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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.
| 1. |
The median of the following marks of `9` students in a class, is `34,32,48,38,24,30,27,21,35`A. 32B. 34C. 38D. 30 |
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Answer» Correct Answer - A Arranging the data in ascending order of magnitude, we have `21,24,27,30,32,34,35,38,48` Since there are 9, and odd number of items. Therefore, median is the value of `((9+2)/(2))^(th)` observation i.e., 32. |
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| 2. |
Compute the geometric mean of 2, 4, 8. |
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Answer» Let G be the required geometric mean. Then, `G=(2xx4xx8)^(1//3) rArr G=(2^(6))^(1//3)=4.` |
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| 3. |
If the mean of a set of observations `x_(1),x_(2), …,x_(n)" is " bar(X)`, then the mean of the observations `x_(i) +2i , i=1, 2, ..., n` isA. `bar(X) +2`B. `bar(X) +2n`C. `bar(X)+(n+1)`D. `X+n` |
| Answer» Correct Answer - C | |
| 4. |
If the arithmetic mean of the observations `x_1,x_2,x_3,.............x_n` is `1`, then the arithmetic mean of `x_1/k,x_2/k,x_3/k,.............x_n/k , (k >0)` isA. greater than 1B. Less than 1C. equal to 1D. none of these |
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Answer» Correct Answer - D We have, `(x_(1)+x_(2)+x_(3)+ ..+x_(n))/(n)=1 rArr x_(1)+x_(2)+x_(3)+ ... +x_(n)=n` `therefore` Required arithmetic mean `((x_(1))/(k)+(x_(2))/(k)+ ...+ (x_(n))/(k))/(n)` `=(1)/(k)((x_(1)+x_(2)+ ... +x_(n))/(n))=(1)/(k)` |
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| 5. |
The A.M. of the series `1, 2, 4, 8, 16,.....,2^n` is -A. `(2^(n)-1)/(n)`B. `(2^(n+1)-1)/(n+1)`C. `(2^(n)+1)/(n)`D. `(2^(n)-1)/(n+1)` |
| Answer» Correct Answer - B | |
| 6. |
The mean of the series `x_1, x_2,...x_n` is `barX`. If `x_2` is replaced by `lambda` then the new mean isA. `bar(X) + x_(2) + lambda`B. `(bar(X)-x_(2)-lambda)/(n)`C. `((n-1)bar(X)+lambda)/(n)`D. `(n bar(X)-x_(2)+lambda)/(n)` |
| Answer» Correct Answer - D | |
| 7. |
The mean of discrete observations `y_(1), y_(2), …, y_(n)` is given byA. `(sum_(i=1)^(n) y_(i)f_(i))/(sum_(i=1)^(n)f_(i))`B. `(sum_(i=1)^(n) y_(i)f_(i))/(n)`C. `(sum_(i=1)^(n) y_(i))/(n)`D. `(sum_(i=1)^(n) y_(i))/(sum_(i=1)^(n)i)` |
| Answer» Correct Answer - C | |
| 8. |
The geometric mean of numbers `7, 7^(2),7^(3),…,7^(n),` isA. `7^(7//4)`B. `7^(4//7)`C. `7^((n-1)/(2))`D. `7^((n+1)/(2))` |
| Answer» Correct Answer - D | |
| 9. |
The mean of the distribution, in which the values of X are 1, 2, ..,n the frequency of each being unity is :A. `(n(n+1))/(2)`B. `(n)/(2)`C. `(n+1)/(2)`D. none of these |
| Answer» Correct Answer - C | |
| 10. |
If the mean of n observations `x_1,x_2,x_3...x_n` is `barx` then the sum of deviations of observations from mean is |
| Answer» Correct Answer - A | |
| 11. |
The averageof 50 numbers is 38. If the numbers 45 and 55 are discarded, then the averageof the remaining numbers is(a)36.5 (b) 37 (c) 37.5 (d) 37.52A. 36B. 36.5C. 37.5D. 38.5 |
| Answer» Correct Answer - C | |
| 12. |
The most stable measure of central tendency isA. the meanB. the medianC. the modeD. none of these |
| Answer» Correct Answer - A | |
| 13. |
The one which is the measure of the central tendency isA. modeB. mean deviationC. standard deviationD. coefficient of correlation |
| Answer» Correct Answer - A | |
| 14. |
For dealing with qualitative data the best average isA. AMB. GMC. ModeD. Median |
| Answer» Correct Answer - D | |
| 15. |
The positional average of central tendency isA. GMB. HMC. AMD. Median |
| Answer» Correct Answer - D | |
| 16. |
Theaverage marks of boys in a class is 52 and that of girls is 42. The averagemarks of boys and girls combined is 50. The percentage of boys in the classis(1)40(2) 20(3) 80(4) 60A. 80B. 60C. 40D. 20 |
