aust written math solution 2018 by khairul alam...written part aust written math solution 2018...

Written Part AUST Written Math Solution 2018

Khairul’s - 1

File wU‡Z cv‡eb: Combined 3 bank So, Sonali Cash, Basic Bank, PKB EO, Karmasansthan bank

Ab¨ †h †Kvb mgvavb †_‡K A‡bK mn‡R Ges e¨vL¨v we‡kølY I mvBW‡bvU mn mgvavb|

Avkv Kwi mevB fv‡jvfv‡e eyS‡Z cvi‡eb| Prepared by: MD Khairul Alam (Writer)

�AUST Gi cixÿvi Rb¨ wb‡Pi UwcK¸‡jv †`L‡Z cv‡ib:

AUST �� ব�� :

1. Tima & Speed (With Train+Boat) �� ব�� । ��

� � �� ব�� ।

2. Equation �� । �� !�� "��# �ব$�� % �� &�� '(�)#*।

3. Time & Work �� &�� + �� । (��& �!, ��, -�� %� ��, �ব�� ,

�"�� , )

4. Partnership: �� - ��। �ব��$ �� + ��-�� %"�; ব�ব�� "��. �%�/ �� -��

%� ��, + "+�� "�� ব�� 01�2� �� +�� ব��, %�� "��# ��&�� ব �� !�,

�� 0 ��*�� + �3��2 ��*��। �ব��/ ��ব� ��0 ��ব� 4�� +�� ব� �� &�� ।

5. Percentage+Profit loss and Interest (�% �� ব��2� &�ব� �� "��# �5��*

�� ব�"� �'(�)#*, �%6�� AUST �'( ��, �ব �� ব�� ।)

6. Probability �� AUST �� ব�� ব� �� । �� 7/8 ব��। �/�"� �� &�� 9�� +: ��/ �� %�� Probability �� "� �� । %�� ব+.� ��+ �"�� !� �� ;��;ব �� "�� "��ব�� 0�< �� 6�� -� �0�< ��2�� /��

Probability �� 2� �"�� 6� ��।

7. AUST ব�&/�# �� "� �� # । �ব �"� �� 6 %� �� ব��।

8. Mensuration ব� ��"� �� 8�� 6+�ব =�0��ব� (��/�� )

9. Solid Geometry �� ব ��> �?�� !�ব %�� "�� 6�� %��,

��@� �� # �� # ��"�� &�� !�ব।

10. ��ব*�� ব�� &�� + �� 6 �%� � !�ব। ;�� '( �ব�� ব�। ( ��9��

��+ব�� A� 8��। �� %� �� )

B0��"�� ব C�� &��।


Khairul’s - 2

1. A train 300m long, overtakes a man walking along the line (in the same direction of the

train) at the speed of 4 km per hour and passed him in 30 sec. The train reached the

station in 15 minutes after it has passed the man. In what time did the man reach the

station? [BSC-Combined exam-(SO-3Banks)-2018-(Written)]

A_©: 300 wgUvi j¤v GKwU †Uªb GKB w`‡K 4 wKwg/N›Uv †e‡M Pjgvb GKRb e¨w³‡K 30 †m‡K‡Û AwZµg K‡i| †UªbwU gvbylwU‡K AwZµg Kivi 15 wgwbU ci †÷k‡b †cŠ‡Q| KLb gvbylwU †÷k‡b †cŠQv‡e?

�Solution: In 30 sec the train goes = 300m

∴ ‘’ 1 ‘’ ‘’ ‘’ ‘’ = 30

300m

∴ ‘’ 3600 ‘’ ‘’ ‘’ ‘’ = 30

3600300×

= 36000m or 36000÷1000 = 36km/hr. [Since, 1000m = 1km]

(GB 36wKwg ïay †Uªb ev ïay gvby‡li MwZ‡eM bv| KviY ‡Uªb I gvbyl GKcv‡k hvIqvi Kvi‡Y MwZ‡eM we‡qvM K‡i 36 n‡q‡Q| Zvn‡j we‡qvM Kivi Av‡M wQj, 36+4 = 40 hv †Uª‡bi MwZ| )

Let, Train’s speed = x Φ

ATQ,

x- 4 = 36 (‡h‡nZz GKB w`‡K †M‡j MwZ‡eM we‡qvM Ki‡j Relative speed ‡ei nq|)

So, x = 40km/hr (GUv †Uª‡bi MwZ‡eM)

Now, in 1hr or 60min the train goes = 40km

∴ in 1min ‘’ ‘’ ‘’ = 60

40km

∴ in 15min‘’ ‘’ ‘’ = 60

1540×km = 10km (A_©vr gv‡Si duvKv RvqMvUv 10wKwg)

[‡Uª‡bi hvIqv 10 wKwg iv ÍvB gvbylUv‡K AwZµg Ki‡Z †h mgq jvM‡e †mUvB DËi| ]

Now,The man goes 4km in = 1 hr.

∴ ’’ ‘’ ‘’ 1 ‘’ ‘’ = 4

1hr.

∴ ’’ ‘’ ‘’ 10 ‘’ ‘’ = 4

10hr.= 2.5 hrs. Ans: 2.5hrs

Bankers Selection Committee (BSC)-Combined Exam

Post name : Senior Officer (3Banks) Exam date: 17-08-2018 Exam taker: Ahsanullah University of Science & Technology (AUST).

e¨vL¨v: gv‡Si ‡h iv Ív †Uª‡bwU 15wgwb‡U hvq, Zv gvbylwU‡K †h‡Z KZ mgq jvM‡e? Zv †ei Ki‡Z n‡e|

gvbyl

‡Uªb = 300wg. 15 wgwb‡Ui iv Ív •

•

‡÷kb


Khairul’s - 3

2. Two boats on opposite banks of a river start moving towards each other. They first pass

each other 1400 meters from one bank. They each continue to the opposite bank,

immediately turn around and start back to the other bank. When they pass each other a

second time, they are 600 meters from the other bank. We assume that each boat travels

at a constant speed all along the journey. What's the width of the river? [BSC-Combined

exam-(SO-3Banks)-2018-(Written)]

A_©: GKwU b`xi y Zxi †_‡K ywU †bŠKv GKwU AciwUi w`‡K PjwQj| ‡bŠKv ywU cÖ_g Zxi †_‡K 1400 wgUvi ~‡i cÖ_gevi wgwjZ nq| Gfv‡e Zx‡i †cŠQvi ci Avevi hvÎvi‡¤¢i ’v‡b wd‡i Avmvvi mgq Zviv Aci Zxi †_‡K 600wgUvi `~‡i wØZxqev‡ii gZ wgwjZ nq| m¤ú~Y© hvÎvc‡_ ‡bŠKv ywU GKB MwZ‡Z Pj‡Z _v‡K| b`xwUi cÖ ^ ’ KZUyKz?

�Solution: (cÖkœwU fv‡jvfv‡e †evSvi Rb¨ wb‡Pi wPÎwU †`Lyb|)

Let the width of the river be ‘w’ meters.

Speed of first boat = x m and speed of second boat = y

According to the 1st condition,

y

1400w

x

1400 −= -------- (i) ( Time

Speed

ceDis=

tan Ges mvÿv‡Zi mgq `yRb B mgvb mgq ‡j‡M‡Q)

According to the 2nd

condition,

y

600w2

x

600w −=

+ -------- (ii) (cÖ_g †bŠKv b`xi G cvk †_‡K Icv‡k wM‡q Avevi 600 wd‡i Avmvi mgq

2q †bŠKvwUI m¤ú~Y© c_ wM‡q wØZxqevi ïiæi RvqMvq wd‡i Avm‡Z 600 wg Av‡MB †`Lv n‡q‡Q)

By, (ii) ÷ (i) we get,

y

1400w

y

600w2

x

1400

x

600w −÷

−=÷

+(‡hvM we‡qvM Ki‡j (x,y) ev` hv‡e bv, ZvB fvM Ki‡Z n‡e)

⇒1400w

y

y

600w2

1400

x

x

600w

−×

−=×

+

⇒ 1400w

600w2

1400

600w

−

−=

+

⇒ w

2 -1400w+600w-840000 = 2800w-840000

⇒ w2 -800w = 2800w (By adding 840000 in both side)

⇒ w2 = 3600w

∴w = 3600 (Dividing by w)

So, width of the river = 3600 m.

river

1st boat 2

nd boat •

•

2nd

meeting 600m

1st meeting 1400m

Zxi-2 Zxi-1

w-1400m

2w-600

cÖkœwU Kíbvq †`L‡Z wVK Ggb


Khairul’s - 4

♦Alternative solution: (eyS‡Z cvi‡j Kg wj‡L mgvavb Kivi Rb¨ GB wbqgwU mnR)

Let the width of the river be ‘w’ meters.

