Bank Math

Question: A person walks at 14 km/h instead of 10 km/h and in the same time covers 20 km more distance. Find the actual distance traveled.

Solution: 
Given that,
Usual speed = 10 km/h
Increased speed = 14 km/h
Extra distance covered = 20 km

Let the actual distance traveled be = d km

Now,
Time = d/10 
And, At 14 km/h, distance covered in the same time = d + 20 
So,
⇒ d/10 = (d + 20)/14
⇒ 14d = 10d + 200
⇒ 14d - 10d = 200
⇒ 4d = 200
⇒ d = 200/4
∴ d = 50 km

The actual distance traveled is 50 km.

Let,
total marks = x.
Then,
(30% of x) + 15 = (40% of x) - 35
⇒ (30x/100) + 15 = (40x/100) - 35
⇒ (x/10) = 50
⇒ x = 500.
So, passing marks = (30% of 500) + 15
= (30/100) × 500 + 15
= 165
∴ Pass percentage = {(165/500) × 100}% = 33%.

X can do in 1 day = 1/20 part 
Y can do in 1 day = 1/(7×5) = 1/35 part
X & Y together can do in 1 day = 1/20 + 1/35 = 11/140
11/140 part of the work is done in 1 day
So, 11/14 part of the work is done in = (11/14) × (140/11) = 10 days

Question: The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?

Solution:
Given,
Average of 10 numbers = 23
⇒ Sum/Total numbers = 23
⇒ Sum/10 = 23
∴ Sum of the 10 numbers = 230

If each number is increased by 4, the total increase = 4 × 10 = 40
New sum = 230 + 40 = 270

Therefore, the new average = 270/10 = 27

Question: If cotθ = 5/12, then secθ = ?

Solution:
এখানে,
cotθ = 5/12 = ভূমি/লম্ব

∴ ভূমি = 5, লম্ব= 12
∴ অতিভুজ = √(5²+ 12²)
= √169 = 13

∴ secθ = অতিভুজ/ভূমি
= 13/5

Question: A hollow cylinder has an internal radius of 8 cm, external radius of 12 cm, and height 15 cm. What is the volume of the material used?

Solution: 
Let, 
Internal radius (r) = 8 cm
External radius (R) = 12 cm
Height (h) = 15 cm

Volume of the the material used, 
V = πh(R² - r²) 
= π × 15 (144 - 64)
= π × 15 × 80
= 1200π

A team of 6 members has to be selected from the 10 players.
This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore,
total ways the selection can be made is = 210 × 6
= 1260

Question: In two identical glasses, one is half full of milk and the other is three-fifth full of milk. Both glasses are then filled with water and poured into a tumbler. What is the ratio of milk to water in the tumbler?

Solution:
In the 1st glass,
Milk = 1/2
Water = 1 − 1/2 = 1/2

In the 2nd glass,
Milk = 3/5
Water = 1 − 3/5 = 2/5

Now total milk in tumbler = (1/2 + 3/5)
= {(5 + 6)/10}
= 11/10

Total water in tumbler = (1/2 + 2/5)
= {(5 + 4)/10}
= 9/10

∴ Ratio of milk and water
= (11/10) : (9/10)
= 11 : 9

Question: Ten years ago, the ratio of ages of S and T was 4 : 3. Now the ratio is 5 : 4. Find the present age of T.

Solution: 
Let the ages of S and T ten years ago be 4x and 3x.
Present ages of
S = 4x + 10 and T = 3x + 10

Now ratio = 5 : 4 
⇒ (4x + 10)/(3x + 10) = 5/4
⇒ 4(4x + 10) = 5(3x + 10)
⇒ 16x + 40 = 15x + 50
⇒ 16x - 15x = 50 - 40
⇒ x = 10

∴ Present age of T = 3x + 10 
= (10 × 3) + 10
= 30 + 10 
= 40

Question: Sadia is currently three times older than Rina. Ten years ago, Sadia’s age was seven times Rina’s age.
What will be the sum of their ages after 8 years?

Solution:
Let Rina’s current age = x years
Sadia’s current age = 3x years

10 years ago:
Sadia’s age = 3x - 10
Rina’s age = x - 10

According to the question,
3x - 10 = 7(x - 10)
⇒ 3x - 10 = 7x - 70
⇒ - 4x = - 60
⇒ x = 15

So, Rina’s age = 15 years, Sadia’s age = 45 years

After 8 years:
Rina = 15 + 8 = 23
Sadia = 45 + 8 = 53

Sum = 23 + 53 = 76

Question: A shirt is being sold for Tk. 720 after a 25% discount. Determine its original price.

