Mathematics Test

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Mathematics Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

A box contains 3 white and 2 black balls. Two balls are drawn at random one after the other. If the balls are not replaced, what is the probability that both the balls are black?

A
2/5

B
1/5

C
1/10

D
None of these

Solution

Total number of balls = 5

Number of black balls = 2

Required probability

$= \frac{n (E)}{n (S)} = \frac{^{3} C_{0} \times^{2} C_{2}}{^{5} C_{2}} = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$
Question 2
1 / -0

For two variables x and y, the two regression coefficients are b_yx= $- \frac{3}{2}$ and b_xy= - $\frac{1}{6}$ .The correlation coefficient between x and y is :

A
-1/4

B
1/4

C
-1/2

D
1/2

Solution

$r = \sqrt{b_{x y} \cdot b_{y x}}$

$= \sqrt{(- \frac{1}{6}) \times (- \frac{3}{2})} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \pm \frac{1}{2}$

b_xyand b_yx both have negative sign. Therefore we have to take negative sign Hence, correlation coefficient (r) = $- \frac{1}{2}$
Question 3
1 / -0

The variance of number $x^{1}, x^{2}, x^{3}, \dots \dots . x^{n}$ ,is V. Consider the following statement:

1.If every $x_{1}$ is increased by 2, the variance of the new set of numbers is V.

2. If the numbers $x_{i}$ is squared, the variance of the set is $V^{2}$ .

Which of the following statement is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

I: Variance is not dependent on change of origin.

Therefore, if every x_iis increased by 2, the variance of the new set of numbers is not changed.

II: Variance is dependent on change of scale.

If the number x_iis squared the variance of the new set is V²

V_xtV_xt= V. V = V²
Question 4
1 / -0

What is the mean of the squares of the first 20 natural numbers?

A
151.5

B
143.5

C
65

D
72

Solution

Mean of the squares of the first 20 natural number

= $= \frac{(n + 1) (2 n + 1)}{6} = \frac{21 \times 41}{6} = 143.5$
Question 5
1 / -0

p, q, r, s, t, are five numbers such that the average of p, q and r is 5 and that of s and t is 10. What is the average of all the five numbers ?

A
7.75

B
7.5

C
7

D
5

Solution

According to question p + q + r = 5 x 3 = 15 ....(i)

s + t = 10 x 2 = 20 ........(ii)

From equations (i) and (ii), p + q + r + s + t = 15+ 20 = 35

Average p, q, r, s and t = $\frac{35}{5}$ = 7
Question 6
1 / -0

The cumulative frequency of the largest observed value must always be:

A
Less than the total number of observations

B
Greater than the total number of observations

C
Equal to total number of observations

D
Equal to mid point of the last class interval

Solution

The cumulative frequency of the largest observed value must always be less then the total number of observations.
Question 7
1 / -0

It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they win alternately, is:

A
5/12

B
5/36

C
19/27

D
5/27

Solution

P(A) $= \frac{6}{12} = \frac{1}{2}, P (B) = \frac{4}{12} = \frac{1}{3}$

Req. probability = $\frac{1}{2} . \frac{1}{3} . \frac{1}{2} + \frac{1}{3} . \frac{1}{2} . \frac{1}{3} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}$
Question 8
1 / -0

Out of 7 consonants and 4 vowels, words are to be formed by involving 3 consonants and 2 vowels. The number of such words formed is:

A
25200

B
22500

C
10080

D
5040

Solution

Number of words = 5! × ⁷C₃× ⁴C₂

$= 120 \times \frac{7!}{4! 3!} \times \frac{4!}{2! 2!} = 25200$
Question 9
1 / -0

Let X denote the number of scores which exceed 4 in 18, tosses of a symmetrical die. Consider the following statements:

1. The arithmetic mean of X is 6.

2. The standard deviation of X is 2.

Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Statement I :

n(X) = 2

p = $\frac{n (X)}{n (S)} = \frac{2}{6} = \frac{1}{3}$

q = 1 - p = 1 - $\frac{1}{3}$ = $\frac{2}{3}$

arithmetric mean of X = np

= 18 × $\frac{1}{3} = 6$

Statement 2 : Standard deviation of

X = $\sqrt{v a r i a n c e o f X}$

= $\sqrt{18 \times \frac{1}{3} \times \frac{2}{3}} = \sqrt{4} = 2$

Hence, statements 1 and 2 both are correct.
Question 10
1 / -0

How many different words can be formed by taking four letters out of the letters of the word ‘AGAIN’ if each word has to start with A?

A
6

B
12

C
24

D
Neither 1 nor 2

Solution

As 'A' must be first letter of each word.

Total number of words = 4! = 24

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 14

Result

Mathematics Test - 14

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SHARING IS CARING

Question 1

2/5

1/5

1/10

None of these

Solution

Question 2

-1/4

1/4

-1/2

1/2

Solution

Question 3

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 4

151.5

143.5

65

72

Solution

Question 5

7.75

7.5

7

5

Solution

Question 6

Less than the total number of observations

Greater than the total number of observations

Equal to total number of observations

Equal to mid point of the last class interval

Solution

Question 7

5/12

5/36

19/27

5/27

Solution

Question 8

25200

22500

10080

5040

Solution

Question 9

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 10

6

12

24

Neither 1 nor 2

Solution

User

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