Bank Math

Question: A committee of 4 members is to be formed from 6 men and 5 women. In how many different ways can the committee be formed if it must contain at least 2 men?

Solution:
To form a committee of 4 members with at least 2 men, we consider the following possible cases:

Case 1: 2 Men and 2 Women
Ways = ⁶C₂ × ⁵C₂
= 15 × 10
= 150

Case 2: 3 Men and 1 Woman
Ways = ⁶C₃ × ⁵C₁
= 20 × 5
= 100

Case 3: 4 Men and 0 Women
Ways = ⁶C₄ × ⁵C₀
= 15 × 1
= 15

Total number of ways = 150 + 100 + 15
= 265

Given that T = R
S.I = (P × R × T)/100
⇒ 686 = (1400 × R × R)/100
⇒ 686 = 14R²
⇒ R² = 49
⇒ R = 7.

Let Siya's age = x
Then Riya's age = 4x
But given Riya's age = 24
=> 4x = 24 
=> x = 6
Hence Siya's age = 6 years
=> The age difference = 24 - 6 = 18 years.

Question: The square root of (5 + 2√7)(5 - 2√7) is:

Solution:
√{(5 + 2√7)(5 - 2√7)
= √{5² - (2√7)²}
= √{25 - (4 × 7)}
= √(25 - 28)
= √(- 3)
= √{3 × (- 1)}
= √(3 × i²)  [ i² = - 1]
= i√3

Question: If 0 ≤ θ ≤ 90° and 4 cos²θ - 4√3 cosθ + 3 = 0 then the value of θ is

Solution:
4 cos²θ – 4√3 cosθ + 3 = 0
⇒ (2cosθ)² – 2 · 2 cosθ · √3 + (√3)² = 0
⇒ (2cosθ – √3)² = 0
⇒ 2 cosθ – √3 = 0
⇒ 2 cosθ = √3
⇒ cosθ = (√3)/2
⇒ cosθ = cos 30°
∴ θ = 30°

Question: A train ticket price increased from Tk. 45 to Tk. 60. What is the approximate percentage increase?

Solution:
Increase in ticket price = 60 - 45 = Tk. 15

Percentage increase = (15/45) × 100%
= (1/3) × 100%
= 33.33%
≈ 33%

∴ The approximate percentage increase is 33%.

Question: If 0 ≤ x ≤ 4 and y < 6, which of the following cannot be the value of xy?

Solution:
0 ≤ x ≤ 4; x = 0, 1, 2, 3, 4 এবং y < 6

এখন,
x = 2 হলে,
a) xy = 2 × (- 1) = - 2   [y = - 1]
b) xy = 2 × 0 = 0          [y = 0]
c) xy = 2 × 5 = 10          [y = 5]

কিন্তু,
d) xy = 24 
= 1 × 24
= 2 × 12
= 3 × 8
= 4 × 6 

সুতরাং, 24 সম্ভব নয়। কারণ, প্রশ্নমতে y < 6.

Question: An outlet pipe can empty a cistern in 2 hours and 30 minutes. In what time will it empty 3/5 of the cistern?

Solution: 
The outlet pipe empties the one complete cistern in 2 hours and 30 minutes or 2.5 hours

∴ Time taken to empty 3/5 Part of the cistern = (3/5) × 2.5
= 1.5 hours

Question: Find the midpoint of the line segment joining the points P₁ = (4, -3) and P₂ = (6, 1).

Solution: 
দেওয়া আছে, 
P₁ = (4, -3) এবং P₂ = (6, 1)

আমরা জানি, 
দুটি বিন্দু (x₁, y₁) এবং (x₂, y₂)-এর জন্য মধ্যবিন্দু,
M = {(x₁​ + x₂)/2​​, (y₁​ + y₂​​)/2}

যেখানে, x₁​ = 4, y₁​ = - 3, x₂ ​= 6, y₂ ​= 1

এখন, 
M_x​ = (4 + 6)/2​ = 10​/2 = 5
এবং,
M_y ​= (- 3 + 1​)/2 = - 2​/2 = - 1 

∴ মধ্যবিন্দু = (5, - 1) 

Let the sum lent to C be x
Simple interest on Tk. 1500 at 8% for 4 years + simple interest on x at 8%for 4 years.
= Tk. 1400
(1500 × 8 × 4)/100 + (x × 8 × 4)/100 = 1400
⇒ {8 × 4(1500 + x)}/100 = 1400
⇒ 1500 + x = 4375
⇒ x = 4375 - 1500
⇒ x = 2875.

