Iba mcq math solution( 2017 2019) by khairul alam [www.itmona.com]

Khairul’s IBA Math Book - 1
IBA
All MCQ Math & Solution
Exam date
01 cvwb Dbœqb †evW© (wnmve KiwYK) 10-08-2018
02 BADC –( Computer-Operator) 11-05-2018
03 Titas Gas Field-Assistant Officer (General) 27-04-2018
04 (GTCL)-Assistant Manager (General) 20-04-2018
05 (DAE)(Officer Assistant cum computer typist) 13-04-2018
06 DAE(Store Keeper) 01-12-2017
07 BADC (Administrative Officer) 27-10-2017
08 BADC (Assistant Cashier) 11-08-2017
09 BADC (Store Keeper) 11-08-2017
10 BREB
11 BSCIC (Extension officer) 10-11-2018
12 City Bank Ltd. (Management Trainee officer) 20-07-2018

1. `yBwU µwgK c~Y© msL¨vi e‡M©i AšÍi 27 n‡j eo msL¨vwU KZ ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 11 B) 13 C) 14 D) 15 E) ‡KvbwUB bq DËi: C
mgvavb: (wjwLZ wbq‡g)
awi,
eo msL¨vwU = x, myZivs †QvU msL¨vwU = x-1
cÖkœg‡Z,
x2
-(x-1)2
= 27
ev, x2
-x2
+2x-1 = 27
ev, 2x = 27+1
x =
2
127 
(GB jvBbwU †_‡KB kU©KvU© myÎ †jLv hvq: eo msL¨vwU=
2
1K¨cv_©iM©e‡qimsL¨vØ‡ 
)
myZivs x ev eo msL¨vwU = 14
2
28
 ( kU©Kv‡U© Kivi Rb¨ mivmwi, 14
2
28
2
127


DËi: 14)
kU©KvU©: eo msL¨vwU =
2
1K¨cv_©iM©e‡qimsL¨vØ‡ 
Ges †QvU msL¨vwU =
2
K¨-1cv_©iM©e‡qimsLvØ‡
2. GKwU †Uwej 10% ÿwZ‡Z wewµ Kiv nj| weµqg~j¨ 51 UvKv †ewk n‡j 7% jvf nZ| †UwejwUi µqg~j¨ KZ UvKv ?
[cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
awi, µqg~j¨ = 100 UvKv|
10% ÿwZ‡Z cÖ_g weµqg~j¨ = 100-10 = 90 UvKv|
Avevi, 7% jv‡f wØZxq weµqg~j¨ = 100+7 = 107 UvKv|
`yB weµqg~‡jï eëavb =107-90 = 17UvKv| (cÖ‡kœ 51 UvKv n‡”Q 2 weµ‡qi cv_©K¨ ZvB cv_©K¨ †ei Kiv n‡jv)
`yB weµqg~‡jï g‡a¨ eëavb 17 UvKv n‡j µqg~j¨ = 100 UvKv
ÕÕ ÕÕ ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ =
17
100
ÕÕ
 ÕÕ ÕÕ ÕÕ ÕÕ 51 ÕÕ ÕÕ ÕÕ =
17
51100 
= 300 UvKv| DËi 300 UvKv|
cvwb Dbœqb †evW©
c‡ì bvg: wnmve KiwYK cixÿvi ZvwiL: 10-08-2018
cixÿvwU wb‡q‡Q: IBA, DU.
cÖgvY: 14 Gi eM© = 142
= 196
14 Gi †_‡K 1 †QvU 13 Gi eM© 132
= 169
myZivs e‡M©i AšÍi : 196-169 = 27|
Note: µwgK c~Y© msL¨v ejv eowU 14 n‡j †QvUwU 13 n‡e|

gy‡L gy‡L Kivi Rb¨:
G ai‡bi As‡Ki ‡ÿ‡Î cÖ_‡gB % Gi †gvU eëavb ‡ei K‡i Zv = eëav‡bi ‡gvU UvKv (hv cÖ‡kœ †`qv _vK‡e Zv) wj‡L
100% Gi gvb Avb‡Z nq|
3. RvwKi, Rwm‡gi †P‡q hZ eQ‡ii †QvU, ewk‡ii ‡_‡K wVK ZZ eQ‡ii eo| Rwmg I ewk‡ii eq‡mi mgwó 56 eQi n‡j,
RvwK‡ii eqm KZ ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 20 B) 28 C) 32 D) 36 E) ‡KvbwUB bq DËi: B
mgvavb:
GB cÖkœwU g~jZ gvbwmK `ÿZv As‡ki|
cv‡ki wPÎwU‡Z RvwKi n‡”Q Rwmg I
ewk‡ii wVK gvSLv‡b| Zvn‡j Rwmg I
ewk‡ii eq‡mi Mo †ei Ki‡j ga¨we›`y †ei
n‡e hv RvwK‡ii eqm| A_©vs RvwK‡ii eqm
= 56 2 = 28 eQi|
wjwLZ mgvavb: (hviv wjwLZ mgvavb bv †`‡L Qvo‡eb bv Zv‡ì Rb¨| )
GLv‡b,
Rwmg-RvwKi = RvwKi-ewki (KviY cÖ_g `yR‡bi eq‡mi eëavb hZ c‡ii `yR‡bi eëavb I ZZ)
ev, Rwmg+ewki = RvwKi + RvwKi
ev, 56 = 2 RvwKi ( †h‡nZz Rwmg I ewk‡ii eq‡mi mgwó = 56 eQi)
ev, 2 RvwKi = 56
myZivs RvwKi = 56  2 = 28 eQi|
4. `yBwU msL¨v Z…Zxq GKwU msL¨v †_‡K h_vµ‡g 40% I 25% Kg| cÖ_g msL¨vwU wØZxq msL¨vwUi Zzjbvq kZKiv KZ
†QvU ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 15% B) 18% C) 20% D) 25% E) ‡KvbwUB bq DËi: C
mgvavb:
awi, Z…Zxq msL¨vwU = 100
myZivs 1g msL¨vwU 100-40 = 60, Ges 2q msL¨vwU 100-25 = 75
cÖ_g msL¨vwU †QvU – 75-60 = 15
kZKiv ‡QvU =
75
10015 
= 20%
5. GKwU e„‡Ëi e¨mva© kZKiv 10% evov‡bv n‡j Gi †ÿÎdj kZKiv KZ e„w× cv‡e ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-
2018]
A) 10% B) 16.5% C) 21% D) 25% E) ‡KvbwUB bq DËi: C
17% = 51 UvKv n‡j (3 ¸Y)
1% =
17
51
(gy‡L gy‡L 3)
100%=
17
10051 
= 300 UvKv
90%
1g weµqg~j¨
107%
2q weµqg~j¨
100%
µqg~j¨
`yB weµqg~‡jï e¨veavb 107-90=17%
Avevi UvKvq eëavb cÖkœvbyhvqx GB 17% = 51 UvKv|
RvwKiRwmg hZ †QvU ZZ eo ewki
56

mgvavb:
awi,
cÖ_‡g †QvU e„‡Ëi evmva© = 100
myZivs †QvU e„ËwUi †ÿÎdj = (100)2
= 10000
10% e„w×i ci bZzb e¨vmva© = 100+10 = 110
bZzb e„‡Ëi †ÿÎdj = (110)2
= 12100
myZivs ‡ÿÎdj e„w× †c‡q‡Q = = 12100 - 10000 = 2100
e„w×i, kZKiv nvi =


10000
1002100
= 21%
6. Lv‡qi Zvi Av‡qi kZKiv 60% LiP K‡i| Zvi Avq 32% e„w× cvIqv‡Z ‡m Zvi LiP AviI 20% evwo‡q w`j| G‡Z
Zvi mÂq kZKiv KZ e„w× cv‡e ev Kg‡e ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 32% e„w× cv‡e B) 50% e„w× cv‡e C) 32% Kg‡e D) 50% Kg‡e E) ‡KvbwUB bq DËi: B
mgvavb:
awi, Lv‡q‡ii eZ©gvb Avq = 100 UvKv, myZivs e¨q = 60 UvKv Ges mÂq = 100-60 = 40 UvKv|
32% e„w×‡Z bZzb Avq = 100+32 = 132| bZzb e¨q = 60 + (60 Gi 20%) = 60+12 = 72 UvKv|
Zvn‡j bZzb mÂq = 132-72 = 60 UvKv| Zvi mÂq evo‡jv 60-40 = 20 UvKv
mÂq e„w×i kZKiv nvi =
40
10020
= 50% (40 UvKvq 20 UvKv evo‡j 100 †Z evo‡j 50UvKv ev 50%)
7. wcZv I cy‡Îi eZ©gvb eq‡mi AbycvZ 3:1| wZb eQi Av‡M Zv‡ì eq‡mi AbycvZ wQj 7:2| wZb eQi ci Zv‡ì eq‡mi
AbycvZ KZ n‡e ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 7:3 B) 5:2 C) 9:5 D) 8:3 E) ‡KvbwUB bq DËi: D
mgvavb:
awi, wcZv I cy‡Îi eZ©gvb eq‡mi AbycvZ = 3x Ges x
cÖkœg‡Z,
2
7
3-x
3-3x
 (Dfq cv‡k 3 eQi Av‡Mi eq‡mi AbycvZ)
ev, 7x-21 = 6x – 6
ev, 7x- 6x = 21- 6 myZivs x = 15
myZivs Zv‡ì eZ©gvb eqm: = 315 = 45 eQi Ges 15 eQi|
3 eQi ci Zv‡ì eq‡mi AbycvZ n‡e (45+3) : (15+3) = 48:18 = 8:3 DËi: 8:3
8. hw` n GKwU †Rvo msL¨v nq Z‡e wb‡Pi †KvbwU †Rvo msL¨v n‡Z cvi‡e bv ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) n2
B) 5(n-1)+1 C) 2n+2 D) 7n+3 E) ‡KvbwUB bq DËi: D
mgvavb:
awi, n = 2
Zvn‡j: A) n2
= 22
=4 = †Rvo| ( jwRKvwj: ‡h †Kvb ‡Rvo msL¨vi eM© †Rvo B n‡e)
B) 5(n-1)+1 = 5 (2-1)+1 = 51+1 = 6 = †Rvo| (‡Rvo †_‡K 1 we‡qvM Ki‡j we‡Rvo Ges 5 we‡Rv‡oi mv‡_
¸Y Ki‡j ¸Ydj I we‡Rvo †k‡l we‡Rv‡oi mv‡_ 1 †hvM Ki‡j †Rvo n‡e| )
C) 2n+2 = 22+2 = 4+2 = 6 = †Rvo| (2 w`‡q †h †Kvb msL¨v‡K ¸Y Ki‡j †Rvo Ges 2 †hvM Ki‡jI †Rvo)
D) 7n+3 = 72+3 = 14+3 = 17 = we‡Rvo| (7 we‡Rvo msL¨vi mv‡_ †h †Kvb †Rvo msL¨v n ¸Y Ki‡j ¸YdjwU
Aek¨B †Rvo n‡q hv‡e Ges †mB †Rv‡oi mv‡_ 3 †hvM Ki‡j †hvMdj we‡Rvo n‡e| ) GUvB DËi: D
gy‡L gy‡L: e„‡Ëi e¨vm ev e¨vmva© Kgv ev evovi K_v ejv n‡j
Zv e‡M©i gZ nq| A_©vr 1 evi evov A_©B 2 evi evov| Zvn‡j
GLv‡b cÖ_‡g 10% evo‡j 100 †_‡K n‡e 100+10 = 110 Ges
cieZ©x‡Z 110 Gi Dci 10% evo‡j n‡e 110+(110 Gi 10%)
= 110+11 = 121 | †gv‡Ui Dci evo‡jv 121-100 = 21% |

9. 60 wjUv‡ii GKwU wgkª‡Y `ya I cvwbi AbycvZ 2:1 | AbycvZ 1:2 Ki‡Z KZ wjUvi cvwb †gkv‡Z n‡e ? [cvwb Dbœqb
†evW©: (wnmve KiwYK)-2018]
A) 20 B)30 C)50 D)60 E) ‡KvbwUB bq DËi: D
mgvavb: ( wjwLZ wbq‡g)
cÖ`Ë Abycv‡Zi †hvMdj = 2+1 = 3
`y‡ai cwigvb: = 60 Gi
3
2
= 40 Ges cvwbi cwigvY = 60 Gi
3
1
= 20
awi,
cvwb †gkv‡Z n‡e, = x wjUvi|
cÖkœg‡Z,
x20
40

=
2
1
(x ‡KwR †hvM K‡i `ycv‡k bZzb AbycvZ)
ev, 20+x = 80
myZivs x = 60 wjUvi|
10.
4
3
,
2
1
Gi kZKiv KZ n‡e ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 150% B) 110% C) 300% D) 400% E) †KvbwUB bq DËi: A
mgvavb:
Dci fMœvskwU †`L‡j A‡b‡Ki AsKUv ïiæ Ki‡Z mgm¨v n‡e| Zvn‡j Gfv‡e hw` fvev hvq, 5 msL¨vwU 10 Gi kZKiv
KZ? Aek¨B A‡a©K ev 50% | wKfv‡e? kZKiv nvi †ei Kivi wbq‡g, Gfv‡e,
10
1005 
= 50% (hvi mv‡_ Zzjbv
Kiv nq Zv‡K wb‡P Ges hv‡K Zzjbv Kiv nq ‡mB msL¨vwU‡K Dc‡i)
GLb cÖ‡kœ cÖ`Ë fMœvskØq kZKivi wbq‡g mvRv‡j,
2
1
100
4
3

= 752 = 150%
11. †Kvb msL¨vi `yB-Z„Zxqvsk H msL¨vi †P‡q 50 Kg n‡j msL¨vwU KZ ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
A) 208 B) 350 C) 250 D) 150 E) ‡KvbwUB bq DËi: D
awi, msL¨vwU = x
cÖkœg‡Z,
x-
3
x2
=50 ev,
3
x2x3 
=50 x = 150
12. 7 Rb †jvK 7 w`‡b 7 wU †Uwej ˆZwi K‡i | 5Rb †jv‡Ki 5 wU †Uwej ˆZwi Ki‡Z Kqw`b jvM‡e ? [cvwb Dbœqb †evW©:
(wnmve KiwYK)-2018]
 kU©KvU©: 10 †m‡K‡Û mgvav‡bi Rb¨ Gfv‡e fveyb
msL¨vwU‡K †gvU 3 fvM Ki‡j ( ni‡K †gvU a‡i)
‡mLvb †_‡K 2 fvM ev` w`‡j _vK‡e Avi 1 fvM| GLb 1 fvM
= 50 n‡j 3 fvM = 503 = 150 |
gy‡L gy‡L Kivi Rb¨:
1+2 = 3 Ask = 60 n‡j 2 Ask = 40 Ges 1 Ask = 20
GLb cvwbi 20 wjUv‡ii mv‡_ wKQz †hvM n‡e wKš‘ `ya Av‡Mi
40 c‡iI 40 B _vK‡e| ZvB GB 40 B 2q Abycv‡Z n‡q
†M‡j 1:2 G 1 Ask| GLb 1 Ask = 40 n‡j cvwbi 2 Ask =
80 n‡Z n‡e| Av‡M †_‡K 20 wjUvi cvwb _vKvq bZzb K‡i
†hvM Ki‡Z n‡e 80-20 = 60 †KwR|

mgvavb: ( wjwLZ wbq‡g)
7 Rb ‡jvK‡K 7 wU †Uwej evbv‡Z mgq jv‡M = 7 w`b| (w`b †ei Ki‡Z n‡e ZvB w`b †k‡l)
1 ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ =
wU7
Rb77 
(‡jvK Kg‡j †ewk w`b ZvB ¸Y, Kg †Uwej n‡j Kg mgq ZvB fvB)
5 ÕÕ ÕÕ 5ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ =
57
577


[‡ewk †jvK = Kg mgq (fvM) Ges ‡ewk †Uwej =‡ewk mgq(¸Y)]
= 7 w`b|
 ev¯Í‡e fve‡j: 5 Rb †jv‡Ki 5wU ‡Uwej evbv‡bvi Rb¨ 5w`b jv‡M| GLb †mLv‡b 7-5 = 2wU ‡Uwej evwo‡q †`qvq †ewk
w`b jvMvi K_v wKš‘ GB `ywU AwZwi³ ‡Uwe‡ji Rb¨ †h‡nZz 2Rb AwZwi³ †jvK cÖ`vb Kiv n‡q‡Q ZvB Av‡M hZ w`b jvMZ
GLb ZZ w`bB jvM‡e|
GB ai‡bi Confusing Question ¸‡jv gvbwmK `ÿZv As‡k cÖPzi Av‡m|
‡`‡L g‡b n‡e GK DËi wKš‘ Avm‡jB DËiwU GKUz wfbœ| ZvB Av‡M eySzb|
13. QqwU msL¨vi Mo 6 | hw` cÖwZwU msL¨v †_‡K 2 we‡qvM Kiv nq Z‡e bZzb msL¨v¸‡jvi Mo KZ n‡e ? [cvwb Dbœqb †evW©:
(wnmve KiwYK)-2018]
6wU msL¨vi Mo =6 n‡j 6wU msL¨vi mgwó = 66 = 36
cÖwZwU msL¨v †_‡K 2 we‡qvM Ki‡j †gvU Kg‡e = 62 = 12
6wU msL¨vi eZ©gvb mgwó = 36-12 = 24
myZivs msL¨v 6wUi bZzb Mo = 246 = 4|
14. 40 dzU j¤^v GKwU euvk Ggbfv‡e †K‡U `yfv‡M Kiv nj †hb †QvU Ask eo As‡ki 2/3 fvM nq | †QvU As‡ki ˆ`N©¨
KZ d~U ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-2018]
mgvavb:
awi, euvkwUi eo AskwU = 3x Ges †QvU AskwU 2x (fMœvsk aivi †_‡K Gfv‡e c~Y© msL¨v ai‡j mnR n‡e|)
cÖkœg‡Z,
3x+2x = 40 (`yÕ As‡ki †hvMdj = 40 dzU|)
5x = 40 x = 8 myZivs †QvU AskwU = 28 = 16 dzU| DËi: 16 dzU|
15. GKwU †Zjc~Y© cv‡Îi IRb 30 †KwR | A‡a©K †Zj mn cv‡Îi IRb 20 †KwR n‡j cvÎwUi IRb KZ †KwR ? [cvwb
Dbœqb †evW©: (wnmve KiwYK)-2018]
awi, cvÎwUi IRb: = x †KwR|
cÖkœg‡Z,
30-x = 2 (20-x) ( ev‡g, †gvU IRb - cvÎ = ‡gvU †Zj |
Avevi Wv‡b †gvU IRb †_‡K cvÎ ev` w`‡j hv _v‡K Zv †gvU
†Z‡ji A‡a©K, ZvB 2 w`‡q ¸Y Ki‡j ev‡gi †gvU †Z‡ji mgvb n‡e )
ev, 30-x = 40-2x
x = 10 myZivs cvÎwUi IRb| = 10 †KwR| `ywU wPÎ Zzjbv Ki‡j †evSv hvq †Zj ev‡`
cv‡Îi IRb n‡e 30-20 = 10 †KwR|
10 †KwR
10 †KwR
Lvwj
10 †KwR
cvÎmn20‡KwR
cvÎmn30‡KwR
wPÎwU †`Lyb: ev‡g †Zj c~Y©
Ges Wv‡b A‡a©K c~Y©
 kU©KvU©: 5 †m‡K‡Û mgvav‡bi Rb¨ Gfv‡e fveyb
cÖwZwU msL¨v †_‡K 2 we‡qvM Ki‡j †h‡nZz cÖwZwU msL¨vB
2 K‡i K‡g hv‡e ZvB Zv‡`i Mo I 2 Kg‡e|
myZivs bZzb Mo n‡e 6-2 = 4|

kU©KvU©: gy‡L gy‡L Kivi Rb¨:
1g wbqg: ‡Zj Kg‡jv 30-20 = 10 †KwR| hv ‡gvU
†Z‡ji A‡a©K | myZivs †gvU †Zj wQj 102 = 20
†KwR| AZGe cv‡Îi IRb = 30-20 = 10 †KwR|
2q wbqg: A‡a©‡Ki IRb †h‡nZz 20 Zvn‡j m¤ú~Y© Gi IRb =
202 = 40‡KwR| wKš‘ †Zj wØ¸Y n‡jI cvÎwU wKš‘ wØ¸Y
n‡e bv| ZvB cv‡Îi IRb n‡”Q 40-30 = 10 †KwR|
 ev¯Íe m¤§Z e¨vL¨v: A‡a©K †Zj †d‡j w`‡j cvÎwU wKš‘ A‡a©K n‡q hvq bv eis ïiæ‡Z hv wQj †k‡lI Zv B _vK‡e|
GRb¨ cÖ_‡g †h 30-20 = 10 †KwR Kg‡jv GUv m¤ú~Y© †Zj| A‡a©K †d‡j w`‡j hw` 10 †KwR †Zj K‡g Zvn‡j m¤ú~Y© †d‡j
w`‡j 20 †KwR K‡g hv‡e| A_©vr GB 20 †KwR †Zj Ges 20 †KwR †d‡j †`qvi ciI AwZwi³ 30-20 = 10 †KwR Av‡Q hv
cv‡Îi IRb|
16. GKwU †Uªb N›Uvq 180 wK‡jvwgUvi †e‡M Pj‡j cÖwZ †m‡K‡Û †UªbwU KZ wgUvi Pj‡e ? [cvwb Dbœqb †evW©: (wnmve KiwYK)-
2018]
mgvavb:
1 N›Uv = 60 wgwbU ev 6060 = 3600 †m‡KÛ Ges 180 wK.wg. = 1801000 = 180000 wg (1wKwg =1000 wg)
GLv‡b, 3600 †m‡K‡Û hvq = 180 wK.wg.
1 †m‡K‡Û hvq,
3600
180000
= 50wgUvi|
================================
==
Gfv‡e cÖwZwU As‡Ki we¯ÍvwiZ wbq‡g mgvav‡bi cv‡k †hŠw³K e¨vL¨vi mv‡_ mv‡_ K‡qK
†m‡K‡Û gy‡L gy‡L mgvavb Kivi ev¯Íe m¤§Z †UKwbK wkL‡Z Avgv‡`i mv‡_ _vKzb| ab¨ev`|

