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MERIT TEST – I FOR TNPSC GROUP – II / II A MENTAL ABILITY SOLUTIONS
1. There are 800 soldiers in an army camp. There is enough provisions for them for 60 days. If 400 more soldiers join the camp for how many days will the provisions last? a. 40 days b. 22 days c. 21 days d. 15 days xU uhZt Kfhkpy; 800 tPuh;fs; ,Uf;fpwhh;fs;. mth;fSf;F 60 ehl;fSf;F NghJkhd kspif rhkhd;fs; cs;sd. me;j Kfhkpw;F NkYk; 400 tPuh;fs; te;J Nrh;e;jhh;fs; vdpy;> vj;jid ehl;fSf;F me;j kspif rhkhd;fs; NghJkhdjhf ,Uf;Fk;? a. 40 ehl;fs; b. 22 ehl;fs; c. 21 ehl;fs; d. 15 ehl;fs;
Solution: M1 x D1 = M2 x D2 800 x 60 = 1200 x D2
D2 = 800 60
1200
D2 = 480
12
D2 = 40 days.
2. A train covers a distance of 195 km in 3 hrs. At the same speed, the distance travelled in 5 hours is _______. a. 195 km b. 325 km c. 390 km d. 975 km xU ,uapy; tz;b 195 fpNyhkPl;lh; J}uj;ij 3 kzp Neuj;jpy; flf;fpd;wJ. mNj Ntfj;jpy;> me;j ,uapy; tz;b 5 kzp Neuj;jpy; flf;Fk; J}uk; a. 195 fp.kP b. 325 fp.kP c. 390 fp.kP d. 975 fp.kP
Solution: km hrs 195 3 x 5
x = 195 5
3
x = 65 x 5
∴ x = 325 km
3. A book contains 120 pages. Each page has 35 lines. How many pages will the book contain if every page has 24 lines per page? a. 173 b. 172 c. 174 d. 175 xt;nthU gf;fj;jpYk; 35 thpfisf; nfhz;l Gj;jfj;jpd; nkhj;j gf;fq;fs; 120. mNj nra;jp xt;nthU gf;fj;jpYk; 24 thpfshf ,Ue;jhy;> Gj;jfj;jpd; nkhj;j gf;fq;fs; vt;tsthf ,Uf;Fk;? a. 173 b. 172 c. 174 d. 175
Solution: pages lines 120 35 x 24
x = 120 35
24
∴ x = 175 pages
4. Ram bought 36 mangoes, 5 mangoes were rotten. What is percentage of the mangoes that were rotten? a. 12.99% b. 13.89% c. 13.79% d. 13.59% uhk; thq;fpa 36 khk;goq;fspy;> 5 khk;goq;fs; mOfptpl;ld vdpy;> mOfpa khk;goq;fspd; rjtPjj;ijf; fhz;f. a. 12.99% b. 13.89% c. 13.79% d. 13.59%
Solution: Ram bought = 36 mangoes Number of rotten = 5 mangoes
Percentage of rotten mangoes = 5
10036
= 500
36
= 13.89%
5. There are 250 students in a school. 55 students like basketball, 75 students like football, 63 students like throw ball, while the remaining like cricket, what percent of students like basketball? a. 32% b. 42% c. 12% d. 22% 250 khztu;fs; cs;s xU gs;spapy;> 55 khztu;fs; $ilg; ge;ijAk; 75 khztu;fs; fhy;ge;ijAk;> 63 khztu;fs; vwpge;ijAk; kPjKs;stu;fs; fpupf;nfl;ilAk; tpUk;Gfpd;wdu; vdpy; $ilg;ge;ij tpUk;Gk; khztu;fspd; rjtPjk; vd;d? a. 32% b. 42% c. 12% d. 22%
Solution:
6. Gain or loss percent is always calculated on _______. a. cost price b. selling price c. gain d. loss ,yhg my;yJ el;l rjtPjk; vg;nghOJk; _______ Nky; fzf;fplg;gLk; a. mlf;ftpiyapd; b. tpw;gid tpiyapd; c. ,yhgj;jpd; d. el;lj;jpd;
Explanation:
7. A man bought an old bicycle for ` 1,250. He spent ` 250 on its repairs. He then sold it for ` 1400. Find his gain or loss% a. 7.2% b. 6.67% c. 5% d. 6% xU egh; xU gioa kjptz;bia `1>250f;F thq;fpdhh;. mjid rPh;g;gLj;j `250 nryT nra;jhh;. mth; mjid `1400f;F tpw;whh;. mthpd; ,yhgj;ij my;yJ el;lj;ijf; fhz;f. a. 7.2% b. 6.67% c. 5% d. 6%
