06. mathematics arts pm t 9-10 genarel scince [onlinebcs.com]

MATHEMATICS 281
Arts KZ©„K 2016-2020 e¨vsK wcÖwj I wjwLZ cixÿvi cÖ‡kœi c~Y©v½ mgvavb
1. Number System
M.C.Q
672. How many integers from 1 to 1000 are divisible
by 30 but not by 16? (1 †_‡K 1000 Gi g‡a¨
KZ¸‡jv c~Y©msL¨v 30 Øviv wefvR¨, wKš‘ 16 Øviv wefvR¨
bq?) [Sonali Bank (S.O.-2018); Bangladesh Bank (Officer-2018);
Agrani Bank (Officer Cash-2017); Bangladesh Bank
(Officer General-2019); Rupali Bank Ltd. (Officer-2019)]
a 29 b 31 c 32 d 38 a
 mgvavb : 30 1000
90
100
90
10
33  1 †_‡K 1000 Gi gv‡S
33wU msL¨v 30 Øviv wefvR¨
GB 33wU msL¨vi gv‡S 16
Øviv wefvR¨ msL¨v¸‡jv n‡e
30 Ges 16 Gi j.mv.¸ ev
240 Gi ¸wYZK 1|
240 1000
960
40
4 1 †_‡K 1000 Gi gv‡S 240
Gi ¸wYZK 4wU|
 30 Øviv wefvR¨ wKš‘ 16 Øviv wefvR¨ bq Giƒc msL¨v
(33 – 4) ev 29wU|
673. If n is an even integer, which of the following
must be an odd integer? (n hw` †Rvo c~Y©msL¨v
nq, Zvn‡j wb‡Pi †KvbwU Aek¨B we‡Rvo msL¨v?)
[Bangladesh Bank (Officer General-2019)]
a n2
– n b n + 2 c 3n – 1 d 3n3
c
 mgvavb : awi, n = 2
a n2
– n = 22
– 2 = 2, †Rvo
b n + 2 = 2 + 2 = 4, †Rvo
c 3n – 1 = 3  2 – 1 = 5, we‡Rvo [mwVK DËi]
d 3n3
= 3  23
= 24, †Rvo
674. If both x and y are prime numbers, which of
the following CANNOT be the product of x
and y? (x I y †gŠwjK msL¨v n‡j wb‡Pi †KvbwUÑ x I y
Gi ¸Ydj n‡Z cv‡i bv?) [Rupali Bank Ltd. (Officer-2019)]
a 6 b 10 c 35 d 27 d
 mgvavb : Ack‡bi msL¨v¸‡jv‡K `ywU †gŠwjK msL¨vi
¸Ydj AvKv‡i cÖKvk K‡i cvB,
(a) 6 = 2  3
(b) 10 = 2  5
(c) 35 = 5  7
(d) 27 = 3  9
†hŠwMK msL¨v
 mwVK Ackb : d|
675. How many integers from 1 to 100 are divisible
by 3 but not by 8? (1 †_‡K 100 ch©šÍ KZ¸‡jv
c~Y©msL¨v Av‡Q hviv 3 Øviv wefvR¨ wKš‘ 8 Øviv wefvR¨ bq?)
[Combined 5 Banks (Officer-2018)]
a 30 b 29 c 31 d 32 b
 mgvavb : 3 100 33
9
10
9
1
 3 Gi ¸wYZK = 33wU
3 I 8 Gi j.mv.¸. = 3  8 = 24
AZGe 24 Øviv wefvR¨ msL¨v¸‡jv 3 I 8 Dfq Øviv wefvR¨|
24 100 4
96
4
 24 Øviv wefvR¨ msL¨v = 4wU
 †gvU msL¨v = 33 – 4 = 29wU
676. After dividing a positive integer y, by 3, the
remainder is 2; but when y is divided by 7,
the remainder is 4. What is the least possible
value of y? (3 Øviv †Kv‡bv c~Y©msL¨v y †K fvM Ki‡j
fvM‡kl _v‡K 2| wKš‘ y †K hLb 7 Øviv fvM Kiv nq,
ZLb fvM‡kl _v‡K 4| y Gi m¤¢ve¨ me©wb¤œ gvb KZ?)
[Sonali & Janata Bank (S.O. IT-2018)]
a 11 b 22 c 18 d 32 a
 mgvavb : Option Check :
(a) 11 – 2 = 9; hv 3 Øviv wefvR¨
11 – 4 = 7; hv 7 Øviv wefvR¨
(b) 22 – 2 = 20; hv 3 Øviv wefvR¨ bq
(c) 18 – 2 = 16; hv 3 Øviv wefvR¨ bq
(d) 32 – 2 = 30; hv 3 Øviv wefvR¨
32 – 4 = 28; hv 7 Øviv wefvR¨
wKš‘ †h‡nZz 11 < 32; ZvB mwVK DËi a
weKí mgvavb : awi, 3 I 7 Øviv y †K fvM Ki‡j
fvMdj nq h_vµ‡g x I z|
fvR¨ = fvRK  fvMdj + fvM‡kl
 y = 3x + 2 Ges y = 7z + 4
x I z me©wb¤œ n‡j y Gi gvbI me©wb¤œ n‡e|
y = 7z + 4 = (7z + 2) + 2 = 3x + 2
 7z + 2 = 3x
 x =
7z + 2
3
z Gi gvb 1 n‡j (me©wb¤œ) x =
7  1 + 2
3
=
9
3
= 3
A_©vr c~Y©msL¨v nq|
 x = 3; z = 1
 y = 3x + 2 = 3  3 + 2 = 11

282 PHENOM’S RECENT BANK SOLUTION
677. If m and p are positive integers and (m + p) m
is even, which of the following must be true?
(hw` m I p abvZ¥K c~Y©msL¨v nq Ges (m + p)m GKwU
†Rvo msL¨v nq, Z‡e wb‡Pi †KvbwU Aek¨B mZ¨?)
[Sonali Bank (S.O.-2018)]
a If m is odd, then p is odd
b If m is odd, then p is even
c If m is even, then p is even
d If m is even, then p is odd a
 mgvavb : (m + p) m †Rvo n‡e hw` G‡`i AšÍZ GKwU
†Rvo nq|
m we‡Rvo n‡j, (m + p) †K †Rvo n‡Z n‡e, †m‡ÿ‡Î
p I we‡Rvo n‡Z n‡e|
 Ackb a mwVK I b fzj|
m †Rvo n‡j p Gi gvb †Rvo ev we‡Rvo hvB †nvK bv
†Kb (m + p)m †Rvo n‡e|
AZGe c I d m¤ú~Y© mwVK e³e¨ bq|
mwVK DËi : a
678. 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 g‡a¨ (10 I 100 mn) †_‡K †Kvb c~Y© msL¨v‡K
ˆ`ePq‡b wbe©vPb Ki‡j Zv 5 A_ev 9 Øviv wefvR¨ nevi
m¤¢ve¨Zv KZ?) [Bangladesh Bank (A.D.-2018)]
a
27
89
b
20
91
c
27
91
d
23
89
c
 mgvavb : 10 †_‡K 100 Gi g‡a¨,
5 Gi ¸wYZK 10, 15, 20, 25, 30, 35, 40, 45, 50, 55,
60, 65, 70, 75, 80, 85, 90, 95 Ges 100 A_©vr 19wU
9 Gi ¸wYZK 18, 27, 36, 45, 54, 63, 72, 81, 90,
99 A_©vr 10wU
9 Ges 5 ¸wYZK 45, 90
†gvU msL¨v = (100 – 10) + 1 = 91wU
 5 A_ev 9 Øviv wefvR¨ nevi m¤¢ve¨Zv =
19
91
+
10
91
–
2
91
=
27
91
679. If * is defined for all positive real numbers a
and b by a * b =
ab
(a + b)
, then 10 * 2 = ? (mKj
abvZ¥K ev¯Íe msL¨vi (a, b) Rb¨ a * b =
ab
(a + b)
n‡j, 10 * 2 = ?) [Bangladesh Bank (Officer-2018)]
a
5
3
b
5
2
c 5 d
20
3
a
 mgvavb : a * b =
ab
a + b
a = 10, b = 2 n‡j,
10 * 2 =
10  2
10 + 2
=
20
12
=
5
3
680. In each expression below, N represents a negative
integer. Which expression could have a negative
value? (wb‡Pi ivwk¸‡jv‡Z N Øviv GKwU FYvZ¥K c~Y©
msL¨v cÖKvwkZ nq| †Kvb ivwkwUi gvb FYvZ¥K n‡Z cv‡i|)
[Bangladesh Bank (Officer-2018); Agrani Bank (Officer Cash-2017)]
a N2
b 6 – N c – N d 6 + N d
 mgvavb : awi, N = – 10
(a) N2
= (– 10)2
= 100 > 0
(b) 6 – N = 6 – (– 10) = 16 > 0
(c) – N = – (– 10) = 10 > 0
(d) 6 + N = 6 + (– 10) = – 4 < 0
 mwVK DËi d
681. Find the value of : 6 (– 3)



1
3
(– 0.25)
[Sonali Bank (Officer-2018)]
a 6 b 4.5 c 1.5 d – 0.5 c
 mgvavb : 6(–3)



1
3
(–0.25)
= (–) (–) 6  3 
1
3

1
4
=
6
4
= 1.5
682. In m and p are positive integers and m + pm
is even, which of the following must be true?
(hw` m Ges p abvZ¥K c~Y©msL¨v nq Ges m + pm
†Rvo msL¨v nq, wb‡Pi †Kvb Z_¨wU mZ¨?)
[Agrani Bank (Officer Cash-2017)]
a If m is odd, then p is odd.
b If m is odd, then p is even.
c If m is even, then p is even.
d If m is even, then p is odd. a
 mgvavb : Avgiv Rvwb,
†Rvo msL¨v  †Rvo msL¨v = †Rvo msL¨v
we‡Rvo msL¨v  we‡Rvo msL¨v = we‡Rvo msL¨v
we‡Rvo msL¨v  †Rvo msL¨v = †Rvo msL¨v
we‡Rvo msL¨v + we‡Rvo msL¨v = †Rvo msL¨v
we‡Rvo msL¨v + †Rvo msL¨v = we‡Rvo msL¨v
†Rvo msL¨v + †Rvo msL¨v = †Rvo msL¨v
m p m + pm
1) †Rvo †Rvo †Rvo + †Rvo  †Rvo = †Rvo
2) †Rvo we‡Rvo †Rvo + †Rvo  we‡Rvo = †Rvo
3) we‡Rvo we‡Rvo we‡Rvo + we‡Rvo  we‡Rvo = †Rvo
4) we‡Rvo †Rvo we‡Rvo + we‡Rvo  †Rvo = we‡Rvo
 a, c Ges d wZbwU mwVK|
wKš‘ c I d `ye©j Ackb| KviY m †Rvo n‡j, p
Gi gvb †Rvo/we‡Rvo hv-B †nvK (m + pm) †Rvo
n‡e| wKš‘ m hw` we‡Rvo nq, Z‡e p †K we‡Rvo n‡ZB
n‡e| ZvB me‡P‡q mwVK DËi a|

MATHEMATICS 283
683. In first 1000 natural numbers, how many integers
exist such that they leave a remainder 4 when
divided by 7 and a remainder 9 when divided
by 11? (cÖ_g 1000 ¯^vfvweK msL¨vi g‡a¨, KZ¸‡jv
c~Y©msL¨v Av‡Q hv‡ì 7 w`‡q fvM Ki‡j 4 fvM‡kl
_v‡K Ges 11 Øviv fvM Ki‡j 9 fvM‡kl _v‡K?)
[B.D.B.L. (S.O.-2017); B.H.B.F.C. (S.O.-2017)]
a 11 b 13 c 15 d 17 b
 mgvavb : 7 Øviv fvM Ki‡j 4 fvM‡kl _v‡K Ggb msL¨v :
N = 7x + 4; x = 0, 1, 2, ............
 N = 4, 11, 18, 25, ............
11 Øviv fvM Ki‡j fvM‡kl 9 _v‡K Ggb msL¨v :
M = 11Y + 9; Y = 0, 1, 2, ............
M = 9, 20, 31, ............
N = 4, 11, 18, ............ d = 7
M = 9, 20, 31, ............ d = 11
awi, N Gi n Zg I M Gi m Zg c` mgvb| GiKg
hZ¸‡jv mgvb c` cvIqv hv‡e, ZZ¸‡jv-B n‡e DËi|
 4 + 7(n – 1) = 9 + 11(m – 1)
 7(n – 1) – 11(m – 1) = 5
GLb, n – 1 = 7 Ges m – 1 = 4 n‡j Dfqcÿ mgvb n‡e|
7  7 – 11  4 = 5
 n – 1 = 7  n = 8
m – 1 = 4  m = 5
G‡ÿ‡Î 1g msL¨vwU n‡e 53
GLb, 53 Gi ci †_‡K we‡ePbv Ki‡j :
53 + 7(n – 1) = 53 + 11  (m – 1)
 7(n – 1) = 11(m – 1)
7  11 = 11  7
 1g avivq 53 Gi c‡i 11 Ni ci ci Ges 2q avivq
53 Gi 7 Ni ci ci msL¨v¸‡jv mgvb n‡e|
n – 1 = 11 m – 1 = 7
n = 12 m = 8
7 1000 142
994
6
 H iƒc †gvU msL¨v =



142
11
+ 1
= 12 + 1 = 13 (Ans.)
684. If n – 5 is an even integer, what is the next
large consecutive even integer? (n – 5 GKwU
†Rvo c~Y©msL¨v, cieZ©x eo µwgK †Rvo msL¨vwU KZÑ)
[B.H.B.F.C. (S.O.-2017)]
a n – 7 b n – 3 c n – 4 d n – 2 b
 mgvavb : (n – 5) Gi cieZ©x µwgK †Rvo msL¨vwU
= n – 5 + 2
= n – 3
2. H.C.F and L.C.M
M.C.Q
685. What is the H.C.F of the numbers 36, 54 and
90? [Bangladesh Bank (Officer General-2019)]
a 6 b 9 c 12 d 18 d
 mgvavb : 36 = 2  2  3  3
54 = 2  3  3  3
90 = 2  3  3  5
 M.mv.¸. = 2  3  3 = 18
686. 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 †KvbwU G‡ì j.mv.¸. n‡Z cv‡i?)
[Rupali Bank Ltd. (Officer-2019); Agrani Bank (Officer Cash-2017)]
a 84 b 128 c 148 d 120 d
 mgvavb : `ywU msL¨vi j.mv.¸. Aek¨B Zv‡ì M.mv.¸.
Øviv wbt‡k‡l wefvR¨ n‡e|
(a)
84
24
= 3
12
24
(b)
128
24
= 5
8
24
(c)
148
24
= 6
4
24
(d)
120
24
= 5
 mwVK Ackb : d|
687. The pair of co-prime numbers is– (mn‡gŠwjK
†Rvo n‡jvÑ) [Rupali Bank (Officer Cash-2018)]
a 2, 3 b 2, 4 c 2, 6 d 2, 110 a
 mgvavb : †h mKj msL¨vi M.mv.¸ 1 A_©vr 1 Qvov
hv‡ì g‡a¨ Avi †Kv‡bv mvaviY Drcv`K †bB, Zv‡ì
ci¯úi mn‡gŠwjK msL¨v e‡j|
2, 3 `ywUB †gŠwjK msL¨v hv‡ì M.mv.¸ = 1
2, 4 Gi M.mv.¸ = 2
2, 6 Gi M.mv.¸ = 2
2, 110 Gi M.mv.¸ = 2
 (2, 3) mn‡gŠwjK †Rvo|
688. What is the largest number that divides 84,
144 or 18 without any remainder? (†Kvb e„nËg
msL¨v Øviv 84, 144 Ges 18 †K fvM Ki‡j †Kvb
fvM‡kl _v‡K bv?) [Sonali Bank (Officer-2018)]
a 6 b 12 c 18 d 24 a
 mgvavb : 84, 144 Ges 18 Gi M.mv.¸-B n‡e wb‡Y©q
e„nËg msL¨v|
84 = 2  2  3  7
144 = 2  2  2  2  3  3
18 = 2  3  3
 wb‡Y©q M.mv.¸ = 2  3 = 6

