# FYBCOM Maths Sem 2 MCQ Questions with Answers: Mumbai University 2021

*FYBCOM Maths Sem 2 MCQ Questions with Answers*

### F.Y.B.Com Sem-II MCQ Mathematical & Statistical Techniques for April,2021

Question | Answer1 | Answer2 | Answer3 | Answer4 | Correct Answer | |

If y = e^{x} + 5x – 10 then dy / dx is —– |
_{e}x |
e^{x} + 5 |
5 | 5x | Answer2 | |

If y = log 6 then dy/dx is | log 6 | 6 | 1/6 | 0 | Answer4 | |

If y= (5x+1)(5x-1) then dy/dx is | 2x | 50 x | 5 x | 25 x | Answer2 | |

If elasticity of demand is 1 , then demand is said to be —– | unelastic | inelastic | perfectly elastic |
directly proportional to |
Answer4 | |

If D is the number of units demanded and the price is p = 50 + 2D, then the average revenue when D = 10 is Rs.—– | 35 | 70 | 100 | 10 | Answer2 | |

If y = u + v , then dy/dx is | v du/dx + u dv/dx –k | du/dx +dv/dx | k. du/dx. dv/dx |
v du/dx + u dv/dx +k |
Answer2 | |

If y = u v , then dy/dx is | [v du/dx + u dv/dx ] | (v du/dx)- (u dv/dx ) |
0 | 1 | Answer1 | |

If y = 2x^{2} + 3x + 20 then dy/dx is |
4x | 4x + 3 | 20 | 3x | Answer2 | |

If D is the number of units demanded and the price is p = 30 + D , then the revenue when D =2 is Rs.—– | 64 | 70 | 60 | 54 | Answer1 | |

A break- even point is the stage when—– | everything breaks down | the market crashes | the demand and supply balance |
Interest is null | Answer3 | |

If C = x^{2} + 10 x + 10 where x is number of units produced, find marginal cost at x = 2 |
10 | 14 | 2 | 20 | Answer2 | |

The demand function is D=1+4p-p^{3}. Then the rateof change of Demand w.r.t. price at p=1, is |
0 | -1 | 1 | 10 | Answer3 | |

The cost of manufacturing x toys is C=x^{2} -5x+7. Then the marginal cost of manufacturing 10 toys is Rs.— |
57 | 15 | 10 | 5 | Answer2 | |

The demand function of a commodity is p=3+5D-D^{2}, where p is the price. Then the rate at which its price is changing when demand is 5 is—- |
10 | 5 | -10 | -5 | Answer4 | |

The point of maxima for the function f(x) =3+4x-x^{2} is x = —– |
2 | 1 | 3 | 4 | Answer1 | |

If the Profit Function in lakhs, for selling x tons of good is p(x)= – x^{2} + 8x+28, then at what value of x, we have max profit in lakhs. |
2 | 4 | 3 | 1 | Answer2 | |

For the demand function D =2 – 3p + p^{2} , the demand when the price p= 5, is —— |
10 | 15 | 12 | 20 | Answer3 | |

The total cost function of producing certain article is C= x^{2} – 5x +100. So the average cost of producing 10 such items is— |
15 | 10 | 20 | 25 | Answer1 | |

The demand for a certain goods in the market is D = 10 – p^{2} , where p is its price . So its total revenue when the price is Rs. 3 per unit is |
3 | 1 | 0 | 100 | Answer1 | |

If elasticity of demand is zero , then demand is said to be —– | perfectly elastic | inelastic | elastic | directly proportional to price |
Answer1 | |

If y= 10 x^{2} + 7x + 1000 then dy/dx is |
20x | 20 x + 7 | 1000 | 7 | Answer2 | |

If y= 6^{3} then dy/dx is— |
216 | 0 | log 6 | 6 | Answer2 | |

If y= 5^{x} + 15 x + log 5 then dy/dx is—– |
15x + log 5 | log 5 | 5^{x} log 5 + 15 |
1 | Answer3 | |

If y= x^{3} + 4x then dy/dx is—– |
3x + 4 | 3x ^{2} + 4 |
4 | 3x^{2} |
Answer2 | |

