cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018 cÖwZiÿv gš¿Yvjq · 2020-04-04 · cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018 Solved by: Ajgar Ali Spotlight Publication 1g bj Øviv

cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018

Solved by: Ajgar Ali Spotlight Publication

cÖwZiÿv gš¿Yvjq c‡ì bvg: mnKvix cwiPvjK; cixÿvi ZvwiL: 02.02.2018

Exam Taker: Department of Management Information Systems, DU

1. e„‡Ëi e¨vm wZb¸Y e„w× †c‡j Gi †ÿÎdj KZ¸Y e„w× cv‡e? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 3 ¸Y (L) 6 ¸Y (M) 9 ¸Y (N) 12 ¸Y mgvavb :

e„‡Ëi e¨vm = 2r Ges e„‡Ëi e¨vmva© = r

e„‡Ëi †ÿÎdj = r2

ewa©Z e¨vm = 6r Ges ewa©Z e¨vmva© = 3r

ewa©Z †ÿÎdj = (3r)2 = 9r2 = 9 r2 myZivs e„‡Ëi †ÿÎdj 9 ¸Y e„w× cv‡e| DËi: M kU©KvU: (hZ¸Y)2 = (3)2 = 9 ¸Y

2. GKwU †PŠev”Pvq 19200 wjUvi cvwb a‡i| Gi MfxiZv 2.56 wgUvi Ges cÖ¯’ 2.5 wgUvi n‡j, ˆ`N©¨ KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 4 wgUvi (L) 6 wgUvi (M) 3 wgUvi (N) 14 wgUvi

mgvavb : ‡PŠev”PvwUi Zjvi ‡¶Îdj = 2.56 wgUvi 2.5 wgUvi = 256 ‡m.wg. 250 ‡m.wg. = 64000 eM© ‡m.wg.

‡PŠev”Pvq 19200 wjUvi ev (19200 1000) Nb ‡m.wg. cvwb a‡i| [ 1000 Nb ‡m.wg. = 1 wjUvi] AZGe, ‡PŠev”PvwUi AvqZb 19200000 Nb ‡m.wg.

‡PŠev”PvwUi MfxiZv = (19200000 64000) ‡m.wg. = 300 ‡m.wg. ev 3 wgUvi| DËi: M

3. 10 eQi c~‡e© wcZv-cy‡Îi eq‡mi AbycvZ wQj 4:1| 10 eQi ci Zv‡ì eq‡mi AbycvZ n‡e 2:1| Zv‡ì eZ©gvb eqm KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 50 I 20 eQi (L) 47 I 20 eQi (M) 54 I 21 eQi (N) 36 I 25 eQi

mgvavb : g‡bKwi, 10 eQi c~‡e© wcZv I cy‡Îi eqm wQj h_vμ‡g 4K eQi Ges K eQi| cÖkœg‡Z,

12

1010K10104K

ev, 4K + 20 = 2K + 40 ev, 2K = 20

K = 10

myZivs wcZvi eZ©gvb eqm (410+10) = 50 eQi Ges cy‡Îi eZ©gvb eqm (10+10) = 20 eQi DËi: K

cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018


4. wb‡Pi fMœvsk¸‡jvi g‡a¨ †KvbwU e„nËg? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 65

(L) 43

(M) 10075

(N) 118

mgvavb :

Ackb (M) †Z 10075

= 43

hv Ackb (L) Gi mgvb| ZvB Ackb (L) I (M) ev` w`‡q evwK `y‡Uv fMœvs‡ki g‡a¨ Zzjbv Ki‡Z

n‡e| GLb (K) Ges (N) Ack‡bi fMœvsk y‡Uvi ni mgvb Ki‡j Zzjbv Ki‡Z mnR n‡e| A_©vr 65,

118

†K Gfv‡e †jLv hvq: 6655

, 6648

| GLv‡b ¯úóZ 3655

fMœvs‡ki je eo| Avi y‡Uv fMœvs‡ki ni mgvb n‡j hvi je eo, †mUvB me‡P‡q eo fMœvsk| ZvB mwVK

DËi Ackb (K)| DËi: K

5. evwl©K 4.5 UvKv nvi gybvdvq KZ UvKv wewb‡qv‡M 4 eQ‡i Zv 826 UvKv n‡e? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 725 UvKv (L) 700 UvKv (M) 702 UvKv (N) 726 UvKv

mgvavb : 4 eQ‡ii my` = 44.5 = 18 UvKv my vmj = 100+18 = 118 UvKv my vmj 118 UvKv n‡j Avmj 100 UvKv

Ó 1 Ó Ó Ó 118100

Ó

Ó 826 Ó Ó Ó 118

826100 Ó = 700 UvKv DËi: L

6. GKwU evBmvB‡K‡ji g~j¨ 10,000 UvKv| Dnv 10% evÆvq μq Kiv nj| wZbgvm eënv‡ii ci μqg~‡jï Dci 15% evÆvq wewμ

Ki‡j weμqg~j¨ KZ wQj? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 8500 UvKv (L) 7650 UvKv (M) 7500 UvKv (N) 8000 UvKv

mgvavb : 10% evÆvq μqg~j¨ = 10000 1000010% = 10000 1000 = 9000 UvKv

15% evÆvq weμqg~j¨ = 9000 900015% = 9000 1350 = 7650 UvKv DËi: L

7. GKwU bj 12 wgwb‡U GKwU Lvwj †PŠev”Pv c~Y© K‡i| Aci GKwU bj cÖwZ wgwb‡U 14 wjUvi cvwb †ei K‡i †`q| †PŠev”PvwU Lvwj _vKv Ae ’vq bj ywU Ly‡j w`‡j 96 wgwb‡U Dnv c~Y© nq| †PŠev”PvwU‡Z KZ wjUvi cvwb a‡i? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 145 wjUvi (L) 155 wjUvi (M) 208 wjUvi (N) 192 wjUvi

mgvavb :

kU©KvU: 1412961296

= 1484

1296

= 14

796

= 192 wjUvi DËi: N

we¯ÍvwiZ mgvavb ‡`Lyb cieZ©x c„ôvq



1g bj Øviv 1 wgwb‡U c~Y© nq = 121

Ask

Avevi ywU bj Øviv 1 wgwb‡U c~Y© nq = 961

Ask

1 wgwb‡U 2q bj Øviv Lvwj nq = 961

121 =

961-8

= 967

Ask

2q bj Øviv 967

Ask Lvwj nq 1 wgwb‡U

2q bj Øviv 1 Ask Lvwj nq 796

wgwb‡U

2q bj Øviv 1 wgwb‡U cvwb †ei nq = 14 wjUvi

2q bj Øviv 796

wgwb‡U cvwb †ei nq =

14

796 = 192 wjUvi

myZivs †PŠev”PvwU‡Z 192 wjUvi cvwb a‡i| DËi: N

8. GKwU eB‡qi ˆ`N©¨ 25 †m.wg. Ges cÖ¯’ 18 †m.wg.| eBwUi c„ôv msL¨v 200 Ges cÖwZ cvZvi cyiZ¡ 0.1 wg.wg. n‡j, eBwUi AvqZb KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 455 Nb †m.wg. (L) 450 Nb †m.wg. (M) 440 Nb †m.wg. (N) 250 Nb †m.wg.

mgvavb : eBwUi AvqZb ej‡Z ˆ`N©¨, cÖ¯’ I D”PZv †ei Ki‡Z ejv n‡q‡Q| eB‡qi ˆ`N©¨ = 25 †m.wg.; cÖ¯’ = 18 †m.wg. Ges c„ôv msL¨v = 200

cvZvi msL¨v = 2002 = 100

cÖwZ cvZvi cyiæZ¡ = 0.1 wg.wg. = (0.110) †m.wg. = 0.01 †m.wg.

eBwUi AvqZb = (25 18 100 0.01) Nb †m.wg. = 450 Nb †m.wg. DËi: L Note: eB‡qi cÖwZ cvZvq 2wU c„ôv _v‡K| A_©vr 2 c„ôv wg‡j 1 cvZv nq|

