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Exam Aid Bank Written Math

Jafar Iqbal Ansary

Shahjalal Islami Bank Ltd.

Management Trainee Officer 2016 [Exam Taker: BIBM]

1. Babu, Arif & Salam started a business investing Tk. 28000 where Arif invested Tk. 4500 more

what Babu invested and Babu invested Tk. 7000 less than what Salam invested. Their business

earned Tk. 5600. What is Babu's share in that profit? Abyev`t

evey, Avwid I mvjvg †gvU 28000 UvKv wb‡q GKwU eëmv ïiæ K‡i| evey hZ UvKv †`q Avwid Zvi †P‡q 4500 UvKv †ewk †`q

Ges mvjvg hZ UvKv †`q evey Zvi †P‡q 7000 UvKv Kg †`q| hw` Zv‡ì †Kv¤úvwbi 5600 UvKv jvf nq Zvn‡j evey KZ UvKv

jf¨vsk cv‡e?

Solution:

Let, the amount of Babu = Tk. x

Then, the amount of Arif = Tk. (x+4500)

And the amount of Salam = Tk. (x+7000)

According to the question,

x + x +4500 + x + 7000 = 28000

Or, 3x + 11500 = 28000

Or, 3x = 16500

x = 5500

So, the amount of Babu = Tk. 5500

The amount of Arif = Tk. (5500 + 7000) = Tk.12500

And the amount of Salam = Tk. (5500 + 4500) = Tk.10000

We get,

Babu: Arif: Salam = 5500 : 12500 : 10000

= 11:25:20

Babu’s profit =Tk. (

×5600) = Tk. 1100 (Ans.)

2. Rahim can do a piece of work in 80 days. After working 10 days he left and rest of the work is

done by Karim in 42 days. If they worked together how many days they need to complete the

work? Abyev`t

iwng GKwU KvR 80 w`‡b Ki‡Z cv‡i| †m 10 w`b KvR Kivi ci Kwig GKvB evwK KvR 42 w`‡b †kl K‡i| Zviv `yÕR‡b

GKmv‡_ KvR Ki‡j m¤ú~b© KvRwU Ki‡Z Zv‡ì KZw`b mgq jvM‡e?

Solution:

Let, the whole work be 1 portion

In 80 days, Rahim can do 1 portion

In 1 day,…………

portion

In 10 days………

portion

=

portion

Rest of the work = (1–

)=

portion


Jafar Iqbal Ansary

Now,

In 42 days, Karim can do

portion

In 1 day, Karim can do

portion

=

portion

In 1 day, Rahim and Karim together can do = (

+

) portion

=

portion

=

portion

So, they can do

part works in 1 day

They can do 1 part works in 30 days. (Ans.)

3. Length of a rectangular field is 1.5 times as large as its width. The field cost Tk. 10260 for

planting grass at Tk. 1.9 per sq. meter. What will be the total cost of providing fence around the

field that costs Tk. 2.5 per meter? Abyev`t

GKwU AvqZKvi gv‡Vi •`N¨© Gi cÖ‡¯’i †`o ¸Y| cÖwZ eM©wgUv‡i 1.9 UvKv wnmv‡e m¤ú~Y© gvVwU‡Z Nvm jvMv‡Z 10,260 UvKv

cÖ‡qvRb| cÖwZ wgUvi 2.5 UvKv wnmv‡e gvVwUi Pvicv‡k †eov w`‡Z KZ UvKv LiP n‡e?

Solution:

Let, the width of the rectangular field be x meters and length = 1.5x meters.

Total area = 1.5x×x = 1.5x2

Total cost = 1.9×1.5x2


1.9×1.5x2 = 10260

Or, 2.85x2 = 10260

Or, x2 = 3600

x = 60

Width = 60m and length = 1.5×60 = 90m

Perimeter = 2(90 + 60) = 300m

Cost = Tk. (300×2.5) = Tk. 750 (Ans.)

4.

+

=

Solution:

+

=

Or,

+

=

Or,

+

=

+

Or,

-

=

-

Or, ( ) ( )

( )( ) = ( ) ( )

( )( )

Or,

( ) =

( ) [Multiplying both sides by (3x-5)]

Or,

( ) =

( )

Or,

( ) =

( ) [Dividing both sides by 5]

Or, 4(2x-5) = 5(x+5)

Or, 8x – 20 = 5x + 25


Jafar Iqbal Ansary

Or, 8x – 5x = 25+20

Or, 3x = 45

x = 15.

Trainee Senior Officer 2016 [Exam Taker: BIBM]

1. A man’s running speed is 3 times of his walking speed. The man runs a distance and comes

back walking. Total time taken by him 2 hours. What is the distance in miles if he runs 9 miles

per hour?

Solution: Running speed = 9 miles per hour.

So, walking speed = 9/3 = 3 miles per hour.

Let, the distance be x km.


+

= 2 [Time =

]

Or,

= 2

Or, 4x = 18

x = 4.5

The distance = 4.5 miles. (Ans.)

2. A borrower pays 8.5% interest per year on the first Tk. 600 he borrows and 7.25% per year

on the part of the loan in excess of Tk. 600. How much interest will the borrower pay on a loan

of Tk. 6500 for one year? Abyev`t

†Kvb FYMÖnxZv F‡Yi 1g 600 UvKvi Rb¨ 8.5% Ges 600 UvKvi AwZwi³ UvKvi 7.25% my` †`q| Zvi F‡Yi cwigvb 6,500

UvKv n‡j wZwb eQ‡i KZ UvKv my` w`‡eb?

Solution: Out of Tk. 6500, the first 600 is charged with 8.5% interest and the rest amount Tk. (6500 – 600) =

Tk.5900 is charged with 7.25% interest.

So, 8.5% interest for first Tk. 600 = 8.5% of 600 = Tk. 51

7.25% interest for rest Tk.5900 is = 7.25% of 5900 = Tk. 427.75

So, total interest for 6500 Tk. for 1 year =Tk. (51 + 427.75)

=Tk. 478.75 (Ans.)

