CONTINUOS MONTHLY ASSESSMENT TEST (CMAT) BATCHES …

CONTINUOS MONTHLY ASSESSMENT TEST (CMAT) FOR

TNPSC GROUP I PRELIMINARY CURRENT BATCHES

MENTAL ABILITY SOLUTION

1. If 48 men working 7 hours a day can do a work in 24 days, then in how many days

will 28 men working 8 hours a day can complete the same work? 48 Mz;fs; xU Ntiyia ehnshd;Wf;F 7 kzp Neuk; Ntiy nra;J 24 ehs;fspy; Kbg;gh; vdpy;> 28 Mz;fs; mNj Ntiyia ehnshd;Wf;F 8 kzp Neuk; Ntiy nra;J vj;jid ehs;fspy; Kbg;gh;?

A. 36 days

B. 24 days

C. 12 days

D. 48 days

Explanation:

M1D1H1 M2D2H2

48 7 24 288 H2

H2

48 7 24 28 8

36 days

2. A is thrice as fast as B. If B can do a piece of work in 24 days, then find the number

of days they will take to complete the work together.

A vd;gth; B vd;gtiuf; fhl;bYk; Ntiy nra;tjpy; %d;W klq;F Ntfkhdth;.

B Mdth; xU Ntiyia 24 ehs;fspy; Kbg;ghh; vdpy;> ,UtUk; ,ize;J me;j Ntiyia Kbf;f vj;jid ehs;fs; vLj;Jf; nfhs;th; vdf; fhz;f.

A. 2 days

B. 3 days

C. 4 days

D. 6 days

Explanation:

A

B

Work Efficiency

3

:

1

Time

1

:

3

A = 24 1 8days 3

Work together = 8 24 6days

32

3. P and Q can do a piece of work in 12 days and 15 days respectively. P started the

work alone and then, after 3 days Q joined him till the work was completed. How

long did the work last?

P kw;Wk; Q MfpNahh; xU Ntiyia Kiwia 12 kw;Wk; 15 ehs;fspy; Kbg;gh;.

PMdth; mej; Ntiyiaj; jdpNaj; njhlq;fpa gpwF> 3 ehs;fs; fopj;J Q

Mdth; mtUld; Nrh;eJ; NtiyahdJ KbAk; NtiyahdJ vj;jid ehs;fs; ePbj;jJ?

tiu mtUld; ,Ue;jhh; vdpy;>

A. 12 days

B. 8 days

C. 14 days

D. 6 days

Explanation:

P work for 3 days = 3 1 12 4

Remaining work = 3 part completed by P & Q.

4

3 4

12 12

15 15

3 1215 5 days 4 27

Total work completed in 5 + 3 = 8 days.

4. A camp had provisions for 490 soldiers for 65 days. After 15 days, more soldiers

arrived and the remaining provisions lasted for 35 days. How many soldiers joined

the camp? xU Kfhkpy; 65 ehs;fSf;F 490 tuP h;fSf;Fg; NghJkhd kspifg; nghUs;fs; ,Ue;jd. 15 ehs;fSf;Fg; gpwF> NkYk; gy tuP h;fs; Kfhkpw;F tej; jhy;> kjP kpUe;j kspifg; nghUs;fshdJ 35 ehs;fSf;F kl;LNk NghJkhdjhf ,Ue;jJ vdpy;> vj;jid tPuuf; s; Kfhkpy; Nrh;eJ; s;sdh;?

A. 700

B. 210

C. 200

d. 500

Explanation:

M1?D1 =M 2 ?D2

490 65 15 490 x 35

490 50 490 x 35

490 x 700

x 210 Men

5. 2 men and 7 boys can do a piece of work in 14 days, 3 men and 8 boys can do the

same in 11 days. 8 men and 6 boys can do 3 times the amount of this work in

A. 21 days

B. 18 days

C. 24 days

D. 36 days

2 Mz;fs; kw;Wk; 7 rpWth;fs; xU Ntiyia 14 ehl;fspy; nra;J Kbf;fpd;wdh;

kw;Wk; 3 Mz;fs; kw;Wk; 8 rpWth;fs; mNj Ntiyia 11 ehl;fspy; nra;J

Kbf;fpd;wdh; vdpy; 8 Mz;fs; kw;Wk; 6 rpWth;fs;> mej; Ntiyiag; Nghy; 3

klqF; Ntiyia vj;jid ehl;fspy; nra;J Kbg;gh;?

A. 21 ehl;fs; B. 18 ehl;fs; C. 24 ehl;fs;

D. 36 ehl;fs;

Explanation:

(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys

 5 men= 10 boys 1man= 2 boys Therefore, Eqn (1) (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys in 14 days Required (8 men + 6 boys) = (8 x 2 +6) boys = 22 boys completed in ? days.

1114 7 days 22

3 Times of work 73 21 days

6. P alone can do 1 of a work in 6 days and Q alone can do 2 of the same work in 4

2

3

days. In how many days working together, will they finish 3 of the work?

4

A. 2 days

B. 8 days

C. 4 days

D. 3 days

P vd;gth; jdpNa xU Ntiyapd; 1 gFjpia 6 ehs;fspYk;> Q vd;gth; jdpNa

2

mNj Ntiyapd; 2 gFjpia 4 ehs;fspYk; Kbg;gh;. ,UtUk; ,ize;J mej;

3

Ntiyapd; 3 gFjpia vj;jid ehs;fspy; Kbg;gh;?