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Answer» Correct Answer - A Let `n_(1) and n_(2)` be the number of boys and girls respectively in the class. We have, `bar(X_(1))=52,bar(X_(2))=42 and bar(X) =50` `therefore bar(X)=(n_(1)bar(X_(1))+n_(2)bar(X_(2)))/(n_(1)+n_(2))` `rArr 50=(52n_(1)+42n_(2))/(n_(1)+n_(2)) rArr 8n_(2)=2n_(1) rArr n_(1)=4n_(2)` `therefore " Percentage of boys "=(n_(1))/(n_(1)+n_(2)) xx 100=(4n_(2))/(5n_(2)) xx 100=80` |
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| 17. |
If the median of 25 observations is 45 and if the observations greater than the median are increased by 4, then the median of the new data isA. 49B. 41C. 45D. none of these |
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Answer» Correct Answer - C As the median is the positional average and by adding 4 to each observation greater than the median the positions of the values are unchanged. Hence the median remains same. |
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| 18. |
The median of a set of a observations is 20.5. If each of the largest 4 observations of the set is increasted by 2, then the median of the new set isA. remains the same as that of origingal setB. is increased by 2C. is decreased by 2D. is two times the original median |
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Answer» Correct Answer - A The observation still remain in the same order. So, the median remains same. |
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| 19. |
The median of 100 observations grouped in classes of equal width is 25. If the median class interval is 20-30 and the number of observations less than 20 is 45, then the frequency of median class isA. 20B. 12C. 10D. 15 |
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Answer» Correct Answer - C We have, `N=100, F=45, l =20, h=10" and Median "=25` Let f be the frequency of the meidan class. Then, Median `=l+((N)/(2)-F)/(f) xx h` `rArr 25=20+(50-45)/(f) xx 10` ` rArr 5=(50)/(f) rArr f=10` |
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| 20. |
The mode of the following distribution is `{:("Class interval :",1-5,6-10,11-15,16-20,21-25),("Frequency :" ,4,7,10,8,6):}`A. 14.5B. 16.5C. 10.5D. 13.5 |
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Answer» Correct Answer - D The given distribution is not a continuous Frequency distribution. So, we first convert it into a continuous frequency distribution as given below: `{:("Class interval :",0.5-5.5,5.5-10.5,10.5-15.5,15.5-20.5,20.5-25.5),("Frequency :" ,4,7,10,8,6):}` Clearly, 10.5-15.5 is the modal class such that `l=10.5, f=10,f_(1)=7,f_(2)=8 and h=5.` `therefore " Mode"=l+(f-f_(1))/(2f-f_(1)-f_(2)) xx h=10.5 + (3)/(5) xx 5=13.5` |
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| 21. |
If `x lt 6` and 17.5 is the mode of the following frequency distribution. `{:("Class-interval:", 0-5,5-10,10-15,15-20,20-25),("Frequency:", 5,2,3,6,x):}` Then, x =A. 3B. 2C. 4D. 5 |
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Answer» Correct Answer - A Clearly, 15-20 is the modal class such that `l=15, h=5, f=6, f_(1)=3 and f_(2)= x.` `therefore " mode"= l+(f-f_(1))/(2f-f_(1)-f_(2)) xx h` `rArr 17.5=15+(6-3)/(12-3-x)xx5 rArr 2.5x =7.5 rArrx=3` |
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| 22. |
The modal class for the following frequency distribution, is `{:("Marks:", 0-10,10-20,20-40,40-50,50-60,60-70,70-90,90-100),("No. of students:",4,6,14,16,14,8,16,5):}`A. 20-40B. 40-50C. 50-60D. 70-90 |
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Answer» Correct Answer - B In the given frequency distribution the class intervals are not equal. So, we first adjust the frequencies taking 10 as minimum class interval as given below : `{:("Marks:", 0-10,10-20,20-30,30-40,40-50,40-50,50-60,60-70,70-80,80-90,90-100),("No. of students:",4,6,7,16,7,16,14,8,8,8,5):}` Since 40-50 is the class of maximum frequency. So, it is the modal class. |
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| 23. |
