Maruti Suzuki Quiz on Algebra with Answers

1/2x +1/4x+1/8x=14.
Take x as common so you will get
(1/2 +1/4+1/8)x=14,
(7/8)x=14,
x=16.

D=0
b^2 -4ac=0
25k2-4(2K)(2)=0
25k2=16k
k=4/5 or 16/25

given, (m⁶ + m⁴ + m² + 1)/m³

(m⁶/m³ + m⁴/m³ + m²/m³ + 1/m³)

(m³ + m + 1/m + 1/m³)

now,

m = (2 - √3)------------( 1 )

m³ = (8 - 3√3 - 6√3(2 - √3)

m³ = (8 - 3√3 - 12√3 + 18)

m³ = (26 - 15√3)-----------( 2 )

1/m = 1/(2 - √3) * (2 + √3)/(2 + √3)

1/m = (2 + √3)/(4 - 3)

1/m = (2 + √3)------------( 3 )

1/m³ = 1/(26 - 15√3) * (26 + 15√3)/(26 + 15√3)

1/m³ = (26 + 15√3)/(676 - 675)

1/m³ = (26 + 15√3)-------------( 4 )

adding------( 1 ) , --------( 2 ) , --------( 3 ) & --------( 4 )

(m + 1/m + m³ + 1/m³)

= (2 - √3 + 2 + √3 + 26 - 15√3 + 26 + 15√3)

= (4 + 52)

= 56

Hence, (m⁶ + m⁴ + m² + 1)/m³ = 56

 

a2+b2-4(a+b)= -8

(a-b)= 4

then it is equal to -8

Here,12+4 is 16 and it is written as 2⁴, then 12/4 is 3,
so now it is 2¹² so answer is 4096.

As this equation should be divisible by a number for all Integral values, Assume n=1
(7^2)-(3^3) = 22
above equation is divisible by 22

Let s+s1+s2 =x;
Then s(x)=9 --(1)
S1(x)=16 -- (2)
S2(x)=144 –(3)
Add above three equations s(x)+S1(x)+S2(x)=9+16+144
x(s+s1+s2 ) = 169
x2=169
x=13
substitute x value in equation(1) so s=9/13

1/18 = 1/18 – 1/9 +x
X=1/9
Ans: Filling the tank in 9 hours

3481

Shortcut : let (x+1/x) = a
Then (x5+1/x5) = a5-5a3+5a
So answer : 6726