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Bangladesh Bank (BB) Officer Written Question Solution 25 May Exam

Bangladesh Bank AD Written Question Solution 2018

 

Bangladesh Development Bank Ltd. (BDBL) Written Math Solution:

Exam Held: 06-01-2018

Post: Senior Officer

 

a. The profit of a company is given in Taka by P = 3x²-35x+50, where x is the amount in Taka spent on advertising. For what values of x does the company make a profit?

Here given,

3x² – 35x + 50 > 0
=> 3x² – 30x – 5x + 50 > 0
=> 3x(x-10) – 5(x-10) > 0
=> (x-10)(3x-5) > 0

Here, the Roots of the equation are: 10 and 5/3

Since, the in-equation is Greater than 0; i.e. positive
So, range of x < 5/3; Or x>10

Ans. x < 5/3; Or x>10

b. An amount of Tk. 7200 is spent to cover the floor of a room by carpet. An amount of Tk. 576 would be saved if the breadth were 3 meters less. What is the breadth of the room?

let, length = x
breadth = y
area = xy
AQ, xy = 7200
or, x = 7200/y…….(1)
3 m less new breadth = y – 3 m
new area = x(y-3)
AQ, x(y-3)=7200-576
=> y = 37.5

c. Find the three digit prime number whose sum of the digits is 11 and each digit representing a prime number. Justify your answer.

Sum of digit to be 11 of a 3-digit number and each digit being a prime number, the number is: 227 or 353

In case of 227, sum of the digits is 2+2+7 = 11.
And 2, 2, 7 all the digits are prime.

Similarly, In case of 353, sum of the digits is 3+5+3 = 11.
And 3, 5, 3 all the digits are prime.

d. If a/(q-r) = b/(r-p) = c/(p-q) then show that, a+b+c = pa+qb+rc

Let, a/(q-r) = b/(r-p) = c/(p-q) = k
So, a = k(q-r); b = k(r-p); and c = k(p-q)

Now, L.H.S. => a+b+c = k(q-r)+ k(r-p)+ k(p-q) = k(q-r+r-p+p-q) = k x 0 = 0
And, R.H.S. => pa+qb+rc = p*k(q-r)+ q*k(r-p)+ r*k(p-q) = kpq-kpr+kqr-kpq+kpr-kqr = 0
So, L.H.S. = R.H.S. (Showed)

e. Prove that a cyclic parallelogram must be a rectangle.

Opposite angles in a parallelogram are congruent (equal), while opposite angles in a cyclic quadrilateral are supplementary (add up to 180°). Congruent supplementary angles are right angles, so opposite angles in a cyclic parallelogram are right angles. Thus all four angles are right angles, and it’s a rectangle.

Let ABCD be the cyclic parallelogram with angles A and C being opposite angles.
A = C
And, A+C = 180°
since A = C
So, A+A = 180°
2A = 180°
A = 90°
if any one angle of parallelogram is 90°, the parallelogram is a rectangle.

Or,

f. After travelling 108 km, a cyclist observed that he would have required 3 hrs less if he could have travelled 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

g. Solve: x/2 + 6/y = 9, x/3 + 2/y = 4

g) Detail solution…

x/2+6/y = 9
or,18y – xy = 12….(1)
x/3+2/y=4
or,12y – xy=6…(2)
(1) – (2)
18y – xy =12
12y – xy =6
………………..
y = 1
putting the value equation (2)
12*1 – x*1 = 6
or, x = 6
so, ( x, y) = (6,1)

See the Written Question Solution:

8. Ans:

a) x < 5/3 or x>10 That means X= 11,12,13……..
b) 37.5
c) 227 or 353
f) 9
g) (6,1)

Bangladesh Development Bank Ltd. (BDBL) Math Question:

Bangladesh Development Bank Ltd. (BDBL) Full Question:

Bangladesh Krishi Bank Written Questions And Math Solution:

Post: Cash Officer

Written Exam date: 13 January 2018

Math Solution:

a.7,10,13
b.15
c.50
d.a=16, B=27
e.100 & 500
f.220
g.500;480
h.(a+1/a) (a+1/a-2).

Detail Solution:

6. a)

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)=318
500+2A.A-40A-318=0
A.A -20A+91=0
A=7,13
A=7
7,10,13and
A=13
13,10,7,ans

6.b)

Atq, 50=35+b-25
b=40
only speck bangla 40-25=15

6. c)

Area of triangle= 1/2 (10*10)= 50 Ans.

