IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: If θ be a positive acute angle satisfying cos²θ + cos⁴θ = 1, then the value of tan²θ + tan⁴θ is?

Solution: 
Given that, 
cos²θ + cos⁴θ = 1
⇒ cos⁴θ = 1 - cos²θ
⇒ cos⁴θ = sin²θ ; [sin²θ = 1 - cos²θ] 
⇒ cos²θ.cos²θ = sin²θ
⇒ cos²θ = sin²θ/cos²θ
⇒ cos²θ = tan²θ

Now, 
cos²θ + cos⁴θ = 1
⇒ cos²θ + (cos²θ)² = 1
⇒ tan²θ + (tan²θ)² = 1
∴ tan²θ + tan⁴θ = 1

Question: An accurate clock shows 2:00 PM. Through how many degrees will the hour hand rotate by the time the clock shows 8:00 PM?

Solution:
We know,
A clock is a full circle = 360°
There are 12 hours on the clock face
So the hour hand moves = 360° ÷ 12 = 30° per hour

From 2 : 00 PM to 8 : 00 PM is exactly 6 hours later.
Therefore, the hour hand rotates,
6 hours × 30° per hour = 180°

So the hour hand will rotate through 180°.

Question: What will come at the place of question mark?
3, 6, 11, 18, 27, ?

Solution:
Look at the differences between consecutive terms.
6 - 3 = 3
11 - 6 = 5
18 - 11 = 7
27 - 18 = 9
The first differences are 3, 5, 7, 9
These increase by + 2 each time (3 ⇒ 5 ⇒ 7 ⇒ 9).
So the next difference should be = 9 + 2 = 11

Therefore next number = 27 + 11 = 38

Question: March 20, 1991 was a Wednesday. What day of the week was March 20, 1992?

Solution:
Given that, 
March 20, 1991 was a Wednesday.

1992 was a leap year which is divisible by 4.
A normal year (non-leap year) has 365 days and 365 ÷ 7 = 52 weeks + 1 day remainder ; day advances by + 1.
A leap year has 366 days and 366 ÷ 7 = 52 weeks + 2 days remainder ; day advances by + 2.

Since the period from March 20, 1991 to March 20, 1992 includes February 29, 1992 (the leap day), it is a full 366-day period.
Therefore Wednesday + 2 days
Wednesday ⇒ Thursday ⇒ Friday

So March 20, 1992 is Friday.

Question: If sec²θ + tan²θ = 7/12, then sec⁴θ - tan⁴θ = ?
 
Solution: 
Given that, 
sec²θ + tan²θ = 7/12

Now, 
sec⁴θ - tan⁴θ
= (sec²θ)² - (tan²θ)²
= (sec²θ + tan²θ)(sec²θ - tan²θ) ; [sec²θ - tan²θ = 1]
= (7/12) × 1
= 7/12

Question: A clock shows 12 : 00. Through how many degrees has the hour hand moved by 25 minutes past 5? 

Solution:
Full circle = 360°

Clock has 12 hours; Hour hand moves per hour, 360°/12 = 30° per hour
And, 1 hour = 60 minutes
Hour hand moves per minute = 30°/60 = 0.5° 

Now, 
From 12 : 00 to 5 : 25 = 5 hours 25 minutes = (5 × 60) + 25 = 325 minutes

∴ Degrees moved by hour hand = 325 × 0.5° = 162.5° 

By 25 minutes past 5, the hour hand has moved through 162.5° from the 12 : 00 position.

Question: What should come in place of the question mark '?' in the following number series.
10, 30, 68, 130, ?

Solution:
Each term is n³ + n
1^st term = 2³ + 2 = 8 + 2 = 10
2^nd term = 3³ + 3 = 27 + 3 = 30
3^rd term = 4³ + 4 = 64 + 4 = 68
4^th term = 5³ + 5 = 125 + 5 = 130
5^th term = 6³ + 6 = 216 + 6 = 222 

∴ The missing number is 222. 

Question: If tomorrow is Friday, what day of the week will it be after 46 days?

Solution: 
Given that,
Tomorrow is Friday, then today is Thursday

Days of the week repeat every 7 days. So, find the remainder when 46 is divided by 7.
So 46 ÷ 7 = 6 weeks + 4 days remainder.
So the day will move 4 days forward. 

Count 4 days from Thursday,
Thursday → Friday → Saturday → Sunday → Monday

Therefore, 46 days from today (Thursday) is Monday.

Question: A 60-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

Solution:
 
Let,
Height, AB = h

Given that, 
AC = 60m                    
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/60
⇒ h = 60 × (1/2)
∴ h = 30 m

So the height of the pole is 30 meters

Question: How many times are the hands of a clock at right angle in a day?
 
Solution: 
In 12 hours, they are at right angles 22 times.
∴ In 24 hours, they are at right angles = 2 × 22 = 44 times.

