**Himanshi Verma**

Vitamin D.

*-Answered by Himanshi Verma* On 10 March 2019:04:41:55 PM

**Question 1.** **Asked on :**09 March 2019:01:54:23 PM

कैल्सीफिरोल किस विटामिन का रासायनिक नाम है ?

*-Added by Rohit rajput* **Mathematics** » **G.k **

**Answer:**

Vitamin D.

*-Answered by Himanshi Verma* On 10 March 2019:04:41:55 PM

**Question 2.** **Asked on :**01 March 2019:04:06:17 PM

Find the value of k for which one root of the quadratic equation kx^{2}-14x+8=0 is 2.

*-Added by Harshita Rathore* **Mathematics** » **G.k **

**Answer:**

Since 2 is one root of the quadratic equation,

We substitute X=2

k(2)^2 - 14(2) +8= 0

4k-28+8=0

4k-20=0

4k=20

k=5

*-Answered by Himanshi Verma* On 10 March 2019:04:48:03 PM

**Question 3.** **Asked on :**01 March 2019:03:26:35 PM

How to prove Pythagorus theorem .

*-Added by Akki chauhan* **Mathematics** » **Maths**

**Answer:**

ABC is a right angle triangle at B and BD is a perpendicular from B on to AC, the diagonal.

Triangles ABC and BDC are similar, as:

. angle DBC = 90 - angle C = angle A

. angle ABC = angle BDC = 90

Ratios of corresponding sides are equal.

AB / BD = BC / DC = AC / BC

So we get BC² = AC * DC --- (1)

Triangle ABC amd ADB are similar, as:

. angle DBA = 90 - (90 - C) = angle C

. angle ABC = 90 = angle ADB

so ratios of corresponding sides are equal

AB / AD = BC / BD = AC / AB

So we get AB² = AD * AC --- (2)

Add (1) and (2) to get:

AB² + BC² = AC * ( AD + DC) = AC * AC = AC²

So the theorem is proved.

*-Answered by Himanshi Verma* On 01 March 2019:03:38:26 PM

**Question 4.** **Asked on :**01 March 2019:03:15:18 PM

*-Added by Akki chauhan* **Mathematics** » **Algebra**

**Answer:**

*-Answered by Shalu Shukla* On 01 March 2019:03:47:38 PM

**Question 6.** **Asked on :**13 February 2019:10:06:57 AM

** Find the value of X**

*-Added by Akki chauhan* **Mathematics** » **Circles**

**Question 7.** **Asked on :**09 February 2019:04:37:21 PM

In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

*-Added by Harshita Rathore* **Mathematics** » **Areas Of Pallelograms And Triangles**

**Answer:**

Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle = 1/2 x Base x Altitude

It is given that DE = BD = EC.

⊥ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that Budhia has divided her field into 3 equal parts.

*-Answered by Himanshi Verma* On 09 February 2019:04:40:18 PM

**Question 8.** **Asked on :**09 February 2019:04:31:17 PM

What is the range of the probability of an event.

*-Added by Harshita Rathore* **Mathematics** » **Probability**

**Answer:**

If an event is impossible its probability is zero. Similarly, if an event is certain to occur, its probability is one. The probability of any event lies in between these values. It is called the range of probability and is denoted as **0** ≤ P (E) ≤ 1.

*-Answered by Himanshi Verma* On 09 February 2019:04:32:27 PM

**Question 9.** **Asked on :**09 February 2019:04:27:51 PM

Water is placed in a cylindrical vessel of 14cm radius. If a spherical ball of radius 7cm is dropped in the vessel completely. How much water will rise to the height of the cylindrical vessel?

*-Added by Himanshi Verma* **Mathematics** » **Volume**

**Answer:**

बेलन के पानी का आयतन = गोले का आयतन

πr^{2}h = 4/3 πr^{3}

14×14×h = 4\3×7×7×7

h = 4/3 × 343/196

h = 7/3cm

*-Answered by Shalu Shukla* On 11 February 2019:05:45:38 PM

**Question 10.** **Asked on :**09 February 2019:03:02:25 PM

In the following,find the area of the shaded portains:

*-Added by Himanshi Verma* **Mathematics** » **Chapter 11.4**

**Question 11.** **Asked on :**07 February 2019:09:36:04 AM

*-Added by SURAJ KUMAR* **Mathematics** » **Trignometry **

**Question 12.** **Asked on :**06 February 2019:04:16:04 PM

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD .If the area of parallelogram is 1470cm^{2 }, AB =35cm and AD=49 *cm ,find the length of BM and DL.*

*-Added by Himanshi Verma* **Mathematics** » **Chapter 11 Perimeter And Area**

**Answer:**

Considering AB = Base;

Area = Base x Height

⇒ 1470 cm2 = AB x DL

⇒ 1470 cm2 = 35cm x DL

Or, DL = 1470 ÷ 35 = 42 cm

Now considering AD = Base;

