Quantitative Aptitude - Arithmetic Ability Questions

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.

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

AC-AB<BC Or AB+BC>AC

BC-AC<AB Or AB+AC>BC

AB-BC<AC Or AC+BC>AB

Sum of 2 sides of triangle is always greater than the third side

So all three statements are true

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

 S.I=Rs.6250*14*25100=Rs.350

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.

Let the edge of the cube bea. Then, 

 6a2 = 1734 

 => a = 17 cm.

  Volume =  a3 = 173 = 4193 cu.cm

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3

Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

Let the average age of the whole team be x years.

11x - (26 + 29) = 9 (x - 1)

=> 11x - 9x = 46

=> 2x = 46

=> x = 23.

So, average age of the team is 23 years.