ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: Two trains of equal length are running on parallel lines in the same direction at 46 km and 36 km per hour. The faster train passes the slower train in 36 seconds. The length of each train is-

Solution:
To cross each other, two trains have to cover a distance equal to the sum of the lengths of the train.
Let the length of the trains be = a m each.

So the distance to be covered = 2a
Now the trains are running int he same direction.
∴ Their relative speed = (46 - 36) km/hr.
= 10km/hr. = 10 × (5/18) km/hr. = (25/9) m/sec.

So, the time taken by the trains to cove 2a m distance
= 2a ÷ (25/9) sec

∴ By the given conditions,
2a ÷ (25/9) = 36
⇒ 2a × (9/25) = 36
⇒ 2a = (36 × 25)/9
⇒ 2a = 100
∴ a = 50

So, the length of each train = 50 m.

Question: X can complete a piece of work in 18 days, Y in 20 days and Z in 30 days, Y and Z together start the work and forced to leave after 2 days. The time taken by X alone to complete the remaining work is:

Solution:
(Y + Z) 2 day's work:
= 2 × (1/20) + (1/30)
= 2 × (3 + 2)/60
= 1/6 part

Remaining work = 1 - (1/6)
= 5/6 part

X's one day's work = 1/18 part

Time taken to complete the work = (5/6)/(1/18) days

∴ Time taken to complete the work = (5/6) × 18
= 15 days

Question: Three runners P, Q and R run a race, with runner P finishing 15 metres ahead of runner Q and 24 metres ahead of runner R, in another race of same type runner Q finished 10 metres ahead of runner R. Each runner travels the entire distance at a constant speed. The length of the race is?

Solution:
Let P finish the race of = x metres
Q finish the race of = (x - 15) meters.
R finish the race of = (x - 24) ............. (1)

In another race of Q & R
Q finish race of = x meters.
R finish race of = (x - 10) .............. (2)

Ratio of speeds of Q & R,
(x - 15)/(x - 24) = x/(x - 10)
⇒ (x - 15)(x - 10) = x(x - 24)
⇒ x² - 15x - 10x + 150 = x² - 24x
⇒ x² - 25x + 150 = x² - 24x
⇒ 150 = x
∴ x = 150 metres

Question: The radius of a circle is same as the diagonal of a square whose area is 36 sq. cm. The area of the circle is-

Solution:
Area of square = 36
Side of square = √36 = 6

Diagonal of square = 6√2

So, the radius of the circle is = 6√2 cm

∴ Area of circle = πr²
= π(6√2)²
= π(36 × 2)
= 72π cm²

Question: Two trains 150 m and 180 m long run at the speed of 72 km/hr and 36 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is:

Solution:
Relative speed = (72 + 36) km/hr
= 108 × (5/18) m/sec
= 30 m/sec

Distance covered in crossing each other = (150 + 180) m
= 330 m.

Required time = 330/30 sec
= 11 sec

Question: If 2x + 3y = 10 and 4x - y = 6, find the value of 5y - 2x.

Solution:
2x + 3y = 10 ...............(1)
4x - y = 6
⇒ 2(4x - y) = 2 × 6
⇒ 8x - 2y = 12 .................(2)

(1) + (2) ⇒
2x + 3y + 8x - 2y = 10 + 12
⇒ 10x + y = 22
⇒ y = 22 - 10x .................(3)

Putting the value of y in (1)
2x + 3(22 - 10x) = 10
⇒ 2x + 66 - 30x = 10
⇒ - 28x = 10 - 66
⇒ - 28x = - 56
∴ x = 2

Putting x = 2 in (3),
y = 22 - 10 · 2
⇒ y = 22 - 20
∴ y = 2

Now,
5y - 2x = (5 × 2) - (2 × 2)
= 10 - 4
= 6

Question: Sunday, Monday, Wednesday, Saturday, Wednesday, Monday, .....

