Bank PO Questions

Number of blocks =160×100×6020×20×20=120

Speed of boat in still water = 1/2 (12 + 6) = 9 kmph.

Longest and shortest phases Of cell cycle:

Shortest Phase - M phase or Mitotic phase.

Longest phase - G1 phase of the Interphase

Longest and shortest phases During Mitosis:

Longest Phase - Prophase

Shortest Phase - Anaphase

Distance travelled by Ramu = 45 x 4 = 180 km

Somu travelled the same distance in 6 hours.

His speed = 180/6 = 30 km/hr

Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.

Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.

Hence difference in the time taken = 0.5 hours = 30 minutes.

Clearly, we may mark such sets of 3 numbers, in which the middle number is 8 and each of the two numbers on both sides of it is a factor of 8, as shown below :

2 8 3 8 2 4 8 2 4 8 6 8 2 8 2 4 8 3 8 2 8 6

So, there are two such 8's. Hence the answer is C.

Let the tin contain 5x litres of liquids

5x - 45 x 455x - 45 x 15 + 45 = 25

=> 5(4x - 36) = 2(x + 36)

=> 20x - 180 = 2x + 72

=> x = 14 litres

Hence, the initial quantity of mixture = 70l

Quantity of liquid B

= 70 - 45 x 15 + 45

= 50 litres.

Water can moderate temperature because of the two properties: high-specific heat and the high heat of vaporization. 

Hence, all the above given options are not the properties of water.

Here the most stable carbocation is the option (2).

Carbocations are always hungry for electrons. The stability of carbocations increases from primary to secondary to tertiary. Here, option (c) shows the most stable carbocation because it can easily withdraw electrons from two other carbon atoms and the more electronegative oxygen atom.

Carbocation stability depends on hyperconjugation. The tertiary carbocation is stabilized by electron releasing inductive effect.