Bank Math

Question: A is twice as good a workman as B and therefore can finish a job 30 days earlier than B. Working together, in how many days can they finish the job?

Solution:

Given,
A is twice as good workman as B. Means,
A = 2B
Let B can finish work in X days, then A will finish same work in (X - 30) days alone
Now,
BX = 2B × (X - 30)
⇒ BX = 2BX - 60B
∴ X = 60 days
B can finish work in 60 days, then A can finish the work in 30 days.
One day work of B = 1/60

One day work of A = 1/30

One day work of (A+B) = (1/60) + (1/30) ⇒ (1+2)/60 ⇒ 3/60 ⇒ 1/20

So, they can finish work together in 20 days.

Question: A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is-

Solution: 
Let the length of the train is x m and speed is s.
ATQ,
s = (x + 800)/100 and,
s = (x + 400)/60

∴ (x + 800)/100 = (x + 400)/60
or, 60x + 48000 = 100x + 40000
or, 40x = 8000
or, x = 200 m

Question: A group of women decided to do a job in 5 days. But since 15 women dropped out every day, the job completed at the end of the 8th day. How many women were there at the beginning?

Solution:
Let X be the initial number of women then,
According to the question,
5X = X + (X - 15) + (X - 30) + (X - 45) + (X - 60) + (X - 75) + (X - 90) + (X - 105)
⇒ 5X = 8X - 420
⇒ 8X - 5X = 420
⇒ 3X = 420
⇒ X = 420/3
∴ X = 140 women

On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
Answer : 3

Greatest number of five digits = 99999.
Required number must be divisible by L.C.M. of 15, 21 and 36, i.e 1260
On dividing 99999 by 1260, we get 459 as a reminder.
∴ Required number = (99999 - 459) = 99540
Answer : 99540

Question: Rina and Shila started a business by investing Tk. 85,000 and Tk. 15,000 respectively. If the business runs for 2 years, what will be the ratio of profit earned by Rina and Shila?
 

Solution:
Given,
Rina's investment = Tk. 85,000
Shila's investment = Tk. 15,000
Time = 2 years for both

The ratio of profit earned after 2 years between Rina and Shila respectively = (85000 × 2) : (15000 × 2)
= 170000 : 30000
= 17 : 3

Question: Which trigonometric ratio is undefined in value?

Solution:
sin 90° = 1 
cos 90° = 0
cosec 90° = 1
cot 0° = ∞ (Undefined)

Question: A boat moves at a speed of 30 km/h in still water. The speed of the current is 5 km/h. How far will the boat travel downstream in 12 minutes?

Solution:
Speed of the boat in still water = 30 km/h
Speed of the current = 5 km/h

∴ Speed downstream = 30 + 5 = 35 km/h

Time = 12 minutes = 12/60 hours = 1/5 hours

∴ Distance travelled downstream = 35 × (1/5) km
= 7 km

∴ The distance travelled downstream is 7 km.

ধরি,
19 টি জিনিসের ক্রয়মূল্য x টাকা
∴ 29 টি জিনিসের বিক্রয়মূল্য x টাকা 
∴ ক্ষতি = (x/19) - (x/29)
= (29x - 19x)/551 = 10x/551 টাকা
∴ শতকরা ক্ষতির পরিমাণ = (10x/551)/(x/19) × 100%
= (10x/551) × (19/x) × 100%
= 34.48%

Question: A certain principal amount, invested at simple interest, grows to Tk. 950 after 3 years and Tk. 1010 after 4 years. What is the original principal amount?

Solution:
Given,
Amount after 3 years = Tk. 950
Amount after 4 years = Tk. 1010
∴ Interest for 1 year = 1010 - 950 = Tk. 60

∴ Interest for 3 years = 60 × 3 = Tk. 180

∴ Principal = 950 - 180 = Tk. 770

Cyclist:Jogger
Ratio of distance→ 2:1
Ratio of time→ 1:2
Ratio of their speed (Jogger:Cyclist)
= (1/2):(2/1)
= 1:4

Question: If the least common multiple of two numbers is twelve times their highest common factor, and their sum (HCF + LCM) equals 403, then what is the other number when one number is 93?

Solution:
Let HCF be h and LCM be l
Then l = 12h and
l + h = 403

∴12h + h = 403
⇒ h = 31

So, l = (403 − 31) = 372

Hence, the other number = (31 × 372)/93 = 124

Question: A bag contains red, blue, and green balls in the ratio 2 : 3 : 5. If there are 15 blue balls, how many total balls are there in the bag?