Solution:
Sunday to Monday = no gap
Monday to Wednesday = One day gap.
Wednesday to Saturday = Two days gap.
Saturday to Wednesday = Three days gap.
Wednesday to Monday = Four days gap.
In the next term there must be five days and six days gap.
So, Next term would be Sunday, Sunday.

Question: Two pipes can fill a tank in 30 and 40 minutes respectively, and a waste pipe can empty 5 gallons per minute. All three pipes working together can fill the tank in 20 minutes. What is the capacity of the tank?

Solution:
Let the waste pipe empty the tank in = x minutes.

According to the question,
(1/30) + (1/40) - (1/x) = (1/20)
⇒ 1/x = (1/30) + (1/40) - (1/20)
⇒ 1/x = (4 + 3 - 6)/120
⇒ 1/x = 1/120
∴ x = 120 minutes

A waste pipe can empty 5 gallons per minute
In 120 minutes it can empty = (5 × 120) = 600 gallons.

∴ Capacity of the tank = 600 gallons.

Question: The factors of the polynomial a³ - 6a² + 12a - 9 are:

Solution:
a³ - 6a² + 12a - 9
= a³ - 3 . a² . 2 + 3 . a . 2² - 2³ - 1
= (a - 2)³ - 1³
= (a - 2 - 1) {(a - 2)² + (a - 2) . 1 + 1²}
= (a - 3) (a² - 4a + 4 + a - 2 + 1)
= (a - 3) (a² - 3a + 3)

Question: Emon takes twice as much time as Rayhan to complete a work and Hemel does it in the same time as Emon and Rayhan together. If all three working together can finish the work in 9 days, then the time taken by Emon to finish the work is-

Solution:
Let,
Rayhan takes x days to complete a work
Then, Emon takes 2x days to complete the work

Rayhan's 1 day's work = 1/x
Emon's 1 day's work = 1/2x
Hemel's 1 day's work = (1/x) + (1/2x) = 3/2x

∴ (Emon + Rayhan + Hemel)'s 1 day's work
= (1/x) + (1/2x) + (3/2x)
= (2 + 1 + 3)/2x
= 3/x

ATQ,
3/x = 1/9
⇒ x = 27

Hence, Emon takes (2 × 27) = 54 days to complete the work.

Question: The area of a trapezium is 84 square cm. The lengths of its parallel sides are 14 cm and 10 cm. What is the distance between the parallel sides?

Solution:
We know,
The area of a trapezium = (1/2) × Sum of the lengths of the parallel sides × Distance between the parallel sides.

Let, the distance between the parallel sides be = d.

Then,
d = (2 × Area of the trapezium)/Sum of the lengths of the parallel sides
= (2 × 84)/(14 + 10)
= 168/24
= 7 cm

Thus, the distance between the parallel sides 7 cm.

Question: If x = b + c - 2a, y = c + a - 2b, z = a + b - 2c, then the value of x² + y² - z² + 2xy is?

Solution:
Given,
x = b + c - 2a,
y = c + a - 2b,
z = a + b - 2c
∴x + y + z = (b + c - 2a) + (c + a - 2b) + (a + b - 2c)
= 0

 Now,
⇒ x² + y² + 2xy - z²
= (x + y)² - z² 
= (x + y - z) (x + y + z)  [(a² - b²) = (a + b) (a - b)]
= (x + y - z) × 0
= 0

Question: Which number will complete the series:
1, 3, 7, 15, 31, 63, 127, 97, 255, __?

Solution:
3 - 1 = 2
7 - 3 = 4 = 2 × 2
15 - 7 = 8 = 4 × 2
31 - 15 = 16 = 8 × 2
63 - 31 = 32 = 16 × 2
127 - 63 = 64 = 32 × 2
225 - 97 = 128 = 64 × 2

∴ The next number of 127 will be = (255 + 128 × 2)
= 255 + 256
= 511

Question: A man starts from a point and travels 12 km in the north direction. Again he went east for 6 km and changed direction and again went north for 8 km. Finally went west for 6 km. What is the direct distance between his destination and journey?

