Janata Bank Limited (JBL) Written Exam Question Solution 2020

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Janata Bank Limited (JBL) Written Exam Question Solution 2020:

Post Name: Assistant Executive Officer (AEO-Teller)

Total Vacancy: 536

MCQ Exam Date: 20 December 2019

Exam Time: 3.00 PM – 4.00 PM

Written Exam Date: 07 February 2020

Written Exam Time: 2.30 PM to 4.30 PM

Total Selected For Written: 9742

Another Exam Notice:

Post Name And Vacancy:

1. Assistant Executive officer (AEO)- 464

MCQ Exam Date: 29 November 2019

Exam Time: 10.00 AM – 11.00 AM

Written Exam Date: 10 January 2020

Written Exam Time: 10.00 AM to 12.00 AM

See/download Janata Bank Limited (JBL) Written Exam Question Solution 2020 From below:

Written Math Solution:

Post Name: Assistant Executive officer (AEO Teller)

1. Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for Point C which is 240 km away from B. The first car arrives at C an hour after the second car. The third car, having reached C, immediately turns back and heads towards B. The first and the third car meet a point that is 80 km away from C. What is the difference between the speed of the first and the third car?

Ans: 60 km/hr

Full Solution:

Let, time taken be 2nd car be x hr
So, 1st takes= (x+1) hr
And 3rd takes= (x-1) hr
Speed of 1st=240/(x+1) km/hr
Speed of 2nd=240/(x-1) km/hr
ATQ,
(240-80)/{240/(x+1)}=(240+80)/{240/(x-1)}
Or, (2x+2)/3 = (4x-4)/3
Or, x=3
Speed of 1st=240/(x+1)=60 km/hr
Speed of 2nd=240/(x-1)=120 km/hr
So, difference=(120-60)=60km/hr

Another Solution:

2. 16 women take 12 days to complete a work which can be completed by 12 men in 8 days. 16 men started working and after 3 days 10 men left and 4 women joined them. How many days will they take to complete the remaining work?

Ans: 6 days

Detail Full Solution:

$\begin{aligned} 16 women \times 12 = 12 men \times 8 \\ 2 women = 1 man \\ Total work \\ = 12 men \times 8 days \\ = 96 units \\ 16 men do work in 3 days \\ = 16 \times 3 \\ = 48 units \\ Work left 96 - 48 = 48 units \end{aligned}$

16 men – 10 men left = 6 men + 4 women join

6men + 2 men = 8 men

8 men will do 48 units in = 48/ 8 = 6 days

Another Solution:

3. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

Ans: 70% $\begin{aligned} = (\frac{295}{420} \times 100) % \\ = \frac{1475}{21} % \\ = 70 % (approximately) \end{aligned}$

Detail Full Solution:

Let C.P.= Tk. 100. Then, Profit = Tk. 320, S.P. = Tk. 420
New C.P. = 125% of Tk. 100 =Tk. 125
New S.P. = Tk. 420
Profit = Tk. (420 – 125) = Tk. 295
∴ Required percentage

= (295/ 420 \times 100) % = 1475/ 21 % = 70 % (approximately)

Ans: 437 taka (approx)

Full Solution:

Ans: 3600 taka

Full Solution:

6. Three bottles whose capacities are as 5 : 3 : 2 are completely filled with milk mixed with water. The ratio of milk to water in the mixture of bottles are as 3 : 2, 2 : 1 and 3 : 1 respectively. Find the percentage of water in the new mixture obtained when 1/3rd of first, 1/2 of second and 2/3rd of the third bottle is taken out and mixed together.

Ans: 33.33 %

Full Solution:

Explanation:

Let the percentage of water be X
The quantity of new mixture = (5X/3) + (3X/2 )+ (4X/3) = 27X/6 = 9X/2.
Percentage of milk = (5X/3) x (3/5) + (3X/2) x (2/3) + (4X/3) x (3/4) = 3X
Percentage of water = (5X/3) x (2/5) + (3X/2) x (1/3) + (X/2) x (2/3) = 3X/2
Percentage of water = (3X/2)/(9X/2) x 100 = 100/3 = 33 (1/3)%

if this math want……………

the percentage of water in the new mixture is…….then answer will 24%

Ans: 5

Full Solution:

Post Name: Assistant Executive officer (AEO)

1. A sum of taka 725 is lent in the beginning of a year at a certain rate of interest. After 8 months , a sum of tk 362.50 more is lent but at the rate of twice the former. At the end of the year Tk 33.50 is earned as interest from both the loans. what was the original rate of Interest ?