At the time of first meeting first boat travelled = 1400m and 2nd

boat travelled w-1400 m,

At the time of 2nd

meeting first boat travelled = w+600m and 2nd

boat travelled 2w-600m.

According to the question,

1400 : w-1400 = w+600 : 2w-600 (‡h‡nZz MwZ‡eM Dfq‡ÿ‡Î Constant wQj ZvB c‡_i AbycvZ mgvb n‡e)

⇒600w2

600w

1400w

1400

−

+=

−

⇒ w2 -1400w + 600w - 840000 = 2800w - 840000

⇒ w2 -800w = 2800w

⇒ w2 = 3600w

∴w = 3600 (Dividing by w)

So, width of the river = 3600 m. Ans: 3600 meters

3. In a mixture of milk and water, their ratio is 4:5 in the first container and the same

mixture has 5:1 in the 2nd container. In what ratio should be extracted from each

container and poured into the 3rd container, so that the ratio of milk and water comes

to 5:4 in the 3rd container? [BSC-Combined exam-(SO-3Banks)-2018-(Written)]

A_©: ywU cv‡Î ya I cvwbi AbycvZ h_vµ‡g 4:5 Ges 5:1| Dfq cvÎ †_‡K KZ Abycv‡Z ya I cvwb Zz‡j 3q GKwU cv‡Î ivL‡j ‡mLv‡b ya I cvwbi AbycvZ 5:4 n‡e?

�Solution: (eyS‡Z mgm¨v n‡j mgvav‡bi †k‡li evsjv e¨vL¨wU co–b ) Let, x liter of mixture taken from 1

st container and poured into 3

rd container

and y liter of mixture taken from 2nd

container and poured into 3rd

container.

Ratio of milk and water in 1st container is 4:5 sum of ratio = 4+5 = 9

So, amount of milk extracted from 1st container = x ×

9

4 =

9

x4(KviY x Gi g‡a¨ ya : cvwb = 4:5)

∴Amount of water extracted from 1st container = x ×

9

5 =

9

x5(KviY x G ya ev‡` evKxUv cvwb)

Ratio of milk and water in 2nd

container is 5:1 sum of ratio = 5+1 = 6

So, amount of milk extracted from 2nd

container = y ×6

5 =

6

y5(KviY y Gi g‡a¨ ya : cvwb = 5:1)

∴Amount of water extracted from 1st container = y ×

6

1 =

6

y(KviY y ‡Z ya ev‡` evKxUv cvwb))

Total milk in 3rd

container = 18

y15x8

6

y5

9

x4 +=+ (cÖ_g I 2q cvÎ †_‡K ‡bqv y‡ai †hvMdj)

Total Water in 3rd

container = 18

y3x10

6

y

9

x5 +=+ (cÖ_g I 2q cvÎ †_‡K ‡bqv cvwbi †hvMdj)


Khairul’s - 5


18

y15x8 +:

18

y3x10 + = 5:4 (Z…Zxq cv‡Î ya I cvwbi cwigv‡Yi AbycvZ = 5:4 )

⇒ 4

5

18

y3x1018

y15x8

=+

+

⇒18

y15x50

18

y60x32 +=

+

⇒32x-50x = 15y-60y (Dividing both side by 18 )

⇒ -18x = -45y

⇒2

5

18

45

y

x=

−

−= (AbycvZ †ei Ki‡Z ejv n‡j Gfv‡e Dc‡i wb‡P wj‡L cÖ_‡g fMœvsk mvRv‡Z nq|)

∴x:y = 5:2 Ans: 5:2

�Confusion clear: x= ya Ges y = cvwb fve‡jB Dëvcvëv jvM‡e| A_P welqUv Ggb †h: aiæb cÖ_g cv‡Î 5x

cwigvY wgkÖY Av‡Q ( ya+cvwb), †mLvb †_‡K me¸‡jvB bv wb‡q eis gvÎ x wjUvi ªeY Z…Zxq cv‡Î ivLv n‡jv| GLb GB x

wjUv‡ii meB wK ya? Aek¨B bv| KviY cÖ_g cv‡Îi †gvU ªe‡Y †hgb ya I cvwb GK‡Î wQj x Gi g‡aÏ wVK ya I cvwb Df‡q Av‡Q| †KvbUv KZUzKz Av‡Q? 4:5 Abycv‡Z ya I cvwb Av‡Q| Gfv‡e †mB x Gi ya I 2q cvÎ †_‡K †bqv y Gi g‡a¨ `y‡ai cwigvY †ei K‡i Zv †hvM Ki‡j 3q cv‡Î ya KZUzKz Av‡Q Zv †ei n‡e GKBfv‡e cvwb I †ei n‡e| Gici AbycvZ mvRv‡bv n‡q‡Q|

4. The number of girls in a school is 160 more than 1/3 of the total enrollment in the

school. The number of boys is 280 more than 1/7 of the total enrollment in the school.

How many pupils in the school are girls and boys? [BSC-Combined exam-(SO-3Banks)-2018-

(Written)]

A_©: GKwU K¬v‡mi †gvU QvÎx‡ì msL¨v me©‡gvU QvÎ-QvÎx‡ì 1/3 As‡ki †_‡KI 160 Rb †ewk| Avevi QvÎ‡ì msL¨v †gvU QvÎ-QvÎx‡ì 1/7 Ask †_‡KI 280 †ewk| H ¯‹z‡j ‡gvU KZRb QvÎx Av‡Q ?

�Solution:

Let, total number of pupils in the school = x

So, total girls = 3

x+ 160, and total number of boys =

7

x+280


3

x+ 160+

7

x+280 = x (QvÎ-QvÎx wg‡j me©‡gvU = x Rb|)

⇒ x = 3

x+

7

x+ 440

⇒ x - 3

x-

7

x= 440

�cÖgvY †`‡L wbb: AvZ¥wek¦vm evo‡e| me©‡gvU QvÎ-QvÎx 840 Gi g‡a¨ QvÎx 440 Ges 840 Gi 1/3 Ask = 280| Zvn‡j 440-280 = 160 Avevi QvÎ = 840-440 = 400, GLb, 840 Gi 1/7 =120 Zvn‡j 400-120 = 280 hv cÖ‡kœ cÖ Ë K¬z¸‡jvi mv‡_ wg‡j hvq|

� †evSvi Rb¨ e¨vL¨: cÖ_g cvÎ †_‡K †h x GKK ‡bqv n‡q‡Q Zv‡Z y‡ai cwigvY KZ? cvwbi cwigvb KZ? GUv †ei Kivi Rb¨ cÖ_g cv‡Îi ya:cvwb = 4:5 Gi mvnvh¨ †bqv n‡q‡Q: GKB fv‡e 2q cvÎ ‡bqv y ‡Z KZUzKz ya Ges cvwb Av‡Q Zv †ei Kiv n‡q‡Q Ges †k‡l yB cvÎ †_‡K †bqv ya I cvwbi ‡gvU cwigv‡Yi AbycvZ = 5:4


