Bank Written Questions Taken By AUST

Bank Written Questions Taken By AUST
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All Bank Written Questions Taken By AUST With Solution

এই পোস্টে AUST কর্তৃক নেয়া All Bank Written Questions Answer (যেগুলো পাওয়া গেছে) হিসেবে রিটেনের ম্যাথ সল্যুশন সহ দেয়া হল।

Sonali Bank (Assistant Engineer IT) – 2016

1. A, B and C enter into partnership with investment in the ratio 5:7:8. If at the end of the year A’s share of profit is 42360, how much is the total profit?

Solution:

Let,A’ investment is 5x tk
B investment is 7x tk
C investment is 8x tkTotal investment= 5x+7x+8x=20xtkSo,5x = 42360
Or, x = 8472Now, total investment=20×8472= 169440 tk

Answer: 169440 tk

2. Two-third of the faculty members of a department are female. Twelve of the male teachers are unmarried, while 60% of them are married. The total number of faculty members in the department is:

Solution:

Let, total members are x then female members are 2x/3

So, male members are = x – 2x/3 = x/3

ATQ,

12 teachers are unmarried
Or, (100-60)% =40% =12
So, 100% = 12*100/40 = 30

Again,

x/3 = 30
Or, x=90

Now, total faculty members are 90

Answer:90

3. A trader marked the price of the T.V. 30% above the cost price of the T.V. and gave the purchaser 10% discount on the marked price, thereby gaining Rs. 340. Find the cost price of the TV?

Solution:

Let,cost price of tv is x tk
Then, the market price = x + x×30/100 = 13x/10 tk
At 10% discount on market price,
Selling price = 13x/10 – 13x×10/10×100 = 117x/100 tk
Profit = 117x/100 – x = 17x/100 tkATQ,

17x/100 = 340
Or,x = 340×100/17
Or, x = 2000
So, the cost price of the TV is 2000 tk

Answer: 2000 tk

4. A circular wheel 28 inches in diameter rotates (moves) the same number of inches per second as a circular wheel 35 inches in diameter. If the smaller wheel makes x revolutions per second, how many revolutions per minute does the larger wheel make in terms of x ?

Solution:

As we know,1 revolution of circle= Circumference of that circle

Again,

Circumference of circle=2πr

Given that,

Diameter = 28 inches
So, radius, r= 28/2= 14 inches

Now, circumference = 28π inches

Hence,

x revolutions per sec = 28πx inches
Then in 60 sec 28πx × 60 inches

ATQ,

28πx × 60 = 35πn
=> n = 48x

Thus the required no of revolutions is 48x

Answer: 48x

PKB SEO- 2018

1. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

Solution:

Let,The duration of the fight be=x hrs

Original distance be=6i00 km

According to the question,

Original speed-Reduced speed=200

600/x-600/(x+1/2)=200
Or, 600/x-1200/(2x+1)=200
Or, 3/x-{6/(2x+1)}=1
Or, (6x+3-6x)/{x(2x+1)}=1
Or, 2x2+3x-2x-3=0
Or, x(2x+3)-1(2x-3)=0
Or, (2x+3)(x-1)=0

Now,

X-1=0
Or, X=1

And

2x+3=0
Or, X=-(3/2)

[neglecting the negative value]

Answer: 1 hour

2. A alone can reap a certain field in 15 days and B in 12 days. If A begins alone and after a certain interval B joins him, the field is reaped in 7.5 days. How long did A and B work together?

Solution:

Let,‘x’ be the number of days that A and B worked together

And

Total work be 1 portion

A’s 1 day’s work =1/15
B’S 1 day’s work = 1/12

According to the question,

(7.5 -x)/1 + x(1/15 + 1/12) = 1
Or, (7.5 -x)/1 + 9x/60 = 1
Or, (30-4x+9x )/60=1
Or, 30 + 5x = 60
Or, x = 6

Hence, A & B work together 6 days

Answer:6 days

3. a, b, c, d, e are 5 consecutive numbers in increasing order, deleting one of them from the set decreased the sum of the remaining numbers by 20% of the sum of 5. Which one of the number is deleted from the set?

Solution:

Since a, b, c, d, e are increasing order consecutive number,b = a + 1,

c = a + 2,
d = a + 3,
e = a + 4

The sum of five numbers = a + a + 1 + a + 2 + a + 3 + a + 4 = 5a + 10

We are given that the sum decreased by 20% when one number was deleted
New sum = (5a + 10) – 20% of (5a + 10) = 4a + 8
Then, 4a + 8 = 5a + 10 – Dropped number

Dropped number = a + 2 = c

4. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time?

