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x+4=9x+4-4=9-4x=5

x/3 =4x/3 ×3=4×3x=12

i don't write answer

i don't write answer

if person use 1kg sugar that price 100 rupees so that raised by 25% then total amount should be 125 rupees per kg . so that person want to reduce .so that situation person reduce 20% cost.then actually he had 100 rupees.that meens answer is (b) 20%

correct awnser is 20%

x^3+2x^2+kx-3 by x=3 ; k=21; (3)^3+2(3)^2+21(3)-3=27+18+63=108

As there is 20 litres of acid, which is fixed. To make a 16 percent acidic solution we add water and now the quantity becomes say x. So 16 % of x = 20..16 * x = 20X = 125.Therefore 25 litres of water is to be added

The answer of this question is 16

6x² - 5x------------ 2²

6x² - 5x------------ 4

3x²/2 - 5x/4

√2 x² +7x +5√2 = 0

⇒√2 x² + 5x + 2x + 5√2 = 0

⇒√2x² + 5x +√2*√2*x + 5√2=0

⇒ x(√2x+5) +√2(√2x+ 5) =0

⇒ (x+√2)(√2x+5)=0

That gives x=-√2 or x= -5/√2

Bro 2nd wala

100x2- 20x + 1 = 0

By splitting of middle terms method,

100x2 - 10x - 10x + 1 = 0

10x ( 10x - 1 ) - 1 ( 10x - 1 ) = 0

( 10x - 1 ) ( 10x - 1 ) = 0

10x -1 = 0 ; 10x - 1 = 0

x = 1/10 ; x = 1/10

Therefore the roots are 1/10 and 1/10

Thnk bro

Given that a, a+b, a+2b are roots of given polynomialx³-6x²+3x+10

From this polynomial,Sum of the roots⇒ a+2b+a+a+b = -coefficient of x²/ coefficient ofx³ ⇒ 3a+3b = -(-6)/1 = 6 ⇒ 3(a+b) = 6 ⇒ a+b = 2 --------- (1) b = 2-a

Product of roots ⇒ (a+2b)(a+b)a = -constant/coefficient of x³ ⇒ (a+b+b)(a+b)a = -10/1

Placing the value of a+b=2 in it ⇒ (2+b)(2)a = -10 ⇒ (2+b)2a = -10 ⇒ (2+2-a)2a = -10 ⇒ (4-a)2a = -10 ⇒ 4a-a² = -5 ⇒ a²-4a-5 = 0 ⇒ a²-5a+a-5 = 0 ⇒ (a-5)(a+1) = 0 a-5 = 0 or a+1 = 0 a = 5 a = -1

a = 5, -1 in (1) a+b = 2

When a = 5, 5+b=2⇒ b=-3 a = -1, -1+b=2⇒ b= 3

∴ If a=5 then b= -3 or If a= -1 then b=3

p-3/p = 3/p=p/12; p^2=12×3=36; p=6

Highest power is 5 so degree of polynomial is 5

Let original price of a unit = Rs. 100Increased price = Rs. 160Reduction on Rs. 160 = Rs. 60Reduction on Rs. 100 = 60/160 * 100 = 37.5 %75/2 %

S.P = Rs 475 Let C.P be x Loss percent = 5% Loss = 5% of x = 5x/100 S.P = × - 5x/100 = 95x/100Here 95x/100 = 475 x = 500C.P = 500Now Profit Percent = 8% Profit = 8% of 500 = 8×500/100 = 40 New S.p = CP + Profit = 500 + 40 = 540

Football is played by rani.

70) option (A) is correct

79 is correct answer

79 is the correct answer

this is book of class 7th

If equation (p-3)x + 3y = pand px + py = 12 have infinite many solutions

Then,(p-3)/p = 3/p p-3 = 3p = 6

Value of p = 6

Let,Old price = xNew price = y

According to questiony = x - (25% of x)y = x - (25/100 × x)y = x - x/4y = 3x/4 ……(1)

Now, in order to restore the original price (x) the new price (y) should be increased by zx = y + z4y/3 = y + z ……[∵ from (1)]z = 4y/3 - yz = y/3

∴ New price should be increased by One-third.

I don't understand how u solved

Let the old price be xso 8% of x = 8/100 × x = 8x/100 = 4x/25So According to questionx + 4x/25 = 129.629x/25 = 129.6x = (129.6×25)/29x = ₹ 111.72Old cost is ₹111.72

Since the 5% discount, the new price is 0.95 of the old price.

If x is the original price, new is .95x.

So just divide by .95 to get x.

1/.95 = 1.053, or 5.3%.

To be exact, it's (20/19)%

1 is the simplified answer of the given question.

good

5-11/4= -6/4= -3/2

5/4+(-11/4)=5/4-11/4=-6/4=-3/2

5-11/4=-6/4=-3/2

is the best answer.

5-11/4-6/4-3/2 is correct answer.

-6/4 the ans of this

condition for equal rootsb²-4ac=0or D=0k^2-576= 0k= √576k= 24

4/11 ÷ 5/22 * 5/4

= (4/11 * 22/5) * 5/4

= 8/5 * 5/4

= 8/4

= 2

(x-y)^-1=(2/3-3/4)^-1=-1/12)^-1=-12andx^-1-y^-1=3/2-4/3=9-8/6=1/6

((4/27)^-2)x=(-2/3)^-3((27/4)^2 )x=(-3/2)^3(729/16)*x=-27/8x=-(2/27)

please like my answer if you find it useful

a) 34^2-6^2= (34+6)×(34-6)= 40×28=1120b) 20^2- 8^2 = 400-64 =336

c,d and e also