Bank Math

জোড় স্থানগুলোতে সর্বদা 3 স্থির রেখে প্রথম সংখ্যা থেকে তৃতীয়, পঞ্চম, সপ্তম সংখ্যায় যথাক্রমে
5; 5 × 2 = 10; 10 × 2 = 20 ......... এভাবে বাড়বে ।
তাই, 4 + 5 = 9
9 + (5 × 2)
= 9 + 10
= 19
19 + (10 × 2)
= 19 + 20
= 39 হবে ।
ANswer: 39.

Question: 

Solution: 

Question: The number of numbers from 1 to 200 which are divisible by neither 3 nor 7 is-

Solution: 
The required number = Number of numbers, which are (divisible by 3 + divisible by 7 - divisible by 21)

Now,
Number of number divisible by 3,
 = {(198 - 3)/3} + 1
= 65 + 1 = 66 

Number of number divisible by 7
= {(196 - 7)/7} + 1
= 27 + 1 = 28 

And,
Number of number divisible by 21,
= {(189 - 21)/21} + 1
= 8 + 1 =  9

Thus, the divisible value = 66 + 28 - 9 = 85 

Thus, number of numbers which are not divisible by 3 or 7
= 200 - 85 = 115

Question: If a team of 5 workers can assemble a motorcycle in 6 hours, how many hours would it take a team of 10 workers to assemble the same motorcycle, working at the same constant rate?

Solution:
Given,
5 workers can assemble a motorcycle in 6 hours
∴ 1 worker can assemble a motorcycle in (6 × 5) = 30 hours
∴ 10 workers can assemble a motorcycle in 30/10 = 3 hours
= 3 × 60 minutes
= 180 minutes

Question: Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6 ?

Solution:
Two fair dice are thrown together.
So total possible outcomes = 6 × 6 = 36

And, 
Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.
Then E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5,  3), (6, 2), (6, 6)}
∴ n (E) = 14.

Hence, P(E) = n(E)/n(S) = 14/36
= 7/18

So the probability that the sum is divisible by 4 or 6 is 7/18.

Question: How much coffee, costing Tk. 100 per kg, should be mixed with 20 kg of cocoa priced at Tk. 300 per kg to get a blend worth Tk. 200 per kg?

Solution:
Ratio in which cocoa and coffee should be mixed
= 300 - 200 : 200 - 100
= 100 : 100
= 1 : 1

Let x be the quantity of coffee at 100/kg.

∴ 1 : 1 = x : 20
⇒ x = 20

Given that, six years ago, ratio of the ages of Ashraf and Rahat = 6:5
Hence we can assume that,
age of Ashraf six years ago = 6x
age of Rahat six years ago = 5x
After 4 years, the ratio of their ages = 11:10
⇒ (6x + 10):(5x + 10) = 11:10
⇒ 10(6x + 10) = 11(5x + 10)
⇒ 5x = 10
⇒ x = 10/5
⇒ x = 2
Rahat's present age
=(5x + 6) = 5 × 2 + 6
= 16.

Question: There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?

Solution: 
As the lecturer can't leave the hall by the door he/she enters.

So, number of ways can a student enter the hall through a door and leave the hall by a different door is = 3 × 2 = 6

Let the length of the train is x m. and its speed is v. m/s.

Distance = Speed × time [S = V × T]
x = v × 9 .........(i).
(x+700) = v × 30 ........(ii).

Dividing the eqn. (i) by (ii).

x/(x+700)= 3/10.
⇒ 10x=3x + 2100.
⇒ 7x=2100.
⇒ x= 2100/7.
⇒ x= 300. m.

putting x = 300 in eqn. (1).
300 = v × 9
⇒ v = 300/9
⇒ v = 100/3 m/s.

Let the train crosses a 800 m. long platform in t seconds.
(x + 800) = v × t .........(iii) [ S = V × T]
⇒ (300 + 800) = (100/3) × t. [putting x= 300. and v= 100/3.]
⇒ t = (1100×3)/100
⇒ t = 33 seconds.

Question: The price of an article is raised by 30% and then two successive discounts of 10% each are allowed. Ultimately, the price of the article is:

Solution: 
let, the price be 100 taka 

after 30% raise = 100 + 30 = 130 taka

after first 10% discount = 130 - 13 = 117 taka

after second 10% discount = 117 - 117 × 0.1 
= 117 - 11.7
= 105.3

Question: What is the value of (¹⁰P₁ × ⁵P₃).

Solution:
¹⁰P₁ = 10!/(10 - 1)!
= 10!/9!
= (10 × 9!)/9! 
= 10

⁵P₃ = 5!/(5 - 3)! 
= 5!/2!
= (5 × 4 × 3 × 2!)/2!
= 60

∴ ¹⁰P₁ × ⁵P₃ = 10 × 60 
= 600  

∴ Total distance of shadow of boy and distance from base of lamp post = 2.4 + 4.5 = 6.9 m
Let the height of lamp post be 'h' m

According to question,
⇒ 1.3/2.4 = h/6.9
⇒ h = (6.9 × 1.3)/2.4
⇒ h = 3.7375 = 3.74m

So, The height of the lamp post is 3.74 meters.