Solution:

Work done by both in 1 minute= (1/20 + 1/60) = 4/60=1/15

Work done by both in 10 minute

=10/15 =2/3 portion

Remaining part

=(1 – 2/3)
= 1/3 portion
Now, 1/60 part is filled in 1 minute
So, 1/3 part will be filled in 20 minute

Answer: 20 m in

5. Two persons are running in 3.6km/h and 7.2 km/h speed. A train passes them in 9 & 9.5 seconds. What is the length of the train and speed of the train?

Solution:

Let,The speed of the train = x kmh

Know that,

Distance = speed × time

ATQ,

(x-3.6)×9 = (x-7.2) × 9.5
Or, 9x – 32.4 = 9.5x – 68.4
Or, 0.5x = 36
Or, x = 72 kmh

Here, speed of the train = 72 kmh

So, distance covered = {(72-3.6) × 9 × (5/18)}=171 meter

Basic Bank (Assistant Manager)-2018

1. If a person invest Tk. 4000 at x% and Tk. 5000 at y%, he will get total Tk. 320 as interest. On the other hand if he invests Tk. 5000 at x% and Tk. 4000 at y%, he will get total Tk. 310 as interest. Find the value of x and y.

Solution:

,ATQ,{4000x/100}+{5000y/100}=320

Or, 4x + 5y = 32 ————-(i)

Again,

{5000x/100}+{4000y/100}= 310

Or,5x + 4y = 31—————-(ii)

From (i)×5 – (ii)×4,

20x + 25y = 160
20x + 16y = 124
—————————–

Or,9y = 36
Or, y = 4

From equation(I)

4x + 5y = 32
x = (32-5*4)/4 = 3

Answer:(3,4)

2. Working together pipe P, Q and T can fill a tank in 5 hours. Working together P and Q can fill it in 7 hours. Find in how many hours T can fill it?

Solution:

Part filled by P, Q and T in 1 hour = 1/5Part filled by P and Q in 1 hour = 1/7

Part filled by T in 1 hour = 1/5 – 1/7 = 2/35
2/35 part is filled by T in 1 hour

So, Whole part is filled by T in 35/2=17.2 hrs

Ans: 17.5 hours

3. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random. What is the probability that they are not of same colour?

Solution:

Total balls = 5 + 4 + 3 = 12No of ways of drawing 3 marbles out of 12 = 12c3 = 220

event of getting 3 marbles of same color = 5c3 + 4c3 + 3c3=10+4+1= 15

probability of getting 3 marbles of same color =15/220 = 3/44
the probability that not getting marbles of same colour

=(1-3/44)=41/44

Answer: 41/44

4. There was a shipment of cars. Out of which half was black in colour. Remaining cars were equally blue, white and red. 70% of black cars, 80% of blue cars, 30% of white cars, 40% of red cars were sold. What percentage of total cars were sold?

Solution:

Let,The total car was x

So,

Black was = x/2
Remaining cars =(x-x/2)=x/2

ATQ,

Other cars are equal colours
So,

Blue, White, and Red are
x/6,x/6,x/6 respectively

Sold cars:
Black cars=x/2 of 70%=7x/20
Blue cars=x/6 of 80%=2x/15
White cars=x/6 of 30%=x/20
Red cars=x/6 of 40%=x/15

Total sold

=(7x/20)+(2x/15)+(x/20)+(x/15)
=180x/300

Required percentage

=(180x × 100)/(300x)
=60%

Answer:60%

5. A train passes a man in 3 second, and another train from opposite direction pass the man 4 second, both train same length. How long time need to pass the train each other?

Solution:

Let,Both train length be x

So,

Total length of two train=(x+x)=2x

Speed Of First train = x/3
Speed Of second train =x/4

We know that,

Time=Distance/speed
Or, T = 2x/(x/3 + x/4)
Or, T= 2x*12/7x
Or, T=24/7

Hence, 3.43 second need to pass each other

Answer:3.43 seconds

6. A man works for certain hours. If his hourly payment increase by 20%, what percent of working hours he may reduce so that total income remain unchanged?

Solution:

Suppose,hourly payment=100tk

After increased 20% then new payment
=(100+20)=120tk

So,

TK 20 reduced from tk 120
Tk 100::::::::::::::::::::::={20*100/120}
={100/6}
=16.67%

Answer:16.67%

7. There were some books of novel and non-fiction. Board discuss 3 times for any novel and 5 times for any non-fiction. During a year they discuss total 52 times. If there were 12 books, how many of them were novel?