6. d) 64x^3 -9ax^2 + 108x -b
=(4x)^3- 3 (4x)^2 (3) + 3 (4x) (3)^2-(3)^3
=(4x-3)^3
so, a= 16 and b= 27

6. h)
a^2 + 1/a^2 +2 -2a – 2/a
=a^2 + 1/a^2 +2*a*1/a -2a -2/a
= (a+1/a)^2 -2a -2/a
= (a+1/a)^2 -2(a+1/a)
= (a+1/a) (a+1/a -2)

See The question in image below:

Uttara Bank Ltd. Written Questions Solution:

Post: Assistant Officer (Cash)

Written Exam date: 02/02/2018

Question-01:

Anik visited his cousin Rowhan during the summer vacation. In the mornings, they both would go for swimming. In the evenings, they would play tennis. They would engage in at most one activity per day, i.e. either they went swimming or played tennis each day. There were days when they took rest and stayed home all day long. There were 32 mornings when they did nothing, 18 evenings when they stayed at home, and a total of 28 days when they swam or played tennis. What duration of the summer vacation did Anik stay with Rowhan?
Solution:
Let
The duration of Anik’s vacation be n days.
Given that
On each day, he had engaged in exactly one of swimming and tennis,
Also given that he was free on 32 mornings and
On 18 evenings and on total 28 days he either went for swimming or tennis.
So,
He was busy on (n−32)mornings
And (n−18) evenings
Now we can write,
(n−32)+(n−18)=28
Or, 2n =28+32+18
Or, n=39

So,39 day’s summer vacation did Anik stay with Rowhan
Answer: 39 day’s

Question-02:

In a three digit number the number in unit place is 75% of tenth digit number, the tenth digit number is greater than hundred digit by 1 & their sum will be 15, find out the number?

Solution:
Let,
Tenth digit be=y
Unit’s digit= 75% of y=3y/4
Hundred’s digit=y-1
So,
The number be
=100(y-1)+10y+3y/4
=100y-100+10y+3y/4
=(443y-400)/4…………(i)
Now From Question,
y-1+y=15
or,2y=16
or,y=8
Putting the value of y=8 in equation(i)
(443*8-400)/4
=(3544-400)/4
=(3144/4)
=786
Hence,the required number is 786
Answer:786
Solution-2:
Let,
Unit digit=x
Tenth=y
And,
Hundred=z
So,
Original number=100z+10y+x
From question condition,
x= 75% of y=3/4*y——-(i)
And,
y=z+1———(ii)
Again from question,
y+z=15
Or,2z+1=15
Or,z=7
From equation(ii)
y=8
From (i)
& x=3/4*8=6
So,
Required number=7*100+8*10+6
=786.
Ans:786

Agrani Bank Written Questions Solution:

Post: Officer (Cash)

Exam date: 20 January 2018

Exam type: Written ( Writing Ability Test )

8)

a) A novelist earned Tk. 100,000 from royalties on her book. She paid 20% income tax on the royalties. She invested Tk. 50,000 at one rate and the rest at a rate that was 1% lower, earning 6,100 taka annual interest on the two investments. What was the lower rate?

Solution:

Remaining amount after deducting tax=100000*(100-20)%=80000

let lower rate=x

Atq,
30000*x%+50000*(x+1)%=6100
or, 300x+500x+500=6100
or, 800x=5600
or, x=7%

Ans: 7%

b) A working couple earned a total of Tk. 43,520. The wife earned Tk. 640 per day, the husband earned Tk. 560 per day. If the total number of days worked by both was 72, formulate a system of equation and solve the system to find the number of days worked by each.

Solution:

let, wife work =x days
husband work=72-x days

ATQ,
640x+(72-x)560=43520
or, 640x+40320-560x=43520
or, 80x=3200
or, x=40

so wife works 40 days
husband works (72-40)=32 days

Ans: 40 and 32 Days

c) A man’s salary in 2015 was Tk. 20,000 per annum and it increased by 10% each year. Find how much he earned in the years 2015 to 2017 inclusive.

Solution:
2015=20000
2016=20000+20000*.10=22000
2017=22000+22000*.10=24200

sum= 66200
Ans: 66200

d) Prove that the sum of the odd numbers from 1 to 125 inclusive is equal to the sum of the odd numbers from 169 to 209 inclusive.

Solution:

Number of odd numbers between -1-125=63

1-3-5-7—+125

sum 63rd series =63/3{2*1+(n-1)*2}
solving =3969

number of odd series in between 169-209=21

169-171—209

sum of 21st series =n/2{2*169+(21-1)*2}

solving =3969

(proved)

e) proved-1 LHS=RHS (Proved)

f) Solution:

x/2+y/3=1—–(I)
x/3+y/2=1—–(ii)
multiply by 3 equation (I) and by 2 equation (ii) then (I)-(ii)
3x/2-2x/3=3-2
or, (9x-4x)/6=1
or, 5x=6
or, x=6/5

putting value in equation (I)
6/10+y/3=1
or, (18+10y)/30=1
or, 10y=12
or, y=6/5

so (x,y)=(6/5,6/5)

f) x=6/5 y=6/5

g) A hemisphere and a right circular cone on equal bases are of equal height. Find the ratio of their volumes.

Solution in image below:

See the Written ( Writing Ability Test ) question in the image below:

 

Agrani Bank Written Questions Solution:

Post: Senior Officer ( Auditor)

Exam date: 19 January 2018

Exam type: Written ( Writing Ability Test )

Math Solution: Full Solution is ongoing……

B.

