Bank Written 90 Math Bank

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1. In a two digit number, the digit in the units place is more than twice the digit in ten's place by

1. If the digits in the units place and the ten's place are interchanged, the difference between the

newly formed number and the original number is less than the original number by 1, what is the

original number? (JBL AEO 15)

Solution>>>

Let, tens digit = x, so units digit = 2x+1, the number is = 10x+2x+1 & the alternate number = 10(2x+1)+x

According to question,

10(2x+1)+x{10x +(2x+1)} = 10x+(2x+1)-1 Or, 20x+10+x-10x-2x-1 = 12x

Or, 9x+9 = 12x

Or, x = 3

So, the number is = 10x3+2x3+1=37 Ans.

2. A, B & C Started a business each investing tk 20,000/=, after Five months A withdraw tk

5000, B withdraw tk 4000 and C invest tk 6000 more. At the end of the year a total profit of tk

69,900/= was recorded. Find the share of each. (JBL AEO 15)

Solution>>>

Let us consider, k=1000tk

then, ratio is, A:B:C=20k*5+15k*7: 20k*5+16k*7:20k*5+26k*7

=205k:212k+282k

=205:212:282

sum of the ratio=205+212+282=699

profit,

A=205*69900/699

=20500

B=212*69900/699

=21200

C=282*69900/699

=28200 Ans.

3. A machine P can print 1 lakh books in 8 hour, Q can same in 10 hour & R can print them in 12

hour. All the machine r started at 9 AM, while machine P in closed at 11 am and the remaining

two machine complete the work. Approximately at what time the work will be finished? (JBL

AEO 15)

Solution>>>

(P + Q + R)'s 1 hour's work = (1/8 + 1/10 + 1/12) = 37/120

Work done by P, Q and R in 2 hours = (37 x 2)/120 = 37/60

Remaining work = (1 - 37/60) = 23/60

(Q + R)'s 1 hour's work = (1/10 + 1/12) = 11 /60.

Now, 11/60 work is done by Q and R in 1 hour.

So, 23/60 work will be done by Q and R in (60 /11x 23/60) = 23/11 hours = 2 hours.

So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M. Ans.

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4. 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? (BB AD 14)

Solution>>>

Let, the consecutive numbers are,

1, 1+1=2, 1+2=3, 1+3=4 & 1+4=5

So, total = 1+2+3+4+5 = 15

Deleting 1 of the 5 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 from the set Ans.

5. Rahim bought 2 varieties of rice costing tk 5 & 6 per kg each. Then he sold the mixture at

tk7/kg, making profit of 20%. What was the ratio of the mixture? (BB AD 14)

Solution>>>

Let, Rahim bought xkg rice at tk 5, so his cost = 5x tk & in same way other variety cost = 6y tk

According to ques.

(5x + 6y) + (5x + 6y)20/100 = 7(x + y)

Or, (5x + 6y)(1 + 1/5) = 7x + 7y

Or, (5x + 6y)6/5 = 7x + 7y

Or, 30x + 36y = 35x + 35y

Or, y = 5x

Or, x/y = 1/5, so the ratio is x:y = 1:5 Ans.

Another Way:

Profit = (7x+7y) (5x+6y)tk = (2x + y)tk According to the question,

2x + y = (5x+6y)20/100

or, 10x + 5y = 5x + 6y

or, 5x = y

Or, x/y = 1/5.

So, X:Y = 1:5 Ans.

6. A team of 2 men and 5 women completed 1/4th of a job in 3ds. After that another man joined

them and they all complete the next 1/4th of the job in 2ds. How many men can complete the

whole job in 4 ds? (BB AD 14, Pubali SO 13)

Solution>>>

Here, in 3ds, 2 men & 5 women do part

Again, in 2ds, 3 men & 5 women do part

or in 1d 3 men & 5 women do 1/8 part

or, in 3d 3 men & 5 women do 3/8 part

so, 1 mans 3 day work = 1/4 3/8 = 1/8 part

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now, 1/8 part work done in 3ds by 1 man

1/8 part work done in 1d by 1x3 men

1(whole) part work done in 1d by 3x8 men

1(whole) part work done in 4ds by 3x8/4 men = 6 men Ans.

7. A trader bought some mangoes for tk150/dozon and equal number of apples for tk 100/doz. If

he sells all the fruits tk 140/dozon, what will be his profit/loss in percentage? (RKUB Senior,

2014)

Solution>>>

Here, cost price of 1dozon mango = 150tk & cost price of 1dozon apple = 100tk

So, cost price (1dozon mango + 1dozon apple) = 150 + 100 = 250tk

Cell price (1dozon mango + 1dozon apple) = 140 x 2 = 280tk

Profit = 280 250 = 30tk Percentage of profit = (30 x 100)/250 = 12% Ans.

8. Two partners A & B have 70% and 30% share in a business. After sometimes, a third partner

C joined by investing tk 10 lakh and thus having 20% share in the business. What is the

percentage of share of A now in the business? (RKUB Senior, 2014)

Solution>>>

Ratio of A & B, A:B = 70%:30% = 7:3

After joining C, he got 20% share, so then total share of A&B = 100-20 = 80%

Now As share is = (7 x 80%)/10 = 56% Ans. Another way:

20% share = 10lakh, 100% share = (10 x 100)/20 = 50 lakh

Here, share of A&B = 50 10 = 40lakh As investment = (7 x 40)/10 = 28 lakh So, As percentage after joining C = (28 x 100)/50 = 56% Ans.

9. The average weight of three men A, B and C is 84kg. Another man D joins the group and the

average becomes 80kg. Another man E, his wt is 3 kg more than that of D, replaces A. Now

average wt of B, C, D & E becomes 79kg. What is the weight of A? (RKUB Senior, 2014)

Solution>>>

A+B+C = 84 x 3 = 252kg

A+B+C+D = 80 x 4 = 320kg

So, D = 320 252 = 68kg, E = 68 + 3 = 71kg B+C+D+E = 4 x 79 = 316

Or, B+C+68+71 = 316 or, B+C = 316 (68+71) = 177 So, A = 252 177 = 75 Ans.

10. A rectangle PQRS inscribed in a circle and PQ = 6. If the area of the rectangular region is 48,

what is the area of the circular region? (RKUB Senior, 2014)

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Solution>>>

Here, area of PQRS = 48, PQ = 6, so, PQ x QR = 48 or, QR = 8

Now, if we connect P to R through a line, a right triangle PQR is formed.

