Bank Math

Let,
7 years ago son’s age was x years old and father’s age was 5x year.
Now,
(x+7)/(5x+7) = 1/3,
so x = 7
Son’s age = 7 + 7 = 14
Father’s age= 5 × 7 + 7 = 42
After 7 years,
Son’s age : Father’s age = 21:49 = 3:7

Question: One morning after sunrise, Sakib was standing facing a pole. The shadow of the pole fell exactly to his left. Which direction was Sakib facing?

Solution:
1. In the morning, the sun is in the east, so all shadows fall toward the west.
2. If the shadow (west) is to Sakib's left, he must be facing north.If he were facing south, the shadow would be to his right

Question: If the difference between the squares of two consecutive natural numbers is 21, what is the sum of the squares of these two numbers?

Solution:
Bigger number = (difference of squares + 1)/2 = (21 + 1)/2 = 11
Smaller number = (difference of squares - 1)/2 = (21 - 1)/2 = 10

So, sum of the squares of these numbers = 11² + 10²
= 121 + 100
= 221 

Question: Person 1 to 4 receive equal shares of an income, while Person 5 receives half of what each of Persons 1 to 4 receives. If the total income is 18000 taka, how much does Person 5 get?

Solution: 
Let the amount each of Persons 1 to 4 receives = x taka.
Then Person 5 receives = x/2 taka.

ATQ,
Total income = amount received by all 5 persons
⇒ x + x + x + x + (x/2) = 18000
⇒ 4x + x/2 = 18000
⇒ (8x + x)/2 = 18000
⇒ 9x/2 = 18000
⇒ 9x = 36000
⇒ x = 36000/9
∴ x = 4000

Therefore, Person 5 receives = x/2 = 4000/2 = Tk. 2000

Question: 40 is subtracted from 60% of a number, the result is 50. Find the number.

Solution:
Let the number be x.

According to the question,
60% of x - 40 = 50

⇒ (60/100)x - 40 = 50
⇒ (3/5)x = 90
⇒ x = (5 × 90)/3
⇒ x = 150

∴ The required number is 150. 

Question: The present ages of three persons are in the ratio of 4 : 7 : 9. Six years ago, the sum of their ages was 62. What was the age of the oldest person six years ago? 

Solution: 
Let their present ages are 4x, 7x and 9x years respectively. 
∴ 6 years ago, their present ages are (4x - 6), (7x - 6) and (9x - 6) respectively. 

ATQ,
(4x - 6) + (7x - 6) + (9x - 6) = 62 
⇒ 20x - 18 = 62
⇒ 20x = 62 + 18
⇒ x = 80/20
⇒ x = 4

∴ Their present ages are (4 × 4) =16, (7 × 4) = 28  and (9 × 4) = 36 years respectively.
∴ Age of the oldest person six years ago is (36 - 6) = 30 years. 

Question: The quadratic equation whose one rational root is 4 + √3 is-

Solution:
ধরি, একটি মূল হলো 4 + √3
যেহেতু এটি একটি দ্বিঘাত সমীকরণ এবং একটি মূলে মূলদ আছে, তাই অপর মূলটি হবে এর অনুবন্ধী (conjugate)।
অতএব, অপর মূলটি হলো 4 - √3

মূলদ্বয়ের যোগফল = (4 + √3) + (4 - √3) = 8

মূলদ্বয়ের গুণফল = (4 + √3)(4 - √3)
= 4² - (√3)²
= 16 - 3
= 13

আমরা জানি,
মূলদ্বয় α এবং β হলে দ্বিঘাত সমীকরণটি হয়:
x² - (α + β)x + α × β = 0

সুতরাং, নির্ণেয় দ্বিঘাত সমীকরণটি হলো,
x² - 8x + 13 = 0

Relative speed= 24-18= 6 km/hr
Time required by faster train to overtake slower train
= 27/6 hr
= 4(1/2) hr
∴ Distance between Q and R:
= 18×4(1/2)
= 81 km

Total weight including teacher = 18 × 90.2 = 1623.6 kg
Total weight of 17 students = 17 × 90 = 1530 kg
So, weight of the teacher = 1623.6 – 1530 = 93.6 kg

Question: There are 45 students in a certain class 2/3 of the students are girls and 1/2 of the girls are blue-eyed. How many blue- eyed girls are there in the class?

Solution:
মোট ছাত্র-ছাত্রীর সংখ্যা = 45 জন
বালিকার সংখ্যা = (45 × 2)/3 জন
= 30 জন

blue- eyed বালিকার সংখ্যা = (30 × 1)/2 জন
= 15 জন

Question: P, Q, and R started a business by investing Tk. 120,000, Tk. 180,000 and Tk. 200,000 respectively. Find the share of Q out of an annual profit of Tk. 125,000.

