aptitude & mental ability

24 Pages·2017·0.69 MB·English

TNPSC GROUP-I A CONSERVATOR OF FORESTS
TEST vi – aptitude & mental ability

1. A shopkeeper marks his goods 30% above the cost price and allows 10%
discount on the marked price. His profit per cent is:
a) 10% b) 12% c) 17% d) 20%
xU filf;fhuh; jd;Dila nghUl;fis mtw;wpd; mlf;f tpiyiatpl 30
tpOf;fhL mjpfkhf tpw;gid nra;a KbT nra;J> me;j eph;za tpiyapy;
,Ue;J 10 tpOf;fhL jsS; gb mspf;fpwhh;. mtUila ,yhgk; vj;jid
tpOf;fhL?
a) 10% b) 12% c) 17% d) 20%
Solution
Marked price = 130% of cost price
Sale price = 130% - 10% of 130 = 117% of cost price
Profit Per cent = 17%
Alternative method:
3010
Profit per cent 3010 17%
100

2. A man bought an old typewriter for ì 1200 and spent ì 200 on its repair. He
sold it for ì 1680. His profit per cent is:
a) 20% b) 10% c) 8% d) 16%
xU egh; xU gioa jl;lrR; g; nghwpia ì 1200 f;F thq;fp mijg; gOJ efP ;fk;
nra;tjw;F ì 200 nryT nra;J> mij ì 1680f;F tpw;gid nra;jhh;. mtUila
,yhg tpOf;fhL:
a) 20% b) 10% c) 8% d) 16%
Solution
Total cost of typerwriter = Rs.(1200+200) = Rs.1400
S.P = Rs.1680
Profit = Rs.(1680-1400) = Rs.280
2800
Hence Profit %  10020%
1400

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3. If 3 toys are sold at the cost price of 4 toys of the same kind, the profit will be:
1 2
a) 25% b) 33 % c) 62 % d) 50%
3 3
4 nghk;ikfspd; mlf;f tpiyf;F mNj tifapyhd 3 nghk;ikfis tpw;gid
nra;jhy;> ,yhgk; vj;jid tpOf;fhlhf ,Uf;Fk;?
1 2
a) 25% b) 33 % c) 62 % d) 50%
3 3
Solution
Cost price of 4 toys = Selling price of 3 toys
Cost price : Selling price = 3 : 4
1 1
Profit%  10033 %
3 3
8
4. If selling price of an article is times its cost price, the profit per cent on it is:
5
a) 120% b) 160% c) 40% d) 60%
8
xU nghUspd; tpw;gid tpiy> mjDila mlf;f tpiyiag; Nghy klqF;
5

vdpy;> me;jg; tpw;gid kPjhd ,yhg tpOf;fhL vtt; sT?
a) 120% b) 160% c) 40% d) 60%
Solution
8 3
Profit  1
5 5
3
 100%60%
5

5. Sourav purchased 30 kg of rice at the rate of ì 10 per kg and 35 kg at the rate of ì
11 per kg. He mixed the two. At what price per kg (in ì ) should he sell the
mixture to make a 30% profit in the transaction?
a) 12.5 b) 13 c) 13.7 d) 14.25
nrsut; xU fpNyh `10 vd;w tpiyapy; 30 fpNyh mhprpiaAk;> xU fpNyh `11
vd;w tpiyapy; 35 fpNyh mhprpiaAk; thq;fpdhh;. mth; mtt; puz;ilAk;
xd;whff; fye;jhh;. mth;> mej; f; fyitia xU fpNyh vd;d tpiyf;F
tpw;why; mej; g; ghpkhw;wj;jpd; thapyhf mtUf;F 30 tpOf;fhL ,yhgk;
fpilf;Fk;.
a) 12.5 b) 13 c) 13.7 d) 14.25
Solution
Total cost price (of 65kg) = 30 10 + 35  11 = 300 + 385 = 685
Total selling price = 685 + 30% of 685 = 890.50
890.50
Selling price per kg = 890.50  13.70
65

