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Asked/Added Questions From NCERT: Mathematics


Question 1. Asked on :21 August 2019:09:36:36 AM

 In a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

-Added by Prince Rawat Mathematics » Triangle

Answer:

Master Purushottam
Theorem 6.8 : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

-Answered by Master Purushottam On 22 August 2019:09:17:36 PM


Answer:

Master Purushottam
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points ,then theother two sides are divided in the same ratio

.Image result for If a line is drawn parallel to one side of a triangle to intersect the other two sides in point, the other two sides are divided in the same ratio.

-Answered by Master Purushottam On 22 August 2019:09:18:51 PM


Question 3. Asked on :19 August 2019:09:49:02 AM

 Who discovered the zero?

-Added by Akki chauhan Mathematics » Number

Answer:

Akki chauhan

 Aryabhatta discovered zero.

-Answered by Akki chauhan On 19 August 2019:09:50:45 AM


Question 4. Asked on :18 August 2019:06:31:17 PM

X-2=7


-Added by Master Purushottam Mathematics » One Variable In Linear Equation

Answer:

Akki chauhan

 X-2=7

 X=7+2

 X=9

-Answered by Akki chauhan On 18 August 2019:08:35:07 PM


Question 5. Asked on :17 August 2019:05:57:38 PM

 Find a quadratic polynomial , The sum and product of whose zeroes are -3 and 2 respectively.

-Added by Khushi Chauhan Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

 Let the quadratic polynomial be ax2+bx+c=0 and its zeroes be α and β.

sum of zeroes     α+β = -3 = -b/a 

product of zeroes      αβ = 2= c/a

If a=1   then b=3,  c=2

So, the quadratic polynomial is x2+3x+2.

-Answered by Himanshi Verma On 18 August 2019:11:00:35 AM


Question 6. Asked on :17 August 2019:05:50:41 PM

 If LM||CB and LN||CD prove that AM/AB =AN/AD.

-Added by Khushi Chauhan Mathematics » Chapter 6

Answer:

Himanshi Verma

 By Thales theorem or basic propositionality Theorem

IN∆ABC
LM||BC so by B.P.T
AL/AC=AM/AB-----------(1)
IN∆ADC
AL/LC=AN/AD------(2)
from equation 1 &2 L.H.S is same so R.H.S will also same.
so
AM/AB=AN/AD

-Answered by Himanshi Verma On 18 August 2019:10:58:35 AM


Question 7. Asked on :17 August 2019:09:38:34 AM

 What is mean , mode and Median?

-Added by Akki chauhan Mathematics » Stastics

Answer:

Himanshi Verma

 Mean, Mode, Median are the measurment of central tendency.  mean and mode cann't be shown on graph but median can be shown on the graph.

-Answered by Himanshi Verma On 17 August 2019:10:34:55 AM


Question 8. Asked on :17 August 2019:09:27:38 AM

 What is the form of Arithmetic Progressions? 

-Added by Akki chauhan Mathematics » Arithmetic Progressions

Answer:

Himanshi Verma

 The general form of an Arithmetic progression is a, a+1, a+2, a+3.....

-Answered by Himanshi Verma On 17 August 2019:10:41:29 AM


Question 9. Asked on :17 August 2019:09:24:51 AM

What is Arithmetic Progressions?

-Added by Akki chauhan Mathematics » G.

Answer:

Himanshi Verma
A sequence of numbers in which each differs from the preceding one by a constant quantity.examples are 1,2,3,4 etc.

-Answered by Himanshi Verma On 18 August 2019:10:41:56 AM


Question 10. Asked on :17 August 2019:09:22:47 AM

 Prove that √2 is irrational?

-Added by Akki chauhan Mathematics » Real Numbers

Answer:

Himanshi Verma
Let √2 be a rational number 

Therefore, √2= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 
                   p²= 2q²           ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p²          [since, 2q²=p²]
⇒ 2 is a factor of p

 Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get 
            p²= 4 m²          ...(2)
From (1) and (2), we get 
           2q² = 4m²      ⇒      q²= 2m²
Clearly, 2 is a factor of 2m²
⇒       2 is a factor of q²          [since, q² = 2m²]
⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

 Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

-Answered by Himanshi Verma On 18 August 2019:11:09:55 AM


Question 11. Asked on :22 July 2019:07:46:58 PM

 S and T are points on sides PR and QR of ΔPQR such that ∠P= ∠RTS. Show that ΔRPQΔRTS.

-Added by Himanshi Verma Mathematics » Traingles

Answer:

Himanshi Verma

-Answered by Himanshi Verma On 23 July 2019:07:52:25 PM


Question 12. Asked on :18 July 2019:07:27:12 PM

 If a line intersects sides AB and AC of a ΔABC at D and E respectively and is a parallel to BC, prove that AD/AB= AE/AC.

-Added by Himanshi Verma Mathematics » Triangles

Answer:

Himanshi Verma

-Answered by Himanshi Verma On 18 July 2019:09:05:33 PM


Question 13. Asked on :10 July 2019:11:49:11 PM

The first term of an A.P. is 6. The sum of first 6 terms is 66. Find it's 6th term.

