Question Paper Quantitative Methods – I (131): July 2006
	· Answer all questions. · Marks are indicated against each question.

1.

The factors of 9x⁴ – 12x³ – 2x² + 4x + 1 are

(a) (x – 1)(x – 2)(x – 3)(x + 1)

(b) (x – 1)²(3x + 1)(2x + 1)

(c) (x – 1)²(3x + 1)²

(d) (3x + 1)(x – 1)(x + 1)(x – 4)

(e) (x – 3)²(x –2)(x + 1).

(2 marks)

< Answer >

2.

In how many ways can the letters of the expression X³Y²Z be arranged when written at full length?

(a) 20 (b) 40 (c) 60 (d) 80 (e) 100.

(1 mark)

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3.

In how many ways can seven girls be seated at a round table so that two particular girls are separated?

(a) 120 (b) 260 (c) 320 (d) 480 (e) 560.

(1 mark)

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4.

If , then find the value of .

(a) 12870 (b) 8008 (c) 5005 (d) 6435 (e) 924.

(1 mark)

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5.

A team of 11 is to be selected out of 14 players of whom 5 are bowlers. Find the number of ways in which this can be done so as to include at least 4 bowlers.

(a) 264 (b) 180 (c) 84 (d) 50 (e) 17.

(1 mark)

< Answer >

6.

How many odd numbers having four digits can be formed from {1, 2, 3, 4, 5, 6, 7, 8, 9}?

(a) 1860 (b) 1680 (c) 8610 (d) 6801 (e) 1068.

(1 mark)

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7.

If then the value of abc is

(a) 0 (b) 1 (c) b – c (d) c – a (e) a – b.

(2 marks)

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8.

The value of is equal to

(a) 0 (b) 1 (c) 1/2 (d) 3 (e) 1/3.

(1 mark)

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9.

The value of is equal to

(a) 0 (b) 1 (c) 20/3 (d) 15/8 (e) 25/7.

(2 marks)

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10.

The sum of a series containing 5 terms in A.P. whose first term is 5 and common difference
2 is

(a) 41 (b) 43 (c) 45 (d) 47 (e) 49.

(1 mark)

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11.

Three terms are in geometric progression. The second and third terms are 54 and 81 respectively, the first term is

(a) 15 (b) 24 (c) 36 (d) 45 (e) 48.

(1 mark)

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12.

The third term of an arithmetic progression (A.P.) is 11 and the sixth term is 20. What is the tenth term of the A.P.?

(a) 8 (b) 17 (c) 20 (d) 29 (e) 32.

(1 mark)

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13.

If 7 times the seventh term of an A.P. is equal to 11 times its eleventh term, then the eighteenth term of an A.P. is

(a) 0 (b) 46 (c) 68 (d) 84 (e) 88.

(1 mark)

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14.

If x + 2y = 13; 3x + y = 14 then find the value of x.

(a) 3 (b) 5 (c) 7 (d) 9 (e) 11.

(1 mark)

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15.

Find the roots of the equation 3x² + 10x – 32 = 0.

(a) 8 or – 16/3 (b) 2 or – 16/3 (c) 2 or – 5/3

(d) 3 or – 7/11 (e) 5 or – 7/11.

(1 mark)

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16.

What is the value of ?

(a) 0 (b) (c) 1 (d) 1/2 (e) 1/4.

(1 mark)

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17.

What is the value of ?

(a) 3/5 (b) 5/3 (c) (d) (e) 0.

(1 mark)

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18.

Find .

(a) (b) (c) (d) (e) .

(1 mark)

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19.

If , then find .

(a) (b) (c)

(d) (e)

(1 mark)

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20.

Which of the following describes a linear function?

(a) (b)

(c) (d)

(e) .

(1 mark)

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21.

Find the value of .

(a) log x + c (b) log(logx) + c

(c) log(log(logx)) + c (d) + c (e) x + c.

(2 marks)

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22.

A function y = f (x) is said to have a relative maxima at a point x = ‘a’ if

(a) (x) = 0 and (x) > 0 (b) (x) = 0 and (x) < 0

(c) (x) = 0 and (x) = 0 (d) (x) = 1 and (x) = 0

(e) (x) = 1 and (x) > 0.

(1 mark)

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23.

Which of the following is not true with regard to a logarithmic function y = log_b x?

(a) The function is not defined for zero or negative values of x

(b) The value of the function is zero when x = 1

(c) The value of the function is negative when x lies between 0 and 1, and b > 1

(d) The value of the function decreases as the value of x increases for b > 1

(e) The value of the function is positive when x > 1.

(1 mark)

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24.

The domain of a function, y = f (x), includes

(a) The values which may be assumed by y

(b) The values of x for which the function is defined

(c) The values of x for which the function is undefined

(d) The values of x which are greater than a pre-specified value

(e) The values of x which are less than a pre-specified value.

(1 mark)

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25.

The derivative of a constant is equal to

(a) 0

(b) 1

(c) –1

(d) A positive number if the constant is positive

(e) A negative number if the constant is negative.

(1 mark)

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26.

f(t) = 112t – t² – 600, then f(t) is maximum at t = ?

(a) 0 (b) 15 (c) 30 (d) 112 (e) 600.

(2 marks)

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27.

Which of the following statements is true with regard to the exponential function y = m.a^x ?

(a) The exponential curve falls from left to right as the values increase along the X-axis if m > 0 and 0 < a < 1

(b) The exponential curve falls from left to right as the values increase along the X-axis if m > 0 and a > 1

(c) The exponential curve is parallel to the X-axis if m > 0 and 0 < a < 1

(d) The exponential curve is parallel to the X-axis if m > 0 and a > 1

(e) If x = 0 then y = a.