From Pythagoras theory we found,

PR^2 = PQ^2 + QR^2

Or, PR = (6^2 + 8^2)^1/2 = 10

PR = Diameter of the circle = 10

So, area of the circle = x (10/2)^2 = 25 Ans.

11. If 5 is added to the sum of two digits of a number consisting of two digits, the sum will be

three times the digits of the tens place. Moreover, if the place of the digits is interchanged, the

number thus found will be 9 less than the original number. Find the number. (RB SO, 2013)

Solution>>>

Let, ten's digit = x & unit's digit = y, the number is = 10x+y.

According to the question,

x+y+5 = 3*x

or, y = 2x-5 (1) again, 10y+x = 10x+y-9

or, 9y = 9x-9

or, y = x-1

or, 2x-5 = x-1 [from (1)],

so, x = 4, & y= 4-1 = 3.

The number is = 10*4+3 = 43 (ans)

12. If a^2 - root3a + 1 = 0, what is the value of a^3 + 1/a^3? (RB SO, 2013)

Solution>>>

a^2 - root3a + 1 = 0

or, a(a+1/a - root3) = 0

or, a+1/a - root3 = 0

or, a + 1/a = root3..............(1)

now, a^3 + 1/a^3

= (a+1/a)^3 - 3a. 1/a (a+1/a)

= (root3)^3 - 3* root3* root3 [from (1)]

= 3root3 - 3root3

= 0. (ans)

13.

3 1200sq.m 50cm ? (SBL SO 14)

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Solution>>>

, = x, = 3x,

, 3x*x = 1200, or, x = 20m & = 60m

= 2(60+20) = 160m =

, , 4A = 160 or, A = 40m & = 40^2 = 1600sq.m

= 50cm = 0.5m, = 0.5x0.5 = 0.25sq.m

= 1600/0.25 = 6400 Ans.

14. &

? (SBL SO 14)

Solution>>>

= =

= / =

= * =

, = =

= ( * )/ = .

. % Ans.

15. 20%

3 2

? (SBL SO 14)

Solution>>>

, = x, = x*20/100 = x/5

3 = x + 3*x/5 = 8x/5

, = y, 2 = y + 2*y/5 = 7y/5

,

=

8x/5 = 7y/5

, x/y = 7/8 Ans.

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16. 13.5% 5 8375 ?

10400 ? (SBL JO 14)

Solution>>>

, P = 100*A/100+rt, A= , r= , t=

, = (100*8375)/100+13.5*5 = 5000

, 10500 = 10400-5000 = 5400

5000 1 = 5000*13.5/100 = 675

675 1

5400 = 5400/675 = 8 Ans.

17.

, ? (SBL JO 14)

Solution>>>

= * = .

= ( + * )*( + * )= * = .

= - = .

= ( * ) = Ans.

18. Mr. Zaman defeated Mr. Younus in a vote where the ratio of their vote was 4:3, total number

of voters was 581 of which 91 didnt vote. Find out the margin of defeat.

Solution>>>

Casting of vote = 581-91=490, vote for Zaman = 490*4/7 = 280, Yonus = 210

So, margin of defeat = 280-210=70 Ans.

19. A mixture of 20kg spirit and water contains 10% water. How much water is to be added to

increase the water up to 25%?

Solution>>>

Here, in 20kg solution amount of water = 20*10/100 = 2kg & amount of spirit = 20-2 = 18kg

Now, in new solution, amount of water will be changed but amount of spirit will be the same.

Percentage of water will be = 25% & percentage of spirit will be = 100-25 = 75%

So, 75% spirit = 18kg, or, 100% spirit = 18*100/75 = 24kg.

Water should be added = 24-20 = 4kg. Ans.

Same problem: % % ? Ans:

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20. 15 years hence A will be twice of his son, 5 years ago he was 4th times of his son, what is

their present age?

Solution>>>

Let us consider, present age of father = x & son = y

5 yrs ago fathers age, x-5 = 4(y-5) or, x = 4y-15

Again, after 15yrs fathers age,

x+15 = 2(y+15)

4y-15+15 = 2y+30

Y = 15, so father x = 4*15-15 = 45 Ans.

21. A series has 3 numbers a, ar, ar^2. In the series, the first term is twice of the second term.

What is the ratio of the sum of the first 2 terms to the sum of the last 2 terms? (BB AD 12)

Solution:

Let the 3rd term, ar^2 = x, so, 2nd term = 2x, 1st term = 4x

So, a + ar = 4x + 2x = 6x

ar + ar^2 = 2x + x = 3x

ratio of first two terms and last two terms = 6x : 3x = 2:1 Ans.

22.Two alloys A and B are composed of two basic elements. The ratios of the compositions of

the two basic elements in the two alloy are 5:3 and 1:2. A new alloy X is formed by mixing the

two alloys A & B in the ratio 4:3. What is the ratio of the composition of the two basic elements

in alloy X? (BB AD 12)

Solution:

Let the amount of A = 4x and amount of B = 3x in alloy X

amount of A in the new alloy X = (5/8)*4x + (1/3)*3x = (7x)/2

amount of B in the new alloy X = (3/8)*4x + (2/3)*3x= (7x)/2

so ratio A to B = [(7x)/2 ] / [ 7x/2] = 1/1 = 1:1 Ans.

23. A trade where selling an item was asking for such a price that would enable him to offer a

10% discount and still make a profit of 20%. If the cost of the product was tk 50, what was his

asking price? (BB Officer 01)

Solution:

If the cost price was 50tk then @20% profit selling price is = 50 + 50*20/100 = 60tk

@ 10% discount

If the selling price is 90 then asking price is 100

If the selling price is 1 then asking price is 100/90

If the selling price is 60 then asking price is (100*60)/90 = 66.67 Ans.

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24. Mr. X pays 10% tax on all income over 60000tk bt he does not pay any tax on interest on

postal saving certificate. In 2000 he paid 7500 as tax & he earned 12000 as interest on postal

savings account. What is his net income in 2000? (BB Officer 01)

solution:

Mr. X gives 10% taxes upon his income, so,

If tax is 10tk then income is 100tk

If tax is 7500tk the income is (100*7500)/10 = 75000tk

His total income in 2000 = 75000 + 60000 + 12000 = 147000tk

So, his net income in 2000 is = 147000 7500 = 139500tk Ans.

25. A candidate answered all 22q in a test & received 63.5 marks. If the total marks were derived

by adding 3.5 marks each correct answer and deducting 1 mark for each incorrect answer, how

many q did the student answer correctly? (BB Officer 01)

Solution:

If all the answers were right he might got = 22*3.5 = 77

So, his deducted number is = 77 63.5 = 13.5 Number deducted for each wrong answer = 1+3.5 = 4.5

Quantity of given wrong answer = 13.5/4.5 = 3

So, the candidate answered correctly = 22 3 = 19 q. Ans.