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Mind

Given: a1 = 6, S= 66 

using Formula: Sn = n2 (a + l

Step by Step Explanation: 

S6 = 62 ( 6 + a6)

66 = 3(6 + 6th term) 

(6 + 6th term) = 663 = 22

6th term = 22 - 6 

6th term = 16 Answer 

-Answered by Master Mind On 11 July 2019:09:08:06 PM


Question 14. Asked on :09 July 2019:11:51:22 PM

Solve the equation 6(7b - 4) = 60 for b. 

-Added by ATP Admin Mathematics » Algebra

Answer:

ATP Admin

Given equation is 6(7b - 4) = 60  Step by Step explanation

7b - 4 = 606 

7b - 4 = 10 

7b = 10 + 4 

7b = 14 

b = 147 = 2 Answer

-Answered by ATP Admin On 09 July 2019:11:53:03 PM


Question 15. Asked on :09 July 2019:07:05:00 PM

If the sum of nth terms of an A.P. is given by 4n2 - 7n. Find 

(i) Sum of 23 terms 

(ii) 15th term

(iii) nth term

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Mind

Given:Sn = 4n2 - 7n 

Step by step explanation: 

(i) Sum of 23 terms 

Sn = 4n2 - 7n ............. (i

Replace n by 23 we have,

S23 = 4(23)2 - 7(23) 

= 4 × 529 - 161

= 2116 - 161 

= 1955 Answer

(ii) 15th term 

a15 = S15 - S14 

S15 = 4(15)2 - 7(15) = 4 × 225 - 105 = 900 - 105 = 795 

S14 = 4(14)2 - 7(14) = 4 × 196 - 98 = 784 - 98 = 686

Now, a15 = S15 - S14 

              = 795 - 686 

              = 109 Answer

(iii) nth term 

an = Sn - Sn-1

Given, Sn = 4n2 - 7n 

Replace n by n-1 we have 

Sn-1 = 4(n-1)2 - 7(n-1) 

       = 4(n2 - 2n + 1) - 7n + 7

       = 4n2 - 8n + 4 - 7n + 7 

       = 4n2 - 15n + 11 ............. (ii) 

Now Applying equation (i) and (ii) in formula an = Sn - Sn-1 

an = (4n2 - 7n) - (4n2 - 15n + 11) 

    = 4n2 - 7n - 4n2 + 15n - 11

    = 8n - 11 

Therefore, an = 8n - 11 Answer  

*******************************

-Answered by Master Mind On 11 July 2019:09:42:19 PM


Question 16. Asked on :09 July 2019:06:56:05 PM

Prove the identity: 


-Added by ATP Admin Mathematics » Trigonometry


Question 17. Asked on :09 July 2019:06:53:37 PM

Find the nature of quadratic equation x2 + 3x + 7.  

-Added by ATP Admin Mathematics » Quadratic Equations

Answer:

Himanshi Verma

 x2+ 3x+7 

This equation is of the form of ax2+bx+c=0

so, a=1,     b=3,    c=7

b2+4ac = 3×3 - 4×1×5

           = 9 - 28 

            = 19 

∴  19 > 0

So, the given equation has two real roots.

-Answered by Himanshi Verma On 18 July 2019:07:35:08 PM


Question 18. Asked on :08 July 2019:10:36:51 PM

x and y are connected parametrically by the equation x = 2at2, y = at4 without eleminating the parameter.  

Find dydx .

-Added by ATP Admin Mathematics » Continuity And Differentiability

Answer:

Master Purushottam

 

-Answered by Master Purushottam On 08 July 2019:10:58:57 PM


Question 19. Asked on :08 July 2019:09:55:52 PM

In the Given fig. DE ‖ AC and DF ‖AE.  

BFFE = BEEC .


-Added by ATP Admin Mathematics » Triangles

Answer:

ATP Admin

-Answered by ATP Admin On 08 July 2019:10:45:29 PM


Question 20. Asked on :08 July 2019:04:56:29 PM

 Find the sum of the first 40 positive integer divisible by 6.

-Added by Himanshi Verma Mathematics » A.p

Answer:

Akki chauhan

 This is an AP a=6,d=6, and n=40

Using formula,  

Sn = n2 [2a+(n-1)d]


S40= 402 [2*6+(40-1)6]

S40=20(12+234)

S40=20*246

S40=4920

-Answered by Akki chauhan On 18 July 2019:10:26:06 AM


Question 21. Asked on :17 June 2019:11:25:45 PM

Show that one and only one out of n, n+2 or n+4 is divisible by 3, where n is any positive integer. 

-Added by ATP Admin Mathematics » Real Numbers

Answer:

ATP Admin

Solution: Using Euclid's division lemma any positive integer can be written in the form of a = bq + r where r = 0, 1, 2 ...... and q is quotients. 