1. GKwU eM©‡ÿ‡Îi ˆ`N©¨ I cÖ¯’ h_vµ‡g 30% I 20% evov‡bv nj| bZzb ˆZwi AvqZ‡ÿÎwUi †ÿÎdj, g~j eM©‡ÿÎwUi
†ÿÎdj †_‡K kZKiv KZ †ewk ? [ BADC –( Computer-Operator)-2018]
a. 72% b. 60% c. 56% d. 44% Ans: c
Solution:
cÖ_‡g 30% evov‡j 100+30 = 130 nq Gici 130 Gi 20% = 26 evov‡j †gv‡U evo‡e 30+26 = 56%|
2. `yRb cyiæl I GKRb gwnjv 4 N›Uvq GKwU KvR Ki‡Z cv‡i| GKB KvR GKRb cyiæl I wZbRb gwnjv wg‡jI 4 N›Uvq
Ki‡Z cv‡i| †mB KvRwU Ki‡Z wZbRb cyiæl I PviRb gwnjvi KZ N›Uv jvM‡e ? [ BADC –( Computer-
Operator)-2018]
a. 3 b. 2.5 c. 2 d. 1 Ans:c
Solution:
(2Rb cyiæl + 1 Rb gwnjv)4 = (1 Rb cyiæl+3Rb gwnjv)  4
A_ev, 8Rb cyiæl + 4 Rb gwnjv = 4 Rb cyiæl +12 Rb gwnjv
A_ev, 4 Rb cyiæl = 8 Rb gwnjv
 1 Rb cyiæl = 2 Rb gwnj|
GLb,
cÖ_‡gi, 2 Rb cyiæl + 1 Rb gwnjvi = 4 w`b
4 Rb gwnjv + 1 Rb gwnj = 4 w`b
5 Rb gwnjv = 4 w`b ( mevB‡K gwnjv‡Z KbfvU© Kiv n‡q‡Q)
Ges †k‡li , ( †hUv †ei Ki‡Z n‡e)
3 Rb cyiæl + 4 Rb gwnjv
6Rb gwnjv + 4 Rb gwnjv
10 Rb gwnjv
5 Rb gwnjv‡K GK‡Î 4 w`b jvM‡j
10 Rb ev wØ¸Y msL¨K gwnjv‡K jvM‡e A‡a©K w`b A_©vr 2 w`b|
3. 10 wU msL¨vi †hvMdj 462| G‡`i cÖ_g 4wUi Mo 52 Ges †kl 5 wUi Mo 38 n‡j cÂg msL¨vwU KZ ? [ BADC –(
Computer-Operator)-2018]
a. 65 b. 64 c. 58 d. 57 Ans: b
Solution:
cÖ_g 4wU Ges †kl wUi †hvMdj = (452)+ (538) = 208+190 = 398
myZivs cÂg msL¨vwU = 462-398 = 64|
4. GKwU Rjvav‡ii `yB-cÂgvsk cvwb Øviv c~Y© Ges G‡Z Av‡iv 25 wjUvi cvwb Xvj‡j Gi 90% cvwbc~Y© nq| RjvaviwUi
aviYÿgZv KZ wjUvi ? [ BADC –( Computer-Operator)-2018]
a. 50 b. 100 c. 150 d. 200 Ans: a
BADC
‡cv‡÷i bvg: Awdm mnKvix Kvg Kw¤úDUvi Acv‡iUi:
cixÿvwU wb‡q‡Q: IBA, DU cixÿvi ZvwiL 11-05-2018

Solution:
2/5 Ask A_© = 40% GLb 25 wjUvi Xvjvi ci 90% c~Y© n‡j bZzb fv‡e c~Y© n‡jv 90-40 = 50%
myZivs 50% = 25 wjUvi n‡j aviY ÿgZvi 100% = 50 wjUvi n‡e|
5. 7% nvi mij my‡` 3000 UvKv KZ eQ‡ii Rb¨ wewb‡qvM Ki‡j †gvU 420 UvKv gybvdv cvIqv hv‡e ? [ BADC –(
Computer-Operator)-2018]
a. 5 b. 4 c. 3 d. 2 Ans:d
Solution:
3000 UvKvi 7% nv‡i GKeQ‡ii gybvdv = 307 = 210 UvKv|
Zvn‡j 420 UvKv †c‡Z mgq jvM‡e 420210 = 2 eQi|
6. hw` ÔKÕ Ges ÔLÕ Dfq †Rvo msL¨v nq, Zvn‡j wb‡Pi †KvbwU Aek¨B we‡Rvo msL¨v n‡e ? [ BADC –( Computer-
Operator)-2018]
a. K+2L b. KL+1 c. K+L d. 2K+L Ans:b
Solution:
‡Rvo msL¨vi mv‡_ †Rvo msL¨v †hvM we‡qvM ev ¸Y Ki‡j Zv †Rvo B _v‡K| ZvB KL = KL = †Rvo|
ZvB G‡`i mv‡_ we‡Rvo msL¨v †hvM Ki‡j Zv we‡Rvo n‡e|
myZivs Ackb B ‡Z cÖ`Ë KL+1 = we‡Rvo|
7. myg‡bi Kv‡Q †h UvKv Av‡Q Zv w`‡q †m 18 wU WvKwUwKU µq Ki‡Z cv‡i| hw` cªwZwU WvKwUwK‡Ui g~j¨ 4 UvKv Kg nZ
Zvn‡j †m Av‡iv `ywU WvKwUwKU †ewk µq Ki‡Z cviZ| Zvi Kv‡Q KZ UvKv Av‡Q ? [ BADC –( Computer-
Operator)-2018]
a. 180 b. 360 c. 540 d. 720 Ans: d
Solution:
awi, myg‡bi Kv‡Q _vKv UvKvi cwigvY, = K
cÖkœg‡Z,
20
K
18
K
 = 4
A_ev,
180
9K10K 
= 4
 K = 720
myZivs myg‡bi Kv‡Q ‡gvU 720 UvKv wQj|
cÖgvY: 720 UvKv w`‡q 18 wU wKb‡j cÖwZwUi `vg = 40 UvKv
Avevi, 720 UvKv w`‡q 20wU wKb‡j cÖwZwUi `vg = 36 UvKv | `yB `v‡gi cv_©K¨ 40-36 = 4 UvKv|
8. 12 dzU ‰`N©¨ Ges 8 dzU cÖ¯’ wewkó GKwU Kv‡c©U Øviv GKwU †g‡Si 60% †gvov‡bv hvq| †g‡SwUi AvqZb KZ eM©dyU ?
[ BADC –( Computer-Operator)-2018]
a. 96 b. 160 c. 64 d. 180 Ans: b
Solution:
60% Gi †ÿÎdj = 12 8 = 96 eM©dzU|
myZivs 1% =
60
96
 100% =
60
10096
= 160 eM©dzU|
gy‡L gy‡L Kivi Rb¨ Gfv‡e fveyb:
18wU Ges 18+2 = 20 wU w`‡q Ack‡bi
msL¨v¸‡jvi g‡a¨ hv‡K fvM Ki‡j fvMdj `ywUi
cv_©K¨ 4 n‡e Zv B DËi: GLv‡b 720 a‡i wn‡me
Ki‡j wg‡j hvq ZvB GUvB DËi|

9. GKwU c¨v‡K‡U 520 wU gv‡e©j Av‡Q| G‡Z Kgc‡ÿ Av‡iv KZ ¸‡jv gv‡e©j †hvM Kiv n‡j †m¸‡jv 3,4 A_ev 6 Rb
Qv‡Îi g‡a¨ mgvbfv‡e fvM K‡i †`qv hv‡e ? [ BADC –( Computer-Operator)-2018]
a. 4wU b. 6wU c. 8wU d. 12wU Ans: c
Solution:
3, 4 Ges 6 Gi j.mv.¸ = 12 | GLb 520 Gi mv‡_ KZwU †hvM Ki‡Z n‡e Zv ‡ei Kivi Rb¨ 520 †K 12
w`‡q fvM Ki‡j fvM‡kl Av‡m 4|Zvn‡j 12w`‡q wb:‡k‡l fvM Kivi Rb¨ Av‡iv gv‡e©j jvM‡e 12-4 = 8 wU|
10. GKwU Mv‡Qi D”PZv cÖwZeQi 20% e„w× cvq| hw` eZ©gv‡b MvQwUi D”PZv 1080 †m.wg. n‡q _v‡K Zvn‡j `yB eQi Av‡M
MvQwUi D”PZv KZ wQj ? [ BADC –( Computer-Operator)-2018]
a. 675 ‡m.wg b. 750 †m.wg c. 775 †m.wg d. 800 †m.wg. Ans: b
Solution:
awi, 2 eQi Av‡M D”PZv wQj = K
2 eQi ci 20% K‡i 2 evi evo‡j nq
K Gi 120% Gi 120% = 1080
K = 1080 
120
100

120
100
= 750 ‡m.wg.
11. wZb eQi Av‡M GKRb wkÿK I GKRb Qv‡Îi eq‡mi AbycvZ wQj 4 : 1 Ges Zv‡`i eq‡mi ¸bdj wQj 196| 8 eQi
ci Zv‡`i eq‡mi AbycvZ KZ n‡e ? [ BADC –( Computer-Operator)-2018]
a. 13.6 b. 6.13 c. 12.5 d. 5.12 Ans: a
Solution:
awi,
3 eQi Av‡M Qv‡Îi eqm = K eQi Ges wkÿ‡Ki eqm = 4K eQi
cÖkœg‡Z,
K4K = 196
ev, 4K2
= 196
ev, K2
= 49 K = 7
myZivs 3 eQi Av‡M QvÎ I wkÿ‡Ki eqm h_vµ‡g 7 eQi Ges 47 = 28 eQi|
eZ©gvb eqm 7+3 = 10 eQi Ges 28+3 = 31 eQi|
8 eQi ci wkÿK I Qv‡Îi eq‡mi AbycvZ n‡e 31+8: 10+8 = 39:18 = 13:6 eQi|
12. 10% jeYhy³ 12 wjUv‡ii GKwU `ªeY †_‡K 2 wjUvi cvwb ev®úxf~Z Kiv n‡j, Aewkó `ªe‡Y KZ kZvsk jeY _vK‡e ? [
BADC –( Computer-Operator)-2018]
a. 6% b. 10% c. 12% d. 14.4% Ans: c
Solution:
12 wjUv‡ii g‡a¨ jeY = 12 Gi 10% = 1.2 Ges cvwb = 12-1.2 = 10.8
2wjUvi cvwb Zz‡j †bqvq eZ©gv‡b cvwb Av‡Q 10.8- 2 = 8.8
eZ©gvb †gvU `ªe‡Yi cwigvY = 12-2 = 10
myZivs 10 wjUvi `ªe‡Y jeY Av‡Q 1.2 Zvn‡j 100 †Z n‡e 12%| ( eyS‡j mivmwi GB jvBb †_‡K ïiæ)
13. ‰`wbK 9 N›U KvR K‡i 5 Rb kÖwgK 3 w`‡b 9 wU ev· evbv‡Z c‡i| ˆ`wbK 10 N›Uv KvR K‡i 8 Rb kªwgK 6 w`‡b
GKB iK‡gi KZwU ev· evbv‡Z cvi‡e ?[ BADC –( Computer-Operator)-2018]
a. 54wU b. 48wU c. 36wU d. 32wU Ans:d

Solution:
9 N›Uv K‡i KvR K‡i 5 Rb kÖwgK 3 w`‡b evbv‡Z cv‡i = 9 wU e·|
1 ÕÕ ÕÕ ÕÕ ÕÕ 1 ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ =
359
9

wU e·|
10 ÕÕ ÕÕ ÕÕ ÕÕ 8 ÕÕ ÕÕ 6 ÕÕ ÕÕ ÕÕ =
359
68109


wU e·| = 32wU|
mn‡R †evSvi Rb¨ e¨vL¨v:
9 N›Uvi cwie‡Z© 1 N›Uv A_©vr Kg KvR Ki‡j Kg e· ˆZix n‡e ZvB fvM, Avei 5 R‡bi cwie‡Z© 1 Rb
KvR Ki‡j Av‡iv Kg e· ˆZix n‡e ZvB Avevi fvM, Ges 6 w`‡b cwie‡Z© 1 w`b KvR Ki‡j Avevi Kg n‡e
ZvB Avevi fvM| Gfv‡e me¸‡jv Kgvi Kvi‡Y fvM n‡j †k‡li jvB‡b me¸‡jv evovi Kvi‡Y ¸Y n‡e | ZvB
mivmwi me¸‡jv msL¨v Dc‡i wj‡L KvUvKvwU|
14. GKwU †Uªb 20 wK.wg/N›Uv †e‡M Pj‡Q| GKRb e¨w³ GKB w`‡K 15 wK.wg/N›Uv ‡e‡M Pj‡Q| †UªbwU hw` e¨w³wU‡K 3
wgwb‡U AwZµg K‡i, Zvn‡j †Uª‡bi ˆ`N©¨ KZ ? [ BADC –( Computer-Operator)-2018]
a. 200 wgUvi b. 220 wgUvi c. 225 wgUvi d. 250 wgUvi Ans: d
Solution:
‡Uªb I e¨w³ GKB w`‡K Pjvq Zv‡`i AvcwÿK MwZ, 20-15 = 5 wKwg,
5
18
5
=
18
25
wg./‡m‡KÛ
1 †m‡K‡Û hvq =
18
25
3 wgwbU ev 180 †m‡K‡Û hvq =
18
25
180 = 250 wgUvi|
15. `yBwU msL¨vi ¸bdj 156 Ges Zv‡`i e‡M©i †hvMdj 313| msL¨v `ywUi †hvMdj KZ ? [ BADC –( Computer-
Operator)-2018]
a. 23 b. 25 c. 26 d. 27 Ans: b
Solution:
awi, msL¨v `yw&U h_vµ‡g K Ges L
cÖkœg‡Z, KL = 156 Ges K2
+L2
= 313
Avgiv Rvwb,
(K+L)2
= K2
+L2
+2KL
(K+L)2
= 313+2156
 (K+L)2
= 313+312
 (K+L)2
= 625
 K+L = 25
gv_v LvuUv‡Z cvi‡j:
msL¨v `ywUi ¸Ydj = 156 †K fv½v‡bv| hvq 1213 Avevi 122
= 144 Ges 132
= 169 Gi †hvMdj
144+169 = 313 myZivs †hvMdj 12+13 =25|
==================================

1. GKwU cixÿvq cvk b¤î 42%| Avwè †m cixÿvq 133 b¤î cvq Ges 35 b¤î Kg †c‡q †dj K‡i| †gvU KZ b¤^‡ii
cixÿv n‡qwQj ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 500 B. 400 C. 300 D. 200 E. ‡KvbwUB bq
mgvavb: (B)
133 b¤î †c‡qI 35 b¤^‡ii Rb¨ †dj Ki‡j †gvU cvk b¤î = 133+35 = 168
‡h‡nZz cvk b¤î 42% ZvB 42% = 168, 1% = 100% = = 400 DËi: 400
2. `ywU msL¨vi AbycvZ 5 : 3 Ges G‡ì j mv ¸ 1815| cÖ_g msL¨vwU KZ ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 55 B. 45 C. 35 D. 25 E. ‡KvbwUB bq
mgvavb: (E)
awi, msL¨v `ywU h_vµ‡g 5K Ges 3K
myZivs msL¨v `ywUi j.mv.¸ = 15K (¸Y K‡i 15K2
wb‡j fzj n‡e|)
cÖkœg‡Z,
15K = 1815
K = = 121
myZivs cÖ_g msL¨wU 5121 = 605 DËi: E (‡KvbwUB bq|)
3. GKwU AvqZ‡ÿ‡Îi cÖ¯’ Zvi ‰`N©¨ A‡cÿv 40% Kg| hw` AvqZ‡ÿ‡Îi cÖ¯’ 36 wgUvi nq, Zvn‡j AvqZ‡ÿ‡Îi †ÿÎdj
KZ ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 2400 eM ©wg. B. 2610 eM ©wg C. 2410eM© wg D. 2160eM© wg E. ‡KvbwUB bq
mgvavb: (D)
awi, AvqZ‡ÿÎwUi ˆ`N¨© = 100 wg. Ges cÖ¯’ 40% Kg A_©vr 60wg.
cÖkœg‡Z,
60% = 36, 1% = 100% = = 60 wg. myZivs ˆ`N©¨ = 60
‡ÿÎdj = ‰`N©¨ cÖ¯’ = 6036 = 2160 eM© wgUvi| DËi: 2160 eM© wgUvi|
4. iæ‡nj 25% ÿwZ‡Z GKwU Nwo weµq K‡i| hw` NwowU 350 UvKv †ewk `v‡g weµq Kiv †hZ Zvn‡j Zvi 10% jvf
nZ| NwowUi µqg~j¨ KZ wQj ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 400 UvKv B. 500UvKv C. 1000 UvKv D. 100 UvKv E. ‡KvbwUB bq
42
168
42
100168
15
1815
60
36
60
10036
Titas Gas Field
Post name: Assistant Officer (General) Exam date: 27-04-2018
Exam taker: IBA.DU.
cÖgvY: cÖ_g msL¨vwU 605 Ges 2q msL¨wU 3121 = 363
GLb 605 Ges 363 Gi j.mv.¸ 1815 |

mgvavb: (C)
awi, µqg~j¨ = 100
myZivs 25% ÿwZ‡Z cÖ_g weµqg~j¨ = 100-25 = 75
Ges 10% jv‡f wØZxq weµqg~j¨ = 100+10 = 110
`yB weµqg~‡jï eëavb = 110-75 = 35
weµqg~j¨ Av‡iv 35 UvKv †ewk n‡j µqg~j¨ = 100 UvKv
 ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ = ÕÕ
 ÕÕ ÕÕ 350 ÕÕ ÕÕ ÕÕ ÕÕ = = 1000 UvKv| DËi: 1000 UvKv|
5. GKwU AvqZ‡ÿ‡Îi cÖ¯’ I cwimxgvi AbycvZ 1 : 5| AvqZ‡ÿ‡Îi ˆ`N©¨ I cÖ‡¯’i AbycvZ KZ ? [wZZvm M¨vm wdì-mn:
Awd:-2018]
A. 1 : 5 B. 5 : 1 C. 3 : 2 D. 2 : 3 E. ‡KvbwUB bq
mgvavb: (C)
awi, AvqZ‡ÿ‡Îi cÖ¯’ = K Ges cwimxgv = 5K
Avgiv Rvwb,
AvqZ‡ÿ‡Îi cwimxgv = 2 (‰`N©¨ + cÖ¯’)
 2 (‰`N©¨ + K) = 5K
‰`N©¨ + K =
 ˆ`N©¨ = - K = =
myZivs ˆ`N©¨ I cÖ‡¯’i AbycvZ = : K = 2 : K2 = 3K : 2K = 3 : 2 DËi: 3:2
6. GKRb LyPiv we‡µZv Zvi c‡Yï wjwLZ g~‡jï Dci 10% Kwgkb †`qvq Zvi 12.5% jvf nq| wjwLZ g~‡jï Dci
20% Kwgkb wb‡j Zvi kZKiv KZ jvf ev ÿwZ n‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 10% ÿwZ B. 12% ÿwZ C. 12% jvf D. jvf ev ÿwZ wKQzB n‡e bv E. †KvbwUB bq
mgvavb: (D) (wjwLZ gyj¨‡K 100 a‡i mvgvavb)
awi,
wjwLZg~j¨ = 100 UvKv
10% Kwgkb w`‡q 1g weµqg~j¨ = 100-10 =90UvKv|
Avevi 12.5% jv‡f weµqg~‡jï nvi =112.5%
cÖkœg‡Z, 112.5% = 90 (Gi g‡a¨ µ‡qi 100% + jv‡fi 12.5% Av‡Q)
 1% =
100% = = 80
myZivs µqg~j¨ 100% = 80UvKv|
GLb 100UvKvi c‡Y¨ 20% Qvo w`‡j weµqg~j¨ I n‡e 80 UvKv| A_©vr 20% Qvo w`‡j jvf ev ÿwZ wKQzB n‡e bv|
35
100
35
350100 
2
5K
2
5K
2
2K5K 
2
3K
2
3K
2
3K
112.5
90
112.5
10090
Shortcut:
25% ÿwZ + 10% jvf = 35% = 350
myZivs 1% =
100% = = 1000UvKv|
35
350
35
100350

