Bank Math

Multiple of 7 between 11 and 200 are 14, 21, 28, 35, 42,....., 189, 196
T_m = 196 ⇒ 14 + (m - 1) × 7 = 196
⇒ (m - 1) × 7 = 182
⇒ (m - 1) = 26
⇒ m = 27.
Multiples of 7 and 3 both i.e. that of 21 are 21, 42, 63,......., 189
T_n = 189
⇒ 21 + (n - 1) × 21 = 189
⇒ (n - 1) × 21 = 168
⇒ (n - 1) = 8
⇒ n = 9.
Required number of terms = (27 - 9) = 18
Answer : 18

Question: The area of the largest triangle that can be inscribed in a semi-circle of radius r, is-

Solution:
 Largest triangle inscribed in a semicircle has,
Base = diameter of semicircle = 2r
Height = radius = r (vertex at the top of the semicircle)

We know,
Area of triangle = (1/2) × base × height
= (1/2) × 2r × r
= r²

Question: Find the value of .

Solution:

Question: If tan(θ + 30°) = √3, what is the value of sinθ?

Solution:
Given that,
tan(θ + 30°) = √3
⇒ tan(θ + 30°) = tan 60°
⇒ θ + 30° = 60°
⇒ θ = 60° - 30°
⇒ θ = 30°

Now,
sinθ 
= sin 30°
= 1/2

Let K be Kalam's age and D be his daughter's age
Five years ago, K-5 = 2(D-5)
K - 2D = -5 ....(1)
and their ages are in the ratio 11:6.
i.e., K/D = 11/6
6K = 11D
⇒ K = 11D/6 ....(2)
Sub. K = 11D/6 in (2)
⇒ (11D/6) - 2D = -5
⇒ 11D - 12D = -30
⇒ -D = -30
⇒ D=30
Thus, the age of his daughter is 30 years.
Now, K = 11 x 30/6 = 55 years
Hence the age of Kalam is 55 years.

Question: P and Q are two positive integers such that PQ = 60. Which of the following cannot be the value of P + Q?

Solution:
60-এর উৎপাদক জোড়াগুলো হল:
1 × 60 = 60 ⇒ P + Q = 1 + 60 = 61
2 × 30 = 60 ⇒ P + Q = 2 + 30 = 32
3 × 20 = 60 ⇒ P + Q = 3 + 20 = 23
4 × 15 = 60 ⇒ P + Q = 4 + 15 = 19
5 × 12 = 60 ⇒ P + Q = 5 + 12 = 17
6 × 10 = 60 ⇒ P + Q = 6 + 10 = 16

সুতরাং, P + Q এর সম্ভাব্য মানগুলো: 16, 17, 19, 23, 32, 61.

তাই, P + Q = 20 হতে পারে না।

Given ratio = 5 : 4
Let the numbers are 5x and 4x
ATQ, 5x - 4x = 10
So, x = 10
∴ Smallest number = 4×10 = 40 

Water in the mixture = 40×(10/100) = 4 litres
And, Milk in the mixture = 40 − 4 = 36 litres
Let x litres of water is mixed
⇒ (4 + x)/(40 + x) = 20/100
⇒ x = 5 litres

A : B = 100 : 75
B : C = 100 : 96

∴ A : C = A/B × B/C
= (100/75) × (100/96)
= 100/72
100 : 72

∴ A beats C by
100 - 72 = 28 m.

Here S = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH}.
Let,
E = event of getting at least two heads
= {THH, HTH, HHT, HHH}.
∴ P(E) = n(E)/n(S)
= 4/8
= 1/2.

Question: A and B invest in a business in the ratio 7 : 3. If 20% of the total profit goes to charity and B's share is TK 12000, the total profit is-

Solution: 
মনে করি,
মোট লাভ 100 টাকা
20% দান করার পর থাকে = (100 - 100 এর 20%) = 80 টাকা

80 টাকার মধ্যে B পাবে = 80 × (3/10) = 24 টাকা

24 টাকা পায় যখন মোট লাভ 100 টাকা
∴ 12000 টাকা পায় যখন মোট লাভ = (12000 × 100)/24 = 50,000 টাকা 

Question: If dividing P(x) = 4x³- 7x² + bx - 5 by (x - 3) results the remainder 10, then find the value of b.