Solution:

Let,Novel books be x

Non Fiction books be(12-x)

According to the question,

3x+5(12-x)=52
Or,3x+60-5x=52
Or -2x=-8
Or, x=4.

So, Nobel books were 4

Answer:4

Sonali Bank (Cash Officer)-2018

1. A, B and C are partners. ‘A’ whose money has been in the business for 4 months claims 1/8 of the profits, ‘B’ whose money has been in the business for 6 months claims 1/3 of the profits. If ‘C’ had Tk. 1560 in the business for 8 months, how much money did A and B contribute to the business?

Solution:

A’s share of profit = 1/8B’s share of profit = 1/3

So, C’s share of profit = 1 – (1/8 + 1/3) = 13/24

Ratio of profit of A, B and C = 1/8:1/3:13/24 = 3:8:13

Let, A contributes Tk. x for 4 months and B contribute Tk. y for 6 months.

Ratio of investments of A, B and C = 4x : 6y : (1560 × 8) = 4x : 6y : 12480

Now,

4x : 12480 = 3 : 13
Or, 4x/12480 = 3/13
Or, 52x = 12480 × 3
Or, x = (12480 × 3) / 52
Or, x = 720

Again,

6y : 12480 = 8 : 13
Or, 6y/12480 = 8/13
Or, 78y = 12480 × 8
Or, y = (2480 × 8) / 78
Or, y = 1280

So, contribution of A and B is Tk. 720 and Tk. 1280 respectively.

Ans: Tk. 720 and Tk. 1280

2. Machine A, working alone at its constant rate, produces x pounds of peanut butter in 12 minutes. Machine B, working alone at its constant rate, produces x pounds of peanut butter in 18 minutes. How many minutes will it take machines A and B, working simultaneously at their respective constant rate, to produce x pounds of peanut butter?

Solution:

Machine A and B produce in 1 minute = (x/12 + x/18) = 5x/365x/36 pounds is produced by both machines in 1 minute

Or, 1 pounds is produced by both machines in 36/5x minute

Or,  x pounds is produced by both machines in (36/5x × x) = 7.2 minute

Ans: 7.2 minutes

3. Two trains running at the rate of 75 km and 60 km an hour respectively on parallel rails in opposite directions, are observed to pass each other in 8 seconds and when they are running in the same direction at the same rates as before, a person sitting in the faster train observes that he passes the other in 31.5 seconds. Find the lengths of the trains.

Solution:

Moving in opposite direction:Relative speed of two trains = (75 + 60) km/hr = 135 km/hr = (135 × 5/18) m/s = 75/2 m/s

Length of the both trains = (75/2 × 8) meters = 300 meters

When moving in the same direction,

Relative speed of two trains = (75 – 60) km/hr = 15 km/hr = (15 × 5/18) m/s = 25/6 m/s
So, Length of the slower train = (25/6 × 31.5) meters = 131.25 meters
Thus, length of the faster train = 300 – 131.25 = 168.75 meters

Ans: 131.25 meters and 168.75 meters.

4. A gardener plants two rectangular gardens in separate regions on his property. The first garden has an area of 600 square feet and a length of 40 feet. If the second garden has a width twice that of the first garden, but only half of the area, what is the ratio of the perimeter of the first garden to that of the second garden?

Solution:

Width of the first garden = 600/40 = 15 feet.ATQ,

width of the second garden = 15 × 2 = 30 feet
and, area of the second garden = ½ × 600 = 300 sq. feet
So, length of the second garden = 300/30 = 10 feet

Now, Perimeter of the first garden : Perimeter of the second garden

= 2(40+15) : 2(10+30)
= 110 : 80
= 11 : 8

5. In a certain class, 1/5 of the boys are shorter than the shortest girls in the class and 1/3 of the girls are taller than the tallest boy in the class. If there are 16 students in the class and no two people have the same height, what percent of the students are taller than the shortest girl and shorter than the tallest boy?

Solution:

Total students = 161/5 implies that no. of boys must be a multiple of 5

And

1/3 implies that the no. of girls must be a multiple of 3.

