Correct

1 km = 1000 m; 1 minute = 60 sec.

Time to run 1000 m = 7 min 10 sec. = 430 seconds

$\therefore Time\:to\:run\:1\:m=\frac{430}{1000}sec.=\frac{43}{100}sec$

$\therefore Time\:to\:run\:400\:m=\frac{43}{100}\times 400sec.=172sec$

= 2 min 52 sec.

Correct

Cost of 3 dozen pencils = Rs. 370

$\therefore Cost\:of\:1dozen\:pencils=Rs.\frac{370}{3}$

$\therefore Cost\:of\:21dozen\:pencils=Rs.\frac{370}{3}\times 21$

= Rs. 2590

Correct

Cost of $8\frac{1}{3}$km of cloth = Rs. 50

Cost of 1 m of cloth = $Rs.50\times \frac{3}{25}=Rs.6$

Cost of $3\frac{1}{4}m\:of\:cloth=Rs.6 \times \frac{3}{14}= Rs.19.50$

Correct

Worth of $\frac{5}{39}$ of cargo = Rs. 1120

Total worth of cargo = Rs. $1120 \times \frac{39}{5}$ = Rs. 8736

Worth of $\frac{1}{8}\:of\:cargo\:=Rs.8736 \times \frac{1}{8}= Rs.1092$

Correct

Quantity of diesel to travel 594 km = 108 l

$\therefore Quantity\: of\: diesel\: travel\: 1 km =\frac{108}{594}l=\frac{2}{11}l$

“1650 km and back means $1650 \times 2=3300"$

$\therefore$ quantity of diesel to travel 3300 km

= $\frac{2}{11}\times 3300\: l=600\: l$

Correct

5 kg 100 g = 5100 gm 1kg 300 kg = 1300 gm

Required ratio = 5100 : 1300 or 51 : 13

Correct

$\frac{1}{3}\times 12.60:\frac{3}{5}\times 21$

= 4 : 2 : 12.6 or 1 : 3

Correct

$\frac{10}{100}x=\frac{20}{100}y\Rightarrow x=2y\Rightarrow \frac{x}{y}=\frac{2}{1}or\:x : y=2 : 1$

Correct

Let the numbers be x and 5x

$\Rightarrow 5x^{2}=320\Rightarrow x=8$

Hence the numbers are 8 and 40

$40^{2}-8^{2}=1600-64=1536$

Correct

Speed of train : Speed of bus

= $\frac{258}{3}:\frac{301}{7}$ = 86 : 43 = 2 : 1

Correct

Total profit = Rs. (12,000 + 16,000) = Rs. 50,000

Profit in March : total profit = 16,000 : 50,000 = 8 : 25

Correct

Total number of members = 25 + 30 + 20 = 75

Total number of adults = 30 + 20 = 50

Adults : total = 50 : 75 = 2 : 3

Correct

$\frac{63}{4}:\frac{315}{16}=\frac{63}{4}:\frac{63}{4}\times \frac{5}{4}$

= $1:\frac{5}{4}=4:5$

Correct

Put x = 2 and y = 3

$3\times 2+2\times 3:9\times 2+5\times 3=12:33=4:11$

Correct

Put x = 7 and y = 5

$6\times 7+7\times 5:12\times 7+5\times 5=77:109$

Correct

$A:B=(3:4)\times 5;\: B:C=(5:7)\times 4$

$\Rightarrow$ A : B = 15 : 20; B : C = 20 : 28

$\Rightarrow$ A : C = 15 : 28

Also B : C = $(5:7)\times 8;\, C:D=(8:9)\times 7$

$\Rightarrow$ B : D = 40 : 56; C: D = 56 : 63

Also A : C = $(15:28)\times 2;\, C:D=(56:63)\times 1$

A : C = 30 : 56; C : D = 56 : 63

$\Rightarrow$ A : D = 30 : 63 = 10 : 21

Hence all these statements I, II and III are correct.

Correct

For A : B if A > B and a > 0 then

A + a : B + a < A : B

Correct

For A : B if A > B and a > 0 then

A – a : B – a > A : B.

Correct

Duplicate ratio of 25 : 9 = $25^{2}.9^{2}$ or 625 : 81.