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6. Given that 10% of A’s income = 15% of B’s income = 20% of C’s income. If sum
of their income is ` 7800, then B’s income is
a. ` 3600 b. ` 3000 c. ` 2400 d. ` 1800
A d; tUkhdj;jpy; 10 rjtPjKk;> B apd; tUkhdj;jpy; 15 rjtPjKk;> C apd;
tUkhdj;jpy; 20 rjtPjKk; rkk.; ,k;%thpd; nkhj;j tUkhdk ; &.7800 vdpy; B
apd; tUkhdk; vtt; sT?
a. ` 3600 b. ` 3000 c. ` 2400 d. ` 1800
Solution
10% of A = 15% of B = 20% of C  2A = 3B = 4C
A B C
Dividing the ratio by LCM of 2, 3 and 4 i.e., by 12, we get :  
6 4 3
 A:B:C = 6:4:3
4
B ‘s income  78002400
13

7. A number, on subtracting 15 from it, reduces to its 80%. What is 40% of the
number?
a. 75 b. 60 c. 30 d. 90
xU vz;zpypUe;J 15 I fopf;Fk; nghOJ mej; vz;zpd; kjpg;G Fiwe;J 80
rjtPjkhfpwJ> vdpy; mej; vz;zpy; 40 rjtPj kjpg;G ahJ?
a. 75 b. 60 c. 30 d. 90
Solution
Number is reduced by 15 or 20% of original number
 40 % of number = 2  20% of number =2  15 = 30

8. If 60% of A’s income is equal to 75% of B’s income, then B’s income is equal to
x% of A’s income. The value of x is:
a. 70 b. 60 c. 80 d. 90
A d; tUkhdj;jpy; 60 rjtPjKk;> B apd; tUkhdj;jpy; 75 rjtPjKk; rkk ; vdpy;
B apd; tUkhdkhdJ A apd; tUkhdj;jpy; ‘x’ rjtPjkhFk;. vdpy; ‘x’ d; kjpg;G
ahJ?
a. 70 b. 60 c. 80 d. 90
Solution
60% of A’s income = 75% of B’s income
A’s income : B’s income = 75 : 60 = 5 : 4
4
 Required percentage  10080
5

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9. If the numerator of a fraction is increased by 20% and its denominator by 25%,
3
then the fraction so obtained is what is the original fraction?
5
3 2 5
a. b. c. d. 1
5 8 8
xU gpd;dj;jpy; njhFjpapd; kjpg;G 20 rjtPjk; mjpfhpf;fpwJ> gFjpapd; kjpg;G 25
3
rjtPjk; FiwfpwJ> gpd;G mg;gpd;dk; khWfpwJ> vdpy; mg;gpd;dj;jpidf; fhz;f.
5
3 2 5
a. b. c. d. 1
5 8 8
Solution
Let the numerator be x and denominator be y.
x
Now, fraction 
y
120x 3
Then, from question  
125y 5
x 3 125
 
y 5 120

x 3 25 5
  
y 5 24 8
5
Original fraction =
8

10. In a group of students, 70% can speak English and 65% can speak Hindi. If 27%
of the students can speak none of the two languages, then what per cent of
group can speak both the languages?
xU khzth;fs; FOtpy;> 70 rjtPjk; Ngh; Mqf; pyk; NgRgth;fs;> 65 rjtjP k;
Ngh; ,e;jp NgRgth;fs;. 27 rjtPjk; khzth;fs; Nkw;fz;l ,uz;L nkhopfisAk;
Ngrhjthf; s; vdpy;> ,uz;L nkhopfisAk; NgRk; khzth;fs; vj;jid rjtPjk;?
A. 38% B. 62% C. 28% D. 23%
Solution
Let total students = 100
Students who know none of the two lanugages = 27
Remaining students = 100 – 27 = 73
Students who know both langugages = 70 + 65 – 73 = 62%

