Written Math-PKB Senior Executive Officer 2018

#Written Math Solution-2018
#PKB SEO -2018
#Exam_Taker:AUST
#Date:02.03.2018

Question-1: 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.[Indian Bix]
[Rupali Bank Written-2013]
Solution:1==================
Let,
The duration of the flight be=x hrs
Original distance be=600 km
According to the question,
Original speed-Reduced speed=200
600/x-600/(x+1/2)=200
Or,600/x-1200/(2x+1)=200
Or,2/x-{6/(2x+1)}=1
Or,(6x+3-6x)/{x(2x+1)}=1
Or,2x^2+3x-2x-3=0
Or,x(2x+3)-1(2x-3)=0
Or,(2x+3)(x-1)=0
Now,
X-1=0
Or,X=1
And
2x+3=0
Or,X=-(3/2)
[neglecting the negative value]
Answer:1 hour
============================
Solution:2
Let,
Original time be= t hrs
Original speed be=S km/hrs
Here,Distance (D)=600 km
According to the first condition,
D=st
Or,s=D/t
Or,s=600/t=======(1)
According to the 2nd condition,
600=(s-200)*(1/2+t)
{(Putting value s=600/t) 2nd condition}
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Or,(2t+3)(t-1)=0
Now,
2t+3=0
Or,t=-(3/2)
[neglecting Negative value]
And
t-1=0
Or,t=1
So Duration of the flight 1 hour
Answer:1 hour

Question-2:A alone can reap a certain field in 15 days and B in 12 days .If A begins alone and after a certain interval B joins him,the field is reaped in 7.5 days .How long did A and B work together.
Solution:
Suppose, 'x' be the number of days that A and B worked together
And Total work be 1 portion
A’s 1 day’s work =1/15
B’S 1 day’s work = 1/12
According to the question,
(7.5-x)/15 + x(1/15 + 1/12) = 1
=»(7.5-x)/15 + 9x/60 = 1
=»(30-4x+9x )/60=1
=»30 + 5x = 60
=»x = 6
Hence, A & B work together 6 days
Answer:6 days

Question-3:a, b, c, d, e are 5 consecutive numbers in increasing order, deleting one of them from the set decreased the sum of the remaining numbers by 20% of the sum of 5. Which one of the number is deleted from the set?
[Bangladesh Bank AD-2014]
[Bangladesh Bank AD-2012]
[PBL SO-2013]
Solution-1:====
Since a,b,C,d,e are increasing order consecutive number
b=a+1
c=a+2
d=a+3
e=a+4
The sum of five numbers
=a+a+1+a+2+a+3+a+4
=5a +10
Now we are given that the sum decreased by 20% when one number was deleted
Hence,
The new sum should be
=(5a+10)-20% of( 5a+10)
=4a+8
Now,
New sum
=old sum- Dropped number
4a+8=5a+10-Dropped number
Dropped number=a+2=C
Answer:C
Solutions-2:====
Let, the consecutive numbers are,
a =1
b = 1 + 1 = 2
c = 1 + 2 = 3
d = 1 + 3 = 4
e = 1 + 4 = 5
So, Total = 1 + 2 + 3 + 4 + 5 = 15
Deleting one of the five numbers from the set then decreased 20% of the sum.
20% of the sum
=(15 x 20)/100
=3
So, the deleted number is the 3rd as c from the set
Answer :C
Solution-3:=====
Let,
C=x. b=x-1. a=x-2
d=x+1
e=x+2
Sum=x-2+x-1+x+x+1+x+2
=5x
Suppose,
The deleted number was=p
According to the question,
P=20% of 5x
P=x
Answer:C

Question-4:A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?
Solution:
Work done by both in 1 minute
= (1/20 + 1/60) = 4/60=1/15
Work done by both in 10 minute
=10/15=2/3 portion
Remaining part
=(1 - 2/3)
= 1/3 portion
Now, 1/60 part is filled in 1 minute.
    So, 1/3 part will be filled in 20 minute.
Answer:20 min