Correct

Triplicate ratio of 3 : 2 is $3^{3}.2^{3}$ or 27 : 8

Correct

Sub duplicate ratio of 225 : 169 is

= $\sqrt{225}:\sqrt{169}=15:13$

Correct

Sub triplicate ratio of 1331 : 729

= $\sqrt[3]{1331}:\sqrt[3]{729}=11:9$

Correct

3x – 8 : 4x – 8 = 2 : 3 $\Rightarrow$ 9x – 24 = 8x – 16 $\Rightarrow$ x = 8

Smaller number = $3\times 8=24$

Correct

x + 4 : 2x + 4 = 2 : 3 $\Rightarrow$ 3x + 12 = 4x + 8

$\Rightarrow$ x = 4

Greater number = $2\times 4=8$

Correct

x : 143 = 315 : 231

X = $\frac{315 \times 143}{231}=195$

Correct

29(2x + 3y) = 18(3x + 5y)

$\Rightarrow$ 58x + 87y = 54x + 90y $\Rightarrow$ 4x = 3y

$\frac{x}{y}=\frac{3}{4}$ or x : y = 3 : 4

Correct

a : b = $(a+c)^{2}:(b+c)^{2}$

$\Rightarrow a(b^{2}+c^{2}+2bc)=b(a^{2}+c^{2}+2ac)$

$\Rightarrow ab^{2}+ac^{2}+2abc=a^{2}b+bc^{2}+2abc$

$\Rightarrow (a-b)c^{2}=ab(a-b)$

$\Rightarrow c^{2}=ab$

Correct

(6 + x) : (10 + x) = (14 + x) : (22 + x)

$\Rightarrow$ (6 + x) (22 + x) = (10 + x) (14 + x)

$\Rightarrow 132+28x+x^{2}=140+24x+x^{2}$

$\Rightarrow$ 4x = 8

$\Rightarrow$ x = 2

Correct

Let the two numbers be x and y

$\sqrt{xy}=24\Rightarrow xy=576$

Also x : y = y : 1536 $\Rightarrow y^{2}=1536x$

$\Rightarrow y^{3}=1536xy=1536 \times 576$

$\Rightarrow y=4 \times 24=96$

$\Rightarrow x=\frac{576}{96}=6$

Sum = x + y = 96 + 6 = 102

Correct

8 : 14 = 16 : X

$\Rightarrow$ X = 28

Correct

Applying $\frac{c}{d}$

$\frac{\sqrt{x+4}}{\sqrt{x-10}}=\frac{7}{3}$

$\Rightarrow \frac{x+4}{x-10}=\frac{49}{9}\Rightarrow 9x+36=49x-490$

Solving x = $\frac{263}{20}$

Correct

Paint : oil = $\left ( \frac{9}{10} \right )^{3}:\left ( 1-(\frac{9}{10})^{3} \right )$

= 729 : 271

Correct

Let bag 1 have x mangoes and bag 2 have y mangoes

Then $\frac{x}{2}:y+\frac{x}{2}=y:x$

Also $\frac{2}{3}x+\frac{2}{3}y=\frac{1}{3}x+\frac{1}{3}y$

Required ratio = $\frac{\frac{2}{3}(x+y)}{\frac{1}{3}(x+y)}=\frac{2}{1}=2:1$

Correct

Let CP of milk be Rs. 1

$\Rightarrow$ SP of milk is Rs. 1/l

Gain % = $16\frac{2}{3}\%$

CP of mixture = Rs. $\frac{6}{7}/l$

$\frac{Quantity\:of\:water}{Quantity\:of\:milk}=\frac{1-\frac{6}{7}}{\frac{6}{7}-0}=\frac{1}{6}$

Quantity of water : quality of milk = 1 : 6

Correct

% of gold = $\frac{75 \times 2+66 \times 3}{2+3}\:\:\%$

= 69.9%

$\approx$ 70%

Correct

SP of 1l of mixture = Rs. 45

Gain % = 12.5%

$\Rightarrow CP\:of\:1l\:of\:mixture\:=Rs.\frac{100 \times 45}{100+12.5}$

= Rs. 40 (we assume cost of water to be Rs. 0/l)

