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Organization Name: Bangladesh Bank
Exam Type: Written
Post: Assistant Director (AD)
Total Candidate: 9837
Exam Taker: Arts Faculty, DU
Written Exam Time: 10.00 AM to 12.00 PM
A.
A senior citizen invest Tk. 50 lac in a fixed deposit scheme at 11.5 % annual interest for six months. In every six months he withdraws Tk. 2 Lac from his principal plus interest earned. What will be his principal amount to invest after two years?
Ans: 42 Lac
B.
After traveling 108 km, a cyclist observed that he would have required 3 hrs less if he could have traveled at a speed 3 km/hr more. At what speed did he travel?
S*T = 108
Again, (S+3)(T-3) = 108
=> ST – 3S + 3T – 9 = ST
=> 3T – 3S = 9
=> T – S = 3
=> T = S+3
Now, S*(S+3) = 108
=> S² + 3S – 108 = 0
=> S² + 12S – 9S – 108 = 0
=> S(S+12) – 9(S+12) = 0
=> (S+12)(S-9) = 0
S = 9 (Ans.)
SHORTCUT:
108 = 9*12 = 12*9
D = S*T
So, S = 9 and T = 12
Ans: 9
C. Three numbers are in A.P. and their sum is 30. Also, the sum of their squares is 308. Find the numbers.
Solution:
Let,
Second Term = a
Common deference = d
So, first term will be = a – d
Third term will be = a + d
A.T.Q.,
a- d + a + a+d = 30
or, 3a = 30
so, a = 10
So, first term = 10 – d
Second term = 10
Third term = 10 + d
Again A.T.Q.,
(10 – d)2 + (10)2 + (10+d)2 = 308
Or, 100 – 20d + d2 + 100 + 100 + 20d + d2 = 308
Or, 2d2 + 300 = 308
Or, 2d2 = 8
Or, d2 = 4
So, d = 2
So first term = 8, second term = 10 , third term = 12
Ans: 8,10,12
OR,
Let the Three numbers: a,a+x,a+2x
Atq, 3a+3x=30
A+X=10
X=10-A
And,
A.A+10.10+(20-A)(20-A)=308
500+2A.A-40A-308=0
A.A -20A+192=0
A=8,12
A=8
8,10,12
Ans: 8,10,12
D.
A square is inscribed inside a circle. What is the area of the square, if the radius of the circle is 10 cm?
Here,
Radius of circle = 10 cm
So, Diameter of circle = 10*2 = 20 cm
Then, Diagonal of square = 20 cm
Therefore, Area of square = d^2/2 = 20^2/2 = 400/2 = 200 cm^2
Ans: 200 CM2
A:
Ans: Write Yourself
B. 50 daily workers can complete a dam project in 40 days. If 30 of them work daily and the rest work in every alternative day, how many more days will be required to complete the project.
Solution:
Total Works: 50*40=2000
30*40=1200
20*20=400
Total =1600
2000………. 1 work
1600…………1600/2000 work or 4/5 work
remain, 1- 4/5=1/5 work
1600….4/5…… 40 days
1600…..1……….(40*5)/4=50
1600…..1/5……..50/5 =10 days
Another Approach:
Total work= 50*40= 2000
In 40 days, daily workers complete: 30*40=1200
Alternate workers complete:20 * (40/2) = 400
Let us assume that they work for x days after 40 days.
Alternate workers will work for x/2 days.
A/Q, 2000 = 1200 + 400 + 30x + 20*(x/2)
Solving, we get x=10
Ans: 10 days
C. 3 coins are tossed random. Construct the sample space and find the probability of getting.
Ans: 3/8, 3/8, 3/8
D. Analyze your Strength and weakness.
Ans: Write Yourself