3. The length of a rectangular field is 30 feet long than its width and the perimeter is 380 feet.

How much would it cost to cover the field with grass at the rate of Tk. 20 per square? Abyev`t

GKwU AvqZKvi gv‡Vi •`N¨© Gi cÖ‡ ’i †P‡q 30 dzU j¤^v Ges AvqZ‡¶ÎwUi cwimxgv 380 dzU| cÖwZ eM©wgUv‡i 20 UvKv LiP n‡j

m¤ú~Y© gv‡V Nvm jvMv‡Z KZ UvKv LiP n‡e?

Solution:

Let, width be x feet and length be (x+30) feet.

We know,

Perimeter =2(width+ length)

=2(x+x+30)

= 2(2x+30)

= 4x+ 60


Jafar Iqbal Ansary


4x+ 60 = 380

Or, 4x = 380-60

Or, 4x = 320

x = 80

Width = 80 feet and length = (80+ 30) feet = 110 feet

Area of the field = (length× width)

= (110×80) sq. feet

= 8800 sq. feet (Ans.)

Total cost to cover the field with grass at rate of Tk. 20 per square feet

= Tk. (8800×20)

= Tk. 176,000 (Ans.)

4. If √x+

√ = a, then x

2+

=?

Solution:

Given that,

√x+

√ = a

Or, (√x+

√ )2 = a

2 [Squaring both sides]

Or, x +

+ 2. √x.

√ = a

2

Or, x +

+ 2 = a

2

Or, x +

= a

2-2

Or, (x +

)2 = (a

2-2)

2

Or, x2 +

+ 2.x.

= a

4 – 4.a

2+ 4

Or, x2 +

+ 2 = a

4 – 4.a

2+ 4

Or, x2 +

= a

4 – 4.a

2+ 4 -2

x2 +

= a

4 – 4.a

2+ 2 (Ans.)

Trainee Officer (Cash) 2013 [Exam Taker: BIBM]

1. Age of three persons is now in the proportion 2: 3: 4 and in 5 years from now, the proportion

will be 5: 7: 9. What is the present age of the youngest person?

Solution:

Let, the age of the persons be 2x, 3x and 4x respectively.

After 5 years, ratio of 1st two persons

(2x+5) : (3x+5) = 5: 7

Or, 5(3x+5) = 7(2x+5)

Or, 15x + 25 = 14x + 35

Or, 15x – 14x = 35- 25

x = 10

The present age of the youngest person = 2×10 = 20 years. (Ans.)


Jafar Iqbal Ansary

2. Mr. X can finish a work in 6 days and Mr. Y can finish the same work in 8 days. How many

days will it take to finish the work it they work together? Abyev`t

X GKwU KvR 6 w`‡b Ges Y GKB KvR 8 w`‡b Ki‡Z cv‡i| Zviv GK‡Î H KvR KZ w`‡b †kl Ki‡Z cvi‡e?

Solution:

In 6 days X can do 1 part of the work

” 1 ” X ” ” 1/6 ” ” ”

In 8 days Y can do 1 part of the work

” 1 ” Y ” ” 1/8 ” ” ”

In 1 day (X+Y) can do = (

+

) part of the work

= (

) ” ” ”

=

” ” ”

part of the work can do in 1 day

1 ” ” ” ” ” ”

days

Ans:

days.

3. A shirt was sold for Tk. 171 and the gain was as much percent as it costs in Taka amount.

What was the purchase price of the shirt?

Solution: Let, the purchase price be Tk. x and profit be x%.


x + x% of x = 171

Or, x +

= 171

Or, 100x + x2 = 17100

Or, x2 + 100x – 17100 = 0

Or, x2 + 190x -90x – 17100 = 0

Or, x(x+190) -90(x+190) = 0

Or, (x+190) (x-90) = 0

x = 90

The purchase price = Tk. 90. (Ans.)

4. If

= 2, then the value of

.

Solution: Given that,

= 2

Or,

= 2

Or, x2 +1 = 2x

Or, x2 – 2x +1 = 0

Or, (x-1)2 = 0

Or, x-1 = 0

x = 1

=

= 1 (Ans.)


Jafar Iqbal Ansary


1. The percentage profit earned by selling an article for Tk. 1920 is equal to the percentage loss

incurred by selling the same article for Tk 1280. At what price should the article be sold to make

25% profit? (South East Bank MTO 13, BB AD(ff) 15, Shahjalal Islami Bank MTO 13) Abyev`t

GKwU cY¨ 1,920 UvKvq weµq Ki‡j kZKiv hZ jvf nq, cY¨wU 1,280 UvKvq weµq Ki‡j kZKiv ZZ UvKv ¶wZ nq| 25%

jvf Ki‡Z n‡j cY¨wU‡K KZ `v‡g weµq Ki‡Z n‡e?

Solution: Let, cost price be Tk. x


SP1-CP = CP-SP2

1920-x = x-1280

Or, 1920+1280 = x+ x

Or, 2x = 3200

Or, x = 3200/2

x = 1600

Cost price is Tk. 1600

At 25% profit,

New selling price = Tk. (1600+25% of 1600) = Tk. 2000

Ans: Tk. 2000.

Alternative Method:

Let, profit and loss both be Tk. x


SP1-Profit = SP2+ Loss

1920-x = 1280+x

Or, 2x = 640

x = 320

Cost price = Selling price – profit

= Tk. (1920-320)

= Tk. 1600

At 25% profit,


Ans: Tk. 2000

Alternative Method:

Let, amount of cost price be Tk. x

Profit percentage =

×100% =

×100%

Loss percentage =

×100% =

×100%


×100% =

×100%

Or, 1920-x = x-1280

Or, 2x = 3200

Or, x = 3200/2

x = 1600

At 25% profit,


Jafar Iqbal Ansary


Ans: Tk. 2000

Alternative Method:

Let, profit and loss both be x %.

( ) =

( )

Or, 128,000+1280x = 192,000-1920x

Or, 1280x+1920x = 192,000-128,000

Or, 3200x = 64,000

x = 20

Cost price = Tk.

