Bank Math

Question: An amount of Tk. 12,000 yields a simple interest of Tk. 2,160 in 4 years. What is the annual rate of interest?

Solution:
Given, Principal, P = 12000
Simple Interest, SI = 2160
Time, n = 4 years
Rate of interest, r = ?

We know, I = Pnr/100
⇒ r = (I × 100)/(P × n)
⇒ r = (2160 × 100)/(12000 × 4)
⇒ r = 216000/48000
⇒ r = 4.5%

So, the annual rate of interest is 4.5%.

Question: If y > 1 and y < 4, then which of the following expressions is positive?
I. (y - 1)(y - 4)
II. (1 - y)(y - 4)
III. (1 - y)(4 - y)

Solution:
Given,
y > 1 and y < 4

For expression I: (y - 1)(y - 4)
Since y > 1, (y - 1) will be positive.
Since y < 4, (y - 4) will be negative.
∴ (y - 1)(y - 4) = positive × negative = negative

For expression II: (1 - y)(y - 4)
Since y > 1, (1 - y) will be negative. 
Since y < 4, (y - 4) will be negative.
∴ (1 - y)(y - 4) = negative × negative = positive

For expression III: (1 - y)(4 - y)
Since y > 1, so (1 - y) will be negative. 
Since y < 4, so (4 - y) will be positive.
∴ (1 - y)(4 - y) = negative × positive = negative

∴ Only expression II is positive.

Question: A is four years older than B, who is thrice as old as C. If the total of the ages of A, B, and C is 46, how old is A?

Solution:
Let
C's age be = a years
Then, B's age = 3a years
A's age = (3a + 4) years

∴ (3a + 4) + 3a + a = 32
⇒ 7a + 4 = 46
⇒ 7a = 42
⇒ a = 6

Hence, A's age = (3 × 6) + 4 = 22 years.

Question: If 8 workers can assemble a car in 9 hours, how long would it take 12 workers to assemble the same car?

Solution: 

Here, M₁ = 8, M₂ = 12, W₁ = W₂ = 1, T₁ = 9, T₂ = ?

(M₁ × T₁)/(M₂ × T₂) = W₁/W₂ 
⇒ (8 × 9)/ (12 × T₂) = 1
⇒ 12 × T₂ = 72
⇒ T₂ = 72/12 
∴ T₂ = 6

Question: Pipe A can fill the tank in 8 hours and pipe B can fill it in 12 hours. If pipe A is opened at 8:00 AM and pipe B is opened at 10:00 AM, then at what time will the tank be full ?

Solution: 
A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 10 A.M. + 3 hours 36 minutes = 1 : 36 P.M.

Question: A box contains 5 green, 3 yellow, and 4 black balls. If one ball is drawn at random, what is the probability that it will not be a green ball?

Solution:
Given that,
Green balls = 5
Yellow balls = 3
Black balls = 4
∴ Total balls = 5 + 3 + 4 = 12

And, number of non-green balls = Yellow + Black = 3 + 4 = 7

We know,
P(not green) = favorable outcomes/total outcomes
= 7/12

∴ The probability of drawing a non-green ball is 7/12

Cost price = Tk 3000
Selling price = [{3600 × 100}/{100 + (10 × 2)}]
= Tk. 3000
Gain = 0%.

Question: Find the product of two consecutive numbers if three times the first number is 5 more than twice the second number.

Solution:
Let the numbers be a and a + 1.

According to the question:
3 × (first number) = 2 × (second number) + 5
⇒ 3a = 2(a + 1) + 5
⇒ 3a = 2a + 2 + 5
⇒ 3a = 2a + 7
⇒ 3a - 2a = 7
⇒ a = 7

∴ The numbers are 7 and 8.
Product = 7 × 8 = 56

Question: A hat contains a total of 30 cards, of which 12 are marked with a star and the remaining 18 are unmarked. If a card is drawn at random from the hat, what is the probability that it will be a card marked with a star?

Solution:

Total number of cards = 12 (marked) + 18 (unmarked) = 30

Number of favorable outcomes (marked with a star) = 12

Probability = (Number of favorable outcomes)/(Total number of outcomes)
= 12/30
= 2/5

Question: The volume of a cone is 300π cubic centimeters. If the radius of its base is 6 cm, what is the height of the cone?

solution:
দেওয়া আছে,
কোণকের আয়তন, V = 300π ঘন সে.মি.
ভূমির ব্যাসার্ধ, r = 6 সে.মি.
ধরি, কোণকের উচ্চতা = h সে.মি.