If y= 5 log x + 15 x then dy/dx is—— | 1/x | 1 | 5/x + 15 | 5x | Answer3 | |

If y = 5 x^{8} + log x then dy/dx is —- |
log x | 40 x | 40x^{7} + 1/x |
40 x^{7} |
Answer3 | |

If 0 < ղ( elasticity of demand) < 1 , the demand is said to be —– | inelastic | perfectly elastic | directly proportional to price |
elastic | Answer1 | |

If ղ( elasticity of demand) > 1 , the demand is said to be —– |
inelastic | perfectly elastic | directly proportional to price |
elastic | Answer4 | |

The cost function of a commodity is given by C = 10x^{2} – 400 x + 100 , x being the quantity produced. Find the quantity x for which the cost is minimum. |
30 | 5 | 10 | 20 | Answer4 | |

The total Revenue function of a commodity is given by R = -5x^{2} + 40 x + 300 , x being the quantity demanded . Find the quantity x for which the revenue is maximum. |
5 | 4 | 3 | 1 | Answer2 | |

If y = 15 x^{2} , then d^{2} y / d^{2} x is —— |
15 | 30 | 30 x | 0 | Answer2 | |

If y = e ^{x} + 10x^{2} , then d^{2}y / d^{2}x is —— |
_{e}x |
10 x | 20 x | e^{x} + 20 |
Answer4 | |

If the demand function is D = 25 – 3p ,where D is demand and p is price, find the elasticity of demand at p = 5 | 0.50 | 1.50 | 2 | 2.50 | Answer2 | |

If the demand function is D = 10 – 2p ,where D is demand and p is price, find the elasticity of demand at p = 2 | 2/3 | 1/3 | 1 | 1/4 | Answer1 | |

If the Revenue function is given by R = D^{2} – D + 10 . Find the marginal revenue if the number of units demanded (D) = 1. |
7 | 6 | 1 | 4 | Answer3 | |

If the Revenue function is given by R = 10 D^{2} + 10 D – 2. Find the marginal revenue if the number of units demanded (D) = 20 |
400 | 10 | 410 | 450 | Answer3 | |

If the demand function is given by p = 20 + 10D. Find the average revenue if the number of units demanded (D) = 4. | 60 | 100 | 65 | 250 | Answer1 | |

If the cost function of manufacturing x items is given by 5x^{2} +10 x+ 20. Find the average cost if 2 items are made( x = 2) |
60 | 30 | 20 | 40 | Answer2 | |

A function is given as f(x) = 10 x^{2} + 5x – 10 , find its value at x = 1 |
-5 | 5 | 0 | 10 | Answer2 | |

A function is given as f(x) = 6x^{2} – 2x + 15 , find its value at x = 2 |
30 | 5 | 35 | 10 | Answer3 | |

The simple interest at 20% on Rs. 60,000 for 5 years is Rs.—- |
60,000 | 1,20,000 | 50,000 | 65,000 | Answer1 | |

A bank promises to double the principal invested by their customers in 10 years. What is the rate of simple interest offered by the |
8% | 10% | 12% | 14% | Answer2 | |

The compound interest for 3 years on Rs. 15,000 at the rate of 8% per year, calculated annually is approx. Rs.—- | 5,000 | 3,896 | 3,500 | 2,950 | Answer2 | |

The difference between the compound interest and the simple interest on a principal P, placed at 10% p.a. , compounded annually for 2 years is Rs. 50, then the principal P is Rs. —– | 300000 | 400000 | 5000 | 600000 | Answer3 | |

A fixed sum kept on compound interest got Rs. 800 in the first year and Rs. 864 in the next year. Find the rate of interest , when compounded annually. | 9% | 7% | 8% | 10% | Answer3 | |

A sum of Rs. 25000 accumulated to Rs. 55000 after 12 years. Find the rate of S.I. p.a. | 30% | 10% | 40% | 50% | Answer2 | |

The Compound interest at 10 % on Rs. 40,000 for 2 years is Rs.— |
8,000 | 9,500 | 8,400 | 12,000 | Answer3 | |

