CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Putting the above values in equation (i) and solving the extreme left and extreme right portion of the equality we get the value of a
2/4 = 3/k
⇒ 2k = 12 ⇒ k = 6

The value of k for which the system of equations has infinitely many solution is k = 6

0Thank You

Sandeep Yadav 6 years, 4 months ago

Given - infinitely many solution =2/4=3/k K=6

0Thank You

ANSWER

300 oranges are distributed equally among A certain number of students had there been 10 more students each would have received one orange less

Posted by Nancy Dsouza 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let the number of students be x, then
{tex}\frac { 300 } { x } - \frac { 300 } { ( x + 10 ) } = 1 \Rightarrow \frac { 1 } { x } - \frac { 1 } { ( x + 10 ) } = \frac { 1 } { 300 }{/tex}
{tex}\Rightarrow \frac { x + 10 - x } { x ( x + 10 ) } = \frac { 1 } { 300 }{/tex}
{tex}\Rightarrow{/tex} x(x + 10) = 3000
{tex}\Rightarrow{/tex} x² + 10x - 3000 = 0
{tex}\Rightarrow{/tex} x² + 60x - 50x - 3000 = 0
{tex}\Rightarrow{/tex} x(x + 60) - 50(x + 60) = 0
{tex}\Rightarrow{/tex} x + 60 = 0 or x - 50 = 0
{tex}\Rightarrow{/tex} x = -60 or x = 50
x = 50 {{tex}\because{/tex} number of students cannot be negative}
Hence the number of students is 50.

0Thank You

ANSWER

If a,b and c be the sums of p,q and r terms respectively of an AP, then show that a/p(q-r) + b/q(r-p) + c/r(p-q) = 0

Posted by Kishor Sahu 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given that, S_p = a, S_q = b and S_r = c
Let A be the first term and d be the common difference. Then,
S_p = {tex}\frac { p } { 2 }{/tex} [2 A + (p - 1)d] = a
{tex}\Rightarrow{/tex} 2 A + (p - 1)d = {tex}\frac { 2 a } { p }{/tex} ...(i)
S_q = {tex}\frac { q } { 2 }{/tex} [2 A + (q - 1)d] = b
{tex}\Rightarrow{/tex} 2 A + (q - 1)d = {tex}\frac { 2 b } { q }{/tex} ...(ii)
and S_r = {tex}\frac { r } { 2 }{/tex} [2 A + (r - 1)d] = c
{tex}\Rightarrow{/tex} 2 A + (r - 1)d = {tex}\frac { 2 c } { r }{/tex} ...(iii)
On multiplying Eq. (i) by (q - r), Eq. (ii) by (r - p) and Eq. (iii) by (p - q), we get
[2 A + (p - 1)d] (q - r) = {tex}\frac { 2 a } { p }{/tex} (q - r) ...(iv)
[2 A + (q - 1)d] (r - p) = {tex}\frac { 2 b } { q }{/tex} (r - p) ...(v)
and [2 A + (r - 1)d] (p - q) = {tex}\frac { 2 c } { r }{/tex} (p - q) ...(vi
On adding Eq. (iv), Eq. (v) and Eqs. (vi), we get
{tex}\frac { 2 a } { p } ( q - r ) + \frac { 2 b } { q } ( r - p ) + \frac { 2 c } { r } ( p - q ){/tex}
= [2 A + (p - 1)d] (q - r) + [2 A + (q - 1)d] (r - p) + [2 A + (r - 1)d] (p - q)
= 2 A (q - r + r - p + p - q) + d[(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]
= 2 A (0) + d[(pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]
= 0 + d (0) = 0
{tex}\therefore{/tex} {tex}\frac { a } { p } ( q - r ) + \frac { b } { q } ( r - p ) + \frac { c } { r } ( p - q ){/tex}= 0
Hence proved.

0Thank You

ANSWER

Why we shall use a value of pi as 22/7

Posted by Arpit Shukla 6 years, 4 months ago

CBSE > Class 10 > Mathematics

4 answers

Ansuman Rathore 6 years, 4 months ago

as we know Aryabhatta ,a great mathematician discovered the perimeter of circle in a rope phenomena by which a discovery of 22/7 asπ came

0Thank You

Krishna Bharadwaj 6 years, 4 months ago

It is easy for calculating except for some problems wherein you have to use 3.14 as the value of π