Let the number which is divisible by 3 be 3q + 0 or 3q + 1 or 3q + 2 where [0 <= r < b] 

Now n = 3q or n = 3q + 1 or n = 3q + 2 

Case I, 

When n = 3q  .......... (i) 

⇒ n = 3(q) where n is divisible by 3

Adding 2 both sides in equ. (i) 

We have,

    n + 2 = 3q + 2 Where n + 2 is not divisible by 3 

Now adding 4 both side in equ. (i) 

We have,

    n + 4 = 3q + 4 Where n + 4 is not divisible by 3 


Case II  

  Taking n = 3q + 1 ........ (ii) where n is not divisible by 3

Adding 2 both sides in equ. (ii) 

We have, 

n + 2 = 3q + 1 + 2 = 3q + 3 

n + 2 = 3(q + 1) where n + 2 is divisible by 3 

Now adding 4 both sides in equ. (ii) 

n + 4 = 3q + 1 + 4 = 3q + 5 where n + 4 is not divisible by 3 


Case III  

taking n = 3q + 2 .... (iii) where n is not divisible by 3 

Adding 2 both sides in equ. (iii) 

We have,

n + 2 = 3q + 2 + 2 = 3q + 4 where n + 2 is not divisible by 3

Now adding 4 both sides in equ. (iii) 

We have, 

n + 4 = 3q + 2 + 4 = 3q +6 

n + 4 = 3(q + 2) where n + 4 is divisible by 3 

Hence in all these three cases we have seen that either one and only one n or n + 2 or n + 4 is divisible by 3. 

 

 



-Answered by ATP Admin On 17 June 2019:11:28:43 PM


Question 22. Asked on :13 May 2019:10:29:43 AM

 7 और  9 के बीच 10 परिमेय संख्या लिखे ?


-Added by ravi varma Mathematics » Maths

Answer:

Himanshi Verma

 The rational numbers between 7 and 9 are 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90. 

-Answered by Himanshi Verma On 14 May 2019:11:11:27 AM


Question 23. Asked on :08 May 2019:10:19:35 AM

x का मान  ज्ञात कीजिए  2x-1=14-x?

-Added by ravi varma Mathematics » Maths

Answer:

Jyoti Srivastva

 2x-1=14-x

2x+x=14+1

3x=15

x=15/3 

x=5

-Answered by Jyoti Srivastva On 07 July 2019:10:05:52 PM


Question 24. Asked on :08 May 2019:10:14:22 AM

 क्या शुन्य परिमेय संख्या है ? 

-Added by Hitesh kumar Mathematics » Mathematics

Answer:

Himanshi Verma

 yes zero is a rational number.

-Answered by Himanshi Verma On 14 May 2019:11:23:16 AM


Question 25. Asked on :08 May 2019:10:15:06 AM

4z+3=6+2z 

-Added by ravi varma Mathematics » Maths

Answer:

Jyoti Srivastva

 4z+3=6+2z

4z-2z=6-3

2z=3

z=3/2

-Answered by Jyoti Srivastva On 07 July 2019:10:11:51 PM


Question 26. Asked on :08 May 2019:09:55:15 AM

5 और 6 के बीच दो  परिमेय संख्या लिखो ?

-Added by Hitesh kumar Mathematics » Math

Answer:

Himanshi Verma

 The rational number are 16, 17.

-Answered by Himanshi Verma On 14 May 2019:11:16:10 AM


Question 27. Asked on :22 April 2019:09:43:52 AM

 Why study of density is important for a science student?

-Added by Nitish kumar Mathematics » ORGANISATION IN THE LIVING WORLD

Answer:

rocki kumar

 Density is important because it affects whether objects will float or sink. It is animportant property to consider when building things like ships and hot air balloons. ... He realized that the amount of water that spilled was equal in volume to the space that his body occupied.

-Answered by rocki kumar On 08 May 2019:08:23:43 AM


Question 28. Asked on :21 April 2019:10:53:53 AM

What is activity series ? 

-Added by Nitish kumar Mathematics » Types Of Chemical Reactions

Answer:

Himanshi Verma

 Arrangement of metals in order of decreasing reactivities in vertical column is called activity series.

-Answered by Himanshi Verma On 21 April 2019:10:58:07 AM


Question 29. Asked on :21 April 2019:10:51:49 AM

 what happens when Mg ribbon is burnt in oxygen?

-Added by Nitish kumar Mathematics » Types Of Chemical Reactions

Answer:

Himanshi Verma

 

When magnesium ribbon is burnt in oxygen it burns with a dazzling flame and gives a white magnesium oxide.


-Answered by Himanshi Verma On 21 April 2019:10:59:16 AM


Question 30. Asked on :21 April 2019:10:45:16 AM

 what is oxidation?

-Added by Nitish kumar Mathematics » Types Of Chemical Reactions

Answer:

Himanshi Verma

 The reaction in which addition of oxygen and removal of hydrogen is called oxidation.

-Answered by Himanshi Verma On 21 April 2019:11:03:26 AM


Question 31. Asked on :19 April 2019:08:34:46 AM

 Given that HCF (306,657) =9 find LCM (306,657).