weKí mgvavb: (µqg~j¨‡K 100 a‡i mgvavb|)
awi µqg~j¨ 100 UvKv| ( †h‡nZz cÖ‡kœ 2wU % Av‡Q ZvB `y cvk †_‡KB aiv hvq)
myZivs 12.5% jv‡f weµqg~j¨ = 100+12.5 = 112.5 UvKv|
GLb, 10% Qvo w`‡q weµgy‡jï % = 100-10 = 90%
Zvn‡j †jLv hvq 90% = 112.5
1% =
100% = = 125 UvKv| A_©vr 10% Qvo †`qvi Av‡M wjwLZ gyj¨ wQj 125 UvKv|
GLb 125 UvKvi cY¨ 20% Qvo w`‡q bZzb weµqg~j¨ n‡e 125-125 Gi 20% = 125-25 = 100 UvKv|
Zvn‡j †`Lv hv‡”Q µqg~j¨ I weµqg~j¨ G‡ÿ‡Î mgvb n‡q hv‡”Q myZivs jvf ev ÿwZ wKQzœB n‡e bv| DËi: E
7. GKRb kÖwgK 25 w`‡b GKwU Kv‡Ri 5/16 Ask †kl Ki‡Z cv‡i| GB nv‡i KvR Ki‡j m¤ú~Y© KvR †kl Ki‡Z Zvi
AwZwi³ Avi KZ w`b jvM‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 80 w`b B. 120 w`b C. 55 w`b D. 45 w`b E. †KvbwUB bq
mgvavb: (C)
ev¯Í‡e ey‡S Ki‡j gy‡L gy‡L n‡e|
aiæb GKUv 16 Zjv wewìs‡qi 5 Zjv evbv‡Z 25 w`b ‡j‡M‡Q, Zvn‡j cÖwZ Zjv evbv‡Z mgq eiv× 255 = 5 w`b|
GLb 16 Zjvi g‡a¨ 5 Zjv evbv‡bv n‡q †M‡j Avi Aewkó 11 Zjv evb‡Z mgq jvM‡e, 115 = 55 w`b|
eB‡qi fvlvq mgvavb:
KvR n‡q †M‡Q Ask| myZivs KvR Aewkó& Av‡Q, 1 - Ask = Ask
Ask KvR Ki‡Z mgq jv‡M = 25 w`b|
1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = 25 w`b& (cÖ_gevi fMœvsk Dwë‡q ¸Y Ki‡Z nq|)
 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = 25  w`b ( 2q evi fMœvsk †K mivmwi ¸Y Ki‡Z nq|)
= 55 w`b| DËi: 55 w`b|
8. 36 wU Kj‡gi µqg~j¨ K wU Kj‡gi weµq g~‡jï mgvb| hw` jv‡fi nvi 20% nq Zvn‡j K Gi gvb KZ ? [wZZvm
M¨vm wdì-mn: Awd:-2018]
A. 30 B. 32 C. 28 D. 24 E. ‡KvbwUB bq
mgvavb: (A)
awi, 36wU Kj‡gi cÖwZwUi µqg~j¨ = 100 UvKv| ‡gvU LiP = 36100 = 3600
myZivs 20% jv‡f cÖwZwU Kj‡gi weµqg~j¨ = 120UvKv| †gvU weµq = 120K
cÖkœg‡Z,
36100 = K120 (36wUi me©‡gvU LiP = 20% jv‡f K wUi me©‡gvU weµqgy‡jï mgvb)
3600 = 120K
K = = 30 DËi: 30wU|
90
112.5
90
100112.5 
16
5
16
5
16
11
16
5
5
16
16
11
5
16
16
11
120
3600
civgk©: †h cvk †_‡KB hv‡K a‡iB wn‡me
Kiæb bv †Kb ‡Kvb % Gi bvg wK? Ges Kvi
mv‡_ †Kvb % jvMv‡j wK n‡e GB welq¸‡jv
gv_vq ivL‡Z n‡e| Zvn‡j K‡qK ‡m‡K‡Û
mgvavb Kiv m¤¢e n‡e|

weKí mgvavb:
awi, 36wUi µqg~j¨ = 36 UvKv| ( cÖwZwUi 1UvKv `‡i)
myZivs K wUi weµqg~j¨ I = 36UvKv ( Kvib 36wUi µqg~j¨ = K wUi weµqg~j¨)
GLb
20% jv‡f 36UvKvi Kj‡gi weµqg~j¨ = 36+36 Gi 20% = 36+7.2 = 43.2 UvKv|
43.2 UvKvq wewµ Ki‡Z n‡e = 36wU
 1 ÕÕ ÕÕ ÕÕ ÕÕ = wU|
 36 ÕÕ ÕÕ ÕÕ ÕÕ = = 30wU| ( 36UvKvq hZwU wewµ Ki‡Q Zvi cwigvY B n‡”Q K)
DËi: 30wU|
9. GKwU cixÿvq 24% QvÎQvÎx weÁv‡b AK…ZKvh© nq Ges 43% QvÎQvÎx MwY‡Z AK…ZKvh© nq| hw` 15% QvÎQvÎx Dfq
wel‡q AK…ZKvh© n‡q _v‡K, Zvn‡j kZKiv KZRb QvÎQvÎx Dfq wel‡q cvm K‡i‡Q ? [wZZvm M¨vm wdì-mn: Awd:-
2018]
A. 52% B. 67% C. 33% D. 48% E. ‡KvbwUB bq
mgvavb: (D) (cv‡ki wPÎwU †`Lyb)
Dfq wel‡q AK…ZKvh© = 15%
ïay weÁv‡b AK…ZKvh© = 24% - 15% = 9%
ïay MwY‡Z AK…ZKvh© - 43-15 = 28%
myZivs †gvU AK…ZKvh©: = 15%+9%+28% = 52%
Zvn‡j Dfq wel‡q cvm K‡i‡Q = 100% - 52% = 48% DËi: 48%
10. ‡Kvb QvÎvev‡m 320 Rb Qv‡Îi 18 w`‡bi Lvevi Av‡Q| 6 w`b ci QvÎvev‡m Av‡iv 160 Rb QvÎ Avm‡j, evKx Lv`¨
Avi KZ w`b Pj‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 12 w`b B. 10 w`b C. 8 w`b D. 6 w`b E. ‡KvbwUB bq
mgvavb: (C)
w`b Aewkó 18-6 = 12 w`b|
160 Rb Avmvq ‡gvU QvÎ= 320+160 = 480
320 R‡bi Lvevi Av‡Q = 12 w`‡bi
1 ÕÕ ÕÕ ÕÕ = 12320 (Kg †jvK †L‡j †ewkw`b hv‡e)
480 ÕÕ ÕÕ ÕÕ = w`b|
= 8 w`b| DËi: 8w`b|
43.2
36
43.2
3636
480
32012 
gy‡L gy‡L 10 †m‡K‡Û mgvavb: 120% = 36,  1% = 100% = = 30wU|
Gfv‡e Kivi Rb¨ wKQz e¨wmK wRwbm Rvbv _vK‡Z n‡e| †hgb: msL¨vevPK As‡Ki †ÿ‡Î memgq weµq msL¨vi
Dci jvf ÿwZ wn‡me nq | GLv‡b 20% jv‡f 100% msL¨v 30 Gi 120% =36| ZvB DËi: 30wU|
120
36
120
10036
weÁvb=24 MwYZ =43
15
%
9 28
48
‡gvU = 100
Shortcut: ey‡S †M‡j mivmwi,
w`b| = 8 w`b|
480
32012 

11. 180 †KwR IR‡bi GKwU mv‡ii wgkª‡Y ˆRe I BDwiqv mv‡ii AbycvZ 2 : 1| Av‡iv KZ †KwR BDwiqv mvi wgkv‡j †mB
AbycvZ 1 : 2 n‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 180 †KwR B. 140 †KwR C. 120 ‡KwR D. 240 ‡KwR E. ‡KvbwUB bq
mgvavb: (A)
Abycv‡Zi †hvMdj = 2+1 = 3
‰Re Av‡Q, 180 Gi = 120 †KwR|
BDwiqv Av‡Q, 180 Gi = 60 †KwR|
awi, bZzb BDwiqv †gkv‡Z n‡e = K ‡KwR|
cÖkœg‡Z,
120:60+K = 1:2 ( `ycv‡kB bZzb AbycvZ)
ev,
ev, 60+K = 240
K = 240-60 = 180 ‡KwR DËi: 180 †KwR|
12. ‰`wbK 14 N›Uv KvR K‡i 9 wU †gwkb 10 w`‡b 2800 wjUvi cvwb D‡Ëvjb Ki‡Z cv‡i| ˆ`wbK 18 N›Uv KvR K‡i 8 wU
†gwkb KZ w`‡b 3200 wjUvi cvwb D‡Ëvjb Ki‡Z cvi‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 9 w`‡b B. 10 w`‡b C. 12 w`‡b D. 16 w`‡b E. ‡KvbwUB bq
mgvavb: (B) (A‡bK eo HwKK wbq‡gi cÖkœ n‡jI jwRK wVK †i‡L Ki‡j mgq Kg jvM‡e+mwVK DËi
Avm‡e)
14 N›Uv K‡i KvR K‡i 9wU ‡gwkb w`‡q 2800 wjUvi cvwb D‡Ëvjb Ki‡Z mgq jv‡M = 10w`b ( ‡hUv Pvq Zv †k‡l)
 1 ÕÕ ÕÕ ÕÕ ÕÕ 1ÕÕ ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = w`b|
 18 ÕÕ ÕÕ ÕÕ ÕÕ 8ÕÕ ÕÕ ÕÕ 3200 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = w`b|
= 10 w`b| DËi: 10 w`b|
e¨vL¨v: e¨vL¨vwU B Avcbvi m¤ú`| KviY e¨vL¨v c‡o †h jwRK †Wfjc n‡e Zv w`‡qB cieZ©x‡Z GiKg AsK
`ªZ Ki‡Z cvi‡eb|
cÖ_‡g, 14 N›Uv KvR bv K‡i 1 N›Uv KvR Ki‡j mgq †ewk jvM‡e, ZvB ¸Y A_©vr Dc‡i, Avevi 9wU ‡gwk‡bi e`‡j
1wU w`‡q KvR Ki‡j Av‡iv †ewk mgq jvM‡e Zvn‡j Avevi ¸Y| wKš‘ 2800 wjUvi cvwbi e`‡j 1 wjUvi cvwb Zzj‡Z Kg
mgq jvM‡e ZvB fvM|
GLb Dc‡ii ¸Y fvM †`‡L Zvi wb‡Pi msL¨v¸‡jv wecixZ cv‡k ewm‡q †hgb 14 Gi wecixZ cv‡k 18, 9 wecix‡Z
8 Ges 2800 wb‡P e‡m‡Q ZvB 3200 †K Dc‡i ewm‡q KvUvKvwU Ki‡jB DËi †ei n‡e| A_ev Avevi †f‡½ †f‡½I
fve‡Z cv‡ib| DËi: 10 w`bB Avm‡e|
3
2
3
1
2
1
K60
120


2800
91410 
8182800
320091410


eyS‡j 15 †m‡K‡Ûi †Ljv: cÖ_‡g ‰Re 180 Ges BDwiqv 60 |
GLb BDwiqv †gkv‡bvi ci BDwiqvi Abycv‡Zi gvb 2 Ges ˆRe Gi
gvb 1| †h‡nZz ˆRe ‡gkv‡bv nq wb ZvB Av‡Mi 120 B 1 Ask n‡q
†M‡Q| Zvn‡j 2 Ask n‡Z BDwiqv jvM‡e 1202 = 240 | Av‡M
‡_‡K 60 Av‡Q, ZvB ‡gkv‡Z n‡e 240-60 = 180|
( GiKg cÖ‡kœ †hUv †gkv‡Z n‡e bv †mUv wb‡q wn‡me Ki‡Z nq)

13. Kwe‡ii evmv BDmy‡di evmvi 4 gvBj cwð‡g| gvRnv‡ii evmv BDmy‡ci evmvi 6 gvBj DË‡i Ges ûgvqy‡bi evmvi 4
gvBj cwð‡g| Kwei Ges ûgvqy‡bi evmvi mij‰iwLK `yiZ¡ KZ ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 14 gvBj B. 10 gvBj C. 8 gvBj D. 5 gvBj E. ‡KvbwUB bq
mgvavb: (B)
cv‡ki wPÎwU †`Lyb Ges e¨vL¨vwU co–b|
cÖ_‡g BDmyd I Kwe‡ii evmvi `~iZ¡ = 4 gvBj
GLb BDmy‡di evmv †_‡K gvRnv‡ii evmv DË‡i = 6
gvRnv‡ii evmvwU BDmy‡di DË‡i Ges GKB mv‡_
ûgvqy‡bi evmvi cwð‡g nIqvq ûgvqy‡bi evmv n‡e
gvRnv‡ii evmvi c~‡e©|
gvRnv‡ii evmv †_‡K ûgvqy‡bi evmvI 4 gvBj c~‡e©|
GLb Kwe‡ii evmv ‡_‡K AvovAvwo ûgvqy‡bi evoxi `~iZ¡ †ei Kivi Rb¨ wc_v‡Mviv‡mi m~Î cÖ‡qvM Ki‡Z n‡e|
`~iZ¡2
= 82
+ 62
ev, `~iZ¡ = = 10 gvBj|
14. †¯ªv‡Zi wecix‡Z GKwU †bŠKv 52 wgwb‡U 13 wKwg †h‡Z cv‡i| †¯ªv‡Zi †eM 4 wKwg/N›Uv| w¯’i cvwb‡Z †bŠKvi †eM KZ ?
[wZZvm M¨vm wdì-mn: Awd:-2018]
A. 19 wKwg/N›Uv B. 23 wKwg/N›Uv C. 13 wKwg/N›Uv D. 11 wKwg/N›Uv E. ‡KvbwUB bq
mgvavb: (A)
52 wgwb‡U hvq = 13 wKwg,
1 ÕÕ ÕÕ = ÕÕ
60 ÕÕ ÕÕ = =15wKwg| (MwZ‡eM †ei Ki‡Z ejvq 60 wgwb‡U KZUzKz hvq Zv †ei Kiv n‡q‡Q)
(Dc‡ii 3 jvB‡bi GB HwKK wbqg 5 ‡m‡K‡Û Gfv‡e Kiv hvq 52 wg = 13 wKwg A_©vr mg‡qi 4 fv‡Mi 1 fvM c_)
myZivs 60 wgwb‡Ui 4 fv‡Mi 1 fvM n‡e 15 ( G‡Z KvUvKvwU Avi wjLv wjwL Ki‡Z n‡e bv| ïay †eª‡bi †Ljv) GLb
‡¯ªv‡Zi wecix‡Z ev cÖwZK~‡j MwZ 15 Ges ‡¯ªv‡Zi MwZ 4 nIqvq w¯’i MwZ = 15+4 = 19wKwg/N›Uv|
15. wb‡Pi †Kvb fMœvsk †_‡K eo ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. B. C. D. E. ‡KvbwUB bq
mgvavb: (B)
†h‡nZz eo fMœvskwU †ei Ki‡Z ejv n‡q‡Q| ZvB †h¸‡jv †QvU n‡e †m¸‡jv ev` w`‡Z n‡e|
cÖ_‡g, Gi mv‡_ AvovAvwo ¸Y K‡i cvB 93  100 myZivs fMœvskwU †QvU|
GLb, Gi mv‡_ AvovAvwo ¸Y K‡i cvB 27  22 myZivs fMœvskwU eo ( GwUB DËi)
GKB fv‡e Ab¨ Ackb `ywUi mv‡_ AvovAvwo ¸Y Ki‡j †m¸‡jv †QvU nIqvq DËi: B
100
52
13
52
6013 
3
2
50
31
11
9
5
2
27
11
50
31
3
2
50
31
11
9
3
2
11
9
3
2
c~e©cwðg
DËi
`wÿY
BDmyd (ïiæ)Kwei 4 gvBj
gvRnvi ûgvqyb
6 gvBj
4 gvBj
4 gvBj
6 gvBj
8 gvBj

16. GK e¨emvqx Zvi Avg`vwbK…Z 240 wU evB‡Ki g‡a¨ 15% evB‡K ÎæwU Ly‡R †cj| cieZ©x‡Z Avg`vwbK…Z 160 wU
evB‡Ki 5% evB‡K ÎæwU Ly‡R †cj| `yev‡i †m kZKiv KZ¸‡jv ÎæwUhy³ evBK †cj ? [wZZvm M¨vm wdì-mn: Awd:-
2018]
A. 20% B. 15% C. 11% D. 10% E. ‡KvbwUB bq
mgvavb: (C)
cÖ_gev‡ii ÎæwUc~Y© evBK =240 Gi 15% = 240 = 36 (gy‡L gy‡L 100 †Z 15 n‡j 200 †Z 30 Ges 10 G 1.5
n‡j 40 G 6 †gvU 30+6 = 36wU)
Avevi , wØZxq ev‡ii ÎæwUc~Y© evB‡Ki msL¨v = 160 Gi 5% = 160 = 8 (gy‡L gy‡L: 100 †Z 5 n‡j 10 G .5
Ges 60 G 3 myZivs 160 G 5+3 = 8)
myZivs ‡gvU 240+160 = 400 wU evB‡Ki g‡a¨ ÎæwU c~Y© evB‡Ki msL¨v = 36+8 = 44wU|
ÎæwUc~Y© evB‡Ki kZKiv nvi = =11% DËi: 11%
17. GKwU eM© †ÿ‡Îi cÖwZ evûi ‰`N©¨ 40% K‡i evov‡bv n‡j Gi †ÿÎdj kZKiv KZUzKy e„w× cv‡e ? [wZZvm M¨vm wdì-
mn: Awd:-2018]
A. 96% B. 80% C. 69% D. 40% E. ‡KvbwUB bq
mgvavb: (A)
eM©‡ÿ‡Îi GKevû‡K 40% evov‡j Zvi ˆ`N©¨ I cÖ¯’ Dfq B 40% K‡i †e‡o hvq|
myZivs cÖ_‡g 100 †_‡K 40 evov‡j nq 140 Gici 140 Gi 40% ev 56 evov‡j †gvU evo‡e 40+56 = 96%| D:
18. ivRy I gy³vi eq‡mi AbycvZ h_vµ‡g 5 : 4| 8 eQi ci ivRyi eqm n‡e 28 eQi| eZ©gvb gy³vi eqm KZ ? [wZZvm
M¨vm wdì-mn: Awd:-2018]
A. 8 B. 12 C. 16 D. 20 E. ‡KvbwUB bq
mgvavb: (C)
awi,
ivRy I gy³vi eqm h_vµ‡g 5K Ges 4K
cÖkœg‡Z,
5K = 28-8 ( †h‡nZz 5K n‡jv eZ©gvb ZvB 28 †_‡K 8 we‡qvM K‡i Avm‡Z n‡e|)
5K = 20
K = 4
myZivs gy³vi eZ©gvb eqm = 44 = 16 eQi| DËi: 16 eQi|
19. `yBwU bj w`‡q GKwU U¨vsK h_vµ‡g 12 I 15 N›Uvq c~Y© nq| Z…Zxq GKwU bj Øviv †mwU 20 N›Uvq Lvwj nq| wZbwU bj
GKmv‡_ Lyy‡j w`‡j U¨vsKwU c~Y© KZ mgq jvM‡e ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 12 N›Uvq B. 10 N›Uvq C. 8 N›Uvq D. 7 N›Uvq E. ‡KvbwUB bq
mgvavb: (B)
cÖ_g bj `ywU 1 N›Uvq c~Y© K‡i h_vµ‡g , Ask Ges Ask|
Avevi 3q bj Øviv 1 N›Uvq Lvwj nq = Ask|
100
15
100
5
400
10044 
12
1
15
1
20
1

myZivs wZbwU bj GKmv‡_ Pvjy _vK‡j 1 N›Uvq|
(c~Y© Ki‡j †hvM Ges Lvwj Kivq we‡qvM Kiv n‡q‡Q)
Ask|
myZivs c~Y© nq = 1 N›Uvq  1 Ask ev m¤ú~Y© Ask c~Y© nq = 110 = 10 N›Uvq| DËi: 10 N›Uv|
20. GKwU _‡j‡Z 25 cqmv, 10 cqmv I 5 cqmvi gy`ªv 3 : 4 : 5 Abycv‡Z Av‡Q| hw` me¸‡jv wgwj‡q 28 UvKv nq, Zvn‡j
10 cqmvi gy`ªv KZwU ? [wZZvm M¨vm wdì-mn: Awd:-2018]
A. 80 wU B. 60 wU C. 100 wU D. 110 wU E. ‡KvbwUB bq
mgvavb: (A)
awi, 25 cqmv, 10 cqmv Ges 5 cqmvi g~`ªvi msL¨v h_vµ‡g 3K, 4K Ges 5K|
GLb, 25 cqmvi g~`ªvi gvb = ev UvKv ( †h‡nZz 28 UvKv †`qv Av‡Q, ZvB UvKv evwb‡q mgxKiY n‡e|)
10 cqmvi g~`ªvi gvb = ev UvKv
Ges 5 cqmvi g~`ªvi gvb = ev UvKv
cÖkœg‡Z,
(3K  ) + (4K  ) + ( 5K ) = 28 UvKv|
 + + = 28 UvKv|
 = 28 UvKv|
28K = 2820 K = 20
myZivs 10 cqmvi g~`ªv = 420 = 80wU|
(‡Kb fMœvsk evbv‡bv n‡jv?? ey‡S wbb:)
aiæb, 3K = 310 ev 30wU 25cqmv wK 30 UvKv n‡e? Aek¨B bv| ZvB 30 Gi mv‡_ ¸Y Ki‡j Zv UvKv n‡e Ges
Gfv‡e me¸‡jv UvKvi cwigv‡Yi †hvMdj = 28 UvKv n‡e| Gfv‡e UvKvq gvb †ei bv K‡i Ki‡j fzj n‡e| )
20
1
15
1
12
1

10
1
120
12
120
6810


10
1
100
25
4
1
100
10
10
1
100
5
20
1
4
1
10
1
20
1
4
3K
5
2K
4
K
20
5K8K15K 
4
1
cÖgvY: 25 cqmvi gy`ªv = 320 = 60wU = 604 = 15UvKv
10cqmvi g~`ªv = 420 = 80wU = 8010 = 8UvKv|
5 cqmvi g~`ªv = 520 = 100wU = 10020 = 5UvKv|
‡gvU UvKv = 15+8+5 = 28UvKv|
D‡jøL¨; 4wU 25cqmvi g~`ªvq 1UvKv, 10wU 10cqmvi g~`ªvq 1UvKv Ges
20wU 5 cqmvi g~`ªv 1UvKv nq weavq 4, 10 Ges 20 w`‡q fvM|

