Bank Math

Question: A jar contains 6 white balls, 4 red balls, and 2 black balls. Two balls are drawn one after the other without replacement. Find the probability that both balls drawn are white.

Solution:
A jar contains 6 white balls, 4 red balls and 2 black balls.
∴ Total balls = 6 + 4 + 2 = 12
Two balls are drawn successively without replacement.

Now, 
Probability first ball is white = 6/12 = 1/2
After drawing one white ball, 5 white balls remain and the total number of balls remaining = 11
Probability second ball is white = 5/11

Since the draws are dependent (without replacement), multiply the probabilities.
∴ P(both white) = (6/12) × (5/11)
= (1/2) × (5/11)
= 5/22

So the probability that both balls are white is 5/22.

Question: The compound interest on Tk. 30000 at 7% per annum is Tk. 4347. The period (in years) is-

Solution:
Given that,
Principal, P = Tk. 30000
Compound Interest, CI = Tk. 4347
Rate, r = 7% per annum
And, Amount, A = P + CI = 30000 + 4347 = 34347

We know,
A = P(1 + r/100)ⁿ
⇒ 34347 = 30000(1 + 7/100)ⁿ
⇒ (107/100)ⁿ = 34347/30000
⇒ (107/100)ⁿ = 11449/10000
⇒ (107/100)ⁿ = (107/100)²
∴ n = 2 

Hence, the period = 2 years.

If 'w₁' work is done by 'm₁' men by working for 'h₁' hours per day in 'd₁' days & 'w₂' is work done by men 'm₂' working for 'h₂' hours per day in 'd₂' days,
then,
m₁d₁h₁/w₁ = m₂d₂h₂/w₂

Since we need to find 'd₂', we can rearrange the formula as
d2 = (m₁d₁h₁w₂)/(m₂d₂h₂w₁)
= (6 x 7 x 7 x 18)/(14 x 9 x 12)
= 3.5 days

ধরি, ক্রয়মূল্য 100 টাকা
12.5% ক্ষতিতে বিক্রয়মূল্য (100 - 12.5) = 87.5 টাকা
ক্রয়মূল্য : বিক্রয়মূল্য = 100 : 87.5 = 8 : 7

Let C = x
Then B = x/2
and A = x/4
A : B : C = 1 : 2 : 4
C's share = Tk. (4/7 × 700) = 400

Surface area of sphere = 4πr² 
If the new radius is 20% increased, then new surface area will be = 4π(1.2)²  = 5.76πr² 

Surface area Increased in percentage = (5.76πr²/4πr² × 100) - 100 =  144 - 100 = 44%

Question: Robi and Bobi together can paint a wall in 16 days. Robi can do it alone in 20 days. How many days would it take Bobi to do this work alone?

Solution:
Robi's 1 day's work 1/20
Robi's and Bobi's 1 day's work 1/16
∴ Bobi's 1 day's work = (1/16) - (1/20)
= (5 - 4)/80
= 1/80

1/80 part of the job done by Bobi in 1 day
∴ Full work done by Bobi in (80/1) days
= 80 days

Question: Rakib can do a piece of work in 15 days, Rakib and Asif together can do in 12 days. If Asif does the work only for half a day daily then in how many days the work will be completed ?

Solution: 

Rakib's 1 day work = 1/15

Since, Rakib and Asif can together complete in 12 days
i.e. (Rakib + Asif)'s 1 day work = 1/12

Then,
Asif's 1 day work,
= (1/12) - (1/15) = (5 - 4)/60 = 1/60

If Asif Works only for half a day daily, then his 1 day work becomes (1/2) × (1/60)
  = (1/120)

Therefore, 1 day work of both Rakib and Asif,
=(1/15) + (1/120) = ( 8 + 1)/120 = 9/120

Hence, the work will be completed in 120/9 = 40/3 days

Question: What will be the result if (4x + 20)/4 is subtracted from (x + 10)?