9. GKwU LvZv 36 UvKvq wewμ Ki‡j hZ ÿwZ nq, 72 UvKv wewμ Ki‡j Zvi wØ¸Y jvf nq| LvZvwUi μqg~j¨ KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 45 UvKv (L) 44 UvKv (M) 42 UvKv (N) 48 UvKv

mgvavb : awi, ÿwZ = K UvKv

jvf = 2K UvKv

cv_©K¨ = 2K ( K) = 2K + K = 3K cÖkœg‡Z,

3K = 72 36 ev, 3K = 36

K = 12 myZivs LvZvwUi μqg~j¨ = weμqg~j¨ + ÿwZ = 36 + 12 = 48 UvKv DËi: N



10. ywU msL¨vi j.mv.¸ 60 Ges M.mv.¸ 10| GKwU msL¨v Aci msL¨vi yB-Z…Zxqvsk n‡j †QvU msL¨vwU KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 20 (L) 30 (M) 10 (N) 40

mgvavb : awi, eo msL¨vwU = 3K

†QvU msL¨vwU = 3K32 = 2K

Avgiv Rvwb,

ywU msL¨vi ¸Ydj = j.mv.¸ M.mv.¸

ev, 3K 2K = 6010 ev, 6K2 = 600 ev, K2 = 100

K = 10

myZivs †QvU msL¨vwU (210) = 20 DËi: K

11. GK bwUK¨vj gvB‡j KZ wgUvi? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 1750.18 wg. (L) 1853.18 wg. (M) 1650.20 wg. (N) 1953.18 wg.

mgvavb : 1 bwUK¨vj gvBj = 1853.18 wg. DËi: L

12. 3a

1a n‡j

33

a

1a Gi gvb KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 9 (L) 18 (M) 27 (N) 36 mgvavb :

‡Ìqv Av‡Q, 3a

1a

33

)3(a

1a

[Nb K‡i]

27a

1a

a

1.a.3

a

1a 3

3

33a

1a

33 = 27

33

a

1a = 27 + 9

36a

1a

33 DËi: N



13. 015x 2 x n‡j 2

2

x

1x Gi gvb KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 1 (L) 5 (M) 3 (N) 52 mgvavb :

‡Ìqv Av‡Q, 1x5x2 = 0

x51x2

x

x5

x

1

x

x2

[Dfq cv‡k x Øviv fvM K‡i]

5x

1x

22

5x

1x

[Dfq cv‡k eM© K‡i]

5x

1.x.4

x

1x

2

45x

1x

2

1x

1x

22

x

1x =

x

1x

x

1x = 51 = 5 DËi: L

14. GKwU cÖK…Z fMœvs‡ki ni, je A‡cÿv 4 †ewk| fMœvskwU eM© Ki‡j †h fMœvsk cvIqv hvq Zvi ni, je A‡cÿv 40 †ewk| fMœvskwU

KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 117

(L) 73

(M) 139

(N) 51

mgvavb : awi, fMœvskwUi je = x

ni = x + 4

fMœvskwU = 4x

x

2

2

)4x(

x

[eM© K‡i]

cÖkœg‡Z, (x + 4)2 x2 = 40

x2 + 8x + 16 – x2 = 40

8x = 40 – 16 8x = 24 x = 3

fMœvskwU = 43

3

=

7

3 DËi: L



15. a I b ywU abvZ¥K msL¨v Ges ax

ab n‡j

b

x KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) a (L) ab (M) b

a (N) a

b

mgvavb :

‡Ìqv Av‡Q, ax

ab x =

a

ab

b

x

ba

ab

= b

1

a

ab =

b

1

a

bbaa

= ab DËi: L

16. 3x3x n‡j x = KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 3 (L) – 3 (M) 0 (N) 3 mgvavb :

GB AsKUv Ackb †U÷ K‡i mn‡R Kiv hvq| Ackb (M) †Z †`Lyb| x = 0 n‡j, 3x = 30 = 3

Ges 3x = 30 = 3

myZivs x = 0 n‡j, 3x3x nq| ZvB Ackb (M) B mwVK| DËi: M

17. GKwU mij‡iLvi Dci Aw¼Z e‡M©i †ÿÎdj H mij‡iLvi GK-PZz_©vs‡ki Dci Aw¼Z e‡M©i †ÿÎd‡ji KZ ¸Y? [cÖwZiÿv

gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 16 (L) 4 (M) 8 (N) 2

mgvavb : awi, m¤ú~Y© mij‡iLv = 4x

GK-PZz_v©sk = 4

x4= x

GK-PZz_©vs‡ki Dci Aw¼Z e‡M©i †ÿÎdj = x2

m¤ú~Y© mij‡iLvi Dci Aw¼Z e‡M©i †ÿÎdj = (4x)2 = 16x2 ev, 16 GK-PZz_©vs‡ki Dci Aw¼Z e‡M©i †ÿÎdj A_©vr GKwU mij‡iLvi Dci Aw¼Z e‡M©i †ÿÎdj H mij‡iLvi GK-PZz_©vs‡ki Dci Aw¼Z e‡M©i †ÿÎd‡ji 16 ¸Y| DËi: K

18. GKwU Mvoxi PvKv cÖwZ wgwb‡U 90 evi Nyi‡j 1.5 †m‡K‡Û PvKvwU KZ wWMÖx Ny‡i? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018] (K) 1800 (L) 2700 (M) 3600 (N) 5400

mgvavb : Mvoxi PvKv e„ËvKvi| ZvB PvKvwU 1 evi Nyi‡j hvq 360 wWwMÖ| Zvn‡j 90 evi Nyi‡j hv‡e (90360) wWwMÖ|

60 †m‡K‡Û PvKvwU †Nv‡i (90360) wWwMÖ

1.5 †m‡K‡Û PvKvwU †Nv‡i

601.536090 wWwMÖ ev 810 wWwMÖ (Ack‡b mwVK DËi †bB)

x x x x



19. 8log 2 = KZ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 1 (L) 2 (M) 3 (N) 4 mgvavb :

8log 2 = 3

22log = 2log3 2 = 31 = 3 DËi: M

20. 324 1x n‡j x = ? [cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018]

(K) 3

2 (L)

5

3 (M)

8

1 (N)

2

3

mgvavb :

324 1x

324.4 1x

4

324x

84x

3x2 22

2x = 3

x = 2

3 DËi: N



cÖwZiÿv gš¿Yvjq c‡ì bvg: Dc-mnKvix cwiPvjK; cixÿvi ZvwiL: 02.02.2018


1. e„‡Ëi e¨vm wØ¸Y e„w× †c‡j Gi cwiwa KZ¸Y e„w× cvq? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 3 ¸Y (L) 6 ¸Y (M) 4 ¸Y (N) 12 ¸Y

mgvavb : e„‡Ëi e¨vm = 2r e„‡Ëi e¨vmva© = r e„‡Ëi †ÿÎdj = r2

bZzb e¨vm = 4r bZzb e¨vmva© = 2r bZzb †ÿÎdj = (2r)2 = 4r2 = 4 r2 myZivs e„‡Ëi †ÿÎdj 4 ¸Y e„w× cv‡e| DËi: M kU©KvU: (hZ¸Y)2 = (2)2 = 4 ¸Y

2. KZ eM©dzU GK kZvsk Rwgi mgvb? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 430.5 eM©dzU (L) 435.6 eM©dzU (M) 434.6 eM©dzU (N) 436 eM©dzU mgvavb :

1 kZvsk = 436 eM©dzU DËi: N

3. ‡Kvb wbw`©ó mg‡qi gybvdv Avm‡j 5600 UvKv Ges gybvdv Avm‡ji 40 kZvsk| gybvdvi evwl©K nvi 8% n‡j, mgq wbY©q Kiæb| [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 7 eQi (L) 6 eQi (M) 5 eQi (N) 4 eQi

mgvavb : g‡bKwi, Avmj 100 UvKv| Zvn‡j gybvdv n‡e 40 UvKv| GLb 8 UvKv gybvdv nq 1 eQ‡i| Zvn‡j 40 UvKv gybvdv nq (408) = 5 eQ‡i| DËi: M

4. wb‡Pi †Kvb msL¨vwU me‡P‡q eo? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 0.2 (L) 0.2 (M) 0.3 (N) 0.3 mgvavb :

Avgiv Rvwb, `kwgK msL¨vi eM©g~j Ki‡j msL¨vwU Av‡iv eo nq| GLv‡b 0.3, 0.2 Gi †P‡q eo| Zvn‡j 0.3 Gi eM©g~j Ki‡j

Zvi gvb wbðqB 0.2 Gi eM©g~‡ji †P‡q eo n‡e| myZivs cÖ`Ë Ackb¸‡jvi g‡a¨ 0.3 n‡jv me‡P‡q eo msL¨v| DËi: M