26. in an organization 30% of all employees live over 10miles away from the place of work &

60% of worker who live who live over 10miles use company transport. If 40% of employees of

the company use company transport, what percent of the employees live 10miles or less from

work and use company transport? (BB Officer 01)

Solution:

Let total employees = 100, employees live over 10 miles = 30

So, employees live in 10 mile or less = 100 30 = 70 Total 40% or 40 employees use transport

Now, employees live over 10miles using transport = 60% of 30 = (60*30)/100 = 18

So, employees live in 10miles or less using transport = 40 18 = 22 Here, total 70 employee live in 10miles or less and among them 22 use company transport

So, percentage of using transport of employee who live in 10miles or less = (22*100)/70 =

31*3/7% Ans.

27. Mr. Reach sold two properties P1 & P2 for tk 50000 each. He sold property P1 for 20% more

then what he paid for it & sold P2 less than 20% what he paid for it. What was his total gain or

loss, if any, on the scale of two properties? (BB Officer 01)

Solution:

In case of P1 @ 20% profit

If selling price is 120tk then buying price 100tk

If selling price is 50000tk then cost price = (50000*100)/120 = 500000/12 tk = 125000/3

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In case of P2 @ 20% loss

If selling price is 80tk then buying price 100tk

If selling price is 50000tk then cost price = (50000*100)/80 = 62500tk

Now, total sell price = 50000*2 = 100000tk

Total cost price = 125000/3 + 62500 = 312500/3 = 104166.67tk

So, loss = 104166.67 100000 = 4166.67tk Ans.

28. Mr. X has a investable amount of tk 100000, he will invest the amount for two years. He has

two options. He can invest at simple interest rate of 12% per annum, alternatively he can invest

at compound rate of 10% (semi annually). Calculate the earnings at two option and advice him.

(BB Officer 01)

Solution:

At simple interest, Mr. X can get = (12*100000*2)/100 [formula, I = rpt/100]

= 24000tk

At 10 compound amount = 100000(1+10/100)^4, [formula, CA = P(1+r/100)^t](t=4 because of

semi annually)

= 121550.6tk

So, compound interest = 121550.6 100000 = 21550.6tk So, simple interest is better option for Mr. X Ans.

29. A total of 50 employees work in a branch. Of them 22 have taken the accounting course, 15

have taken finance, 14 marketing. 9 of them taken exactly 2 of the courses, 1 of them has taken

all. How many of the 50 employees have taken none of this? (BB AD 01)

Solution:

Here, 1 of the employee has taken all of the courses

9 of the employee have taken exactly 2 of the courses

Number of employee have taken only Accounting = 22 (9+1) = 12 Numbers of employee have taken only Finance = 15 (9+1) = 5 Numbers of employee have taken only Marketing = 14 1 = 13

Numbers of total employee have taken 1, 2 or 3 courses = 12+5+13+9+1 = 40

So, employees who have not taken any course = 50 40 = 10 Ans.

30. 3 partners A, B & C start a business. Twice the investment of A is equal to thrice the capital

of B is 4 times the capital of C. Find the share of each out of a profit of tk. 297000. (Rupali SO

13)

Solution

let, 2A = 3B = 4B = x.

A = x/2,

B = x/3 &

C = x/4.

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Ratio = A:B:C = x/2 : x/3 : x/4

= 1/2 : 1/3 : 1/4 = 6:4:3.

A's share of profit = 297000*6/13

= tk. 137077.

B's share of profit = 297000*4/13

= tk. 91385.

C's share of profit = 29000*3/13

= tk. 68538. Ans.

31. 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 & the time of flight increased by 30 minutes. Find out the

duration of the flight. (Rupali SO 13)

solution

let, usual speed of the flight is X km/hr.


[600/(x-200)] - 600/x = 30mnts = 1/2hr,

12000/(x^2 - 200x) = 1/2,

x^2 - 200x = 24000,

x^2 - 200x + 1000 = 25000,

(x-100)^2 = (500)^2,

x-100 = 500,

x = 600.

Duration of the flight

= (600/600 hr + 1/2 hr)

= (1+1/2)hr = 3/2 hour Ans.

32. 2 pipes A & B can fill a tank in 36 minutes & 45 minutes respectively. Waste pipe C can

empty the tank in 30 minutes. 1st A & B are opened. After 7 minutes, C is also opened. In how

much time, the tank is full? (Rupali O 13)

solution:

In 1 min,A & B together can fill =(1/36+1/45)=1/20 part.

In 7 min's they can fill=7/20 part.

Remaining =(1-7/20)=13/20 part.

IN 1 min C can empties=1/30 part.

After 7 min's,every 1 min tank fill=(1/20-1/30)=1/60 part.

SO, Tank fill 1/60 part in 1 min.

Tank fill remaining 13/20 part in=60*13/20=39 min's.

so total time require to fill tank=(7+39)=46 minutes. Ans.

33. If 5 is added to the sum of two digits of a number consisting of two digits, the sum will be

three times the digits of the tens place. Moreover, if the places of the digits are interchanged, the

number thus found will be 9 less than the original number. Find the number. (Rupali SO 13)

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Solution

Let, ten's digit = x

& unit's digit = y

The number is = 10x+y.


x+y+5 = 3*x,

y = 2x-5..........(1)

& 10y+x = 10x+y-9,

9y = 9x-9,

y = x-1,

2x-5 = x-1 [from (1)],

x = 4, & y= 4-1 = 3.

The number is = 10*4+3 = 43 Ans.

34. If a^2 - root3a + 1 = 0, what is the value of a^3 + 1/a^3 ? (Rupali SO 13)

Solution

a^2 - root3a + 1 = 0,

a(a+1/a - root3) = 0,

a+1/a -root3 = 0,

a + 1/a = root3..............(1)

now, a^3 + 1/a^3

= (a+1/a)^3 - 3a. 1/a (a+1/a)

= (root3)^3 - 3* root3* root3 [from (1)]

= 3root3 - 3root3

= 0. Ans.

35. Akti ayotokar ghor er doyrgho prostho opekkha 14 m beshi. Ghortir porishima 72 m hole

ghortir porishima r soman porishimar kono borgokhetrer khetrofol koto? (Sonali SO 13)

Solution

Ayotokar ghor er porishima = borgokhetrer porishima = 72 m.

