Bangladesh Bank Cash Officer Written 24/02/2017

1) Shakil started a business investing Tk. 25000 in 2009. In 2010, he invest ed an additional amount of Tk. 10000 and Raihan joined him with an amount of Tk. 35000. In 2011, Shakil invested another additional amount of Tk. 10000 and Jafor joined them with an amount of Tk.35000. What will be Raihan's share in profit of Tk. 150000 earned at the en d of 3 years from the start of the business in 2009? 

Equivalent amount of Shakil= 25000*3+10000*2+1000 =105000
 " Raihan = 35000*2= 70000
" Jafor = 35000*1= 35000
now S:R:J = 3:2:1
sum of ratio= 6
Raihan 's share = 150000*2/6 = 50000


2) If 9 engines consume 24 metric tonnes of coal, when each is working 8 hou rs a day, how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter t ype? 

4 later type = 3 former type
8 later type = 6 former type
consuderating consuming rate same
 x/6*13 = 24/9*8
 x= 26


3) Dawood invested certain amount in three different schemes A, B and C with the rate of 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interes t accrued in one year was Tk. 3200 and the amount invested in Scheme C was 150% of the amount in vested in Scheme A and 240% of the amount invested in Scheme B, what was th e amount invested in Scheme B ?

100% of C = 150% of A = 240% of B
so, 100C = 150 A
A: C = 2:3
 and 150% A = 240% of B
A: B = 8:5
A:B=( 8:5)*2= 16:10
A:C = (2:3)*8=16:24
A:B:C =16:10:24=8:5:12
 let, A investment = 8x
 B.. = 5x
C. ............... = 12x
 now. 8x*10/100 + 5x*12/100 + 12*15/100 = 3200
80x+60x+180x = 3200*100
 32x= 32000
5x = 32000*5/32 =5000

ONE BANK LTD Special Cadre Officer – 2017

ONE BANK LTD Special Cadre Officer -2017 

WRITTEN MATH QUESTION SOLUTION

1. Length of a rectangular field is 1.5 times as large as its width. The field cost Tk 10260 for planting grass at 1.9TK per sq.meter. What will be the total cost of providing fence around the field that costs 2.5TK per meter?

 Let the width of the rectangular field = x m and length = 1.5x m
 According to the question
1.5x * x*1.9 = 10260
O r, x^2 = 3600
Therefore, x = 60
So, width = 60m, length = 90m
Perimeter = 2(60 + 90) = 350m
Cost = 350*2.5 = Tk750 (ans)



2. Salam used a part of TK 100,000 to purchase a TV of the remaining portion he invested 1/3 of it at 4% simple annual interest and 2.3 of it at 6% simple annual interest, If after a year the income from two investment totaled Tk 320. What was the purchase price of the TV?

 Let, the purchase price of the TV = tk x
 and remaining amount of money = tk(100,0 0 0 – x)
According to the question
 (100,000 – x) * 1/3 *4% + ( 100,000 – x)*2/3 * 6% = 320
Or, 1000/3 * 4 + 12000/3 – 4x/300 – 12x/300 = 320
Or, 4000/3 + 12000/3 – 4x/300 – 12x/300 = 320
Or, 400000 + 1200000 - 4x – 12x = 320*300
Or, 16x = 1600000 – 96000
Or, 16x = 1504000
 Or, x = 94,000(ans )


3. Solve the equation : 4/(2x+3) + 15/(5x+4) = 35/(7x+6) 

4/( 2x+3) + 15/(5x+4) = 35/(7x+6)
Or, 4/(2x+3) + 15(5x+4) = 21/(7x+6) + 14/(7x+6)
Or, 4/(2x+3) - 14/(7x+6) = 21/(7x+6) – 15/(5x+4)
Or, (28x+24 - 28x - 42)/ (7x+6)( 2x+3) = (105x+84 - 105x - 90) / (5x+4)(7x+6)
Or, - 18/( 2x+3) = - 6/(5x+4)
Or, 9/(2x+3) = 3/(5x+4)
Or, 45x+36 = 6x+9
Or, 39x = - 27
Or, x = - 27/39 = - 9/13(ans) 







4. A man has to go 10 km to catch a bus. He walks part of the way at 7 km/h and runs rest of the way at 12 km/h. If he takes 1 hr and 15 mins to complete his journey find the distance he covers by walking?

 Let, he walks x km and run (10 – x) km
According to the question x/7 + (10 - x)/12 = 1hr 15mins
Or, (12x + 70 – 7x)/84 = 1+15/60
Or, 70 + 5x = 1.25*84 = 105
Or, 5x = 35
Or, x = 7 km (ans)