5. UvKvq 3wU wRwbm μq K‡i UvKvq 2wU K‡i weμq Ki‡j kZKiv KZ jvf n‡e? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 30% (L) 15% (M) 50% (N) 30% mgvavb :

G RvZxq AsK m~‡Îi mvnv‡h¨ `ªæZ mjf Ki‡Z cv‡ib|

m~Î = 100msL¨v weμqK¨cv_©

= 1002

2-3 = 100

21 = 50%

gy‡L gy‡L mgvavb: 2wU †Z jvf nq 1wU A_©vr 50%| DËi: M



6. 1wU eM©vKvi Rwgi f‚wgi ˆ`N© 20% evov‡j Ges cÖ ’ 10% Kgv‡j †ÿÎdj KZUzKz n«vm ev e„w× cv‡e? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 2% e„w× (L) 3% e„w× (M) 10% e„w× (N) 8% e„w×

mgvavb :

100

ABBA =

100

)10(201020

= 8% DËi: N

7. GKwU bj 12 wgwb‡U GKwU Lvwj †PŠev”Pv c~Y© K‡i| Aci GKwU bj cÖwZ wgwb‡U 15 wjUvi cvwb †ei K‡i †`q| †PŠev”PvwU Lvwj

_vKv Ae ’vq bj ywU Ly‡j w`‡j 48 wgwb‡U Dnv c~Y© nq| †PŠev”PvwU‡Z KZ wjUvi cvwb a‡i? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 220 wjUvi (L) 230 wjUvi (M) 240 wjUvi (N) 250 wjUvi

mgvavb :

kU©KvU: 1512481248

= 1536

1248

= 1615 = 240 wjUvi DËi: M

we¯ÍvwiZ mgvav‡bi Rb¨ ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 7 bs cÖkœ †`Lyb| 8. hw` GKwU KvR 9 Rb †jvK 12 w`‡b Ki‡Z cv‡i, AwZwi³ 3 Rb †jvK wb‡qvM Ki‡j KvRwU KZw`‡b †kl n‡e? [cÖwZiÿv gš¿Yvj‡qi

Dc-mnKvix cwiPvjK-2018] (K) 7 w`b (L) 9 w`b (M) 8 w`b (N) 10 w`b

mgvavb : 9 Rb †jv‡K KvRwU K‡i 12 w`‡b

12 Rb †jv‡K KvRwU K‡i

12912 w`‡b = 9 w`‡b DËi: L

9. GKwU LvZv 36 UvKvq wewμ Ki‡j hZ ÿwZ nq 72 UvKvq wewμ Ki‡j Zvi wØ¸Y jvf nq| LvZvwUi μqg~j¨ KZ? [cÖwZiÿv

gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 45 UvKv (L) 44 UvKv (M) 42 UvKv (N) 48 UvKv

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 9 bs cÖkœ †`Lyb| DËi: N

10. ywU msL¨vi j.mv.¸ 60 Ges M.mv.¸ 10| GKwU msL¨v Aci msL¨vi `yB-Z…Zxqvsk n‡j ‡QvU msL¨vwU KZ? [cÖwZiÿv gš¿Yvj‡qi

Dc-mnKvix cwiPvjK-2018] (K) 20 (L) 30 (M) 10 (N) 40

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 10 bs cÖkœ †`Lyb| DËi: K

11. log327 = KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 1 (L) 2 (M) 3 (N) 4

mgvavb : log327 = log333 = 3log33 = 31 = 3 DËi: M



12. wb‡Pi †KvbwU g~j` msL¨v? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 9 (L) 11 (M) 3 (N) 2

mgvavb : ‡h me ¯^vfvweK msL¨vi eM©g~j Kiv hvq †m¸‡jv g~j` msL¨v| cÖ`Ë Ackb¸‡jvi g‡a¨ ïay 9 Gi eM©g~j Kiv hvq ( 9 = 3)|

ZvB 9 g~j` msL¨v| DËi: K

13. 3a

1a n‡j

33

a

1a Gi gvb KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 9 (L) 18 (M) 27 (N) 36


14. 01x5x 2 n‡j 2

2

x

1x Gi gvb KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 1 (L) 5 (M) 3 (N) 52

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 13 bs cÖkœ †`Lyb| DËi: L

15. GKwU cÖK…Z fMœvs‡ki ni, je A‡cÿv 2 †ewk| fMœvskwU eM© Ki‡j †h fMœvsk cvIqv hv‡e Zvi ni, je A‡cÿv 40 †ewk| fMœvskwU KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 97

(L) 53

(M) 119 (N)

1513

mgvavb : awi, fMœvskwUi je = x

ni = x + 2

fMœvskwU = 2x

x

2

2

)2x(

x

[eM© K‡i]

cÖkœg‡Z,

(x + 2)2 x2 = 40 x2 + 4x + 4 – x2 = 40 4x = 36 x = 9

fMœvskwU = 11

9

29

9

DËi: M

16. a I b ywU abvZ¥K msL¨v Ges ax

ab n‡j x = KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) a (L) ab (M) b

a (N) a

b

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 15 bs cÖkœ †`Lyb| DËi: L



17. 3x3x n‡j x = KZ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 3 (L) 3 (M) 0 (N) 3

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 16 bs cÖkœ †`Lyb| DËi: M

18. GKwU mij‡iLvi Dci Aw¼Z e‡M©i †ÿÎdj H mij‡iLvi GK-PZz_©vs‡ki Dci Aw¼Z e‡M©i †ÿÎd‡ji KZ ¸Y? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 16 (L) 4 (M) 8 (N) 2

mgvavb : ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 17 bs cÖkœ †`Lyb| DËi: K

19. GKwU Mvoxi PvKv cÖwZ wgwb‡U 90 evi Nyi‡j 1.0 †m‡K‡Û PvKvwU KZ wWwMÖ Ny‡i? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018] (K) 1800 (L) 2700 (M) 3600 (N) 5400

mgvavb : 60 †m‡K‡Û PvKvwU †Nv‡i (90360) wWwMÖ

1.0 †m‡K‡Û PvKvwU †Nv‡i

6036090 wWwMÖ ev 540 wWwMÖ DËi: N

20. 324 1x n‡j x = ? [cÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018]

(K) 3

2 (L)

5

3 (M)

8

1 (N)

2

3


ICB Cashier-2018


Investment Corp. Bangladesh (ICB) Post Name: Cashier; Exam Date: 16.03.2018


1| 0.30.6 = KZ? [ICB Cashier-2018] (K) 1.8 (L) 0.18 (M) 0.018 (N) 0.0018

mgvavb : 0.30.6 = 0.18 DËi: L

2| Zij c`v‡_©i AvqZb cwigv‡ci GKK ‡KvbwU? [ICB Cashier-2018] (K) MÖvg (L) dzU (M) wjUvi (N) Gqi

mgvavb : DËi: M Zij c`v‡_©i AvqZb cwigv‡ci g~j GKK wjUvi| IRb cwigv‡ci g~j GKK MÖvg| ˆ`N©¨ cwigv‡ci g~j GKK wgUvi|

3| dvinv‡bi Rb¥w`b 29 †deªæqvwi| Zvi Rb¥MÖn‡Yi mvjwU n‡Z cv‡i? [ICB Cashier-2018]

(K) 1900 (L) 2002 (M) 2008 (N) 2010 mgvavb :

wjcBqvi ev Awae‡l© †deªæqvwi gvm 29 w`‡b Ges dvêyb gvm 31 w`‡b nq| ZvB dvinv‡bi Rb¥mvjwU Aek¨B wjcBqvi n‡e| cÖ`Ë Ackb¸‡jvi g‡a¨ ïay 2008 mvjwU wjcBqvi, hv 4 Øviv wbt‡k‡l wefvR¨| D‡jøL¨, 1900 mvj wjcBqvi n‡Z n‡j 400 Øviv wbt‡k‡l wefvR¨ n‡Z n‡e (kZvãxi †ÿ‡Î); 4 Øviv bq| ZvB mwVK DËi Ackb (M)| DËi: M

