Nine engines consume 24 metric tonnes of coal, when each is working 8 hours day. How much coal is required for 8 engines, each running 13 hours a day, if 3 engines of former type consume as much as 4 engines of latter type?

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