Bank Math Master

Question: Using all the letters of the word 'GIFT' how many distinct words can be formed?

Solution:
The word 'GIFT' has 4 letters.
The number of distinct words using all the letters can be calculated as 4! = 4 × 3 × 2 × 1 = 24.

So, there are 24 distinct words that can be formed from the letters in the word 'GIFT'.

Question: What would be the mirror image of the clock when the time is 01 : 40?

Solution:
Hence,
11 : 60 -  01 : 40
= 10 : 20

∴ The mirror image of 1 : 40 would be 10 : 20.

Question: A 80 m long ladder is leaning on a wall. If the ladder makes an angle of 45° with the ground, find the distance of the ladder from the wall.

Solution:

Here,
cosθ = Base/Hypotenuse
⇒ cos45° = Base/80
⇒ Base = 80cos 45° = 80 × (1/√2) = 40√2 
∴ Distance of the ladder from the wall = 40√2 m

Question: If sin3A = x, then value of x is-

Solution:
sin3A = sin(2A + A)
⇒ sin3A = sin2A.cosA + cos2A.sinA
⇒ sin3A = 2sinA.cosA.cosA + (1 - 2sin²A).sin A
⇒ sin3A = 2sinA(1 - sin²A) + sinA - 2sin³A
⇒ sin3A = 2sinA - 2sin³A + sinA - 2sin³A
∴ sin3A = 3sinA - 4sin³A

Question: In how many different ways can five friends sit for a photograph of five chairs in a row?

Solution:
We have to find total number of arrangements of 5 persons seated in a row.
We know that arrangement of n different things can be done in n! ways.

So, arrangements of 5 persons can be done in 5! = 120 ways.

Question: When you looked at a clock, it was showing 6 : 00 in the morning. By how much angle will the hour’s hand rotate when you again look at the clock at 12 : 00 in the noon?

Solution:
In 12 hours, the hour’s hand turns 360°

Hence, the difference between time 12 : 00 - 6 : 00 = 6 hours

Therefore, the required angle = (360°/12) × 6 = 180°

Question: There are two poles, one on each side of the road. The higher pole is 54 m high. From the top of this pole, the angle of depression of the top and bottom of the shorter pole is 30° and 60° respectively. Find the height of the shorter pole.

Solution:

Let AB and CD be the two poles.
Let AC = x m
CD = h m

Now, in triangle ABC,
tan60° = AB/AC
⇒ √3 = 54/AC
∴ AC = 18√3 m

Clearly, AC = DE = 18√3 m

In triangle BED,
tan30° = BE/DE
⇒ BE = DE tan 30
⇒ BE = 18 √3 / √3 m
⇒ BE = 18 m
⇒ CD = AE = AB - BE
⇒ CD = 54 - 18 = 36 m

Therefore, the height of the shorter pole = 36 m.

Question: If sinθ + cosθ = √3, then what is tanθ + cotθ equal to?

Solution:
sinθ + cosθ = √3
⇒ (sinθ + cosθ)² = (√3)²
⇒ sin²θ+ 2sinθcosθ + cos²θ  = 3

We know,
sin²θ + cos²θ = 1

∴ 1 + 2sinθcosθ = 3
⇒ 2sinθcosθ = 2
∴ sinθcosθ = 1

Now,
tanθ + cotθ
= sinθ/cosθ + cosθ/sinθ
= (sin²θ + cos²θ)/(sinθcosθ)
= 1/1
= 1

Question: If today is Monday. After 57 days, it would be:

Solution:
57/7 = 8 ( Remainder = 1)

It is well known that each day of the week is repeated after 7 days.
So we can say that after multiples of 7 days, the day will repeat itself as monday.
Hence after 56 days Monday will occur again.

So we calculate after after 57 days = Monday + 1day = Tuesday.

Question: The angle of elevation of the sun, when the length of the shadow of a tree 1/√3 times the height of the tree, is:

Solution:

Let,
Height of the tree BC = x
Length of the shadow AB = (1/√3)x
The angle of elevation of the sun ∠A = θ

Now,
tanθ = BC/AB 
⇒ tanθ = x/{(1/√3)x}
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°
 

Question: January 2, 2007 was Tuesday. What will be the day on January 2, 2008?

Solution:
January 2, 2007 was Tuesday
∴ January 1, 2007 was Monday

2007 is not a leap year so the last day of the year is as same as first day.
∴  December 31, 2007 will be Monday.

January 1, 2008 will be Tuesday
January 2, 2008 will be Wednesday.