Solution:
Given,
Let the original price be x Tk.

Discount = 25% of x
= 25x/100
= x/4

∴ Selling Price = Original Price - Discount
= x - (x/4)
= 3x/4

Now,
3x/4 = 720
⇒ 3x = 720 × 4
⇒ 3x = 2880
∴ x = 960 

∴ Original price = 960 Tk

⇒ x + 1/x = 13/6
⇒ (x² + 1)/x = 13/6
⇒ 6x² - 13x + 6 = 0
⇒ 6x² - 9x - 4x + 6 = 0
⇒ 3x(2x - 3) -2(2x - 3) = 0
⇒ (3x - 2)(2x - 3) = 0
⇒ x = 2/3 or 3/2

(A +B)'s 20 day's work = 20 × 1/30 = 2/3
Remaining work = 1 − 2/3 = 1/3
Now,1/ 3 work is done by A in 20 days
∴ The whole work will be done by A
in 20×3 = 60days

Question: How many ways can the letters of the word 'LETTER' be arranged so that the two T’s are together?

Solution:
The word LETTER has 6 letters.
L, E, T, T, E, R
Here, 2 T’s are identical, 2 E’s are identical, and L and R appear once each.

Now,
Treat the two T’s as a single unit (bundle them together).
So we now have 5 units to arrange: (TT), L, E, E, R

Total units = 5 and E is repeated 2 times
∴ Number of distinct arrangements = 5!/2!
= 120/2
= 60
Inside the (TT) bundle, the two T’s are identical, so there is only 1 way to arrange them inside the bundle (TT = TT).

Therefore, total number of arrangements where the two T’s are together 60.

Question: While on a holiday, X persons have decided to rent a van. The rent of the van is Tk D and each person is to pay an equal share. If Y persons cancel their trip, which of the following represents the additional amount of Tk per person that each remaining person must pay in order to still rent the van?

Solution: 
Total rent = Tk. D
Original number of persons = X
∴ Original share per person = D/X

And, 
If Y persons cancel then, 
∴ Remaining persons = X - Y
∴ New share per person = D/(X - Y)

∴ Additional amount each remaining person must pay = {D/(X - Y)} - (D/X)
= D{1/(X - Y)} - (1/X)}
= D{(X - X + Y)/X(X - Y)}
= YD/(X(X - Y))

∴ The additional amount per remaining person is YD/(X(X - Y)).

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want black OR red pebble
There are 6 black and 8 red pebbles
Total pebbles = 2 + 6 + 4 + 8 = 20

So, Probability = 6/20 + 8/20 = 14/20 = 7/10.

Let, Rahim donates x taka each.
And, Karim donates = x + 1 taka each
ATQ,
100/x – 100/(x + 1) = 5
Or, x² + x - 20 = 0
Or, x² + 5x - 4x - 20 = 0
Or, (x - 4)(x + 5) = 0
So, x = 4; x ≠ -5
∴ Rahim donates to = 100/4 = 25 people
Kaim donates to = 100/5 = 20 people
Therefore, number of people in the charity = 20 + 25 = 45

Question: If m is an even integer, which of the following must be an odd integer?

Solution:
ধরি m = 2

ক) m² + m = 2² + 2 = 4 + 2 = 6 [যা জোড়]

খ) m - 2 = 2 - 2 = 0 [যা জোড়]

গ) 5m + 3 = 5 × 2 + 3 = 10 + 3 = 13 [যা বিজোড়]

ঘ) 2m + 4 = 2 × 2 + 4 = 8 [যা জোড়]

∴ উত্তর: গ) 5m + 3

Question: The difference of two numbers is 20% of the larger number. If the smaller number is 12, the larger one is-

Solution:
Let,
The larger number = x

According to the question,
x - 12 = 20% of x
⇒ x - 12 = 20x/100
⇒ x - 12 = x/5
⇒ 5x - 60 = x
⇒ 5x - x = 60
⇒ 4x = 60
⇒ x = 60/4
∴ x = 15

So, the larger number is 15

Question: A mixture contains two liquids 'A' and 'B' in the ratio 5 : 2. If 14 litres of mixture is withdrawn and replaced with 14 litres of 'B', then the ratio becomes 3 : 4. What was the initial quantity of A?

