Arts faculty all mcq math & solution (2019 2017) by khairul alam [www.onlinebcs.com]

[All MCQ Math & Solution of Arts Faculty (2019-2017)] By Khairul Alam
Khairul’s Bank Recent Math - 1
Prepared By: Khairul Alam For BB Off and AD exams
Arts faculty
All MCQ Math & Solution
Exam date Page
01 Rupali Bank Limited Officer 08-11-2019 01
02 Combine Exam (5 bank) Officer 20-07-2018 08
03 Bangladesh Bank Assistant Director 06-07-2018 15
04 Sonali Bank Limited Senior Officer 01-06-2018 22
05 Bangladesh Bank Officer, (General) 27-04-2018 30
06 Rupali Bank Limited Officer-(Cash) 09-03-2018 36
07 Agrani Bank Limited Senior Officer (Auditor) 15-12-2017 43
08 Agrani Bank Limited Officer ( Cash) 15-12-2017 51
09 Bangladesh Development Bank Senior Officer 24-11-2017 55
10 Bangladesh Krishi Bank Officer-(Cash) 24-11-2017 62
11 Bangladesh House Building Finance Corporation
Senior Officer – 2017
2017 67
Written cixÿvi cÖkœ¸‡jv c‡i †kqvi Kiv n‡e | mv‡_ _vKzb|

1. If both x and y are prime numbers, which of the following cannot be product of x and
y? (x I y ‡gŠwjK msL¨v n‡j wb‡Pi †KvbwU x I y Gi ¸Ydj n‡Z cv‡i bv?) [Rupali Bank Officer-2019]
(A) 6 (B) 10 (C) 35 (D) 27 Ans: D
Solution: 6 = 32, 10 = 5  2, 35= 5  7, 27= 3  9
Ack‡bi g‡a¨ 6, 10, 35 Gi Drcv`K ¸‡jv ‡gŠwjK msL¨v wKš‘ 27 †K fv½‡j 39 nq
†hLv‡b 9 ‡gŠwjK msL¨v bq| ZvB DËi 27|
2. How many integers from 1 to 1000 are divisible by 30 but not by 16? (1 ‡_‡K 1000 ch©šÍ
KZ¸‡jv c~Y©msL¨v 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq?) [Agrani Bank – (Cash)-2017] + [BB-(officer)-
2018] & [Sonali Bank –(SO)-2018]+[Rupali Bank Officer-2019]
(A) 29 (B) 31 (C) 32 (D) 38 Ans: A
Solution:
Divisible by 30=
30
1000
= 33.33 = 33 Numbers (30,60,90-----990 = 33wU)
L.C.M of 30 and 16 = 240,
So, the numbers from 1-1000 divisible by 30 and 16 is 240 is =
240
1000
= 4.
So, the number divisible by 240 is also divisible by 30 and 16,
So, the numbers only divisible by 30 is 33-4 = 29. Ans : 29
evsjvq e¨vL¨v:
30 w`‡q wefvR¨ †h msL¨v¸‡jv Av‡Q †hgb: 30, 60, 120, 240 - - - - 990 = 33wU | Gi g‡a¨ wKQz msL¨v Av‡Q
hv‡`i‡K 16 w`‡qI fvM Kiv hvq|
GLb GB msL¨v¸‡jv †_‡K 30 w`‡q fvM Kiv hvq wKš‘ 16 w`‡q fvM Kiv hvq bv Ggb msL¨v †ei Kivi Rb¨ 16 I 30
Dfq w`‡q fvM Kiv hvq †mB msL¨v¸‡jv ‡ei K‡i Av‡Mi 33wU †_‡K ev` w`‡Z n‡e|
wK Ki‡j 16 I 30 Gi wgjb ¯’j Lyu‡R cvIqv hv‡e? Aek¨B j.mv,¸ Ki‡j| ZvB 16 Ges 30 Gi j.mv.¸ 240 †ei
K‡i 240 w`‡q 30 w`‡q wefvR¨ 33wU msL¨vi g‡a¨ †h 4wU msL¨v ( 240,480,720,960) †K 16 w`‡qI fvM Kiv hvq
†m¸‡jv ev` w`‡jB 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq Ggb msL¨v¸‡jv †ei n‡e| DËi: 33-4 = 29wU|
3. P and Q are brothers. R and S are sisters. The son of P is brother of S. Q is related to R
as- (P I Q `yB fvB| R I S `yB †evb| P Gi mšÍvb S Gi fvB| Q I R Gi m¤úK© Kx ?) [Rupali Bank Officer-
2019]
(A) Son (B) Brother (C) Uncle (D) Father Ans: C
Solution:
cv‡ki wPÎwU †`Lyb, ïiæ‡Z P Ges Q `yB fvB| GLb awi P
Gi cyÎ X n‡jv S Gi fvB| Zvn‡j R Ges S, n‡jv P Gi
`yB †g‡q| A_©vr R,S Ges X n‡jv 3 fvB †evb hv‡`i evev P
Gi fvB Q n‡e R Gi PvPv ev Uncle.
Rupali Bank Limited
Post name : Officer. Exam date: 08-11-2019
Exam taker: Arts Faculty, DU
3 fvB ‡evb
P(+)
R(-)
X(+)
Q(+)
`y,fvB
S(-)
?

4. If w is 10% less than x, and y is 30% less then z, than wy is what percent less then xz ?
(hw` w, x Gi †_‡K 10% †QvU Ges y, z Gi †_‡K 30% ‡QvU nq Zvn‡j wy, xz Gi †_‡K kZKiv KZ Kg?)
[Sonali Bank –(SO)-2018]+[Rupali Bank Officer-2019]
(A) 10% (B) 20% (C) 37% (D) 40% Ans: C
Solution:
If x = 100 then w = 90
if z = 100 then y = 70
so, xz = 100100 = 10000
and wy = 9070 = 6300
xz – wy = 10000-6300 = 3700 less % =
10000
1003700
= 37%
5. Every 3 minutes, 4 litres of water are poured into a 2000 litre tank After 2 hours, what
percent of the tank is full ? (2000 wjUvi aviY ÿgZvi GKwU U¨vs‡K cÖwZ 3 wgwb‡U 4 wjUvi cvwb Xvjv nq| 2
N›Uv ci U¨vsKwU KZ kZvsk c~Y© n‡e?) [Rupali Bank Officer-2019]
(A) 0.4% (B) 4% (C) 8% (D) 12% Ans: C
Solution:
In 3 min the tank will be filled by = 4 liters
In 120 min the tank will be filled by =
3
1204
= 160 liters
Required % =
2000
100160
= 8%
6. The next number in the sequence 3, 6, 11, 18, 27,--- is – (3, 6, 11, 18, 27,--- wmwiRwUi cieZ©x
msL¨v KZ?) [Sonali Bank –(SO)-2018]+ [Rupali Bank Officer-2019]
(A) 34 (B) 36 (C) 38 (D) 40 Ans: C
Solution:
Gap between the numbers is 3, 5, 7, 9 (difference of the gap is 9-7 = 2)
So next gap should be 9+2 = 11 and the number should be 27+11 = 38
7. The H.C.F. of two numbers is 24. The number which can be their L.C.M. is (`ywU msL¨vi
M.mv.¸ 24| wb‡Pi †Kvb msL¨vwU H msL¨vØ‡qi j.mv.¸ n‡Z cv‡i?) –[Agrani Bank – (Cash)-2017]+ [Rupali
Bank Officer-2019]
(A) 84 (B) 128 (C) 148 (D) 120 Ans: D
Solution:
The HCF is 24.
So LCM should be the numbers which are divisible by 24 .
from the given number only 120 divisible by 24 Ans: 120
g‡b ivLyb, †h †Kvb `ywU msL¨vi j.mv.¸ H msL¨vØ‡qi M.mv.¸ Øviv wb:‡k‡l wefvR¨ nq|
8. The average of eight numbers is 14 and the average of six of these numbers is 16. What is the
average of the remaining numbers?(AvUwU msL¨vi Mo 14, G‡`i g‡a¨ 6wU msL¨vi Mo 16 n‡j Aewkó
msL¨v¸‡jvi Mo KZ?) (Agrani Bank Off. (Cash)-2013)+[Rupali Bank Officer-2019]
(A)4 (B) 8 (C) 16 (D) Data inadequate Ans: B
Shortcut: 10% †QvU n‡j 100 †_‡K 90 nq | Gici 90 Gi
30% †QvU n‡j 27 ‡QvU nq Ges `ywU †QvU GKmv‡_ n‡q eo Gi
mv‡_ Zzjbv Ki‡j †gv‡Ui Dci †QvU nq 10%+27% = 37%

Solution: Sum of eight numbers = 8 14 = 112 Sum of first six numbers = 616 = 96
So, Sum of two numbers = 112-96=16 So average of last two numbers is 16÷2 = 8
9. If A = {1,2,3,4,5}, then the number of proper subsets of A is – (hw`, A = {1,2,3,4,5} nq
Zvn‡j A Gi cÖK…Z Dc‡mU KZwU?) [Agrani Bank – (Cash)-2017]+ [Sonali Bank –(SO)-2018]+[Rupali
Bank Officer-2019]
(A) 120 (B) 30 (C) 31 (D) 32 Ans: C
Solution:
Subset = 2n
= Here n = number of elements.
And proper subset = 2n
-1 = 25
– 1 = 32-1 = 31
e¨vL¨v: Subset n‡jv Dcv`vb¸‡jv wb‡q hZ¸‡jv bZzb †mU MVb Kiv hvq|
Avi proper subset = me¸‡jv Dcv`vb wb‡q MwVZ Dc‡mU wU ev` w`‡q ‡h¸‡jv Dc‡mU _vK‡e †m¸‡jvB cÖK…Z
Dc‡mU|
A= {1,2,3,4,5} Gi GKwU Dc‡mU {12345} wKš‘ Bnv cÖK…Z Dc‡mU bq| GRb¨ m~ÎwU 2n
-1
10. Two numbers are in the ratio 2 : 5. If 16 is added to both the numbers, their ratio
becomes 1 : 2. The number are- (`ywU msL¨vi AbycvZ 2:5| hw` Dfq msL¨vi mv‡_ 16 †hvM Kiv nq Zvn‡j
Zv‡`i AbycvZ 1:2 n‡q hvq| msL¨v `ywU KZ?) [Rupali Bank Officer-2019]
(A) 16, 40 (B) 20, 50 (C) 28, 70 (D) 32, 80 Ans: D
Solution:
Let, the numbers be 2x and 5x
ATQ,
16x5
16x2


=
2
1
 5x + 16 = 4x + 32  x = 16
So, the numbers are 216 = 32 and 516 = 80
11. If 1 -2x  3, then x = ? [Agrani Bank – (Cash)-2017] & [BB-(officer)-2018]+[Rupali Bank
Officer-2019]
(A) x  - 2 (B) x  - 2 (C) x  - 1 (D) x  - 1 Ans: D
Solution:
1-2x  3
 - 2x  3 -1
 - 2x  2  x  - 1 (Dfq cÿ‡K FYvZ¥K msL¨v w`‡q ¸Y Ki‡j gv‡Si wPýwU D‡ë hvq|)
12. If
3
1
x
y
 and x + 2y = 10 then x is- [Rupali Bank Officer-2019]
(A) 2 (B) 3 (C) 4 (D) 6 Ans: A
Solution:
Given,
3
1
x
y
 x = 3y……(i)
And x + 2y = 10 3y + 2y = 10 5y = 10 y = 2 putting this value in (i) we get x = 32=6
13. How many real roots does the polynomial 2x2
+ 8x - 7 have? [Agrani Bank – (Cash)-2017]+
[Rupali Bank Officer-2019]
(A) None (B) One (C) two (D) three Ans: B

Solution:
Here, f(x) = 2x3
+8x -7
And two positive sign and one negative sign, one sign changes.
so according to Descartes’ rule of signs One real root possible Ans: One
14. If xy = 2 and xy2
= 8, what is the value of x? [Rupali Bank Officer-2019]
(A) 4 (B) 2 (C)
2
1
(D) 8 Ans: C
Solution:
xy = 2 ---- (i) and xy2
= 8---- (ii)
by (ii)(i) We get
2
8
xy
xy2
 y = 4 putting this value in (i) We get x =
2
1
15. If 2
a
1
a  . What is 3
3
a
1
a  ? [Rupali Bank Officer-2019]
(A) 16 (B) 10 (C) 14 (D) 12 Ans: C
Solution: 3
3
a
1
a  =
3
a
1
a 





 + 3a
a
1







a
1
a = 23
+ (3  2)= 8 + 6 = 14
16. The three sides of a triangle are x+1, 2x-1 and 3x+1 respectively and the perimeter is 25
cm. The length of the smallest side is (GKwU wÎfz‡Ri wZbwU evûi ˆ`N©¨ h_vµ‡g x+1, 2x-1 Ges 3x+,
wÎf~RwUi cwimxgv 25 †m.wg. n‡j ÿz`ªZg evûi ˆ`N©¨ KZ?)- [Agrani Bank – (Cash)-2017]+ [Rupali Bank
Officer-2019]
(A)5 cm (B) 6 cm (C) 4 cm (D) 7 cm Ans: B
Solution:
Here, (x+1) + ( 2x-1) + (3x+1) = 25 Or, 6x = 24  x = 4
So the sides are 4+1 = 5, 24 – 1 = 8-1 = 7 and 34+1 = 12+1 = 13  Smallest side = 5
17. There are 10 true-false questions in an examination. These questions can be answered in
(hw` GKwU cixÿvq 10wU true – false cÖkœ _v‡K Zvn‡j H cÖkœ¸‡jvi DËi KZ fv‡e †`qv hv‡e? ) [Rupali Bank
Officer-2019]
(A) 20 ways (B) 100 ways (C) 102
ways (D) 1024 ways Ans: D
Solution:
GiKg cÖ‡kœi DËi †`qvi wbqg n‡jv: nr
GL‡b, n = number of ways (true or false)
Ges r = total number of questions
myZivs DËi †`qvi Dcvq n‡e, 210
= 1024
wKQz welq †R‡b ivLyb, Zvn‡j GiKg Ab¨ cÖkœ¸‡jvi DËi †`qv mnR n‡e|
[ - - - - - - - - - -] Gfv‡e †gvU 10wU cÖ‡kœi Rb¨ 10wU M¨vc| cÖwZwU M¨vc c~Y© Kiv hvq true ev False `yfv‡e| Zvn‡j
2222222 . . . Gfv‡e 10wU 2 ¸Y Kiv A_© n‡jv 210
= 1024| fzj K‡i 102
†`qv hv‡e bv| KviY n‡jv GB
cÖkœwU n‡jv mgv‡ek ev Combination Gi cÖkœ| 10wU B mZ¨ ev 10wU wg_¨v Gfv‡e †f‡e 102
w`‡j Zv w`‡q mgv‡ek
†evSvq bv| eis welqUv Ggb, 5wU mZ¨ 5wU wg_¨v| ev 4wU mZ¨v 6wU wg_¨v, 3wU mZ¨ 7wU wg_¨v| Gfv‡e wn‡me Pj‡Z
_vK‡e|

18. If two fair coins are flipped. What is the probability that one will come up heads and the
other tails? (`ywU gy`ªv wb‡ÿc Kiv n‡jv| GKwU‡Z †nW Ges Ab¨wU‡Z †Uj Avmvi m¤¢vebv KZ?) [Agrani Bank –
(Cash)-2017]& [BSC-Combined Exam –(Officer)-2018]+ [Rupali Bank Officer-2019]
A.
2
1
B.
3
1
C.
4
1
D. 1 Ans: A
Solution:
total sample(4)= TT, HH, HT, TH.
So one heads and one tails =HT, & TH
So probability =
2
1
4
2

Confusion Clear: cÖ‡kœ hw` ejv _vK‡Zv First one is heads and next one is tails ZLb cÖ_gevi
wbw`©ó nIqvq DËi n‡Zv =
4
1
KviY ZLb cÖ_gevi †nW Qvov Ab¨ wKQz †bqv hv‡e bv|
wKš‘ cÖ_gevi bv e‡j GKwU‡Z †nW ejvq GLv‡b wbw`©ó K‡i †evSvq wb| ZvB †h †Kvb GKwU †nW Ges Ab¨wU †Uj wn‡m‡e
AbyK~‡j djvdj = 2wU| ZvB DËi:
2
1
4
2

19. If the diagonal of a square measures 216 cm, what is the area of the square in sq.cm?
(hw` GKwU e‡M©i K‡Y©i ˆ`N©¨ 216 ‡m.wg. n‡j eM©wUi ‡ÿÎdj KZ eM© †m.wg.) [Rupali Bank Officer-2019]
(A) 232 (B) 246 (C) 128 (D) 256 Ans: D
Solution: e‡M©i KY© = a2 = 216 myZivs e‡M©i GKevû a = 16 e‡M©i †ÿÎdj = a2
= 162
= 256
20. If 16
9
xLog =
2
1
 the value of the base is- [Rupali Bank Officer-2019]
(A)
9
16
(B)
16
9
(C)
81
256
(D)
256
81
Ans: C
Solution:
16
9
xLog =
2
1