689. Find the least number of six digits which is
exactly divisible by 15, 21 and 28. (Qq A‡¼i
ÿz`ªZg †Kvb msL¨v 15, 21 Ges 28 Øviv wbt‡k‡l
wefvR¨Ñ) [B.H.B.F.C. (S.O.-2017)]
a 100480 b 100270
c 100380 d 100340 c
 mgvavb : 15, 21 I 28 Gi j.mv.¸ = 420
6 A‡¼i ÿz`ªZg msL¨v = 100000
420 100000 238
840
1600
1260
3400
3360
40
 6 A‡¼i wb‡Y©q ÿz`ªZg msL¨v = 100000 + (420 – 40)
= 100380
Written
690. Find the H.C.F. of x3
 16x, 2x3
+ 9x2
+ 4x,
2x3
+ x2
 28x. (x3
 16x, 2x3
+ 9x2
+ 4x Ges
2x3
+ x2
 28x Gi M.mv.¸ KZ?)
[Agrani Bank Ltd. (S.O. Auditor-2018)]
 Solution: Given,
1st expression = x3
 16x
= x(x2
 42
)
= x (x + 4) (x  4)
2nd expression = 2x3
+ 9x2
+ 4x
= x(2x2
+ 9x + 4)
= x{2x2
+ 8x + x + 4}
= x {2x (x + 4) + 1(x + 4)}
= x (x + 4) (2x + 1)
3rd expression = 2x3
+ x2
 28x
= x(2x2
+ x  28)
= x(2x2
+ 8x  7x  28)
= x{2x (x + 4)  7(x + 4)}
= x(x + 4)) (2x  7)
 H.C.F (Highest common factor) of given
expressions = x(x + 4)
3. Algebra
M.C.Q
691. If
a2
+ 1
a
= 3 what is
a3
+ 1
a3 ?
a 24 b 7 c 30 d 18 d
 mgvavb : †`Iqv Av‡Q,
a2
+ 1
a
= 3

a2
a
+
1
a
= 3
 a +
1
a
= 3




a +
1
a
3
= 33
 a3
+
1
a3 + 3. a.
1
a 


a +
1
a
= 27
 a3
+
1
a3 + 3  3 = 27
 a3
+
1
a3 = 27 – 9 = 18
692. If a –
1
a
= 2 what is a3
–
1
a3?
[Rupali Bank Ltd. (Officer-2019)]
a 16 b 10 c 14 d 12 c
 mgvavb : cÖ`Ë ivwk = a3
–
1
a3
=



a –
1
a
3
+ 3. a.
1
a 


a –
1
a
‹ x3
– y3
= (x – y)3
+ 3xy (x – y)
= 23
+ 3  2
= 8 + 6
= 14
693. The values of p for equation 2x2
– 4x + p = 0
to have real roots is– (2x2
– 4x + p = 0
mgxKi‡Yi g~j¸‡jv ev¯Íe n‡e hw` p Gi gvbÑ)
a p  – 2 b p  2 c p  2 d p  – 2 c
 mgvavb : 2x2
– 4x + p = 0 mgxKiYwU GKwU wØNvZ
mgxKiY hv ax2
+ bx + c = 0 Gi mgZzj¨
 a = 2, b = – 4, c = p
ev¯Íe g~‡ji Rb¨ wbðvqK : b2
– 4ac  0
 (– 4)2
– 4(2) (p)  0
 16 – 8p  0
 8 (2 – p)  0
 2 – p  0
 – (2 – p)  0
[– 1 Øviv ¸Y Kivq AmgZvi w`K cwieZ©b]
 – 2 + p  0
 p  2
694. If a +
1
a
= 2 what is a3
+
1
a3 ? (a +
1
a
= 2 n‡j a3
+
1
a3 Gi gvb KZ?) [Combined 5 Banks (Officer-2018)]
a
1
2
b 7 c 2 d
3
2
c
 mgvavb : a +
1
a
= 2
 a3
+
1
a3 =



a +
1
a
3
– 3.a.
1
a 


a +
1
a
[‹ x3
+ y3
= (x + y)3
– 3xy(x + y)]
= 23
– 3  2 = 8 – 6 = 2

MATHEMATICS 285
695. Bangladeshi supporters in a stadium double
every match. In the eighth match, there were
48000 supporters which was the full capacity of
the stadium. In which match did the Bangladeshi
supporter fill up half the capacity of the stadium?
(GKwU †÷wWqv‡g evsjv‡`kx mg_©K‡ì msL¨v cÖwZ g¨vP
AšÍi wØ¸Y nq| Aóg g¨v‡P 48000 Rb mg_©K Dcw¯’Z wQj
hv †÷wWqv‡gi aviY ÿgZvi mgvb| †Kvb g¨v‡P evsjv‡`kx
mg_©K‡ì msL¨v †gvU aviY ÿgZvi A‡a©K wQj?)
a 2nd
match b 4th
match
c 6th
match d 7th
match d
 mgvavb : 8g g¨v‡P Dcw¯’Z wQj 48000 Rb `k©K hv
†÷wWqv‡gi aviY ÿgZvi mgvb|
 A‡a©K aviY ÿgZv =
48000
2
= 24000 Rb
Avevi, cÖwZ g¨vP AšÍi `k©K Dcw¯’wZ wØ¸Y nq|
A_©vr c~‡e©i g¨v‡Pi Dcw¯’wZ =
cieZ©x g¨v‡Pi Dcw¯’wZ
2
 7g g¨v‡Pi Dcw¯’wZ =
8g g¨v‡Pi Dcw¯’wZ
2
=
48000
2
= 24000 Rb
myZivs 7g g¨v‡P Dcw¯’wZ A‡a©K aviY ÿgZvi mgvb wQj|
696. You bought 11 pencils and erasers worth
BDT 80. If erasers cost half that of a pencil
and you bought one extra eraser, how much
is the eraser worth? (Avcwb 80 UvKv Li‡P †gvU
11wU †cwÝj I B‡iRvi wKb‡jb| cÖwZwU B‡iRv‡ii g~j¨
cÖwZwU †cwÝ‡ji g~‡jï A‡a©K Ges Avcwb B‡iRvi GKwU
†ewk wKb‡jb (†cwÝj A‡cÿv)| cÖwZwU B‡iRv‡ii g~j¨
KZ?) [Sonali & Janata Bank (S.O. IT-2018)]
a 3 b 4 c 5 d 6 c
 mgvavb : aiv hvK, wZwb †cwÝj wKb‡jb NwU
 B‡iRvi (N + 1)wU
 N + (N + 1) = 11
 2N = 10
 N = 5
myZivs Zuvi µqK…Z †cwÝj = 5wU
B‡iRvi = (5 + 1) = 6wU
awi, cÖwZwU B‡iRv‡ii g~j¨ x UvKv
†cwÝ‡ji 2x UvKv
cÖkœg‡Z, 6x + 5  2x = 80
 6x + 10x = 80  16x = 80
 x =
80
16
 x = 5
 cÖwZwU B‡iRv‡ii g~j¨ 5 UvKv
697. Jashim buys 10 CDs for BDT 200. If DVDs
cost BDT 20 more, how many DVDs can he
buy for the same amount? (Rwmg 200 UvKv e¨‡q
10wU CD wKbj| hw` cÖwZwU DVD Gi `vg cÖwZwU CD
Gi `v‡gi Zzjbvq 20 UvKv †ewk nq, Z‡e H UvKvq †m
KqwU DVD wKb‡Z cvi‡e?)
a 4 b 5 c 6 d 10 b
 mgvavb : 10wU CD Gi g~j¨ 200 UvKv
 1wU CD
200
10
= 20 UvKv
AZGe, cÖwZwU DVD Gi g~j¨ = (20 + 20) = 40 UvKv
40 UvKvq cvIqv hvq 1wU DVD
 1
1
40
DVD
 200
200
40
DVD
= 5wU DVD
698. The value of k, if (x – 1) is a factor of 4x3
+
3x2
– 4x + k, is– (hw` 4x3
+ 3x2
– 4x + k ivwkwUi
GKwU Drcv`K (x – 1) nq, Z‡e k Gi gvbÑ)
[Sonali Bank (S.O.-2018); Rupali Bank (Officer Cash-2018)]
a 1 b 2 c – 3 d 3 c
 mgvavb : (x – 1) hw` 4x3
+ 3x2
– 4x + k Gi GKwU
Drcv`K nq, Z‡e x = 1 n‡e 4x3
+ 3x2
– 4x + k =
0 mgxKi‡Yi GKwU g~j|
 4  13
+ 3  12
– 4  1 + k = 0
 4 + 3 – 4 + k = 0
 k = – 3
weKí mgvavb :
awi, (x) = 4x3
+ 3x2
– 4x + k
(x – 1), (x) Gi GKwU Drcv`K n‡j,
(1) = 0
 4(1)3
+ 3(1)2
– 4(1) + k = 0
 4 + 3 – 4 + k = 0
 k + 3 = 0
 k = – 3
699. If x +
1
x
= 3, then x –
1
x
= ? (x +
1
x
= 3 n‡j, x –
1
x
= ?)
a 5 b 13 c 7 d 0 a
 mgvavb : x +
1
x
= 3




x +
1
x
2
= 32




x –
1
x
2
+ 4x 
1
x
= 9
[‹ (a + b)2
= (a – b)2
+ 4ab]




x –
1
x
2
+ 4 = 9 



x –
1
x
2
= 5
 x –
1
x
= 5

700. The factors of x2
– 5x – 6 are : (x2
– 5x – 6 †K
Drcv`‡K we‡kølY Ki :) [Sonali Bank (S.O.-2018)]
a (x – 6) (x + 1) b (x + 6) (x – 1)
c (x – 3) (x + 2) d (x – 3) (x – 2) a
 mgvavb : cÖ`Ë ivwk = x2
– 5x – 6
= x2
+ x – 6x – 6
= x(x + 1) – 6(x + 1)
= (x – 6) (x + 1)
701. The roots of the equation 9x2
– bx + 81 = 0
will be equal, if the value of b is (9x2
– bx +
81 = 0 mgxKiYwUi g~j¸‡jv mgvb n‡j b Gi gvbÑ)
[Rupali Bank (Officer Cash-2018)]
a  9 b  18 c  27 d  54 d
 mgvavb : 9x2
– bx + 81 = 0 GKwU wØNvZ mgxKiY hv
ax2
+ px + c = 0 Gi mgZzj¨
a = 9; p = – b; c = 81
wbðvqK = p2
– 4ac
g~j¸‡jv mgvb nevi kZ© :
p2
– 4ac = 0
 (– b)2
– 4  9  81 = 0
 b2
= 4  9  81
 b =  54
702. How many cases do you need if you have to
pack 112 pairs of shoes into cases that each
hold 28 shoes? (28wU Ry‡Zv avib Ki‡Z cv‡i Ggb
KZ¸‡jv †Km eënvi K‡i 112 †Rvov Ry‡Zv ivLv hv‡e?)
[Bangladesh Bank (Officer-2018)]
a 8 b 10 c 12 d 14 a
 mgvavb : 28wU Ry‡Zv =
28
2
= 14 †Rvov Ry‡Zv|
14 †Rvov Ry‡Zv ivLv hvq 1wU †K‡m
 1
1
14
wU
 112
112
14
wU
= 8wU †K‡m
703. A person needs to pay Tk. 500 to buy pencils
and Tk. X for any additional unit of pencil.
If the customer pays a total of Tk. 4,700 for
1200 pencils, what is the value of X? (GKRb
e¨w³i Kv‡Q 500 UvKv Av‡Q wKQz †cwÝj µq Kivi Rb¨
Ges AviI wKQz †cwÝj †Kbvi Rb¨ x UvKv Av‡Q| hw`
e¨w³wU †gvU 1200 †cwÝj wKb‡Z 4700 UvKv e¨q K‡i,
Zvn‡j cÖwZwU †cwÝ‡ji g~j¨?) [Sonali Bank (Officer-2018)]
a 4.0 b 3.91 c 3.85 d 3.5 b
 mgvavb : cÖwZwU †cwÝ‡ji g~j¨ =
4700
1200
= 3.91 UvKv
704. If P = 5 + 2 then the value of P2
is–
a 5 + 10 2 b 20 + 5 2
c 27 + 10 2 d 27 c
 mgvavb : †Ìqv Av‡Q, p = 5 + 2
 p2
= (5 + 2)2
= 52
+ 10 2 + 2
= 25 + 2 + 10 2
 p2
= 27 + 10 2
705. If x –
1
x
= – 3, then x4
+
1
x4 = ?
[Agrani Bank (S.O. Auditor-2018)]
a 23 b 27 c 3 d 9 a
 mgvavb : †Ìqv Av‡Q, x –
1
x
= – 3
cÖ`Ë ivwk = x4
+
1
x4
= (x2
)2
+



1
x2
2
=



x2
+
1
x2
2
– 2  x2

1
x2
=









x –
1
x
2
+ 2  x 
1
x
2
– 2
= { }(– 3)2
+ 2 2
– 2
= {5}2
– 2 = 25 – 2 = 23
706. How many real roots does the polynomial
2x3
+ 8x – 7 have? (2x3
+ 8x – 7 eûc`xi KZwU
ev¯Íe g~j Av‡Q?) [Agrani Bank (Officer Cash-2017)]
a None b One c Two d Three b
 mgvavb : awi, (x) = 2x3
+ 8x – 7
x –3 –2 –1 0 1 2 3
(x) –85 –39 –17 –7 3 25 71
x = 0 †_‡K x = 1 ch©šÍ (x) Gi gvb –7 †_‡K 3 nq|
A_©vr, 0 †_‡K 1 Gi gv‡S eûc`xwUi GKwU ev¯Íe g~j
i‡q‡Q|
 eûc`xwUi 1wU ev¯Íe g~j Av‡Q|
weKí mgvavb :
– 3 – 2 – 1 1 2 3– 7
– 17
– 39
– 85
25
71
†jLwPÎ n‡Z †`Lv hvq, x = 0 Ges x = 1 Gi g‡a¨
†jLwU x-Aÿ‡K GKevi †Q` K‡i| A_©vr x = 0 Ges x
= 1 Gi g‡a¨ GKwU ev¯Íe g~j we`¨gvb|

MATHEMATICS 287
707. The solutions of 2x2
+ 3x – 2 = 0 are
a x = – 3 and x = 2 b x =
1
2
and x = – 2
c x = – 1 and x = 2 d x = 1 and x = – 2 b
 mgvavb : 2x2
+ 3x – 2 = 0
 2x2
+ 4x – x – 2 = 0
 2x (x + 2) – 1 (x + 2) = 0
 (x + 2) (2x – 1) = 0
 x + 2 = 0 ev, 2x – 1 = 0
 x = – 2  x =
1
2
708. A grocer buys some eggs at Tk. 3 each. He finds
that 12 of them are broken, but he sells the
others at Tk. 4 each and makes profit of Tk. 96.
How many eggs did he buy? (GKRb †`vKvb`vi
cÖwZwU wWg 3 UvKv nv‡i wKQz wWg wK‡b| Zvi gv‡S
12wU wWg fv½v wQj, evwK¸‡jv †m cÖwZwU wWg 4 UvKv
K‡i weµq K‡i 96 UvKv jvf K‡i| †m KZwU wWg
wK‡bwQj?) [B.K.B. (Officer Cash-2017)]
a 140 b 142 c 144 d 150 c
 mgvavb : g‡b Kwi, †`vKvb`vi xwU wWg wK‡bwQj
†`vKvb`vi wewµ K‡i (x – 12)wU wWg
cÖkœg‡Z, 4(x – 12) – 3x = 96
 4x – 48 – 3x = 96
 x = 96 + 48 = 144
709. A man buys doughnuts at the rate of Tk. 35 per
100 pieces and sells them at Tk. 7.20 per dozen.
If the profit is Tk. 30, how many doughnuts
did he buy? (GKRb e¨w³ cÖwZ 100 wcm †WvbvU 35
UvKv nv‡i wK‡b Ges cÖwZ WRb 7.20 UvKv nv‡i weµq
K‡i| hw` jvf 30 UvKv nq, Zvn‡j KZ¸‡jv †WvbvU
wK‡bwQj?)
[B.K.B. (Officer Cash-2017)]
a 60 b 120 c 180 d 210 b
 mgvavb : 1 WRb ev 12wU †Wvbv‡Ui weµqg~j¨ 7.20 UvKv
 1wU
7.20
12
 100wU
7.20  100
12
= 60 UvKv
 100wU †Wvbv‡U jvf = 60 – 35 = 25 UvKv
25 UvKv jvf nq 100wU †Wvbv‡U
 30
100
25
 30wU †Wvbv‡U
= 120wU †Wvbv‡U
710. If x2
– 7xy + y2
is divided by x – 2y, the result is–
[B.D.B.L. (S.O.-2017)]
a 3x + 2y b 3x – 2y
c 2x – 3y d 2x + 3y c
 mgvavb : cÖkœwU‡Z x2
– 7xy + y2
Gi cwie‡Z©
2x2
– 7xy + 6y2
n‡e|
2x2
– 7xy + 6y2
= 2x2
– 4xy – 3xy + 6y2
= 2x (x – 2y) – 3y (x – 2y)
= (x – 2y) (2x – 3y)

2x2
– 7xy + 6y2
x – 2y
=
(x – 2y) (2x – 3y)
(x – 2y)
= (2x – 3y)
711. If x –
1
x
= 3 then x +
1
x
= ?
[B.H.B.F.C. (S.O.-2017); B.D.B.L. (S.O.-2017)]
a 3 3 b 7 c 2 3 d 7 b
 mgvavb : †`Iqv Av‡Q, x –
1
x
= 3
Gme, 


x +
1
x
2
=



x –
1
x
2
+ 4. x.
1
x
= ( 3)2
+ 4 = 3 + 4




x +
1
x
2
= 7
 x +
1
x
= 7
Written
712. Factorize: x3
– 21x + 20.
 Solution: x3
– 21x + 20
= x3
– x2
+ x2
– x – 20x + 20
= x2
(x – 1) + x (x – 1) – 20(x – 1)
= (x – 1) (x2
+ x – 20)
= (x – 1) (x2
+ 5x – 4x – 20)
= (x – 1) {x(x + 5) – 4 (x + 5)}
= (x – 1) (x + 5) (x – 4)
713. If
a
q  r
=
b
r  p
=
c
p  q
then show that, a + b + c
= pa + qb + rc [Bangladesh Development Bank Ltd. (SO)-2018]
 Solution: Given,
a
q  r
=
b
r  p
=
c
p  q
Let,
a
q  r
=
b
r  p
=
c
p  q
= k
 a = k(q  r), b = k(r  p) and c = k(p  q)
L.H.S = a + b + c
= k(q r)+ k(r p) + k(p q)
= k{q  r + r  p + p  q}
= k  0
= 0
R.H.S = pa + qb + rc
= pk(q  r) + qk((r  p) + rk (p  q)
= k{pq  pr + qr  pq + pr  qr}
= k  0 = 0
 L.H.S = R.H.S

714. What should be the values of a and b for
which 64x3
 9ax2
+ 108x  b will be a perfect
cube? [Bangladesh Krishi Bank Ltd. (Officer Cash-2018)]
 Solution: We know,
(p  q)3
= p3
 3p2
q + 3pq2
 q3
   
Given expression = 64x3
 9ax2
+ 108x  b
comparing,
p3
= 64x3
 p3
= (4x)3
 p = 4x
Ges 3p2
q = 9ax2
 3  (4x)2
 q = 9ax2
 q =
9ax2
48x2 =
3a
16
Ges 3pq2
= 108x  3  4x 



3a
16
2
= 108x
 12x 
9a2
256
= 108x  a2
=
108x  256
12x  9
= 256
 a = 16
Ges b = q3
=