1. A 60 litre mixture of sugar and water contains sugar and water in the ratio of 2:3. How
many litre of the mixture should be replaced by sugar so that the ratio of sugar and
water becomes 1:1?(60 wjUv‡ii GKwU wgkÖ‡Y wPwb I cvwbi AbycvZ 2:3|G‡Z KZ wjUvi wgkÖY, wPwb Øviv
cÖwZ¯’vwcZ Ki‡j bZzb wgkÖ‡Y wPwb I cvwbi AbycvZ 1:1 n‡e?) [GTCL (AM-General)-2018]
a. 12 liters b. 15 liters c. 10 liters d. 20 liters Ans: c
Solution:
Given that, Total mixture = 60 litres
Ratio of Sugar: Water = 2:3
Sum of Ratio = 5
So, Sugar =
5
260
=24 litres and water =
5
360
= 36 litres
Suppose, x litres mixture should be replaced by sugar, (cÖ_gevi x Zz‡j wb‡j Zv‡Z wPwb I cvwb Df‡q
Av‡Q, wKš‘ c‡i Avevi x wPwb w`‡j Zv‡Z ‡Kvb cvwb †bB|)
then the new mixture = (60-x) litres
So,the new ratio of sugar and water = {(60-x)
5
2
}:{(60-x) 
5
3
)}
According to the question,
{(60-x)
5
2
)+x)}:{(60-x) 
5
3
)} =1:1 [m¤úyY© wgkÖ‡Yi mv‡_ wPwb cÖwZ ¯’vcb Ki‡j bZzb wgkÖ‡Y wPwb: cvwb=1:1]
Or,
5
x5x2120 
:
5
x3180 
=1:1
Or, 120+3x=180 -3x Or, 6x = 60 x = 10 Ans:10 liters
cÖgvY:
60 Gi g‡a¨ wPwb I cvwb 24 I 36 Av‡Q,| ‡gvU wgkÖY †_‡K 10 Zz‡j wb‡j Aewkó 60-10 = 50 G wPwb _v‡K 20 Ges
cvwb _v‡K 30 ( KviY AbycvZ Av‡Mi gZB 2:3 Av‡Q) GLb GB 50 G 10wjUvi wPwb w`‡j bZzb =50+10 = 60 wjUvi
wgkÖ‡Y wPwb _vK‡e 20+10 = 30 I cvwbI _vK‡e 30 hvi AbycvZ 30:30 ev 1:1|
2. If
z
x
is 1 more than
z
y
,than =? [GTCL (AM-General)-2018]
a. x+z b. x-y c. y-z d. x-z Ans: d
Solution:
z
x
-
z
y
=1 (more A_© †ewk| Avi †ewk n‡j e¨eavb 1 n‡e|)
Or,
z
yx 
=1 Or, x-y = z y = x-z Ans:x-z
Gas Transmission Company Ltd, (GTCL)
Post Name: Assistant Manager (General) Exam Date: 20-04-2018
Exam taker : IBA, DU.

3. Arif bought 17 pens of three colors –black ,green and red. they cost TK . 5, TK. 10 and
TK. 25 each respectively . The total amount that Arif paid was TK .205.If Arif bought
twice as many green pens as red pens , how many black pens did he buy? (Avwid Kv‡jv,
meyR I jvj i‡Oi 17 wU Kjg wKb‡jv hvi g~j¨ h_vµ‡g 5 UvKv, 10 UvKv Ges 25 UvKv| †m †gvU 205 UvKv cwi‡kva
K‡i| hw` Avwid jvj i‡Oi wØ¸Y meyR i‡Oi Kjg wK‡b _v‡K, Zvn‡j †m KZwU Kv‡jv i‡Oi Kjg wK‡bwQj?) [GTCL
(AM-Gen)-2018]
a. 5 b. 6 c. 7 d. 8 Ans: a
Solution:
Let, red pens = x So, green pens =2x
And black pens =17 – (x+2x) = 17 -3x (me©‡gvU †_‡K Ab¨ `ywU we‡qvM|)
25x+2x10 + (17- 3x) 5 = 205 (me¸‡jv Kj‡gi †gvU `vg = 205 UvKv|)
Or, 25x+20x+85-15x = 205
Or, 30x = 205 – 85 Or, 30x = 120 x = 4
So, number of black pens = (17 – 3  4) =17 – 12 = 5 Ans:5
4. Alam starts working on a job and works on if for 12 days and completes 40% of the
work. Then Babu joins alam and together the complete the rest of the work in 12 days .
How long (in days) will it taka Babu to complete the job if he works alone? (Avjg GKwU
KvR Ki‡jv †m 12 w`‡b KvRwUi 40% †kl Kivi ci evey mv‡_ ‡hvM †`q Ges Df‡q Aewkó KvRUzKz 12 w`‡b †kl
K‡i, evey GKvKx KvRwU Ki‡j m¤ú~Y© KvRwU †kl Ki‡Z Zvi KZ mgq jvM‡e? [GTCL (AM-General)-2018]
a. 45 days b. 50 days c. 55days d. 60 days Ans:d
Solution:
Alam completes alone 40% in 12 days.
So, Alam can complete 1% in
40
12
days.
, Alam can complete 100% in
40
10012 
= 30days.(A_©vr Avjg GKv m¤ú~Y© KvRUv 30w`‡b Ki‡Z cv‡i)
Again, Alam and Babu both completes rest 100% - 40% = 60% work in = 12 days.
Alam and Babu = 1% work in =
60
12
days.
, Alam can complete 100% in
60
10012 
= 20days (2 R‡b wg‡j m¤ú~Y© KvRwU 20 w`‡b Ki‡Z cv‡i)
Now, Alam and Babu’s 1 day’s work =
20
1
And Alam’s 1 day’s work =
30
1
So, Babu’s 1 day work =
20
1
-
30
1
=
60
1
[2 R‡bi 1 w`‡bi Kiv KvR -1 R‡bi KvR = Ab¨ Rv‡bi KvR]
So, Babu can complete
60
1
part in = 1 day
Babu completes the whole work in = 160 = 60 days . Ans:60 days

5. If Rahim walks at 14 km/hr instead of 10 km/hr for a certain time, he would have
walked 20 km more .If Rahim walks at a speed of 10 km/hr ,the distance travelled by
him within that time is –(hw` iwng ‡Kvb wbw`©ó mg‡q 10 wK.wg./N›Uv Gi cwie‡Z© 14 wK.wg/N›Uv †e‡M nvu‡U,
Zvn‡j †m 20 wK.wg. c_ †ewk †h‡Z cv‡i| wKš‘ hw` †m 10 wK.wg./N›U †e‡M nuv‡U, Zvn‡j H mg‡qi g‡a¨ ‡m KZUyKz
c_ AwZµg Ki‡e?) [GTCL (AM-General)-2018]
a. 42km b. 50km c. 60km d. 64km Ans: b
Solution:
Speed difference = 14-10 = 4,
If difference is 4 then actual distance travelled = 10 km
Then when difference is 20km ‘’ ‘’ = 50km
4
2010


[Note: 10 wKwgi cwie‡Z© 14 wKwg †M‡j GiKg n‡Zv| wKš‘ Avm‡j †m 10wKwg MwZ‡ZB wM‡qwQj| ZvB DËi 50]
6. A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at
18% profit. He gains 14% on the whole. The quantity sold at 18% profit is: (GKRb e¨emvqx
Gi Kv‡Q 1000 †KwR wPwb Av‡Q| Zvi ga¨ †_‡K wKQz wPwb 8% jv‡f wewµ Ki‡jb Ges Aewkó wPwb 18% jv‡f wewµ
Ki‡jb| †gv‡Ui Dci Zvi 14% (1000 ‡KwR wewµ‡Z) jvf n‡j 18% jv‡f KZ †KwR wewµ K‡iwQ‡jb?) [BD
House Building FC (OF)-2017] + [Pubali Bank (TAT)-2017] & [GTCL (AM-General)-2018]
a. 600 kg b. 560 kg c. 400 kg d. 640 kg Ans: a
 Written Solution:
Let the sugar of 18% profit is = x
So, the sugar of 8% profit = 1000-x
ATQ,
18% of x + 8% of ( 1000-x) = 14% of 1000  x = 600
Solution, By the rule of allegation, we have :
Ratio of 1st
and 2nd
parts = 4 : 6 = 2 : 3
 Quantity of 2nd
kind = kg1000
5
3






 = 600 kg
e¨vL¨v: cÖ_‡g Zv‡`i wgkÖ‡Yi ev wewµi AbycvZ †ei Kiv n‡q‡Q| A_©vr †Kvb ai‡Yi wPwb KZ Abycv‡Z wewµ Ki‡j
‡gv‡Ui Dci jvf 14% n‡e| Zv †ei Kivi ci †gvU 1000 †KwR †_‡K 18% jv‡f wewµ Kiv wPwbi cwigvY †ei Kiv
n‡q‡Q|
7. The sum of three consecutive odd integers is 40 more than the first of the number. What
is the middle number? (3 wU avivevwnK we‡Rvo msL¨vi mgwó cÖ_g msL¨viwUi †P‡q 40 †ewk| gv‡Si msL¨vU
KZ?) [GTCL (AM-General)-2018]
a. 16 b. 18 c. 19 d. 24 Ans: c
Profit on 1st
part
8%
4 6
18%
14%
Mean
profit
Profit on 2nd
part

Solution:
Let, the numbers are x, x+2, x+4 (we‡Rvo msL¨v ZvB e¨eavb: 2 K‡i)
x + x + 2 + x + 4 = x + 40 (3 wU mgwó = cÖ_g msL¨v +40)
Or, 3x – x + 6 = 40
Or, 2x=34 x = 17
So, the middle number is = x +2 =17+2 =19 Ans:19
gy‡L gy‡L: 3Uv msL¨v-1g msL¨v = 40 A_©: 2q+3q msL¨i ‡hvMdj = 40 n‡j Mo n‡e 20 Ges msL¨v `ywU n‡e GKwU
20 Gi †_‡K 1 Kg I Ab¨wU 20 Gi †_‡K 1 †ewk| ( KviY msL¨v¸‡jv we‡Rvo)
gv‡Si msL¨vwU n‡e 20-1 = 19|
8. When the positive integer n is divided by 5,the remainder is 2. Which of the following
must be true? (n GKwU abvZ¥K c~Y©msL¨v hv 5 Øviv fvM Ki‡j fvM‡kl 2 n‡e| Zvn‡j wb‡Pi ‡KvbwU Aek¨B mZ¨
n‡e?) [GTCL (AM-General)-2018]
i. n is odd
ii. n+1 can not be a prime number
iii. (n+2) divided by 7 has remainder 2
a. none b. i only c. i and ii only d. ii and iii only Ans: a
Solution:
Let, n = 12 (5 w`‡q fvM Ki‡j 2 Aewkó _v‡K Ggb msL¨v )
now, (i) n = not an odd number
(ii) n+1 = 12+1 = 13 is a prime number
(iii) n+2 = 12+2 = 14 is divided by 7 has no reminder,
So, all the given information is not true, thus ans is none.
9. When x is divided by 7, the remainder is 6.Which of the following must be an even
number? (x ‡K 7 Øviv fvM Ki‡j fvM‡kl 6 nq, Zvn‡j wb‡Pi †KvbwU Aek¨B ‡Rvo msL¨v n‡e ?) [GTCL (AM-
General)-2018]
a. x+6 b. x3
+x2
+x c. x2
+x d. x- 4 Ans:c
Solution:
if x= 13 then,
(a) X+6 = 13+6 = 19 = Odd number
(b) x3
+x2
+x = (13)3
+(13)2
+13 = 2197+169+13=2379 = Odd number
(c) x2
+x = (13)2
+13 = 169+13 =182 = Even number
(d) x- 4 = 13 – 4 = 9 = Odd number
GLv‡b option (c) QvWv evwK me¸‡jv we‡Rvo msL¨v| myZivs mwVK DËi: c
10. Pens that used to cost TK .150 for a package of 3 now cost TK.300 for a package of 5.
What is the present increase in the price of these pens? (3 wU Kj‡gi GKwU c¨v‡K‡Ri g~j¨ wQj
150 UvKv, eZ©gv‡b 5 wU Kj‡gi GKwU c¨v‡K‡Ri g~j¨ 300 UvKv| eZ©gv‡b Kj‡gi g~j¨ KZ e„w× †c‡q‡Q ?) [GTCL
(AM-General)-2018]
a. 5% b. 10% c. 15% d. 20% Ans: d

Solution:
Old price of 1 pen =
3
150
= Tk.150 New price of 1 pen =
5
300
= Tk.60
Price increased = 60-50 = Tk.10
 increased % =
50
10010 
= 20% Ans: 20%
11. A year ago the price of a toothbrush and the price of a comb were both TK. 50.the price
of the toothbrush was increased by 20% while the price of the comb was decreased by
10%.what is the difference in taka between the current price of the toothbrush and the
comb?(GK eQi c~‡e© eªvk Ges wPiæwb cÖwZwUi g~j¨ 50 UvKv wQj| cieZ©v‡Z eªv‡ki g~j¨ 20% e„w× †cj Ges wPiæwbi
g~j¨ 10% n«vm †cj| eªvk I wPiæwbi eZ©gvb g~‡j¨i cv_©K¨ KZ UvKv?) [GTCL (AM-General)-2018]
a. 12 b. 14 c. 15 d. 20 Ans: c
Solution:
Current price of a toothbrush = 50+20% of 50 = 50+10 = 60
Current price of a comb = 50 – 10% of 50 = 50-5 = 45
Difference of two prices = 60-45 = Tk. 15 Ans:15
12. If 7 > x > 2 and 3 < x < 8, which of the following best describes x? [GTCL (AM-General)-
2018]
a. 2 < x < 8 b. 2 < x < 7 c. 3< x < 5 d. 3 < x < 6 Ans: a
Solution:
From the given data, 7 > x > 2 and 3 < x < 8, we can write 7 > x and 8 > x means 8> x
Again, x > 2 and x > 3 means x > 2
So, we can write 2 < x < 8
evsjvq e¨vL¨:
x Gi gvb hw` 7 †_‡K †QvU nq Zvn‡j Aek¨B 8 ‡_‡KI ‡QvU n‡e|
Avevi x Gi gvb hw` 3 ‡_‡K eo nq Zvn‡j Aek¨B 2 ‡_‡KI eo n‡e|
myZivs ejv hvq 2 < x < 8 Zvn‡j mwVK DËi n‡e a.
13. Apu took 3/5 of the marbles kept in a box. His younger took another 3/5 of the
remaining marbles. Then his sister took another 3/5 of remaining marbles. What
fractions of the marbles are left in the box?(Acy GKwU ev· †_‡K 3/5 Ask gv‡e©j wbj| Gici Zvi
†QvU fv&B Aewkó gv‡e©j †_‡K 3/5 Ask gv‡e©j wbj| Avevi Zvi †evb Aewkó gv‡e©‡ji 3/5 Ask gv‡e©j wbj| ev‡·
Avi KZ fvM gv‡e©j _vK‡jv?) [GTCL (AM-General)-2018]
a.
125
8
b.
125
20
c.
125
45
d.
125
25
Ans:a
Solution:
Let, box contain = 125 marbles [5 Gi Nb 53
=125 KviY wZbevi 5 fvM Ki‡Z n‡e]
Apu took =125
5
3
= 75 marbles
remaining marbles = 125-75 = 50
Then, younger brother took = 50
5
3
= 30 marbles
and now remaining = 50-30 = 20
Tips:
[x a‡iI Kiv hvq wKš‘& wcÖwji cixÿvq Aí
mg‡q fMœvs‡ki †ewk †ewk wn‡me Ki‡Z A‡bK
mgq jvM‡e| ZvB ni w`‡q wefvR¨ †h †Kvb
msL¨v a‡i wn‡me Kiv DËg|]

Again, his sister took = 20
5
3
= 12
So, at last remaining marbles = 20-12 = 8
 require fraction is =
125
8
(‡gvU gv‡e©j= 125wU Ges Aewkó Av‡Q 8 wU ) Ans:
125
8
14. The average daily wages of female workers in a factory is TK. 30 and that of male
workers is Tk.42. if the average wages of all the workers is TK.37, what is the ratio of
male workers? (GKwU KviLvbvi gwnjv Kg©Pvix‡ì ˆ`wbK Mo gRywi 30 UvKv Ges cyiæl Kg©Pvix‡ì ˆ`wbK Mo
gRyix 42 UvKv| mKj Kg©Pvixi Mo gRyix 37 UvKv n‡j cyiæl I gwnjv Kg©Pvixi AbycvZ KZ ?
a. 6:5 b. 5:7 c. 5:6 d. 7:5 Ans:d
Solution:
Let, number of male = x and number of female = y
ATQ,
42x+30y = 37(x+y) [cyiæl‡ì †gvU Avq+gwnjv‡ì †gvU Avq = cyiæl gwnjv mevi GKwÎZ Avq]
 42x+30y = 37x+37y
 42x-37x = 37-30y
 5x = 7y
5
7
y
x

 x: y = 7:5
So, the ratio of male to female workers is 7:5 Ans:7:5
Alternative solution:
Rule of Allegation Gi gva¨‡g GB AsKwU K‡qK †m‡K‡Û mgvavb Kiæb|
So, ratio of male and female is 7:5
================================
==
Male wages
42
7 5
30
37
Mean
profit
Female wages

1. iv‡mj, Avmv` I ivRy‡K 315 UvKv fvM K‡i †Ìqv n‡jv | G‡Z iv‡m‡ji UvKv Avmv‡ì UvKvi 3/5 Ges Avmv‡ì
UvKv ivRyi UvKvi 2 ¸Y n‡jv| ivRy KZ UvKv †cj? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. 60 UvKv L. 90 UvKv M. 150 UvKv N. 75 UvKv DËi:N
Solution: (fMœvsk Qvov mnR wbq‡g mgvavb)
g‡bKwi , ivRy cvq = 3K UvKv Zvn‡j Avmv` cv‡e = 5K UvKv (j‡ei UvKv iv‡mi Ges n‡ii UvKv Avmv`)
myZivs ivRy cv‡e, 5K  2 = 2.5K (Avmv‡ì UvKv ivRyi UvKvi wØ¸Y n‡j ivRyi UvKv Avmv‡ì UvKvi A‡a©K n‡e)
cÖkœg‡Z,
3K+5K+2.5K = 315 (mevi UvKvi †hvMdj = 315 UvKv)
ev, 10.5K = 315  K = = 30UvKv|
Zvn‡j ivRy cv‡e, 2.530 = 75 UvKv|
fMœvsk a‡i mgvavb Kivi Rb¨, ivRy = K n‡j Avmv` = 2K Ges iv‡mj = = UvKv|
cÖkœg‡Z,
K+2K+ =315  = 315  K =  K = 75 A_©vr ivRy cv‡e, 75 UvKv|
[
2. x, y Ges z wZbwU cyY© msL¨v | hw` x  y  z Ges y  2 nq Z‡e wb‡Pi †KvbwU Aek¨B fyj? [K…wl m¤úªmviY Awa`ßi-
(Awd: mn:+UvBwc÷)-2018]
K. xyz  0 L. xy –z  0 M. y-xz  0 N. †KvbwU bq DËi:M
Solution:
‡Ìqv Av‡Q , x  y  z
awi, x = 2, y =3, z = 4
GLb Ackb ¸‡jv‡Z GB gvb ewm‡q †`Lv hvK |
(K) xyz  0 234  0  24  0 GUv n‡Z cv‡i|
(L) xy – z  0  23 - 4 0 2 0 GUvI n‡Z cv‡i|
(M) y- xz  0  3 - 24  0  - 5  0 = Bnv KL‡bvB mwVK bq| KviY FYvZ¥K msL¨v memgq 0 Gi †_‡K
†QvU| myZivB GUvB DËi|
3. evey I Rvgv‡ji gvwmK †eZ‡bi bycvZ 7:5 Ges `yR‡bi gvwmK †eZb GK‡G 24,000 UvKv | GK eQi c‡i eveyi †eZb
500UvKv Ges Rvgv‡ji †eZb 350UvKv e„w× †cj | GK eQi c‡i Zv‡ì gvwmK †eZ‡bi AbycvZ KZ n‡e ? [K…wl
m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. 290:207 L. 145:103 M. 500:350 N. 7:5 DËi:K
10.5
315
5
3
2K 
5
6K
5
6K
5
21K
21
5315
Department of Agricultural Extension (DAE)
Post name : Officer Assistant cum computer typist
Exam Date : 13-04-2018 Exam taker: IBA, DU.