= Tk. 1600

At 25% profit,

Selling Price = Tk. (1600+25% of 1600) = Tk. 2000

Ans: Tk. 2000

2. The ratio between the perimeter and the breadth of rectangle is 5:1. If the area of the

rectangle is 216 sq. cm, what is the length of the rectangle?(Shahjalal Bank MTO 13,South East

Bank MTO 13) Abyev`t

GKwU AvqZKvi †¶‡Îi cwimxgv Ges cÖ‡¯’i AbycvZ 5 t 1| AvqZ‡¶ÎwUi †¶Îdj 216 eM© †m.wg n‡j, Gi •`N©¨ KZ?

Solution: Given that, perimeter: breadth = 5:1.

Let, the perimeter be 5x and breadth be x.

Here,

2(length +breadth) = 5x

Or, 2(length +x) = 5x

Or, 2×length+2x = 5x

Or, 2×length = 5x-2x = 3x

Length = 1.5x.


(1.5x)×(x) = 216 [Area = Length ×Breadth]

Or, x2 = 216/1.5 = 144

x = 12

Length = 1.5×12 = 18 cm.

Ans: 18 cm

Alternative Method:

Let, the length be x and breadth be y

So, area of the rectangle, xy = 216 …………. (1)

Now, the perimeter = 2(x + y)

According to question,

2(x+ y): y = 5:1

Or, 2x + 2y = 5y [Cross multiplication]

Or, 2x = 3y

x = 1.5y

From equation (1)=>

y×1.5y = 216

Or, y2 = 144

y = 12


Jafar Iqbal Ansary

x = 1.5×12 = 18

Length = 18 cm.

Ans: 18 cm

Alternative Method:

Let, Length = L and Breadth = B


2(L+B) : B = 5:1

Or, ( )

= 5

Or, 2(L+B) = 5B [Cross multiplication]

Or, 2L+2B = 5B

Or, 5B-2B = 2L

Or, 3B = 2L

B =

L

Again,

Length × Breadth = Area

Or, L×B = 216

Or, L×

L = 216 [ B =

L]

Or, L2 = 216 ×

Or, L2 = 324

L = 18

Length = 18 cm. (Ans.)

3. A, B and C enter into a partnership in the ratio 7/2:4/3:6/5. After 4 months, A increases his

share by 50%. If the total profit at the end of one year is Tk. 21,600, then what is B's share in the

profit? (Shahjalal Islami Bank MTO 13, South East Bank MTO 13) Abyev`t

A, B I C

t

t

Abycv‡Z GKwU Askx`vix Kvievi ïiæ K‡i| 4 gvm ci A Zvi †kqvi 50% e„w× K‡i| hw` 1 eQi c‡i

†gvU jv‡fi cwigvb 21,00 UvKv nq Zvn‡j B Gi jf¨vsk KZ n‡e?

Solution:

Given ratio,

A:B:C = (

:

:

)

= 105: 40: 36 [LCM of 2,3 and 5 = 30; so, multiplying by 30]

Let, the initial investments of A = 105x, B = 40x and C = 36x.

Total investment of

A = (105x×1) + 105x×50%×(

) [Here, 8 months =

year]

= 105x+ 35x

= 140x

B = 40x×1 = 40x and C = 36x×1 = 36x

A: B: C = 140x: 40x: 36x = 35: 10: 9

Sum of the ratio = 35+10+9 = 54

B's share = Tk. 21600×(

) = Tk. 4000

Ans: Tk. 4000


Jafar Iqbal Ansary

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

of the male and 60% of female citizen are eligible to cast their vote. What fraction of citizens

voted during their election?( Janata Bank EO 12, South East Bank MTO 13, Shahjalal Islami

Bank MTO 13, Midland Bank MTO 15) Abyev`t

GKwU †`‡ki 60% cyiæl Ges 70% gwnjv †fvU w`‡Z cvi‡eb| G‡`i g‡a¨ 70% cyiæl Ges 60% gwnjv †fvU `vb K‡ib|

KZRb bvMwiK wbe©vP‡bi mgq †fvU cÖ`vb K‡ib?

Solution:

Let, the number of male be x and female be y.

So, male voted =60% of 70% of x

= 0.6×0.7× x = 0.42x

And female voted = 70% of 60% of y

= 0.7×0.6× y = 0.42y

Total citizen = x+y

Total citizen voted = 0.42x +0.42y = 0.42(x+y)

Fraction of the citizen voted = ( )

( ) = 0.42 =

=

Ans:

Alternative Method:

Let, the number of male be 100 and number of female be 100

Total citizen = 100+100 = 200

So, eligible Male voter = 60% of 100 = 60 and Female = 70% of 100 = 70

Male voted = 70% of 60 = 42 and female voted = 60% of 70 = 42

Total voted= 42+42= 84

Fraction of the citizen voted =

=

Ans:


1.The total income of Mr. Atiq in the years 2008, 2009 and 2010 was Tk. 36,40,000. His income

increased by 20% each year. What was his income in 2010? (Shahjalal Islami Bank TSO 13) Abyev`t

Rbve AvwZ‡Ki 2008, 2009 Ges 2010 mv‡ji †gvU Avq 36,40,000 UvKv| Zvi Avq cÖ‡Z¨K eQi 20% e„w× cvq| 2010

mv‡j Zvi Avq KZ wQj?

Solution: Let, in 2008 income was Tk. 100x

So, in 2009 income = Tk. (100x+20% of100 x) =Tk. 120x

And in 2010 income = Tk. (120x+20% of 120x) =Tk. 144x


100x+120x+144x = 36,40,000

Or, 364x = 36,40,000

Or, x = 36,40,000/364

Or, x = 10000

100x = 10,00,000

In 2010 income = Tk. (10,00,000×144) =Tk 14,40,000

Ans: Tk. 14,40,000.


Jafar Iqbal Ansary

2. In a set of three numbers, the average of first two numbers is 2, the average of the last two

numbers is 3, and the average of the first and last numbers is 4. What is the average of three

numbers? (Shahjalal Islami Bank TSO 13) Abyev`t

wZbwU msL¨vi g‡a¨ 1g msL¨v ywUi Mo 2, †kl msL¨v ywUi Mo 3 Ges 1g I †kl msL¨vi Mo 4| msL¨v wZbwUi Mo KZ?