(1 mark)

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28.

If x² – xy + y² = 1 then at (1, 1) is = ?

(a) (b) –1 (c) –3 (d) (e) .

(2 marks)

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29.

If then the value of is equal to

(a) (b) (c) (d) (e) .

(2 marks)

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30.

A finance company offers to pay a lump sum of Rs.16,000 at the end of 6 years to investors who deposit Rs.2000 annually for 6 years. What is the implicit interest rate?

(a) 11.00% (b) 11.43% (c) 11.78% (d) 11.81% (e) 12.00%.

(2 marks)

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31.

Which of following statements is/are false regarding interpolation?

I. It is a statistical technique.

II. It is a method of statistical estimation.

III. The figures obtained by interpolation are fairly correct as long as underlying assumptions hold good.

IV. It allows us to make insertions.

V. It allows us to forecast a value for some future date.

(a) Only (II) above (b) Only (IV) above

(c) Only (V) above (d) Both (I) and (III) above

(e) Both (II) and (V) above.

(1 mark)

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32.

Which of the following is a method of solving interpolation problems?

(a) Simplex method (b) Dual simplex method

(c) Graphical method (d) Census method

(e) Revised simplex method.

(1 mark)

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33.

A project requires an initial outlay of Rs.35 lakh and has the following cost flow projections:

Year	0	1	2	3	4
Cash flows (Rs.lakh)	35	15	10	10	20

Using interest factor tables and interpolation techniques find out the IRR of the project.

(a) 19.82% (b) 35.10% (c) 34.90% (d) 10.06% (e) 15.73%.

(2 marks)

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34.

Which of the following is/are true?

I. Linear approximation method is the only method, which can be used for interpolation.

II. Interpolation allows us to anticipate a value for some future date.

III. Extrapolation helps in completing the incomplete, lost or destroyed records.

IV. There must exist a functional relationship between an independent and dependent variable for interpolation to work.

(a) Only (I) above (b) Only (IV) above

(c) Both (I) and (II) above (d) Both (III) and (IV) above

(e) All (I), (II), (III) and (IV) above.

(1 mark)

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35.

Which of the following statements is/are false?

I. Graphical method of estimation is some times very accurate.

II. Graphical method can be used for both interpolation and extrapolation.

III. Graphical method of interpolation is fully reliable.

(a) Only (I) above (b) Only (II) above

(c) Only (III) above (d) Both (I) and (II) above

(e) Both (II) and (III) above.

(1 mark)

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36.

A company manufactures two products (A and B) and the profit per unit sold is Rs.300 and Rs.500 respectively. Each product has to be assembled on a particular machine, each unit of product A takes 12 minutes of assembly time and each unit of product B takes 25 minutes of assembly time. The company estimates that the machine used for assembly has an effective working week of only 30 hours (due to maintenance/breakdown). Technological constraints mean that for every five units of product A produced at least two units of product B must be produced. The company wants to maximize the profit per week. In the framework of linear programming problem, which of the following would be a true representative of the objective function? [Assume the production of A is x units and that of B is y units]

(a) 5x + 3y (b) 12x + 25y (c) 3x + 5y (d) 12x + 30y (e) 30x + 25y.

(1 mark)

< Answer >

37.

The CSTC, Calcutta State Transport Corporation, is considering the purchase of additional buses to expand its service. Two different models are being considered. A small model would cost Rs.10,00,000, carry 45 passengers, and operate at an average speed of 25 kilometers per hour over the existing bus routes. A larger model would cost Rs.15,00,000, carry 55 passengers, and operate at an average speed of 30 kilometers per hour. The transport authority has Rs.3,00,00,000 in its capital budget for purchasing new buses during the forthcoming year. However, the authority is also restricted in its expansion program by limitations imposed on its operating budget. Specifically, a hiring freeze is in effect and only 25 drivers are available for the foreseeable future to operate any new buses that are purchased. The transport authority wishes to determine how many buses of each model to buy to maximize additional capacity measured in passenger-kilometers-per hour while satisfying these constraints. Using the linear-programming framework, let be the number of small buses purchased and the number of large buses purchased. The constraint with regard to capital budgeting of the above mentioned problem would be

(a) (b)

(c) (d)

(e)

(2 marks)

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38.

In a linear programming problem the feasible region is bounded by the following system of inequalities:

Suppose the objective function is defined as P = 6x + 8y. The maximum value for P, subject to the constraints of the region of feasibility, would be

(a) 36 (b) 46 (c) 56 (d) 66 (e) 76.

(2 marks)

< Answer >

39.

In which of the following conditions, the graphical solution to a linear programming problem will not have multiple optimal solutions?

(a) One of the edges of the feasible region coincides with the horizontal axis

(b) One of the edges of the feasible region is parallel to the horizontal axis

(c) One of the edges of the feasible region coincides with the vertical axis

(d) One of the edges of the feasible region is parallel to the vertical axis

(e) None of the edges of the feasible region is parallel to the objective function.

(1 mark)

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40.

When all the constraints of a linear programming problem are expressed as equalities, then the problem is said to be in

(a) Quadratic form

(b) Non-linear form

(c) Standard form

(d) Dual form

(e) Classical form.

(1 mark)

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41.

For a company profit maximization may not be the only objective. It may have other objectives also viz., sound ecological management, networking in the neighborhood and maximizing market share. Which of the following approaches is best suited to find out the optimal solution for a company having multiple objectives?

(a) Linear programming problem in primal form

(b) Linear programming problem in dual form

(c) Integer programming problem

(d) Goal programming problem

(e) Using differential calculus to arrive at the optimal solution.

(1 mark)

< Answer >

42.

