tnpsc deo main premodel exam mental ability & aptitude

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TNPSC DEO MAIN PREMODEL EXAM MENTAL ABILITY & APTITUDE SOLUTION

SECTION A: 3 MARK

1. Find the rate percent at which a sum of money becomes 7

6times in 3 years.

xU mryhdJ 3 tUlj;jpy; 7

6klq;fhf MFnkdpy; mjd; tl;b tpfpjk; vt;tsT?

Solution

N= 3year P = x A = 7

6x SI =

1

6x

1 3

6 100

50 55 %

9 9

x Rx

R

2. In a simultaneous throw of two dice, what is the probability of getting a total of

10 or 11? ,U gfilfs; xNu rkaj;jpy; tPrg;gLk; NghJ> $Ljy; 10 my;yJ 11 fpilf;f epfo;jfT ahJ? Solution:

Total No.of events = 6 6 36n s

Sum of 10 = (4, 6) (6, 4) (5, 5) Sum of 11 = (5, 6) (6, 5) n(E)=5

Probability = 5

36

3. What is relative-cumulative frequency distribution?

njhlh;G FtpT epfo;ntz; guty; vd;why; vd;d? The relative cumulative frequency is defined as the ratio of the cumulative frequency to the total frequency. The relative cumulative frequency is usually expressed in terms of a percentage. The arrangement of relative cumulative frequencies against the respective class boundaries is termed as relative

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cumulative frequency distribution or percentage cumulative frequency distribution.

4. Find the range of the following distribution. nfhLf;fg;gl;l gutypd; tPr;R fhz;f

Age (in years) taJ (tUlq;fspy;)

16 – 18 18-20 20-22 22-24 24-26 26-28

Number of students khzth;fspd; vz;zpf;if

0 4 6 8 2 2

5. A garden is in the form of a trapezium. The parallel sides are 40 m and 30 m. The perpendicular distance between the parallel side is 25 m. Find the area of the garden. xU Njhl;lkhdJ rhptfk; tbtpy; cs;sJ. mjd; ,izg;gf;fq;fs; 40kP> 30kP. ,izg;gf;fq;fSf;F ,ilNa cs;s njhiyT 25 kP. Njhl;lj;jpd; gug;gsT fhz;f.

Solution:



6. Raju bought a motorcycle for ` 36,000 and then bought some extra fittings to make it perfect and good looking. He sold the bike at a profit of 10% and he got `

44,000. How much did he spend to buy the extra fittings made for the motorcycle?

,uh[P ` 36>000f;F xU Nkhl;lhh; irf;fpis thq;fp> mjd; Njhw;wg; nghypT ed;F mikaTk; NkYk; ed;Kiwapy; ,aq;fTk; rpy ,ju ghfq;fisg; nghUj;jpdhh;. gpd;G mk;Nkhl;lhh; irf;fpis ` 44>000f;F 10% ,yhgj;jpy; tpw;fpd;whh; vdpy; ,ju ghfq;fs; thq;f vt;tsT nryT nra;jhh;?

Directions (7): Study the following graph carefully and answer the questions given below: mwpTiufs; (tpdh 7): gpd;tUk; tiuglj;ijf; ftdkhf Ma;T nra;Jtpl;L> mjw;Ff; fPNo nfhLf;fg;gl;Ls;s tpdhf;fSf;F tpilaspf;fTk;: DISTRIBUTION OF CANDIDATES WHO WERE ENROLLED FOR MBA ENTRANCE EXAM AND THE CANDIDATES (OUT OF THOSE ENROLLED) WHO PASSED THE EXAM IN DIFFERENT INSTITUTES



MBA EioTj; Njh;Tf;fhf gjpT nra;j tpz;zg;gjhuh;fspd; guty; tptuk; kw;Wk; me;jj; Njh;tpy; ntt;NtW epWtdq;fspy; ,Ue;J Njh;T ngw;wth;fs; (gjpT nra;j tpz;zg;gjhuh;fspy; ,Ue;J) tptuk; Candidates Enrolled = 8550 Candidates who passed the Exam = 5700 gjpT nra;j tpz;zg;gjhuh;fs; = 8550 Njh;r;rp mile;j tpz;zg;gjhuh;fs; = 5700