Quantity of spirit for Rs. 40 = $\frac{1}{50}\times 40\: l=0.8\: l$

Hence quantity of water in each liter of mixture = 0.2 l or 200 ml

Correct

Let us take cost of water = Rs. 0/l

$\frac{Quantity\: of\: water}{14000}=\frac{10.80-8.40}{8.40-0}$

$\frac{Quantity\: of\: water}{14000}=\frac{2.4}{8.4}$

Quantity of water = $\frac{24}{84}\times 14000\:l$

= 4000 l

Or 4 kl

Correct

(Using Quicker Method-2): Quantity of water to be added =

$\frac{40(20-10)}{100-20}\: lt=5l.$

Correct

$\frac{Quantity\:of\:water}{Quantity\:of\:milk}=\frac{12-8}{8-0}=\frac{4}{8}=\frac{1}{2}$

Quantity of water : Quantity of milk = 1 : 2

Correct

Initial % of milk = 80%

Find % of milk = 90%

Quantity of milk to be added = $\frac{125(90-80)}{(100-90)}\: l.$

= 125 l

Correct

$\frac{1}{t}=\frac{1}{20}+\frac{1}{30}=\frac{5}{30}\Rightarrow t=\frac{60}{5}=12hrs$

$\Rightarrow$ time = 12 hours

Correct

$\frac{1}{t}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12};t=2.4$

Time = 2.4 hrs or 2 hours 24 minutes

Correct

$\frac{1}{t}=\frac{1}{12}+\frac{1}{18}+\frac{1}{9}=\frac{1}{4}$

$\Rightarrow$ time = 4 min.

Correct

$\frac{1}{8}-\frac{1}{16}=\frac{1}{16}$

$\frac{1}{16}$ of tank is emptied in 1 min.

$\Rightarrow \frac{1}{2}$ of tank will be emptied in 8 min.

Correct

$\frac{1}{12}+\frac{1}{15}-\frac{1}{x}=\frac{1}{20}$

$\Rightarrow \frac{1}{x}=\frac{1}{12}+\frac{1}{15}-\frac{1}{20}=\frac{1}{10}$

$\Rightarrow$ x = 10

$\Rightarrow$ time = 10 min.

Correct

$\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

$\frac{5}{12}$ of the tank is filled in every 2 hours

$\Rightarrow tank\: will\: be\: full\: in\: 2 \times \frac{12}{5}$

= $\frac{24}{5}\: hours$

= 4.8 hous or 4 hours 48 minutes

Correct

Vol of tank = $12 \times 13.5=162\:l$

$9l \times n=162\:l$

$\Rightarrow$ n = 18

18 buckets are needed.

Correct

$\frac{1}{t}=\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$

Time = 24 hours = $24 \times 60\: min$

= 1440 min.