 2
1
x

=
16
9

2
1
x
1
=
16
9

x
1
=
16
9

x
1
=
256
81
[eM© K‡i]  x =
81
256
21. If cosA + cos2
A = 1, then the value of the expression (cos2
A + cosA) is- [Rupali Bank
Officer-2019]
(A) 1 (B)
2
1
(C) 2 (D) 3 Ans: A
Solution: (GUv †Kvb cÖkœB bv, hv †`qv Av‡Q Zv B †ei Ki‡Z ejv n‡q‡Q)
cosA + cos2
A = 1, cos2
A + cosA = 1

cÖgvY + hyw³:
(1) x = -8 n‡j, y = (-2 -8) - 8 = 16-8 = 8
(2) x = -9 n‡j y = (-2-9) -8 = 18-8 = 10
(3) x = 40 n‡j y = (-240) -8 = -80-8 = -88
GLv‡b y Gi gvb 9 Gi †_‡K Kg 1 b¤^‡i| hwÌ 3 b¤^‡i -
88 I 9 Gi †_‡K Kg wKš‘ G‡Z x Gi gvbwU ÿz`ªZg wQj bv
eis e„nËg n‡q †M‡Q| ZvB DËi: x = -8
22. The difference in taka between simple and compound interest at 5% annually on a sum
of Tk. 5000 after 2 years is- (5000 UvKvi Dci 5% nv‡i 2 eQ‡ii mij my` I Pµe„w× my‡ì cv_©K¨ KZ?)
[Rupali Bank Officer-2019]
(A) 12.5 (B) 25 (C) 50 (D) 500 Ans: A
Solution: (Formal way)
Simple interest = (5% of 5000)  2 = 500
Compound interest = (105% of 105% of 5000 ) – 5000 = 5512.5 – 5000 = 512.5
Required difference = 512.5-500 = 12.5
Alternative Solution: (Smart way)
5% of 5000 = 250 then 5% of 250 = 12.5 (cÖ_g eQ‡ii †hUv my` ‡mB my‡ì my`B n‡jv cv_©K¨ )
‡evSvi Rb¨ wPÎ|
23. If x is an integer and y = -2x-8, what is the least value of x for which y is less than 9 ?
(hw` x GKwU c~Y© msL¨v nq Ges y = -2x-8 nq, Zvn‡j x Gi me©wb¤œ †Kvb gv‡bi Rb¨ y Gi gvb 9 Gi †_‡K Kg
n‡e?) [Sonali Bank –(SO)-2018] & [BSC-Combined Exam –(Officer)-2018]+ [Rupali Bank Officer-
2019]
(A) - 9 (B) - 8 (C) - 7 (D) - 6 Ans: B
Solution:
y  9 Since y is less than 9
or, -2x-8 9 since y = -2x-8 [ y Gi gvb ewm‡q|]
or, -2x17
or, x  - 8.5 (- w`‡q ¸Y Ki‡j wPý D‡ë hvq|)
A_©vr x Gi gvb -8.5 Gi †_‡K eo c~Y© msL¨v = -8
24. A group of 7 members having a majority of boys is to be formed out of 7 boys and 4
girls. The number of ways the group can be formed is- (7 Rb evjK I 4 Rb evwjKv †_‡K
AwaKvsk evjK †i‡L 7 R‡bi `j MVb Kiv n‡e| KZ Dcv‡q `jwU MVb Kiv hv‡e?) [Rupali Bank Officer-2019]
(A) 80 (B) 100 (C) 295 (D) 110 Ans: C
Solution:
evjK‡ì msL¨v AwaKvsk A_© n‡jv 7 R‡bi `‡j evjKiv me mgq †ewk n‡e|
Zvn‡j evj‡Ki msL¨v 4, 5, 6 A_ev, 7 R‡bi mevB evjK n‡e| A_©vr 4 Gi †_‡K Kg bv n‡jB n‡jv|
Way no Boys Girls Group Total result
1 no way 4 3 7
C44
C3 = 354 140
2 no way 5 2 7
C54
C2 = 216 126
3 no way 6 1 7
C6  4
C1 = 74 28
7 0 7
C7 = 1 1
Total number of ways is 140+126+28+1= 295
5000
250
250 12.5+
gyjab
1g eQi
2q eQi cÖ_g eQ‡ii my‡ì my` 12.5 B n‡jv cv_©K¨
ïay Avm‡ji Dci my` 250+250 = 500 B n‡jv mij my`

1. The values of p for equation 2x2
-4x+p = 0 to have real roots is -[BSC-Combined Exam –
(Officer)-2018]
a. p ≤ - 2 b. p ≥ 2 c. p ≤ 2 d. p ≥ - 2 Ans: c
Solution:
To have one or more real roots, b2
-4ac ≥ 0 (GK ev GKvwaK real roots ‡ei Kivi wbqg b2
-4ac ≥ 0)
or, (-4)2
- 4×2×p ≥ 0 [GLv‡b, cÖ`Ë ivwk 2x2
-4x+p †Z, b = 4 xGi mnM, a= x2
Gi mnM 2 Ges c = p]
or, 16 - 8p ≥ 0
or, -8p ≥ -16
or, p ≤ 2
2. How many integers from 1 to 100 are divisible by 3 but not by 8? (1 ‡_‡K 100 Gi g‡a¨
KZ¸‡jv c~Y©msL¨v Av‡Q hv‡`i‡K 3 w`‡q fvM Kiv hvq wKš‘ 8 w`‡q fvM Kiv hvq bv?) [BSC-Combined Exam –
(Officer)-2018]
a. 30 b. 29 c.31 d.32 Ans: b
Solution:
Divisible by 3 =
3
100
= 33 integers
Divisible by 8&3, or 24 =
24
100
= 4 (j.mv,¸ w`‡q fvM Ki‡j me¸‡jv w`‡q fvM Kiv hvq|)
Divisible by 3 but not by 8 = 33-4 = 29.
evsjvq e¨vL¨v: 1 †_‡K 100 ch©šÍ 100 3 = 33wU msL¨v‡K 3 w`‡q fvM Kiv hvq| †hgb: 3, 6, 9, 12, 15,...24, .
. . 48, , ,72,,,96,,,99
wKš‘ GB 33 wU msL¨vi g‡a¨ wKQz msL¨v Av‡Q hv‡`i‡K 8 w`‡qI fvM Kiv hvq| cÖ‡kœ †h‡nZz 8 w`‡q fvM Kiv hvq bv
Ggb msL¨v wb‡Z ejv n‡q‡Q ZvB 8 w`‡q wefvR¨ msL¨v¸‡jv H 33 wU msL¨v †_‡K ev` w`‡Z n‡e|
3 I 8 Gi j.mv.¸ = 24 ( j.mv.¸ w`‡q hv‡K fvM Kiv hvq 3 I 8 w`‡qI Zv‡K fvM Kiv hvq)
GLb 3 I 8 Dfq msL¨v w`‡q wefvR¨ 100 ch©šÍ msL¨v¸‡jv n‡j 24, 48, 72 Ges 96 = 4wU|
myZivs 3 w`‡q wefvR¨ wKš‘ 8 w`‡q wefvR¨ bq Ggb msL¨v n‡e 33-4 = 29wU|
n‡e?) [Sonali Bank –(SO)-2018] & [BSC-Combined Exam –(Officer)-2018]
a. -9 b. -8 c. 38 d. 40 Ans: b
Bankers Selection Committee (BSC)-Combined Exam
Post name : Officer. (5 banks) Exam date: 20-07-2018
Exam taker: Arts Faculty, DU

cÖgvY + hyw³:
(1) x = -8 n‡j, y = (-2 -8) - 8 = 16-8 = 8
(2) x = -9 n‡j y = (-2-9) -8 = 18-8 = 10
(3) x = 40 n‡j y = (-240) -8 = -80-8 = -88
Solution:
or, -2x-8 9 since y = -2x-8 [ y Gi gvb ewm‡q|]
or, -2x17
4. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) = ? [Sonali Bank –(SO)-2018] & [BSC-Combined Exam
–(Officer)-2018]
a. 5 : 11 b. 6 : 5 c. 5:6 d. 3 : 8 Ans: a
Solution:
x : y = 5 : 3 or,
3
5
y
x
 or, 3x =5y
Now, (8x – 5y) : (8x + 5y) = (8x – 3x) : (8x + 3x) [Since 3x = 5y] = 5x:11x = 5:11
5. If a+
a
1
=2 what is a3
+ 3
a
1
? [BSC-Combined Exam –(Officer)-2018]
a. 1/2 b. 7 c.2 d.3/2 Ans: c
Solution:
a3
+ 3
a
1
= 












a
1
a
a
1
a3
a
1
a
3
.. = 23
-32 = 8-6 = 2
6. If 10% of x is equal to 25% of y, and y=16, what is the value of x? (x Gi 10% Ges y Gi
25% ci¯úi mgvb | hw` y =16 nq| Zvn‡j x Gi gvb KZ? ) [BSC-Combined Exam –(Officer)-2018]
a. 4 b. 6.4 c.24 d.40 Ans: d
Solution:
x×
100
10
= y×
100
25

4
y
10
x
 
4
16
10
x
 x = 40
gy‡L gy‡L:
‡k‡li w`K †_‡K 16 Gi 25% = 4 fv‡Mi 1 fvM A_©vr 4 GLb x Gi 10% ev 10 fv‡Mi 1 fvM = 4 n‡j m¤ú~Y© Ask ev
100% n‡e 4 Gi 10¸Y A_©vr 40|
7. If sinA + sin2
A = 1, then the value of the expression cos2
A + cos4
A is — [Rupali Bank Off-
(Cash)-2018] & [BSC-Combined Exam –(Officer)-2018]
a. 1
b.
2
1 c.2 d.3 Ans: a
Solution:
sinA + sin2
A = 1
sinA = 1-sin2
A

or, sinA = cos2
A
or, cos2
A = sinA or, cos4
A = sin2
A or, cos4
A = 1 – cos2
A So, cos2
A + cos4
A =1
8. A pole casts a m3 shadow on a ground on an elevation of 600
. The height of the pole is?
(GKwU LyuwU f~wg‡Z m~‡h©i mv‡_ 600
†KvY Drcbœ Ki‡j LuywUwUi ˆ`N©¨ KZ?) [BSC-Combined Exam –(Officer)-
2018]
a. 60° b. 45° c. 30° d. 90° Ans: a
Solution:
‡`qv Av‡Q,  = 60°, Ges f~wg/ Qvqv = 3 , LyuwUi ˆ`N©¨ = ?
awi, j¤^ = h
we know,
tan =
wg~f
^j¤
[f~wg †`qv Av‡Q, Ges j¤^ †ei Ki‡Z n‡e ZvB tan]
myZivs, tan60° =
3
h
ev, 3 =
3
h
h = 3
9. If the difference between the circumference and diameter of a circle is 90 cm ,then the
radius approximately is? (GKwU e„‡Ëi cwiwa I e¨v‡mi cv_©K¨ 90‡m.wg. n‡j H e„‡Ëi e¨vmva© KZ?) [BSC-
Combined Exam –(Officer)-2018]
a. 21cm b. 19 cm c.20 cm d.22 cm Ans: a
Solution:
Let the radius of the circle = r so the diameter is 2r and the circumference is 2r
Here,
2r - 2r = 90 or, 2r (-1) = 90 or, 





 1
7
22
r = 45 or, 




 

7
722
r = 45 or, r = 45 21
15
7

10. If the length of a side of a regular pentagon is 4 cm ,the area of the pentagon is
approximately- (GKwU mylg cÂf~‡Ri cÖwZwU evûi ‰`N©¨ 4 †m.wg. n‡j Zvi †ÿÎdj Gi m¤¢ve¨ cwigvY KZ?)
[BSC-Combined Exam –(Officer)-2018]
a. 25cm2
b. 27cm2
c.29cm2
d.32cm2 Ans: b
Solution:
Area of pentagon = 2
a5255
4
1
 )( [GUv m~Î]
= 2
451025
4
1
 )(
= 451025  )(
= 27.5 cm2
 27 cm2
B C
A
60
Pole = ?
Qvqv= 3
m~q©

11. The height of an equilateral triangle with a side 2 cm is (GKwU mgevû wÎf~‡Ri GKevûi ˆ`N©¨ 2
†m.wg. n‡j Zvi D”PZv KZ?) [BSC-Combined Exam –(Officer)-2018]
a. 3 cm b. 32 cm c. 23 m d. 5 cm Ans: a
Solution:
cv‡ki wPÎwU †`Lyb: mgevû wÎfzR ABC Gi ga¨gv AD  BC
GLb ABC mgevû wÎfz‡R, AB=BC=AC=2
BD = DC = 1 (‡h‡nZz D, BC evûi ga¨we›`y| )
GLb,
ADC mg‡KvYx wÎfz‡R
AD2
+DC2
= AC2
ev, AD2
+12
= 22
ev, AD= 3
Shortcut: Height of an equilateral triangle = a
2
3
(a = GK evû) = 2
2
3
= 3
12. The next number in the sequence 3, 4, 8, 17, 33,--- is? [BSC-Combined Exam –(Officer)-2018]
a. 54 b. 56 c.58 d.60 Ans: c
Solution:
Given 3, 4, 8, 17, 33
Differences: 1 4 9 16
Clues: 12
22
32
42
GLv‡b cieZ©x msL¨vwUi mv‡_ cv_©K¨ n‡e 52
= 25| myZivs msL¨vwU n‡e 33+25 = 58.
13. The second and third terms of a geometric series are 9 and 3 respectively, The fifth term
of the series is- ? (GKwU ¸‡YvËi avivi wØZxq I Z…Zxq c` h_vµ‡g 9 Ges 3 n‡j H avivwUi cÂgc` KZ?)
[BSC-Combined Exam –(Officer)-2018]
a. 1/9 b. 2/9 c.1/3 d.1/4 Ans: c
Solution:
avivwUi 2q c` = 9 Ges Z…Zxq c` = 3 myZivs avivwUi mvaviY AbycvZ q =
3
1
9
3
 (A_©vr cÖwZevi
3
1
w`‡q ¸Y )
avivwUi 4_© c` = 3 1
3
1
 myZivs avivwUi cÂg c` n‡e 1
3
1
=
3
1
Ans: c
wmwiRwU n‡e: 27, 9, 3, 1 †k‡l
3
1
A_©vr cÖwZevi 3 w`‡q fvM n‡”Q ,3 w`‡q fvM Kiv A_© hv,
3
1
w`‡q ¸Y KivI GK|
14. If two fair coins are flipped. What is the probability that one will come up heads and the
other tails? (`ywU gy`ªv wb‡ÿc Kiv n‡jv| GKwU‡Z †nW Ges Ab¨wU‡Z †Uj Avmvi m¤¢vebv KZ?) [Agrani Bank –
(Cash)-2017]& [BSC-Combined Exam –(Officer)-2018]
a.
2
1
b.
3
1
c.
4
1
d. 1 Ans: a
11
?
22
A
B D C

Solution:
total sample(4)= TT, HH, HT, TH.
So one heads and one tails =HT, & TH
So probability =
2
1
4
2

Confusion Clear: cÖ‡kœ hw` ejv _vK‡Zv First one is heads and next one is tails ZLb cÖ_gevi wbw`©ó
nIqvq DËi n‡Zv =
4
1
KviY ZLb cÖ_gevi †nW Qvov Ab¨ wKQz †bqv hv‡e bv|
wKš‘ cÖ_gevi bv e‡j GKwU‡Z †nW ejvq GLv‡b wbw`©ó K‡i †evSvq wb| ZvB †h †Kvb GKwU †nW Ges Ab¨wU †Uj wn‡m‡e
AbyK~‡j djvdj = 2wU| ZvB DËi:
2
1
4
2

15. 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 (1,3 Ges 5 wWwRU¸‡jv‡K eënvi K‡i hZ¸‡jv
msL¨v MVb Kiv hvq Zv‡ì g‡a¨ †_‡K ‰èfv‡e GKwU msL¨v wb‡j Zv 5 w`‡q wefvR¨ nIqvi m¤¢vebv KZ?) [BSC-
Combined Exam –(Officer)-2018]
a. 1/9 b. 2/9 c.1/3 d.1/4 Ans:
Solution:
Total number of 3 digit numbers that can be formed by 1, 3, 5 is = 333= 33
= 27 (g‡b ivL‡Z
n‡e, All possible A_© hZfv‡e msL¨v evbv‡bv hvq A_©vr G‡ÿ‡Î wWwRU wiwcU K‡i ˆZix nIqv msL¨v¸‡jv‡KI ai‡Z
n‡e|) 27 wU msL¨v = 135,115,315,351,551,555,333 Gfv‡e hZ¸‡jv evbv‡bv hvq me¸‡jv wb‡j 27wU n‡e|
Keeping 5 at the units position Fixed
Total number of 3 digit numbers that can be formed by 1, 3, 5 is = 33= 9 (GLv‡b ‡k‡li wWwRwU
5 wbw`©ó Ki‡j 5 †K 3 evi †bqv hvq bv, ZvB 5 ev‡` Ab¨ `ywU 3evi K‡i †bqv hvq ZvB 3 w`‡q ¸Y|)
Probability =
3
1
27
9

16. The line perpendicular to y=x-2 is ? [BSC-Combined Exam –(Officer)-2018]
a. y=2x+1 b.2y = -2x-5 c.2y=x+7 d.y=3x+1 Ans:
Solution:
Here, y = x-2
the slope of the line = 1 (KviY Slope ‡ei Kivi myÎvbymv‡i, y=mx+c n‡j y Gi mnM B Slope )
so, the slope of the line perpendicular to y = 1
1
1
 (j‡¤î Slope ‡ei Ki‡Z n‡j FYvZ¥K wecixZ
ivwk †ei Ki‡Z nq|)
GLb Ackb¸‡jvi g‡a¨ †h ivwki Slope Gi gvb (-1) n‡e Zv B DËi
Ackb b ‡Z cÖ`Ë, 2y = -2x-5 = y = -x -
2
5
nq †hLv‡b x Gi mnM ev Xvj (-1) ZvB GUvB DËi|

17. If logx
2
16
9
=
2
1
 the value of the base is -[BSC-Combined Exam –(Officer)-2018]
a. 16/9 b.9 /16 c.256/81 d.81/256 Ans: c
Solution:
logx
2
16
9
=
2
1
    2
1
2
x

=
16
9
 1
x =
16
9

x
1
=
16
9
x =
9
16
‡h‡nZz cÖ‡kœ value of base ‡ei Ki‡Z ejv n‡q‡Q, Avevi †em †`qv wQj logx
2
ZvB x2
=
2
9
16






