All Bank Written Math Solution 2016 bd

Pubali bank 1. A certain sum of money amounts to Tk. 1008 in 2 years and to Tk.1164 in 3 ½ years. Find the sum and rate of interest. (SO/O 16) Solution: In 3.5 years sum of money amounted to= Tk. 1164 In 2 years sum of money amounted to = Tk. 1008  in (3.5-2)=1.5 years interest = (1164-1008)= Tk. 156  in 1 year interest =Tk. 156/1.5 = Tk. 104  Sum = 1008-(104×2) = Tk. 800 We know, I=pnr/100 100×104  r =100I/pn = 800×1 % =13 % [Or,

Interest×100 Principal

=

104×100 800

= 13%]

 Rate of interest= 13% Ans: Sum Tk. 800 and rate of interest 13%. 2. A room is half as long as it is broad. The cost of carpeting the room at Tk. 5 per sq. m is Tk. 270 and the cost of papering the four walls at Tk. 10 per sq. m is Tk. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room. (SO/O 16) Solution: Let, breadth = x meters, length =(x+x/2)= 3x/2 meters and height = h meters. Area of the floor=Total cost of carpeting/Rate =(270/5) sq. m =54 sq. m Now, Breadth ×Length= Area x×(3x/2) = 54 54×2 Or, x2 = 3 Or, x2 =36  x = 6. So, breadth = 6 m and length = (3/2)×6 = 9 m. Now, papered area = (1720/10) sq. m = 172 sq. m Area of one door and two windows = 8 sq. m  Total area of 4 walls = (172 + 8) sq. m = 180 sq. m According to the question, 2(9+ 6)×h = 180 Or, 30h= 180 Or, h=180/30  h= 6  Height 6 m. Ans: breadth = 6 m, length = 9 m and height 6 m.

3. Four milkmen rented a pasture. A grazed 24 cows for 3 months; B 10 for 5 months; C 35 cows for 4 months and D 21 cows for 3 months. If A's share of rent is Tk. 720, find the total rent of the field. (SO/O 16) Solution: Ratio of shares of A: B: C: D = (24×3) : (10×5) : (35×4) : (21×3) = 72 : 50 : 140 : 63. Sum of the ratios= 72+50+140+63= 325 Let, total rent be Tk. x. 72x Then, A’s share = Tk. 325 According to the question, 72x =720 Tk. 325 720×325

Or, x= 72 Tk.  x= 3250  Total rent 3250 Tk. (Ans.) 4. The simple interest on a certain sum of money for 2.5 years at 12% per annum is Tk. 40 less than the simple interest on the same sum for 3.5 years at 10% per annum. Find the sum. (JO Cash 16) Solution: We know, Interest (I)= principle(p) × Time(n)× rate of interest(r) Here, Sum=p, n1=3.5years, r1=10%, n2=2.5 and r2= 12% According to the question, pn1r1-pn2r2=40 p×3.5×10%-p×2.5×12%=40 35p 30p Or, 100 − 100 = 40 Or, (35p-30p) 4000 Or, 5p= 4000  p= 8000  The sum is Tk. 800. (Ans.) 5. A, B and C started a business by investing Tk. 1,20,000, Tk. 1,35,000 and Tk. 1,50,000 respectively. Find the share of each out of an annual profit of Tk. 56,700. (JO Cash 16) Solution: Ratio of shares A: B:C = 120000 : 135000 : 150000 = 8: 9: 10 Sum of the ratios= 8+9+10=27 8 A’s share = ( 56700 × 27 ) =Tk. 16800 9

B’s share = ( 56700 × 27 ) =Tk. 18900 10

C’s share = ( 56700 × 27 ) =Tk. 21000 Ans: Tk. 16800, Tk. 18900 and Tk. 21000.

6. A retailer buys 40 pens at the market price of 36 pens from a wholesaler. If he sells these pens giving a discount of 1%, what is the profit percent? (JO Cash 16) Solution: Let, the market price of each pen to be Tk 10, So, the market price of 36 pens = Tk. (36×10)= Tk. 360 Buying price of 40 pens =Tk. 360  Buying price of each pen = 360/40 =Tk. 9 In 1% discount, Selling price = 99% of 10 = Tk. 9.90  Profit = 9.90 – 9.00 =Tk. 0.90 0.90×100  Profit percentage = 9.00 % = 10%  Profit is 10% (Ans.) Alternative Method: Let, 40 pens buy at Tk. p  1 ” ” Tk. p/40. 36 pens sell at Tk. p  1 ” ” Tk. p/36. In 1% discount, Selling price = 99% of 99

= 100 of 11p

= 11p

400 p

p p

36

36

Tk.