Solution:
awi, eZ©gv‡b eveyj I Rvgv‡ji gvwmK †eZb h_vµ‡g 7K I 5K UvKv|
cÖkœg‡Z,
7K + 5K =24000 ev, 12K = 24000 K = 2000 UvKv
myZivs eveyj I Rvgv‡ji †eZb h_vµ‡g 14,000 UvKv I 10,000 UvKv|
GK eQi c‡i Zvì †eZ‡bi AbycvZ n‡e =(14000+500) :(10000+ 350) =14500 : 10,350 = 290 : 207
4. XvKv ‡_‡K Kwi‡gi evwoi `„iZ¡ 355 wK. wg. | †m ev‡m XvKv †_‡K evwo iIqvbv n‡jv | 319 wK. wg hvIqvi ci evmwU
bó n‡q ‡M‡j Kwig evwK c_ wi·vq †Mj| ev‡mi MwZ‡eM 22 wK. wg./ N›Uv I wiKkvi MwZ‡eM 6 wK. wg ./N›Uv n‡j
evwo †cxuQv‡Z Kwi‡gi †gvU KZ mgq jvM‡e ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. 20 N›Uv L. 20 N›Uv 30 wgwbU M. N›Uv 50 wgwbU N. 21 N›Uv DËi:L
Solution:
ev‡m hvq = 319 wK. wg.c_ Ges wi·vvq †M‡Q = 355-319 = 36 wK. wg.|
ev‡m †h‡Z mgq jv‡M = = 14.5 N›Uv
wi·vq mgq jv‡M= = 6 N›Uv
‡gvU mgq jv‡M = 14.5 + 6 = 20.5 N›Uv ev 20 N›Uv 30 wgwbU | DËi: 20 N›Uv 30 wgwbU
5. 8 Rb †jvK GKwU KvR 6 w`‡b Ki‡Z cv‡i | KvRwU wZb w`‡b Ki‡Z n‡j KZR‡b bZzb †jvM wb‡qvM Ki‡Z n‡e ?
[K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.6 L. 8 M.7 N.12 DËi:L
Solution:
6 w`‡b Ki‡Z cv‡i = 8 Rb †jvK|
 1 Ó Ó Ó = 8 6 Ó Ó (Kg w`‡b Ki‡Z †ewk †jvK jvM‡e ZvB ¸Y|)
 3 Ó Ó Ó = Ó Ó = 16 Rb myZivs AwZwi³ †jv‡Ki msL¨v = 16-8 = 8 Rb|
gy‡L gy‡L: 6 w`‡bi KvR 3 w`‡b A_©vr A‡a©K mg‡qi g‡a¨ Ki‡Z PvB‡j Av‡Mi 8 Rb †jv‡Ki wØ¸Y †jvK jvM‡e|
8 Rb †h‡nZz Av‡M †_‡KB Av‡Q ZvB Av‡iv bZzb †jvK jvM‡e 8 Rb| DËi:8
6. GKwU Kjg I GKwU eB‡qi g~j¨ GK‡Î 95 UvKv | KjgwUi g~j¨ 15 UvKv †ewk Ges eBwUi g~j¨ 14 UvKv Kg n‡j
KjgwUi g~j¨ eBwUi g~‡jï wØMyY n‡Zv | eBwUi g~j¨ KZ UvKv? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.49 L. 46 M.50 N.40 DËi:L
Solution:
g‡b Kwi, eBwUi g~j¨ = K UvKv
 Kj‡gi g~j¨ = (95- K) UvKv
cÖkœg‡Z,
95 - K + 15 = 2(K - 14) ev, 110 - K = 2K - 28 ev, 3K = 138  K = 46
myZivs, eBwUi g~j¨ 46 UvKv| DIi : 46
7. GKwU c‡Yï weµqg~j¨ wØ¸Y n‡j we‡µZvi gybvdv †e‡o 3 ¸Y n‡e | g~jë„w× bv K‡i cY¨wU weµq Ki‡j we‡µZv
kZKiv KZ gybvdv Ki‡e ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.50 L. 100 M.150 N. †KvbwU bq DËi:L
22
319
6
36
3
68

Solution:
g‡b Kwi, cY¨wUi weµqg~j¨ = 100 UvKv| hv‡Z jvf Av‡Q K UvKv|
weµqg~j¨ wØ¸Y n‡j bZzb weµqg~j¨ = 200UvKv hv‡Z jvf Av‡Q 3K UvKv|
cÖkœg‡Z,
3K-K = 200-100 ( `yB jv‡fi cv_©K¨ = `yB weµqg~‡jï cv_©K¨)
ev, 2K = 100 myZivs K = 50 UvKv|
Zvn‡j cÖ_gevi 100 UvKvq jvf n‡qwQj 50UvKv Ges µqg~j¨ = 100-50 = 50 UvKv|
50UvKvq jvf 50 UvKv n‡j jv‡fi nvi = 100% DIi : 100%
8. GKwU †Kv¤úvwbi 46 kZvsk Kg©KZ©v cyiæl | hw` 60 kZvsk Kg©KZv© †jevi BDwbqb K‡i Ges Zv‡ì g‡a¨ 70 kZvsk
cyiyl nq, BDwbqb K‡i bv Ggb Kg©KZ©v‡ì g‡a¨ kZKiv KZ Rb gwnjv? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-
2018]
K.90 L. 87.5 M.80 N.75 DËi:K
Solution:
‡gvU Kg©KZv© 100 Rb n‡j cyiæl 46 Rb Ges gwnjv = 100-46 = 54 Rb|
‡jevi BDwbqb K‡i 60 ( cyiæl+gwnjv) Rb Ges K‡i bv = 100-60 = 40 (cyiæl+gwnj) Rb|
GLb †jevi BDwbDb Kiv cyiæ‡li msL¨v = 60 Gi 70% = 42 Rb|
‡jevi BDwbqb K‡i bv Ggb cyiæl = ‡gvU cyiæl-BDwbqb Kiv cyiæl = 46-42 = 4 Rb|
Zvn‡j †gvU ‡jevi BDwbqb bv Kiv 40 R‡bi g‡a¨ cyiæl 4 ev` w`‡j gwnjv _v‡K 40-4 = 36 Rb|
40 R‡bi g‡a¨ 36 Rb gwnjv n‡j 1 R‡b gwnjv = Ges 100 R‡bi g‡a¨ gwnjv = =90%
9. GKRb weµZv 17 wU Kjg 720 UvKvq wewµ K‡i ‡h †jvKmvb Ki‡jv Zv 5 wU Kj‡gi µqg~‡jï mgvb |GKwU Kj‡gi
µqg~j¨ KZ UvKv ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.52 L. 54 M.60 N. †KvbwU bq DËi: M
Solution:
‡h‡nZz ÿwZ n‡q‡Q ZvB
17wUi µqg~j¨ - 17wUi weµqg~j¨ = 5wUi µqg~j¨ [µqg~j¨ eo Ges Zv †_‡K weµqgyj¨ we‡qvM Ki‡j ÿwZ †ei n‡e]
ev, 17wUi µqg~j¨ - 5wUi µqg~j¨ = 17wUi weµqg~j¨|
ev, 12wUi µqg~j¨ = 720 UvKv ( ‡h‡nZz 17wUi weµqg~j¨ 720 UvKv cÖ‡kœB †`qv Av‡Q)
myZivs 1wUi µqgyj¨ = = 60 UvKv
10. `kwU msLvi Mo x Ges G‡ì cvuPwU ksL¨vi Mo y | hw` evwK cvuPwU msL¨vi Mo z nq Z‡e wb‡Pi †KvbwU mwVK ? [K…wl
m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.2x = y + z L. 2x =y + z M. x = 2y + 2x N. †KvbwU bq DËi:L
Solution:
10 wU msL¨vi mgwó = 10x
cÖ_g cvuPwU msL¨vi mgwó = 5y
evwK 5 wU msL¨vi msL¨vi mgwó = 5z
cÖkœg‡Z,
10x = 5y+5z (cÖ_g 10wUi †hvMdj = 5+5wUi †hvMdj)
ev, 10x = 5(y+z)  2x = y+z
40
36
40
10036
12
720

11. hw` 7,11,15 Ges x Gi Mo y nq Z‡e x Gi gvb KZ ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. 4y -33 L. (y – 26)/4 M. y -33 N. †KvbwU bq DËi:K
Solution:
cÖkœg‡Z,
[`y cv‡ki mgwó mgvb]
ev, 33+x = 4y x = 4y -33
12. hw` 15,28,33 GB wZbwU msL¨vi ¸bdj z nq,Z‡e wb‡Pi ‡KvbwU GKwU c~Y© msL¨v n‡e bv ? [K…wl m¤úªmviY Awa`ßi-
(Awd: mn:+UvBwc÷)-2018]
K. L. M. N. me¸wjB c~Y© msL¨v DËi:L
Solution:
‡`Iqv Av‡Q, z = 152833 GLb Ackb¸‡jvi ni †`qv Av‡Q, 21,24 Ges 55| Avgiv 152833 ‡K Ggbfv‡e
K‡qKwU Drcv`K evbv‡e hv‡Z GB Drcv`K¸‡jv Av‡m wK bv Zv †`Lv hvq| hw` Av‡m Zvn‡j Zv c~Y© msL¨v Avi bv
Avm‡j fMœvsk|
152833 = 53722311 = GLvb †_‡K 73 = 21 evbv‡bv hvq, Avevi, 511 w`‡q 55 evbv‡bv hvq|
wKš‘ †Kvbfv‡e 24 evbv‡bv hv‡”Q bv| Zvn‡j z Gi g‡a¨ 24 Drcv`KwU bv _vKvq eo msL¨vwU 24 w`‡q wefvR¨ n‡e bv
ZLb Zv GKwU fMœvsk n‡e|
13. hw` -2  x  2 and 3  y  8 nq, Z‡e †KvbwU mwVK ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.1  y – x  10 L. 1  y - x  5 M.5  y – x  6 N.‡KvbwUB bq DËi:M
Solution:
‡`qv Av‡Q, 3  y 8
-2  x 2
[we‡qvM K‡i cvB] 5  y - x 6 (Ackb M Gi mv‡_ wg‡j †M‡Q|)
14. GKwU ¯‹zv‡ji 70% QvÎ dzUej, 75% nwK Ges 80% QvÎ wµ‡KU †Lj‡Z cQ›` K‡i H ¯‹z‡ji kZKiv KZ Rb QvÎ
wZbwU ‡LjvB ‡Lj‡Z cQ›` K‡i ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. 25% L. 30% M. 35% N. †KvbwU bq DËi:K
Solution:
dzUej cQ›` K‡i bv = 100-70% = 30%
nwK cQ›` K‡i bv = 100-75% = 25%
wµ‡KU cQ›` K‡i bv = 100-80% = 20%
 †Kvb †LjvB †Lj‡Z cQ›` K‡i bv = {100-(30+25+20)}% = (100-75)% = 25%
15. Kvjv‡gi †eZb x UvKv n‡j, hv mvjv‡gi †eZ‡bi A‡a©K Ges Avwi‡~di †eZ‡bi Pvi ¸Y| Zv‡`i wZb R‡bi †eZ‡bi
†hvMdj KZ ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K. L. M. 3x N. †KvbwU bq DËi:K
Solution:
Kvjv‡gi †eZb = x UvKv, mvjv‡gi †eZb =2x UvKv Ges Avwi‡di ‡eZb = UvKv|
 wZb R‡bi ‡gvU †eZb = x+2x+ = = UvKv
4yx15117 
21
z
24
z
55
z
4
x13
4
73
4
x
4
x
4
xx8x4 
4
x13

16. iwd‡Ki IRb hw` 17 †KwR K‡g hvq Z‡e Zvi IRb Avwi‡di IR‡bi A‡a©K n‡q hv‡e | Zv‡ì `yR‡bi IR‡bi
†hvMdj 140 †KwR n‡j iwd‡Ki IRb KZ †KwR ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.55 L. 58 M.61 N.‡KvbwU bq DËi:L
Solution:
awi, iwd‡Ki IRb = x †KwR Ges Avwi‡di IRb = y †KwR|
1g kZ©g‡Z,
x+y = 140 ev, y = 140-x
2q kZ©g‡Z,
2(x-17) = y (hvi IRb K‡g †m A‡a©K n‡q hvq, ZvB Zv‡K 2 w`‡q ¸Y Ki‡j Ab¨R‡bi mgvb n‡e|)
ev, 2x -34 = 140-x [y = 140-x ewm‡q|]
ev, 3x = 174
 x = 58
myZivs iwd‡Ki IRb = 58 †KwR |
♦ gy‡L gy‡L Kivi wbqg:
GKRb Av‡iKR‡bi A‡a©K A_© GKRb GK¸Y n‡j Ab¨Rb 2 ¸Y|
140 †KwR †_‡K 17 ev` w`‡j hv _v‡K Zv‡K 3 fvM K‡i ‡h A‡a©K Zv‡K 1 ¸Y Ges Ab¨Rb‡K 2 ¸Y w`‡j
gy‡L gy‡L DËi †ei n‡e, 140-17 = 1233 = 41, †h‡nZz hvi IRb Kg ZviUv †ei Ki‡Z e‡j‡Q ZvB DËi
n‡e 41+17 = 58| (1fv‡Mi mv‡_ Zvi IRb †_‡K ev` †`qv 17 †hvM Kiv n‡q‡Q|)
17. `ywU abvZœK msL¨vi cv_©K¨ 6 Ges Zv‡ì e‡M©i cv_©K¨ 108| msL¨v `ywUi †hvMdj KZ ? [K…wl m¤úªmviY Awa`ßi- (Awd:
mn:+UvBwc÷)-2018]
K.6 L. 8 M.12 N.18 DËi:N
Solution:
GLv‡b, x-y = 6 Ges x2
-y2
= 108
GLb, x2
-y2
= 108
 (x+y)(x-y) = 108
 (x+y)6 = 108  x+y = 18
myZivs msL¨v `ywUi †hvMdj = 18 | DËi: 18
18. hw` xy  0 Ges y > 0 nq, Z‡e wb‡Pi †KvbwU Aek¨B fzj? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.  6 L.  6 M.  6 N. me¸‡jv mwVK DËi:L
Solution:
‡Ìqv Av‡Q, xy 0 GLv‡b `ywU msL¨vi ¸Ydj 0 Gi †_‡K †QvU ev FYvZ¥K ZLwb nq hLb Zv‡ì g‡`¨ †h ‡Kvb
GKwU msL¨v FYvZ¥K nq| Avevi †`qv Av‡Q, y  0 A_©vr y Gi gvb 0 Gi †_‡K eo myZivs y GKwU abvZ¥K msL¨v|
GLb Ackb¸‡jv †_‡K cÖgvb Ki‡Z n‡e †KvbUv Aek¨B fzj|
 6 A_©vr L AckbwU fzj| KviY x Gi gvb FYvZ¥K Ges FYvZ¥K msL¨v †_‡K †Kvb wKQz we‡qvM Ki‡j Zv
Avevi FYvZ¥K B _v‡K| Avevi FYvZ¥K msL¨v w`‡q Dc‡ii 2y+3 abvZ¥K msL¨v‡K fvM Ki‡j fvMdj FYvZ¥K B
Avm‡e, Ges FYvZ¥K msL¨v 6 Gi †_‡K eo n‡Z cv‡i bv| ZvB GUvB fzj|
x2
3y2


2x
3y2


x2
1x2


2x
3y2



19. GKwU K¬v‡ei 80% m`m¨ cyiæl | cyiæl m`m¨‡ì 50% †ckvq Wv³vi ,30% †ckvq BwÄwbqvi Ges evwK 40 Rb
eëmvqx | H K¬v‡ei †gvU m`m¨ msL¨v KZ? [K…wl m¤úªmviY Awa`ßi- (Awd: mn:+UvBwc÷)-2018]
K.100 L. 200 M.250 N. ‡KvbwU bq DËi: M
Solution:
g‡b Kwi, K¬v‡ei †gvU m`m¨ msL¨v = 100 Rb |
Gi g‡a¨ cyiæl = 80%
GLb, cyiæl‡ì g‡a¨ Wv³vi I BwÂwbqvi = 50+30 = 80% myZivs eëmvqxi msL¨v =100-8 20% |
GLb, 20% e¨vemvqx = 40 Rb n‡j 1% = Ges 100% = = 200 Rb|
myZivs, me©‡gvU cyiæl = 200 Rb =wKš‘ †ei Ki‡Z n‡e cyiæl gwnjv mevB‡K|
‡h‡nZz cyiæ‡liv †gvU m`‡mï 80%|
Zvn‡j 80% = 200 Rb n‡j 1% = Rb = Ges 100% = = 250 Rb|
 K¬v‡ei †gvU m`m¨ msL¨v = 250 Rb|
================================
20
40
20
10040 
80
200
80
100200

1. GK e¨vw³ GKwU `ªe¨ µq K‡i 10% ÿwZ‡Z weµq Ki‡jv| hw` †m `ªe¨wU 20% K‡g µq K‡i 55UvKv †ekx‡Z weµq
Ki‡Zv Zvn‡j 40% jvf nZ| `ªe¨wUi µqg~j¨ KZ? [DAE-(Store Keeper)-2017]
a. 200 b. 220 c. 250 d. 300 Ans: c
Solution:
cÖ_g µqg~j¨ 100UvKv n‡j 10 ÿw‡Z cÖ_g weµq g~j¨ = 90UvKv|
Avevi 20% K‡g 2q µqg~j¨ hLb 80UvKv ZLb 40% jv‡f 2q weµqg~j¨ 80UvKvi 140% = 112UvKv|
GLb `yB weµqg~‡jï g‡a¨ cv_©K¨ 112-90 = 22UvKv n‡j µqg~j¨ = 100 UvKv|
myZivs `yB weµqg~‡jï g‡a¨ cv_©K¨ 55 UvKv n‡j µqg~j¨ =
22
55100
= 250UvKv|
2. `ywU msL¨vi AbycvZ 3:4 Ges Zv‡ì j.mv.¸ 180 n‡j cÖ_g msL¨vwU KZ? [DAE-(Store Keeper)-2017]
a. 30 b. 45 c. 50 d. 60 Ans:b
Solution:
awi, msL¨v `ywU h_vµ‡g 3K Ges 4K|
Ges Zv‡ì j.mv.¸ = 12K|
cÖkœg‡Z,
12K = 180 K = 15 myZivs cÖ_g msL¨wU = 315 = 45|
3. A I B Gi Av‡qi AbycvZ 3:2 Ges Zv‡ì e¨v‡qi AbycvZ 5:3| hw` Zviv cÖ‡Z¨‡K 1000 UvKv K‡i mÂq K‡i Z‡e
A Avq KZ? [DAE-(Store Keeper)-2017]
a. 6000 UvKv b. 7000UvKv c. 8000UvKv d. 10000 UvKv Ans: a
Solution:
A I B Gi Avq h_vµ‡g = 3K Ges 2K
Ges Zv‡ì e¨vq h_vµ‡g = 5L Ges 3L
cÖkœg‡Z,
3K-5L = 1000 - - - - (1)
2K-3L = 1000 - - - - (2)
-----------------------
1 bs mgxKi‡K 3 w`‡q ¸Y K‡i Ges 2bs mgxKiY‡K 5 w`‡q ¸Y K‡i cvIqv hvq
9K-15L = 3000 - - - - (1)
10K-15L = 5000 - - - - (2)
------------------------------------
K = 2000
myZivs A Gi Avq = 32000=6000UvKv| DËi: 6000 UvKv|
4. 1 eQi c~‡e© ¯§„wZ I cÖxwZi eq‡mi AbycvZ wQj 4:3| GK eQi c‡i Zv‡ì eq‡mi AbycvZ n‡e 5:4| Zv‡ì eZ©gvb
eq‡mi mgwó KZ eQi? [DAE-(Store Keeper)-2017]
Shortcut: 22% = 55
n‡j 100% = 250|
Department OF Agricultural Extension (DAE)
Post name: Store Keeper Exam Date: 01-12-2017

a. 15 b. 16 c. 17 d. 19 Ans: b
Solution:
awi,
1 eQi c~‡e© ¯§„wZ I cÖxwZi eqm h_vµ‡g 4K Ges 3K eQi|
myZivs Zv‡ì eZ©gvb eqm = 4K+1 Ges 3K+1
Ges 1 eQi ci Zv‡ì eqm n‡e 4K+2 Ges 3K+2
cÖkœg‡Z,
2+3K
2+4K
=
4
5
ev, 16K+8 = 15K+10 K = 2
myZivs 1 Zv‡ì eZ©gvb eqm = (42+1) = 9 Ges (32+1) = 7 eQi|
Zv‡ì eq‡mi mgwó = 9+7 = 16 eQi|
5. Rwbi ¯‹z‡j hvIqvi mgq †eM N›Uvq 3 wKwg Ges evmvq †divi mgq †eM N›Uvq 2 wKwg| ¯‹z‡j hvIqv-Avmv‡Z hw` Zvi †gvU
5 N›Uv mgq jv‡M, Zvn‡j ¯‹zj I evmvi g‡a¨ `~iZ¡ KZ wKwg? [DAE-(Store Keeper)-2017]
a. 6 b. 7 c. 8 d. 9 Ans: a
Solution:
awi, `~iZ¡ = K wKwg
cÖkœg‡Z,
2
K
3
K
 = 5
ev, 5
6
3K2K


ev, 5K = 30
K = 6wKwg myZivs `~iZ¡ 6 wKwg|
6. ‡mvbv cvwbi ‡P‡q 19 ¸Y fvix Ges Zvgv cvwbi †P‡q 9 ¸Y fvix| avZz `y‡Uv wK Abycv‡Z wgwkÖZ Ki‡j D³ wgkÖY cvwbi
†P‡q 15¸Y fvix n‡e? [DAE- (Store Keeper)- 2017]
a. 2:3 b. 4:3 c. 3:2 d. 3:4 Ans: c
Solution:
awi, †mvbv cvwbi †P‡q 19x Ges Zvgvi †P‡q 9y ¸Y fvix|
cÖkœg‡Z,
19x+9y= 15(x+y)
 19x + 9y = 15x + 15y
 19x – 15x = 15y – 9y  4x= 6y 
4
6
y
x
 x : y = 3:2 DËi: 3:2
Shortcut: by rule of allegation
Shortcut: G ai‡Yi cÖkœ `ªæZ mgvavb Kivi Rb¨ MwZ‡eMØ‡qi j.mv.¸
†K `~iZ¡ a‡i wn‡me Ki‡Z nq| GLv‡b 2 I 3 Gi j.mv.¸ 6 B DËi|
1g Ask
6: 4 = 3:2
19
15-9=6 19-15=4
9
15
Mo
2q Ask

