Bank Math

Question: If the product of three consecutive integers is 720, then the sum of the two largest integers is-

Solution: 
720 =  2 × 2 × 2 × 2 × 3 × 3 × 5
       = (2 × 2 × 2) ×  (3 × 3) × (2 × 5)
       = 8 × 9 × 10 

 The sum of the two largest integers is = 9 + 10 = 19

Let numbers be A and B
∴ (A - B)² = 9
∴ A² - 2AB + B² = 9
Further, A² + B² = 225
∴ 225 - 2AB = 9
∴ AB = 108 = product of the two numbers.

Net part filled in 1 hour
= (1/x − 1/y)
= (y−x)/xy
∴ The tank will be filled in
= xy/(y−x) hours

√6 x √27
= √(6 x 27)
= √(2 x 3 x 3 x 3 x 3)
= (3 x 3)√2
= 9√2

Question: A circle can have how many parallel tangents at a single time?

Solution:
- একটি বৃত্তের স্পর্শক (Tangent) হলো এমন একটি সরলরেখা যা বৃত্তকে কেবল একটি বিন্দুতে স্পর্শ করে।
- একটি বৃত্তে একটি নির্দিষ্ট দিকে সর্বোচ্চ এক জোড়া বা ২টি সমান্তরাল স্পর্শক থাকা সম্ভব। এই সমান্তরাল স্পর্শক দুটি সর্বদা বৃত্তের ব্যাসের (Diameter) বিপরীত প্রান্তবিন্দুতে অবস্থিত থাকে।
- যদি ব্যাসের দুই প্রান্ত ছাড়া অন্য কোনো বিন্দুতে তৃতীয় একটি সমান্তরাল রেখা আঁকা হয়, তবে সেটি হয় বৃত্তকে দুই বিন্দুতে ছেদ করবে (Secant) অথবা বৃত্তকে স্পর্শই করবে না।

∴ একটি বৃত্তে একসাথে ঠিক ২টি সমান্তরাল স্পর্শক থাকতে পারে (যেকোনো নির্দিষ্ট দিকের জন্য)।

Question: The average age of 12 children is 15 years. If another child comes, the average age comes to 13. What is the age of the new child?

Solution: 
Sum of 12 children ages = 12 × 15 = 180
Sum of the 13 children ages = 13 × 13 = 169
So age of new children = 180 - 169
= 11

[বাস্তবিকভাবে নতুন একজনের বয়স যুক্ত হওয়ার পর বয়সের গড় কম-বেশি হতে পারে তবে ১২ জনের মোট বয়স (১৮০ বছর) অপেক্ষা ১৩ জনের মোট বয়স (১৬৯ বছর) কম হতে পারে না।  এই প্রশ্নটি যেহেতু জব সল্যুশনের প্রশ্ন তাই গাণিতিক নিয়ম অনুযায়ী ১১ বছর উত্তর রাখা হয়েছে] 

Question: What percent of 500 is 1.5?

Solution:
ধরি,
500 এর x% = 1.5
⇒ 500 এর (x/100) = 1.5
⇒ 5x = 1.5
⇒ x = 1.5/5
⇒ x = 0.3

​সুতরাং, 500 এর 0.3% হলো 1.5

At p% profit, selling price = 100 + p
at p% loss, new selling price = (100 + p) - (100 + p) × p/100
= (100 + p) - {(100p + p²)/100}
= 10000 + 100p - 100p - p²
= (10000 - p²)/100

so, (10000 - p²)/100 is equal to 1
1 is equal to 1/(10000 - p²)/100
100 is equal to 1 × (100/ 10000 - p²) × 100
= 10000/(10000 - p²)

অপশন গ) (10000/10000-p^2) হওয়ার কথা ছিলো। সেক্ষেত্রে উত্তর ঠিক হয়।
লাইভ পরীক্ষার প্রশ্নে অপশন ভুল থাকায় উত্তর হিসাবে ঙ) None of above কে ধরা হয়েছে।

Question: A motorcyclist covers the first 40 km in 20 minutes and the second 60 km in 30 minutes. Between these two segments, the motorcyclist stopped for 10 minutes for a rest. What is the average speed of the motorcycle in km/h?

Solution:
মোট দূরত্ব = 40 কিমি + 60 কিমি = 100 কিমি।

প্রথম অংশের সময় = 20 মিনিট।
দ্বিতীয় অংশের সময় = 30 মিনিট।
বিশ্রামের জন্য বিরতি = 10 মিনিট।
মোট সময় = 20 + 30 + 10 মিনিট = 60 মিনিট = 1 ঘন্টা

গড় গতিবেগ = 100 কিমি/1 ঘন্টা
= 100 কিমি/ঘন্টা
∴ মোটরসাইকেলটির গড় গতিবেগ হলো 100 কিমি/ঘন্টা।

Question: For what percentage of profit per annum by which Tk. 12000 will be Tk. 17760 as profit-principal in 6 years?