Solution:
ধরি, মিশ্রণের প্রাথমিক পরিমাণ = 7x লিটার

A এর পরিমাণ = 5x লিটার
B এর পরিমাণ = 2x লিটার

∴ 14 লিটার মিশ্রণ তুলে নেওয়ার পর,
A এর পরিমাণ = 5x - (5/7) × 14 = 5x - 10 লিটার
B এর পরিমাণ = 2x - (2/7) × 14 = 2x - 4 লিটার

আবার, B তে 14 লিটার যোগ করার পর,
B এর পরিমাণ = 2x - 4 + 14 = 2x + 10 লিটার

∴ প্রদত্ত অনুপাত,
⇒ (5x - 10)/(2x + 10) = 3/4
⇒ 4(5x - 10) = 3(2x + 10)
⇒ 20x - 40 = 6x + 30
⇒ 20x - 6x = 30 + 40
⇒ 14x = 70
⇒ x = 5

∴ A এর পরিমাণ = 5 × 5 = 25 লিটার

Speed of the first train
= 120/10 m/sec.
= 12 m/sec.

Speed of the second train
= 120/15 m/sec.
= 8 m/sec.

Relative speed = 12 + 8 = 20 m/sec.

∴ Required time
= (120 + 120)/20 sec.
= 240/20 sec.
= 12 sec.

Question: A merchant has 1500kg of wheat, part of which he sells at 10% profit and the rest at 20% profit. He gains 14% overall. The quantity sold at 20% profit is-

Solution:
ধরি, 20% লাভে বিক্রি করা গমের পরিমাণ = x কেজি
∴ 10% লাভে বিক্রি করা গমের পরিমাণ = (1500 - x) কেজি

প্রশ্নমতে,
10% of (1500 - x) + 20% of x = 14% of 1500
⇒ 10(1500 - x)/100 + 20x/100 = (14 × 1500)/100
⇒ {10(1500 - x) + 20x}/100 = 210
⇒ 15000 - 10x + 20x = 21000
⇒ 10x = 6000
⇒ x = 600

সুতরাং, 20% লাভে বিক্রি করা গমের পরিমাণ হলো 600 কেজি।

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 train running at the speed of 60 km/hr crosses a pole in 9 seconds. Find the length of the train.

Solution: 
Given that, 
Speed of train, v = 60 km/h
Time to cross a pole, t = 9 seconds

We know, 
Length of the train = speed × time
= (50/3) × 9
= (50 × 9)/3
= 450/3
= 150 meters

So the length of the train is 150 meters.

Distance covered by the first plane till 2 P.M.
i.e., in 2 hrs = (500 × 2) miles
= 1000 miles.
Distance covered by the second plane till 2 P.M.
i.e., in 1(1/2) hrs = (400 × 3/2) miles
= 600 miles.
∴ Required distance
= AB =√{(1000)² + (600)²}
=√(1000000 + 360000)
=√1360000
= 200√34 miles
= 200 × 5.83
= 1166 miles.

Question: In how many ways can a group of 3 teachers and 4 students be formed from 6 teachers and 9 students?

Solution:
We have 6 teachers and 9 students.
We need to choose 3 teachers from 6 and 4 students from 9.

∴ Number of ways = ⁶C₃ × ⁹C₄
= {6!/(3!(6 - 3)!)} × {9!/(4!(9 - 4)!)}
= {6!/(3! × 3!)} × {9!/(4! × 5!)}
= {(6 × 5 × 4)/(3 × 2 × 1)} × {(9 × 8 × 7 × 6)/(4 × 3 × 2 × 1)}
= 20 × 126
= 2520 ways

Question: A photographer who is 1.6 meters tall is standing 20 meters away from a tower. If the angle of elevation from his eye to the top of the tower is 45°, what is the height of the tower?

Solution:

এখানে,
ফটোগ্রাফারের উচ্চতা, CD = 1.6 মিটার
এখানে, CD = EB
টাওয়ারের উচ্চতা, = AB

এখন,
tan∠C = AE/CE
⇒ tan45° = AE/20
⇒ 1 = AE/20
∴ AE = 20

∴ AB = AE + BE
= 20 + 1.6
= 21.6 m

Question: If sec²θ - tan²θ = 1 and tan²θ = 3, then the value of θ when 0° ≤ θ ≤ 90° is?

Solution:
Given,
tan²θ = 3
⇒ tanθ = √3
⇒ tanθ = tan60°

∴ θ = 60°