If Ratul age is x,
then Sahin age is x - 7,
so, (x - 7)/x = 7/9
9x - 63 = 7x
2x = 63
x = 31.5
So, Sahin age is = 31.5 - 7
= 24.5

Question: In a code, 'BREAD' is written as 12345 and 'MOUSE' is written as 67893. How is 'DREAM' written in the same code?

Solution:
BREAD is written as 12345
B = 1, R = 2, E = 3, A = 4, D = 5

MOUSE is written as 67893
M = 6, O = 7, U = 8, S = 9, E = 3

So, DREAM will be 
D = 5, R = 2, E = 3, A = 4, M = 6

∴ DREAM = 52346

Question: If log⁡_m243 + log⁡_m81 = 9, find the value of m.
​
​Solution:
​Given that,
​log⁡_m243 + log⁡_m81 = 9
​⇒ ​​log⁡_m(243 × 81) = 9
⇒ ​​log⁡_m19683 = 9
⇒ ​m⁹ = 19683
​⇒ ​m⁹ = 3⁹
∴ m = 3​

Cost of 21 pencils and 9 clippers = Tk. 819
Cost of 7 pencils and 3 clippers = 819/3 = Tk. 273

r = 50
So consumption should reduced = (r/100+r) × 100
= (50/150) × 100
= 33.33% that means 1/3

Question: If the 10th number in a series of 10 consecutive integers has the value n + 5, find the 1st number in the series expressed in terms of n?

Solution:
এখানে, একটি সিরিজে পরপর 10টি পূর্ণসংখ্যা আছে।
ধরা যাক, সিরিজের প্রথম সংখ্যাটি হলো x
তাহলে, সিরিজটি হবে: x, (x + 1), (x + 2), ..., (x + 9)
প্রশ্ন অনুযায়ী, সিরিজের 10ম সংখ্যাটি হলো n + 5
সুতরাং,
x + 9 = n + 5
⇒ x = n + 5 - 9
⇒ x = n - 4
সুতরাং, সিরিজের প্রথম সংখ্যাটি হলো n - 4

Question: Find the largest number of 5-digits which, when divided by 16, 24, 30 or 36, leaves the same remainder 10 each case.

Solution: 
১৬ = ২ × ২ × ২ × ২
২৪ = ২ × ২ × ২ × ৩ 
৩০ = ২ × ৩ × ৫ 
৩৬ = ৩ × ৩ × ২ × ২

১৬, ২৪, ৩০, ৩৬ এর ল সা গু = ২ × ২ × ২ × ২  × ৩ × ৩ × ৫ 
= ১৬ × ৯ ×৫ 
= ৭২০ 

পাঁচ অঙ্কের বৃহত্তম সংখ্যা ৯৯৯৯৯ কে ৭২০ দ্বারা ভাগ করলে ৬৩৯ অবশিষ্ট থাকে। 
অতএব ৭২০ দ্বারা বিভাজ্য ৫ অঙ্কের বৃহত্তম সংখ্যা = ৯৯৯৯৯ - ৬৩৯ 
= ৯৯৩৬০ 

∴ নির্ণেয় সংখ্যাটি হবে = ৯৯৩৬০ + ১০ 
= ৯৯৩৭০ 
 

Question: If 5 workers can harvest 60 kg of wheat in 3 days, how many kilograms of wheat will 8 workers harvest in 5 days?

Solution: 
5 workers 3 days harvest = 60 kg
1 worker 1 day harvest = (60/15) kg
8 workers 5 days harvest = ( 60 × 40 ) / 15 kg
= 160 kg

Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or, 20×(10/60) = 8/(60(20 + x))
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph.

Question: If a circle has a radius of 5 units, what is its circumference?