Khairul’s - 6

⇒21

x3x7x21 −−= 440

⇒21

x11= 440

⇒ 11x- 440×21

⇒ x= 11

21440×

∴ x=840

So, number of girls = 3

840+ 160 = 280 + 160 = 440

and number of girls = 7

840+ 280 = 120 + 280 = 400 Ans: 440 and 400

5. A can do a piece of work in 120 days and B can do it in 150 days. They work together

for 20 days then B leaves and A continues the work alone. 12 days after C joins in place

of B and the work is completed in 48 days more. In how many days can C do it if he

works alone? [BSC-Combined exam-(SO-3Banks)-2018-(Written)]

A_©: A GKwU KvR 120 w`‡b Ges B H KvRwU 150 w`‡b Ki‡Z cv‡i| Zviv GK‡Î 20 w`b KvR Kivi ci B P‡j †Mj Ges A GKvKx KvR Ki‡Z _vK‡jv| A 12 w`b KvR Kivi ci H KvRwU‡Z B Gi ¯’v‡b C Zvi mv‡_ †hvM w`j| Ges Av‡iv 48 w`‡b m¤ú~Y© KvRwU †kl n‡jv| m¤ú~Y© KvRwU C GKvKx KZw`‡b †kl Ki‡Z cvi‡e?

�Solution: (GB wbqgUv eo g‡b n‡jI GUv Ki‡Z Kg mgq jvM‡e, KviY GUv cÖPwjZ wbqg)

In, 120 days A can do = 1 part

∴ ‘’ 1 day A ‘’ ‘’ = 120

1 part.

In, 150 days B can do = 1 part

∴ ‘’ 1 day B ‘’ ‘’ = 150

1 part.

So, in 1 day A and B together can do = 200

3

600

9

600

45

150

1

120

1==

+=+ part.

∴ ‘’ 20 days A and B ‘’ ‘’ ‘’ = 10

3

200

203=

×part.

Again, In, 12 days A can do = = 10

1

120

12= part.

In first 20+10 = 30 days total work done = 5

2

10

4

10

13

10

1

10

3==

+=+ part,

Work left after 30days = 5

3

5

25

5

21 =

−=− part.

Now, 5

3Part is done by A and C = 48 days,


Khairul’s - 7

So, 1 ‘’ ‘’ ‘’ ‘’ A and C = 3

548× = 80 days

In, 80 days A and C can do = 1 part.

∴ ‘’ 1 day A and C ‘’ ‘’ = 80

1 part. (GLvb †_‡K C Gi jvMv w`b‡K x a‡iI mvRv‡bv hvq)

In 1 day C can do = {(A+C)’s 1 day’s work - A’s 1 day’s work} = 240

1

240

23

120

1

80

1=

−=−

240

1 part is done by C in = 1 day

∴ 1 part is done by C in = 240 days.

�Alternative solution: (Kv‡Ri †gvU cwigvY‡K BDwbU a‡i)

Suppose, Total Work be 600 units [LCM of 120 and 150]

So, 1 day work of A =120

600 = 5 units

And, 1 day work of B =150

600 = 4 units

Again, Let 1 day work of C = C units


20×(5+4) + (12×5) + 48×(5+C) = 600

⇒48×(5+C) = 600 - 240

⇒ C = 7.5 - 5 = 2.5

So, C alone can complete the work in 2405.2

600= days Ans: 240 days

6. A manufacturing company uses two machines A and B with different production

capacities. Where working alone machine A can produce a production lot in 5 hours

and machine B can produce the same lot in x hours.when the two machine 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? [BSC-

Combined exam-(SO-3Banks)-2018-(Written)]

A_©: Drc`vbKvix GKwU cÖwZôv‡bi `ywU ‡gwkb A Ges B Gi wfbœ wfbœ Drcv`b ÿgZv i‡q‡Q| †gwkb A, 5 N›Uvq hv Drcv`b Ki‡Z cv‡i †gwkb B, †K GKB cwigvY cY¨ Drcv`b Ki‡Z x N›Uv mgq jv‡M| hLb ywU ‡gwkb GK‡Î KvR K‡i ZLb GKB cY¨ Drcv`b Ki‡Z 2 N›Uv mgq jv‡M| ‡gwkb B GKvKx KvR Ki‡j Zv Drcv`b Ki‡Z me©‡gvU KZ N›Uv mgq jvM‡e?

�Solution: ( cÖPwjZ wbq‡g mnR mgvavb) In, 5hours machine A can produce = 1 part

∴ ‘’ 1 hr ‘’ A ‘’ ‘’ = 5

1 part.


Khairul’s - 8

In, x hours machine B can produce = 1 part

∴ ‘’ 1 hr ‘’ B ‘’ ‘’ = x

1 part.

Again,

In, 2hours machine A and B together can produce = 1 part

∴ ‘’ 1 hr ‘’ A and B ‘’ ‘’ ‘’ = 2

1 part.


2

1

x

1

5

1=+ (A Ges B Gi 1 N›Uvq Kiv †gvU Drcv`‡bi cwigv‡bvi †hvMdj = †gvU Drcv`‡bi

2

1 Ask)

⇒5

1

2

1

x

1−=

⇒10

25

x

1 −=

⇒10

3

x

1=

⇒3x = 10 ∴x = 3

10 So, B alone takes =

3

10hrs or 3.33hrs or 3

3

1hr or 3hrs 20mins

Ans: 3hr 20 min

�Alternate Solution: (‡gvU Kv‡Ri cwigvY‡K BDwbU a‡i|)

Let, the total work be = 10 units. [LCM of 5 and 2]

So, in 1 hr, A can produce = 5

10= 2 units.

And, in 1 hr, A & B together can produce = 2

10= 5 units

Then, in 1 hr, B can produce = (5-2) = 3 units ( yR‡bi †_‡K 1 R‡bi Uv ev` w`‡j Ab¨ R‡biUv †ei n‡e)

So, B alone can produce the lot in 3

10hr or 3

3

1hr or 3hr. 20 min Ans: 3hr 20 min

� ‡`‡L ivLyb Ab¨ A‡bK As‡K Kv‡R w`‡e|

3

10hr wKfv‡e 3 N›Uv 20 wgwbU n‡q †Mj?

3

10hr×60 =200 wgwbU (‡h †Kvb N›Uv‡K 60 w`‡q ¸Y Ki‡j wgwbU nq)= 180+20 wg= 3N›Uv 20 wgwbU|

or 3

10hr = 3

3

1hr = 3hr +

3

1×60min = 3hr 20min

======�=======�=======�=======�=======


Khairul’s - 9

01. A, B and C are Partners. ‘A’ whose money has been in the business for 4 months claims

1/8 of the profits, ‘B’ whose money has been in the business for 6 months claims 1/3 of

the profits. If ‘C’ had Tk.1560 in the business for 8 months, how much money did A and

B contribute to the business? [Sonali Bank (Cash)-2018-(Written)]

A_©: GKwU †hŠ_ eëmvq A,B I C wZbRb Askx`vi| H eëmvq A 4 gv‡mi Rb¨ wewb‡qvM K‡i gybvdvi 1/8 Ask Ges B, 6 gv‡mi Rb¨ wewb‡qvM K‡i gybvdvi 1/3 Ask `vex K‡i ev cvq| hw` C 8 gv‡mi Rb¨ 1560 UvKv wewb‡qvM K‡i, Z‡e AI B cÖ‡Z¨†K eëmvq KZ UvKv wewb‡qvM K‡iwQj ?

�Solution: Let,

A invests Tk. x for 4 months B invests Tk,y for 6months and C invested Tk.z for 8 months

Invest Ratio Of A:B:C = 4x :6y:8z

Here, C’s Monthly investment =1560×8 =12480.

Profit of A = 8

1, Profit of B =

3

1 & Profit of C =

+−

3

1

8

11 =

24

8324 −−=

24

13

Profit ratio of, A:B:C =8

1: 3

1 :

24

13

= =

8

1×24:

3

1×24:

24

13×24 =

3:8:13 or, 3x:8x:13x

Here,C’s profit 13x = Tk. 12480.