3a
16
3
=



3  16
16
3
= 27
715. Resolve into factors : a2
+
1
a2 + 2  2a 
2
a
[Bangladesh Krishi Bank Ltd. (Officer Cash-2018)]
 Solution: a2
+
1
a2 + 2  2a 
2
a
=



a +
1
a
2
 2.a.
1
a
+ 2  2



a +
1
a
=



a +
1
a
2
 2 + 2  2



a +
1
a
=



a +
1
a
2
 2



a +
1
a
=



a +
1
a 


a +
1
a
 2 (Ans.)
716. Simplify :
5x + 2
x2
 x  20
+
2x  1
x2
 4x  5
[Agrani Bank Ltd. (S.O. Auditor-2018)]
 Solution:Given expression
5x + 2
x2
 x  20
+
2x  1
x2
 4x  5
=
5x + 2
x2
 5x + 4x  20
+
2x  1
x2
 5x + x  5
=
5x + 2
x(x  5) + 4(x  5)
+
2x  1
x(x  5) + 1(x  5)
=
5x + 2
(x  5) (x + 4)
+
2x  1
(x  5) (x + 1)
=
(5x + 2) (x + 1) + (2x  1) (x + 4)
(x  5) (x + 4) (x + 1)
=
5x2
+ 5x + 2x + 2 + 2x2
+ 8x  x  4
(x  5) (x + 4) (x + 1)
=
7x2
+ 14x  2
(x  5) (x + 4) (x + 1)
(Ans.)
717. If a = xyp  1
, b = xyq  1
, c = xyr  1
and p + q + r
= 3, then prove that aq  r
× br  p
× cp  q
= 1
[Agrani Bank Ltd. (Officer Cash-2018);
Rupali Bank Ltd. (Officer Cash-2018)]
 Solution: Given, a = xyp  1
, b = xyq  1
, c = xyr  1
and p + q + r = 3
L.H.S = aq  r
 br  p
 cp  q
= ( )xyp  1 q  r
 ( )xyq  1 r  p
 ( )xyr  1 p  q
= xq  r
.y(p  1) (q  r)
 xr  p
. y(q  1) (r  p)
 xp  q
.y(r  1) (p  q)
= xq  r + r  p + p  q
 ypq  pr  q + r
 yqr  pq  r + p
 yrp  rq  p + q
= x0
 ypq  pr  q + r + qr  pq  r + p + rp  rq  p + q
= x0
 y0
= 1 = R.H.S (Showed)
718. Find the value of x6
+
1
x6
,
if x +
1
x
= 3
[Sonali Bank Ltd. (Officer)-2018]
 Solution: Given, x +
1
x
= 3
Given expression
x6
+
1
x6
= (x3
)2
+



1
x3
2
=



x3
+
1
x3
2
 2.x3
.
1
x3
=









x +
1
x
3
 3.x.
1
x 


x +
1
x
2
 2
= {(3)3
 3  3}2
 2 = {27  9}2
 2
= 182
 2 = 324  2 = 322 (Ans.)
719. Simplify :
x  1
x2
 x  20
+
4  x
x2
 4x  5
[Combined 5 Bank (Officer)-2018]
 Solution:Given expression,
x  1
x2
 x  20
+
4  x
x2
 4x  5
=
x  1
x2
 5x + 4x  20
+
4  x
x2
 5x + x  5
=
x  1
x(x  5) + 4(x  5)
+
4  x
x(x  5) + 1(x  5)
=
x  1
(x  5) (x + 4)
+
4  x
(x  5) (x + 1)
=
(x  1) (x + 1) + (4  x) (x + 4)
(x  5) (x + 4) (x + 1)
=
x2
 1  (x  4) (x + 4)
(x  5) (x + 4) (x + 1)
=
x2
 1  (x2
 42
)
(x  5) (x + 4) (x + 1)
=
x2
 1  x2
+ 16
(x  5) (x + 4) (x + 1)
=
15
(x  5) (x + 4) (x + 1)

MATHEMATICS 289
4. Equation
M.C.Q
720. If xy = 2 and xy2
= 16, what is the value of x?
a 4 b 2 c
1
4
d 8 c
 mgvavb : †Ìqv Av‡Q, xy = 2  x2
y2
= 4 ......... (i)
Ges xy2
= 16 ...................... (ii)
(i) (ii) 
x2
y2
xy2 =
4
16
 x =
1
4
721. If
y
x
=
1
3
and x + 2y = 10 then x is–
a 2 b 3 c 4 d 6 d
 mgvavb :
y
x
=
1
3
 y =
x
3
x + 2y = 10
 x +
2
3
x = 10

5x
3
= 10
 x =
3  10
5
 x = 6
722. How many real roots does the equation 2x3
+ 8x
– 7 = 0 have? (2x3
+ 8x – 7 = 0 eûc`x mgxKi‡Y
ev¯Íe gyj KZwU?) [Rupali Bank Ltd. (Officer-2019)]
a None b One c Two d Three b
 mgvavb : †h‡nZz mgxKiYwU wÎNvZ, ZvB G‡Z ev¯Íe g~j
_vK‡e 1wU ev 3wU| †Kbbv RwUj g~j¸‡jv AbyeÜx AvKv‡i
_v‡K| ZvB RwUj g~‡ji msL¨v 2wU ev GKwU I bq|
awi, g~j¸‡jv r1, r2, r3
 r1 + r2 + r3 = 0
r1 r2 + r2 r3 + r3 r1 =
8
2
= 4
(r1 + r2 + r3)2
= r1
2
+ r2
2
+ r3
2
+ 2 (r1 r2 + r2 r3 + r3 r1)
 r1
2
+ r2
2
+ r3
2
+ 2  4 = 0
 r1
2
+ r2
2
+ r3
2
= – 8
wKš‘ KZ¸‡jv ev¯Íe msL¨vi e‡M©i †hvMdj KL‡bvB ïY¨
(0) A‡cÿv †QvU n‡Z cv‡i bv| ZvB r1, r2 I r3 wZbwUB
ev¯Íe bq|
kZ©vbyhvqx, G‡ì g‡a¨ 2wU RwUj g~j|
 ev¯Íe g~‡ji msL¨v =1|
723. If xy = 2 and xy2
= 8, what is the value of x?
(xy = 2 I xy2
= 8 n‡j x Gi gvb KZ?)
a 4 b 2 c
1
2
d 8 c
 mgvavb : xy2
= 8

(xy)2
x
= 8 
22
x
= 8 [ xy = 2]
 x =
4
8
 x =
1
2
724. If 3x – 7y = 0 and x + 2y = 13 then y is– (hw`
3x – 7y = 0 Ges x + 2y = 13 nq, Z‡e y Gi gvbÑ)
a 2 b 3 c 4 d 7 b
 mgvavb : 3x – 7y = 0 ......... (i)
x + 2y = 13 .......... (ii)
(ii)  3 – (i) 
3x + 6y = 39
3x – 7y = 0
(–) (+) (–)
13y = 39
 y =
39
13
 y = 3
725. A bag and a book costs BDT 1100. If the bag
costs 1000 more than the book, how much does
the book cost? (GKwU e¨vM I GKwU eB‡qi g~‡jï
mgwó 1100 UvKv| hw` e¨v‡Mi g~j¨ eB‡qi g~j¨ A‡cÿv
1000 UvKv †ewk nq, Z‡e eB‡qi g~j¨ KZ?)
a 100 b 150 c 50 d 200 c
 mgvavb : awi, eB‡qi g~j¨ = x UvKv
 e¨v‡Mi g~j¨ = (x + 1000) UvKv
cÖkœg‡Z, eB‡qi g~j¨ + e¨v‡Mi g~j¨ = 1100
 x + (x + 1000) = 1100
 2x = 100  x = 50
 eB‡qi g~j¨ 50 UvKv
726. If a + 2b = 6 and ab = 4 what is
2
a
+
1
b
? (hw` a
+ 2b = 6 Ges ab = 4 nq, Z‡e
2
a
+
1
b
Gi gvb KZ?)
[Sonali Bank (S.O.-2018); Sonali Bank (Officer-2018)]
a
1
2
b 1 c
3
2
d 2 c
 mgvavb : a + 2b = 6 ......... (i)
ab = 4 .............. (ii)
(i)  (ii) 
a + 2b
ab
=
6
4
=
3
2

1
b
+
2
a
=
3
2

2
a
+
1
b
=
3
2

727. If
y
x
=
3
7
and x + 2y = 13 then y is–
[Bangladesh Bank (A.D.-2018)]
a 2 b 3
c 4 d 7 b
 mgvavb :
y
x
=
3
7
 x =
7
3
y
Avevi, x + 2y = 13

7
3
y + 2y = 13

7y + 6y
3
= 13
 13y = 13  3
 y = 3
728. If Kabir loses 8 pounds, he will weigh twice
as much as his sister. Together they now weigh
278 pounds. What is Kabir’s present weight,
in pounds? (hw` Kwei 8 cvDÛ IRb nvivq, Zvi
IRb Zvi †ev‡bi IR‡bi wØ¸Y n‡e| GLb `yB R‡bi
GK‡Î IRb 278 cvDÛ| Kwe‡ii eZ©gvb IRb)
a 147 b 188 c 135 d 139 b
 mgvavb : awi, Kwe‡ii eZ©gvb IRb = x cvDÛ
 Kwe‡ii †ev‡bi eZ©gvb IRb = (278 – x) cvDÛ
cÖkœg‡Z, x – 8 = 2 (278 – x)
 x – 8 = 556 – 2x
 3x = 564  x = 188
729. A man spent
1
2
of his money and then lost
1
4
of the remainder. He was left with Tk. 3,600.
How much did he start with? (GKRb e¨w³ Zvi
UvKvi
1
2
Ask LiP K‡i Ges evwK UvKvi
1
4
Ask nvwi‡q
†d‡j| Zvi Kv‡Q 3600 UvKv _vK‡j †m KZ UvKv wb‡q
ïiæ K‡iwQjÑ) [Sonali Bank (Officer-2018)]
a Tk. 7,200 b Tk. 8,800
c Tk. 9,600 d Tk. 10,400 c
 mgvavb : g‡b Kwi, e¨w³wUi Kv‡Q x UvKv wQj|
 LiP K‡i = x Gi
1
2
=
x
2
UvKv
 evwK UvKv = x –
x
2
=
x
2
UvKv
 nvwi‡q †d‡j =
x
2
Gi
1
4
=
x
8
UvKv
 evwK UvKv =
x
2
–
x
8
=
4x – x
8
=
3x
8
UvKv
cÖkœg‡Z,
3x
8
= 3600
 3x = 3600  8  x = 9600 UvKv
730. The population of a certain town increases
by 50% every 50 years. If the population in
1950 was 810, in what year was the population
160? (†Kv‡bv kn‡ii RbmsL¨v 50 eQ‡i 50% e„w×
cvq| 1950 mv‡j RbmsL¨v 810 n‡j, †Kvb mv‡j
RbmsL¨v 160 wQj?) [Agrani Bank (Officer Cash-2017)]
a 1650 b 1800 c 1700 d 1750 d
 mgvavb : g‡b Kwi, x mv‡j RbmsL¨v 160 wQj
 (x + 50) mv‡j RbmsL¨v = 160 + 160 Gi 50%
= 160 + 80 = 240
(x + 50 + 50) ev (x + 100) mv‡j RbmsL¨v
= 240 + 240 Gi 50%
= 240 + 240 
50
100
= 360
(x + 100 + 50) ev (x + 150) mv‡j RbmsL¨v
= 360 + 360 Gi 50%
= 360 + 360 
50
100
= 540
(x + 150 + 50) ev (x + 200) mv‡j RbmsL¨v
= 540 + 540 Gi 50% = 810
cÖkœg‡Z, x + 200 = 1950
 x = 1750
731. 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? (†Kv‡bv K¬v‡mi ïiæ‡Z
A‡a©K wkÿv_x© jvB‡eªix‡Z hvq| K¬v‡mi gv‡S evwK
wkÿv_x©‡`i A‡a©K Kw¤úDUvi j¨v‡e hvq| Ae‡k‡l 8
Rb wkÿv_x© K¬v‡m _v‡K| †gvU wkÿv_x© msL¨v KZ wQj?)
a 12 b 16 c 24 d 32 d
 mgvavb : awi, wkÿv_x© msL¨v wQj x Rb
K¬v‡mi ïiæ‡Z jvB‡eªix‡Z hvq = x Gi
1
2
Rb =
x
2
Rb
 evwK wkÿv_x© = x –
x
2
=
x
2
Rb
c‡i Kw¤úDUvi j¨v‡e hvq =
x
2
Gi
1
2
=
x
4
Rb
cÖkœg‡Z, x –



x
2
+
x
4
= 8
 x –
2x + x
4
= 8

4x – 3x
4
= 8
 x = 32 Rb

MATHEMATICS 291
Written
732. 2n – 1
+ 2n + 1
= 320 n‡j n Gi gvb KZ?
[Bangladesh Bank (Recruitment Test-2020)]
 Solution: 2n – 1
+ 2n + 1
= 320
 2n – 1
+ 2n – 1 + 2
= 320
 2n – 1
+ 2n – 1
 22
= 320
 2n – 1
+ 4  2n – 1
= 320
 5  2n – 1
= 320
 2n – 1
= 64
 2n – 1
= 26
 n – 1 = 6
 n = 7
733. Solve :
x
2
+
6
y
= 9,
x
3
+
2
y
= 4
[Bangladesh Development Bank Ltd. (SO)-2018;
Rupali Bank Ltd. (Officer Cash) Cancelled-2018]
 Solution:
x
2
+
6
y
= 9 ................... (i)
x
3
+
2
y
= 4 (ii)
(i)  (ii)  3 
x
2
+
6
y
= 9
x +
6
y
= 12
() () ()
x
2
 x =  3

x  2x
2
=  3 
 x
2
=  3  x = 6
putting value of x in equation .... (i)
6
2
+
6
y
= 9
 3 +
6
y
= 9 
6
y
= 6  y =
6
6
= 1
 (x, y)  (6, 1) (Ans.)
734. Price of 3 tables and 5 chairs is Tk. 2,000. Again,
price of 5 tables and 7 chairs is Tk. 3,200.
What is the price of 1 table and 1 chair? (3wU
†Uwej Ges 5wU †Pqv‡ii g~j¨ 2000 UvKv| Avevi 5wU
†Uwej Ges 7wU †Pqv‡ii g~j¨ 3200 UvKv| 1wU †Uwej
Ges 1wU †Pqv‡ii g~j¨ KZ?)
[Bangladesh Krishi Bank Ltd. (Officer Cash-2018);
 Solution: Let, Price of 1 Table = x Tk
and Price of 1 chair = y Tk
According to question,
3x + 5y = 2000 ... ... (i)
5x + 7y = 3200 ... ... (ii)
(i)  5  (ii)  3 
15x + 25y = 10000
15x + 21y = 9600
() () ()
4y = 400
 y = 100
Putting value of y in equation ... ... (i)
3x + 5  100 = 2000
 3x =1500  x = 500
 Price of 1 table = 500 Tk
and Price of 1 chair = 100 Tk
735. A working couple earned a total of Tk.
43,520. The wife earned Tk. 640 per day, the
husband earned Tk. 560 per day. If the total
number of days worked by both was 72,
formulate a system of equation and solve the
system to find the number of days worked by
each. (GK Kg©Rxex `¤úwZ †gvU 43520 UvKv DcvR©b
K‡i| ¯¿x 640 UvKv cÖwZw`b Avq K‡i Ges ¯^vgx 560
UvKv cÖwZw`b Avq K‡i| hw` Df‡q wg‡j 72 w`b KvR
K‡i| GKwU mgxKiY MVb K‡i †K KZw`b KvR K‡i
Zv wbY©q Ki?) [Agrani Bank Ltd. (Officer Cash-2018)]
 Solution: Total working days = 72
Let, husband worked for x days
 wife worked for (72  x) days
560  x + (72  x)  640 = 43520
 560x + 46080  640x = 43520
 46080  43520 = 80x
 80x = 2560
 x = 32 days
 husband worked for 32 days
 wife worked for (72  32) = 40 days
736. Solve :
x
2
+
y
3
= 1,
x
3
+
y
2
= 1
[Agrani Bank Ltd. (Officer Cash-2018)]
 Solution:
x
2
+
y
3
= 1 .................... (i)
x
3
+
y
2
= 1 ................... (ii)
(i)  2  (ii)  3 
x +
2y
3
= 2
x +
3y
2
= 3
() () ()
2y
3

3y
2
=  1

4y  9y
6
=  1   5y =  6  y =
6
5

Putting value of y in equation ... (i)
x
2
+
6
5
3
= 1

x
2
+
6
5

1
3
= 1

x
2
+
2
5
= 1 
x
2
= 1 
2
5

x
2
=
3
5
 x =
6
5
 (x, y) 



6
5

6
5
737. Amin has 12 pieces of Tk. 10 and Tk. 5 notes
in his wallet. If the total value of all the notes is
less than Tk. 95, what is the maximum
number of Tk. 10 notes that he has? (Avwg‡bi
Kv‡Q 10 UvKv Ges 5 UvKvi †gvU 12wU †bvU Av‡Q| hw`
†gvU UvKvi cwigvY 95 Gi †P‡q Kg nq| m‡e©v”P
KZwU 10 UvKvi †bvU _vK‡Z cv‡i?)
 Solution: As, total notes = 12
Let, Number of 10 Tk notes = x
 ,, ,, 5 Tk ,, = (12  x)
According to the question
10x + 5(12  x) < 95
 10x + 60  5x < 95
 5x < 35  x < 7
 Maximum number of 10 taka notes will be = 6
738. Solve the equation :
3
x + 2
+
x  1
x  5
= 2
 Solution: Given equation,
3
x + 2
+
x  1
x  5
= 2

3
x + 2
= 2 
x  1
x  5
=
2x  10  x + 1
x  5

3
x + 2
=
x  9
x  5
 3(x  5) = (x + 2) (x  9)
 3x  15 = x2
 9x + 2x  18
 3x  15 = x2
 7x  18
 x2
 7x  18  3x + 15 = 0
 x2
 10x  3 = 0
We know,
Solve of this equation ax2
+ bx + c = 0
x =
 b  b2
 4ac
2a
Comparing this equation with ax2
+ bx + c = 0
a = 1, b =  10, c =  3
 x =
 ( 10)  ( 10)2
 4  1  ( 3)
2  1
=
10  100 + 12
2
=
10  112
2
=
10  16  7
2
=
10  4 7
2
=
2(5  2 7)
2
 x = 5  2 7 (Ans.)
739. Four students contributed to a charity drive
and the average amount contribution by
each student was BDT 20. If no student gave
more than BDT 25, what is the minimum
amount that any student could have
contributed? (4 Rb QvÎ GKwU `vZe¨ ms¯’vq `vb
Ki‡jv Ges M‡o mevB 20 UvKv K‡i w`‡jv| †Kv‡bv
QvÎB 25 UvKvi †ewk †`q bv| GKRb QvÎ me©wb¤œ KZ
UvKv w`‡q‡Q?) [Sonali & Janata Bank Ltd. (SO) IT/ICT-2018]
 Solution: 1 student contributes 20 Tk
 4 ,, ,, (4  20) Tk
= 80 Tk
No student gives, more than 25 Tk
 Let, out of 4 students, 3 students give 25 Tk
 Total contribution of 3 students = 3  25
= 75 UvKv
 minimum amount that any student can
contribute = 80  75 = 5 Tk
740. A school has 40 rooms that can sit 600 people.
Some rooms can sit 10 people and some can
sit 20 people. What is the ratio of the number
of 10 person rooms to the number of 20
person rooms? (GKwU ¯‹z‡j 600 Rb e¨w³i Rb¨
40wU Kÿ Av‡Q| wKQz K‡ÿ 10 Rb Ges wKQz K‡ÿ 20
Rb em‡Z cv‡i| 10 R‡bi Kÿ Ges 20 R‡bi K‡ÿi
AbycvZ KZ?) [Sonali & Janata Bank Ltd. (SO) IT/ICT-2018]
 Solution:Let, Number of 10 seated room = x
 ,, ,, 20 ,, ,, = (40  x)
According to the question,
10x + 20 (40  x) = 600
 10x + 800  20x = 600
 200 = 10x
 x = 20
 Number of 10 seated room = 20
 ,, ,, 20 ,, ,, = 40  20 = 20
 Ratio = 20 : 20 = 1 : 1

