Q:

A bag contains 4 white and 6 blue balls. If a ball is drawn at random, the probability that the drawn ball to blue is:

A:

3/5

Q:

If a coin is tossed twice, the probability of getting at least one head is

A:

3/4

Q:

The distance of a point (3, 3) from x-axis is

A:

3 units

Q:

If an angle is 3 times its complementary angle, then the measure of the angle is

A:

67.5°

Q:

If an angle of a parallelogram is four times of its adjacent angle, then these two angles of the parallelogram are

A:

144^{°}, 36°

Q:

Two sides of a triangle are 5 cm and 9 cm. Which of the following length can be the length of the third side?

A:

13 cm

Q:

A:

Positive rational number

Q:

Which of the following expressions is 'NOT' a polynomial?

A:

2x^{4 }+ (3/x^{2}) - 1

Q:

The length of the chord of a circle is 30 cm and its distance from the centre is 8 cm. Then, the radius of the circle is

A:

17 cm

Q:

Mode of data 24, 17, 13, 24, 26, 20, 26, 30, 8, 41, 24 is

A:

24

Q:

Let the sides of a triangle be 4 cm, 7 cm and 11 cm, then area of the triangle is_____ cm^{2}.

A:

Q:

In geometrical figures, sides decide sizes and _____ decide shapes.

A:

In geometrical figures, sides decide sizes and __angles__ decide shapes.

Q:

A:

Q:

An algebraic expression in which the variables involved have only non-negative integral powers is called a _________.

A:

An algebraic expression in which the variables involved have only non-negative integral powers is called a __polynomial__.

Q:

In the given figure, the measure of angle ABC is

A:

40°

Q:

On comparing 4x – 17 = 23 y with ax + by + c = 0, the value of a + b + c is _____.

A:

The given equation, 4x – 17 = 23 can be written as 4x – 23y – 17 = 0.

On comparing this equation with ax + by + c = 0, we

get, a = 4, b = – 23 and c = – 17

Then, a + b + c = 4 + (– 23) + (– 17)

= 4 – 40

= – 36

On comparing 4x – 17 = 23y with ax + by + c = 0, the value of a + b + c is __–36__.

Q:

A:

Standard form of the equation in two variable is ax+by+c = 0.

Therefore given equation can be written as 2x - 3y - 5 = 0.

Therefore given equation can be written as 2x - 3y - 5 = 0.

Q:

If the graph of a polynomial y = p(x) intersects the *x*-axis at three points, then what is the number of zeros of p(x)?

A:

The number of zeros is 3, as the graph intersects the *x*-axis at three points.

Q:

A:

x^{3} = 27

x = 3

x is a rational number.

x = 3

x is a rational number.

Q:

A:

The data which are arranged in ascending or descending order is called arrayed data.

Q:

A:

No, is non-terminating and non-recurring so it is not rational number.

Q:

A:

The frequency of an item is the number of occurence of that item in the given data.

Q:

A:

There are 4 kings in a deck of cards. Let E be the event of the card which is a king.

The number of outcomes favourable to E = 4

The number of possible outcomes = 52

Therefore,

P(a king) = (4/52)

= 1/13.

Q:

A:

Q:

A:

Q:

A:

Q:

If POR : ROQ = 5 : 7

A:

Q:

Find the zeros of polynomial 2x^{2} - 8.

A:

2x^{2 }- 8 = 0

⇒ 2x^{2} = 8

⇒ x^{2} = 4

⇒ x = 2 and -2

Hence, 2 and -2 are the zeros of the polynomial.

Q:

A:

Let f(x) = 2x^{3} - 3x^{2} + 7x - 6

then f(1) = 2(1)^{3} - 3(1)^{2} + 7(1) - 6

= 2 - 3 + 7 - 6

= 0

Hence, x = 1 is a zero of polynomial f(x)_{.}

then f(1) = 2(1)

= 2 - 3 + 7 - 6

= 0

Hence, x = 1 is a zero of polynomial f(x)

Q:

A:

Q:

A:

Inner radius of well (r) = 8/2 m

= 4 m

Width of embankment = 3 m

Radius of well with embankment (R) = 4 + 3 = 7 m

Depth of well = 21 m

Volume of earth taken out from well = r^{2}h

= (22/7)(4)^{2}21

= 22163

= 1056 m^{3}

Area of embankment = (R^{2} - r^{2})

= (22/7){(7)^{2} - (4)^{2}}

= (22/7)113

= (726/7) m^{2}

Height of embankment = __(Volume of earth taken out from well)__

(Area of embankment)

= 1056 / (726/7)

= __(10567)__

726

= 10.2 approx.

Q:

The cost of papering four walls of a room at 90 paise per square metre is 157.50. The height of the room is 5 metres. Find the length and the breadth of the room if they are in the ratio 4:1.