The probability of drawing either an ace or spade or both from a deck of cards is

(a) 1/13 (b) 2/13 (c) 3/13 (d) 4/13 (e) 1/4.

(1 mark)

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43.

Which of the following is not a major characteristic of linear programming?

(a) There must be alternative courses of action among which to decide

(b) An objective for the firm must exist

(c) The problem must be of maximization type

(d) Resources must be limited

(e) There should be a linear relationship in the objective function.

(1 mark)

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44.

Subject to the constraints

The feasible region to the above linear programming problem is

(a) (0,3) (b) (1,0)

(c) Data is insufficient (d) It has no feasible solution

(e) Unbounded solution.

(2 marks)

< Answer >

45.

In the following linear programming problem, the value of the objective function is

Max. Z = 3x₁ + 2x₂ + 5x₃

Subject to constraints

x₁ + 2x₂ + x₃ < 430

3x₁ + 2x₃ < 460

x₁ + 4x₂ < 420

x₁, x₂, x₃ > 0.

(a) Max. Z = 1220 (b) Max. Z = 1350

(c) Max. Z = 1480 (d) Max. Z = 1640

(e) Max. Z = 1820.

(2 marks)

< Answer >

46.

In solving the linear programming problem, if the type of constraint is greater than or equal to (>) then

(a) The extra variable called slack variable is added

(b) The extra variable called surplus variable is subtracted, and an artificial variable is added

(c) Only an artificial variable is added

(d) The coefficient of extra variables in the objective function of type Max.z is +M

(e) The coefficient of extra variables in the objective function of type Min.z is –M.

(1 mark)

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47.

Which of the following is not true with regard to the graphical procedure of solving a linear programming problem?

(a) The optimal solution always occurs at one of the corner points

(b) Any point within the feasible region satisfies all the constraints

(c) Only two decision variables can be used

(d) There may be multiple optimal solutions

(e) The objective function will have different values in case of multiple optimal solutions.

(1 mark)

< Answer >

48.

The pie chart has the advantage that

(a) The circle area is proportional to the largest data value

(b) It is a simple method of conveying approximate information about a small number of categories

(c) Information is conveyed accurately in the diagram

(d) It readily lends itself to a comparison of several different variables

(e) It is easy to present a set of observations through a pie diagram.

(1 mark)

< Answer >

49.

Which of the following is not the correct step for arranging data in the form of class intervals?

I. The lowest value should be included in the first class and the highest value should be included in the last class.

II. The adjacent classes should not have any interval in between.

III. The adjacent classes should overlap.

IV. Every item of data should be included in one and only one class.

(a) Only (I) above (b) Only (II) above

(c) Only (III) above (d) (I), (II) and (III) above

(e) All (I), (II), (III) and (IV) above.

(1 mark)

< Answer >

50.

For an asymmetric frequency distribution the mean and empirical mode are 5 and 8 respectively. The value of the median is

(a) 5.0 (b) 6.0 (c) 7.0 (d) 8.0 (e) 8.5.

(1 mark)

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51.

The average of a sample consisting of 65 items is 8 and the sum of the squares of the items is 4560. The sample standard deviation is

(a) 2.50 (b) 6.25 (c) 64 (d) 400 (e) 570.

(1 mark)

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52.

The coefficient of variation for the following frequency distribution is(approximately):

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70
Series A	10	16	34	38	24	18

(a) 33% (b) 44% (c) 55% (d) 66% (e) 77%.

(2 marks)

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53.

Which of the following is/are computed average(s)?

(a) Mean (b) Median

(c) Mode (d) Quartile Deviation

(e) Both (b) and (c) above.

(1 mark)

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54.

The larger the spread of scores around the mean

(a) The smaller the interquartile range

(b) The smaller the standard deviation

(c) The smaller the coefficient of variation

(d) The larger the standard deviation

(e) The smaller the variance.

(1 mark)

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55.

The coefficient of variation cannot be meaningfully used to compare the variability of two or more sets of data, when

(a) The standard deviation is zero for one or more sets of data

(b) The standard deviation is 1 for one or more sets of data

(c) The mean is zero for one or more sets of data

(d) The mean is 1 for one or more sets of data

(e) The mean and standard deviation are equal for one or more sets of data.

(1 mark)

< Answer >

56.

As study was conducted on the daily wages earned by the unskilled labourers in an industrial area. The observations from the survey are given below:

Daily wages (in Rs.)	Number of labourers
25 – 35	15
35 – 45	25
45 – 55	32
55 – 65	28
65 – 75	18
75 – 85	12

The median daily wage earned by the unskilled labourers is

(a) Rs.35.50 (b) Rs.45.25 (c) Rs.52.66 (d) Rs.60.56 (e) Rs.65.25.

(1 mark)

< Answer >

57.

The observations from a population are presented in the following frequency distribution:

Class interval	Number of observations
10 – 20	5
20 – 30	15
30 – 40	17
40 – 50	25
50 – 60	18
60 – 70	10
70 – 80	6
80 – 90	4
Total number of observations in the population	100

What is the standard deviation of the population?

(a) 14.25 (b) 17.23 (c) 20.18 (d) 22.16 (e) 24.36.

(2 marks)

< Answer >

58.

The geometric mean between two given quantities is equal to

(a) The sum of the arithmetic mean and harmonic mean between the two quantities

(b) The difference between the arithmetic mean and harmonic mean between the two quantities

(c) The geometric mean of the arithmetic mean and the harmonic mean between the two quantities

(d) The product of the arithmetic mean and the geometric mean between the two quantities

(e) The ratio of the arithmetic mean to the geometric mean between the two quantities.