4| 1 n‡Z KZ we‡qvM Ki‡j we‡qvMdj 0 (k~b¨) n‡e? [ICB Cashier-2018] (K) 2 (L) -1 (M) 1 (N) 4

mgvavb : 1 (1) = 0 DËi: L

5| x + y = x – y n‡j, Gi gvb wb‡Pi †KvbwU? [ICB Cashier-2018] (K) -1 (L) 0 (M) 1 (N) 2

mgvavb : ‡Ìqv Av‡Q, x + y = x – y

Or, 2y = 0 y = 0 DËi: L

6| 1 wewjqb = KZ †KvwU? [ICB Cashier-2018] (K) 10 (L) 100 (M) 1000 (N) 10000

mgvavb : 100 jv‡L 1 †KvwU| 100 ‡KvwU‡Z 1 wewjqb| DËi: L

ICB Cashier-2018


7| GKwU Kjg 11 UvKvq weμq Ki‡j 10% jvf nq| KjgwUi μqg~j¨ KZ? [ICB Cashier-2018] (K) 9 UvKv (L) 10 UvKv (M) 11 UvKv (N) 12 UvKv

mgvavb :

KjgwUi μqg~j¨ = 11110100

= 10 UvKv DËi: L

8| hw` x = – 3 Ges y = 2 n‡j, xy2 = KZ? [ICB Cashier-2018]

(K) 36 (L) -36 (M) -12 (N) 12 mgvavb :

xy2 = – 3.(2)2 = – 12 DËi: M

9| x2 – x – 12 = 0 mgxKi‡Yi g~jØq wb‡Pi †KvbwU? [ICB Cashier-2018]

(K) 3, 4 (L) 3, -4 (M) -3, 4 (N) -3, -4 mgvavb :

‡Kvb mgxKi‡Yi g~j †ei Kiv ej‡Z mgxKiY mgvavb Kiv‡K eySvq| x2 – x – 12 = 0 Or, x2 – 4x + 3x – 12 = 0 Or, x(x – 4) + 3(x – 4) = 0 Or, (x – 4) (x + 3) = 0

x = (– 3, 4) DËi: M 10| ‡KvbwU wÎfy‡Ri †ÿÎdj? [ICB Cashier-2018]

(K) D”PZvf‚wg21

(L) D”PZvf‚wg (M) f‚wg^j¤ (N) D”PZv^j¤21

mgvavb :

wÎfz‡Ri †ÿÎdj = D”PZvf‚wg21

DËi: K

11| x + y = 3, xy = 2 n‡j, x3 + y3 Gi gvb KZ? [ICB Cashier-2018]

(K) 9 (L) 18 (M) 19 (N) 27 mgvavb :

x3 + y3 = (x + y)3 – 3xy(x + y) = (3)3 – 3.2.3 = 27 – 18 = 9 DËi: K

12| 9x2 – 16y Gi Drcv`‡K we‡kølY wb‡Pi †KvbwU? [ICB Cashier-2018]

(K) (3x+4y)(3x-4y) (L) (4y-3x)(4y+3x)

(M) (-4x-3x)(4y-3x) (N) (3x+4y)(-4y-3x) mgvavb :

cÖ‡kœ fzj Av‡Q| 16y Gi cwie‡Z© 16y2 n‡j, DËi n‡e Ackb (K)| 9x2 – 16y2 = (3x)2 – (4y)2 = (3x + 4y)(3x – 4y)

ICB Cashier-2018


13| Avwid I AvwK‡ei eq‡mi AbycvZ 5 : 3| Avwi‡di eqm 20 eQi n‡j, KZ eQi ci Zv‡ì eq‡mi AbycvZ 7 : 5 n‡e? [ICB

Cashier-2018] (K) 5 eQi (L) 6 eQi (M) 8 eQi (N) 10 eQi

mgvavb : Avwid I AvwK‡ei eq‡mi AbycvZ 5 : 3 = (54) : (34) = 20 : 12 g‡bKwi, ÔKÕ eQi c‡i Zv‡ì eq‡mi AbycvZ 7 : 5 n‡e| cÖkœg‡Z,

57

K12K20

ev, 100 + 5K = 84 + 7K ev, 2K = 16 K = 8 eQi myZivs 8 eQi c‡i Zv‡ì eq‡mi AbycvZ 7 : 5 n‡e| DËi: M

14| av‡b Pvj I Zz‡li AbycvZ 7 : 3 n‡j, G‡Z kZKiv Kx cwigvY Pvj Av‡Q? [ICB Cashier-2018] (K) 70 (L) 30 (M) 7 (N) 3

mgvavb : Abycv‡Zi †hvMdj a‡i HwKK wbq‡g Kiæbt 10 †KwR‡Z Pvj Av‡Q 7 †KwR

1 †KwR‡Z Pvj Av‡Q

107 †KwR

100 †KwR‡Z Pvj Av‡Q

101007 †KwR = 70 †KwR DËi: K

15| x + y = 6 Ges 2x = 4 n‡j, y Gi gvb KZ? [ICB Cashier-2018]

(K) 2 (L) 4 (M) 6 (N) 8 mgvavb :

‡Ìqv Av‡Q, 2x = 4 or x = 2

x + y = 6 or 2 + y = 6 or y = 4 DËi: L 16| GKwU mvgvšÍwi‡Ki yBwU mwbœwnZ evûi ˆ`N©¨ h_vμ‡g 7 †m.wg. I 5 †m.wg. n‡j, Gi cwimxgvi A‡a©K KZ? [ICB Cashier-

2018] (K) 12 (L) 20 (M) 24 (N) 28

mgvavb : cwimxgv = 2 (‰`N© + cÖ ’) = 2(7+5) = 24 ‡m.wg.

Aa©-cwimxgv = 242 = 12 ‡m.wg. DËi: K

7

7

5 5

A

B C

D

ICB Cashier-2018


17| 20% wWmKvD‡›Ui c‡i GKwU eB‡qi LiP 400 UvKv `vuovq| eBwUi cÖK…Z g~j¨ KZ? [ICB Cashier-2018] (K) 500 UvKv (L) 480 UvKv (M) 320 UvKv (N) 333 UvKv

mgvavb : awi, eBwUi cÖK…Z g~j¨ 100 UvKv| 20% Qv‡o, weμqg~j¨ (100 - 20) ev 80 UvKv n‡j cÖK…Z g~j¨ = 100 UvKv

weμqg~j¨ 1 UvKv n‡j cÖK…Z g~j¨ =

80100 UvKv

weμqg~j¨ 400 UvKv n‡j cÖK…Z g~j¨ =

80400100 UvKv = 500 UvKv DËi: K

18| 68, 81, 100, 121,......, ......, k~b¨¯’v‡b wmwi‡Ri cieZ©x msL¨v¸‡jv KZ n‡e? [ICB Cashier-2018]

(K) 142, 163 (L) 144, 169 (M) 145, 168 (N) 140, 160 mgvavb :

cÖ‡kœ fzj Av‡Q| avivi cÖ_g msL¨vwU 68 Gi cwie‡Z© 64 n‡e| Zvn‡j DËi n‡e Ackb (L)| 64 + 17 = 81 81 + 19 = 100 100 + 21 = 121 121 + 23 = 144 144 + 25 = 169

19| 43

,21

,41

Gi Mo †KvbwU? [ICB Cashier-2018]

(K) 45 (L)

32 (M)

21 (N)

43

mgvavb :

343

21

41

=

31

46 =

21 DËi: M

20| fvRK fvMd‡ji 10 ¸Y| fvRK 0.5 n‡j, fvR¨ KZ? [ICB Cashier-2018]

(K) 0.025 (L) 0.25 (M) 25 (N) 2.5 mgvavb :

fvMdj = 0.510 = 0.05| myZivs, fvR¨ = fvRKfvMdj = 0.50.05 = 0.025 DËi: K

ICB Cashier-2018


21| GKwU K¬v‡m 9% wkÿv_©x A †MÖW †c‡q‡Q| hw` K¬v‡m wkÿv_©xi msL¨v 300 nq, Z‡e KZRb cixÿv_©x A †MÖW †cj? [ICB