11. If the price of gold increased by 20% on Monday and decreased by 20% on
Tuesday then by what percentage is Tuesday's price higher or lower than the
initial price of that week?
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jqf; j;jpd; tpiy jpq;fl;fpoik 20% mjpfhpf;fpwJ nrt;tha;fpoikapy; 20%
FiwfpwJ vdpy; nrt;thad;W jq;fj;jpd; tpiy thuj;jpd; njhlf;ftpiyiatpl
Fiwthd rjtPjk; my;yJ mjpfkhd rjtPjk; vtt; sT?

a. 4% lower b. 12.5% higher c. 10% lower d. No change
Solution
I×D 2020
Formula= I-D- = 2020 4%
100 100

12. The price of a loaf of bread was increased by 25%. How many loaves can be
purchased for the amount that used to buy 300 loaves?
nuhl;bj; Jz;bd; tpiy 25% Mf cah;ej; hy;> 300 nuhl;bj; Jz;Lfs; thq;fpa
njhifapy; vj;jid nuhl;bj; Jz;Lfs; thq;f KbAk;?
A. 240 B. 250 C. 260 D. 275
Solution
1 bread cost = Rs.1
300 loaves = 300 Rs.
1 bread 25% increased
Price = 1.25
300 nuhl;bj; Jz;Lfs ; thq;fpa njhifapy;
No of loaves in purchased for the amount buy 300 loaves
300 300
  100  240

1.25 125

13. If x% of y is 100 and y% of z is 200, then the relation between x and z.
yd; x rjtPjk; 100 kw;Wk; zd; y rjtPjk; 200 vdpy; xf;Fk; zf;Fk; cs;s
njhlh;G.
x x
a. z  b. z 2x c. z  d. z 4x

2 4
Solution
x y
y100------ (1) z 100------ (2)
Given
100 100
x 1
eqn (1)(2) 
z 2
z = 2x

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14. If A exceeds B by 40%, B is less than C by 20%, then A : C is :
a. 28 : 25 b. 26 : 25 c. 3 : 2 d. 3 : 1
A d; kjpg;G B ia tpl 40% mjpfk;> B apd; kjpg;G C I tpl 20% FiwT
vdpy; A : C tpfpjk; ahJ?
a. 28 : 25 b. 26 : 25 c. 3 : 2 d. 3 : 1
Solution
A : B = 140 : 100
B : C = 80 : 100
A A B 140 80 28
     
C B C 100 100 25

15. Price of cloth having been raised by 75%, by how much per cent a householder
must reduce his consumption of cloth so as not to increase his expenditure?
6 1
a. 42 % b. 57 % c. 75% d. 50%
7 7
Jzpapd; tpiy 75% mjpfhpf;fpwJ> ,Ugg; pDk;> mth;fsJ FLk;g tuT nrytpy;
ve;jtpj khw;wKk; ,y;iy vdpy;> Jzp cgNahfpf;Fk; msit vt;tsT rjtjP k;
Fiwf;f Ntz;Lk;?
Solution:
100R 10075 6
Formula:  42 %
100R 175 7

16. The population of town 2 years ago was 62,500. Due to migration to big cities, it
decreases every year at the rate of 4%. The present population of the town is:
A. 56,700 B. 57,600 C. 58,800 D. 60,000
xU efuj;jpd; kf;fs;njhif ,uz;L Mz;LfSf;F Kd;G 62>500 nghpa
efuq;fSf;F ,lk;ngah;tjhy; mee; fufj;jpd; kf;fs;njhif Mz;Lf;F 4%
FiwfpwJ. me;efuj;jpd; jw;Nghija kf;fs;njhif vtt; sT?
A. 56,700 B. 57,600 C. 58,800 D. 60,000
Solution
2
 4   24 24
Present population 62500 1  62500  57600
   