Solution: Let, the numbers be a, b, c.

= 2

a+ b = 4 ….(i)

= 3

b+ c = 6 ….(ii)

and

= 4

a+ c = 8 …..(iii)

(i)+ (ii)+(iii)=>

a+ b+ b+ c+ a+ c = 4+6+8

Or, 2(a+ b+ c) = 18

Or, (a+ b+ c) = 18/2 = 9

Or, (a+ b+ c) = 9

Or,

=

= 3

Average of three numbers = 3

Ans: 3.

Alternative Method:

Let, the numbers be a, b, c.

= 2

Or, a+b = 4

b = 4-a ….(i)

= 3

Or, b+c = 6

Or, 4-a+c = 6

c = 2+a ….(ii)

and

= 4

Or, a+c = 8

Or, a+2+a = 8

Or, 2a = 8-2

Or, 2a = 6

a = 3 …..(iii)

Putting the value of ‘a’ in equation (i) and (ii)

b = 4-3 = 1

c = 2+3 = 5

Now,

=


Jafar Iqbal Ansary

= 3

Average of three numbers = 3

Ans: 3.

3. (√ √ )

√ √ - √

√ =? (Shahjalal Islami Bank TSO 13)

Solution: (√ √ )

√ √ - √

√

= (√ √ )(√ √ )

(√ √ )(√ √ ) – ( √ )( √ )

( √ )( √ )

= (√ √ )

(√ ) (√ ) –

( √ )

( ) (√ )

= ( √ √ )

– ( √ )

= ( √ )

– (7+4√ )

= (8+2.2√ ) - (7+4√ )

= 8+4√ – 7- 4√

= 1

4. What is the area of the equilateral triangle if the base BC = 6? (Shahjalal Islami Bank TSO

13)

Solution:

Given that,

AB = BC = CA = 6

We know,

Area of Equilateral Triangle = √

×(6)

2 = 9√

Ans: 9√

Trainee Officer 2013 [Exam Taker: BIBM]

1. In a certain Accounting class, the ratio of the number of Accounting majors to the under of

students who are not Accounting major is 2 to 5. If 2 more Accounting majors were to enter the

class, the ratio would be 1 to 2. How many students are in the class? (Shahjalal Islami Bank TO

13) Abyev`t

†Kvb GKwU A¨vKvDw›Us K¬v‡m hv‡ì A¨vKvDw›Us †gRi Ges hv‡ì A¨vKvDw›Us †gRi bq Zv‡ì AbycvZ 2 t 5| 2 Rb bZzb

A¨vKvDw›Us †gRi K¬v‡m cÖ‡ek Kivq AbycvZ nq 1 t 2| K¬v‡m KZRb wk¶v_©x Av‡Q?

Solution: Let, number of Accounting major be 2x and non Accounting major be 5x

B

A

C


Jafar Iqbal Ansary


=

Or, 5x = 4x+4

x = 4

Total students = 2×4 + 5×4 = 28

Ans: 28.

Alternative Method:

Let, total students be x

Given that,

Accounting major : Accounting non major = 2:5

Sum of the ratio = 2+5 = 7

Number of Accounting major students =

and Accounting non major students =


(

+2) :

= 1:2

Or, 2(

+2) =

Or,

+4 =

Or, 4 =

-

Or, 4 =

Or, 4 =

x = 28

Ans: 28.

2. A retailer buys a sewing machine at discount of 15% and sells it for Tk. 1955. Then he makes

a profit of 15%. What is the amount of discount? (Shahjalal Islami Bank TO 13) Abyev`t

†Kvb LyPiv we‡µZv GKwU †mjvB †gwkb 15% wWmKvD‡›U µq K‡ib Ges 1,955 UvKvq weµq K‡ib| G‡Z Zvi 15% jvf nq|

wWmKvD‡›Ui UvKvi cwigvb KZ wQj?

Solution: Let, marked price = Tk. 100x

At 15% discount, cost price = Tk. (100x- 15% of 100x) = Tk. 85x


85x+ 15% of 85x = 1955

Or, 85x+ 12.75x = 1955

Or, 97.75 x = 1955

Or, x = 1955/97.75

100x = 2000

Amount of discount = 15% of 2000 = Tk. 300

Ans: Tk. 300.

3. If x =

, then what is the value of

√ √

√ √ =? (Shahjalal Islami Bank TO 13)


x =

Or,

=


Jafar Iqbal Ansary

Or,

=

Or,

= 9

Or, √

√ = √ [Square on both sides]

Or, √

√ = 3

Or, √ √

√ √ =

Or, √ √

√ √ =

√ √

√ √ = 2 (Ans.)

4. What is perimeter of ∆ABC show in the figure? (Shahjalal Islami Bank TO 13)

Solution: Given that, AC = 2

B = x = y, C = y = z

x = y = z and B = C

And A = 2z

Now,

A + B + C = 180°

Or, 2z + z+ z = 180° [ B = C = z]

Or, 4z = 180°

z = 45°

B = C = 45° and A = 2×45° = 90°

So, it is a right angled triangle.

Pythagoras theorem,

BC2 = AB

2 + AC

2

Or, BC2 = 2

2 + 2

2 [AC = AB = 2]

Or, BC2 = 8

BC = 2√

Perimeter = AB + AC + BC = 2+ 2+ 2√ = 4+ 2√ (Ans.)

2

y0

C B

y0

Z

0

2z A

x0


Jafar Iqbal Ansary

Trainee Officer (Cash) 2013 [Exam Taker: BIBM]

1. If a number of two digits is divided by the product of its digits, the quotient is 3. When 18 is

added to the number, the digits of the number change their places. Find the number. (Shahjalal

Islami Bank TO cash 13) Abyev`t

`yB A¼wewkó GKwU msL¨v‡K Gi A¼Øi‡qi ¸Ydj Øviv fvM Ki‡j fvMdj 3 nq| hw` msL¨vwUi mv‡_ 18 †hvM Kiv nq Zvn‡j

A¼Øq ’vb wewbgq K‡i| msL¨vwU wbY©q Kiæb|

Solution: Let, the unit digit be y and tenth digit be x

The number = (10x + y).

