# HBFC Senior Officer Exam Question Solution [English+Math]

Bangladesh House Building Finance Corporation complete there job exam for the post of Senior Officer. HBFC Written Exam Full Question Solution 2017 published by ejobscircular.com . Let`s Check carefully below this Job Exam Question in Bangladesh.

## HBFC Senior Officer Exam Question Solution

Many Candidate want to know about HBFC Job Question Solution from Bangladesh. Now in this post, I am going to share Exam Question of 11 November 2017 (Bangladesh House Building Finance Corporation) . Be Carefully when check this job Question correction.  a) 8:3
b) all 3 are 3/8
c) 6(2/15) days
d) 500; 800
e) 100
f) 12

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

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.

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,
L = x
W = 2x-10

Atq,
2x^2-10x= 600
2x^2-10x-600 =0
2(x^2-5x-300) = 0
x^2-20x+15x-300 = 0
(x-20)(x+15) = 0
x= 20

L = 20
W = 20*2-10 = 20

Perimeter, 2(20+30) = 100 Ans.

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.