(1 mark)

< Answer >

59.

The quartile deviation for the following frequency distribution is :

Class interval	5-6	6-7	7-8	8-9	9-10	10-11
Frequency	40	56	60	96	84	68

(a) 0.682 (b) 1.262 (c) 2.846 (d) 2.103 (e) 3.874.

(2 marks)

< Answer >

60.

Which of the following statements is/are true?

I. The semi-interquartile range may be used when the median is the measure of central tendency.

II. The semi-interquartile range may be used when the middle 50% of the scores are of primary importance.

III. The semi-interquartile range may be used when the distribution has extreme scores, which would affect the Standard Deviation disproportionately.

(a) Only (I) above (b) Only (II) above

(c) Only (III) above (d) Both (I) and (III) above

(e) All (I), (II) and (III) above.

(1 mark)

< Answer >

61.

Which of the following is /are the measures of dispersion?

I. Range.

II. Mean deviation.

III. Median.

IV. Deciles.

V. Standard deviation.

(a) Only (II) above (b) Only (V) above

(c) (I), (II) and (V) above (d) (II), (IV) and (V) above

(e) All (I), (II), (III), (IV) and (V) above.

(1 mark)

< Answer >

62.

Which of the following statement is not true?

(a) The arithmetic mean can often be misleading if the data fall in a homogeneous group

(b) The arithmetic mean can often be misleading if the data don’t fall in a homogeneous group

(c) Arithmetic mean does not convey any information about the spread or trend of data

(d) Arithmetic mean is not a suitable measure of central value in case of highly skewed distribution

(e) Arithmetic mean cannot be calculated for frequency distribution with open end classes.

(1 mark)

< Answer >

63.

Which of the following measures of dispersion enables meaningful comparison of variability of different sets of data?

(a) Range (b) Mean absolute deviation

(c) Standard deviation (d) Coefficient of variation (e) Variance.

(1 mark)

< Answer >

64.

There are three sets of data. The first set contains four quantities and their geometric mean is 1.252. The second set contains eight quantities and their geometric mean is 1.164. The third set contains five quantities and their geometric mean is 1.206. Each of the three sets of data is expanded by including one more quantity as follows:

The quantity 1.20 is included into the first set, the quantity 1.18 is included into the second set and the quantity 1.24 is included into the third set. What is the combined geometric mean of all the three sets of data after the expansion?

(a) 1.166 (b) 1.212 (c) 1.241 (d) 1.198 (e) 1.389.

(2 marks)

< Answer >

65.

A data set includes some quantities. The sum of reciprocals of the quantities in the data set is . The harmonic mean of the data set is . The data set is expanded by including the quantities 5 and 10 into it.

What is the harmonic mean of the expanded data set?

(a) (b) (c) (d) (e) .

(2 marks)

< Answer >

66.

A multiple choice test consists of three problems. For each problem, there are five choices, one of which is correct. One student comes totally unprepared and decides to answer by sheer guessing. What is the probability that he will answer at least one problem correctly?

(a) (b) (c) (d) (e) .

(2 marks)

< Answer >

67.

The first box contains one green ball and three blue balls. A second box contains two green balls and four blue balls. A third box contains three green balls and one blue ball. One of the three boxes is selected at random and a ball is randomly taken out of it.

What is the likelihood that the ball is green?

(a) (b) (c) (d) (e) .

(1 mark)

< Answer >

68.

From twenty tickets marked with the first twenty numerals, one ticket is drawn at random. What is the probability that the numeral marked on it is a multiple of 3 or 5?

(a) (b) (c) (d) (e) .

(1 mark)

< Answer >

69.

If six fair dice are rolled, what is the probability that each of the six numbers will appear exactly once?

(a) 0.0065 (b) 0.0154 (c) 0.0198 (d) 0.0215 (e) 0.0285.

(1 mark)

< Answer >

70.

One shot is fired from each of the three guns, where G₁, G₂ and G₃ denote the events that the target is hit by the first, second and third guns respectively. If P(G₁) = 0.5, P(G₂) = 0.6 and P(G₃) = 0.8 and G₁, G₂ and G₃ are independent of each other. The probability that at least two hits to be registered is

(a) 0.50 (b) 0.55 (c) 0.60 (d) 0.65 (e) 0.70.

(2 marks)

< Answer >

71.

If then the value of is

(a) 0 (b) 1 (c) 1/3 (d) 1/4 (e) 1/5.

(1 mark)

< Answer >

72.

A pair of fair dice is thrown. Find the probability of getting a sum of 7, when it is known that the digit in the first die is greater than that of the second.

(a) 5/12 (b) 1/5 (c) 1/12 (d) 5/8 (e) 7/10.

(2 marks)

< Answer >

73.

The manufacturing process of an article consists of two parts A and B. The probabilities of defect in parts A and B are 8% and 12% respectively. What is the probability that the assembled product will not have any defect?

(a) 80.9% (b) 76.5% (c) 74.02% (d) 70.25% (e) 68.57%.

(1 mark)

< Answer >

74.

If and then the value of is

(a) 0.60 (b) 1.93 (c) 0.10 (d) 1.15 (e) 2.32.

(1 mark)

< Answer >

75.

When five coins are tossed, Heads and tails show up on the coins. Find the probability of showing exactly two heads on the coins.

(a) 1/2 (b) 2/5 (c) 3/8 (d) 5/16 (e) 7/20.

(1 mark)

< Answer >

76.

If the occurrence of some events are dependent on the occurrence of an event A then the sum of all the joint probabilities in which the occurrence of event A is considered gives the

(a) Subjective probability of event A

(b) Classical probability of event A

(c) Conditional probability of event A

(d) Marginal probability of event A

(e) Relative frequency of occurrence of event A.