Solution: 
circumference = 2πr
=  2π × 5
= 10π units

Question: Speed of a boat in standing water is 18 kmph and the speed of the stream is 3 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is-

Solution:
Given,
Speed of a boat in standing water = 18 kmph
The speed of the stream = 3 kmph

∴ Speed upstream = (18 - 3) kmph
= 15 kmph.

∴ Speed downstream = (18 + 3) kmph
= 21 kmph.

So, Total time taken = (105/15 ) + (105/21) hours
= (7 + 5) hours
= 12 hour

Let the numerator be x.
The denominator of a fraction is 2 more than the numerator. Therefore, denominator = x + 2
Now,
Numerator/Denominator = 8/10
x/(x + 2) = 8/10

The numerator and denominator are increased by 4.

Therefore,
(x + 4)/{(x + 2) + 4} = 8/10
⇒ 10 (x + 4) = 8 ( x + 6)
⇒ 10 x + 40 = 8x + 48
⇒ 2x = 8
⇒ x = 4

Hence, the fraction is = 4/6
= 2/3.

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
n(E)= 10 + 8 - 1 (since 2 is common in both item)
= 17
P(E U P) = n(E)/n(s)
= 17/20 + 8/20 - 1/20
= 17/20

Question: 'A' began a business with Tk. 85000. He was joined afterwards by 'B' with Tk. 42500. For how much period does 'B' join, if the profits at the end of the year are divided in the ratio of 3 : 1 ?

Solution: 
Suppose 'B' joined for x months.

Now,
A's capital = Tk. 85000, time = 12 months
B's capital = Tk. 42500, time = x months

According to the question,
A's share : B's Share = 3 : 1
⇒ (85000 × 12)/(42500 × x) = 3/1
⇒ x = (85000 × 12)/(42500 × 3)
⇒ x = 8
∴ 'B' joined for 8 months.

Since both trains are of equal length and cross each other in 10 seconds with relative speed, we get,
∴ Distance = 2x (because both trains together cover each other’s lengths)
∴ Time = distance/speed
⇒ 10 = 2x/50
⇒ x = 250 m
So, length of the first train = 250 meters.

Therefore, the time taken to cross the platform by the first train,
⇒ t = (250 + 350)/[72 × (5/18)]
∴ t = 600/20 = 30 sec.

Let the radius of the smaller circle be r cm
Then, radius of bigger circle = (r + 14) cm
Given, π(r+14)² - π(r)² = 1056
⇒ π(r² + 28r + 196) – πr² = 1056
⇒ 28πr + 196π = 1056
⇒ r + 7 = 1056/28π = 12
∴ r = 5

Question: A box contains 5 green, 3 yellow, and 2 black balls. If one ball is drawn at random, what is the probability that it will not be yellow?

Solution:
Green balls = 5
Yellow balls = 3
Black balls = 2

∴ Total balls = 5 + 3 + 2 = 10
Number of non-yellow balls = Green + Black = 5 + 2 = 7

We know,
Probability,
P(not yellow) = favorable outcomes/total outcomes
= 7/10

∴ The probability that the drawn ball will not be yellow is 7/10.

Question: In a class of 120 students, the average score in science is 75. If the 70 girls scored an average of 78, what is the average score of the remaining boys?

Solution:
ধরি, ছাত্রদের গড় নম্বর = x

120 জন শিক্ষার্থীর মোট নম্বর = 120 × 75 = 9000
70 জন ছাত্রীর মোট নম্বর = 70 × 78 = 5460

প্রশ্নমতে,
5460 + (120 - 70) × x = 9000
⇒ 5460 + 50x = 9000
⇒ 50x = 9000 - 5460
⇒ 50x = 3540
⇒ x = 3540/50
⇒ x = 70.8

∴ 50 জন ছাত্রের গড় নম্বর = 70.8

Question: What is the reflex angle between the hands of a clock at 10:20?

Solution:
The angle between the hands of the clock is |11M - 60H|°/2
= |(11 × 20) - (60 × 10)|°/2
= |220 - 600|°/2
= |- 380|°/2
= 190°

∴ Reflex angle is 190°

Question: Fahim is travelling to City B. He calculated that if he travels at 20 km/h, he will reach there at 3 : 00 p.m., but if he travels at 30 km/h, he will reach there at 1 : 00 p.m. At what speed must he travel to reach City B exactly at 2 : 00 p.m.?