Solution:
Expression = (x + 10) - {(4x + 20)/4}
= (x + 10) - {4(x + 5)/4}
= (x + 10) - (x + 5)
= x + 10 - x - 5
= 5

Question: A square and a circle have the same perimeter. The side length of the square is 11 cm. What is the area of the circle?

Solution:
দেওয়া আছে,
বর্গক্ষেত্রের এক বাহুর দৈর্ঘ্য, a = 11 সে.মি.
∴ বর্গক্ষেত্রের পরিসীমা = 4 × a
= 4 × 11
= 44 সে.মি.

প্রশ্নমতে,
বৃত্তের পরিধি = বর্গক্ষেত্রের পরিসীমা
∴ 2πr = 44
⇒ 2 × (22/7) × r = 44
⇒ (44/7) × r = 44
⇒ r = 44 × (7/44)
∴ r = 7 সে.মি.

এখন,
বৃত্তের ক্ষেত্রফল = πr²
= (22/7) × 7²
= (22/7) × 49
= 22 × 7
= 154 বর্গ সে.মি.

অতএব, বৃত্তের ক্ষেত্রফল = 154 বর্গ সে.মি.

Question: A ladder leans against a vertical wall making an angle of 60° with the ground. If the foot of the ladder is 4.6 m away from the base of the wall, find the length of the ladder.

Solution:
 
Let AB be the wall and BC be the ladder.

 Then, ∠ACB = 60°
and AC = 4.6 m

We know,
cos∠ACB = AC/BC
⇒ cos60° = AC/BC
⇒ AC/BC= 1/2
⇒ BC = 2 × AC
⇒ BC = 2 × 4.6
∴ BC = 9.2 m

Question:

Solution:

Question: What is the least number which when divided by 3, 5, 6, 8, 10, and 12 leaves a remainder 2 in each case, but when divided by 22 leaves no remainder?

Solution:
The number, when divided by 3, 5, 6, 8, 10, and 12 leaves a remainder of 2. This means the number is 2 more than a multiple of their L.C.M.
LCM of 3, 5, 6, 8, 10, 12 = 120

The required number, N, must be in the form:
N = 120K + 2, where K is a positive integer.

N divisible by 22 ⇒ (120K + 2) ÷ 22 has remainder 0

120K + 2 = 22 × m ⇒ Check for smallest K

Try K = 2 ⇒ 120 × 2 + 2 = 242, divisible by 22

Hence, required number = 242

Question: An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of copper and lead per kg in new alloy will be-

Solution:
Ratio of Zinc, Copper and Tin is given as, Z : C : T = 2 : 3 : 1 = 4 : 6 : 2
Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Tin)
Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3 (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead)

Alloys are mixed together to form third alloy. Then the ratio of content in it,
Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3 = 4 : 11 : 6 : 3

 Weight of third alloy = 12 + 12 = 24 Kg.
∴ Weight of Copper = 11/24 
 And, weight of Lead = 3/24 
= 1/8

Question: A train takes 8 seconds to cross a pole and 18 seconds to cross a platform of length 120m. What is the length of the train?

Solution:
Let the length of the train be L meters.

Now,
Time to cross a pole = 8 s
Distance covered = length of train = L
∴ Speed = L/8 m/s

Again, 
Platform length = 120 m
Distance covered = L + 120
Time taken = 18 s
∴ Speed = (L + 120)/18 m/s

ATQ, 
L/8 = (L + 120)/18
⇒ 18L = 8L + 960
⇒ 18L - 8L = 960
⇒ 10L = 960
⇒ L = 960/10
∴ L = 96 m

So the length of the train is 96 meters.

If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required greatest number.
34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2

Question: The difference between two positive numbers is 8 and the difference of their squares is 160. What is the smallest number?

