Bank Math Master

Question: In How many different ways can the letters of the word 'PRESENT' be arranged?

Solution:
The word 'PRESENT' contains 7 letters, with 2 E and all other 5 different.

Ways = 7!/2!
= 2520

Question: How many diagonals can be drawn in a pentagon?

Solution: 
A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs.
Total number of sides and diagonals,
= ⁵C₂
= 10
This includes its 5 sides also.

∴ Diagonals = 10 – 5 = 5

Question: There are five women and six men in a group. From this group, a committee of 4 is to be chosen. How many different ways can a committee be formed that contains three women and one man?

Solution: 
there are five women and six men in the group.
a committee of 4 that contains 3 women and 1 man can be formed in
= (⁵C₃) × (⁶C₁)
= 60

Question: When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?

Solution: 
outcomes in which 0 coins turn heads are,
⁶C₀ = 1 outcome.
outcomes in which 1 coin turn head are,
⁶C₁ = 6 outcomes.
outcomes in which 2 coins turn heads are,
⁶C₂ = 15 outcomes.
outcomes in which 3 coins turn heads are,
⁶C₃ = 20 outcomes.
Therefore, the total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.

Question: The angle of depression of a point situated at a distance of 70m from the base of a tower is 60°. The height of the tower is-

Solution:

Length of the tower AB = h meter.
∠DAC = ∠ACB = 60°
BC = 70 meter
In △ABC,
tan 60° = AB/BC
⇒√3 = h/70
⇒ h=70√3 meter

Question: How many 3-digit numbers can be built by using 1, 2, 3, 4, and 5, if repetition is allowed?

Solution:
total individual  digits = 5
every number will have 3 digits.
so, total 3-digits numbers = 5³ = 125.

Question: A team of 8 students goes on an excursion. In two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

Solution: 
There are 8 students and the maximum capacity of the cars together is 9.
We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second
Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in ⁸C₃ ways.
Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in ⁸C₄ ways.

Therefore, the total number of ways in which 8 students can travel is:
⁸C₃+⁸C₄ = 56 + 70 = 126

Question: A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

Solution:
A team of 6 members (5 players and 1 captain) has to be selected from 10 players.
this can be done in = ¹⁰C₆ = 210 ways.
A captain can be selected from these 6 in 6 ways.
∴ total number of ways = 210 × 6 = 1260 ways

Question: The ratio of the length of a man and its shadow is 1 : √3. The angle of elevation of the sun is-

Solution:

Let AB be a man and BC be its shadow
So that AB : BC = 1 : √3

Let, θ be the angle of elevation

∴tanθ = AB/BC = 1/√3 = tan30°

∴θ = 30°

Question: How many words can be formed from the letters of the word 'EXTRA' so that the vowels are never together?

Solution:
EXTRA has 5 letters.
total number of words = 5! = 120

putting the vowels together we get XTE(EA) = 4! = 24 ways.
EA can be arranged in = 2! = 2 ways.

total arrangements while putting the vowels together is = 24 × 2 = 48 ways.

∴ Total arrangements where the vowels are never together is = 120 - 48 = 72 ways

Question: Today is Sunday. After 61 days, it will be -

Solution: 
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Sunday.
∴ After 61 days, it will be Friday.

Question: If ⁶P_r = 360 and ⁶C_r = 15, find the value of r.

Solution:
We know that,
ⁿP_r = ⁿC_r × r!
360 = 15 × r!
r! = 24
r! = 4!
r = 4

Question: How many days are there in x weeks 3x days?

Solution: 
x weeks 3x days = (7x + 3x) days = 10x days.

Question: What is the angle formed between the hour hand and the minute hand of a clock at half past two in the afternoon?

Solution: 
দুপুর আড়াইটায় ঘড়ির ঘণ্টার কাঁটা ও মিনিটের কাঁটার ব্যবধান
= ।(১১M - ৬০H)/২।°
= ।{(১১ × ৩০) - (৬০ × ২)}/২।°
= ।(৩৩০ - ১২০)/২।°
= ১০৫°

Question: How many leap years are there in 100 years?

Solution: 
Given year is divided by 4, the quotient gives the number of leap years.
Here, 100 ÷ 4 = 25

But, as 100 is not a leap year
⇒ 25 - 1 = 24 leap years.

Question: In an examination paper, there are two groups each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?

Solution: 
5 questions can be selected in the following ways,
2 questions from the first group and 3 questions from the second group Or 3 questions from the first group and 2 questions from the second group.
= (⁴C₂ × ⁴C₃) + (⁴C₃ × ⁴C₂)
= 24 + 24
= 48

Question:  If we consider an anticlockwise direction, what is the time difference between 2 am and 10:30 pm?

Solution: 


Since we're going anticlockwise (backward in time), we need to count from 2 am back to 10:30 pm of the previous day
Breaking this down:
From 2:00 am back to midnight (12:00 am) = 2 hours
From 12:00 am back to 10:30 pm = 1 hour and 30 minutes

Total time difference = 2 hours + 1 hour and 30 minutes
= 3 hours and 30 minutes

Therefore, going anticlockwise from 2 am to 10:30 pm, the time difference is 3 hours and 30 minutes.

Question: Which of the following is not a leap year?

Solution: 
The century divisible by 400 is a leap year. [2000, 800 and 1200 is divisible by 400]
∴ The year 700 is not a leap year. [700 is not divisible by 400]

[Rules for leap years:
• Must be divisible by 4,
• If it's a century year (divisible by 100), it must also be divisible by 400.]

Question: In how many different ways can the letters of the word 'DETAIL' be arranged so that the vowels occupy only the odd positions?

Solution:
there are 6 letters, where there are 3 vowels and 3 consonants.

3 vowels in 3 odd positions can be arranged in = ³P₃ = 3! = 6 ways
3 consonants in 3 even positions can be arranged in = ³P₃ = 3! = 6 ways

total ways = 6 × 6 = 36 ways

Question: A box contains 10 balls, of which 3 are red and the rest are blue. How many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all 6 balls of the same color?

Solution: 
Six balls can be selected in the following ways,
One red ball and 5 blue balls
Or Two red balls and 4 blue balls.
Total number of ways,
= (³C₁ × ⁷C₅) + (³C₂ × ⁴C₇)
= 63 + 105
= 168