Sunday, September 13, 2015

Janata Bank Limited Assistant Executive Officer (Teller) 2015 Question With Complete Solution

Janata Bank Limited
Assistant Executive Officer (Teller) 2015
Mathematics

  1. What will be the compound interest on a sum of TK. 25,000 after 3 years at the rate of 12 p.c.p.a.?
Solution: Amount=[25000×(1+12/100)]
=(25000×28/25×28/25×28/25)
=35123.20
So, C.I.=( 35123.20-25000)
=10123.20
  1. In a school, 10% of the boys are some in number as ¼ th of the girls. What is the ratio of boys to girls in the school?
Solution: 10% of boy=10/100 of boy =1/4 th of girl
So, boy/girl=100/(10*4)=5/2
  1. If a number is decreased by 4 and divided by 6, the result is 8. What would be the result if 2 is subtracted from the number and then it is divided by 5?
Solution: Let, the number be x
x-4/6=8
or,x=6*8+4=48+4=52
Now, (52-2)/5=50/5=10
  1. A 70 cm long wire is to be cut into two pieces such that one piece will be 2/5 as long as the other. How many centimeters will the shorter piece be?
Solution: Let the other piece be of length x cm, so the shorter piece = 2/5 of x =2x/5
Now, 2/5 x + x =70
Or, 7/5 x=70
Or, x=50 and 2/5 x=2/5 * 50 =20 cm
  1. The Product of two numbers is 4107. If the H.C.F. of the these numbers is 37, then the greater number is :
Solution: Let the numbers be 37a and 37b
Then, 37a × 37 b = 4107
Or, ab = 3
Now, Co-primes with product 3 are (1,3)
So, the required numbers are (37 * 1, 37 *3) = (37, 111)
So, Greater number = 111
  1. What will be the ratio of sample interest earned by certain amount at the same rate of interest for  6 years and that for 9 years?
Solution: Let the principal be P and rate of interest be R%
So, Required ratio = (P×R×6/100)/(P×R×9/100)
=6 PR / 9 PR
=2/3
=2:3
  1. The average of six number is x and the average of three of these is y. If the average of the remaining three is z, then:
Solution: Average of 6 numbers = x
Sum of 6 numbers = 6 x
Average of the 3 numbers = y
Sum of the 3 numbers = 3y
Average of the remaining 3 numbers = z
Sum of the remaining 3 “  =3z
Now we know that 6x= 3y + 3z
Or, 2x = y + z
  1. In what ratio must a grocer mix two varieties of pulses costing TK. 15 and TK. 20 per kg respectively so as to get a mixture worth TK. 16.50 per Kg?
Solution: By the rules of allegation:
Cost of Cheaper                  15 TK                                                            Cost of dearer 20 TK

                                                                                Mean Price
                                                                                   16.50 TK


              (20-16.50) or, 3.50                                                                              (16.50-15) or,1.50