=
81
256
18. n
C1+ n
C2+ n
C3+ …………….. .
n
Cn = ? [BSC-Combined Exam –(Officer)-2018]
a. 2n
b. 2n-1
c. d.2n
-1 Ans: d
Solution:
n
C1+ n
C2+ n
C3+ …………….. .
n
Cn = 2n
-1
cÖgvY:
suppose, n = 4
so, 4
C1 + 4
C2 + 4
C3 + 4
C4 = 4+6+4+1 = 15 Avevi = 24
-1 = 16-1 = 15
suppose, n = 5
so, 5
C1 + 5
C2 + 5
C3 + 5
C4 + 5
C5 = 31 Avevi = 25
– 1 = 32-1 = 31 A_©vr `y cv‡kB mgvb|
so, n
C1+ n
C2+ n
C3+ …………….. .
n
Cn = 2n
-1
Shortcut: GKwU myÎ Av‡Q, n
C0+ n
C1+ n
C2+ n
C3+ …………….. .
n
Cn = 2n
wKš‘ †h‡nZz cÖ`Ë cÖkœwU‡Z avivwUi ïiæ‡Z n
C0 ‡`qv †bB Avevi n
C0=1 ZvB m~‡Îi †hvMdj †_‡K 1 we‡qvM K‡i DËi:
n
C1+ n
C2+ n
C3+ …………….. .
n
Cn = 2n
-1
19. The solution of the inequality 23x-7  is-[BSC-Combined Exam –(Officer)-2018]
a. -3 x
3
5
b.3  x 
3
5
c.-3 x
2
5
d.-3 x -
3
5 Ans:
Solution:
The solution of the inequality |7-3x |  2
if (7-3x) is non negative, then,
7-3x  2 or, -3x  - 5 or, 3x  5 [-1 Øviv ¸Y Ki‡j wPý D‡ë hvq|]
x 
3
5
if (7-3x) is negative, then,
- (7-3x)  2
or, 7-3x  -2
or, -3x  -9
 x  3
Solution: 3  x 
3
5
(mgvavb †jLvi mgq x Gi Dfq cv‡k GKB wPý w`‡q mgvavb K‡i cÖvß `ywU DËi †_‡K
GKUv ûeû wjL‡Z nq Ges Ab¨Uv wecixZ cv‡k w`‡Z nq|)

20. The difference in take between simple and compound interest at 5% annually on sum of
TK 2000 after 2 years is- (kZKiv evwl©K 5 UvKv nvi my‡` 2000 UvKvi 2 eQ‡ii Pµe„w× my` Ges mij my‡ì
cv_©K¨ KZ UvKv?) [BSC-Combined Exam –(Officer)-2018]
a. 5 b. 50 c.20 d.200 Ans:
Solution:
Compound Interest = 2000
100
105
100
105
 = 2205Tk (myÎ cÖ‡qvM Ki‡jI 2205 B Avm‡e|)
Simple Interest: = 200
100
252000


Difference of interest = 205-200 = 5
Shortcut: mij my‡` ïay Avm‡ji Dci my` cvIqv hvq Avevi Pµe„w× my‡ì †ÿ‡Î my‡ì Dci AwZwi³ my` cvIqv
hvq| GB AwZwi³ my` B n‡”Q `y ai‡Yi my‡ì cv_©K¨|
myZivs cv_K¨© n‡e: 2000 Gi 5% = 100 UvKv GB my‡ì my` 100 Gi 5% = 5 UvKv| DËi: 5|
21. If 3x – 7y = 0 and x+2y =13 then y is -[BSC-Combined Exam –(Officer)-2018]
a. 2 b. 3 c.4 d.7 Ans: b
Solution:
If 3x-7y = 0 ---- (i), and x+2y = 13 ------ (ii) multiply (ii) by 3 we get
3x-7y = 0
3x+6y = 39
------------------
(-) -13y = - 39  y = 3
22. The sum of squares of 3 consecutive integers is less than 97,What is the greatest possible
value of the smallest one? (3 wU avivevwnK msL¨vi e‡M©i †hvMdj 97, G‡ì g‡a¨ ÿy`ªZg msL¨vwUi m‡e©v”P gvb
KZ n‡Z cv‡i?) [BSC-Combined Exam –(Officer)-2018]
a.4 b. 5 c.6 d.7 Ans: a
Solution:
22
+32
+42
= 4+9+16 = 29  97, 32
+42
+52
= 9+16+25 = 50  97,
42
+52
+62
= 16+25+36 = 77  97, 52
+62
+72
= 25+36+49 = 110  97,
‡`Lv hv‡”Q 4 w`‡q ïiæ n‡j cici wZbwU msL¨vi e‡M©i †hvMdj 97 Gi †_‡K †QvU Ges 97 Gi me‡_‡K KvQvKvwQ
n‡”Q| Gi †_‡K Kg wb‡q A_©vr 3 w`‡q ïiæ Ki‡jI ‡hvMdj 97 Gi †_‡K Kg nq, wKš‘ ‡m‡ÿ‡Î ÿy`ªZg msL¨vwUi
m‡e©v”P gvb Avm‡Q bv| Avevi 4 ev‡` 5 w`‡q ïiæ Ki‡j †hvMdj 97 Gi †_‡K †ewk n‡q hvq| ZvB DËi: 4|
=======================================

1. Shonghoti and Shouhardo Clubs consists of 200 and 270 members respectively. If the
total member of the two clubs is 420 then how many members belong to both clubs?
(msnwZ Ges †mŠnv`© K¬v‡ei h_vµ‡g 200 Ges 270 Rb m`m¨ Av‡Q| hw` Dfq K¬v‡ei me©‡gvU 420 Rb m`m¨ _v‡K
Zvn‡j KZRb m`m¨ Dfq K¬v‡ei mv‡_ hy³ Av‡Qb?)[BB-(AD)-2018]
a. 30 b.40 c. 50 d. 60 Ans: c
Solution:
Let, P(A) = 200 and P(B) =270
We know that, P(A)  P(B) = P(A) + P(B) – P(A  B)
or, 420 = 200 + 270 – P(A  B)
or, P(A  B) = 470-420
 P(A  B) = 50
GK jvB‡b Kivi Rb¨: 200+270 – 420 = 470- 420 = 50
2. The one third of the complementary angle to 600
is – (600
†Kv‡Yi c~iK †Kv‡Yi GK Z…Zxqvs‡ki gvb
KZ?) [BB-(AD)-2018]
a. 1500
b. 1000 c. 400
d. 100
Ans: d
Solution:
Complementary angle 600
is (900
– 600
) = 300
So, one third of 300
is 300

3
1
= 100
3. If the area of a rhombus is 96 sq. cm and the length of one of the diagonals is 16 cm.The
length of the other diagonal is (GKwU i¤^‡mi †ÿÎdj 96 eM© ‡m.wg. hw` i¤^mwUi GKwU K‡Y©i ‰`N©¨ 16
†m.wg. nq Zvn‡j Aci K‡Y©i ‰`N©¨ KZ?) [BB-(AD)-2018]
a. 18 b. 12 c. 9 d. 6 Ans: b
Solution:
Let ,length of other diagonal is = x (myÎ: i¤^‡mi †ÿÎdj =
2
1
 KY©Ø‡qi ¸Ydj)
Area of rhombus is
2
1
 x ×16= 96 or 8x = 96  x = 12
4. The ratio of two numbers is 3:4 and their sum is 630. The smaller one of the two
numbers is (`ywU msL¨vi AbycvZ 3:4| Zv‡`i †hvMdj 630 n‡j ÿy`ªZg msL¨vwU KZ?) [BB-(AD)-2018]
a. 360 b. 270 c. 180 d. 120 Ans:b
Solution:
If the number is 3x and 4x
then, 3x+ 4x = 630 or 7x = 630  x = 90 So the smaller number is 390 = 270
Bangladesh Bank
Post name: Assistant Director (General) Exam date: 06-07-2018
Exam taker : Arts Faculty, DU.
200 50 270
400

5. If 42x+1
=32, then x =? [BB-(AD)-2018]
a. 2 b. 3 c.
4
3
d.
3
4
Ans: c
Solution:
4x+1
=32
Or, 22(2x+1)
=25
or, 4x+2 = 5 or, 4x = 3 x =
4
3
6. What will be the difference between simple and compound interest at 10% on a sum of
Tk. 1000 after 4 years? (kZKiv evwl©K 10 UvKv nvi gybvdvq 1000 UvKvi 4 eQ‡ii mij my` I Pµe„w× my‡`i
cv_©K¨ KZ UvKv?) [BKB – (CASH ) -2017] & [BB-(AD)-2018]
a. Tk.31.90 b. Tk. 32.10 c. 44.90 d. Tk. 64.10 Ans: d
Solution:
Simple interest =
100
4101000 
= 400tk
Compound interest = (110% of 110% of 110% of 110% of 1000 ) -1000
Or, (1000×
100
110
×
100
110
×
100
110
×
100
110
) – 1000 = 1464.1-1000 = 464.1
So difference = 464.1- 400 = 64.1
gy‡L gy‡L:
cÖ_g eQ‡ii my` = 1000Gi 10% = 100 UvKv|
2q eQ‡ii my` = (1000+100) = 1100 Gi 10% = 110 UvKv|
Z…Zxq eQ‡ii my` = (1100+110) = 1210 UvKvi 10% 121UvKv|
4_© eQ‡ii my` (1210+121) = 1331 UvKvi 10% = 133.1 UvKv|
†gvU = 100+110+121+133.1 = 464.1
Zvn‡j cv_©K¨ = 464.1-400 = 64.1 UvKv|
cv‡ki wPÎwUi e¨vL¨v: ïay 1000 UvKv Avm‡ji Dci 4 eQ‡ii my` 400 UvKv B n‡”Q mij my`| Gici cÖ_g eQi
cvIqv 100 UvKv my‡`i Dci c‡ii 3 eQi cvIqv my` 10+10+10 = 30| Avevi 2q eQi cvIqv 100+10 = 110
UvKvi Dci c‡ii 2 eQi A_©vr 3q I 4_© eQ‡i my` 11+11 = 22 UvKv Avevi 3q eQ‡ii 100+10+11 = 121 UvKvi
Dc‡i 4_© eQ‡i cvIqv my` 12.1 UvKv|GLv‡b cÖ_g 400UvKv Avmj †_‡K cvIqv ZvB GUv mij my` Ges evKx¸‡jv
Pµe„w×i Kvi‡YB AwZwi³ cvIqv hvq e‡j cv_©K¨ 30+22+12.1 = 64.1 UvKv| (Pµe„w× my`400+64.1=464.1)
7. In a series of 6 consecutive odd numbers, If 15 is the 6th
number. what is the 4th
number
in the series? (6wU avivevwnK †e‡Rvo msL¨vi g‡a¨ 6ô msL¨vwU 15 n‡j 4_© msL¨vwU?) [BKB – (CASH ) -2017]
& [BB-(AD)-2018]
a. 7 b. 9 c. 11 d. 13 Ans: c
Solution:
The series is 5, 7, 9, 11, 13 and 15
So 4th
number is 11
1000
100
100
100
100
10
10
10
11
11 12.1
+
+
+
+
+
30 + 22 +12.1 = 64.1
Easiest way
1g
2q
3q
4_©

8. If x = ya
, y = zb
and z = xc
then the value of abc is- [BD House Building FC (SO)-2017] &
[Rupali Bank Off- (Cash)-2018] & [BB-(AD)-2018]
a. 1 b. 0 c. 0.5 d. Infinity Ans:a
Solution:
x = ya
Or, x=zab
(Since y= zb
) Or, x = xabc
(Since z= xc
) Or, xabc
=x1
Or, abc = 1
Alternative way, y = zb
or, y = xbc
or, y = yabc
or, abc = 1 (A_©vr †h ‡KvbUv a‡iB DËi 1 )
9. If 1+sin = xcos then tan is - ? [BB-(AD)-2018]
a.
x
1x2

b.
x
1x2

c.
x2
1x2

d.
x2
1x2
 Ans:d
Solution:
Given, 1+sin = xcos



cos
sinθ
cosθ
1
= x (cos w`‡q fvM K‡i|)
 sec + tan = x ------ (i)
Now, We know that, sec2
 - tan2
 = 1
(sec + tan ) (sec - tan) = 1
x(sec- tan) = 1 [from equation (i)]
or,(sec- tan) =
x
1
- - - - - (ii)
Again, sec + tan = x - - - - - (iii)
by (iii)-(ii)
2tan = x-
x
1
or, 2tan =
x
1x2

tan =
x2
1x2

10. The difference between two number is 5 and the difference between their squares is 65.
What is the larger number ? (`ywU msL¨vi cv_©K¨ 5 Ges Zv‡`i e‡M©i cv_©K¨ 65 n‡j e„nËg msL¨vwU KZ?)
[BD House Building FC (SO)-2017]+ [BDBL – (SO ) -2017] & [BB-(AD)-2018]
a. 13 b. 11 c. 8 d. 9 Ans: d
Solution:
let two number a and b
a-b = 5…. (i) here a > b
and. a2
-b2
= 65
or, (a+b)(a-b) = 65 or a+b =
5
65
=13 ---(ii)
by adding (i) and (ii) we get 2a = 18  a = 9 So, the larger number is 9
11. A 240 m long train passed a pole in 24 seceond. How long will it take to pass a 650 m
long platform ? (240 wgUvi j¤^v GKwU †Uªb 24 †m‡K‡Û GKRb gvbyl‡K AwZµg Ki‡Z cv‡i| H GKB †UªbwU
650wgUvi j¤^v GKwU cøvUdg© AwZµg Ki‡Z KZ mgq jvM‡e?) [BD House Building FC (SO)-2017]+ [BDBL –
(SO ) -2017] & [BB-(AD)-2018]
a. 65 sec b. 89 sec c. 100 sec d. 130 sec Ans: b

Solution:
train speed in 1 second is = 24024 = 10m/s
The train has to go = 240+650 = 890m
So Total time taken = 89010 = 89 seconds
[ g‡b ivLyb: hLb ‡UªbwU 650 wgUvi cøvUdg© AwZµg K‡i ZLb Zvi wb‡Ri ˆ`N©¨ 240 wg mn AwZµg K‡i|]
12. The slope of the line perpendicular to the line y = -5x+ 9 is –[Agrani Bank – (Cash)-2017] &
[Sonali Bank –(SO)-2018]
a. 5 b. -5 c.
5
1
d. -
5
1 Ans: c
Solution:
Since y = mx+c (Slope ‡ei Kivi m~Î)
Here slope m= -5 (x Gi mnM †h‡nZz -5 ZvB Zvi slope ev Xvj I -5 )
(A‡bK cÖ‡kœ hLb ïay slope ‡ei Ki‡Z ejv nq ZLb ïay y = mx+c Gfv‡e mgxKiY †K G‡b m Gi mnM Uv B
slope nq Ges †mUvB DËi| )
GLb perpendicular ev j¤^ †iLvi Slope ‡ei Kivi wbqg n‡jv slope Gi wecixZ fMœvsk ‡ei K‡i FYvZ¥K gvb
w`‡q ¸Y Kiv|
So, line perpendicular will have slope m = -
m
1
= -
5
1
5
1



Ans:
5
1
GKB iKg Av‡iKUv †`Lyb: The equation of a perpendicular line to y = 3x − 9
Must have a slope that is the negative reciprocal of the original slope. m = -
m
1
=
3
1

13. If
7
3
x
y
 = and x + 2y= 13 then y is- [BB-(AD)-2018]
a.2 b. 3 c. 4 d. 7 Ans: b
Solution:
7
3
x
y
 or, 3x = 7y or, or, 3x -7y = 0 -------- (i) and x + 2y= 13 --------(ii)
By, (ii)  1 & (ii)  3 we get
3x -7y = 0
3x + 6y = 39
----------------------------
-13y = -39 (we‡qvM K‡i|)
 y = 3
14. 1-3x  4, Then [BB-(AD)-2018]
a. x -2 b. x -2 c. x -1 d. x -1 Ans: c
Solution:
1-3x  4 or, -3x  3 or, -x  1 (3 w`‡q ¸Y K‡i) or, x  -1 (-1 Øviv ¸Y Ki‡j wPý D‡ë hvq|)
3x = 7y Gici x =
3
y7
c‡ii As‡k ewm‡q Ki‡jI n‡e|

15. A pole 6 m high casts a shadow 32 m long on the ground, then the Sun's elevation is ?
(GKwU LuywUi D”PZv 6 wg. Ges LuywUi Qvqvi ˆ`N©¨ 32 wgUvi n‡j m~‡h©i DbœwZ †KvY KZ?) [Rupali Bank Off-
(Cash)-2018] +[Sonali Bank –(SO)-2018] & [BB-(AD)-2018]
a. 60° b. 45° c. 30° d. 90° Ans: a
Solution:
Avgiv Rvwb, tan =
fzwg
^j¤
tan =
BC
AB
( cv‡ki wPÎ Abyhvqx j¤^ Ges f~wg †`qv Av‡Q ZvB)
or, tan =
32
6
=
2
3
= 3
3
33

.
or, tan = tan600
[Since tan60
= 3 ]
  = 600
(Ans)
16. If a, b and c are the lengths of the three sides of a triangle, then which of the following is
true? (hw` GKwU wÎfz‡Ri 3wU evû h_vµ‡g a,b Ges c nq Zvn‡j wb‡Pi †KvbwU mwVK) [Rupali Bank Off-
(Cash)-2018] & [BB-(AD)-2018]
a. a+b < c b. a-b <c c. a+b = c d. a+b  c Ans:b
Solution:
`ywU ¸iæZ¡c~Y© Abywm×všÍ:
 wÎfz‡Ri †h †Kvb `yB evûi mgwó Zvi Z…Zxq evû A‡cÿv e„nËi| Ges Gi wecix‡Z
 wÎfz‡Ri †h †Kvb `yB evûi AšÍi ev e¨eavb ev we‡qvMdj Z…Zxq evû A‡cÿv ÿz`ªZi|
2q wbqg Abymv‡i Ackb B. †Z cÖ`Ë a-b < c mwVK| A_©vr `ywU evû we‡qvM Ki‡j Zv Z…Zxq evû †_‡K †QvU n‡e|
17. A football team is to be consisted out of 14 boys. In how many ways the team can be
chosen so that the owner of the ball is always in the team? (14 Rb evj‡Ki ga¨ †_‡K GKwU
dzUej `j KZfv‡e MVb Kiv hv‡e †hLv‡b e‡ji gvwjK me©`v `‡ji g‡a¨ _vK‡e?) [BB-(AD)-2018]
a. 200 b. 201 c. 210 d. 286 Ans: d
Solution:
one is fixed So, total boys remain 14-1 = 13 boys
and the number of boys should be selected = 11-1 = 10
Total number of team = 13
C10=
123
111213


