A train travelling at a speed of 75 mph enters a tunnel 312 miles long. The train is 14 mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?

According to question ,we can say that
From equation ?. x⁴ = 398 + 227 = 625
? x⁴ = 5⁴

From equation ?. y² = 346 - 321 = 25
? y² = 5²

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

∴ The day on 6^th March, 2005 will be 1 day beyond the day on 6^th March, 2004.

Given that, 6^th March, 2005 is Monday.

∴ 6^th March, 2004 is Sunday (1 day before to 6^th March, 2005).

Couldn't be determined, since the total amount of money is not given in either of the case

Given that a /b = b/c
? b² = ac
? a⁴ : b⁴ = a⁴ : a² c² = a² : c²

Given, A = 6 km/h, B = 3 km/ h
According to the formula, Average speed = 2AB/(A + B)
? Required average speed = 2 x 6 x 3/(6 + 3)
= 36/9 = 4 km/h

Given that, W + D = 6 ...(i)
[ w = Time taken while walking and
D = Time taken while driving ]
From Eq. (i)
5 +D = 6
? D = 1
2D = 2 x 1 = 2
? He will take 2 h to drive both ways.

175760 = 2⁴ x 5 x 13 ³
= 2³ x 2 x 5 x 13³
To make perfect cube, it must be divided by 2 x 5 = 10

On comparing it with ax² + bx + c = 0 , we get a = 1, b = p, c = q
The roots of the equation x² + px + q = 0 are equal if
b² - 4ac = 0
? p² - 4q = 0 ? p² = 4q.
Thus , required answer is option B .

Let the volume be x³ and 27x³
? Their edges are x and 3x

Now Ratio of their surface area = 6x² : 54x² = 1 : 9