Solution:
The ratio of investments of P, Q, and R is:
P : Q : R = 120,000 : 180,000 : 200,000
= 12 : 18 : 20
= 6 : 9 : 10

Sum of the ratios = 6 + 9 + 10 = 25
Total annual profit = Tk. 125,000

Q's share of profit = (9/25) × 125,000
= 9 × 5,000
= 45,000 Tk.

Let, the length of the rectangle is l and breadth is b.
Given that the breadth of the rectangular field is 60% of its length.
b = 60l/100
=3l/5
Perimeter of the field 800 m
⇒ 2(l + b) = 800
⇒2{l + (3l/5)} = 800
⇒ l + (3l/5) = 400
⇒ 8l/5 = 400
⇒ l = 250 m.
b = 3l/5
= (3 × 250)/5
= 150 m
Area = lb
= (250 × 150)
= 37500 m².

Question: Which of the following describes all values of x for which 1 - x² ≥ 0?

Solution:
1 - x² ≥ 0
⇒ - x² ≥ - 1
⇒ x² ≤ 1
⇒ x² ≤ 1²
∴ - 1 ≤ x ≤ 1

Question: The marked price of a t-shirt was Tk. 800. A man bought the same for Tk. 420 after getting two successive discounts. the first being 25%. then the second discount rate is-

Solution:
Marked price = 800
Actual price = 420
First discount = 25%
Let the second discount be x%

Then, we can write

after 25% discount,
discounted amount = 800 × (25/100)
= 200

New price = 800 - 200
= 600 Tk

Again, second discount,
discounted amount = 600 × (x/100)
= 6x

New price = 600 - 6x

ATQ,
600 - 6x = 420
⇒ 600 - 420 = 6x
⇒ 6x = 180
⇒ x = 30

∴ Second discount = 30%

n < 0 ⇒ 2n < 0,
again, - n > 0 and n² = (-n)2 > 0.
Thus, out of the numbers 0, -n, 2n and n²
We find that 2n is the least number here. 
Answer: 2n

By investing tk. 1552, income = tk. 128.
By investing tk. 97, income = tk.128/1552 x 97= tk. 8.
Dividend = 8%

নাম্বার বেশি পায় = 83 - 63 = 20
প্রশ্নমতে,
0.5 অংশ বৃদ্ধি = 20
∴ 1 অংশ বৃদ্ধি = 20/0.5 = 40
∴ ছাত্র-ছাত্রী সংখ্যা = 40

Question: Find the difference between the compound profit and the simple profit on Tk. 12,000 for 2 years at 25% per annum.

Solution:
Given that,
P = 12000 Tk
r = 25%
= 25/100
= 1/4

t = 2 years

We know,
The compound profit = P (1 + r)ⁿ - P
= 12000 (1 + 1/4)²- 12000
= (12000 × 25/16)- 12000
= 18750 - 12000
= 6750 Tk

and
Simple profit I = Prn
= 12000 × 1/4 × 2
= 6000

∴ The different between compound profit and simple profit = (6750 - 6000) Tk
= Tk 750

Question: A shirt is sold for Tk. 1500 at a profit of 20%. What would have been the actual profit or loss percentage if it had been sold for Tk. 1200?

সমাধান:
ধরি, শার্টের ক্রয়মূল্য = x টাকা

20% লাভে বিক্রয়মূল্য = x + x এর 20%
= x + (x × 20/100)
= x + x/5 = 6x/5

প্রশ্নমতে,
6x/5 = 1500
⇒ 6x = 1500 × 5
⇒ 6x = 7500
⇒ x = 7500/6
⇒ x = 1250

∴ শার্টের ক্রয়মূল্য = Tk. 1250

বিক্রয়মূল্য = Tk. 1200 ক্রয়মূল্য = Tk. 1250

যেহেতু বিক্রয়মূল্য < ক্রয়মূল্য, তাই ক্ষতি হবে।

ক্ষতি = ক্রয়মূল্য - বিক্রয়মূল্য
= 1250 - 1200 = 50 টাকা

শতকরা ক্ষতি = (ক্ষতি/ক্রয়মূল্য) × 100%
= (50/1250) × 100%
= 5000/1250
= 4%

∴ 4% ক্ষতি হবে

বাগানের পরিসীমা = 5300/26.5
= 200 মিটার
এখন ধরি, বাগানের প্রস্থ x মিটার
∴ বাগানের দৈর্ঘ্য (x + 20) মিটার
প্রশ্নমতে, 2(x + x + 20) = 200
⇒ 2(2x + 20) = 200
⇒ 2x + 20 = 100
⇒ 2x = 100 - 20
⇒ 2x = 80
⇒ x = 40
∴ বাগানের দৈর্ঘ্য = 40 + 20 = 60 মিটার

Question: A cistern is filled by Pipe A and Pipe B together in 2 hours. Pipe A alone can fill the cistern at the rate of 100 litres per hour. Pipe B alone can fill the cistern in 4 hours. What is the capacity of the cistern?