Solution: Cost Price of the bicycle = `1,250
Repair Charges = `250
Total Cost Price = 1250 + 250 = `1,500
Selling Price = `1,400
C.P > S.P., there is a Loss
Loss = Cost Price – Selling Price
= 1500 – 1400
= 100
Loss = `100
Percentage of Loss = 100Pr
Loss
Cost ice
= 100
1001500
= 20
3 =
26
3
Loss % = 6.67%
8. A trader mixes two kinds of oil, one costing Rs. 100 per kg and the other costing Rs. 80 per kg in the ratio 3 : 2 and sells the mixture at Rs. 101.20 per kg. Find his profit or loss percent. a. 8% b. 7% c. 9% d. 10% fpNyh &.100> &.80 cs;s ,uz;L tifahd vz;izia xU tpw;gidahsh; 3 : 2 vd;w tpfpjj;jpy; fye;J mf;fyitia fpNyh &. 101.20f;F tpw;gidr; nra;fpd;whh;. mtUila ,yhg my;yJ el;l rjtPjj;ij fhz;f. a. 8% b. 7% c. 9% d. 10%
Solution: Let, Quantity of oil is 3 kg and 2 kg. Total Quantity = 5 kg The cost of first oil = ` 100 3 = 300 The cost of second oil = ` 80 2 = 160 Total Cost Price= ` 460. Selling the Mixture = 101.2 5 = 506 Profit = 506 – 460 = 46
Profit % = 46
100 10%460
9. Sathish sold a camera to Rajesh at a profit of 10%. Rajesh sold it to John at a
loss of 12%. If John paid Rs. 4,840, at what price did Sathish buy the camera? a. Rs. 5,000 b.Rs. 4000 c. Rs. 5500 d. Rs. 5200 rjP~; xU Nfkuhit uhN[~plk; 10% ,yhgj;jpw;F tpw;whh;. uhN[~; mjid [hdplk; 12% el;lj;jpw;F tpw;whh;. [hd; &.4,840f;F thq;fpapUe;jhy;. me;j Nfkuhit rjP~; vd;d tpiyf;F thq;fpdhh;? a. Rs. 5,000 b.Rs. 4000 c. Rs. 5500 d. Rs. 5200
Solution: Satish sold a camera to Rajesh at profit 10%. Rajesh sold the same camera to John at loss of 12%. Let Cost Price be x.
100 10 100 12
4840100 100
x
110 88
4840100 100
x
100 100
4840110 88
x
∴ x = 5000
10. Find the amount when `2,500 is invested for 146 days at 13% per annum. a. ` 2,650 b. ` 2,620 c. ` 2,630 d. ` 2,625 `2>500 I 13% tUl tl;b tPjk; itg;G epjpahf nrYj;jpdhy;> 246 ehl;fspy; ngWk; njhifiaf; fhz;f. a. ` 2,650 b. ` 2,620 c. ` 2,630 d. ` 2,625
Solution:
Given: P = ` 2,500 n = 146 days = 146
365 r = 13%
Simple Interest = I = 100
Pnr
= 13 146
2500100 365
= 25 13
2.5
= 250 13
25
= ` 130
Amount = Principal + Interest = 2,500 + 130
∴ Amount = ` 2,630
11. Find the S.I. on ` 12,000 from May 21st 1999 to August 2nd 1999 at 9% per annum. a. ` 126 b. ` 2226 c. ` 216 d. ` 2216 `12>000 f;F 9% tUl tl;b tPjk; 21 Nk 1999 ypUe;J 2 Mf];L 1999 tiu fpilf;Fk; jdptl;bia fhz;f. a. ` 126 b. ` 2226 c. ` 216 d. ` 2216
Solution: Given: P = ` 12,000 n = May 21st 1999 to August 2nd 1999
73 days =
73 1
365 5year
Simple Interest = I = 100
Pnr
= 12,000 x 9 1
100 5
= 24 x 9 = ` 216
12. A sum of money doubles itself at 12½% per annum over a certain period of time. Find the number of years. a. 7 years b. 8 years c. 9 years d. 6 years xU njifahdJ 12½% tUl tl;b tPjj;jpy; xU Fwpg;gpl;l tUlq;fspy; ,ul;bg;ghfpwJ. tUlq;fspd; vz;zpf;ifiaf; fhz;f. a. 7 Mz;Lfs; b. 8 Mz;Lfs; c. 9 Mz;Lfs; d. 6 Mz;Lfs;
Solution:
x=2 Rate of Interest = 1 25
12 %2 2
N=?
