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33051.

What is mode

Answer» Highest frequent value
33052.

356

Answer» 356
33053.

Find the sum of odd numbers between 0 and 50.

Answer» {tex}\\text {Series of odd is }1, 3, 5, ..., 49.{/tex}{tex}\\therefore \\text { first number } a = 1 \\text { last number } l = 49{/tex}{tex}a_n=a+(n-1)d{/tex}{tex}\\Rightarrow 49=1+(n-1)(2){/tex}{tex}\\Rightarrow n=25{/tex}{tex}\\text {Total odd numbers from 0 to 50 are}{/tex}{tex}\\text {Now, sum of odd numbers from 1 to 50 } S_n \\text { is }{/tex}{tex}S_n = \\cfrac {n}{2}(a+l){/tex}{tex}\\Rightarrow S_n = \\cfrac {25}{2} (1+49){/tex}{tex}\\Rightarrow S_n = 25 \\times 25=625{/tex}{tex}\\text {Thus, required sum is 625.}{/tex}\xa0
33054.

When was class 10 science compartment exame

Answer» 19 jul
16 july
33055.

x+y=0x-y=0

Answer» {tex}\\text {Given two equations are } x+y=0 \\text { and } x-y=0{/tex}{tex}\\text {On solving these two equations, algebrically with any of methods knows, } i.e., \\text { substitution, elimination, etc., we get,}{/tex}{tex}x=0 \\text { and } y = 0{/tex}
33056.

|x-3|+2|x|+3|x-2|

Answer»
33057.

|x-1|=3

Answer» In simple words we can also write it as X-1 = 3 X-1+1= 3+1X = 4Because there is sign of absolute value the answer could also be -2
33058.

LCM and HCF of fraction..formula

Answer» H.C.F\xa0=\xa0H.C.F of numerators\xa0/\xa0L.C.M of denominatorsL.C.M\xa0=\xa0L.C.M of numerators\xa0/\xa0H.C.F of denominators
33059.

How can we determine where to use lcm in word problem

Answer» Whenever you are required to find a number that is\xa0divisible\xa0by more than one number, you have to find LCM.\xa0
33060.

n(T)=85%, n(E)=40%, n(H)=20%, n(EnT)=32%, n(TnH)=13%, n(EnH)=10%

Answer» I gon\'t no
33061.

Find the lowest number which is less by 9 to be divided by 21,35,and 49

Answer» 3,5,7
33062.

2x/x_4+2x-5/x-3=25/3

Answer»
33063.

let p be a prime number then prove that square root of it is an irrationak number

Answer»
33064.

3 x minus y is equals to 3 upon 9 x minus 3 Y is equals to 9 in linear equation

Answer»
33065.

Solve the x+y=a+b and ax_by=a^2+b^2 by method of elimination

Answer»
33066.

Using euclid division lemna find the hcf of 72 , 126 and 156

Answer»
33067.

Solve a/(x-b)+b/(x-a)=2

Answer»
33068.

Father of geometry

Answer» Euclid
33069.

What do you mean by Euclid division Lemma

Answer» a=bq +r it is quite sufficient
33070.

Why 4 is called blackhole number?

Answer»
33071.

In the roots of quadratic equation of 6x2-6-2=0

Answer» 6x2- x - 2 = 0or, 6x2 + 3x - 4x - 2 = 0or, 3x(2x + 1) -2(2x + 1) = 0or, (2x + 1)(3x - 2) = 0{tex}\\Rightarrow{/tex} either 3x - 2 = 0 or 2x + 1 = 0{tex}\\therefore \\quad x = \\frac { 2 } { 3 } \\text { or } x = - \\frac { 1 } { 2 }{/tex}Therefore, Roots of equation are\xa0{tex}\\frac { 2 } { 3 } \\text { and } - \\frac { 1 } { 2 }{/tex}.
33072.

What square of 3256

Answer» 10601536
33073.

From A quadratic polynomial whose zeroes are 7+2*1/2 and 7-2*1/2

Answer»
33074.

From

Answer» From
33075.

12(((9/3)-(4/2))*2+(2-4-3))

Answer» 84
33076.

Use euclid\'s division algorithm to find hcf of 1190and1445. Express the hcf in the form 1190m+1445n

Answer» Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.1445 = 1190 ×\xa01 + 2551190 = 255 ×\xa04 + 170255 = 170 ×\xa01 + 85170 = 85 ×\xa02 + 0So, now the remainder is 0, then HCF is 85Now,85 = 255 - 170= (1445 - 1190) - (1190 - 255 ×\xa04)= 1445 - 1190 - 1190 + 255 ×\xa04= 1445 - 1190 ×\xa02 + (1445 - 1190) ×\xa04= 1445 - 1190 ×\xa02 + 1445 × 4 - 1190 ×\xa04= 1445 ×\xa05 - 1190 ×\xa06= 1190 ×\xa0(- 6) + 1445 ×\xa05= 1190m + 1445n , where m = - 6 and n = 5
33077.