Solution:
Let the capacity of the cistern = x litres.

Pipe A fills at 100 litres per hour
∴ Time taken by A alone = x/100 hours
Pipe B alone fills the cistern in 4 hours
∴ Pipe B's rate = x/4 litres per hour
And Combined rate of A + B = 100 + (x/4) litres per hour

They together fill the cistern in 2 hours, so:
Combined rate = x/2 litres per hour

Therefore, 100 + (x/4) = x/2
⇒ (x/2) - (x/4) = 100
⇒ (2x - x)/4 = 100
⇒ x = 4 × 100
∴ x = 400

So the capacity of the cistern is 400 litres.

Question: What word can be formed by arranging the letters of 'AEEHLPTN'?

Solution:
By rearranging the letters 'AEEHLPNT', the name of an animal (ELEPHANT ⇒ Elephant) can be formed.

Man's/Boat's Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X - Y

Downstream Speed = Distance covered/Time taken
= 20/1
= 20 km/hr
Upstream Speed = 20/2 = 10 km/hr

X + Y = 20 km/hr
and, X - Y = 10 km/hr

Adding them we get,
X + Y + X - Y = 30 km/hr
∴ X=15 km/hr = Speed of swimmer in still water

∴ Y = 20 - 15 = 5 km/hr = Speed of river.

Given speed = 63 km/hr = 63 × 5/18 = 35/2 m/s
Let the length of the bridge = x mts
Given time taken to cover the distance of (170 + x) mts is 30 sec.
We know speed = distance / time 
⇒ 35/2 = (170+x)/30
⇒ 340 + 2x = 1050
⇒ 2x = 710
∴ x = 355

Question: If 5 students run a mile in 5 minutes, how much time will 50 students take to run a mile?

Solution:
5 students run a mile in 5 minutes
1 students run a mile in 5 minutes
50 students run a mile in 5 minutes

Question: If A can do 1/4 of a work in 3 days and B can do 1/9 of the same work in 4 days, how much will A get if both work together and paid Tk 800 in all?

Solution:
Whole work is done by A in (3 × 4) = 12 days
∴ A's 1 day's work = 1/12 part
Whole work is done by B in (4 × 9) = 36 days
∴ B's 1 day's work = 1/36 part

A's 1 day's work : B's 1 day's work
= A's wages : B's wages
= 1/12 : 1/36
= 3 : 1

∴ A's share = (800 × 3/4) Tk
= 600 Tk

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

We want a grey card OR ENGLISH card
There are 30 - 17 = 13 grey cards
There are 4 + 5 = 9 ENGLISH cards
Total cards = 30
Also, 5 grey cards are ENGLISH cards.

So Probability = 13/30 + 9/30 - 5/30 = (13 + 9 - 5)/30
= 17/30 [This subtraction is needed a grey card gets counted twice - once in 13 grey cards and once again in 9 ENGLISH cards.]

Question: The difference between simple interest for 4 years and 6 years at 5.5% per annum is BDT 220. Find the principal amount.

Solution:
Given that,
Rate of interest, r = 5.5%
Difference in simple interest for 6 years and 4 years = 220
Time difference, n = 6 - 4 = 2 years

We know that,
I = (P × r × n)/100
⇒ 220 = (P × 5.5 × 2)/100
⇒ 11 × P = 22000
⇒ P = 22000/11
∴ P = 2000
∴ The sum is Tk. 2000.

Question: What is the least number which when doubled is exactly divisible by 8, 14, 18, and 24?

Solution:
Let the number be x.
Double the number is 2x.

8 = 2 × 2 × 2
14 = 2 × 7
18 = 2 × 3 × 3
24 = 2 × 2 × 2 × 3

∴ LCM = 2 × 2 × 2 × 3 × 3 ×7
= 504

∴ x = 504/2 = 252

Question: The value of 1 + {(tan 30° - tan 45°)/(cot 45° - cot 60°)} is -

Solution:
1 + (tan 30° - tan 45°)/(cot 45° - cot 60°)
= 1 + (tan 30° - tan 45°)/{cot (90° - 45°) - cot (90° - 60°)}
= 1 + (tan 30° - tan 45°)/(tan 45° - tan 30°)
= 1 + (tan 30° - tan 45°)/(-1)(tan 30° - tan 45°)
= 1 - 1
= 0