Bank Exams Questions

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P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) 

 = 1-16C1× 14C1×12C1×10C116C4=1539

The given puzzle follows a pattern that,

Number - <last digit> <sum of all digits>.

111 - 13

112 - 24

113 - 35

So 

117 - 79

1.12.91 is the first Sunday of December 91.
So, 3.12.91 is the first Tuesday of the month.
Clearly, 10.12.91, 17.12.91, 24.12.91 and 31.12.91 are also Tuesdays.
So, 24.12.91 is the fourth Tuesday.

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅
= (10 x 6)/252 = 5/21

Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 5780
= x= 5780/2.89 = 2000

Therefore, amount invested by Raghu is Rs. 2000.

Given that the month begins on a Friday and has 31 days

Sundays = 3rd, 10th, 17th, 24th, 31st
=> Total Sundays = 5

Every second & fourth Saturday is holiday.
2nd & 4th Saturday in every month = 2

Total days in the month = 31

Total working days = 31 - (5 + 2) = 24 days.

The given series follows a pattern of +19, x19, +17, x17, +15, x15,...

Here the number which is to be replaced in ? is

41310+15=41325.