ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

A number divisible by 4 formed using the digits 1, 2, 3, 4 and 5 has to have the last two digits 12 or 24 or 32 or 52.
In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways.
A number divisible by 4 can be formed in 6 × 4 = 24 ways.
Total number that can be formed using the digits 1, 2, 3, 4 and 5 without repetition
= 5! = 120
Required probability,
= 24/120
= 1/5

Probability of getting a tail when a single coin is tossed 
=1/2
Probability of getting 4 when a die is thrown 
=1/6
Required probability,
=1/2 × 1/6
=1/12

Here, n(S) = 52
There are 26 red cards (including 2 kings) and there are 2 more kings.
Let E = event of getting a red card or a king.
Then, n(E) = 28
∴P(E)=n(E)/n(S)
=28/52
=7/13

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting a total of 10 or 11
= [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)]
∴P(E) = n(E)/n(S) = 5/36

In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
∴P(E)=n(E)/n(S)
=4/36
=1/9

The letters of the word 'BAKERY' be arranged in 6! ways 
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

The digits to be used are 0,6 and 9
The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 3⁵
But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 3⁵ - 1
= 243 - 1
= 242

Required number of ways,
= ⁶C₄
= 6×5×4! /2!4!
= 15 ways

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

100 years contain 5 odd days.
∴ Last day of 1st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

The year 2008 is a leap year. So, it has 2 odd days.
1st day of the year 2008 is Tuesday (Given)
So, 1st day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.

Angle traced by hour hand in 13/ 3 hrs
=( 360/12 × 13/3)^∘ =130^∘
Angle traced by min. hand in 20 min
=( 360/60 ×20)^∘=120^∘
∴Required angle
=(130−120)^∘ =10^∘

In 12 hours, they are at right angles 22 times.
∴ In 24 hours, they are at right angles 44 times.

Angle traced by the minute hand in 5 minutes. =(360/60×5)^∘ = 30^∘