7. The number of candidates passed from institutes S and P together exceeds the

number of candidates enrolled from institutes T and R together by: S kw;Wk; P epWtdq;fspy; ,Ue;J Njh;r;rp ngw;w nkhj;j tpz;zg;gjhuh;fspd; vz;zpf;if> T kw;Wk; R epWtdq;fspy; gjpT nra;j nkhj;j tpz;zg;gjhuh;fspd; vz;zpf;ifiatpl vj;jid mjpfk;?

Explanation: Required difference = [(16% + 18%) of 5700] - [(8% + 10%) of 8550] = [(34% of 5700) - (18% of 8550)] = (1938 - 1539) = 399.

8. The LCM of 2 3 4 9

, , ,3 5 7 13

is

2 3 4 9, , ,

3 5 7 13- d; kPg;ngU nghJ klq;fhdJ

Explanation:

LCM of fractions =

LCM 2,3,4,9LCMof numerotors 3636

HCFof denominators HCF 3,5,7,13 1

njhFjpapYs;s vz; fspd; kP.rp.k

gpd;d vz; fspd; k.P rp.kgFjpapYs;s vz; fspd; kP.ng.t

=



9. Convert (1265)8 to equivalent Decimal number (1265)8 jrk vz;zhf khw;wTk;

Solution:

10. fPNo nfhLf;fg;gl;Ls;s epidtfq;fspd; tphpthf;fj;ij juTk; m. CD-ROM M. EEPROM ,. DRAM Expand the following memories: a. CD-ROM b. EEPROM c. DRAM

Explanation:

CD-ROM - "Compact Disc Read-Only Memory." EEPROM - Electrically Erasable Progammable Read-only Memory DRAM - Dynamic Random-Access Memory



SECTION B: 8 MARK

11. Divide Rs. 1586 into 3 parts so that the respective amounts at 5 per cent in 2, 3, 4 years respectively, same interest in all three cases. &.1>586 %d;W ghfq;fshf gphpf;f> mit KiwNa 2> 3> 4 Mz;Lfspy; 5 rjtPjk; tl;b tpfpjj;jpy; rkkhd jdptl;b njhifia ju Ntz;Lk;.

Solution: Let Sum of money Rs. x, y and z.

x + y + z = 1568 SI = PNR

100

S.I1 = S.I2 = S.I3

x×2×5 y×3×5 z×4×5= =

100 100 100

10x = 15y = 20z LCM of 10,15,20=60

10x=60 15y=60 20z=60

x= 6 y=4 z=3

12. Answer the following questions gpd;tUk; tpdhf;fSf;F tpilasp

a) Mention the different kinds of Flip –flop circuits. What is the use of flip-flop

circuits? /gpsg;/g;shg;Gfspy; cs;s tiffis vOJf? fzpg;nghwpapd; xUq;fik Rw;Wfspy; Kf;fpa gq;fhw;WtJ vJ? Solution:

There are several kinds of flip-flop circuits, with designators such as D, T, J-K, and R-S. Flip- flop circuits are interconnected to form the logic gates that comprise digital integrated circuits (ICs) such as memory chips and microprocessors.

/gpspg;/g;shg;Gfspy;> R-S, J-K, D, Tvd gy tiffs; cz;L kw;w Rw;WfSld; ,it ,ize;J nrayhw;Wk;. njhlh; Rw;Wfspy; xU Neuf;$wpy; tUk; ntspaPl;bid epidtpy; itj;J mLj;J Neuf;$wpy; nfhLf;Fk; epidtfkhf



,J nray;gLfpwJ. vdNt fzpg;nghwpapd; xUq;fik Rw;Wfspy; ,J xU Kf;fpa gq;fhw;WfpwJ.

b) Tabulate the different memory sizes in which the computer memory is measured. ntt;NtW epidtf msTfs; nfhz;l fzpdp epidtf msTfis gl;baypLf.