If the mean of 26 , 19 , 15 , 24 , and x is x , then find the median of the data .A. 23B. 22C. 20D. 21 |
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Answer» Correct Answer - D We have, `(26+19+15+24+x)/(5) =x rArr 4x=84 rArr x=21` Thus, the given data is `26,19,15,24,21.` Arranging the data in ascending the data in ascending order, we get `15,19,21,24,26` Clearly, 21 is its median. |
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| 24. |
The mean of a certain number of observations is m. If each observation is divided by `x(ne 0)` and increased by y, thenA. `mx+y`B. `(mx+y)/(x)`C. `(m+xy)/(x)`D. `m+xy` |
| Answer» Correct Answer - C | |
| 25. |
If the mode of observations `5,4,4,3,5,x,3,4,3,5,4,3,5 ` is 3, then median of the observation isA. 3B. 4C. 5D. 3.5 |
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Answer» Correct Answer - B It is given that the mode of given observations is 3. So, 3 must have maximum frequency. `therefore x=3` Arranging the observation in ascending order, we have `3,3,3,3,3,4,4,4,4,5,5,5,5` Clearly, median = 4. |
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| 26. |
The mean of a set of observations is `a` . If each observation is multiplied by `b` and each product is decreased by `c`, then the mean of new set of observations isA. `(a)/(b)+c`B. `ab-c`C. `(a)/(b)-c`D. `ab+c` |
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Answer» Correct Answer - B Let `x_(1),x_(2),x_(3), …, x_(n)` be n observations. Let `y_(i)=bx_(i) -c, i=1,2, …, n.` Then, `bar(Y)=b bar(X)-c rArr bar(Y) =ab-c.` |
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| 27. |
If the median of the data `x_(1), x_(2), x_(3), x_(4), x_(5),x_(6), x_(7), x_(8) " is " alpha and x_(1) lt x_(2) lt x_(3) lt x_(4) lt x_(5) lt x_(6) lt x_(7) lt x_(8)`, then the median of `x_(3), x_(4), x_(5), x_(6)` isA. `alpha`B. `(alpha)/(2)`C. `(alpha)/(3)`D. `(alpha)/(4)` |
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Answer» Correct Answer - A We have, `alpha = (x_(4)+x_(5))/(2)` Median of `x_(3),x_(4), x_(5), x_(6) " is " (x_(4)+x_(5))/(2)= alpha` |
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| 28. |
If a , b, c, d are non-zero integers such that `a lt b lt c lt d` and median of a, b, c are both equal to zero, then which one of the following is correct?A. a = -cB. a = -dC. both (a) and (b)D. none of these |
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Answer» Correct Answer - C It is given that both the mean and the median of given data are zero. `therefore (a+b+c+d)/(4)=0and (b+c)/(2)=0` `rArr a+b+c+d=0 and b+c=0` `rArr a= -d and b= -c` |
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| 29. |
if `x < y < 2x` and the mean and median of `x,y,2x` are `15,12` respectively then `x`A. 7B. 11C. 10D. 8 |
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Answer» Correct Answer - B We have, Mean = 15 and, Median = 12 `rArr (x+y+2x)/(3) =15 and y=12 rArr x=11, y= 12` |
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| 30. |
The mean of 16 observations is 16 . If one observation 16 is deleted and three observations 5 , 5 and 6 are included , then find the mean of the final observations .A. 16B. 15.5C. 13.5D. none of these |
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Answer» Correct Answer - D We have, `("Sum of 16 observations")/(16) =16` `rArr ` Sum of 16 observations = 256 Mean of new observations `=(256-16+5+5+6)/(18) = 14.22` |
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| 31. |
The mean and median of the data a, b and c are 50 and 35 , where `a lt b lt c` . If `c - a = 55` , then find (b - a) .A. 8B. 7C. 3D. 5 |
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Answer» Correct Answer - D We have, mean = 50, Median = 35 `rArr (x+y+z)/(3)=50 and y=35` `rArr x+y+z=150 and y=35 rArr x+z=115` But , `z-x=35-30=5` |
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| 32. |