Question: From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Solution:

Let AB be the tower.
Then, ∠APB = 30° and AB = 100 m.

AB/AP = tan 30° = 1/√3
⇒ AP = (AB × √3) m
= 100 × √3 m
= (100 × 1.73) m
= 173 m.

Question: If tan53° = 4/3, then, what is the value of tan8°?

Solution:
We know,
tan(A - B) = (tanA - tanB)/(1 + tanAtanB)

Now,
tan8° = tan(53° - 45°)
⇒ tan8° = (tan53° - tan45°)/(1 + tan53° tan45°)
⇒ tan8° = (4/3 - 1)/{1 + (4/3) × 1}
⇒ tan8° = (1/3)/(7/3)
∴ tan8° = 1/7

Question: How many different words can be formed from the alphabets of the word 'SCISSORS'?

Solution:
The word SCISSORS consists of 8 alphabets in which S repeat 4 times and remaining alphabets are C, I, O, R and occuring only once.

Number of words can be formed 8!/4! = 1680

Question: For what year will the calendar be the same as for the year 2009?

Solution:
For the year to have the same calendar as 2009 you need the sum of the number of odd days. When this sum is divisible by 7 than that year will have the same calendar as 2009.

Year : 2009, 2010, 2011, 2012, 2013, 2014
Odd day : 1, 1, 1, 2, 1, 1

Sum of odd days = 7 odd days [divisible by 7]

 ∴ Calendar for the year 2015 will be the same as for the year 2009.

Question: The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

Solution:

Let AB be the wall and AC be the ladder.
 
Then,  ∠ACB = 60°  and BC = 4.6 m

BC/AC = cos60° [ cosθ = ভূমি/অতিভুজ ]
⇒ BC/AC = 1/2
⇒ AC = 2BC
⇒ AC = 2 × 4.6
∴ AC = 9.2m

∴ The length of the ladder is 9.2 m.

Question: If sec²θ + tan²θ = 5/3, then what is the value of tan2θ?

Solution:
We know that,
sec²θ = 1 + tan²θ

Given that,
sec²θ + tan²θ = 5/3
⇒ 1 + tan²θ + tan²θ = 5/3
⇒ 2tan²θ = 5/3 - 1
⇒ 2tan²θ = 2/3
⇒ tan²θ = 1/3
⇒ tanθ = 1/√3
∴ θ = 30°

Now,
tan2θ = tan(2 × 30°) = tan60° = √3

Question: In a word jumble, there are 8 consonants and 5 vowels given. Find out in how many ways can we form a 5-letter word having three consonants and 2 vowels?

Solution:
Number of ways of selecting 3 consonants from 8 is ⁸C₃
Number of ways of selecting 2 vowels from 5 is ⁵C₂

Number of ways of selecting 3 consonants from 8 and 2 vowels from 5 is ⁸C₃ × ⁵C₂
= {(8 × 7 × 6)/(3 × 2 × 1)} × {(5 × 4)/(2 × 1)} 
= 56 × 10 
= 560
It means we can have 560 groups where each group contains total 5 letters (3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves  5! = 5 × 4 × 3 × 2 × 1 = 120

∴ Total number of words will be formed = 560 × 120
= 67200

∴ Required number of ways is 67200

Question: If the second-hand moves 480 times then how much space (in terms of degrees) will the minute hand move?

Solution:
Second hand moves 480 times = 480/60 min = 8 min

In 60 min the minute hand moves 360°
∴ In 8 min the minute hand moves (360°/60) × 8
= 48°

Question: Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Solution:

ধরি,
বাতিঘরের উচ্চতা AB = 100 মিটার 
C ও D হলো জাহাজের অবস্থান 

ΔABC এ 
tan∠ACB = AB/BC
⇒ tan 30° = 100/BC
⇒ 1/√3 = 100/BC
∴ BC = 100√3 

ΔABC এ 
tan∠ADB = AB/BD
⇒ tan45° = 100/BD
⇒ 1 = 100/BD
∴ BD = 100

∴ CD = 100√3  + 100
= 173.205 + 100 
= 273.205
≈ 273 m

Question: If sinθ = 3/5 then tanθ =?

Solution:
sinθ = 3/5

We know,
cosθ = √(1 - sin²θ)
= √(1 - 9/25)
=√(16/25)
= 4/5

∴ tanθ = sinθ/cosθ
= (3/5)/(4/5)
= (3/5) × (5/4)
= 3/4