Solution:
Let the numbers be x and (x + 8)

According to the question,
(x + 8)² - x² = 160
⇒ x² + 16x + 64 - x² = 160
⇒ 16x + 64 = 160
⇒ 16x = 96
⇒ x = 6

∴ The smallest number is = 6

The ratio of speeds of A, B, C = 6 : 3 : 1
The ratio of time taken by A, B, C = 1/6 : 1/3 : 1
To simplify it, we will multiply it by the LCM of ratio of speeds given.
Hence, the ratio of time taken by A, B, C = 1 : 2 : 6

[Speed is inversely proportional to time, meaning if speed increases, time decreases. So, ratio of time would be reciprocal of the ratio of speed given. ]

Time taken by C to covered given distance = 78 = 6 × 13
The ratio of time of A and C = 1 : 6
Thus, time taken by A = 13 min.

Alternative method:
Let C's speed be x metres/min
Let time taken by A be y min
Then B's speed = 3x metres/min
And, A's speed = 6x metres/min
Ratio of speed of A and C = Ratio of time by C and A
6x : x = 78 : y
6x/x = 78/y
y = 13 min

Just too much information is given in the question to confuse. This is a straight and simple question
Market Value of Company X (his selling price) = Tk. 30
Total shares sold = 4000
The amount he gets = Tk. (4000 × 30)
He invests this amount in ordinary shares of Company Y
Market Value of Company Y(His purchasing price) = 15
Number of shares of company Y which he purchases = (4000 × 30)/15
= Tk. 8000.

Question: The volume of a right circular cylinder is 25π cubic units and its height is 4 units. What is the circumference of its base?

Solution:
আমরা জানি, একটি সিলিন্ডারের আয়তন = πr²h
যেখানে, r হলো ভূমির ব্যাসার্ধ এবং h হলো উচ্চতা।

প্রশ্নমতে,
πr² × 4 = 25π
⇒ 4r² = 25
⇒ r² = 25/4
⇒ r = √(25/4)
⇒ r = 5/2 = 2.5 একক

সিলিন্ডারের ভূমির পরিধি = 2πr
= 2π × 2.5
= 5π একক

∴ সিলিন্ডারটির ভূমির পরিধি হলো 5π একক।

Question: If x and y are whole numbers such that x^y = 64, then the value of (x - 2)^{y + 1} is = ?

Solution:
দেওয়া আছে,
x^y = 64
⇒ x^y = 8²
এখানে, x = 8 এবং y = 2

এখন,
(x - 2)^{y + 1} = (8 - 2)^{2 + 1}  [মান বসিয়ে]
= 6³
= 216

Question:  If 1 + sinθ = mcosθ than what is the value of cotθ? 

Solution:
Given that,
1+ sinθ = m cos θ
⇒ (1 + sinθ)/cosθ = m
⇒ (1/cosθ) + (sinθ/cosθ) = m
∴ secθ + tanθ = m ...............(i)

We know,
(secθ + tanθ) (secθ - tanθ) = 1
⇒ m(secθ - tanθ) = 1
⇒ secθ - tanθ = 1/m .................(ii)

Now, (i) - (ii) ⇒
secθ + tanθ - (secθ - tanθ) = m - (1/m)
⇒ secθ + tanθ - secθ + tanθ = (m² - 1)/m
⇒ 2tanθ = (m² - 1)/m
⇒  tanθ = (m² - 1)/2m
⇒  1/cotθ = 1/{(m² - 1)/2m}
∴ cotθ = 2m/(m² - 1)

Let, distance from A to B = x km
ATQ, 
x/40 + x/30 + 1 = 15
Or, (3x + 4x)/120 = 15 - 1
Or, 7x = 14 × 120
Or, x = 240
∴ distance = 120 km

Question: A 30-meter pole has fractured and bent over to make a 30° angle with the ground, remaining partially attached. How high from the base did it break?

Solution:

ধরি,
খুটিটি x মিটার উচুতে ভেঙ্গেছিল।
∴ অপর ভাঙ্গা অংশের দৈর্ঘ্য = (30 - x) মিটার

এখন, 
sin θ = লম্ব/অতিভুজ
বা, sin θ = x/(30 - x)
বা, sin 30° = x/(30 - x)
বা, 1/2 = x/(30 - x)
বা, 2x = 30 - x
বা, 2x + x = 30
বা, 3x = 30
⇒ x = 10

∴ খুটিটি ভূমি থেকে 10 মিটার উচুতে ভেঙ্গেছিল।

Greatest number of four digits = 9999
LCM of 15, 25, 40 and 75 = 600
9999 ÷ 600 = 16, remainder = 399
Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75 = 9999 - 399 = 9600

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 and Days 1 : 2 } :: 8:x
=> (4 × 1 × x) = (3 × 2 × 8)
=> x = 12

p < 1
⇒ 1/p > 1
⇒ 2/p > 2
2/p - p > 2 - p > 0
[∵ p < 1]
Hence, (2/p - p) is a positive number.