 100  25 25

7
17. Find the rate percent at which a sum of money becomes times in 3 years.
6
7
xU mryhdJ 3 tUlj;jpy; klqf; hf MFnkdpy; mjd; tl;b tpfpjk;
6
vtt; sT?
5 5
a. 12% b. 5 % c. 6 % d. 24%
9 9
Solution
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Principal = P
7P
Amount 
6
7P P
SI= -P=
6 6
SI×100 P×100 50
R= = 
P×T 6×P×3 9
5
R = 5 %
9

18. A lent ì 450 to B for 2 years, and ì 500 to c for 3 years, at a certain rate of simple
interest. If he received altogether from both ì 120 as interest, find the rate
interest.
(a) 3% (b) 4% (c) 5% (d) 6%
A vd;gth; B f;F 2 Mz;LfSf;F &.450 I fldhf nfhLj;jhh;> A vd;gth; C
f;F 3 Mz;LfSf;F &.500 I fldhf nfhLj;jhh;> ,t;tpUtUk; &.120
jdptl;bahf nfhLj;jdh;> vdpy; Mz;L tl;b tjP k; vtt; sT?
(a) 3% (b) 4% (c) 5% (d) 6%
Solution
4502x 5003x
  120
   
 100   100 
9x + 15x = 120
x = 5%

19. The Simple Interest on Rs.3,500 at 8% from 4th February, 1996 to 24th April,
1996 is...
a. Rs.61.37 b. Rs.22.40 c. Rs.224 d. Rs.70.20
&.3500 I 8% jdptl;bapy; gpg;uthp 4> 1996 ypUe;J Vg;uy; 24> 1996 tiu
fpilf;Fk; jdptl;b vtt; sT?
a. Rs.61.37 b. Rs.22.40 c. Rs.224 d. Rs.70.20
Solution
P = Rs.3500 R = 8% N = (Feb 25 + Mar 31 + April 24) = 16/73
3500168
SI  61.37
10073

20. A sum of money 5 times of itself in 22 years. In how many years would it
become 4 times of itself?
a. 16 years and 3 months b. 16 years and 2 months
c. 16 years and 9 months d. 16 years and 6 months
xU Fwpg;gpl;l mry; njhif 22 Mz;Lfspy; 5 klq;fhfpwJ vdpy;> mNj mry;
njhif vj;jid Mz;Lfspy; 4 klq;F njhifahFk;?
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a. 16 Mz;Lfs; kw;Wk; 3 khjk; b. 16 Mz;Lfs; kw;Wk; 2 khjk;
c. 16 Mz;Lfs; kw;Wk; 9 khjk; d. 16 Mz;Lfs; kw;Wk; 6 khjk;
Solution
R
400=100×22×
100
400
R= %
22

100×N×400×1 66
300= =
22×100 4
1
N=16 years
2

21. A man borrowed Rs.12,000 from two persons. He paid 5% interest to one and
8% per annum to the other. In one year he paid total interest Rs.840. How
much did he borrow at 5% rate?
a. Rs.7500 b. Rs.4500 c. Rs.8000 d. Rs.4000
xUth; &.12000 I ,uz;L egh;fsplk; ,Ue;J fldhf ngw;whh;. ,jpy; Kjy; egh;
5% tl;bAk; kw;Wk; ,uz;lhk; egh; 8% tl;bAk; ngw;wdh;. ,t;tpUtUf;Fk;
Nrh;j;J xU Mz;L Kbtpy; &.840 tl;bahf nfhLj;jhh; vdpy;> 5% tl;bapy;
ngw;w fld; njhif ahJ?
a. Rs.7500 b. Rs.4500 c. Rs.8000 d. Rs.4000
Solution
x15 12000x18
  Rs.840
100 100
5x960008xRs.84000
3xRs.12000
x = 4000