সমাধান:
ধরি,
ফাহিমের অতিক্রান্ত মোট দূরত্ব হলো x কিমি।

20 কিমি/ঘন্টা গতিতে এবং 30 কিমি/ঘন্টা গতিতে পৌঁছানোর সময়ের পার্থক্য = 3 : 00 p.m. - 1 : 00 p.m. = 2 ঘন্টা।

প্রশ্নমতে,
x/20 - x/30 = 2
⇒ (3x - 2x)/60 = 2
⇒ x/60 = 2
⇒ x = 120 কিমি।

20 কিমি/ঘন্টা গতিতে 120 কিমি যেতে সময় লাগে = 120/20 = 6 ঘন্টা।

যেহেতু এই গতিতে সে 3 : 00 p.m. এ পৌঁছায়, তাই যাত্রা শুরুর সময় ছিল:
3 : 00 p.m. - 6 ঘন্টা = 9 : 00 a.m.

9:00 a.m. এ শুরু করে 2:00 p.m. এ পৌঁছানোর জন্য প্রয়োজনীয় সময় = 5 ঘন্টা।

∴ প্রয়োজনীয় গতিবেগ = দূরত্ব/প্রয়োজনীয় সময়
= 120 কিমি 5 ঘন্টা
= 24 কিমি/ঘন্টা।
∴ ফাহিমকে গড়ে 24 কিমি/ঘন্টা গতিতে যেতে হবে।

Question: A garrison of 600 men has provisions to last 30 days. After 3 days, 300 additional men arrive. How many days will the remaining food now last?

Solution:
After 3 days, food having for = (30 - 3) = 27 days
After arriving 300 men, total men = (600 + 300) = 900 men

600 men can eat the food for 27 days
∴ 1 man can eat the food for (27 × 600) days
∴ 900 men can eat the food for (27 × 600)/900 days
= 18 days

Question: A train travels between X and Y in 3 hours. When the speed of train is increased by 6 km/hr, then it covers the same distance in 2 hours. What is the original speed of train?

Solution: 
Let the original speed of the train be a km/hr.
The distance between X and Y is the same in both cases.

Now, given that, 
Speed = a km/hr
Time = 3 hours
∴ Distance = a × 3 = 3a km

And,
Speed = (a + 6) km/hr
Time = 2 hours
∴ Distance = (a + 6) × 2 = 2(a + 6) km

Since distance is the same. Then we get,
⇒ 3a = 2(a + 6)
⇒ 3a = 2a + 12
⇒ 3a - 2a = 12
∴ a = 12

So the original speed of the train is 12 km/hr

Question: A passenger travels from Dhaka to Cumilla at a speed of 30 kmph and return with a speed of 60 kmph. What is the average speed?

Solution:
Let the total distance be 60 km [L.C.M. of 30 kmph and 60 kmph]

Time to cover 60 km distance at the speed of 30 km/hr = 60/30 = 2 hrs
Time to cover 60 km distance at the speed of 60 km/hr = 60/60 = 1 hr

Total distance = 60 + 60 = 120 km

Total time = 2 hrs + 1 hrs = 3 hrs

∴ Average speed = 120/3 = 40 km/hr

Let p be the number
p ÷ 44 = 432, remainder = 0
⇒ p = 432 × 44 + 0 = 19008
p ÷ 31
= 19008 ÷ 31
= 613, remainder = 5.

Question: If α and β are the roots of the equation 4x² - 25x + 36 = 0, then the value of αβ equals:

Solution:
4x² - 25x + 36 = 0
⇒ 4x² - 16x - 9x + 36 = 0
⇒ 4x(x - 4) - 9(x - 4) = 0
⇒ (x - 4)(4x - 9) = 0
হয়, x - 4 = 0 ⇒ x = 4
অথবা, 4x - 9 = 0 ⇒ x = 9/4

অর্থাৎ, α = 4 এবং β = 9/4
∴ αβ = 4 × (9/4) = 9

Shortcut:
আমরা জানি, ax² + bx + c = 0 সমীকরণের মূলদ্বয়ের গুণফল αβ = c/a
এখানে, a = 4 এবং c = 36
∴ αβ = 36/4 = 9

Question: A lead ore mine contains 50% metal. Of this metal, 4/5% is gold and the remainder is lead. If 6000 kg of ore is extracted, what is the mass of the lead?