-Added by Prince Rawat Mathematics » Real Number

Answer:

Himanshi Verma
Product of LCM and HCF= Product of two numbers

9×LCM = 306×657

9×LCM = 201042

LCM   = 22338

-Answered by Himanshi Verma On 21 April 2019:10:32:51 AM


Question 32. Asked on :19 April 2019:08:23:46 AM

 Check whether the first polynomials is a factors of the second polynomials by dividing the second polynomials by the first polynomials.

-Added by Prince Rawat Mathematics » POLYNOMIALS


Question 33. Asked on :19 April 2019:08:21:00 AM

Prove that 5
 is irrational.

-Added by Prince Rawat Mathematics » Real Number

Answer:

Himanshi Verma

Let us assume that √5 is rational 

Then √5 =  

(a and b are co primes, with only 1 common factor and b≠0) 

⇒ √5 =  

(cross multiply) 

⇒ a = √5b 

⇒ a² = 5b² -------> α

⇒ 5/a² 

(by theorem if p divides q then p can also divide q²) 

⇒ 5/a ----> 1 

⇒ a = 5c 

(squaring on both sides) 

⇒ a² = 25c² ----> β 

From equations α and β 

⇒ 5b² = 25c²

⇒ b² = 5c² 

⇒ 5/b² 

(again by theorem) 

⇒ 5/b---> 2 


we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong. 

This contradiction arises because we assumed that √5 is a rational number 

∴ our assumption is wrong 

∴ √5 is irrational number.

-Answered by Himanshi Verma On 21 April 2019:11:24:50 AM


Question 34. Asked on :19 April 2019:08:18:18 AM

 Check whether 6n can end digit 0 for any nature number n.

-Added by Prince Rawat Mathematics » Real Number

Answer:

Akki chauhan

 prime factors of 6n 

                = (2*3)n 

                = 2n*3n 

While the number ending with zero have prime factor as 2n*5n 

       Hence, 6n is not end with zero.

-Answered by Akki chauhan On 28 June 2019:09:30:20 AM


Question 35. Asked on :18 April 2019:08:18:35 AM

 Use euclid's division alogrithan to find the HCF of :

(i) 135 and 225 .

-Added by Nitish kumar Mathematics » Real Number

Answer:

Himanshi Verma

  Use Euclid division alogrithan 

 a=bq+r

225=135×1+90

135=90×1+45

90=45×2+0

HCF=45 .

-Answered by Himanshi Verma On 18 April 2019:04:26:31 PM


Answer:

Himanshi Verma

 Suppose, at first the side of the square was a

the area was a^2

now, the side of the square is (a+a★50/100)=a+a/2=3a/2

the area will be, 9a^2/4

so, 9a^2/4-a^2=5a^2/4

we now tell that, after increasing 50%of the side, the area will 5/4 times of the older(a^2).

-Answered by Himanshi Verma On 06 April 2019:06:44:55 PM


Question 37. Asked on :31 March 2019:01:06:03 PM

If the length of a rectangle is increased by 60% by what per cent would the width have to be decreased to maintain the same area?

-Added by Master Purushottam Mathematics » Mensuration

Answer:

Himanshi Verma




% Decrease

-Answered by Himanshi Verma On 06 April 2019:06:46:28 PM


Question 38. Asked on :31 March 2019:12:59:06 PM

 If the length and breadth of a rectangle plot are increased by 50% and 20% respectively, then the new area is how many times the original area

-Added by Master Purushottam Mathematics » Mensuration

Answer:

Hitesh kumar

  LET THE ORIGINAL LENGTH BE L

AND

ORIGINAL BREATH BE B.

ORIGINAL AREA = LB

LENGTH INCREASED BY 50% = 150 L / 100

BREATH INCREASED BY 20% = 120 B / 100

INCREASED AREA = 150 L X 120 B / 100 X 100

= 18000 LB / 10000

= 1.8 LB

NOW,

1.8 LB / LB = 1.8

CHANGING 1.8 INTO FRACTION.

1.8 = 18/10 = 9/5 

SO,

THE INCREASED AREA WILL BE 9/5 TIMES OF THE ORIGINAL AREA.

-Answered by Hitesh kumar On 13 April 2019:08:31:51 AM


Question 39. Asked on :31 March 2019:12:53:32 PM

The length of diagonal of the square whose area is 16900m2 is: 

-Added by Master Purushottam Mathematics » Mensuration

Answer:

Himanshi Verma

 so let us first consider a square with area x² unit²

so the side is of length √x² = x unit 

so square has angles of 90° so using Pythagoras we  get the length of the diagonal as  = 

√2x

so for this problem the area is 16900  m² so length of the side is √16900 = 130 m

so the length of the diagonal is √2 x 130 m = 130√2 m ≈ 183.82 m ANSWER

-Answered by Himanshi Verma On 06 April 2019:06:51:54 PM


Question 40. Asked on :31 March 2019:12:49:03 PM

 The perimeter of the rectangle field is 480m and the ratio between the length and breadth is 5 : 3. the area of the field  in square meters is :

-Added by Master Purushottam Mathematics » Mensuration

Answer:

Himanshi Verma

 The perimeter is given by twice the sum of length and breadth.