Borgokhetrer bahur doyrgho = 72/4 m = 18 m.

Borgokhetrer khetrofol = (18)^2

= 324 sq.m Ans.

36. Sud-Asol e 5 years e tk. 525 & 8 years e tk. 660 hoi. Shotkora sud er har & asol koto? (Sonali

SO 13)

Solution

(8-5) = 3 years e sud = (660-525) = tk. 135

1 ,, ,, ,, = tk. 135/3

5 ,, ,, ,, = tk.(135*5)/3 = tk. 225.

Asol = tk. (525-225) = tk. 300.

Tk. 300 er 5 years e sud tk. 225

tk. 1 ,, 1 ,, ,, =tk. 225/300*5

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tk. 100 ,, 1 ,, ,, = tk. (225*100)/1500 = 15%.

Suder har 15% & asol tk. 300 Ans.

37. A simple interest rate of a bank was reduced to 5% from 7%. As a consequences Mr. Bs income was reduced by tk 2100 in 5 yrs. How much is Mr. Bs initial deposit in the bank? (BB AD 01)

Solution:

Here, reduced interest rate = 7% - 5% = 2%

In 5 yrs interest reduced = 2100 tk

In 1 yr interest reduced = 2100/5 = 420 tk

Now, if interest reduced 2 tk then deposit is 100 tk

If interest reduced 1 tk then deposit is 100/5 tk

If interest reduced 2100 tk then deposit is (100*2100)/5 tk

= 21000 tk Ans.

38. Mr. A purchased a house for tk 1000000 tk in 1995, he spent 100000 tk for routine

maintenance & upkeep of the house. 1n 1999 he sold the house for 25% of more then what he

paid for it. He paid 5% of the proceeds as gain tax & he has to pay 50% of his net profit to the

broker, what is his net income? (BB AD 01)

Solution:

Here, purchasing cost + routine maintenance & upkeep cost

= (1000000 + 100000) = 1100000 tk

Profit gained @ 25% of total cost = (1100000*25)/100 = 275000 tk

5% gain tax = (5*275000)/100 = 13750 tk

So, net profit after deducting gain tax = (275000 13750) tk = 261250 tk

Net income after deducting brokers commission = 261250/2

= 130625 tk Ans.

39. A trader sells on an average 18 pencils and 12 pens per day. The profit comes from pencil is

1/3rd of the profit made from selling a pen. If he makes profits of tk 900 in a month by selling

pencils, how much profit does he make per month by selling pens? The trader sells 30ds in a

month. (BB AD 01)

Solution:

Let, Profit from 1 pen = x tk so, 12*30 pen = 360x tk.

Profit from 1 pencil = x/3 tk, 18*30 pencil = 540x/3 = 180x tk.

According to ques. 180x = 900 or, x = 5 tk

So, profit from pen = 360*5 = 1800 tk. Ans.

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40. A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together

for 5 days and the rest of the work is done by C in 2 days. If they get Tk.450 for the whole work,

how should they divide the money? (BASIC-2014)

Solution:

A's 1 day work=1/10, his 5 days work = 5/10 = 1/2 of the work

B's 1 day work =1/15, his 5 days work = 5/15 = 1/3 of the work

1 day work of (A+B) =1/10+1/15 = 1/6 so, 5 days work = 5/6

Remaining work=1-5/6=1/6

So, C has done 1/6 of the work

So, As money = 450/2 = 225 tk, B = 450/3 = 150 tk, C = 450/6 = 75 tk Ans.

41. A sum of Tk. 1260 is borrowed from a money lender at 10% p.a compounded annually. If the

amount is to be paid in two equal annual installments, find the annual installments. (BASIC-

2014)

Solution:

We know, compound amount = P(1+r/100)^t

= 1260(1+10/100)^2

= 1260(1.1) ^2

= 1260*1.21 =1524.6

So, annual installment = 1524.62=762.3tk. Ans.

42. Rahim, Karim & Gazi 3jon 1ti kaj korte pare jothakrome 15, 6 & 10 days e. Tahara akotre

3jon oi kajti koto din e sesh korbe? (SBL O 13)

Solution:

/

/ /

= / + / + / = ( + + )/ = /

, = /

, , = Ans.

43. How much interest will tk 1000 earn in 1 year @ an annual interest rate of 20% if interest

rate is compounded every 6 months? (BB AD 10, Rupali SO 13)

Solution:

We know, compound amount = P(1+r/100)^t

= 1000 (1+10/100)^2 [t=2 bcz interest rate is compounded

in every 6 months]

= 1000 * (1 + 1/10) ^2

= 1000*(1.1)^2 = 1000*1.21 = 1210 tk.

So, interest = 1210 1000 = 210 tk. Ans.

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44. A train went 300 miles from X to city Y @ an average speed of 80mph. At what speed did it

travel on the way back if it its average speed for the whole trip was 100mph?

Solution:

The train, started its journey from X to Y, than came back to X again

So, total distance covered by it = 300*2 = 600 mile

Here, average speed of total journey is 100 mph, so, total time spent = 600/100 = 6 hr.

Time req. to go from X to Y at 80mph = 300/80 = 3.75 hrs.

Time req. to comeback from Y to X = 6 3.75 = 2.25 hrs. So, speed of travelling on way back = 300/2.25 = 133.33 mph Ans.

45. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the

same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be?

Solution :

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y = 1/10

and 26x + 48y = 1/2.

Solving these two equations, we get: x =1/ 100 and y = 1/200 .

(15 men + 20 boy)'s 1 day's work =15/100 + 20/200 = 1/ 4

15 men and 20 boys can do the work in 4 days. Ans.

46. An article is sold at 20% profit. Its cost price is increased by tk. 50 and at the same time if its

selling price is also increased by tk. 30, the percentage of profit decreases by 3.33333. find the

cost price? (Pubali SO 13)

Solution: Cost = p

Profit = 0.2p

Selling price = 1.2p

New:

Cost = p+50 [cost increases by 50 tk]

Profit = (20 - 3.33)% of (p+50) = 16.67% of (p+50)= 0.1667p + 8.33

Selling price = 1.2p + 30 [s.p. increases by 30 tk]

According to the question:

Cost + profit = selling price

p+50 +( 0.1667p + 8.33) = 1.2p + 30

0.0333p = 28.33

p = 850 tk Ans.

47. A picnic was attended by 240 persons. There were 20 more men than women and 20 more

adults than children. How many men were there in the picnic?