MATHEMATICS 293
741. A teacher has 3 hours to grade all the papers
submitted by the 35 students in her class. She
gets through the first 5 papers in 30 minutes.
How much faster does she have to work to
grade the remaining papers in the allotted
time? (GKRb wkÿ‡Ki 35 Rb wkÿv_x©i LvZv †`Lvi
Rb¨ 3 NÈv mgq Av‡Q| cÖ_g 5wU LvZv †`L‡Z 30 wgwbU
mg‡q jv‡M| eivÏK…Z mg‡q LvZv †`Lv †kl Ki‡Z KZ
`ªæZ LvZv †`L‡Z n‡e?) [Sonali Bank Ltd. (SO) IT/ICT-2018]
 Solution: To grade first 5 papers the teacher
takes 30 minutes or,
30
60
hours or
1
2
hours
Now, The teacher takes
1
2
hours to grade 5 papers
 ,, ,, ,, 1 ,, ,, ,, 10 ,,
 speed of grading = 10 papers/hour
Remaining alloted time = 3 
1
2
= 2.5 hours
Remaining papers to grade = 35  5 = 30
In 2.5 hours the teacher has to grade 30 papers
 ,, 1 ,, ,, ,, ,, ,, ,,
30
2.5
,,
= 12 papers
speed has to be = 12 papers/hour
 percentage of speed has to be increased
` =
12  10
10
 100% = 20%
742. Solve : x2
 yx = 7, y2
+ xy = 30
[Sonali Bank Ltd. (SO) IT/ICT-2018]
 Solution: x2
 yx = 7 ................ (i)
y2
+ xy = 30 ............. (ii)
(i) + (ii)  x2
+ y2
= 37 ........... (iii)
From (i), x2
 7 = yx
 y =
x2
 7
x
=
x2
x

7
x
 y = x 
7
x
............................. (iv)
Putting the value of y in equation (iii)
x2
+



x 
7
x
2
= 37
 x2
+ x2
 2.x.
7
x
+



7
x
2
= 37
 2x2
 14 +
49
x2 = 37
 2x2
+
49
x2 = 51

2x4
+ 49
x2 = 51
 2x4
+ 49 = 51x2
 2x4
 51x2
+ 49 = 0
 2x4
 49x2
 2x2
+ 49 = 0
 x2
(2x2
 49)  1 (2x2
 49) = 0
 (x2
 1) (2x2
 49) = 0
x2
 1 = 0 or, 2x2
 49 = 0
 x2
= 1  x2
=
49
2
 x =  1  x = 
7
2
From equation (iv),
if x = 1, y = 1 
7
1
=  6
if x = 1, y = (1) 
7
(1)
= 6
if x =
7
2
, y =
7
2

7
7
2
=
7
2
 7 
2
7
=
7
2
 2
=
7  2
2
=
5
2
if x = 
7
2
, y = 
7
2

7




7
2
= –
7
2
+ 7 
2
7
= –
7
2
+ 2 =
 7 + 2
2
= 
5
2
So, (x, y)  (1,  6), (1, 6),



7
2

5
2
,



 7
2

 5
2
809. A car owner buys petrol at Tk. 75, Tk. 80
and Tk. 85 per liter for three successive years.
What approximately is the average cost per
liter of petrol if he spends Tk. 40,000 each year
in this concern? (GKRb Mvoxi gvwjK cici wZb
eQi 75 UvKv, 80 UvKv Ges 85 UvKv w`‡q †c‡Uªvj wK‡b|
†m cÖwZ eQi †gvU 40000 UvKv LiP Ki‡j, M‡o cÖwZ
wjUv‡i KZ LiP n‡qwQj?) [Combined 5 Bank (Officer)-2018]
 Solution:
Amount of petrol used in 1st year =
40000
75
=
1600
3
liter
,, ,, ,, ,, ,, 2nd year =
40000
80
= 500 liter
,, ,, ,, ,, ,, 3rd year =
40000
85
=
8000
17
liter

 Total amount of petrol used in 3 years
=



1600
3
+ 500 +
8000
17
liter
=
27200 + 25500 + 24000
51
liter
=
76700
51
liter
And, total cost in 3 years = 3  40000 = 120000
 Average cost =
120000
76700
51
= 120000 
51
76700
= 79.79 Taka
743. A family has 480kg of rice for X number of
weeks. If they need to use the same amount
for 4 more weeks, they need to cut down their
weekly assumption of rice by 4 kg. What is
the value of X? (GKwU cwiev‡ii X mßv‡ni Rb¨
480 kg Pvj Av‡Q| hw` H Pvj Av‡iv 4 mßvn †ewk
eënvi Ki‡Z Pvq, Zv‡ì mvßvwnK Lvev‡ii cwigvY 4
kg Kgv‡Z n‡e| X Gi gvb KZ?)
[Rupali Bank Ltd. (Officer Cash-2018)]
 Solution: As, X is number of weeks and 480 kg
is quantity of rice
 per week consumption =
480
x
kg
If 4 weeks is to be increased than (x + 4) is
number of weeks
Again, per week consumption =
480
x + 4
kg
480
x

480
x + 4
= 4

480 (x + 4)  480x
x(x + 4)
= 4

480x + 1920  480x
x(x + 4)
= 4
 1920 = 4x2
+ 16x
 x2
+ 4x  480 = 0
 x2
+ 24x  20x  480 = 0
 x(x + 24)  20(x + 24) = 0
 (x  20) (x + 24) = 0
So, x  20 = 0 or, x + 24 = 0
 x = 20  x =  24
but number of weeks can't be negative.
744.
1
2x
+
6
y
= 3,
5
x
+
3
y
= 11
[Rupali Bank Ltd. (Officer Cash-2018)]
 Solution: Given equations,
1
2x
+
6
y
= 3 ......... (i)
5
x
+
3
y
= 11 ........ (ii)
(i)  (ii)  2 
1
2x
+
6
y
= 3
10
x
+
6
y
= 22
() () ()
1
2x

10
x
=  19

1  20
2x
=  19 
 19
2x
=  19  2x = 1
 x =
1
2
Putting value of x in equation (i)
1
2 
1
2
+
6
y
= 3
 1 +
6
y
= 3 
6
y
= 2  y = 3
 (x, y) 



1
2
 3
5. Series
M.C.Q
745. The sum of first 17 terms of the series 5, 9,
13, 17, — is– (5, 9, 13, 17 — avivwUi cÖ_g 17
c‡ì mgwó—) [Bangladesh Bank (Officer General-2019);
Sonali Bank (S.O.-2018); Agrani Bank (Officer Cash-2017)]
a 529 b 462 c 629 d 523 c
 mgvavb : d = 9 – 5 = 13 – 9 = 4
 avivwU mgvšÍi aviv hvi cÖ_g c`, a = 5
mvaviY AšÍi, d = y
 17wU c‡ì mgwó =
n
2
{2a + (n – 1)d}
=
17
2
{2  5 + (17 – 1) 4}
=
17
2
{10 + 64} =
17
2
 74
= 629
746. The next number in the sequence 3, 6, 11, 18,
27, — is– [Sonali Bank (S.O.-2018); Bangladesh Bank (Office
General-2019); Rupali Bank Ltd. (Officer-2019)]
a 34 b 36 c 38 d 40 c
 mgvavb : cÖ`Ë avivwU :
3, 6, 11, 18, 27 27 + 11
+ 3 + 5 + 7 + 9 9 + 2 ev + 11
ev 38
+ 2 + 2 + 2 + 2
 cieZ©x c` = 27 + 11 = 38

MATHEMATICS 295
747. The next number in the sequence 3, 4, 8, 17,
33, ... is (3, 4, 8, 17, 33, ......... avivwUi cieZ©x c`
KZ?) [Combined 5 Banks (Officer-2018)]
a 54 b 56 c 58 d 60 c
 mgvavb : 8 17 33 58
42
52
3 4
22
32
12
748. The second and third terms of a geometric
series are 9 and 3 respectively. The fifth term
of the series is– (†Kv‡bv R¨vwgwZK ev ¸‡YvËi avivi
wØZxq I Z…Zxq c` h_vµ‡g 9 I 3| avivwUi cÂg c`Ñ)
a 1 b
1
9
c
1
3
d
1
27
c
 mgvavb : awi, cÖ_g c` = a
mvaviY AbycvZ = r
 n Zg c`, Tn = arn – 1
wØZxq c`, T2 = ar2 – 1
= ar
Z…Zxq c`, T3 = ar3 – 1
= ar2
cÖkœg‡Z, ar = 9 ................... (i)
ar2
= 3 ................. (ii)
(ii)  (i) 
ar2
ar
=
3
9
 r =
1
3
(i) bs G r Gi gvb ewm‡q,
a 
1
3
= 9
 a = 27
 cÂg c`, T5 = ar4
 T5 = 27 



1
3
4
 T5 =
27
34 =
33
34
 T5 =
1
3
749. n
C1 + n
C2 + n
C3 + ... n
Cn = ?
a 2n
b 2n – 1
c
n(n – 1) (n2
+ 1)
2
d 2n
– 1 d
 mgvavb : (1 + x)n
= n
C0 + n
C1x + n
C2x2
+ .... + n
Cnxn
 (1 + x)n
= 1 + n
C1x + n
C2x2
+ n
C3x3
+ .... + n
Cnxn
x = 1 n‡j,
(1 + 1)n
= 1 + n
C1 + n
C2 + n
C3 + .... + n
Cn
 2n
= 1 + n
C1 + n
C2 + .... + n
Cn
 n
C1 + n
C2 + .... + n
Cn = 2n
– 1
750. A person buys a TV worth BDT 3,90,000
with a down payment of 40,000, including
Tk. 5000 as first month’s installment. How
many more installments does he have to pay
if his installments had to double after each
successive payment? (GKRb e¨w³ cÖ_g gv‡mi
5000 UvKv wKw¯Ímn GKKvjxb 40000 UvKv cÖ`vb K‡i
390000 UvKvi GKwU TV µq K‡ib| hw` cÖ‡Z¨K gvm
AšÍi wKw¯Íi cwigvY wØ¸Y nq, Z‡e Zuv‡K AviI KZwU
wKw¯Í cwi‡kva Ki‡Z n‡e?)
a 6 b 7 c 8 d 10 a
 mgvavb : TV wUi g~j¨ = 390000 UvKv
cÖ_g gv‡mi wKw¯Ímn cwi‡kvaK…Z g~j¨ = 40000 UvKv
 evwK UvKv = (390000 – 40000) UvKv
= 350000 UvKv
cÖkœg‡Z, cÖ_g gv‡mi wKw¯Í 5000 UvKv n‡j cieZ©x
gv‡m 10000, Zvi c‡ii gv‡m 20000 Ges Gfv‡e
Pj‡Z _vK‡e|
 10000 + 20000 + 40000 + ......... = 350000
evgcv‡ki avivwUi cÖ_g c`, a = 10000
mvaviY AbycvZ, r = 2 [wØ¸Y n‡”Q]
hw` cieZ©x n gvm a‡i wKw¯Í cwi‡kva Ki‡Z nq, Z‡e
H n gv‡mi wKw¯Íi mgwó =
a(rn
– 1)
r – 1
cÖkœg‡Z,
a(rn
– 1)
r – 1
 350000

10000  (2n
– 1)
2 – 1
 350000
 2n
– 1  35
 2n
 36
n = 5 n‡j, 25
= 32 < 36
n = 6 n‡j, 26
= 64 > 36
 Zuv‡K AviI 6 gvm wKw¯Í cwi‡kva Ki‡Z n‡e|
weKí mgvavb : cÖ_g gv‡mi wKw¯Ímn 40000 UvKv cÖ`v‡bi
ci evwK UvKv = (390000 – 40000) = 350000 UvKv
cieZ©x n gv‡m H UvKv †kva Ki‡Z n‡j,
cÖkœg‡Z, 5000  2 + 5000  2  2 + .... = 350000
a = 2  5000 = 10000
r = 2

arn
– 1
r – 1
 350000
 10000  (2n
– 1)  350000
 2n
– 1  35
 2n
 36
 2n
 32 + 4
 2n
 25
+ 4
 n > 5; AZGe n = 5 + 1 = 6 gvm

751. 6, 7, 9, 13, __, __. What are the two missing
numbers in the series?
a 21, 37 b 17, 21
c 21, 39 d 17, 19 a
 mgvavb : 9 13 21 37
23
=8 24
=16
6 7
2=21
4=22
1=20
752. How many terms of Arithmetic Progression
(A. P.) 21, 18, 15, 12, ... must be taken to give
the sum zero? (21, 18, 15, 12, ...... mgvšÍi
avivwUi KZ msL¨K c‡ì mgwó k~Y¨ n‡e?)
a 10 b 15 c 22 d 27 b
 mgvavb : mgvšÍi avivwUi cÖ_g c`, a = 21
mvaviY AšÍi, d = 18 – 21
= – 3
awi, n msL¨K c‡ì mgwó = 0
n msL¨K c‡ì mgwó =
n
2
{2a + (n – 1)d}
cÖkœg‡Z,
n
2
{2a + (n – 1)d} = 0
 {2a + (n – 1)d} = 0 [n  0]
 2  21 + (n – 1)  (– 3) = 0
 3(n – 1) = 2  21
 n – 1 =
2  21
3
 n – 1 = 14
 n = 15
753. After being dropped a certain ball always
bounces back to
2
5
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
we‡kl ej f‚wg‡Z covi ci me©`v Gi c~‡e©i D”PZvi
2
5
¸Y D”PZvq jvwd‡q I‡V| cÖ_g evi f‚wg‡Z cZ‡bi ci
GwU 125 BwÂ I‡V| PZz_© evi cZ‡bi ci GwU KZ
D”PZvq DV‡e?) [Sonali Bank (S.O.-2018)]
a 20 b 8 c 5 d 3.2 b
 mgvavb : cÖ_g evi cZ‡bi ci I‡V 125 BwÂ
 wØZxq Ó Ó Ó Ó
2
5
 125 BwÂ
= 50 BwÂ
Z…Zxq Ó Ó Ó Ó
2
5
 50 BwÂ
= 20 BwÂ
PZz_© Ó Ó Ó Ó
2
5
 20 BwÂ
= 8 BwÂ
weKí mgvavb :
cÖ‡Z¨KwU cZ‡bi ci ejwUi DÌvb c~‡e©i
2
5
¸Y nq|
ZvB GB DwÌZ D”PZv¸‡jv GKwU ¸‡YvËi aviv hviÑ
cÖ_g c`, a = 125
mvaviY AbycvZ, r =
2
5
 PZz_© c` = arn – 1
= 125 



2
5
4 – 1
= 125 



2
5
3
= 125 
8
125
= 8
 PZz_© cZ‡bi ci ejwU 8 BwÂ DV‡e|
754. If a + 1, 2a + 1, 4a – 1 are in Arithmetic
Progression then the value of ‘a’ is : (hw` a +
1, 2a + 1, 4a – 1 mgvšÍi avivq _v‡K, Z‡e 'a' Gi
gvbÑ) [Rupali Bank (Officer Cash-2018)]
a 1 b 2 c 3 d 4 b
 mgvavb : mgvšÍi avivi c`¸‡jvi exRMvwYwZK eëavb
mgvb _v‡K hv‡K mvaviY AšÍi e‡j|
cÖkœg‡Z,
mvaviY AšÍi = (2a + 1) – (a + 1) = (4a – 1) – (2a + 1)
 a = 2a – 2
 a = 2
755. In a series of 6 consecutive odd numbers if
15 is the 6th
number, what is the 4th
number
in the series? (6wU µwgK we‡Rvo msL¨vi 6th
msL¨vwU 15 n‡j, H avivi 4th
msL¨vwU KZ?)
[Bangladesh Bank (A.D.-2018); B.K.B. (Officer Cash-2017)]
a 7 b 9 c 11 d 13 c
 mgvavb : awi, cÖ_g we‡Rvo msL¨vwU a
 msL¨v¸‡jv a, a + 2, a + 4, a + 6, a + 8, a + 10|
6th
msL¨vwU = a + 10
 a + 10 = 15
 a = 5
 4th
msL¨vwU = a + 6 = 5 + 6 = 11
756. Which number logically follows the sequence?
4 6 9 6 14 6 ... (†Kvb msL¨vwU 4, 6, 9, 6, 14, 6 ...
avivwUi cieZx© msL¨v?) [Bangladesh Bank (Officer-2018)]
a 6 b 17 c 19 d 21 c
 mgvavb : †Rvo ¯’v‡bi msL¨v¸‡jv A_©vr wØZxq, PZz_©,
lô, .... c`¸‡jv 6|
Avgv‡ì mßg c` †ei Ki‡Z n‡e, hv we‡Rvo ¯’v‡bi
c` we‡Rvo ¯’vbxq c`¸‡jv:
myZivs, mwVK c`wU 19|
cÖ_g Z…Zxq cÂg mßg
4
+ 5
9
+ 5
14
+ 5 19

MATHEMATICS 297
757. The next number of the sequence is–
4 3 9 3 19 3 ... [Bangladesh Bank (A.D.-2018)]
a 31 b 32 c 39 d 49 c
 mgvavb : 4 3 9 3 19 3 39
5 52=10 102=20
758. 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 bvU¨kvjvi GK mvwi Avmb‡K cici T1 †_‡K
T50 ch©šÍ µwgK EaŸ©µ‡g wPwýZ Kiv Av‡Q| mygb I mRxe
h_vµ‡g T17 I T39 bs Avm‡b e‡m‡Q| `yR‡bi gv‡S
KZ¸‡jv Avmb Av‡Q?) [Bangladesh Bank (Officer-2018)]
a 23 b 21 c 22 d 20 b
 mgvavb : T17 I T39 Gi gaëZx© Avmb¸‡jv n‡jv: T18,
T19, T20, .... T37, T38
†gvU Avmb msL¨v = (38 – 18 + 1)wU = 21wU
759. In the given series, which number will come
next : 91, 86, 76, 61, —? [Sonali Bank (Officer-2018)]
a 31 b 36 c 41 d 46 c
 mgvavb : 91 86 76 61
– 5 –10 –15
 cieZx© msL¨v = 61 – 20 = 41
760. If the first and sixth term of a geometric series
are respectively
1
2
and
1
64
, then the common
ratio is– (†Kv‡bv ¸‡YvËi avivi cÖ_g I lô c`
h_vµ‡g
1
2
Ges
1
64
n‡j, mvaviY AbycvZÑ)
[B.D.B.L. (S.O.-2017)]
a
1
4
b
1
2
c 1 d 2 b
 mgvavb : ¸‡YvËi avivi n Zg c` = arn–1
†hLv‡b, a = cÖ_g c`, r = mvaviY AbycvZ
GLb, cÖ_g c` a =
1
2
n = 6 Zg c` =
1
2
r6 – 1
=
1
2
r5
cÖkœg‡Z,
1
2
r5
=
1
64
 r5
=
1
32
 r5
=