Area = Base x Height

⇒ 1470 cm2 = AD x BM

⇒ 1470 cm2 = 49 cm x BM

Or, BM = 1470 ÷ 49 = 30 cm

Hence, DL = 42 cm and BM = 30 cm

*-Answered by Himanshi Verma* On 09 February 2019:04:21:59 PM

**Question 13.** **Asked on :**05 February 2019:05:27:35 PM

*-Added by Himanshi Verma* **Mathematics** » **G.k **

**Answer:**

8 8 8

8 8

8

8

+ 8

= 1000

*-Answered by Himanshi Verma* On 05 February 2019:05:36:22 PM

**Question 14.** **Asked on :**05 February 2019:05:12:59 PM

If x = , what is the value of

*-Added by Harshita Rathore* **Mathematics** » **Number System **

**Question 15.** **Asked on :**05 February 2019:05:11:52 PM

*-Added by Himanshi Verma* **Mathematics** » **Perimeter And Area**

**Answer:**

Given

Ratio of side =3:2

Perimeter =60cm

Altitude =5cm

So let take one side as 3x and other is 2x

Then perimeter of parallelogram

=3x+2x+3x+2x

=10x

According to question perimeter =60cm

Then

10x=60cm

X=60/10=6cm

Then one side is =6 x 3=18

And other is =2x6=12

So base=18cm

Area of parallelogram =bxh

=18x5=90cm²

Altitude corresponding to smaller side

90=12xh

H=7.5cm

*-Answered by Himanshi Verma* On 05 February 2019:05:26:39 PM

**Question 16.** **Asked on :**05 February 2019:09:14:22 AM

Which is the largest chord in circle.

*-Added by Akki chauhan* **Mathematics** » **Circles**

**Answer:**

Diameter is the greatest chord of circle.

*-Answered by Himanshi Verma* On 05 February 2019:04:58:06 PM

**Question 17.** **Asked on :**03 February 2019:05:07:16 PM

Prove that :- 777+888+999=999

*-Added by Himanshi Verma* **Mathematics** » **Addition**

**Answer:**

9+8+8+7+7=39,

9+9+8+3=29,

7+2=9

We write last nine of the three solution so, we get 777+888+999=999

Hence, proved.

*-Answered by Nishant Verma* On 03 February 2019:05:23:34 PM

**Question 18.** **Asked on :**03 February 2019:05:02:10 PM

Prove that :- 777+888+999=999

*-Added by Shalu Shukla* **Mathematics** » **Addition**

**Answer:**

9+8+8+7+7=39,

9+9+8+3=29,

7+2=9

We write last nine of the three solution so, we get 777+888+999=999

Hence, proved.

*-Answered by Himanshi Verma* On 05 February 2019:04:51:38 PM

**Question 19.** **Asked on :**30 January 2019:02:19:19 PM

Why is the area of circle is πr^{2}?

*-Added by Himanshi Verma* **Mathematics** » **Circle**

**Answer:**

The usual definition of pi is the ratio of the circumference of a **circle** to its diameter, so that the circumference of a **circle** is pi times the diameter, or 2 pi times the radius. ... This give a geometric justification that the **area** of a **circle** really is "pi r squared".

*-Answered by Himanshi Verma* On 02 February 2019:01:33:01 PM

**Question 20.** **Asked on :**28 January 2019:11:33:19 AM

tanA+secA-1/tanA+secA+1=cosA/1-sinA .

Proove that.

*-Added by SURAJ KUMAR* **Mathematics** » **Trignometry **

**Answer:**

(secA+tanA-(sec^2 A - tan ^2A)) )/tanA-secA+1) as sec^2 A= 1 + tan ^2 A so 1= sec^2 A - tan ^2 A

= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1)

= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1)

= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1)

= (sec A + tan A)

= 1/cos A + sin A/ cos A

= (1+ sin A)/ cos A

= (1 + sin A )(1- sin A)/(cos A (1- sin A))

= (1- sin ^2 A/(cos A (1- sin A))

= cos ^2 A / (cos A (1- sin A))

= cos A /(1- sin A)

proved

*-Answered by Himanshi Verma* On 02 February 2019:01:40:14 PM

**Question 22.** **Asked on :**28 January 2019:10:53:52 AM

A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. find the area of the garden in hectare.

*-Added by praful kumar* **Mathematics** » **Perimeter And Area **

**Answer:**

Length (l) of garden = 90 m

Breadth (b) of garden = 75 m

Area of garden = l × b = 90 × 75 = 6750 m2

From the figure, it can be observed that the new length and breadth of the garden, when path is also included, are 100m and 85m respectively.

Area of the garden including the path = 100 × 85 = 8500 m2

Area of path = Area of the garden including the path − Area of garden

= 8500 − 6750 = 1750 m2

1 hectare = 10000 m2

Therefore, area of garden in hectare

=6750/10000

=0.675

*-Answered by Himanshi Verma* On 02 February 2019:01:49:33 PM

**Question 23.** **Asked on :**28 January 2019:10:49:14 AM

Find the area of a square park whose perimeter is 320m.

*-Added by praful kumar* **Mathematics** » **Perimeter And Area **

**Answer:**

A=6400m^{2}

*-Answered by Himanshi Verma* On 03 February 2019:04:19:42 PM

**Question 24.** **Asked on :**27 January 2019:11:08:39 AM

(4-k)x^{2}+(2k+4)x+(8k+1)=0.

k=?

एक पुर्ण वर्ग है ?

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

The given quadratic will be a perfect square if it has two real and equal roots. To have two real and equal roots the discriminant must be zero.

i.e., b²-4ac=0

or, (2k+4)²-4(4-k)(8k+1)=0

or, 4k²+16k+16-4(32k-8k²+4-k)=0

o², 4k²+16k+16-128k+32k²-16+4k=0

or, 36k²-108k=0

or, 36k(k-3)=0

either 36k=0

or, k=0

or, k-3=0

or, k=3

∴, k=0,3 Ans.