আমরা জানি,
কোণকের আয়তন, V = 1/3 × π × r² × h
∴ 300π = 1/3 × π × 6² × h
⇒ 300 = 1/3 × 36 × h (π উভয় পক্ষ থেকে বাদ দিয়ে)
⇒ 300 = 12h
⇒ h = 300 / 12
∴ h = 25 সে.মি.

অতএব, কোণকটির উচ্চতা = 25 সে.মি.

Question: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Solution:
Here, S = {1, 2, 3, 4, ...., 19, 20}
Let E = event of getting a multiple of 3 or 5 = {3, 6, 9, 12, 15, 18, 5, 10, 20}

∴ P(E) = n(E)/n(S)
= 9/20

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
= ⁷C₃× ⁴C₂
= {(7 × 6 × 5)/(3 × 2 × 1)} × {(4 × 3)/(2 × 1)}
= 210.

Question: When bent in the form of a circle, a wire has a radius of 28 cm. If the same wire is bent into the shape of a square, what will be its area in cm²?

Solution:
প্রদত্ত, বৃত্তের ব্যাসার্ধ, r = 28 cm

অতএব, পরিধি = 2πr
= 2 × (22/7) × 28 = 176 cm

ধরি, বর্গের বাহু = a cm
তাহলে, বর্গের পরিসীমা = 4a

এখন,
বৃত্তের পরিধি = বর্গের পরিসীমা
⇒ 176 = 4a
⇒ a = 176/4
= 44 cm

∴ বর্গের ক্ষেত্রফল = a²
= 44²
= 1936 cm² 

Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 5 x 4 = 20),
so there will be 6 significant digits to the right of the decimal point.

Question: If A = {p, q, r, s, t}, then how many proper subsets does A have?

Solution:
Given that, 
A = {p, q, r, s, t}
The number of elements in set A is 5.

We know that,
Number of proper subsets = 2ⁿ - 1  ; [where n = number of elements in the set]
∴ Number of proper subsets of A = 2⁵ - 1
= 32 - 1
= 31

Question: What number should come next in the series:
3, 7, 15, 31, 63, .......?

Solution: দেওয়া আছে,
সিরিজটি হলো: 3, 7, 15, 31, 63, .......

প্রতিটি পার্থক্য আগের পার্থক্যের 2 গুণ।

3 থেকে 7 পর্যন্ত পার্থক্য: 4
7 থেকে 15 পর্যন্ত পার্থক্য: 8 (4 × 2)
15 থেকে 31 পর্যন্ত পার্থক্য: 16 (8 × 2)
31 থেকে 63 পর্যন্ত পার্থক্য: 32 (16 × 2)
সুতরাং, পরবর্তী পার্থক্যটি হবে: 32 × 2 = 64

পরবর্তী সংখ্যাটি হবে শেষ সংখ্যা এবং এই পার্থক্যের যোগফল: 63 + 64 = 127

অতএব, পরবর্তী সংখ্যাটি হলো 127

Question: The mean weight of 100 students in a class is 46 kg. The mean weight of boys is 50 and of girls is 40 kg. Therefore, the number of boys is-

Solution:
Given that, 
Total students = 100
Mean weight of all students = 46 kg
∴ Total weight of all students = 100 × 46 = 4600 kg.

Let,
The number of boys = x. Then, the number of girls = 100 - x 
Mean weight of boys = 50 kg,
∴ total weight of boys = 50x
And, 
Mean weight of girls = 40 kg,
∴ total weight of girls = 40(100 - x)

ATQ,
50x + 40 × (100 - x) = 4600
⇒ 50x + 4000 - 40x = 4600
⇒ 10x = 4600 - 4000
⇒ x = 600/10
∴ x = 60

So the number of boys is 60.

AB = BC
⇒ ∠C = ∠A = (2x - 20)°.
∠A+ ∠B + ∠C =180°
⇒ (2x - 20) + x + (2x - 20 ) = 180
⇒ 5x - 40 = 180
⇒ 5x = 220
⇒ x = 44.
∴ ∠B = 44°

Question: A dealer buys dry fruits at Tk. 100, Tk. 80, and Tk. 60 per kilogram. He mixes them in the ratio 3 : 4 : 5 by weight and sells at a profit of 50%. At what price per kilogram does he sell the dry fruits?

Solution: 
Let the dealer buy 3 kg, 4 kg and 5 kg.
∴ Price of total dry fruits = (3 × 100) + (4 × 80) + (5 × 60) = Tk. 920

At 50% Profit,
Selling Price, SP = 920 + 50% of 920
= 920 + (50/100) × 920
= 1380

Hence,
Price of dry fruits per kg = 1380/12 = 115 Tk.

