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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Let a business-man have sold rice of 15 quintal , 19 quintal and 20 quintal in three consecutive months. Then how many quintal of rice had he sold per month as a average ? |
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Answer» Here , let x is the variable of quantity of rice . Then the three values of x and `x_(1), x_(2) and x_(3)` the unit being quintal each. Then the arithemetic mean of these values of x is `bar x=(x_(1)+x_(2)+x_(3))/(3)=(15+19+20)/(20)` quintals `=(54)/(3)` quintals =18 quintals Hence , he had sold 18 quintals of rice per month. |
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| 52. |
A shop keeper bougth rice of Rs. 30 at a rate of Rs. 1.5 per kg and of Rs. 30 at a rate of Rs. 4 per kg. Then what is the value of mixed rice per kg ? |
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Answer» Here , the total amont of buying price of the shopkeeper = Rs. (30+30)=Rs.60 The quantity of the total rice boughtb `=(30)/(1.5)+(30)/(4)kg =2+7(1)/(2) kg =27(1)/(2) kg ` `therefore ` the price of the mixed rice per kg =Rs. `(60)/(27(1)/(2))` `=Rs. (60)/((55)/(2))=Rs. (24)/(11)` =Rs. 2.18 (approx.) `therefore ` The required rate of price =Rs. 2.18 per kg . We can also find the result by using harmonic mean. According to the formula, the H.M.`=(2)/((1)/(15)+(1)/(4))=(2)/((1)/(1(1)/(2))+(1)/(4))=(2)/((2)/(3)+(1)/(4))` `=(2)/((8+3)/(12))=(2xx12)/(11)=(24)/(11)=2.18` `therefore ` the average price of rice =Rs. 2.18 per kg. |
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| 53. |
There are two groups A and B in a shop . In the group A, 50 toys have been sold the average price of each toy being Rs. 15 . In group B there also have been sold 30 toys, the average price of which is Rs. 20 each . Find the average of average selling price of the total 80 toys sold in the shop. |
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Answer» Here , the total selling price of 50 toys in group A =Rs. 50`xx`50 =Rs. 750 The total selling price =Rs. (750+600)=Rs. 1350 `therefore ` the required average =Rs.`(1350)/(50+30)=Rs. (1350)/(80)=Rs. 16.87` (approx.) |
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| 54. |
A statistical measure which cannot be determind graphically isA. medianB. modeC. harmonic meanD. mean |
| Answer» Correct Answer - C | |
| 55. |
The weighted means of of first n natural numbers whose weights are equal to the squares of corresponding numbers isA. `(n+1)/(2)`B. `(3n(n+1))/(2(2n+1))`C. `((n+1)(2n+1))/(6)`D. `(n(n+1))/(2)` |
| Answer» Correct Answer - B | |
| 56. |
If a variable takes value 0,1,2,3,....,n with frequencies `1`,`C_1`,`C_2`,.....,`C_n`A. nB. `(2^(n))/(n)`C. `n+1`D. `(n)/(2)` |
| Answer» Correct Answer - D | |
| 57. |
A group of 10 items has arithmetic mean 6. If the arithmetic mean of 4 of these items is 7.5, then the mean of the remaining items isA. 6.5B. 5.5C. 4.5D. `5.0` |
| Answer» Correct Answer - D | |
| 58. |
The mean weight of 9 items is 15. If one more item is added to the series the mean becomes 16. The value of 10th item is (a) `35` (b) `30` (c) `25` (d) `20`A. 35B. 30C. 25D. 20 |
| Answer» Correct Answer - C | |
| 59. |
The mode of the data `6,4,3,6,4,3,4,6,3,x` can beA. only 5B. both 4 and 6C. both 3 and 6D. 3, 4 or 6 |
| Answer» Correct Answer - D | |
| 60. |
Median of these numbers : 3,5,7,9,12 isA. 3B. 6C. 7D. 12 |
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Answer» Correct Answer - C because 7 is situated in middle of series |
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| 61. |
`Q_(1)` is known as upper quartile of the series. |
| Answer» False because he first quartile (also called the lower quartile) is the number below which lies the 25 percent of the bottom data. | |
| 62. |
The arithmetic mean , the mode and the median of a group of 75 observations were calculate to be 27, 34 and 29 respectively. It was later discovered that one observation was wrongly read as 43 instead of the correct value 53. Examine to what extent the calculate values of the averages will be affected by the error. |
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Answer» Given : N `=75,overline(X)=27` we know , `overline(X)= (sumX)/(N) or sumX= Noverline(X)` `:. sumX=75xx27=2,025` Corrected `overline(X)=2,025-43+53=2,035` Corrected `overline(X)=(2,035)/(75)=27.13` Thus , the value of mean is affected ,. But the values of median and mode will be affected by this error because these are positional averages and the median value is 29 which is far away from the values 43 and 53 . Similarly , the value of mode would not be affected by the error since the modal value 34 is far away from the values 43 and 53. |
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| 63. |
If in an asymmetrical distribution , median is 280 and mean is 310 , what will be the mode ? |