Solution:

First condition : Men+Women+Children =240

Second Condition: Men=Women+ 20

15

Women= Men-20-----(1)

Third Condition: (Men+Women)=Children+ 20

Men+Women-20=Children

Men+(Men-20)-20=Children [putting the value of eq (1)]

2Men-40=Children....(2)

Putting the value of eq. (1) and (2) in condition 1, we get ,

Men+(Men-20)+(2Men-40)=240

Men= 75 Ans.

48. The ratio of the number of boys and girls in a school is 3:2. If 20% of the boys and 25% of

girls are scholarship holders, what % of the students does not get scholarship?

Solution:

Let total Student 100

Boys = 100*3/5= 60

Scholarship holder = 60*20%= 12

Girls = 100*2/5=40

Scholarship Holder = 40*25%= 10

Total Scholarship holder = 12+10= 22

Does not get Scholarship: 100-22=78% Ans.

49. A salesman's commission is 5% on all sales up to tk 10000 and 4% on all sales exceeding

this. He remits tk 31100 to his parent company after deducting his commission. Find the total

sales.

Solution :

let total sales =s

First condition : 10000x5%= 500

Second Condition : (s-10000)x4%=0.04s- 400

According to the question ,

S-(500+0.04S-400)= 31100

S-0.04S-100=31100

0.96S= 31100+100=31200

S=31200/0.96

S=32500 tk Ans.

50. A,B and C do a job alone in 20,30,and 60 days respectively. In how many days can A do the

job if he is assisted by B & C?(BB AD 13)

Solution:

(A+B+C) work 1 days=1/20+1/30+1/60 = 1/10

With help of B & C A do 1/10 part work in 1 day

,, ,, ,, ,, ,, ,, 1 (whole) part in 10 days Ans

16

51. A bus is traveling with 52 passengers. When it arrives at a stop, Y passengers get off and 4

get I at the next stop one-third of the passengers get off and 3 get on. There are 25 passengers.

Find out how many passengers got off at the first stop? (BB AD 13)

Solution:

According to q,

(52-Y+4) *2/3 + 3 = 25

Or, Y= 23 Ans.

52. An Eskimo leaves its igloo and travels 3 km north, then 8 km east and finally 3 km north to

reach the north pole. How many km does he have two travels to return to his igloo in a straight

line? (BB AD-13)

Solution:

Here, 1st go A to B -3 Km

Then go B To C -8Km

Finally, go C to D -3km

A to D = ?

Now using the formula of Pythagoras, we get

AD^2 = AE^2 + ED^2

Or, AD^2 = 8^2 + 6^2

Or, AD=10 Ans.

53. If sugar price reduced 25/4%, then one can buy 1kg more sugar at 120tk. Find the rate of

original and reduced price. (JBL Cash 15)

Solution:

Let, the original rate = xtk/kg, so in 120 tk sugar can be found = 120/x kg

Now, 25/4% discount in x tk = 25x/4*100 = x/16 tk

So, discount rate = x - x/16 = 15x/16 tk per kg and in 120 tk we can found = 120*16/15x kg

sugar


120*16/15x - 120/x = 1 Or, 120/15x = 1, or x = 8 tk/kg

So, discount price is = 15*8/16 = 7.5 tk/kg Ans.

54. If 2 men and 3 boys can do a piece of work in 10 days; and if 3 men and 2 boys can do the

same piece of work in 8 days, then 2 men and 1 boy can do that work in how many days? (JBL

Cash 15)

Solution:

In I day

2 men and 3 boy do = 1/10 part (1) 3 men and 2 boy do = 1/8 part .. (2) _____________________________________

17

5 men and 5 boy do = 9/40 part

So, 1 man and 1 boy do = 9/200 part in 1 day . (3)

We find from (2) (3) 2 men and 1 boy do = 1/8 9/200 = (25 9)/200 = 2/25 part So, 2 men and 1 boy do 2/25 part work in 1 day

2 men and 1 boy do 1 part or whole work in 25/2 days = 12.5 days Ans.

Alternate way:

(2M+3B)10=(3M+3B)8

So, 4M=14B

We can say 2 men = 7 boy

2 men and 3 boys or 10 boys can do the work in 10 days

So 1 boys can do the work in 10*10 days

So 2 men and 1 boy or 8 boys can do the work in 10*10/8 days

So 12.5 days Ans.

55. A total amount of 1550 tk was invested in two parts. One part is 8% rate and the other part is

6% rate. If the annual income is tk 106, then how much money was invested in each part? (JBL

Cash 15)

Solution:

Let, 8% invest is = x

6% invest is =1550 x The interest on tk x = 8x/100 tk

And interest on tk (1550 x) = 6(1550 x)/100 tk


(1550 - x)*6/100 + 8x/100 = 106

=> (1550-x)*6 + 8x = 106*100

=>9300 - 6x + 8x = 10600

=>2x = 10600 - 9300

=>x = 650

8% invest is = 650

6% invest is = 1550 650 = 900 Ans.

56. A and B together can do a piece of work in 12 days, which B and C together can do in 16

days. A has been working at it for 5 Days and B for 7 Days, C finishes in 13 days. In how many

days C alone will do the work?

Solution:

5 days work of A and B=5/12 so B's (7-5) or 2 days remain which is done with C so 2 days work

of B&C=2/16=1/8 Now C's remain (13-2) or 11 days Now remain part=1-(5/12+1/8)=11/24

which is done by C in 11 days so whole part needs for C=11*24/11=24 days

18

57. A,B and C enter into partnership by investing in the ratio of 3:2:4.After one year, B invests

another tk 270000.At the end of three years profits are shared in the ratio of 3:4:5. Find the initial

investment of each.

Solution:

Say, A,B,C initial investment=3x,2x,4x

ATQ,

3x*3:[2x*1+(2x+270000)2]:4x*3=3:4:5

9x:2x+4x+540000:12x=3:4:5

9x:6x+540000:12x=3:4:5

Now,

9x/6x+540000=3/4

36x=18x+1620000

18x=1620000

x=90000

so A's initial investment=3*90000=270000 tk

B's " " =2*90000=180000 "

C's " " =4*90000=360000 " (ans)

58. A & B can do a work in 12 days where B & C can do the job in 16 days. A worked for 5

days, B worked for 7 days and C did the remaining job in 13 days. How many days would

require for C to do the job alone? (BCBL 2015)

Solution:

A&B do in 12 days =1 work

" " " 1 day =1/12 "

" " " 5 days =5/12 "

Again B&C do in 16 days= 1 work

" " " 1 day =1/16 "

" " " 2 days=2/16=1/8 work

Work remaining=1-5/12-1/8=11/24 work

11/24 work completed by C in = 11 days

1 " " " C " =11*24/11~24 days (ans)

59. Mr. Nader drove for Mymensingh to Dhaka @ 60 miles/hr. Returning on the same route

there was a lot of traffic and he was only able to drive @ 40 miles/hr. If the return trip took 1hr

longer, what is the distance between Dhaka & Mymensingh? (BASIC AM 12)

Solution:

Let the distance between Dhaka & Mymensingh = X km

We know, Speed = Distance/Time or, Time = Distance/Speed


x/40 x/60 = 1 or, (60x 40x)/2400 = 1 or, 20x = 2400

or, x = 120 miles Ans.