= 286
18. The next number of the sequence is? (wmwi‡Ri cieZ©x msL¨vwU KZ?) [BB-(AD)-2018]
4, 3, 9, 3, 19, 3………….?
a. 31 b. 32 c. 39 d. 49 Ans: c
Solution:
wmwiRwUi 2q, 4_©, 6ô Uvg©¸‡jv‡Z †Kej 3 Av‡Q|
A_©vr GLv‡b `ywU wfbœ wfbœ wmwiR Av‡Q †hgb: 4, 9 , 19 Ges 3 , 3 ,3
B C
A
60
Pole=6
Qvqv= 32
m~q©

3q ivwk -1g ivwk = 9-4 = 5 Avevi 5g ivwk – 3q ivwk = 19-9 = 10| A_©vr ivwk¸‡jvi gv‡S cv_©K¨ 5 Gi ci 10
Zvn‡j 7g I 5g ivwki cv_©K¨ n‡e 10 Gi wØ¸Y A_©vr 20 Ges ivwkwU n‡e 19+20 = 39|
19. Which of the following can be arranged into an English word? [BB-(AD)-2018]
a. ANSLAIT b. LSNIT c. OTATM d. WQRGS Ans: a
Solution:
ANSLAIT ‡K mvRv‡j nq LATINAS nq| hvi A_© A‡ckv`vi|
20. October 1985 corresponds to Bangla year - ? [BB-(AD)-2018]
a. 1392 b. 1391 c. 1394 d. 1390 Ans: a
Solution:
‡h †Kvb Bs‡iRx mvj †_‡K 593 we‡qvM Ki‡j H eQi evsjv KZ mvj wQj Zv †ei nq|
GLv‡b Bs‡iRx mvj 1985 - 593 = 1392 mvj|
21. if 21215120 represents ‘bloat’ then 6121135 represents? [BB-(AD)-2018]
a. voice b. bald c. flame d. castle Ans: c
Solution:
GLv‡b Bs‡iRx eY©gvjvi eY©µg‡K msL¨vq cÖKvk Kiv n‡q‡Q| GUv‡K g~jZ Coding ejv nq hv msL¨v I eY©gvjvi auvauv|
Given, 21215120 = ‘bloat’
Where, 2 = b, 12 = l, 15 = o, 1 = a and 20 = t (Bs‡iRx e‡Y©i wmwiqvj bv¤^vi)
Then, 6121135 = ?
Following the given code, 6 = f, 12 = l, 1 = a, 13 = m, and 5 = e
So, 6121135 = flame
22. What is the probability that an integer selected at random from those between 10 and
100 inclusive is a multiple of 5 or 9? (10 ‡_‡K 100 Gi ga¨ †_‡K (10 I 100 mn) ‡h †Kvb GKwU c~Y©
msL¨v ˆ`efv‡e wbe©vPb Ki‡j Zv 5 A_ev 9 Gi ¸wYZK nIqvi m¤¢vebv KZ?) [BB-(AD)-2018]
a.
89
27
b.
91
20
c.
91
27
d.
89
23 Ans: c
Solution:
Multiple of 5 from 10 to 100 = 19 numbers such as [ 10, 15, 20, .. . 45,….. 90, 95,100]
(kU©Kv‡U©, 1 †_‡K 100 ch©šÍ 1005 = 20wU msL¨v‡K 5 w`‡q fvM Kiv hvq wKš‘ GLv‡b ïiæ‡Z 5 ev` w`‡q †gvU 19 wU|)
Multiple of 9 from 10 to 100 = 10 numbers such as [ 18,27,36, 45,….. 90,99 ]
(kU©Kv‡U©: 1 †_‡K 100 ch©šÍ 9 w`‡q wefvR¨msL¨v 119 = 99 A_©vr 11wU| wKš‘ ïiæi 9 ev` w`‡q GLv‡b 10wU )
Total numbers from 10 to 100 = 100-10+1 = 91
Multiple of 5 is 19numbers and multiple of 9 is 10 numbers
But, 2 numbers are common such as 45 and 90 (Dfq †ÿ‡Î 45 I 90 _vKvq GB 2wU msL¨v ev` )
So, total multiple of 5 or 9 from 10 to 100 is 19+10-2 = 27 numbers

Probability of selecting an integer =
91
27
‡UKwbK: `ywU wfbœ msL¨vi GKB ¸wYZK †ei Kivi Rb¨ msL¨v `ywUi j.mv.¸ †ei K‡i wn‡me Kiv mnR|
‡hgb: GB cÖ‡kœ 5 I 9 Gi j.mv.¸ 45 Zvn‡j 10 †_‡K 100 Gi g‡a¨ 5 I 9 Gi mvavib ev Kgb ¸wYZK n‡”Q 45 Ges
45 Gi wØ¸Y 90| GB `ywU
23. What does make ‘you’ young? (you ‡K young Ki‡Z wK jv‡M?) [BB-(AD)-2018]
a. Adding 2 velars
c. Eating sweet fruits
b. Drinking energy beverage
d. Changing outfits
Ans: a
Solution:
velar A_© aŸwb| you k‡ãi mv‡_ n I g `ywU velar ev aŸwb ‡hvM K‡i young n‡e| ZvB DËi n‡e a.
24. The sum of 3 consecutive integers is less then 75. What is the greatest possible value of
the smallest one? (wZbwU avivevwnK c~Y© msL¨vi †hvMdj 75 †_‡K Kg n‡j ÿz`ªZg msL¨vwUi m‡ev©”P gvb KZ n‡Z
cv‡i?) [BB-(AD)-2018]
a. 16 b. 19 c. 22 d. 23 Ans: d
Solution:
let the smallest number be x
so, 2nd
number is x+1 and 3rd
number is x+2
ATQ,
x+ x+1 + x+ 2  75 (‡h‡nZz wZbwU msL¨vi †hvMdj 75 Gi †_‡K Kg|)
3x+3 75
3x 72
x  24
A_©vr ÿz`ªZg msL¨vwU n‡e 24 Gi †_‡K †QvU wKš‘ e„nËg| GLb Ack‡bi g‡a¨ 23 B n‡”Q 24 Gi †_‡K †QvU m‡e©v”P
msL¨v| ZvB DËi: 23
==================================
GB eB‡qi †h †Kvb cÖkœ eyS‡Z mgm¨v n‡j Qwe Zz‡j A_ev wj‡L †dmey‡K
†cv÷ Kiæb| Avgv‡`i G·cvU©iv Avcbv‡K eywS‡q w`‡Z mn‡hvwMZv Ki‡e|
Facebook group: Khairul’s Basic Math

1. If w is 10% less than x, and y is 30% less then z, than wy is what percent less then xz ?
(hw` w, x Gi †_‡K 10% †QvU Ges y, z Gi †_‡K 30% ‡QvU nq Zvn‡j wy, xz Gi †_‡K kZKiv KZ Kg?)
a. 10% b. 20% c. 37% d. 40% Ans: c
Solution:
if x = 100 then w = 90
if z = 100 then y = 70
so, xz = 100100 = 10000
and wy = 9070 = 6300
xz – wy = 10000-6300 = 3700
less % = %37
10000
1003700


n‡e?) [Sonali Bank –(SO)-2018]
a. -9 b. -8 c. 38 d. 40 Ans: b
Solution:
or, -2x-8 9 since y = -2x-8
or, -2x17
3. The next number in the sequence 3, 6, 11, 18, 27,--- is – (3, 6, 11, 18, 27,--- wmwiRwUi cieZ©x
msL¨v KZ?) [Sonali Bank –(SO)-2018]
a. 34 b. 38 c. -7 d. - 6 Ans: b
Solution:
Gap between the numbers is 3, 5, 7, 9 (difference of the gap is 9-7 = 2)
So next gap should be 9+2 = 11 and the number should be 27+11 = 38
4. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) = ? (Agrani Bank Ltd. Seni Offi-2013) & [Sonali Bank
–(SO)-2018]
a. 5 : 11 b. 6 : 5 c. 5:6 d. 3 : 8 Ans: a
Solution:
Sonali Bank Limited
Post name : Senior Officer. Exam date: 01-06-2018
Exam taker: Arts Faculty, DU.
Shortcut: 10% †QvU n‡j 100 †_‡K 90 nq | Gici 90 Gi
30% †QvU n‡j 27 ‡QvU nq Ges `ywU †QvU GKmv‡_ n‡q eo Gi
mv‡_ Zzjbv Ki‡j †gv‡Ui Dci †QvU nq 10%+27% = 37%
cÖgvY + hyw³:
(1) x = -8 n‡j, y = (-2 -8) - 8 = 16-8 = 8
(2) x = -9 n‡j y = (-2-9) -8 = 18-8 = 10
(3) x = 40 n‡j y = (-240) -8 = -80-8 = -88

x : y = 5 : 3 or,
3
5
y
x
 or, 3x=5y
Now, (8x – 5y) : (8x + 5y) =(8x – 3x) : (8x + 3x) [Since 3x=5y] = 5x:11x =5:11
5. The sum of first 17 terms of the series 5, 9, 13, 17…..( 5, 9, 13, 17 - -- wmwiRwUi cÖ_g 17wU
msL¨vi †hvMdj KZ?) [Sonali Bank –(SO)-2018]
a. 529 b. 462 c. 629 d. 523 Ans: c
Solution:
5+9+13+17 . . . . sum of first 17 terms?
Here first terms a = 5 , difference d = 9-5 = 4 and total terms n = 17
Sum = })({ d1na2
2
n
 = })({ 411752
2
17
 = )( 6410
2
17
 = 1737 = 629
gy‡L gy‡L cvivi Rb¨ e¨vwmK AvBwWqv jvM‡e: wmwi‡Ri ivwk¸‡jvi Mo  ivwki msL¨v = mgwó|
wmwiRwUi 17 Zg c` n‡e 5+16wU cv_©K¨ = 5+164 = 5+64 = 69
wmwiRi Mo
2
569
= 37| myZivs mgwó n‡e 3717 = 629|
6. If A = {1,2,3,4,5}, then the number of proper subsets of A is – (hw`, A = {1,2,3,4,5} nq
Zvn‡j A Gi cÖK…Z Dc‡mU KZwU?) [Sonali Bank –(SO)-2018]
a. 120 b. 30 c. 31 d. 32 Ans: c
Solution:
Subset = 2n
= Here n = number of elements.
And proper subset = 2n
-1 = 25
– 1 = 32-1 = 31
e¨vL¨v: Subset n‡jv Dcv`vb¸‡jv wb‡q hZ¸‡jv bZzb †mU MVb Kiv hvq|
Avi proper subset = me¸‡jv Dcv`vb wb‡q MwVZ Dc‡mU wU ev` w`‡q ‡h¸‡jv Dc‡mU _vK‡e †m¸‡jvB cÖK…Z
Dc‡mU|
A= {1,2,3,4,5} Gi GKwU Dc‡mU {12345} wKš‘ Bnv cÖK…Z Dc‡mU bq| GRb¨ m~ÎwU 2n
-1
7. How many terms of Arithmetic Progression (A.P) 21,18,15,12,… must be taken to give
the sum zero ? (21,18,15,12 wmwiRwUi KZwU msL¨vi †hvMdj 0 n‡e?) [Sonali Bank –(SO)-2018]
a. 10 b. 15 c. 22 d. 27 Ans: b
Solution:
0d1na2
2
n
 )((
or,
2
n
{221+(n-1) -3} = 0
or,
2
n
(42-3n+3) = 0
or, 45n-3n2
= 0
or, 15-n = 0  n = 15
wj‡L wj‡L `ªæZ mgvavb:
21+18+15+12+9+6+3+0+
(-3)+(-6)+(-9)+(-12)+(-15)+(-
18)+(-21) = 0 A_©vr 0 Gi AvM
ch©šÍ 7wU abvZ¥K msL¨v Ges 0 Gi ci
H 7wU msL¨vB FYvZ¥K Ges 0 mn †gvU
15wU msL¨vi †hvMdj 0 n‡e|
me‡_‡K `ªæZ mgvav‡bi Rb¨:
21 w`‡q ïiæ cÖwZevi 3 K‡i Kg‡j
0 †h‡Z ivwk 213 = 7wU| Zvn‡j
0 Gi ci FYvZ¥K I 7wU 7+7 =
14wUi mv‡_ 0 mn †gvU 15wU|

8. Which of the numbers below is not equivalent to 4%? (wb‡Pi †Kvb msL¨vwU 4% Gi mgvb bq?)
a. 1/25 b. 4/100 c. 0.40 d. 0.04 Ans:c
Solution:
4% =
100
4
or
25
1
or .04
wKš‘ 4% = .40 †jLv hvq bv KviY 0.40 Avi 0.4 GKB | Avevi % Gi Kvi‡Y wb‡P 100 Avm‡j `yÕNi Av‡M `kwgK
em‡e A_©vr 0.04 n‡e|
9. After being dropped, a certain ball always bounces back to
5
2
of the height of its
previous bounce. After the first bounce it reaches a height of 125 inches. How high (in
inches) will it reach after its fourth bounce? (GKwU ej cÖ_g bounce G hZUzKz D”PZvq D‡V cieZ©x
bounce G Zvi c~e©eZ©x bounce Gi
5
2
fvM D‡V| cÖ_g G 125 BwÂ D”PZvq DV‡j PZz_© bounce G KZUzKz DV‡e?)
(Mercantile Bank MTO 2013) & [Sonali Bank –(SO)-2018]
a. 20 b. 8 c. 5 d. 3.2 Ans: b
Solution:
The ball reaches after first bounce125 inches given
After 2nd
bounce= 125
5
2
=50
After 3rd
bounce = 50
5
2
=20
After 4th
bounce =20
5
2
= 8
KZ¸‡jv c~Y©msL¨v 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq?) [Agrani Bank – (Cash)-2017] + [BB-(officer)-
2018] & [Sonali Bank –(SO)-2018]
a. 29 b. 31 c. 32 d. 38 Ans: a
Solution:
Divisible by 30=
30
1000
= 33.33 = 33 Numbers (30,60,90-----990 = 33wU)
L.C.M of 30 and 16 = 240,
240
1000
= 4.
 GK jvB‡b Gfv‡e:
125Gi
5
2

5
2

5
2
= 8
(4 evi ejvq cÖ_gev‡ii 125 ev‡` 3 evi
5
2
)

evsjvq e¨vL¨v:
30 w`‡q wefvR¨ †h msL¨v¸‡jv Av‡Q †hgb: 30,,60,120,240 - - - - 990 Gi g‡a¨ wKQz msL¨v Av‡Q 16 w`‡q fvM
Kiv hvq|
Avevi 16 w`‡q fvM Kiv hvq Ggb msL¨v¸‡jv 16,32,48 ,240 Ggb A‡bK ¸‡jv msL¨vi g‡a¨ ïay†h¸‡jv 30 w`‡q fvM
Kiv hvq †m¸‡jvi mv‡_ †h¸‡jv wg‡j hv‡e Zv ev` w`‡Z n‡e| (KviY 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq|)
wK Ki‡j 16 I 30 Gi wgjb ¯’j Lyu‡R cvIqv hv‡e? Aek¨B j.mv,¸ Ki‡j| ZvB j.mv.¸ 240 †ei K‡i 240 w`‡q
30 Gi wmwiqv‡ji †h 4wU msL¨v ( 240,480,720,960) †K fvM Kiv hvq †m¸‡jv ev` w`‡jB 30 w`‡q wefvR¨ wKš‘ 16
w`‡q wefvR¨ bq Ggb msL¨v¸‡jv †ei n‡e|
11. The slope of the line perpendicular to the line y = -5x+ 9 is –[Agrani Bank – (Cash)-2017] &
a. 5 b. -5 c.
5
1
d. -
5
1 Ans: c
Solution:
w`‡q ¸Y Kiv|
m
1
= -
5
1
5
1



Ans:
5
1
must have a slope that is the negative reciprocal of the original slope. m = -
m
1
=
3
1

12. If m and p are positive integers and (m+p)m is even, which of the following must be true
? (hw` m Ges p `ywU abvZ¥K c~Y© msL¨v nq Ges (m+p)m Gi gvb †Rvo nq Zvn‡j wb‡Pi †KvbwU Aek¨B mZ¨?)
a. If m is odd, then p is odd
c. If m is even, then p is even
b. If m is odd, then p is even
d. If m is even, then p is odd
Ans: a
Solution:
(m+p)m = even
GLv‡b `yÕ‡Uv cvU© Av‡Q, (m+p) Ges m Ges `yÕ‡Uv ¸Y K‡i ¸Ydj = even ev †Rvo n‡q‡Q|
GLb, `ywU ivwki ¸Ydj †Rvo n‡Z n‡j `y‡Uv ivwkB †Rvo A_ev Zv‡`i AšÍZ GKwU †Rvo n‡Z n‡e|
ïw× cixÿv:
a.If m is odd, then p is odd, = evB‡ii m we‡Rvo n‡j (m+p) †Rvo n‡Z n‡e| ( bvn‡j (m+p)m = even
n‡e bv|) GLb (m+p) Gi g‡a¨I m ‡h‡nZz we‡Rvo ZvB m+p ‡K †Rvo evbv‡Z n‡j p= Aek¨B we‡Rvo n‡Z n‡e|
cÖkœvbyhvqx GUvB mwVK DËi|

b. If m is odd, then p is even = m = we‡Rvo n‡j †fZ‡ii (m+p) = Aek¨B ‡Rvo n‡Z n‡e| bvn‡j ¸Ydj
†Rvo n‡e bv| GLb , (m+p) = †Rvo n‡j Ges m = we‡Rvo nIqvq (m+p) ‡K †Rvo evbv‡bvi Rb¨ ïay p ‡K
we‡Rvo n‡Z n‡e| wKš‘ Ack‡b p is even ejvq Zv fzj| †hgb: m=3 Ges p = 2 n‡j (m+p)m = (3+2)3 =
15 = we‡Rvo| hv mwVK bq|
c. If m is even, then p is even = m = ‡Rvo n‡j p = ‡Rvo ev we‡Rvo †h †Kvb wKQz n‡jI ¸Ydj †Rvo B
n‡e| KviY `ywUi g‡a¨ Avgv‡ì 1wU †Rvo cÖ‡qvRb| cÖgvY: m=2 Ges p = 4 n‡j (m+p)m = (2+4)2 = 12 =
‡Rvo| Avevi m=2 Ges p = 3 n‡jI (m+p)m = (2+3)2 = 10 = ‡Rvo| A_©vr `yfv‡eB †Rvo n‡”Q| ZvB
GwUI Must n‡”Q bv|
d. If m is even, then p is odd = m = ‡Rvo n‡j c Gi e¨vL¨vi gZB p = ‡Rvo ev we‡Rvo †h †Kvb wKQzB n‡Z
cv‡i ZeyI Zv‡ì ¸Ydj †Rvo B n‡e| GRb¨ GUvI ‡bqv hv‡e bv|
Logic::
G ai‡Yi cÖkœ¸‡jv‡Z †h jwRKwU me‡_‡K †ewk Kv‡R jv‡M, ¸Y Kivi mgq †h †Kvb GKwU msL¨v †Rvo n‡j Ab¨
msL¨vwU ‡Rvo ev we‡Rvo hv B †nvK bv †Kb ¸YdjwU †Rvo n‡e| †hgb: 2x ev 32x G¸‡jv memgq B †Rvo| G‡ÿ‡Î x
= ‡Rvo ev we‡Rvo hv B †nvK| wKš‘ 3x ev, 31x = †Rvo bvwK we‡Rvo Zv wbf©i K‡i x Gi gv‡bi Dci x = ‡Rvo n‡j
3x ev, 31x Df‡q †Rvo| wKš‘ x = we‡Rvo n‡j 3x ev, 31x Df‡q we‡Rvo|
GB e¨vwmK AvBwWqv¸‡jv _vK‡j †h †Kvb gvb ewm‡qB wnme Kiæb bv †Kb DËi mn‡R wg‡j hv‡e|
(Cash)-2018] & [Sonali Bank –(SO)-2018]
a. 60° b. 45° c. 30° d. 90° Ans: a
Solution:
Avgiv Rvwb, tan =
fzwg
^j¤
tan =
BC
AB
( cv‡ki wPÎ Abyhvqx j¤^ Ges f~wg †`qv Av‡Q ZvB)
or, tan =
32
6
=
2
3
= 3
3
33

