Govt bank exam [www.chakritips.com]

1. Recent Publication 257
Bangladesh Bank
Recruitment Test for Officer (General) – 27.12.2019
01. If n is an even integer, which of the following must be an odd integer?
a) n2
– n b) n + 2
c) 3n – 1 d) 3n3
ans. C
Shortcut: cÖ‡kœ ejv n‡”Q †h, n hw` GKwU †Rvo msL¨v nq, Zvn‡j Ack‡bi †KvbwU Aek¨B we‡Rvo n‡e?
†h‡nZz, n n‡jv even ev †Rvo GB AsK Dfq †_‡K Back Solve Gi gva¨‡g Kiv hv‡e|
†m‡nZz, Dˇi n = 2 emv‡j cvB,
Option n = even (†Rvo) = 2 n‡j Remark
a) n2
– n 22
– 2 = 4 – 2 = 2 Even (†Rvo), ZvB ev`
b) n + 2 2 + 2 = 4 Even (†Rvo), ZvB ev`
c) 3n – 1 (3  2) – 1 = 6 – 1 = 5 Odd (we‡Rvo), ZvB GwU n‡e
d) 3n3
3  23
= 3  8 = 24 Even (†Rvo), ZvB ev`
02. How many integers from 1 to 1000 are divisible by 30 but not by 16?
a) 29 b) 31
c) 32 d) 38 ans. A
Solution: cÖ‡kœ ejv n‡”Q, 1 †_‡K 1,000 Gi g‡a¨ Ggb KZwU c~Y© msL¨v Av‡Q, hv‡K 30 w`‡q fvM Kiv hvq, wKš‘ 16
w`‡q fvM Kiv hvq bv|
AsKwU RwUj| GwU Avcwb, Step by Step wPš—v K‡i mgvavb Ki‡eb|
1 †_‡K 1,000 ch©š— †gvU msL¨v Av‡Q 1,000 wU|
GB 1,000 wU msL¨vi g‡a¨ 30 w`‡q fvM Kiv hvq Ggb msL¨v Av‡Q =
1000
30
= 33 wU|
GLb, GB 33 wU msL¨vi g‡a¨ †_‡K Avcbv‡K H msL¨v¸‡jvB †ei Ki‡Z n‡e, hv‡`i‡K 16 w`‡q fvM Kiv hvq bv|
GwU Kivi Rb¨ Avcwb 30 Ges 16 Gi j.mv.¸. †ei Ki‡j cv‡eb 4wU msL¨v|
Avi GB 4 wU msL¨v‡K 30 Ges 16 Dfq msL¨v w`‡q fvM Kiv hvq| ZvB 1 †_‡K 1,000 Gi g‡a¨ 30 w`‡q fvM Kiv
hvq, wKš‘ 16 w`‡q fvM Kiv hvq bv Ggb msL¨v n‡e = 30 w`‡q fvM Kiv hvq Ggb msL¨v – 30 Ges 16 Dfq msL¨v
w`‡q fvM Kiv hvq Ggb msL¨v = 33 – 4 = 29 wU|
03. Bus fares were recently increased from Taka 1.70 to Taka 2.00. What was the approximate
percentage of increase?
a) 18% b) 15%
c) 0.15% d) 0.18% ans. A
Solution: cÖ‡kœ ejv n‡”Q, m¤cÖwZ evm fvov 1.70 UvKv †_‡K †e‡o 2.00 UvKv n‡jv| evm fvov kZKiv KZ evo‡jv?
evm fvov = 2.00 – 1.70 = 0.30 UvKv
 evm fvov kZKiv ev‡o =



0.30
1.70
 100 % = 17.65%  18%
04. Which of the numbers below is not equivalent to 20%?
a)
1
5
b)
20
100
c) 0.5 d) 0.2 ans. C
Solution: cÖ‡kœ ejv n‡”Q, 20% Gi mgZzj¨ fMœvsk †KvbwU?
20% =
20
100
=
1
5
= 0.2
Ackb c) †Z Av‡Q 0.5 =
50
100
= 50%
05. What is the H.C.F. of the numbers 36, 54 and 90?
a) 6 b) 9
c) 12 d) 18 ans. D

2. Recent Publication 258
Solution: cÖ‡kœ ejv n‡”Q, Dˇii †Kvb eo Ackb w`‡q 36, 54, 90 †K fvM Kiv hvq?
H.C.F gv‡b n‡jv Highest Common Factor A_©vr M.mv.¸.
Ackb b) Gi 9 Ges d) Gi 18 w`‡q hvq|
wKš‘ 18 eo n‡j M.mv.¸. nq 18.
06. If x is an integer and y = – 2x – 8, what is the least value of x for which y is less than 9?
a) – 9 b) – 8
c) – 7 d) – 6 ans. B
Solution: cÖ‡kœ ejv n‡”Q, hw` x GKwU c~Y© msL¨v nq Ges y = – 2x – 8 nq, Z‡e x Gi †mB me©wbgœ gvb ev Least
value †ei Ki“b hvi Rb¨ y Gi gvb 9 Gi †P‡q Kg nq?
†`Lyb, Ackb¸wj‡Z me©wbgœ gv‡bi µg n‡”Q, – 9 < – 8 < – 7 < – 6.
GLb GB PviwU msL¨vi g‡a¨ x Gi gvb †KvbwU n‡j y Gi gvb 9 A‡c¶v †QvU n‡e, †mUvB †ei Ki‡Z n‡e|
Gevi Avmyb Ack‡bi gvb¸wj y = – 2x – 8 mgxKi‡Y emvB
x = – 9 n‡j y = (– 2)  (– 9) – 8 = 18 – 8 = 10 hv 9 Gi †P‡q eo, ZvB Ackb a) ev`|
Gevi, x = – 8 n‡j y = (– 2)  (– 8) – 8 = 16 – 8 = 8 hv 9 Gi †P‡q †QvU, ZvB Ackb b) DËi n‡e|
GLb Avgiv Ackb c) Gi – 6 Ges Ackb d) Gi – 7 w`‡q †Póv Ki‡ev bv, KviY – 6 Ges – 7 DfqB – 8 A‡c¶v
eo| myZivs mwVK DËi n‡e b).
07. If a +
1
a
= 3, what is a3
+
1
a 3 ?
a) 24 b) 7
c) 30 d) 18 ans. D
Solution: a3
+
1
a3 =



a +
1
a
3
– 3  a 
1
a 


a +
1
a
= (3)3
– 3  3 = 27 – 9 = 18.
08. If sec + tan = x, then tan is
a)
x2
+ 1
x
b)
x2
- 1
x
c)
x2
+ 1
2x
d)
x2
- 1
2x
ans. D
Solution: Avgiv Rvwb, sec2
 – tan2
 = 1
 (sec + tan) (sec – tan) = 1
 x(sec – tan) = 1
 sec – tan =
1
x
......... (i)
†`qv Av‡Q, sec + tan = x ......... (ii)
(ii) bs n‡Z (i) bs mgxKiY we‡qvM K‡i cvB,
sec + tan = x
sec – tan =
1
x
2tan = x –
1
x
 2tan =
x2
- 1
x
 tan =
x2
- 1
2x
 tan =
x2
- 1
2x
Clipboard
Q. x and y are positive integers. If (xy + x) is odd, then
which of the following must be even?
a) x b) y c) x2
– y
d) y – x e) None of these ans. B
Solution: Abyev`t x Ges y n‡jv abvZ¥K c~Y© msL¨v| hw` (xy + x)
we‡Rvo msL¨v nq Z‡e wb‡Pi †KvbwU Aek¨B †Rvo msL¨v n‡e?
†`Iqv Av‡Q, (xy + x) is odd
Zvi gv‡b xy I x ivwk `ywUi †h †Kvb GKwU Odd n‡e e‡j
†hvMdj Odd n‡q‡Q| GLv‡b x Aek¨B Odd n‡e Ges y Aek¨B
Even n‡e|
ZLb xy + x = Even + Odd = Odd nq|
myZivs GLv‡b Equation wUi Rb¨ y Aek¨B Even n‡Z n‡e|

3. Recent Publication 259
09. If the length of a rectangle is increased by 20% and width is decreased by 20%, what is the
change in area of the rectangle?
a) unchanged b) decreased by 4%
c) increases by 4% d) increases by 5% ans. B
Solution: cÖ‡kœ ejv n‡”Q, GKwU Rectangle Gi ˆ`N©¨ cÖ_‡g hw` 20% ev‡o Ges c‡i 20% K‡g, Z‡e †¶Îd‡ji wK
cwieZ©b n‡e?
awi, Rectangle wUi ˆ`N©¨ wQj = 100
20% evo‡j ˆ`N©¨ nq = 100 + 100 
20
100
= 120
Avevi, 20% Kg‡j ˆ`N©¨ nq = 120 – 120 
20
100
= 120 – 24 = 96
 †¶Îdj Kg‡e = 100 – 96 = 4%
10. The sum of first 17 terms of the series 5, 9, 13, 17, ........ is
a) 529 b) 462
c) 629 d) 523 ans. C
Solution: cÖ‡kœ ejv n‡”Q 5, 9, 13, 17 ......... GB avivi cÖ_g 17 wU c‡`i †hvMdj KZ?
avivwUi cÖ_g c` a = 5;
mvaviY Aš—i d = 9 – 5 = 4 Ges c`msL¨v n = 17.
 cÖ_g 17 wU msL¨vi mgwó =
n
2
{2a + (n – 1)d}
=
17
2
{(2  5) + (17 – 1)4} =
17
2
{10 + (16  4)}
=
17
2
{10 + 64} =
17
2
 74 = 17  37 = 629
11. All possible three digit numbers are formed by 1, 3, 5. If one number is chosen randomly, the
probability that it would be divisible by 5 is
a) 0 b)
2
9
c)
1
3
d)
1
4
ans. C
Solution: cÖ‡kœ ejv n‡”Q 1, 3 Ges 5 m¤¢ve¨ mKj msL¨v MVb K‡i †h‡Kvb GKwU msL¨v ˆ`efv‡e Pqb Ki‡j msL¨vwU 5
Øviv wefvR¨ nIqvi m¤¢ve¨Zv KZ?
1, 3 Ges 5 GB wZbwU msL¨v Øviv |3 = 6 Dcv‡q msL¨v MVb Kiv hvq|
GLb, †k‡li msL¨v 5 †K me©`v w¯’i †i‡L evwK 2 wU msL¨v‡K |2 = 2 Dcv‡q mvRv‡bv hvq|
 msL¨vwU 5 Øviv wefvR¨ nevi m¤¢vebv =
2
6
=
1
3
12. The three sides of a triangle are x + 1, 2x – 1 and 3x + 1 respectively and the perimeter is 25cm.
The length of the smallest side is
a) 5cm b) 3cm
c) 4cm d) 7cm ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, GKwU wÎfz‡Ri wZbwU evûi ˆ`N©¨ h_vµ‡g x + 1, 2x – 1 Ges 3x + 1. wÎfzRwUi cwimxgv
25cm n‡j ¶z`ªZg evûi ˆ`N©¨ KZ?
†h‡nZz, wÎfz‡Ri cwimxgv n‡jv wÎfz‡Ri wZb evûi mgwó| Kv‡RB,
x + 1 + 2x – 1 + 3x + 1 = 25
 6x + 1 = 25
 6x = 25 – 1 = 24
 x =
24
6
= 4

4. Recent Publication 260
AZGe 1g evû n‡e = x + 1 = 4 + 1 = 5cm.
2q evû n‡e = 2x – 1 = 2  4 – 1 = 8 – 1 = 7cm.
Ges 3q evû n‡e = 3x + 1 = 3  4 + 1 = 12 + 1 = 13cm.
A_©vr m‡e©v”P †QvU evû 5cm hv Ackb a) †Z Av‡Q|
13. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) =?
a) 5 : 11 b) 6 : 5
c) 5 : 6 d) 3 : 8 ans. A
Solution: GLv‡b, x : y = 5 : 3  x = 5 Ges y = 3
GLb, (8x – 5y) : (8x + 5y) = (8  5 – 5  3) : (8  5 + 5  3) = 25 : 55 = 5 : 11
14. The difference in Taka between simple and compound interest at 5% annually on a sum of Tk.
5,000 after 2 years is
a) 12.5 b) 25
c) 50 d) 500 ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, 5% nvi my‡` 5,000 UvKvi 2 eQ‡ii mij I Pµe„w× my‡`i nvi KZ?
5% nvi my‡` 5,000 UvKvi 2 eQ‡ii my` =
5000  5  2
100
= 500 UvKv
 Pµe„w× my` =



5000



1 +
5
100
2
- 5000
=



5000



1 +
1
20
2
- 5000
=



5000



21
20
2
- 5000 =



5000  441
400
- 5000
= 5,512.5 – 5,000 = 512.5 UvKv
AZGe, mij I Pµe„w×my‡`i cv_©K¨ = 512.5 – 500 = 12.5 UvKv|
15.
Log 36
Log 6
=
a) 5 b) 8
c) 3 d) 2 ans. D
Solution: †`qv Av‡Q,
Log 36
Log 6
=
Log 6 2
Log 6
=
2 Log 6
Log 6
= 2
16. If A = {1, 2, 3, 4, 5} then the number of proper subsets of A is
a) 120 b) 30
c) 31 d) 32 ans. C
Solution: cÖ‡kœ ejv n‡”Q, A = {1, 2, 3, 4, 5} †m‡Ui cÖK…Z Dc‡mU KZwU?
G‡¶‡Î Dc‡mU †ei Kivi wbqg n‡jv 2n
– 1.
GLv‡b, n = †m‡Ui Dcv`vb msL¨v
 cÖK…Z Dc‡mU msL¨v = 25
– 1 = 32 – 1 = 31 wU|
17. P and Q are brothers. R and S are sisters. The father of P is brother of S. Q is related to R as
a) son b) brother
c) uncle d) aunt ans. D
Solution: cÖkœ ejv n‡”Q P Ges Q ci¯úi fvB| R Ges S ci¯úi †evb| P Gi wcZv S Gi fvB n‡j Q I R Gi m¤úK©
wK?
Avm‡j cÖkœUv cÖ_‡g co‡j GKUz KwVb g‡b n‡Z cv‡i| wKš‘ mnR K‡i wPš—v Ki‡j RU Lyj‡Z Lye †eM †c‡Z n‡e
bv| GLv‡b P I Q `yB fvB| R Ges S `yB †evb| †m‡nZz P I Q Gi evev R Ges S Gi fvB| ZvB R n‡jv Q Gi
Aunt.