1
100x
NR
2 1
10025 2
=
2100
25
∴ N = 8 years
May = 11 days June = 30 days July = 31 days August = 1 days Total = 73 days
13. A certain sum of money amounts to ` 6,500 in 3 years and ` 5,750 in 1½ years respectively. Find the principal and the rate percent? a. ` 5,100; 10% b. ` 5,000; 10% c. ` 4,000; 10% d. ` 5,000; 15% xU Fwpg;gpl;lj; njhifahdJ 3 Mz;Lfspy; `6>500 MfTk; 1½ Mz;Lfspy; `5>750 MfTk; khWfpwJ. mry; kw;Wk; tl;b tPjj;ijf; fhz;f. a. ` 5,100; 10% b. ` 5,000; 10% c. ` 4,000; 10% d. ` 5,000; 15%
Solution: The amount in 3 years = 6500
The amount in 1
12
years = 5750
Interest in 1
12
years = 750 = Simple interest
Before 1
12
years the principal = P = A – I
= 5,750 – 750 P = 5,000
Rate of Interest (r) = . 100S I
P n
[ n =
11
2 year =
3
2 years]
= 750 100 75 2
3 1550002
∴ r = 10%
14. If the angles of a triangle are in the ratio 1 : 3: 5 Then the angles are a. 20°, 100°, 60° b. 20°, 60°, 100° c. 30°, 60°, 90° d. 60°, 30°, 90° xU Kf;Nfhzj;jpd; Nfhzq;fs; 1:3:5 vd;w tpfpjj;jpy; cs;sd vdpy; mf;Nfhzq;fspd; kjpg;Gfs; KiwNa a. 20°, 100°, 60° b. 20°, 60°, 100° c. 30°, 60°, 90° d. 60°, 30°, 90°
Solution: Sum of three angles of triangle = 180° Given ratio = 1 : 3 : 5
= 180 180
201 3 5 9
Required Angle = (20 x 1) : (20 x 3) : (20 x 5) = 20°, 60°, 100°
15. If A is 1/3 of B and B is 1/2 of C then A:B:C is : a. 1:3:6 b. 2:3:6 c. 3:2:6 d. 3:1:2
A vd;gJ B-y; 1/3> B vd;gJ C-y; 1/2 vdpy; A:B:C
a. 1:3:6 b. 2:3:6 c. 3:2:6 d. 3:1:2
Solution: A B B C 1 : 3 1 : 2 A B C
1 3
1 2
∴ Required Ratio = 1 : 3 : 6
16. In a proportion, the product of extremes = ______. a. product of means b. Sum of Means c. Subtract of Means d. Sum of Extremes tpfpjrkj;jpy; <w;nwz;fspd; ngUf;Fj;njhif = ________. a. ,il vz;fspd; ngUf;Fj;njhif b. ,il vz;fspd; $Ljy; c. ,il vz;fspd; tpj;jpahrk; d. <w;nwz;fspd; $Ljy;
17. If a plant grows 3
5cm in an hour, than its height after
12
2days is ____.
a. 6
10cm b. 30 cm c. 35 cm d. 36 cm
xU jhtuk; 1 kzp Neuj;jpy; 3
5nr.kP. cauk; tsh;fpwJ vdpy;
12
2ehl;fSf;F
gpwF mj;jhtuj;jpd; cauk; _______.