Ans: 3.46% (Approximately)

Detail Full Solution:

Let, Initial Rate = r%
Final rate = 2r%
ATQ,
725*1* r/100+ (362.5)*(1/3)*2r/100= 33.50
=> 725r+ 241.67r= 3350
=> r= 3.46%

2. If x+1/x=2; find the value of x^17+1/(x^19)=?

Ans: 2

Detail Full Solution:

x+ 1/x = 2
=> x^2+1= 2x
=> x^2-2x+1= 0
=> x^2-x-x+1= 0
=> x(x-1)-1(x-1)= 0
=> x= 1

Now,
x^17+ 1/x^19
= (1)^17+1/(1)^19
= 1+1
= 2

3. Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

Ans: 93,300

Detail Full Solution:

X salary for first year = 12*300= 3600 for first year
Salary up to 1959 = (3600+3960+———–+n)
Here, n= 10
a = 3600
d= 360
Total salary = 10/2*{2*3600+(10-1)*360}
= 5{7200+9*360}= 52,200

Y salary up to 1959
= 1200 for 6 month
= 1200+ 15*6= 1290 for next six month

Total salary of Y = 20/2*{2*1200+19*90}
= 41,100
Total Salary = (52,200+41,100)
= 93,300

4. 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?

Ans: 1 Hour

Detail Full Solution:

Let the duration of the flight be x hours.

x(2x + 1) = 3

2x² + x – 3 = 0

(2x + 3)(x – 1) = 0

x = 1 hr. [neglecting the -ve value of x]

or,

Flight Distance= 600 km
Let, Avg. Time = t
New time = (t+1/2) hr
ATQ,
600/t – 600/(t+1/2)= 200
=> 600/t – 600/(2t+1)/2= 200
=> 600/t -1200/(2t+1)= 200
=> (1200t+600-1200t)= 200*t(2t+1)
=> 600= 200(2t^2+t)
=> 600= 400t^2+200t
=> 400t^2+200t-600= 0
=>4t^2+2t-6= 0
=> 2t^2+t-3= 0
=> 2t^2+3t-2t-3= 0
=> t(2t+3)-1(2t+3)= 0
So, t= 1hr

5. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. How many hours will take C alone to fill the tank ?

Ans: 14 hours

Detail Full Solution:

(A+B) can fill = 2/3 part in = 7 hr
1 part in = 21/2 hr
(A+B+C) can 1 part in = 6 hr
Eff. (A+B+C):(A+B)= 7:4
Eff. Of C = 7-4= 3
ATQ,
42= 3*T
=> T= 14 hr

6. A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is?

Ans: 5

Detail Full Solution:

Let, Length and Breadth L& B
Area, L*B= 100
L= 100/B
And,
2L+ B= 30
=> 200/B + B= 30
=> 200+B^2-30B= 0
=> B^2-30B+200= 0
=> B^2-20B-10B+200= 0
=> B= 20,10
If B = 10
L= 10
And, if perimeter, 2B+L= 30
Then, 2B+ 100/B= 30
=> 2B^2-30B+100= 0
=> B^2-15B+50= 0
=> B= 10,5
If B = 5, L = 20
Since, it is rectangular L should be greater than Breadth
So, L= 20
B= 5

7. P does a work in 12 days for 8 hours and Q does a Work in 8 days for 10 hours.If both work for 8 hours then how many days it needed to complete the work ?

Ans: 60/11 days

Detail Full Solution:

12*8*P= 8*10*Q
=> 12P= 10Q
=> P:Q= 5:6
ATQ,
96*5= 8*(5+6)*D
=> 60= 11 D
=> D= 60/11

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Mirpur 1, Dhaka, Bangladesh.

Janata Bank Limited (JBL) Written Exam Question Solution 2020