Question: Kalam earns Tk. 7.50 per hour on days other than Friday and twice the rate on Friday. Last week he worked a total of 60 hours, including 8 hours on Friday. What is his earnings for the week?

Solution: 
During the week, Kalam worked a total of 60 - 8 = 52 hours at a rate of Tk. 7.50 per hour.
On Friday, he worked 8 hours at a rate of Tk. 7.50 × 2 = Tk. 15.00 per hour.

Therefore, his total earnings for the week were (52 × 7.50 + 8 × 15) = Tk. 510

x:2x:3x
x³+8x³+27x³=4500
36x³=4500
x³= 4500/ 36 =125
x= 5
Smallest number is 5

cost price = 80000 + 5000 + 1000
= 86000
profit = 25%
Selling price = 86000 + 86000 × (1/4)
= Tk. 107500

AC = 40 × 4
= 160 km
BD = 35 × 4
= 140 km
BC = 200 - 160
= 40 km
AD = 200 - 140
= 60 km
∴ CD = 200 - BC - AD
= 200 - 40 - 60
= 100 km
Hence, 100 km apart will the two cars be after four hours of continuous travelling.

Question: If tan 53° = 4/3, then, what is the value of tan8°?

Solution:
Given that,
tan 53° = 4/3

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

Now,
8° = 53° - 45°
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: In the figure below, AB is perpendicular to BC and DB = DC. If AD = √7 cm and AC = 5 cm, what is the value of BC?

Solution:

ΔABD-এ, পিথাগোরাসের উপপাদ্য অনুযায়ী,
BD² + AB² = AD²
⇒ BD² + AB² = (√7)²
⇒ BD² + AB² = 7 ... (1)

আবার, ΔABC-এ,
BC² + AB² = AC²
⇒ (BD + DC)² + AB² = 5²
⇒ (2DC)² + AB² = 25 [যেহেতু BD = DC, তাই BC = BD + DC = 2DC]
⇒ 4DC² + AB² = 25
⇒ 3DC² + (DC² + AB²) = 25 [যেহেতু BD = DC, তাই BD² = DC²]
⇒ 3DC² + 7 = 25 [সমীকরণ (1) থেকে মান বসিয়ে]
⇒ 3DC² = 25 - 7
⇒ 3DC² = 18
⇒ DC² = 6
⇒ DC = √6

অতএব, BC = 2DC (যেহেতু BD = DC)
= 2√6 cm

∴ BC এর মান 2√6 cm

Question: The ages of five people are as follows 36, 48, 52, 40, and 44 years. What should be the age of a sixth person so that the average age of all six becomes 45 years?

Solution:
Let the age of the sixth person be x.
The average of six persons = Total age of 6 persons ÷ 6.

ATQ, 
(36 + 48 + 52 + 40 + 44 + x)/6 = 45
⇒ (220 + x)/6 = 45
⇒ 220 + x = 270
⇒ x = 270 - 220
⇒ x = 50

∴ The age of the sixth person should be 50 years.

Let x /1776 = 111/ x
Then,
⇔x²=111×1776
⇔x²=111×111×16
⇔x= √ {(111)²×(4)²}
⇔x=111×4
⇔x=444

Let A is to be added them 2x² + 3x - 5 + A = x² - x + 1
A = x² - x + 1 - (2x² + 3x - 5)
A = x² - x + 1 - 2x² - 3x + 5
A = -x² - 4x + 6.

sec (x − 30°) = 2/√3
Or, sec (x - 30°) = sec 30°
Or, x - 30° = 30°
Or, x = 60°
∴ tan 60° = √3

Question: If α, β are the roots of the equation x² - 9x + 20 = 0, then αβ equals:

Solution:
x² - 9x + 20 = 0
⇒ x² - 5x - 4x + 20 = 0
⇒ x(x - 5) - 4(x - 5) = 0
⇒ (x - 5)(x - 4) = 0
⇒ x = 5, 4

Hence, α = 5, β = 4

Hence, The value of α × β = 5 × 4 = 20
∴ αβ = 20

Shortcut:
দ্বিঘাত সমীকরণ ax² + bx + c = 0 এর মূলদ্বয় α এবং β হলে,
αβ = c/a [যেখানে, a হলো x² এর সহগ এবং c ধ্রুবক পদ]
∴ αβ = 20/1 = 20

Question: A bus was supposed to travel 240 km at its normal speed. But because of heavy traffic, it had to slow down by 10 km/h and therefore reached the destination 2 hours later than the scheduled time. What was the bus’s original (normal) speed?