*-Answered by Himanshi Verma* On 02 February 2019:01:43:20 PM

**Question 25.** **Asked on :**27 January 2019:10:59:13 AM

abx^{2}_{+(b}^{2}_{ac)x-bc=0.सरल करे|}

_{}

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

Consider, abx +(b-ac)x-bc = 0

⇒ abx + bx – acx – bc = 0⇒ bx (ax + b) – c(ax + b) = 0⇒ (ax + b)(bx – c) = 0*-Answered by Himanshi Verma* On 02 February 2019:01:44:13 PM

**Question 26.** **Asked on :**27 January 2019:10:54:59 AM

ax^{2}_{+4ax+(a}^{2}_{-b}^{2}_{)=0}

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Question 27.** **Asked on :**27 January 2019:10:51:05 AM

1/a+b+x=1/a+1/b+1/x,a+b not equal 0.सरल करे |

*-Added by SURAJ KUMAR* **Mathematics** » **POLYNOMIALS**

**Answer:**

ON solving the above equation we have a quadratic equation

(a+b)x^2+(a+b)^2x+ab(a+b)

on solving the above equation by formula -b+root over(b^2-4ac) and -b-root over(b^2-4ac)

we got -a and -b as roots.

According to me this is much easier then others when compared.

*-Answered by Himanshi Verma* On 10 March 2019:04:49:29 PM

**Question 29.** **Asked on :**27 January 2019:10:19:50 AM

1,2,3 ,...,33,34,35 में 7 का गुणज आने

*-Added by praful kumar* **Mathematics** » **Prayikta**

**Question 30.** **Asked on :**21 January 2019:10:00:50 AM

__+__+__=30 (use this number and solve question, you also reapet this number )

[1,3,5,7,9,11,13,15,]

*-Added by Akki chauhan* **Mathematics** » **G.k **

**Answer:**

: (11 + 9) + (3) + (7) = 30

ANOTHER: (13 + 15) + (1) + (1) = 30

OR: (7 + 9) + (1) + (13) = 30

*-Answered by Himanshi Verma* On 21 January 2019:04:52:39 PM

**Question 31.** **Asked on :**20 January 2019:10:13:42 AM

cotΦ-1/sinΦ=1

*-Added by Rohit Rajput* **Mathematics** » **Maths**

**Answer:**

Cscϕ/secϕ = (1+cotϕ)/(1+tanϕ)

LHS = (1+cotϕ)/(1+tanϕ)

where cotϕ=cosϕ/sinϕ and tanϕ = sinϕ/cosϕ

= (1+(cosϕ/sinϕ))/(1+(sinϕ/cosϕ))

= ((cosϕ+sinϕ)/sinϕ)/((cosϕ+sinϕ)/cosϕ)

= cosϕ/sinϕ

wkt cosϕ=1/secϕ

and sinϕ=1/cosecϕ

= cscϕ/secϕ

LHS = RHS

Hence Proved

*-Answered by Himanshi Verma* On 22 January 2019:05:15:48 PM

**Question 32.** **Asked on :**19 January 2019:10:22:48 AM

On tossing a coin 1000 times head comes 425 times find the probability of getting a tails in this event.

*-Added by Prince Rawat* **Mathematics** » **STATISTICS**

**Answer:**

Total no. of times tossing a coin= 1000

no. of head comes= 425

no. of tail comes= 1000-425= 575

probability p(e)= 575/1000

*-Answered by Himanshi Verma* On 20 January 2019:04:27:24 PM

**Question 33.** **Asked on :**19 January 2019:10:07:46 AM

Three spheres having radii 2 cm ,3 cm and 5 cm are melted together to from a single find the radius of the new sphere.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

Let three spheres are S1, S2 & S3

having radii r₁ = 3cm, r₂ = 4cm & r₃ = 5cm respectively.

Let the radius of new Big sphere S is R.

A/Q,

Volume of new Sphere S = Sum of volumes three Spheres S1, S2 & S3

⇒ 4/3πR³ = 4/3π(r₁)³ + 4/3π(r₂)³+4/3π(r₃)³

⇒ 4/3πR³ = 4(r₁³ +r₂³ + r₃³)/3π

⇒ R³ = r₁³ +r₂³ + r₃³

⇒ R³ = 3³ +4³ + 5³ = 27 + 64 + 125

⇒ R³ = 216

⇒ R = ∛216

⇒ R = 6cm

Therefore radius of new Sphere is 6 cm.

*-Answered by Himanshi Verma* On 20 January 2019:05:15:36 PM

**Question 34.** **Asked on :**19 January 2019:10:05:05 AM

The slant height of a cone is 10 cm and its base radius is 8 cm find total surface area of the cone.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

slant height of cone is : 10

Radius of the base is : 8 cm

Total surface area of cone is : πr(l+r)

227 ×8(10+8)

227 ×144

= 452.57

*-Answered by Himanshi Verma* On 20 January 2019:04:27:57 PM

**Question 35.** **Asked on :**18 January 2019:05:21:24 PM

Write the formula of cylinder open from the top ?