a. 6
10cm b. 30 cm c. 35 cm d. 36 cm
Solution:
1 hour → 3
5 cm ∴
12
2 days = 60 hours
60 hours → 3
605 = 36 cm
18. If Ramya needs Rs.9,00,000 after ten years, how much should she invest in a
bank pays 20% simple interest. a. Rs. 2,00,000 b. Rs. 3,00,000 c. Rs. 4,00,000 d. Rs.5,00,000 gj;J Mz;bw;Fg; gpwF uk;ahtpw;F &.9>00>000 Njitg;gLfpwJ vdpy; Mz;bw;F 20% jdptl;b mspf;Fk; tq;fpapy;> uk;ah vt;tsT mryhf nrYj;j Ntz;Lk;? a. Rs. 2,00,000 b. Rs. 3,00,000 c. Rs. 4,00,000 d. Rs.5,00,000
Solution: Let, Investment be Rs. P P + S.I = A
P + 10 20
100
P = 9 lakhs
P + 2P = 9 lakhs 3P = 9
∴ P = 3 lakhs
19. 40% of the population of a town are men and 35% are women. If the number of children are 20000, then the number of men will be : a. 3200 b. 80000 c. 32000 d. 3,20,000 xU efuj;jpd; kf;fs; njhifapy; 40% Ngh; Mz;fs;> 35% Ngh; ngz;fs;> rpWth;fspd; vz;zpf;if 20,000 vdpy;> me;efuj;jpYs;s Mz;fspd; vz;zpf;if vt;tsT? a. 3200 b. 80000 c. 32000 d. 3,20,000
Solution: No. of children = 100% - [40 + 35]% = 100 – 75 = 25%
25% → 20,000 40% → x
x = 20,000 40
25
= 32,000 men
20. A garden is in the form of a triangle. Its base is 26 m and height is 28 m. Find
the cost of levelling the garden at `5 per m2. xU Njhl;lkhdJ Kf;Nfhz tbtpy; cs;sJ. mjd; mbg;gf;fk; 26 kP> cauk; 28 kP. Njhl;lj;ijr; rkd;nra;a rJu kPl;lUf;F Rs. 5 tPjk; MFk; nkhj;j nryitf; fhz;f. a. ì 1,800 b. ì 1,900 c. ì 1,820 d. ì 2,000
Solution: Base of the garden = 26 m = b Height of the garden = 28 m = h
Area of the garden = 1
.2
b h sq units
= 1
26 282
= 13 x 28 = 364 m2 Cost of levelling the garden at m2 = ` 5 Cost of levelling 364 m2 = 364 x 5 = ` 1,820
21. A diagonal of a quadrilateral is 25 cm, and perpendicular on it from the opposite vertices are 5 cm and 7 cm. Find the area of the quadrilateral. xU ehw;fuj;jpd; %iytpl;lk; 25 nr.kP vjph; cr;rpfspy; ,Ue;J %iytpl;lj;jpd; Nkyike;j nrq;Fj;jpd; ePsq;fs; 5 nr.kP> 7 nr.kP vdpy; ehw;fuj;jpd; gug;gsT ahJ? a. 150 cm2 b. 145 cm2 c. 155 cm2 d. 125 cm2
Solution:
Diagonal of a quadrilateral = 25 cm = d Perpendicular distance from the opposite vertices h1 = 7 cm h2 = 5 cm
Area of the quadrilateral = 1 2
1( ) .
2d h h sq units
= 1
25(7 5)2
= 1
25(7 5)2
= 25 x 12
2 = 25 x 6
= 150 cm2
22. A ground is in the form of a parallelogram. Its base is 324 m and its height is 75 m. Find the area of the ground. xU tpisahl;Lj; jply; ,izfuk; tbtpy; cs;sJ. mjd; mbg;gf;fk; 324 kP kw;Wk; Fj;Jauk; 75 kP vdpy; tpisahl;Lj; jplypd; gug;gsT vd;d? a. 23,300 m2 b. 24,300 m2 c. 22,400 m2 d. 23,500 m2
Solution: Base of a parallelogram = 324 m = b Height of a parallelogram = 75 cm = h
∴ Area of the ground = b x h sq.units = 324 x 75 = 24,300 m2.
23. A field is in the form of a rhombus. The diagonals of the fields are 50 m and 60 m. Find the cost of levelling it at the rate of `2 per sq. m. xU tayhdJ rha;rJu tbtpy; cs;sJ. taypd; %iytpl;l msTfs; 50kP> 60kP. me;j taiyr; rkd;nra;a rJu kPl;lUf;F `2 tPjk; MFk; nryitf; fhz;f. a. `2900 b. `2800 c. `3000 d. `2700
Solution:
24. The area of a trapezium is 102 sq. cm and its height is 12 cm. If one of its parallel sides is 8 cm. Find the length of the other side. xU rhptfj;jpd; gug;gsT 102 r.nr.kP> nrq;Fj;Jj; njhiyT 12 nr.kP. rhptfj;jpd; ,izg;gf;fq;fspy; xU gf;fj;jpd; ePsk; 8 nr.kP vdpy; kw;nwhU gf;fj;jpd; ePsnkd;d? a. 7 cm b. 8 cm c. 11 cm d. 9 cm
Solution:
25. A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel. xU kfpOe;jpd; rf;fuk; 66 kP. njhiyT flf;f 20 Rw;Wfs; Rw;wpdhy; mr;rf;fuj;jpd; tpl;lk; fhz;f. a. 1.02 m b. 1.10 m c. 1.03m d. 1.05 m
Solution: A wheel covers a distance in 20 revolution = 66 m
∴ A wheel covers a distance in one revolution = 66 33
20 10 = 3.3 m
πd = 3.3
22
7d = 3.3
d = 3.3 x 7
22
d = 0.3 7
2
=
2.1
2 = 1.05 m
∴ The diameter of the wheel = 1.05 m
26. The circumference of a circular field is 44 m. A cow is tethered to a peg at the centre of the field. If the cow can graze the entire field, find the length of the rope used to tie the cow. xU tay; ntspapd; Rw;wsT 44 kP. tay; ntspapd; ikaj;jpy; mbf;fg;gl;Ls;s Kidapy; Jk;Gf;fapW nfhz;L xU gRkhL fl;lg;gl;Ls;sJ. tay;ntsp KOtJk; gRkhL Nka KbAkhdhy; gRkhL fl;lg;gl;Ls;s Jk;Gf; fapw;wpd; ePsnkd;d?