Solution: Given ratio = Red : Blue : Green = 2 : 3 : 5
Let the numbers be:
Red = 2x, Blue = 3x, Green = 5x

Given,
Blue balls, 3x = 15
∴ x = 5

Now calculate total balls:
= 2x + 3x + 5x
= (2 + 3 + 5)x 
= 10x
= 10 × 5 
= 50

Question: The total of three successive multiples of 3 is 117. Determine the greatest number.

Solution:
Let,
First multiple: 3x
Second multiple: 3(x + 1) = 3x + 3
Third multiple: 3(x + 2) = 3x + 6

ATQ,
3x + (3x + 3) + (3x + 6) = 117
⇒ 9x + 9 = 117
⇒ 9x = 108
⇒ x = 108/9
∴ x = 12

∴ The largest number = 3x + 6
= 3 × 12 + 6 
= 36 + 6
= 42

Question: A sum increases by 80% in 8 years at simple interest. If Tk 2500 is invested at the same rate for 2 years under compound interest, what is the final amount?
 
Solution:
let
the principal = Tk. 100
Simple Interest = Tk 80 (since 80% increase in 8 years)

We know,
Simple interest = Pnr/100
⇒ 80 = (100 × r × 8)/100
⇒ 80 = 8r
∴ r = 10%

The compound interest = 2500 {1 + (10/100)}² - 2500
= 2500 × {1 + (1/10)}² - 2500
= 2500 × (11/10)² - 2500
= (2500 × 121)/100 - 2500
= 3025 - 2500
= Tk. 525

So the compound interest on Tk 2500 after 2 years is Tk 525.

Question: A tradesman fixed his selling price of goods at 25% above the cost price. He sells half the stock at this price, one-quarter of his stock at a discount of 20% on the original selling price, and the rest at a discount of 40% on the original selling price. Find the gain percentage altogether?

Solution:
Let C.P = 100; 
then S.P = 100 + 25 = 125 Tk

Now, revenue
= [(1/2) × 125] + [(1/4) × (80/100) × 125] + [(1/4) × (60/100) × 125]
= 62.5 + 25 + 18.75
= 106.25

∴ Gain = 106.25 - 100
= 6.25 Tk

Gain percentage = (6.25 × 100)/100
= 6.25% 

Question: Three bells ring at intervals of 9, 12, and 15 minutes respectively. If they all ring together at 3:00 PM, when will they ring together again?

Solution:
তিনটি ঘণ্টা বিকাল ৩ টায় একত্রে বাজলে 9, 12, 15 এর ল.সা.গুর সমান সময়ের পর ঘণ্টাগুলো পুনরায় একত্রে বাজবে।

সংখ্যা গুলোর মৌলিক উৎপাদক:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5

9, 12, 15 এর ল.সা.গু. = 2² × 3² × 5
= 4 × 9 × 5
= 180 অর্থাৎ 180 মিনিট

সুতরাং, ঘণ্টা গুলো একবার বিকাল 3 টায় বাজার পর পুনরায় বাজবে = 3 টা + 180 মিনিটে
= 3 টা + (60 + 60 + 60) মিনিটে
= 3 টা + 3 ঘণ্টা
= 6 টা
∴ তারা আবার 6:00 PM-এ একত্রে বাজবে।

Question: A team of workers can finish a project in 20 days. However, when 5 of them were absent, it took 25 days to complete the same work. How many workers were originally assigned to the project?

Solution:
Let,
the total number of people working originally = x
When 5 people were absent,
Total present workers = x - 5

x workers can complete the work in 20 days
∴ 1 worker can complete it in 20x days
∴ (x - 5) workers can complete it in 20x/(x - 5) days

ATQ,
20x/(x - 5) = 25
⇒ 4x/(x - 5) = 5
⇒ 5x - 25 = 4x
∴ x = 25

∴ The total number of people working originally = 25

HCF = 17
Let numbers are = 17x, 17y
LCM = 17xy = 714 (given)
xy = 42
Possible pairs are (1, 42), (2, 21), (3, 14), (6, 7)
Possible numbers are (17, 714), (34, 357), (51, 238), (102, 119)
but given that both numbers are of three digits
∴ numbers are = (102, 119)
∴ sum of numbers = 102 + 119 = 221

Suppose pipe A alone takes x hours to fill the tank in 144 min.
Then,
pipes B and C will take x/2 and x/4 hours respectively to fill the tank.
∴ 1/x + 2/x + 4/x = 1/5
⇒ 7/x = 1/5
⇒ x = 35.