*-Added by khushi chauhan* **Mathematics** » **Maths**

**Answer:**

Total surface area= 2πr(r+h) -πr^{2}

curved surface area = 2πrh+πr^{2}

*-Answered by Himanshi Verma* On 03 February 2019:04:44:16 PM

**Question 36.** **Asked on :**17 January 2019:05:34:31 PM

ग्राफ में आमने सामने की भुजाओ को समान्तर बताने के लिए सूत्र लिखो?

*-Added by Sumitra Mahajan* **Mathematics** » **G.k **

**Question 37.** **Asked on :**17 January 2019:08:47:48 AM

path 1m wide built along the border and inside a square garden of side 30m. find ;

[i] The area of the path.

[ii] the cost of planting grass in the remaining portion of the garden at the rate of Rs.40 per m^{2}

*-Added by Nitish kumar* **Mathematics** » **Perimeter And Area **

**Answer:**

Let ABCD, the square garden of side= 30 m.

PQRS is the region inside the garden.

so,

PQ = (30 -1-1 ) = 28 m

also

PS = ( 30 – 1 – 1 ) = 28 m

(i) Area of the path = Area of ABCD – Area of PQRS

= [( 30 x 30) – (28 x 28)]

= 900 – 784

= 116 sq.m

(ii) Area of the remaining portion in which the grass is planted = Area of square PQRS

= 28 x 28

= 784 sq.m

cost of planting the grass in the region PQRS = area x cost = 764 x 40

=31,360

*-Answered by Himanshi Verma* On 18 January 2019:05:16:10 PM

**Question 38.** **Asked on :**16 January 2019:06:13:49 PM

An integer that is divisible by 2 is called ?

*-Added by Shalu Shukla* **Mathematics** » **G.k**

**Answer:**

An integer that is divisible by 2 is called Even Number.

*-Answered by Himanshi Verma* On 17 January 2019:05:28:03 PM

**Question 39.** **Asked on :**16 January 2019:06:14:35 PM

Counting number are kept under _______ number.

*-Added by Himanshi Verma* **Mathematics** » **G.k **

**Answer:**

**Counting numbers** are **numbers** you can use to **count** whole items. You can use the positive integers to **count** marbles, for example. Whether 0 is a **counting number** or not, that's a big debate you may or may not be interested in

*-Answered by Harshita Rathore* On 17 January 2019:05:31:45 PM

**Question 40.** **Asked on :**16 January 2019:05:34:22 PM

(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}(a-b)^{3}+(b-c)^{3}+(c-a)^{3}Simplify this Equation:

*-Added by Shalu Shukla* **Mathematics** » **Ch-2**

**Answer:**

We know,

---> if a + b + c = 0 ; a³ + b³ + c³ = 3abc

We observe,

( a² - b² ) + ( b² - c² ) + ( c² - a² ) = 0

=> ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ = 3( a² - b² )( b² - c² )( c² - a² )

Similarly,

( a - b ) + ( b - c ) + ( c - a ) = 0

=> ( a - b )³ + ( b - c )³ + ( c - a )³ = 3( a - b )( b - c )( c - a )

Now, [ ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ ] / [ ( a - b )³ + ( b - c )³ + ( c - a )³ ]

== 3( a² - b² )( b² - c² )( c² - a² ) / 3( a - b )( b - c )( c - a )

= ( a + b )( b + c )( c + a ) <---- Answer..

*-Answered by Himanshi Verma* On 03 February 2019:04:45:43 PM

**Question 41.** **Asked on :**16 January 2019:05:39:35 PM

Among the following which natural numbers had no Prodecessor?

*-Added by Himanshi Verma* **Mathematics** » **G.k **

**Answer:**

1

*-Answered by Himanshi Verma* On 17 January 2019:05:27:08 PM

**Question 42.** **Asked on :**16 January 2019:05:34:43 PM

How many digits are there in Hindu Arabic System?

*-Added by Himanshi Verma* **Mathematics** » **G.k **

**Answer:**

The Hindu-Arabic numerals are the **ten digits** (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). They are descended from Indian numerals, and the Hindu-Arabic numeral system by which a sequence of digits such as "406" is read as a whole number was developed by Indian mathematicians.

*-Answered by Himanshi Verma* On 17 January 2019:05:27:36 PM

**Question 43.** **Asked on :**15 January 2019:09:48:55 AM

The lateral surface area of a cube is 256cm^{2}. Find its volume.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

Lateral Surface Area of cube=256 cm²

4(side)² =256side²=256÷4

side²=64

side=√64=8 cm

Volume of the cube=(side)³=8×8×8=512 cm³.

*-Answered by Himanshi Verma* On 16 January 2019:06:04:58 PM

**Question 44.** **Asked on :**15 January 2019:09:31:00 AM

The length, breadth and height of room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of s. 7.50perm^{2}.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

l=5m b=4m h=3m

The total area of cuboid = 2(lb+bh+hl)+l×b

= 2(5×4+4×3+3×5)+5×4

= 2(20+12+15)+20

= 2(47) +20

= 94+20

= 114m^{2}

The cost of white washing the wall and the ceiling of the room = 7.50 per m^{2}

The cost of 114m^{2} of the walls and ceiling of the room= (7.50×114)Rs

= 855.00Rs.