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Answer» Mode = 3 Median -2 Mean Mode` =3xx280-2xx310` `=840-620` `=220` `{:("Whan Mode is 220 and Median is","Whan Mode is 220 and Mean is 310, "),("280 , Mean would be :","Median would be:"),(220=280xx3-2 "Mean",220 =3 "Median" -2xx310),(220 =840-2 "Mean",220=3 "Median"-620),(2 " Mean" =840 -220,3 "Median" =220 +620),(2 "Mean"=620,3 "Mean"=840),("Mean"=(620)/(2)=310,"Medina"=(840)/(3)=280):}` |
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| 64. |
The mode of the following data : `120,110,130,110,120,140,130,120,140,120,` isA. 110B. 120C. 130D. 115 |
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Answer» Correct Answer - B The frequency table for the given data is as given below: `{:("Value"(x_(i)):,110 ,120,130,140),("Frequency"(f_(i)):, 2,4,2,2):}` We observe that the value 120 has the maximum frequency. Hence, the mode or modal value is 120. |
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| 65. |
The arithmetic mean of the squares of first n natural numbers isA. `(n+1)/(6)`B. `((n+1)(2n+1))/(6)`C. `(n^(2)-1)/(6)`D. none of these |
| Answer» Correct Answer - B | |
| 66. |
Mode of a certain series is x. If each score is decreased by 3, then mode of the new series isA. `x`B. `x-3`C. `x+3`D. `3x` |
| Answer» Correct Answer - B | |
| 67. |
The arthmetic mean of the numbers `1,3,3^(2), … ,3^(n-1),` isA. `(3^(n)-1)/(2)`B. `(3^(n)-1)/(2n)`C. `(3^(n)+1)/(2)`D. `(3^(n)+1)/(2n)` |
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Answer» Correct Answer - B Let M be the required mean. Then, `M=(1+3+3^(2)+ ... +3^(n-1))/(n)` `rArr M=(1xx((3^(n)-1)/(3-1)))/(n)=(3^(n)-1)/(2n)` |
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| 68. |
The arithmetic mean of first `n` odd natural numbers, isA. nB. `(n)/(2)`C. `(n-1)/(2)`D. `(n+2)/(2)` |
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Answer» Correct Answer - A First n odd natural numbers are `1,2,3,5, … , (2n-1).` So, required mean `=(1+3+5+ …+(2n-1))/(n)` `rArr " Required Mean "=((n)/(2){1+2n-1})/(n)=n`. |
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| 69. |
The arithmetic mean of first `n` natural numbers, isA. `(pi)/(2)`B. `(n+1)/(2)`C. `(n(n+1))/(2)`D. `(n-1)/(2)` |
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Answer» Correct Answer - B Clearly, Required mean `=(1+2+ ... +n)/(n)=(n(n+1))/(2n)=(n+1)/(2)` |
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| 70. |
If the arithmetic mean of the first n natural numbers is 15 , then n is `"__________"` .A. 15B. 20C. 14D. 29 |
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Answer» Correct Answer - D We have, `(1+2+3+...+n)/(n)=15 rArr (n(n+1))/(2n)=15 rArr n = 29` |
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| 71. |
The arithmetic mean of `""^(n)C_(0),""^(n)C_(1),""^(n)C_(2), ..., ""^(n)C_(n)`, isA. `(2^(n))/(n)`B. `(2^(n)-1)/(n)`C. `(2^(n))/(n+1)`D. `(2^(n-1))/(n+1)` |
| Answer» Correct Answer - C | |
| 72. |
The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102,126,85,91,107 Find the average of the amount of rupees sold per day. |
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Answer» Here, the values of the variable x , is near the values 100 of x If we transfer the origin A to 100 , then the values of u are 15-2,2,26,-15,-8,7. `therefore baru=1/7(15-2+2+26-15-9+7)=(50-26)/(7)=(24)/(7)=3.43` (approx.) `therefore baru=100+3.43 =103.43` (approx) `therefore ` the required daily seling amount =Rs. 103.43 Now , let x is a descrete variable. The possible values of `x_(1), x_(2),.............., x_(n)` . Frequency distribution of x are made from n values of x. Let the frequencies of `x_(1), x_(2), .........x_(n)` be `f_(1),f_(2), ..............f_(n)` i.e. `x_(1)` occurs `f_(1)` times , `x_(2)` occurs `f_(2)` times, .......`f_(n)` occurs n-times and `f_(1)+f_(2) +........+f_(n)=n` Now , `underset(i=1)overset(n)sum f_(i)x_(i)=x_(i)xxf_(i)+x_(2)xxf_(2) + ........+x_(n)xxf_(n)` and `=f_(1)+f_(2)+............+f_(n)=n` Arithmetic mean of `x=barx=(x_(1)f_(1)+x_(2)f_(2)+.....+x_(n)f_(n))/(f_(1)+f_(2)+.......+f_(n))` `=(underset(i=1)overset(n)sumx_(i)f_(i))/(underset(i=1)overset(n)sum f_(i)=(1)/(n)underset(i=1)overset(n)sum x_(i)f_(i)................(""^(**)4)` This `barx` is the weighted mean of the variable x. |
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| 73. |
The median can graphically be found fromA. ogiveB. histogramC. frequency curveD. none of these |
| Answer» Correct Answer - A | |
| 74. |
If y = f(x) be a monotonically increasing or decreasing function of x and M is the median of variable x, then the median of y isA. `f(M)`B. `M//2`C. `f^(-1)(M)`D. none of these |
| Answer» Correct Answer - A | |
| 75. |
If the first item is increased by 1, second by 2 and so on, then the new mean isA. `bar(X) +n`B. `bar(X)+(n)/(2)`C. `bar(X)+(n+1)/(2)`D. none of these |
| Answer» Correct Answer - C | |
| 76. |
The weighted mean of first n natural numbers when their weights are equal to corresponding natural number, isA. `(n+1)(2)`B. `(2n+1)/(2)`C. `(2n+1)/(3)`D. `((2n+1)(n+1))/(6)` |
| Answer» Correct Answer - A | |