1Thank You

Rani Mishra ??? 6 years, 4 months ago

For easy calculation. The original value of pi is 3.14

1Thank You

Ádítî Chàûdhäry 6 years, 4 months ago

Because it's universal value of π

1Thank You

ANSWER

how to get 500 / 500 in 10 th board exam

Posted by Gurwinder Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

5 answers

Ansuman Rathore 6 years, 4 months ago

study with great lagan

1Thank You

Krishna Bharadwaj 6 years, 4 months ago

Do hardwork

2Thank You

Ronak Choudhary 6 years, 4 months ago

Focus and hardwork for any particular aim bring success

3Thank You

Ayushi Rawat 6 years, 4 months ago

to study hard in the exam

2Thank You

?????? ?? 6 years, 4 months ago

Do hard work

2Thank You

ANSWER

Prove /5 is irrational

Posted by Santy Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 6 years, 4 months ago

let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 × q = p
squaring on both sides
=> 5×q×q = p×p ------> 1
p×p is divisible by 5
p is divisible by 5
p = 5c [c is a positive integer] [squaring on both sides ]
p×p = 25c×c --------- > 2
sub p×p in 1
5×q×q = 25×c×c
q×q = 5×c×c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational

0Thank You

ANSWER

For commerc we have to choose standard

Posted by Agampreet Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Shivam Gupta 6 years, 4 months ago

Yes

0Thank You

ANSWER

Is there any simpler way or trick to solve 8.4 exercise

Posted by Agampreet Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Ronak Choudhary 6 years, 4 months ago

No never but everything is possible try to solve tough questions like fourth ,fifth , sixth as much u can make it shorter by using each and every thing u know to solve aproblem

0Thank You

?????? ?? 6 years, 4 months ago

U searh on youtube chanel in truemath

0Thank You

ANSWER

Happt independence day ??????

Posted by ????? ?????? 6 years, 4 months ago

CBSE > Class 10 > Mathematics

5 answers

? ? 6 years, 4 months ago

Haanji pata hai ap thik nhi ho aditi ji..take care nd get well soon☺

2Thank You

#Aditi~ Angel???? 6 years, 4 months ago

Gungun im not fine...& take care

1Thank You

? ? 6 years, 4 months ago

Same 2 youuu

2Thank You

Pk . 6 years, 4 months ago

Same to you sis

1Thank You

#Aditi~ Angel???? 6 years, 4 months ago

Happy Independence Day & Rakhsha Bandhan

1Thank You

ANSWER

Prove that sin theeta - 2sin^3 theeta /2cos^3 theeta -cos theeta is equal to tan theeta

Posted by Khushi ?? 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

LHS = (sin{tex}\theta{/tex} - 2sin³{tex}\theta{/tex})
= sin{tex}\theta{/tex}(1 - 2sin²{tex}\theta{/tex})
RHS = (2cos³{tex}\theta{/tex} - cos{tex}\theta{/tex})tan{tex}\theta{/tex}
= cos{tex}\theta{/tex}(2cos²{tex}\theta{/tex} - 1){tex}\frac { \sin \theta } { \cos \theta }{/tex}
= [2(1 - sin²{tex}\theta{/tex}) - 1)sin{tex}\theta{/tex}
= (2 - 2sin²{tex}\theta{/tex} -1)sin{tex}\theta{/tex}
= (1 - 2sin²{tex}\theta{/tex})sin{tex}\theta{/tex}
{tex}\Rightarrow{/tex} LHS = RHS
{tex}\therefore{/tex} (sin{tex}\theta{/tex} - 2sin³{tex}\theta{/tex}) = (2cos³{tex}\theta{/tex} - cos{tex}\theta{/tex})tan{tex}\theta{/tex}

1Thank You

ANSWER

I want a brief details about the two types of mathematics in 10th class.

Posted by Aakash Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

6 answers

Rani Mishra ??? 6 years, 4 months ago

Cbse has released this news a very long time ago.

1Thank You

Ritika Vyas 6 years, 4 months ago

1st is Basic. In which in future we cannt go in maths field and if u will take medicle in that instead of physics u have to choose agriculture and u cannot take non medicle, commerce etc. And paper will come from only text book. 2nd is standard in which u can choose maths field in future . And the exam will come from text book and any other resource book. The come little tough as compare to basic.