So, x = 13

12480= 960

So, A,s profit 3x = 3 ×960 = 2880

and B’s profit 8x = 8×960 = 7680

Again,

4x = 2880 (ïiæ‡Z †jLv 4 gv‡mi wewb‡qv‡Mi cwigvY 4x) ∴x= 4

2880= Tk. 720.

and 6y =7680. ∴y= 6

7680= Tk. 1280. Ans: Tk. 720 and Tk.1280

�Alternative Solution:

Let, the investment of A, B and C is x, y and z respectively.

They were in the business for 4, 6 and 8 months.

So, their monthly investment = 4x, 6y and 8z (‡gvU wewb‡qvM‡K gvm w`‡q ¸Y Ki‡j gvwmK wewb‡qvM nq)

Now,

Profit of A = 8

1, Profit of B =

3

1 & Profit of C =

+−

3

1

8

11 =

24

8324 −−=

24

13

Sonali Bank Ltd.

Post name: Officer (Cash) Exam Date: 18-05-2018

Exam taker: Ahsanullah University of Science & Technology (AUST)


Khairul’s - 10

Profit ratio of, A:B:C =8

1: 3

1 :

24

13

= =

8

1×24:

3

1×24:

24

13×24 =

3:8:13

ATQ,

4x: 6y: 8z = 3:8:13 (wewb‡qv‡Mi AbycvZ = jv‡fi AbycvZ A_© hv, jv‡fi AbycvZ = wewb‡qv‡Mi AbycvZ GKB)

Here, y6

x4=

8

3 (GiKg wZbUv AbycvZ Avm‡j Zvi ga¨ †_‡K †h †Kvb ywU wb‡q wn‡me Kiv hvq|)

⇒ y

x=

32

18

∴ x : y = 18: 32

Again. 13

8

z8

y6=

⇒

78

64

z

y=

⇒

39

32

z

y= ∴ y : z = 32 : 39

Now,

x:y = 18:32

and y:z = 32:39

So, x:y:z = 18:32:39 ( yyB Abycv‡ZB y Gi gvb 32 Gi mgvb ZvB GKUv ev` †`qv hvq| )

So, the investment ratio of A,B & C is 18:32:39 (G¸‡jv n‡”Q †gvU wewb‡qv‡Mi AbycvZ KviY Avgiv aivi mgq x,y,z ‡K †gvU wewb‡qvM a‡iwQjvg|)

Let, Investment of A,B and C is 18x, 32x and 39x


39x = 1560 (C Gi ‡gvU wewb‡qvM = 1560 UvKv)

or, x = 39

1560 = 40

So, investment of A = 18×40 = Tk. 720and Investment of B = 32×40 = Tk. 1280

Ans: Tk. 720 and Tk.1280

�Alternative Solution: (MCQ cixÿvq mn‡R Kivi Rb¨)

A's profit = 8

1 and B's profit =

3

1So, C's profit = 1 – (

8

1+

3

1) = 1-

24

13

24

11=

Then, Ratio of profits, A : B: C =8

1:

3

1:

24

13= 3 : 8 : 13

Now, Ratio of investment = timeof Ratio

profit of Ratio(Gfv‡e Ki‡j ªæZ n‡e) =

4

3:

6

8:

8

13= 18 : 32 : 39

So. A's contribution = Tk. 1560 ×39

18 = Tk. 720 (1560 †K 39 w`‡q fvM K‡i 18 w`‡q ¸Y )

And B's contribution = Tk.1560 × 39

32= Tk.1280 Ans: Tk. 720 and Tk.1280

� fzj n‡Z cv‡i †hfv‡e fve‡j:

A : B: C =8

1:

3

1:

24

13= 3 : 8 : 13 or 3x:8x:13x

Here, 13x = 1560 so, x = 120


Khairul’s - 11

Therefore, 3x = 3×120 = 360

And 8x = 8×120 = 960

Gfv‡e mn‡R Ki‡Z wM‡q Avm‡j A,B & C Gi jv‡fi cwigvY KZ Zv †ei n‡q‡Q| hv mwVK bq| ( mvaviY MCQ cixÿvq Gfv‡e mn‡R Pvq, ZvB A‡b‡KB Gfv‡eB Af¨¯’)

wKš‘ GB cÖ‡kœ †h‡nZz wewb‡qv‡Mi cwigvY †ei Ki‡Z ejv n‡q‡Q ZvB cÖ_‡g †`qv wbqg¸‡jv Abymv‡i Ki‡Z n‡e|

2. Machine A, working alone at its constant rate, produces x pounds of peanut butter in 12

minutes. Machine B, working alone at its constant rate, produces x pounds of peanut

butter in 18 minutes. How many minutes will it take machine A and B, working

simultaneously at their respective constant rate, to produce x pounds of peanut butter?

[Sonali Bank (Cash)-2018-(Written)]

A_©: †gwkb A ‡h MwZ‡Z 12 wgwb‡U x cvDÛ ev`v‡gi gvLb ˆZix Ki‡Z cv‡i| GKB MwZ‡Z mgcwigvY ev`v‡gi gvLb ˆZix Ki‡Z †gwkb B Gi 18 wgwbU mgq jv‡M| ‡gwkb A Ges †gwkb B GK‡Î KvR Ki‡j x cvDÛ ev`v‡gi gvLb ˆZix Ki‡Z KZ mgq †b‡e?

♦Solution: In 12 minutes machine A produces = x pounds.

∴ ‘’ 1 ‘’ ‘’ A ‘’ = 12

x ‘’

Again,

In 18 minutes machine B produces = x pounds.

∴ ‘’ 1 ‘’ ‘’ B ‘’ = 18

x ‘’

Machine A and B together produce in 1 minute 18

x

12

x+ =

36

x3x2 +=

36

x5pounds

36

x5

pounds is produced by A & B in = 1 minute

∴1 ” ” ” ” A & B ” = x5

361× ”

∴ x ” ” ” ” A & B ” = x5

x36×

= 7.2 minutes Ans: 7.2 minutes

03. Two trains running at the rate of 75 km. and 60 km. an hour respectively on parallel

rails in opposite directions, are observed to pass each other in 8 seconds and when they

are running in the same direction at the same rates as before, a person sitting in the

faster train observes that he passes the other in 31.5 seconds. Find the lengths of the

trains? [Sonali Bank (Cash)-2018-(Written)]

A_©: wecixZ w`K †_‡K Avmv ywU †Uªb mgvšÍivj jvB‡b h_vµ‡g 75 wKwg/N›Uv Ges 60 wKwg/N›Uv †e‡M ci¯úi‡K 8 †m‡K‡Û AwZµg K‡i| Avevi, c~‡e©i b¨vq GKB ‡e‡M GKB w`‡K Pj‡j `ªæZMwZi †UªbwU‡Z emv GK e¨w³ jÿ¨ Ki‡jv †h,Zvi †UªbwU (`ªZZi †Uªb) axiMwZi †UªbwU‡K 31.5 †m‡K‡Û AwZµg K‡i‡Q| †Uªb ywUi ˆ`N © †ei Kiæb ?


Khairul’s - 12

♦Solution: Here, opposite direction moving time,

Relative speed=75+60 = 135km or 135×1000 = 135000m

In 1 hr or 3600 sec both the train go = 135000m

∴ in 1 sec both the train go = 3600

135000= 37.5m

∴ in 8sec both the train go = 37.5×8 = 300 meters

Length of both train = (37.5×8) = 300 meters (GB 8 †m‡K‡Û hvIqv c_wUB n‡”Q ywU †Uª‡bi ˆ`N© )

Now, same direction moving time,

Relative speed = (75-60) km./hr = 15 km,/hr or, 15×1000 = 15000 m

In 1 hr or 3600 sec the train goes = 15000m

∴ in 1 sec the train goes = =3600

15000=

6

25m/sec.