Cashier-2018] (K) 21 Rb (L) 30 Rb (M) 33 Rb (N) 28 Rb

mgvavb :

cÖ`Ë Ackb¸‡jvi ‡KvbwUB mwVK bq| mwVK DËi n‡e 27 Rb| 3001009 = 27 Rb

22| ywU †gwkb GKmv‡_ N›Uvq 4wU †Ljbv ˆZix K‡i| 6wU †gwkb 2 N›Uvq KZwU †Ljbv ˆZix Ki‡e? [ICB Cashier-2018] (K) 32 (L) 16 (M) 12 (N) 24

mgvavb : 2wU †gwkb 1 N›Uvq ˆZix K‡i 4wU †Ljbv

1wU †gwkb 1 N›Uvq ˆZix K‡i

24 wU †Ljbv

6wU †gwkb 2 N›Uvq ˆZix K‡i

2264 wU †Ljbv = 24wU †Ljbv DËi: N

23| GKwU †cwÝ‡ji `vg 10 UvKv I GKwU Kj‡gi `vg 20 UvKv| hw` wgRv‡bi Kv‡Q 100 UvKv _v‡K, Z‡e †m KZ¸‡jv †cwÝj I

Kjg wKb‡Z cvi‡e? [ICB Cashier-2018] (K) 4wU †cwÝj I 4wU Kjg (L) 4wU †cwÝj I 2wU Kjg (M) 2wU †cwÝj I 4wU Kjg (N) 4wU †cwÝj I 3wU Kjg

mgvavb : (102) (204) = 20 80 = 100 UvKv DËi: M

24| mKvj QqUvi mgq N›Uvi KvuUv I wgwb‡Ui KvuUvi g‡a¨ †Kv‡Yi cwigvc KZ? [ICB Cashier-2018]

(K) 900 (L) 1800 (M) 1200 (N) 600 mgvavb :

Drcbœ †Kv‡Yi gvb = 2

H60M11 =

2

660011=

2

360= 180 DËi: L

we: ª: Drcbœ †Kv‡Yi gvb FYvZ¥K n‡jI abvZ¥K ai‡Z n‡e|

ICB Computer Operator-2018


Investment Corp. Bangladesh (ICB) Post Name: Computer Operator; Exam Date: 25.05.2018


1. hw` nq, Z‡e KZ? [ICB Computer Operator-2018]

(K) 4 (L) 2 (M) 1 (N) 0 mgvavb:

‡Ìqv Av‡Q, x4 – x2 + 1 = 0

x4 + 1 = x2

[Dfq cv‡k¦© x2 Øviv fvM K‡i]

DËi: M

2. 300 UvKvi 4 eQ‡ii mij gybvdv I 400 UvKvi 5 eQ‡ii mij gybvdv GK‡Î 148 UvKv n‡j, kZKiv gybvdvi nvi KZ? [ICB

Computer Operator-2018] (K) 10% (L) 15% (M) 4% (N) 6%

mgvavb:

gybvdvi nvi = = = 4.625% DËi: M

we ÍvwiZ mgvav‡bi Rb¨ ÔKarmasangsthan Bank DEO-2018Õ Gi 2 bs cÖkœ †`Lyb|

3. ‰`wbK 8 N›Uv cwikÖg K‡i 50 Rb †jvK GKwU KvR 12 w`‡b Ki‡Z cv‡i| ˆ`wbK KZ N›Uv cwikÖg K‡i 60 R‡b 16 w`‡b H KvRwU Ki‡Z cvi‡e? [ICB Computer Operator-2018] (K) 10 N›Uv (L) 5 N›Uv (M) 15 N›Uv (N) 7 N›Uv

mgvavb: Applying MDH Method:

=

Or, =

Or, H2 =

H2 = 5 N›Uv DËi: L

01xx 24 2

2

x

1x

2

2

22

4

x

x

x

1

x

x

1x

1x

22

5)(4004)(300100148

3200100148

1

111

W

HDM

2

222

W

HDM

1

81250 1

H1660 2

1660

81250



4. GKwU eB‡qi g~j¨ 24.00 UvKv| GB g~j¨ cÖK…Z g~‡jÏ 80%| evwK g~j¨ miKvi fZz©wK w`‡q _v‡Kb| miKvi cÖwZ eB‡q KZ UvKv fZz©wK †`b? [ICB Computer Operator-2018] (K) 10 UvKv (L) 5 UvKv (M) 6 UvKv (N) 15 UvKv

mgvavb: 80% = 24 UvKv

1% = UvKv

20% = = 6 UvKv

myZivs cÖwZ eB‡q miKvi 6 UvKv fZ©zwK †`q| DËi: M

5. 2x – y = 8 Ges x – 2y = 4 n‡j, x + y = KZ? [ICB Computer Operator-2018] (K) 0 (L) 4 (M) 8 (N) 12

mgvavb: ‡Ìqv Av‡Q, 2x – y = 8 x – 2y = 4 x + y = 4 [we‡qvM K‡i] DËi: L

6. GKwU mgevû wÎfz‡Ri evûi ˆ`N©¨ 6 †m.wg. n‡j, Gi †ÿÎdj KZ eM© †m.wg.? [ICB Computer Operator-2018]

(K) (L) (M) (N) mgvavb:

Avgiv Rvwb, mgevû wÎfz‡Ri †ÿÎdj =

GLv‡b, GKwU evû, a = 6 ‡m.wg.

wÎfzRwUi ‡ÿÎdj = = = DËi: N

7. Avwid I AvwK‡ei eq‡mi AbycvZ h_vμ‡g 5:3| Avwi‡di eqm 20 eQi n‡j, KZ eQi ci Zv‡ì eq‡mi AbycvZ 7:5 n‡e?

[ICB Computer Operator-2018] (K) 5 eQi (L) 6 eQi (M) 8 eQi (N) 10 eQi

mgvavb: Avwid I AvwK‡ei eq‡mi AbycvZ = 5 : 3 = 20 : 12

myZivs Avwid I AwK‡ei eZ©gvb eqm h_vμ‡g 20 eQi I 12 eQi| awi, x eQi c‡i Zv‡ì eq‡mi AbycvZ 7:5 n‡e| cÖkœg‡Z,

ev, 100 + 5x = 84 + 7x ev, 2x = 16 x = 8

myZivs 8 eQi c‡i Zv‡ì eq‡mi AbycvZ 7:5 n‡e| DËi: M

8024

802024

33 34 36 39

2a4

3

2)6(4

336

4

3 39

5

7

x12

x20



8. x2 – 7x + 6 Gi Drcv`‡K we‡køwlZ iƒc wb‡Pi †KvbwU? [ICB Computer Operator-2018] (K) (x – 2) (x – 3) (L) (x – 1)(x + 8) (M) (x – 1)(x – 6) (N) (x + 1)(x + 6)

mgvavb: x2 – 7x + 6 = x2 – 6x – x + 6 = x(x – 6) – 1(x – 6) = (x – 1)(x – 6) DËi: M

9. x + y = 3, xy = 2 n‡j, x3 + y3 Gi gvb KZ? [ICB Computer Operator-2018] (K) 9 (L) 18 (M) 19 (N) 27

mgvavb: x3 + y3 = (x + y)3 – 3.xy(x + y) = (3)3 – 323 = 27 – 18 = 9 DËi: K

hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018


hye Dbœqb Awa`ßi c‡ì bvg: Rywbqi Awdmvi; cixÿvi ZvwiL: 04.05.2018; ‡mU- 8161


1. ‡Kvb cixÿvq cixÿv_©x‡ì 40% Bs‡iwR‡Z, 30% evsjvq †dj K‡i‡Q| hw` 20% Dfq wel‡q †dj K‡i _v‡K Z‡e kZKiv KZRb Dfq wel‡q cvk K‡i‡Q? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 5 B. 20 C. 50 D. 35

mgvavb: ïay Bs‡iwR‡Z †dj K‡i‡Q = (40 20)% = 20%

ïay evsjvq †dj K‡i‡Q = (30 20)% = 10% ‡gvU †dj K‡i‡Q = (20 + 10 + 20)% = 50%

Dfq wel‡q cvk K‡i‡Q = (100 50)% = 50% DËi: C 2. evwl©K gybvdv 12.5% †_‡K K‡g 10.75% nIqvq Rvwgj mv‡n‡ei Avq 4 eQ‡i 280 UvKv K‡g †M‡Q| Zvi g~jab KZ UvKv? [hye

Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 3500 B. 1600 C. 2500 D. 4000

mgvavb: gybvdv K‡g‡Q = 12.50% - 10.75% = 1.75% 4 eQ‡i Avq K‡g ‡M‡Q 280 UvKv

1 Ó Ó Ó Ó (2804) Ó = 70 UvKv A_©r gybvdvi nvi 1.75% K‡g hvIqvq Rvwgj mv‡n‡ei evrmwiK Avq 70 UvKv K‡g‡Q| GLb, 1.75% = 70 UvKv