5. Recent Publication 261
18. Every 3 minutes, 4 litres of water are poured into a 2,000 litre tank. After 2 hours, what
percent of the tank will be full?
a) 0.4% b) 4%
c) 8% d) 12% ans. C
Solution: cÖwZ 3 wgwbU Aš—i 4 wjUvi cvwb 2,000 wjUvi aviY ¶gZv m¤úbœ GKwU U¨v‡¼ Xvjv nq| 2 NÈv ci U¨v‡¼ KZ
Percent c~Y© n‡e Zv †ei Ki‡Z n‡e|
2 NÈv = 2  60 = 120 wgwbU|
3 wgwb‡U Xvjv nq = 4 wjUvi cvwb
 120 wgwb‡U Xvjv nq =
4  120
3
= 160 wjUvi cvwb
GLb, 2000 wjUv‡ii g‡a¨ U¨v¼ c~Y© nq = 160 wjUvi
 100 wjUv‡ii g‡a¨ U¨v¼ c~Y© nq =
160  100
2000
= 8%
19. The next number in the sequence 3, 6, 11, 18, 27, ........ is
a) 34 b) 36
c) 38 d) 40 ans. C
Solution: cÖ‡kœ ejv n‡”Q, 3, 6, 11, 18, 27 .......... avivwUi cieZ©x msL¨v KZ?
c‡ii msL¨v n‡Z Av‡MiwU we‡qvM Ki‡j cvB,
6 – 3 = 3; 11 – 6 = 5; 18 – 11 = 7; 27 – 18 = 9.
A_©vr cÖ_g msL¨vi mv‡_ avivevwnK fv‡e Odd number †hvM Ki‡j c‡ii msL¨vwU cvIqv hvq|
9 Gic‡i Odd 11.
 cieZ©x msL¨vwU = 27 + 11 = 38.
20. The average of six numbers is 14. The average of four of these numbers is 15. The average of
the remaining two numbers is
a) 4 b) 8
c) 12 d) 16 ans. C
Solution: cÖ‡kœ ejv n‡”Q, 6 wU msL¨vi Mo 14. G‡`i g‡a¨ 4 wU msL¨vi Mo 15 n‡j Aci `ywUi Mo KZ?
6 wU msL¨vi mgwó = 6  14 = 84
Avevi, 4 wU msL¨vi mgwó = 4  15 = 60
 Aci `ywUi mgwó = 84 – 60 = 24
 Zv‡`i Mo =
24
2
= 12
21. If 2x – 1 > – 3, then
a) x < – 2 b) x > – 2
c) x < – 1 d) x > – 1 ans. D
Solution: cÖ`Ë AmgZvwU n‡jv 2x – 1 > – 3
 2x – 1 + 1 > – 3 + 1
 2x > – 2

2x
2
>
- 2
2
[Dfq c¶‡K 2 Øviv fvM K‡i]
 x > – 1
Note: AmgZvi Dfq cv‡k GKB msL¨v †hvM, we‡qvM, ¸Y, fvM Ki‡j AmgZvi wP‡ýi cwieZ©b nq bv|
22. If xy = 2 and xy2
= 16, what is the value of x?
a) 4 b) 2
c)
1
4
d) 8 ans. C

6. Recent Publication 262
Solution: †`qv Av‡Q,
xy2
xy
=
16
2
 y = 8
GLb, xy = 2
 x  8 = 2
 x =
2
8
=
1
4
23. There are 10 true-false questions in an examination. These questions can be answered in
a) 20 ways b) 100 ways
c) 210
ways d) 1024 ways ans. C & D
Solution: cÖ‡kœ ejv n‡”Q, GKwU cix¶vq 10wU mZ¨-wg_¨v Ackb msewjZ cÖkœ i‡q‡Q| KZfv‡e DËi cÖ`vb Kiv †h‡Z cv‡i?
Avgiv Rvwb, 1 wU mZ¨-wg_¨v cÖ‡kœi DËi 2 fv‡e †`qv †h‡Z cv‡i|
Gfv‡e 10wU cÖkœ = 2  2  2  2  2  2  2  2  2  2 = 210
= 1,024 Dcv‡q †`qv †h‡Z cv‡i|
24. The perimeter of a circle measures 16cm, what is the area of the circle in sq.cm?
a) 32 2 b) 64
c) 256 d) 128 ans. B
Solution: cÖ‡kœ ejv n‡”Q, GKwU e„‡Ëi cwiwa 16cm n‡j e„‡Ëi †¶Îdj KZ?
awi, e„‡Ëi e¨vmva© r cm
 Gi †¶Îdj r2
cm2
cÖkœg‡Z, 2r = 16
 r =
16
2
= 8
 †¶Îdj = r2
cm2
=   (8)2
cm2
= 64 cm2
Investment Corporation of Bangladesh
Recruitment Test for Capital Management (AP) – 02.03.2019
01. my‡`i nvi `kwgK 75 kZvsk n«vm cvIqv‡Z GKRb AvgvbZKvixi Avgvb‡Zi Dci 4 eQ‡ii cÖvß Avq 750 UvKv
K‡g hvq| Zvi Avgvb‡Zi †gvU cwigvY KZ?
a) 25,000 UvKv b) 18,750 UvKv
c) 30,000 UvKv d) 1,00,000 UvKv ans. A
Solution: †`qv Av‡Q, my‡`i nvi n«vm †c‡q‡Q 0.075; mgq n = 4 eQi; my` I = 750 UvKv|
Avgiv Rvwb, I =
prn
100
ev, p =
100  I
r  n
=
100  750
0.075  4
= 25,000 UvKv
02. GKwU UvKvi _‡j‡Z 1 UvKv, 50 cqmv I 25 cqmv g~j¨gv‡bi 342 wU gy`ªv i‡q‡Q| gy`ªv¸‡jvi g~j¨gv‡bi AbycvZ
11 t 9 t 5 n‡j 50 cqmv g~j¨gv‡bi gy`ªvi msL¨v KqwU?
a) 99 wU b) 180 wU
c) 150 wU d) 162 wU ans. Note
Solution: 1 UvKv, 50 cqmv I 25 cqmv Gi g~j¨gv‡bi AbycvZ h_vµ‡g = 11 : 9 : 5
 1 UvKv, 50 cqmv I 25 cqmv Gi msL¨vi AbycvZ h_vµ‡g =
11
1 :
9
1
2
:
5
1
4
= 11 : 18 : 20
50 cqmv g~j¨gv‡bi gy`ªvi msL¨v =
18
11 + 18 + 20  342 =
18
49  342 = 125.63wU

7. Recent Publication 263
03. cyjK Avwe‡ii †P‡q wظY `¶ Ges GKB KvR Avwe‡ii †P‡q 60 w`b Kg mg‡q m¤úbœ Ki‡Z cv‡i| Zviv Df‡q
GK‡Î KZw`‡b KvRwU †kl Ki‡Z cvi‡e?
a) 90 w`b b) 60 w`b
c) 40 w`b d) 30 w`b ans. C
Solution: kZ© bv †f‡O aiv hvq, Avwei K‡i KvRwU 120 w`‡b Ges cyjK KvRwU K‡i 60 w`‡b|
Avwei I cyjK GK‡Î 1 w`‡b K‡i =



1
120 +
1
60 =



1 + 2
120 =
3
120 =
1
40 Ask
A_©vr Zviv GK‡Î
1
40 Ask K‡i = 1 w`‡b
 Zviv 1 ev m¤ú~Y© Ask K‡i =
40  1
1 = 40 w`‡b
04. a, b, c, d PviwU µwgK ¯^vfvweK msL¨v n‡j wb‡Pi †KvbwU c~Y©eM© msL¨v?
a) abcd b) ab + cd
c) abcd + 1 d) abcd – 1 ans. C
Solution: g‡b Kwi, a, b, c I d gvb h_vµ‡g 1, 2, 3 Ges 4
 a  b  c  d = 1  2  3  4 = 24
GLv‡b 24 c~Y©eM© msL¨v bq| 24 Gi mv‡_ 1 †hvM Ki‡j A_©vr 24 + 1 = 25 nq c~Y©eM©|
myZivs abcd + 1 n‡jv c~Y© eM© msL¨v
05. a = 3 + 2 n‡j a3
+
1
a 3 =?
a) 18 3 b) 5 3
c) 6 3 d) 12 3 ans. A
Solution: †`qv Av‡Q, a = 3 + 2

1
a
=
1
3 + 2
=
3  2
( )3 + 2 ( 3  2)
=
3  2
( 3)
2
 ( 2)
2 = 3  2
 a +
1
a
= 3 + 2 + 3  2 = 2 3
 a3
+
1
a3 =



a +
1
a
3
 3. a.
1
a 


a +
1
a
= ( )2 3
3
 3( )2 3 = 23
 ( )3
3
 6 3 = 8  3 3  6 3 = 24 3  6 3 = 18 3
06. GKwU Mvwoi PvKvi e¨vm 1.67 wgUvi| 21 wK‡jvwgUvi c_ †h‡Z PvKvwU KZevi Ni‡e?
a) 1,000 evi b) 3,000 evi
c) 25,000 evi d) 4,000 evi ans. D
Solution: †`qv Av‡Q, PvKvi e¨vm, 2r = 1.67
 PvKvi cwiwa = 2r = 1.67 
22
7
= 5.25 wgUvi [Here, 2r = 1.67]
5.25 wgUvi †h‡Z cv‡i 1 evi
21 km ev 21,000 wgUvi †h‡Z †Nv‡i =
21000
5.25
= 4,000 evi
07. GKwU cY¨ weµq K‡i cvBKvix we‡µZv 20% Ges LyPiv we‡µZv 20% jvf Z‡i| hw` `ªe¨wUi LyPiv weµqg~j¨ 576
UvKv nq, Z‡e cvBKvix we‡µZvi µqg~j¨ KZ?
a) 250 UvKv b) 300 UvKv
c) 400 UvKv d) 480 UvKv ans. C
Solution: g‡b Kwi, cvBKvwi weµqg~j¨ = 100 UvKv
 20% jv‡f, cvBKvwi we‡µZvi weµqg~j¨ = 100 + 20 = 120 UvKv

8. Recent Publication 264
cvBKvwi we‡µZvi weµqg~j¨ = LyPiv we‡µZvi µqg~j¨ = 120 UvKv
 20% jv‡f, LyPiv we‡µZvi weµqg~j¨ = 120 + 120  20% = 144 UvKv
weµqg~j¨ 144 UvKv n‡j µqg~j¨ = 120 UvKv
weµqg~j¨ 576 UvKv n‡j µqg~j¨ =
120
144  576 = 480 UvKv
LyPiv we‡µZvi µqg~j¨ = cvBKvwi we‡µZvi weµqg~j¨ = 480 UvKv
cvBKvwi we‡µZvi weµqg~j¨ 120 UvKv n‡j µqg~j¨ = 100 UvKv
cvBKvwi we‡µZvi weµqg~j¨ 480 UvKv n‡j µqg~j¨ =
100  480
120 = 400 UvKv
08. 10 wU msL¨vi †hvMdj 462, cÖ_g 4wUi Mo 52 I †kl 5wUi Mo 38 n‡j 5g msL¨vwU KZ?
a) 62 b) 72
c) 64 d) 54 ans. C
Solution: cÖ_g 4wU Ges †kl 5wUi †hvMdj = (4  52) + (5  38) = 208 + 190 = 398
myZivs cÂg msL¨vwU = 462  398 = 64
09. 28 x + 1,426 = three-fourths of 2984, find X.
a) 659 b) 694
c) 841 d) 859 ans. C
Solution: 28 x + 1426 =
3
4
 2984
 28 x + 1426 = 2238
 28 x = 2238  1426
 28 x = 812
 x =
812
28
 ( )x
2
= (29) 2
 x = 841
10. 4x + 1
= 32 n‡j x = ?
a)
2
3
b)
3
5
c)
1
8
d)
3
2
ans. D
Solution: †`Iqv Av‡Q, 4x + 1
= 32
 2 2x + 2
= 25
 2x + 2 = 5
 x =
5 - 2
2
 x =
3
2
Probashi Kallyan Bank Ltd.
Recruitment Test for Programmar – 20.10.2019
01. If 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8
women do the same piece of work?
a) 4 days b) 5 days
c) 3 days d) 2 days ans. C
Clipboard
Q. If
n
41
is 1 more than
m
41
, then n = ?
a) m – 41 b) m + 41 c) m + 40
d) m – 40 e) None of these ans. B
Solution: cÖkœg‡Z,
n
41
–
m
41
= 1

n - m
41
= 1
 n – m = 41
 n = m + 41

9. Recent Publication 265
Solution: cÖ‡kœ ejv n‡”Q †h, 3 Rb cyyi“l ev 6 Rb ¯¿x‡jvK GKwU KvR 16 w`‡b K‡i| 12 Rb cyi“l I 8 Rb ¯¿x‡jvK
KZ w`‡b KvRwU Ki‡Z cvi‡e?
†`qv Av‡Q, 3 Rb cyi“l = 6 Rb ¯¿x‡jvK
 12 Rb cyi“l =
6  12
3
= 24 Rb ¯¿x‡jvK
 12 Rb cyi“l + 8 Rb ¯¿x‡jvK = 24 + 8 = 32 Rb ¯¿x‡jvK
GLb, 6 Rb ¯¿x‡jvK KvRwU K‡i = 16 w`‡b
 1Rb ¯¿x‡jvK KvRwU K‡i = (16  6) w`‡b
 32Rb ¯¿x‡jvK KvRwU K‡i =
16  6
32
= 3 w`‡b
02. A milkman bought 15kg of milk and mixed 3kg of water in it. If the price per kg of the mixture
becomes Tk. 22, what is cost price of the milk per kg?
a) Tk. 26.40 b) Tk. 28.00
c) Tk. 22.60 d) Tk. 24.00 ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, GK e¨w³ 15 †KwR `ya wK‡b Zvi mv‡_ 3 †KwR cvwb †gkv‡jv| cvwb wgwkªZ cÖwZ †KwR
`y‡ai `vg 22 UvKv n‡j cÖwZ †KwR `y‡ai `vg KZ?
awi, cÖwZ †KwR `y‡ai `vg x
GLb, wgkªYwewa Abymv‡i,
cÖkœg‡Z,
22
x - 22
=
15
3

22
x - 22
= 5
 5x – 110 = 22
 5x = 22 + 110 = 132
 x =
132
5
= 26.40
03. If the rate of interest is 10% per annum and is compounded half yearly, the principal of Tk.
400 in
3
2
years will amount to
a) Tk. 463.00 b) Tk. 463.05
c) Tk. 463.15 d) Tk. 463.20 ans. B
Solution: cÖ‡kœ ejv n‡”Q †h, 10% Aa©evwl©Kx nvi Pµe„w× my‡` 400 UvKv
3
2
eQ‡ii my`vmj KZ?
awi, my`vmj C; Avmj, P = 400 UvKv; nvi, r = 10%; eQi, n =
3
2
eQi; m = eQ‡i †h Kqevi my` Count nq|
GLb, Pµe„w× my‡`i †¶‡Î Avgiv Rvwb, C = P



1 +
10
100  m
nm
 C = 400



1 +
10
100  2
2 
3
2 = 400



1 +
1
20
3
= 400



21
20
3
= 400 
21  21  21
20  20  20
= 463.05 Tk.
04. A dealer offers a discount of 10% on the marked price of an article and still makes a profit of
20%. If its marked price is Tk. 800, then the cost price of the article is
a) Tk. 900 b) Tk. 800
c) Tk. 700 d) Tk. 600 ans. D
Solution: cÖ‡kœ ejv n‡”Q †h, ZvwjKvg~‡j¨i Dci 10% discount w`‡qI GK e¨w³ GKwU cY¨ weµq K‡i 20% jvf
K‡i| ZvwjKvg~j¨ 800 UvKv n‡j cY¨wUi µqg~j¨ KZ?
†`qv Av‡Q, ZvwjKvg~j¨ 800 UvKv
 10% discount G weµqg~j¨ = 90% of 800 =
90
100
 800 = 720 UvKv
Mo `vg
cvwbi `vg 0`y‡ai `vg x
x – 2222 UvKv

10. Recent Publication 266
GLb, 20% jvf
weµqg~j¨ 120 UvKv n‡j µqg~j¨ = 100 UvKv
 weµqg~j¨ 720 UvKv n‡j µqg~j¨ =
100  720
120
= 600 UvKv
05. Pipe A alone can fill a tank in 8 hours. Pipe B can fill it in 6 hours. If both the pipes are opened
and after 2 hours pipe A is closed, then the other pipe will fill the tank in
a) 6 hours b) 4 hours
c) 2
1
2
hours d) 3
1
2
hours ans. C
Solution: cÖ‡kœ ejv n‡”Q †h, A cvBcwU 8 NÈvq GKwU U¨vsK c~Y© K‡i Ges B GKv 6 NÈvq c~Y© K‡i| cvBc `ywU
GKmv‡_ 2 NÈv Pjvi ci A eÜ K‡i w`‡j B cvBcwU KZ mg‡q U¨vsKwU c~Y© Ki‡e?
(A + B) GK‡Î 1 NÈvq c~Y© K‡i =
1
8
+
1
6
=
3 + 4
24
=
7
24
Ask
 (A + B) GK‡Î 2 NÈvq c~Y© K‡i =
7  2
24
=
7
12
Ask
A_©vr (A + B) GK‡Î 2 NÈvq P‡j U¨vsKwU
7
12
Ask c~Y© K‡i|
 evwK _v‡K = 1 –
7
12
=
12 - 7
12
=
5
12
Ask
GLb GB
5
12
Ask B cvBc Øviv c~Y© n‡e|
GLb B cvBcwU 1 ev m¤ú~Y© Ask c~Y© K‡i = 6 NÈvq
 B cvBcwU
5
12
ev m¤ú~Y© Ask c~Y© K‡i =
6  5
12
=
5
2
= 2
1
2
NÈvq
A_©vr evwK Ask B cvBcwU 2
1
2
NÈvq c~Y© Ki‡e|
06. One tap can fill a drum three times as fast as another tap. Both the taps together can fill the
drum in 36 minutes. How much time would the slower tap take to fill the drum?
a) 144 minutes b) 108 minutes
c) 124 minutes d) 160 minutes ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, GKwU bj Aci GKwU b‡ji †P‡q 3 ¸Y †ewk nv‡i GKwU U¨vsK c~Y Ki‡Z cv‡i| `ywU bj
GK‡Î 36 wgwb‡U U¨vsKwU c~Y© Ki‡Z cvi‡j axi MwZi bjwU KZ mg‡q U¨vsKwU c~Y© Ki‡e?
awi, axiMwZi bj Øviv x wgwb‡U U¨vsKwU c~Y© Kiv nq|
 `ª“ZMwZi bj Øviv
x
3
wgwb‡U U¨vsKwU c~Y© n‡e|
cÖkœg‡Z,
1
x
+
3
x
=
1
36

1 + 3
x
=
1
36
 x = 4  36
 x = 144
A_©vr 144 wgwb‡U c~Y© n‡e|
07. 8 taps through which water flows at the same rate can fill a tank in 30 minutes. If two taps go
out of order, how long will the remaining taps take to the fill the tank?
a) 35 minutes b) 40 minutes
c) 37 minutes d) 38 minutes ans. B
Solution: cÖ‡kœ ejv n‡”Q †h, GKB iK‡gi 8wU bj Øviv GKwU U¨vsK 30 wgwb‡U c~Y© nq| `ywU bj A‡K‡Rv n‡q co‡j
evwK bj¸‡jv Øviv KZ mg‡q U¨vsKwU c~Y© n‡e?
Clipboard
Q. Given a =
x
12
and b =
y
27
, if a > b, which of
the following must be true?
a) x > y b) x > 2 c) 9x > 4y
d) x > 3y e) None of these ans. C
Solution: †`Iqv Av‡Q, a =
x
12
I b =
y
27
, †hLv‡b
kZ© n‡jv a > b

x
12
>
y
27

x
4
>
y
9
 9x > 4y.
Clipboard
Q. What is the radius of the in-circle of the triangle
whose sides measure 5, 12 and 13 units?
a) 2 b) 12 c) 6.5
d) 6 e) 7.5 ans. A
Shortcut: †h‡nZz wÎfy‡Ri evû wZbwU h_vµ‡g 5, 12, Ges 13 ZvB
GwU Aek¨B mg‡KvYx wÎfyR|
Radius of in circle
=
2(area of triangle)
sum of sides
=
2 
1
2
 5  12
5 + 12 + 13
=
60
30
= 2 Unit.