a. 3.5 m b. 14 m c. 10.5 m d. 7 m
Solution:
27. A silver wire when bent in the form of a square encloses an area of 121 sq. cm.
If the same wire is bent in the form of a circle. Find the area of the circle. xU nts;spf; fk;gp tisf;fg;gl;L rJukhf khw;Wk; NghJ> mjdhy; milgLk; gFjpapd; gug;gsT 121 r.nr.kP. mNj nts;spf;fk;gp tl;lkhf tisf;fg;gLfpwJ vdpy; tl;lj;jpd; gug;gsT vd;d?
a. 77 sq.cm b. 154 sq.cm c. 38.5 sq.cm d. 308 sq.cm
Solution:
28. The length of a building is 20 m and its breadth is 10 m. A path of width 1 m is
made all around the building outside. Find the area of the path. xU nrt;tf tbtf; fl;llj;jpd; ePsk; 20 kP> mfyk; 10 kP. fl;llj;ijr; Rw;wp ntspg;Gwkhf 1 kP mfyKs;s ghij mikf;fg;gl;Ls;sJ. ghijapd; gug;gsT fhz;f.
a. 264 sq.m b. 200 sq. m c. 32 sq.m d. 64 sq.m
Solution:
29. Which of the following statement(s) is/are true statements?
i. All squares are rhombuses also ii. All parallelograms are rectangles also iii. All parallelograms are quadrilaterals also iv. All trapeziums are rhombuses also a. (i)only b. (ii), (iii) c. (i), (ii), (iii) d. (i), (iii) fPo;f;fz;l $w;WfSs; nka;ahd $w;W vJ/vit? i. vy;yh rJuq;fSk; rha;rJuq;fs; MFk; ii. vy;yh ,izfuq;fSk; nrt;tfq;fs; MFk; iii. vy;yh ,izfuq;fSk; ehw;fuq;fs; MFk; iv. vy;yh rhptfq;fSk; rha;rJuq;fs; MFk; a. (i) kl;Lk; b. (ii), (iii) c. (i), (ii), (iii) d. (i), (iii)
30. Which of the following statement is false in a Parallelogram? a. The opposite sides are parallel b. The opposite angles and sides are equal c. The diagonals are equal d. The diagonals bisect each other Xh; ,izfuj;jpy; vJ jtwhd $w;W?
a. vjph;g; gf;fq;fs; ,izahFk; b. vjpnujph; Nfhzq;fs; kw;Wk; gf;fq;fs; rkkhFk; c. %iy tpl;lq;fspd; ePsq;fSk; rkkhFk; d. %iy tpl;lq;fs; xd;iwnahd;W ,U rkf;$wpLk;
31. The diagonals of a rhombus bisect each other at rha;rJuj;jpd; %iytpl;lq;fs; xd;iwnahd;W ve;j Nfhzj;jpy; ,Urkf;$wpLk;?