Tk.

Tk.

 Profit= ( 400 - 40 ) Tk. p

= 400 Tk. p

p

In TK. 40, profit is 400 Tk.  ” Tk. 1  ” Tk. 100

p×40

” ” 400×p Tk. ” ”

p×40×100 400×p

= 10 Tk.  Profit is 10% (Ans.)

Tk.

NRB Bank 1. A certain sum of money is Tk.900 (including principal) in 5 years and Tk.1200 (including principal) in 8 years. What's the rate of simple interest? (MTO 16) Solution: In 8 years, Principal amount + Interest= 1200 Tk. In 5 years, Principal amount + Interest= 900 Tk. [By Subtracting] In 3 years interest= 300 Tk.  In 1 year interest=Tk. 300/3= Tk. 100  In 8 years interest = Tk. (100×8) = Tk. 800 So, Principal amount= Tk. (1200-800) = Tk. 400 In 1 year, Tk. 400 gives interest Tk. 100 Tk. 1 ” ” Tk. 100/400 100×100 Tk. 100 ” ” Tk. 400 = Tk. 25  Interest rate 25%. Ans: 25% 2. A is 10 miles west to B, B is 30 miles north to C, C is 20 miles east to D. What's the distance from A to D? (MTO 16) Solution:

According to the Pythagoras theorem, AD2 = DE2 + AE2 Or, AD= 102 + 302 Or, AD2 = 1000 Or, AD= √1000  AD= 10√10 (Ans.)

Standard Bank 1. A person deposited money 6% and 7% respectively. If total Tk. 4000 money earns Tk. 250 as interest in 1 year. How much money was invested at 7% interest rate? (TAO 16) Solution: Let, at 7% invested x Tk. and at 6% invested (4000-x) Tk. According to the question, 7% of x+ 6% of (4000-x)= 250 7 6 Or,100 ×x + 100 ×(4000-x) = 250 7x+6(4000−x)

Or, = 250 100 Or, 7x+ 24000-6x= 25000 Or, 7x-6x=25000-24000  x= 1000  1000Tk. invested at 7% interest. (Ans.) 2. A trader, while selling an item, was asking for such a price that would enable him to offer a 20% discounts and still make a profit of 30% on cost. If the cost of the item was Tk.50 what was his asking price? (TAO 16) Solution: Given that, cost = 50 Tk. At 30% profit, Price becomes =130% of 100 = (50×130)/100Tk. = 65 Tk. Or, 50+ (50×30%) = 50+15 = 65TK Now, Let, cost price= 100 Tk. At 20% discount price =(100-20)= 80 Tk. When asking price is 80 Tk cost is 100 Tk. When asking price is 1 Tk cost is 100/80 Tk. When asking price is 65 Tk cost is (100×65)/80 Tk. = 81.25 Tk (Ans.) Alternative method: At 30% profit, Selling price= 130% of 50= 65 Tk. Let, asking price be Tk. x At 20% discount, Selling price=80% of x= 0.80x According to the question, 0.80x= 65 Or, x= 65/0.80  x= 81.25  Asking price is 81.25 Tk. (Ans.)

3. A sales person earns 250 Tk. fixed salary. He also gets 15% commission on total sales. What will be sales volume, if he wants to earn total 1000 Tk? (TAO 16) Solution: Given that, fixed salary= 250 Tk.  Commission= (1000-250) Tk. = 750 Tk. According to the question, 15% = 750 Tk.  1% =750/15 Tk. 750×100  100% = 15 Tk. = 5000 Tk.  Sales volume is 5000 Tk. (Ans.) Alternative method: Let, sales volume be p Tk. 15% of p +250=1000 Or, 0.15p= 1000-250 Or, 0.15p= 750  p= 5000  Sales volume is 5000 Tk. (Ans.) 4. The length of a rectangular field is 3 feet long than its breadth. If the perimeter of rectangular field is 380 feet. What will be area of the field in square feet? (TAO 16) Solution: Let, breadth be x feet and length be (x+3) feet. We know, Perimeter=2(breadth+ length) =2(x+x+3) = 2(2x+3) = 4x+ 6 According to the question, 4x+ 6=380 Or, 4x= 380-6 Or, 4x= 374  x= 93.5  Breadth= 93.5 feet and length= (93.5+ 3) feet= 96.3 feet  Area of the field= (length× breadth) = (96.5×93.5) sq. feet = 9022.75 sq. feet (Ans.)