Vol = $1440 \times 6\:l=8640\:l$

Correct

Correct

$\angle AOC=\angle AOB-\angle COB=68^{\circ}-42^{\circ}=26^{\circ}$

Correct

$\angle XOZ=\angle XOY +\angle YOZ=56^{\circ}+34^{\circ}=90^{\circ}$

Correct

Correct

Let angle = $x^{\circ}$

Then $x^{\circ}+98^{\circ}$ = 180

$\Rightarrow x^{\circ}=82^{\circ}$

Correct

Let angles be 8x and x

8x + x = 180

X = $20^{\circ}\:then\:angles\:are\:160^{\circ}\:and\:20^{\circ}$

Correct

6x + 7x + 5x = $180^{\circ}$

X = $10^{\circ}$

Correct

$58^{\circ}+42^{\circ} + z = 180^{\circ}$

Z = $180^{\circ}-100^{\circ}=80^{\circ}$

$\frac{1}{2}\: z=50^{\circ}$

Correct

$x+y+2\times 50^{\circ}=360^{\circ}$

$x+y=360^{\circ}-100^{\circ}=260^{\circ}$

Correct

$\angle POC=\frac{1}{2}\angle AOC$

$\angle COQ=\frac{1}{2}\angle COB$

$\angle POC+\angle COQ=\frac{1}{2}(\angle AOC+\angle COB)$

$\angle POQ=\frac{1}{2}(180^{\circ})$

$\angle POQ=90^{\circ}$

Correct

$\angle F=50$

$\angle F+x=180^{\circ}$

X = $130^{\circ}$

Correct

Correct

Correct

$\angle BHE=30^{\circ}$

$\angle EHB=180^{\circ}-30^{\circ}=150^{\circ}$

$\angle HEC=180^{\circ}-(150+10)=20^{\circ}$

$\angle CFE=180^{\circ}-20=160^{\circ}$

Correct

$\angle COD=20^{\circ}$

$\angle OCD=60^{\circ}$

$\angle ODC=180^{\circ}-(20^{\circ}+60^{\circ})=100^{\circ}$

Correct

y = $180^{\circ}1-103^{\circ}=77^{\circ}$

$x+y+35^{\circ}=180^{\circ}$

X = $180^{\circ}-35-77=68^{\circ}$

Correct

$\angle PRS=69^{\circ}+25^{\circ}=94^{\circ}$

Correct

Let angles be 2x, 3x and 4x then

2x + 3x + 4x = $180^{\circ}$

9x = 180

X = $20^{\circ}\:measure\:of\:angle\:is\:40^{\circ},60^{\circ},80^{\circ}$

Correct

Correct

Let $\angle 1=3x\:and\:\angle 2=2x$

3x + 2x = $180^{\circ}$

X = $36^{\circ}$

$\angle 2=72^{\circ}\Rightarrow \angle 3=72^{\circ}$

Correct

$\angle BOC=180^{\circ}-\left ( \frac{1}{2}\angle COCB+\frac{1}{2}\angle OBC \right )$