19

60. A businessman before closing his shop, counted the money kept in the cash box and found

there were X number 50 paisa coin, X number of 1 tk notes, X number of 2 tk notes and X

number of 5 tk notes. Apart from this there is nothing in the box. The next day when he opened

the shop he founds that the cash box had been stolen. As he was insured, he got tk 1615 which is

95% of the stolen money from the insurance company. How many 2 tk notes were in the box?

(BASIC AM 12)

Solution:

Here, 95% of stolen money = 1615 tk

So, 100% of the stolen money = (1615*100)/95 = 1700tk


0.5x + 1x + 2x + 5x = 1700

Or, x(0.5 + 1 +2 + 5) = 1700

Or, x = 1700/8.5 = 200 Ans.

61. %

? (BKB CO 12)

Solution:

%, = = %

= + =

, % =

% = ( * )/ = Ans.

62. % %

? (BKB CO 12)

Solution:

= + =

, =

%

=

= ( * )/ =

%

=

20

= ( * )/ =

, = + =

= = Ans.

63. A person spends 1/3rd of the money with him on food, 1/5th of the remaining on education,

1/4th of the remaining on treatment. Now he is left with tk 200. How much did he have with him

in the beginning? (One Bank Officer 12)

Solution:

Let us consider initial money = x tk

1/3rd of the money is spent for food, so remaining money = x x/3 = 2x/3 1/5th of the remaining is spent for education = 1/5*2x/3 = 2x/15,

Remaining money = 2x/3 2x/15 = 8x/15 1/4th of the remaining money spent on treatment = * 8x/15 = 2x/15

Remaining money = 8x/15 2x/15 = 2x/5 Acq. to ques. 2x/5 = 200 or, x = (200*5)/2 = 500 tk Ans.

64. Suppose 81p + 62q = 138 and 62p + 81q = 5, find out the value of p & q. (One Bank Officer

12)

Solution:

Here,

81p + 62q = 138 (1) 62p + 81q = 5. (2) We find from (1)*62 and 2*81

5022p + 3844q = 8556 . (4) 5022p + 6561q = 405 (5) Now, we can find from (4) (5) - 2717q = 8151 or, q = -3

So, p = 4 Ans.

65. in a business Piku invested tk 6500 for 6 months, Qazi invested 8400 for 5 months, Raj

invested 10000 for 3 months. Piku wants to be working member for which he will receive 5% of

profit. If the total profit earned is tk 7400, what is the share of Qazi in profit? (Jamuna O 12)

Solution:

Ratio of Piku, Qazi & Rajs investment = (6500*6) : (8400*5) : (10000*3) = 39000 : 42000 : 30000 = 13:14:10

As a working member Piku will receive = 5% of 7400 = (5*7400)/100 = 370 tk

So, remaining = 7400 370 = 7030 tk So, Qazis share in profit = (7030*24)/37 = 2660 tk Ans.

66. A bag contains tomatoes that are green or red. The ratio of green tomatoes in the bag is 4:3.

When 5 green and 5 red tomatoes are removed, ratio becomes 3:2. How many red tomatoes were

originally in the bag? (Jamuna MTO 14)

21

Solution:

Let, Green tomatoes = 4x, Red tomatoes = 3x

According to Q, 4x-5/3x-5=3/2 or, 8x 10 = 9x 15 or, x=5 Red tomatoes originally in the bag are 15 Ans.

67. Kalim is asked to write a study guide for a text book. For his work the publishing company is

giving him a choice of a onetime payment of tk 13375 or tk 2000 plus 10% royalty per copy

sold. If the proposed royalty rate of tk 3.25 per copy sold, how many study guide to be sold for

the total income received by Kalim to be the same from either choice? (Jamuna MTO 14)

Solution:

Here, 10 % of royalties =3.25tk/copy

Amount of royalties kalim has to receive = 13375-2000 =11375tk

So, copy to be sold = 13375/3.25 = 3500 nos. Ans.

68. In a certain store, the profit was 320% of the cost. If the cost increases by 25% but the selling

price remains constant, approximately what percentage of the selling price is the profit now?

(Jamuna O 12)

Solution:

Let, cost price is = 100tk, profit = 320 tk, so selling price = 100+320 = 420tk

Then, new cost price = 125 tk, selling price = 420 tk, so profit = 420 125 = 295 tk Profit is the percentage of new selling price = (295*100)/420 = 1475/21 = 70% approximately

Ans.

69. A can do a work in 10 days, B can do it in 15 days. They work together for 5 days, rest of the

work is done by C in 2 days. If they get tk 4500 for whole work, how should they divide money?

(Basic Officer 14)

Solution:

Here, A do in 1 day = 1/10, so in 5 days =

Similarly, B don in 5 days = 1/3

(A+B) 5 days work = + 1/3 = 5/6

So, 2 days work of C = 1 5/6 = 1/6 So, share of A = 4500*1/2 = 2250 tk

Share of B = 4500*1/3 = 1500 tk

Share of C = 4500*1/6 = 750 tk Ans.

70. There are 2 examination room A and B. if 10 students are sent from A to B, then the number

of students in each room is the same. If 20 candidates are sent B to A, then the number of

students in A is double of the number of the students in B. Number of the students in A? (NRB O

14)

Solution:

Let us consider, there are x students in the room A and y students in the room B.


22

x 10 = y + 10 or, y = x 20 . (1) Again, x + 20 = 2(y 20) . (2) Solving for x on (2)

X + 20 = 2(x 40) Or, x + 20 = 2x 80 Or, x = 100

So, there are 100 students in room A Ans.

