Job Roles

Take nearest values
√580×∛510+ 49.999 x 3.999=?
24x 8 + 200 = 392

x = ±23

And, y = 23

hence, x ≤ y

x^2+ 30x + 81 = 0
x^2+ 27x + 3x + 81 = 0
x = -3, -27
y^2-9y -162 = 0
y^2-18y + 9y -162 = 0
y = -9, 18
Hence, no relationship can be established between x and y

5x^2+ 28x = -15x = (-3/5, -5)

II. 3y^2+ 11y + 6 = 0y = (-3, -2/3)

So Relationship cannot be established

Let the breadth of rectangle be x m.

Then, the length of rectangle = (x+ 6) m

Perimeter of rectangle = 2 (x+x+ 6) m

Therefore, 2 (x+x+ 6) = 84 m

4x+ 12 = 84

4x= 84 –12
4x = 72/4 = 18
Therefore, length of rectangle = 18 + 6 = 24 m = height of triangle

Diagonal of rectangle = √(324+576) = 30 m = base of triangle

Therefore, are of triangle =1/2× base × height =1/2× 24 × 30 = 360 sq. m

Let length of platform be ‘y’metres.

Then, at the platform,Distance travelled = y + (length of train) = (y + 480)m

Then, Speed of train = (y + 480)m / (3x60) sec = (y+480)/180 ...(1)

Also, at the pole,Distance travelled = length of train = 480m

Then, Speed of train = 480m / 30sec = 16 m/s ...(2)

Equating eq.(1) & eq.(2), we get,

(y+480)/180 = 16(y+480) = 16 x 180 = 2880

y = 2880 –480 = 2400m or 2.4 km long platform.

Let time period of S.I. be T years.

Then for a principal amount, say P,

ATQ, as, S.I. = C.I. for rate =10%p.a. and

time for C.I. = 2(P x 10 x T)/100 = P{[ (100+10)/100 ]2-1}T/10 = [ 110/100 ]2–1 = [(11/10)2–1] = (121-100)/100T/10 = 21/100T = 21/10 = 2.1 years