∴ in 31.5 sec both the train goes = 5316

25.× meters

length of the slower train = 5316

25.× meters = 131.25 meters

And length of the faster train = 300-131.25 = 168.75 meters Ans: 168.75m And 131.25m.

�Alternative question: (MCQ Gi Rb¨)

Total distance covered of length of both trains = (75+60)×18

5×8 = 300 m.

Length of the slower train = (75-60)×18

5×31.5 = 131.25m.

So, length of the faster train = (300-131.25) = 168.75 m. Ans: 168.75m And 131.25m.


Khairul’s - 13

04. A gardener plants two rectangular gardens in separate regions on his property. The

first garden has an area of 600 square feet and a length of 40 feet. If the second garden

has a width twice that of the first garden, but only half of the area, what is the ratio of

the perimeter of the first garden to that of the second garden? [Sonali Bank (Cash)-2018-

(Written)]

A_©: GKRb evMv‡bi gvwjK Zvi Rwg‡Z wfbœ RvqMvq y‡Uv AvqZvKvi evMvb ˆZix K‡i‡Qb| cÖ_g evMv‡bi †ÿÎdj 600 eM©dyU Ges ˆ`N © 40 dzU | hw` wØZxq evMv‡bi cÖ ’ cÖ_g evMv‡bi cÖ‡¯’i wØ¸Y Ges ‡ÿÎdj cÖ_g evMv‡bi ‡ÿÎd‡ji A‡a©K nq, Zvn‡j cÖ_g I wØZxq evMv‡bi cwimxgvi AbycvZ KZ ?

�Solution: Given that, area of first garden = 600 sq.m and length = 40m.

So width of first garden = 40

600=15 m.

Perimeter of the first garden = 2 (40+15) = 2×55 = 110m

So, width of 2nd garden =15×2=30 (Since the width of 2nd garden double of 1st garden)

and area of the 2nd

garden 2

600=300 sq.m (Since the area is half of 1st garden)

So length of 1st garden 30

300=10m

Perimeter of the 2nd

garden = 2 (30+10) = 2×40 = 80m

So ratio of perimeter of two garden is = 110:80 = 11:8 Ans:11: 8

05. In a certain class, 1/5 of the boys are shorter than the shortest girls in the class and 1/3

of the girls are taller than the tallest boy in the class. If there are 16 students in the class

and no two people have the same height, what percent of the students are taller than the

shortest girl and shorter than the tallest boy? [Sonali Bank (Cash)-2018-(Written)]

A_©: GKwU †kÖYx‡Z GK cÂgvsk †Q‡j H K¬v‡mi me‡P‡q Luv‡Uv †g‡qi †P‡qI Lvu‡Uv Ges GK Z…Zxqvsk †g‡q me‡P‡q j¤v †Q‡ji †P‡qI j¤v| hw` H †kÖYx‡Z 16 Rb QvÎQvÎx _v‡K Ges 2 Rb QvÎQvÎxi D”PZv GKB bv nq, Zvn‡j kZKiv KZRb QvÎ- QvÎx me‡P‡q Luv‡Uv †g‡q‡`i †P‡q j¤v Ges me‡P‡q j¤v †Q‡ji †P‡q Luv‡Uv ?

♦Solution: GB AsKUvi fvlv¸‡jv eyS‡Z A‡b‡KiB mgm¨v n‡Z cv‡i ZvB Av‡M evsjvq eySzb Ges cÖ‡kœi fvlvi mv‡_ wb‡Pi wPÎwU GKevi wgwj‡q wbb| 16 R‡bi g‡a¨ QvÎ = 10 Rb Ges QvÎx 6 Rb ( KviY 5w`‡q QvÎ msL¨v‡K Ges 3 w`‡q QvÎx msL¨v‡K fvM Kiv †h‡Z n‡e, bv n‡j GK cÂgvsk ev GK Z…Zxqvsk evbv‡bv hv‡e bv|

me‡_‡K Lvu‡Uv †g‡q me‡_‡K j¤v †Q‡j

Luv‡Uv †g‡qi †_‡KI Lvu‡Uv ‡Q‡j 2 Rb

j¤v †Q‡ji †_‡KI j¤v †g‡q 2 Rb `ycv‡k 3+3 = 6 Rb n‡j gv‡S 16-6 = 10 GB 10 RbB n‡jv me‡_‡K Lvu‡Uv †g‡qi PvB‡Z

j¤v Avevi me‡_‡K j¤v †Q‡ji †P‡q Lvu‡Uv| hviv †gvU 16 R‡bi 62.5% GUvB DËi|

wb‡Pi wPÎwU ‡evSvi †Póv Kiæb


Khairul’s - 14

� cixÿvq Gfv‡e wjL‡Z cv‡ib Total student = 16

5

1of the boys means the number of boys must be a multiple of 5.

3

1of the girls means that the no. of girls must be a multiple of 3.

So, the number of boys can be = 5,10 or 15 and girls = 3,6,9,12 or 15

from this number of boys 10 and number of girls 6 satisfy the condition;

(KviY QvÎ 5 Rb wb‡j Aewkó 16-5 = 11 Rb QvÎx‡K 3 w`‡q fvM Kiv hvq bv ZvB QvÎ 10 wb‡j Aewkó 16-10 = 6 ‡K 1/3 Ask evbv‡bv hvq|)

So, Total boys and girls =10+6 = 16

Shorter boys = 5

110 × = 2 and shortest girl = 1

Taller girls = 63

1× =2 and tallest boy = 1

Taller than shortest girls and shorter than tallest boy = [16- (2+2+1+1)] (me‡_‡K Lvu‡Uv †g‡qi †_‡K Lvu‡Uv 2 Rb + me‡_‡K j¤v †Q‡ji †_‡KI j¤v 2 Rb + me‡_‡K Lvu‡Uv 1 Rb + me‡_‡K j¤v †g‡q 1Rb mn †gvU = 6 Rb | hv me©†gvU 16 Rb †_‡K we‡qvM Ki‡j hviv _v‡K Zviv 10 Rb n‡”Q Av‡Mi 6 R‡bi gv‡S _vK‡e )

= 16 – 6 = 10

∴Required Percentage = 10016

10× = 62.5% Ans: 62.5%

======�=======�=======�=======�=======


Khairul’s - 15

1. wcZv I cy‡Îi eZ©gvb eq‡mi mgwó 50 eQi| hLb cy‡Îi eqm wcZvi eZ©gvb eq‡mi mgvb n‡e ZLb Zv‡ì eq‡mi mgwó 102 eQi n‡e| wcZv I cy‡Îi eqm †ei Kiæb| [Kg©ms ’vb e¨vsK-(WvUv Gw›Uª Acv‡iU)-2018-(wjwLZ)] [31 Zg wewmGm-(wjwLZ)]

�mgvavb: awi,

cy‡Îi eZ©gvb eqm = x

wcZvi eZ©gvb eqm = 50 - x

Zv‡ì eq‡mi eëavb = (50 – x) - x = 50 - 2x

cyÎ hLb wcZvi mgvb n‡e ZLb cy‡Îi eqm = x + 50 - 2x eQi ci (cy‡Îi eZ©gvb eq‡mi mv‡_ eëavb †hvM)

A_©vr cy‡Îi fwel¨r eqm = 50 - x

Avevi ZLb wcZvi eqm n‡e = (50 - x) + (50 - 2x) (cy‡Îi eqm hZUzKz evo‡e wcZvi I ZZUzKz evo‡e)

A_©vr wcZvi fwel¨r eqm = 100 - 3x

cÖkœg‡Z, 50 - x + 100 - 3x = 102 [ yÕR‡bi fwel¨r eq‡mi mgwó = 102]

ev, -4x = 102 - 150

ev, - 4x = - 48

∴ x = 12

myZivs cy‡Îi eZ©gvb eqm = 12 Ges wcZvi eZ©gvb eqm = 50 - 12 = 38 Ans. 38 I 12 eQi|

2. 300 UvKvi 4 eQ‡ii mij gybvdv I 400 UvKvi 5 eQ‡ii mij gybvdv GK‡Î 148 UvKv n‡j, kZKiv gybvdvi nvi KZ

? [Kg© ms ’vb e¨vsK-(WvUv Gw›Uª Acv‡iU)-2018-(wjwLZ)]

�mgvavb: 300 UvKvi 4 eQ‡ii my` hZ 300×4 = 1200 UvKvi 1 eQ‡ii my` ZZ|

400 UvKvi 5 eQ‡ii my` hZ 400×5 = 2000 UvKvi 1 eQ‡ii my` ZZ|

(1200+2000) = 3200 UvKvi 1 eQ‡ii my` = 148 UvKv

∴1 UvKvi 1 eQ‡ii my` = 3200148

UvKv

∴100 UvKvi 1 eQ‡ii my` = 3200

100148 × UvKv = 4.625% DËi: 4.625%

Karmasangsthan Bank Ltd.