The length and  breadth are on ratio 5:3.Let the common ratio be x.
length = 5x
breadth = 3x
Perimeter = 2(length + breadth)
480 = 2(5x + 3x)
480 = 16x
x=30
length = 5x=5*30 = 150
breadth = 3x = 3*30 =90
Area = length * breadth
Area = 150 * 90
Area = 13500 sq. m 

-Answered by Himanshi Verma On 06 April 2019:06:53:50 PM


Question 41. Asked on :31 March 2019:12:34:26 PM

 A child walk 5m to cross a rectangular field  diagonally, if breadth of the field in 3m, its length is: 

-Added by Master Purushottam Mathematics » Mensuration

Answer:

Himanshi Verma



In the figure, ABCD is a rectangular field.
AC is the diagonal.
Triangle ABC, 
AC2=AB2+BC2
202=162+BC2
BC2=202162
=(20+16)(2016)=36×4
BC=6×2 = 12 m.
The breadth of the field is 12 m.

-Answered by Himanshi Verma On 06 April 2019:06:43:16 PM


Question 42. Asked on :29 March 2019:04:53:40 PM

 Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

-Added by Rohit rajput Mathematics » Real Numbers

Answer:

Himanshi Verma

 (i)135 and 225

  a = 225, b = 135               {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

  a = bq + r (then)

  225 = 135 ×1 + 90

 135 = 90 ×1 + 45

 90 = 45 × 2 + 0                  {when we get r=0, our computing get stopped}

 b = 45 {b is HCF}

Hence:  HCF = 45

 (ii)196 and 38220

 a = 38220, b = 196          {Greatest number is ‘a’ and smallest number is ‘b’}

 Using Euclid’s division algorithm

 a = bq + r (then)

 38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

 b = 196 {b is HCF}

Hence:  HCF = 196

(iii)867 and 255

a = 867, b = 255               {Greatest number is ‘a’ and smallest number is ‘b’}

Using Euclid’s division algorithm

a = bq + r (then)

38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}

b = 196 {b is HCF}

Hence:  HCF = 196


-Answered by Himanshi Verma On 29 March 2019:04:55:16 PM


Question 43. Asked on :09 March 2019:01:54:23 PM

कैल्सीफिरोल किस विटामिन का रासायनिक नाम है ?

-Added by Rohit rajput Mathematics » G.k

Answer:

Himanshi Verma

 Vitamin D.

-Answered by Himanshi Verma On 23 March 2019:05:59:38 PM


Question 44. Asked on :01 March 2019:04:06:17 PM

 Find the value of k for which one root of the quadratic equation kx2-14x+8=0 is 2.

-Added by Harshita Rathore Mathematics » G.k

Answer:

Himanshi Verma

 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 45. Asked on :01 March 2019:03:26:35 PM

 How to prove Pythagorus theorem .

-Added by Akki chauhan Mathematics » Maths

Answer:

Himanshi Verma

 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


Answer:

Shalu Shukla


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


Question 47. Asked on :27 February 2019:07:34:58 AM

 

-Added by praful kumar Mathematics » P


Question 48. Asked on :13 February 2019:10:06:57 AM

   Find the value of X

-Added by Akki chauhan Mathematics » Circles


Answer:

Himanshi Verma

 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 50. 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:

Himanshi Verma

 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


Answer:

Shalu Shukla

 बेलन के पानी का आयतन = गोले का आयतन 

   πr2h   = 4/3 πr3

   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 52. 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 53. Asked on :07 February 2019:09:36:04 AM

tanA-conA /sinAcosA =tan2A-cot2

-Added by SURAJ KUMAR Mathematics » Trignometry


Question 54. 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 1470cm2 , AB =35cm and AD=49 cm ,find the length of BM and DL. 


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

Answer:

Himanshi Verma

 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 55. Asked on :05 February 2019:05:27:35 PM

सिद्ध कीजेये की 8 को आठ बार जमा (Plus+) करने पर 1000 आएगा |

-Added by Himanshi Verma Mathematics » G.k

Answer:

Himanshi Verma

  8 8 8

     8 8

        8

        8

+      8

= 1000 

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


Question 56. Asked on :05 February 2019:05:12:59 PM

  

Rationalise & find value of cubic expression

If x = 1860, what is the value of 

-Added by Harshita Rathore Mathematics » Number System


Answer:

Himanshi Verma

 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 58. Asked on :05 February 2019:09:14:22 AM

 Which is the largest chord in circle.

-Added by Akki chauhan Mathematics » Circles

Answer:

Himanshi Verma

 Diameter is the greatest chord of circle.

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


Question 59. Asked on :03 February 2019:05:07:16 PM

 Prove that :- 777+888+999=999

-Added by Himanshi Verma Mathematics » Addition

Answer:

Nishant Verma

 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 60. Asked on :03 February 2019:05:02:10 PM

 Prove that :- 777+888+999=999

-Added by Shalu Shukla Mathematics » Addition

Answer:

Himanshi Verma

 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 61. Asked on :30 January 2019:02:19:19 PM

 Why is the area of circle is πr2?