So, Boys = 10 and Girls = 6 which satisfy the no. of total students (10 + 6) = 16

Shorter boys = (1/5 × 10) = 2 and shortest girl = 1
Taller girls = (1/3 × 6) =2 and tallest boy = 1
Taller than shortest girls and shorter than tallest boy = (16 – 2 – 2 – 1 – 1) = 10

So, Required percentage = (10/16 × 100) = 62.5%

Ans: 62.5%

5 bank combined officer written – 2018

1. A car owner buys petrol at Tk.75, tk. 80 and tk. 85 per liter for three successive years. What approximately is the average cost per liter of petrol if he spends Tk. 40000 each year in this concern?

Solution:

Among the three successive years,

1st year, he buys = 40000/75 = 533.33 litres of petrol
2nd year, he buys = 40000/80 = 500 litres of petrol
3rd year, he buys = 40000/85 = 470.58 litres of petrol

He buys total = (533.33+500+470.58) = 1503.91 litres of petrol

Total cost = (40000×3) = 120000 tk.

So, average cost of per litre of petrol = (120000/1503.91) = 79.79 tk. = 80 tk (Approx)

2. P is a working and Q is an investing partner. P puts in Tk. 340000 and Q puts Tk. 650000. P receives 20% of the profits for managerial works. The rest is distributed in proportion to their capitals. Out of a total profit of Tk. 99000, how much does P get?

Solution:

For managerial works, P gets from profit = 20% of 99000 = 19800 tk.
Profit remains = (99000 – 19800) = 79200 tk.

ATQ,

P:Q = 340000: 650000 = 34:65
Sum of ratio = (34+65) = 99

From remaining profit P gets = 34/99 of 79200 = 27200 tk

So, P gets total = (19800 + 27200) = 47000 tk.

3. A lawn is in the form of a rectangle having its sides in the ratio 2:3. The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn.

Solution:

Let,

Length = 3x meter
And breadth = 2x meter

Given,

Area = 1/6 hectares

= (10000/6) meter2
= 5000/3 meter2

ATQ,

3x × 2x = 5000/3

Or, 6x2 = 5000/3
Or, x2 = 5000/18
Or, x2 = 2500/9
Or, x = ±√(2500/9)
Or, x = ±50/3
Or, x = 50/3 (negative is not acceptable)

So, length = 3x = 3 × (50/3) meter = 50 meter

breadth = 2x = 2 × (50/3) meter = 100/3 meter = 33.33 meter

4. Consider an example from a maintenance shop. The inter-arrival times of tools at that shop are exponential with an average time of 10 minutes. The length of the service time is assumed to be exponentially distributed, with mean 6 minutes. Estimate the fraction of the day that an operator will be idle.

5. Simplify: (x-1)/(x2 – x – 20) + (4 – x)/(x2 – 4x – 5)

Solution:

Given,

(x-1)/(x2 – x – 20) + (4 – x)/(x2 – 4x – 5)
= (x-1)/(x2 – 5x + 4x – 20) + (4 – x)/(x2 – 5x + x – 5)
= (x-1)/{(x(x – 5) + 4(x – 5)} + (4 – x)/{(x(x – 5) + 1(x – 5)}
= (x-1)/(x – 5)(x + 4) + (4 – x)/(x – 5)(x + 1)
= 1/(x – 5){ (x-1)/(x + 4) – (x – 4)/(x + 1)}
= 1/(x – 5){(x2 – 1 – x2 + 16)/(x + 4)(x + 1)}
= 1/(x – 5){15/(x + 4)(x + 1)}
= 15/(x – 5)(x + 4)(x + 1)

6. Prove that a parallelogram inscribed in a circle must be a rectangle.

Solution:

 

7. The angle of elevation of a hot air balloon, climbing vertically from 25 degrees to 60 degrees at 10:00 am and at 10.02 am respectively. The point of observation of the angle of elevation is situated 300 meters away from the take-off point. What is the upward speed (m/sec), assumed constant for the balloon?

Solution:

At 10:00 am, height of the ballon is,
tan250 = height1/300
or, height1 = 300 tan250 = 139.89 meter

Again,

At 10:02 am, height of the ballon is,
Tan600 = height2/300
or, height2 = 300 tan600 = 519.62 meter

Difference of height = (height2 – height1)
= (519.62 – 139.89) meter
= 379.73 meter

Time difference = (10:02 – 10:00) am = 2 minute

So, upward speed (m/sec) of ballon = height/time

= 379.73/120 (2 minute = 120 sec)
= 3.16 m/sec (Approx)

Bank Written Questions Taken By AUST ছাড়া আরও পড়ুনঃ

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