Post name: Data Entry Operator Exam Date: 27-04-2018

Exam Taker:


Khairul’s - 16

3. hš¿vs‡k 40%, `vjv‡j 25%, KuvPvgv‡j 15% Ges Avmevec‡Î 5% UvKv LiP Kivi ci nvmv‡bi nv‡Z 1305 UvKv _v‡K| Zvi Kv‡Q KZ UvKv wQj ? [Kg© ms ’vb e¨vsK-(WvUv Gw›Uª Acv‡iU)-2018-(wjwLZ)]

�mgvavb: g‡b Kwi, cÖ_‡g nvmv‡bi Kv‡Q wQj = 100 UvKv| Zvi ‡gvU LiP = 40+25+15+5=85 UvKv| Aewkó _v‡K = 100-85 = 15UvKv nvmv‡bi Kv‡Q eZ©gv‡b 15 UvKv _vK‡j cÖ_‡g wQj = 100 UvKv

∴ ÕÕ ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ = 15100

UvKv

∴ ÕÕ ÕÕ ÕÕ 1305 ÕÕ ÕÕ ÕÕ ÕÕ = 15

1305100× UvKv = 8700 UvKv DËi: 8700 UvKv

4. GKwU eB 65 UvKvq wewµ Ki‡j 30% jvf K‡i| 10% jv‡f wewµ Ki‡Z PvB‡j bZzb wewµi g~j¨ KZ n‡e? [Kg© ms ’vb e¨vsK-(WvUv Gw›Uª Acv‡iU)-2018-(wjwLZ)]

�mgvavb: 30% jv‡f, weµqg~j¨ = 100+30 = 130 UvKv| eBwUi weµqg~j¨ 130 UvKv n‡j µqg~j¨ = 100 UvKv

eBwUi weµqg~j¨ 1 UvKv n‡j µqg~j¨ = 130100

UvKv

eBwUi weµqg~j¨ 65 UvKv n‡j µqg~j¨ = 130

65100× UvKv = 50 UvKv

Avevi 10% jv‡f, weµqg~j¨ = 100+10 = 110 UvKv|

µqg~j¨ 100 UvKv n‡j weµqg~j¨ = 110 UvKv

∴ µqg~j¨ 1 UvKv n‡j weµqg~j¨ =100110

UvKv

µqg~j¨ 50 UvKv n‡j weµqg~j¨ = 100

50110 × UvKv = 55 UvKv DËi: 55 UvKv

======�=======�=======�=======�=======


Khairul’s - 17

1. If a person Invest Tk. 4000 at x% and Tk. 5000 at y%. He will get total Tk. 320 as

interest. On the other hand if he invest Tk. 5000 at x% and Tk. 4000 at y% , he will get

total Tk. 310 as interest. Find the value of x and y. [Sonali Bank (Officer)-2018-

(Written),[Dhaka Bank(TACO)-2018] & [Basic Bank – (AM)-2018- (written)]

A_©: GK e¨w³ x% gybvdvq 4,000 UvKv Ges y% gybvdvq 5,000 UvKv Rgv K‡ib| eQi †k‡l wZwb 320 UvKv gybvdv AR©b K‡ib| wZwb hw` x% my‡` 5,000 UvKv Ges y% my‡` 4,000 UvKv Rgv ivL‡Zb Zvn‡j eQi †k‡l 310 UvKv gybvdv †c‡Zb| x Ges y Gi gvb KZ ?

♦Solution:

According to the first condition

x% of 4000 + y% of 5000 = 320 [2 Avmj †_‡K cÖvß †gvU my‡`i cwigvY = 320UvKv| ]

Or, 100

x × 4000 +

100

y × 5000 = 320

Or, 40x+50y = 320

∴ 4x+5y = 32 ……………(i)

According to the second condition

x% of 5000 + y% of 4000 = 310 ( ywU wfbœ Avmj †_‡K cÖvß †gvU my‡`i cwigvY = 310 UvKv|)

Or, 100

x× 5000+

100

y× 4000 = 310

Or, 50x+40y = 310

∴ 5x+4y = 31……………(ii)

Now, (i) × 5 - (ii) × 4

⇒ 25y- 16y =160 -124

Or, 9y = 36

∴ y = 4

By putting the of y into the equation (ii) we get,

5x + 4×4 =31

Or, 5x =31-16

Or, 5x =15

∴ x = 3 Ans: x = 3 and y = 4

Basic Bank Ltd.

Post name: Assistant Manager-(General) Exam Date: 16-03-2018

Exam Taker : Ahsanullah University of Science & Technology


Khairul’s - 18

2. Working together pipe P, Q and T can fill a trunk in 5 hours. Working together P and

Q can fill it in 7 hours. Find in how many hours T can fill it? [Basic Bank- (AM)-2018-

(Written)]

A_©: wZbwU cvBc P, Q I T GK‡Î GKwU U¨vsK 5 N›Uvq c~Y© Ki‡Z cv‡i| Avevi P I Q GK‡Î 7 N›Uvq c~Y© Ki‡Z cv‡i| m¤ú~Y© U¨vsKwU c~Y© Ki‡Z T Gi GKvKx KZ mgq jvM‡e?

♦Solution: Part filled by P, Q and T in 5hours = 1 part

Part filled by P, Q and T in 1 hour = 5

1 part

Part filled by P and Q in 1 hour = 1 part

Part filled by P and Q in 1 hour =7

1 part

Part filled by T in 1 hour = 5

1-

7

1=

35

57 −=

35

2(3 wU cvB‡ci KvR †_‡K 2wU ev` w`‡j 1wUi KvR †ei n‡e)

Now, 35

2 part is filled by T in = 1 hour

∴ Whole part is filled by T in = 2

35hour = 17.5 hours. Ans: 17.5 hours

3. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at random.

What is the probability that they are not of same color?[Basic Bank- (AM)-2018-(Written)]

A_©: GKwU ev‡· 5wU meyR, 4wU njy` Ges 3wU mv`v ej Av‡Q| ev· †_‡K ‰`efv‡e 3wU ej †Zvjv n‡jv| ej¸‡jv GKB is‡qi bv nIqvi m¤¢vebv KZ?

