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Sia ? 6 years, 3 months ago
The numbers divisible by 9 between 400 and 800 are:
405, 414, 423,.................792
Here a=414, d=9 and l=792
Let the number of these terms be n, then
{tex}\mathrm{We}\;\mathrm{know}\;\mathrm{that}\;{\mathrm a}_{\mathrm n}=\mathrm a+(\mathrm n-1)\mathrm d{/tex}
an=792
Or, a + (n-1)d=792
{tex}\Rightarrow{/tex} 405 + (n-1){tex}\times{/tex}9=792
{tex}\Rightarrow{/tex} 9(n - 1) = 387
{tex}\Rightarrow{/tex} (n - 1) = 43
{tex}\Rightarrow{/tex} n = 44
So, S44 = {tex}\frac{n}{2}{/tex}(a+l)
= {tex}\frac{{44}}{2}{/tex}(405 + 792)
= 22 {tex}\times{/tex} 1197 = 26334
Hence, Sn=26334
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Tribhuwan Singh 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
No there is no additional content. The previous year's difficulty level was a standard.
Aditi Shukla 6 years, 3 months ago
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Aaliya ? 6 years, 3 months ago
Ragini Agrawal 6 years, 3 months ago
Posted by Aman Tiwari 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Suppose, the digit at units and tens place of the given number be x and y respectively.
{tex}\therefore{/tex} the number is {tex}10y + x{/tex}
After interchanging the digits, the number becomes {tex}10x + y{/tex}
Given: The sum of the numbers obtained by interchanging the digits and the original number is 66.
Thus, {tex}(10x + y) + (10y + x) =66{/tex}
{tex}\Rightarrow{/tex} {tex}10x + y + 10y + x = 66{/tex}
{tex}\Rightarrow{/tex} {tex}11x +11y =66{/tex}
{tex}\Rightarrow{/tex} {tex}11(x + y) = 66{/tex}
{tex}\Rightarrow x + y = \frac{{66}}{{11}}{/tex}
{tex}\Rightarrow{/tex} {tex}x + y = 6{/tex} .....(i)
Also given, the two digits of the number are differing by 2.
{tex}\therefore{/tex} we have {tex}x - y = ±2{/tex}....(ii)
So, we have two systems of simultaneous equations,
{tex}x - y = 2, \;x + y = 6{/tex}
{tex}x - y = -2, \;x + y = 6{/tex}
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
- First, we solve the system
{tex}x - y = 2{/tex}
x + y = 6
Adding the two equations,
{tex}\Rightarrow(x - y) + (x + y) = 2 + 6{/tex}
{tex}\Rightarrow{/tex} {tex}x - y + x + y = 8{/tex}
{tex}\Rightarrow{/tex} {tex}2x = 8{/tex}
{tex}\Rightarrow x = \frac{8}{2} {/tex}
{tex}\Rightarrow{/tex} {tex}x = 4{/tex}
Substituting the value of x in the first equation, we have
{tex}4 - y = 2{/tex}
{tex}\Rightarrow{/tex} {tex}y = 4 - 2{/tex}
{tex}\Rightarrow{/tex} {tex}y = 2{/tex}
Hence, the number is 10 {tex}\times{/tex} 2 + 4 = 24 - Now, we solve the system
{tex}x - y= -2{/tex}
{tex}x + y = 6{/tex}
Adding the two equations, we have
{tex}(x - y) + (x + y) = -2 + 6{/tex}
{tex}\Rightarrow{/tex} {tex}x - y + x + y = 4{/tex}
{tex}\Rightarrow{/tex} {tex}2x = 4{/tex}
{tex}\Rightarrow x = \frac{4}{2} {/tex}
{tex}\Rightarrow{/tex} x = 2
Substituting the value of x in the first equation,
{tex}\Rightarrow2 - y = -2{/tex}
{tex}\Rightarrow{/tex} {tex}y = 2 + 2{/tex}
{tex}\Rightarrow{/tex} y = 4
Hence, the number is 10 {tex}\times{/tex} 4 + 2 = 42
Thus, the two numbers are 24 and 42.