(1 mark)

< Answer >

77.

If events B and C are dependent on event A and P(A and B) = 0.30, P (A and C) = 0.20, and the dependent events B and C are mutually exclusive and collectively exhaustive, then P(C/A) is equal to

(a) 0.20 (b) 0.40 (c) 0.50 (d) 0.60 (e) 0.80.

(1 mark)

< Answer >

78.

A bin in a hardware store, contains 125 bolts and 200 nuts. One-fifth of the bolts and three-fourth of the nuts are defective. One item is picked randomly from the bin. What is the probability that the item is either a defective item or it is a nut?

(a) 92.3% (b) 84.6% (c) 69.2% (d) 53.8% (e) 38.5%.

(2 marks)

< Answer >

Suggested Answers
Quantitative Methods – I (131): July 2006

1.

Answer : (c)

Reason : If we substitute x = 1 in 9x⁴–12x³–2x²+4x+1 , the value of the expression will be 9(1)⁴ – 12(1)³ – 2(1)² + 4(1) + 1 = 0

That is, x = 1 or x–1 = 0 is one of the factors of this expression. The other factors are obtained by dividing the expression with x–1.

x-1) 9x⁴–12x³–2x²+4x+1 ( 9x³–3x²–5x–1

(–1) 9x⁴–9x³

------------------

–3x³–2x²

(–1) –3x³+3x²

--------------------

–5x²+4x

(–1) –5x²+5x

--------------------

–x+1

(–1) –x+1

--------------

0

------------

The other factor is 9x³–3x²–5x–1. To get other factors we substitute x = 1 in the above factor. The value of the expression will be

9(1)³–3(1)²–5(1)–1 = 0

That is, x–1 = 0 or x=1 is again a factor of 9x³–3x²–5x–1. We again divide

x-1) 9x³–3x²–5x–1 (9x²+6x+1

(–1) 9x³–9x²

-------------------

6x²–5x

(–1) 6x²–6x

------------

x–1

(–1) x–1

----------

0

----------

That is, 9x²+6x+1 is another factor. We find 9x²+6x+1 is identical to a²+2ab+b². That is, for a=3x and b=1, we have 9x²+6x+1 = (3x+1)²

Therefore the factors of 9x⁴–12x³–2x²+4x+1 are (x–1)(x–1)(3x+1)² or (x–1)²(3x+1)²

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2.

Answer : (c)

Reason : X³Y²Z at full length = XXX YY Z . In this there are 3X’s, 2Y’s and Z.

Therefore the no. of arrangements = .

Therefore (c) is the correct answer.

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3.

Answer : (d)

Reason : Let the two particular girls be taken together as one unit. Then the persons will be 6.

They can sit round the table in (6 – 1)! =5! ways .

For each of this arrangement, the two can be interchanged in 2! Ways.

Therefore, the total arrangement = 5!2!.

The arrangements that the two persons are separated = 6! – 5!2! = 480.

Therefore (d) is the correct answer.

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4.

Answer : (a)

Reason : implies n = 6 + 10 = 16.

.

Therefore (a) is the correct answer.

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5.

Answer : (a)

Reason : The only possibilities regarding the selections are

Bowlers others

i. 4 7

ii. 5 6

in case (i) 4 bowlers can be selected out of 5 bowlers in ways and 7 others can be selected out of 9 other players in ways.

Therefore the team 11 can be selected in.ways.

Similarly in the second case a team of 11 can be selected in . ways.

Therefore total number of ways of forming the team 11 = . + .=264.

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6.

Answer : (b)

Reason : Given {1, 2, 3, 4, 5, 6, 7, 8, 9}

Odd number set is {1, 3, 5, 7, 9}=5. If we place one of these five numbers first from right then only three positions are left i.e., they can be arranged in ways

Therefore number of odd four digit numbers can be formed in 5=1680 ways. Therefore (b) is the correct answer.

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7.

Answer : (b)

Reason :

Now add 1, 2 and 3 above

log a + log b + log c = k(b – c) + k(c – a) + k(a – b)

log a + log b + log c = k(b – c + c –a + a – b)

log a + log b + log c = k(0)

log a + log b + log c = 0

Þ log (abc) = 0

Þ log (abc) = log1

Þ abc = 1

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8.

Answer : (b)

Reason :

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9.

Answer : (d)

Reason :

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10.

Answer : (c)

Reason :

Sum = {2a + (n – 1) d} ´ = {2 ´ 5 + (5 – 1) ´ 2} x = {10 + 8} ´ = = 45

Where a = first term of an AP series n = total number of terms in AP series d = common difference of the series

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11.

Answer : (c)

Reason : The nth term in a geometric progression (G.P.) is given

t_n = ar^{n – 1}

Given: t₂ = ar = 54

t₃ = ar² = 81

\ = = r = = 1.5

First term of the G.P. = a = = = 36

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12.

Answer : (e)

Reason : t₃ = a + (3 – 1)d i.e., 11 = a + 2d (A)

t₆ = a + (6 – 1)d i.e., 20 = a + 5d (B)

Subtracting equation (A) from equation (B)

(20 – 11) = (a + 5d) – (a + 2d)

or 9 = 3d

or d = 9/3 = 3

\ a = 11 – (2 ´ 3) = 5

\ t₁₀ = a + 9d = 5 + (9 ´ 3) = 32.

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13.

Answer : (a)

Reason :

= a+ 10d ---à is the 11^th term.

Given 7 times the 7^th term of an A.P. is equal to 11 times. Its 11^th term

=>7a+42d=11a+110d

=> –68d =4a

=> a = –17d

=a+17d

= –17d+17d=0

\

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14.