1
2
5
 r =
1
2
761. The sum of fourth and twelfth term of an
arithmetic progression is 20. What is the sum
of the first fifteen terms of that arithmetic
progression? (GKwU mgvšÍi avivi 4_© Ges 12Zg
c‡ì †hvMdj 20| avivwUi cÖ_g 15 c‡ì mgwóÑ)
[B.H.B.F.C. (S.O.-2017)]
a 300 b 120 c 150 d 130 c
 mgvavb : mgvšÍi avivi 1g c` a Ges mvaviY AšÍi d n‡j,
PZz_© c` = a + (4 – 1)d = a + 3d
Ges 12 Zg c` = a + (12 – 1)d = a + 11d
cÖkœg‡Z,
a + 3d + a + 11d = 20
 2a + 14d = 20 ...................(i)
GLb, cÖ_g 15 c‡ì mgwó =
15
2
{2a + (15 –1)d}
=
15
2
{2a + 14d} =
15
2
 20 [(i) n‡Z gvb ewm‡q]
= 150
Written
762. `yB A¼wewkó †hmKj msL¨v 3 Øviv wefvR¨ Zv‡ì
†hvMdj wbY©q Kiæb|
[Bangladesh Bank (Recruitment Test-2020)]
 Solution: 3 Øviv wefvR¨ `yB As‡Ki msL¨v¸‡jv n‡jv
: 12, 15, 18, 21, 24, ......... 90, 93, 96, 99
†hvMdj, S = 12 + 15 + 15 + 18 + ....... + 96 + 99
hv GKwU mgvšÍi avivi mgwó †hLv‡bÑ
cÖ_g c`, a = 12
†kl c` = 99
mvaviY AšÍi, d = 3 [15 – 12 =18 – 15 = 96 – 99 = 3]
†gvU Giƒc msL¨v n n‡j
99 = a + (n – 1)d
 99 = 12 + 3(n – 1)
 3(n – 1) = 87  n – 1 = 29
 n = 30
 s =
n
2
 (cÖ_g c` + †kl c`)
=
30
2
 (12 + 99) = 15  111 = 1665
763. Two numbers x, y and in G.P and their sum
is 30. Also, the sum of their squares is 468.
Find the numbers. (`yBwU msL¨v x Ges y ¸‡YvËi
aviv‡Z Av‡Q Ges Zv‡ì mgwó 30| msL¨vØ‡qi e‡M©i
mgwó 468| msL¨vØq KZ?) [Rupali Bank Ltd. (Officer-2019)]
 Solution: Given, x + y = 30  x = 30 – y .... (i)
Again, x2
+ y2
= 468
 (30 – y)2
+ y2
= 468
 900 – 60y + y2
+ y2
= 468
 2y2
– 60y + 900 – 468 = 0
 2y2
– 60y + 432 = 0
 y2
– 30y + 216 = 0
 y2
– 18y – 12y + 216 = 0
 y (y – 18) – 12(y – 18) = 0
 (y – 18) (y – 12) = 0
 y = 12, 18
If, y = 12, x = 30 – 12 = 18
y = 18, x = 30 – 18 = 12
 Numbers are 12, 18.

764. The sum of three numbers in Arithmetic
Progression is 30, and the sum of their
squares is 318. Find the numbers. (GKwU
mgvšÍi avivi 3wU msL¨vi mgwó 30 Ges Zv‡`i e‡M©i
†hvMdj 318 msL¨v wZbwU KZ?)
 Solution:
Let, first term of series = a
 2nd term = a + d d is common difference
and, 3rd term = a + 2d
a + a + d + a + 3d = 30
 3a + 3d = 30  a + d = 10
 a = 10  d ... (i)
Again, a2
+ (a + d)2
+ (a + 2d)2
= 318
 (10  d)2
+ (10  d + d)2
+ (10  d + 2d)2
=
318 From (i)
 100  20d + d2
+ 100 + (10 + d)2
= 318
 200  20d + d2
+ 100 + 20d + d2
= 318
 2d2
+ 300 = 318  2d2
= 18  d2
= 9
 d =  3
From equation (i)
If d = 3, a = 10  3 = 7
d =  3, a = 10  (3) = 13
For, d = 3 and a = 7
1st term = 7
2nd term = 7 + 3 = 10
3rd term = 7 + 3  2
= 13
For d =  3 and a = 13
1st term = 13
2nd term = 13  3 = 10
3rd term = 13  3  2
= 7
 Three numbers are 7, 10, 13
765. Prove that the sum of the odd numbers from
1 to 125 inclusive is equal to the sum of the
odd numbers from 169 to 209 inclusive. (cÖgvY
Ki †h, 1 †_‡K 125 Gi g‡a¨ we‡Rvo msL¨v¸‡jvi
†hvMdj 169 †_‡K 209 Gi g‡a¨ we‡Rvo msL¨v¸‡jvi
†hvMd‡ji mgvb|) [Agrani Bank Ltd. (Officer Cash-2018)]
 Solution:
Sum of odd numbers from 1 to 125
1 + 3 + 5 + ... + 125
First term a = 1, common difference d = 3  1 = 2
Last term or nth
term = 125
 a + (n  1)d = 125
 1 + (n  1)2 = 125
 1 + 2n  2 = 125  2n = 126
 n = 63
Now,
Sum of n terms =
n
2
{2a + (n  1)d}
 ,, ,, 63 ,, =
63
2
{2  1 + (63  1) 2}
=
63
2
{2 + 124}
= 63  63 = 3969
Sum of odd numbers from 169 to 209
169 + 171 + 173 + ...... + 209
Here, first term a = 169, common difference
d = 171  169 = 2
nth
term = 209
 a + (n  1)d = 209
 169 + (n  1) 2 = 209
 2n  2 = 40
 2n = 42
 n = 21
Sum of 4 terms =
21
2
{2  169 + (21  1)  2}
=
21
2
{338 + 40} =
21
2
 378 = 3969
 1 + 3 + 5 + .... + 125 = 169 + 171 + ... + 209
6. Set
M.C.Q
766. If A = {1, 2, 3, 4, 5}, then the number of proper
subsets of A is– (A = {1, 2, 3, 4, 5} Zvn‡j A Gi
cÖK…Z Dc‡mU KZwUÑ) [Bangladesh Bank (Officer
General-2019); Rupali Bank Ltd. (Officer-2019);
Sonali Bank (S.O.-2018); Agrani Bank (Officer Cash-2017)]
a 120 b 30 c 31 d 32 c
 mgvavb : A = {1, 2, 3, 4, 5}
 A †m‡U Dcv`vb msL¨v n = 5
Avgiv Rvwb, n m`m¨wewkó †Kv‡bv †m‡Ui cÖK…Z Dc‡mU
msL¨v = 2n
– 1|
 cÖK…Z Dc‡mU = 2n
– 1 = 25
– 1 = 31
767. In a group of 60 people, 27 like coke and 42
like borhani and each person likes at least
one of the two drinks. How many people like
both coke and borhani? (60 Rb e¨w³i GKwU
MÖæ‡c 27 Rb †KvK I 42 Rb †evinvwb cQ›` K‡i Ges
cÖ‡Z¨‡KB `ywUi AšÍZ GKwU cQ›` K‡i| KZRb e¨w³
†KvK Ges †evinvwb `y‡UvB cQ›` K‡i?)
a 9 b 129 c 69 d 15 a
 mgvavb : †gvU e¨w³, n(S) = 60
†KvK cQ›` K‡i Ggb e¨w³i msL¨v, n(C) = 27
†evinvwb cQ›` K‡i Ggb e¨w³i msL¨v, n(B) = 42
†KvK I †evinvwb Dfq cQ›` K‡i, Ggb e¨w³i msL¨v,
n(C  B) = ?
†h‡nZz cÖ‡Z¨‡KB †Kv‡bv bv †Kv‡bv cvbxq cQ›` K‡i
ZvB, AšÍZ GKwU cQ›` K‡i Ggb e¨w³i msL¨v,
n(C  B) = †gvU e¨w³i msL¨v, n(S) = 60
n(C  B) = n(C) + n(B) – n(C  B)
 n(C  B) = n(C) + n(B) – (C  B)
 n(C  B) = 27 + 42 – 60
 n(C  B) = 9

MATHEMATICS 299
768. In a room there are six Bengali, twelve engineers
and fifteen football players. Only one of them
was a Bengali Engineer who played football.
Two were Bengali Engineers but did not play
football and two were Bengali football players
and were not engineers. If there were 24 people
in the room, and at least one of them were
Bengali, engineer or a football player, how
many were engineers and played football but
not Bengali? (GKwU K‡ÿ 6 Rb evOvwj, 12 Rb
cÖ‡KŠkjx I 15 Rb dzUejvi Av‡Qb| Zv‡ì g‡a¨ †_‡K
gvÎ GKRb evOvwj cÖ‡KŠkjx whwb dzUej †Lj‡Zb| `yRb
wQ‡jb evOvwj cÖ‡KŠkjx hviv dzUej †Lj‡Zb bv Ges
`yRb wQ‡jb evOvwj dzUejvi huviv cÖ‡KŠkjx bb| hw` H
K‡ÿ †gvU 24 Rb †jvK _v‡Kb Ges Zv‡ì g‡a¨ AšÍZ
GKRb evOvwj, cÖ‡KŠkjx ev dzUejvi, Z‡e Zv‡ì g‡a¨
KZRb cÖ‡KŠkjx I dzUejvi wKš‘ evOvwj bb?)
a 1 b 9 c 6 d 3 d
 mgvavb : †gvU e¨w³i msL¨v, n(B  E  F) = 24
evOvwji msL¨v, n(B) = 6
cÖ‡KŠkjxi msL¨v, n(E) = 12
dzUejv‡ii msL¨v, n(F) = 15
evOvwj cÖ‡KŠkjx huviv dzUej †Lj‡Zb Zuv‡ì msL¨v,
n(B  E  F) = 1
evOvwj cÖ‡KŠkjx wKš‘ dzUejvi bb Ggb e¨w³i msL¨v,
n((B  E)F) = 2
 n(B  E) – n (B  E  F) = 2
 n(B  E) – 1 = 2
 n(B  F) = 3
evOvwj dzUejvi wKš‘ cÖ‡KŠkjx bb Ggb e¨w³i msL¨v,
n((B  F)E) = 2
 n(B  F ) – n (B  F  E) = 2
 n(B  F) – 1 = 2
 n(B  E) = 3
cÖ‡KŠkjx I dzUejvi wKš‘ evOvwj bb Ggb e¨w³i msL¨v,
= n((E  F)B)
= n(E  F) – n(E  F  B)
= n(E  F) – 1
Avgiv Rvwb, n(B  E  F) = n(B) + n(E) + n(F) –
n(B  E) – n(E  F) – n(B  F) + n(B  E  F)
 24 = 6 + 12 + 15 – 3 – n(E  F) – 3 + 1
= 28 – n(E  F)
 n(E  F) = 28 – 24 = 4
 cÖ‡KŠkjx I dzUejvi wKš‘ evOvwj bb Ggb e¨w³i
msL¨v = n(E  F) – 1 = 4 – 1 = 3
weKí mgvavb : †fbwPÎ †_‡K
B(6) E(12)
F(15)
6–(2+2+1)
= 1
12–(2+1+x)
= 9–x
15–(2+1+x)
= 12–x
24
GLb, 24 = 1 + 9 – x + 12 – x + 2 + 1 + 2 + x
 24 = 27 – x  x = 3
769. Shonghoti and Shouhardo Clubs consist 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 I
†mŠnv`©¨ K¬v‡ei †gvU m`m¨ msL¨v h_vµ‡g 200 I 270
Rb| hw` `ywU K¬v‡ei †gvU m`m¨ msL¨v 420 nq, Z‡e
KZRb Dfq K¬v‡ei m`m¨?) [Bangladesh Bank (A.D.-2018)]
a 30 b 40 c 50 d 60 c
 mgvavb : msnwZ K¬v‡ei m`m¨ msL¨v, N(A) = 200
†mŠnv`©¨ Ó Ó Ó , N(B) = 270
†gvU m`m¨ msL¨v N(A  B) = 420
N(A  B) = N(A) + N(B) – (A  B)
 N(A  B) = N(A) + N(B) – (A  B)
= 200 + 270 – 420 = 50
 50 Rb Dfq K¬v‡ei m`m¨|
770. In a class of 78 students 41 are taking French,
22 are taking German. Of the students taking
French or German, 9 are taking both courses.
How many students are not enrolled in either
course? (†Kv‡bv K¬v‡m 78 Rb wkÿv_x©i gv‡S 41 Rb
†d«Â, 22 Rb Rvg©vb wbj| 9 Rb `ywU †Kvm©B †bB|
KZRb wkÿv_x© †Kvb †Kv‡m©B AskMÖnY K‡i wbÑ)
a 6 b 15 c 24 d 33 c
 mgvavb :
Avgiv Rvwb, N = (F  G) = N(F) + N(G) – n (F  G)
= 41 + 22 – 9
= 54
 †Kvb †Kvm©B †bq wb = 78 – 54 = 24 Rb
771. Club A has 20 members and Club B has 27.
If a total of 42 people belong to the two clubs,
how many people belong to both clubs? (A
K¬v‡e 20 Rb m`m¨ Ges B K¬v‡e 27 Rb m`m¨ Av‡Q|
hw` `yB K¬ve wgwj‡q †gvU 42 Rb m`m¨ _v‡K Zvn‡j
KZRb `yB K¬v‡eB Av‡QÑ) [B.K.B. (Officer Cash-2017)]
a 3 b 4 c 5 d 6 c
F G
(F  G)

 mgvavb :
Avwg Rvwb,
n(A  B) = n(A) + n(B) – n (A  B)
 42 = 20 + 27 – n(A  B)
 n(A  B) = 47 – 42 = 5
772. If 61% of Bangladeshi people like coffee and
74% like tea, how many like both? (hw` 61%
evsjv‡`kx gvbyl Kwd cQ›` K‡i Ges 74% Pv cQ›`
K‡i| KZRb DfqwU cQ›` K‡iÑ)
a 13% b 16% c 26% d 35% d
 mgvavb :
Avgiv Rvwb,
n(C  T) = n(C) + n(T) – n(C  T)
 100% = 61% + 74% – n(C  T)
 n(C  T) = 135% – 100% = 35%
 35% †jvK DfqwUB cQ›` K‡i|
Written
773. Among 50 people, 35 can speak English, 25
can speak both English and Bangla and each
can speak at least in one of the two languages.
How many of them can speak only Bangla?
(50 Rb gvby‡li g‡a¨, 35 Rb Bs‡iwR, 25 Rb Bs‡iwR
I evsjv Ges c‡Z¨‡K †h‡Kvb GKwU fvlv ej‡Z cv‡i|
Zv‡`i g‡a¨ ïaygvÎ evsjv ej‡Z cvi‡e KZ Rb?)
 Solution:
Let, set of Bangla speaking people = B
and ,, ,, English ,, ,, = E
We know,
n(B  E) = n(B) + n(E)  n(B  E)
 50 = n(B) + 35  25
 n(B) = 40
So, 40 people can speak Bangla
So, Number of people who can speak only
Bangla = 40  25 = 15
774. In a survey at an airport, 55 travelers said that
last year they had been to Spain, 53 to France
and 79 to Germany, 18 had been to Spain and
France, 17 to Spain and Germany and 25 to
France and Germany, while 10 had to all three
countries. How many travelers took part in
the survey? (wegvbe›`‡i GKwU Rwic †`Lv hvq, 55
Rb hvÎx MZ eQi †¯ú‡b, 53 Rb d«v‡Ý Ges 79 Rb
Rvg©vwb ågY K‡i‡Qb| 18 Rb †¯úb Ges d«vÝ, 17 Rb
†¯úb Ges Rvg©vwb Ges 25 Rb d«vÝ Ges Rvg©vwb‡Z
ågY K‡i‡Qb| 10 Rb me¸‡jv †`‡kB ågY K‡i‡Qb|
KZRb Rwic AskMÖnY K‡iwQj?) [Agrani Bank Ltd.
(S.O. Auditor-2018); Rupali Bank Ltd. (Officer Cash-2018)]
 Solution:
Number of travellers who had been to spain n(s) = 55
,, ,, ,, ,, ,, ,, ,, France n(F) = 53
,, ,, ,, ,, ,, ,, ,, Germany n(G) = 79
,, ,, ,, ,, ,, ,, ,, both spain
and France n(S  F) = 18
,, ,, ,, ,, ,, ,, ,, both spain and
Germany n(S  G) = 17
,, ,, ,, ,, ,, ,, ,, both France and
Germany n(F  G) = 25
,, ,, ,, ,, ,, ,, ,, all three countries
n(S  F  G) = 10
Total Number of travellers n(S  F  G)
= n(S) + n(F) + n(G)  n(S  F)  n(S  G) 
n(F  G) + n(S  F  G)
 n(S  F  G) = 55 + 53 + 79  18  17  25 + 10 = 137
775. 70 students are studying physics, mathematics
and chemistry. 40 students study mathematics,
35 study physics and 30 study chemistry, 15
students are studying all the subjects. How
many students are studying exactly two of the
subjects? (70 Rb QvÎ c`v_©, MwYZ Ges imvqb GB
wZbwU welq co‡Q| 40 Rb QvÎ MwYZ c‡o, 35 Rb
c`v_© c‡o Ges 30 Rb imvqb c‡o| 15 Rb wZbwU
welqB c‡o| KqRb `yBwU welq c‡o?) [Rupali Bank Ltd.
(Officer Cash) Cancelled-2018; Sonali Bank Ltd. (Officer)-2018]
A B
A  B
Coffce (C)
C  T
Tea (T)
only English
only Bangla
B E
Both Bangla and English [n(B  E)]
50
S F
G