11. Recent Publication 267
8wU bj Øviv U¨vsKwU c~Y© nq = 30 wgwb‡U
 1wU bj Øviv U¨vsKwU c~Y© nq = (30  8) wgwb‡U
 6wU bj Øviv U¨vsKwU c~Y© nq =
30  8
6
= 40 wgwb‡U
08. ‘A’ can do a piece of work in ‘x’ days and ‘B’ can do the same work in 3x days. To finish the
work together they take 12 days. What is ‘x’ equal to?
a) 8 b) 10
c) 12 d) 16 ans. D
Solution: cÖ‡kœ ejv n‡”Q †h, ‘A’ GKwU K‡i ‘x’ w`‡b Ges ‘B’ KvRwU K‡i 3x w`‡b| Zviv GK‡Î 12 w`‡b KvRwU
K‡i| ‘x’ Gi gvb KZ?
‘A’ Gi 1 w`‡bi KvR =
1
x
Ask
 ‘B’ Gi 1 w`‡bi KvR =
1
3x
Ask
 (A + B) Gi 1 w`‡bi KvR =
1
x
+
1
3x
=
3 + 1
3x
=
4
3x
Ask
GLb †`qv Av‡Q, (A + B) Gi 1 w`‡bi KvR =
1
12
Ask
ZvB,
4
3x
=
1
12
 3x = 4  12
 x =
4  12
3
= 16
09. In a business partnership among A, B, C and D the profit is shared by A and B, B and C, C
and D in 1 : 3 ratio respectively. If the total profit is Tk. 4,00,000 the share of C is
a) Tk. 1,12,500 b) Tk. 1,37,500
c) Tk. 90,000 d) Tk. 2,70,000 ans. C
Solution: cÖ‡kœ ejv n‡”Q A, B, C Ges D GKwU e¨emvq †hvM w`j| A I B, B I C Ges C I D Gi g‡a¨ gybvdv 1 :
3 Abycv‡Z fvM K‡i wbj| †gvU gybvdv 4,00,000 UvKv n‡j C Gi †kqvi KZ?
†`qv Av‡Q, A : B = B : C = C : D = 1 : 3
GLv‡b,
A
B
=
1
3
A_©vr A Gi jvf 1 ¸Y n‡j B Gi jvf 3 ¸Y|

B
C
=
1
3
=
3
9
A_©vr B Gi jvf 3 ¸Y n‡j C Gi jvf 9 ¸Y|

C
D
=
1
3
=
9
27
A_©vr C Gi jvf 9 ¸Y n‡j D Gi jvf 27 ¸Y|
AZGe, A : B : C : D = 1 : 3 : 9 : 27
 C Gi †kqvi = 400,000 Gi
9
1 + 3 + 9 + 27
= 400,000 
9
40
= 90,000 UvKv
10. A tradesman marks his goods 10% above his cost price. If he allows his customers 10%
discount on the marked price, how much profit or loss does he make, if any?
a) 1% loss b) 1% gain
c) 5% gain d) No gain, no loss ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, GK we‡µZv µqg~‡j¨i †P‡q 10% †ewk‡Z ZvwjKvg~j¨ wba©viY Ki‡jv| †jvKwU GLb
ZvwjKvg~‡j¨i Dci 10% discount w`‡j Zvi kZKiv KZ jvf/¶wZ n‡e?
awi, †jvKwUi µqg~j¨ = 100 UvKv
 ZvwjKvg~j¨ = 100 + 10 = 110 UvKv
GLb, 10% discount G weµqg~j¨ = 110 – 110 Gi 10% = 110 – 11 = 99 UvKv
 ¶wZ = 100 – 99 = 1%
Clipboard
Q. If m =
4n
(x + n)
, then n =
a)
m(x + 1)
4
b)
mx + m
(4 - m)
c)
mx
(4 - m)
d)
mx
4
e) None ans. C
Solution: m =
4n
(x + n)
 mx + mn = 4n
 mx = 4n – mn
 n(4 – m) = mx
 n =
mx
(4 - m)

12. Recent Publication 268
11. 2 men and 1 woman can complete a piece of work in 14 days, while 4 women and 2 men can do
the same work in 8 days. If a man gets Tk. 90 per day, what should be the wages per day of a
woman?
a) Tk. 48 b) Tk. 60
c) Tk. 72 d) Tk. 135 ans. B
Solution: cÖ‡kœ ejv n‡”Q †h, 2 Rb cyi“l I 1 Rb ¯¿x‡jvK GKwU KvR K‡i 14 w`‡b| Avevi, 4 Rb ¯¿x‡jvK I 2 Rb
cyi“l KvRwU K‡i 8 w`‡b| GKRb cyi“l w`b 90 UvKv gRyix †c‡j GKRb ¯¿x‡jv‡Ki ˆ`wbK gRyix KZ nIqv DwPr?
awi, cyi“l M Ges ¯¿x‡jvK W
cÖkœg‡Z, 14(2M + 1W) = (4W + 2M)  8
 28M + 14W = 32W + 16M
 2M  6 = 3W  6
 2M = 3W
AZGe, 1W =
2
3
men
GLb, GKRb cyi“l ˆ`wbK gRyix = 90 UvKv n‡j
 GKRb ¯¿x‡jv‡Ki ˆ`wbK gRyix n‡e = 90 Gi
2
3
= 60 UvKv
12. A, B and C subscribe together Tk. 50,000 for a business. A subscribes Tk. 4,000 more than B
and B subscribes Tk. 5,000 more than C. Out of total profit of Tk. 35,000. A receives
a) Tk. 8,500 b) Tk. 11,998
c) Tk. 12,600 d) Tk. 14,700 ans. D
Solution: cÖ‡kœ ejv n‡”Q A, B Ges C wg‡j 50,000 UvKv w`‡q GKwU e¨emvq ïi“ Ki‡jv| e¨emvq B A‡c¶v A 4,000 UvKv
†ewk Ges C A‡c¶v B 5,000 UvKv †ewk wewb‡qvM Ki‡jv| e¨emvq †gvU 35,000 UvKv gybvdv n‡j A KZ cv‡e?
awi, C wewb‡qvM K‡iwQj x UvKv; B wewb‡qvM K‡iwQj (x + 5,000) UvKv
Ges A wewb‡qvM K‡iwQj (x + 5,000 + 4,000) = (x + 9,000) UvKv
cÖkœg‡Z, x + x + 5,000 + x + 9,000 = 50,000
 3x + 14,000 = 50,000
 3x = 50,000 – 14,000 = 36,000
 x =
36000
3
= 12,000 UvKv
AZGe A, B Ges C Gi wewb‡qv‡Mi AbycvZ
= (12,000 + 9,000) : (12,000 + 5,000) : 12,000 = 21,000 : 17,000 + 12,000 = 21 : 17 : 12
 A gybvdv wn‡m‡e cv‡e =



35000 
21
50
= 14,700 UvKv
13. If the average of 10% of a number, 25% of that number, 50% of that number and 75% of that
is 24, then the number will be
a) 50 b) 70
c) 60 d) 80 ans. C
Solution: cÖ‡kœ ejv n‡”Q †h, GKwU msL¨vi 10%, 25%, 50% Ges 75% Gi Mo 24 n‡j msL¨vwU KZ?
awi, msL¨vwU x
cÖkœg‡Z,
(10% of x) + (25% of x) + (50% of x) + (75% of x)
4
= 24

0.1x + 0.25x + 0.5x + 0.75x
4
= 24
 1.6x = 4  24 = 96
 x =
96
1.6
= 60
Clipboard
Q. If 40% of all women are voters and 52% of
the population is women, what percent the
population are women voters?
a) 18.1 b) 20.8
c) 26.4 d) 40 ans. B
Shortcut:
A  B
100
=
40  52
100
= 20.8

13. Recent Publication 269
14. A trader sells two cycles at Tk. 1,188 each and gains 10% on the first and loses 10% on the
second. What is the profit or loss percent on the whole?
a) 1% loss b) 1% gain
c) 2% loss d) No gain, no loss ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, GKRb e¨emvqx `ywU mvB‡Kj 1,118 UvKv `‡j weµq K‡i GKwU‡Z 10% jvf I Ab¨wU
10% ¶wZ n‡jv| †gv‡Ui Dci Zvi KZ jvf/¶wZ n‡jv?
GKB `v‡g weµq K‡i GKwU‡Z †h Percent jvf I Ab¨wU‡Z †mB Percent ¶wZ n‡j wb‡gœv³ m~Î e¨envi Ki‡Z
nqt ¶wZi nvi =
mgvb mgvb jvf ev ¶wZi Percent
10 % =



10
10
2
% = (1)2
% = 1%
15. A and B invest in a business in the ratio 5 : 4. If 10% of the total profit goes to charity and A’s
profit share is Tk. 7,500 then the total profit is
a) Tk. 7,500 b) Tk. 15,000
c) Tk. 12,000 d) Tk. 10,000 ans. B
Solution: cÖ‡kœ ejv n‡”Q †h, A Ges B GKwU e¨emvq 5 : 4 Abycv‡Z wewb‡qvM Ki‡jv| jv‡fi 10% `vb Kiv nj| A
jvf wn‡m‡e 7,500 UvKv †c‡j †gvU jvf KZ n‡qwQj?
awi, †gvU gybvdvi cwigvY 100 UvKv
`v‡bi ci Aewkó _v‡K = 100 – 10 = 90 UvKv
 A gybvdv wn‡m‡e cvq = 90 Gi
5
4 + 5
= 90 Gi
5
9
= 50 UvKv
GLb, gybvdv wn‡m‡e A 50 UvKv †c‡j †gvU gybvdv = 100 UvKv
 gybvdv wn‡m‡e A 7,500 UvKv †c‡j †gvU gybvdv =
100  7500
50
= 15,000 UvKv
16.
1
48
of a work is completed in half a day by 5 persons. Then,
1
40
of the work can be completed by
6 persons in how many days?
a) 1 b) 2
c) 3 d)
1
2
ans. D
Solution: cÖ‡kœ ejv n‡”Q, GKwU Kv‡Ri
1
48
Ask 5 Rb †jvK Øviv
1
2
w`‡b m¤úbœ n‡j KvRwUi
1
40
Ask 6 Rb †jvK KZ
w`‡b Ki‡e?
G‡¶‡Î mn‡R DËi †ei Kivi cÖPwjZ m~Î n‡jvt
M1 D1
W1
=
M2 D2
W2
GLv‡b, M1 = 1g †¶‡Î †jvK; M2 = 2q †¶‡Î †jvK; D1 = 1g †¶‡Î w`b; D2 = 2q †¶‡Î w`b;
W1 = 1g †¶‡Î KvR; W2 = 2q †¶‡Î KvR|
awi, D2 = x w`‡b
cÖkœg‡Z,
5 
1
2
1
48
=
6  x
1
40

5
2
1
48
=
6x
1
40

5
2

48
1
= 6x  40
 x =
5  48
2  6  40
=
1
2
w`‡b
Clipboard
Q. Tina rented a rickshaw for Tk. 18 plus Tk. 0.10 per km.
Ruby rented another rickshaw for Tk. 25 plus Tk. 0.05
per km. If each moves d km and paid exactly the same
amount of rental, then d = ?
a) 100 b) 120
c) 135 d) 140 ans. D
Solution: cÖ_g Rickshaw Gi †¶‡Î †gvU fvov
= 18 + (d  0.10) = 18 + 0.10d UvKv
wØZxq Rickshaw Gi †¶‡Î †gvU fvov
= 25 + (d  0.05) = 25 + 0.05d UvKv
cÖkœg‡Z, 18 + 0.10d = 25 + 0.05d  0.05d = 7
 d =
7
0.05
=
7
5
100
= 7 
100
5
= 140 wKwg

14. Recent Publication 270
17. A and B are partners in a business. A contributes
1
4
of the capital for 15 months and B receives
2
3
of the profit. Find for how long B’s money was used.
a) 6 months b) 8 months
c) 10 months d) 12 months ans. C
Solution: cÖ‡kœ ejv n‡”Q †h, GKwU Kviev‡i A Ges B n‡jb Askx`vi| A e¨emvi g~ja‡bi
1
4
Ask 15 gv‡mi Rb¨
LvUv‡jb Ges B gybvdv wn‡m‡e
2
3
Ask †c‡jb| B Gi UvKv KZ gv‡mi Rb¨ †L‡UwQj?
awi, H e¨emvq †gvU g~ja‡bi cwigvY wQj x UvKv
 A Gi wewb‡qv‡Mi cwigvY wQj
x
4
Ask Ges B Gi wewb‡qv‡Mi cwigvY wQj =



x -
x
4
=
3x
4
Ask
†h‡nZz †gvU gybvdvi
2
3
Ask cvq B
 A cv‡e =



1 -
2
3
=
3 - 2
3
=
1
3
Ask
AZGe A Ges B Gi gybvdvi AbycvZ =
1
3
:
2
3
= 1 : 2
GLb awi, B Gi
3x
4
UvKv n gv‡mi Rb¨ LvUvq|
Zvn‡j cÖkœg‡Z,
x
4
 15
3x
4
 n
=
1
2