Answer : (a)

Reason :

We have x+2y = 13 (1)

3x+y = 14 (2)

Express equation (1) in terms of x.

It will be x = 13–2y. Substitute this in equation (2)

We have 3(13-2y) + y = 14

39–6y+y = 14

–5y = –25

Y = 5

Substitute the value of ‘y’ in equation (1)

x + 2(5) = 13

x + 10 = 13

x = 3

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15.

Answer : (b)

Reason :

In the given quadratic equation a = 3, b = 10 and c = –32. We now substitute these values in the formula

or

x = 2 or

That is, the roots of the equation 3x²+10x–32=0 are

x= 2;

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16.

Answer : (e)

Reason :

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17.

Answer : (c)

Reason :

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18.

Answer : (a)

Reason :

.

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19.

Answer : (a)

Reason :

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20.

Answer : (c)

Reason : Linear function is of the form

So (c) is the correct answer.

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21.

Answer : (b)

Reason : let log x =t

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22.

Answer : (b)

Reason : At the point of relative maxima i.e at x = a, the tangent to the curve of the function is horizontal; hence, the slope of the function i.e.(x), is equal to zero. Further, at the point of relative maxima the second derivative of the function i.e. is less than zero.

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23.

Answer : (d)

Reason : According to the properties of logarithmic functions all the alternatives (a) through (d) are true. Alternative (d) is not true because the value of the logarithmic function increases as the value of x increases, for b > 1.

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24.

Answer : (b)

Reason : The domain of a function, y = f(x), includes the values of x for which the function is defined.

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25.

Answer : (a)

Reason : (a) The derivative of a constant is equal to 0.

(b), (c), (d) and (e) are all incorrect conclusions with regard to the derivative of a constant.

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26.

Answer : (c)

Reason : f (t) = 112t – t² – 600

(t) = 112 – ´ 2 t = 112 –

For maxima or minima f ¢(t) = 0

(t) = 0 Þ 112 – t = 0 or t = = 30

(t) = – which is always negative

Hence, f (t) is maximum at t = 30.

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27.

Answer : (a)

Reason : The exponential function falls from left to right as the values increase along the X-axis if m > 0 and 0 < a < 1. Hence (a) is true. (b), (c ) (d) and (e) are not true because they are not the characteristics of the exponential function.

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28.

Answer : (b)

Reason :

x² – xy + y² = 1

=

at (1,1) is = =

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29.

Answer : (c)

Reason :

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30.

Answer : (b)

Reason : We know that 16,000 = 2000 × FVIFA(K, 6)

Where FVIFA is the future value interest factor of an annuity, and K is the implicit rate.

FVIFA(K, 6) = 16000/2000 = 8

If we refer to the FVIFA tables,

At K = 11%, FVIFA(11, 6) = 7.913 and

At K = 12%, FVIFA(12, 6) = 8.115

So, K must be greater than 11% but lower than 12%.

For an ascent of (8.115 – 7.913), the ascent in rate is 1. for a required ascent of (8.000 – 7.913), the ascent in rate is

The implicit rate of interest =

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31.

Answer : (c)

Reason : Options (I), (II), (III), (IV) are true with regard to the interpolation. Option (v) is false because Extrapolation is used for forecasting purposes. Hence (c) is the answer.

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32.

Answer : (c)

Reason : The correct answer is (c) . Therefore variety of methods that can be used for interpolation such as graphical method, linear approximation, etc.,

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33.

Answer : (a)

Reason : let the IRR of the project be ‘r’

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34.

Answer : (b)

Reason : The following statements are true.

1. There are a variety of methods that can be used for interpolation such as graphical method, linear approximation, etc.

2. Interpolation helps in completing the incomplete lost or destroyed records.

3. Extrapolation allows us to forecast or anticipate a value for some future date.

4. There must exist a functional relation between an independent and dependent variable for interpolation to work.

Therefore (b) is the correct answer.

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35.

Answer : (c)

Reason : Graphical method of interpolation is simple, non-algebraical and not fully reliable. Therefore (c) is the correct answer.

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36.

Answer : (c)

Reason : The profit per unit sold is Rs.300 and Rs.500 for product A and product B respectively. The production of A is x units and that of B is y units. So, the profit function or objective function of the problem is 300x + 500y or .

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37.

Answer : (d)

Reason : A small bus costs Rs.10,00,000 and a large bus costs Rs.15,00,000. Total budget for this purpose is Rs.3,00,00,000. As X₁ number of small buses and X₂ number of large buses are purchased at optimum solutions so total cost of purchase should limit within budget. This constraint can be represented as

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38.

Answer : (d)

Reason : The constraints are

and the terminal line of feasibility is …….(i)

and the terminal line of feasibility is …….(ii)

So, the vertices are

(6, 0), (0, 7), and (3, 6).

The value of P at different vertices are

(6, 0):

(0, 7):

(3, 6):

As the value of P is maximum at (3, 6) so, optimum point is (3, 6) and maximum value of P is 66.

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39.

Answer : (e)

Reason : A graphical solution to a linear programming problem will have multiple optimal solutions if the objective function is parallel to an edge of the feasible region which is in the direction of the optimal movement of the objective function. If none of the edges of the feasible region is parallel to the objective function then, the possibility of multiple optimal solution does not arise.

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40.

Answer : (c)

Reason : When all the constraints of a linear programming problem are expressed as equalities, then the problem is said to be in standard form.

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41.

Answer : (d)

Reason : The company has more than one goal. Hence it has to formulate a single objective function by taking all the goals into consideration as per their importance. This type of model is called goal programming problem.

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42.