MATHEMATICS 301
 Solution:
From venndiagram,
Number of students studying only mathematics
= 40  (x + y + 15)
Number of students studying only physics
= 35  (x + z + 15)
Number of students studying only chemistry
= 30  (y + z + 15)
Total students = 70
From diagram
40  (x + y + 15) + 35  (x + z + 15) + 30  (y
+ z + 15) + x + y + z + 15 = 70
 120  (2x + 2y + 2z + 45) + x + y + z = 70
 75 – x – y – z = 70
 x + y + z = 5 (Ans.)
776. Out of 85 football players, 42 have scored a
goal and 54 have received a yellow card. If 5
players did not do either, what fraction of
the players secored a goal and received a
yellow card as well? (85 Rb dzUej †L‡jvqv‡oi
gv‡S 42 Rb †Mvj K‡i‡Q Ges 54 Rb njy` KvW©
†c‡q‡Q| hw` 5 Rb †KvbwUB bv K‡i Z‡e KZ Ask
†L‡jvqvo †Mvj K‡i‡Q Ges njy` KvW© †c‡q‡Q?)
[Sonali & Janata Bank Ltd. (SO) IT/ICT-2018]
 Solution:
we know,
n(G  Y) = n(G) + n(Y)  n(G  Y)
 85  5 = 42 + 54  n(G  Y)
 80 = 96  n(G  Y)
 n(G  Y) = 6
 Fraction =
16
85
7. Logarithm
M.C.Q
777.
log 36
log 6
= [Bangladesh Bank (Officer General-2019);
BangladeshBank (Officer-2018);AgraniBank (OfficerCash-2017)]
a 5 b 8
c 3 d 2 d
 mgvavb :
log 36
log 6
=
log 62
log 6
=
2 log 6
log 6
‹ logaxn
= nlogax = 2
778. If logx
9
16
=
1
2
the value of the base is– (logx
9
16
=
1
2
n‡j x Gi gvbÑ) [Rupali Bank Ltd. (Officer-2019)]
a
16
9
b
9
16
c
256
81
d
81
256
d
 mgvavb : logx



9
16
=
1
2
 x
1
2 =
9
16
 x =



9
16
2
=
81
256
779. If logx2
9
16
= –
1
2
the value of the base is– (logx2
9
16
= –
1
2
n‡j, jMvwi`‡gi wfwËi gvbÑ)
a
16
9
b
9
16
c
256
81
d
81
256
c
 mgvavb : GLv‡b wfwË = x2
logx2
9
16
= –
1
2
 (x2
)
– 1
2 =
9
16
 [ ](x2
)
– 1
2
– 2
=



9
16
– 2
[Dfqc‡ÿ – 2 NvZ wb‡q]
 x2
=



16
9
2
=
256
81
780. If logx
1
4
= – 2, the x = ? [Rupali Bank (Officer Cash-2018)]
a
–1
2
b
1
2
c 2 d 3 c
 mgvavb : logx



1
4
= – 2
 x– 2
=
1
4

1
x2 =
1
4
 x2
= 4  x = 4
 x = 2
Here, x + y + z =
Number of students
studying exactly two
subjects
G
n(G  Y)
Y 85
5
M P
C
x
y z
15
70

781. If x = ya
, y = zb
and z = xc
then the value of
abc is– [RupaliBank(Officer Cash-2018);AgraniBank(S.O.Auditor-2018);
Bangladesh Bank (A.D.-2018); B.H.B.F.C. (S.O.-2017)]
a 1 b 0
c 0.5 d infinity a
 mgvavb : x = ya
 log x = log ya
= a log y
 a =
log x
log y
.................. (i)
y = zb
 log y = log zb
 log y = b log z
 b =
log y
log z
......................... (ii)
z = xc
 log z = log xc
= c log x
 c =
log z
log x
........................ (iii)
 (i)  (ii)  (iii) 
abc =
log x
log y

log y
log z

log z
log x
 abc = 1
weKí mgvavb :
x = ya
, y = zb
, z = xc
 z = xc
= (ya
)c
= yac
= (zb
)ac
= zabc
 z1
= zabc
 abc = 1
782. If 42x + 1
= 32, then x = ? [Bangladesh Bank (A.D.-2018)]
a 2 b 3 c
3
4
d
4
3
c
 mgvavb : 42x + 1
= 32
 (22
)2x + 1
= 25
 24x + 2
= 25
 4x + 2 = 5  4x = 3
 x =
3
4
783. If x
–
7
2
=
1
128
then the value of x is–
a 8 b – 4 c 4 d 2 c
 mgvavb : x
–
7
2 =
1
128
 x
7
2 = 128 [e¨¯ÍKiY K‡i]
 x
7
2 = 27
 x = ( )27
2
7 = 2
7 
2
7
 x = 22
 x = 4
784. (2x–1
)2
 x–5
is equal to : [Sonali Bank (Officer-2018)]
a 2x2
b 4x c 4x2
d 4x3
d
 mgvavb : (2x–1
)2
 x–5
= 22
 (x–1
)2
 x–5
= 22

1
x2 
1
x5
= 22

1
x2  x5
= 4x3
785. Log 3 81 = ?
[Agrani Bank (S.O. Auditor-2018); B.K.B. (Officer Cash-2017)]
a 9 b 7 c 6 d 8 d
 mgvavb : log 3
81 = log 3
( 3)8
= 8 log 3
3
= 8  1 [loga
a
= 1]
= 8
786. If 4x + 1
= 32, then x = ? [B.K.B. (Officer Cash-2017)]
a 2 b 3 c
3
2
d
2
3
c
 mgvavb : 4x + 1
= 32
 (22
)x + 1
= 25
 22x + 2
= 25
 2x + 2 = 5  2x = 3  x =
3
2
787. If logx
1
9
= – 2, the x = ? [B.D.B.L. (S.O.-2017)]
a
– 1
3
b
1
3
c – 3 d 3 d
 mgvavb : logx



1
9
= – 2
 x–2
=
1
9

1
x2 =
1
9
 x2
= 9  x = 9 = 3
8. Inequality
M.C.Q
788. If x is an integer and y = – 2x – 8, what is the
least value of x for which y is less than 9? (x
GKwU c~Y©msL¨v Ges y = – 2x – 8 n‡j x Gi †Kvb
me©wb¤œ gv‡bi Rb¨ y Gi gvb 9 Gi Kg n‡e?)
[Combined 5 Banks (Officer-2018); Sonali Bank (S.O.-2018);
Bangladesh Bank (Officer General-2019);
Rupali Bank Ltd. (Officer-2019)]
a – 9 b – 8 c – 7 d – 6 b
 mgvavb : y < 9
 – 2x – 8 < 9
 – 2x – 8 + 8 < 9 + 8
 – 2x < 17

– 2x
– 2
>
17
– 2
 x > – 8.5
 c~Y©msL¨v wn‡m‡e x Gi me©wb¤œ gvb – 8 n‡Z n‡e|

MATHEMATICS 303
789. If 2x – 1  – 3, then– [Bangladesh Bank (Officer General-2019)]
a x  – 2 b x  – 2 c x  – 1 d x  – 1 d
 mgvavb : 2x – 1  – 3
 2x – 1 + 1  – 3 + 1
2x  – 2
2x
2

– 2
2
 x  – 1
790. If 1 – 2x  3, then– [Rupali Bank Ltd. (Officer-2019);
Bangladesh Bank (Officer-2018), Agrani Bank (Officer Cash-2017)]
a x  – 2 b x  – 2
c x  – 1 d x  – 1 d
 mgvavb : 1 – 2x  3
 – 2x  3 – 1  – 2x  2
 x  – 1 [– 2 Øviv fvM K‡i]
791. The solution of the inequality | 7 – 3x | < 2 is–
(| 7 – 3x | < 2 AmgZvwUi mgvavbÑ)
a – 3 < x <
5
3
b 3 > x >
5
3
c – 3 < x <
5
2
d – 3 < x < –
5
3
b
 mgvavb : | 7 – 3x | < 2
 – 2 < 7 – 3x < 2
 – 2 – 7 < 7 – 3x – 7 < 2 – 7
[AmgZvwUi cÖ‡Z¨K As‡k – 7 †hvM K‡i]
 – 9 < – 3x < – 5
 9 > 3x > 5
[– 1 Øviv ¸Y Kivq AmgZvi w`K cwieZ©b]

9
3
> x >
5
3
 3 > x >
5
3
792. The sum of squares of 3 consecutive integers
is less than 97. What is the greatest possible
value of the smallest one? (wZbwU µwgK msL¨vi
e‡M©i mgwó 97 A‡cÿv †QvU| †QvU msL¨vwUi m‡e©v”P
gvb KZ?) [Combined 5 Banks (Officer-2018)]
a 4 b 5 c 6 d 7 a
 mgvavb : awi, msL¨vÎq x – 1, x, x + 1
cÖkœg‡Z, (x – 1)2
+ x2
+ (x + 1)2
< 97
 x2
+ {(x + 1)2
+ (x – 1)2
} < 97
 x2
+ 2  (x2
+ 1) < 97
 3x2
+ 2 < 97
 3x2
< 95
 x2
<
95
3
 x2
< 31.66
 x < 31.66
 x < 25
31.66 Gi wbKUZg I ÿz`ªZi c~Y©eM© 25
 x < 5
 wb‡Y©q m‡e©v”P gvb 4.
793. In a graph there are two curves, y1 = 2x1 – 5
and y2 = – x2 + 10. y2 will be greater than y1
when– (GKwU †jLwP‡Î y1 = 2x1 – 5 I y2 = – x2 +
10 `ywU †iLv Av‡Q| y1 Gi Zzjbvq y2 e„nËi n‡e hw`Ñ)
a x > 5 b x < 5 c – 1 < x d x < 9 b
 mgvavb : GKB †j‡L D‡jøwLZ GKvwaK †iLvi g‡a¨ Zzjbvi
mgq ¯^vaxb PjK (x1, x2) `yB †ÿ‡Î GKB (x) n‡Z n‡e|
 y1 = 2x – 5
y2 = – x + 10
y2 > y1
 – x + 10 > 2x – 5
 – 2x – 5 < – x + 10
 2x + x < 15  3x < 15
 x < 5
794. In 1 – 3x  4, then– [Bangladesh Bank (A.D.-2018)]
a x  – 2 b x  – 2
c x  – 1 d x  – 1 d
 mgvavb : 1 – 3x  4
 1 – 3x – 1  4 – 1 [Dfqcÿ †_‡K 1 we‡qvM K‡i]
 – 3x  3
 3x  – 3 [– 1 Øviv ¸Y Kivq AmgZvi w`K e`j]
 x 
– 3
3
 x  – 1
795. The sum of 3 consecutive integers is less than
75. What is the greatest possible value of the
smallest one? (3wU µwgK c~Y©msL¨vi mgwó 75 A‡cÿv
Kg| ÿz`ªZg msL¨vwUi m¤¢ve¨ m‡e©v”P gvb KZ?)
[Bangladesh Bank (A.D.-2018)]
a 16 b 19 c 22 d 23 d
 mgvavb : awi, msL¨vÎq x, x + 1 I x + 2
cÖkœg‡Z, x + (x + 1) + (x + 2) < 75
 3x + 3 < 75
 3(x + 1) < 75  x + 1 <
75
3
 x + 1 < 25  x < 25 – 1
 x < 24
24 Gi Zzjbvq ÿz`ª m‡e©v”P c~Y©msL¨v = 23
 wb‡Y©q m‡e©v”P gvb = 23
Written
796. Find the maximum value of z = 6x + 2y, subject
to the conditions x  0, y  0, x + y = 5, x  2, y  0
[Bangladesh Bank (Officer)-2018]
 Solution:
Given, conditions
x  0 ... (i)
y  0 ... (ii)
x + y = 5 ... (iii)
x  2 .... (iv)
y  4 ... (v)
Required function,
zmax = 6x + 2y

From equation (i) and (iv) we get,
minimum value of x is 0
and maximum value of x is 2
To get zmax we have to maximize value of x and y
From equation (ii) and (v) we get,
minimum value of y is 0
and maximum value of y is 4
If we take both maximum value,
then x + y = 2 + 4 = 6 which does not satisfy
equation .... (iii)
So, to get maximum value and to satisfy equation
(iii) we can use x = 1, y = 4 or x = 2, y = 3
As our function is 6x + 2y, so x must be
maximzed as x is multiplied by 6
So, x = 2, y = 3
 zmax = 6  2 + 2  3
= 12 + 6 = 18
797. x2
 12x + 27 < 0
 Solution:
x2
 12x + 27 < 0
 x2
 9x  3x + 27 < 0
 x(x  9)  3(x  9) < 0
 (x  9) (x  3) < 0 ..... (i)
sign of
(x  9)
sign of
(x  3)
sign of
(x  9) (x  3)
x < 3   +
3 < x < 9  + 
x > 9 + + +
From equation (i) we can see (x  9) (x  3) <
0, that means (x  9) (x  3) has to be negative
From the chart, Answer is 3 < x < 9
798. x2
 13x + 40  0
 Solution:
x2
– 13x + 40  0
 x2
– 8x – 5x + 40  0
 x(x  8)  5(x  8)  0
 (x  5) (x  8)  0 .... (i)
sign of
(x  5)
sign of
(x  8)
sign of
(x  5) (x  8)
x  5   +
5 < x  8  + 
x  8 + + +
From equation (i) we can see (x  5) (x  8) 
0, that means (x  5) (x  8) must be positive.
From the chart, Answer is x  5 or x  8
9. Average
M.C.Q
799. The average of six numbers is 14. The average
of four of these numbers is 15. The average
of the remaining two number is– (6wU msL¨vi
Mo 14| G‡`i 4wU msL¨vi Mo 15| evwK `ywU msL¨vi
Mo KZÑ) [Bangladesh Bank (Officer General-2019)]
a 4 b 8 c 12 d 16 c
 mgvavb : 6wU msL¨vi mgwó = 14  6 ev 84
4wU ” ” = 15 4 ev 60
(–) K‡i,  2wU ” ” = 24
 2wU msL¨vi Mo =
24
2
= 12
800. The average of eight numbers is 14. The average
of six of these numbers is 16. The average of
the remaining two numbers is– (AvUwU msL¨vi
Mo 14| G‡`i g‡a¨ 6wU msL¨vi Mo 16| Aewkó `ywU
msL¨vi MoÑ) [Rupali Bank Ltd. (Officer-2019);
Agrani Bank (Officer Cash-2017)]
a 4 b 8
c 16 d data inadequate b
 mgvavb : AvUwU msL¨vi mgwó = 8  14 = 112
QqwU msL¨vi mgwó = 6  16 = 96
 Aewkó `ywU msL¨vi mgwó = 112 – 96 = 16
 wb‡Y©q Mo =
16
2
= 8
801. Consider that w + x = – 4, x + y = 25 and y + w
= 15. Then the average of w, x, y is– (w + x =
– 4, x + y = 25 Ges y + w = 15 n‡j, w, x, y Gi
MoÑ) [Rupali Bank (Officer Cash-2018);
B.K.B. (Officer Cash-2017); B.D.B.L. (S.O.-2017)]
a 3 b 4 c 5 d 6 d
 mgvavb : w + x = – 4 ........... (i)
x + y = 25 ............ (ii)
y + w = 15 .......... (iii)
(i) + (ii) + (iii)
 (w + x) + (x + y) + (y + w) = – 4 + 25 + 15
 2(w + x + y) = 36
 w+ x + y = 18
 Mo =
w + x + y
3
=
18
3
= 6
802. The numbers 2, 3, 5 and x have an average equal
to 4. What is the value of x? (2, 3, 5 Ges x Gi
Mo 4 n‡j x Gi gvb?) [B.K.B. (Officer Cash-2017)]
a 4 b 6 c 8 d 10 b
 mgvavb :
2 + 3 + 5 + x
4
= 4
 2 + 3 + 5 + x = 16
 x = 16 – 10 = 6

MATHEMATICS 305
803. If 12a + 3b = 1 and 7b – 2a = 9, what is the
average of a and b? [Sonali Bank (Officer-2018)]
a 0.1 b 0.5 c 1 d 2.5 b
 mgvavb : †Ìqv Av‡Q, 12a + 3b = 1 ............... (i)
7b – 2a = 9 ................ (ii)
a Ges b Mo =
a + b
2
(i) Ges (ii) †hvM K‡i cvB,
10a + 10b = 10
 a + b = 1 
a + b
2
=
1
2
= 0.5
804. The average of the smallest and largest primes
between 60 and 80 is– (60 †_‡K 80 Gi gv‡S
e„nËg I ÿz`ªZg †gŠwjK msL¨vi MoÑ)
[B.D.B.L. (S.O.-2017)]
a 60 b 70 c 80 d 77 b
 mgvavb : 60 †_‡K 80 Gi gv‡S e„nËg †gŠwjK msL¨v
= 79 Ges ÿz`ªZg †gŠwjK msL¨v 61; msL¨v `ywUi
Mo =
msL¨vØ‡qi mgwó
2 =
61 + 79
2
=
140
2
= 70
805. Which of the following is the average of first five
prime numbers? (cÖ_g 5wU †gŠwjK msL¨vi MoÑ)
[B.H.B.F.C. (S.O.-2017)]
a 4.5 b 5.6 c 7.5 d 8.6 b
 mgvavb : cÖ_g 5wU †gŠwjK msL¨v 2, 3, 5, 7, 11
cuvPwU msL¨vi Mo =
2 + 3 + 5 + 7 + 11
5
=
28
5
= 5.6
806. The average of ten numbers is 7. What will
be the new average if each of the numbers is
multiplied by 8? (10wU msL¨vi Mo 7| hw` cÖwZ
msL¨v‡K 8 Øviv ¸Y Kiv nq Z‡e bZzb MoÑ)
[B.H.B.F.C. (S.O.-2017)]
a 45 b 52 c 56 d 55 c
 mgvavb : †Ìqv Av‡Q, 10wU msL¨vi Mo 7
 10wU msL¨vi mgwó = 7  10 = 70
cÖwZwU msL¨v‡K 8 Øviv ¸Y Ki‡j mgwó n‡e = 70  8
= 540
 Mo n‡e =
560
10
= 56
10. Problem on Numbers
M.C.Q
807. Find the largest fraction from the following :
(e„nËi fMœvskwU †ei Kiæb :) [Sonali Bank (S.O.-2018)]
a –
5
11
b –
8
13
c –
7
19
d –
15
97
d
 mgvavb : cÖwZwU fMœvs‡ki mvg‡b FYvZ¥K wPý Av‡Q|
ZvB †h fMœvs‡ki mvswL¨K gvb me©wb¤œ, †mB fMœvs‡ki
gvb n‡e m‡e©v”P|
(a)
5
11
= 0.45 (b)
8
13
= 0.615
(c)
7
19
= 0.368 (d)
15
97
= 0.154