.
or, tan = tan600
[Since tan60
= 3 ]
  = 600
(Ans)
14. All possible three digit numbers are formed by 1, 2, 3. If one number is chosen
randomly, the probability that it would be divisible by 111 is (1,2 Ges 3 ‡K eënvi K‡i 3
A‡¼i hZ¸‡jv msL¨vi MVb Kiv hvq Zv‡ì g‡a¨ †h †Kvb GKwU msL¨v‡K wb‡j Zv 111 w`‡q wefvR¨ nIqvi m¤¢vebv KZ?
) [Sonali Bank –(SO)-2018]
a. 0 b.
9
2
c.
3
1
d.
4
1
Ans:
B C
A
60
Pole=6
Qvqv= 32
m~q©

Solution: (no answers)
Three digit numbers by using 1,2 and 3 is 33
= 27
Multiple of 111 in this number is = (111). (222) and (333) = 3
So, the probability is
27
3
=
9
1
Confusion clear : cÖ‡kœi g‡a¨ All possible ejvq 1,2,3 w`‡q MVb Kiv hvq Ggb me¸‡jv msL¨vB wb‡Z
n‡e| A_©vr wiwcU Ki‡jI mgm¨v †bB| wKš‘ hw` ejv n‡Zv cÖwZwU msL¨v GKevi gvÎ e¨envi K‡i MwVZ msL¨v ZLb †gvU
msL¨v n‡Zv 3! = 6 wU hvi g‡a¨ †Kvb msL¨vB 111 w`‡q wefvR¨ bv nIqvq DËi: 0 n‡Zv| Ack‡bi mv‡_ DËi
†gjv‡bvi Rb¨ 0 DËi w`‡jI cÖK…Zc‡ÿ DËi n†e
9
1
15. If sec + tan = x, then tan is? [Rupali Bank Off- (Cash)-2018] & [Sonali Bank –(SO)-2018]
a.
x
1x2

b.
x
1x2

c.
x2
1x2

d.
x2
1x2
 Ans: d
Solution:
We know that, sec2
 - tan2
 = 1
(sec + tan ) (sec - tan) = 1
 x(sec- tan) = 1
x
1
- - - - - (i)
again, sec + tan = x - - - - - (ii)
by (ii)-(i)
2tan = x-
x
1
or, 2tan =
x
1x2

tan =
x2
1x2

16. The area of a triangle with sides 3 cm, 5 cm. and 6 cm. is- (GKwU wÎfz‡Ri wZbwU evû 3 †m.wg 5
†m.wg Ges 6 †m.wg. n‡j wÎfzRwUi †ÿÎdj KZ?) [Sonali Bank –(SO)-2018]
a. 2 3 cm2
b. 2 14 cm2
c. 5 12 cm2
d. 4 14 cm2
Ans: b
Solution:
cimxgv 2S = 3+5+6=14  S = 7
2
14
 Rules: area = ))()(( csbsass 
Area of the triangle = ))()(( 6757377  = 1421441247 
17. The value of k, if (x-1) is a factor of 4x3
+3x2
- 4x+ k, is - [Sonali Bank –(SO)-2018]
a. 1 b. 2 c. -3 d. 3 Ans: c
Solution:
(x-1) is a factor of 4x3
+3x2
- 4x+ k
So, x-1 = 0 or x = 1 (Drcv`K n‡j Gfv‡e Drcv`K = 0 wjL‡Z nq|)
Now,
4x3
+3x2
- 4x+ k = 0 (KviY x=1 emv‡j m¤ú~Y© ivwkwUi gvb 0 n‡e|)
 4.13
+3.12
-4.1+k=0 4+3-4+k = 0 k = -3

18. If the radius of cylinder is halved (A‡a©K) and height is doubled, the what will be the
curved surface area ? (GKwU †ej‡bi e¨vmva© A‡a©K K‡i Zvi D”PZv wØ¸Y Ki‡j H †ej‡bi eµZ‡ji †ÿÎdj
wKiƒc n‡e?) [Sonali Bank –(SO)-2018]
a. increase by 1 b. the same c. double d. triple Ans: b
Solution:
Rules: Curved Surface Area of a cylinder = 2rh
Let, radius = 2r and height = h (2r aij A‡a©K I wØ¸Y Kiv mnR n‡e Ges fMœvsk Avm‡e bv|)
So, Curved Surface Area of the cylinder = 2.2r.h = 4rh
New radius = 2r2 = r and new height = h2 = 2h
So, new Curved Surface Area of the cylinder = 2..r2h = 4rh
Both time the area is same.
[Logic: ‡ej‡bi eµc„‡ôi †ÿÎd‡ji 2rh Gi r I h Gi †h †Kvb GKUv wØ¸Y Ges Ab¨Uv A‡a©K n‡j Dfq‡ÿ‡ÎB
†ÿÎdj mgvb n‡e| KviY GKUv hLb evo‡Q ZLb Ab¨Uv Kg‡Q|]
19. If a+2b = 6 and ab = 4 , what is
a
2
+
b
1
? [Sonali Bank –(SO)-2018]
a. 2
1
b. 1 c.
2
3
d. 2 Ans: c
Solution:
a
2
+
b
1
=
ab
ab2 
=
ab
b2a 
=
4
6
=
2
3
20. The number of parallelograms (mvgšÍwiK) that can be formed from a set of four parallel
(mgvšÍivj) lines intersecting (†Q`) another set of three parallel lines is (PviwU mgvšÍivj †iLv Ab¨
wZbwU mgvšÍivj †iLv‡K †Q` Ki‡j KZ¸wj mvgšÍwiK ‰Zwi n‡e?) (Pubali Bank Ltd. SO 2013)+[Sonali Bank So-
(2018) ]
a. 6 b. 9 c. 12 d. 18 Ans:d
Solution:
wb‡¤œv³ wPÎwU‡Z PviwU mgvšÍivj †iLv Aci wZbwU mgvšÍivj †iLv‡K †Q` Kivq †gvU 18wU mvgvšÍwiK ‰Zix n‡q‡Q|
Super Shortcut: cvkvcvwk (1+2) Dc‡i wb‡Pi (1+2+3) = 36 = 18 wU|
K‡qKwU wÎfzR Ges PZzf~©R wg‡j A‡bK K‡qKwU bZzb bZzb wÎf~R Ges PZzf~©R MwVZ n‡j Zv wKfv‡e Lye mn‡R †ei Ki‡Z
nq G wel‡q we¯ÍvwiZ Av‡jvPbv cv‡eb: Khairul’s Mental Ability eB‡q| wb‡R c‡o wb‡RB wkLyb|
1 †_‡K 6 ch©šÍ = 6wU h_v: 1, 2, 3, 4, 5 I 6
2 A¼ w`‡q 7wU = (1+2), (3+4), (5+6), (1+3), (3+5), (2+4) I
(4+6)
3 A¼ w`‡q =2wU| h_v: (1+3+5), (2+4+6)
4 A¼ w`‡q 2wU h_v: (1+2+3+4) Ges (3+4+5+6)
Ges me¸‡jv msL¨v e¨venvi K‡i 1wU = ( 1+2+3+4+5+6)
me©‡gvU mvgšÍwiK 6+7+2+2+1 =18wU| (Mfxifv‡e bv fve‡j fzj n‡Z cv‡i)
1 2
3
5 6
4
21
1
2
3

21. There are 5 red and 3 black balls in a bag. probability of drawing a black ball is (GKwU
e¨v‡M 5wU jvj Ges 3wU Kv‡jv ej Av‡Q| Zv †_‡K GKwU ej wb‡j Zv Kv‡jv nIqvi m¤¢vebv KZ?) [Sonali Bank –
(SO)-2018]
a.
8
5
b.
2
1
c.
8
3
d.
4
1
Ans: c
Solution:
Total balls = 5+3 = 8 and Black balls = 3 So probability of taking black ball =
8
3
22. Find the largest fraction from the following : (wb‡Pi fMœvsk¸‡jv †_‡K e„nËg fMœvskwU †ei Kiæb|)
a.
11
5
 b.
13
8
 c.
19
7
 d.
97
15
 Ans: d
Solution:
‡h‡nZz cÖwZwU fMœvs‡ki c~‡e© ( - ) we‡qvM wPý Av‡Q Ges e„nËg fMœvsk †ei Ki‡Z ejv n‡q‡Q ZvB ÿz`ªZg fMœvskwUi
†ei Ki‡j B Zv FYvZ¥K wn‡m‡e e„nËg n‡q hv‡e|
a.
11
5
 = 11 Gi A‡a©K 5.5 wKš‘ Dc‡i Av‡Q 5 A_©vr A‡a©‡Ki †_‡K GKUz Kg|
b.
13
8
 = 13 Gi A‡a©K 6.5 wKš‘ Av‡Q 8 A_©vr A‡a©‡Ki †_‡K GKUz †ewk|
c.
19
7
 = 19 Gi A‡a©K 9.5 wKš‘ 7 _vKvq Zv A‡a©‡Ki GKUz Kg|
d.
97
15
 = 97 Gi 6 fv‡Mi 1 fvM = 15 Gi KvQvKvwQ A_©vr 16 Gi GKUz †ewk | Av‡Mi ¸‡jvi †_‡K GUv A‡bK
Kg| KviY Av‡Mi ¸‡jv 0.5 Gi Av‡k cv‡kB wQj| Avevi ïiæ‡Z we‡qvM wPý _vKvq GB fMœvskwUB e„nËg|
mvaviY wbq‡g Ki‡Z †M‡j AvovAvwo ¸Y K‡i †ei Ki‡Z n‡e| G‡Z A‡bK †ewk mgq jvM‡e|
23. If x+
x
1
= 3, then x -
x
1
= ? [Sonali Bank –(SO)-2018]
a. 5 b. 13 c. 7 d. 0 Ans: a
Solution:
x
1
x4
x
1
x
x
1
x
22
..











 
2
x
1
x 





 = 32
-4 
x
1
x  = 5
24. The factors of x2
- 5x - 6 are: (x2
- 5x – 6 Gi Drcv`K KZ?) [Sonali Bank –(SO)-2018]
a. (x – 6)(x + 1) b. (x + 6)(x - 1) c. (x – 3)(x + 2) d. (x – 3)(x - 2) Ans: a
Solution:
x2
- 5x – 6 = x2
-6x+x-6 = x(x-6) +1 (x-6) = (x-6)(x+1)
==================================

1. How many integers from 1 to 1000 are divisible by 30 but not by 16?(1 ‡_‡K 1000 ch©šÍ
KZ¸‡jv c~Y©msL¨v 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq?) [Agrani Bank – (Cash)-2017] & [BB-(officer)-
2018]
a. 29 b. 31 c. 32 d. 38 Ans: a
Solution:
Divisible by 30=
30
1000
= 33.33 = 33 Numbers (30,60,90 ----- 990 = 33wU)
L.C.M of 30 and 16 = 240,
240
1000
= 4
evsjvq e¨vL¨v:
30 w`‡q wefvR¨ †h msL¨v¸‡jv Av‡Q †hgb: 30,,60,120,240 - - - - 990 Gi g‡a¨ wKQz msL¨v Av‡Q 16 w`‡q fvM
Kiv hvq| Avevi 16 w`‡q fvM Kiv hvq Ggb msL¨v¸‡jv 16,32,48 ,240 Ggb A‡bK ¸‡jv msL¨vi g‡a¨ ïay†h¸‡jv
30 w`‡q fvM Kiv hvq †m¸‡jvi mv‡_ †h¸‡jv wg‡j hv‡e Zv ev` w`‡Z n‡e| (KviY 30 w`‡q wefvR¨ wKš‘ 16 w`‡q
wefvR¨ bq|)
wK Ki‡j 16 I 30 Gi wgjb ¯’j Lyu‡R cvIqv hv‡e? Aek¨B j.mv,¸ Ki‡j| ZvB j.mv.¸ 240 †ei K‡i 240 w`‡q
30 Gi wmwiqv‡ji †h 4wU msL¨v ( 240,480,720,960) †K fvM Kiv hvq †m¸‡jv ev` w`‡jB 30 w`‡q wefvR¨ wKš‘ 16
w`‡q wefvR¨ bq Ggb msL¨v¸‡jv †ei n‡e|
2. If * is defined for all positive real numbers a and b by a * b = ab/(a+b) then 10*2 ? (hw` *
w`‡q me abvZ¥K msL¨v‡K †evSvq Ges a I b w`‡q a * b = ab/(a+b) †evSvq Zvn‡j 10*2 w`‡q wK †evSv‡e?)-
[BB-(officer)-2018]
a. . 5/3 b. 5/2 c. 5 d. 20/3 Ans:a
Solution:
Given a*b =
ba
ab

(Dc‡i ¸Ydj Ges wb‡P †hvMdj) So,10*2 =
210
210


=
3
5
3. 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?(hw` GKwU AvqZ‡ÿ‡Îi ˆ`N©¨‡K 20% e„w× Kivi ci Zvi cÖ¯’‡K 20%
n«vm Kiv‡bv nq Zvn‡j AvqZ‡ÿÎwUi †ÿÎdj kZKiv wK iKg cwieZ©b n‡e?) [Agrani Bank – (Cash)-2017] &
[BB-(officer)-2018]
a. unchanged b.decreases by 4% c. increases by 4% d. increases by 5% Ans: b
Solution:
Let the length = 100 and the width 100.
So area = 100 100 = 10000
Bangladesh Bank
Post name: Officer (General) Exam date: 27-04-2018

New length = 120 and new width = 80
New area = 12080 = 9600
Area decreased = 10000 – 9600 = 400 Decrease % =
10000
100400 
= 4%
 gy‡L gy‡L Kivi Rb¨ : cÖ_‡g 100 †_‡K 20% evov‡j n‡e 120 Gici 120 ‡_‡K 120 Gi 20% = 24 Kgv‡j n‡e
120-24 = 96| †gv‡Ui Dci Kg‡jv 100-96 = 4%|
myÎ cÖ‡qv‡M: Shortcut: 20 - 20 +
 
100
2020 
= - 4
4. The dimensions of a box are 2, 3 and 4 meters. The cost of Painting the outer sides of the
box, at the rate of Tk. 3 per square meter is? (GKwU e‡·i wZbwU evû h_vµ‡g 2, 3 Ges 4 wgUvi|
e·wUi evB‡ii Ask cÖwZ eM© wgUv‡i 3UvKv nv‡i is Ki‡Z ‡gvU KZ UvKv LiP n‡e?) [Agrani Bank – (Cash)-
2017] & [BB-(officer)-2018]
a. Tk.156 b. Tk. 220 c. Tk. 136 d. Tk. 160 Ans:a
Solution:
Total surface area 2 (ab+bc+ca) (evB‡ii †ÿÎdj †ei Kivi myÎ)
2(23+34+42) = 2(6+12+8) = 52 Sq. meter
Cost = 523 = Tk.156
g‡b ivLyb: is Kivi Rb¨ ïay evB‡ii As‡ki †ÿÎdj is Ki‡Z n‡e|
5. In each expression below, N represents a negative integer. Which expression could have
a negative value? (wb‡Pi expression ¸‡jvi g‡a¨ N n‡”Q GKwU FbvZ¥K c~Y© msL¨v| wb‡Pi †Kvb expression
gvb FbvZ¥K n‡Z cv‡i? ) [Agrani Bank – (Cash)-2017] & [BB-(officer)-2018]
a. N2
b. 6-N c. -N d. 6+N Ans: d
Solution:
a) N2
= Positive number KviY abvZ¥K ev FYvZ¥K †h †Kvb msL¨vi eM© memgq abvZ¥K B nq|
b) 6 - N= Positive number KviY - (negative integer) A_©vr FYvZ¥K msL¨v‡K Avevi (-) w`‡q ¸Y Ki‡j
Zv abZ¥K n‡q hvq Ges 6 Gi mv‡_ †hvM Ki‡j Zv Avevi aYvZ¥K B _v‡K|
c) - N = Positive number ‡h‡nZz ïay N GKwU FYvZ¥K msL¨v|
d) 6 + N = Positive number / Negative number (abvZ¥K Ges FYvZ¥K †hvM Ki‡j Zv abvZ¥K ev FYvZ¥K
nq|) ‡hgb: N = -2 n‡j 6 + N = 6 + (-2) = 4 A_©vr FYvZ¥K msL¨vwU 6 Gi †_‡K †QvU n‡j 6 + N abvZ¥K|
wKš‘ N = -8 n‡j 6 + N = 6 + (-8) = -2 A_©vr FYvZ¥K msL¨vwU 6 Gi †_‡K eo n‡j 6 + N = FYvZ¥K n‡Z
cv‡i|
Zvn‡j †`Lv hv‡”Q Dc‡ii 3wU Ackb †Kvbfv‡eB FbvZ¥K nq bv, wKš‘ †k‡li AckbwU FYvZ¥K n‡Z cv‡i| ZvB GUvB
DËi|
civgk©: N = -4 A_ev -8 A_ev Ab¨ †h †Kvb FYvZ¥K msL¨v a‡i Ackb¸‡jv cÖgvY Kiv hvq|
6. 1f the length of rectangle A is one-half the length of rectangle B, and the width of
rectangle A is one-half the width of rectangle B, what is the ratio of the area of rectangle
A to the area of rectangle B? (AvqZ‡ÿÎ A Gi ˆ`N©¨ AvqZ‡ÿÎ B Gi A‡a©K Avevi AvqZ‡ÿÎ A Gi cÖ¯’
AvqZ‡ÿÎ B Gi A‡a©K n‡j A Gi †ÿÎdj I B Gi †ÿÎd‡ji AbycvZ KZ?)- [BB-(officer)-2018]
a.
4
1
b.
2
1
c.
1
1
d.
1
2
Ans: a
4m
3m
2m