Answer : (d)

Reason : Let E₁ be the drawing an ace and E₂ be the drawing a spade , then E₁ and E₂ are not mutually exclusive since the ace of spade can be drawn.

P{E₁ + E₂ } = P{ E₁}+ P{ E₂ } – P{ E₁ E₂ } = 4/52 + 13/52 – 1/52 = 16/52

= 4/13.

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43.

Answer : (c)

Reason : In case of linear programming the problem may be maximization or minimization type.

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44.

Answer : (d)

Reason :

there is no feasible region for the above problem.

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45.

Answer : (b)

Reason :

			3	2	5	0	0	0
Basic variables	CB	XB	X1	X2	X3	S1	S2	S3	Min Ratio (XB/Xt)
S1	0	430	1	2	1	1	0	0	430/1 = 430
¬ S2	0	460	3	0	2	0	1	0	460/2 = 230 ¬
S3	0	420	1	4	0	0	0	1	–
X1=X2=X3=0	Z = 0		–3	–2	–5^*	0	0 ¯	0	¬Dj
¬ S1	0	200	– ½	2	0	1	½	0	200/2 = 100 ¬
® X3	5	230	3/2	0	1	0	½	0	–
S3	0	420	1	4	0	0	0	1	420/4 = 105
X1=X2=S2=0	Z = 1150		9/2	–2^*	2	0 ¯	5/2	0	¬ Dj
X2	2	100	– ¼	1	0	½	– ¼	0
X3	5	230	3/2	0	1	0	½	0
S3	0	20	2	0	0	–2	1	1
X1=S2=S3=0	Z = 1350		4	0	0	1	2	0	¬Dj > 0

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46.

Answer : (b)

Reason : In solving the linear programming problem, if the type of constraint is greater than or equal to (>) then the extra variable called surplus variable is subtracted, and an artificial variable is added. Therefore (b) is the correct answer. The coefficient of extra variable in the objective function of type Max.z is –M and the coefficient of extra variables in the objective function of type Min.z is +M.

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47.

Answer : (e)

Reason : The objective function will have the same value even when there are multiple optimal solutions. All other alternatives are true

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48.

Answer : (b)

Reason : It is a simple method of conveying approximate information about a small number of categories

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49.

Answer : (c)

Reason : Adjacent classes should not overlap. Hence (III) is not correct.

Alternatives (I), (II) and (IV) are correct.

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50.

Answer : (b)

Reason : Mode = 3 ´ Median – 2 ´ Mean

Let the median be x

\ 8 = 3x – 2 ´ 5

or 3x = 18

or x = 6

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51.

Answer : (a)

Reason : Sample standard deviation =

=

=

=

= 2.50.

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52.

Answer : (a)

Reason :

C.I	xi	ui=(xi-A)/c	fi	fiui	fiui^2
10-20	15	-3	10	-30	90
20-30	25	-2	16	-32	64
30-40	35	-1	34	-34	34
40-50	45=A	0	38	0	0
50-60	55	1	24	24	24
60-70	65	2	18	36	72
			140	-36	284

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53.

Answer : (a)

Reason : Mean is a computed average, while others are positional averages.

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54.

Answer : (d)

Reason : The smaller the spread of scores around the mean the smaller will be the interquartile range, the smaller the standard deviation, and smaller will be the coefficient of variation.

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55.

Answer : (c)

Reason : Coefficient of variation = (standard deviation / mean) × 100

(a) A standard deviation equal to zero implies that there is no deviation in the data set. The same will be reflected by the c.v. provided mean is not equal to zero.

(b) Even when the standard deviation is 1 the c.v. can be meaningfully used for comparison of variability provided mean is not equal to zero.

(c) Hence it cannot be meaningfully used for comparison of variability when mean of one or more data sets is zero.

(d) When the mean is equal to 1, the c.v. can be meaningfully used for comparison of variability.

(e) When the mean and standard deviation are equal for one or more sets of data, the c.v. can be meaningfully used for comparison of variability.

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56.

Answer : (c)

Reason : Median =

Class	Frequency (f)	Cumulative frequency
25 – 35	15	15
35 – 45	25	40
45 – 55	32	72
55 – 65	28	100
65 – 75	18	118
75 – 85	12	130
	130

Position of the median = = 65.5th item

From above can be seen that the median class is 45 – 55.

\ N = 130

L_m = 45

f_m = 32

F = 40

W = 10

\ Median wage = 10 + 45 = Rs.52.66 (approx.)

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57.

Answer : (b)

Reason : Standard deviation of population, s =

Class	Mid-point(x)	Frequency (f)	fx
10-20	15	5	75	–31	961	4805
20-30	25	15	375	–21	441	6615
30-40	35	17	595	–11	121	2057
40-50	45	25	1125	–1	1	25
50-60	55	18	990	9	81	1458
60-70	65	10	650	19	361	3610
70-80	75	6	450	29	841	5046
80-90	85	4	340	39	1521	6084
		100	4600			29700

Mean,

Standard Deviation = = 17.23.

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58.

Answer : (c)

Reason : The geometric mean between two given quantities is equal to the geometric mean of the arithmetic mean and the harmonic mean between the two given quantities. (a), (b), (d) and (e) are all incorrect with regard to the geometric mean between two given quantities.

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59.

Answer : (b)

Reason :

C.I	Frequency	Cumulative frequency
5-6	40	40
6-7	56
7-8
8-9	96
9-10
10-11	68	404
	N=404

Here N=404

This observation will fall in class (7-8)

\

C = 1

=7+0.08=7.08

Similarly

=9 + 0.607

=9.607

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60.

Answer : (e)

Reason : I. This is true. The semi-interquartile range may be used when the median is the measure of central tendency.