1% =

1.7570 UvKv

100% =

1.7510070 UvKv

= 4000 UvKv DËi: D

3. 1, 4, 7, 10... ... ... avivi 29Zg c`wU KZ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 79 B. 82 C. 85 D. 88

mgvavb: avivi cÖ_g c` (a) = 1 Ges mvaviY AšÍi (d) = 4 1 = 3 Avgiv Rvwb, n Zg c` = a + (n – 1)d

= 1 + (29 – 1)3 [where, n = 29]

= 1 + (283) = 1 + 84 = 85 DËi: C



4. ywU msL¨vi ¸Ydj 1536, msL¨v ywUi j.mv.¸ 96 n‡j M.mv.¸ KZ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 16 B. 24 C. 32 D. 12

mgvavb: Avgiv Rvwb, msL¨vØ‡qi ¸Ydj = j.mv.¸ M.mv.¸

ev, 1536 = 96 M.mv.¸

ev, M.mv.¸ = 96

1536

M.mv.¸ = 16 DËi: A

5. 30 dzU j¤^v GKwU evuk Ggb fv‡e †K‡U yÕfvM Kiv nj †hb †QvU Ask eo As‡ki yB-Z…Zxqvsk nq, †QvU As‡ki ˆ`N©¨ KZ dzU? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 10 B. 8 C. 6 D. 12

mgvavb:

awi, eo Ask = 3K| Zvn‡j †QvU Ask = 3K32 = 2K

cÖkœg‡Z, 3K + 2K = 30

ev, 5K = 30 K = 6 dzU

myZivs †QvU As‡ki ˆ`N©¨ (26)=12 dzU| DËi: D

6. 10wU msL¨vi †hvMdj 462, cÖ_g 4wUi Mo 52 Ges †kl 5wUi Mo 38 n‡j 5g msL¨vwU KZ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 62 B. 72 C. 64 D. 54

mgvavb: cÖ_g 4wU msL¨vi †hvMdj = 452 = 208

‡kl 5wU msL¨vi †hvMdj = 538 = 190

5g msL¨vwU = 10wU msL¨vi ‡hvMdj 9wU msL¨vi †hvMdj = 462 (208+190) = 64 DËi: C 7. 20 R‡b †h mg‡q GKwU KvR Ki‡Z cv‡i, Zvi 37.5 kZvsk Kg mg‡q KvRwU †kl Ki‡Z n‡j Rbej KZ¸‡Y DbœxZ Ki‡Z

n‡e? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 20 B. 32 C. 16 D. 48

mgvavb: awi, ¯^vfvweKfv‡e KvRwU 100 w`‡b †kl nq|

GLb 37.5% Kg mg‡q †kl Ki‡Z n‡j KvRwU (10037.5) ev 62.5 w`‡b †kl Ki‡Z n‡e| KvRwU 100 w`‡b †kl Ki‡Z Rbej jv‡M 20 Rb

Ó 1 Ó Ó Ó Ó Ó (20100) Ó

Ó 62.5 Ó Ó Ó Ó Ó

62.510020 Ó ev 32 Rb

A_©vr Rbej DbœxZ Ki‡Z n‡e

2032 ev 1.6 ¸Y| (Ack‡b mwVK DËi †bB)



8. ywU msL¨vi j.mv.¸ 96 Ges M.mv.¸ 16| GKwU msL¨v Aci msL¨vi 1.5 ¸Y n‡j eo msL¨vwU KZ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 20 B. 32 C. 16 D. 48

mgvavb: awi, †QvU msL¨vwU = K

eo msL¨vwU = 1.5K

Avgiv Rvwb, msL¨vØ‡qi ¸Ydj = j.mv.¸ M.mv.¸

ev, 1.5K K = 96 16

ev, 1.5K2 = 96 16

ev, K2 = 1.5

1696

ev, K2 = 1024

K = 32

myZivs eo msL¨vwU (1.532) = 48 DËi: D

9. x + y = 6, xy = 8, then (x – y)3 = ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 3 B. 8 C. 9 D. 12

mgvavb: Given, x + y = 6 Or, y = 6 – x Again, xy = 8

x(6 – x) = 8 [ y = 6 – x]

x2 – 6x + 8 = 0

x2 – 4x – 2x + 8 = 0

x(x – 4) – 2(x – 4) = 0

(x – 4) (x – 2) = 0 Now, x – 4 = 0 Or, x – 2 = 0

x = 4 x = 2 So, y = 6 – x Again, y = 6 – x = 6 – 4 [when, x = 4] = 6 – 2 [when x = 2] = 2 = 4 Now, (x – y)3 = (2 – 4)3 [when, x = 2 and y = 4] = (–2)3 = –8 Again, (x – y)3 = (4 – 2)3 [when, x = 4 and y = 2] = (2)3 = 8

A_©vr (x – y)3 = 8 n‡e| wKš‘ Ack‡b †h‡nZz 8 †`qv Av‡Q| ZvB GUvB mwVK DËi| DËi: B



10. cici wZbwU msL¨vi ¸Ydj 120, Zv‡ì †hvMdj KZ? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 20 B. 32 C. 16 D. 48

mgvavb: mwVK DËi 15 (Ack‡b †bB)| ÔGKwU evwo GKwU Lvgvi cÖK‡íi wdì mycvifvBRvi-2018Õ Gi 16 bs cÖkœ †`Lyb|

11. If 102y = 25, then 10-y equals: [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018]

A. 50

1 B.

625

1 C.

2

1 D.

5

1

mgvavb: 102y = 25

(10y)2 = (5)2

10y = 5

5

1

10

1y

10-y = 5

1 DËi: D

12. What is the difference between the interests earned on Tk. 10000 for five years on yearly

compounding basis and simple interest basis? [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. Tk. 1770 B. Tk. 1760 C. Tk. 1750 D. Tk. 1700

mgvavb: cÖ‡kœ fzj Av‡Q| KviY my‡ì nvi †`qv †bB| ZvB K¨vjKz‡jkvb Kiv hv‡e bv|

13. If the radius of a circle is increased by 20% then the area is increased by (e„‡Ëi cwiwa 20% e„w× †c‡j †ÿÎdj kZKiv KZ e„w× cv‡e?) [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 44% B. 400% C. 40% D. 120%

mgvavb: Area of circle = r2 New radius = 120% of r = 1.2r

New area = (1.2r)2

Area increased = (1.2r)2 – r2

= 1.44r2 – r2

= 0.44r2

% increased =

100r

r44.02

2

= 44%

Shortcut: 100

ABBA =

100

20202020

= 44% DËi: A



14. What is the effective rate of interest of a lump sum (GKKvjxb cÖ`Ë UvKv) that is compounded quarterly

at 12% annually? (eQ‡i kZKiv 12 UvKv ˆÎgvwmK Pμe„w× my` nv‡i GKKvjxb cÖ`Ë †Kvb UvKvi (Avm‡ji) Kvh©Kix my` KZ?) [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 10.25% B. 10.32% C. 10.36% D. 10.38%

mgvavb: Effective rate of interest Gi †ÿ‡Î-

r = 1n

i1

n

= 14

31

4

= 14

74

= 9.378 – 1 = 8.378 (Ack‡b mwVK DËi †bB)

15. A retailer sets 220% of invoice price as catalogue price and offers 40% off on catalogue price in sales. What is his profit percentage? (GKRb LyPiv we‡μZv Pvjv‡bi g~‡jï ‡P‡q 220% †ewk g~‡j¨ †Kvb c‡Yï ZvwjKvg~j¨ wba©viY K‡i Ges ZvwjKvg~‡jï Dci 40% wWmKvD›U †`q| G‡Z H we‡μZvi kZKiv KZ jvf n‡e?) [hye Dbœqb Awa`ß‡ii Rywbqi Awdmvi-2018] A. 24% B. 28% C. 32% D. 36%

mgvavb: GLv‡b invoice price UvB n‡jv we‡μZvi μqg~j¨| Let, invoice price = Tk. 100 Catalogue price = 220% of 100 = Tk. 220 Selling price after discount = 220 – 40% of 220 = Tk. 132 Profit % = 132 – 100 = 32% DËi: C

Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018


Jibon Bima Corporation Post Name: Junior Officer; Exam Date: 20.04.2018


1. GKwU cY¨ weμq K‡i cvBKvix we‡μZv 20% Ges LyPiv we‡μZv 20% jvf K‡i| hw` ªe¨wUi LyPiv weμqg~j¨ 576 UvKv nq, Z‡e

cvBKvix we‡μZvi μqg~j¨ KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 250 UvKv B. 300 UvKv C. 400 UvKv D. 480 UvKv

mgvavb: g‡bKwi, cvBKvix we‡μZvi μqg~j¨ 100 UvKv| 20% jv‡f cvBKvix we‡μZvi weμqg~j¨ (100+20) = 120 UvKv| [GwU LyPiv we‡μZvi μqg~j¨]

Avevi 20% jv‡f LyPiv we‡μZvi weμqg~j¨ (120 + 12020%) = 120 + 24 = 144 UvKv LyPiv we‡μZvi weμqg~j¨ 144 UvKv cvBKvix we‡μZvi μqg~j¨ 100 UvKv

Ó Ó Ó 1 Ó Ó Ó Ó

144100 Ó

Ó Ó Ó 576 Ó Ó Ó Ó

144576100 Ó = 400 UvKv

myZivs cvBKvix we‡μZvi μqg~j¨ 400 UvKv| DËi: C 2. hw` ap = b, bq = c Ges cr = a nq, Zvn‡j pqr KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

A. 0 B. 1 C. 2 D. 3 mgvavb:

‡Ìqv Av‡Q, ap = b bq = c cr = a

p = a

b q =

b

c r =

c

a

pqr = a

b

b

c

c

a = 1 DËi: B

3. GKwU Mvoxi PvKvi e¨vm 1.67 wgUvi| 42 wK‡jvwgUvi c_ †h‡Z PvKvwU KZevi Nyi‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-

2018] A. 1000 evi B. 3000 evi C. 25000 evi D. 6000 evi

mgvavb: PvKvi e¨vm, 2r = 1.67 wgUvi| PvKvi cwiwa, 2r = (1.673.1416) = 5.25 wgUvi

42 wK.wg. ev 42000 wgUvi †h‡Z PvKvwU Nyi‡e = 5.25

42000 = 8000 evi (Ack‡b mwVK DËi †bB) DËi:



4. ÔKÕ kn‡i RbmsL¨v 68,000 Rb Ges cÖwZeQi Gi RbmsL¨v 1200 Rb K‡i n«vm cv‡”Q| ÔLÕ kn‡ii RbmsL¨v 42,000 Rb Ges cÖwZeQi Gi RbmsL¨v 800 Rb K‡i e„w× cv‡”Q| KZ eQ‡i Dfq kn‡ii RbmsL¨v mgvb n‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 9 eQi B. 10 eQi C. 13 eQi D. 15 eQi

mgvavb: cÖwZ eQi Dfq kn‡i RbmsL¨v n«vm-e„w×i cv_©K¨ = 800 (1200) = 800 + 1200 = 2000 Rb

Dfq kn‡ii †gvU RbmsL¨vi cv_©K¨ = 68000 42000 = 26000 Rb

Dfq kn‡ii RbmsL¨v mgvb n‡e = 260002000 = 13 eQ‡i DËi: C

5. 1426x28 = three-fourths of 2984, find x. [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 659 B. 694 C. 841 D. 859

mgvavb:

Given, 1426x28 = 29844

3

1426x28 = 2238

x28 = 2238 – 1426

x28 = 812

x = 29

x = 841 [squaring both sides] DËi: C 6. w`‡b iv‡Z 24 N›Uvq GKwU Nwoi KvuUv ywU 90 wWwMÖ †KvY m„wó K‡i- [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

A. 23 evi B. 44 evi C. 24 evi D.48 evi mgvavb:

Nwoi KuvUv 12 N›Uvq 90 wWwMÖ †KvY m„wó K‡i 22 evi| Zvn‡j 24 N›Uvq 90 wWwMÖ †KvY m„wó K‡i (222) ev 44 evi| DËi: B

7. my‡ì nvi `kwgK 75 kZvsk n«vm cvIqv‡Z GKRb AvgvbZKvixi Avgvb‡Zi Dci 4 eQ‡ii cÖvß Avq 750 UvKv K‡g hvq| Zvi

Avgvb‡Zi cwigvY KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 25,000 UvKv B. 18,750 UvKv C. 30,000 UvKv D. 1,00,000 UvKv

mgvavb: 4 eQ‡i K‡g 750 UvKv| Zvn‡j 1 eQ‡i K‡g (7504) ev 187.5 UvKv| GLb, 0.75% = 187.5 UvKv

1% = 0.75187.5

UvKv

100% =

0.75100187.5 UvKv = 25000 UvKv

myZivs H AvgvbZKvixi Avgvb‡Zi cwigvY 25000 UvKv| DËi: A



8. GKwU UvKvi _‡j‡Z 1 UvKv, 50 cqmv I 25 cqmvi g~j¨gv‡bi 342wU gy`ªv i‡q‡Q| gy ªv¸‡jvi g~j¨gv‡bi AbycvZ 11 : 9 : 5 n‡j 50 cqmv g~j¨gv‡bi gy`ªvi msL¨v KqwU? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 19 wU B. 180 wU C. 150 wU D. 162 wU

mgvavb:

50 cqmv = 21

UvKv Ges 25 cqmv = 41

UvKv

awi, gy`ªv¸‡jvi g~j¨gvb h_vμ‡g 11x, 9x Ges 5x cÖkœg‡Z,

4

1x5

2

1x9

1

x11 = 342

ev, 11x + (9x2) + (5x4) = 342

ev, 11x + 18x + 20x = 342

ev, 49x = 342

x = 49

342

50 cqmv g~j¨gv‡bi gy ªvi msL¨v = 249

3429 = 126wU (Ack‡b mwVK DËi †bB) DËi:

9. cyjK Avwe‡ii †P‡q wØ¸Y `ÿ Ges GKB KvR Avwe‡ii †P‡q 90 w`b Kg mg‡q m¤úbœ Ki‡Z cv‡i| Zviv Df‡q GK‡Î KZw`‡b

KvRwU †kl Ki‡Z cvi‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 90 w`b B. 60 w`b C. 45 w`b D. 30 w`b

mgvavb: cyjK Avwe‡ii †P‡q wØ¸Y `ÿ| Zvn‡j cyjK hw` KvRwU 90 w`‡b Ki‡Z cv‡i, Avwei KvRwU Ki‡Z cvi‡e 180 w`‡b|

Zviv GK‡Î 1 w`‡b K‡i = 1801

901 =

18012

= 1803

= 601

Ask

Zvn‡j Zviv Df‡q GK‡Î m¤ú~Y© KvRwU 60 w`‡b †kl Ki‡Z cvi‡e| DËi: B

10. a, b, c, d PviwU μwgK ¯^vfvweK msL¨v n‡j wb‡Pi †KvbwU c~Y©eM© msL¨v? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

A. abcd B. ab + cd C. abcd + 1 D. abcd – 1 mgvavb:

a, b, c, d PviwU μwgK ^vfvweK msL¨v| awi, a = 1, b = 2, c = 3, d = 4

abcd = 1234 = 24

abcd + 1 = 24 + 1 = 25 = (5)2 myZivs abcd + 1 GKwU c~Y©eM© msL¨v| DËi: C



11. 23a n‡j ?a

1a

33 [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

A. 318 B. 35 C. 36 D. 312

mgvavb: ‡Ìqv Av‡Q, 23a

23

1

a

1

= 2323

23

= 22

23

23

=

23

23

= 23

a

1a = 2323 = 32

33

a

1a =

a

1a

a

1.a.3

a

1a

3

= 32.3323

= 36324

= 318 DËi: A

12. GKwU mgevû wÎfz‡Ri me¸‡jv evû 20% e„w× cvIqv‡Z wÎfzRwUi †ÿÎdj KZ kZvsk e„w× cv‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi

Awdmvi-2018] A. 44% B. 40% C. 20% D. 8%

mgvavb: ÔmgvR‡mev Awa`ß‡ii Aaxb Z…Zxq †kÖYxi wewfbœ c`-2018Õ Gi 6 bs cÖkœ †`Lyb| DËi: A

13. 3x3x n‡j x = KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 3 B. 3 C. 0 D. 3

mgvavb: ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 16 bs cÖkœ †`Lyb| DËi: C

14. 4x+1 = 32 n‡j x = ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

A. 3

2 B.

5

3 C.

8

1 D.

2

3

mgvavb: ÔcÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018Õ Gi 20 bs cÖkœ †`Lyb| DËi: D