Let distance travelled by cat before dog catches it be D
We know, time for which Dog and Cat ran is same
∴ T = T
∴ D/5 = (D + 80)/7 [D = S x T]
∴ D = 200 m

Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) =8000 x 1000/320 x 250= 100
 x = 10 cm = 1 dm.

Question: The difference between the length and breadth of a rectangle is 23 m. if its area is 2520 m², then its perimeter is:

Solution:
Let,
The breadth is x m
The length is x + 23 m 

ATQ,
x(x + 23) = 2520
⇒ x² + 23x - 2520 = 0
⇒ x² + 63x - 40x - 2520 = 0
⇒ x(x + 63) - 40(x + 63) = 0
⇒ (x + 63)(x - 40) = 0
∴ x = - 63 Or x = 40 
We ignore the negative value.
so, x = 40

∴ The breadth is 40 m
∴ The length is 40 + 23 m = 63 m

∴ The perimeter is 2(63 + 40)m = 2 × 103 m = 206 m 

Question: If 4x² - 6x + 1 = 0, then the value of 8x³ + 1/8x³ is- 

Solution:

Question: The sum of five consecutive multiples of 3 is 165. What is the largest number?

Solution:
ধরি, পাঁচটি ক্রমিক 3 এর গুণিতক যথাক্রমে x, (x + 3), (x + 6), (x + 9) এবং (x + 12).

প্রশ্নমতে,
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 165
⇒ 5x + 30 = 165
⇒ 5x = 165 - 30
⇒ 5x = 135
⇒ x = 135/5
⇒ x = 27

∴ বৃহত্তম সংখ্যা = x + 12
= 27 + 12 = 39

Question: The average of five numbers is 32. The sum of the first three numbers is 75. If the fourth number is 28, what is the fifth number?

Answer:
Average of five numbers = 32
Total sum = 5 × 32
= 160

Sum of first three numbers = 75.
Fourth number = 28.

Sum of first four numbers = 75 + 28 = 103.

∴ Fifth number = total sum - sum of first four 
= 160 - 103
= 57

So the fifth number is 57.

The Distance covered by the motorcycle with speed 60km/hr in 3 hours = (60 x 3)
= 180 km
Now,
Speed = Distance/Time
Since the train covers the same 180 km in 3/2 hours (we can write 1 and half hours as 3/2hours)
Then the speed of the train = 180/ (3/2)
= 180 × (2/3)
= 120 km/hr.
Hence, the train travelled at the speed of 120km/hr to cross 180km in 1 and half hour.

Radius of incircle
=a/2√3
=42/ 2√3
=7√3
Area of incircle
=22/7×49×3
=462 cm²

a² + b²
= 1/2{(a + b)² + (a - b)²}
= 1/2(13² + 3²)
= 1/2(169 + 9)
= 89

Question: A pipe can fill a cistern in 20 hours. Once the cistern is half full, three additional identical pipes are opened. How long will it take to fill the cistern completely?

Solution:

Work done by 1 pipe in 1 hour = 1/20
∴ Time to fill half the cistern with 1 pipe = (1/2) ÷ (1/20) = 10 hours

After cistern is half full,
three additional identical pipes are opened, 
total pipes = 4
Work done by 4 pipes in 1 hour = 4 × (1/20) = 1/5
Time to fill remaining half = (1/2) ÷ (1/5)
= 2.5 hours
= 2 hours 30 minutes

∴ Total time to fill cistern = 10 + 2.5
= 12.5 hours = 12 hours 30 minutes