7. wgbv ivRy I wgVz GKwU KvR h_vµ‡g 15w`b, 10 w`b I 6 w`‡b m¤úbœ Ki‡Z cv‡i| Zviv GK‡Î KvR Ki‡j H KvRwU m¤úbœ
n‡Z †gvU KZw`b mgq jvM‡e? [DAE-(Store Keeper)-2017]
a. 2w`b b. 3w`b c. 4w`b d. 6w`b Ans: b
Solution:
wgbv, ivRy Ges wgVz GK w`‡b Ki‡Z cv‡i,
3
1
30
10
30
532
6
1
10
1
15
1


 Ask|
GLb,
3
1
Ask Ki‡Z 1 w`b jvM‡j 1 Ask ev m¤ú~Y© KvRwU Ki‡Z mgq jvM‡e = 3 w`b|
8. ‡Kvb cixÿvq 35% QvÎ GK wel‡q 42% QvÎ Ab¨ wel‡q ‡dj Ki‡jv Ges 15% QvÎ Dfq wel‡q †dj Kij| hw` 2500
Rb QvÎ cixÿvq AskMÖnY K‡i Zvn‡j KZ Rb QvÎ †h †Kvb GKwU‡Z †dj K‡i‡Q? [DAE-(Store Keeper)-2017]
a. 1000 b. 1100 c. 1175 d. 1250 Ans: c
Solution:
ïay GK wel‡q †dj = 35% -15% = 20%
ïay Ab¨ wel‡q †dj = 42% - 15% = 27%|
Zvn‡j ïay GK wel‡q †dj Kiv QvÎ = 20%+27% = 47% (wP‡Î Kv‡jv Ask)
GK ‡h †Kvb GK wel‡q †dj Kiv Qv‡Îi msL¨v = 2500 Gi 47% = 1175 Rb|
(D‡jøL¨ †h, hviv ïay GK wel‡q †dj ZvivB ïay GK wel‡q cvk|
KviY `ywU wel‡qi g‡a¨ ïay GKwU wel‡q ‡dj Kiv A_© Ab¨ welqwU‡Z cvk|)
9. kni A ‡_‡K kni B Gi 2wU wU‡KU Ges kni A ‡_‡K kni C Gi 3wU wU‡K‡Ui g~j¨ 77 UvKv| Avevi kni A ‡_‡K B
Gi 3wU wU‡KU Ges kni A ‡_‡K C Gi 2wU wU‡K‡Ui g~j¨ 73 UvKv| kni A ‡_‡K B Ges kni A ‡_‡K C Gi wU‡K‡Ui
g~j¨ h_vµ‡g KZ? [DAE-(Store Keeper)-2017]
a. 13 UvKv, 17UvKv b. 17 UvKv, 13 UvKv c. 9UvKv, 11UvKv d. 11UvKv, 9UvKv Ans: a
Solution:
awi,
kni A †_‡K kni B ‡Z hvIqv wU‡K‡Ui g~j¨ = x UvKv|
Ges kni A †_‡K kni C ‡Z hvIqv wU‡K‡Ui g~j¨ = y UvKv|
cÖkœg‡Z,
2x+3y=77…………(i)
3x+2y=73…………(ii)
(i) bs mgxKiY‡K 2 Ges (ii) bs mgxKiY‡K 3 w`‡q ¸Y K‡i cvIqv hvq|
4x+6y=154
9x+6y=219 ………………………………….
5x=65 (we‡qvM K‡i|)
 x=13
Avevi 2x+3y=77  213+3y=77  3y = 51 y = 17
myZivs A †_‡K B Gi fvov 13UvKv Ges A †_‡K kni C Gi fvov 17 UvKv|
10. GKRb K…lK Zvi eM©vK…wZ evMv‡bi Pvicv‡k †eov w`j| ‡m cÖwZ cv‡k^© 27wU K‡i LyuwU emv‡j, Zvi me©‡gvU KZwU LyuwUi
cÖ‡qvRb n‡qwQj? [DAE-(Store Keeper)-2017.
a. 108 b. 107 c. 106 d. 104 Ans:d
20% 15% 27%
Shortcut: G ai‡Yi cÖkœ mgxKiY mvwR‡q mgvavb bv K‡i
Ackb a‡i wn‡me Ki‡j mn‡R DËi †ei n‡e|

Solution:
GiKg cÖkœ¸‡jv `y fv‡e Kiv hvq|
cÖ_‡g me¸‡jv KY©v‡i GKwU K‡i LyuwU emv‡j †gvU 4wU|
Ges cÖwZ jvB‡b 27wU †gjv‡bvi Rb¨ `yÕ KY©v‡ii `ywUi gv‡S 25wU K‡i Pvicv‡k †gvU
100wU LyuwU jvM‡e| Zvn‡j †gvU LyuwU = 100+4 = 104 wU|
A_ev cÖ_‡gB 274 = 108wU †_‡K Pvi‡Kv‡Yi 4wU wiwcU n‡q hvIqvq 4 we‡qvM Ki‡j 108-4 = 104wU|
11. ‡kvfv cÖwZwU 70UvKv `‡i wKQz LvZv Ges cÖwZwU 30UvKv `‡i wKQz Kjg wKbj| †m †gvU 810 UvKv LiP K‡i Ges
m‡e©v”P hZ¸‡jv LvZv †Kbv m¤¢e ZZ¸‡jvB LvZv µq K‡i | Zvi µxZ LvZv I Kj‡gi AbycvZ KZ? [DAE-(Store
Keeper)-2017]
a. 4:3 b. 2:3 c. 3:4 d. 3:2 Ans: d
Solution:
m‡e©v”P msL¨K LvZv wKb‡Z n‡e|
GLb me¸‡jvB LvZv wKb‡j UvKv Awekó Zv‡K 81070 =11wU †Kbvi ci Aewkó = 40UvKv|
40UvKv Aewkó †_‡K GKwU Kjg †Kbvi ciI 10UvKv _vK‡e|
Zvn‡j Av‡iv 20UvKv jvM‡e bZzb GKwU Kjg †Kbvi Rb¨|
GLb GKwU LvZv Kg wKb‡j Zv w`‡q `ywU Kjg †Kbvi ciI 10UvKv Aewkó _vK‡e|
‡h‡nZz 20UvKv jvM‡e Zvn‡j `ywU LvZv bv wK‡b †mB UvKv w`‡q Kjg wKb‡j †mLvb †_‡K ‡h 20UvKv Aewkó _vK‡e
Zv Av‡Mi 10UvKvi mv‡_ †hvM K‡i Av‡iv GKwU Kjg †Kbv hv‡e|
Zvn‡j LvZv wK‡bwQj 9wU = 970 = 630 UvKv Ges Aewkó UvKv 810-630 = 180 UvKv w`‡q 18030 = 6wU|
LvZv I Kj‡gi AbycvZ = 9:6 ev 3:2|
12. x I y Dfq we‡Rvo msL¨v nq, Zvn‡j wb‡Pi †KvbwU Aek¨B †Rvo n‡e? [DAE-(Store Keeper)-2017]
a. xy+2 b. x+y c. xy d. x+y+1 Ans: b
Solution:
`ywU we‡Rvo msL¨v †hvM Ki‡j Zv Aek¨B †Rvo n‡e| †hgb: 3 I 5 Gi †hvMdj 3+5 = 8|
myZivs cÖ`Ë Ack‡bi g‡a¨ x+y B mwVK|
13. GK e¨vw³‡K Zvi †gvU F‡Yi cÖ_g 600 UvKvi Rb¨ 8% nv‡i my` w`‡Z nq Ges 600 UvKvi AwaK AskwUi Rb¨ 7%
nv‡i my` w`‡Z nq| hw` †Kvb eQi Zvi F‡Yi cwigvY 6000 UvKv nq, †m eQi Zv‡K KZ UvKv my` w`‡Z n‡e? [DAE-
(Store Keeper)-2017]
a. 480 b. 420 c. 378 d. 426 Ans: d
Solution:
6000 UKvi g‡a¨ cÖ_g 600 UvKvi my` = 600 Gi 8% = 48UvKv|
Avevi c‡ii 6000-600 = 5400 UvKvi my` = 5400 Gi 7% = 378UvKv|
Zvn‡j †gvU my`: 48+378 = 426 UvKv|
gy‡L gy‡L GK jvB‡b Kivi Rb¨: 6000UvKviB my` 7% n‡j n‡Zv 420UvKv Gi mv‡_ cÖ_g 600UvKvi Rb¨ 1UvKv K‡i
6UvKv AwZwi³ n‡j †gvU my` n‡e 420+6 = 426UvKv|
====================================

1. GKRb wUwf we‡µZv 45% jv‡f wUwf wewµ KiZ| g›`vi Kvi‡Y †m Zvi jv‡fi nvi 40% K‡i Ges G‡Z Zvi weµq
10% †e‡o hvq| Zvi bZzb jvf I Av‡Mi jv‡fi AbycvZ KZ? [BADC (AO)-2017]
a. 9:8 b. 11:10 c. 45:44 d. 44:45 Ans: d
Solution:
awi, Av‡Mi wewµ 10wU wUwf Ges cÖwZwU‡Z jvf = 45UvKv| Zvn‡j †gvU jvf = 1045 = 450UvKv|
Avevi, 10wUi wewµ evo‡j bZzb wewµ 11wU Ges cÖwZwU‡Z 40UvKv K‡i bZzb †gvU jvf = 1140 = 440UvKv|
bZzb jvf I Av‡Mi jv‡fi AbycvZ = 440:450 ev 44:45 UvKv|
2. hw` K2
+ L2
= 4 Ges K2
- L2
= - 4 nq, Zvn‡j K 4
+ L4
Gi gvb KZ n‡e ? [BADC (AO)-2017]
a. 16 b. -16 c. 18 d. -8 Ans: a
Solution:
K2
+ L2
= 4
K2
- L2
= - 4
--------------------------------
2K2
= 0 [ †hvM K‡i]
myZivs K2
= 0 K = 0
Avevi,
2L2
= 8 [we‡qvM K‡i]
ev, L 2
= 4 L = 2
myZivs
K 4
+ L4
= 04
+24
= 16 [(a+b)(a-b) = a2
-b2
myÎ cÖ‡qvM K‡iI Kiv hvq|]
3. wibv gx‡gi †_‡K 10 eQ‡ii eo| 7 eQi ci, wibvi eqm gx‡gi eq‡mi wØ¸Y n‡e| wibvi eZ©gvb eqm KZ ? [BADC
(AO)-2017]
a. 17 b. 16 c. 13 d. 11 Ans: c
Solution:
awi, wibvi eqm = x eQi|
Zvn‡j gx‡gi eqm = x-10 eQi|
cÖkœg‡Z,
x+7 = 2(x-10+7) [gxg ‡QvU ZvB gx‡gi mv‡_ 2¸Y]
ev, x+7 = 2x-6
 x = 13 Ans:
4. wZbwU avivevwnK c~Y©msL¨vi ¸Ydj 120 n‡j G‡ì †hvMdj KZ ? [BADC (AO)-2017]
a. 9 b. 14 c. 12 d. 15 Ans: d
Solution:
120 ‡K fv½‡j cvIqv hvq 22235 GLb wZbwU avivevwnK msL¨v evbv‡bvi Rb¨ G‡ì‡K evbv‡Z n‡e 456
myZivs †hvMdj: 4+5+6 = 15|
Shortcut:
`ywU eM© msL¨vi †hvMdj 4 n‡j Zv Aek¨B 2 I 2 A_ev 0 I 2 n‡Z cv‡i|
wKš‘ Zv‡ì e‡M©i we‡qvMdj -4 n‡j Aek¨B GKwU msL¨v 0 Ges AciwU 2 n‡Z n‡e|
ZvB cvIqvi 4 w`‡j 04
+24
= 16 B n‡e|
Bangladesh Agricultural Development Corporation (BADC)
Post name: Administrative Officer Post: Exam date: 27-10-2017
Exam Taker: IBA, DU

5. 8 R‡bi GKwU `‡j 65 †KwR IR‡bi GKR‡bi cwie‡Z© bZzb GKRb †hvM †`qvq Zv‡ì Mo IRb 2.5 †KwR †e‡o
hvq| bZzb e¨w³i IRb KZ †KwR ? [BADC (AO)-2017]
a. 45 b. 76 c. 80 d. 85 Ans: d
Solution:
8 R‡bi M‡o 2.5 †KwR K‡i evo‡j †gvU IRb evo‡e 82.5 = 20 †KwR|
Zvn‡j bZzb e¨vw³i IRb = 65+20=85 †KwR|
6. ˆ`wbK 8 N›Uv KvR K‡i 3wU cv¤ú 1wU Rjvavi‡K 2w`‡b Lvwj Ki‡Z cv‡i| Rjvavi‡K 1w`‡b Lvwj Ki‡Z n‡j 4wU
cv‡¤úi ˆ`wbK KZ N›Uv KvR Ki‡Z n‡e ? [BADC (AO)-2017]
a. 10 b. 12 c. 15 d. 16 Ans:b
Solution:
3wU cv¤ú w`‡q 2 w`‡b Lvwj Ki‡Z n‡j KvR Ki‡Z n‡e = 8 N›Uv
1 ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = 832 (Kg cv¤ú w`‡q †ewk mgq Ges Kg w`‡b Av‡iv
†ewk| )
4 ÕÕ ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ =
4
238 
=12N›Uv|
7. 1056 Gi mv‡_ me©wb¤œ KZ †hvM Ki‡j †hvMdj 23 Øviv wb:‡k‡l wefvR¨ n‡e ? [BADC (AO)-2017]
a. 2 b. 3 c. 18 d. 21 Ans: a
Solution:
23 w`‡q 1056 †K fvM Ki‡j fvM‡kl _vK‡e = 21 Zvn‡j ‡`Lv hv‡”Q 23-21 = 2 NvUwZi Kvi‡Y Av‡iKevi 23 w`‡q
fvM Kiv hv‡”Q bv| myZivs 1056 Gi mv‡_ †hvM Ki‡Z n‡e 2|
8. GKwU Aswk`vix eëmv‡qi †gvU gybvdv †_‡K ÔKÕ 40%, ÔLÕ 25%, ÔMÕ 20%, ÔNÕ 10%, Ges ÔOÕ 5% cvq| ÔKÕ Gi
gybvdv ÔLÕ Gi gybvdvi kZKiv KZ Ask ? [BADC (AO)-2017]
a. 62.5% b. 120% c. 160% d. 175% Ans: c
Solution:
GLv‡b K Gi 40% Ges L Gi 25%
Zvn‡j L Gi 25G hw` K Gi 40 nq
myZivs kZKiv nvi n‡e
25
10040 
=160%
9. ‡Kvb GK KviLvbvq Kgx©‡ì RbcÖwZ ˆ`wbK gRywi wQj 100 UvKv| g›`vi Kvi‡Y Zv‡ì ˆ`wbK gRywi 50% Kgv‡bv
n‡qwQj| m¤úªwZ ˆ`wbK gRywi 60% evov‡bv n‡q‡Q| eZ©gv‡b RbcÖwZ ˆ`wbK gRywi KZ ? [BADC (AO)-2017]
a. 160 b. 110 c. 80 d. 60 Ans: c
Solution:
100 UvKv †_‡K 50% gRywi Kgv‡j bZzb gRywi 50 UvKv| GLb GB 50 UvKvi 60% ev 30UvKv evo‡j bZzb gRywi n‡e
50+30 = 80 UvKv|
10. GKwU wngvMvi‡K bevqb Kivi mgq Gi ˆ`N©¨ 30% I cÖ¯’ 50% evov‡bv nj Ges Gi D”PZv 20% Kgv‡bv nj|
bevqbK…Z wngvMv‡ii AvqZb cy‡iv‡bv wngvMv‡ii †_‡K kZKiv KZ Ask †ewk ? [BADC (AO)-2017]
a. 56% b. 50% c. 45 % d. 65% Ans: a
Solution:
cÖ_‡g ˆ`N¨©, cÖ¯’ I D”PZv 10, 10 Ges 10 n‡j AvqZb = 101010 = 1000
bZzb ‰`N©¨, cÖ¯’ I D”PZv = 13,15 Ges 8 myZivs bZzb AvqZb = 13158 = 1560

AvqZb e„w× cvq 1560-1000 = 560 | e„w×i kZKiv nvi = 56%
1000
100560


11. 36 msL¨vwUi †gvU KZ¸‡jv fvRK i‡q‡Q ? [BADC (AO)-2017]
a. 6 b. 8 c. 9 d. 10 Ans: c
Solution:
36 †K fv½‡j cvIqv hvq
49 = 22
32
GLb GKB msL¨vi Dc‡ii cvIqvi¸‡jvi mv‡_ 1 †hvM K‡i Zv‡ì ¸Ydj nj 33 = 9|
12. GKwU QvÎvev‡m hZRb QvÎ _v‡K, Zv‡ì cÖ‡Z¨‡Ki gvwmK LiP Zv‡ì †gvU msL¨vi `k¸Y| H QvÎvev‡mi †gvU gvwmK
LiP 6,250 UvKv n‡j H QvÎvev‡m KZRb QvÎ _v‡K ? [BADC (AO)-2017]
a. 15 b. 25 c. 35 d. 45 Ans: b
Solution:
awi, QvÎ msL¨v = x Rb|
Zvn‡j Zv‡ì cÖ‡Z¨‡Ki gvwmK LiP = 10x UvKv|
cÖkœg‡Z,
x10x = 6250 ev, x2
= 625  x = 25
13. 13, 17, 25, 41 ....... avivwUi cieZx© msL¨vwU KZ ? [BADC (AO)-2017]
a. 73 b. 89 c. 101 d. 145 Ans: a
Solution:
13, 17, 25, 41 - - - avivwUi gv‡Si eëavb ¸‡jv n‡jv 4,8,16 myZivs Gi c‡ii msL¨vi mv‡_ eëavb n‡e 32 Ges
msL¨vwU n‡e 41+32 = 73|
14. ‡Kv‡bv K¬v‡m evjK‡ì Mo eqm evwjKv‡ì msL¨vi wØ¸Y| 36 R‡bi H K¬v‡m evjK I evwjKv‡ì msL¨vi AbycvZ 5.1 |
H K¬v‡mi evjK‡ì †gvU eqm KZ eQi ? [BADC (AO)-2017]
a. 300 b. 320 c. 360 d. 400 Ans: c
Solution:
5+1 = 6 As‡ki g‡a¨ ‡gvU evjK = 36
6
5
= 30 Rb Ges evwjKv = 36-30 = 6 Rb|
evjK‡ì Mo eqm = 12eQi | myZivs evjK‡ì †gvU eqm = 3012 = 360eQi|
15. ivRy GKwU eB nvmv‡bi Kv‡Q 10% ÿwZ‡Z wewµ Kij| ivRy hw` eBwU 20% Kg `v‡g wKbZ Ges 44 UvKv †ewk `v‡g
wewµ KiZ, Zvn‡j Zvi 40% jvf nZ| ivRy eBwU KZ `v‡g wK‡bwQj? [BADC (AO)-2017]
a. 50 b. 100 c. 200 d. 500 Ans: c
Solution:
awi, cÖ_g µqg~j¨ = 100UvKv|
Zvn‡j cÖ_g weµqg~j¨ = 100-10 =90UvKv|
Avevi 2q µqg~j¨ = 100-20 = 80UvKv|
Avevi 40% jv‡f 2q weµqg~j¨ = 80+80 Gi 40% = 112 UvKv|
GLb `yB weµqg~‡jï e¨veavb = 112-90=22UvKv
e¨veavb 22UvKv n‡j µqg~j¨ = 100UvKv
myZivs eëavb 44UvKv n‡j µqg~j¨ n‡e =200 UvKv|
Shortcut: 22% = 44 n‡j 100% = 200

16. 240 Rb †jvK GKwU eb‡fvR‡b hvq| †mLv‡b hZRb gwnjv wQj Zvi †_‡K 20 Rb cyiæl †ewk wQj| Avevi hZRb wkï
wQj Zvi †_‡K 20 Rb cÖvßeq®‹ †ewk wQj| eb‡fvR‡b KZRb cyiæl wQj? [BADC (AO)-2017]
a. 100 b. 140 c. 145 d. 75 Ans: d
Solution:
g‡b Kwi, cyiæl = x
gwnjv = x - 20 (cyiæl †_‡K gwnjv 20 Rb Kg|)
Ges wkï = (x + x – 20) – 20 (cyiæl gwnjv wg‡j cÖvß eq¯‹ hv‡`i †_‡K wkï 20 Rb Kg|)
= 2x– 40
cÖkœg‡Z,
x + (x – 20) + (2x – 40) = 240 (cyiæl+gwnjv+wkï = 240)
4x – 60 = 240  4x = 300  x = 75  cyiæl = 75
17. GKRb wVKv`vi 1920 wgUvi `xN© iv¯Ív 120 w`‡b wbg©vY K‡i †`qvi Rb¨ 160 Rb kÖwgK wb‡qvM Kij| 24 w`b ci, gvÎ
8
1
Ask KvR m¤úbœ nj| wba©vwiZ mg‡qi g‡a¨ KvR †kl Ki‡Z n‡j AwZwi³ KZRb †jvK wb‡qvvM Ki‡Z n‡e?
[BADC (AO)-2017]
a. 110 b. 160 c. 180 d. 120 Ans: d
Solution:
wbg©vY n‡q †M‡Q 1920 Gi
8
1
= 240wgUvi|
wbg©vY Aewkó = 1920-240 = 1680wgUvi Ges w`b Aewkó 120-24 = 96
24 w`‡b 240wgUvi wbgvY© Ki‡Z †jvK jv‡M = 160 Rb|
1 ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ =
240
24160 
(Kg w`‡b †ewk †jvK ZvB ¸Y Ges Kg Kv‡R Kg †jvK ZvB fvM)
96 ÕÕ 1680 ÕÕ ÕÕ ÕÕ ÕÕ =
96240
168024160


