2 | Page PH: 24339436, 42867555, 9840226187

(A)369 (B)2169 (C)21627 (D)2 Solution: 43 33 = 6427 82 72 = 6449 33 13 = 271 62 32 = 369

4. Find the sum of the first 20 terms of the geometric series 5 5 5

....2 6 18

+ + +

(A) 20

15 11

4 3

−

(B) 18

15 11

4 3

−

(C) 16

15 11

4 3

−

(D) 14

15 11

4 3

−

5 5 5

....2 6 18

+ + + vd;w ngUf;Fj; njhlhpd; Kjy; 20 cWg;Gfspd; $Ljiyf;

fhz;f.

(A) 20

15 11

4 3

−

(B) 18

15 11

4 3

−

(C) 16

15 11

4 3

−

(D) 14

15 11

4 3

−

Solution:

1

1

n

n

rS a

r

−=

−

5

2a =

n=20,

5 2 1

6 5 3r = × =

20 201 1

1 15 53 3

1 22 21

3 3

nS

− −

= =

− −

= 20

15 11

4 3

−

5. If 40% of 1640 + x = 35% of 980 + 150% of 850, find the value of x

(A)1052 (B) 842 (C) 962 (D) 372 40% of 1640 + x = 35% of 980 + 150% of 850, x-d; kjpg;ig fhz;f (A)1052 (B) 842 (C) 962 (D) 372 Solution:

40 35 150

1640 980 250100 100 100

x

× + = × + ×

656+x= [343+1275] x = 962

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6. Three different containers contain 396 litres, 432 litres, and 612 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly? A) 36 B) 32 C) 28 D)24 %d;W nfhs;fyd;fspy; ghy; kw;Wk; jz;zPh; fyit KiwNa 396 ypl;lh;> 432 ypl;lh; kw;Wk; 612 ypl;lh; cs;sJ. ,k;%d;W msTfisAk; msf;ff;$ba kpfg;nghpa msit vJ? A) 36 B) 32 C) 28 D)24 Solution: HCF (396, 432, 612) = 36

7. Find the wrong number in the series. 2, 9, 28, 65, 126, 216, 344 a. 2 b. 28 c. 65 d. 216 gpd;tUk; njhlhpy; jtwhd vz;iz Njh;T nra;f. 2, 9, 28, 65, 126, 216, 344 a. 2 b. 28 c. 65 d. 216 Solution Cube Nummber +1 13 + 1=2, 23 + 1=9, 33 + 1=28, 43 + 1=64, 53 + 1=216, 63 + 1=217, 73 + 1=344 Wrong Number = 216

8. The GCD and LCM of two polynomials are (x+1) and x6-1 respectively. if one of the polynomial is (x3+1), find the other. A. (x3-1)(x+1) B. (x6-1)(x+1) C. (x3-1)(x3+1) D. (x3+1)(x-1) ,uz;L gy;YWg;Gf;Nfhitapd; kP.ng.t kw;Wk; kP.rp.k KiwNa (x+1) kw;Wk; x6-1 ,jpy; xU gy;YWg;Gf;Nfhitapd; kjpg;G (x3-1) vdpy; kw;nwhd;iwf; fz;Lgpb

A. (x3-1)(x+1) B. (x6-1)(x+1) C. (x3-1)(x3+1) D. (x3+1)(x-1) Solution:

LCM HCF

OtherNumberGivenNumber

×=

3 3

3

( 1)( 1) ( 1)

1

x x xOtherNumber

x

+ − × +=

+

∴(x3-1)(x+1)

6 | Page PH: 24339436, 42867555, 9840226187

40 Ngh; 300 kP Rtiu fl;l xU ehisf;F 6 kzpNeuk; tPjk; Ntiynra;J 12 ehl;fs; Kbf;fpd;wdh; vdpy; 30 Ngh; 200 kP Rtiu fl;l xU ehisf;F 8 kzpNeuk; Ntiy nra;jhy; mt;Ntiyia vj;jid ehl;fspy; Kbg;gh;? A. 4 ½ ehl;fs; B. 8 ehl;fs; C. 10 ½ ehl;fs; D. 11 ehl;fs; Solution

1 1 2 2

1 2

240 12 6 30 8

300 200

M D M D

W W

D

=

× × × ×=

2

D = 8 day

14. The compound interest on Rs.15,625 at 8% per annum is Rs.4058. The period (in

years) is .... &.15>625f;F Mz;Lf;F 8 rjtPjk; vd;w mbg;gilapy; $l;L tl;b &.4058 vdpy;> fhyk; vj;jid Mz;Lfs;?

a. 3 years b. 2 1

2 years c. 2 years d. 4 years

Solution Amount = Principal + C.I Amount = Rs.19683

1

19683 15625 18

n

= +

19683 27

15625 25

n

=

327 27

25 25

n

=

n = 3 years

15. Two years ago the average age of a family of 8 members was 20 years. If the present average age of the family after increasing one child remains same. What is the present age of the child ?