= $180^{\circ}-(32^{\circ}+27^{\circ})$

= $180^{\circ}-59$

= $121^{\circ}$

Correct

x = $50^{\circ}$

X + y = $125^{\circ}$

Y = $125^{\circ}-50^{\circ}$

Y = $75^{\circ}$

Correct

$\angle PQR=180^{\circ}-110^{\circ}=70^{\circ}$

$\angle QPR=180^{\circ}-135^{\circ}=45^{\circ}$

$\angle PQR=180^{\circ}-(70^{\circ}+45^{\circ})$

= $65^{\circ}$

Correct

Let angle be $x^{\circ}$

$x^{\circ}+(3x+2)=90^{\circ}$

4x = 90 – 2 = $88^{\circ}$

X = $22^{\circ}$

Correct

Let angles be $2x^{\circ}\:and\:3x^{\circ}$

2x + 3x = 180

X = 36

Correct

$\angle BOD=180-70^{\circ}=110^{\circ}$

$\angle 2=\frac{1}{2}\angle BOD$

= $55^{\circ}$

Correct

Let angle be $x^{\circ}$

X + 2x = $90^{\circ}$

X = $30^{\circ}$

Correct

$\angle AOP=180^{\circ}-40^{\circ}=140^{\circ}$

$\angle AOQ=\frac{1}{2}(140)$

= $70^{\circ}$

Correct

2x + x = $180^{\circ}$

X = $60^{\circ}$

$\angle ACD=120^{\circ}$

$\angle ABC+50^{\circ}=120^{\circ}$

$\angle ABC=70^{\circ}$

Correct

$\angle C=180-(80+40)$

$\angle C=60^{\circ}$

$\angle OCB=30^{\circ}$

$\angle BOC=180-(\angle OBC+\angle OCB)$

= 180 - $(20^{\circ}+30^{\circ})$

= $130^{\circ}$

Correct

Correct

Because angle opposite to BC is largest

Correct

Because side opposite to $\angle X$ is the longest

Correct

$\angle ACD=y+30^{\circ}$

180 - $(x+30^{\circ})$ = y + 30

$120^{\circ}=y+x$

Correct

180 – x + 180 + 180 – z = $180^{\circ}$

X + y + z = $360^{\circ}$

Correct

x = 55 + 30 + 50

= $135^{\circ}$

Correct

$\angle PRQ=180-(46+24^{\circ})$

= 110

$\angle SRP=180-(110+40)=30^{\circ}$

X= (30 + 40) = 70

Correct

$\angle x=\angle ADC$

$\angle x +\angle ADC + 100^{\circ}=360^{\circ}$

$2\angle x=260^{\circ}$

$\angle x=130^{\circ}$

Correct

$\angle CPD=(35+30)$

$\angle CPD=65$

$\angle BPC=2\times 65=130^{\circ}$

Correct

x = 180 – 40 = $140^{\circ}$

Correct

$\angle q=55^{\circ}$

$\angle BOQ=180-(65+55)$

= $60^{\circ}$

X = 180 – 60 = $120^{\circ}$

Correct

$\angle PRT=63^{\circ}$

$\angle QRP=180-(63^{\circ}+57^{\circ})$

= $60^{\circ}$

Correct

$\angle BAC=35^{\circ}$

$108^{\circ}=\angle B+35$

$\Rightarrow \angle B=73^{\circ}$

Correct

x = $(180^{\circ}-120^{\circ})+(180^{\circ}-145^{\circ})$

X = 60 + 35

= $95^{\circ}$

Correct

$\angle ACD=62$

$x+62+58^{\circ}=180$

X = $60^{\circ}$

Y = 62

X + y = 60 + 62 = $122^{\circ}$

Correct

x = 180 – 38 = $142^{\circ}$

$78^{\circ}=y+38$

Y = $40^{\circ}$

X + y = $182^{\circ}$

Correct

Correct

Correct

Correct

$\angle ABC=90$

Y – 90 + 50 = 180

Correct

Number of revolutions = $\frac{Distance}{Circumstance\:of\:wheel}$

Circumstance of wheel = $2\times \frac{22}{7}\times 28=176\: cm=1.76\: m$

No. of revolutions = $\frac{1.1\times 1000}{1.76}=625$

Correct

A = $13.86\times 10000\:m^{2}$

= $138600 \:m^{2}$

$\pi r^{2}=138600$

$r^{2}=\frac{138600}{22}\times 7=6300\times 7$

R = 210

Length of fence = $2\pi r=2\times \frac{22}{7}\times 210=1320$

Cost = $1320 \times 0.60= Rs.792$

Correct

a + a + 8 = 18

$\Rightarrow$ a = 5

DC = $\frac{1}{2}\times 8=4$

$\Delta ADC$ is a right triangle. Hence

$h^{2}|4^{2}=5^{2}$

$\Rightarrow$ h = 3

Area of $\Delta ABC=\frac{1}{2}\times 8\times 3\:cm^{2}$

= $12\:cm^{2}$

Correct

Side of each square of chess board = $\sqrt{9}$

= 3 cm

Side of board = $8\times side\:of\:square+2 \times width\:of\:margin.$

Side of board = $8\times 3+2 \times 3$

= 24 + 6 = 30 cm

Correct

$2\pi r=22\: cm$

$\Rightarrow r=22 \times \frac{7}{24}\: cm=\frac{7}{2}\: cm$

Area quadrant = $\frac{1}{4}\times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\:cm^{2}=9.625\:cm^{2}$