3nx
4
=
30x
4
 n =
30x
4

4
3x
 n = 10
A_©vr B Zvi g~jab 10 gv‡mi Rb¨ LvwU‡qwQj|
18. The ratio between the ages of A and B is 2 : 5. After 8 years, their ages will be in the ratio 1 : 2.
What is the difference between their present ages?
a) 24 years b) 25 years
c) 22 years d) 20 years ans. A
Solution: cÖ‡kœ ejv n‡”Q †h, A Ges B Gi eq‡mi AbycvZ 2 : 5. 8 eQi ci Zv‡`i eq‡mi AbycvZ 1 : 2 n‡j Zv‡`i
eZ©gvb eq‡mi cv_©K¨ KZ?
awi, eZ©gv‡b A Gi eqm 2x eQi Ges B Gi eqm 5x eQi|
cÖkœg‡Z,
2x + 8
5x + 8
=
1
2
 5x + 8 = 4x + 16
 x = 8
AZGe, A Gi eZ©gvb eqm = 2  8 = 16 eQi|
Ges B Gi eZ©gvb eqm = 5  8 = 40 eQi|
 eq‡mi cv_©K¨ = 40 – 16 = 24 eQi|
19. A person invested part of Tk. 45,000 at 4% and the rest at 6%. If his annual incomes from
both are equal, then what is the average cost of interest?
a) 4.6% b) 4.8%
c) 5.0% d) 5.2% ans. B
Solution: cÖ‡kœ ejv n‡”Q GK e¨w³ 45,000 UvKvi wKQz Ask 4% Ges evwK Ask 6% nvi my‡` wewb‡qvM Kij| `y‡Uv
Ask n‡Z cÖvß Av‡qi cwigvY mgvb n‡j Mo my‡`i n‡i KZ wQj?
awi, 4% nvi my‡` wewb‡qvM K‡iwQj x UvKv
Clipboard
Q. The average of a set of 12 numbers, which
includes 34, is A. If 34 is removed from the set
and 36 is included to the set, what is the average
of the new set of numbers in terms of A?
a) A + 4 b)
(A + 38)
12
c) 12A + 4
d) A +
1
3
e) None ans. D
cÖ‡kœ ejv n‡”Q, 34 mn 12 wU msL¨vi Mo n‡”Q A. hw` H
msL¨v¸‡jvi MÖ“c n‡Z 34 †K mwi‡q 38 †K †bqv nq Z‡e A Gi
mv‡c‡¶ msL¨v¸‡jvi Mo KZ n‡e?
Shortcut: 12A + (38 – 34) = 12A + 4
New average =
12A + 4
12
= A +
1
3

15. Recent Publication 271
 6% nvi my‡` wewb‡qvM K‡iwQj (45,000 – x) UvKv
cÖkœg‡Z, 4% of x = 6% of (45,000 – x)
 4x = 6(45,000 – x)
 4x = 2,70,000 – 6x
 10x = 2,70,000
 x = 27,000
A_©vr 4% nv‡i wewb‡qvM K‡iwQj = 27,000 UvKv
 6% nv‡i wewb‡qvM K‡iwQj = 45,000 – 27,000 = 18,000 UvKv
AZGe †gvU my` = 4% of 27,000 + 6% of 18,000 = 1,080 + 1,080 = 2,160 UvKv
 Mo my‡`i nvi =
2160  100
45000
% = 4.8%
20. A person sold an article for Tk. 136 and made a loss at 15%. Had he sold it for Tk. x, he would
have made a profit of 15%. Which one of the following is correct?
a) 190 < x < 200 b) 170 < x < 170
c) 160 < x < 170 d) 180 < x < 190 ans. D
Solution: cÖ‡kœ ejv n‡”Q †h, GK e¨w³ 136 UvKvq GKwU `ªe¨ weµq K‡i 15% ¶wZ w`‡jb| `ªe¨wU x UvKvq weµq
Ki‡j wZwb 15% jvf Ki‡Z cvi‡Zb| wb‡Pi †KvbwU mZ¨?
A_©vr x Gi m¤¢ve¨ Value KZ n‡Z cv‡i, Zv †ei Ki‡Z n‡e|
15% ¶wZ‡Z
weµqg~j¨ 85 UvKv n‡j µqg~j¨ = 100 UvKv
 weµqg~j¨ 136 UvKv n‡j µqg~j¨ =
100  136
85
= 160 UvKv
GLb, 15% jv‡f `ªe¨wU weµqg~j¨ n‡e = 160 + 160 Gi 15% = 160 +
160  15
100
= 160 + 24 = 184 UvKv
GLb, x = 184 e‡j 180 < x < 190 B mwVK|
Joint Recruitment for 4 Banks
Recruitment Test for AP; Held On: 25.09.2020
01. How many feet are equal to 1 nautical mile?
a) 5220 b) 7080
c) 6250 d) 6076 ans. D
Solution: bwUK¨vj gvBj †bŠ c_ ev AvKvk c‡_i ˆ`N©¨ wbY©‡q e¨eüZ cwigv‡ci GKK| 1 bwUK¨vj gvBj mgvb 1,852
wgUvi ev 6076.12 dzU| GwU wm‡÷g Ad B›Uvib¨vkbvj (Gm AvB) GKK bq (hw`I Avš—R©vwZKfv‡e e¨env‡ii Rb¨
we AvB wc Gg KZ…©K ¯^xK…Z)| Avi 1 bwUK¨vj gvBj = 2025 MR| 1 bwUK¨vj gvBj = 1.853 wK‡jvwgUvi (cÖvq)|
02. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) = ?
a) 3 : 12 b) 8 : 12
c) 5 : 11 d) 5 : 15 ans. C
Solution: GLv‡b, x : y = 5 : 3
 x = 5 Ges y = 3
GLb, (8x – 5y) : (8x + 5y)
= (8 × 5 – 5 × 3) : (8 × 5 + 5 × 3)
= 25 : 55 = 5 : 11
03. The average of 5 consecutive number integers starting with m as the first integer is n. Then n
=?
a) 5m b) m + 3
c) m + 2 d) nm + 2 ans. C

16. Recent Publication 272
Solution: cÖ‡kœ ejv n‡”Q, m Øviv ïi“ nIqv 5wU µwgK msL¨vi Mo n n‡j n =?
†`qv Av‡Q, cÖ_g msL¨vwU = m
cÖkœg‡Z,
m + m + 1 + m + 2 + m + 3 + m + 4
5
= n

5m + 10
5
= n
 n =
5(m + 2)
5
= m + 2
04. The difference of two numbers is 20% of the large number. If the smaller number is 20 then
the larger number is-
a) 25 b) 65
c) 40 d) 60 ans. A
Solution: cÖ‡kœ ejv n‡”Q, `ywU msL¨v- GKwU ¶z`ªZi Avi Av‡iKwU e„nËi| msL¨v `ywUi cv_©K¨ n‡jv e„nËi msL¨vi
20%| ¶z`ªZi msL¨vwU 20 n‡j e„nËi msL¨vwU KZ?
awi, e„nËi msL¨vwU x
cÖkœg‡Z, x – 20 = 20% of x
 x – 20 =
20x
100
 x – 20 =
x
5
 5x – 100 = x
 4x = 100
 x = 25
05. A mother said to her daughter “I was as old as you are at present at the time of your birth”. If
the mother’s age is 38 years now, the daughters age five years back was-
a) 19 b) 15
c) 14 d) 33 ans. C
Solution: cÖ‡kœ ejv n‡”Q, gv Zvi †g‡q‡K ej‡Q ÒGLb †Zvgvi eqm hv, †Zvgvi R‡b¥i mgq Avgvi eqm ZvB wQj|Ó
GLv‡b AZxZ Ges eZ©gvb GKB n‡q †M‡Q| Kvib ai“b, GLb †g‡qi eqm 20 eQi| Zvn‡j †g‡qwU hLb Rb¥MÖnY
K‡i‡Q A_©vr 20 eQi Av‡M gv‡qi eqm wQj 20 eQi| Zvn‡j eZ©gv‡b gv‡qi eqm n‡e 20 + 20 = 40 eQi| Z‡e
wK `vov‡”Q? gv‡qi eqm †g‡qi eq‡mi wظY| Ab¨fv‡e †g‡qi eqm gv‡qi eq‡mi A‡a©K| Gevi Avmyb g~j cÖkœwU
mgvavb Kwit
†`qv Av‡Q, gv‡qi eZ©gvb eqm 38 eQi|
 †g‡qi eZ©gvb eqm = 38  2 = 19 eQi|
 5 eQi Av‡M †g‡qi eqm wQj = 19 – 5 = 14 eQi|
GwU x a‡iI Ki‡Z cv‡ib wb‡Pi gZ K‡i,
awi, †g‡qi eZ©gvb eqm = x eQi|
 gv‡qi eZ©gvb eqm (x + x) = 2x eQi
kZ©g‡Z, 2x = 38
 x = 19
 5 eQi c~‡e© †g‡qi eqm wQj = 19 – 5 = 14 eQi|
06. The price of a pen is 25% more than the price of a book. The price of a pen holder is 50%
more than the price of the book. How much is the price of the pen holder more than the price
of the pen?
a) 20% b) 25%
c) 35% d) 15% ans. A
Clipboard
Q. A ferry can carry 24 buses or 36 cars at a
time. If there are 18 buses on the ferry, how
many cars can be loaded onto it?
a) 6 b) 8 c) 9
d) 12 e) None ans. C
24 Bus = 36 Car
18 Bus =
36
24
 18= 27 Car
evKx 36 – 27 = 9wU Car a‡i|
Clipboard
Q. If (2x – 1)2
= 100, then which one of the following
could equal x?
a) –
11
2
b) –
9
2
c)
11
2
d)
13
2
ans. C
Solution: †`qv Av‡Q, (2x – 1)2
= 100
 2x – 1 = 10
 2x = 11
 x =
11
2

17. Recent Publication 273
Solution: cÖ‡kœ ejv n‡”Q, eB‡qi †P‡q Kj‡gi g~j¨ 25% †ewk| Avevi, eB‡qi †P‡q Kjg`vwbi g~j¨ 50% †ewk n‡j
Kj‡gi †P‡q Kjg`vwbi g~j¨ kZKiv KZ †ewk?
awi, Book Gi `vg = 100 Tk.
Pen Gi `vg = 100 + 25 = 125 Tk.
Pen holder Gi `vg = 100 + 50 = 150 Tk.
Pen †_‡K Pen holder Gi `vg †ewk = 150 – 125 = 25 Tk.
Pen †_‡K Pen holder kZKiv †ewk =
25  100
125
= 20%
07. A sum of Tk. 312 was divided among 100 boys and girls in such a way that each boy gets Tk.
3.6 and each girl Tk. 2.4. The number of girls is:
a) 35 b) 40
c) 55 d) 50 ans. B
Solution: cÖ‡kœ ejv n‡”Q, 312 UvKv 100 evjK evwjKvi g‡a¨ Ggbfv‡e fvM K‡i w`‡Z n‡e hv‡Z cÖ‡Z¨K evjK 3.60
UvKv K‡i Ges cÖ‡Z¨K evwjKv 2.40 UvKv K‡i cvq| Zvn‡j evwjKvi msL¨v †ei Ki‡Z n‡e|
awi, evwjKvi msL¨v = x Rb
 evj‡Ki msL¨v = (100 – x) Rb
cÖ‡Z¨K evjK 3.60 UvKv K‡i cvq
 (100 – x) Rb evjK cvq = 3.60 (100 – x) UvKv K‡i
cÖ‡Z¨K evwjKv cvq = 2.40 UvKv
 x Rb evwjKv cvq = 2.40x UvKv K‡i
cÖkœg‡Z, 3.60(100 – x) + 2.40x = 312
 360 – 3.60x + 2.40x = 312
 x = 40
 evwjKvi msL¨v = 40
08. A nurse has to record temperature of a Covid-19 patients in Celsius but her thermometer
reads Fahrenheit. The patient’s temperature is 100.7 Fahrenheit. What is the temperature in
Celsius?
a) 32 C b) 36.5
c) 37.5 C d) 38.2 C ans. D
Solution: cÖ‡kœ ejv n‡”Q, GKRb bvm© GKRb †ivMxi ZvcgvÎv †g‡c †`L‡jb 100.7 dv‡ibnvBU †mjwmqvm| H †ivMxi
ZvcgvÎv KZ?
Avgiv Rvwb,
c
5
=
F - 32
9
[GLv‡b, c = †mjwmqvm; F = dv‡ibnvBU]

c
5
=
100.7 - 32
9
 c = 38.2 †mjwmqvm
09. If the volume of a cube is 27 cubic meters, find the surface area of the cube?
a) 9 square meter b) 18 square meter
c) 54 square meter d) None of them ans. C
Solution: cÖ‡kœ ejv n‡”Q, GKwU Nb‡Ki AvqZb 27 NbwgUvi n‡j c„ôZ‡ji †¶Îdj KZ?
awi, Nb‡Ki GKevû a
 Volume = a3
Ges Surface area = 6a2
cÖkœg‡Z, a3
= 27
 a = 3
 Surface area = 6a2
= 6  32
= 54
Clipboard
Q. The side length of a square inscribed in a circle is
2. What is the area of the circle?
a) 2 b) 
c) r2
d) 2 2 ans. A
Solution: Abyev`t e„‡Ëi wfZ‡i e‡M©i GK evûi ˆ`N©¨ 2. e„‡Ëi
AvqZb KZ?
GLv‡b, e‡M©i KY© = e„‡Ëi e¨vm = 2  2 = 2 2
 e„‡Ëi e¨vmva© =
2 2
2
= 2
 e„‡Ëi †¶Îdj = ( )2
2
= 2

18. Recent Publication 274
10. The value of 0.1  0.1 is
a) 0.1 b) 0.1
c) 0.01 d) 0.001 ans. C
Sonali & Janata Bank Ltd.
Recruitment Test for Officer IT; Held On: 02.10.2020
01. `yywU msL¨vi ¸Ydj 120 Ges Zv‡`i e‡M©i †hvMdj 289| msL¨v؇qi †hvMdj KZ?
a) 409 b) 469
c) 529 d) 589 ans. Note
Solution: awi, GKwU msL¨v x Ges AciwU y
cÖkœg‡Z, xy = 120 ............. (i)
x2
+ y2
= 289 ................... (ii)
mgxKiY (ii) bs Gi mv‡_ 2xy †hvM Kwi
x2
+ y2
= 289
 x2
+ y2
+ 2xy = 289
 (x + y)2
= 289 + (2  120) = 289 + 240 = 529
 (x + y)2
= 529
 x + y = 529 = 23
A_©vr msL¨v `ywUi †hvMdj 23.
02. GKwU †cbWªvBf weµ‡q †`vKvb`vi 5% wWmKvD›U †`q| 7% wWmKvD‡›U Zvi 15 UvKv jvf nq| †cbWªvBfwUi
ZvwjKv g~j¨ KZ UvKv?
a) 697 b) 712
c) 780 d) 750 ans. D
Solution: awi, †cbWªvBfwUi ZvwjKv g~j¨ x UvKv
GLv‡b 5% wWmKvD›U A_© ZvwjKv g~‡j¨i 95% G weµq|
Avi 7% wWmKvD›U A_© ZvwjKv g~‡j¨i = (100 - 7)% = 95% G weµq|
cÖkœg‡Z, x Gi 95% - x Gi 93% = 15

95x
100 -
93x
100 = 15

95x - 93x
100 = 15
 2x = 1,500
 x =
1500
2 = 750 UvKv
03. 300 wgwjwjUv‡ii wgkª‡Y 12% jeY Av‡Q, hw` Gi mv‡_ 200 wgwjwjUvi cvwb wgkv‡bv nq bZyb wgkÖ‡Y je‡Yi
cwigvY KZ kZvsk?
a) 7.2% b) 7.5%
c) 6.9% d) 6.8% ans. A
Solution: 300 wgwjwjUv‡ii wgkª‡Y jeY Av‡Q = 300 Gi 12% = 300 
12
100 = 36 wgwjwjUvi
 H wgkª‡Y 200 wgwjwjUvi cvwb wgkv‡j †gvU wgkªY n‡e = 200 + 300 = 500 wgwjwjUvi|
AZGe, bZzb wgkª‡Y je‡Yi cwigvY =



36
500  100 % = 7.2%
Clipboard
Q. The average of 5 numbers is 40. If 2 more
numbers, with an average of 21, are added to
these numbers, what will be the average of
the combined 7 numbers?
a) 8.73 b) 30.17
c) 30.35 d) 34.57 ans. D
Shortcut:
(5  40) + (2  21)
5 + 2
= 34.57  34.60
Clipboard
Q. A class of 50 girls and 70 boys sponsored a party. If 40% of the
girls and 50% of the boys attended the party, approximately
what percent of the class attended?
a) 40 b) 42
c) 44 d) 46 ans. D
Shortcut: Attend% =
(50  40%) + (70  50%)
50 + 70
 100% =
20 + 35
120
 100%
=
55
120
 100% = 45.83  46