(A) 2 years (B) 6 years (C) 4 years (D) 8 years ,uz;L Mz;LfSf;F Kd;ghf 8 cWg;gpdh;fisf; nfhz;l xU FLk;gj;jpd; ruhrhp taJ 20 Mz;Lfs;. me;jf; FLk;gj;jpy; xU Foe;ij Nrh;e;j gpwF me;jf; FLk;gj;jpd; jw;Nghija ruhrhp taJ Ke;ija ruhrhp vz;zpf;ifahfNt cs;sJ. me;jf; Foe;ijapd; jw;Nghija taJ vd;d? (A) 2 Mz;Lfs; (B) 6 Mz;Lfs; (C) 4 Mz;Lfs; (D) 8 Mz;Lfs; Solution

Total age of 8 persons, 2 years ago = 8×20 = 160 years

Total age of 8 persons at present = 160 + 8×2 = 176 years

7 | Page PH: 24339436, 42867555, 9840226187

Total age of 9 members = 20×9 = 180 years Present age of the child = 180-176=4 years

16. A lends Rs.3000 to B and a certain sum to C at the same time at 7% p.a. simple interest. If after 5 years, A altogether receives Rs.1400 as interest from B and C, then the sum lent to C is...

a. Rs.1000 b. Rs.1500 c. Rs.750 d. Rs.700 A &.3000I B-f;Fk;> xU Fwpg;gpl;l njhifia C-f;Fk; xNu rkaj;jpy; 7 tpOf;fhL tl;b tPjj;jpy; nfhLf;fpwhh;. 5 Mz;Lfs; fopj;J> B kw;Wk; C-aplk; ,Ue;J nkhj;jkhf &.1400I A tl;bahfg; ngWfpwhh;. vdpy;> mth; C-f;Ff; nfhLj;j njhif vt;tsT? a. &.1000 b. &.1500 c. &.750 d. &.700 Solution Let the Sum lent to C be Rs.x S.I. for Rs.3000 at 7%p.a. for 5 years + S.I. for Rs.x, at 7% p.a. for 5 years = Rs. 1400

3000 7 5 7 51400

100 100

71050 1400

20

21000 7 28000

x

x

x

× × × ×⇒ + =

⇒ + =

⇒ + =

7x = 28000 – 21000

x = 1000

17. How many metres of cloth 10 m wide will be required to make a conical tent with base radius of 14 m and height is 48 m?

a. 110 m b. 55 m c. 77 m d. 220 m 14 kP mbg;gf;f Muk; kw;Wk; 48 kP cauk; nfhz;l xU $k;G tbt $lhuk; mikg;gjw;F 10 kP mfyk; nfhz;l Jzp vj;jid kPl;lh;fs; Njitg;gLk;? a. 110 kP b. 55 kP c. 77 kP d. 220 kP Solution

Area of cloth = πrl = 22

7×14×50=2200 m2

Hence length of cloth = 2200/10 = 220m

18. The cost of cultivating a square field at the rate of ì 250 per hectare is ì 1000. the cost of putting a fence around it at ì 12 per m is a. ì 8400 b. ì 9200 c. ìììì 9600 d. ì 9400

8 | Page PH: 24339436, 42867555, 9840226187

xU rJu tbt epyj;jpy; gaph; nra;tjw;F xU n`f;NlUf;F &.250 tPjk; &.1000 MfpwJ. me;j epyj;ijr; Rw;wpYk; Ntyp mikg;gjw;F xU kPl;lUf;F &.12 tPjk; vt;tsT MFk;? a. ì 8400 b. ì 9200 c. ìììì 9600 d. ì 9400 Solution

Area = 1000

4250

hectare=

a2 = 40,000 m2

a = 200 m

Perimetre = 4×200 = 800m

Cost of Putting a fence = 800×12=9600Rs.

19. The area (in sq cm) of the regular hexagon whose perimeter is 12 cm, is

a. 18 3 b. 15 3 c. 12 3 d. 6 3 12 nrkP Rw;wsitf; nfhz;l xU xOq;F mWNfhzj;jpd; gug;gsT vj;jid rnrkP?

a. 18 3 b. 15 3 c. 12 3 d. 6 3 Solution Perimetre = 6a = 12, a = 2cm

= 23

64

a×

= 23

6 4 6 34

cm× × =

20. A sphere of copper with radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire would be a. 18 m b. 12 m c. 36 m d. 24 m 3 nrkP Muk; nfhz;l xU gpj;jisf; Nfhsk; 0.2 nrkP tpl;lk; nfhz;l xU fk;gpahf mbj;J ePl;lg;gLfpwJ. mt;thW cUthf;fg;gLk; fk;gpapd; ePsk; a. 18 kP b. 12 kP c. 36 kP d. 24 kP Solution Volume of sphere = Volume of Cylinder

D = 0.2, r=0.1cm=1

10cm

3 24

3

4 1 127

3 10 10

r r h

h

π π=

× = × ×

h = 3600cm = 36m 21. Five children A, B, C, D and E are sitting in a row. D is sitting next to A but not E,

B is sitting next to C who is sitting on the extreme left and E is not sitting next to B. Who are sitting adjacent to D.