*-Answered by Himanshi Verma* On 19 January 2019:05:10:07 PM

**Question 45.** **Asked on :**15 January 2019:09:27:40 AM

Find the ratio of total surface area of sphere and a hemisphere of same radius.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

The total surface area of sphere= 4πr

The total surface area of hemisphere= 3πr^{2}

The ratio of total surface area of sphere and hemisphere= 4πr^{2}/3πr^{2}

= 4:3

*-Answered by Himanshi Verma* On 20 January 2019:04:38:18 PM

**Question 46.** **Asked on :**15 January 2019:09:24:30 AM

The total surface area of a cube is 150sq. cm. Find the perimeter of any one of its faces.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

Total surface area of cube is 150sq. cm

Total surface area of cube : 6a^{2}

^{To find perimeter of its one faces : ?}

Total surface area of cube =150sq.cm^{}

6a^{2 }=150sq.cm

= 1506

= 25cm

=a^{2} = √25cm

=a = 5 cm

Perimeter of its one face is = 4 × a

= 4 × 5 cm

= 20 cm **Answer**

*-Answered by Himanshi Verma* On 19 January 2019:04:39:34 PM

**Question 47.** **Asked on :**15 January 2019:09:20:19 AM

The surface area of the cuboid is 1372 sq. cm. If its dimensions are in the ratio of 4:2:1. Then find length.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

The surface area of the cuboid= 1372 sq. cm

Let the dimensions =4x, 2x, 1x

The total surface area of the cuboid = 2(lb+bh+hl)=1372

= 2(4x × 2x + 2x × 1x + 4x × 1x )=1372

=2(8x^{2}+2x^{2}+4x^{2}) =1372

=2(14x^{2})= 1372

=28x^{2}=1372

= x^{2}=1372/28

= 49

x=7m

the dimensions are 4x=4×7=28m

2x=2×7=14m

1x=1×7=7m

*-Answered by Himanshi Verma* On 19 January 2019:05:17:58 PM

**Question 48.** **Asked on :**14 January 2019:10:35:56 AM

The ratio of height of two cylinder is 5:3 as well as the ratio of their radii is 2:3. find the ratio of the volume of the cylinder.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

Let the radii be 2x and 3x,

let the height be 5y and 3y.

so, ratio of volume =r²h/R²H =(2x)²×5y/(3x)²×3y =20x²y/27x²y =20/27,

The ratio is = 20:27

*-Answered by Himanshi Verma* On 20 January 2019:05:11:58 PM

**Question 49.** **Asked on :**13 January 2019:05:13:09 PM

show that if sum of the two angles of a triangles is equal to the third angle is right triangle.

*-Added by khushi chauhan* **Mathematics** » **Lines And Angles**

**Answer:**

Let the first and second angle be a and b respectively

Third angle = a+b

By angle sum property of triangle

a +b + a+b = 180

=> 2a +2b = 180

=> 2(a+b) = 180

=> a+b = 90°

Third angle = 90°

*-Answered by Himanshi Verma* On 20 January 2019:05:07:15 PM

**Question 50.** **Asked on :**13 January 2019:04:58:17 PM

In a ΔABC, ∠A+∠B=125º And ∠B+∠C=150º. Find all the angles of ΔABC.

*-Added by Himanshi Verma* **Mathematics** » **Lines And Angles **

**Question 51.** **Asked on :**13 January 2019:01:13:28 PM

Find the median of11,12,2x+2,3x,21,23 where x is the mean of 5,10,3.

*-Added by Akki chauhan* **Mathematics** » **Probability**

**Answer:**

We know,

if the number of observation (n) is even

then,

1. first of all find the value at the position

2. and find the value at the position

3. now find the average of two value to get the median .

e.g.,

Given, 11, 12, 14, 18, (x + 2), (x + 4) , 30, 32 , 35 , 41 are in ascending order .

number of terms = 10 {even}

so, median = {(n/2)th + (n/2 + 1) th }/2

24 = (5th + 6th)/2

24 = {(x + 2) + (x + 4)}/2

24 = (x + 3)

x = 21

hence, x = 21

*-Answered by Jyoti Srivastva * On 30 January 2019:04:32:47 PM

**Question 52.** **Asked on :**13 January 2019:12:42:04 PM

Fill in the blanks :

(i) The probability of an impossible event is .............

(ii) The probability of happening of a possible event is always lies between ........... and...........

(iii) On tossing a coin, We get a total of ............. outcomes.

*-Added by Akki chauhan* **Mathematics** » **Probability**

**Answer:**

Ans 1. Zero(0)

Ans 2. 0 and 1

Ans 3. 2

*-Answered by Akki chauhan* On 15 January 2019:09:06:44 AM

**Question 53.** **Asked on :**13 January 2019:12:13:10 PM

If the ratio of altitude and area of the parallelogrm is 2:1 then find the length of the base of parallelogram.

*-Added by Himanshi Verma* **Mathematics** » **Areas Of Parallelogram And Triangles**

**Question 54.** **Asked on :**13 January 2019:11:53:09 AM

A bag contain slips bearing from 1-89. A slip is drawn at random from the bag. Find the probability that the slip drawn has:

(a) A number written on it such that sum of its digit is 11.

(b) A number greater than 63 written on it.

*-Added by Akki chauhan* **Mathematics** » **Probability**

**Answer:**

Let 'x' be the number of boys and 'y' be the number of girls.