19. Recent Publication 275
04. 150 wgUvi `xN© GKwU †Uªb 450 wgUvi `xN© c­vUdg© 20 †m‡K‡Û AwZµg K‡i| †UªbwUi MwZ‡eM NÈvq KZ
wK‡jvwgUvi?
a) 25 b) 54
c) 45 d) 75 ans. Note
Solution: Avgiv Rvwb, 1 NÈv = 60  60 = 3,600 †m‡KÛ
 †Uªb + c­vUd‡g©i ˆ`N©¨ = 150 + 450 = 600 wgUvi
20 †m‡K‡Û hvq =
600
1000 wK‡jvwgUvi [1,000 wgUvi = 1 wK‡jvwgUvi]
 1 †m‡K‡Û hvq =
600
1000  20
wK‡jvwgUvi
 1 †m‡K‡Û hvq =
600  3600
1000  20
= 108 wK‡jvwgUvi
Shortcut: †gvU AwZµvš— `~iZ¡ = (150 + 450)m = 600m; AwZµvš— mgq = 20 seconds.
 wb‡Y©q MwZ‡eM =
600
20
ms – 1
= 30 ms – 1
=
30  18
5
ms – 1
= 108 kmph
05. GKwU wmwjÛvi I GKwU e„ËvKvi †gvPvi e¨vmva© Ges AvqZb mgvb| wmwjÛv‡ii D”PZv I †gvPvi D”PZvi AbycvZ
KZ?
a) 3 t 5 b) 2 t 5
c) 1 t 3 d) 3 t 3 ans. C
Solution: awi, wmwjÛvi I e„ËvKvi †gvPvi e¨vmva© r
Ges wmwjÛvi I e„ËvKvi †gvPvi D”PZv h_vµ‡g x Ges y
Avgiv Rvwb, wmwjÛv‡ii AvqZb = r2
h NbGKK
Ges e„ËvKvi †gvPvi AvqZb =
1
3 r2
h NbGKK
cÖkœg‡Z, r2
h =
1
3 r2
h
 r2
x =
1
3 r2
y
 x =
y
3

x
y =
1
3
 x : y = 1 t 3
06. log5625 = KZ?
a) 1 b) 2
c) 3 d) 4 ans. D
Solution: †`qv Av‡Q, log5625 = log554
= 4log55 = 4  1 = 4
07. xy †jLwPÎ mgxKiY y = –
1
2
x + 3 Gi †iLv hw` (c, – c) †jLwe›`yi ga¨ w`‡q †M‡j c Gi gvb KZ?
a) – 6 b) – 3
c) 2 d) 3 ans. A
Solution: †`qv Av‡Q, x(A_©vr fyR) hw` c Gi mgvb nq Zvn‡j y = –
1
2
x + 3 mgxKi‡Yi Xvj, y = – c n‡e|
Clipboard
Q. A sales representative will receive a 15%
commission on a sale of Tk. 28,000. If she has
already received Tk. 1500 as an advance of
that commission, how much more is due on
the commission?
a) Tk. 1200 b) Tk. 2700
c) Tk. 3200 d) Tk. 4200 ans. B
Solution: Abyev`t GKRb e¨w³ 28,000 UvKv weµ‡qi
Dci 15% Kwgkb cvq| †m hw` H Kwgk‡bi g‡a¨
1,500 UvKv AwMÖg †c‡q _v‡K Z‡e Avi KZ UvKv
Kwgkb e‡Kqv _vK‡jv?
15% nv‡i Total Commission
= 28,000  15% = 4,200 UvKv
 1,500 UvKv AwMÖg MÖnY Kivq Commission evwK
= 4,200 – 1,500 = 2,700 UvKv

20. Recent Publication 276
A_©vr – c = –
1
2
x + 3
 c =
x
2
– 3 [– 1 Øviv ¸Y K‡i]
 c =
c
2
– 3
 c –
c
2
= – 3

c
2
= – 3
 c = – 6
08. If logx
144
= 4 then find value of x.
a) 16 b) 3 2
c)2 3 d) 6 ans. C
Solution: †`qv Av‡Q, logx
144
= 4
 x4
= 144
 (x2
)
2
= (12)2
 x2
= 12
 x = 12 = 3  4 = 2 3
09. `ywU msL¨vi ¸Ydj 1,536| msL¨v `ywUi j.mv.¸. 96 n‡j M.mv.¸. KZ?
a) 16 b) 18
c) 32 d) 12 ans. A
Solution: Avgiv Rvwb, `ywU msL¨vi ¸Ydj = j.mv.¸  M.mv.¸
 1536 = 96  M.mv.¸  M.mv.¸ =
1536
96 = 16
10. If the radius of a circle is decreased by 20% then the area is decreased by:
a) 20% b) 30%
c) 36% d) 40% ans. C
Solution: cÖ‡kœ ejv n‡”Q, GKwU e„‡Ëi e¨vmva© 20% Kg‡j †¶Îdj KZ Kg‡e?
Avgiv Rvwb, †¶Îdj n«vm e„w×i m~Î
=



A + B +
AB
100
% =



- 20 - 20 +
- 20  - 20
100
%
= (– 40 + 4)% = – 36% [(–) n«vm cvIqv]
Janata Bank Ltd.
Recruitment Test for Officer (Cash); Held On: 31.10.2020
01. There is 60% increase in an amount in 6 years at simple interest. What will be the compound
interest of taka 12,000 after 3 years at the same rate?
a) 6,240 b) 3,972
c) 3,420 d) 4,972 ans. B
Solution: cÖ‡kœ ejv n‡”Q, mij my‡` 6 eQ‡i †Kv‡bv Avm‡ji 60% e„w× cvq| GKB nvi my‡` 12,000 UvKvi 3 eQ‡ii
Pµe„w× KZ n‡e?
awi, †gvU UvKv 100
 6 eQi c‡i †gvU UvKv n‡e = 100 + 100  60% = 100 + 60 = 160 UvKv
my` (Simple Interest) = 160 – 100 = 60
Clipboard
Q. A worker union contract specifies a 6 percent salary increase plus
a Tk 450 bonus for each worker. For a worker, this is equivalent
to 8 percent salary increase. What was this worker's salary before
the new contract?
a) Tk 21,500 b) Tk 22,500
c) Tk 23,500 d) Tk 24,300 ans. B
Solution: g‡b Kwi, c~‡e© Salary wQj x UvKv
cÖkœg‡Z, (x  6%) + 450 = x  8% 
6x
100
+ 450 =
8x
100

6x + 45000
100
=
8x
100
 6x + 45,000 = 8x  2x = 45,000  x = 22,500

21. Recent Publication 277
mvaviY my‡`i nvi =
60
6
= 10%
Avgiv Rvwb, Compound Amount = p(1 + r)n
[GLv‡b, p = g~jab = 12000, r = rate = 10%, n = mgq = 3 eQi]
= 12,000  (1 + 10%)3 = 12,000  (1.1)3 = 15,972
 Compound Interest = 15,972 – 12,000 = Tk. 3,972.
02. A system of equations is shown below: x + l = 6; x – m = 5; x + p = 4; x – q = 3
What is the value of l + m + p + q?
a) 3 b) 2
c) 6 d) 5 ans. B
Solution: †`qv Av‡Q, x + l = 6 ............. (i)
x – m = 5 ............................ (ii)
x + p = 4 ............................ (iii)
x – q = 3 ............................ (iv)
mgxKiY (i) †_‡K (ii) bs we‡qvM K‡i cvB
x + l – x + m = 6 – 5
 l + m = 1
Avevi, mgxKiY (iii) †_‡K (iv) bs we‡qvM K‡i cvB
x + p – x + q = 4 – 3
 p + q = 1
AZGe, l + m + p + q = 1 + 1 = 2
03. What would be the measure of the perimeter of a square whose area is equal to 256 sq cm?
a) 16cm b) 36cm
c) 64cm d) 256cm ans. C
Solution: cÖ‡kœ ejv n‡”Q, †h e‡M©i †¶Îdj 256 eM©‡mtwgt, H e‡M©i cwimxgv KZ?
awi, eM©‡¶‡Îi GKevûi ˆ`N©¨ a †mw›UwgUvi
 eM©‡¶‡Îi GKevûi †¶Îdj a2
eM© †mw›UwgUvi
cÖkœg‡Z, a2
= 256
 a2
= 16 2
 a = 162
 a = 16
Avgiv Rvwb, e‡M©i cwimxgv = 4a GKK = 4  16 = 64 †mw›UwgUvi|
04. In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD =?
a) 12 b) 11
c) 13 d) 17 ans. D
Solution: wP‡Î A, B, C wÎfy‡R B, C & D GKB mij †iLvq Aew¯’Z|
GLb wÎfyR ABD –G BD = BC + CD = 6 + 9 = 15
GLb wÎfyR ABD mg‡KvYx wÎfyR e‡j wc_v‡Mviv‡mi m~Î cÖ‡qvM K‡i cvBt
AD = AB2
+ BD2
= 82
+ 152
= 64 + 225 = 289 = 17.
A
8
B
6 C 9
D
10
Shortcut
†¶Îdj 256cm2
e‡j e‡M©i GK evû n‡e 16cm.
 cwimxgv n‡e = 4  16 = 64 †mw›UwgUvi|
Clipboard
Q. Two trains are moving in opposite directions
at 60 km/hr and 90 km/hr. Their lengths are
1.10 km and 0.9 km respectively. The time
taken by the slower train to cross the faster
train in seconds is -
a) 36 b) 45 c) 48
d) 49 e) None ans. C
Shortcut: Time =
Distance
Speed
=
2 hour
150 hour
=
2  60  60
150
= 48 †m‡KÛ|

22. Recent Publication 278
05. Mr. Ronaldo can finish a task in 20 days and Mr. Messi can do it in 30 days. With help of Mr.
Salah, they did the job in 10 days only. Then, how many days are necessary to complete the
task by Mr. Salah alone?
a) 80 days b) 60 days
c) 50 days d) 70 days ans. B
Solution: cÖ‡kœ ejv n‡”Q, †ivbvì GKwU KvR K‡i 20 w`‡b| †gwm KvRwU K‡i 30 w`‡b| †ivbvì, †gwm Ges mvjvn
GK‡Î KvRwU K‡i 10 w`‡b| mvjvn GKv KZ w`‡b KvRwU Ki‡e?
†ivbvì 20 w`‡b K‡i = 1 wU KvR
 †ivbvì 1 w`‡b K‡i =
1
20
Ask
†gwm 30 w`‡b K‡i = 1 wU KvR
 †gwm 1 w`‡b K‡i =
1
30
Ask
Avevi, †ivbvì + †gwm + mvjvn GK‡Î 10 w`‡b K‡i = 1 wU KvR
 †ivbvì + †gwm + mvjvn GK‡Î 1 w`‡b K‡i =
1
10
Ask
Avevi, †ivbvì + †gwm GK‡Î 1 w`‡b K‡i =
1
20
+
1
30
=
3 + 2
60
=
5
60
=
1
12
Ask
 mvjvn GKv 1 w`‡b K‡i =
1
10
–
1
12
=
6 - 5
60
=
1
60
Ask
A_©vr mvjvn
1
60
Ask KvR K‡i = 1 w`‡b
 mvjvn 1 ev m¤ú~Y© Ask KvR K‡i = 1  60 = 60 w`‡b
Shortcut: (R + M) Gi 1 w`‡bi KvR =
1
20
+
1
30
=
3 + 2
60
=
5
60
=
1
12
Ask
(R + M + S) Gi 1 w`‡bi KvR =
1
10
Ask
 S Gi 1 w`‡bi KvR =
1
10
–
1
12
=
6 - 5
60
=
1
60
Ask
A_©vr S GKv KvR Ki‡e 60 w`‡b|
06. When 4 is added to
1
2
of a number, the result is 14. What is the number?
a) 20 b) 21
c) 27 d) 35 ans. A
Solution: cÖ‡kœ ejv n‡”Q, †Kvb msL¨vi
1
2
Gi mv‡_ 4 †hvM Ki‡j †hvMdj 14 nq| msL¨vwU KZ?
awi, Number wU = x
cÖkœg‡Z,
x
2
+ 4 = 14

x
2
= 10
 x = 20
07. If x = 10, which of the following has the minimum value?
a)
x
2
b)
2
x
c) 2 – x d) (2 – x) (2 – x) ans. C
Solution: cÖ‡kœ ejv n‡”Q, x = 10 n‡j wb‡Pi †KvbwUi gvb me©wbgœ n‡e?
†`qv Av‡Q, x = 10

23. Recent Publication 279
GLb Avmyb DËi †_‡K Back Solve Kwit
Ackb a)
x
2
=
10
2
= 5;
Ackb b)
2
x
=
2
10
= 0.2;
Ackb c) 2 – x = (2 – 10) = – 8;
Ackb d) (2 – x)(2 – x) = (2 – 10) ( 2 – 10) = (– 8  – 8) = 64
 me©wbgœ gvb = – 8 hv Ackb c) †Z Av‡Q|
08. Kowshik earns taka 11 for each ticket that he sells and a bonus of taka 2 per ticket for each
ticket he sells over 100. If Kowshik was paid taka 2400, how many tickets did he sell?
a) 200 b) 180
c) 220 d) 250 ans. A
Solution: cÖ‡kœ ejv n‡”Q, Kowshik cÖwZwU wU‡KU weµq K‡i 11 UvKv K‡i jvf K‡i| wKš‘ 100wU wUwKU wewµi ci
†m cÖwZ wUwK‡U Av‡iv 2 UvKv K‡i †ewk jvf K‡i| Kowshik †gvU 2,400 UvKvi wUwKU weµq Ki‡j †m KZwU
wUwKU weµq K‡iwQj?
awi, Kowshik wU‡KU sell K‡i‡Q = x wU
 Bonus Qvov all wU‡K‡Ui g~j¨ = 11x UvKv
 Bonus wU‡K‡Ui msL¨v = (x  100) wU
cÖkœvbymv‡i, 11x + 2(x  100) = 2,400
 11x + 2x  200 = 2,400
 13x = 2,600
 x = 200
 Kowshik †gvU wU‡KU sell K‡iwQj 200 wU|
09. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers.
a) 28 and 17 b) 28 and 18
c) 28 and 19 d) 28 and 16 ans. A
Solution: cÖ‡kœ ejv n‡”Q, `ywU msL¨vi Aš—i 11 Ges mgwó GK cÂgvs‡ki gvb 9 n‡j msL¨vØq KZ?
msL¨vØq x Ges y n‡j cÖkœg‡Z,
1
5
(x + y) = 9  x + y = 45 ....... (i)
Ges x – y = 11 ....... (ii)
GLb, {(i) + (ii)}  2x = 56  x = 28
Ges {(i) – (ii)}  2y = 34 = y = 17
 msL¨vØq 28 Ges 17
10. If ax
= b, by
= c, and cz
= a, then the value of xyz is
a) – 1 b) 1
c)
1
abc
d) abc ans. B
Solution: GLb, x, y, z wZbwUB wKš‘ Power. ZvB Avcbv‡K (Something)x y z
= (Something)*
Ki‡Z n‡e
A_©vr `ywU Base GKB Ki‡Z n‡e Ges ZLb Avcwb `yB cv‡ki Power †K mgvb AvKv‡i wj‡L x, y, z Gi gvb †ei
Ki‡Z cvi‡eb|
Avmyb †kl †_‡K ïi“ Kwi: cz
= a
 (by
)z
= a [†h‡nZz c = by
]
 byz
= a
 (ax
)yz
= a [†h‡nZz b = ax
]
 axyz
= a1
Shortcut
(i) †_‡K mgwó 45 †ei Kivi ci Av‡Mi m~ÎwU
Ki‡j msL¨vØq = (
45 + 11
2
,
45 - 11
2
) = (28, 17)