-Added by Himanshi Verma Mathematics » Circle

Answer:

Himanshi Verma

 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 62. 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:

Himanshi Verma

 (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 63. Asked on :28 January 2019:11:31:30 AM

 

-Added by SURAJ KUMAR Mathematics » Trigmometpy


Question 64. 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:

Himanshi Verma

 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 65. 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:

Himanshi Verma

 A=6400m2

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


Question 66. Asked on :27 January 2019:11:08:39 AM

 (4-k)x2+(2k+4)x+(8k+1)=0.

k=? 

एक पुर्ण वर्ग है ?

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

 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 67. Asked on :27 January 2019:10:59:13 AM

 abx2+(b2ac)x-bc=0.सरल करे|


-Added by SURAJ KUMAR Mathematics » POLYNOMIALS

Answer:

Himanshi Verma

 Consider, abx2 +(b2-ac)x-bc = 0

⇒ abx2 + b2x – 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 68. Asked on :27 January 2019:10:54:59 AM

 ax2+4ax+(a2-b2)=0

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS


Question 69. 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:

Himanshi Verma

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 70. Asked on :27 January 2019:10:46:14 AM

 

-Added by SURAJ KUMAR Mathematics » POLYNOMIALS


Question 71. 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 72. 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:

Himanshi Verma

 : (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 73. Asked on :20 January 2019:10:13:42 AM

 cotΦ-1/sinΦ=1

-Added by Rohit Rajput Mathematics » Maths

Answer:

Master Purushottam

 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 Master Purushottam On 24 August 2019:12:17:05 AM


Question 74. 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:

Master Purushottam

 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 Master Purushottam On 24 August 2019:12:18:13 AM


Question 75. 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:

Master Purushottam

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 Master Purushottam On 24 August 2019:12:18:19 AM


Question 76. 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:

Master Purushottam

 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 Master Purushottam On 24 August 2019:12:15:34 AM


Question 77. 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:

Master Purushottam

 Total surface area= 2πr(r+h) -πr2

curved surface area = 2πrh+πr2

-Answered by Master Purushottam On 24 August 2019:12:15:48 AM


Answer:

Master Purushottam

 vb

-Answered by Master Purushottam On 24 August 2019:12:14:15 AM


Answer:

Master Purushottam

  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 Master Purushottam On 24 August 2019:12:14:42 AM


Question 80. 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:

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

-Answered by Master Purushottam On 24 August 2019:12:15:00 AM


Question 81. Asked on :16 January 2019:06:14:35 PM

 Counting number are kept under _______ number.

-Added by Himanshi Verma Mathematics » G.k

Answer:

Master Purushottam
 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 Master Purushottam On 24 August 2019:12:15:06 AM


Question 82. Asked on :16 January 2019:05:34:22 PM

(a2-b2)3+(b2-c2)3+(c2-a2)3(a-b)3+(b-c)3+(c-a)3Simplify this Equation: 

-Added by Shalu Shukla Mathematics » Ch-2

Answer:

Master Purushottam

 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 )

-Answered by Master Purushottam On 24 August 2019:12:09:53 AM


Question 83. 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:

Master Purushottam

 2

-Answered by Master Purushottam On 24 August 2019:12:11:09 AM


Question 84. 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:

Master Purushottam
 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 Master Purushottam On 24 August 2019:12:11:22 AM


Question 85. Asked on :15 January 2019:09:48:55 AM

 The lateral surface area of a cube is 256cm2. Find its volume.

-Added by Prince Rawat Mathematics » SURFACE AREAS AND VOLUMES

Answer:

Master Purushottam

  Lateral Surface Area of cube=256 cm²

4(side)² =256
side²=256÷4
side²=64
side=√64=8 cm
Volume of the cube=(side)³=8×8×8=512 cm³.

-Answered by Master Purushottam On 24 August 2019:12:03:45 AM


Answer:

Master Purushottam

 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

                                  =     114m2

The cost of white washing the wall and the ceiling of the room = 7.50 per m2

      The cost of 114m2 of the walls and ceiling of the room= (7.50×114)Rs

                                                                                   = 855.00Rs.

-Answered by Master Purushottam On 24 August 2019:12:04:01 AM


Question 87. 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:

Master Purushottam
The total surface area of sphere= 4πr2

The total surface area of hemisphere= 3πr2

The ratio of total surface area of sphere and hemisphere= 4πr2/3πr2

                                                                                = 4:3

-Answered by Master Purushottam On 24 August 2019:12:04:08 AM


Question 88. 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:

Master Purushottam

   Total surface area of cube is 150sq. cm

Total surface area of cube : 6a2

To find perimeter of its one faces :  ?