Question: A can complete a work in 20 days, B in 30 days, and C in 60 days. A stops working 4 days before the completion of the work, and B stops 6 days before completion. C continues working alone till the end. What was the total number of days taken to complete the entire work?

Solution:
Let the total work be completed in y days.

∴ A worked for (y - 4) days
So his contribution = (y - 4)/20

B worked for (y - 6) days
So his contribution = (y - 6)/30

C worked full y days, so his contribution = y/60

Therefore,
(y - 4)/20 + (y - 6)/30 + y/60 = 1
⇒ 3(y - 4) + 2(y - 6) + y = 60
⇒ 3y - 12 + 2y - 12 + y = 60
⇒ 6y - 24 = 60
⇒ 6y = 84
⇒ y = 14

∴ The total work was completed in 14 days.

Question: Weekly incomes of two persons are in the ratio of 7 : 3 and their weekly expenses are in the ratio of 5 : 2. If each of them saves Tk. 300 per week, find their weekly incomes.

Solution:
Let the incomes of the two persons be 7x and 3x 
And their expenses be 5y and 2y respectively.

Then we get,
7x - 5y = 3x - 2y
⇒ 4x = 3y
∴ y = 4x/3

Now, 7x - 5y = 300
⇒ 7x - 5(4x/3) = 300
⇒ (21x - 20x)/3 = 300 
∴ x = Tk. 900

∴ Weekly income of first person = (7 × 900) = Tk. 6300
∴ Weekly income of second person = (3 × 900) = Tk. 2700

So weekly incomes are Tk. 6300 and Tk. 2700

Question: A train running at the speed of 90 km/h crosses a pole in 10 seconds. What is the length of the train?

Solution:
Speed = 90 km/h
= [90 × (5/18)] m/sec
= 25 m/sec

∴ Length of the train = (25 × 10) m
= 250 m

So, the length of the train is 250 meters.

Number of bricks
=Courtyard area/1 brick area
=(2500×1600 / 20×10)=20000

Question: The greatest value of sin⁴2θ + cos⁴2θ is?

Solution:
sin²2θ + cos²2θ = 1

(sin²2θ + cos²2θ)² = 1²
⇒ sin⁴2θ + cos⁴2θ + 2 sin²2θ cos²2θ = 1
⇒ sin⁴2θ + cos⁴2θ = 1 − 2 sin²2θ cos²2θ [2 sin²2θ cos²2θ = 0 (when θ = 0° or 90°)]
∴ sin⁴2θ + cos⁴2θ = 1 

∴ greatest value = 1

Question: A train 360 m long passes a pole in 30 seconds. How long will it take to pass a platform 540 m long?

Solution: 
সমাধান:
ট্রেনটির মোট দূরত্ব অতিক্রম করতে হবে = (360 + 540) মিটার = 900 মিটার 

ট্রেনটি 360 মিটার অতিক্রম করতে সময় নেয় = 30 সেকেন্ড 
ট্রেনটি1 মিটার অতিক্রম করতে সময় নেয় = 30/360 সেকেন্ড 
ট্রেনটি 900 মিটার অতিক্রম করতে সময় নেয় = (30 × 900)/360 সেকেন্ড 
= 75 সেকেন্ড

Suppose,
A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then,
(1/x) + (2/x) + (3/x) = 1/2
⇒ 6/x = 1/2
⇒ x = 12.
So, B takes 6 days to finish the work.

Question: rsinθ = 1, rcosθ = √3 then the value of (√3tanθ + 1) = ?

Solution:
rsinθ = 1
rcosθ = √3

Now,
rsinθ/rcosθ = 1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1
⇒ √3tanθ + 1 = 1 + 1
∴ √3tanθ + 1 = 2

84 × 59 × 13 × 76 = 4896528
Without the use of calculator, to count the unit digit = 4 × 9 × 3 × 6 = 36 × 18 = 648

So, 8 is the unit digit

Question: Three unbiased coins are tossed simultaneously. What is the probability of getting at most one tail?

Solution: Total outcomes = 2³ = 8 (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)

Favorable outcomes = cases with at most one tail (zero tails or exactly one tail) = (HHH, HHT, HTH, THH)

At most one tail refers to zero or one tail only.

Therefore, Probability = Favorable outcomes / Total outcomes = 4/8 = 1/2

He sells 40% of oranges and still there are 420 oranges remaining.
=> 60% of oranges = 420
=> Total oranges × 60 /100 = 420
=> Total oranges = 420 × 100/ 60 = 700