Question of Solved Question

# Question Fifty seeds were selected at random from each of 5 bags A, B, C, D, E of seeds, and were kept under standardized conditions equally favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follow: Bag A B C D E Number of seeds germinated 40 48 42 39 41 What is the probability of germination of (i) more than 40 seeds in a bag? (ii) 49 seeds in a bag? (iii) more than 35 seeds in bag?

Represent Root 9 point 3 on the number line

Solution:

Explanation:

Step 1: Draw a line segment AB of length 9.3 units.

Step 2: Now, Extend the line by 1 unit more such that BC=1 unit .

Step 3: Find the midpoint of AC.

Step 4: Draw a line BD perpendicular to AB and let it intersect the semicircle at point D.

Step 5: Draw an arc DE such that BE=BD.

Hence, Number line of √ 9.3 is attached below.

Which one of the following statement is true

A: Only one line can pass through a single point.

B: There are an infinite number of lines which pass through two distinct points.

C: Two distinct lines cannot have more than one point in common.

D: If two circles are equal, then their radii are not equal.

Solution:

Explanation:

From one point there is an uncountable number of lines that can pass through.

Hence, the statement “ Only one line can pass through a single point” is false.

We can draw only one unique line passing through two distinct points.

Hence, the statement “There are an infinite number of lines which pass through two distinct points” is false.

Given two distinct points, there is a unique line that passes through them.

Hence, the statement “Two distinct lines cannot have more than one point in common” is true.

If circles are equal, which means the circles are congruent. This means that circumferences are equal and so the radii of two circles are also equal.

Hence, the statement “If two circles are equal, then their radii are not equal” is false.

The correct option is (C) Two distinct lines cannot have more than one point in common.

The class mark of the class 90-120 is

A: 90

B: 105

C: 115

D: 120

Solution:

Explanation:

To find the class mark of a class interval, we find the sum of the upper limit and lower limit of a class and divide it by 2 Thus,

Class -mark=Upper limit + Lower limit/2

Here, the lower limit of 90-120=90

And the upper limit of 90-120=120

So,

Class -mark=120+90/2

=210/2

=105

Hence, the class mark of the class 90-120 is 105

That is, option (B) is correct.

Option (B) 105  is correct.

ABCD is a parallelogram and AP

and CQ are perpendiculars from vertices A and C on diagonal BD Show that i)ΔAPBΔCQD ii) AP = CQ

Solution:

Find the roots of the following equation

A: - 1, - 2

B: - 1, - 3

C: 1,3

D: 1,2

Solution:

Explanation:-