x = 30x$

and C.P. of books = $\frac{100}{125} \times 63x = 50.4x$

=> C.P. of DVDs = $100x – 30x – 50.4x = 19.6x$

=> Loss made on DVDs = $19.6x – 15x = 4.6x$

$\therefore$ Loss % on DVDs = $\frac{4.6 x}{19.6 x} \times 100 = 23.46 \%$

13) Answer (B)

Selling price of first watch = 300
Profit = 10%
cost price = $\frac{300}{1.1}$
Selling price of second watch = 300
Loss = 10%
cost price = $\frac{300}{0.9}$

Total selling price of transaction= 600
Total cost price of transaction = $300(\frac{10}{11} + \frac{10}{9}) = 600 (\frac{100}{99})$
Loss = $600 (\frac{100}{99} – 1)$
%loss = $(600 (\frac{100}{99} – 1)) \div (600(\frac{100}{99})) \times 100 = 1$

14) Answer (C)

Let the principal amount = $P$ and rate of interest = $r \%$

Interest accumulated from 2004 to 2007 is Rs.10,000 and from 2004 to 2010 is Rs.25,000

Using, $C.I. = P[(1 + \frac{R}{100})^T – 1]$

=> $P[(1 + \frac{r}{100})^3 – 1] = 10,000$ ———-Eqn(I)

and $P[(1 + \frac{r}{100})^6 – 1] = 25,000$ ———–Eqn(II)

Dividing eqn(II) from (I), we get :

=> $\frac{P[(1 + \frac{r}{100})^6 – 1]}{P[(1 + \frac{r}{100})^3 – 1]} = \frac{5}{2}$

Let $(1 + \frac{r}{100})^3 = x$

=> $\frac{x^2 – 1}{x – 1} = \frac{5}{2}$

=> $2x^2 – 5x + 3 = 0$

=> $(2x – 3) (x – 1) = 0$

=> $x = \frac{3}{2} , 1$    $(x \neq 1)$ because then, r = 0

=> $(1 + \frac{r}{100})^3 = \frac{3}{2}$

Substituting it in eqn(I)

=> $P[\frac{3}{2} – 1] = 10,000$

=> $P = 10,000 \times 2 = 20,000$

15) Answer (C)

Amount man gets after 1 year

= $6000 + (\frac{6000 \times 5 \times 1}{100}) – 1200$

= $6000 + 300 – 1200 = 5100$

$\therefore$ Amount at the beginning of third year, i.e. after 2 years

= $5100 + (\frac{5100 \times 5 \times 1}{100}) – 1200$

= $5100 + 255 – 1200 = 4155$

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