1st condition,

3

10x+y = 3xy …..(i)

2nd condition,

(10x + y) +18 = 10y + x

Or, 10x-x +18 = 10y-y

Or, 9x+18 = 9y

Or, 9y-9x = 18

Or, y-x = 2

y = x+2 ……(ii)

Putting the value of y in equation (i)

10x + (x+2) = 3x (x+2)

Or, 10x+x+2 = 3x2 +6x

Or, 3x2 +6x-10x-x-2 =0

Or, 3x2 -5x-2 = 0

Or, 3x2 -6x+x-2 = 0

Or, 3x(x-2) +1(x-2) = 0

Or, (x-2)(3x+1) = 0

x = 2 [x = - 1/3, not acceptable]

Putting the value of x in equation (ii)

y = x+2 = 2+2 = 4

The number = 10×2+4 = 24

Ans: 24.

2. An old man distributed all the gold coins he had to his two sons into two different numbers

such that the difference between the squares of the two numbers is 36 times the difference

between the two numbers. How many coins did the old man have? (Shahjalal Islami Bank TO

cash 13) Abyev`t

GKRb e„× †jvK Zvi mg Í ¯^Y©gy`ªv Zvi `yB cy‡Îi g‡a¨ Ggb msL¨vq fvM K‡i †`b †hb msL¨v ywUi e‡M©i AšÍi, msL¨v `ywUi

AšÍ‡ii †P†q 36 ¸Y eo nq| e„× ‡jvKwUi Kv‡Q KZwU ¯^Y©gy`ªv wQj?

Solution: Let, the two numbers be x and y.

So, total number of coins = x+ y


Jafar Iqbal Ansary


x2-y

2 = 36(x-y)

Or, (x+y)(x-y) = 36(x-y)

(x+y) = 36

Total number of gold coins is 36.

Ans: 36.

3. Two equal amount of money are deposited in two banks, each at 15% per annum, for 3.5

years and 5 years. If difference between their profits is Tk. 144, what is the each amount of

money deposited? (Shahjalal Islami Bank TO cash 13) Abyev`t

15% my‡` mgcwigvb UvKv 3.5 eQi Ges 5 eQ‡ii Rb¨ 2wU Avjv`v e¨vs‡K ivLv n‡jv| hw` G‡`i gybvdvi cv_©K¨ 144 UvKv nq

Zvn‡j cÖ‡Z¨K e¨vs‡K KZ UvKv Rgv ivLv n‡qwQj?

Solution: Let, initial deposit be Tk. 100x


(100x×15%×5)-(100x×15%×3.5) = 144 [Here, Interest (I) = Prn]

Or, 75x- 52.5x = 144

Or, 22.5x = 144

Or, 100x =

100x = 640

Ans: Tk. 640.

4. In the following diagram, if BC = CD = BD = 1 and angle ADC is a right angle, what is the

perimeter of triangle ABD? (Shahjalal Islami Bank TO cash 13)


BC = CD = BD = 1

So, BCD is an equilateral triangle.

BCD = CDB = CBD = 60°

BDA = 90°-60° = 30° and ABD = 180°-60° = 120°

So, BAD = 180°-120°-30° = 30°

BAD is an Isosceles triangle.

Therefore, AB = BD = 1 and AC = AB + BC = 1+1 = 2

Pythagoras theorem,

AC2 = CD

2 + AD

2

Or, 22 = 1

2 + AD

2

Or, AD2 = 4-1 = 3

AD = √

Perimeter of ABD = 1+1+√ = 2+√ (Ans.)

A

B

D C


Jafar Iqbal Ansary


1. A train travelling at 48 km/h completely crosses another train having half its length and

travelling in opposite direction at 42 km/h, in 12 second. It also passes a railway platform in 45

second. What is the length of the platform? (Shahjalal Islami Bank MTO 11) Abyev`t

48 wK.wg/N›Uv †e‡Mi GKwU †Uªb wecixZ w`K †_‡K 42 wK.wg/N›Uv †e‡M Avmv Gi A‡a©K •`‡N©¨i GKwU †Uªb‡K 12 †m‡K‡Û m¤ú~Y©

iƒ‡c AwZµg K‡i| GwU †ijI‡q cøvUdg©wU‡K 45 †m‡K‡Û AwZµg K‡i| cøvUd‡g©i •`N©¨ KZ?

Solution: Let, length of the smaller train = x meters

So, length of the longer train = 2x meters

Total length of two trains = x+2x = 3x meters

Actual speed of two trains = (48+42) km/hr = 90 km/hr

= 90×(

) m/s

= 25 m/s [Here, 1km/h =

m/s]

We know,

Distance = Speed ×Time

3x = 25×12

x = 100

So, length of the longer train = 2×100 = 200 meters

Let, length of the platform be y meters.


200 + y = 45 × (48×

) [Distance = Time ×Speed]

Or, 200 + y = 600

y = 400

So, the length of the platform = 400 meters.

Ans: 400 meters.

2. The cost price of two watches taken together is Tk. 840. If by selling one at a profit of 16%

and the other at a loss of 12%, there is no loss or gain in the whole transaction, find the cost

price of the two watches. (Shahjalal Bank MTO 11, Social Islami Bank PO 17) Abyev`t

GKm‡½ †Kbv 2wU Nwoi µqg~j¨ GK‡Î 840 UvKv| hw` GwU‡K 16% jv‡f Ges Ab¨wU‡K 12% ¶wZ‡Z weµq Ki‡j †gv‡Ui Dci

†Kvb jvf ev ¶wZ nq bv Zvn‡j Nwo 2wUi cÖwZwUi µqg~j¨ wbY©q Ki|

Solution: Let, 1st cost price = Tk. x and 2nd cost price = Tk. (840-x)


16% of x = 12% of (840-x)

Or, 0.16x = 0.12(840-x)

Or, 0.16x = 100.8-0.12x

Or, 0.16x+0.12x = 100.8

Or, 0.28x = 100.8

Or, x = 100.8/0.28

x = 360.