71. Amin and Sajal are friends. Each has some money. If Amin gives Tk 30 to Sajal, then Sajal

will have twice the money left with Amin. But, if Sajal gives Tk 10 to Amin, then Amin will

have thrice as much as is left with Sajal. How much money does each have? (NRB SEO 14)

Solution:

Let, Amin has Tk x Sajal has Tk y Sajals money will be twice of Amin, if Amin gives Tk 30 to Sajal. So according to the question, 2(x-30)=y+30

2x-60= y+30

2x-y=90.. (i) If Sajal gives taka 10 to Amin, according to the question,

3(y-10)=x+10

3y-30=x+10

3y-x=40... (ii) -x+3y=40

-2x+6y=80. (iii)

Adding equation (i) & (iii) we can get,

2x-y =90

-2x+6y=80

5y=170

y=34

Substituting the value of y in equation (i), we can get 2x-34=90

2x=124

x=62

Amin has Taka 62 & Sajal has Taka 34. (Ans)

72. A contract is to be completed in 46 days and 117 men were set to work, each working 8

hours a day. After 33 days, 4/7 of the work is completed. How many additional men may be

employed so that the work may be completed in time, each man now working 9 hours a day?

(NRB SEO 14)

Solution:

After 33 days time remains = (46-33) = 13 days

4/7th of the work is completed, so works remains=1-4/7=(7-4)/7=3/7

23

In 33 days 8 hours 4/7th of work is done by 117 men

1 1 1 (117x7x33x8)/4 13 9 3/7 (117x7x33x8x3)/(4x19x9x7)=198 Men Additional fund required = 198-117 =81 men. (Ans)

73. Mr. Jones gave 40% of the money he had, to his wife. He also gave 20% of the remaining

amount to each of his three sons. Half of the amount now left was spent on miscellaneous items

and the remaining amount of Tk. 12000 was deposited in the Bank. How much money did Mr.

Jones have initially? (NRB SEO 14)

Solution:

Jones gave 40% of money to his wife,

Remaining amount (100-40) = 60%

Each son got 20% of the remaining (60x20%)=(60x20/100)=12%

Three Son get= (12+12+12) =36%

Money spent on miscellaneous items was half of the amount left.

Amount left=(60-36)=24%

Spent of miscellaneous items=24%/2 =12%

Remaining amount is deposited to bank =24%-12%=12%

Accordingly to the question,

12%=12000

1%=12000/12

100%=(12000x100)/12

=10000

Ans: 10000 Taka.

74. Robi was 4 times as old as his son 8 years ago. After 8 years, Robi will be twice as old as his

son. What are their present ages? (NRB O 14)

Solution:

Robi was 4 times old than his son 8 years ago

Let, 8 years ago sons age was x years So, Robis age was 4x years At present sons age = (x+8) years

Robis age = (4x+8) years After 8 years sons age will be= x+8+8 = (x+16) years

8 Robis =4x+8+8 = (4x+16) years

After 8 years Robi will be twice of his son

So, 4x+16=2(x+16)

4x+16= 2x+32

2x=16

X=8

At present sons age = (8+8) = 16 years

24

Robis age = (4x8)+8 =40 years Ans: Robis age 40 years, sons age 16 years

75. In an examination, 80%of the students passed in English, 85% in Mathematics, and 75% in

both English and Mathematics. If 40 students failed in both the subjects, find the total number of

students? (NRB O 14)

Solution:

Let us consider, total students = 100

Passed in English, E = 80

Passed in Mathematics, M = 85

Passed in both subject, EM = 75

According to Venn theory we find,

Total students = E + M - EM + EuM(Both Failed)

Or,100 = 80 + 85 -75 + Both Failed

Or, Both Failed = 100 90 = 10%

If 10 students failed in both subjects, so total students =100

1 100/10 40 (100/10)x40 = 400 Ans.

76. A, B and C enter into a partnership by investing in ratio 3:2:4. After one year, B invests

another Tk 270000. At the end of three year profits are shared in the ratio of 3:4:5. Find the

initial investment of each. (NRB O 14)

Solution:

Let us consider, invest of A, B & C for 1 year is 3x, 2x & 4x

3 year invest of A = 3x*3 = 9x, B= 3*2x + 270000*2 = 6x + 540000, C = 3x*4 = 12x

Now, according to the question,

6x + 540000/9x = 4/3

Or, 36x = 18x + 1620000

Or, 18x = 1620000

Or, x = 90000

So, investment of A = 3*90000 = 270000 tk

Investment of B = 2*90000 = 180000 tk

Investment of C = 4*90000 = 360000 tk Ans.

77. In a country, 60% of the male citizen and 70% of the female citizen are eligible to vote. 70%

of the male & 60% of female citizen is eligible to cast their vote. What fraction of citizens voted

during their election? (JBL EO 12).

Solution:

Let, total number of male = x

Total number of female = y

So, male voted = x*(60/100)*(70/100)

And female voted = y*(70/100)*(60/100)

25

So, fraction of the citizen voted = x*(60/100)*(70/100) + y*(70/100)*(60/100)

x+y

= (x+y)(60*70)/100*100

(x+y)

= 60*70/10000

= 21/50 Ans.

78. A boy purchased some chocolates from a shop for tk 120. In the next shop he found that the

price of per piece chocolate is tk3 less than that charged @the previous shop, as such he could

have purchased 2 more chocolates. How many chocolates did he buy from the first shop? (JBL

EO 12).

Solution:

Let, the price of one piece chocolate in 1st shop = x

So, total chocolate = 120/x

Price of one piece chocolate in 2nd shop = x 3 Total chocolate 120/(x 3) According to the question,

120/(x 3) - 120/x = 2 Or, (120x 120x + 360)/x(x 3) = 2 Or, 360/x^2 3x = 2 Or, 2x^2 6x 360 = 0 Or, x^2 3x 180 = 0 Or, x^2 15x + 12x 180 = 0 Or, (x + 12)(x 15) = 0 So, x = 15

Number of chocolates bought in the first shop = 120/15 = 8 nos. Ans.

79. A borrower pays 8% interest/yr on the first 600tk he borrows and 7%/yr on the part of the

loan in excess of tk 600. How much interest will the borrower pay on a loan of tk 6,000 for one

year? (PJO Cash 12)

Solution:

Out of tk 6000, the first 600 is charged with 8% interest & the rest amount is charged with 7%

interest.

So, 8% interest for first 600 tk = 600*8/100 = 48 tk

And, 7% interest for next (6000 600) = 5400tk is = 5400*7/100 = 378tk So, total interest for 6000tk for 1 year = 48 + 378 = 426 tk Ans.