11110000+111145567

Answer» 122255567
122255567
33078.

prove that Sec theta (1-sin theta)(sec theta+tan theta)=1

Answer»
33079.

formules of ap chpter

Answer» Get formulae in revision notes :\xa0https://mycbseguide.com/cbse-revision-notes.html
33080.

in an ap if tje common diffrenceis equal to -4and the seventh term id 4then find the first term

Answer» an\xa0= a + (n - 1)da7\xa0= a + (7 - 1)da + 6d = 4a + 6(-4) = 4{tex}\\Rightarrow{/tex} a = 28
33081.

3cos55°/7sin35° - 4(cos70° cosec20°)/ 7(tan5° tan25° tan45° tan65° tan85°)

Answer»
33082.

Euclid\'s division algorithm to find the HCF of 135 and 225

Answer» 225 = 135×1+90, 135=90×1+45, 90=45×2+0 hence the hcf is = 45
225=135 × 1 +90135=90 × 1 +4590=45 × 2 + 0hence hcf = 45 did you got that vaishali
33083.

In an ap if d=-4 and seventh term is 4 then find first term

Answer» Here -20 is correct
I hope it may help u
d=4a7=4a=?a7=a +(n-1)d4=a+6(4)A=-20
16
24
33084.

Prove sin30 geometrically

Answer»
33085.

If x-3 is one root of quactic equation x-2k-6=0 find value of X

Answer» Firstly check the question... Since x-2k-6=0 is not quadratic..It should be x²-2kx-6=0....In that case put value of x=3 and find the value of k...
33086.

Sec thetaa=5by4 verify that tan thetaby 1+tan square theta = sin theta by sectheta

Answer» Thanks bolne ke lia mat bhulna
Sec theta 5/4so ,p=3 by pythagoros theormBy using foormula u find value of l.h.s &RHSLHS=RHS
Sorry you write correct question
Sec theta=5/4 ya 4/5 please bolo
33087.

What must be subtracted from 4x^4+2x^3-8x^2+3x-7 so that it is easily divisible by 2x^2+x-2?

Answer»
33088.

10+12×34

Answer» I get answer 418 but how.......414
414
No answer
33089.

99x + 101y = 499 : 101x + 89y =501

Answer» Given, 99x + 101y = 499 ....(i)101x + 99y = 501... (ii)Adding eqn. (i) and (ii),( 99x + 101y ) + (101x + 99y ) = 499 + 50199x + 101y + 101x + 99y = 1000200x + 200y = 1000x + y = 5 ...(iii)Subtracting eqn. (ii) from eqn. (i), we get( 99x + 101y ) - (101x + 99y ) = 499 - 50199x + 101y - 101x - 99y = -2-2x + 2y = - 2or, x - y= 1 ........ (iv)Adding equations (iii) and (iv)x + y + x - y = 5 + 12x = 6{tex}\\therefore {/tex}\xa0x = 3Substituting the value of x in eqn. (iii), we get3 + y = 5y = 2Hence the value of x and y of given equation are 3 and 4 respectively.
33090.

A (3,5) B (7,2 ) C (-3,7) AB=BC coordinate geometry

Answer»
33091.

Find the 19th term from the last term of the AP: 3,8,13,....,253.

Answer» a = 3, d = 5Now, 253 = a + (n + 1) d⇒ 253 = 3 + (n -1) x 5⇒ 253 = 3 + 5n – 5 = – 2⇒ 5n = 253 + 2 = 255⇒ n = 255/5 = 51Therefore, 19th term from the last term = 51 – 18 = 33a33 = a + 32d= 3 + 32 x 5= 3 + 160 = 163Thus, required term is 163
33092.

Find the equilateral of an equilateral triangle side 8cm

Answer»
33093.

6:222::8:

Answer» 296
33094.

usefulness of mathematics in the field of Defence

Answer»
33095.

usefulness of mathematics in the field of Aeronautics

Answer»
33096.

Prove that root 2 is irrational no.

Answer» Please yogita prove √5 is an iirational
Given √2 is irrational number.Let √2 = a / b wher a,b are integers b ≠ 0we also suppose that a / b is written in the simplest formNow √2 = a / b⇒ 2 = a2 / b2\xa0⇒ 2b2 = a2∴ 2b2 is divisible by 2⇒ a2 is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2 = 4c2⇒ 2b2 = 4c2⇒\xa0b2 = 2c2∴ 2c2 is divisible by 2∴ b2 is divisible by 2∴ b is divisible by 2∴ a are b are divisible by 2 .This contradicts our supposition that a/b is written in the simplest form.Hence our supposition is wrong∴ √2 is irrational number.
33097.

Factorise x*2+5√2+12

Answer» (x+2√2)(x+3√2)
33098.

What is the hcf of first prime and composition no.

Answer» 2
33099.

If hcf of two number is 2and their product is 120 find their lcm

Answer»
33100.

Solve for x and y ax/b-by/a=a+b;ax-by=2ab

Answer»