Solution:
According to the Remainder Theorem, if a polynomial P(x) is divided by (x - c), then the remainder = P(c).
Here divisor is (x - 3).
So remainder = P(3).

Now,
P(3) = 4(3)³ - 7(3)² + b(3) - 5

= 4 × 27 - 7 × 9 + 3b - 5

= 108 - 63 + 3b - 5

= 40 + 3b

According to the question, the remainder is 10.
So, 40 + 3b = 10

⇒ 3b = 10 - 40
⇒ 3b = - 30
⇒ b = - 10

Let the speed of the trains be 7 x and 8 x respectively.
Speed of second train = 400 /4 = 100 km/hr
⇒ 8 x = 100
⇒ x = 100/8 = 12.5
Speed of the first train = 7x = 7 × 12.5
= 87.5 km/hr

3/4 = 0.75
5/6 = 0.83
2/3 = 0.667
4/5 = 0.80
9/10 = 0.9
1/2 = 0.50

So, the fraction 4/5 is greater than 3/4 and less than 5/6

Sum of 6 numbers = (6 × 25) = 150.
Sum of 3 additional numbers = (3 × 22) = 66.
Sum of (6 + 3) =9 numbers = (150 + 66) = 216
∴ average of the combined 9 numbers = 216/9 = 24

Let,
S be the sample space and
E be the event of select.
Then,
n(S) = number of ways of selecting 3 students out of 25
= ²⁵C₃
= (25 × 24 × 23)/(3 × 2 × 1)
= 2300.
And,
n(E) = (¹⁵C₂ × ¹⁰C₁)
= {(15 × 14)/(2 × 1)} × 10
= 1050.
∴ P(E) = n(E)/n(S)
= 1050/2300
= 21/46.

Number of games played = 40
Number of won games = 24
Percentage of games played = (24/40)× 100
= 60%

Question: The speed of three cars is in the ratio of 3 : 4 : 5. The ratio of the times taken by these cars to travel the same distance is-

Solution:
যেহেতু দূরত্ব একই থাকে, তাই গতিবেগ সময়ের সাথে ব্যস্তানুপাতিক হয়।
⇒ গতিবেগ ∝ (1/সময়)
⇒ s ∝ (1/t)

∴ সময়ের অনুপাত = 1/3 : 1/4 : 1/5
= (1/3 × 60) : (1/4 × 60) : (1/5 × 60)
= 20 : 15 : 12

Question: Which number will complete the series 4, 25, 151, —, 5443.

Solution:
Given,
4
(4 × 6 + 1) = 25
(25 × 6 + 1) = 151
(151 × 6 + 1) = 907
(907 × 6 + 1) = 5443

Question: The average age of a group of 20 employees is 26 years. When 4 more employees join the group, the average age increases by 4 years. What is the average age of the new employees?

Solution:
Average age of 20 employees = 26 years
∴ Total age of 20 employees = 20 × 26 = 520 years

After 4 more employees join:
Total number of employees = 20 + 4 = 24 employees
New average age = 26 + 4 = 30 years
Total age of 24 employees = 24 × 30 = 720 years

∴ Total age of 4 new employees = 720 - 520 = 200 years

∴ Average age of 4 new employees = 200 ÷ 4 = 50 years

Question: Rahim's regular pay is Tk 40 per hour up to 40 hours. Overtime is paid at twice the regular rate. If he was paid Tk 2400 in total, how many hours of overtime did he work?

Solution:
Rahim’s regular wage for 40 hours = (40 × 40) = 1600 Taka.
Amount earned from overtime = (2400 - 1600) Taka = 800 Taka.
Since the overtime rate is twice the regular hourly wage,
Total overtime hours worked = 800 ÷ (40 × 2) hours
= 800 ÷ 80
= 10 hours

Question: A hall is 30m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to of the areas of four walls, the volume of the hall is-

Solution:
Let,
The height of the hall = h

Given that, 
Length of the room = 30 m
Breadth of the room = 10 m

According to the question,
2 × (30 × 10) = 2 × (30 + 10) × h
⇒ 2 × 40 × h = 2 × (30 × 10)
⇒ 40h = 300
⇒ h = 300/40
⇒ h = 30/4
∴ h = 15/2

We know,
 Volume = 30 × 10 × (15/2)
= 2250 m³

Question: In how many different ways can the letters of the word 'SOFTWARE' be arranged in such a way that the vowels always come together?