24. Recent Publication 280
 xyz = 1 [Dfqc¶ †_‡K Base ev` w`‡q]
Note: GUv Avcwb †h‡Kv‡bv RvqMv †_‡K ïi“ Ki‡Z cv‡ib: †hgb gvSLvb †_‡K ïi“ Ki‡j n‡e:
c = by
= (ax
)y
= axy
= (cz
)xy
= cxyz
 xyz = 1
11. The length of the sides of a triangle are in the ratio of 3 to 5 to 6. If the perimeter of the
triangle is 70, what is the length of the longest side?
a) 10 b) 15
c) 25 d) 30 ans. D
Solution: cÖ‡kœ ejv n‡”Q, wÎfz‡Ri wZbevûi ˆ`N©¨i AbycvZ 3 : 5 : 6 GKK| wÎfz‡Ri cwimxgv 70 GKK n‡j e„nËg
evûi ˆ`N©¨ KZ?
†h‡nZz wÎfz‡Ri wZbevûi ˆ`N©¨i AbycvZ 3 : 5 : 6, ZvB
awi, wÎfz‡Ri evû wZbwUi ˆ`N©¨ h_vµ‡g 3x, 5x Ges 6x GKK|
cÖkœg‡Z, 3x + 5x + 6x = 70
 14x = 70
 x =
70
14
= 5
AZGe, e„nËg evûi ˆ`N©¨ = 6x = 6  5 = 30 GKK
12. How many ‘7’ will you pass on the way when you count from 1 to 100?
a) 18 b) 19
c) 20 d) 21 ans. C
Solution: cÖ‡kœ ejv n‡”Q, 1 †_‡K 100 ch©š— KZ¸‡jv 7 cvIqv hv‡e?
1 †_‡K 100 ch©š— †gvU 7 cvIqv hv‡e = 7, 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79,
87, 97 = †gvU 20 wU
13. P sells a product to Q and makes a profit of 25%. Q sells in to R at a loss of 20%. If R buys it
for Tk. 140, what price did P pay for it?
a) 211 b) 169
c) 140 d) 139 ans. C
Solution: cÖ‡kœ ejv n‡”Q P, Q Gi wbKU GKwU cY¨ weµq K‡i 25% jvf Ki‡jv| Q H cY¨wU R Gi wbKU weµq K‡i
20% ¶wZ Ki‡jv| hw` R cY¨wU 140 UvKv w`‡q wK‡b Z‡e P Gi µqg~j¨ KZ wQ‡jv?
awi, P Gi µqg~j¨ x UvKv
 P Gi weµqg~j¨ = 125% Gi x =
125x
100
=
5x
4
UvKv
A_©vr Q Gi µqg~j¨ =
5x
4
UvKv
Avevi, Q Gi weµqg~j¨ = 80% Gi
5x
4
=
80
100

5x
4
= x UvKv
A_©vr R Gi µqg~j¨ = x UvKv = 140 UvKv
14. A garden of 100 meter length and 60 meter width has a walkway of 2 meter width on every
side. What is the area of the garden, in square meters, excluding the walkway?
a) 5,376 b) 5,576
c) 2,556 d) 7,874 ans. A
Solution: cÖ‡kœ ejv n‡”Q, 100 wgUvi `xN© Ges 60 wgUvi cÖ‡¯’i GKwU evMv‡bi †fZ‡i 2 wgUvi PIov GKwU iv¯—v Av‡Q|
H iv¯—v ev‡` evMv‡bi †¶Îdj KZ?
iv¯—v ev‡` evMv‡bi ˆ`N©¨ = {100 – (2  2)} = 100 – 4 = 96 wgUvi
 iv¯—v ev‡` evMv‡bi cÖ¯’ = {60 – (2  2)} = 60 – 4 = 56 wgUvi
AZGe, iv¯—v ev‡` evMv‡bi †¶Îdj = 96  56 = 5,376 eM© wgUvi|
Shortcut
eo evû =
70
3 + 5 + 6
 6 = 30

25. Recent Publication 281
15. Which one is true for the mathematical expression
2n
+ 2n - 1
2n + 1
- 2n =?
a)
3
2
b)
2
3
c) 1 d) 2n
ans. A
Solution:
2 n
+ 2 n - 1
2 n + 1
- 2 n =
2 n
+ 2 n
 2- 1
2 n
 2 1
- 2 n =
2 n
+ 2 n

1
2
2 n
 2 1
- 2 n =
2n



1 +
1
2
2 n
(2 - 1)
= 1 +
1
2
=
2 + 1
2
=
3
2
16. 230
+ 230
+ 230
+ 230
= ?
a) 8120
b) 830
c) 232
d) 230
ans. C
Solution: 230
+ 230
+ 230
+ 230
= 4  230
= 22
 230
[†h‡nZz †gvU 230
Av‡Q 4wU]
= 22 + 30
= 232
Alternate Solution:
230
+ 230
+ 230
+ 230
= 230
(1 + 1 + 1 + 1) [230
†K Common wb‡q]
= 230
 4
= 230
 22
= 230 + 2
= 232
17. 40% of 200 is what percent of 160?
a) 20 b) 80
c) 60 d) 50 ans. D
Solution: 40% of 200 = 200  .40 = 80
cÖkœg‡Z, x% of 160 = 80

x
100
 160 = 80
 x = 50%
18. What is the value of ‘n’ if (8 – 3) (8 – n) = 40?
a) – 3 b) 8
c) 0 d) – 5 ans. C
Solution: cÖ‡kœ ejv n‡”Q, (8 – 3) (8 – n) = 40 n‡j n =?
†`qv Av‡Q, (8 – 3) (8 – n) = 40
 5(8 – n) = 40
 8 – n =
40
5
= 8
 n = 8 – 8 = 0
 n = 0
19. If
1
y
= 3
1
2
then
1
y + 2
= ?
a)
7
16
b)
2
7
c)
2
11
d)
2
16
ans. A
Solution: †`qv Av‡Q,
1
y
= 3
1
2
=
7
2

1
y
=
7
2
 y =
2
7
Clipboard
Q. How many squares of 2 inch dimension will
be required to cover a rectangle of 8 inch
breadth and 6 inch length?
a) 7 b) 24 c) 12
d) 48 e) None ans. C
Shortcut: †gvU Squares =
6  8
22 = 12wU|

26. Recent Publication 282
AZGe,
1
y + 2
=
1
2
7
+ 2
=
1
2 + 14
7
=
1
16
7
=
7
16
20. Mr. X lost a wallet containing Tk. 120. Incidentally, he had only notes of Tk. 2, Tk. 3 and Tk. 5
denominations in the wallet. If the total number of notes was 30, how many Tk. 5
denomination notes did he has?
a) 15 b) 16
c) 18 d) 20 ans. Note
Solution: cÖ‡kœ ejv n‡”Q, Rbve X Gi wbKU 120 UvKv wQj †hLv‡b 2 UvKv, 3 UvKv Ges 5 UvKvi †bvU wQj| †gvU
†bv‡Ui msL¨v 30 n‡j 5 UvKv †bvU wQj KqwU?
GwU IQ UvB‡ci A¼| cÖ_‡g Ackb a‡i Ki‡j `ª“Z mgvav‡b †cuŠ‡Q hv‡eb|
15wU 5 UvKvi †bvU _vK‡j †gvU UvKvi cwigvY n‡e = 15  5 = 75 UvKv
d‡j evwK 15wU 2 Ges 3 UvKvi †bvU w`‡q 45 UvKv evbv‡bv hvq bv| ZvB GwUI ev`|
Avevi, 16wU 5 UvKvi †bvU _vK‡j †gvU UvKvi cwigvY n‡e = 16  5 = 80 UvKv|
d‡j evwK 14wU 2 Ges 3 UvKvi †bvU = 2  2 = 4 Ges 12  3 = 36 w`‡q 4 + 36 = 40 nq|
ZvB 40 + 80 = 120 nq|
A_©vr 16wU DËi n‡Z cv‡i|
Avevi, 18wU 5 UvKvi †bvU _vK‡j = 5  18 = 90 UvKv
evwK 30 UvKv 2 I 3 UvKvi †bv‡Ui g‡a¨ = 2  6 = 12
Ges 3  6 = 18 Gfv‡e fvM K‡i †`qv hvq| Kv‡RB GwUI DËi n‡Z cv‡i|
Avevi, 5 UvKvi †bvU 20wU n‡j †gvU nq 5  20 = 100 UvKv|
d‡j evwK 20 UvKv 2 Ges 3 UvKvi 10wU †bv‡Ui g‡a¨ fvM K‡i †`qv hvq bv| ZvB GwU DËi n‡e bv|
ZvB mwVK DËi n‡Z cv‡i 16 Ges 18 wU|
Sonali, Janata & Rakub Bank Ltd.
Recruitment Test for AE (IT); Held On: 31.10.2020
01. If x is a whole number greater than 1, then x2
(x2
– 1) is always divisible by?
a) 8 b) 10
c) 12 d) 16 ans. C
Solution: cÖ‡kœ ejv n‡”Q, x GKwU c~Y© msL¨v hv 1 Gi †P‡q eo| Zvn‡j x2
(x2
– 1) me©`v wb‡Pi †KvbwU Øviv wefvR¨
n‡e?
1 Gi †P‡q eo msL¨v 2 e‡j x = 2 awi|
cÖ`Ëivwkgvjv = x2
(x2
– 1) = 22
(22
– 1) = 4(4 – 1) = 4  3 = 12
A_©vr x2
(x2
– 1) ivwkwU me©`v 12 Øviv wefvR¨ n‡e|
02.
a
b + c
=
b
c + a
=
c
a + b
= k, then the value of k is
a)
1
2
b) +
1
2
c) 1 d) – 1 ans. A
Solution: †`qv Av‡Q,
a
b + c
=
b
c + a
=
c
a + b
= k
GLb,
a
b + c
= k,
b
c + a
= k Ges
c
a + b
= k n‡e|
Zvn‡j, a = k(b + c) ............ (i)
b = k(c + a) ................ (ii)
c = k(a + b) ................ (iii)

27. Recent Publication 283
GLb (i), (ii) Ges (iii) bs mgxKiY †hvM Kwi
a + b + c = k(b + c) + k(c + a) + k(a + b) = k(b + c + c + a + a + b)
 a + b + c = k(2a + 2b + 2c)
 a + b + c = 2k(a + b + c)
 2k =
a + b + c
a + b + c
= 1
 k =
1
2
03. 80% of a number is equal to the
4
5
th of the other number. What is the ratio between the first
number and the second number respectively?
a) 1 : 1 b) 3 : 4
c) 3 : 5 d) 5 : 3 ans. A
Solution: cÖ‡kœ ejv n‡”Q, GKwU msL¨vi 80% Aci msL¨vi
4
5
As‡ki mgvb| 1g Ges 2q msL¨vi AbycvZ KZ?
awi, GKwU msL¨v x Ges Aci msL¨v y.
cÖkœg‡Z, 80% of x =
4
5
of y

80x
100
=
4y
5

4x
5
=
4y
5
 x = y [Dfqc¶‡K
5
4
Øviv ¸Y K‡i]

x
y
= 1 =
1
1
 x : y = 1 : 1
04. If x is the length of a median of an equilateral triangle, then the area is
a) x2
+ 2 b) x – 2
c)
x2
3
3
d)
1
2
x ans. C
Solution: cÖ‡kœ ejv n‡”Q, GKwU mgevû wÎfy‡Ri ga¨gv x n‡j wÎfyRwUi †¶Îdj KZ?
mgevû wÎfyR n‡jv †mB wÎfyR hvi wZb evû-B mgvb|
H wÎfy‡Ri GKevûi ˆ`N©¨ a GKK n‡j ga¨gv n‡e
x =
3
2
a.
 a =
2x
3
GLb Avgiv Rvwb, mgevû wÎfy‡Ri †¶Îdj =
3
4
(a)2
=
3
4 


2x
3
2
=
3
4

4x2
3
=
x2
3
3
eM© GKK|
05. Which one of the following is the minimum value of the sum of two integers whose product is
36?
a) 37 b) 20
c) 15 d) 12 ans. D
Solution: cÖ‡kœ ejv n‡”Q, `ywU msL¨vi ¸Ydj 36 n‡j msL¨v `ywUi †hvMdj Minimum KZ n‡e?
36 = 1  36
36 = 2  18
36 = 3  12
36 = 4  9
A
CB
a
aa
x
Clipboard
Q. (1000)7
 1018
=?
a) 10 b) 100
c) 1000 d) 10000 ans. C
Solution: (1000)7
 1018
= (10)3  7
 1018
=
1021
1018 = 1021 – 18
= 103
= 1000. [1000 = 103
]

28. Recent Publication 284
36 = 18  2
36 = 6  6
AZGe †`Lv hv‡”Q †k‡li = 6 + 6 = 12
06. If 0 < x < 1, then which one of the following is the maximum value of (x – 1)2
+ x?
a) – 1 b) – 2
c) 0 d) 1 ans. D
Solution: cÖ‡kœ ejv n‡”Q, 0 < x < 1 n‡j (x – 1)2
+ x Gi m‡e©v”P gvb KZ n‡e?
0 < x < 1 n‡j eySv hvq x Gi gvb 0 A‡c¶v eo wKš‘ 1 Gi †P‡q †QvU ev mgvb|
GLb, (x – 1)2
+ x Gi m‡e©v”P gvb †ei Kivi Rb¨ x = 1 ai‡Z n‡e|
Zvn‡j, (x – 1)2
+ x = (1 – 1)2
+ 1 = 02
+ 1 = 0 + 1 = 1
07. A product is sold at a profit of 20%. If the cost price is increased by 10% and sale price by Tk.
26, then the percentage of profit reduce by 5%, cost price is
a) 200 b) 250
c) 320 d) 400 ans. D
Solution: cÖ‡kœ ejv n‡”Q, GKwU cY¨ 20% jv‡f weµq Kiv nq| µqg~j¨ 10% †e‡o †M‡j weµqg~j¨ 26 UvKv ev‡o|
Gi d‡j gybvdv 5% K‡g hvq| cY¨wUi µqg~j¨ KZ?
awi, cY¨wUi µqg~j¨ x UvKv
20% jv‡f weµqg~j¨ = 120% of x =
120x
100
=
6x
5
UvKv
10% LiP e„wׇZ bZzb µqg~j¨ = 110% of x =
110x
100
=
11x
10
UvKv
AZGe, bZzb weµqg~j¨ =



6x
5
+ 26 UvKv
†m‡¶‡Î bZzb jvf n‡e =



6x
5
+ 26 –
11x
10
=
6x
5
–
11x
10
+ 26 =
12x - 11x
10
+ 26 =



x
10
+ 26 UvKv
cÖkœg‡Z, 


x
10
+ 26 
11x
10
 100 = 15 [jvf 20% †_‡K K‡g 15% n‡q‡Q KviY cÖ‡kœ ejv n‡q‡Q jvf 5% K‡g hvq]




x + 260
10
1000
11x
= 15
 (x + 260) 
100
11x
= 15
 100x + 26,000 = 165x
 65x = 26,000
 x =
26000
65
= 400
08. If (x + 3) is a factor of 3x2
+ ax + 6, then what is the value of a?
a) 3 b) 8
c) 11 d) 18 ans. C
Solution: cÖ‡kœ ejv n‡”Q, (x + 3) hw` (3x2
+ ax + 6) Gi GKwU Drcv`K nq Z‡e a =?
(x + 3) ivwkwU (3x2
+ ax + 6) Gi GKwU Drcv`K n‡j x + 3 = 0 n‡e|
 x = – 3
A_©vr x = – 3 n‡j (3x2
+ ax + 6) ivwkwUi gvb 0 n‡e|
cÖkœg‡Z, 3x2
+ ax + 6 = 0
 3  (– 3)2
+ (– 3)a + 6 = 0
 (3  9) – 3a + 6 = 0
 27 – 3a + 6 = 0
 3a = 33
 a =
33
3
= 11
Clipboard
Q. If a number x is 10% less than another
number y and y is 10% more than 125, then x
is equal to:
a) 140.55 b) 123.75
c) 143 d) 150 ans. B
Solution: x + 10% of 137.5 = 137.5
Avevi, x + 10% of 137.5 = 137.5
 x +
10  137.5
100
= 137.5
 x + 13.75 = 137.5  x = 123.75

29. Recent Publication 285
09. A man’s salary was reduced by 50%, again the reduce salary was increased by 50%. Find the
loss of in terms of percentage.
a) 25% b) 50%
c) 75% d) No loss ans. A
Solution: cÖ‡kœ ejv n‡”Q, GK e¨w³i †eZb cÖ_‡g 50% Kwg‡q Avevi c‡i 50% evov‡bv n‡j G‡Z Zvi kZKiv KZ
¶wZ n‡e?
awi, †eZb wQj 100% UvKv
50% K‡g †eZb n‡e = 100 – 100 Gi 50% = 100 – 50 = 50 UvKv
Avevi, †eZb 50% evov‡bv n‡j eZ©gvb †eZb n‡e = 50 + 50 Gi 50% = 50 + 25 = 75 UvKv
 kZKiv ¶wZ n‡e = 100 – 75 = 25 UvKv
10. What is the weight of 1 cubic meter of water?
a) 10kg b) 100kg
c) 500kg d) 1,000kg ans. D
Solution: cÖ‡kœ ejv n‡”Q, GK Nb wgUvi cvwbi IRb KZ?
GwU m~‡Îi A¼|
Avgiv Rvwb, GK NbwgUvi cvwbi IRb 1,000 †KwR|
Sonali & Bangladesh Development Bank Ltd.
Recruitment Test for Senior Officer (IT) – 16.10.2020
01. How many different selections of 4 books can be made from 10 different books, if 2 particular
books are never selected?
a) 20 b) 45
c) 70 d) 210 ans. C
Solution: cÖ‡kœ ejv n‡”Q, 10wU eB †_‡K me©`v 2wU eB bv wb‡q KZ fv‡e †mLvb †_‡K 4wU eB evQvB Kiv hv‡e?
10wU n‡Z 2wU ev` w`‡j eB _v‡K = 10 – 2 = 8wU
8wU eB n‡Z 4wU eB 8c4
Dcv‡q evQvB Kiv hvq|
GLv‡b, 8c4
=
8!
4! (8 - 4)!
=
8  7  6  5  4!
4  3  2  1  4!
= 70 Dcv‡q|
02. What is the distance of the origin from the line 12x – 5y + 26 = 0?
a) 2 b) 3
c) 4 d) 5 ans. A
Solution: cÖ‡kœ ejv n‡”Q, g~jwe›`y †_‡K 12x – 5y + 26 = 0 †iLvi `~iZ¡ KZ?
Avgiv Rvwb, g~jwe›`y ¯’vbv¼ n‡jv (0, 0)
cÖ`Ë mij †iLvi mgxKiY n‡jv 12x – 5y + 26 = 0
 wb‡Y©q `~iZ¡ =