(55+53+79)-(18+17+25)+10=x
=)x=197-60
=)x=137
d.
X^2+4^2=5^2
=)x^2=9
=)x=3

C)

1st cp=100×100/110=1000/11

2nd cp=100×100/90=1000/9

total cp of two=1000/11+1000/9=20000/99.
Sp=100+100=200

proft/loss=200-20000/99=19800-20000/99=-200/99
% of loss=200×99×100/99×20000=1%

E.

Ans: x(x+4)

See the Written ( Writing Ability Test ) question in the image below:

 HBFC  Written Math Solution:

Post: Senior Officer

Exam held: 11/11/2017

Math the answer with below questions in image:

a) 8:3
b) all 3 are 3/8
c) 6(2/15) days
d) 500; 800
e) 100
f) 12

Download Written math solution in pdf: BHBFC Senior Officer Written Maths

Detail Solution :

a) A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and the speed of the water current respectively?

Solution:
Suppose, the speed of the boat in still water and the speed of the water current are x km/hr and
y km/hr.
According to the question:
(8 + 48/60)(x-y) = 4(x+y)
(44/5)(x-y) = 4(x+y)
11x-11y = 5x+5y
6x = 16y
x : y = 8 : 3

Answer: 8 : 3

b)
3 coins are tossed at random. Show the sample space and find the probability of getting:
– (i) one head two tails
– (ii) One tail
– (iii) One tail and two heads
Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT)
– (i) Probability of getting one head and two tails:
In the sample space we can see, a total of 8 types of outcome is possible.
Among these 8 types of outcomes, the combinations with one head and two tails are : HTT, THT,
TTH, i.e: 3 outcomes.

So, the required probability is 3/8.
– (ii) Probability of getting one tail:
In the sample space we can see, a total of 8 types of outcome is possible.
Among these 8 types of outcomes, the combinations with one tail are : HHT, HTH, THH, i.e: 3
outcomes.
So, the required probability is 3/8.
– (iii) Probability of getting one tail and two heads:
In the sample space we can see, a total of 8 types of outcome is possible.
Among these 8 types of outcomes, the combinations with one tail and two heads are : HHT, HTH,
THH, i.e: 3 outcomes.
So, the required probability is 3/8.
Answer: Sample Space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT), (i) 3/8, (ii) 3/8, (iii) 3/8.

c) A, B and C can complete a work in 12, 15 and 25 days respectively. A and B started working together whereas C worked with them in every third day. Find the number of days required to complete the work.

Solution:
A and B together can complete in 1 day = 1/12 + 1/15 = 9/60 = 3/20 of the work.
A and B with the help of C can complete in 1 day = 3/20 + 1/25 = 19/100 of the work.
So, their 3 days’ work = 2×3/20 + 19/100 = (30+19)/100 = 49/100 of the work.
So, their 3×2 = 6 days’ work = 49×2/100 = 49/50 of the work.
Remaining work = 1 – 49/50 = 1/50 of the work.
on the 7th day, A and B will take (1/50)/(3/20) or 20/150 or 2/15 days.
Therefore, the required number of days = 6 2 15 days.
Answer: 6 2 15 days.

d)
The price of a shirt and a pant together is Tk. 1300. If the price of the shirt increases by 5% and that of the pant by 10%, it costs Tk. 1405 to buy those two things. Find the respective price of
a shirt and a pant.

Solution:
Suppose, the prices of a shirt and a pant are Tk. x and Tk. y respectively.
According to the question:
x+y = 1300 ———— (i)
1.05x + 1.1y = 1405 ———- (ii)
Subtracting (ii) from 1.1 times the value of (i) we get,
0.05x = 25
x = 500
Substituting the value of x in equation (i) we get,
500+y = 1300
y = 800.
Answer: Shirt = Tk. 500, Pant = Tk. 800.

e) Twice the width of a rectangle is 10 meters more than its length. If the area of the region
enclosed by the rectangle is 600 square meters, then find its perimeter.

Solution:

Suppose, the width of the rectangle is = x meters.
So, the length of the rectangle is = 2x-10 meters.
According to the question,
x(2x-10) = 600
x² – 5x – 300 = 0
x² – 20x + 15x – 300 = 0
(x-20)(x+15) = 0
x = 20 or -15
As the width of a rectangle cannot be negative, x = -15 is not acceptable.
So, width = x = 20 meters
and length = 2x-10 = 2×20 – 10 = 30 meters
Perimeter = 2(30+20) = 100 meters.
Answer: 100 meters.

f)
A customer bought 5 pencils and 6 erasers at Tk. 80. Next week, the price of each pencil increases by 20%, but the price of erasers remains unchanged. Now, the customer buys 2 pencils and 3 erasers at Tk 39. Find the new price of each pencil.

Solution:
Suppose, the original prices of a pencil and an eraser are Tk. x and Tk. y respectively.
According to the question,
5x+6y = 80 ———– (i)
&
2×1.2x + 3y = 39
or, 2.4x+3y = 39 ——— (ii)
Subtracting the double of (ii) from (i) we get,
0.2x = 2
x = 10.
So, the new price of a pencil is = 1.2×10 = 12.

Answer :12

Solved by: Sultan Mahmud Cxilvii

Image of Question:

 

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Previous Bank Written Questions Solution:

 

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