After how many years, a sum of Rs. 6400 accumulated to an amount of Rs. 9280 at 9% p.a. rate of simple Interest? | 12 years | 13 years | 5 years | 10 years | Answer3 | |

Ajay opened a recurring deposit account of Rs. 2,000 at the end of each year for 2 years. What is the accumulated amount if the interest is compounded at 10% p.a.? | Rs. 3,800 | Rs.2,500 | Rs. 4,200 | Rs.3,545 | Answer3 | |

The approximately present value of an immediate annuity of Rs. 500/- per year for 4 years with interest compounded at 11 % p.a. | Rs.2,500 | Rs.1,551 | Rs.2,840 | Rs.3,400 | Answer2 | |

Priya took some money from his friend at simple interest of 6 % per annum. She returned her friend Rs.1,180. After how much time did Priya return the money if she borrowed Rs. 1,000? | 3 years | 4years | 2 years | 6years | Answer1 | |

Manisha took a loan from a bank for 4 years at 12 % p.a. rate of simple interest. She paid Rs.1,440 as interest on a loan taken by her. What was the amount she took as loan? |
Rs.3,000 | Rs.5,500 | Rs.6,400 | Rs.3,880 | Answer1 | |

Suresh invested Rs. 500 in SBI for 2 years. He also invested Rs. 300 in ICICI for 4 years. At the end he received Rs. 220 in all from banks as total simple interest. What must have been rate of interest? | 10% | 11% | 12% | 9% | Answer1 | |

An amount of Rs.5,000 was borrowed by Anjali , at a simple interest of 15 % p.a. . She returned the amount with interest after 4 years. Calculate the total amount returned by her. |
Rs.5,000 | Rs.6,000 | Rs.8,000 | Rs.8,500 | Answer3 | |

The simple interest at 14 % on Rs. 200 for 3 years is Rs—– | 58 | 84 | 45 | 100 | Answer2 | |

The simple interest at 9 %, for a fixed deposit at the end of 6 years was Rs. 2,700, then the principal of the fixed deposit was Rs.—- | 5,000 | 5,500 | 4,800 | 3,400 | Answer1 | |

At what rate a principal of Rs. 5,000 be put for 4 years, so as to get Rs.2,400 as interest on Maturity ? | 6% | 12% | 10% | 8% | Answer2 | |

When the interest is compounded annually , the maturity amounts of a fixed deposit obtained at the end of one year using the compound interest as well as the simple interest are identical. | TRUE | FALSE | Partially true | Depends on the principal | Answer1 | |

I keep Rs. 100 for simple interest in a bank at 10% for 2 years. My friend keeps Rs. 100 for compound interest at 10% for 2 years. If his interest is compounded annually, what is the difference in our interests on maturity? | Rs.100 | Rs.1 | Rs.10 | Rs.1000 | Answer2 | |

A person borrowed Rs 15,000 at 12% interest p.a. If he is supposed to return the money within 1 year, then his EMI using Flat Rate of Interest will be Rs.— |
Rs.2,500 | Rs.1,800 | Rs.2,000 | Rs.1,400 | Answer4 | |

Ruby borrowed Rs. 60,000 at 8 % interest p.a. If she is supposed to return the money within 2 years, then her EMI using Flat Rate of Interest will be Rs.— |
3,600 | 2,000 | 2,900 | 900 | Answer3 | |

The future value of Rs. 25,000 kept in a bank, after 1 year at a 15% rate of compound interest p.a. is Rs.—- | 28,750 | 30,000 | 32,400 | 25,300 | Answer1 | |

The future value of Rs.45,000 kept in a fixed deposit account, after 2 years at 12 % rate of compound interest p.a. is Rs.—- | 40,000 | 56,448 | 42,000 | 35,000 | Answer2 | |

The difference between the compound interest and the simple interest on a principal P, placed at 10% p.a. , compounded annually for 4 years is Rs. 1282, then the principal P is Rs. —- | 20,000 | 10,000 | 15,000 | 25,000 | Answer1 | |