  1. When a commodity is sold for TK. 34.80, there is a loss of 2%. What is the cost price fo the commodity?
Solution: Sell price = 34.80; loss = 2%
Cost price = (100 × 34.80)/98 = 35.51
[Note: Correct answer not available in the question]
  1. How long will a boy take to run round a square field of side 35 meters, if he runs at the rate of 9 km/hr?
Solution: Total length to cover = 4 sides of the square  = 4 × 35 meters
9 km/hr = 9000/(60×60 sec)
Time taken = (4 × 35 × 60 × 60)/9000 sec
=56 sec
  1. The smallest 5-digit number exactly divisible by 41 is:
Solution: The smallest 5-digit number = 10000
If we divide 10000 by 41, 37 will be remainder
So, required number = 10000+ (41-37)= 10004
  1. A, B, C hired a car for TK. 520 and used it for 7, 8 and 11 hours respectively. Hire charges paid by B were:
Solution: Hire charges paid by B = 520 × (8/7+8+11) = 160 TK.
  1. The square root of 64009 is :
Solution: Square root of 64009 is 253
  1. Which of the following is not a leap year?
Solution: The century divisible by 400 is a leap year
So, 700 is not a leap year
  1. The ratio of the father’s age to his son’s age is 7:3. The product of their ages is 756. The ratio of their ages after 6 years will be:
 Solution: Let their age be 7x and 3x
Now, 7x × 3x =756
Or, x2 = 756/(7×3)= 36
Or, x = 6
      After 6 year the ratio would be  = (7x+6)/(3x+6)
      = (7*6+6)/(3*6+6)
      = 2/1
      = 2 : 1
  1. Two Pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full?
Solution: (A+B)’s 2 hour’s work when open A = 1/6 + 1/4   = 5/12
                “              1              “              “              “              = 5/12  * 2 = 5/6
Remaining work = 1-5/6 = 1/6
Now, it’s A’s turn in 5th hour
1/6 work will be done by A in 1 hour
Total time = 4 + 1 = 5 hours
  1. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with the remaining meal?
Solution: There is a meal for 200 children, 150 children have taken the meal.
Remaining meal is to be catered to 50 children
Now, 200 children = 120 men
            50       “              =(120/200 * 50) = 30 men
  1. (1000)7 ÷ 1018
Solution: (1000)7 / 1018 = {(10)3}7 / 1018 = 1021  / 1018 = 1021-18 = 103 = 1000
  1. In a 200 meters race A beats B by 35 m or 7 seconds. A’s time over the course is:
Solution: B runs 35 m in 7 sec
B covers 200 m in (7/35 * 200) = 40 sec
B’s time over the course = 40 sec
A’s time over the course = (40-7) = 33 sec
  1. 617 + 6.017 + 0.617 + 6.0017 = ?
Solution:  617 + 6.017 + 0.617 + 6.0017 = 629.6357
  1. The 15% of 40 is greater than 25% of a number by 2, then the number is :
Solution: 15% of 40 = 25% of number + 2
Or, (15*40)/100 -2 = 25/100 * number
Or, 6-2 = 25/100 * number =
Or, Number = (4*100)/25 = 16
  1. A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day?
Solution: Given that B alone can complete the same work in day  = half the time taken by A = 9 days
A’s one day’s work = 1/18
B’s “       “              “     = 1/9
(A + B)’s one day’s work = 1/18 + 1/9 = 1/6
  1. At 3.40, the hour hand and the minute hand of a clock form an angle of :
Solution: we use the formula= |11m – 60h|0 = |11*40 – 60*3|0 = 1300
  1. The length of a rectangular hall in 5 m more than it’s breadth. The area of the hall is 750 m2 . The length of the hall is :
Solution: Let, the breadth be b meter
b(b  +  5) = 750
or, b2 + 5b – 750 = 0
or, b = 25, -30
so, breadth = 25 meter and length = 25 + 5 = 30 meter
  1. What comes next in the series given?
                                    

?

Solution:
In Every Picture, one side is added. So after triangle, rectangle, pentagon there will be hexagon
 



BCS Math Solution, Worker, Time related Math Shortcut Method

Worker Labor Related Shortcut Method-1
Worker Labor Related Shortcut Method-1
https://www.youtube.com/watch?v=WoX5tiLmQeo
Worker Labor Related Shortcut Method-2
Worker Labor Related Shortcut Method-2
Shortcut Method Worker, Time Related Math

BCS Math Solution, Tank Related Math Shortcut Technic

Tank Related Shortcut Method-1
Tank Related Shortcut Method-1
Tank Related Shortcut Method-2
Tank Related Shortcut Method-2

Shortcut Method Tank Related Math

Thursday, September 10, 2015

Middle Ages of Bengali Literature Part-2

Middle Ages of Bengali Literature part-2-1
Middle Ages of Bengali Literature part-2-1
Middle Ages of Bengali Literature part-2-2
Middle Ages of Bengali Literature part-2-2

Middle Ages of Bengali Literature part-2-3
Middle Ages of Bengali Literature part-2-2

History of Bengali Literature, Ages and Charjapad

History of Bangla Literature-1
History of Bangla Literature-1

History of Bangla Literature-2
History of Bangla Literature-2

History of Bangla Literature-3
History of Bangla Literature-3

History of Bangla Literature-4
History of Bangla Literature-4

History of Bangla Literature-5
History of Bangla Literature-5

History of Bangla Literature-6
History of Bangla Literature-6

Bengali Literature of the Middle Ages

Bengali literature of the Middle Ages

Bengali Literature of the Middle Ages-1
Bengali Literature of the Middle Ages-1
Bengali Literature of the Middle Ages-2
Bengali Literature of the Middle Ages-2
Bengali Literature of the Middle Ages-3
Bengali Literature of the Middle Ages-3
syllabus-for-english-from-35th BCS
 

Sample text

Sample Text

Sample Text

 
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