(24Gi wecix‡Z 96 Ges 240 Gi wecixZ cv‡k 1680)
= 280 Rb|
AwZwi³ †jvK jvM‡e 280-160= 120Rb|
weKí mgvavb:
fMœvsk a‡iI GB cÖkœwU mgvavb Kiv hvq| Gfv‡e;
wbgvY n‡q †M‡Q,
8
1
Ask| Zvn‡j Aewkó KvR Av‡Q 1-
8
1
=
8
7
Ask|
GLb HwKK wbq‡g mvRv‡bv hvq,
24 w`‡b
8
1
Ask wbgvY© Ki‡Z †jvK jv‡M = 160 Rb|
1 ÕÕ 1 ÕÕ ÕÕ ÕÕ ÕÕ ÕÕ = 160248 (Kg w`‡b †ewk †jvK ZvB ¸Y Ges 1 Ask Av‡iv †ewk ZvB
Avevi Dwë‡q ¸Y)
96 ÕÕ
8
7
Ask| ÕÕ ÕÕ ÕÕ ÕÕ =
896
7824160


(24 Gi wecix‡Z 96 Ges
8
7
mivmwi ¸Y|)
= 280 Rb|
AwZwi³ †jvK jvM‡e 280-160= 120Rb|

‡kLvi gZ welq n‡jv ‡h:
cÖ_g evi fMœvsk Avm‡j Dwë‡q ¸Y nq Avi wØZxqevi fMœvsk Avm‡j mivmwi ¸Y nq| cÖkœwU A‡b¸‡jv wjwLZ cixÿvq
ûeû Avmvi Kvi‡Y A‡bK †ewk ¸iZ¡c~Y© |
18. GKRb e¨w³ GKwU eM©‡ÿÎ Ry‡o Avov Avwofv‡e ‡nu‡U wM‡qwQj| cÖvšÍ eivei bv nvuUvi Kvi‡Y KZ kZvsk Kg nvuU‡Z
n‡qwQj ? [BADC (AO)-2017]
a. 20% b. 25% c. 32% d. ‡KvbwUB bq Ans: d
Solution:
awi, eM©‡ÿÎwUi GK evû = 10wgUvi Zvn‡j cÖvšÍ eivei †M‡j nvU‡Z n‡e = 10+10 = 20wg.
Avevi KY© eivei †h‡Z n‡j nvU‡Z n‡e 14.14102 
Zvn‡j Kg nvuU‡Z n‡e, 20-cÖvq 14 = 6 kZKiv nvi 30%
20
1006


cÖvq|
19. GKwU †kÖYxi cÖwZ †e‡Â 4 Rb K‡i QvÎ em‡j 3wU †eÂ Lvwj _v‡K| wKš‘ cÖwZ †e‡Â 3 Rb K‡i em‡j 6 Rb Qv‡Îi
`vuwo‡q _vK‡Z nq| G †kÖYxi QvÎ msL¨v KZ ? [BADC (AO)-2017]
a. 50 b. 60 c. 70 d. 80 Ans:b
Solution:
g‡b Kwi, QvÎmsL¨v = x
cÖkœg‡Z,
4
x
+3 =
3
6x 
[KviY †eÂ msL¨v mgvb]
ev
3
12x 
=
3
6x 
ev 4x-24=3x+36  x = 60
20. hw` n GKwU †Rvo msL¨v nq Z‡e wb‡Pi †KvbwU †Rvo msL¨v n‡Z cvi‡e bv ? [BADC (AO)-2017]
a. n2
b. 3 (n+1)+3 c. 2n+2 d. 2n+3 Ans: d
Solution:
Avgiv Rvwb †h †Kvb msL¨vi mv‡_ †Rvo msL¨v ¸Y Ki‡j Zv †Rvo n‡q hvq| ZvB GLv‡b 2n Aek¨B ‡Rvo msL¨v| Avevi
hw` 2n ‡Rvo msL¨v nq Zvn‡j Zvi mv‡_ Av‡iKwU †Rvo msL¨v †hvM Ki‡j †RvoB _vK‡e| wKš‘ we‡Rvo msL¨v †RvM
Ki‡j Zv we‡Rvo n‡q hv‡e| ZvB 2n+3 Aek¨B we‡Rvo n‡e|
====================================
4 Ges 3 Dfq w`‡qB fvM Kiv hvq Ggb
msL¨v gvÎ 1wUB Av‡Q hv n‡jv 60|

1. GK e¨vw³ 40 w`‡b Zvi `vjv‡bi KvR †kl Kivi Rb¨ 25 Rb †jvK wb‡qvM w`‡jb| 20 w`b ci wZwb Av‡iv 15 Rb
†jvK wb‡qvM w`‡jb Ges KvRwU 5w`b Av‡M †kl n‡q †Mj| AwZwi³ †jvK wb‡qvM bv w`‡j wZwb wba©vwiZ mg‡qi KZw`b
c‡i KvRwU †kl Ki‡Zb? [BADC (AC)-2017]
K. 2 w`b L. 3 w`b M. 4 w`b N.5 w`b O. †KvbwUB bq DËi: 4w`b|
mgvavb: (mvaviY HwKK wbq‡g mgvavb|)
w`b Aewkó Av‡Q = 40-20= 20 w`b
5 w`b Av‡M †kl nq = 20- 5 = 15 w`‡b
AwZwi³ †jvK wb‡qvM = 25+15 = 40 Rb|
40 Rb KvRwU K‡i = 15 w`‡b (Av‡Mi 25 + bZzb 15 Rb mn KvRwU Ki‡j 15 w`‡b †kl nq)
1 Ó Ó Ó = 1540 ÕÕ
25 Ó Ó Ó =
25
4015 
(bZzb 15 Rb wb‡qvM bv w`‡q Av‡Mi 25 Rb B Ki‡j 24 w`b jvM‡Zv|)
= 24 w`b
AwZwi³ †jvK wb‡qvM bv w`‡j mgq †ewk jvMZ = 24 - 20 = 4 w`b DËi t 4 w`b
weKí mgvavb: (Kv‡Ri BDwbU a‡i GgwmwKD cixÿvi Rb¨ 20 †m‡K‡Û mgvavb|)
KvRwU‡Z jvMv †gvU gRy‡ii cwigvb †ei n‡e Gfv‡e,
25 Rb cÖ_g 20 w`‡b K‡i 2520 = 500 BDwbU
Avevi 20 w`b ci 15 Rb Avmvq 25+15 = 40 Rb hLb K‡i ZLb KvRwU †kl nq 40-5 = 35 w`‡b|
myZivs 40 Rb Ki‡jv 35-20 = 15 w`‡b †gvU 4015 = 600 BDwbU|
A_©vr †gvU KvRwU 500+600 = 1100 BDwb‡Ui|
Zvn‡j 25 Rb GKUvbv KvR Ki‡j jvM‡Zv 110025 = 44 w`b|
myZivs KvRwU †kl n‡Z †jU n‡Zv 44-40 = 4 w`b|
weKí mgvavb: (K a‡i mgxKiY mvwR‡q mgvavb)
awi, KvRwU †kl Ki‡Z ‡gvU mgq jvM‡e K w`b|
Zvn‡j 25 Rb‡K †gvU jvM‡e = 25K w`b|
cÖkœg‡Z,
25K = (2025) +(1540) =(25 Rb 20w`b KvR Kivi ci 25+15 = 40 Rb K‡i‡Q 15w`b, KviY †gvUw`b 40-
5=35)
ev, 25K = 1100
myZivs K = 44 w`b|
AwZwi³ †jvK jvMv‡bv bv n‡j mgq †ewk jvM‡Zv: 44-40 = 4 w`b|
Post name: Assistant Cashier. Exam Date: 11-08-2017
Exam Taker: IBA, DU

2. 40 ‡KwR ˆRe I BDwiqv mv‡ii wgkÖ‡Y ˆRe mv‡ii cwigvY 10%| KZ †K‡wR mvi wgkv‡j bZzb wgkÖ‡Y ˆRe mv‡ii
cwigvY 20% n‡e? [BADC (AC)-2017]
K. 4 L.5 M.6 N.7 O. †KvbwUB bq
mgvavb: (L)
cÖ_‡g ˆRe I BDwiqv mv‡ii cwigvY = 4 I 36 †KwR
awi, ˆRe mvi †gkv‡Z n‡e = K †KwR
cÖkœg‡Z,
4+K = (40+K) Gi 20%
ev, 20+5K = 40+K
ev, 4K = 20 K = 5 †KwR|
‡hfv‡e †f‡e Kbwd‡W›Uwj fzj Ki‡Z cv‡ib:
40 Gi 10% = 4 n‡j 20% = 8 n‡e Zvn‡j Av‡M wQj 4 ‡gkv‡Z n‡e 8-4 = 4 (wbwðZ fzj DËi)
‡Kb fzj??
KviY hLb ˆRe †gkv‡bv n‡e ZLb †gvU cwigvY I evo‡e| ZvB ‡gkv‡bvi ci 20% Avi †gkv‡bvi Av‡Mi 20% GK bv|
‡hgb: cÖ_‡g ‡gvU mvi = 40 Gi g‡a¨ ˆRe = 40 Gi 10% = 4 Ges BDwiqv = 40-4 = 36|
GLb 5 †KwR ˆRe mvi †gkv‡bvi ci †gvU mvi n‡e 40+5 = 45
Ges ˆRe mvi n‡e GB 45 Gi 20% = 9 (Av‡Mi 40+bZzb 5)|
wKš‘ hw` 4 †gkvb Zvn‡j †gvU cwigvY n‡e 40+4 = 44 hvi 20% = 8 nq bv eis: 8.8 nq|
Solution By rule of allegation in 10sec:
Ratio of 1st
and 2nd
parts = 20 :10 = 2 : 1
2 part = 10 So, 1 part = 5 Ans: 5
3. ‡Kvb msL¨vi wZb PZy_©vs‡ki GK cÂgvs‡ki gvb 60| msL¨vwU KZ? [BADC (AC)-2017]
K.300 L.400 M.500 N.600 O. †KvbwUB bq
mgvavb: (L)
GiKg cÖkœ †k‡li ‡_‡K wn‡me Ki‡j gy‡L gy‡L DËi ejv hvq|
‡k‡li 5 fv‡Mi 1 fv‡Mi gvb 60 n‡j 5 fv‡Mi gvb = 605 = 300|
GLb GB 300 n‡j 4 fv‡Mi 3 fv‡Mi gvb| Zvn‡j 1 fv‡Mi gvb n‡e 100 Ges 4 fvM ev msL¨vwU n‡e 400
gy‡L gy‡L: 10 †m‡K‡Û:
BDwiqv: 80% = 36 n‡j ˆRe 20% = n‡e BDwiqvi
cwigv‡Yi 4 fv‡Mi 1 fvM| A_©vr 36/4 = 9 †KwR|
bZzb †gkv‡Z n‡e 9-4 = 5 †KwR|
Profit on 1st
part
10%
20-0 20-10
0%
20%
Mean profit
Profit on 2nd
part

4. GKwU `ªe¨ 420 UvKvq µq K‡i 15% jv‡f weµq Kiv nj| weµqg~j¨ µqg~j¨ A‡cÿv KZ UvKv ‡ewk? [BADC
(AC)-2017]
K.42 UvKv L.21UvKv M.84UvKv N.63UvKv O. †KvbwUB bq
mgvavb: (N)
µqg~j¨ = 420
Ges weµqg~j¨ = 420+420 Gi 15% = 420+63 = 483 UvKv|
weµqg~j¨ I µqg~‡jï cv_©K¨ = 483-420 = 63 UvKv| (jv‡fi 63 UvKv B cv_©K¨)
5. `yBwU msL¨vi AbycvZ 5:7 Ges G‡ì †hvMdj 108| e„nËg msL¨vwU KZ? [BADC (AC)-2017]
K.42 L.49 M.56 N.63 O. †KvbwUB bq
mgvavb: (N)
5+7 = 12 As‡ki gvb 108 n‡j 1 As‡ki gvb 10812 = 9
Zvn‡j eo msL¨vwU = 97 = 63
6. gvZv I Kb¨vi eq‡mi mgwó 60 eQi| 5 eQi Av‡M gvZvi eqm Kb¨vi eq‡mi 4 ¸Y wQj| 7 eQi ci gvZvi eqm KZ
n‡e? [BADC (AC)-2017]
K.35 L.40 M.42 N.47 O. †KvbwUB bq
mgvavb: (O)
eZ©gv‡b gvZv I Kb¨vi mgwó 60 eQi n‡j 5 eQi Av‡M mgwó wQj = 60-10 = 50 eQi| ZLb gvZvi eqm Kb¨vi 4
¸Y| Ges 7 eQi ci gvZvi eqm n‡e 45+7 = 52 eQi|
7. GKwU cvBc Øviv GKwU U¨vsK 3 N›Uvq c~Y© nq| wØZxq cvBcwU Øviv U¨vsKwU c~Y© n‡Z 6 N›Uv mgq jv‡M| `yBwU cvBc
GKmv‡_ †Q‡o ‡`qv n‡j U¨vsKwU c~Y© n‡Z KZ wgwbU jvM‡e? [BADC (AC)-2017]
K.60 L.90 M.120 N.140 O. †KvbwUB bq
mgvavb: (M)
cÖ_g I wØZxq cvBc w`‡q 1 N›Uvq c~Y© nq: =
2
1
6
3
6
12
6
1
3
1


 Ask|
Zvn‡j m¤ú~Y© Ask c~Y© Ki‡Z mgq jvM‡e 2 N›Uv ev 120 wgwbU|
8. K, L Ges M GKwU e¨vemvq h_vµ‡g 36,000UvKv, 42,000 UvKv Ges 72,000 UvKv wewb‡qvM K‡i| GKeQi ci
g~jab Abycv‡Z jvf ewÚZ nq Ges L 1400 UvKv jvf cvq| K I M Gi jv‡fi mgwó KZ? [BADC (AC)-2017]
K.2500 L.3000 M.3300 N.3600 O. †KvbwUB bq
mgvavb: (N)
Zv‡ì wewb‡qv‡Mi AbycvZ = K:L:M = 36,000 : 42,000 : 72,000 ev 6:7:12
GLb L Gi 7 Ask = 1400 n‡j 1 As‡ki gvb = 14007 = 200
myZivs K+M Gi 6+12 = 18 As‡ki gvb n‡e 18200 = 3600UvKv|
9. wZbwU msL¨vi AbycvZ 4:5:6 Ges ga¨g msL¨vwUi eM© 225| e„nËg msL¨vwU KZ? [BADC (AC)-2017]
K.18 L.20 M.22 N.24 O. †KvbwUB bq
mgvavb: (K)
awi, msL¨v wZbwU h_vµ‡g, 4K, 5K Ges 6K

cÖkœg‡Z,
(5K)2
=225 ev, 5K = 15 ev, K = 3, myZivs e„nËg msL¨vwU = 63 = 18
10. ‡Kvb cixÿvq 80% MwY‡Z I 70% evsjvq cvk Kij| Dfq wel‡q cvk Kij 60%| Dfq wel‡q kZKiv KZRb †dj
Kij? [BADC (AC)-2017]
K. 5% L.10% M.15% N.20% O. 25%
mgvavb: (L)
Dfq wel‡q †dj = [100- (80+70- 60)]% = 10%
11. `ywU msL¨vi ¸Ydj 1536| msL¨v `ywUi j.mv.¸ 96 n‡j Zv‡ì M.mv.¸ KZ? [BADC (AC)-2017]
K.16 L.12 M.24 N.18 O. 32
mgvavb: (K)
`ywU msL¨vi j.mv.¸  msL¨v `ywUi M.mv.¸ = msL¨v `ywUi ¸Ydj|
myZivs msL¨v `ywUi M.mv.¸ = ¸Ydj  j.mv.¸ ev 153696 = 16
( GLv‡b m¤ú~Y© fvM bv K‡i ïay †k‡li msL¨v 6 I 6 ‡`‡L 5 †m‡K‡Û †gjv‡bv hvq|)
12. hw` 330 UvKv 3 eQi c‡i my`-Avm‡j 429 UvKv nq, Zvn‡j 650 UvKv my‡` Avm‡j 5 eQi c‡i KZ UvKv n‡e?
[BADC (AC)-2017]
K. 825 L.875 M.900 N.975 O. †KvbwUB bq
mgvavb: (N)
cÖ_‡g 3 eQ‡ii ‡gvU my` = 429-330 = 99UvKv, myZivs 1 eQ‡ii my` = 993 = 33 UvKv|
Avevi 330 UvKvi 1 eQ‡ii my` = 33UvKv (10 fv‡Mi 1 fvM n‡j )
100 UvKvi 1 eQ‡ii my` n‡e 100UvKvi 10 fv‡Mi 1 fvM ev 10UvKv A_©vr my‡ì nvi 10% |
GLb 650 UvKvi 1 eQ‡ii myÌ n‡e 10 fv‡Mi 1 fvM = 65UvKv Zvn‡j 5 eQ‡ii my` n‡e 655 = 325 UvKv|
myZivs my`-Avmj n‡e 650+325 = 975 UvKv|
13. GKwU eM©vKvi Rwgi GK cvk¦© gvcvi mgq fz‡j ˆ`‡N©¨ 10 kZvsk †ewk gvcv nq| †ÿÎd‡ji †ÿ‡Î fz‡ji cwigvY KZ
kZvsk? [BADC (AC)-2017]
K. 10% L.10.25% M.21% N.25% O. †KvbwUB bq
mgvavb: (M)
mvaviY fv‡e e‡M©i †ÿÎdj cwigv‡ci Rb¨ GK evû cwigvc K‡iB ‡ÿÎdj †ei Kiv nq| GLb GKwU e‡M©i GK evû
100 n‡j Zvi ‡ÿÎdj = 1002
ev 10000 | GLb hw` fzj K‡i hw` GK evû 10% evov‡bv nq A_©vr †h Rwg ‡g‡c‡Q
†m 100wgUvi Gi cwie‡Z© 110 wgUvi ej‡j †h wn‡me Ki‡e †m 1102
= 12100 †ÿÎdj †ei Ki‡e| G‡Z e‡M©i ‡ÿÎdj
evo‡e 10000 G 2100 ev 100 †Z 21%|
gy‡L gy‡L: e‡M©i GKevû evov gv‡bB `y evi evov‡bv| Zvn‡j cÖ_‡g 100 †_‡K 10 ‡e‡o 110 nq| Ges c‡i 110 Gi
10% A_©vr 11 ev‡o| Zvn‡j †gvU evo‡e 10+11 = 21%|
90%
10%
80-60
=20% 60%
BM
70-60
=10%
Total =100%