22. A sum of money amounts to Rs.7,250 after 2 years and Rs.8,500 after 4 years at
the same rate of simple interest. What is the rate percent?
a. 14.02% b. 8.42% c. 10.42% d. 11.42%
xU Fwpg;gpl;l njhif 2 tUlq;fspy; &. 7250 MfTk; NkYk; 4 tUlq;fspy;
&. 8500 MfTk; Kjph;T milfpwJ. vdpy; tl;b tjP k;?
a. 14.02% b. 8.42% c. 10.42% d. 11.42%
Solution
P4SI8500 (2)
P2SI7250(1)
2SI 1250
SI = 625
P + 2 (625) = 7250
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P = 6000
1250100
R  10.42%
60002

23. Find out compound interest on 16,000 at 15% per annum, compounded yearly
for 3 years.
a. Rs.7200 b. Rs.9000 c. Rs.8343 d. Rs.8334
&.16000 I 15% $l;Ltl;bapy; 3 Mz;LfSf;F fpilf;fg;ngUk; tl;bj;njhif
vtt; sT?
a. Rs.7200 b. Rs.9000 c. Rs.8343 d. Rs.8334
Solution
n
 R 
CI=P 1+ -P
 
 100
3
 15 
16000 1+ -16000
 
 100
23 23 23

=16000   16000
20 20 20
2433416000
8334

24. The compound interest on Rs.15,625 at 8% per annum is Rs.4058. The period (in
years) is ....
a. 3 years b. 2 ½ years c. 2 years d. 4 years
&.15,625I 8 rjtjP $l;Ltl;bapy; vj;jid Mz;Lfspy; tl;bj;njhif
&.4058 fpilf;Fk;?
Solution
 R n 
CI P1  1
 100 
n
 8 
405815625 1 1
 
 100
n
 4058  27
1 
   
 15625 25
3
19683 27

   
15625 25
N = 3 years

25. The C.I. on Rs.24,000 at 10% per annum for 1 ½ years where interest being
compounded half-yearly is _____
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6 khjj;jpw;F xU Kiw tl;b fzf;fplg;gl;lhy; &.24>000f;F 1 ½ Mz;L
fhyj;jpw;F 10% tl;b tPjj;jpy; fpilf;Fk; $l;L tl;b vd;d?
A. Rs. 3783 B. Rs.3873 C. Rs.3378 D. Rs.3837
Solution
 105 105 105
C.I 24000   24000 27783240003783
 
 100 100 100

26. A sum on compound interest becomes three times in 4 years. How many years
will it take to become 27 times the original if the interest is calculated at the
same rate?
xU njhifahdJ $l;L tl;b %yk; ehd;F tUlq;fspy; 3 klq;fhfpwJ. ,Nj
tl;b tPjk; %yk; mej; njhifahdJ vj;jid tUlq;fSf;Fg; gpwF 27
klqf; hFk;?
a. 8 years b. 12 years c. 24 years d. 36 years
Solution
3T  4years
9T  8 years
27T 12years

27. A man invested Rs. 25,000 at 4% per annum in compound interest and received
the amount Rs. 27,040 after n years, then value of n is
xUth;> mry; &. 25>000 &ghia Mz;Lf;F $l;L tl;b 4% tjP k; KjyLP nra;J n
Mz;Lfs; fopj;J & 27,040 I njhifahfg; ngWfpwhh; vdpy; n-d; kjpg;G ahJ?
a. 2 yrs b. 3 yrs c. 2½ yrs d. 3½ yrs
Solution
n n 2
104 26 2704 26
25000 27040   
     
100 25 2500 25
 n = 2 years

28. The least number of complete years in which a sum of money put out at 20%
compound interest will be more than doubled is
20% $l;L tl;b tpfpjj;jpy; itf;fg;gl;l xU njhif ,UklqF; f;F Nky; Mtjw;F
krP ;rpW KO Mz;Lfs; vd;d?
A. 3 B. 4 C. 5 D. 6
Solution
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