So, 1st cost price = Tk. 360 and 2nd cost price = Tk. (840-360) = Tk. 480.

Ans: Tk. 360 and Tk. 480.


Jafar Iqbal Ansary

3. A, B and C can do a piece of work in 16, 32 and 48 days respectively. They started working

together but C left after working 4 days and B left 2 days before the completion of work. How

many days it took to complete the work? (Shahjalal Islami Bank MTO 11, Social Islami Bank

PO 17) Abyev`t

A, B I C GKwU KvR h_vµ‡g 16, 32 Ges 48 w`‡b Ki‡Z cv‡i| Zviv GKmv‡_ KvR Ki‡Z ïiæ K‡i wKš‘ KvR ïiæ nIqvi 4

w`‡b ci C KvR †Q‡o †`q Ges KvR †kl nIqvi 2 w`b Av‡M B KvR †Q‡o †`q| m¤ú~Y© KvRwU †kl n‡Z KZw`b mgq jv‡M?

Solution: Let, the total work will be completed in x days.

In x days A’s work =

parts

Since left 2 days before the completion of work,

B work in (x-2) days =

parts

And C works in 4 days = 4×

=

parts


+

+

= 1

Or,

= 1 -

Or,

=

Or, 3x-2 = (

)×32

Or, 3x =

+ 2 =

x = 94/9 = 10

So, the total work will be completed in 10

days

Ans: 10

days.

4. Two Cans have the same height equal to 21 m. One Can is cylindrical; the diameter of whose

base is 10 cm. The other can has square base of side 10 cm. What is the difference in their

capacities? (Shahjalal Islami Bank MTO 11) Abyev`t

`ywU cv‡Îi cÖwZwUi D”PZv 21 wgUvi| GKwU cvÎ wmwjÛvi AvK…wZi hvi f~wgi e¨vm 10 †m.wg| Ab¨wU eM©vK…wZi hvi GK av‡ii

•`N¨© 10 †m.wg| cvÎ ywUi avib¶gZvi cv_©K¨ KZ?

Solution: Height, h = 21m = 2100cm [1 meter = 100 cm]

Radius, r = 10/2cm = 5cm [Radius =

×Diameter]

We know,

Volume of cylindrical Can = πr2h = 3.1416×5

2×2100 =164934 cm

3

Volume of square box Can = side2×height = 10

2×2100 = 2,10,000 cm

3

Difference = (210,000-164934) cm3 = 45066 cm

3

Ans: 45066 cm3


Jafar Iqbal Ansary


1. If

=

, then

= ? (Shahjalal Islami Bank TSO 11)

Solution:

=

Or,

=

[Multiplying both sides by

]

Or,

=

Or,

=

= -3 (Ans.)

2. The base of a rectangle exceeds three times the height by 8. Find the dimensions of the

rectangle if- (a) the semi-perimeter is 32 (b) the perimeter is 100. (Shahjalal Islami Bank TSO

11)

Solution: Let, the base be x.

So, height = 3x+8

a) Semi- perimeter = Base +Height = x+ 3x+8 = 4x +8


4x +8 = 32

Or, 4x = 32-8

Or, 4x = 24

x = 6

Base = 6 and height = 3×6+8 = 26 (Ans.)

b) Perimeter = 2(Base+ Height) = 2(x+3x+8) = 8x+16


8x +16 = 100

Or, 8x = 100-16

Or, 8x = 84

x = 10.5

Base = 10.5 and height = 3×10.5+8 = 39.5 (Ans.)

3. A club has 20 members. They are electing a principal and a vice president. How many

different outcome of the election are possible? (Assume the president and the vice president

must be different members of the club.) (Shahjalal Islami Bank TSO 11)

Solution: A President can be elected from 20 members by 20C1 = 20 ways.

A vice President can be elected from 19 members by 19C1 = 19 ways.

Possible outcome of the election = 20×19 = 380.

Ans: 380.


Jafar Iqbal Ansary

4. Tk. 1500 is invested at a rate of 10% simple interest and interest is added to the principal

after every 5 years .in how many years will it amount to Tk. 2500? (Shahjalal Islami Bank TSO

11) Abyev`t

hw` 10% mij gybvdvq 1,500 UvKv wewb‡qvM Kiv nq Ges cÖ‡Z¨K 5 eQi ci hw` gybvdv Avm‡ji mv‡_ †hvM Kiv nq Zvn‡j KZ

eQ‡i UvKvi cwigvb 2,500 n‡e?

Solution: At 10 % simple interest,

Interest of 5 years = 1500×5×10% = Tk. 750

After 5 years Total amount = Tk. (1500+750) = Tk. 2250.

Interest = Tk. (2500-2250) = Tk. 250

We know,

Interest = principal× Time× Rate of Interest

Or, 250 = 2250× Time× 10%

Or, 250 = 225× Time

Time = 250/225=1.11

Total time = 5+1.11 = 6.11 years.

Ans: 6.11years.

Trainee Officer 2011 [Exam Taker: BIBM]

1. The average income of A for 15 days is Tk. 70. The average for first five days is Tk. 60 and

that for the last nine days is Tk. 80. What is the income for the sixth day? (Shahjalal Bank

Trainee Officer 11)

Solution: Total income of 15 days = Tk. (70×15) = Tk. 1050

Income of 1st 5 days = Tk. (60×5) = Tk. 300

Income of last 9 days = Tk. (80×9) = Tk. 720

Income of sixth day = Tk. (1050-300-720) = Tk. 30

Ans: Tk. 30.

2. Nipu, Nila and Dipa formed a partnership with investments of Tk. 75000, Tk. 60000 and Tk.

40000 respectively. After 3 years of operation, the partnership had a net profit of Tk. 26250.

What was the share of Dipa in the profit? (Shahjalal Bank Trainee Officer 11)

Solution: Ratio of the capitals of Nipu, Nila and Dipa

Nipu: Nila: Dipa = 75000 : 60000 : 40000

= 75 : 60 : 40

= 15 : 12 : 8.