Question: If (x/y) + (y/x) = √7 then what is the value of [(x⁴/y⁴) + (y⁴/x⁴)]³ ?
(Janata RC 22 এর অনুরূপ)

Solution:
দেওয়া আছে,
(x/y) + (y/x) = √7

∴ x⁴/y⁴ + y⁴/x⁴
= (x/y)⁴ + (y/x)⁴
= {(x/y)²}² + {(y/x)²}²
= {(x/y)² + (y/x)²}² - 2.(x²/y²).(y²/x²)
= {(x/y)² + (y/x)²}² - 2
= [{(x/y) + (y/x)}² - 2.(x/y).(y/x)]² - 2
= {(√7)² - 2}² - 2
= (7 - 2)² - 2
= 5² - 2
= 25 - 2
= 23

[(x⁴/y⁴) + (y⁴/x⁴)]³ = (23)³ = 12,167

Question: Find the volume of a cylindrical-shaped water container that has a height of 15 cm and a diameter of 14 cm.

Solution:
দেওয়া আছে,
 উচ্চতা (h) = 15 cm এবং ব্যাস (d) = 14 cm

আমরা জানি, ব্যাস = 2 × ব্যাসার্ধ
⇒ 2 × ব্যাসার্ধ = 14 cm
⇒ ব্যাসার্ধ (r) = 14/2 = 7 cm

আমরা জানি,
সিলিন্ডারের আয়তন, V = πr²h ঘন একক
= 22/7 × (7)² × 15
= 22/7 × 49 × 15
= 22 × 7 × 15
= 2310 ঘন সেমি।

সুতরাং, সিলিন্ডার আকৃতির পানির পাত্রটির আয়তন হল 2310 cm³।

Question: The value ofis:
(Officer General 22 এর অনুরূপ) 

Solution:
1 - cosx
≈ x²/2 for small x

Again,
1 - cosx/x²

= (x²/2)/x²

= 1/2

In 3 L solution sugar is 60%
So, Sugar Amount = 3 × 3/5 = 1.8 L
Now, 1 L water is added.
So, Total solution = (3 + 1)
= 4 L
% sugar = (1.8 × 100)/4
= 180/4
= 45% [Answer.]

Question: A sum of Tk. 1750 is divided into two parts such that the interest on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. The interest on each part is-

Solution: 
Principal = Tk. 1750
Let the first part = x
Hence, the second part = 1750 - x

x × (8/100) × 1 = (1750 − x) × (6/100) × 1
⇒ 4x = 5250 − 3x
⇒ 7x = 5250
⇒ x = 750

First part = Tk. 750
∴ Second part = Tk. (1750 − 750)
= Tk. 1000

∴ Required interest = 750 × (8/100) 
= Tk. 60

Given,
S.I = 3 Principal Amount
=> 3A = A x 16 x R/100
By solving, we get
=> R = 18.75%

Question: Apu was assigned to do 2 similar tasks. He completes the second task in two-thirds the time it takes him to complete the first task. If it took him an hour to complete both the tasks, in how many minutes did he complete the second task?

Solution:
Let the time taken for the first task be t minutes.
Then, time for the second task = 2t/3 minutes.

Total time for both tasks = 1 hour = 60 minutes.

ATQ, 
t + (2t/3) = 60 
⇒ (3t + 2t)/3 = 60
⇒ 5t = 60 × 3
⇒ t = (60 × 3)/5
∴ t = 36

∴ Second task = 2t/3 = (2 × 36)/3 = 24 minutes

So Apu completed the second task in 24 minutes.

Question: Two pipes X and Y can fill a tank in 30 minutes and 45 minutes, respectively. Both pipes are opened together. After how many minutes should pipe Y be turned off so that the tank is filled in 20 minutes?

Solution:
Pipe X can fill 1 / 30 part of tank in one minute.
Pipe Y can fill 1 / 45 part of tank in one minute.
Both pipes can fill (1/30 + 1/45) part of tank in one minute.
= (3+2) / 90 = 5/90 = 1/ 18

Let, after time t minutes, we turned off the pipe Y.
according to question, [দুটি পাইপ t সময় পর্যন্ত একসাথে চলতে থাকে, বাকি সময় X পাইপটি চলে এবং সম্পূর্ণ বা 1 অংশ ট্যাংক পূর্ণ করে ]
∴ t/18 + (20-t) / 30 = 1 
⇒ (5t + 60- 3t) / 90 = 1
⇒ 5t + 60 - 3t = 90
⇒ 2t = 30
⇒ t = 15

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36
Let E = event of getting a total of 10 or 11
= [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)]
∴P(E) = n(E)/n(S) = 5/36

Question: If Taka 200 is invested at 10% interest for 3 years, what is the difference between the simple and compound interest?