5. A pipe can fill a tank in 30 minutes and other can drain out the tank in 40 minutes. If both pipes are started same time, how much time will be required to fill half of the tank? (TAO 16) Solution: First pipe, In 30 min can fill 1 part of the tank ” 1 ” ” ” 1/30 ” ” ” Second pipe, In 40 min can empty 1 part of the tank ” 1 ” ” ” 1/40 ” ” ” 1 1  Both in 1 min can fill = (30 - 40) part of the tank = ( 120 )

4−3

”

” ” ”

1

= (120)

”

” ” ”

1

part of the tank will be filled in 1 min  1 ” ” ” ” ” ” ” ” 120 min 1 1  2 ” ” ” ” ” ” ” ” (120 ×2) min = 60 min Ans: 60 min. 120

6. A casher received a total amount of Tk. 10,000 from depositor in a total of 140 notes of Tk. 50 and Tk. 100 denominations. How many notes of Tk. 50 denominations did he receive? (TAO Cash 16) Solution: Let, the number 50 Tk. notes be x and the number of 100 Tk. note be (140-x). According to the question, 50x+100(140-x) =10000 Or, 50x + 14,000 – 100x = 10,000 Or, 50x- 100x= 10,000-14,000 Or, -50x= -4,000 Or, 50x= 4,000 x=80  The number of Tk. 50 notes is 80. (Ans.) 7. A can dig a pond in 30 days and B can dig the same pond in 20 days. In how many days A and B can dig the pond if they work together? (TAO Cash 16) Solution: In 30 days A can dig 1 part of the pond  ” 1 ” A ” ” 1/30 ” ” ” In 20 days B can dig 1 part of the pond  ” 1 ” B ” ” 1/20 ” ” ” 1 1 In 1 day (A+B) can dig= (30 + 20) part of the pond

1

= ( 60 )

2+3

”

”

”

= 60

5

”

”

”

= 12

1

”

”

”

part of the pond can dig in 1 day 1 ” ” ” ” ” ” 12 days Ans: 12 days. 12

8. A garden is 60 meter long and 20 meter wide. Inside the garden there is a 5 meter wide path around it. What is the area of the path in square meter? (TAO Cash 16) Solution: Area of the garden without path= (60×20) sq. m. = 1200 sq. m. Area of the garden with path= (60-5-5)×(20-5-5) sq. m. = 50×10 sq. m. = 500 sq. m.  Area of the path= (1200-500) sq. m. = 700 sq. m. (Ans.)

South East Bank 1. A person invests Tk. 24000 at 7.5% interest annually. How much additional money needs to invest at 10% interest to earn overall interest at 9.25% on entire amount? (TO 16) Solution: Let, additional amount be Tk. x According to the question, 7.5% of 24,000+ 10% of x= 9.25% of (24,000+x) 7.5 10 9.25 Or, 100× 24,000 + 100× x = 100 × (24,000+ x) Or, 7.5×24,000+ 10x= 9.25(24,000+x) [Multiplying both sides by 100] Or, 180,000 + 10x= 222000+ 9.25x Or, 10x-9.25x= 222000-180,000 Or, 0.75x= 42000 Or, x= 42000/0.75  x= 56,000  Additional amount is Tk. 56,000. (Ans.) Alternative Method: Let, the total amount be Tk. x. So, additional amount= Tk. (x-24,000) According to the question, 10% of (x-24,000) + 7.5% of 24000= 9.25% of x 10(x−24,000) 7.5×24000 9.25x Or, + 100 = 100 100 Or, 10(x-24,000)+ 7.5×24000= 9.25x Or, 10x- 240,000 + 180,000= 9.25x Or, 10x- 9.25x= 240,000- 180,000 Or, 0.75x= 60,000 Or, x= 60,000/0.75  x= 80,000  Additional amount=Tk. (80,000-24,000) = Tk. 56,000 (Ans.)

