HBFC Senior Officer Exam Question Solution 2017 - [English+Math]

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.

HBFC Senior Officer Written Math Solution

HBFC Senior Officer Written Solution 2017

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

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,
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.

Answer :12

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