Solution:
Let, the original speed = x km/hr
Distance = 240 km

Time taken at original speed = 240 / x
Time taken at reduced speed = 240 / (x - 10)

According to the question:
240 / (x - 10) - 240 / x = 2

LCM: x(x - 10)
Now,
240x - 240(x - 10) = 2x(x - 10)
⇒ 240x - 240x + 2400 = 2x² - 20x
⇒ 2x² - 20x - 2400 = 0
⇒ x² - 10x - 1200 = 0

Solve quadratic: x² - 10x - 1200 = 0

Factors: (x - 40)(x + 30) = 0
x = 40 or -30 (speed cannot be negative)
So the speed is 40 km/hr.

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

According to the question
The cost price of 12 oranges = The sale price of 9 oranges
So profit% = (12 C.P. - 9 C.P.)/9 C.P × 100 = 33.33%

Then it is said that,
5 S.P. - 5 C.P. = 10 M.P. -10 S.P.
From that
we get the relation between M.P. and S.P., that is,
27 S.P. = 24 M.P.(With help of 12 C.P. = 9 S.P.)
Then Discount%
= M.P. - S.P./M.P
= (27 M.P. - 24 M.P.)/27 M.P × 100
= 11.11%
So, % point discount
= 33.33% - 11.11%
= 22.22%

Let the rate of interest be R%
Amount due in 10 months
= 9 + simple interest on Tk. 9 for ten months
= 9 + {9 × R × (10/12)}/100
= 9 + (3R/40)
With the formula mentioned,
1 = 100{9 + 3R/40)}/[(100 × 10) + {R × 10(10 - 1)/(2 × 21)}]
900 + (15R/2) = 1000 + (15R/4)
15R/4 = 100
R = 26.67.
Hence the interest rate is 26.67%.

Sum of the present ages of husband, wife and child
= (27 x 3 + 3 x 3) years
= 90 years
Sum of the present ages of wife and child
= (20 x 2 + 5 x 2) years
= 50 years
∴ Husband's present age
= (90 - 50) years
= 40 years

দেওয়া আছে,
sin2(A + B + C) = 1
⇒ sin 2(A + B + C) = sin 90°
⇒ 2(A + B + C) = 90°
∴ (A + B + C) = 45°............(i)
আবার, tan (B + C - A) = √3
⇒ tan (B + C - A) = tan60°(tan 60° = √3)
B + C - A = 60°............(ii)
এখন, (i) ও (ii) নং সমীকরণ যোগ করে পাই,
(A + B - C) + (B + C - A) = 45° + 60°
⇒2B = 105°
⇒ B = 52(1/2)°

Let us assume that he answered x question correctly. Marks scored by him in x question = 2x
Then, wrong answer would be = 60 – x
Marks lost by him in (60 – x) questions = (60 – x)×1
ATQ,
2x – (60 – x) = 39
Or, 3x = 99
∴ x = 33

Question: If the simple interest on Tk. X at X% per annum for 5 years is Tk. X, what is the value of X?

Solution:
Given,
P = X
r = X% = X/100
n = 5 years
I = X

We know,
I = Prn
⇒ X = X × (X/100) × 5
⇒ X = X × (X/20)
∴ X = 20

Question: A rectangular floor of dimensions 18 m × 12 m is to be covered with a carpet 60 cm wide. Calculate how many metres of carpet are required.

Solution:
Given that,
Floor dimensions = 18 m × 12 m
Carpet width = 60 cm = 0.6 m [1m = 100cm]

Now, Area of floor = length × breadth = 18 × 12 = 216m²

And,
Width of carpet = 0.6m Length of carpet required = L m 
Area covered by L m of carpet = 0.6 × L  m²

This must equal the area of the floor, 0.6 × L = 216
L = 216/0.6 = 360 m

So 360 metres of carpet will be required.

odd + even = odd (4 + 5 = 9)
odd × even = even (4 × 5 = 20)
even × odd = even × even = even (2 × 5 + 4 = 14).

Question:

Solution: 

Let y be the amount of flavour.
(0.2 × 20) + 1 × y = 0.25 (20 + y)
Or, 4 + y = 5 + 0.25y
Or, 0.75y = 1
So, y = 1.33

Question: Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months, and 7 months respectively. What was the ratio of their investment?

Solution:
Let their investments be
x Tk for 14 months,
y Tk for 8 months and
z Tk for 7 months respectively.

Then,
14x : 8y : 7z = 5 : 7 : 8

Now, 
14x/8y = 5/7
⇒ 98x = 40y
⇒ y = 49x/20

And,
14x/7z = 5/8
⇒ 112x = 35z
⇒ z = 16x/5

∴ x : y : z
= x : 49x/20 : 16x/5
= 20 : 49 : 64

Question: A sum of money is distributed equally among 15 persons, but if 5 more persons were included, each person would get Tk. 50 less. What was the total sum?

Solution:
Let the total sum be Tk. x.

When the sum is distributed among 15 persons, each person gets x/15.
If 5 more persons are included, making it 20 persons, each person would get x/20.

According to the question,
(x/15) - (x/20) = 50

⇒ (4x - 3x)/60 = 50
⇒ x/60 = 50
⇒ x = 50 × 60
⇒ x = 3000

∴ The total sum of money is Tk. 3000.