Total surface area of cube =150sq.cm

                           6a2  =150sq.cm

 = 1506 

= 25cm

=a2 = √25cm

=a = 5 cm

Perimeter of its one face is = 4 × a

                                        = 4 × 5 cm

                                        = 20 cm Answer

-Answered by Master Purushottam On 24 August 2019:12:01:30 AM


Question 89. 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:

Master Purushottam

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(8x2+2x2+4x2) =1372

=2(14x2)= 1372

=28x2=1372

= x2=1372/28

       = 49

x=7m

the dimensions are 4x=4×7=28m

                           2x=2×7=14m

                           1x=1×7=7m

-Answered by Master Purushottam On 24 August 2019:12:01:38 AM


Question 90. 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:

Master Purushottam

 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 Master Purushottam On 23 August 2019:11:57:01 PM


Question 91. 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:

Master Purushottam

 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 Master Purushottam On 23 August 2019:11:41:59 PM


Question 92. 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 93. 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:

Master Purushottam

 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 Master Purushottam On 23 August 2019:11:36:15 PM


Answer:

Master Purushottam

  Ans 1. Zero(0)

Ans 2. 0 and 1

Ans 3. 2

-Answered by Master Purushottam On 23 August 2019:11:36:37 PM


Question 95. 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


Answer:

Master Purushottam

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 Master Purushottam On 23 August 2019:11:37:23 PM


Answer:

Master Purushottam

 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 Master Purushottam On 23 August 2019:11:37:42 PM


Question 98. 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:

Master Purushottam

   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 Master Purushottam On 23 August 2019:11:37:53 PM


Question 99. 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:

Master Purushottam

r= 32


Volume of hemisphere= 23 πr3


                                = 23 ×22/7×3/2×3/2×3/2

                                 = 7.071

-Answered by Master Purushottam On 23 August 2019:11:33:27 PM


Question 100. Asked on :13 January 2019:11:15:32 AM

 Define concentric circles.

-Added by Akki chauhan Mathematics » Circles

Answer:

Master Purushottam
 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 Master Purushottam On 23 August 2019:11:34:12 PM


Question 101. 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:

Akki chauhan

 √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 102. Asked on :12 January 2019:05:54:25 PM

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

-Added by Himanshi Verma Mathematics » Number System


Question 103. Asked on :12 January 2019:05:52:26 PM

prove that the cyclic parallelogram is rectangle. 

-Added by khushi chauhan Mathematics » Circles

Answer:

Akki chauhan

  ∠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 104. Asked on :12 January 2019:10:52:27 AM

 Find the zero of 2x-5.

-Added by Akki chauhan Mathematics » Number System

Answer:

Master Purushottam

2x-5=0

2x=5

x= 5/2

-Answered by Master Purushottam On 22 August 2019:09:11:57 PM


Question 105. Asked on :12 January 2019:10:49:51 AM

 Factorise:

x3-23x2+142x-120

-Added by Akki chauhan Mathematics » Polynomials

Answer:

Master Purushottam

 x-23x2+142x-120 

=x3-x2-22x2+22x+120-120

=x2(x-1)-22x(x-1)+120(x-1)

=(x-1)(x2-22x+120)

=(x-1)(x2-12x-10x+120)

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

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

-Answered by Master Purushottam On 22 August 2019:09:12:12 PM


Question 106. Asked on :12 January 2019:10:45:37 AM

 If the polynomials ax3+3x2-13 and 2x3-5x+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:

Master Purushottam

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 Master Purushottam On 22 August 2019:09:12:19 PM


Question 107. Asked on :12 January 2019:10:38:50 AM

If a = 9-4√5, then find the value of

a21a2 

-Added by Akki chauhan Mathematics » Polynomials

Answer:

Master Purushottam

 (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 Master Purushottam On 22 August 2019:09:08:56 PM


Question 108. Asked on :11 January 2019:02:55:18 PM

 Find the median of first 10 prime numbers.


-Added by Prince Rawat Mathematics » STATISTICS

Answer:

Master Purushottam

 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 Master Purushottam On 22 August 2019:09:04:22 PM


Question 109. Asked on :11 January 2019:10:18:43 AM

 Define frequency of the observation.

-Added by Prince Rawat Mathematics » STATISTICS

Answer:

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

-Answered by Master Purushottam On 22 August 2019:09:04:32 PM


Question 110. Asked on :11 January 2019:10:10:33 AM

 The volume of sphere is 310.4 cm3. Find its radius.

-Added by Prince Rawat Mathematics » SURFACE AREAS AND VOLUMES

Answer:

Master Purushottam

volume of the sphere = 43 πr3= 310.4cm3

                                4/3×22/7×r3=310.4

                                88/21×r3= 310.4

                                   r3=   310.4×21/88

                                      =   6518.4/88

                                   r3  = 74.0727272cm

-Answered by Master Purushottam On 22 August 2019:09:04:44 PM


Question 111. Asked on :11 January 2019:10:04:29 AM

 What is Degree of the Polynomials?