Solution:
The given word contains 8 different letters.
We keep the vowels (OAE) together and treat them as 1 letter.
Thus, we have to arrange the 6 letters SFTWR(OAE)
These can be arranged in 6! = 6 5 × 4 × 3 × 2 × 1 = 720 ways

And,
The vowels (OAE) can be arranged among themselves in 3! = 3 × 2 × 1 = 6 ways.

∴ Required number of ways = (720 × 6) = 4320

Bloat এরঃ
B হলো 2^nd Alphabet.
L হলো 12^th Alphabet.
O হলো 15^th Alphabet.
A হলো 1^st Alphabet.
T হলো 20^th Alphabet.
Flame এরঃ
F হলো 6^th Alphabet.
L হলো 12^th Alphabet.
A হলো 1^st Alphabet.
M হলো 13^th Alphabet.
E হলো 5^th Alphabet.

Question: How many years will it take for an investment of Tk. 10000 to earn Tk. 3000 in simple interest at a rate of 6%?

Solution:
Given that,
Principal, P = 10000
Simple Interest, I = 3000
Rate of interest, r = 6%
Time, n = ?

We know,
I = Pnr/100
⇒ 3000 = (10000 × n × 6)/100
⇒ 3000 = 100 × n × 6
⇒ 3000 = 600n
⇒ n = 3000/600
⇒ n = 5

So, it will take 5 years for the investment to earn Tk. 3000 at 6% simple interest.

A : B : C = 7 : 8 : 11.
Hire charges paid by B = Tk. (520 × 8/26)
= Tk. 160.

Question: An investment becomes Tk. 6,720 in 2 years and Tk. 7,392 in 3 years at compound interest. Find the rate of interest per annum.

Solution:
Let the principal = P, rate = R%.

Compound Interest:
A = P[1 + 100/R​]^T

Amount after 2 years = 6,720
6,720 = P[1 + 100/R​]²

Amount after 3 years = 7,392
7,392 = P[1 + 100/R​]³

Divide the 3rd year amount by 2nd year amount:
7,392/6,720 = P(1 + R/100)³/ P(1 + R/100)² 
⇒ 1.1 = 1 + R/100 [7,392/6,720 = 1.1]
⇒ 1.1 - 1 = R/100
⇒ 0.1 = R/100
⇒ R = 0.1 × 100
⇒ R = 10%

∴ R = 10%

Product of 1^st and 4^th terms (extremes) = product of 2^nd and 3^rd terms (means)
⇒ 2.5x = 40
⇒ x = 40/2.5 = 16

We know,
Total = n(A) + n(B) - both + none
⇒ 36 = 20 + 28 - both + 0
⇒ 36 = 48 - both
⇒ both = 48 - 36
⇒ both = 12

Question: In a certain code language, 'APPLE' is coded as '50', and 'BANANA' is coded as '33'. How will 'ORANGE' be coded in that code language?

Solution:

এখানে প্রতিটি অক্ষরের Alphabetical Position (A = 1, B = 2, …, Z = 26) যোগ করা হয়েছে।

APPLE = A(1) + P(16) + P(16) + L(12) + E(5)
= 1 + 16 + 16 + 12 + 5
= 50 

BANANA = B(2) + A(1) + N(14) + A(1) + N(14) + A(1)
= 2 + 1 + 14 + 1 + 14 + 1
= 33 

অতএব,
ORANGE = O(15) + R(18) + A(1) + N(14) + G(7) + E(5)
= 15 + 18 + 1 + 14 + 7 + 5
= 60

Question: The average weight of 12 students is 52 kg. If a new student joins, the average becomes 53 kg. What is the weight of the new student?

Solution:
Average weight of 12 students = 52 
Total weight = 12 × 52
= 624 kg.

Let the weight of the new student = x kg.
New total number of students = 13

New average = 53 
Total weight = 13 × 53
= 689 kg.

Weight of new student = new total - original total
= 689 - 624
= 65 kg

∴ The weight of the new student is 65 kg.

Let daughter’s age be = A
and mother’s age be = B

ATQ, 2A + B = 70 ..... (i)
and, A + 2B = 95 ..... (ii)
(ii)×2 - (i) ⇒ 3B = 120
⇒ B = 40