Solution:
Let, Length of A = x  Length of B = 2x
Width of A = y  Width of B = 2y
Area of A = xy And area of B = 2x2y = 4xy
So ratio of A:B = xy :4xy = 1:4 or
4
1
Shortcut: GKevi A‡a©K Kivi ci Av‡iKevi A‡a©K Ki‡j †gv‡Ui Dci 4 fv‡Mi 1 fvM ev GK PZz_©vsk n‡q hvq|
7. Log 36 / Log 6 --- ? [Agrani Bank – (Cash)-2017] & [BB-(officer)-2018]
a. 5 b. 8 c. 3 d. 2 Ans: d
Solution:
2
6
62
6
6
6
36 2

log
log
log
log
log
log
[mivmwi KvU‡j 6 Avm‡e †hUv fz„j| KviY Av‡M myÎ ‡d‡j Zvici KvU‡Z n‡e|]
8. If 1 -2x  3, then [Agrani Bank – (Cash)-2017] & [BB-(officer)-2018]
a. x  -2 b. x  - 2 c. x  -1 d. x  - 1 Ans: d
Solution:
1-2x  3
 - 2x  3 -1
 - 2x  2  x  - 1 (Dfq cÿ‡K FYvZ¥K msL¨v w`‡q ¸Y Ki‡j gv‡Si wPýwU D‡ë hvq|)
9. If sinA + sin2
A + cos4
A is [Rupali Bank Off-
(Cash)-2018] & [BB-(officer)-2018]
a. 1 b. 1/2 c. 2 d. 3 Ans: a
Solution:
sinA + sin2
A = 1
sinA = 1-sin2
A
or, sinA = cos2
A [Since, 1-sin2
A = cos2
A (myÎ)]
or, cos2
A = sinA
or, cos4
A = sin2
A (eM© K‡i) or, cos4
A = 1 – cos2
A [ 1- cos2
A=sin2
A (myÎ)] So, cos2
A + cos4
=1
10. A median of a triangle divides it into two (GKwU ga¨gv wÎfzRwU‡K wef³ K‡i-?)[Rupali Bank Off-
(Cash)-2018] & [BB-(officer)-2018]
a. Congruent triangles c. Isosceles triangles Ans: b
b . triangles of equal area d. right angle triangles
Solution:
‡h †Kvb wÎfz‡Ri ga¨gv wÎfzR †ÿÎwU‡K mgvb †ÿÎdj wewkó `ywU bZzb wÎfz‡R wef³ K‡i| G‡ÿ‡Î Zv mg‡KvYx ev
mgwØevû nIqvUv ¸iæZ¡c~Y© bq| ZvB DËi: B Triangles of equal area.
11. Which number logically follows the sequence? (wmwi‡Ri cieZ©x msL¨vwU KZ?) [BB-(officer)-
2018] 4, 6, 9, 6, 14, 6………….?
a. 6 b. 17 c. 9 d. 19 Ans: d

Solution:
wmwiRwUi 2q, 4_©, 6ô Uvg©¸‡jv‡Z †Kej 6 Av‡Q|
A_©vr GLv‡b `ywU wfbœ wfbœ wmwiR Av‡Q †hgb: 4, 9 , 14 Ges 6 , 6 ,6
1g Uvg© †_‡K 3q Uvg© Ges 3q Uvg© †_‡K 5g Uv‡g©i cv_©K¨ = 5
‡h‡nZz me©‡kl msL¨v 6 †`qv Av‡Q Zvn‡j †k‡li msL¨vwU n‡e 14+ 5 = 19
12.How many cases (e·) do you need if you have to pack 112 pairs of shoes into cases that
each hold 28 shoes? (112 †Rvov my ivLvi Rb¨ cÖwZ e‡· 28 wU my aviY Ki‡Z cv‡i Ggb KZwU e· jvM‡e? )
[BB-(officer)-2018]
a. 8 b. 10 c. 12 d. 14 Ans: a
Solution:
2 shoes = 1 pair
28 shoes =
2
28
=14 pairs. (112 †Rvov‡K 112 wU a‡i wn‡me Ki‡j fzj n‡e|)
 Cases needs =
14
112
= 8
13. In a row in the theatre the seats are numbered consecutively from T1 to T50. Sumon is
sitting in seat T17 and Shajib is sitting in seat T39. How many seats are there between
them? (GKwU w_‡qUv‡ii GK jvB‡bi Avmb¸‡jv‡K T1 ‡_‡K T50 ch©šÍ avivevwnKfv‡e bv¤^vwis Kiv n‡q‡Q| mygb
T17 Avm‡b Ges mRxe T39 G e‡m‡Q| Zv‡ì gv‡S Avi KZwU Avmb Av‡Q?) [BB-(officer)-2018]
a. 23 b. 21 c. 22 d. 20 Ans: b
Solution:
Total number of seats = (39-17) -1=21 (Ans)
‡evSvi Rb¨ e¨vL¨v:
mygb I mRx‡ei g‡a¨ T18 ‡_‡K T38 ch©šÍ Avmb i‡q‡Q | GLb T1 ‡_‡K T38 ch©šÍ Avmb msL¨v 38 wU|
Avevi T1 ‡_‡K T17 ch©šÍ Avmb msL¨v 17 wU| Zvn‡j Zv‡ì `yÕR‡bi g‡a¨ ‡gvU (38-17) = 21 wU Avmb i‡q‡Q|
14. Which of the following can be arranged into a 5-letter English word? (wb‡Pi kã¸‡jvi g‡a¨
†KvbwU‡K mvRv‡j GKwU 5 Aÿiwewkó Bs‡iRx kã n‡e?) [BB-(officer)-2018]
a. HRGST b. RILSA c. TOOMT d. WQRGS
a. a & c b. b & c c. c & d d. a & d Ans: b
Solution:
Given that; a. HRGST b. RILSA c. TOOMT d. WQRGS
Option a and d Cannot become an English word, because it has no vowels.
option b Can be arranged to spell : Lairs(‡jqvi), Liars(wg_¨vev`x), Rails(‡ij)
option c Can be arranged to spell : Motto (g~jgš¿)
15. Siddik has a new set of golf clubs. Using a club 8, 7 and 6 the ball travels an average
distance of 100 m, 108 m, 114 meters respectively. How far will the ball go-if he uses a
club 5? (wmwÏ‡Ki Kv‡Q _vKv bZzb Mjd jvwVi g‡a¨ 8wg. j¤^v jvwV w`‡q 100 wg, 7wg. w`‡q 108 wgUvi Ges 6 wg.
w`‡q 114 wg Mo `~i‡Z¡ ej cvVv‡Z cv‡i| hw` †m 5 wg. j¤^v GKwU jvwV eënvi K‡i Zvn‡j ejwU KZ`~‡i hv‡e?) [BB-
(officer)-2018]
a. 122m b. 120m c. 118m d. 116m Ans: c

Solution:
Given that;
(cÖ‡kœ cÖ`Ë K¬z¸‡jv‡K †Uwej AvKv‡i mvRv‡j )
Club Club 8 Club 7 Club 6 Club 5
Distance 100 m 108 m 114 m ?
GLv‡b, 100+8 = 108, Gici 108+6 = 114
A_©vr jvwVi ˆ`N©¨ hZ Kg‡Q ej hvIqvi `~iZ¡ ZZ evo‡Q | cÖ_‡g 8 evovi ci 6 ‡e‡o‡Q| (e„w×i cwigvY I 2 K‡i
Kg‡Q) Zvn‡j 5 wg. jvwV w`‡q ejwU hv‡e Av‡iv 4wg. †ewk| A_©vr 114+4 = 118 wgUvi|
16. The day after the day after tomorrow is 4 days before Monday. What day is it today?
[BB-(officer)-2018]
a. Monday b. Tuesday c. Wednesday d. Thursday Ans: a
Solution:
The day after the day after tomorrow = A_© AvMvgx ciïi c‡ii w`b| GLb †mvgev‡ii 4 w`b Av‡M wQj
e„n¯úwZevi| A_v©r AvMvgx ciïi c‡ii w`b e„n¯úwZevi n‡e| Avi AvMvgx ciïi c‡ii w`b e„n¯úwZevi n‡Z n‡j AvR
Aek¨B †mvgevi n‡e|
M, T, W, Th, F, Sa, S, M, GLv‡b wmwiqvj wgwj‡q wbb| AvMvgx ciïi c‡ii w`b e„n¯úwZevi n‡j D‡ëvcv‡k MZ
ciïi Av‡Mi w`b †mvgevi n‡e|
17. 6121135 represents 'flame' then 21215120 represents? [BB-(officer)-2018]
a. Voice b. bald c. bloat d. castle Ans: c
Solution:
GLv‡b Bs‡iRx eY©gvjvi eY©µg‡K msL¨vq cÖKvk Kiv n‡q‡Q| GUv‡K g~jZ Coding ejv nq hv msL¨v I
eY©gvjvi auvauv|
Given, 61121135 = flame
Where, 6 = f, 12 = l, 1 = a, 13 = m and 5 = e
Then, 21215120 = ?
Following the given code, 2 = b, 12 = l, 15 = o, 1 = a and 20 = t
So, 21215120 = bloat
18. My successor's father is my father's son and I don't have any brothers or sons. Who is
my successor? (Avgvi DËivavKixi evev Avgvi evevi cyÎ, Avgvi †Kvb fvB ev cyÎ †bB| Avgvi DËivwaKvix †K?|)
[BB-(officer)-2018]
a. Nephew b. Niece c. Daughter d. Myself Ans: c
Solution:
Avgvi DËivwaKvixi evev = Avgvi evevi cyÎ ( †h‡nZz Avgvi †Kvb fvB †bB) A_©vr Avwg Avgvi evevi GKgvÎ cyÎ|
GLb cÖ_g Ask †_‡K Avgvi evevi cyÎ A_©vr Avwg njvg Avgvi DËivwaKvixi wcZv Zvn‡j Avgvi Aek¨B cyÎ ev Kb¨v
Av‡Q| ( wKš‘ cÖ‡kœ Av‡Q Avgvi fvB ev cyÎ †KvbUv B †bB)
‡h‡nZz Avwg GKR‡bi evev Zvn‡j Avgvi †h DËivwaKvix n‡e †m Aek¨B Avgvi Kb¨ n‡e|

19. At the end of a banquet 10 people shake hands with each other. How many handshakes
will there be in total? (Lvevi LvIqvi ci 10 Rb †jvK cÖ‡Z¨‡K‡ cÖ‡Z¨‡Ki mv‡_ n¨vÛ‡kK Ki‡j me©‡gvU KZwU
n¨vÛ‡kK n‡e? ) [BB-(officer)-2018]
a. 100 b. 20 c. 45 d. 90 Ans: c
Solution:
Total Handshake =10
C2 =
!!
!
)!(!
!
82
8910
2102
10




=
2
910
= 45 (A hw` B Gi mv‡_ n¨vÛ‡kK K‡i
Zvn‡j B GiI A Gi mv‡_ n¨vÛ‡kK n‡q hvq e‡j n¨vÛ‡k‡Ki cÖkœ¸‡jv‡Z mgv‡e‡ki myÎ cÖ‡qvM Ki‡Z nq| )
kU©KvU©: G ai‡Yi cÖ‡kœ hZRbB n¨vÛ‡kK KiæK bv †Kb Zvi Av‡Mi msL¨vi mv‡_ ¸Y K‡i wb‡P 2 w`‡q fvM Ki‡Z
n‡e|
2
910
= 45 ( GKB fv‡e 6 Rb n¨vÛ‡kK Ki‡j †gvU n¨vÛ‡kK = 15
2
56


)
20. In a crime, three suspects X, Y and Z were caught and questioned. Each person said,
"One of the other two stole it. I did not do it." Later on the police found out that Z was
lying and there was only one thief. Who was the thief? (GKwU Aciv‡ai Rb¨ X, Y Ges Z ‡K
†MÖdZvi Kiv n‡q‡Q| Zv‡ì‡K Revew`wnZv Kivi mgq cÖ‡Z¨‡K ej‡jv, Avwg Pzwi Kwi wb, Ab¨ `yR‡bi GKRb K‡i‡Q|
cieZ©x‡Z cywjk †`L‡jv †h Z wg_¨v e‡j‡Q, Avevi ‡mLv‡b GKRb gvÎ †Pvi Av‡Q| ‡K †mB †Pvi?) [BB-(officer)-
2018]
a. X b. Y c. Z d. Someone else Ans: c
Solution:
GLv‡b †`Lv hv‡”Q †h, 3 R‡bi g‡a¨ Z wg_¨v ej‡Q, A_©vr Zvi K_v Abymv‡i Ab¨ 2 Rb †Pvi GB K_vwU mZ¨ bq|
myZivs †m wb‡RB †Pvi|
21. Two people are born in the same moment, but they don’t have the same birthdays.
Because they might be born in different-(`yÕBRb e¨w³ GKB gyn~‡Z© Rb¥MÖnY Kij wKš‘ Zv‡ì Rb¥ ZvwiL
GKB bq| KviY Zviv `ywU wfbœ-----------Rb¥MÖnY K‡i‡‡Q | ) [BB-(officer)-2018]
a. Centuries b. Culture c. Countries d. Casts Ans: c
Solution:
`yÕRb e¨w³ GKB gyn~‡Z© Rb¥MÖnY Ki‡jI Zviv hw` Ggb `ywU †`‡k Rb¥MÖnY K‡i hv‡ì mg‡qi cv_©K¨ 12 N›Uv
Zvn‡j GK‡`‡k †h ZvwiL n‡e Ab¨ †`‡k 1 w`b †ewk ev Kg n‡e| †h Kvi‡Y `y R‡bi Rb¥w`b GKB ZvwiL bv I n‡Z
cv‡i| myZivs wfbœ †`‡k Rb¥ wb‡jB †Kej GgbwU nIqv m¤¢e|
==================================

1. If sinA + sin2
A + cos4
A is — [Rupali Bank Off-
(Cash)-2018]
a. 1 b.
2
1
c.2 d.3 Ans: a
Solution:
sinA + sin2
A = 1
sinA = 1-sin2
A
or, sinA = cos2
A
or, cos2
A = sinA or, cos4
A = sin2
A or, cos4
A = 1 – cos2
A So, cos2
A + cos4
A =1
(Cash)-2018]
a. 60° b. 45° c. 30° d. 90° Ans: a
Solution:
Avgiv Rvwb, tan =
fzwg
^j¤
tan =
BC
AB
( cv‡ki wPÎ Abyhvqx)
or, tan =
32
6
=
2
3
= 3
3
33

.
or, tan = tan600
[Since tan60
= 3 ]
  = 600
(Ans)
3. If a + 1, 2a + 1, 4a - 1 are in Arithmetic Progression, then the value of ‘a’ is (GKwU mgvšÍi
avivi c`¸‡jv a + 1, 2a + 1, 4a – 1 n‡j a Gi gvb KZ?) [Rupali Bank Off- (Cash)-2018]
a. 1 b. 2 c. 3 d. 4 Ans:b
Solution:
(2a+1)-(a+1) = (4a-1) - (2a+1) (‡h‡nZz avivwU GKwU mgvšÍi aviv ZvB me c‡`i e¨veavb mgvb|)
or, 2a+1-a-1 = 4a-1-2a-1
or, a= 2a-2 a = 2 Ans: 2
cÖgvY: a = 2 emv‡j avivwU nq 2+1 , 22+1 Ges 42-1 A_©vr 3, 5 Ges 7 ( gv‡Si e¨veavb 2 K‡i)
Ackb a‡iI Gfv‡e cÖgvY Kiv hvq|
4. Suppose today is Friday. What day of the week will it be 65 days from now? (AvR ïµevi|
AvR †_‡K 65 w`b ci wK evi n‡e?) [Rupali Bank Off- (Cash)-2018]
a. Saturday b. Monday c. Tuesday d. Friday Ans: a
Rupali Bank Ltd.
Post name: Officer- (Cash) Exam date: 09-03-2018
B C
A
60
Pole=6
Qvqv= 32
m~q©