II. This is true. The semi-interquartile range may be used when the middle 50% of the scores are of primary importance.

III. This is true. The semi-interquartile range may be used when the distribution has extreme scores, which would affect the Standard Deviation disproportionately.

Therefore (e) is the correct answer.

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61.

Answer : (c)

Reason : Range, mean deviation and standard deviation are the measures of dispersion.

Median and deciles are the measures of central values.

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62.

Answer : (a)

Reason : The options (b), (c), (d), (e) are the demerits of arithmetic mean. Therefore the are correct and option (a) is wrong.

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63.

Answer : (d)

Reason : Coefficient of variation is a relative measure of dispersion which expresses standard deviation as a percentage of mean. Hence it is used to compare different sets of data having unequal means and standard deviations. All others are absolute measures of dispersion and are not comparable across different sets of data.

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64.

Answer : (d)

Reason : The combined geometric mean of the three sets of data after the expansion,

=

=

= 1.19802 @ 1.198

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65.

Answer : (e)

Reason : Harmonic mean of a set of ‘n’ quantities, H =

Given: = H = =

From above, n = H.

\ n = = 3

Harmonic mean after the expansion = =

= .

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66.

Answer : (b)

Reason : Here, the probability of choosing the right one is given by 1/5 and the probability of at least one is correct is given by the sum of the probability one correct, probability of two correct and the probability of three correct. As it is problem of Binomial distribution, it may given as:

³C₁ ´ p ´ (1 – p)² + ³C₂ ´ p² ´ (1 – p) + ³C₃ ´ p³

= .

OR,

Probability of answering a problem correctly =

Probability of answering a problem wrongly =

Probability of answering all three problems wrongly =

Probability of answering at least one correctly =

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67.

Answer : (c)

Reason : The box is selected at random. So each box is equally likely to be selected.

\ Probability of selecting a green ball from the first box = =

Probability of selecting a green ball from the second box = =

Probability of selecting a green ball from the third box = =

\The likelihood that the ball is green = + + = .

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68.

Answer : (d)

Reason : Probability that the numeral is a multiple of 3 (event A) =

Probability that the numeral is a multiple of 5 (event B) =

Probability that the numeral is a multiple of both 3 and 5 (event A and B) =

\ Probability that the numeral is a multiple of 3 or 5,

P(A or B) = P(A) + P(B) – P(A and B) = .

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69.

Answer : (b)

Reason : In this experiment the sample space contains 6⁶ elements, because each die can take any one of the six possible values from 1 through 6. For all numbers to be different, the first die can take 6 values, but the second can take only five values and so on until the sixth die has only one possible value that it can take. Hence, there are 6! Ways to get all six different values.

Thus, P(Each of the six number will appear exactly once) = 6!/6⁶ = 720/46656 = 0.0154

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70.

Answer : (e)

Reason : We are given that P(G₁) = 0.5, P(G₂) = 0.6 and P(G₃) = 0.8 and hence their complementary events are given by:

.

The event of exactly two hit may be the combinations of the following mutually exclusive events:

(i) happens (ii) happens

(iii) happens

or (iv) happens.

Therefore, from the addition theorem of probability, the required probability may be calculated as:

= 0.5 ´ 0.6 ´ 0.2 + 0.5 ´ 0.4 ´ 0.8 + 0.5 ´ 0.6 ´ 0.8 + 0.5 ´ 0.6 ´ 0.8 = 0.06 + 0.16 + 0.24 + 0.24

= 0.70

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71.

Answer : (c)

Reason :

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72.

Answer : (b)

Reason : The sample space of the experiment of throwing a pair of fair dice contains 6 X 6 = 36 equally likely event points.

Let A denote the event that the sum of the digits in the two dice is 7. again, let B be the event that the digit in the first die is greater than that of the second. Then,

A = { (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

i.e., event A contains 6 equally likely event points

and B = { (2, 1), (3, 1), (4,1), (5, 1), (6, 1), (3, 2), (4, 2), (5, 2), (6,2), (4, 3), (5, 3), (6, 3), (5, 4),

(6, 4), (6, 5)}

i.e., event B contains 15 equally likely event points.

Now we have to find the value of .

i.e., the event contains 3 equally likely event points.

Therefore, we have,

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73.

Answer : (a)

Reason :

Let A and B denote the events that an article is defective due to defect in parts P and Q respectively. We have ,

and

Therefore

Therefore the probability that the assembled product will not have any defect

( since and are independent )

.

Therefore, the probability that the assembled product will not have any defect is 80.9%.

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74.

Answer : (d)

Reason :

Given

therefore (d) is the correct answer.

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75.

Answer : (d)

Reason :

In a throw of five coins, total number of ways of showing up of heads, tails=

Let E be the event of getting exactly two heads on the 5 coins i.e., getting two heads, 3 tails on the coins.
So the number of favorable cases to E = m =

.

Therefore (d) is the correct answer

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76.

Answer : (d)

Reason : If events B and C are dependent on event A then P(A and B) + P(A and C) = P(A).

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77.

Answer : (b)

Reason : P(A and B) = 0.30 P(A and C) = 0.20

\ P(A) = P(A and B) + P(A and C) = 0.30 + 0.20 = 0.50

P(A and C) = P(A) . P(C/A) or P(C/A) = = = 0.40

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78.

Answer : (c)

Reason : Probability that the chosen item is either defective or it is nut

= P (item is defective) + P( item is a nut) – P( item is a defective nut)

Total number of items in the bin = 125 + 200 = 325

Total number of defective items = = 175

Number of nuts = 200

Number of defective nuts = = 150

P (Item is defective or item is a nut) =

i.e., 69.2%.

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