A sum of Rs. 12,000 accumulated to Rs. 17,280 at 20%p.a.compound interest. Find the period. | 1year | 3year | 2year | 4year | Answer3 |

A sum of Rs.24,000 accumulated to Rs. 34,560 after a certain period , at 20 % p.a. rate of compound interest p.a.. Find the period | 1 year | 4 year | 3 year | 2 year | Answer4 |

Find the approx present value of Rs.15,000 payable 3 years hence ,if the interest is compounded annually at 9 % p.a. | Rs.15,700 | Rs.16,450 | Rs.11,583 | Rs.16,000 | Answer3 |

Preeti kept a certain amount in a fixed deposit in a bank which offered 10% rate of interest , compounded annually. At the time of maturity , after 4 years she received Rs. 73,205. Find the Principal amount | Rs.50,000 | Rs.80,000 | Rs.75,000 | Rs.85,000 | Answer1 |

Ramesh deposited Rs. 8000 at the end of every year in a bank for 4 years. How much amount of annuties he received at the end if the rate of interest being 8 % p.a. | Rs. 32,000 | Rs.30000 | Rs.25,000 | Rs.36,050 | Answer4 |

Radhika opened a recurring deposit in a bank for 6 years with payment of Rs. 5000. Find the total amount of annuties she will receive at the end of period with 9 % p.a. | Rs.37,617 | Rs.30,000 | Rs.25,000 | Rs.27,000 | Answer1 |

Find the present value of an annuity of Rs. 2,000,paid at the end of each year for 4 years , at 11 % compounded annually. | Rs.8,000 | Rs.6,205 | Rs.8,200 | Rs.9,000 | Answer2 |

_is an interest earning fund , in which equal deposits are made at equal interval of time . | Mutual fund | Sinking Fund | EMI | Loan | Answer2 |

Find the approx. present value of an annuity of Rs. 2,000 paid at the end of each year for 3 years , at 6 % compounded annually | Rs.8,760 | Rs.5,346 | Rs.6,700 | Rs.7,500 | Answer2 |

Amita deposited Rs. 5000 at the end of every year in a R. D. account for 3 years. How much amount of annuties she received at the end if the rate of interest being 12 % p.a. | Rs.10,300 | Rs.14,000 | Rs.12,200 | Rs.16,872 | Answer4 |

Puja took a loan of Rs.40,000 to be repaid in 2 year at 6 % p.a. . Find the approx. EMI by Flat interest rate method. |
Rs.3,500 | Rs.4,400 | Rs.2,500 | Rs.1,867 | Answer4 |

Mansi takes a loan of Rs. 2,40,000 from a bank for a period of 10 months. Compute the EMI at 10 % p.a.by Flat interest rate method. | Rs. 18,200 | Rs.30,000 | Rs.26,000 | Rs.25,500 | Answer3 |

Rajesh takes a loan of Rs.10,000 from a financial institution for a period of 4 months. Compute the EMI at 12 % p.a.by Flat interest rate method. |
Rs.2,600 | Rs.5,000 | Rs.3,500 | Rs.7,200 | Answer1 |

For EMI , rate of interest should be —- | per annum | per quarter | per month | per half year | Answer3 |

The sum invested /borrowed is called the —— | Interest | accumulated amount |
EMI | principal | Answer4 |

A series of equal payments made at equal interval of time is called the —– |
Principal | Amount | Annuity | Interest | Answer3 |

If there exit a relation between a pair of variables ( x and y) then it is called |
correlation | A.M. | G.M. | Median | Answer1 |

A statistical data consisting of two variables is called a |
multivariate | bivariate | univariate | polynomial | Answer2 |

If both variables increase or decrease together then there is correlation |
negative | no | positive | perfect negative | Answer3 |

The correlation coefficient lies between | ± 1 | ± 2 | ± 3 | 0 & 1 | Answer1 |

A pictorial representation of the correlation is called |
scatter diagram | bar diagram | histogram | pie chart | Answer1 |

For Perfectly positive correlation r = | -1 | -2 | 1 | -3 | Answer3 |

If the value of Pearson’s coefficient of correlation is – 0.93, it can be concluded that there is | High degree of Positive correlation | Perfect negative correlation |
No correlation | High degree of negative correlation | Answer4 |