-Added by Prince Rawat Mathematics » POLYNOMIALS

Answer:

Prince Rawat

 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 112. 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:

Master Purushottam

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 Master Purushottam On 22 August 2019:08:59:15 PM


Question 113. Asked on :09 January 2019:09:37:10 AM

 Ques.1. prove that 2+2= 5


-Added by Master Purushottam Mathematics » Number System

Answer:

Master Purushottam

  Start with: -20 = -20

Which is the same as: 16-36 = 25-45
Which 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 Master Purushottam On 22 August 2019:09:00:11 PM


Question 114. Asked on :09 January 2019:09:36:20 AM

What is Euclid's division algorithm. 

-Added by Master Purushottam Mathematics » Number System

Answer:

Master Purushottam

The basis of the Euclidean division algorithm isEuclid's division lemma. To calculate the Highest Common Factor (HCF) of two positive integers a and b we use Euclid's division algorithm. HCF is the largest number which exactly divides two or more positive integers.


-Answered by Master Purushottam On 22 August 2019:09:00:20 PM


Question 115. 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:

Master Purushottam

  21b - 32 + 7b - 20b

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

=8b - 32 Answer

-Answered by Master Purushottam On 22 August 2019:09:00:32 PM


Question 116. Asked on :07 January 2019:02:26:38 PM

Find the sum : 3p2q2 - 4pq + 5, - 10p2q2, 15 + 9pq + 7p2q2

-Added by Master Purushottam Mathematics » Algebra Expreesion

Answer:

Master Purushottam

 3p2q2 - 4pq + 5 + (-10p2q2) + 15 + 9pq + 7p2q2 

= 3p2q2 + 7p2q2 - 10p2q2 + 9pq - 4pq + 5 + 15 

= 10p2q2 - 10p2q2 + 5pq + 20 

= 5pq + 20

-Answered by Master Purushottam On 22 August 2019:09:00:43 PM


Question 117. 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

Answer:

Himanshi Verma

 ( x + 1/x)² = x² + 1/x² + 2

=> 2² = x² + 1/x² + 2

=> x² + 1/x² = 4–2 = 2 . . . . . . . .(1)

(x - 1/x)² = x² + 1/x² -2

=> ( x-1/x)² = 2 - 2 = 0

=> (x-1/x) = 0 . . . . . . . . . .(2)

Now, x^4 - 1/x^4 = ( x² + 1/x²) ( x² - 1/x²)

= 2 ( x + 1/x) ( x-1/x)

= 2 * 2 * 0 ( by (1) & (2)

> x^4 - 1/x^4 = 0 

-Answered by Himanshi Verma On 22 March 2019:05:16:46 PM


Question 118. Asked on :07 January 2019:02:16:52 PM

If a = 2, b = -2, find the value of a2 + b2.

-Added by Master Purushottam Mathematics » Algebra Expreesion

Answer:

Master Purushottam
The value of a2= 22

                   =4

The value of b2= (-22) = 4

 a2+b2= 4+4

         =  8

-Answered by Master Purushottam On 22 August 2019:08:55:06 PM


Question 119. 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:

Master Purushottam

 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 Master Purushottam On 22 August 2019:08:55:13 PM


Question 120. Asked on :07 January 2019:02:14:21 PM

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

2(a2 + ab) + 3 - ab

-Added by Master Purushottam Mathematics » Algebra Expreesion

Answer:

Master Purushottam

 a=5,      b=-3

2(52+5×(-3))+3-5×(-3)

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

2(25-15)+18

2×10+18

20+18

38

-Answered by Master Purushottam On 22 August 2019:08:55:23 PM


Question 121. 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

Answer:

Master Purushottam

 3a-2b-ab-(a-b+ab)+3ab+b-a


3a-2b-ab-a+b-ab+3ab+b-a
take same terms like this
3a-a-a -2b+b+b -ab-ab+3ab
3a-a-a=a
-2b+b+b=0
-ab-ab+3ab= -2ab+3ab=ab
=a+0+ab
=a+ab
=a(1+b)

-Answered by Master Purushottam On 22 August 2019:08:55:31 PM


Question 122. Asked on :07 January 2019:12:03:07 PM

 What is mean? 

  

-Added by ATP Admin Mathematics » Mean

Answer:

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

-Answered by Master Purushottam On 22 August 2019:08:55:50 PM


Question 123. Asked on :31 December 2018:02:20:01 PM

factorise x2 + 2x + 1 

-Added by ATP Admin Mathematics » Factorisation

Answer:

Akki chauhan

    x2 + 2x + 1 

= x2 + 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 124. Asked on :31 December 2018:09:20:26 AM

Find the median of 11, 12, 2x + 2, 3x, 21, 23 where x is the mean of 5, 10, 3.

-Added by ATP Admin Mathematics » Mean Median

Answer:

Master Purushottam

 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 Master Purushottam On 22 August 2019:08:53:15 PM


Question 125. Asked on :29 December 2018:01:00:20 AM

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Purushottam

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 Master Purushottam On 22 August 2019:08:54:07 PM


Question 126. Asked on :28 December 2018:11:00:50 PM

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Purushottam

 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 Master Purushottam On 22 August 2019:08:54:23 PM


 

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