Solution:

The different memory sizes in which the computer memory is measured are:

Name Abbreviation Size (Bytes)

Kilo K 2^10

Mega M 2^20

Giga G 2^30

Tera T 2^40

Peta P 2^50

Exa E 2^60

Zetta Z 2^70

Yotta Y 2^80

13. Prepare a frequency distribution for the following observations:

The marks obtained by 25 students in a test are given as follows: 10, 20, 20, 30, 40, 25, 25, 30, 40, 20, 25, 25, 50, 15, 25, 30, 40, 50, 40, 50, 30, 25, 25, 15 and 40. gpd;tUk; tptuq;fspypUe;J Xu; miyntz; guty; ml;ltizia jahupf;fTk;.: xU Nju;tpy; 25 khztu;fs; ngw;w kjpg;ngz;fs;: 10, 20, 20, 30, 40, 25, 25, 30, 40, 20, 25, 25, 50, 15, 25, 30, 40, 50, 40, 50, 30, 25, 25, 15 kw;Wk; 40 MFk;.




a. The least number of 4 digit which is divisible by 8, 10 and 12 is

vz;fs; 8, 10 kw;Wk; 12 My; tFgLk; kpfr; rpwpa ehd;F ,yf;f vz; Solution LCM (8, 10, 12) = 120 1000 divisible by 120 get the remainder 40 Required Number = 1000 – 40 + 120 = 1080

b. A bag contains ì 114 in the form of 1 rupee, 50 paisa and 10 paisa coins in the ratio 3: 4: 10. What is the number of 50 paisa coins?

xU igapy; &.1> 50 igrh kw;Wk; 10 igrh ehzaq;fspd; vz;zpf;if tpfpjq;fs; KiwNa 3:4:10. ,itfspd; nkhj;j kjpg;G &. 114. vdpy; 50 igrh ehzaj;jpd; vz;zpf;if vt;tsT?

Solution: Denomination of coins 100p 50p 10p Ratio of coins 3 4 10 Value of coins 300 200 100 Total value = 300+200+100=600paise =Rs.6

Number of 50 paise coins 114

4 766

15. A right circular cylindrical container of base radius 6 cm and height 15 cm is full

of ice cream. The ice cream is to be filled in cones of height 9 cm and base radius 3 cm, having a hemispherical cap. Find the number of cones needed to empty the container. 6 nr.kP Muk; kw;Wk; 15 nr.kP cauk; nfhz;l Xu; cUis tbtg; ghj;jpuj;jpy; KOtJkhf gdpf;$o; (Ice cream) cs;sJ. me;jg; gdpf;$ohdJ> $k;G kw;Wk; miuf;Nfhsk; ,ize;j tbtj;jpy; epug;gg;gLfpwJ. $k;gpd; cauk; 9 nr.kP kw;Wk; Muk; 3 nr.kP vdpy;> ghj;jpuj;jpy; cs;s gdpf;$io epug;g vj;jidf; $k;Gfs; Njit?



SECTION C: 15 marks


a. In the Annual sports meet, among the 260 students in XI standard in the

school, 90 participated in Kabadi, 120 participated in Hockey, and 50 participated in Kabadi and Hockey. A Student is selected at random. Find the probability that the student participated in

i. Either Kabadi or Hockey ii. Neither of the two tournaments

iii. Hockey only iv. Kabadi only v. Exactly one of the tournaments.

Mz;L tpisahl;L Nghl;bapy; 11 Mk; tFg;gpy; gbf;fpd;w 260 khzth;fspy;> 90 Ngu; fgb Nghl;bapYk;> 120 Ngu; `hf;fp Nghl;bapYk; kw;Wk; 50 Ngu; ,uz;L Nghl;bapYk; fye;Jnfhs;fpd;wdu;. xU khztd; rktha;g;G Kiwapy; Nju;e;njLf;fg;gLfpd;whd;. Nju;e;njLf;Fk; khztd;.