So the hands of a clock are at a right angle 44 times in a day.

Question: In a certain code language, 'BACK' is coded as '89', 'LEGS' is coded as '63'. How will 'CHEST' be coded in that code language?

Solution:
 
Logic: Sum of reverse alphabetical values - 2 = Number.

Now, BACK
B = 25, A = 26, C = 24, K = 16
∴ BACK = 25 + 26 + 24 + 16
= 91
= 91 - 2
= 89 

LEGS
L = 15, E = 22, G = 20, S = 8
∴ LEGS = 15 + 22 + 20 + 8
= 65
= 65 - 2
= 63 

Now for CHEST
C = 24, H = 19, E = 22, S = 8, T = 7
∴ CHEST = 24 + 19 + 22 + 8 + 7
= 80
= 80 - 2
= 78

So the code for CHEST is 78. 

Question: Which year is not a leap year?

Solution:
A leap year is divisible by 4.

Here, 
1996 ÷ 4 = 499, divisible by 4 , so leap year.
1994 ÷ 4 = 498.5, not divisible by 4 , not a leap year.
2000 ÷ 4 = 500, divisible by 4 , so leap year.
2004 ÷ 4 = 501, divisible by 4, so leap year.
1988 ÷ 4 = 497, divisible by 4, so leap year.

Question: If x = a secθ.cosφ, y = b secθ.sinφ and z = c tanθ. then the value of (x²/a²) + y²/b²) - (z²/c²) is?

Solution: 
Given that, 
x = a secθ.cosφ
∴ x/a = secθ.cosφ

y = b secθ.sinφ
∴ y/b = secθ.sinφ

And, z = c tanθ
z/c = tanθ

Now, 
(x²/a²) + y²/b²) - (z²/c²)
= (x/a)² + (y/b)² - (z/c)²
= (secθ.cosφ)² + (secθ.sinφ)² - (tanθ)²
= sec²θ.cos²φ + sec²θ.sin²φ - tan²θ
= sec²θ(cos²φ + sin²φ) - tan²θ ; [cos²θ + sin²θ = 1]
= sec²θ - tan²θ
= 1

Question: At what angle are the hands of a clock inclined at 10 minutes past 5?

Solution: 
The standard formula for the smaller angle θ between the hour and minute hands is.
θ = {|60H - 11M|}/2 ; [where H = hour, M = minutes]
= {|(60 × 5) - (11 × 10)|}/2
= |300 - 110|/2
= 190/2
= 95°

So the hands of the clock are inclined at 95° at 10 minutes past 5.

Question: In a code, 'BREAD' is written as 12345 and 'MOUSE' is written as 67893. How is 'DREAM' written in the same code?

Solution:
BREAD is written as 12345
B = 1, R = 2, E = 3, A = 4, D = 5

MOUSE is written as 67893
M = 6, O = 7, U = 8, S = 9, E = 3

So, DREAM will be 
D = 5, R = 2, E = 3, A = 4, M = 6

∴ DREAM = 52346

Question: Some months have 30 days and some have 31. How many months have at least 28 days? 

Solution:
Every month has at least 28 days. [February has 28 days in a common year and 29 in a leap year].

We know,
January, March, May, July, August, October, December is 31 days
April, June, September, November is 30 days
And February is 28 or 29 days

So all 12 months have at least 28 days.

Question: If tan(θ + 45°) = 1, then what is the value of sin θ?

Solution: 
Given that, 
tan(θ + 45°) = 1
We know that, tan 45° = 1

Therefore, 
tan(θ + 45°) = tan 45°
⇒ θ + 45° = 45°
⇒ θ = 45° - 45°
∴ θ = 0°

So, sinθ = sin0° = 0

Question: A watch becomes fast by 5 minutes everyday. By what percent does it become fast ?
 
Solution: 
Number of minutes in a day = (24 × 60) = 1440

∴ Required percentage =(5/1440) × 100%
= (50/144)%
= (25/72)%

Question: Anika is taller than Bristi but shorter than Kabir. Rafi is shorter than Bristi. Shawon is taller than Anika but shorter than Kabir. Who is the smallest among them?

Solution: 
From the given statements,
Anika > Bristi
Kabir > Anika
∴ Kabir > Anika > Bristi

Rafi < Bristi
∴ Bristi > Rafi

And
Shawon > Anika
Kabir > Shawon
∴ Kabir > Shawon > Anika

Full order from tallest to smallest,
Kabir > Shawon > Anika > Bristi > Rafi

Therefore, Rafi is the smallest among them.

Question: If the day before yesterday was Thursday, when will Sunday be?

Solution:
Day before yesterday = Thursday
Yesterday = Friday
Today = Saturday
Tomorrow = Sunday

Therefore, Sunday will be tomorrow.