Posted by Aaliya ? 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
We have, tan 2A = cot (A - 18°)
{tex}\Rightarrow{/tex} tan 2A = tan {90° -(A -18°)}
{tex}\Rightarrow{/tex} tan 2A = tan (108° - A)
{tex}\Rightarrow{/tex} 2A = 108° - A
{tex}\Rightarrow{/tex} 3A = 108°
{tex}\Rightarrow{/tex} A = 36°
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Gaurav Seth 6 years, 3 months ago
The existing Mathematics examination is the Standard Level Examination. Standard-Level and Basic-level Question papers shall be based on the same syllabus. However the Standard-Level Mathematics assesses higher Mathematical abilities compared to Basic-Level. Accordingly, the difficulty level of the Mathematics – ‘Basic’ is less than that of Mathematics-‘Standard’.
for more info : Click the link;
<a href="http://cbse.nic.in/newsite/circulars/2019/03_CircularFAQ_2019.pdf">http://cbse.nic.in/newsite/circulars/2019/03_CircularFAQ_2019.pdf</a>
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Harsh Mishra 6 years, 3 months ago
Q. Find the Perpendicular Distance Of A(6,4) from the X-axis ????
Shreya Aapne http://clay6.com se copy toh kar liya lekin Question theek se read naheen kiya......Haina Miss Darshini....
Jaisa kee Aapne likha hai :- Consider a Point P(x,y) on the coordinate plane Perpendicular distance of the point P(x,y) from the x axis=ordinate of the point P(x,y)=y Perpendicular distance of the point P(x,y) from the y axis=
ordinateof the point P(x,y)=x Perpendicular distance from y-axis = x coordinate of the point Given x coordinate of point is 6, Therefore, perpendicular distance of point A(6,4) from y-axis is 6......toh word language toh lagvag theek hai lekin ending galat hai...... I'm also not going to write by myself but, I'll include some of the changes that in my view're necessary.......Consider a point A(x,y) on the coordinate plane.
Perpendicular distance of the point A(x,y) from the x axis=ordinate of the point A(x,y)=y. [Since, An ordered pair contains abscissa (value of x) and ordinate (value of y) respectively]
Perpendicular distance of the point A(x,y) from the y axis=abscissa of the point A(x,y)=x.
Perpendicular distance from x-axis = y coordinate of the point
Given y coordinate of point is 4,
Therefore, perpendicular distance of point A(6,4) from x-axis is 4. That's it.... :) Sorry, For showing Off.... But As I've heard that peoples learn from their mistakes, so I hope you too'll do the same..... :) Once again, Sorry....
Shreya .... 6 years, 3 months ago
Posted by All In One Pratham 6 years, 3 months ago
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Unnati Dabasi 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
)
Given, sides other than the hypotenuse of a right triangle is 16 cm and 8 cm.
So, Let AC = 16 cm and BC = 8 cm.
Let PQCR be the largest square which can be inscribed in the right triangle ABC. Thus,
Let AC = x cm, So, AQ = 16 - x cm
In {tex}\triangle A P Q{/tex} and {tex}\triangle A BC{/tex}
{tex}\angle A=\angle A{/tex} (common angle)
{tex}\angle A Q P=\angle A C B{/tex} (each has 90o)
{tex}\Rightarrow \triangle A P Q \approx \triangle A B C{/tex} (By AA similarity)
So,
{tex}\frac{A Q}{A C}=\frac{P Q}{B C}{/tex}
{tex}\Rightarrow \frac{16-x}{16}=\frac{x}{8}{/tex}
{tex}\Rightarrow 16-x=2 x{/tex}
{tex}\Rightarrow x=\frac{16}{3} \mathrm{cm}{/tex}
Hence, the length of largest square which can be inscribed in the right triangle ABC is 16/3 cm.
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