MCQ Part
[Bangladesh Bank- (AD) - 2018]
Bangladesh Bank
Post name: Assistant Director (General)
Exam date: 06-07-2018
Exam taker : Arts Faculty, DU
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]
c. 50
d. 60
a. 30
b.40
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)
400
or, 420 = 200 + 270 – P(A ∩ B)
or, P(A ∩ B) = 470-420
∴ P(A ∩ B) = 50
200
50
270
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]
c. 400
d. 100
a. 1500
b. 1000
Ans:
Solution:
Complementary angle 600 is (900 – 600) = 300
1
So, one third of 300 is 300 × = 100
3
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 =
1
× KY©Ø‡qi ¸Ydj)
2
1
× x ×16= 96 or 8x = 96 ∴ x = 12
2
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
Area of rhombus is
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
Khairul’s Bank Recent Math - 1
MCQ Part
5. If 4
2x+1
[Bangladesh Bank- (AD) - 2018]
=32, then x =? [BB-(AD)-2018]
a. 2
c.
b. 3
3
4
d.
4
3
Ans: c
Solution:
4x+1=32
Or, 22(2x+1) =25
or, 4x+2 = 5
or, 4x = 3
∴x =
3
4
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:
1000 × 10 × 4
= 400tk
100
Compound interest = (110% of 110% of 110% of 110% of 1000 ) -1000
110 110 110 110
Or, (1000×
×
×
×
) – 1000 = 1464.1-1000 = 464.1
100 100 100 100
So difference = 464.1- 400 = 64.1
Simple interest =
1000
♦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|
Easiest way
1g 100
2q 100 +
10
3q 100 +
10
+ 11
4_© 100 +
10
+ 11
12.1
30 + 22 +12.1 = 64.1
♦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
Solution:
The series is 5, 7, 9, 11, 13 and 15
So 4th number is 11
Khairul’s Bank Recent Math - 2
d. 13
Ans: c
MCQ Part
[Bangladesh Bank- (AD) - 2018]
a
b
c
8. If x = y , y = z and z = x 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+sinθ = xcosθ then tanθ is - ? [BB-(AD)-2018]
x2 +1
x 2 −1
x2 +1
a.
b.
c.
x
x
2x
d.
x 2 −1
2x
Ans:d
Solution:
Given, 1+sinθ = xcosθ
1
sinθ
+
= x (cosθ w`‡q fvM K‡i|)
⇒
cosθ cos θ
⇒ secθ + tanθ = x ------ (i)
Now, We know that, sec2θ - tan2θ = 1
(secθ + tanθ ) (secθ - tanθ) = 1
x×(secθ- tanθ) = 1 [from equation (i)]
1
or,(secθ- tanθ) =
- - - - - (ii)
x
Again, secθ + tanθ = x - - - - - (iii)
by (iii)-(ii)
1
x2 −1
2tanθ = xor, 2tanθ =
x
x
∴tanθ =
x2 −1
2x
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
65
=13 ---(ii)
5
by adding (i) and (ii) we get 2a = 18 ∴ a = 9
or, (a+b)(a-b) = 65 or a+b =
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
Khairul’s Bank Recent Math - 3
d. 130 sec
Ans: b
MCQ Part
[Bangladesh Bank- (AD) - 2018]
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.
1
5
d. -
1
5
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|
1
−1 1
1
=
So, line perpendicular will have slope m = =Ans:
m
5
−5 5
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 = -
y 3
= = and x + 2y= 13 then y is- [BB-(AD)-2018]
x 7
c. 4
a.2
b. 3
1
1
=−
m
3
13. If
d. 7
Ans: b
Solution:
y 3
=
or, 3x = 7y or, or, 3x -7y = 0 -------- (i)
and x + 2y= 13 --------(ii)
x 7
By, (ii) × 1 & (ii) × 3 we get
7y
3x = 7y Gici x =
c‡ii As‡k ewm‡q Ki‡jI n‡e|
3x -7y = 0
3
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|)
Khairul’s Bank Recent Math - 4
MCQ Part
[Bangladesh Bank- (AD) - 2018]
15. A pole 6 m high casts a shadow 2 3 m long on the ground, then the Sun's elevation is ?
(GKwU LuywUi D”PZv 6 wg. Ges LuywUi Qvqvi ˆ`N©¨ 2 3 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 =
j¤ ^
fzwg
m~q©
AB
∴tanθ =
( cv‡ki wPÎ Abyhvqx j¤^ Ges f~wg †`qv Av‡Q ZvB)
BC
6
3. 3
3
= 3
or, tanθ =
=
=
2
2 3
3
or, tanθ = tan60
∴ θ = 600 (Ans)
0
ο
[Since tan60 =
A
Pole=6
60°
B
3 ]
C
Qvqv= 2 3
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
13 × 12 × 11
Total number of team = 13C10=
= 286
3 × 2 ×1
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
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
Khairul’s Bank Recent Math - 5
Ans: c
MCQ Part
[Bangladesh Bank- (AD) - 2018]
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 wظY 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]
c. flame
d. castle
a. voice
b. bald
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]
Ans: c
27
20
27
23
a.
b.
c.
d.
89
91
91
89
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
Khairul’s Bank Recent Math - 6
MCQ Part
[Bangladesh Bank- (AD) - 2018]
Probability of selecting an integer =
27
91
‡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 wظY 90| GB `ywU
23. What does make ‘you’ young? (you ‡K young Ki‡Z wK jv‡M?) [BB-(AD)-2018]
b. Drinking energy beverage
a. Adding 2 velars
c. Eating sweet fruits
d. Changing outfits
Ans: a
Solution:
velar A_© aŸwb| you k‡ãi mv‡_ n I g `ywU aŸwb ‡hvM K‡i young n‡e|
ZvB DËi n‡e a.
24. The sum of 3 consecutive integers is less than 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:
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
=======×======×========×========×=======
Khairul’s Bank Recent Math - 7