Solution:
AvR 1g w`b ïµevi n‡j Gici 8g, 15Zg, 22Zg, 29Zg, 36Zg, 43Zg, 50Zg, 57Zg Ges 64Zg w`bI
ïµevi n‡e| Zvn‡j 65Zg w`b n‡e ïµev‡ii c‡ii w`b A_©vr kwbevi|
kU©KvU©: †h ‡Kvb w`‡bi e¨veavb‡K 7 w`‡q fvM K‡i 1 Aewkó _vK‡j GKB evi nq| A_©vr 7 Gi ¸wYZ‡Ki mv‡_ 1 †hvM
Ki‡j GKB evi n‡e| GLv‡b 65 Gi Av‡M 7 Gi ¸wYZK 63 Ges 63+1 = 64 Zg w`bI ïµevi| myZivs 65 Zg w`b
n‡e kwbevi| (Uvwb©s c‡q›U: from now = GLb †_‡K A_©vr AvR mn ai‡Z n‡e|)
5. A median of a triangle divides it into two (wÎfz‡Ri ga¨gv wÎfzRwU `ywU bZzb wÎfz‡R wef³
Ki‡j...)[Rupali Bank Off- (Cash)-2018]
a. congruent triangles
c. isosceles triangles
b. triangles of equal area
d. right triangles Ans: b
Solution:
‡h †Kvb wÎfz‡Ri ga¨gv wÎfzR †ÿÎwU‡K mgvb †ÿÎdj wewkó `ywU bZzb wÎfz‡R wef³ K‡i| G‡ÿ‡Î Zv mg‡KvYx ev
mgwØevû nIqvUv ¸iæZ¡c~Y© bq| ZvB DËi: B
6. Which of the following angle can be constructed with the help of a ruler and a pair of
compasses? (wb‡Pi †Kvb †KvYwU GKwU ‡¯‹j Ges `ywU K¤úv‡mi gva¨‡g A¼b Kiv m¤¢e?) [Rupali Bank Off-
(Cash)-2018]
a. 35° b. 40° c. 37.5° d. 47.5
Ans:c
Solution:
bZzb av‡Pi GB cÖkœwUi A_© Av‡M eyS‡Z n‡e|
mvaviYZ †KvY A¼‡bi Rb¨ Pvu`v e¨venvi Ki‡Z nq| wKš‘ ïay ‡¯‹j Ges `ywU K¤úv‡mi mvn‡h¨ †KvY A¼b Ki‡Z n‡j
GKwU mij†iLv Uvbvi ci GKB e¨vmva© wb‡q mij‡iLvi GKB cv‡k K‡qKwU e„ËPvc Av©K‡Z nq| wPÎvbymv‡i GKwU
mij‡iLvi Dci Aa©e„Ë A¼b Kivi ci GKB e¨vmva© wb‡q `ywU e„ËPvc A¼b Ki‡j GKwU mg‡KvY Aw¼Z n‡e| Gici
mg‡KvY Ges 60wWwMÖ †Kv‡Yi evûi mv‡_ e„ËPv‡ci †Q` we›`y‡K †K›`ª K‡i 75wWwMÖ ‡KvY A¼Y Kiv hv‡e| Gfv‡e me¸‡jv
†KvY‡K evi evi A‡a©K Kiv hv‡e|
cÖ‡kœ cÖ`Ë Ackb Abyhvqx 0 I 75 wWwMÖ‡K ‡K›`ª a‡i
G‡`i A‡a©K 37.5 wWwMÖ †KvY AvuKv m¤¢e| Gfv‡e cÖwZevi 7.5 †hvM
Ki‡j †h bZzb bZzb †KvY n‡e Zv A¼b Kiv m¤¢e|
7. If a, b and c are the lengths of the three sides of a triangle, then which of the following is
true? (hw` GKwU wÎfz‡Ri 3wU evû h_vµ‡g a,b Ges c nq Zvn‡j wb‡Pi †KvbwU mwVK) [Rupali Bank Off-
(Cash)-2018]
a. a+b < c b. a-b <c c. a+b = c d. a+b  c Ans:b
Solution:
`ywU ¸iæZ¡c~Y© Abywm×všÍ:
 wÎfz‡Ri †h †Kvb `yB evûi mgwó Zvi Z…Zxq evû A‡cÿv e„nËi| Ges Gi wecix‡Z
 wÎfz‡Ri †h †Kvb `yB evûi AšÍi ev e¨eavb ev we‡qvMdj Z…Zxq evû A‡cÿv ÿz`ªZi|
2q wbqg Abymv‡i Ackb B. †Z cÖ`Ë a-b < c mwVK| A_©vr `ywU evû we‡qvM Ki‡j Zv Z…Zxq evû †_‡K †QvU n‡e|
600
900
750
37.50

8. Which line is parallel to y = x -2 ? (wb‡Pi †KvbwU y = x -2 Gi mgvšÍivj? ) [Rupali Bank Off- (Cash)-
2018]
a. y = 2x+1 b. 2y = 2x -6 c. 2y = x+7 d. y = 3x+1 Ans:b
Solution:
y = x -2
Since y = mx+c
Here slope = 1 (x Gi mnM †h‡nZz 1 ZvB Zvi slope ev Xvj I 1 )
GLb †`L‡Z n‡e Ack‡bi g‡a¨ †Kvb mgxKiYwUi Xvj 1 nq| KviY mgvšÍivj n‡Z n‡j `ywUi Xvj B mgvb n‡Z n‡e|
Ack‡bi ga¨ †_‡K
2x = 2y - 6
x = y - 3 y-3= x  y = x+3 So, Here slope = 1 (GLv‡bI x Gi mnM 1 )
9. The area of a triangle with sides 3 cm, 5 cm and 6 cm is (3 ‡m.wg, 5‡m.wg. I 6 †m.wg. evû wewkó
wÎfz‡Ri †ÿÎdj KZ?)-[Rupali Bank Off- (Cash)-2018]
a. 2 3 cm2
b. 4 14 cm2
c. 2 14 cm2
d. 2 5 cm2 Ans:c
Solution:
Aa© cwimxgv, S = 7
2
653


= )67()57()37(7  = 1.2.4.7 = 2 14 cm2
(Ans)
10. The pair of co-prime numbers is (wb‡Pi †Kvb msL¨vØq mn - †gŠwjK msL¨v?) [Rupali Bank Off-
(Cash)-2018]
a. 2, 3 b. 2, 4 c. 2, 6 d. 2, 110 Ans:a
Solution:
If the H.C.F of two numbers is 1 then they said to be Co-Prime numbers. A_©vr hw` `ywU msL¨vi
M.mv.¸ 1 nq Zvn‡j Zv‡`i‡K ci¯úi mn‡gŠwjK ev Co- Prime msL¨v e‡j| cÖ‡kœ cÖ`Ë Ackb¸‡jvi g‡a¨ ïaygvÎ A
‡Z cÖ`Ë 2 Ges 3 Gi M.mv.¸ 1 ZvB DËi a.
11. The value of k, if (x- 1) is a factor of 4x3
+ 3x2
-4x + k, is (hw` x- 1, 4x3
+ 3x2
-4x + k ivwkwUi
GKwU Drcv`K nq Zvn‡j k Gi gvb KZ? ) [Rupali Bank Off- (Cash)-2018]
a. 1 b. 2 c. - 3 d. 3 Ans: c
Solution:
g‡b ivLyb: †Kvb ivwk Ab¨ ivwki Drcv`K n‡j †hgb: x- 1 GKwU Drcv`K Zvn‡j x-1 = 0 ev x = 1 emv‡j Zv
H m¤ú~Y© ivwkwU‡K 0 evwb‡q w`‡e|
myZivs,
f(x) = 4x3
+3x2
-4x+k
f(1) = 4(1)3
+3(1)2
-4.1+k = 4+3-4+k = k+3
Now, k+3 = 0 ( ‡h‡nZz x- 1 ivwkwUi Drcv`K ZvB x = 1 emv‡bvi ci hv gvb Avm‡e Zv 0 Gi mgvb n‡e| )
 k= -3 (Ans)

12. There are 5 red and 3 black balls in a bag. Probability of drawing a black ball is (GKwU
e¨v‡M 5wU jvj Ges 3wU Kv‡jv ej Av‡Q| †mLvb †_‡K ˆ`efv‡e GKwU ej wb‡j Zv Kv‡jv nIqvi m¤¢vebv KZ?)
[Rupali Bank Off- (Cash)-2018]
a.
8
5
b.
2
1
c.
8
3
d.
4
1 Ans: c
Solution:
Total balls = 5+3 = 8 and black ball = 3
So, probability of getting black ball is =
8
3
(wb‡P †gvU dj Ges Dc‡i †h ej wb‡Z n‡e Zvi msL¨v|)
13. The total surface area of a hemisphere of radius r is (GKwU Aa© †Mvj‡Ki mgMÖ Z‡ji †ÿÎdj KZ?)
–[Rupali Bank Off- (Cash)-2018]
a. 4r2
b. r2
c. 2r2
d. 3r2
Ans:d
Solution:
total surface area of sphere is = 4r2
but hemisphere surface is half of sphere surface =2r2
+ surface area of a circle = r2
So total surface area of hemisphere is 2r2
+ r-2
= 3r2
Av‡iKUz wk‡L ivL‡j Dc‡ii welqwUI Av‡iv wK¬qvi n‡e|
mgMÖ c„ôZj ej‡Z Pvicv‡ki Avei‡Yi †ÿÎdj|
GLb †MvjK ev Sphere A_©vr GKwU dzUe‡ji Pvicv‡ki Avei‡Yi †ÿÎdj = 4r2
Zvn‡j dzUejUv‡K gvSLvb w`‡q †K‡U †dj‡j Zv GKUv cvDiæwUi gZ n‡e| ZLb †mB cvDiæwUi evKv‡bv As‡ki
†ÿÎdj n‡e dzUe‡ji mgMÖ Z‡ji †ÿÎd‡ji A‡a©K A_©vr 4r2
 2 = 2r2
wKš‘ ZLb H cvDiæwUi euvKv‡bv Ask
ev‡`I wb‡Pi mgZj GKwU e„Ë ˆZix n‡e hvi †ÿÎdj n‡e e„‡Ëi †ÿÎd‡ji gZ, r2
Avi †m Rb¨ hemisphere ev
Aa© ‡Mvj‡Ki mgMÖ Z‡ji †ÿÎdj n‡e 2r2
+ r-2
= 3r2
14. The roots of the equation 9x2
- bx + 81 = C will be equal, if the value of b is ( b Gi gvb KZ
n‡j 9x2
- bx + 81 = C mgxKiYwUi gyjØq mgvb n‡e?) [Rupali Bank Off- (Cash)-2018]
a. ±9 b. ±18 c. ± 27 d. ± 54 Ans: d
Solution:
9x2
- bx + 81 = C
we know that,
ax2
+bx+c = 0
and b2
- 4ac = 0 ( GUv gyj mgvb nIqvi myÎ|)
or b2
= 4ac
or, b2
= 4981 (ax2
+bx+c = 0 myÎvbymv‡i a=9 Ges c=81)
or, b = 8194 
b = ±( 239) = ± 54
Alternative solution:
9x2
- bx + 81

ivwkwU‡K (a+b)2
A_ev (a-b)2
GB my‡Î †dj‡Z n‡e| Zvn‡j mgxKiYwUi gyjØq mgvb n‡e|
GLb,
9x2
- bx + 81
= (3x)2
-bx+92
= ( 3x)2
-2.3x.9 +92
+54 (gv‡Si bx ¯’‡j myÎ ‡gjv‡Z 2ab evbv‡Z n‡e)
= (3x-9)2
+54
Avevi GB 54 hw` + bv wb‡q -54 ‡bqv nq Zvn‡j (a+b)2
my‡Îi mv‡_ wg‡j hv‡e|
myZivs DËi: ± 54
15. If sec + tan = x, then tan is? [Rupali Bank Off- (Cash)-2018]
a.
x
1x2

b.
x
1x2

c.
x2
1x2

d.
x2
1x2
 Ans:d
Solution:
We know that, sec2
 - tan2
 = 1
 (sec + tan ) (sec - tan) = 1
 x(sec- tan) = 1
x
1
- - - - - (i)
 again, sec + tan = x - - - - - (ii)
by (ii)-(i) we get,
2tan = x-
x
1
or, 2tan =
x
1x2

tan =
x2
1x2

16. Consider that w + x = -4, x + y = 25 and y + w = 15. Then the average of w, x, y is —
[BKB – (Cash ) -2017] & [BDBL – (SO ) -2017] [Rupali Bank Off- (Cash)-2018]
a. 3 b. 4 c. 5 d. 6 Ans: d
Solution:
w+x + x+y + y+w = -4 + 25 + 15
2w+2x+2y = 36 2(w+x+y) =36
 w+x+y=362=18 So, average of w,x,y=183 = 6 Ans:
17. What is the original price of a T-shirt, if the sale price after 15% discount is 272? (GKwU
kvU© 15% Qvo w`‡q 272 UvKvq weµq Kiv n‡j kvU©wUi cÖK…Z g~j¨ KZ wQj?)? [BDBL – (SO ) -2017] & [Rupali
Bank Off- (Cash)-2018]
a. 300 b. 280 c. 320 d. 314 Ans: c
Solution:
85% = 272 So,1% =
85
272
100% =
85
100272
= 320 Ans:

18. TK. 500 is deposited in a savings account which pays 7% annual interest compounded
semi-annually. To the nearest Taka, how much is in the account at the end of the year?
(kZKiv evwl©K 7% nv‡i lvb¥vwmK Pµe„w×‡Z 500 UvKv GKwU e¨vs‡K ‡mwfs GKvD‡›U Rgv ivL‡j eQi †k‡l Zv cÖvq
KZ UvKv n‡e? ) [Rupali Bank Off- (Cash)-2018]
a. 542 b. 536 c. 512 d. 524 Ans: b
Solution:
cÖ_‡g 7% my`‡K 2w`‡q fvM K‡i 3.5% evbv‡Z n‡e| (KviY 7% nj evrmwiK wKš‘ ‡ei Ki‡Z n‡e 6gvwmK|)
Gici my`vmj = 500 Gi 103.5% Gi 103.5% = 536 ( cÖvq)
Gfv‡e mivmwi wn‡me Kiv ( 103.5% †ei Kiv ) KwVb g‡b n‡j ‡f‡½ †f‡½ wb‡Pi wbq‡g †ei Kiv Lye mnR|
weKí mgvavb:
500 (6 gvm K‡i 2 evi) interest of interest Total interest Amount
17.5 (Avm‡ji my`) Gi 3.5%
17.5 (Avm‡ji my`) 0.6125 17.5+17.5+.6125=35.61 or 36 536
19. 4
1
xlog = -2, the x =? [Rupali Bank Off- (Cash)-2018]
a.
2
1
 b.
2
1
c. 2 d. 17 Ans:c
Solution:
4
1
xlog = -2 or, x-2
=
4
1
or, x-2
= 2-2
x = 2
20. If 5% is gained by selling an article for BDT 350 than selling it for BDT 340, the cost of
the article is (GKwU `ªe¨ 340 UvKvi cwie‡Z© 350 UvKvq wewµ Ki‡j 5% jvf nq| `ªe¨wUi µqg~j¨ KZ?) [BD
House Building FC- (SO)-2017] & [BDBL - (SO ) -2017] & [Rupali Bank Off- (Cash)-2018]
a. BDT 180 b. BDT 150 c. BDT 200 d. BDT 150 Ans: c
Solution:
Selling price difference =350-340=10 tk. And % difference is = 5%
5% of cost =10tk, Or, 1% of cost = tk
5
10
So 100% of cost =
5
10010
= 200 tk
21. If x = ya
, y = zb
and z = xc
then the value of abc is[BD House Building FC (SO)-2017] &
[Rupali Bank Off- (Cash)-2018]
a. 1 b. 0 c. 0.5 d. Infinity Ans:a
Solution:
x = ya
Or, x=zab
(Since y= zb
) Or, x = xabc
(Since z= xc
) Or, xabc
=x1
Or, abc = 1
Alternative way, y = zb
or, y = xbc
or, y = yabc
or, abc = 1 (A_©vr †h ‡KvbUv a‡iB DËi 1 )

22. If x is 30% greater than y, what percent of y is x? (hw` x Gi gvb y Gi †_‡K 30% †ewk nq Zvn‡j
y Gi kZKiv KZ x n‡e?) [Rupali Bank Off- (Cash)-2018]
a. 70 b. 77 c. 120 d. 130 Ans: d
Solution:
if, y = 100 then x = 100+30 = 130
then, x is 130% of y
evsjvq eySzb:
hvi mv‡_ Zzjbv Kiv nq Zv‡K wb‡P wjL‡Z n‡e Ges hv‡K Zzjbv Kiv n‡e Zv‡K Dc‡i wj‡L kZKivi wn‡me Ki‡Z nq|
GLv‡b y = 100 then x = 130
GLb, kZKiv nvi †ei Kivi Rb¨,
100
100130
= 130% (Avi ey‡S †Mj †Kvb wKQz bv wj‡LB 130% ejv hvq|)
23. The lengths of two sides of a right angle triangle are 13cm and 5cm respectively. The
length of the third side is (GKwU wÎfz‡Ri `ywU evûi ‰`N¨© h_vµ‡g 13 I 5| Z…Zxq evûi ‰`N©¨-) [BDBL –
(SO ) -2017] & [Rupali Bank Off- (Cash)-2018]
a. 13 b. 17 c. 11 d. 12 Ans:d
Solution:
cx_v‡Mviv‡mi myÎvbyhvqx
AwZfz‡Ri2
= j¤^2
+ fzwg2
n‡Z n‡j GKwU
wÎfz‡Ri AwZfzR 13 I j¤^ 5 n‡j Zvi Z…Zxq evû Aek¨B 12 Gi mgvb n‡e|
mg‡KvYx wÎfz‡Ri evû¸‡jvi AbycvZ 12:5: 13 nq|
24. The present age of Habib and Shikha are in the ratio of 6 : 4. Five years ago their ages
were in the ratio of 5 : 3. How old is Habib now? (nvwee Ges wkLvi eZ©gvb eq‡mi AbycvZ 6:4| 5
eQi Av‡M Zv‡`i eq‡mi AbycvZ 5:3 wQj | eZ©gv‡b nvwe‡ei eqm KZ?) [Rupali Bank Off- (Cash)-2018]
a. 24 b. 30 c. 36 d. 42 Ans:b
Solution:
Let,
Present age of Habib and Shikha is 6x and 4x
ATQ,
3
5
5x4
5x6



 20x -25 = 18x-15  2x = -15+25 = 10 mx= 5
Habib = 6  5 = 30 years old now. Ans: 30
==================================
A
B C
5
13
12