For positive correlation if x increases then y increases or if x decreases then —– | y increases | y decreases | no change in y | y increases and decreases both | Answer2 |

If changes in x does not affect the changes in y then there is correlation | positive | negative | no | perfect positive | Answer3 |

The relationship between number of strikes of employees of a company and its production is a example of —— correlation |
positive | negative | perfect positive | no | Answer2 |

The relationship between height and weight of a student is a example of correlation | negative | perfect positive | no | positive | Answer4 |

The relationship between I. Q. of a student and price of petrol is a example of correlation | positive | No | negative | Perfect positive | Answer2 |

If the value of Pearson’s coefficient of correlation is 1 , it can be concluded that there is correlation |
Imperfect negative | Perfect negative | perfect positive | imperfect positive | Answer3 |

If the value of Pearson’s coefficient of correlation is 0 , it can be concluded that there is —correlation |
negative | positive | no | perfect positive | Answer3 |

The coefficient of rank correlation( R ) between marks given by judge A and judge B to a series of one act play in a drama competition is 0.75. If ∑ d ^{2} = 30 then number of the play group(n) is—- |
9 | 11 | 12 | 13 | Answer1 |

If n= 8 and ∑ d^{2} = 30 then Spearman’s rankcoefficient of correlation( R ) is…… |
0.85 | 1 | 0.64 | 2 | Answer3 |

If n= 7 , ∑ d^{2} = 45 and ∑ c.f. = 1 then Spearman’s rank coefficient of correlation( R ) is…… |
0.4 | 0.18 | 0.5 | 0.8 | Answer2 |

If n = 4 , ∑xy = 88 , ∑x = 16 , ∑y = 20, ∑x^{2} = 80and ∑ y ^{2} = 116 then Karl Pearson’s coeff. of |
0.4 | 1 | 0.2 | 0.5 | Answer4 |

If a rank is repeated twice then value of correction factor (c.f.) is —– |
1 | 2.00 | 0.5 | 1.5 | Answer3 |

For the data x : 12 , 15 , 11 , 10 ; y: 25,30,28,29 . The rank coeff. of correlation is —- – |
0.2 | 0.3 | 0.4 | 0.5 | Answer1 |

If two variables of x & y are highly correlated then y can be estimated for a given value of x using regression equation of | y on y | y on x | x on x | x on y | Answer2 |

x can be estimated for a given value of y using regression equation of If the two variables of x & y are highly correlated. | x on x | y on y | x on y | y on x | Answer3 |

If the two regression coefficients are positive then the value of the correlation coefficient must be | positive | negative | zero | -1 | Answer1 |

The correlation coefficient is always If the two regression coefficients are negative. |
1 | 0 | positive | negative | Answer4 |

The Pearson’s correlation is mean of both the regression coefficient. | Arithmetic | Geometric | Harmonic | cube | Answer2 |

The regression equation is inter-relation between variables. | 5 | 4 | 2 | 1 | Answer3 |

If one of the regression coefficient is negative then the another regression coefficient is . |
negative | positive | 2 | 3 | Answer1 |

If one of the regression coefficient is > 1 then the another regression coefficient has to be | > 1 | < 1 | 1 | -1 | Answer2 |

If byx = 1.5 & bxy = 0.5 then r = | 0.566 | 0.666 | 0.766 | 0.866 | Answer4 |

If the two regression lines are perpendicular with each other, then the value of correlation coefficient is | 5 | 4 | 0 | 3 | Answer3 |

If the regression equation of x on y is x + 2y = 40, the estimated value of x when y = 10 is | 5 | 10 | 15 | 20 | Answer4 |

The estimated value of y when x = 9 is if the regression equation of y on x is 4x + 3y = 51. |
5 | 4 | 3 | 2 | Answer1 |

The value of the Correlation coefficient is if the regression coefficients are 0.4 and 0.9 | 0.5 | 0.6 | 0.7 | 0.8 | Answer2 |