i. fgb my;yJ `hf;fp ii. ,uz;L Nghl;bapYk; fye;J nfhs;tjw;fhd

iii. `hf;fpapy; kl;Lk; iv. fgbapy; kl;Lk; v. rupahf xd;wpy; kl;Lk; fye;J nfhs;tjw;fhd epfo;jfT fhz;f.



b. The table below shows the status of twenty residents in an apartment

If one of the residents is chosen at random, find the probability that the chosen resident will be

(i) a female (ii) a college student (iii) a female student (iv) a male employee

fPo;f;fz;l ml;ltizahdJ xU FbapUg;gpy; jq;fpAs;Nshhpd; epiyiaf; fhl;LfpwJ.

tif ghypdk; fy;Y}hp khzth; gzpahsh;

Mz; 5 3 ngz; 4 8

FbapUg;Nghh; xUtiu rktha;g;G Kiwapy; Njh;e;njLf;Fk; NghJ mth; (i) ngz;zhf ,Uf;f (ii) fy;Y}hp khztu; / khztpahf ,Uf;f (iii) fy;Y}hp khztpahf ,Uf;f (iv) Mz; gzpahsuhf ,Uf;f epfo;jfT vd;d?

Solution

(i) P (a female) 4 8 12 3

5 3 4 8 20 5

(ii) P (a college student) 5 4 9

20 20

(iii) a female student 4 1

20 5

(iv) P (a male employee) 3

20


32. Answer the following questions

gpd;tUk; tpdhf;fSf;F tpilasp

a. If 5 men with 7 boys can earn Rs. 3825 in 6 days and 2 men with 3 boys can earn Rs. 1050 in 4 days, 7 men with 6 boys will earn Rs. 22,500 in time of: xU Ntiyia Ie;J Mz;fs;> 7 rpWth;fs; Nrh;e;J 6 ehl;fspy; &. 3825-I Cjpakhf ngWfpd;wdh; kw;Wk; 2 Mz;fs;> 3 rpWth;fs; Nrh;e;J 4 ehl;fspy; &.1050–I Cjpakhf ngWfpd;wdh; vdpy; 7 Mz;fs;> 6 rpWth;fs; Nru;e;J vj;jid ehl;fspy; &.22>500-I Cjpakhf ngw KbAk;.



Solution: Let, Men = M, Boys = B

Given: (5M + 7B) earn Rs. 3,825 in 6 days (2M + 3B) earn Rs. 1,050 in 4 days Required: 7M + 6B earn Rs. 22,500 in ? days

Formulae: 1 1 2 2

1 2

M ×D M ×D=

W W

(5 7 ) 6 (2 3 ) 4

3825 1050

M B M B

(5 7 ) 2 (2 3 ) 2

1275 525

M B M B

(5 7 ) (2 3 )

17 7

M B M B

35M + 49B = 34M + 51B 1M = 2B

Equation 1: 5M + 7B (5x2B) + 7B 10B + 7B = 17B Required: 7M + 6B = (7 x 2B) + 6B = 14B + 6B = 20B

217 6 20

3825 22500

B B D

D2 =

17 6 22500

3825 20 = 30 days.

b. A, B, and C can do a piece of work in 10, 12, and 15 days respectively. They

began the work together but B leaves after two days, how long would it take A and C to finish the Remaining work?

XU Ntiyia A> B kw;Wk; C Kbf;f vLj;Jf;nfhs;Sk; ehl;fs; KiwNa 10> 12 kw;Wk; 15 MFk;. ,k;%tUk; Nrh;e;J mt;Ntiyia njhlq;fp ,uz;L ehl;fSf;F gpwF B vd;gth; tpyFfpwhh; vdpy;> kPjKs;s Ntiyia A kw;Wk; C Nrh;e;J Kbf;f vt;tsT ehl;fs; MFk;. Solution:

(A, B, C) 1 day work = 1 1 1

10 12 15

= 6 5 4

60

=

15

60 =

1

4

∴ 2 days work = 1

4 x 2 =

1

2



Remaining work completed by A and C = 1 1

12 2

= 10 151

2 25

= 1

62 = 3 days.

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tnpsc deo main premodel exam mental ability & aptitude

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