2. If 7% of selling price is equal to 8% of cost price and 9% of selling price is 1 greater than 10% of cost price. What is cost price and how much money is earn by selling the product? (TO 16) Solution: Let, the cost price be Tk. A and selling price be Tk. B According to the question, 7% of B= 8% of A Or, 7B/100= 8A/100 Or, 7B=8A  B= 8A/7 And, 9% of B- 10% of A=1 Or, 9B/100- 10A/100= 1 Or, 9B-10A= 100 8A Or, 9× 7 -10A= 100 72A−70A

Or, = 100 7 Or, 2A= 700  A= 350  Cost price= Tk. 350 8  Selling price= Tk. (7 ×350) = Tk. 400  Profit =Tk. (400-350) = Tk. 50 (Ans.)

Janata Bank Ltd 1. In 2005, the number of pairs of the shoes that a company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent from that of the previous year. The company's revenue from sales of the shoes in 2005 was taka 300000. What was the revenue from the sale of the shoes in previous year? (IT Officer 16) Solution: Let, previous year, Shoes sold = x numbers and price per shoes=y Tk.  Total revenue= xy Tk. In 2005, Shoes sold = (x- 0.20x) numbers and price per shoes=(y+0.20y) Tk. According to the question, (x- 0.20x)× (y+0.20y)= 300000 Or, 0.80x ×1.20y=300000  xy = 312500  Total revenue= 3,12,500 Tk. (Ans.)

Alternative Method: Suppose, the company sold 10 pair of shoes in 2004 and price was Tk. 10 per pair. So, revenue was Tk. 10×10 = Tk. 100. In 2005, price increased to Tk. 12 and selling unit decreased to 8. So, revenue in 2005 was Tk. 12×8 = Tk. 96. Thus, actual revenue in 2004 was Tk. (3,00,000×100)/96 = Tk. 3,12,500 (Ans.) 2. A manufacturing company uses two machines A and B with different production capacities. When working alone, machine A can produce a production lot in 5 hours and machine B can produce the same lot in X hour. When the two machines operate simultaneously to fill the same production lot, it takes them 2 hours to complete the job. How many hours will the machine B take to produce the production lot alone? (IT Officer 16) Solution: Let, in x hours B can complete 1 part  in 1 hour B can complete 1/x part. In 5 hours A can complete 1 part,  in 1 hour A can complete 1/5 part. If they both worked together they can do in 1 hour=1/2 part According to the question, 1/2-1/5=1/x Or, (5-2)/10=1/x Or, 3/10=1/x Or, 3x=10  x = 10/3 = 3.33 hours or 3 hours 20 min. (Ans.) Alternative Method: Let, A can produce y products in 5 hours and B can produce y products in x hours. According to the question, y y y + = 5 x 2 1

1

y

Or, y( 5 +x) = 2 1

1

1

Or, ( 5 +x) = 2 Or, (5-2)/10=1/x Or, 3/10=1/x Or, 3x=10  x = 10/3 = 3.33 hours or 3 hours 20 min. (Ans.) 3. Interest on 500 Tk. for 4 years and 600 Tk. for 5 years is 500 Tk. What is the interest on Tk. 1000 for 3 years? (IT Officer 16) Solution: We know, Interest= Principal× Time× Rate of interest Let, rate of interest= r According to the question, 500×4×r% + 600×5×r% = 500 r r Or, 500×4×100 + 600×5×100 = 500 Or, 20r+30r= 500

Or, 50r= 500  r= 10  Rate of interest 10%  Interest= 1000×3×10% = 300 Tk. Ans: 300 Tk. 4. Sumon and Shimul two friends started a business with 5000 Tk and 4000 Tk respectively. After 3 months Mr. Sumon added 1000 Tk and simultaneously Dilip joined with them with 7000 Tk. What is the share of profit among them after one year if profit is 36000 Tk? (IT Officer 16) Solution: Equivalent amount of Sumon= 5000×3+ (5000+1000)× 9= 69,000 Tk. Equivalent amount of Shimul= 4000×12= 48, 000 Tk. Equivalent amount of Dilip= 7000×9= 63,000 Tk.  Sumon: Shimul : Dilip= 69,000: 48,000 : 63,000 = 69: 48: 63 = 23: 16 : 21 Sum of the ratios= 23+ 16 + 21= 60  Profit share of, 23 Sumon= (36000× 60 ) Tk. = 13800 Tk. 16

Shimul= (36000× 60 ) Tk. = 9600 Tk. 21

Dilip= (36000× 60 ) Tk. = 12600 Tk. Ans: 13800 Tk, 9600 Tk and 12600 Tk.

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