[2017 mv‡j Arts faculty KZ…K †bqv GB cÖkœwU‡Z †_‡K cieZ©x‡Z wewfbœ cixÿvq A‡bK cÖkœ
ûeû wiwcU n‡q‡Q| ZvB GiKg wKQz cÖkœ GB As‡k †`qv n‡jv]
1. How many real roots does the polynomial 2x3
+8x -7 have? [Agrani Bank – (Cash)-2017]
a. None b. One c. Two d. Three Ans: b
Solution:
Here, f(x) = 2x3
+8x -7
And two positive sign and one negative sign, one sign changes.
so according to Descartes’ rule of signs One real root possible Ans: One
KZ¸‡jv c~Y©msL¨v 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq?) [Agrani Bank – (Cash)-2017]
a. 29 b. 31 c. 32 d. 38 Ans: a
Solution:
Divisible by 30=
30
1000
= 33.33 = 33 Numbers (30,60,90-----990 = 33wU)
L.C.M of 30 and 16 = 240,
240
1000
= 4.
evsjvq e¨vL¨v:
30 w`‡q wefvR¨ †h msL¨v¸‡jv Av‡Q †hgb: 30, 60, 120, 240 - - - - 990 = 33wU | Gi g‡a¨ wKQz msL¨v Av‡Q
hv‡`i‡K 16 w`‡qI fvM Kiv hvq|
GLb GB msL¨v¸‡jv †_‡K 30 w`‡q fvM Kiv hvq wKš‘ 16 w`‡q fvM Kiv hvq bv Ggb msL¨v †ei Kivi Rb¨ 16 I 30
Dfq w`‡q fvM Kiv hvq †mB msL¨v¸‡jv ‡ei K‡i Av‡Mi 33wU †_‡K ev` w`‡Z n‡e|
wK Ki‡j 16 I 30 Gi wgjb ¯’j Lyu‡R cvIqv hv‡e? Aek¨B j.mv,¸ Ki‡j| ZvB 16 Ges 30 Gi j.mv.¸ 240 †ei
K‡i 240 w`‡q 30 w`‡q wefvR¨ 33wU msL¨vi g‡a¨ †h 4wU msL¨v ( 240,480,720,960) †K 16 w`‡qI fvM Kiv hvq
†m¸‡jv ev` w`‡jB 30 w`‡q wefvR¨ wKš‘ 16 w`‡q wefvR¨ bq Ggb msL¨v¸‡jv †ei n‡e| DËi: 33-4 = 29wU|
3. What is the slope of the line perpendicular to the line y = -5x+9 ? [Agrani Bank – (Cash)-
2017]
a. 5 b. -5 c.
5
1
d.
5
1
 Ans:c
Solution:
Agrani Bank Limited
Post name: Officer (Cash) Exam date: 15-12-2017
Exam taker: Arts Faculty, Du.

w`‡q ¸Y Kiv|
m
1
= -
5
1
5
1



Ans:
5
1
must have a slope that is the negative reciprocal of the original slope. m = -
m
1
=
3
1

4. If m and p are positive integers and m+pm is even, which of the following must be true
? (hw` m Ges p `ywU abvZ¥K c~Y© msL¨v nq Ges m+pm Gi gvb †Rvo nq Zvn‡j wb‡Pi †KvbwU Aek¨B mZ¨?)
[Agrani Bank –(Cash)-2018]
a. If m is odd, then p is odd
c. If m is even, then p is even
b. If m is odd, then p is even
d. If m is even, then p is odd
Ans: a
Solution:
m+pm = even
or m(1+p) = even
we know, eveneven = even
Or, oddeven = even.
so if m is odd then 1+p must even to make it even.
since 1+p = even
so p also must be odd.
ïw× cixÿv:
Avgv‡`i‡K ‡`Lv‡Z n‡e †h, m(1+p) = even
a. If m is odd, then p is odd = m = we‡Rvo n‡j 1+p ‡Rvo Zvn‡j ïa p I we‡Rvo| GLv‡b p
we‡Rvo ev‡` Ab¨ wKQz nIqvi my‡hvM †bB| ZvB GUvB DËi| ( †h‡nZz cÖ‡kœ must ejv n‡q‡Q)
b. If m is odd, then p is even = m = we‡Rvo n‡j ïay p = †Rvo n‡Z cv‡i bv| KviY ïay p = ‡Rvo n‡j
1+p = we‡Rvo n‡q hv‡e ZLb we‡Rvo m Gi mv‡_ ¸Y Ki‡j ¸Ydj I we‡Rvo n‡q hv‡e| wKš‘ ¸Ydj †Rvo n‡Z n‡e|
c. If m is even, then p is even = m = ‡Rvo n‡j (1+p) ‡Rvo, A_ev we‡Rvo †h †Kvb wKQz nIqvi my‡hvM
Av‡Q| KviY †Rvo msL¨vi mv‡_ †Rvo we‡Rvo hv B ¸Y Kiv †nvK ¸Ydj †Rvo n‡e| GLb (1+p) hw` †Rvo nq Zvn‡j
ïay p = we‡Rvo| Avevi (1+p) = we‡Rvo n‡j ïay p = ‡Rvo| wKš‘ GB DËi G Rb¨ †bqv hv‡e bv †h GLv‡b Aek¨B
K_vUv wVK _vK‡Q bv| A_©vr m= even n‡jI, p = even I n‡Z cv‡i Avevi odd I n‡Z cv‡i|
d. If m is even, then p is odd = Ackb c ‡Z cÖ`Ë hyw³ Abymv‡i m = even n‡jI p = odd ev even ‡h
†Kvb wKQz n‡Z cv‡i| ZvB GUv must n‡”Q bv|
Logic::
¸Y Kivi mgq †h †Kvb GKwU msL¨v †Rvo n‡j Ab¨ msL¨vwU
‡Rvo ev we‡Rvo hv B †nvK bv †Kb ¸YdjwU †Rvo n‡e| GB
hyw³wUB GB cÖ‡kœ Kv‡R jvMv‡bv n‡q‡Q|

5. The population of a certain town increases by 50 percent every 50 years. If the
population in 1950 was 810, in what year was the population 160? (GKwU kn‡ii RbmsL¨v cÖwZ
50 eQ‡i 50% K‡i e„w× cvq| hw` 1950 mv‡j H kn‡ii RbmsL¨v 810 Rb _v‡K Zvn‡j KZ mv‡j H kn‡ii
Rb¨msL¨v 160 Rb wQj?) (Pubali Bank Ltd. SO 2013) & [Agrani Bank – (Cash)-2017]
a. 1650 b. 1700 c. 1750 d. none Ans:c
Solution:
50% e„w× cvIqv A_© A‡a©K e„w× cvIqv A_©vr RbmsL¨v 100% †_‡K 50% †e‡o 150% ev
2
3
fvM nIqv |
Zvn‡j 50 eQi Av‡M wQj = wZb fv‡Mi `y fvM ev
3
2
KviY Av‡M †jvK Kg wQj| c‡i †e‡o‡Q|
1950 mv‡j 810 Rb _vK‡j 1900 mv‡j wQj = 810 Gi
3
2
= 540
GLb cÖwZevii mvj wj‡L wj‡L †ei bv K‡i 540 Gi mv‡_
3
2
¸Y Ki‡Z Ki‡Z †hLv‡b 160 Avm‡e †mUvB DËi |
GLv‡b 540
3
2
= 360
3
2
= 240
3
2
=160
A_©vr wZbevi ¸Y Kivi ci 160 n‡q‡Q Zvn‡j mvjwU n‡e 1900-150 = 1750 mvj|
Alternative Solution:
r = 50% p =160 CA= 810
So 160(1+ 0.50)x
= 810
Or 1.5x
=
16
81
Or (
2
3
)x
= (
2
3
)4
Or x = 4
50% per 50 years. So total years = 450 = 200 years.
So the time was =1950-200 =1750 Ans: none
6. If ‘a’ and ‘b’ are integers greater than 100 such that a + b = 300, which of the following
could be the exact ratio of a to b? (hw` a Ges b n‡jv 100 Gi †P‡q eo c~Y© msL¨v Ges a + b = 300
nq Z‡e a I b Gi cÖK…Z AbycvZ KZ n‡e?) (Pubali Bank Ltd. SO 2013) & [Agrani Bank – (Cash)-2017]
a. 9 to 1 b. 5 to 2 c. 5 to 3 d. 3 to 2 Ans:d
Solution:
GLv‡b 300 †K Ggb fv‡e `ywU As‡k fv½‡Z n‡e hv‡Z `ywUB 100 Gi †_‡K eo msL¨v nq Ges Ack‡b cÖ`Ë Abycv‡Zi
mv‡_ wg‡j hvq|
cÖ_g Ackb a‡i mgvavb Ki‡j a = 270 Av‡m wKš‘ b = 30 hv 100 †_‡K †QvU| 2q I 3q Ackb a‡i fv½v‡bvB
hv&q bv| 4_© Ackb a‡i mgvavb Ki‡j a = 180 Av‡m Ges b = 120 Av‡m| hv cÖ‡kœi mv‡_ wg‡j hvq ZvB DËi

7. The three sides of a triangle are x+1, 2x-1 and 3x+1 respectively and the perimeter is 25
cm. The length of the smallest side is (GKwU wÎfz‡Ri wZbwU evûi ˆ`N©¨ h_vµ‡g x+1, 2x-1 Ges 3x+,
wÎf~RwUi cwimxgv 25 †m.wg. n‡j ÿz`ªZg evûi ˆ`N©¨ KZ?)- [Agrani Bank – (Cash)-2017]
a. 5cm b. 3cm c. 4cm d. 7cm Ans: a
Solution:
Here, (x+1) + ( 2x-1) + (3x+1) = 25 Or, 6x = 24  x = 4
So the sides are 4+1 = 5, 24 – 1 = 8-1 = 7 and 34+1 = 12+1 = 13  Smallest side = 5
8. 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? (hw` GKwU AvqZ‡ÿ‡Îi ˆ`N©¨‡K 20% e„w× Kivi ci Zvi cÖ¯’‡K
20% n«vm Kiv‡bv nq Zvn‡j AvqZ‡ÿÎwUi †ÿÎdj kZKiv wK iKg cwieZ©b n‡e?) [Agrani Bank – (Cash)-
2017] & [BB-(officer)-2018]
a. unchanged b.decreases by 4% c. increases by 4% d. increases by 5% Ans: b
Solution:
 gy‡L gy‡L Kivi Rb¨ : cÖ_‡g 100 †_‡K 20% evov‡j n‡e 120 Gici 120 ‡_‡K 120 Gi 20% = 24 Kgv‡j n‡e
120-24 = 96| †gv‡Ui Dci Kg‡jv 100-96 = 4%|
myÎ cÖ‡qv‡M: Shortcut: 20 - 20 +
 
100
2020 
= - 4 Ans: decreases by 4%
9. In how many ways a team of 11 members can be formed from a group of 15 students if a
student who is the owner of the ball is always considered a member of the team? (15 Rb
QvÎ-QvÎx‡ì ga¨ †_‡K 11 Rb m`‡mï GKwU `j KZfv‡e evQvB Kiv hv‡e †hLv‡b ‡h e‡ji gvwjK †m memgq `‡ji
g‡a¨ _vK‡e?) [Agrani Bank – (Cash)-2017]
a. 200 b. 201 c. 210 d. 1001 Ans:d
Solution:
one is fixed.
11-1= 10 should be selected from out of 15-1 = 14 (GKRb wbw`©ó nIqvq 15 Rb †_‡K 1 Rb Kg‡e Ges
11 R‡bi ¯’‡j Av‡M †_‡K 1 Rb‡K wb‡q †bqvq Aewkó †jvK wb‡Z n‡e 11-10 Rb| A_©vs 14 †_‡K 10 Rb|)
14
C10=
123410
1011121314
101410
14



 !
!
)!(!
!
=1001 Ans: 1001
10. At the beginning of a class period, half of the students in a class go to the library. Later
in the period, half of the remaining students go to the computer lab. If there are 8
students remaining in the class, how many students were originally in the class? (K¬v‡ki
ïiæ‡Z A‡a©K QvÎ-QvÎx jvB‡eªix‡Z †Mj| wKQzÿY c‡i Aewkó QvÎ-QvÎx‡ì A‡a©K Kw¤úDUvi j¨v‡e †Mj| hw` K¬v‡k
eZ©gv‡b 8 Rb QvÎ-QvÎx _v‡K Zvn‡j cÖ_‡g KZ Rb QvÎ-QvÎx wQj?) (Pubali Bank SO 2013) & [Agrani Bank –
(Cash)-2017]
a. 12 b. 16 c. 24 d. 32 Ans: d
Solution:
Before going to computer lab students were 82 = 16 (GLv‡b ïiæi msL¨v †`qv ‡bB ZvB †kl †_‡K)
So, at the beginning of the class students were 162 = 32 Ans: 32

11. The dimensions of a box are 2, 3 and 4 meters. The cost of Painting the outer sides of the
box, at the rate of Tk. 3 per square meter is? (GKwU e‡·i wZbwU evû h_vµ‡g 2, 3 Ges 4 wgUvi|
e·wUi evB‡ii Ask cÖwZ eM© wgUv‡i 3UvKv nv‡i is Ki‡Z ‡gvU KZ UvKv LiP n‡e?) [Agrani Bank – (Cash)-
2017] & [BB-(officer)-2018]
a. Tk.156 b. Tk. 220 c. Tk. 136 d. Tk. 160 Ans:a
Solution:
Total surface area 2 (ab+bc+ca) (evB‡ii †ÿÎdj †ei Kivi
myÎ)
2(23+34+42) = 2(6+12+8) = 52 Sq. meter
Cost = 523 = Tk.156
g‡b ivLyb: is Kivi Rb¨ ïay evB‡ii As‡ki †ÿÎdj is Ki‡Z n‡e|
12. If
3
1
x
y
 = and x + 2y= 10 then x is- [Agrani Bank – (Cash)-2017]
a. 2 b. 3 c. 4 d. 6 Ans: d
Solution:
3
1
x
y
 or, 3y = x y =
3
x
now, x+2y=10 Or, x+
3
x2
=10 Or,
3
x5
=10 Or, x=
5
310
 x= 6 Ans:6
13. A coin is tossed twice. What is the probability of getting head on first toss and tail on
second toss? (GKwU g~`ªv `yÕ evi wb‡ÿc Kiv n‡jv| cÖ_gevi †nW Ges c‡iievi †Uj Avmvi m¤¢vebv KZ?)
[Agrani Bank – (Cash)-2017]
a.
2
1
b.
3
1
c.
4
1
d. 1 Ans: c
Solution:
total sample(4)= TT,HH,HT,TH.
So first head and second tail is on only 1=HT
So probability=
4
1
Or: head probability=
2
1
and tail probability=
2
1
So probability=
2
1

2
1
=
4
1
Ans:
4
1
(g‡b ivLyb : cÖ_gevi hLb †nW DV‡e ZLb †Uj DV‡e bv| Avevi cÖ_gevi †Uj DVvi ci c‡ii evi ‡nW DV‡j †mB
djvdj GLv‡b †bqv hv‡e bv| KviY cÖ_gevi ‡nW DV‡Z n‡e hv Specific fv‡e e‡j †`qv Av‡Q)
4m
3m
2m

14. The average of eight numbers is 14 and the average of six of these numbers is 16. What
is the average of the remaining numbers? (AvUwU msL¨vi Mo 14, G‡`i g‡a¨ 6wU msL¨vi Mo 16 n‡j
Aewkó msL¨v¸‡jvi Mo KZ?) [Agrani Bank Off. (Cash)-2013)] & [Agrani Bank – (Cash)-2017]
a. 4 b. 8 c. 16 d. 6 Ans: b
Solution:
Sum of eight numbers = 8 14 = 112 Sum of first six numbers = 616 = 96
So, Sum of two numbers = 112-96=16
So average of last two numbers is 16÷2 = 8 Ans: 8
15. A motorist travels to a place 150 km away at an average speed of 50 km and returns at
30 km per hour. What is his average speed for the whole journey in km per hour?(GKRb
‡gvUi mvB‡Kj Av‡ivnx 50wKwg MwZ‡Z 150 wKwg c_ AwZµg K‡i 30wKwg MwZ‡Z wd‡i Av‡m| m¤úyY© hvÎvq Zvi Mo
MwZ‡eM KZ?) (Agrani Bank Ltd. Seni Offi-2013) & [Agrani Bank – (Cash)-2017]
a. 35 b. 37 c. 37.5 d. 40 Ans: c
Solution:
Here total travel = 150+150 = 300 and first time 15050 = 3 hours
and return time is 15030 = 5hr So the average speed is
53
150150


= 537
8
300
. km.
16. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) = ? [Agrani Bank – (Cash)-2017]
a. 5 : 11 b. 6 : 5 c. 5:6 d. 3 : 8 Ans: a
Solution:
x : y = 5 : 3 or,
3
5
y
x
 or, 3x =5y
Now, (8x – 5y) : (8x + 5y) =(8x – 3x) : (8x + 3x) [Since 3x = 5y] 5x:11x  5:11
17. The sum of first 17 terms of the series 5, 9, 13, 17…..( 5, 9, 13, 17 - -- wmwiRwUi cÖ_g 17wU
msL¨vi †hvMdj KZ?) [Agrani Bank – (Cash)-2017]
a. 529 b. 462 c. 629 d. 523 Ans: c
Solution:
5+9+13+17 . . . . sum of first 17 terms?
Here first terms a = 5 , difference d = 9-5 = 4 and total terms n = 17
Sum = })({ d1na2
2
n
 = })({ 411752
2
17
 = )( 6410
2
17
 = 1737 = 629
gy‡L gy‡L cvivi Rb¨ e¨vwmK AvBwWqv jvM‡e: wmwi‡Ri ivwk¸‡jvi Mo  ivwki msL¨v = mgwó|
wmwiRwUi 17 Zg c` n‡e 5+16wU cv_©K¨ = 5+164 = 5+64 = 69
wmwi‡Ri Mo
2
569
= 37| myZivs mgwó n‡e 3717 = 629| Ans:629

Arts faculty all mcq math & solution (2019 2017) by khairul alam [www.onlinebcs.com]

Arts faculty all mcq math & solution (2019 2017) by khairul alam [www.onlinebcs.com]

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