Bank PO Questions

How to find the years which have the same Calendars : 

 Leap year calendar repeats every 28 years. 

Here 28 is distributed as 6 + 11 + 11.

Rules:

a) If given year is at 1st position after Leap year then next repeated calendar year is Given Year + 6.

 b) If given year is at 2nd position after Leap year then next repeated calendar year is Given Year + 11.

 c) If given year is at 3rd position after Leap year then next repeated calendar year is Given Year + 11. 

Now, the given year is 2018

We know that 2016 is a Leap year.

2016  2017  2018  2019  2020

 LY    1st     2nd    3rd    LY

Here 2018 is at 2 nd position after the Leap year.

According to rule ( b) the calendar of 2018 is repeated for the year is 2018 + 11 = 2029.

The quadratic equation whose roots are reciprocal of 2x2+5x+3=0 can be obtained by replacing x by 1/x.

Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0

=> 3x2+5x+2=0

55min spaces are covered in 60min

60 min. spaces are covered in (60/55 x 60) min= 65+5/11 min.

loss in 64min=(65+5/11)- 64 =16/11

Loss in 24 hrs (16/11 x 1/64 x 24 x 60) min= 32 8/11.

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.

Let the length of the rope be r mts 

then, πr2=9856 ⇒ r2=9856×722=3136 ⇒r =56m

P = Rs. 6250, R = 14 %  & T = (146/365) years = 2/5 years .

 S.I=Rs.6250*14*25100=Rs.350

Let the cost price of a ball is Rs.x
Given, on selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls
The equation is :
17x - 720 = 5x
Solving the equation
we get x = 60
Therefore, cost price of a ball is Rs. 60.

Sumo wrestling is the Japan origin game in which the two heavy weight wrestlers fight in a marked circle to throw the opponent out of the circle to win the game. And the two participants in the game are called Sumos.

Hence, sumo wrestling is the national sport of Japan.

Let the speed of the boat = p kmph

Let the speed of the river flow = q kmph

From the given data,

2 x 28p + q = 28p - q

=> 56p - 56q -28p - 28q = 0

=> 28p = 84q

=> p = 3q.

Now, given that if

$$\begin{gathered} 283q + 2q + 283q - 2q = 67260 \\ => 285q + 28q = 67260 \\ => q = 3 kmph \\ => x =3q = 9 kmph \end{gathered}$$

Hence, the speed of the boat = p kmph = 9 kmph and the speed of the river flow = q kmph = 3 kmph.