Sum of the ratio = 15+12+8 = 35

Nipu’s share = Tk. (26250 ×

) = Tk.11,250

Nila's share = Tk. (26250×

) = Tk. 9000

Dipa's share = Tk. (26250×

) = Tk. 6000.

Ans: Tk. 11,250, Tk. 9000 and Tk. 6000


Jafar Iqbal Ansary

3. A man sells articles at 5% profit. If had bought it at 5% less and sold it for Tk. 1 less, he

would have gained 10%. What is the cost price of the article? (Shahjalal Bank Trainee Officer

11) Abyev`t

†Kvb e¨w³ 5% jvf cY¨ weµq K‡ib| hw` wZwb 5% Kg `v‡g wK‡b 1 UvKv K‡g weµq Ki‡Zb Zvn‡j Zvi 10% jvf n‡Zv|

cY¨wUi µqg~j¨ KZ?

Solution:

Let, cost Price = Tk. 100x.

At 5% profit, selling price = (100x+5% of 100x) = Tk. 105x

When cost price 5% reduced,

New cost = (100x- 5% of 100x) = Tk. 95x,

New selling price = (95x+10% of 95x) = Tk. 104.5x


105x-104.5x = 1 [Difference of two selling price]

Or, 0.5x = 1

Or, x =

Or, 100x =

x = 200

Cost price = Tk. 200 (Ans.)

4. The length of each side of a triangle is 12 cm. Find the height of the triangle. (Shahjalal Bank

Trainee Officer 11)

Solution:

Let, height of triangle be h cm.

Given that,

AB = BC = AC = 12 cm

We know,

Area of triangle =

×BC×AD =

×12×h = 6h

Again,

Area of Equilateral Triangle = √

×(12)

2 = 36√


6h = 36√

h = 6√

Height of the triangle = 6√ cm. (Ans.)

A

B C D


Jafar Iqbal Ansary


1. A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the

bridge in 20 seconds but him in 8 seconds. Find the length of the train. (Sonali Officer IT-16,

Shahjalal Islami Bank MTO 10) Abyev`t

†Kvb e¨w³ 180 wg. j¤^v GKwU †ijI‡q eªx‡Ri Dci `vwo‡q Av‡Q| wZwb †`L‡jb †h GKwU †Uªb 20 †m‡K‡Û eªxRwU AwZµg K‡i

wKš‘ Zv‡K 8 †m‡K‡Û AwZµg K‡i hvq| †Uª‡bi ‣`N © wbY©q Ki|

Solution:

Let, length of the train be x m


=

[Speed=

]

Or, 20x = 8(x+180)

Or, 20x = 8x + 1440

Or, 20x-8x = 1440

Or, 12x = 1440

x = 120

Length of the train = 120 m

Ans: 120 m.

2. A trader bought some mangoes for Tk. 150 per dozen and equal number of apples for Tk. 100

per dozen. If he sells all the fruits Tk. 140 per dozen, what will be his profit/loss in percentage?

(BASIC Bank Officer 1999, RAKUB SO 14, Shahjalal Islami Bank MTO 10) Abyev`t

GKRb e¨emvqx c ÖwZ WRb 150 UvKv `‡i wKQz Avg Ges cÖwZ WRb 100 UvKv `‡i mgvb msL¨K Av‡cj µq K‡ib| hw` †m

me¸‡jv dj 140 UvKv WRb wn‡m‡e weµq K‡i Zvn‡j Zvi kZKiv KZ jvf ev ¶wZ n‡e?

Solution:

Let, he bought x dozen mangoes and x dozen apples.

So, buying price = 150x + 100x = Tk. 250x

Selling price = Tk. (140×2x) = Tk. 280x

Profit = Tk. (280x-250x) = Tk. 30x

So, percentage of profit =

% = 12 %

Ans: Profit 12%

Alternative Method:

Given, cost price of per dozen mango = Tk. 150

and cost price of per dozen apple = Tk. 100

So, cost price (per dozen mango + per dozen apple) =Tk. (150+100)

= Tk. 250

Selling price (per dozen mango+ per dozen apple) = Tk. (140×2)

=Tk. 280

Profit = Tk. (280–250) = Tk. 30

Percentage of profit =

% = 12% (Ans.)


Jafar Iqbal Ansary

3. In a school, there are equal number of boys and girls. Among the students, 1/8 of the girls and

5/6th of the boys are residing in the hostel. What percent of the Students consists of boys who do

not reside in the hostel among all students? (Modhumoti Bank MTO 16, Shahjalal Islami Bank

MTO 10) Abyev`t

GKwU we`¨vj‡q †Q‡j Ges †g‡qi msL¨v mgvb| wk¶v_©x‡`i g‡a¨ 1/8 Ask †g‡q Ges 5/6 Ask †Q‡j †nv‡÷‡j evm K‡i| †gvU

wk¶v_©xi g‡a¨ kZKiv KZRb †Q‡j †nv‡÷‡j evm K‡i bv?

Solution:

Let, the number of boys be ‘x’ and girls be ‘x’.

Total students = 2x students.

Non-residing boys = x-

of x = x-

=

Required percentage =

× 100% =

% = 8.33% (Ans.)

Alternative Method:

L.C.M. of 8 and 6 = 48

Let, total student be 48x

So, boys = 24x and girls = 24x

Boys residing in the hostel = 24x×(

) = 20x

Boys do not reside in the hostel = 24x-20x = 4x

Required percentage = (

×100) % = 8.33 % (Ans.)

4. Two partners A and B have 70% and 30% shares respectively in a business. After sometimes,

a third partner C joined the business by investing Tk. 10 lakh and thus having 20% share in the

business. What percentage is the A’s share now in the business? (Shahjalal Islami Bank MTO

10, RAKUB SO 14)

Solution:

Given that,

A: B = 70%: 30% = 7:3

Sum of the ratios = 7+3 = 10


20% share = 10 lakh,

100% share =

lakh

= 50 lakh

Share of A and B = (50–10) lakh = 40 lakh

A’s share = (

×40) lakh = 28 lakh

Now A’s share percentage =

×100 % = 56% (Ans.)