Ax + By + c
A2
+ B2 =



(12  0) - (5  0) + 26
(12)2
+ (- 5)2 =



26
169
=
26
13
= 2.
GLv‡b, A = 12; B = – 5; C = 26
03. If ‘+’ means ‘’, ‘’ means ‘–’ means ‘’ and ‘’, means ‘+’ then find the value of 36 + 16 – 9 
15 ÷ 3
a) 7 b) 16
c) 12 d) 30 ans. A
Solution: g~jwU A¯úó _vKvi Kvi‡Y cÖkœwUi Abyiƒc GKwU cÖkœw`‡q Zvi DËi cÖ`vb Kiv n‡jvt

30. Recent Publication 286
Q. If ‘+’ means ‘’, ‘’ means ‘–’, ‘–’ means ‘’ and ‘’ means ‘+’ then find the value of 36 + 18 –
17  16 ÷ 3
Solution: 36 + 18 – 17  16  3
= 36  18  17 – 16 + 3
= (2  17) – (16 + 3)
= 34 – 16 – 3
= 37 – 16
= 21
04. If the product of two numbers is 560 and greatest common factor is 4. Then what is the least
common multiple?
a) 70 b) 140
c) 120 d) 49 ans. B
Solution: cÖ‡kœ ejv n‡”Q, `ywU msL¨vi ¸Ydj 560. msL¨v `ywUi M.mv.¸. 4 n‡j Zv‡`i j.mv.¸. KZ?
Avgiv Rvwb, `ywU msL¨vi ¸Ydj = msL¨v `ywUi M.mv.¸.  j.mv.¸.
 560 = 4  j.mv.¸.
 j.mv.¸. =
560
4
= 140
05. The sum of squares of two numbers is 80 and the square of difference between the two
numbers is 36. Find the product of two numbers.
a) 11 b) 22
c) 33 d) 26 ans. B
Solution: cÖ‡kœ ejv n‡”Q, `ywU msL¨vi e‡M©i mgwó 80 Ges Zv‡`i Aš—‡ii eM© n‡”Q 36. msL¨v `ywUi ¸Ydj KZ?
awi, msL¨v `ywU h_vµ‡g a I b
 cÖkœg‡Z, a2
+ b2
= 80 ............. (i)
Ges (a – b)2
= 36
 a2
– 2ab + b2
= 36
 a2
+ b2
– 2ab = 36
 80 – 2ab = 36 [(i) bs n‡Z]
 – 2ab = 36 – 80
 – 2ab = – 44
 ab =
44
2
= 22
06. Find the x intercept of this equation: 2x + 3y = 12
a) (6, 0) b)
c) d) ans. A
Solution: 2x + 3y = 12 mgxKiY x Gi †Q`vsk †ei Ki‡Z n‡e|
†`qv Av‡Q, 2x + 3y = 12
x Gi ¯’v‡b A_©vr f~wg A‡¶ y = 0 e‡j Avgiv cvB,
2x + (3  0) = 12
 2x + 0 = 12
 2x = 12
 x =
12
2
= 6
answer: (6, 0)
0, 4
Y
X
0, 6O
Y
X
Clipboard
Q. How many prime numbers are there between 45 and
72?
a) 5 b) 6
c) 8 d) 7 ans. B
Solution: cÖ‡kœ ejv n‡”Q, 45 n‡Z 72 ch©š— †gŠwjK msL¨v i‡q‡Q?
45 †_‡K 72 ch©š— †gŠwjK msL¨v¸‡jv n‡jv 47, 53, 59, 61, 67,
71. A_©vr †gvU 6 wU †gŠwjK msL¨v i‡q‡Q|

31. Recent Publication 287
07. Which of the following three side of the triangle?
a) 5, 6, 7 b) 5, 7, 14
c) 3, 2, 1 d) 2, 4, 8 ans. A
Solution: cÖ‡kœ ejv n‡”Q, wb‡Pi †KvbwU GKwU wÎfy‡Ri evû wb‡`©k K‡i?
Avgiv Rvwb, wÎfy‡Ri †¶‡Î †h‡Kvb `yB evûi mgwó Z…Zxq evû A‡c¶v e„nËi n‡e|
Ackb a) †Z Av‡Q 5, 6, 7. GwU DËi KviY 5 + 6 > 7
wKš‘, Ackb b) †Z Av‡Q 5, 7, 14. GwU DËi bv KviY 5 + 7 < 14
Ackb c) †Z Av‡Q 4, 5, 12. GwU DËi bv KviY 4 + 5 < 12
Ackb d) †Z Av‡Q 2, 4, 8. GwU DËi bv KviY 2 + 4 < 8
08. What is the value of tan 40° tan 50° tan 60°?
a) 1 b)
1
2
c) 3 d) – 3 ans. C
Solution: †`qv Av‡Q, tan 40° tan 50° tan 60°
=
1
cot 40°
 tan 50°  tan 60°
=
1
tan




2
- 40
 tan 50°  tan 60°
=
1
tan (90 - 40)°
 tan 50°  tan 60°
=
1
tan 50°
 tan 50°  tan 60°
= tan 60°
= 3 answer
09. The cube root of 1,331 is?
a) 13 b) 11
c) 19 d) 17 ans. B
Solution: GLv‡b Cube root A_© Nbg~j|
†`qv Av‡Q,
3
1331 =
3
11 3
[KviY 113
= 1331]
= 11
3 
1
3 = 11 (answer)
10. The area of the right triangle is 184cm2
. One of its leg is 16cm long. Find the length of the other
leg.
a) 23cm b) 22cm
c) 24cm d) 20cm ans. A
Solution: cÖ‡kœ ejv n‡”Q, GKwU mg‡KvYx wÎfy‡Ri †¶Îdj 184cm2
. wÎfyRwUi GK evûi ˆ`N©¨ 16cm n‡j Aci evû KZ?
GLv‡b, ABC GKwU mg‡KvYx wÎfyR
hvi BC = 16cm
AB =?
Avgiv Rvwb, mg‡KvYx wÎfy‡Ri †¶Îdj =
1
2
 mg‡KvY msjMœ evû؇qi ¸Ydj
cÖkœg‡Z, ABC Gi †¶Îdj =
1
2
 AB  BC
 184 =
1
2
 16  AB
 AB =
2  184
16
= 23 (answer)
A
CB
16
Clipboard
Q. What is the difference between 150 and 54 ?
a) 2 6 b) 16 6
c) 9 6 d) 6 2 ans. A
Shortcut: 150 – 54 = 6  25 – 6  9
= 6  (5)2
– 6  (3)2
= 5 6 – 3 6 = 2 6

32. Recent Publication 288
Janata Bank Ltd.
Recruitment Test for Senior Officer (Engineering-Textile) – 07.11.2020
01. The present worth of Tk. 169 due in 2 years at 4% per annum compound interest is
a) 144.75 b) 148.5
c) 15.25 d) 162.25 ans. Note
Solution: cÖ‡kœ ejv n‡”Q, 4% nvi my‡` 2 eQ‡i Pµe„w× my`vmj 169 UvKv n‡j Avmj KZ?
†`qv Av‡Q, my`vmj A = 169 UvKv; mgq n = 2 eQi; my‡`i nvi, r = 4% Ges Avmj, p =?
Avgiv Rvwb, A = p



1 +
r
100
n
 p =
A



1 +
r
100
n =
169



1 +
4
100
2 =
169
(1 + 0.04)2 =
169
(1.04)2 =
169
1.0816
= 156.25
A_©vr Avmj 156.25 UvKv|
02. When the speed is increased by 4 kmph, it takes 4 hours less to cover a distance of 32km. Find
the previous speed.
a) 2km b) 4km
c) 8km d) 12km ans. B
Solution: cÖ‡kœ ejv n‡”Q, MwZ‡eM 4 kmph evov‡j 32km c_ AwZµg Ki‡Z 4 NÈv mgq Kg jv‡M| cÖK…Z ev Avmj
MwZ‡eM KZ?
awi, cÖK…Z MwZ‡eM wQj x kmph
MwZ‡eM 4 kmph evov‡j, bZzb MwZ‡eM n‡e = (x + 4) kmph
cÖkœg‡Z,
32
x
–
32
x + 4
= 4 [ mgq =
`~iZ¡
MwZ‡eM ]

32x + 128 - 32x
x(x + 4)
= 4
 (x2
+ 4x)4 = 128
 x2
+ 4x =
128
4
= 32
 x2
+ 4x – 32 = 0
 x2
+ 8x – 4x – 32 = 0
 x(x + 8)(x – 4) = 0
 x – 4 = 0 Avevi, x + 8 = 0
 x = 4  x = – 8 wKš‘, FYvZ¥K MwZ‡eM MÖnY‡hvM¨ bq e‡j GwU ev`
mwVK DËi n‡e 4 kmph
03. The radius of a circle is increased so that in circumference by 5%. The area of the circle will
be increased by
a) 5% b) 10%
c) 10.25% d) 10.5% ans. C
Solution: cÖ‡kœ ejv n‡”Q, GKwU e„Ë Ggbfv‡e evov‡bv n‡jv hv‡Z cwiwa 5% †e‡o †Mj| e„ËwUi †¶Îdj KZ evo‡e?
awi, e„ËwUi cÖv_wgK e¨vmva© r GKK
 e„ËwUi cÖv_wgK cwiwa 2r GKK
5% e„wׇZ e„ËwUi bZzb cwiwa = 2r + 5% of 2r = 2r + (0.05  2r) = 2(r + 0.05r) = 2  1.05r GKK
A_©vr e„ËwUi byZb e¨vmva© R = 1.05r GKK
 †¶Îdj ev‡o = R2
– r2
= (R2
– r2
) = [(1.05r)2
– r2
] = (1.1025r2
– r2
) = 0.1025r2
eM© GKK
 †¶Îdj kZKiv ev‡o =



0.1025r2
r2  100 % = 10.25%
Clipboard
Q. A gas tank is
1
5
th full and requires 32 gallons
more to make it
3
7
th full. What is the
capacity of the tank?
a) 115 b) 120
c) 135 d) 140 ans. D
Shortcut: GLv‡b
3
7
–
1
5
=
8
35
As‡ki cwigvY 32 M¨vjb
 m¤ú~Y© ev 1 As‡ki cwigvY = 32 
35
8
= 140 M¨vjb

33. Recent Publication 289
Shortcut: e„‡Ëi e¨vmva© Ges cwiwa mgvbycv‡Z ev‡o e‡j cwiwa 5% evo‡j e¨vmva©I 5% evo‡e|
ZvB, cÖv_wgK e¨vmva© r = 1 n‡j ewa©Z e¨vmva© n‡e 1.05r
 †¶Îdj kZKiv ev‡o =



(1.05r)2
 -   r2
  12  100 % =



0.1025r2
r2  100 % = 10.25%
04. A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square,
then what will be its area?
a) 7,744 b) 7,759
c) 1,456 d) None of these ans. A
Solution: cÖ‡kœ ejv n‡”Q, GKwU Zvi‡K e„‡Ëi AvKvi w`‡j Zvi e¨vmva© nq 56cm. H ZviwU‡K e‡M©i AvKvi w`‡j e‡M©i
†¶Îdj KZ n‡e?
†`qv Av‡Q, e„‡Ëi e¨vmva© r = 56cm
 cwiwa, 2r = 2 
22
7
 56 = 352cm.
 e‡M©i GKevûi ˆ`N©¨ n‡e =
352
4
= 88cm.
AZGe, e‡M©i †¶Îdj = (evû)2
= (88)2
cm2
= 7,744 cm2
05. In a railway compartment 6 seats are vacant on a bench. In how many ways can 3 passengers
sit on them?
a) 120 b) 130
c) 160 d) 180 ans. A
Solution: cÖ‡kœ ejv n‡”Q, 6wU mx‡U 3 Rb hvÎx‡K KZfv‡e emv‡bv †h‡Z cv‡i?
6wU mx‡U 3 Rb hvÎx‡K 6
P3 = 6  5  4 = 120 Dcv‡q emv‡bv hv‡e|
06. The distance from the point P to two vertices A and B of an equilateral triangle are |PA| = 2
and |PB| = 3. What is the greatest possible value of |PC|?
a) 5 b) 9
c) 6 d) 12 ans. A
Solution: cÖ‡kœ ejv n‡”Q, GKwU mgevû wÎfy‡Ri `ywU kxl© we›`y A Ges B n‡Z P we›`yi `~iZ¡ n‡jv |PA| = 2 Ges |PB|
= 3. |PC| Gi m¤¢ve¨ me‡P‡q eo gvb KZ n‡Z cv‡i?
awi, |PA| = a = 2; |PB| = b = 3; |PC| = x
Aa© cwimxgv S =
a + b + x
2
=
2 + 3 + x
2
=
5 + x
2
GLv‡b, ABC =
1
2 


3
4
(a2
+ b2
+ x2
) + 3 S(S - a)(S - b)(S - x)
=
1
2 


3
4
(4 + 9 + x2
) + 3



5 + x
2 


5 + x
2
- 2



5 + x
2
- 3



5 + x
2
- x
=
1
2 


3
4
(13 + x2
) + 3



5 + x
2 


5 + x - 4
2 


5 + x - 6
2 


5 + x - 2x
2
=
1
2 


3
4
(x2
+ 13) + 3





(5 + x) (5 - x)
2  2 




(x + 1) (x - 1)
2  2
=
1
2 


3
4
(x2
+ 13) + 3
(25 - x 2
) (x 2
- 1)
16
=
1
2 


3
4
(x2
+ 13) +
3
4
(25 - x2
) (x 2
- 1)
C
BA a = 2
x
P
b = 3

34. Recent Publication 290
 †¶Î ABC Gi gvb cvIqv hvv‡e hLb, (25 – x2
) (x2
– 1) > 0 n‡e|
GLb, 25 – x2
> 0 A_ev, x2
– 1 > 0
1g †¶‡Î 25 – x2
> 0 mZ¨ n‡e hLb
x = + 1, + 2, + 3, + 4, + 5
 – 5 < x < 5
 x Gi m‡e©v”P m¤¢ve¨ gvb n‡e 5
 |PC| = x = 5
Shortcut: †h‡nZz PA = 2
PB = 3
myZivs PC Gi m‡e©v”P `~iZ¡ 5 n‡Z cv‡i|
07. The average 10 integers is 16. If the sum of 6 of them is 100. What is the average of other 4?
a) 21 b) 44
c) 66 d) None of these ans. Note
Solution: cÖ‡kœ ejv n‡”Q, 10wU msL¨vi Mo 16. 6wU msL¨vi mgwó 100 n‡j evwK 4wU msL¨vi Mo KZ?
10wU msL¨vi mgwó = 16  10 = 160
†`qv Av‡Q, 6wU msL¨vi mgwó = 100
 4wU msL¨vi mgwó = 60
AZGe, 4wU msL¨vi Mo =
60
4
= 15
08. If p and n are integers such that p > n > 0 and p2
– n2
= 12, which of the following value of p – n?
a) – 1 b) 2
c) 8 d) 18 ans. B
Solution: cÖ‡kœ ejv n‡”Q, p I n c~Y© msL¨v Ges p > n > 0 I p2
– n2
= 12 n‡j (p – n) =?
GLv‡b, †h‡nZz p > n > 0 I p2
– n2
= 12 Kv‡RB
p = 4 Ges n = 2 n‡j p2
– n2
= 12 mgxKiYwU wm× nq|
KviY, 42
– 22
= 16 – 4 = 12
AZGe, p – n = 4 – 2 = 2
09. Income tax is raised from 4% to 5% but the revenue is increased by 10% only. Find the
decreased percentage in the amount taxed.
a) 8 b) 10
c) 12 d) 15 ans. C
Solution: cÖ‡kœ ejv n‡”Q, AvqKi 4% †_‡K evwo‡q 5% Kiv n‡j ivR¯^ 10% †e‡o hvq| AvqKi kZKiv KZ K‡g
wQ‡jv?
awi, cÖ_‡g AvqKi wQj x UvKv
bZzb AvqKi wQj y UvKv
Ges awi, cÖ_g ivR¯^ wQj 100 UvKv
Zvn‡j 4% of x = 100