Solution: 
Principal, P = Tk. 12000
Total Amount, A = Tk. 17760
Time, n = 6 years
Simple Interest SI = A - P = 17760 - 12000 = Tk. 5760

We know, 
SI = Prn/100
⇒ 5760 = (12000 × r × 6)/100
⇒ r × 720 = 5760
⇒ r = 5760/720
∴ r = 8

So the profit percentage of per annum is 8%.

Question: A man completes a journey in 8 hours. He travels the first half of the journey at the rate of 40 km/hr and the second half at the rate of 60 km/hr. Find the total distance of the journey in km. 

Solution:
Let the total distance of the journey be d km.
Then, the first half of the journey = d/2 km and the second half = d/2 km.

Time taken for the first half,
= (d/2) / 40 hours
= d/80 hours

And,
Time taken for the second half,
= (d/2) / 60 hours
= d/120 hours

According to the question,
(d/80) + (d/120) = 8
⇒ (3d + 2d)/240 = 8
⇒ 5d/240 = 8
⇒ 5d = 8 × 240
⇒ 5d = 1920
⇒ d = 1920/5
⇒ d = 384 km

∴ The total distance of the journey is 384 km.

Let the sum be P
then,
P(1 + r/100)³ = 6690 .......(i)
and
P(1 + r/100)⁶ = 10035 .......(ii)
On dividing (ii) by )i),
(1 + r/100)³ = 10025/6690 = 3/2
Substituting this value in (i) we find, P×(3/2) = 6690
⇒ P = (6690×2/3) = 4460

Question: At what percentage profit must an article be sold such that selling it at two-thirds of that price results in a 20% loss?
​
​Solution:
Let,
Cost Price be x .
Selling Price be y.
​
​Selling at 2/3 of y, causes 20% loss,
So, 2y/3 = x - 20% of x
⇒ 2y/3 = x - (20x/100)
⇒ 2y/3 = x{(1 - (20/100)}
⇒ 2y/3 = x × (80/100)
∴ y = 6x/5
​
​Profit = Selling Price - Cost Price
= (6x/5) - x
= (6x - 5x)/5
= x/5
​
​∴ Profit Percentage = (Profit/Cost Price) × 100%
= {(x/5)/x} × 100%
= 20%

ATQ, ar^2-1 = ar = 9 ......(i)
ar^3-1 = 3 ..... (ii)  
(ii)/(i) = r = 1/3
from (i) we get, a = 9×3 = 27
∴ ar^5-1 = 27×(1/3)4 
= 27×(1/81)
= 1/3 

Question: If cosecθ - cotθ = 1/5, then find the value of sinθ + 5cosθ.

Solution:
cosecθ - cotθ = 1/5
⇒ (1/sinθ) - (cosθ/sinθ) = 1/5
⇒ (1 - cosθ)/sinθ = 1/5
⇒ sinθ = 5(1 - cosθ)
∴ sinθ = 5 - 5cosθ

∴ sinθ + 5cosθ = (5 - 5cosθ) + 5cosθ
= 5 - 5cosθ + 5cosθ
= 5

Question: The equation  is only possible when?

Solution:
cos²θ = (x + y)²/4xy

Maximum value of cos²θ = 1. So,
⇒ 1 = (x + y)²/4xy
⇒ 4xy = (x + y)²
⇒ 4xy = x² + y² + 2xy
⇒ 0 = x² + y² - 2xy
⇒ 0 = (x - y)²
⇒ 0 = x - y
∴ x = y

Question: What least number must be added to 1056, so that the sum is completely divisibly by 23?

Solution:
1056 কে 23  দ্বারা ভাগ করলে 21 অবশিষ্ট থাকে।

এখানে
23 - 21 = 2
2, 1056 এর সাথে যোগ করলে যোগফল 23 দ্বারা বিভাজ্য ।

Question: Find the greatest number that exactly divides each of the numbers 48, 72, and 108.

Solution:
We know,
The HCF (Highest Common Factor) of two or more numbers is the greatest number that divides each of them exactly.

Now,
Prime factorization of 48 = 2 × 2 × 2 × 2 × 3

Prime factorization of 72 = 2 × 2 × 2 × 3 × 3

Prime factorization of 108 = 2 × 2 × 3 × 3 × 3

∴ HCF of 48, 72, and 108 = 2 × 2 × 3 = 12

Therefore, the greatest number is 12.