2Thank You

Pk . 6 years, 4 months ago

The label of standard is slightly tough than that of basic

3Thank You

Pk . 6 years, 4 months ago

if you want to continue your 11 without math then you are free to choose any one of them

2Thank You

Pk . 6 years, 4 months ago

If you want to continue your math in class 11 also then you haave to choose math

2Thank You

Pk . 6 years, 4 months ago

Yar vijay it is basic level and standard label . You have to choose any one of them

2Thank You

ANSWER

Two circles touch each other externally at P.AB is a common tangent to the circle touching them at A,B.Find APB

Posted by Angela Malik 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If sec∅+tan∅=x find the value of sec∅

Posted by Shobha Upadhyay 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given: sec x + tan x = p
As we know,
{tex}sec^2 x - tan^2 x = 1{/tex}
{tex}\therefore{/tex} {tex}(sec\ x + tan\ x)(sec\ x - tan\ x) = 1 {/tex}
{tex}\Rightarrow{/tex} {tex}p(sec\ x - tan\ x) = 1 {/tex}
{tex}\Rightarrow{/tex} {tex}sec\ x - tan\ x ={/tex} {tex}\frac{1}{p}{/tex}
Thus, we have
{tex}sec\ x + tan\ x = p{/tex} and sec x - tan x = {tex}\frac{1}{p}{/tex}
{tex}\Rightarrow{/tex} (sec x + tan x) + (sec x - tan x) = p + {tex}\frac{1}{p}{/tex} and (sec x + tan x) - (sec x - tan x) = p - {tex}\frac{1}{p}{/tex}
{tex}\Rightarrow{/tex} 2 sec x = p + {tex}\frac{1}{p}{/tex} and 2 tan x = -{tex}\frac{1}{p}{/tex} {tex}\Rightarrow{/tex} sec x = {tex}\frac{p^{2}+1}{2 p}{/tex} and tan x = {tex}\frac{p^{2}-1}{2 p}{/tex}

0Thank You

ANSWER

If the HCF of 657 and963 is 657x +963x -15, find x

Posted by Fiza Siddiqui 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given numbers are 657 and 963 .

Here, 657 < 963

By using Euclid's Division algorithmm , we get
963 = (657 × 1) + 306

Here , remainder = 306 .

So, On taking 657 as new dividend and 306 as the new divisor and then apply Euclid's Division lemma, we get
657 = (306 × 2) + 45

Here, remainder = 45

So, On taking 306 as new dividend and 45 as the new divisor and then apply Euclid's Division lemma, we get
306 = (45 × 6) + 36

Here, remainder = 36

So, On taking 45 as new dividend and 36 as the new divisor and then apply Euclid's Division lemma, we get
45 = (36 × 1) + 9
Here, remainder = 9

So, On taking 36 as new dividend and 9 as the new divisor and then apply Euclid's Division lemma, we get

36 = (9 × 4) + 0
Here , remainder = 0 and last divisor is 9.

Hence, HCF of 657 and 963 = 9.

∴ 9 = 657x + 963(-15)
⇒ 9 = 657x - 14445
⇒ 657x = 9 + 14445
⇒ 657x = 14454
⇒x = 14454/657
⇒ x =22

0Thank You

ANSWER

√5 is irrational number prove it

Posted by Aman Patidar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Rani Mishra ??? 6 years, 4 months ago

Check on necrt solutions on maths dashboard in my cbseguide app

1Thank You

Fiza Siddiqui 6 years, 4 months ago

√5=p/q (both sides squaring)p/q (hole square) 5 divisible by Psq Let p=5m 5Qsq=(5m)sq

1Thank You

ANSWER

Show that the progression √18,√50,√98......... is an A.P . Find its first term and the common difference

Posted by Piyush Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Atiqur Rahman 6 years, 4 months ago

Common difference is 2√2

3Thank You

Atiqur Rahman 6 years, 4 months ago

Its first term is√18and its common differece

3Thank You

ANSWER

3x-5y=4 9x=2x+5y

Posted by Wasi Ahamad 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

(6x-4y)+(5x6y)Distance formula

Posted by Yuvraj Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Piyush Singh 6 years, 4 months ago

D=t/s

1Thank You

ANSWER

Solve the following pair of equations graphically 2x+3y=12, x-y-1=0. Shade the region between the two lines representd by the above equation and the x-axis.

Posted by Ujjwal Kumar 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

2x + 3 y = 12
3y = 12 - 2x
or, {tex}y = \frac { 12 - 2 x } { 3 }{/tex}
If x = 0, we have y = 4
If x = 6,we have y = 0
If x = 3 , we have y = 2
The solution table is as follows:

0	6	3
4	0	2

x - y = 1
or, y = x - 1
If x = 0, we have y = -1
If x = 1, we have y = 0
If x = 3, we have y = 2
The solution table is as follows:

0	1	3
-1	0	2

Plotting the above points and drawing the lines joining them, we get the following graph.

The two lines intersect each other at point (3, 2).
Hence, x = 3 and y = 2.
{tex}\triangle A B C{/tex} is the region between the two lines represented by the given equations and the X-axis.