4x
100
= 100
 x = 2,500
Clipboard
Q. 10 minutes after a plane leaves the airport;
it is reported to be 40 miles away. What
average speed in miles per hour of the
plane?
a) 560 b) 660
c) 400 d) 240 ans. D
1g 10 minutes G Plane wU hvq = 40 miles
 1g 60 minutes G Plane wU hvq =
40  60
10
= 240m
P 2 A
B C
3

35. Recent Publication 291
Avevi, ivR¯^ 10% e„wׇZ bZzb ivR¯^ n‡e = 100 + 10 = 110 UvKv
AZGe, 5% of y = 110

5y
100
= 110
 y =
100  110
5
= 2,200
 AvqKi kZKiv K‡g‡Q =



2500 - 2200
2500
 100 % =



300
2500
 100 % = 12%
10. Find the area of rhombus one side of which measures 20cm one diagonal 24cm.
a) 281cm2
b) 320cm2
c) 384cm2
d) 404cm2
ans. C
Solution: cÖ‡kœ ejv n‡”Q, GKwU i¤^‡mi GK evûi ˆ`N©¨ 20cm Ges KY© 24cm n‡j †¶Îdj KZ?
awi, i¤^‡mi GKevû n = 20cm
GKwU KY©, p = 24cm
 Aci KY©, q = 4a2
- p2
= 4  (20)2
- (24)2
= 1600 - 576 = 1024 = (32)2
= 32cm
A_©vr Aci KY© q = 32cm
AZGe, i¤^‡mi †¶Îdj =
1
2
 p  q =
1
2
 24  32 = 384cm2
Rupali Bank Ltd.
Recruitment Test for Financial analysts – 20.11.2020
01. Of the 50 researcher in a workgroup, 40% will be assigned to team A and the remaining 60%
to team B. However 70% of the researcher prefers team A and 30% prefer team B. What is
the least possible number of researchers who will not be assigned to the team they prefer?
a) 15 b) 20
c) 25 d) 30 ans. A
Solution: cÖ‡kœ ejv n‡”Q, 50 R‡bi GKwU M‡elK‡`i `‡ji 40% m`m¨‡K A `‡ji Ges Aewkó 60% m`m¨‡K B
`‡ji `vwqZ¡ †`qv n‡e| Z‡e 70% M‡elK A `j‡K Ges 30% M‡elK B `j‡K cQ›` K‡ib| KZ kZvsk
M‡elK‡K Zv‡`i cQ›` Abyhvqx `j‡K †`qv nqwb?
`j A †Z eivÏK…Z M‡elK‡`i msL¨v =



50 
40
100
= 20 Rb
 `j B †Z eivÏK…Z M‡elK‡`i msL¨v =



50 
60
100
= 30 Rb
`j A cQ›` Kiv M‡elK‡`i msL¨v =



50 
70
100
= 35 Rb
 `j B cQ›` Kiv M‡elK‡`i msL¨v =



50 
30
100
= 15 Rb
cQ›`gZ `‡j eivÏ cvIqv M‡elK‡`i msL¨v = 20 + 15 = 35 Rb
webv cQ‡›`i `‡j eivÏ cvIqv Kgc‡¶ †gvU M‡elK‡`i msL¨v = 50 – 35 = 15 Rb|
a
aa
a
2
p = 24

36. Recent Publication 292
02. a, b, c, d and e are five consecutive integers in increasing order of size. Which one of the
following expression is not odd?
a) a + b + c b) ab + c
c) ac + e d) ac + d ans. C
Solution: cÖ‡kœ ejv n‡”Q, †QvU †_‡K eo µgvbymv‡i a, b, c, d Ges e n‡jv cuvPwU µwgK c~Y©msL¨v| wb‡Pi †KvbwU
we‡Rvo bq?
1g †¶‡Î,
awi, a = 1; b = 2; c = 3; d = 4; e = 5
2q †¶‡Î,
awi, a = 2; b = 3; c = 4; d = 5; e = 6
Ackb a) a + b + c
= 1 + 2 + 3
= 6 (†Rvo)
Ackb b) ab + c
= (1  2) + 3
= 3 + 2 = 5 (we‡Rvo)
Ackb a) a + b + c
= 2 + 3 + 4
= 9 (we‡Rvo)
Ackb b) ab + c
= (2  3) + 4
= 6 + 4 = 10 (†Rvo)
Ackb c) ac + e
= (1  3) + 5
= 3 + 5 = 8 (†Rvo)
Ackb d) ac + d
= (1  3) + 4
= 3 + 4 = 7 (we‡Rvo)
Ackb c) ac + e
= (2  4) + 6
= 8 + 4 = 14 (†Rvo)
Ackb d) ac + d
= (2  4) + 5
= 8 + 5 = 13 (we‡Rvo)
Ackb c) wU me‡¶‡ÎB †Rvo| ZvB GwU DËi|
Alternative: awi, a = 1; b = 2; c = 3; d = 4; e = 5 n‡j
Ackb c) ac + e = (1  3) + 5 = 3 + 5 = 8 (†Rvo)|
Avevi awi, a = 2; b = 3; c = 4; d = 5; e = 6 n‡j
Ackb c) ac + e = (2  4) + 6 = 8 + 6 = 14 (†Rvo)|
wKš‘ evwK Ackb¸‡jv‡Z GB gvb¸wj ewm‡q we‡Rvo msL¨vi cvIqv hvq| ZvB mwVK DËi n‡e Ackb c) Gi ac + e
03. Selling 12 pen at a price of Tk. 10 yields a loss a%. Selling 12 pen at a price of Tk. 12 yields a
profit of a%, what is the value of a?
a) 10 b) 11
c)
100
11
d)
11
100
ans. C
Solution: cÖ‡kœ ejv n‡”Q †h, 10 UvKv `‡i 12 wU Kj‡gi wewµ Ki‡j a% ¶wZ nq| wKš‘ 12 UvKv `‡i H 12 wU Kjg
wewµ Ki‡j a% jvf nq| Avcbv‡K a Gi gvb †ei Ki‡Z n‡e|
†`qv Av‡Q, kZKiv jvf = kZKiv ¶wZ = a
cÖkœg‡Z,
µqg~j¨ - 10
µqg~j¨  100 =
12 - µqg~j¨
µqg~j¨  100
 µqg~j¨ – 10 = 12 – µqg~j¨ [Dfq c¶‡K
µqg~j¨
100 Øviv ¸Y K‡i]
 2 µqg~j¨ = 22
 µqg~j¨ =
22
2
= 11
AZGe, a =
µqg~j¨ - 10
µqg~j¨  100 =
11 - 10
11
 100 =
1
11
 100 =
100
11
04. The base of an isosceles triangle is 6cm and one of the equal sides is 12cm. Find the radius of
the circle through the vertices of the triangle?
a)
7 13
3
b)
7 5
6
c)
8 15
5
d)
5 5
3
ans. C
Solution: cÖ‡kœ ejv n‡”Q, e„‡Ë Aš—wjwL©Z GKwU mgwØevû wÎfy‡Ri f~wgi ˆ`N©¨ 6cm Ges mgvb mgvb evûi ˆ`N©¨ 12cm.
wÎfyRwUi kxl©we›`y a‡i e„ËwUi e¨vmva© wbY©q Ki“b|

37. Recent Publication 293
awi, e„‡Ëi e¨vmva© r †mw›UwgUvi|
wÎfy‡Ri D”PZv h = (r + x) †mw›UwgUvi|
GLv‡b, AB = 6 †mw›UwgUvi
 AD =
6
2
= 3 †mw›UwgUvi|
myZivs, ACD G, D = 90
wc_v‡Mviv‡mi m~Î cÖ‡qvM K‡i cvBt
AC2
= CD2
+ AD2
 122
= h2
+ 32
 h = 144 – 9 = 135
Avevi, ADE G, D = 90
wc_v‡Mviv‡mi m~Î cÖ‡qvM K‡i cvBt
AE2
= AD2
+ DE2
 r2
= 9 + x2
 r2
= 9 + (h – r)2
 r2
= 9 + h2
– 2hr + r2
 0 = 9 + h2
– 2hr
 2hr = h2
+ 9
 r =
(9 + h2
)
2h
=
9 + ( )135
2
2 135
=
9 + 135
2 135
=
144
2 135
=
144  135
2 135  135
=
144 135
270
=
8 15
5
05. A jar contains milk and water in the ratio 5 : 1. If the quantity of milk is more than that of
water by 8 liters, then what is the quantity of water?
a) 1.5 liter b) 2 liter
c) 6 liter d) 8 liter ans. B
Solution: cÖ‡kœ ejv n‡”Q, GKwU cv‡Î `ya Ges cvwb Av‡Q 5 : 1 Abycv‡Z| cvwbi †P‡q `ya 8 wjUvi †ewk n‡j cvwbi
cwigvY KZ?
`ya t cvwb = 5 : 1
awi, `ya Av‡Q = 5x wjUvi Ges cvwb Av‡Q = x wjUvi
kZ©g‡Z, 5x – x = 8
 4x = 8
 x =
4
8
= 2
 cvwb Av‡Q 2 wjUvi|
06. A sum of money amounts of Tk. 460 in 3 years and to Tk. 500 in five years. Find the rate
percent per annum
a) 5 Tk b) 10 Tk
c) 15 Tk d) 25 Tk ans. A
Solution: cÖ‡kœ ejv n‡”Q, †Kv‡bv Avmj 3 eQ‡i my‡` Avm‡j 460 UvKv Ges 5 eQ‡i my‡` Avm‡j 500 UvKv n‡j
kZKiv my‡`i nvi n‡e
5 eQ‡ii my` + Avmj = 500 UvKv
3 eQ‡ii my` + Avmj = 460 UvKv
 2 eQ‡ii my` = 40 UvKv
 3 eQ‡ii my` =
40  3
2 = 60 UvKv
 Avmj = (460 - 60) UvKv = 400 UvKv
C
BA
D
E
p
r
x
Shortcut
Step-1: eo my` =
500 - 460
5 - 3
 5 = 100
Step-2: Avmj, P = 500 – 100 = 400
Step-3: 100 =
400  r  5
100
 r = 5

38. Recent Publication 294
400 UvKvq 3 eQ‡ii my` 60 UvKv
 1 UvKvq 1 eQ‡ii my` =
60
400  3
UvKv
 100 UvKvq 1 eQ‡ii my` =
60  100
400  3
= 5 UvKv
07. Sum of three different positive integers is the same as their product. What is the smallest of
these 3 integers?
a) 0 b) 1
c) – 1 d) 2 ans. B
Solution: cÖ‡kœ ejv n‡”Q, wZbwU c~Y©msL¨vi †hvMdj hv Zv‡`i ¸YdjI Zv| msL¨v wZbwUi g‡a¨ ¶z`ªZg msL¨v †KvbwU?
awi, cÖ_g msL¨v = x; wØZxq msL¨v = x + 1 Ges Z…Zxq msL¨v = x + 2
kZ©g‡Z, x + x + 1 + x + 2 = x(x + 1)(x + 2)
 3x + 3 = (x2
+ x)(x + 2)
 3x + 3 = x3
+ x2
+ 2x2
+ 2x
 3x + 3 = x3
+ 3x2
+ 2x
 x3
+ 3x2
+ 2x – 3x – 3 = 0
 x3
+ 3x2
– x – 3 = 0
 x3
– x2
+ 4x2
– 4x + 3x – 3 = 0
 x2
(x – 1) + 4x(x – 1) + 3(x – 1) = 0
 (x – 1) (x2
+ 4x + 3) = 0
nq, x – 1 = 0 A_ev, x2
+ 4x + 3 = 0
 x = 1
cÖ_g msL¨v x = 1; wØZxq msL¨v, x + 1 = 2 Ges Z…Zxq msL¨v, x + 2 = 3
me‡P‡q ¶z`ªZg msL¨v 1.
08. What is the rate of discount is a car that had a regular price of Tk. 30,00,000 is sold for Tk.
27,90,000?
a) 7% b) 8%
c) 9% d) 10% ans. A
Solution: cÖ‡kœ ejv n‡”Q 30,00,000 UvKvi Mvwo 27, 90, 000 UvKv weµq Ki‡j Discount Gi nvi KZ wQj?
wjwLZ g~j¨ = 30,00,000 UvKv
weµq g~j¨ = 27, 90, 000 UvKv
Qvo = 2,10,000 UvKv
 Qv‡oi nvi =



210000
3000000
 100 % = 7%
09. The average of a, b, c is 6 and a – b = 4, ab = 21, what is the value of c?
a) 6 b) 7
c) 8 d) 9 ans. C
Solution: cÖ‡kœ ejv n‡”Q a, b, c Gi Mo 6 Ges a – b = 4 I ab = 21 n‡j c Gi gvb KZ?
†`qv Av‡Q,
a + b + c
3
= 6
 a + b + c = 18 ................. (i)
a – b = 4 ......................... (ii)
ab = 21 ........................... (iii)
Shortcut
msL¨v wZbwU 1, 2, 3 n‡j 1 + 2 + 3 = 6
Ges 1  2  3 = 6
GLv‡b ¶z`ªZg msL¨v n‡jv 1.

39. Recent Publication 295
mgxKiY (i) bs n‡Z (ii) bs †hvM K‡i cvB,
a + b + c + a – b = 22
 2a + c = 22
 c = 22 – 2a ................. (iv)
ab = 21
 b =
21
a
........................... (v)
mgxKiY (ii) bs n‡Z cvB,
a –
21
a
= 4
 a2
– 21 = 4a
 a2
– 21 – 4a = 0
 a2
– 4a – 21 = 0
 a2
–7a + 3a – 21 = 0
 a(a – 7) + 3(a – 7) = 0
 (a – 7) (a + 3) = 0
nq, a – 7 = 0 A_ev, a + 3 = 0
 a = 7  a = – 3
a = 7 Gi gvb mgxKiY (iv) G ewm‡q cvB,
c = 22 – (2  7) = 22 – 14 = 8
10. If x is an even number, what is the difference between the smallest even number grater than
(5x + 6) and the largest even number less than (3x + 9)?
a) 2x b) 2x – 3
c) 7x + 15 d) 2x + 3 ans. A
Solution: cÖ‡kœ ejv n‡”Q, x n‡jv GKwU †Rvo msL¨v| Avcbv‡K (5x + 6) †_‡K eo †Rvo msL¨v Ges (3x + 9) †_‡K
†QvU †Rvo msL¨vi cv_©K¨ †ei Ki‡Z n‡e?
†h‡nZz x †Rvo msL¨v,
ZvB (5x + 6) I †Rvo msL¨v n‡e|
 (5x + 6) Gi eo ¶z`ªZg †Rvo msL¨vwU n‡e (5x + 6) Gi †P‡q 2 †ewk|
A_©r (5x + 6) + 2 = 5x + 8.
GKBfv‡e, x †Rvo msL¨v nIqvq (3x + 9) Aek¨B we‡Rvo msL¨v n‡e|
 (3x + 9) Gi †P‡q †QvU Largest †Rvo msL¨vwU n‡e (3x + 9) Gi †P‡q 1 Kg A_©vr (3x + 9) – 1 = 3x + 8
 msL¨v؇qi cv_©K¨ = (5x + 8) – (3x + 8) = 5x + 8 – 3x – 8 = 2x
Shortcut
†`q Av‡Q,
a + b + c
3
= 6
 a + b + c = 18
Avevi, †`qv Av‡Q a – b = 4 Ges ab = 21
GLv‡b, a = 7 Ges b = 3 n‡j ab = 21 nq
Ges a – b = 4 nq|
7 + 3 + c = 18
 c = 18 – 10 = 8