The regression equations are given by x + 3y = 88 & 2x + y = 71 then mean values of x & y are respectively. | 20 & 22 | 23 & 26 | 25 & 21 | 24 & 27 | Answer3 |

The regression equation of y on x is 100y-45x- 1400=0 & that of x on y is 4y-5x+200=0 then r = | 0.6 | 0.7 | 0.8 | 0.9 | Answer1 |

bxy = 1/3 and mean values of x & y are 15 & 20 respectively then regression equation of x on y is |
3x – y = 24 | 3x – y = 25 | 3x – y = 26 | 3x – y = 27 | Answer2 |

byx = -4/5 and mean values of x & y are 20 & 25 respectively then regression equation of y on x is | 4x + 5y = 202 | 4x + 5y = 203 | 4x + 5y = 204 | 4x + 5y = 205 | Answer4 |

If r = 0.4 and standard deviation of x & y are 15 & 9 respectively then regression coefficient of y on x is |
0.19 | 0.22 | 0.24 | 0.26 | Answer3 |

The Regression coefficient of x on y is If the standard deviation of x & y are 7 & 4 respectively and r = – 0.8 | -0.9 | -1.4 | -1.6 | -1.9 | Answer2 |

If r = 1/3 & bxy = 2/5 then byx = | 5/16. | 5/17. | 5/18. | 5/19. | Answer3 |

An orderly set of data arranged in accordance with their time of occurrence is called series | Arithmetic | Geometric | Harmonic | Time | Answer4 |

An increase in the number of patients in the hospital due to heatstroke is variation |
Regular | Irregular | Seasonal | Non-Seasonal | Answer3 |

In time series seasonal variations can occur within a period of year | Four | Three | One | Nine | Answer3 |

The general tendency of the time series data is represented using the following component | Trend | random | constant | variable | Answer1 |

In a straight-line equation Y = a + bX; a is | X-intercept | Y-intercept | Slope | Variable | Answer2 |

In a straight-line equation Y = a + bX; b is | Slope | Variable | X-intercept | Y-intercept | Answer1 |

Moving average method is used for measurement of trend when it is | variable | constant | non-linear | linear | Answer4 |

The most commonly used mathematical method for measuring the trend is by method | quarterly | semi-annual | least square | Variable | Answer3 |

Damages due to floods, earthquakes, strikes, fires and political disturbances are variation | Regular | Irregular | Seasonal | Cyclical | Answer2 |

The Additive time series model can be expressed as |
O = T + S + C + I | O = T x S x C x I | O = T x S x C | O = T x S | Answer1 |

The multiplicative time series model can be expressed as | O = T + S + C + I | O = T x S x C x I | O = T + S + C | O = T + S | Answer2 |

Index numbers are expressed in ratio | Squares | cubes | Percentage | Permutation | Answer3 |

Index for base period is always taken as | 100 | 200 | 300 | 400 | Answer1 |

In chain base method, the base period is | Four | Three | Two | Not Fixed | Answer4 |

Price relatives are a percentage ratio of current year price and | Base year quantity | Previous year quantity |
Base year price |
Current year quantity |
Answer3 |

Another name of consumer’s price index number is | Wholesale index | Cost of living index |
Sensitive index |
Composite index | Answer2 |

Cost of living index numbers are obtained by formula | Laspeyre’s | Paasche’s | Fisher | Marshall Edgeworth |
Answer1 |

When the current price is divided by the price of the preceding year, we get relative | non-link | Compound | Complex | link | Answer4 |

The method of link relatives is used to calculate index | Paasche’s | Fisher | Chain Base | Laspeyre’s | Answer3 |

A number that measures a relative change in a single variable with respect to a base is Number | Good Index | Great Index | Simple Index | Quantity Index | Answer3 |

The Trend value data for 3 consecutive years is 58, 62 & 78 then 3 yearly moving average is | 55 | 66 | 77 | 88 | Answer2 |

The Trend value data for 5 consecutive years is 73, 82, 95, 102 & 103 then 5 yearly moving average is | 89 | 90 | 91 | 92 | Answer3 |