80. Ripon, Liton and Pintu started a business jointly with a total amount of taka 280. Ripon paid

taka 45 more than Liton & Liton paid tk 70 less than Pintu. If the company made a profit of tk

56, how much profit should Liton receive? (PJO Cash 12)

Solution:

Let investment of Liton = xtk

26

So, investment of Ripon = (x + 45) tk & Pintu = (x +70) tk


x + (x + 45) + (x +70) = 280

or, 3x + 115 = 280

or, 3x = 165

or, x = 55

so, share of Liton = 45 tk, Ripon = 55 + 45 = 100 tk, Pintu = 55 + 70 = 125 tk

L:R:P = 55:100:125 = 11:20:25

Profit will be received by Liton = (11/56)56 = 11 tk Ans.

81. A worker is paid taka x per hour for the first 5 hrs he works each day. He is paid tk y per hour for each hr he works in excess of 5 hrs. During one week, he works 8 hrs on Saturday, 11

hrs on Sunday, 12 hrs on Monday, 10 hrs on Tuesday, 9 hrs on Wednesday. What is the average

daily wage for five days of week? (PJO Cash 12)

Solution:

Day wise earning of the worker:

Saturday = 5x + 3y

Sunday = 5x + 6y

Monday = 5x + 7y

Tuesday = 5x + 5y

Wednesday = 5x + 4y

Sum of five days = 5x*5 + y(3 + 6 + 7 + 5 + 4) = 25x + 25y

Average earning = 25(x + y)/5 = 5(x + y) Ans.

82. Reena took a loan of tk 1200 at simple interest for as many years as the rate of interest. If she

paid tk 432 as interest at the end of the loan period, what was the rate of interest? (IFIC O 12)

Solution:

We know in case of simple interest, I = rpt/100

Here, interest: I = 432tk, principal: P = 1200tk, t = time, r = rate of interest

According to question, rate = time or, r = t

So, I = rpt/100

Or, rt = 100*I/p

Or, r^2 = (100*432)/1200 = 36

Or, r = 6

So, rate of interest is 6% Ans.

83. Two students Rahim & Karim, appeared at an examination. Rahim scored 9 marks more than

Karim did and Rahims mark was 56% of the sum of their marks. What was Karims score? (IFIC O 12)

Solution:

Let, Karims marks = x So, Rahims mark = x + 9 Their total marks = x + x + 9 = 2x + 9

27

According to question

(2x + 9)56/100 = x + 9

Or, 112x + 504 = 100x + 900

Or, 112x 100x = 900 504 Or, 12x = 396

Or, x = 33 Ans.

84. Abu started a business investing tk 70,000. Robu joined him after 6 months with an amount

of tk 1,05,000 and Sabu joined them with tk 1,40,000 after another 6 months. In what ratio the

amount of profit earned should be distributed among them after 3 years? (DBBL PO 12)

Solution:

Here, ratio of investment of Abu, Robu & Sabu is

(70,000*36):(1,05,000*30):(1,40,000*24)

= 25,20,000:31,50,000:33,60,000

= 252:315:336

= 12:15:16

So, the profit earned by Abu, Robu & Sabu should be distributed to the following proportion

12:15:16 Ans.

85. A trader mixes 26kg of rice priced at tk20/kg with 30kg rice of another variety priced at

tk36/kg & sells the mixture at tk30/kg. What is the percentage of profit or loss? (DBBL PO 12)

Solution:

Here, profit = {30*(26 + 30)} (26*20 + 30*36) = 1680 1600 = 80 tk

So, percentage of profit = (80*100)/1600 = 5% Ans.

86. If zakir loses 8 pounds, he will weight twice as much his sister. Together they are now

weighs 278 pounds. What is zakirs present weight? (SEBL PO 12)

Solution:

Let, zakirs wt. = x & his sisters wt. = y According to question, x + y = 278 .. (1) Again, x 8 = 2y or, x = 2y + 8 . (2) So, we can get from equation (1) by solving (2)

2y + 8 + y = 278

Or, 3y = 278 8 Or, y = 270/3 = 90

So, zakirs wt. = 278 90 = 188lb Ans.

87. A seller incurs a loss of 15% when a table is sold @ tk10,200. At what price the table should

be sold to make a profit of 35% (STBL AO 12)

28

Solution:

At 15% loss,

If tk85 is the sold price then the buying price is tk100

tk1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100/85

10200 ,, ,, ,, ,, ,, ,, ,, ,, ,, 100*10200/85 = 12,000tk

Now, at 35% profit,

If tk100 is the buying price then the sold price is tk135

tk1 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135/100

12,000 ,, ,, ,, ,, ,, ,, ,, ,, ,, 135*12,000/100 = 16,200tk Ans.

88. You want to make a garden in front of your house. The length of the rectangular region is

greater than its breadth by 20 meters. If the perimeter of the land is 200m and gardening cost is

tk20 for each square meter of land, how much will be the total gardening cost? (STBL AO 12)

Solution:

Let, the breadth = x meter & length = x + 20

According to question, the perimeter

2(x + 20 + x) = 200

Or, 4x = 200 40 Or, x = 160/40 = 40

So, breadth = 40m & length = 40 + 20 = 60m

Area of the garden = 40*60 = 2400sqm

So, money required to make the garden = 2400*20 = 48000 tk Ans.

89. A basketball team has won 15 games and lost 9. If these games represent 50/3% of the total

games to be played, then how many more games must the team win to average 75% for the

season? (Premier MTO 12)

Solution:

Here, games already played = 15 + 9 = 24

These are the 50/3% of the total games have to be played

So, total games = (24*3*100)/50 = 144

And 75% of 144 = (144*75)/100 = 108

So, the team has to win more = 108 15 = 93 games Ans.

90. A person earns yearly interest of tk 920 by investing tk x at 4% & tk y at 5% simple interest.

If he invest tk x at 5% & tk y at 4% his yearly interest earning will be reduced by tk 40. Find out

his investment. (Premier MTO 12)

Solution:

The reduced interest due to rate change = 920 40 = 880 According to 1st condition,

(x*4%*1)+(y*5%*1) = 920

Or, 0.04x+0.05y=920 (1) Again, 2nd condition,

29

(X*5%*1)+(y*4%*1) = 880

Or, 0.05x+0.04y = 880 (2) By solving this two equation with (1)*50 - (2)*40 we find

2x + 2.5y = 46,000 .. (3) 2x + 1.6y = 35,200 .. (4) 2.5y 1.6y = 10,800 Or, 0.9y = 10,800

Or, y = 12,000

We can put this figure of y in equation (3) 2x + 2.5*12000 = 46000

Or, 2x = 46000 30000 = 16000 Or, x = 16000/2 = 8000

So, his investment at 4% rate is 8000tk and 12000tk at 5% rate of interest Ans.

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