5. Taxi fare is described by the following relationship:

Total taxi Fare = Fixed Charge: Tk. A up to 2 km + Tk. B per km run exceeding 2 km + Tk. 60

for per hours waiting

A person paid Tk. 432 for running 52 km and 2 hours of waiting charge. The same person paid

Tk. 732 for running 102 km and 2 hours of waiting charge. Find the value of A and B. (Shahjalal

Islami Bank MTO 10)

Solution:

Given that,

Rate of 1st 2 km = Tk. A, per km after exceeding 2 km = Tk. B and per hour waiting charge = Tk. 60


Jafar Iqbal Ansary

1st Condition,

A + (52-2)×B +60×2 = 432

Or, A + 50B = 432- 120

A + 50B = 312 ....(i)

2nd Condition,

A + (102-2)×B +60×2 = 732

Or, A + 100B = 732- 120

A + 100B = 612 ....(ii)

Now, (ii) – (i) =>

100B – 50B = 612 – 312

Or, 50B = 300

B = 6

Putting the value of B in equation (i)

A + 50×6 = 312

Or, A = 312 -300

A = 12

Ans: A = 12 and B = 6.

6. In the figure below, AB is perpendicular to BC and BD = DC. If AD = √10 cm and AC = 4 cm,

then what is the value of BC? (Shahjalal Islami Bank MTO 10)

A

B D C

Solution:

Given that, AC = 4 cm and AD = √10

Let, BD = DC = x and AB = h

Now, according to the Pythagoras theorem

AD2 = AB

2 + BD

2

Or, (√10)2 = h

2 + x

2

Or, 10 = h2 + x

2

h2 = 10 – x

2 ......(i)

Again,

AC2 = AB

2 + BC

2 [BC = BD+DC = x+ x = 2x]

Or, (4)2 = h

2 + (2x)

2

Or, 16 = 10 – x2 + 4x

2 [From equation (i)]

Or, 16-10 = 3x2

Or, 3x2 = 6

Or, x2 = 2

Or, x = √2

2x = 2√2

The value of BC = 2√2 cm.


Jafar Iqbal Ansary


1. The sum of square of two numbers is 80 and the square of their difference is 16. Determine

the product of the two numbers.

Solution:

Let, the numbers be x and y.

1st Condition,

x2 + y

2 = 80 ....(i)

2nd Condition,

(x-y)2 = 16

Or, x2 + y

2 - 2xy = 16

Or, 80 - 2xy = 16

Or, 80-16 = 2xy

Or, 2xy = 64

xy = 32

The product of the two numbers = 32 (Ans.)

2. A and B are two stations 195 km. apart. A train starts from A at 8 a.m. and travels towards B

at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what

time do they meet? Abyev`t

2wU †÷kb A I B Gi `~iZ¡ 195 wK.wg | GKwU †Uªb †ejv 8 Uvi mgq A †_‡K hvÎv ïiæ K‡i Ges B Gi w`‡K 60 wK.wg / N›Uv

†e‡M hvq| Aci GKwU †Uªb B †_‡K †ejv 9 Uvi mgq hvÎv ïiæ K‡i A Gi w`‡K 75 wK.wg / N›Uv †e‡M †h‡Z _v‡K| KLb Zviv

gy‡LvgyLx wgwjZ n‡e?

Solution:

Let, they meet x hours after 8 a.m.

So, first train travelled = x hr and second train travelled = (x-1) hr


60x + 75(x-1) = 195 [Distance = Speed× Time]

Or, 60x + 75x – 75 = 195

Or, 135x = 195 + 75

Or, 135x = 270

x =

= 2 hrs.

So, they meet = (8 a.m. + 2 hrs) = 10 a.m.

Ans: 10 a.m.

3. If 2a + 3

b = 17 and 2

a+2 – 3

b+1 = 5, then determine the values of a and b.


2a + 3

b = 17

2a = 17- 3

b .... (i)

2a+2

– 3b+1

= 5

Or, 2a.2

2 – 3

b.3

1 = 5

Or, 4.2a. – 3.3

b = 5

Or, 4(17- 3b) – 3.3

b = 5

Or, 68-4.3b -3.3

b = 5

Or, -7.3b = 5-68 = -63

Or, 3b =

= 9


Jafar Iqbal Ansary

Or, 3b = 3

2

b = 2

Putting the value of b in equation (i)

2a = 17- 3

2 = 17-9 = 8

Or, 2a = 2

3

a = 3

Ans: a = 3 and b = 2.

4. A number consists of two digits. The sum of the digits is 15. If 27 is subtracted from the

number, its digits are interchanged. Find the number. Abyev`t

GKwU msL¨v `ywU A¼ wb‡q MwVZ| A¼Ø‡qi †hvMdj 15| hw` msL¨vwU †_‡K 27 we‡qvM †`qv nq Zvn‡j A¼Øq ¯’vb wewbgq K‡i|

msL¨vwU wbY©q Kiæb|

Solution: Let, the ten's digit be x.

So, unit's digit = (15 - x).

Number = l0x + (15 - x) = 9x + 15.

Number obtained by reversing the digits = 10 (15 - x) + x = 150 - 9x.


(9x + 15) - 27 = 150 - 9x

Or, 9x +15 -27 = 150-9x

Or, 9x+9x = 150-15+27

Or, 18x = 162

x = 9

The ten's digit = 9 and unit's digit = (15 - 9) = 6.

Number = 9×9+15 = 96. (Ans.)

5. Masum has twice as much money as Selim and Selim has 50% more money than what Badal

has. The average money with them is Tk. 110, then determine the amount of Masum’s money?

Solution: Let, Badal has Tk. 2x.

So, Selim has 150% of Tk.2x = Tk. 3x and Masum has Tk. 3x×2 = Tk. 6x


= 110

Or, 11x = 330

Or, 6x =

6x = 180

Masum has Tk. 180. (Ans.)


Jafar Iqbal Ansary

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