To obtain 5 yearly moving average for a year the corresponding 5 yearly moving total is to be divided by | 5 | 4 | 3 | 2 | Answer1 |

The 4 yearly moving average for a year whose corresponding 4 yearly centered total is 496 has to be | 60 | 61 | 62 | 63 | Answer3 |

The 3 yearly moving average for a year whose corresponding 3 yearly moving total is 546 has to be | 182 | 183 | 184 | 185 | Answer1 |

The 5 yearly moving average for a year whose corresponding 5 yearly moving total is 265 has to be | 50 | 51 | 52 | 53 | Answer4 |

If a = 5 & b = 7 then the Trend line equation is given by | y = 7 + 5x | y = 5 + 7x | x = 5 + 7y | x = 7 + 5y | Answer2 |

If n = 7 then to find trend line equation y = a + bx by least square method x takes the values as | ,-3,-2,-1,0,1,2,3, | ,1,2,3,4,5,6,7, | ,1,3,5,7,9,11,1 3, |
,2,4,6,8,10,12,14, | Answer1 |

If n = 6 then to find trend line equation y = a + bx by least square method x takes the values as | ,1,2,3,4,5,6, | ,-5,-3,-1,1,3,5, | ,1,3,5,7,9,11, | ,2,4,6,8,10,12, | Answer2 |

If ∑y = 240 & n = 12 then in trend line equation y = a + bx we have a = | 17 | 18 | 19 | 20 | Answer4 |

If ∑P0 = 80 & ∑P1= 100 then simple index number by aggregative method is | 115 | 120 | 125 | 130 | Answer3 |

If ∑i = 480 & n = 6 then simple index number by average of price relative method is | 50 | 60 | 70 | 80 | Answer4 |

If ∑P0w = 303 & ∑P1w = 500 then index number by weighted aggregative method is | 145.02 | 155.02 | 165.02 | 175.02 | Answer3 |

If ∑iw = 4200 & ∑w = 20 then cost of living index number by family budget method is | 200 | 210 | 220 | 230 | Answer2 |

If ∑p1q0 = 400 & ∑p0q0 = 210 then Laspeyre’s Index number = | 190.48 | 192.48 | 194.48 | 196.48 | Answer1 |

If ∑p1q1 = 1610 & ∑p0q1 = 1090 then Paasche’s Index number = | 137.71 | 147.71 | 157.71 | 167.71 | Answer2 |

If Laspeyre’s & Paasche’s Index number are 80 & 90 respectively then Fisher’s Index number is | 64.85 | 74.85 | 84.85 | 94.85 | Answer3 |

If Laspeyre’s & Paasche’s Index number are 64 & 72 respectively then Dorbish-Bowley’s Index number is | 62 | 64 | 66 | 68 | Answer4 |

If ∑p1q0 = 400, ∑p0q0 = 210, ∑p1q1 = 490 & ∑p0q1 = 300 then Marshal Edgeworth’s Index number = | 164.51 | 174.51 | 184.51 | 194.51 | Answer2 |

Dorbish-Bowley’s Index number is of Laspeyre’s & Paasche’s Index number | Harmonic Mean | Geometric Mean | Difference | Arithmetic mean | Answer4 |

The mean of the binomial distribution is given by |
n | np | npq | p+q | Answer2 |

The variance of the binomial distribution is given by | p+q | n | np | npq | Answer4 |

The variance of binomial distribution is always mean. | thrice to | more than | less than | twice the | Answer3 |

Each trial in the binomial distribution has possible outcomes. |
1 | 2 | 3 | 4 | Answer2 |

In binomial distribution we always have p + q = | 1 | 3 | 5 | 7 | Answer1 |

The mean of Poisson’s distribution is always variance. | more than | less than | twice the | equal to | Answer4 |

The Poisson distribution has a parameter. | 3 | 2 | 1 | 0 | Answer3 |

The mean of the Poisson’s distribution is given by | m | n | pq | p+q | Answer1 |

The variance of the Poisson’s distribution is given by | pq | m | p+q | n | Answer2 |

We use Poisson’s distribution when ‘n’ is large & probability of success (p) is | large | not known | fixed | small | Answer4 |

