# Free Objective Test 01 Practice Test - 11th and 12th

The range of the observations in 3, 8, 4, 9, 16, 19 is

A.

16

B.

19

C.

8

D.

7

#### SOLUTION

Solution : A

Range = Highest observation - Lowest observation
= 19 − 3
= 16

If the coefficient of variation and standard deviation of a distribution are 2 and 0.4 respectively, then mean of the distribution is

A.

40

B.

20

C.

10

D.

8

#### SOLUTION

Solution : B

C.V=2;σ=0.4(given)C.V=100×s.dmean2=100×0.4meanMean=402=20

The mean of 10 numbers is 12.5; the mean of the first six is 15 and the last five is 10. The sixth number is

A. 15
B. 12
C. 18
D. none of these

#### SOLUTION

Solution : A

Let the mean of the last four be A2. Then by the formula for combinedmean,12.5=6×15+4×A26+4;or 125=90+4 A2;  A2=354Let the sixth number=x; then taking the sixth number as a collection,the combined mean of this collection and the collection of the last five is 10, by question. By definition of combined mean10=1×x+4×3541+4;50=x+35;    x=15 sixth number=15

If the mean of the set of number x1,x2...xn is ¯x ,then the mean of the number xi+2i,1n is

A. ¯x+2n
B. ¯x+n+1
C. ¯x+2
D. ¯x+n

#### SOLUTION

Solution : B

¯x=ni=1xinni=1xi=n¯x    ni(xi+2i)n=ni=1xi+2(1+2+...+n)n=n¯x+2n(n+1)2n=¯x+(n+1)

Mean deviation of the series a, a + d, a + 2d, a + 2nd from its mean is

A. (n+1)d(2n+1)
B. (nd)(2n+1)
C. n(n+1)d(2n+1)
D. 2(n+1)dn(n+1)

#### SOLUTION

Solution : C

¯x=2n+12(a+a+2nd))(2n+1)=a+nd|x¯x|=2d(1+2+.....+n)=n(n+1)d   M.D=n(n+1)d(2n+1)

The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are:

A.

2 and 9

B.

3 and 8

C.

4 and 7

D.

5 and 6

#### SOLUTION

Solution : C

Let the other two numbers be
α and β.

¯x=4,N=5 and (x¯x)2N=5.2 (x¯x)2=(5.2)5  (x¯x)2=26   (14)2+(24)2+(64)2+(α4)2+(β4)2=26   (α4)2+(β4)2=9Also,1+2+6+α+β5=4    α+β=209=11Clearly 4,7 only satisfy the above equation in α,β.Hence required numbers are 4,7.

The mean deviation from the mean for the set of observations – 1, 0, 4 is

A.

less than 3

B.

less than 1

C.

greater than 2.5

D.

greater than 4.9

#### SOLUTION

Solution : A

¯x=1+0+43=1. Mean Deviation=13[|11|+|01|+|41|]=2

If the S.D of a set of observations is 4 and if each observation is divided by 4, the S.D of the new set of observations will be

A.

4

B.

3

C.

2

D.

1

#### SOLUTION

Solution : D

We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of "k2".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4×14 = 1

The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is

A.

36

B.

36.5

C.

37.5

D.

38.5

#### SOLUTION

Solution : C

we have,xi50=38,      xi=1900New value of xi=19005545=1800 and n=48 new mean=180048=45012=2256=37.5.

If the mean of 3, 4, x, 7, 10 is 6, then the value of x is

A. 4
B. 5
C. 6
D. 7

#### SOLUTION

Solution : C

6=3+4+x+7+105
30=24+x
x=6

Consider the frequency distribution of the given number
Value:1234Frequency:546f
​If the mean is known to be 3, then the value of f is

A. 3
B. 7
C. 10
D. 14

#### SOLUTION

Solution : D

Mean=1×5+2×4+3×6+4×f15+f
i.e., 3=5+8+18+4f15+f45+3f=31+4f
4531=ff=14

A school has four sections of chemistry in class XII having 40, 35, 45 and 42 students.  The mean marks obtained in chemistry test are 50, 60, 55 and 45 respectively for the four sections, the over all average of marks per students is

A. 53
B. 45
C. 55.3
D. 52.25

#### SOLUTION

Solution : D

Total number of students =40+35+45+42=162
Total marks obtained
=(40×50)+(35×60)+(45×55)+(42×45)
=8465
Overall average of marks per students =8465162=52.25

The mean of 5 numbers is 18. If one number is excluded, their mean becomes 16. Then the excluded number is

A.

18

B.

25

C.

26

D.

30

#### SOLUTION

Solution : C

Sum of total number =18×5=90
After one number excluded
Sum of total number =16×4=64
Then, excluded number is 9064=26

The S.D. of scores 1, 2, 3, 4, 5 is:

A. 2
B. 3
C. 25
D. 35

#### SOLUTION

Solution : A

Here ¯x=1+2+3+4+55=3
xx¯x(x¯x2)124211300411524 (x¯x)2=10

S.D=σ=(x¯x)N=105=2

The coefficient of variation of two series are 58% and 69%.  If their standard deviations are 21.2 and 15.6, then their A.Ms are

A.

36.6,22.6

B.

34.8, 22.6

C.

36.6, 24.4

D.

none of these

#### SOLUTION

Solution : A

We know thatC.V=σ×100¯x  or  ¯x=σc.v×100 Mean of first series=21.2×10058=36.6Mean of second series=15.6×10069=22.6