14. 300 wgUvi `xN© GKwU gvjevnx †Uªb N›Uvq 72wK‡jvwgUvi †e‡M 25 †m‡K‡Û GKwU ‡mZz AwZµg K‡i | †mZzwUi ‰`N¨©
KZ? [BADC (AC)-2017]
K.200 wgUvi L.220 wgUvi M.250wgUvi N.300wgUvi O. †KvbwUB bq
mgvavb: (K)
‡Uª‡bi MwZ‡eM = 72 wKwg/N›Uv ev 72
18
5
=20wgUvi/‡m‡K‡Û| Zvn‡j 25 †m‡K‡Û ‡gvU hvq 2025 = 500wg.
‡h‡nZz †Uª‡bi ˆ`N©¨ = 300wgUvi Zvn‡j †mZzwUi ˆ`N¨© n‡e = 500-300 = 200 wgUvi|
15. `yBwU msL¨vi AbycvZ 5:8| cÖwZwU msL¨vi mv‡_ 15 †hvM Ki‡j G‡`i AbycvZ nq 10:13| msL¨vØ‡qi †hvMdj KZ?
[BADC (AC)-2017]
K.31 L.33 M.36 N.39 O. †KvbwUB bq
mgvavb: (N)
awi, msL¨v `ywU: = 5K Ges 8K
cÖkœg‡Z,
5K+15 : 8K + 15 = 10:13
mgvavb K‡i cvB K = 3
myZivs msL¨v `ywUi †hvMdj = 5K+8K = 13K = 133 = 39| DËi: 39
(gy‡L gy‡L: Ack‡bi g‡a¨ ïay 39 †K AbycvZ Ø‡qi †hvMdj 8+5 = 13 w`‡q fvM Kiv hvq| )
16. GKwU AvqZ‡ÿ‡Îi ˆ`N©¨ A‡cÿv cÖ¯’ 4 wgUvi Kg| Gi †ÿÎdj 192eM© wgUvi n‡j, cwimxgv KZ? [BADC (AC)-
2017]
K.48wgUvi L.64wgUvi M.52wgUvi N.56wgUvi O. †KvbwUB bq
mgvavb: (N)
awi, ‰`N¨© = K myZivs cÖ¯’ = K-4
cÖkœg‡Z, K  (K-4) = 192 ev, K2
- 4K -192 = 0 ev, K-2
-16K+12K -192 = 0 myZivs K = 16
‰`N©¨ 16 n‡j cÖ¯’ = 16-4 =12 I cwimxgv = 2(16+12) = 56
{gy‡L gy‡L : 1612 = 192 nq| †hLv‡b 16-12=4 myZivs DËi 2(16+12) = 56}
17. 12 Rb kÖwgK 3 w`‡b 720 UvKv Avq K‡i| 9 Rb kÖwgK mgcwigvY UvKv Avq K‡i KZw`‡b? [BADC (AC)-2017]
K.2w`‡b L.3w`‡b M.4 w`‡b N.5w`‡b O. †KvbwUB bq
mgvavb: (M)
12 Rb Avq Ki‡Z jv‡M = 3 w`b|
1 Rb Avq Ki‡Z jv‡M = 312w`b|
9Rb Avq Ki‡Z jv‡M =
9
123
= 4 w`b|
18. cvwbc~Y© GKwU Wªv‡gi IRb 20 †KwR| hw` Gi GK PZz_©vsk cvwbc~Y© _v‡K Zvn‡j Gi IRb nq 8 †KwR| Lvwj Wªv‡gi
IRb KZ †KwR? [BADC (AC)-2017]
K.3 ‡KwR L.4 †KwR M. 5 †KwR N.6 †KwR O. †KvbwUB bq
gy‡L gy‡L: 10 †m‡K‡Û:
cÖ_g `ywU msL¨v ¸Y K‡i c‡ii msL¨vwU w`‡q fvM|
(12 Rb 720 UvKv Avq K‡i, Avevi 9 RbI 720 UvKv
Avq K‡i,ZvB 720UvKv wjL‡Z n‡e bv| )

mgvavb: (L)
4
3
Ask = 12 n‡j
4
1
Ask = 4 n‡e| eZ©gv‡b cvwb Av‡Q 4 †KwR| myZivs Wªv‡gi IRb 8-4 = 4 †KwR|
19. GKwU Kv‡Ri
23
1
Ask †kl nq 3 w`‡b| H Kv‡Ri 3 ¸Y KvR Ki‡Z KZw`b mgq jvM‡e? [BADC (AC)-2017]
K.69 L.207 M.138 N.23 O. 175
mgvavb: (L)
23 Zjv wewìs‡qi 1 Zjv evbv‡Z 3 w`b jvM‡j, cyiv wewìs evbv‡Z mgq jvM‡e 233=69w`b Ges GiKg 3wU wewìs
evbv‡Z mgq jvM‡e 693 = 207 w`b|]
20. GKwU ¯‹z‡ji †gvU QvÎQvÎx‡`i A‡a©K dzUej †L‡j Ges Aewkó‡`i GK Z…Zxqvsk †Uwbm †L‡j | Aewkó 300 Rb `ywU
†Ljvi †KvbwUB †L‡j bv| ¯‹z‡j KZRb QvÎ-QvÎx Av‡Q? [BADC (AC)-2017]
K.450 Rb L.600Rb M. 900 Rb N.1200Rb O. †KvbwUB bq
mgvavb: (M)
awi, †gvU QvÎ-QvÎx = K Rb
dzUej †L‡j =
2
K
,Aewkó =
2
K
, †Uwbm †L‡j =
2
K
Gi
3
1
=
6
K
Aewkó =
2
K
-
6
K
=
6
K3K 
=
6
2K
=
3
K
cÖkœg‡Z,
3
K
= 300 K = 900 Rb| DËi: 900 Rb|
gy‡L gy‡L †k‡li w`K †_‡K: 3 fv‡Mi 1 fvM wµ‡KU †Lj‡j wµ‡KU †L‡j bv 3 fv‡Mi 2 fv‡Mi gvb = 300 myZivs
wµ‡KU †L‡j 3 fv‡Mi 1 fvM = 150| GLvb wµ‡KU+‡KvbwUB bv = 150+300 = 450 n‡jv †gv‡Ui A‡a©K myZivs
†gvU QvÎ-QvÎx = 4502 = 900 Rb| ) DËi:
====================================

1. cÖwZwU 3600 UvKv K‡i `ywU †Uwej weµq Kiv nj| GKwU 20% jv‡f Ges Ab¨wU 20% ÿwZ‡Z weµq Kiv nj|
me©‡gvU KZ jvf ev ÿwZ n‡q‡Q? [BADC-(Store Keeper)-2017]
a. 200UvKv jvf b. 300UvKv jvf c. 600UvKv ÿwZ d. 300UvKv ÿwZ Ans: b
Solution:
wjwLZ mgvavb:
jv‡fi †ÿ‡Î ,
3600 UvKvi ‡Pqv‡ii µqg~j¨ n‡e
20% jv‡f 100 UvKvi †Pqv‡ii `vg = 120 UvKv|
GLb 120 UvKv weµqg~j¨ n‡j µqg~j¨ = 100 UvKv |
Avevi 1 ÕÕ ÕÕ ÕÕ ÕÕ =
120
100
Avevi 3600 ÕÕ ÕÕ ÕÕ ÕÕ =
120
3600100  = 3000 UvKv|
Avevi, ÿwZi ‡ÿ‡Î,
20% ÿwZ‡Z 100 UvKvi ‡Pqv‡ii weµqg~j¨ =80 UvKv
GLb weµqg~j¨ 80UvKv n‡j µqg~j¨ = 100UvKv
ÕÕ 1 ÕÕ ÕÕ ÕÕ
80
100
ÕÕ 3600 ÕÕ ÕÕ ÕÕ
80
3600100 
= 4500 UvKv|
GLb †gvU µqg~j¨ = 3000+4500 = 7500 Ges †gvU weµqg~j¨ = 3600+3600 = 7200 myZivs †gvU ÿwZ 7500-
7200 = 300UvKv| A_ev jv‡fi mgq 600 jvf wKš‘ ÿwZi mgq 900 ZvB †gv‡U ÿwZ 900-600=300UvKv
Learning points: GgwmwKD cixÿvi Rb¨ Dc‡ii AsKwU `ªæZ mgvavb Kiv hvq Gfv‡e,
GiKg GKB `v‡g `ywU cY¨ weµq Kivq GKB nv‡i jvf ev ¶wZ n‡j, †gv‡Ui Dci memgq ¶wZ nq|
KZUzKz ¶wZ nq Zv †ei Kivi Rb¨ wb‡Pi m~ÎwU cÖ‡qvM Kiæb:
¶wZi nvi = %
10
nviÿwZievjvfmgvbmgvb
2






20% K‡i jvf- -ÿwZ n‡j †gv‡Ui Dci ÿwZ = %
2
10
20






= 4% (Dc‡ii m~Î cÖ‡qvM K‡i)
GLb 3600 UvKv K‡i 2wU 7200 UvKvq wewµ Kivq 4% ÿwZ n‡j †gvU ÿwZ n‡e 4% Gi mgvb|
96% = 7200 1% =
96
7200
4% = 300
96
47200


UvKv [ †h‡nZz †gvU ÿwZ 4%]
Post Name: Store Keeper Exam Date:11-08-2017
Exam Taker: IBA, DU

2. wZbwU avivevwnK we‡Rvo msL¨vi †hvMdj 177| ga¨g msL¨vwU KZ? [BADC-(Store Keeper)-2017]
a. 47 b. 59 c. 65 d. 69 Ans: b
Solution:
msL¨v wZbwUi Mo = 1773 = 59 B n‡jv ga¨vg msL¨v| KviY msL¨v wZbwU n‡e 58,59 Ges 60| †hLv‡b 59B ga¨g
3. ‡Kvb fMœvskwU ÿz`ªZg? [BADC-(Store Keeper)-2017]
a. 5/6 b. 12/15 c. 11/14 d. 17/21 Ans: c
Solution:
K I L Gi g‡a¨
30
25
6
5
 Ges
30
24
15
12
 myZivs cÖ_g fMœvskwU eo Ges ZvB †mwU ev` w`‡q c‡ii fMœvskwU ivwL|
Avevi, M I N Gi g‡a¨
42
33
14
11
 Ges
42
34
21
17
 GLv‡b Ackb N eo ZvB Zv ev` w`‡q
14
11
ivwL|
GLb:
15
12
Ges
14
11
Gi g‡a¨ cÖ_g fMœvskwU eo|
KviY `ywU fMœvs‡ki gv‡Si cv_©K¨ mgvb n‡j †h fMœvs‡ki ni eo †mwUB eo
nq| Avevi †h‡nZz †k‡li Ack‡bi †k‡li fMœvskwU GKwU AcÖK…Z fMœvsk ZvB DËi
14
11
|
(GLv‡b A‡bK¸‡jv wbq‡gi K_v ejv n‡jI G¸‡jv cÖ_‡g GKevi eyS‡Z n‡e|
A_ev AvovAvwo ¸Y K‡iI mn‡R †ei Kiv hvq)
4. ‡mŠif 8% nvi my‡` †gv‡gb‡K Ges 12% nvi my‡` iæ‡ej‡K mgcwigvY UvKv avi w`j| †m wZb eQi ci `yR‡bi KvQ
†_‡K me©‡gvU 720 UvKv my` †cj| †m cÖ‡Z¨K‡K KZ UvKv avi w`‡qwQj? [BADC-(Store Keeper)-2017]
a. 1200 UvKv b. 1400 UvKv c. 1600 UvKv d. 1800 UvKv Ans: a
Solution:
`yRb‡K 100 UvKv K‡i w`‡j 1 eQi ci my` cvIqv hv‡e 8+12 = 20UvKv
3 eQi ci cvIqv hv‡e 203 =60UvKv|
‡gvU my` 60 UvKv n‡j GKRb‡K w`‡qwQj = 100UvKv
‡gvU my` 720UvKv (60 Gi †_‡K 12¸Y †ewk) n‡j Avmj = 10012 =1200UvKv|
5. wcZv I cy‡Îi eq‡mi AbycvZ 4:1 Ges Zv‡ì eq‡mi ¸Ydj 256| 4 eQi ci Zv‡ì eq‡mi AbycvZ n‡e-[BADC-
(Store Keeper)-2017]
a. 3:1 b. 4:1 c. 5:1 d. 5:2 Ans: a
Solution:
awi, wcZv I cy‡Îi eqm h_vµ‡g, 4K I K eQi|
cÖkœg‡Z
4KK = 256
ev, 4K2
= 256 ev, K2
= 64 K = 8 myZivs Zv‡ì eqm = 48 = 32 eQi Ges 8 eQi|
4 eQi c‡i Zv‡ì eq‡mi AbycvZ n‡e 32+4:8+4 = 36:12 = 3:1
6. GKwU K‡j‡R 70% cwiÿv_©x Bs‡iRx‡ZI 80% cwiÿv_©x evsjvq cvm K‡i‡Q| wKš‘ 10% wkÿv_x© Dfq wel‡q †dj
K‡i‡Q| hw` Dfq wel‡q 300 Rb cwiÿv_©x cvm K‡i _v‡K, Z‡e H K‡j‡R KZ Rb cwiÿv_©x cixÿv w`‡q‡Q?
[BADC-(Store Keeper)-2017]
a. 300 b. 400 c. 500 d. 550 Ans: c

Solution:
cÖ_‡g GKwU wel‡q †d‡ji nvi †ei Kiæb, ‡Kbbv Dfq wel‡q †d‡ji nvi †`qv Av‡Q 10%|
ZvB ïay Bs‡iRx †dj 100-70 = 30% Ges ïay evsjvq †dj = 100-80=20%
‡gvU †dj = 30+20 -10 = 40%, GB 40% wK??? GwU n‡jv †gvU †dj ( nq GK wel‡q A_ev Dfq wel‡q
†dj) Zvn‡j Dfq wel‡q cvm n‡jv 100-40% = 60% | GLb GB 60% = 300 Rb †Kbbv 300 Rb nj
Dfq wel‡q †gvU cvk Kiv wk¶v_©xi msL¨v| GLb 100% Gi gvb n‡e, 300
60
100
= 500 Rb
cv‡ki wPÎwU eyS‡j GK jvB‡b Gfv‡e mgvavb Kiv hvq|
{100% - (30+20-10)} = 60%
GLb 60%= 300 n‡j 100% = 500
7. GKwU _‡j‡Z 25 cqmv, 10 cqmv, I 5 cqmvi gy`ªv 3:4:5 Abycv‡Z Av‡Q| hw` me¸‡jv wgwj‡q 28UvKv nq, Zvn‡j 10
cqmvi gy`ªv KZwU? [BADC-(Store Keeper)-2017]
a. 80wU b. 60wU c. 100wU d. 110wU Ans: a
Solution:
awi,
_‡jwU‡Z 25 cqmvi gy`ªv Av‡Q 3K, 10 cqmvi gy`ªv Av‡Q 4K Ges 5 cqmvi gy`ªv Av‡Q 5KwU|
cÖkœg‡Z,
253K + 104K + 55K = 28100 [me¸‡jv †h‡nZz cqmv ZvB 28 UvKv‡K cqmv evbv‡bv n‡jv|]
75K+40K+25K = 2800
140K = 2800
K = 20
GLb 10 cqmvi gy`ªv Av‡Q = 420 = 80wU|
8. `yB AsK wewkó GKwU msL¨v, msL¨vwUi AsKØ‡qi †hvMd‡ji 4 ¸Y| msL¨vwUi mv‡_ 27 †hvM Ki‡j AsKØq ¯’vb cwieZ©b
K‡i| msL¨vwU KZ? [BADC-(Store Keeper)-2017]
a. 27 b. 36 c. 39 d. 45 Ans: b
Solution:
G ai‡Yi cÖkœ x a‡i mgvavb Kiv hvq| wKš‘ Zv‡Z A‡bK mgq jv‡M|
ZvB †h †Kvb wcÖwji mnR AsK¸‡jv Ackb a‡i mgvavb KivB DËg|
GLv‡b cÖ`Ë Ackb¸‡jvi g‡a¨ ïaygvÎ L †Z †`qv 36 Gi 3+6 = 9 Ges 36 n‡jv 9 Gi 4¸Y| GLb Ab¨¸‡jv a‡i bv
†f‡e GUv a‡i B wØZxq kZ© wgwj‡q †bB 36 Gi mv‡_ 27 †hvM Ki‡j 63 nq hv‡Z msL¨vwUi AsKØq ¯’vb wewbgq K‡i|
ZvB DËi: 36|
9. GKwU evm cÖwZw`b GKwU wbw`©ó MwZ‡Z 60 gvBj c_ AwZµg K‡i| GKw`b hvwš¿K ÎæwUi Kvi‡Y ev‡mi MwZ 10
gvBj/N›Uv K‡g hvq Ges H c_ AwZµg Ki‡Z 3 N›Uv †ewk jv‡M| evmwUi ¯^vfvweK MwZ KZ? [BADC-(Store
Keeper)-2017]
a. 20gvBj/N›Uv b. 25gvBj/N›Uv c. 32gvBj/N›Uv d. 45gvBj/N›Uv Ans:a
gy‡L gy‡L:
Ackb ‡_‡K 10 cqmvi gy`ªvi AbycvZ 4 Ges 4 w`‡q wefvR¨
msL¨vi g‡a¨ 80 wb‡jB 80wU 10 cqmvi gy`ªv 8UvKv Ges
Ab¨ `ywU wg‡j 20 UvKv n‡j †gvU 28UvKv wg‡j hvq|
40%
60%
30-10 =20% 10%
BE
20-10 =10%
Total =100%

Solution:
awi, evmwUi MwZ‡eM = K
cÖkœg‡Z,
K
60
-
10K
60

= 3 [Av‡Mi mgq - MwZ Kgvi c‡ii mgq = 3 N›Uv]
mgvavb K‡i cvIqv hvq, K = 20 gvBj|
10. wZbwU msL¨vi †hvMdj 110| cÖ_g msL¨vwU wØZxq msL¨vi wØ¸Y Ges Z…Zxq msL¨vwU cÖ_g msL¨vi GK Z…Zxqvsk| wØZxq
msL¨vwU KZ? [BADC-(Store Keeper)-2017]
a. 30 b. 40 c. 50 d. 60 Ans: a
Solution:
awi, Z…Zxq msL¨vwU = 2K (Gfv‡e ai‡j fMœvsk Avm‡e bv)
myZivs cÖ_g msL¨vwU = 6K
Ges wØZxq msL¨vwU = 3K
cÖkœg‡Z,
6K+3K+2K = 110
ev, 11K = 110  K = 10 myZivs wØZxq msL¨vwU = 30
11. `ywU msL¨vi e‡M©i †hvMdj 68 Ges G‡ì cv_©‡Kï eM© 36| msL¨vØ‡qi ¸Ydj KZ? [BADC-(Store Keeper)-
2017]
a. 16 b. 20 c. 24 d. 28 Ans: „
Solution:
awi, GKwU msL¨v = x Ges Aci msL¨vwU = y
cÖkœg‡Z,
x2
+y2
= 68 Ges (x-y)2
= 36
Avgiv Rvwb,
x2
+y2
= (x-y)2
+2xy ev, 68 = 36+2xy
ev, 68 - 36 = 2xy ev, 32 = 2xy xy = 16
12. ‡Kvb msL¨v H msL¨vi 25% A‡cÿv 60 †ewk| msL¨vwU KZ? [BADC-(Store Keeper)-2017]
a. 72 b. 80 c. 100 d. 120 Ans: b
Solution:
msL¨vwU 100% Ges Zv †_‡K 25% we‡qvM Ki‡j _v‡K 75% Avi GB 75% ev 4 fv‡Mi 3 fvM = 60 n‡j 1 fvM n‡e
20 Ges m¤ú~Y© msL¨vwU n‡e 4 fvM A_©vr 204 = 80|
13. 9wU msL¨vi Mo 12| Gi g‡a¨ cÖ_g 7wU msL¨vi Mo 10| evKx msL¨v `yBwUi Mo KZ? [BADC-(Store Keeper)-
2017]
a. 17 b. 18 c. 19 d. 20 Ans: c
Solution:
9wU msL¨vi mgwó = 912 = 108
7wUi mgwó = 710 = 70
gy‡L gy‡L:
`ywU msL¨vi e‡M©i †hvMdj 68 †`‡L ‡evSv Svq 64+4 = 68
A_©vr 82
+22
= 68 nq Ges G‡ì e‡M©i cv_©K¨ (8-2)2
=
36 nq| Zvn‡j ¸Ydj n‡e 82 = 16
gy‡L gy‡L: 7wU msL¨vq K‡g †Mj 72 = 14
GLb GB 14 Ab¨ `ywU msL¨vq fvM K‡i w`‡j n‡e 12+7 = 19
gy‡L gy‡L:
Ack‡bi g‡a¨ ïay 20 w`‡qB 60 ‡K fvM Kiv hvq|
Zvn‡j 60gvBj †h‡Z cÖ_‡g mgq jv‡M 6020 = 3 N›Uv
Ges MwZ‡eM 10 gvBj Kg‡j mgq jv‡M 6010 = 6 N›Uv|
hv Av‡Mi mg‡qi †_‡K 3 N›Uv †ewk|

evKx `ywUi mgwó = 108-70 = 38
myZivs Mo = 382 = 19
14. GKwU mgevû wÎfz‡Ri cÖwZwU evûi ˆ`N©¨ 2 wgUvi eov‡j †ÿÎdj 3 3 eM© wgUvi †e‡o hvq| mgevû wÎfzRwUi cÖwZwU
evûi ˆ`N©¨ KZ wgUvi? [BADC-(Store Keeper)-2017]
a. 1 b. 2 c. 3 d. 4 Ans: b
Solution:
mgevû wÎfz‡Ri †ÿÎdj =
4
3
a2
,
Avevi mgevû wÎfz‡Ri bZzb evûi ˆ`N©¨ = a + 2 wgUvi myZivs mgevû wÎfz‡Ri bZzb †ÿÎdj =
4
3
(a + 2)2
kZ©g‡Z,
ev,
4
3
(a + 2)2
-
4
3
a2
= 3
ev,
4
3
(a2
+ 4a + 4 - a2
) = 3
ev,
4
3
(4a + 4) = 3 ev,
4
3
4 (a + 1) = 3 ev, a + 1 = 3  a = 2
15. ‰`wbK 7 N›Uv KvR K‡i 16 Rb †jvK 60 w`‡b GKwU `vjvb ˆZix Ki‡Z cv‡i| ˆ`wbK 15 N›Uv KvR K‡i 14 Rb ‡jvK
H `vjvb KZ w`‡b ˆZix Ki‡Z cvi‡e? [BADC-(Store Keeper)-2017]
a. 24 b. 30 c. 32 d. 42 Ans: c
Solution:
7 N›Uv KvR K‡i 16 Rb †jvK `vjvbwU ‰Zix K‡i = 60 w`‡b
1 N›Uv KvR K‡i 1 Rb †jvK `vjvbwU ‰Zix K‡i =60716 (Kg N›Uv Ki‡j †ewk w`b +Kg ‡jvK Ki‡j Av‡iv
†ewk )
15 N›Uv KvR K‡i 14 Rb †jvK `vjvbwU ‰Zix K‡i =
1415
16760


(†ewk N›Uv I †ewk †jvK KvR Ki‡j Kg w`b
jvM‡e)
= 32 w`b|
16. GKwU QvÎvev‡m 200 Rb Qv‡Îi 120 w`‡bi Lvevi Av‡Q| 30 w`b ci 20 Rb QvÎ QvÎvevm †Q‡o P‡j †Mj| Aewkó
Lv`¨ QvÎ‡`i KZw`b Pj‡e? [BADC-(Store Keeper)-2017]
a. 90 w`b b. 95 w`b c. 100 w`b d. 120 w`b Ans: c
Solution:
‡gvU w`b = 120-30 = 90 Ges †gvU QvÎ = 200-20 = 180
200 R‡bi Lvevi Pj‡e = 90 w`b| 1 R‡bi Pj‡e = 90200 w`b|
myZivs 180 R‡bi Pj‡e
180
20090
=100 w`b|
3
3
3 3

Iba mcq math solution( 2017 2019) by khairul alam [www.itmona.com]

Iba mcq math solution( 2017 2019) by khairul alam [www.itmona.com]

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Iba mcq math solution( 2017 2019) by khairul alam [www.itmona.com]