To find the total marks

The total marks of boys = 42x

The total marks of girls = 45y

Total number of boys and girls = x + y

To find the ratio

The total marks of boys and girls = 44(x + y)

Then we can write

44(x + y) = 42x + 45y

44x + 44y = 42x + 45y

2x = y

x/y = 1/2

Therefore the ratio of the number of boys to the number of girls is 1:2

*-Answered by Himanshi Verma* On 20 January 2019:05:29:47 PM

**Question 55.** **Asked on :**13 January 2019:11:49:45 AM

In a class test, the mean marks of boys and girls are respectively 42 and 45. If the mean marks of boys and girls are 44, Find the ratio of the number of boys to the number of girls.

*-Added by Akki chauhan* **Mathematics** » **Statistics**

**Answer:**

Let 'x' be the number of boys and 'y' be the number of girls.

To find the total marks

The total marks of boys = 42x

The total marks of girls = 45y

Total number of boys and girls = x + y

To find the ratio

The total marks of boys and girls = 44(x + y)

Then we can write

44(x + y) = 42x + 45y

44x + 44y = 42x + 45y

2x = y

x/y = 1/2

Therefore the ratio of the number of boys to the number of girls is 1:2

*-Answered by Himanshi Verma* On 22 January 2019:05:24:43 PM

**Question 56.** **Asked on :**13 January 2019:11:44:06 AM

THe slant height of cone is 10 cm and its base radius is 8 cm. Find the total surface area of cone.

*-Added by Akki chauhan* **Mathematics** » **Surface Ares And Volumes.**

**Answer:**

l=10cm r=8cm

The total surface area of cone=πr(l+r)

=22/7×8(10+8)

=22/7×8(18)

=22/7×144

=3168/7cm

*-Answered by Akki chauhan* On 15 January 2019:09:06:10 AM

**Question 57.** **Asked on :**13 January 2019:11:37:50 AM

Find the volume of hemisphere whose radius is

32 cm.

*-Added by Akki chauhan* **Mathematics** » **Surface Ares And Volumes.**

**Answer:**

r= 32

Volume of hemisphere= 23 πr^{3}

^{}

= 23 ×22/7×3/2×3/2×3/2

= 7.071

*-Answered by khushi chauhan* On 22 January 2019:05:23:49 PM

**Question 58.** **Asked on :**13 January 2019:11:15:32 AM

Define concentric circles.

*-Added by Akki chauhan* **Mathematics** » **Circles**

**Answer:**

The concentric circle are the circles with a common center the region between two concentric circle of different radii is called an annulus.

*-Answered by Akki chauhan* On 15 January 2019:09:05:48 AM

**Question 59.** **Asked on :**12 January 2019:05:55:36 PM

add √125 + 2√27 and -5√5 - √3

*-Added by khushi chauhan* **Mathematics** » **Number System **

**Answer:**

√125 +2√27 +(-5√5 -√3)

=√5×5×5 +2√3×3×3 -5√5 -√3

=5√5 +6√3 -5√5 -√3

=5√3

*-Answered by Akki chauhan* On 15 January 2019:09:05:29 AM

**Question 60.** **Asked on :**12 January 2019:05:54:25 PM

Rationalise the denominator = 1√3+√5+√7

*-Added by Himanshi Verma* **Mathematics** » **Number System **

**Question 61.** **Asked on :**12 January 2019:05:52:26 PM

prove that the cyclic parallelogram is rectangle.

*-Added by khushi chauhan* **Mathematics** » **Circles**

**Answer:**

∠A + ∠C = 180 ....1

But ∠A = ∠C

So ∠A = ∠C = 90

Again

∠B + ∠D = 180 ....2

But ∠B = ∠D

So ∠B = ∠D = 90

Now each angle of parallelogram ABCD is 90.

Hense ABCD is a rectangle

*-Answered by Akki chauhan* On 15 January 2019:09:05:05 AM

**Question 62.** **Asked on :**12 January 2019:10:52:27 AM

Find the zero of 2x-5.

*-Added by Akki chauhan* **Mathematics** » **Number System**

**Answer:**

2x-5=0

2x=5

x= 5/2

*-Answered by Akki chauhan* On 15 January 2019:09:04:26 AM

**Question 63.** **Asked on :**12 January 2019:10:49:51 AM

Factorise:

x^{3}-23x^{2}+142x-120

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

x^{3 }-23x^{2}+142x-120

=x^{3}-x^{2}-22x^{2}+22x+120-120

=x^{2}(x-1)-22x(x-1)+120(x-1)

=(x-1)(x^{2}-22x+120)

=(x-1)(x^{2}-12x-10x+120)

=(x-1)[x(x-12)-10(x-12)]

=(x-1)(x-10)(x-12)

*-Answered by Himanshi Verma* On 16 January 2019:05:57:48 PM

**Question 64.** **Asked on :**12 January 2019:10:45:37 AM

If the polynomials ax^{3}+3x^{2}-13 and 2x^{3}-5x+a , are divided by x+2 if the remainder in each case is the same, then find the value of 'a'

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

*-Answered by Himanshi Verma* On 20 January 2019:05:29:25 PM

**Question 65.** **Asked on :**12 January 2019:10:38:50 AM

If a = 9-4√5, then find the value of

a^{2} + 1a^{2}

*-Added by Akki chauhan* **Mathematics** » **Polynomials**

**Answer:**

(a+1/a)2=a2+1/a2+2

a2+1/a2=(a+1/a)2-2→ I

Now,

a=9-4root5

1/a=1/9-4root5

=(1/9-4root5)×9+4root5/9+4root5

=(9+4root5)

a+1/a=9-4root5+9+4root5

=81

Now,

a2+1/a2=(a+1/a)2-2 →(from I)

=(81)2-2

=6561-2

=6559

*-Answered by Himanshi Verma* On 22 January 2019:05:31:51 PM

**Question 66.** **Asked on :**11 January 2019:02:55:18 PM

Find the median of first 10 prime numbers.