Question: Which of the following is a leap year?

Solution:
অধিবর্ষ বা লিপ ইয়ার নির্ণয়ের দুটি প্রধান নিয়ম রয়েছে:
১. সাধারণ বছর: বছরটি 4 দ্বারা নিঃশেষে বিভাজ্য হতে হবে।
২. শতাব্দী বছর (100 দ্বারা বিভাজ্য): বছরটি 400 দ্বারা নিঃশেষে বিভাজ্য হতে হবে।

এখন,
1982 ÷ 4 = 495.5 → বিভাজ্য নয় → Leap year নয়। 
1984 ÷ 4 = 496 → বিভাজ্য → Leap year. 
1998 ÷ 4 = 499.5 → বিভাজ্য নয় → Leap year নয়। 
2002 ÷ 4 = 500.5 → বিভাজ্য নয় → Leap year নয়। 

অতএব, 1984 সালটি অধিবর্ষ।

Question: A box contains 24 electric bulbs, out of which 6 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is:

Solution:
Given that,
Total bulbs = 24
Defective bulbs = 6
Non-defective bulbs = 24 - 6 = 18
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (18/24) × (17/23)
= 306/552
= 51/92

And,
∴ P(at least one defective)
= 1 - P(both non-defective)
= 1 - (51/92)
= (92 - 51)/92
= 41/92

Question: In how many different ways can the letters of the word 'EXTRA' be arranged so that the vowels are never together?

Solution:
Taking the vowels (EA) as one letter, the given word has
the letters X T R (EA) = 4 letters.
∴ These letters can be arranged in = 4! = 24 ways.
∴ The letters EA may be arranged amongst themselves in = 2! = 2 ways.

∴ Number of arrangements having vowels together = (24 × 2) = 48 ways.

∴ Total arrangements of all letters = 5! = (5 × 4 × 3 × 2 × 1) = 120.

∴ Number of arrangements not having vowels together = (120 - 48) = 72

So number of arrangements where vowels are never together = 72

Let the required number of working hours per day be x
More pumps, Less working hours per day (Indirect proportion)
Less days, More working hours per day (Indirect proportion)
Pumps 4:3 & Days 1:2 } :: 8:x
∴ 4×1×x = 3×2×8
⇔ x = (3×2×8)/(4)
⇔ x = 12

n(S) = number of ways of choosing 4 persons out of 9
= ⁹C₄ = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 126

n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) = ⁴C₂ × ⁵C₂ = {(4 × 3)/(2 × 1) × (5 × 4)/(2 × 1)} = 60

∴P(E) = n(E)/n(S) = 60/126 = 10/21.

প্রশ্ন: ৮টি ধারাবাহিক বিজোড় সংখ্যার ৬ষ্ঠ সংখ্যাটি হল - ১৭, সবগুলো সংখ্যার সমষ্টি কত?

সমাধান:
৮টি ধারাবাহিক বিজোড় সংখ্যার সিরিজটি হলো: - ২৭, - ২৫, - ২৩ - ২১, - ১৯, - ১৭, - ১৫, - ১৩
সবগুলো সংখ্যার সমষ্টি = - (২৭ + ২৫ + ২৩ + ২১ + ১৯ + ১৭ + ১৫ + ১৩)
= - ১৬০

আবার
৮টি ধারাবাহিক বিজোড় সংখ্যার সিরিজটি হলো: - ৭, - ৯, - ১১, - ১৩, - ১৫, - ১৭, - ১৯, - ২১
সবগুলো সংখ্যার সমষ্টি = - (৭ + ৯ + ১১ + ১৩ + ১৫ + ১৭ + ১৯ + ২১)
= - ১১২

এখানে দুইভাবে সমাধান করলে উত্তর আসে - ১১২ এবং - ১৬০
অপশনে যেহেতু  - ১১২ আছে তাই সঠিক উত্তর হিসেবে  - ১১২ নেওয়া হয়েছে।

Question: A man spends 45% of his salary on rent, 25% on food, and saves the rest. If his salary is Tk. 60000. how much does he save?
 
Solution: 
Given that, 
Total salary = Tk. 60000

Now, 
Spends on rent = 45% of 60000
= (45/100) × 60000
= Tk. 27000 

And spends on food = 25% of 60000
= (25/100) × 60000
= Tk. 15000

∴ Total expenditure = rent + food
= 27000 + 15000
= Tk. 42000

∴ Savings = Total salary - Total expenditure
= 60000 - 42000
= Tk. 18000

He saves Tk. 18000

Shortcut method:
Percentage saved = 100% - (45% + 25%)
= 100% - 70%
= 30%

∴ Savings = 30% of 60000
= 0.30 × 60,000
= Tk. 18000

Question: In a class average age of 15 boys is 11. If 5 boys each of age 9 years are added, what would be the new average?