15
97
Gi mvswL¨K gvb me©wb¤œ
 –
15
97
fMœvskwU m‡e©v”P
808. The difference between two numbers is 5 and
the difference between their squares is 65.
What is the larger number?– (`ywU msL¨vi cv_©K¨
5 Ges G‡ì e‡M©i cv_©K¨ 65| e„nËi msL¨vwU KZ?)
[B.D.B.L. (S.O.-2017); B.H.B.F.C. (S.O.-2017); Bangladesh Bank (A.D.-2018)]
a 13 b 11 c 8 d 9 d
 mgvavb : awi, e„nËi msL¨vwU x
 ÿz`ªZi msL¨vwU (x – 5)
cÖkœg‡Z, x2
– (x – 5)2
= 65
 x2
– [x2
– 10x + 25] = 65
 10x – 25 = 65
 10x = 90
 x = 9
809. Find the largest fraction from the following?
(wb‡Pi †Kvb fMœvskwU e„nËg) [Sonali Bank (Officer-2018)]
a –
15
30
b –
8
64
c –
7
98
d –
15
90
c
 mgvavb : cÖwZwU fMœvs‡ki mvg‡b FYvZ¥K wPý Av‡Q|
ZvB †h fMœvs‡ki mvswL¨K gvb me©wb¤œ, †mB fMœvs‡ki
gvb n‡e m‡e©v”P|
GLb,
–15
30
= –
1
2
–
8
64
= –
1
8
–
7
98
= –
1
14
–
15
90
= –
1
6
fMœvsk¸‡jvi me¸‡jvi je 1 A_©vr ni hvi me‡P‡q eo
†m fMœvskwUi mvswL¨K gvb me©wb¤œ|

1
14
Gi mvswL¨K gvb me©wb¤œ
 m‡e©v”P fMœvsk = –
7
98
810. 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, 100
Gi †P‡q eo c~Y©msL¨v Ges a + b = 300 nq Z‡e a
Ges b Gi AbycvZ KZ?) [Agrani Bank (Officer Cash-2017)]
a 9 to 1 b 5 to 2
c 5 to 3 d 3 to 2 d

 mgvavb : †`Iqv Av‡Q,
a > 100 Ges b > 100
Avevi, a + b > 300 [a Ges b c~Y©msL¨v]
Option check
(a) 9 : 1
300 G a = 300 
9
10
= 270
b = 300 
1
10
= 30 < 100
(b) 5 : 2
300 G a = 300 
5
7
 c~Y©msL¨v
(c) 5 : 3
300 G a = 300 
5
8
 c~Y©msL¨v
(d) 3 : 2
300 G a = 300 
3
5
= 180
b = 300 
2
5
= 120
811. If one number exceeds another number by 14
and the larger number is
3
2
times the smaller
number, then the smaller number is– (hw` GKwU
msL¨v Aci GKwU msL¨vi †P‡q 14 †ewk nq Ges e„nËi
msL¨vwU ÿz`ªZi msL¨vi
3
2
¸Y nq, Zvn‡j †QvU msL¨vwU?)
a 13 b 26 c 28 d 31 c
 mgvavb : g‡b Kwi, †QvU msL¨vwU = x
 eo msL¨vwU = x + 14
cÖkœg‡Z, x + 14 =
3
2
 x
 2x + 28 = 3x  x = 28
812. The ratio of two numbers is 3 : 4 and their sum
is 420. The greater one of the two numbers is–
(`ywU msL¨vi AbycvZ 3 : 4 Ges msL¨v `ywUi mgwó
420| eo msL¨vwU KZ?) [B.K.B. (Officer Cash-2017)]
a 360 b 240 c 180 d 120 b
 mgvavb : g‡b Kwi, †QvU msL¨vwU = 3x
 eo msL¨vwU = 4x
cÖkœg‡Z, 3x + 4x = 420
 7x = 420  x = 60
 msL¨vwU = 4  60 = 240
Written
813. A two digit number is six times the sum of
the two digits. If the digits are reversed, the
number so obtained is 9 less than the
original number. What is the original
number? (`yB A¼ wewkó GKwU msL¨v Zvi A¼Ø‡qi
†hvMd‡ji 6 ¸Y| A¼Øq ¯’vb cwieZ©b Ki‡j cÖvß
msL¨v cÖK…Z msL¨vi †P‡q 9 Kg nq| cÖK…Z msL¨vwU
KZ?) [Rupali Bank Ltd. (Officer-2019)]
 Solution: Let, unit digit is x
and tenth disit is y
  The number is = 10y + x
10y + x = 6 (x + y) .......... (i)
If the digits are reversed, the new number is
10x + y
(10y + x) – (10x + y) = 9
 9y – 9x = 9
 y – x = 1
 y = x + 1 .............. (ii)
From (i) 10 (x + 1) + x = 6 (x + x + 1)
 10x + 10 + x = 6(2x + 1)
 11x + 10 = 12x + 6
 10 – 6 = 12x – 11x
 x = 4
From (ii), y = 4 + 1 = 5
 Original number is = 10  5 + 4
= 54
814. Find the three-digit prime number whose sum
of the digits is 11 and each digit representing
a prime number. Justify your answer. (wZb
A‡¼i Ggb GKwU †gŠwjK msL¨v wbY©q Kiæb hvi
A¼¸‡jvi mgwó 11 Ges cÖwZwU A¼B †gŠwjK msL¨v|)
[Bangladesh Development Bank Ltd. (SO)-2018;
 Solution: Sum of three prime is 11
possible options are, 2 + 2 + 7 = 11
3 + 3 + 5 = 11
Numbers that can be formed by 2, 2 and 7 =
227, 272, 722
Within these numbers only 227 is prime number
and satisfies all the conditions of question.
Now, numbers that can be formed by 3, 3 and 5
= 335, 353 and 533 within these numbers only
353 is prime number and satisfies all the
conditions of question.
 227 and 353 are both correct.

MATHEMATICS 307
815. The sum of the digits of a two-digit number
is subtracted from the number. How many
such two-digit numbers can be formed so
that the digit in the unit place of the
resulting number is 6? (`yB A‡¼i †Kv‡bv msL¨v
n‡Z †mB msL¨vi A¼Ø‡qi mgwó we‡qvM Kiv nq| Ggb
KZ¸‡jv msL¨v MVb Kiv m¤¢e †hb we‡qvMd‡ji GKK
¯’vbxq A¼ 6 nq?) [Bangladesh Bank (Officer)-2018]
 Solution: Let, Unit place of the number = Y
and tenth ,, ,, ,, ,, = X
 Number = 10X + Y
Sum of the digit = X + Y
Subtracting we get = (10X + Y)  (X + Y)
= 9X
To have 6 in the unit place value of X must be
4 as 9  4 = 36
As, x = 4 so tenth place = 4
So, possible numbers are
40, 41, 42, 43, 44, 45, 46, 47, 48, 49
So, 10 number can be formed.
816. The sum of 15 consecutive integers is 88.
What is the largest of these integers? (15wU
ch©vqµwgK c~Y©msL¨vi mgwó 88| c~Y©msL¨v mg~‡ni
gv‡S e„nËg msL¨v †KvbwU?)
 Solution: Let, 15 consecutive numbers are x, x + 1,
x + 2, x + 3, x + 4, x + 5, x + 6, x + 7, x + 8, x + 9,
x + 10, x + 11, x + 12, x + 13 and x + 14
x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x +
6 + x + 7 + x + 8 + x + 9 + x + 10 + x + 11 + x
+ 12 + x + 13 + x + 14 = 88
 15x + 105 = 88
 15x =  17  x =
17
15
 Largest number =
17
15
+ 14
=
 17 + 210
15
=
193
13
which is not integer
So, data in the question is not correct.
817. A two digit number is four times the sum of the
two digits. If the digits are reversed, the number
so obtained is 18 more than the original number.
What is the original number? (`yB A‡¼i GKwU
msL¨v Gi A¼Ø‡qi †hvMd‡ji 4 ¸Y| A¼ `ywU ¯’vb
cwieZ©b Ki‡j cÖvß msL¨v g~j msL¨vi †P‡q 18 †ewk|
g~j msL¨vwU KZ?) [Sonali Bank Ltd. (SO) IT/ICT-2018]
 Solution: Let, Unit digit of the number is x
and tenth ,, ,, ,, ,, ,, y
 Number = 10y + x
Sum of the digits = x + y
According to question, 4(x + y) = 10y + x ... (i)
If the digits are reversed, obtained number = 10x + y
10x + y = 10y + x + 18
 9x = 9y + 18
 x = y + 2 ... ... (ii)
From (i) we get,
4(y + 2 + y) = 10y + y + 2
 4(2y + 2) = 11y + 2
 8y + 8 = 11y + 2
 6 = 3y  y = 2
From (ii), x = 2 + 2 = 4
 The number = 10  2 + 4 = 24
818. Three numbers x, y and z are in A.P. and their
sum is 30. Also, the sum of their squares is 308.
Find the numbers. (GKwU mgvšÍi avivi wZbwU msL¨v
x, y Ges z Gi mgwó 30 Ges msL¨v¸‡jvi e‡M©i mgwó
308| msL¨v wZbwU KZ?) [Bangladesh Bank (AD)-2018]
 Solution: According to question,
Common difference of the series is
y  x = z  y
 2y = x + z ......................... (i)
Again, according to question,
x + y + z = 30
 y + 2y = 30 [From (i)]
 y = 10
If, common difference = d
then first number x = 10  d
and second number z = 10 + d
Again, x2
+ y2
+ z2
= 308
 (10  d)2
+ 102
+ (10 + d)2
= 308
 100  20d + d2
+ 100 + 100 + 20d + d2
= 308
 2d2
+ 300 = 308
 d2
= 4  d =  2
If d = 2, x = 10  2
= 8
z = 10 + 2
= 12
If d =  2
x = 10  (2)
= 12
z = 10 + (2)
= 8
So, Three numbers are 8, 10, 12
11. Problem on Ages
M.C.Q
819. The present age of Habib and Shikhan 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 I wkL‡bi eZ©gvb eq‡mi AbycvZ 6 : 4| cuvP
eQi c~‡e© Zv‡`i eq‡mi AbycvZ wQj 5 : 3| nvwe‡ei
eZ©gvb eqm KZ?) [Rupali Bank (Officer Cash-2018);
a 24 b 30 c 36 d 42 b

 mgvavb : nvwee I wkL‡bi eZ©gvb eq‡mi AbycvZ
= 6 : 4 = 3 : 2
awi, nvwe‡ei eZ©gvb eqm = 3x
wkL‡bi Ó Ó = 2x
5 eQi c~‡e© Zv‡ì eq‡mi AbycvZ wQj (3x – 5) : (2x – 5)
cÖkœg‡Z, (3x – 5) : (2x – 5) = 5 : 3

3x – 5
2x – 5
=
5
3
 10x – 25 = 9x – 15
 x = 10
 nvwe‡ei eZ©gvb eqm = 3  10 = 30 eQi
820. Today is Aziz’s 12th birthday and his father’s
40th birthday. How many years from today
will Aziz’s father be twice as old as Aziz’s at
that time? (AvR‡K AvwR‡Ri 12 Zg Rb¥w`b Ges
Zvi wcZvi 40 Zg Rb¥w`b| GLb †_‡K KZ eQi ci
AvwR‡Ri wcZvi eqm AvwR‡Ri eq‡mi wØ¸Y n‡e?)
[Agrani Bank (S.O. Auditor-2018);
a 12 b 24 c 18 d 16 d
 mgvavb : g‡b Kwi, x eQi ci AvwR‡Ri wcZvi eqm,
AvwR‡Ri eq‡mi wØ¸Y n‡e|
cÖkœg‡Z, 2(12 + x) = 40 + x
 24 + 2x = 40 + x
 x = 16
821. Mira is 30 times older than her son. 18 years
later she will be thrice as old as her son. What
is Mira’s present age? (wgivi eqm Zvi †Q‡ji 30
¸Y| 18 eQi ci wgivi eqm Zvi †Q‡ji eq‡mi wZb¸Y
n‡e| wgivi eZ©gvb eqmÑ) [Agrani Bank (Officer Cash-2017)]
a 36 b 40 c 52 d 86 b
 mgvavb : g‡b Kwi,
eZ©gv‡b wgivi †Q‡ji eqm = x eQi
 eZ©gv‡b wgivi eqm = 30x eQi
18 eQi ci †Q‡ji eqm = (x + 18) eQi
18 ,, ,, wgivi ,, = (30x + 18) eQi
cÖkœg‡Z, 30x + 18 = 3(x + 18)
 30x + 18 = 3x + 54
 27x = 36  x =
4
3
 wgivi eZ©gvb eqm = 30 
4
3
= 40 eQi
822. If a man was r years old s years ago, how many
years old will he be t years from now? (hw`
GKRb e¨w³i eqm s eQi Av‡M r wQj, Zvn‡j t eQi
c‡i Zvi eqm KZ n‡eÑ) [B.H.B.F.C. (S.O.-2017)]
a s + r + t b rs + t
c s – r + t d r – s + t a
 mgvavb : s eQi Av‡M e¨w³wUi eqm wQj r eQi
 e¨w³wUi eZ©gvb eqm (r + s) eQi
GLb †_‡K t eQi ci e¨w³i eqm (r + s + t) eQi
823. Samir is twice as old as Babul was two years
ago. If the difference between their ages is 2
years, how old is Samir now? (evey‡ji `yB eQi
Av‡Mi eq‡mi wØ¸Y mvwg‡ii eZ©gvb eqm| hw` Zv‡ì
eq‡mi cv_©K¨ 2 eQi nq, Zvn‡j mvwg‡ii eqmÑ)
[B.H.B.F.C. (S.O.-2017)]
a 8 years b 6 years c 10 years d 12 years a
 mgvavb : g‡b Kwi, evey‡ji `yB eQi Av‡Mi eqm = x eQi
 mvwg‡ii eZ©gvb eqm = 2x eQi
evey‡ji eZ©gvb eqm = x + 2 eQi
Avevi, 2x – (x + 2) = 2
 x – 2 = 2
 x = 4
 mvwg‡ii eZ©gvb eqm = 2  4 = 8 eQi
Written
824. Shakib is twice as old as Fahim. Four years
ago, Shakib was six years younger than three
times of Fahims age at that time. How old will
they be in two years from now? (mvwK‡ei eqm
dvwn‡gi eq‡mi wØ¸Y| Pvi eQi Av‡M, mvwK‡ei eqm
dvwn‡gi eq‡mi 3 ¸Y A‡cÿv 6 eQi Kg wQj| 2 eQi
ci Zv‡ì eqm KZ n‡e?)
 Solution: Let, Fahim's age is x years
 Shakib's age is 2x years
4 years age, Fahim's age was (x  4)
and Shakib's age was (2x  4)
3(x  4)  6 = 2x  4
 3x  12  6 = 2x  4
 x = 14
So, after 2 years
Fahim's age will be = 14 + 2 = 16 years
and shakib's age will be = 2  14 + 2 = 30 years
12. Percentage
M.C.Q
825. Bus fares were recently increased from Taka
1.70 to Taka 2.00. What was the approximate
percentage of increase? (m¤úªwZ evm fvov 1.7
UvKv †_‡K 2 UvKv Kiv n‡q‡Q| kZKiv e„w×?)
a 18% b 15%
c 0.15% d 0.18% a
 mgvavb : kZKiv e„w× =
†gvU fvov e„w×
Avw` fvov  100%
=
2 – 1.70
1.70
 100% =
0.3
1.7
 100%
> 18%

MATHEMATICS 309
826. Which of the numbers below is not equivalent
to 20%? (wb‡Pi †KvbwU 20% Gi mgZzj¨ bq?)
a
1
5
b
20
100
c 0.5 d 0.2 c
 mgvavb : 20% =
20
100
=
1
5
= 0.2
 mwVK DËi: 0.5|
827. Every 3 minutes, 4 litres of water are poured
into a 2000 litre tank, After 2 hours, what
percent of the tank will be full? (GKwU 2000
wjUvi U¨v‡¼ cÖwZ 3 wgwb‡U 4 wjUvi cvwb Xvjv nq| 2
N›Uv ci U¨v‡¼i kZKiv KZ Ask fwZ© nq?)
[Bangladesh Bank (Officer General-2019); Rupali Bank Ltd. (Officer-2019)]
a 0.4% b 4% c 8% d 12% c
 mgvavb : 2 N›Uv = 2  60 ev 120 wgwbU
3 wgwb‡U Xvjv nq 4 wjUvi
 120 ” ” ”



4
3
 120 ”
= 160 wjUvi
 U¨v¼wU c~Y© nq kZKiv =
160
2000
 100% = 8%
828. If w is 10% less than x, and y is 30% less than
z, then wy is what percent less than xz? (W
Gi gvb x Gi gvb A‡cÿv x Gi 10% cwigvY Kg
Ges y Gi gvb z Gi gvb A‡cÿv z Gi 30% cwigvY
Kg| Zvn‡j wy Gi gvb xz Gi gvb A‡cÿv xz Gi
KZ kZvsk Kg?) [RupaliBankLtd.(Officer-2019);SonaliBank(S.O.-2018)]
a 10% b 20% c 37% d 40% c
 mgvavb : w = x – x Gi 10%
 w = x –
10
100
x
 w =
9
10
x
y = z – z Gi 30%
y = z –
30
100
z
 y =
7
10
z
 wy =
9
10
x 
7
10
z
 wy =
63
100
z
 wy =
100 – 37
100
z
 wy =



1 –
37
100
z
 wy = z –
37
100
z
 wy = z – z Gi 37%
829. If 10% of x is equal to 25% of y, and y = 16,
what is the value of x? (hw` x Gi 10% Gi gvb
y Gi 25% Gi mgvb nq Ges y = 16 nq, Z‡e x Gi
gvb KZ?) [Combined 5 Banks (Officer-2018)]
a 4 b 6.4 c 24 d 40 d
 mgvavb : x Gi 10% =
10
100
x =
x
10
y Gi 25% =
25
100
y =
y
4
cÖkœg‡Z,
x
10
=
y
4
 x =
10
4
y  x =
5
2
y  x =
5
2
 16  x = 40
830. Which of the numbers below is not equivalent
to 4%? (wb‡Pi †Kvb msL¨vwU 4% Gi mgZzj¨ bq?)
a
1
25
b
4
100
c 0.40 d 0.04 c
 mgvavb : 4% =
4
100
=
1
25
Avevi, 4% =
4
100
=
1
10

