Showing posts with label CLASS 10 NCERT Solution. Show all posts
Showing posts with label CLASS 10 NCERT Solution. Show all posts

Problems Dated Nov 05 2021. 




Problem No 09 

First Number = x 

Second Number =y 

Case 1 
(x+2)/(y+2) = 1/2

Case 2 
(x-4)/(y-4) = 5/11

From 1 

2x - y = -2 ...... 3 

from 2 

11x - 5y = 24 .... 4 

Multiplying Eqn 1 by 5 

10 x - 5y = -10   .......5

Eqn 4 -Eqn 5 

11x - 5y -10 x + 5y = 24 + 10 

x = 34 

From Eqn 1 

2 (34) - y = -2 
or y = 68 + 2 = 70  Answer. 


 

Exercise 3E R S Agarwal 

Problem No 01 

Chair = x 

Tables = y 


5 x + 4 y = 5600 ..... 1 

4 x + 3 y = 4340 ....... 2 

1 * 4  &  2 * 5 

20x + 16 y = 22400.......... a 

20x + 15y = 21700............b 

a-b 

y = 700

5x + 2800 = 5600 

5x = 2800 

x = 560 Ans 


Problem No 02

 

MISC

Maths Problems dated 03 Nov 2021


Problem No 01

In the given figure APB and AQB are two semi circle. (Given) 

Perimeter of the given shape will be equal to 

= Circumference of half semi circle (APB) + Circumference of quarter semicircle AQO + QO RADIUS + OB RADIUS 

= ( PI * R ) + (0.5 * PI * R) + R + R 

But this perimeter is given equal to the 47 cm.

Hence, 

( PI * R ) + (0.5 * PI * R) + R + R  = 47 cm 

R * ( 22/7 + 11/7 + 1 + 1 ) = 47 CM 

R * ( 47/7) = 47 CM 

It implies that R = 7 cm 

Area of Shaded Region 


Area of total Circle - Area of Quarter CIrcle 

(PI * R * R ) - (0.25 * PI * R* R) = (0.75 * PI * R* R) 

= 0.75 * 22/7 * 7 * 7 = 115.50 CM2 ANS 
It seems that answer given is incorrect. 

 Chapter 02 Polynomials 


Problem No 01 :- Find the Zeros of the Polynomials and verify the relationship between the zeros and coefficients :-


x2 + 7x + 12


To find the zeros of polynomial, It should be equated to zero first

Hence,


x2 + 7x + 12 = 0


Or, x2 + 4x + 3x + 12 = 0


Or, x (x + 4)+ 3 (x + 4 ) = 0


Or, (x + 3) * (x + 4 ) = 0


It implies that x = - 3 & -4 Ans ............................................... (1)


Comparing the Given Eqn with Stnd Eqn i.e. ax2 + bx + c = 0


We Get, a=1 , b = 7 & c = 12


α + β = -3 + ( - 4) = - 7................................... (2)


-b /a = -7 / 1 = -7 .............................................(3)


Also α * β = -3 * -4 = 12 .................................(4)


c / a = 12 / 1 ......................................................(5)


From (2), (3) (4) & (5) relation between the zeros and coefficients can be verified.