Solution:
Sum of ages of 15 boys = 15 × 11= 165
Sum of ages of 5 boys = 5 × 9 = 45
Total age of 20 boys = 165 + 45 = 210
∴ Average of ages of 20 boys = 210/20 = 10.5 years

Question: A sample of 50 liters of glycerine is found to be adulterated to the extent of 20%. How much pure glycerine should be added to it so as to bring down the percentage of impurity to 5%?

Solution: 
Initially, the 50 liters of glycerine is 20% adulterated. So, the pure glycerine in it is 80% of 50 liters.

Amount of pure glycerine initially = 80% of 50 liters
= 0.8 × 50
= 40 liters

Let 'x' represent the amount of pure glycerine that needs to be added to reduce the percentage of impurity to 5%.

The total volume of glycerine after adding pure glycerine will be 50 + x liters

After adding 'x' liters of pure glycerine, the total amount of pure glycerine in the mixture will be 40  + x liters (added).

Now, this total amount of pure glycerine should be 95% of the new total volume (50 liters original + x liters added), as the impurity percentage is to be reduced to 5%.

So, we can set up an equation:

Total amount of pure glycerine = 95% of total volume after adding pure glycerine

40 + x = 0.95 × 50 + 0.95 × x
⇒ 40 + x = 47.5 + 0.95x
⇒ x - 0.95x = 47.5 - 40
⇒ 0.05x = 7.5
⇒ x = 7.5 / 0.05
⇒ x = 150 liters

Question: If one-fifth of one-half of a number is 12, then two-thirds of that number is: 

Solution:
Let the number be X.
(1/5) × (1/2) × X = 12
⇒ X = 12 × 5 × 2
⇒ X = 120
Two-thirds of the number:
(2/3) × 120 = 80

Questions: Three taps P, Q, and R can fill a tank in 6 hours. After working together for 2 hours, tap R is closed, and P and Q can fill the rest of the tank in 7 hours. The number of hours taken by R alone to fill the tank is –

Solution:
Three taps P, Q and R can fill a tank in 6 hours.
Three taps can fill in one hour 1/6 part of the tank
Three taps can fill in 2 hours 1/3 part of the tank.

Rest part 1 - 1/3 = 2/3 part
2/3 part can be filled in 7 hours by P and Q

∴ In 1 hour P and Q can fill 2/21 part
∴ In 1 hour P, Q and R can fill 1/6 part

∴ in 1 hour R can fill (1/6 - 2/21) = 1/14 part
Hence, R alone fills the tank in 14 hours.

Question: The speed of P and Q are in the ratio 5 : 8. Q takes 24 minutes less than P to reach a destination. Time in which Q reaches the destination?

Solution:
Given, speed of P and Q = 5 : 8

So, ratio of time taken = 8 : 5 [Time ∝ 1/Speed]

Let time taken by P and Q be 8x and 5x minutes respectively.

According to the question,
8x - 5x = 24
⇒ 3x = 24
⇒ x = 8

Hence, time taken by Q = 5 × 8 = 40 minutes

Question: If log_m(81) = 4, what is the value of m?

Solution:
log_m(81) = 4
⇒ m⁴ = 81 [log_b(a) = c ⇒ b^c = a]
⇒ m⁴ = 3⁴
∴ m = 3

Question: An 84 kg metal sphere is melted down and reshaped into 4,000 nails of equal size. Find the weight of one nail in grams.

Solution:
দেওয়া আছে,
ধাতুর বলের ওজন = 84 কেজি = 84 × 1000 = 84000 গ্রাম
পেরেকের সংখ্যা = 4000 টি 

এখন,
4000 পেরেকের ওজন = 84000 গ্রাম
∴ 1 টি পেরেকের ওজন = (84000/4000) গ্রাম = 21 গ্রাম

We know, Area of rhombus = 1/2 × x × y [Here, x and y are two diagonals of the rhombus]
Or, x = (91 × 2) / 14 = 13 cm

Question: Which of the following is the largest?

Solution: 
ক) 9/12 = 0.75

খ) 10/15 = 0.66

গ) 5/8 = 0.62

ঘ) 7/10 = 0.70

So 9/12 = 0.75 is clearly the biggest.