4
10
=
1
10
 0.4 = 0.04
 4%  0.40
 mwVK DËi : C
weKí mgvavb :
Ackb¸‡jv †PK K‡i †`Lv hvq :
(a)
1
25
=
1
25
 100% = 4%
(b)
4
100
=
4
100
 100% = 4%
(c) 0.40 = 0.40  100% = 40%
(d) 0.04 = 0.04  100% = 4%
 mwVK DËi : C
831. What is the original price of a T-shirt, if the
sale price after 15% discount is 272? (15% g~j¨
Qv‡o †Kv‡bv wU-kvU© Gi weµqg~j¨ 272 n‡j cÖK…Z g~j¨
KZ?) [Rupali Bank (Officer Cash-2018); B.D.B.L. (S.O.-2017)]
a 300 b 280 c 320 d 314 c
 mgvavb : 15% g~j¨Qvo Gi A_© n‡jv :
cÖK…Z g~j¨ 100 UvKv n‡j weµqg~j¨ (100 – 15) = 85 UvKv
weµqg~j¨ 85 UvKv n‡j cÖK…Zg~j¨ 100 UvKv
Ó 1 Ó Ó Ó
100
85 Ó
Ó 272 Ó Ó Ó
100
85
 272 Ó
= 320 UvKv
832. If x is 30% greater than y, what percent of y
is x? (x hw` y Gi Zzjbvq 30% eo nq, Z‡e x, y
Gi KZ kZvsk?) [Agrani Bank (S.O. Auditor-2018);
Rupali Bank (Officer Cash-2018)]
a 70 b 77 c 120 d 130 d

 mgvavb : x = y + y Gi 30%
 x = y +
30
100
y
 x = y +
3
10
y  x =
13
10
y

x
y
=
13
10
=
13
10
 100% = 130%
833. As the price of mango has reduced 20%, it is
now possible to buy 2 more mangoes at Tk. 12.
What is the current price of 50 mangoes?
(Av‡gi `vg 20% K‡g †M‡j, GLb 12 UvKvq 2wU †ewk
Avg cvIqv hvq| 50wU Av‡gi eZ©gvb g~j¨?)
a Tk. 50 b Tk. 40 c Tk. 30 d Tk. 60 d
 mgvavb : g‡b Kwi, c~‡e© 100 UvKvq cvIqv †hZ xwU Avg
eZ©gvb xwU Avg cvIqv hvq (100 – 20) ev 80 UvKvq
cÖwZwU Av‡gi c~e©g~j¨ =
100
x
UvKv
Ges ,, ,, eZ©gvb g~j¨ =
80
x
UvKv
c~‡e©,
100
x
UvKvq cvIqv †hZ 1wU Avg
 12 ,, ,, ,,
12x
100
,, ,,
eZ©gv‡b,
80
x
UvKvq cvIqv hvq 1wU Avg
 12 ,, ,, ,,
12x
80
,, ,,
cÖkœg‡Z,
12x
80
–
12x
100
= 2

60x – 48x
400
= 2  12x = 800
 x =
200
3
200
3
wU Av‡gi eZ©gvb g~j¨ 80 UvKv
1wU ,, ,, ,,
80  3
200
UvKv
50wU ,, ,, ,,
80  3  50
200
UvKv
= 60 UvKv
834. If the salary of an employee is reduced by 10
percent for his late attendance and then increased
by 10 percent on a pardon, how much does
he lose? (hw` †Kvb Kg©Pvixi †eZb 10% K‡g hvq
†`wi K‡i Awd‡m Avmvi Rb¨ Ges c‡i 10% e„w× Kiv
nq ÿgv PvIqvi Rb¨, Zvn‡j Zvi †eZb kZKiv KZUzKz
n«vm †cj) [Sonali Bank (Officer-2018)]
a 2% b 1% c –
1
2
% d 9% b
 mgvavb : awi, Kg©Pvixi eZ©gvb †eZb = 100 UvKv
cÖ_‡g 10% K‡g †M‡j †eZb = 100 – 10 = 90 UvKv
Avevi, 10% evo‡j †eZb = 90 + 90 Gi 10%
= 90 + 90 
10
100
= 99 UvKv
 100 UvKvq †eZb K‡g = (100 – 99) = 1 UvKv
 kZKiv 1% n«vm †cj
weKí mgvavb :
kZKiv n«vm = 10 – 10 –
10 10
100
= – 1%
 1% n«vm cvq|
835. 35% of Nabila’s income is equal to 25% of
Nuru’s income. The ratio of their income is–
(bvwejvi 35% Avq byiæi 25% Av‡qi mgvb| Zv‡`i
Av‡qi AbycvZÑ) [Sonali Bank (Officer-2018)]
a 5 : 7 b 4 : 7 c 7 : 3 d 4 : 3 a
 mgvavb : awi, bvwejvi Avq x UvKv
Ges byiæi Avq y UvKv
cÖkœg‡Z, x Gi 35% = y Gi 25%
 x 
35
100
= y 
25
100

x
y
=
25
100

100
35

x
y
=
5
7
 x : y = 5 : 7
836. 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`
AvqZ‡ÿ‡Îi ˆ`N©¨ 20% e„w× Kiv nq Ges cÖ¯’ 20%
Kgv‡bv nq Zvn‡j AvqZ‡ÿ‡Îi †ÿÎd‡ji kZKiv wK
cwieZ©b n‡e?) [Agrani Bank (Officer Cash-2017)]
a Unchanged b decreases by 4%
c increases by 4% d increases by 5% b
AvqZ‡ÿ‡Îi ˆ`N©¨ x GKK
Ges ,, cÖ¯’ y GKK
 AvqZ‡ÿ‡Îi †ÿÎdj = xy eM© GKK
ˆ`N©¨ 20% evov‡bv n‡j, bZzb ˆ`N©¨ = x + x Gi 20%
= 1.2x
cÖ¯’ 20% Kg‡j, bZzb cÖ¯’ = y – y Gi 20%
= .8y
 bZzb †ÿÎdj = 1.2x  .8y
= 0.96 xy
 cwiewZ©Z †ÿÎdj = xy – 0.96 xy
= 0.04 xy
 †ÿÎd‡ji kZKiv cwieZ©b =
0.04xy
xy
 100%
= 4%
 †ÿÎdj 4% n«vm cv‡e|

MATHEMATICS 311
weKí mgvavb:
†ÿÎdj cwieZ©‡bi nvi =



20 + (–20) +
20  (–20)
100
%
=



20 – 20 –
400
100
%
= – 4%
 †ÿÎdj 4% n«vm cvq|
837. What is the original price of a T-shirt, if the
sale price after 16% discount is 264? (16% Qv‡o
wU kv‡U©i g~j¨ 264 UvKv n‡j, wU kv‡U©i cÖK…Z g~j¨ KZ?)
a 300 b 214 c 320 d 314 d
 mgvavb : kv‡U©i cÖK…Z g~j¨ x n‡j,
16% Qv‡o weµq g~j¨ x – x Gi 16%
= x – x 
16
100
=
84x
100
=
21x
25
cÖkœg‡Z,
21x
25
= 264
 x =
264  25
21
= 314.28  314 UvKv
838. If y% of x = 29, then x = ? [B.K.B. (Officer Cash-2017)]
a
2900
y
b
29x
y
c
29y
x
d 29xy a
 mgvavb : x Gi y% = 29
 x 
y
100
= 29  x =
2900
y
839. If A’s income is 25%less than that of B, then
what percent is B’s income more than that of
A? (hw` A Gi Avq B Gi †P‡q 25% Kg nq Zvn‡j
B Gi Avq A Gi †_‡K kZKiv KZ †ewk?)
a 33.33 b 66.67
c 11.67 d 31.50 a
 mgvavb : awi, B Gi Avq = 100 UvKv
 A Gi Avq = (100 – 25) UvKv
= 75 UvKv
 B Gi Avq †ewk = (100 – 75)
= 25 UvKv
A Gi Avq 75 UvKv n‡j B Gi †ewk 25 UvKv
 A ,, ,, 100 ,, ,, B ,, ,,
25
75
 100 ,,
= 33.33 UvKv
840. After a 20% price decrease, a computer monitor
is on sale for Tk. 7200. What is its orginal price?
(20% g~j¨ n«v‡m GKwU Kw¤úDUvi gwbUi 7200 UvKvq
wewµ Kiv nq| cÖK…Z g~j¨ KZ?) [B.K.B. (Officer Cash-2017)]
a Tk. 9000 b Tk. 10000
c Tk. 11000 d Tk. 12000 a
gwbU‡ii cÖK…Z g~j¨ = x UvKv
cÖkœg‡Z, x – x Gi 20% = 7200
 x – x 
20
100
= 7200

4x
5
= 7200
 4x = 5  7200
 x =
5  7200
4
= 9000
841. A cricket team has won 40 games out of 60
played. It has 32 more games to play. How
many of these must the team win to make it
record 70% win for the season? (GKwU wµ‡KU
wUg 60wU g¨v‡P 40wU‡Z wR‡Z‡Q| wUg‡K AviI 32wU
g¨vP †Lj‡Z n‡e| Avi KZ g¨vP wRZ‡j GB wmR‡b
70% R‡qi †iKW© _vK‡e?) [B.D.B.L. (S.O.-2017)]
a 20 b 25
c 23 d 32 b
 mgvavb : †gvU g¨vP = 60 + 32 = 92wU
70% R‡qi †iKW© ivL‡Z,
†gvU g¨vP wRZ‡Z n‡e = 92 Gi 70%
= 92 
70
100
= 64.4wU
g¨vP wR‡Z‡Q 40wU
 AviI wRZ‡Z n‡e = 64.4 – 40
= 24.4
> 25wU
842. Three workers X, Y and Z are paid a total of
Tk. 5,500 for a particular job. X is paid 133.33%
of the amount paid to Y and Y is paid 75%
of amount paid to Z. How much is paid to Z?
(wZbRb kÖwgK x, y Ges z †K GKwU wbw`©ó Kv‡Ri Rb¨
5500 UvKv †Ìqv nq| x †K y Gi 133.33% Ges y
†K z Gi 75% †Ìqv nq| z †K KZ UvKv †Ìqv nq?)
[B.D.B.L. (S.O.-2017)]
a Tk. 1780 b Tk. 1890
c Tk. 1975 d Tk. 2000 d
 mgvavb : cÖkœg‡Z,
x = y Gi 133.33%
= y 
133.33
100
x = 1.33y ... ... (i)
y = z Gi 75%
= z 
75
100
y =
3z
4

y Gi gvb (i) G ewm‡q cvB,
x = 1.33 
3z
4
= 0.9975 z
cÖkœg‡Z, x + y + z = 5500
 0.9975 z +
3z
4
+ z = 5500
 2.7475 z = 5500
 z =
5500
2.7475
=
5500
2.75
= 2000 UvKv
Written
843. A teacher has 6 hours to grade all the papers
submitted by the 35 students in his class. He
gets through the first 5 papers in 1 hour. How
much faster should he work to grade the
remaining papers in the allotted time? (GKRb
wkÿ‡Ki 35 Rb wkÿv_©xi LvZv †MÖwWs Kivi Rb¨ 6
NÈv mgq †`Iqv nj| cÖ_g 5wU LvZv †MÖwWs Ki‡Z 1
NÈv mgq jv‡M| KZ `ªæZ LvZv †MÖwWs Ki‡j wbw`©ó
mg‡q †kl Ki‡Z cvi‡eÑ) [Rupali Bank Ltd. (Officer-2019)]
 Solution: Time left = (6 – 1) or 5 hours
Papers left for grading = (35 – 5) or 30 papers
Now, the teacher has to grade
30
5
or 6 papers
per hour
  The teacher has to be



6 – 5
5
 100% faster
= 20% faster
844. A man’s salary in 2015 was Tk. 20,000 per
annum and it increased by 10% each year.
Find how much he earned in the years 2015
to 2017 inclusive. (2015 mv‡j GK e¨w³i evwl©K
Avq wQj 20000 Tk Ges cÖwZ eQi †eZb 10% K‡i
e„w× cvq| 2015 mvj †_‡K 2017 mvj ch©šÍ H e¨w³i
†gvU Avq KZ?) [Agrani Bank Ltd. (Officer Cash-2018);
 Solution:
In 2015,
Man's annual salary = 20000 Tk
In 2016,
annual salary = 20000 + 20000 
10
100
= 22000 Tk
In 2017,
annual salary = 22000 + 22000 
10
100
= 24200 Tk
So, From 2015 to 2017,
Total earnings = 20000 + 22000 + 24200
= 66200 Tk (Ans.)
845. A shopkeeper sells two shirts at the same
price. He makes 10% profit on one and loses
10% on the other. How much in percentage
does he gain or lose? (GKRb †`vKvb`vi `ywU kvU©
GKB `v‡g wewµ K‡i| GKwU‡Z 10% jvf K‡i Ges
Ab¨wU‡Z 10% ÿwZ nq| †gv‡Ui Dci kZKiv KZ
jvf ev ÿwZ nq?) [Agrani Bank Ltd. (S.O. Auditor-2018);
 Solution:
Shopkeeper sells two shirts in same price.
for 10% profit he sells 1st shirt = (100 + 10) Tk
sell again for 10% loss he sells 2nd shirt
= (110  110 of 10%)
= (110  11) Tk
= 99 Tk.
Loss percentage =
cost price  selling price
cost price
 10%
=
100  99
100
 100%
=
1
100
 100% = 1%
846. The price of a shirt and a pant together is
Tk. 1,300. If the price of the shirt increases by
5% and that of the pant by 10%, it costs Tk.
1,405 to buy those two things. Find the
respective price of a shirt and a pant. (GKwU
kvU© Ges GKwU c¨v‡›Ui g~j¨ GK‡Î 1300 UvKv| hw`
kv‡U©i g~j¨ 5% e„w× cvq Ges c¨v‡Èi g~j¨ 10% e„w×
cvq Z‡e kvU© I c¨v‡›Ui g~j¨ nq 1405 UvKv| Z‡e kvU©
I c¨v‡›Ui g~j¨ KZ?)
[Bangladesh House Building Finance Corporation (SO) Written-2017]
 Solution:
Let, price of the shirt = x Tk
So, ,, ,, ,, pant = (1300  x) Tk
If price of shirt increases by 5%
then price of shirt = x + 5% of x
= x + x 
5
100
=
21x
20
If price of pant increases by 10%
The price of pant = (1300  x) + 10% of (1300  x)
= 1300  x +
1300  x
10
=
13000  10x + 1300  x
10
=
14300  11x
10

MATHEMATICS 313
21x
20
+
14300  11x
10
= 1405

21x + 28600  22x
20
= 1405
  x + 28600 = 28100
 28600  28100 = x
 x = 500
 Price of shirt = 500 Tk
and ,, ,, pant = 1300  500 = 800 Tk
847. A customer bought 5 pencils and 6 erasers at
Tk. 80. Next week, the price of each pencil
increases by 20%, but the price of erasers
remains unchanged. Now, the customer buys
2 pencils and 3 erasers at Tk. 39. Find the new
price of each pencil.? (GKRb †µZv 80 UvKv w`‡q
5wU †cwÝj Ges 6wU B‡iRvi wKbj| c‡ii mßv‡n
cÖwZwU †cwÝ‡ji g~j¨ 20% †e‡o hvq, wKš‘ B‡iRv‡ii
g~j¨ cwieZ©b nq bv| Ggb †µZv 2Uv †cwÝj Ges 3Uv
B‡iRvi 39 UvKvq wK‡b| cÖwZwU †cwÝ‡ji bZzb g~j¨
KZ?) [Bangladesh House Building Finance Corporation (SO) Written-2017]
 Solution:
Let, previous price of each pencil was x Tk
and ,, ,, ,, ,, eraser ,, y Tk
5x + 6y = 80 ... (i)
Next week after 20% increment,
new price of the pencil = x + 20% of x
= x + x 
20
100
=
6x
5
Price of eraser doesn’t change
Now,
2 
6x
5
+ 3y = 39

12x
5
+ 3y = 39 ... (ii)
(i)  (ii)  2 
5x + 6y = 80
24x
5
+ 6y = 78
() () ()
5x 
24x
5
= 2

x
5
= 2  x = 10
 New price of pencil =
6  10
5
= 12 (Ans.)
13. Ratio and Proportion
M.C.Q
848. If x : y = 5 : 3, then (8x – 5y) : (8x + 5y) = ?
[Bangladesh Bank (Officer General-2019); Sonali Bank (S.O.-
2018); Agrani Bank (Officer Cash-2017)]
a 5 : 11 b 6 : 5
c 5 : 6 d 3 : 8 a
 mgvavb : †`Iqv Av‡Q, x : y = 5 : 3

x
y
=
5
3

x
y

8
5
=
5
3

8
5

8x
5y
=
8
3

8x – 5y
8x + 5y
=
8 – 3
8 + 3
=
5
11
 (8x – 5y) : (8x + 5y) = 5 : 11
849. Two numbers are in the ratio 2 : 5. If 16 is
added to both the numbers, their ratio becomes
1 : 2. The numbers are– (`ywU msL¨vi AbycvZ 2 :
5| Df‡qi mv‡_ 16 †hvM Ki‡j G‡`i AbycvZ nq 1 :
2| msL¨vØqÑ) [Rupali Bank Ltd. (Officer-2019)]
a 16, 40 b 20, 50
c 28, 70 d 32, 80 d
 mgvavb : awi, msL¨vØq 2x I 5x
cÖkœg‡Z,
2x + 16
5x + 16
=
1
2
 5x + 16 = 4x + 32
 x = 16
msL¨vØq 2  16 = 32 I 5  16 = 80
850. If x : y = 5 : 3, then (8x + 5y) : (8x – 5y) = ?
(hw` x : y = 5 : 3 nq, Z‡e (8x + 5y) : (8x – 5y) = ?)
a 5 : 11 b 6 : 5 c 5 : 6 d 11 : 5 d
 mgvavb : x : y = 5 : 3

x
y
=
5
3

8
5

x
y
=
8
5

5
3
; [Dfqcÿ‡K
8
5
Øviv ¸Y K‡i]

8x
5y
=
8
3

8x + 5y
8x – 5y
=
8 + 3
8 – 3
; [†hvRb-we‡qvRb K‡i]

8x + 5y
8x – 5y
=
11
5
 (8x + 5y) : (8x – 5y) = 11 : 5

06. mathematics arts pm t 9-10 genarel scince [onlinebcs.com]

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