*-Added by Prince Rawat* **Mathematics** » **STATISTICS**

**Answer:**

The first 10 prime numbers are : 2,3,5,7,11,13,17,19,23 and 29.

n=10

Median if n is even numbers

n2 th term = 5th term = 11

n2 + 1 = 5 + 1 = 6th term = 13

Median = 5th term + 6th term2

Median = 11 + 132

= 242 = 12 **Answer**

*-Answered by Akki chauhan* On 15 January 2019:09:03:48 AM

**Question 67.** **Asked on :**11 January 2019:10:18:43 AM

Define frequency of the observation.

*-Added by Prince Rawat* **Mathematics** » **STATISTICS**

**Answer:**

The number of times an observation occurs in the given data is called frequency of the observation.

*-Answered by Rohit Rajput* On 12 January 2019:08:59:05 AM

**Question 68.** **Asked on :**11 January 2019:10:10:33 AM

The volume of sphere is 310.4 cm^{3}. Find its radius.

*-Added by Prince Rawat* **Mathematics** » **SURFACE AREAS AND VOLUMES**

**Answer:**

volume of the sphere = 43 πr^{3}= 310.4cm^{3}

4/3×22/7×r^{3}=310.4

88/21×r^{3}= 310.4

r^{3}= 310.4×21/88

= 6518.4/88

r^{3} = 74.0727272cm

*-Answered by Himanshi Verma* On 19 January 2019:04:56:34 PM

**Question 69.** **Asked on :**11 January 2019:10:04:29 AM

What is Degree of the Polynomials?

*-Added by Prince Rawat* **Mathematics** » **POLYNOMIALS**

**Answer:**

Highest power of x in the algebraic expression is called the degree of the polynomials.

*-Answered by Prince Rawat* On 11 January 2019:10:08:00 AM

**Question 70.** **Asked on :**09 January 2019:09:54:18 AM

Prove that if the chords are equal then the subtend angle at the center are equal?

*-Added by Master Purushottam* **Mathematics** » **Circles**

**Answer:**

Given : In a circle C(O,r) , ∠AOB = ∠COD

To Prove : Chord AB = Chord CD .

Proof : In △AOB and △COD

AO = CO [radii of same circle]

BO = DO [radii of same circle]

∠AOB = ∠COD [given]

⇒ △AOB ≅ △COD [by SAS congruence axiom]

⇒ Chord AB = Chord CD [c.p.c.t]

*-Answered by Akki chauhan* On 13 January 2019:01:04:33 PM

**Question 71.** **Asked on :**09 January 2019:09:37:10 AM

Ques.1. prove that 2+2= 5

*-Added by Master Purushottam* **Mathematics** » **Number System**

**Answer:**

Start with: -20 = -20

Which is the same as: 16-36 = 25-45Which can also be expressed as: (2+2) 2 (9 X (2+2) = 52) 9 X 5

Add 81/4 to both sides: (2+2) 2 (9 X (2+2) + 81/4 = 52) 9 X 5 + 81/4

Rearrange the terms: ({2+2}) 9/2) 2 = (5-9/2) 2

Ergo: 2+2 - 9/2 = 5

Hence: 2 + 2 = 5

*-Answered by Akki chauhan* On 15 January 2019:09:03:13 AM

**Question 72.** **Asked on :**09 January 2019:09:36:20 AM

*-Added by Master Purushottam* **Mathematics** » **Number System**

**Question 73.** **Asked on :**07 January 2019:02:34:53 PM

Q. simplify combining like terms

(1) 21b - 32 +7b - 20b*-Added by Master Purushottam* **Mathematics** » **Algebra Expression**

**Answer:**

21b - 32 + 7b - 20b

=(21b+7b -20b) - 32

=8b - 32 Answer

*-Answered by Akki chauhan* On 10 January 2019:10:54:03 AM

**Question 74.** **Asked on :**07 January 2019:02:26:38 PM

Find the sum : 3p^{2}q^{2} - 4pq + 5, - 10p^{2}q^{2}, 15 + 9pq + 7p^{2}q^{2}

*-Added by Master Purushottam* **Mathematics** » **Algebra Expreesion**

**Answer:**

3p^{2}q^{2} - 4pq + 5 + (-10p^{2}q^{2}) + 15 + 9pq + 7p^{2}q^{2}

= 3p^{2}q^{2} + 7p^{2}q^{2} - 10p^{2}q^{2} + 9pq - 4pq + 5 + 15

= 10p^{2}q^{2} - 10p^{2}q^{2} + 5pq + 20

= 5pq + 20 **Answer**

*-Answered by Akki chauhan* On 10 January 2019:10:54:27 AM

**Question 75.** **Asked on :**07 January 2019:02:17:56 PM

Putting x = 2 (i) In x + 4, we get the value of x + 4, i.e., x + 4 = 2 + 4 = 6

*-Added by Master Purushottam* **Mathematics** » **Algebra Expreesion**

**Question 76.** **Asked on :**07 January 2019:02:16:52 PM

If a = 2, b = -2, find the value of a^{2} + b^{2}.