Solution: 
এখানে,
মূলধন, P = 200 টাকা
সুদের হার, r = 10% = 10/100 = 1/10
সময়, n = 3 বছর

আমরা জানি,
সরল মুনাফার ক্ষেত্রে,
SI = P × r × n
= 200 × (1/10) × 3
= 60 টাকা

আবার,
চক্রবৃদ্ধি মুনাফায়,
C = P(1 + r)ⁿ
= 200 × [1 + (1/10)]³
= 200 × (11/10)³
= 200 × (11/10) × (11/10) × (11/10)
= 200 × (1331/1000)
= 266200/1000
= 266.2 টাকা

∴ চক্রবৃদ্ধি মুনাফা = C - P
= (266.2 - 200) টাকা
= 66.2 টাকা

∴ সরল ও চক্রবৃদ্ধি মুনাফার পার্থক্য = 66.2 - 60 = 6.2 টাকা

Question: The ratio between the speeds of two buses is 5 : 6. If the second bus runs 450 km in 5 hours, then the speed of the first bus is:

Solution:
Given that,
Ratio of speeds of first bus : second bus = 5 : 6
Second bus covers 450 km in 5 hours

Now, Speed of second bus = distance/time
= 450/5
= 90 km/h

Let speed of first bus = 5x km/h
Speed of second bus = 6x km/h
We know speed of second bus = 90 km/h
So, 6x = 90
⇒ x = 90/6
∴ x = 15

∴ Speed of first bus = 5x = 5 × 15 = 75 km/h

So the speed of the first bus is 75 km/h.

Question: If 2^{n - 1} + 2^{n + 1} = 320, then the value of n is = ?

Solution:
Given that, 
2^{n - 1} + 2^{n + 1} = 320
⇒ 2^{n - 1} + 2^{n - 1} . 2² = 320
⇒ 2^{n - 1}(1 + 2²) = 320
⇒ 2^{n - 1} . 5 = 320
⇒ 2^{n - 1} = 320/5 = 64
⇒ 2^{n - 1} = 26
⇒ n - 1 = 6
⇒ n = 6 + 1
∴ n = 7

So the value of n is 7.

Question: A ladder is leaning against a wall. It makes a 60° angle with the wall. If the distance between the foot of the ladder and the wall is 7.5 meters, find the length of the ladder.

Solution:

 
Let BC be the wall and AC be the ladder.
∠BAC = 60° and AB = 7.5 meter
In ΔABC,
cos60° = AB/AC
⇒ 1/2 = 7.5/AC
⇒ AC = 7.5 × 2
∴ AC = 15

Original Building height = h = MQ
New building height = PQ

In △MQN, tan 30° = 1/√3 = MQ/NQ
MQ = 300/√3

In △PQN, tan 60° = √3 = PQ/NQ
PQ = 300√3
Building grew = PQ - MQ

∴ Building grew = 300√3 - (300/√3 )
= 3 × 300 × (2/√3)
= 200√3

Number of pages typed by Robi in 1 hour = 32/6 = 16/3 .
Number of pages typed by Kamal in 1 hour = 40/5 = 8
Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3
Therefore, Time taken by both to type 110 pages = 110/(40/3) hours
= 110 × 3/40 hours (or) 8 hours 15 minutes.

Therefore, the length of the ladder is 16 m.

Question: After selling a saree for Tk. 3360 a shopkeeper suffers a loss of 16%. If he wants to earn 15% profit after giving the discount of 8%, what will be the marked price of the saree?

Solution:
At 16% loss,
Selling price Tk. 84 when cost price Tk. 100
Selling price Tk. 1 when cost price Tk. 100/84
Selling price Tk. 3360 when cost price Tk. (100 × 3360)/84
= Tk. 4000

At 15% profit,
cost price Tk. 100 when selling price Tk. 115
cost price Tk. 1 when selling price Tk. 115/100
cost price Tk. 4000 when selling price Tk. (115 × 4000)/100
= Tk. 4600

At 8% discount,
Selling price Tk. 92 when marked price Tk. 100
Selling price Tk. 1 when marked price Tk. 100/92
Selling price Tk. 4600 when marked price Tk. (100 × 4600)/92
= Tk. 5000

∴ The marked price of the saree should be Tk. 5000.