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A. B & A B. A & C C. Only A D. A & E A, B, C, D, E vd;w Ie;J Foe;ijfs; xU thpirapy; mkh;e;jpUe;jdh;. D vd;gth; A-f;F gf;fj;jpYk; mkh;e;J Mdhy; E-f;F gf;fj;jpy; mkuhYk; ,Ue;jhh;. B vd;gth; ,lJ Nfhbapy; mkh;e;jpUf;Fk; C-f;F gf;fj;jpy; mkh;e;jpUe;jhh; kw;Wk; E vd;gth; B-f;F gf;fj;jpy; mkuhky; ,Ue;jhh; vdpy; D-f;F gf;fj;jpy; mkh;e;jpUg;gth;fs; ahth;? A. B & A B. A & C C. A kl;Lk; D. A & E

Solution C B D A E

22. Raju starts walking towards south. After walking 20 m he turns towards North and Walks 8 m. Again he turns towards East and walks 5 m. How far and in which direction is he from his starting point. A. 15 m, South B. 17 m, North-West C. 7 m, East D. 13 m, South-East uh[_ vd;gth; njw;F Nehf;fp elf;fj; njhlq;Ffpwhh;. 20kP J}uk; ele;j mth; tlf;F Nehf;fp jpUk;gp 8 kP J}uk; elf;fpwhh;. kWgbAk; fpof;F Nehf;fp jpUk;gp 5 kP J}uk; elf;fpwhh; vdpy;> Muk;g ,lj;jpypUe;J vt;tsT J}uk; ve;jj; jpirapy; mth; ,Ug;ghh;?

A. 15 kP njw;F B. 17kP tlNkw;F C. 7kP fpof;F D. 13kP njd;fpof;F Solution 12m South East 5m 8m x2 = 144+25=169 Ans: 13m South East

23. If the price of a book is first decreased by 20% and then increased by 20% then the net change in the price will be

(A)No change (B) 5% increase (C)4% decrease (D) 10% decrease xU Gj;jfj;jpd; tpiyahdJ Kjypy; 20% Fiwf;fg;gl;L gpd; 20% tpiy Vw;wg;gl;lhy; Kbtpy; mjd; tpiyapy; Vw;gLk; khw;wj;jpd; rjtPjk; (A)xU khw;wKk; ,y;iy (B) 5% Vw;wk; (C)4% FiwTFiwTFiwTFiwT (D) 10% FiwT Solution

100

I DI D

×− −

10 | Page PH: 24339436, 42867555, 9840226187

400

20 20 4%100

− − =−

24. If in a certain language, COUNSEL is coded as BITIRAK, how is GUIDANCE

written in that code ? A. EOHYZJBB B. FOIYZJBB C. FOHYZJBB D. HGRMSCHI

xU FwpaPl;L nkhopapy;;, “COUNSEL” vDk; thh;j;ij “BITIRAK” vd;W khw;wp vOjg;gLfpwJ. ,ijg;NghyNt “GUIDANCE” vDk; thh;j;ijia vt;thW khw;Wtha;?

A. EOHYZJBB B. FOIYZJBB C. FOHYZJBB D. HGRMSCHI

Explanation: The letters at odd positions are each moved one step backward, while the letters at even positions are respectively moved six, five, four, three, two,... steps backward to obtain the corresponding letters of the code.

25. In a row of students, the place of Rahul from right is 12th and from left is 4th. How many students should be added to make the total number of students 28?

A. 14 B. 13 C. 20 D. 18 fpilthpirapy; mkh;e;Js;s khzth;fspy; uhFy; mkh;e;Js;s ,lk; tyJGwkpUe;J 12-tjhfTk; ,lJGwkpUe;J 4-tjhfTk; cs;sJ. vj;jid khzth;fis Nrh;j;jhy; nkhj;j khzth;fspd; vz;zpf;if 28-Mf ,Uf;Fk;? A. 14 B. 13 C. 20 D. 18

L + R – 1 = Total 4 + 12 – 1 = 15 15 + 13 = 28

Therefore, 13 students added to make the total number of students 28