Correct

BD = $\sqrt{5^{2}+12^{2}}\;cm=13\:cm$

Also BD is diameter

$\Rightarrow$ r = 6.5 cm

Required = Area of circle – area of rectangle

= $(3.14\times 6.5^{2}-12 \times 5)\:cm^{2}=72.665\:cm^{2}$

Correct

Required area = area semicircle PS + area semicircle PQ – area semicircle QS

= $\frac{1}{2}\pi (6^{2}+2^{2}-4^{2})\: cm^{2}$

= $\frac{1}{2}\pi (36+4-16)\: cm^{2}$

= $12\pi \:cm^{2}$

Correct

$l \times b=150$

And 2(l + b) = 50 $\Rightarrow$ l + b = 25

Clearly l = 15, b = 10

Correct

Area of path = $(50 \times 40-46 \times 36)m^{2}$

= $344 \:m^{2}$

Cost = Rs. $344 \times 50$

= Rs. 17,200

Correct

$\frac{1}{2}ab=6\Rightarrow ab=12$

$a^{2}+b^{2}=5^{2}=25$

Use a + b = $\sqrt{a^{2}+b^{2}+2ab}\:\:and\:a-b=\sqrt{a^{2}+b^{2}-2ab}$

Correct

13x + 12 x + 5x = 450

X = 15

A = 195

B = 180

C = 75

Area = $\sqrt{225 \times 30 \times 45 \times 150}\:\:=6750\:m^{2}$

Correct

$A_{1}=100A_{2}$

$\Rightarrow \pi r\: _{1}\: ^{2}=100\pi r\: _{2} \: ^{2}$

$\Rightarrow r\: _{1}\: ^{2}=100r\: _{2}\: ^{2}$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow \frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{r_{1}}{r_{2}}=\frac{10}{1}$

$\Rightarrow C_{1}:C_{2}=10:1$

Correct

Join centres to obtain a square of side 14 cm.

Correct

Area of foot path = $(30 \times 20-28 \times 18)m^{2}=96\:m^{2}$

Number of tiles = $\frac{960000}{20 \times 20}=2400$

Correct

Original area = $l \times b$

New area = $\frac{3}{2}l \times \frac{3}{2}b=\frac{9}{4}lb$

Increase % = $\frac{\frac{9}{4}lb-lb}{lb}\times 100\%$

= 125%

Correct

$A_{1}=\pi r^{2},A_{2}=\pi \left ( \frac{r}{2} \right )^{2}=\frac{\pi r^{2}}{4}$

$A_{3}=A_{1}-A_{2}=\frac{3}{4}\pi r^{2}$

Correct

Area of road = 3.14 $\left ( 30^{2}-26^{2} \right )\: m^{2}$

= 3.14 (30 + 26) (30 – 26) $m^{2}$

= 703.36 $m^{2}$

Correct

$d_{1}=3x,d_{2}=4x$

$\frac{1}{2} \times 3x \times 4x=2400$

$\Rightarrow x^{2}=400\Rightarrow x=20$

Use : side = $\frac{1}{2}\sqrt{d_{1}^{2}+d_{2}^{2}}$

Correct

Required area = Area of smaller circle + $2 \times area\:of\:minor\:segment\:AB$

= $\pi \times \left ( \frac{7}{2} \right )^{2}+2\times \frac{7^{2}}{2}\left ( \frac{90 \times \pi }{180}-sin\: 90 \right )$

= $66.5\:cm^{2}$

Correct

AC = 54 cm; BC = 10 cm

$\Rightarrow$ AB = (54 – 10) cm = 44 cm

Radius of larger circle = 27 cm

Radius of smaller circle = 22 cm

Required area = $\pi (27^{2}-22^{2})\: cm^{2}$

= $\frac{22}{7}(27+22))(27-22)\:cm^{2}=770\:cm^{2}$

Correct

Area of star = $Area\: of\: \Delta ABC+3 \times area\: \Delta DEF.\: \Delta DEF$

is an equilateral triangle with each side equal to 1 cm.

Correct

$Side^{2}\times \frac{\sqrt{3}}{4}=36\sqrt{3}$

$\Rightarrow$ side = 12 cm

Semi-Perimeter = $\frac{12\times 3}{2}=18\:cm$

Use r = $\frac{\Delta }{S}$

Correct

Let the length of the longer diagonal be L.

$\therefore$ Length of shorter diagonal = 80% of L = $\frac{4}{5}L$

Area of rhombus = $\frac{1}{2} \times L \times \frac{4}{5}L$

$\Rightarrow Area\:of\:rhombus =\frac{2}{5}L^{2}$

Correct

OA = OB = OC = AB = BC = 10 cm

Clearly $\Delta OBA\:and\:\Delta OBC$ are both equilateral and also congruent.

Then area Rhombus = $2 \times area\:equilateral\:\Delta OAB=2 \times \left ( 10^{2}\times \frac{\sqrt{3}}{4} \right )\: cm^{2}$

= $50\sqrt{3}\: cm^{2}$

Correct

$\Delta ABC\:is\:an\:equilateral\Delta$

$\Rightarrow each\:interior\:angle\:is\:60^{\circ}$

$Side^{2}\times \frac{\sqrt{3}}{4}=17300$

Side = 200 m

Radius of each circle = 100 m

Shaded area = area of $\Delta$ - 3(area of any sector).