a. 30° b. 45° c. 60° d. 90°
32. A line segment joining any two points on the circle is called _____. a. Diameter b. Radius c. Chord d. None tl;lj;jpd; NkYs;s VNjDk; ,U Gs;spfisr; Nrh;f;Fk; Nfhl;Lj; Jz;bd; ngah; a. tpl;lk; b. Muk; c. ehz; d. xd;Wkpy;iy
33. Match the following:
A. Square 1. 1
2 × b × h sq. units
B. Parallelogram 2. All sides equal
C. Quadrilateral 3. Opposite sides equal D. Right triangle 4. Closed figure bounded by four line segments
A B C D a. 1 2 3 4 b. 4 1 3 2 c. 2 1 4 3 d. 2 3 4 1 gpd;tUtdtw;iwg; nghUj;Jf
A. rJuk; 1. 1
2 × b × h r. myFfs;
B. ,izfuk; 2. midj;Jg; gf;fq;fSk; rkk; C. ehw;fuk; 3. vjpnujph; gf;fq;fs; rkk; D. nrq;Nfhz Kf;Nfhzk; 4. ehd;F Nfhl;Lj; Jz;Lfshy;
#og;gl;l xU tbtk; A B C D
a. 1 2 3 4 b. 4 1 3 2 c. 2 1 4 3 d. 2 3 4 1
34. Match the following: A. Perimeter 1. Surface enclosed by closed figure B. Area 2. d1 = d2 C. Rhombus 3. Boundary of the closed figure D. Square 4. d1 d2
A B C D a. 1 2 3 4 b. 3 1 4 2 c. 2 1 4 3 d. 2 3 4 1 gpd;tUtdtw;iwg; nghUj;Jf: A. Rw;wsT 1. %lg;gl;l tbtj;jpdhy; #og;gl;l gFjp B. gug;gsT 2. d1 = d2 C. rha; rJuk; 3. %lg;gl;l tbtj;jpd; vy;iy D. rJuk; 4. d1 d2
A B C D a. 1 2 3 4 b. 3 1 4 2 c. 2 1 4 3 d. 2 3 4 1
35. 7 men can complete a work in 52 days. In how many days will 13 men finish the same work?
a. 20 days b. 22 days c. 28 days d. 15 days 7 Ml;fs; xU Ntiyia 52 ehl;fspy; nra;J Kbf;fpd;wdh;. mNj Ntiyia 13 Ml;fs; vj;jid ehl;fspy; nra;J Kbg;ghh;fs;?
a. 20 days b. 22 days c. 28 days d. 15 days
Solution:
36. The perimeter of a square is 120 m. Find its area.
a. 800 m2 b. 900 m2 c. 1000 m2 d. 700 m2
xU rJuj;jpd; Rw;wsT 120 kP vdpy;> mjd; gug;gsitf; fz;Lgpbf;fTk;. a. 800 m2 b. 900 m2 c. 1000 m2 d. 700 m2
Solution: Perimeter of a square = 4a 4a = 120 m
a = 120
304
m
Area of a square = a2
= 30 x 30 = 900 m2 37. Find the area of the following figures:
fPo;f;fhZk; glj;jpd; gug;gsitf; fhz;f.
a. 345 cm2 b. 260 cm2 c. 365 cm2 d. 225 cm2
Solution: Area of the figure = Area of square + Area of rectangle 15 = 3 + x + 5 x = 15 – 8 x = 7 = (15 x 15) cm2 + (7 x 20) cm2 = 225 cm2 + 140 cm2 = 365 cm2 38. Find the area of the following figures:
fPo;f;fhZk; glj;jpd; gug;gsitf; fhz;f.
a. 750 cm2 b. 400 cm2 c. 80 cm2 d. 1200 cm2
Solution: Area of the figure = Area of rectangle + Area of right triangle
= (10 x 30) cm2 + 1
( 30 30)2 cm2
= 300 cm2 + (15 x 30) cm2 = 300 cm2 + 450 cm2 = 750 cm2
39. The radii of 2 concentric circles are 56 cm and 49 cm. Find the area of the pathway. nghJ ika tl;lq;fspd; Muq;fs; 56 nr.kP.> 49 nr.kP.> vdpy;> tl;lg; ghijapd; gug;G ahJ?
a. 2300 sq.cm b. 2310 sq.cm c. 2315 sq.cm b. 2210 sq.cm
Solution: Area of the circular pathway = π(R + r) (R – r) R = 56 cm r = 49 cm
= 22
(56 49)(56 49)7
= 22
105 77
= 2,310 cm2 40. Find the cost of fencing a square flower garden of side 50 m at the rate of `10
per metre. rJu tbtg; g+e;Njhl;lj;jpd; gf;fk; 50 kP. g+e;Njhl;lj;ijr; Rw;wp kPl;lUf;F &.10 tPjk; NtypNghl MFk; nryitf; fhz;f. a. `1900 b. `2000 c. `2100 d. `2200
Solution:
APPOLO STUDY CENTRE
NO: 25, NANDHI LOOP STREET
WEST CIT NAGAR
CHENNAI – 35
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