4

A. 2 ehl;fs; B. 8 ehl;fs;

C. 4 ehl;fs;

D. 3 ehl;fs;

Explanation:

P 1 6 days P 12 days 2

Q 2 4 days Q 6 days 3

3 4

12 18

6

3

days

7. A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how

many days will it produce 14400 soaps working 3 hours more a day?

A. 20 days

B. 7 1 days

C. 40 days

D. 30 days

2

xU Nrhg;Gj; njhopw;rhiyahdJ> ehnshd;Wf;F 15 kzp Neuk; Ntiy nra;J 6

ehs;fspy; 9600 Nrhg;Gfisj; jahhpf;fpwJ. ehnshd;Wf;F $Ljyhf 3 kzp Neuk;

Ntiy nra;J 14400 Nrhg;Gfs; jahhpf;f mjw;F vj;jid ehs;fs; MFk?;

A. 20 ehl;fs; B. 7 1 ehl;fs;

2

C. 40 ehl;fs;

D. 30 ehl;fs;

Explanation:

D1H1 = M2H2

W1

W2

6?15 = D2?18 9600 14400

D2

6?15 14400 9600 18

7

1 2

days

8. X alone can do a piece of work in 6 days and Y alone in 8 days. X and Y undertook

the work for . 4800. With the help of Z, they completed the work in 3 days. How

much is Z's share?

X vd;gth; jdpNa xU Ntiyia 6 ehs;fspYk;> Y vd;gth; jdpNa mNj

Ntiyia 8 ehs;fspYk; Kbg;gh;. X kw;Wk; Y MfpNahh; ,e;j Ntiyia

.

4800f;F xg;Gf; nfhz;ldh;. Z vd;gthpd; cjtpAld;> mth;fs; mej; Ntiyia

3 ehs;fspy; Kbj;jdh; vdpy;> njhifapd; Z,d; gq;F vtT; sT?

A. . 600

B. . 800

C. . 700

D. . 500

Z 4800 1 600 8

9. A and B together can do a work in 10 days. They worked together for 4 days and

then B leaves off. A finished the remaining works in 18 days. In how many days,

can A alone finish the whole work

a. 15 days

b. 20 days

c. 24 days

d. 30 days

A kw;Wk; B Mfpa ,UtUk; Nrh;e;J xU Ntiyia 10 ehspy; Kbf;f ,aYk;.

mth;fs; ,UtUk; Nrh;e;J 4 ehl;fs; Ntiy nra;jdh;. gpd;dh; B mjpypUe;J

tpyfp tpLfpwhh;. kjP pAs;s Ntiyia A 18 ehl;fspy; Kbj;jhh;. mej; nkhj;j

NtiyiaAk; A kl;Lk; vj;jid ehl;fspy; Kbf;f KbAk;?

a. 15 ehl;fs; b. 20 ehl;fs; c. 24 ehl;fs; d. 30 ehl;fs;

Explanation:

A & B Work for 4 days = 4 1 2

10 5

Remaining Work = 3 A=18 A=30days

5

10.Find the greatest number consisting of 6 digits which is exactly divisible by

24,15,36?

24, 15 kw;Wk; 36My; tFgLk; 6 ,yf;f kpfg;nghpa vz;?

A. 999999

B. 999790

C. 999720

d. 999240

Explanation:

L.C.M of 24, 15 and 36 = 360

Now, 999999 divide by 360

Remainder = 279

Therefore, the remainder is 279. Hence the required number is = 999999 ? 279 = 999720 Hence, 999720 is the greatest number of 6 digits exactly divisible by 24, 15 and 36.

11. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(1) 2025

(2) 5220

(3) 5025

(4) 2520

Explanation:

The least number divisible by all the numbers from 1 to 10 will be the LCM of these

numbers.

We have,

1 = 1

2 = 2 ? 1

3 = 3 ? 1

4 = 2 ? 2

5 = 5 ? 1

6 = 2 ? 3

7 = 7 ? 1

8 = 2 ? 2 ? 2

9 = 3 ? 3

10 = 2 ? 5

So, LCM of these numbers = 1 ? 2 ? 2 ? 2 ? 3 ? 3 ? 5 ? 7 = 2520

Hence, least number divisible by all the numbers from 1 to 10 is 2520

12.The sum of two numbers is 2000 and their LCM is 21879. Find the numbers

a. 1993, 7

b. 1991, 9

c. 1989, 11

d. 1987, 13

,U vz;fspd; $Ljy; 2000 mtw;wpd; k.P ngh.k 21879 vdpy; mej; vz;fisf; fhz;f.

a. 1993, 7

b. 1991, 9

c. 1989, 11

d. 1987, 13

13.L.C.M. of x3 ? 27, (x ? 3)2 and x2 ? 9 is a. (x ? 3)2 (x + 3)2 c. (x ? 3)2 (x + 3) (x2 + 3x + 9)

b. (x ? 3)2 (x2 + 6x + 9) d. (x + 3)2 (x ? 3)2 (x2 + 3x + 9)

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