Correct

$r_{1} =r,r_{2}=r+50\%\:of\:r=\frac{3}{2}r,$

$h_{1}=h;h_{2}=\frac{3h}{2}$

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi \left ( \frac{3}{2}r \right )^{2}\left ( \frac{3}{2}h \right )}{\frac{1}{3}\pi r^{2}h}=\frac{27}{8}=3.375$

Volume increasing by $(3.375-1)\times 100\%=237.5\%$

Correct

$\frac{V_{2}}{V_{1}}=\frac{\frac{1}{3}\pi (2r)^{2}(2h)}{\frac{1}{3}\pi r^{2}h}$

$V_{2}:V_{1}=8:1$

Correct

Radius of ball : r = ?; Radius of tube : R = 12 cm

H = 6.75 cm

Volume of ball = Volume of cylinder ABCD

$\frac{4}{3}\pi r^{3}=\pi R^{2}h$

$\frac{4}{3}\times r^{3}=12 \times 12 \times \frac{675}{100}$

$\Rightarrow$ r= 9 cm

Correct

h = 42 cm; r = 30 cm, n = 700

Area leveled = $700 \times 2 \times \frac{22}{7} \times 30 \times 42\:cm^{2}$

= $5544000\:cm^{2}$

= $554.4\:m^{2}$

Cost = Rs. $554.4 \times 0.45=RS.\: 249.48\approx Rs.\: 250$

Correct

L = 24 cm, B = 12 cm, H = 12 cm

V = $24 \times 12 \times 12\: cm^{3}=3456\: cm^{3}$

Correct

r = $\frac{105}{2}$ m; l = 53 m

H = 3 m

Area of canvas

= C.S.A. cone + C.S.A. of cylindrical

= $\pi rl+2\pi rh$

= $\pi r(l+2h)$

= $\frac{22}{7}\times \frac{105}{2}(53+10)\: m^{2}$

= $10395\:m^{2}$

Correct

r = $\frac{7}{2}\: cm$

H + 2r = 19 cm

$\Rightarrow$ h = 12 cm

Surface area = $2\pi rh+2 \times 2\pi r^{2}$

= $2\pi r(h+2r)$

= $2 \times \frac{22}{7} \times \frac{7}{2} \times 19\:cm^{3}$

= $418\:cm^{2}$

Correct

$r_{1}=r,r_{2}=2r$

$\frac{A_{1}}{A_{1}}=\frac{(2r)^{2}}{r^{2}}=4$

Increase in area = 300%

Correct

Volume of wire (cylinder) = Volume of sphere

$\pi r^{2} \times 72=\frac{4}{3}\pi \times 3^{3}$

$\Rightarrow r=\frac{1}{\sqrt{2}}\: cm$

$\Rightarrow diameter=\frac{2}{\sqrt{2}}\: cm$

= $\sqrt{2}\: cm$

$\approx 1.4\: cm$

Correct

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

Side of new cube =24 cm

Correct

$2(l+b)h \times rate=100$

$2(2l+2b)\frac{h}{4} \times rate$

$\Rightarrow$ required cost = Rs. 50

Correct

$\frac{22}{7}\times r^{2}\times 28=6358$

$\Rightarrow r=\frac{17}{2}\: cm$

C.S.A. = $2\times \frac{22}{7}\times \frac{17}{2}\times 28=1496\:cm^{2}$

Correct

r = 5 cm

R = (5 + 2.1) cm = 7.1

Volume of material = $\frac{2}{3}\times \frac{22}{7}(7.1^{3}-5^{3})\:cm^{3}$

= $488.004\: cm^{3}$

$\approx 488\: cm^{3}$

Correct

$\frac{V_{2}}{V_{1}}=\frac{\frac{4}{3}\pi \left ( \frac{r}{2} \right )^{3}}{\frac{4}{3}\pi \left ( r \right )^{3}}=\frac{1}{8}$