♦Solution: Total balls = 5+4+3 = 12

Total cases = 12

C3 =123

101112

××

××= 220 (A_©vr 12 Uv ej †_‡K 3 Uv †bqv hv‡e me©‡gvU 220 fv‡e)

Total cases of drawing same colour =5C3+

4C3+

3C3=15

Probability of same color =220

15=

44

3 (A_©vr 3wU ejB GKB nIqvi m¤¢vebv GUv|) Ans:

44

41

Probability of not same colur =1-44

3=

44

41 (me©‡gvU djvdj †_‡K GKwU ev` w`‡j Ab¨wU †ei nq|)

�Alternative solution: The probability that they all are of same color =

44

3

10

1

11

2

12

3

10

2

11

3

12

4

10

3

11

4

12

5=××+××+××

the probability that they are not of same colour = 1- 44

41

44

3= Ans:

44

41


Khairul’s - 19

4. There was a shipment of cars. Out of which half was black in colour. Remaining cars

were equally blue, white and red. 70% of black. 80% of blue.30% of white and 40% of

red cars were sold. What % of total cars were sold? [Basic Bank- (AM)-2018-(Written)]

A_©: GKwU Mvoxi Pvjvb wQj| Gi g‡a¨ A‡a©K Mvox Kv‡jv is‡qi Ges evwK A‡a©K Mvoxi mgvb msL¨K bxj, mv`v I jvj is‡qi| 70% Kv‡jv Mvox, 80% bxj Mvox, 30% mv`v Mvox Ges 40% jvj Mvox wewµ Kiv n‡jv| †gv‡Ui Dci kZKiv KZ fvM Mvox wewµ Kiv n‡jv?

♦Solution: Let, total cars 600 cars. (Ggb msL¨v ai‡Z n‡e hv‡Z †k‡l cÖ‡kœi % ¸‡jvi wn‡me Lye mn‡R Kiv hvq|)

So black cars = 2

600= 300 cars.

Remaining cars= 600-300 = 300 cars.

So blue cars = 3

300= 100, white cars = 100 and yellow cars = 100

Cars sold = (300×70%+100×80%+100×30%+100×40%) = 210+80+30+40 = 360 cars

So car sold %=600

100360×= 60% Ans: 60%

�Alternative solution: (x a‡i mgvavb:)

Let, blue cars = x white = x and red cars = x and black color cars = 3x (‡h‡nZz †gv‡Ui A‡a©K Kv‡jv) Total number of cars = 3x+x+x+x = 6x

Total number of cars sold = 70% of 3x + 80% of x + 30% of x + 40% of x

= 2.1x+0.8x+0.3x+0.4x ( †QvU `kwgK msL¨v fMœvs‡ki †_‡K mnR|)

= 3.6x

Percentage of total car sold = x6

100x63 ×. = 60%

5. A train passes a man in 3 second, and another train from opposite direction pass the

man 4 second, both train same length. How long time need to pass the train each other?

[Basic Bank- (AM)-2018-(Written)]

A_©: mgvb ˆ`‡N© i ywU †Uª‡bi g‡a¨ GKwU †Uªb GK e¨w³‡K 3 †m‡K‡Û Ges wecixZ w`K †_‡K Avmv Av‡iKwU †Uªb GKB e¨w³‡K 4 †m‡K‡Û AwZµg K‡i| †Uªb ywUi GKwU Av‡iKwU‡K AwZµg Ki‡Z KZ mgq jvM‡e?

GKUz mnR fvlvq : GKUv †jvK `y cvk †_‡K Avmv mgvb ˆ`‡N© i ywU †Uª‡bi gv‡S c‡i †M‡Q| mvg‡bi w`K †_‡K Avmv †UªbUv Zv‡K 3 †m‡K‡Û Avi †cQb †_‡K Avmv †UªbwU Zv‡K 4 †m‡K‡Û AwZµg K‡i | cÖkœ n‡”Q †h †Uªb `ywU Zv‡K AwZµg bv K‡i G‡K Aci‡K ( GK †Uªb Av‡iK †Uªb‡K) AwZµg Ki‡Z †M‡j KZ mgq jvM‡Zv?


Khairul’s - 20

♦Solution: Let the length of each train = x meter

So speed of 1st train=

3

x= 4 meter/per second

And speed of 2nd

train=4

x= 3m/s

So relative speed =3

x+

4

x

(since opposite direction)

=12

x7

12

x3x4=

+ m/s

Total distance to cross the train = length of two train=x+x = 2x meter

So time needed =

12

x7

x2seconds (`~iZ¡ ÷ MwZ = mgq) =

7

24 or,

7

33 seconds Ans:

7

33 sec

6. A man works for certain hours. If his hourly payment increase by 20%, what percent of

working hours he may reduce so that total income remain unchanged ? [Basic Bank-

(AM)-2018-(Written)]

A_©: GK e¨w³ wbw`©ó N›Uv KvR K‡i| hw` Zvi cÖwZ N›Uvi gRyix kZKiv 20 UvKv e„w× cvq, Z‡e Kg© N›Uv kZKiv KZ Kgv‡j †gvU Avq AcwiewZ©Z _vK‡e?

♦Solution: Let, his hourly pay 10 Tk. and total work hour 10 hours

So overall payment=10×10=100 Tk.

Now, his hourly pay=10× 120% =12 Tk.

So no of hours ×12=100

Or no of hours=12

100=

3

25hours

So hours reduction=10-3

25=

3

5hours

Hours reduction in percentage = 10

1003

5×

= 30

500= 16.67% Ans: 16.67%

♦Alternative Solution: Applying rule of proportionality,

A:B = 5:6 (Wage)

A:B = 6:5 (Hour)

So hour reduces = (6-5) = 1 ∴ Required percentage =

×100

6

1 = 16.67 % Ans: 16.67%

♦wKfv‡e ? &gvbyl‡K AwZµg Kivi mgq †Uªb ywU wb‡Pi ˆ`N© B AwZµg K‡i Zvn‡j ‡Uªb ywU G‡K Aci‡K AwZµg Kivi mgq †h‡Z n‡e Zv‡`i ˆ`‡N© i †hvMdj| GB ˆ`N© ‡K Zv‡`i yR‡bi MwZ‡e‡Mi †hvMdj w`‡q fvM Ki‡jB DËi ‡ei n‡e|

eyS‡j A‡bK mn‡R †jLv: 100 UvKv‡Z 20 UvKv evo‡j 120 nq Gici 120 †_‡K, Avevi 100 †Z Avb‡Z PvB‡j 20 Kgv‡Z n‡e| 120 G 20 Kgv‡j 100 †Z 16.67%|

g‡b ivLyb: GKevi evov‡bvi ci †hUv nq Kgv‡bvi mgq †mB ewa©Z cwigvY †_‡K Kgv‡Z nq|


Khairul’s - 21

7. There were some books of novel and nonfiction. Board discuss 3 times for any novel and

5 times for any nonfiction. During a year they discuss total 52 times. If there were 12

books. How many of them were novel? [Basic Bank- (AM)-2018-(Written)]

A_©: wKQz Dcb¨vm I bb wdKk‡bi eB Av‡Q| †ev‡W©i m`m¨iv †Kvb Dcb¨vm wb‡q 3 evi Ges bb wdKkb wb‡q 5 evi Av‡jvPbv K‡i| Gfv‡e GK eQ‡i Zviv †gvU 52 evi Av‡jvPbv K‡i| †mLv‡b hw` †gvU 12wU eB _v‡K, Zvn‡j Gi g‡a¨ Dcb¨vm KZwU?

� Solution: (GK PjK wewkó mgxKiY mvwR‡q mgvavb|)

Let the number of novel = x

Then, the number of fiction =12 - x

According to question,

3x + 5(12- x) = 52 (Dcb¨v‡mi Av‡jvPbvi 3¸Y + wdKk‡bi Av‡jvPbvi 5 ¸Y = ‡gvU Av‡jvPbv 52)

Or, 3x + 60 – 5x = 52

Or, 2x = 8

∴ x = 4

So, number of novel is 4. Ans: 4

♦Alternative Solution: (`ywU mgxKiY mvwR‡q mgvavb|)

Let number of novels book x and nonfiction books y

So x+y =12 ………….(i)

And 3x+5y=52 ……..(ii)

Now (i)×5- ii

5x+5y = 60

3x+5y = 52

2x = 8 [by subtracting]

∴ x = 4

So, number of novel is 4. Ans: 4

======�=======�=======�=======�=======


Khairul’s - 22

1. In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed

for the trip was reduced by 200km/hr and the time of flight increased by 30 minutes.