A:

Since ratio of length and breadth of room are 4:1.

So, let length of room = 4x

and breadth of room = x

Rate of papering = 90 paise

= 0.90

Then,

Area of 4 walls of room = 2(l + b)h

= 2(4x + x)5

= 50x m^{2}

Cost of papering wall = 50x0.90

= 45x

But cost of papering wall= 157.50

So, 45x = 157.50

x = 157.50 / 45

= 3.5

Hence,

length of the room = 4x

= 4(3.5)

= 14 m

breadth of the room = x

= 3.5 m

Q:

A:

Q:

A:

Q:

A:

Q:

A:

Q:

Factorise:

_{}

A:

_{}

Q:

A:

(i) = 4.79583 ……. [By square root method]

Which is non- terminating and non -recurring.

SO, by the decimal - expansion property of irrationals, is irrational.

(ii) Which is terminating.

SO, by the decimal expansion property of rational, is rational.

(iii) By the decimal expansion property of rational, is rational.

Q:

A:

To Construct:-ABC

Step 1. Draw a line segment AB 5.8 cm

Q:

A:

Since 1 kg.=1000 gm

So, the weight of one cubic cm of wood is 62.89 gm and one cubic cm of sand is 101.01 gm.

Q:

A:

Q:

A:

Let the cost of 1 kg apples be

According to the question, linear equation is

5x = 150 +7y

5x – 7y =150

It is the required equation.

Since the cost of one kg grapes is

By putting y = 70 in the above equation,

we get

5x – 7(70) = 150

5x = 490 + 150

x = 640/5

= 128

Thus, the cost of 1 kg apples is

Q:

A:

Since D is the mid point of AB hence CD is the median

ar(BCD) = ar(ACD)= 1/2ar(ABC) ...(i) (Median divides the triangle into two triangles of equal areas.)

ar(BCD) = ar(ACD)= 1/2ar(ABC) ...(i) (Median divides the triangle into two triangles of equal areas.)

ar(DPQ) = ar(PDC) (Two triangles having same base and between same parallel lines are equal in area.)

ar(DPQ) + ar(BDP) = ar(PDC)) + ar(BDP)

ar(BPQ)= ar(BCD)

ar(BPQ) = 1/2ar(ABC) [ from (i)]

Q:

A:

Construction: Produce BD to E so that BD = DE. Join EC

In ADB and CDE,

AD = DC (Given) ..(1)

BD = DE (By construction) ..(2)

Also, ADB = CDE ..(3)

So, ABD ACD (Using (1), (2), (3) and SAS rule).

This gives EC = AB and CED = ABD (CPCT)

CE AB ( Alternate angles CED = ABD)

ABC + ECB = 180°

90°^{ }+ ECB = 180°

ECB = 90°

Thus, In ABC and ECB,

AB = EC (Given) ..(1)

BC = CB (By construction) ..(2)

Also, ABC = ECB ..(3)

So, ABC ECB (Using (1), (2), (3) and SAS rule)

This gives AC = BE (CPCT)

(1/2)AC = (1/2)BE

or BD = (1/2)AC

or BD = (1/2)AC

Q:

A:

To Prove:-QTR=(1/2)QPR

Proof:-In PQR,

Ext.PRS=PQR+QPR

2TRS=2TQR+QPR [Since TQ and TR are angle bisectors]

2(TRS-TQR)=QPR

TRS-TQR=(1/2)QPR ...(1)

In TQR,

Ext.TRS=TQR+QTR

TRS-TQR=QTR ...(2)

from equation (1) and equation (2), we have

QTR =(1/2) QPR

Q:

A circular park of radius 20 m is situated in a colony. Three boys

Ankur, Syed and David are sitting at equal distance on its

boundary each having a toy telephone in his hands to talk to

each other. Find the length of the string of each telephone.

A:

It is given that AS = SD = DA

Therefore, ASD is an equilateral triangle.

OA (radius) = 20 m

Medians of equilateral triangle pass through the circumcentre (O) of the

equilateral triangle ASD. We also know that medians intersect each other

in the ratio 2: 1. AB is the median of equilateral triangle ASD,

(OA/OB) = 2/1

(20/OB) = 2/1

OB = 10 m

AB = OA + OB = (20 + 10) m = 30 m

In ABD, ABD=90, So by Pythagoras Theorem

AD^{2} = AB^{2} + BD^{2 }

AD^{2} = (30)^{2} + (AD/2)^{2 }

AD^{2} = 900 + (1/4)AD^{2 }

(3/4)AD^{2} = 900

AD^{2} = 1200

AD = 203

Therefore, the length of the string of each phone will be 203 m.

Q:

A:

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