$V_{2}=\frac{1}{8}V_{1}$

Correct

$x^{3}=12^{3}+16^{3}+20^{3}$

$x^{3}=13824=24^{3}$

$\therefore$ Side of new cube = 24 cm

$\therefore diagonal=24\sqrt{3}\:cm$

Correct

Circumference of cylinder = 88 cm

$\Rightarrow 2\pi r=88$

$\Rightarrow r=14$

Also h = 8

$\therefore Volume\: of\: cylinder=\frac{22}{7}\times 14\times 14\times 8\:cm^{3}$

= $4928\: cm^{3}$

Correct

r = $\frac{42}{2}\: cm=21\: cm$

$\frac{1}{3}\times \frac{22}{7}\times 21^{2}\times h=12936$

$\Rightarrow h=28\:cm$

L = $\sqrt{28^{2}+21^{2}}=35$

C.S.A. = $\frac{22}{7}\times 21\times 35\: cm^{2}$

= $2310\:cm^{2}$

Correct

$\frac{V_{2}}{V_{1}}=\frac{(3x)^{3}}{x^{3}}=\frac{27x^{3}}{x^{3}}=\frac{27}{1}$

Increse in vol = $V_{2}-V_{1}$ = 27 -1 = 26

$\therefore$ increase % = 2600%

Correct

Number of small balls = $\frac{Volume\:of\:original\:ball}{Volume\:of\:small\:ball}$

= $\frac{\frac{4}{3}\times \pi \times r^{3}}{\frac{4}{3}\times \pi \times \left ( \frac{r}{2} \right )^{3}}=8$

$\frac{CSA\: of\: original\: ball}{Sum\: of\: CSA\: of\: 8\: small\: balls}=\frac{4\pi r^{2}}{8\times \pi \times \left ( \frac{r}{2} \right )^{2}}=\frac{2}{1}$

Or 2 : 1

Correct

Volume wire (cyl) =Vol. of sphere

$\Rightarrow \pi \times (0.2)^{2}h=\frac{4}{3}\times \pi \times 3^{3}$

$\Rightarrow h=900\:cm=9\:m$

Correct

$r_{1}=4x,r_{2}=3x$

$h_{1}=3y;h_{2}=4y$

$\frac{V_{2}}{V_{1}}=\frac{\pi (4x)^{2}(3y)}{\pi (3x)^{2}(4y)}=\frac{4}{3}$

Correct

$\frac{\pi r^{2}h}{2\pi rh}=\frac{924}{264}$

$\Rightarrow \frac{r}{2}=\frac{7}{2}$

$\Rightarrow r=7$

$2\times \frac{22}{7}\times 7\times h=264$

$\Rightarrow h=6\:m$

Correct

$2\pi r=132$

$\Rightarrow r=21\:cm$

Volume = $\frac{2}{3}\times \frac{22}{7}\times 21^{3}\: cm^{3}$

= $19404\:cm^{3}$

Correct

$\frac{CSA_{1}}{CSA_{2}}=\frac{2\pi (2x)^{2}(5y)}{2\pi (3x)^{2}(4y)}=\frac{5}{9}$

Correct

r = 3 cm

H = (11 – 3) cm = 8 cm

Capacity of vessel = $\frac{2}{3}\pi r^{3}+\pi r^{2}h$

= $\pi r^{2}\left ( \frac{2}{3}r+h \right )$

= $3.14\times 3^{2}\left ( \frac{2}{3}\times 3+8 \right )cm^{3}$

= $3.14\times 9 \times 10\: cm^{3}$

= $282.6\: cm^{3}$

Correct

Width of interval is different.

Correct

Correct

27 is median as it is appearing in middle of data and

Mean = $\frac{25+26+27+28+29}{5}=\frac{135}{7}=27$

Correct

Correct

1900, because its frequency is the highest

Correct

Mean = $\frac{2+4+6+8+10+12+14+16+18+20}{10}$

=$\frac{110}{10}=11$

Correct

Correct

G.M. = $(2 \times 6 \times 24 \times 72)^{\frac{1}{4}}$

= $(2\times 2\times 3\times 2\times 2\times 2\times 3\times 2\times 2\times 2\times 3\times 3)^{\frac{1}{4}}$

= $2 \times 3=6$