15. log327 = KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] A. 1 B. 2 C. 3 D. 4

mgvavb: ÔcÖwZiÿv gš¿Yvj‡qi Dc-mnKvix cwiPvjK-2018Õ Gi 19 bs cÖkœ †`Lyb| DËi: C

Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi (wjwLZ)-2018


Jibon Bima Corporation Post Name: Junior Officer; Exam Date: 20.04.2018


1. GKwU AvqZvKvi Rwgi †ÿÎdj 12 †n±i Ges K‡Y©i ˆ`N©¨ 500 wgUvi| H Rwgi ˆ`N©¨ I cÖ‡¯’i m‡½ Aci GKwU Rwgi ˆ`N©¨ I

cÖ‡ ’i AbycvZ h_vμ‡g 3 : 4 I 2 : 3| cÖ`Ë AvqZvKvi RwgwUi †ÿÎdj KZ eM©wgUvi I Aci RwgwUi †ÿÎdj KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb : 12 ‡n±i = (1210,000) eM©wgUvi = 1,20,000 eM©wgUvi [1 †n±i = 10,000 eM©wgUvi] awi, cÖ Ë RwgwUi ˆ`N©¨ I cÖ ’ h_vμ‡g 3x wg. Ges 2y wg. Ges Aci RwgwUi ˆ`N©¨ I cÖ ’ h_vμ‡g 4x wg. Ges 3y wg.

cÖ Ë Rwgi †ÿÎdj = 3x 2y = 6xy

Aci Rwgi †ÿÎdj = 4x 3x = 12xy

cÖkœg‡Z, 6xy = 120000 xy = 20000

12xy = 1220000 = 240000 eM©wgUvi myZivs cÖ`Ë AvqZvKvi RwgwUi †ÿÎdj 120000 eM©wgUvi Ges Aci RwgwUi †ÿÎdj 240000 eM©wgUvi| DËi: 120000 eM©wgUvi Ges 240000 eM©wgUvi

2. 20,000 UvKv Avm‡j 2 eQ‡ii eëav‡b Pμe„w× my` I mij my‡ì cv_©K¨ 392 UvKv n‡j my‡ì nvi KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb : Avgiv Rvwb, mij my‡ì †ÿ‡Î:

I = pnr = [‡hLv‡b, I = mij my`, p = Avmj, n = mgq Ges r = my‡ì nvi]

Avevi Pμe„w× my‡ì †ÿ‡Î:

C = P(1 + r)n = = = = 2(100 + r)2

[‡hLv‡b, C = my`vmj, P = Avmj, n = mgq Ges r = my‡ì nvi] Pμe„w× my` = my`vmj Avmj = 2(100 + r)2 – 20000 = 2(r2 + 200r + 10000) – 20000 = 2r2 + 400r +

20000 – 20000 = 2r2 + 400r cÖkœg‡Z, 2r2 + 400r – 400r = 392 Or, 2r2 = 392

r = 14 myZivs my‡ì nvi 14%| DËi: 14%

r400100

r220000

2

100

r1P

2

100

r10020000

10000

)r100(20000

2

3x

2y 3y

4x

cÖ Ë Rwg Aci Rwg



3. GKwU eB‡qi ˆ`N©¨ 25 †m.wg. I cÖ¯’ 18 †m.wg.| eBwUi c„ôv msL¨v 200 Ges cÖwZ cvZvi cyiæZ¡ 0.1 wg.wg. n‡j eBwUi AvqZb KZ? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb : eB‡qi ˆ`N©¨ = 25 †m.wg.; cÖ¯’ = 18 †m.wg. Ges c„ôv msL¨v = 200

cvZvi msL¨v = 2002 = 100

cÖwZ cvZvi cyiæZ¡ = 0.1 wg.wg. = (0.110) †m.wg. = 0.01 †m.wg.

eBwUi AvqZb = (25 18 100 0.01) Nb †m.wg. = 450 Nb †m.wg. Note: eB‡qi cÖwZ cvZvq 2wU c„ôv _v‡K| A_©vr 2 c„ôv wg‡j 1 cvZv nq|

4. 12 †m.wg. e¨v‡mi GKwU cyiæ avZe †MvjK‡K Mwj‡q GKB e¨vm I fi wewkó wmwjÛv‡i iƒcvšÍi Kiv n‡j, wmwjÛviwUi D”PZv KZ n‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb : ‡Mvj‡Ki e¨vm = 12 †m.wg.| e¨vmva©, r = 12 2 = 6 †m.wg.|

‡Mvj‡Ki AvqZb = = = 288

cÖkœg‡Z, r2h = 288

ev, r2 h = 288

ev, 62 h = 288

ev, h =

h = 8 myZivs wmwjÛviwUi D”PZv 8 ‡m.wg.| DËit 8 †m.wg.

5. In a code language ‘mok dan sil’ means ‘big dad wresler’; ‘fit kon dan’ means ‘wrestler is good’; ‘cold tir fit’ means ‘he is new’. Which word stands for ‘is’ in this language? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb : Common Code Common Word

1st statement: mok dan sil big dad wresler

2nd statement: fit kon dan wresler is good

3rd statement: cold tir fit he is new In the first and second statements, the common code word is ‘dan’ and the common word is ‘wresler’. So, ‘dan’ means ‘wrestler’. In the second and third statements, the common code word is 'fit' and the common word is ‘is’. So ‘fit’ stands for ‘is’. Ans: fit

3πr3

4 3)6(π3

4

36

288

wmwjÛv‡ii AvqZb = ‡Mvj‡Ki AvqZb

Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi (wjwLZ)-2018


6. f(x) = x3 – 6x2 + 11x – 6 n‡j, x Gi †Kvb gv‡bi Rb¨ f(x) = 0 n‡e? [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018] mgvavb :

f(x) = x3 – 6x2 + 11x – 6

x3 – 6x2 + 11x – 6 = 0

x3 – 3x2 – 3x2 + 9x + 2x – 6 = 0

x2(x – 3) – 3x(x – 3) + 2(x – 3) = 0

(x – 3) (x2 – 3x + 2) = 0

(x – 3) (x2 – 2x – x + 2) = 0

(x – 3) {x(x – 2) – 1(x – 2)} = 0

(x – 1) (x – 2) (x – 3) = 0 Now, x – 1 = 0 or, x = 1 x – 2 = 0 or, x = 2 x – 3 = 0 or, x = 3 Ans: 1, 2, 3

7. n‡j, Gi gvb wbY©q Kiæb| [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb :

Avgiv Rvwb, = = (3)2 – 2 = 9 – 2 = 7

Avevi, = = (3)3 – 3.3 = 27 – 9 = 18

GLb, = 718

ev, = 126

ev, = 126

ev, = 126

ev, = 126 – 3

= 123

DËit 123

3x

1x

55

x

1x

22

x

1x

x

1.x.2

x

1x

2

33

x

1x

x

1x

x

1.x.3

x

1x

3

33

22

x

1x

x

1x

52

3

3

25

x

1

x

x

x

xx

55

x

1x

x

1x

55

x

13x

55

x

1x

55

x

1x



8. mgvavb Kiæbt [Rxeb exgv K‡c©v‡ik‡bi Rywbqi Awdmvi-2018]

mgvavb :

ev,

ev,

ev,

ev, (12x + 45) (3x – 5) = 18 (2x – 5) (x + 5)

ev, 36x2 – 60x + 135x – 225 = 18(2x2 + 10x – 5x – 25)

ev, 36x2 + 75x – 225 = 36x2 + 180x – 90x – 450

ev, 75x – 180x + 90x = 225 – 450

ev, 15x = 225

x = 15 DËit 15

53x

18

5x

1

52x

10

5x3

18

5x

1

5x2

10

5x3

18

)5x)(5x2(

)5x2(1)5x(10

5x3

18

)5x)(5x2(

5x250x10

5x3

18

)5x)(5x2(

45x12

Solved by: Ajgar Ali

mv¤úªwZK g¨v_ mjykvb †c‡Z mv‡_B _vKzb|

Ajgar Ali

Documents

cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018 cÖwZiÿv gš¿Yvjq · 2020-04-04 · cÖwZiÿv gš¿Yvj‡qi mnKvix cwiPvjK-2018 Solved by: Ajgar Ali Spotlight Publication 1g bj Øviv