Solution:
Given mass of lead ore = 6000 kg 

The mass of metal is 50% of 6000 = 3000 kg 
∴ Mass of gold is 4/5% of 3000
= (4 × 3000)/(5 × 100)
= 24 kg 

∴ Mass of lead = Mass of metal - Mass of gold
= 3000 - 24 
= 2976 kg

Let the total capacity of the tank is 30 units.

The efficiency of Leakage(Pipe A) will be = 30/10 = 3

And the efficiency of the leakage (Pipe A) and another Pipe (B) which is filling the tank will be = 30/15 = 2
Pipe A is emptying at 3 units/hr and when filling pipe B started then the emptying rate will come down to 2 units/hr.

∴ Filling Pipe B efficiency is 3 - 2 = 1unit/hr
Pipe B will be fill the tank in = 30/1 = 30 hrs.
Filling rate of Pipe B per minute is 4 litter

∴ Total Capacity of tank will be = (4 × 60) × 30 = 7200 litters.

Take the LCM of days = LCM of (20, 30, and 60) = 60
Let the total work = 60

1 Man's one day work = 60/20 = 3 unit
1 Woman's one day work = 60/30 = 2 unit
1 The boy's one day work = 60/60 = 1 unit........................ (i)

ATQ, (2Men + 8Women + x Boy) can finish a work in 2 days.

Or, (2Men + 8women)'s 1 day work = (2 × 3) + (8 × 2) = 22

Or, (2Men + 8women)'s 2 day work = 22 × 2 = 44
Remaining work = 60 - 44 = 16
That means the remaining 16 work is done by X boys in 2 days
Or, X boys' one day work = 16/2 = 8

By equation 1, 1 boy completes 1 unit of work per day
So, 8 boys are required to complete 8 units of work per day.

Question: A boat goes 20 km upstream and 44 km downstream in 8 hours. In 5 hours, it goes 15 km upstream and 22 km downstream. Determine the speed of the boat in still water.

Solution: 
Let,
Upstream speed = U km/h
Downstream speed = D km/h

Then we get speed of boat = (U + D)/2

Now,
According to the question,
20/U + 44/D = 8 ....… (i)
15/U + 22/D = 5 ....… (ii)

Now, multiply by 2 the equation (ii) then subtract from equation (i) we get
20/U + 44/D = 8
30/U + 44/D = 10
⇒ - 10/U = - 2
⇒ 2U = 10
⇒ U = 10/2 = 5
∴ U = 5 km/hr

Putting the value in equation (i), we get
20/5 + 44/D = 8
⇒ 44/D = 8 - 4
⇒ 4D = 44
⇒ D = 44/4
∴ D = 11

So, the speed of boat = (U + D)/2 = (5 + 11)/2 = 8 km/hr

So the speed of the boat in still water is 8 km/h.

Question: If θ = 60°, then sec²θ - tan²θ = ?

Solution: 
Given, θ = 60°

Now,
sec²θ - tan²θ
= (sec60°)² - (tan60°)²
= 2² - (√3)²
= 4 - 3
= 1

Let,
The son's present age is x years.
Then, the man's present age is = (x + 24)
∴ (x + 24) + 2 = 2(x+ 2)
⇒ x + 26 = 2x + 4
⇒ x = 22

Question: A certain sum of money becomes 2.5 times itself in 5 years at simple interest. What is the rate of interest per annum? 

Solution: 
Let the principal amount be P.
The sum becomes 2.5 times itself in 5 years.

So, A = 2.5P
∴ Simple Interest, I = 2.5P - P = 1.5 P

We know, I = Pnr/100
⇒ 1.5 P = (P × 5 × r)/100
⇒ 5r = 150
∴ r = 30%