In Normal distribution, Total area under the normal curve is | 4 | 3 | 2 | 1 | Answer4 |

We use normal distribution when x is | small | continous | not known | fixed | Answer2 |

In Normal distribution, Shape of normal curve can be related to | square | rectangle | triangle | bell | Answer4 |

In case of symmetrical normal distribution we always have |
mean = mode = median |
mean < mode | mean > mode | mean < median | Answer1 |

The normal distribution has parameters. | 0 | 1 | 2 | 3 | Answer3 |

In Normal distribution, the shape of normal curve depends upon deviation | Geometric | Standard | Quartile | Quadratic | Answer2 |

In Normal distribution, the normal curve is symmetrical about | y = 2x | x = 2y | y = mean | x = mean | Answer4 |

In Normal distribution, area under the normal curve on either side of ordinate is | 0.5 | 0.6 | 0.7 | 0.8 | Answer1 |

In Normal distribution, the Quartile deviation is standard deviation | more than | less than | equal to | thrice the | Answer2 |

In Normal distribution, the Standard deviation is Mean deviation | less than | equal to | more than | twice the | Answer3 |

In a binomial distribution, If mean = 4 and n = 5 then p = | 0.8 | 0.7 | 0.6 | 0.3 | Answer1 |

In a binomial distribution, If n = 8 and mean = 2.4 then q = | 0.2 | 0.4 | 0.6 | 0.7 | Answer4 |

In a binomial distribution, If mean = 3 and p = 0.5 then n = | 8 | 7 | 6 | 5 | Answer3 |

In a binomial distribution, If n = 6 and q = 0.7 then mean = | 1 | 1.8 | 2 | 3 | Answer2 |

In a binomial distribution, If p = 0.4 and n = 7 then variance = | 3 | 2 | 1.68 | 1 | Answer3 |

In a binomial distribution, If variance = 0.8 and q = 0.8 then n = | 1 | 2 | 3 | 5 | Answer4 |

In a binomial distribution, If mean = 3 and variance = 2.4 then p = | 0.2 | 0.3 | 0.4 | 0.5 | Answer1 |

In a binomial distribution, If variance = 2 and mean = 3 then n = | 8 | 9 | 10 | 12 | Answer2 |

In a binomial distribution, If n = 5 and p = 0.6 then mode = | 2 | 4 | 3 | 6 | Answer3 |

In a binomial distribution, If n = 4 and q = 2/3 then P(X=4) = | 1/21. | 1/31. | 1/41. | 1/81. | Answer4 |

In a Poisson’s distribution, If n = 300 and p = 1% then mean = | 0 | 3 | 6 | 9 | Answer2 |

In a Poisson’s distribution, If variance = 2 and p = 4% then n = | 50 | 60 | 70 | 80 | Answer1 |

In a Poisson’s distribution, If mean = 3.8 then mode = | 0 | 2 | 3 | 5 | Answer3 |

In a Poisson’s distribution, If mean = 4 then standard deviation = | 1 | 2 | 3 | 5 | Answer2 |

In a Poisson’s distribution, If P(X=1) = P(X=2) then m = | 7 | 5 | 3 | 2 | Answer4 |

In a Poisson’s distribution, If standard deviation= 3 and n = 300 then p = % | 0 | 2 | 3 | 4 | Answer3 |

In a Poisson’s distribution, If P(X<1) = 0.83 then P(atleast one) = | 0.17 | 0.18 | 0.19 | 0.21 | Answer1 |

In a Poisson’s distribution, If P(X = 0) = 0.14, P(X= 1) = 0.17 & P(X = 2) = 0.23 then P(X < 3) = | 0.24 | 0.32 | 0.41 | 0.54 | Answer4 |

In a Normal distribution, If P(0 < z < 2) = 0.4772 then P(-2 < z < 2) = | 0.6255 | 0.9544 | 0.7255 | 0.8255 | Answer2 |

In a Normal distribution, If P(0 < z < 1) = 0.3413 then P(z > 1) = | 0.1355 | 0.1455 | 0.1587 | 0.1655 | Answer3 |

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