If theratio of mean and median of a certain data is 2:3, then find the ratio of itsmode and mean.A. `4:3`B. `7:6`C. `7:8`D. `5:2` |
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Answer» Correct Answer - D We have, `("Mean")/("Median")=(2)/(3) rArr "Mean" =2lambda " and Median"=3lambda` `therefore " Mode"=3" Median"-2" Mean"` `rArr "Mode"=9lambda -4lambda=5lambda` `therefore "Mode": "Mean"=5lambda:2lambda=5:2` |
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| 33. |
The sum of squares of the deviation of the values of the variable is minimum when taken about their arithmetic meanA. AMB. GMC. HMD. median |
| Answer» Correct Answer - A | |
| 34. |
The arithmetic mean of first `n` odd natural numbers, isA. nB. `(n+1)/(2)`C. `n-1`D. none of these |
| Answer» Correct Answer - A | |
| 35. |
If the meanof `x+2, 2x+3, 3x+4, 4x+5 i s x+2,`find `x` |
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Answer» Correct Answer - C We have, `rArr ((x+2)+(2x+3)+(3x+4)+(4x+5))/(4) = x+2` `rArr 10x+14=4x+8 rArr x= -1` |
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| 36. |
If the mean of the following distribution is 2.6, then the value of `y`isVariable `(x)`: 1 2 3 4 5Frequency `(y)`: 4 5 `y`1 2(a) 3 (b) 8 (c) 13 (d) 24A. 3B. 8C. 13D. 24 |
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Answer» Correct Answer - B We have, Mean`=2.6` `rArr (1xx4+2xx5+3y+4xx1+5xx2)/(4+5+y+1+2)=2.6` `rArr (3y+28)/(12+y)=2.6` `rArr 3y+28=31.2+26y rArr 0.4y=3.2 rArr y=8` |
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| 37. |
If the mean of the following frequency distribution is 5, then b = `{:(x_(i) " :",3,5,7,4),(f_(i)" :",2,a,5,b):}`A. 10B. 6C. 8D. none of these |
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Answer» Correct Answer - B We have, `(3xx2+5xxa+7xx5 +4xxb)/(2+a+5+b)=5` `rArr 5a+4b+41+35+5a+5b rArrb=6` |
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| 38. |
Thearithmetic mean and mode of a data are 24 and 12 respectively, then find the medina of the data.A. 20B. 18C. 21D. 22 |
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Answer» Correct Answer - A We have, Mean = 24 and Mode = 12. `therefore "Mode"=3 " Median"-2 " Mean" ` `rArr 12=3 " Mean" - 2xx24 rArr "Median"=20` |
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| 39. |
If a mode exceeds a mean by 12 , then the mode exceeds the median by `"__________"`.A. 4B. 8C. 6D. 10 |
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Answer» Correct Answer - B We have, Mode - Mean = 12 `therefore ` Mode = 3 Median - 2 Mean `rArr ` Mode - Mean = 3(Median - Mean) `rArr` 12 = 3(Median - Mean) `rArr ` Median - Mean = 4 Again, Mode = 3 Median - 2 Mean `rArr` Mode - Median = 2( Median - Mean) = 2 `xx `4=8 |
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| 40. |
The arithmetic mean of the following frequency distribution: `{:("Variate(X)" ,:,0,1,2,3, ..., n),("Frequency" (f),:,""^(n)C_(0),""^(n)C_(1),""^(n)C_(2),""^(n)C_(3),...,""^(n)C_(n)):}` isA. `(2^(n))/(n)`B. `(2^(n))/(n+1)`C. `(n)/(2)`D. `(2)/(n)` |
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Answer» Correct Answer - C Let `bar(X)` denote the required mean, Then, `bar(X)=(sum_(r=0)^(n)r.""^(n)C_(r))/(sum_(r=0)^(n)""^(n)C_(r))=(sum_(r=1)^(n)r.(n)/(r)""^(n-1)C_(r-1))/(sum_(r=0)^(n)r""^(n)C_(r))=(n sum_(r=1)^(n)""^(n-1)C_(r-1))/(2^(n))` ` rArr bar(X)=(n xx 2^(n-1))/(2^(n))=(n)/(2) " "[ therefore sum_(r=1)^(n)""^(n-1)C_(r-1)=2^(n-1)]` |
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| 41. |
The arithmetic mean of `""^(n)C_(0),""^(n)C_(1),""^(n)C_(2), ..., ""^(n)C_(n)`, isA. `(1)/(n)`B. `(2^(n))/(n)`C. `(2^(n-1))/(n)`D. `(2^(n+1))/(n)` |
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Answer» Correct Answer - B Clearly, Required mean `=(""^(n)C_(0)+""^(n)C_(1)+ ... +""^(n)C_(n))/(n) =(2^(n))/(n)` |
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| 42. |
If theratio of mode and median of a certain data is 6:5, then find the ratio of itsmean and median.A. `8:9`B. `9:10`C. `9:7`D. `8:11` |
