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

Question 1

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

SOLUTION

Solution : A

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

Question 2

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

SOLUTION

Solution : B

$C . V = 2; σ = 0.4 (g i v e n) C . V = 100 \times \frac{s . d}{m e a n} 2 = \frac{100 \times 0.4}{m e a n} M e a n = \frac{40}{2} = 20$

Question 3

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

$L e t t h e m e a n o f t h e l a s t f o u r b e A_{2} . T h e n b y t h e f o r m u l a f o r c o m b i n e d m e a n, 12.5 = \frac{6 \times 15 + 4 \times A_{2}}{6 + 4}; o r 125 = 90 + 4 A_{2}; ∴ A_{2} = \frac{35}{4} L e t t h e s i x t h n u m b e r = x; t h e n t a k i n g t h e s i x t h n u m b e r a s a c o l l e c t i o n, t h e c o m b i n e d m e a n o f t h i s c o l l e c t i o n a n d t h e c o l l e c t i o n o f t h e l a s t f i v e i s 10, b y q u e s t i o n . ∴ B y d e f i n i t i o n o f c o m b i n e d m e a n 10 = \frac{1 \times x + 4 \times \frac{35}{4}}{1 + 4}; 50 = x + 35; ∴ x = 15 ∴ s i x t h n u m b e r = 15$

Question 4

If the mean of the set of number $x_{1}, x_{2} . . . x_{n}$ is $¯ x$ ,then the mean of the number $x_{i} + 2 i, 1 \leq\leq n$ is

¯ x + 2 n

¯ x + n + 1

¯ x + 2

¯ x + n

SOLUTION

Solution : B

$¯ x = \frac{\sum_{i = 1}^{n} x_{i}}{n} \Rightarrow \sum_{i = 1}^{n} x_{i} = n ¯ x ∴ \frac{\sum_{i}^{n} (x_{i} + 2 i)}{n} = \frac{\sum_{i = 1}^{n} x_{i} + 2 (1 + 2 + . . . + n)}{n} = \frac{n ¯ x + 2 \frac{n (n + 1)}{2}}{n} = ¯ x + (n + 1)$

Question 5

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

\frac{(n + 1) d}{(2 n + 1)}

\frac{(n d)}{(2 n + 1)}

\frac{n (n + 1) d}{(2 n + 1)}

\frac{2 (n + 1) d}{n (n + 1)}

SOLUTION

Solution : C

$¯ x = \frac{\frac{2 n + 1}{2} (a + a + 2 n d))}{(2 n + 1)} = a + n d \sum | x - ¯ x | = 2 d (1 + 2 + . . . . . + n) = n (n + 1) d ∴ M . D = \frac{n (n + 1) d}{(2 n + 1)}$

Question 6

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:

2 and 9

3 and 8

4 and 7

5 and 6

SOLUTION

Solution : C

Let the other two numbers be
$α a n d β .$

$¯ x = 4, N = 5 a n d \frac{\sum (x - ¯ x)^{2}}{N} = 5.2 \Rightarrow \sum (x - ¯ x)^{2} = (5.2) 5 ∴ \sum (x - ¯ x)^{2} = 26 ∴ (1 - 4)^{2} + (2 - 4)^{2} + (6 - 4)^{2} + (α - 4)^{2} + (β - 4)^{2} = 26 ∴ (α - 4)^{2} + (β - 4)^{2} = 9 A l s o, \frac{1 + 2 + 6 + α + β}{5} = 4 ∴ α + β = 20 - 9 = 11 C l e a r l y 4, 7 o n l y s a t i s f y t h e a b o v e e q u a t i o n i n α, β . H e n c e r e q u i r e d n u m b e r s a r e 4, 7.$

Question 7

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

less than 3

less than 1

greater than 2.5

greater than 4.9

SOLUTION

Solution : A

$¯ x = \frac{- 1 + 0 + 4}{3} = 1. ∴ Mean Deviation = \frac{1}{3} [| - 1 - 1 | + | 0 - 1 | + | 4 - 1 |] = 2$

Question 8

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

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 " $k^{2}$ ".
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 $\times \frac{1}{4}$ = 1

Question 9

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

36.5

37.5

38.5

SOLUTION

Solution : C

$w e h a v e, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900 N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800 a n d n = 48 ∴ n e w m e a n = \frac{1800}{48} = \frac{450}{12} = \frac{225}{6} = 37.5.$

Question 10

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

$6 = \frac{3 + 4 + x + 7 + 10}{5}$
$\Rightarrow 30 = 24 + x$
$\Rightarrow x = 6$

Question 11

Consider the frequency distribution of the given number
$\begin{matrix} V a l u e : & 1 & 2 & 3 & 4 F r e q u e n c y : & 5 & 4 & 6 & f \end{matrix}$
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

$M e a n = \frac{1 \times 5 + 2 \times 4 + 3 \times 6 + 4 \times f}{15 + f}$
$i . e ., 3 = \frac{5 + 8 + 18 + 4 f}{15 + f} \Rightarrow 45 + 3 f = 31 + 4 f$
$\Rightarrow 45 - 31 = f \Rightarrow f = 14$

Question 12

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
$= (40 \times 50) + (35 \times 60) + (45 \times 55) + (42 \times 45)$
$= 8465$
Overall average of marks per students $= \frac{8465}{162} = 52.25$

Question 13

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

SOLUTION

Solution : C

Sum of total number $= 18 \times 5 = 90$
                After one number excluded
                Sum of total number $= 16 \times 4 = 64$
                Then, excluded number is $90 - 64 = 26$

Question 14

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

\sqrt{2}

\sqrt{3}

\frac{2}{5}

\frac{3}{5}

SOLUTION

Solution : A

$H e r e ¯ x = \frac{1 + 2 + 3 + 4 + 5}{5} = 3$
$\begin{matrix} x & x - ¯ x & (x - {¯ x}^{2}) 1 & - 2 & 4 2 & - 1 & 1 3 & 0 & 0 4 & 1 & 1 5 & 2 & 4 \end{matrix} ∴ \sum (x - ¯ x)^{2} = 10$

$∴ S . D = σ = \sqrt{\frac{(x - ¯ x)}{N}} = \sqrt{\frac{10}{5}} = \sqrt{2}$

Question 15

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

36.6,22.6

34.8, 22.6

36.6, 24.4

none of these

SOLUTION

Solution : A

$We know that C . V = \frac{σ \times 100}{¯ x} o r ¯ x = \frac{σ}{c . v} \times 100 ∴ Mean of first series = \frac{21.2 \times 100}{58} = 36.6 Mean of second series = \frac{15.6 \times 100}{69} = 22.6$

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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