Solution:

ঐ ব্যক্তির যাত্রাস্থান A এবং গন্তব্য স্থান = E
∴ তার গন্তব্য স্থান ও যাত্রা স্থানের সরাসরি দূরত্ব, AE = (12 + 8) কি.মি.
= 20 কি.মি.

Question: The equation 12x² + 4kx + 3 = 0 has real and equal roots, if-

Solution:
Here a = 12, b = 4k, c = 3
Since the given equation has real and equal roots
∴ b² - 4ac = 0
⇒ (4k)² - 4 × 12 × 3 = 0
⇒ 16k² - 144 = 0
⇒ 16k² = 144
⇒ k² = 9
⇒ k = ± 3

Question: How many triangles are there?


Solution:

The simplest triangles are AEH, EHI, EBF, EFI, FGC, IFG, DGH and HIG.
so, there are 8 triangles

The triangles composed of two components each are HEF, EFG, HFG and EFG
so, there are 4 triangles

Thus, there are = 8 + 4
= 12 triangles in the figure.

Question: REF, SGH, TIJ, ____, VMN.

Solution:
There are two alphabetical series here.
The first series is with the first letters only: R, S, T, U, V.
The second series involves the remaining letters: , EF, GH, IJ, KL, MN.

∴ Answer will be = UKL

Question: A and B can do a work in 12 days. B and C can do it in 15 days. A and C can do it in 20 days. If all of them work together, in how many days can they finish the work?

Solution:
(A + B), 1days works = 1/12,
(B + C), 1days work = 1/15,
(C + A), 1 days work = 1/20

Now,
2(A + B + C) 1days work = (1/12) + (1/20) + (1/15)
= (5 + 3 + 4)/60
=12/60
=1/5

 (A + B + C) 1 days works = 1/10
∴ A, B and C together can do it in = 10 days

Question: The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 Km/hr. At what time do they meet?

Solution:
Suppose they meet x hrs after = 8 a.m

Then,
[Distance moved by first in x hrs] + [Distance moved by second in (x - 1) hrs] = 330

Therefore, 60x + 75(x - 1) = 330
⇒ 60x + 75x - 75 = 330
⇒ 135x = 405
∴ x = 3

So,they meet at = (8 + 3)
= 11 a.m.

Question: Two pipes A and B together can fill a tank in 6 hours. If pipe A can fill 5 hours faster than pipe B, in how many hours pipe B alone can fill the tank?

Solution:
Pipe A and B can fill the tank = 6 hours

Let
Pipe B can fill the tank in = x hours
Pipe A will take = (x - 5) hours

Now
(1/x) + {1/(x - 5)} = 1/6
⇒ (x - 5 + x)/{x(x - 5)} = 1/6
⇒ 6(x - 5 + x) = x² - 5x
⇒ 12x - 30 = x² - 5x
⇒ x² - 17x + 30 = 0
⇒ x² - 15x - 2x + 30 = 0
⇒ x(x - 15) - 2(x - 15) = 0
⇒ (x -15)(x - 2) = 0
x = 15 and x = 2

If x = 2, A = - 3 and time cannot be negative
So x = 15 hours

Question: 3x + (1/2x) = 5, then the value of 8x³ + (1/27x³)  is?

Solution:

Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:

ধরি,
খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC = 20√3 m

এখন,
ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
⇒ tanθ = AB/BC
⇒ tanθ = 20/(20√3)
⇒ tanθ = 1/√3
⇒ tanθ = tan30°
∴ θ = 30°

Question: The speed of a boat in still water is 15 km/hr. If it can travel 40 km downstream and 20 km upstream at the same time, the speed of the stream is-

Solution:
Let the speed of the stream be = x km/hr.
Then speed downstream = (15 + x) km/hr.
Speed upstream = (15 - x)km/hr.

ATQ,
40/(15 + x) = 20/(15 - x)
⇒ 600 - 40x = 300 + 20x
⇒ - 40x - 20x = 300 - 600
⇒ - 60x = - 300
⇒ 60x = 300
∴ x = 5 km/hr