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Answer» Correct Answer - B We have, `"Mode" :"Median"=6.5 rArr Mode =6lambda, "Median"=5lambda` Now, ` "Mode"=3 " Median"-2 " Mean" ` `rArr 6lambda =3 xx 5 lambda -2" Mean" rArr "Mean"=(9)/(2)lambda` `therefore "Mean" : "Median"=(9lambda)/(2) : 5lambda =9:10` |
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| 43. |
There are 50 workers in a small industry. The dialy income of them is Rs. 9 Amongst them, the daily income of the 30 workers, working in the morning is Rs. 8 . Then find the daily average income of workers working in the evening. |
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Answer» Here, the total number of workers, n =50 The number of workers in the morning `n_(1)=30` `therefore ` The numbers of workers in the eveing `n_(2)=n-n_(1)=50-30=20` The average daily income of the total workers =Rs. 9 The average daily income of the workers in the morning , `barx_(1)=Rs. 8` According to the law of collective average, `barx=(n_(1)barx_(1)+n_(2)barx_(2))/(n_(1)+n_(2)) therefore 9=(30xx8+20barx_(2))/(30+20)` or , `9=(240+20barx_(2))/(50)` or , `450=240 + 20 barx_(2) ` or , `20 barx_(2)=210` or , `barx_(2)=(210)/(20)=10.50` `therefore ` THe daily income of the workers in the evening =Rs. 10.50. |
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| 44. |
If each of n numbers `x_(i)=i`, is replaced by `(i+1)x_(i)`, then the new mean isA. `((n+1)(n+2))/(n)`B. `n+1`C. `((n+1)(n+2))/(3)`D. none of these |
| Answer» Correct Answer - D | |
| 45. |
The mean age of a combined group of men and women is 25 yrs . If mean age of men is 26 and that of women is 21 , then percentage of men and women in the group , is ,A. 60, 40B. 80, 20C. 20, 80D. 40, 60 |
| Answer» Correct Answer - B | |
| 46. |
The median of the data 5 , 6 , 7, 8 , 9 , 10 is `"__________"`.A. 7B. 8C. 7.5D. 8.5 |
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Answer» Correct Answer - C There are six observations. `therefore " Median"=AM " of " 3^(rd) and 4^(th) ` observation `rArr " Median"=(7+8)/(2)=7.5` |
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| 47. |
The AM of n observations is M. If the sum of `n-4` observations is a, then the mean of remaining 4 observations isA. `(nM-a)/(4)`B. `(nM+a)/(2)`C. `(nM-a)/(2)`D. `nM+a` |
| Answer» Correct Answer - A | |
| 48. |
The number of observations in a group is 40. If the average of first 10 is 4.5 and that of the remaining 30 is 3.5 then the average of the whole group isA. `(15)/(4)`B. `(1)/(5)`C. 8D. 4 |
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Answer» Correct Answer - A We have, `n_(1)=10,n_(2)=30,bar(X)_(1)=4.5, X_(2)=3.5` Let `bar(X)` be the average of the whole group. Then, `bar(X)=(n_(1)bar(X)_(1)+n_(2)bar(X)_(2))/(n_(1)+n_(2))=(10xx45+30xx35)/(40)=(45+105)/(40)=(15)/(4)` |
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| 49. |
A car complete the first half of its journey with speed `v_1`, and the rest half with speed `v_2`. The average speed of the car in whole journey:A. `(2v_(1)v_(2))/(v_(1)+v_(2))`B. `(v_(1)+v_(2))/(2)`C. `sqrt(v_(1)v_(2))`D. none of these |
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Answer» Correct Answer - A The average velocity is the Harmonic mean of `v_(1) and v_(2)`. `therefore "Average Velocity" =(2v_(1)v_(2))/(v_(1)+v_(2))` |
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| 50. |
Let the velocity of a bicycler in 10 consecutive hours are 17,25,30,32,28,24,20,18,16,10 in kilometers.Find the average velocity. |
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Answer» Then the average velocity , `barx=(x_(1)+x_(2) + …………..+x_(10))/(10) =(17+25+30+………+10)/(10)km=(220)/(10) km =22 km ` Hence , the average velocity of him , =22 km/h . Now, if the person travels 10 hours with the same velocity , he would travel a distance of `22xx10km=220` km in 10 hours . |
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