Find out the duration of the flight? [Rupali Bank Ltd.(JO)-2013(written)] & [PKB-(SEO)-2018-

(Written)]

A_©: 600 wK.wg Gi hvÎv c‡_ GKwU D‡ovRvnvR Lvivc AvenvIqvi Kvi‡Y Gi MwZ Kwg‡q †`q| GZ H hvÎvq D‡ovRvnvRwUi Mo †eM 200 wK.wg/N›Uv n«vm cvq Ges mgq 30 wgwbU e„w× cvq| H hvÎvi mgq wbY©q Kiæb|

�Solution: Let. normal speed of the flight be x km/hr.

So. normal time to travel 600km be x

600hour

If 30 min or 2

1 hr more takes due to slow down.


2

1

x

600

200x

600=−

−

Or, 2

1

200xx

120000x600x600=

−

+−

)(

Or, x2-200x=240000

Or, x2 -200x-240000 = 0

Or, x2 -600x+400x-240000 = 0

∴(x-600) (x+400) = 0

∴x= 600 and x = - 400 [distance can never be native]

∴Duration of the flight= h1hr600

600

x

600== Ans:1 hrs.

(Since the duration of the flight is fixed, extra time will not be added)

2. A alone can reap a certain field in 15 days and B in 12 days .If A begins alone and after

a certain interval B joins him, the field is reaped in 7.5 days .How long did A and B

work together. [PKB-(SEO)-2018-(Written)]

A_©: GKwU Rwgi avb KvU‡Z A GKvKx 15 w`b Ges B GKvKx 12 w`b mgq †bq| hw` A GKvKx avb KvUv ïiæ K‡i Ges wKQzw`b ci B Zvi mv‡_ †hvM †`q , Gid‡j †gvU 7.5 w`‡b Zviv m¤ú~Y© Rwgi avb KvU‡Z cvi‡jv|AI B

GKmv‡_ KZw`b KvR K‡iwQj?

Probasi Kollyan Bank

Post name: Senior Executive Officer Exam date: 02-03-2018

Exam taker: AUST


Khairul’s - 23

♦Solution: Let, B works for x days.

In 1 day A can do = 15

1 part

In 7.5 day A can do = 2

1

15

57=

. part ( †h‡nZz A KvRwUi ïiæi w`b †_‡K †gvU 7.5 w`b KvR K‡i‡Q|)

In 1 day B can do =12

1 part

In x days B can do =12

x part


15

57.+

12

x=1 (© yR‡bi Kiv Kv‡Ri As‡ki †hvMdj = m¤ú~Y© KvR ev 1 Ask|)

Or, 2

1+

12

x=1

Or, 12

x6+=1

Or,6+x= 12

∴x = 6 days

Thus, A and B work 6 days together. (KviY B ‡h 6 w`b KvR K‡i‡Q H 6 w`b A I KvR K‡i‡Q|)

Ans: 6 days

3. 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)-2014(Written)] & [PKB-(SEO)-2018-(Written)]

A_©: A, B, C, D, E 5wU D×©µ‡g mvRv‡bv avivevwnK msL¨v hv‡`i g‡a¨ GKwU msL¨v‡K evwZj Ki‡j msL¨v¸‡jvi †hvMdj 20% n«vm cvq| †Kvb msL¨vwU‡K evwZj Kiv n‡q‡Q?

♦Solution: Let, the consecutive numbers are,

A = 1, B = 2, C = = 3, D = 4 & E = 5 [Gfv‡e msL¨vi gvb a‡i wn‡me Kiv hvq ]

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 = 15× 3100

20=

So, the deleted number is the 3rd as C from the set Ans. C


Khairul’s - 24

♦Alternative solution: Since a,b,c,d,e are increasing order consecutive number

a, b= a+1, c= a+2, d= a+3, e = a+4

The sum of five numbers

=a+ (a+1) + (a+2) + (a+3) + (a+4) = 5a +10

20% of 5a+10 = 5(a+2)×100

20 = a+2

if 20% a number is deleted then 20% of sum or a+2 is reduced.

So, the number is a+2 = c Ans: c

4. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the

taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank

will be completely filled in what time ?[ Basic Bank (AO)-2009- (Written)] &[PKB-(SEO)-2018-

(Written)]

A_©: GKwU bj w`‡q GKwU U¨vsK 20 wgwb‡U c~Y© nq, Ges Ab¨ GKwU bj w`‡q 60 wgwb‡U c~Y© nq| `ywU bjB GKmv‡_ Pvjy Kivi 10 wgwbU ci cÖ_g bj‡K eÜ K‡i †`qv n‡jv| Gi KZÿY ci m¤ú~Y© U¨vsKwU c~Y© n‡e?

� Solution: First Tap in 20 minutes can fill = 1 portion

∴ First Tap in 10 minutes can fill 2

1

20

10= portion

2nd

Tap in 60 minutes can fill = 1 portion

2nd

Tap in 10 minutes can fill 6

1

60

10= portion

∴ In 10 minutes tank filled = 6

1

2

1+ =

6

13 +=

3

2

6

4= portion

Remaining = 1-3

2 =

3

1 portion (GB Aewkó AskwU 2q bj‡K c~Y© Ki‡Z n‡e| KviY cÖ_g bj eÜ n‡q †M‡Q|)

2nd

Tap can fill 1 portion in 60 minutes

∴ 2nd

Tap can fill 3

1 portion in 60 ×

3

1 = 20 minutes Ans: 20 min.

5. Two persons are running in 3.6km/h and 7.2 km/h speed, A train passes them in 9 & 9.5

seconds. What is the length of the train and speed of the train? [PKB-(SEO)-2018-

(Written)]

A_©: yB Rb e¨w³ h_vµ‡g 3.6 wK.wg ./N›Uv Ges 7.2 wKwg/ N›Uv ‡e‡M hv‡”Q | GKwU †Uªb Zv‡`i yRb‡K h_vµ‡g 9 †m‡KÛ Ges 9.5 †m‡K‡Û AwZµg K‡i | †UªbwUi ‰`N © Ges MwZ‡eM KZ?

� Solution:

3.6km or 3.6× 1000 = 3600 meters


Khairul’s - 25

in 1 hr or 3600 sec first person goes = 3600meter

So, in 1 ‘’ ‘’ ‘’ ‘’ = 3600

3600 = 1 mps

Similarly,

7.2km or 7.2× 1000 = 7200 meters

in 1 hr or 3600 sec 2nd

person goes = 7200meter

So, in 1 ‘’ ‘’ ‘’ ‘’ = 3600

7200 = 2 m

Let, length of train be x meter & speed y mps.

Relative speed for first person = y – 1 ( †Uªb Ges gvbyl GKB w`‡K hv‡”Q| hw`I cÖ‡kœ w`‡Ki K_v ejv †bB wKš‘ Kg w¯ú‡W Pjv †jvK‡K AwZµg Ki‡Z †h mgq jvM‡Q †ewk MwZ‡Z Pjv †jvK‡K AwZµg Ki‡Z Zvi †_‡K †ewk mgq ‡j‡M‡Q, †`‡LB †evSv hv‡”Q GKB w`‡K hv‡”Q me| bv n‡j D‡ëvUv n‡Zv|)

First condition,

1y

x

−= 9

Or, x= 9y-9……(i)

2nd

Condition,

2y

x

− = 9.5

Or, x = 9.5y-19

Or, 9y-9 = 9.5y-19 [x=9.5-9y ewm‡q|]

Or, 0.5y=10

∴y= 20mps

Putting the value of x in equation (i) we get

x= 9×20-9

or, x=180-9 ∴x =171 meters

if the train goes in 1 sec = 20m

then it goes in 3600 sec = 20×3600 = 72000 m or 72km/hr

So, the length of the train is 171 meters and speed of the train is 72km/hr

Ans: 171 meters and 72km/hr

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aust written math solution 2018 by khairul alam...written part aust written math solution 2018...

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