*-Added by Master Purushottam* **Mathematics** » **Algebra Expreesion**

**Answer:**

The value of a

=4

The value of b^{2}= (-2^{2}) = 4

a^{2}+b^{2}= 4+4

= 8

*-Answered by Himanshi Verma* On 22 January 2019:05:38:40 PM

**Question 77.** **Asked on :**07 January 2019:02:15:12 PM

. Simplify the expressions and find the value if x is equal to 2

(i) x + 7 + 4 (x – 5)

*-Added by Master Purushottam* **Mathematics** » **Algebra Expreesion**

**Answer:**

x+7+4(x-5)=2

x+7+4x-20=2

5x-13=2

5x=2+13

5x=15

x=3

So, The value of x is 3.

*-Answered by Akki chauhan* On 14 January 2019:10:15:24 AM

**Question 78.** **Asked on :**07 January 2019:02:14:21 PM

simplify the expression and find its value when a = 5 and b = -3.

2(a^{2} + ab) + 3 - ab

*-Added by Master Purushottam* **Mathematics** » **Algebra Expreesion**

**Answer:**

a=5, b=-3

2(5^{2}+5×(-3))+3-5×(-3)

2(25+(-15))+3+15

2(25-15)+18

2×10+18

20+18

38

*-Answered by Himanshi Verma* On 10 March 2019:04:58:49 PM

**Question 79.** **Asked on :**07 January 2019:02:07:12 PM

Q1. 3a - 2b - ab - (a-b+ab)+ 3ab +b - a

*-Added by Master Purushottam* **Mathematics** » **Algebra**

**Question 80.** **Asked on :**07 January 2019:12:03:07 PM

What is mean?

*-Added by ATP Admin* **Mathematics** » **Mean**

**Answer:**

Mean is an average of some data which is calculated by dividing ∑Sum of data by total numbers of data

*-Answered by Akki chauhan* On 19 January 2019:10:46:10 AM

**Question 81.** **Asked on :**31 December 2018:02:20:01 PM

factorise x^{2} + 2x + 1

*-Added by ATP Admin* **Mathematics** » **Factorisation**

**Answer:**

x^{2} + 2x + 1

= x^{2} + x + x + 1

= x(x + 1) + 1(x + 1)

= (x + 1) (x + 1)

*-Answered by Akki chauhan* On 10 January 2019:10:57:10 AM

**Question 82.** **Asked on :**31 December 2018:09:20:26 AM

*-Added by ATP Admin* **Mathematics** » **Mean Median**

**Answer:**

For finding the value of x first we find mean of 5, 10, 3

sum of observation = 5 + 10 + 3 = 18

mean = 18 / 3 = 6

Therefore, x = 6

Now we have data for median 11, 12, 14, 18, 21, 23

n = 6 { n is an even number}

n/2 th trem = 14

and (n/2 + 1)th trem = 18

Median = (14 + 18)/2 = 32/ 2 = 16 Answer

*-Answered by Akki chauhan* On 14 January 2019:10:14:53 AM

**Question 83.** **Asked on :**29 December 2018:01:00:20 AM

*-Added by ATP Admin* **Mathematics** » **Arithmetic Progressions**

**Answer:**

S7 = 7/2(2a + 6d)

=> 49 = 7a + 21d

=> 7 = a + 3d ------------ (i)

S17 = 17/2 (2a + 16d)

=> 289 = 17a + 136

=> 17 = a + 8d -------------- (ii)

Substracting equation (i) from (ii)

we have ...

17 - 7 = a - a + 8d - 3d

10 = 5d

d = 10/5

d = 2

Putting the value in Equation (i)

7 = a + 3d

7 = a + 3 x 2

a = 7 - 6

a = 1

Sum of first n terms = n/2 [2a + (n - 1) d]

= n/2 [2x 1 + (n - 1)2]

= n/2 [2 + 2n - 2 ]

= n/2 (2n)

= nxn = n2

*-Answered by Akki chauhan* On 14 January 2019:10:14:29 AM

**Question 84.** **Asked on :**28 December 2018:11:00:50 PM

*-Added by ATP Admin* **Mathematics** » **Arithmetic Progressions **

**Answer:**

A.p= 9,17,25

a=9

d=17-9

=8

Sum of n terms=n÷2[2a+(n-1)d

636=n÷2[2(9)+(n-1)8]

636=n÷2[18+8n-8]

636=n÷2[10+8n]

1272=10n+8n^2

8n^2+10n-1272=0

2[4n^2+5n-636]=0

4n^2+5n-636=0

4n^2+53n-48n-636=0

n(4n+53)-12(4n+53)=0

(n-12)(4n+53)=0

n-12=0

n=12

12 terms must be given to sum of 636

*-Answered by Himanshi Verma* On 20 January 2019:05:32:22 PM

You can see here all the solutions of this question by various user for **NCERT Solutions**. We hope this try will help you in your study and performance.

This Solution may be usefull for your practice and **CBSE Exams** or All label exams of **secondory examination**. These solutions or answers are user based solution which may be or not may be by expert but you have to use this at your own understanding of your **syllabus**.