Question: If 0.13 ÷ p² = 13, then p equals:

Solution:
0.13 ÷ p² = 13
⇒ p² = 0.13/13
⇒ p² = 1/100
⇒ p = √1/100
⇒ p = 1/10
⇒ p = 0.1

A:: - G:S = 40:60
B:: - G:S:C = 35:40:25
New,
G:S= (1×40 + 4×35) : (40×4 + 1×60)
= 180:220
= 18:22
= 9:11

Question: A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Tk. 1000 more than D, what is B's share?

Solution: 
let, they get 5x, 2x, 4x, 3x 

4x - 3x = 1000 
⇒ x = 1000 

∴ B's share is = 2x 
= 2 × 1000 
= 2000 taka 

Question: 2³⁰ + 2³⁰ + 2³⁰ + 2³⁰ = ?

Solution
2³⁰ + 2³⁰ + 2³⁰ + 2³⁰
= 4 × 2³⁰
= 2² × 2³⁰
= 2^{2 + 30}
= 2³²

Question: Alif is shorter than Meena but taller than Zara. Pial is taller than Alif. Meena is the second-tallest person among them. Sumon is shorter than Zara. Who is the third-tallest person among them?

Solution:
First statement: Meena > Alif > Zara

Second statement: Pial > Alif

Third statement: Meena is the second-tallest, meaning one person is taller than Meena. Since Pial is taller than Alif and Meena is taller than Alif, Pial must be taller than Meena.
Therefore, Pial > Meena > Alif

Fourth statement: Zara > Sumon

Putting everyone together: Pial > Meena > Alif > Zara > Sumon

∴ The third-tallest person is Alif.

Question: Tina works four times as fast as Rina. If Rina can complete a work in 20 days alone, how many days will Tina and Rina together take to complete the work?

Solution:
Tina : Rina work ratio = 4 : 1
∴ Time taken ratio = 1 : 4

Since Rina takes 20 days, Tina takes = 20 / 4 = 5 days

∴ (Tina + Rina)'s 1 day work = (1/5) + (1/20)
= (4 + 1)/20
= 5/20
= 1/4

∴ Total days to finish the work = 1 ÷ (1/4) = 4 days

∴ Tina and Rina together can complete the work in 4 days.

log₁₀4+log₁₀25
= log₁₀(4×25)
= log₁₀100
= log₁₀10²
= 2log₁₀10
= 2

Question: Currently, a mother’s age is five times that of her son. After 10 years, her age will be three times her son’s age. What is the son’s present age?

Solution:
Let,
The son’s present age = x years
Then, mother’s present age = 5x years

According to the question,
5x + 10 = 3(x + 10)
⇒ 5x + 10 = 3x + 30
⇒ 5x - 3x = 30 - 10
⇒ 2x = 20
⇒ x = 10

∴ The son’s present age is 10 years.

Question: If a coin is tossed once, what is the probability of getting a tail?

Solution: Here, the total outcome is 2 (Head and Tail) 
The favorable outcome is 1 (Tail)

Therefore, Probability = Favorable outcome/Total outcome
= 1/2
= 0.5

Question: If the price of a commodity is decreased by 25% and its consumption is increased by 25%, what will be the increase or decrease in expenditure on the commodity?

Solution:
Let,
The initial expenditure on the commodity be Tk. 100.
Now, the price decreases by 25%,
∴ Current Price = (100 - 25% of 100) = Tk. 75

Same time due to decrements in price 25% consumption has been increased.
So, Current expenses on commodity = (75 + 25% of 75) = Tk. 93.75

Here,
The initial expenditure was Tk. 100 which became Tk. 93.75 at the end, it means there is 6.25% decrements in the expenditure of the commodity.

So the expenditure decreases by 6.25%

Question: If p + q + r = 10 and p² + q² + r² = 38, then what is the value of pq + qr + rp?

Solution:
We know the identity:
(p + q + r)² = p² + q² + r² + 2(pq + qr + rp)
⇒ 10² = 38 + 2(pq + qr + rp)
⇒ 100 = 38 + 2(pq + qr + rp)
⇒ 2(pq + qr + rp) = 100 - 38 = 62
⇒ pq + qr + rp = 62/2
∴ pq + qr + rp = 31

Question: If the sum of two numbers is 20 and the sum of their squares is 208, then what is the product of the two numbers?

Solution:
ধরি, সংখ্যা দুটি যথাক্রমে x এবং y।
দেওয়া আছে,
x + y = 20
x² + y² = 208

আমরা জানি,
(x + y)² = x² + y² + 2xy
⇒ (20)² = 208 + 2xy
⇒ 400 = 208 + 2xy
⇒ 2xy = 400 - 208
⇒ 2xy = 192
⇒ xy = 192 / 2
∴ xy = 96

সুতরাং, সংখ্যা দুটির গুণফল হলো 96.