200% of x = 90 $\Rightarrow$ x = 45

$\Rightarrow$ 50% of x = $\frac{45}{2}$ = 22.5

A gets 65% of votes $\Rightarrow$ B gets 35% of votes as no votes are invalid.

Let total number of votes be x

Then (65-35)% of x = 300

$\Rightarrow$ x = 1000

Gross value – 10% of gross value = Rs. 18000

$\Rightarrow$ 90% of gross value = Rs. 18000

$\Rightarrow$ gross value = Rs. 20,000

90% of population = 27000

$\Rightarrow$ Total population = $27000\times \frac{100}{90}=30,000$

50% of x + 100 = x

$\Rightarrow \frac{x}{2}+100=x$

$\Rightarrow$ x = 200

x + 62.5% of x = 78

$\Rightarrow x+\frac{5}{8}x=78$

$\Rightarrow \frac{13}{8}x=78$

$\Rightarrow$ x = 48

x – 31% of x = 552

$\Rightarrow$ 69% of x = 552

$\Rightarrow x = 552\times \frac{100}{69}=800$

$4000\times 1.23=4920$

$(1-0.028)\times 850=0.72\times 850=612$

Inc% = $\frac{6305-4850}{4850}\times 100\%=30\%$

$x \times 1.24+(11000-x)\times 0.8=11000$

$\Rightarrow$ 0.44x + 8800 = 11000

$\Rightarrow$ x = 5000

See important fact 1 of theory

Loss % = $\frac{10 \times 10 }{100} =1\%$

$x \times 1.1+(22000-x)0.88=22000$

$\Rightarrow$ 19360 + 0.22x = 22000

$\Rightarrow$ x = 12000

Costs of the scooters are Rs. 12000 and Rs. 10,000

SP of 12 items = CP of 15 items = $12 \times 15$

SP = 15 and CP = 12

Gain % = $\frac{3}{12} \times 100\%=25\%$

SP of 23 items = CP of 20 items = $20 \times 23$

$\Rightarrow$ SP = 20 and CP = 23

$\Rightarrow Loss\%= \frac{23-20}{20} \times 100\%=13 \frac{1}{23}\%$

SP of 20 items = CP of 23 items = $20 \times 23$

$\Rightarrow$ SP = 23 and CP = 20

$\Rightarrow Gain\%= \frac{23-20}{20} \times 100\%=15\%$

SP of 20 items = CP of 16 items = $20 \times 16$

Let

$\Rightarrow$ SP = 16; CP = 20

Loss % = $\frac{4}{20} \times 100\%=20\%$

Let MP of 1 pen be Rs. 1

$\therefore$ CP = 15; SP = 20

Profit % = $\frac{5}{15} \times 100\%=33 \frac{1}{3}\%$

SP = $600 \times \frac{100-10}{100} \times \frac{100-5}{100}$

= $600 \times 0.9 \times 0.95 = Rs.513$

Let marked price of each pen be Rs. 1

CP = 16; SP = 20

Profit % = $\frac{4}{16} \times 100\%=25\%$

SP of 2nd purchaser = $840 \times \frac{100+50}{100} \times \frac{100-25}{100}$

= $840 \times 1.5 \times 0.75= Rs.945$

(10% + 10%) of CP = 80

20% of CP = 80

$\Rightarrow CP=80 \times 5 =Rs.400$

(5% + 5%) of CP = 6.72

10% of CP = 6.72

CP = Rs. 67.20

(2.5% + 5%) of CP = 6

7.5% of CP = 6

CP = $\frac{6}{7.5} \times 100 = Rs.80$

CP = 1 quintal = Rs. 150

Spoilage = 10% of 1 quintal = 0.1 quintal

Qty. available for sale = 0.9

Desired SP of 0.9 quintals = $150 \times 1.2 = Rs. 180$

$\Rightarrow$ SP of 1 quintal = Rs. 200

MP = $70 \times 4 = 280$

D% = $\frac{280-238}{280} \times 100\%=15\%$

$\frac{280-238}{280} \times 100\%=15\%$

MP = $80 \times 4 = Rs. 320$

D% = $\frac{320-272}{320} \times 100\%=15\%$

SP = $\frac{100-25}{100} \times \frac{100-10}{100} \times 2,500$

= $0.75 \times 0.9 \times 2,500= Rs. 1,687.50$

SP = $\big\{ \frac{100-20}{100} \big\} \big( \frac{100-15}{100} \big) \times 2,300$

= $0.8 \times 0.85 \times 2,300$

= Rs. 1,564

SP = $\big( \frac{100-15}{100} \big) \big( \frac{100-10}{100} \big) \big( \frac{100-8}{100} \big) \times 35,000$

= $\big( \frac{85}{100} \big) \big( \frac{90}{100} \big) \big( \frac{92}{100} \big) \times 35,000$

= Rs. 24,633

SI = $\frac{20,000 \times 4 \times 5}{100} = Rs. 4,000$

A = 20,000 + 4,000 = Rs. 24,000

T = 2 year 3 m = $2 \frac{1}{4}\:year=\frac{9}{4}\:year$

SI = $\frac{40,000 \times 3 \times 9}{100 \times 4}= Rs. 2,700$

A = 40,000 + 2,700 = Rs. 42,700

: T = 2 year 6 m = $2 \frac{1}{2}\:year= \frac{5}{2}\:year$

SI = $\frac{35000 \times 12 \times 5}{100 \times 2}=Rs. 10,500$

A = 35,000 + 10,500 = Rs. 45,500

T = $\frac{100}{365}year= \frac{20}{73}years$

SI = $\frac{3650 \times 5.5 \times 20}{100 \times 73}=Rs. 55$

A = 3,650 + 55 = Rs. 3,705

P = $\frac{100 \times 2450}{7 \times 7}=Rs. 5,000$

A = 5,000 + 2,450 = Rs. 7,450

T = $\frac{120}{365}years= \frac{24}{73}years$

I = $\frac{7665 \times 5.5 \times 24}{100 \times 73 }$

= Rs. 138.6

A = 7,665 + 138.6 = Rs. 7,803.60

P = $\frac{100 \times 4949}{7 \times 7}= Rs. 10,100$

A = 10,100 + 4,949 = Rs. 15,049

T = $\frac{219}{365}= \frac{3}{5}yrs$

I = 52,100 – 50,000 = Rs. 2,100

r = $\frac{100 \times 2100}{50000 \times \big( \frac{3}{5} \big) }=7\%pa$

P = 45,000 – 5,000 = Rs. 40,000

T = $\frac{100 \times 5000}{40000 \times 2.5}=5 years$

SI for 10 years = Rs. 600

SI for 1 year = Rs. 60

SI of for 5 years = $60 \times 5 = Rs. 300$

After 5 years principal is tripled

Hence SI for next 5 years = $300 \times 3 = Rs. 900$

Total SI for 10 years = 900 + 300 = Rs. 1,200

P = 42,500 – 2,500 = Rs. 40,000

T = $\frac{100 \times 2500}{40000 \times 2.5} = 2.5 years$

I = $\frac{24000 \times 12 \times 3}{100}= Rs. 8,640$

A = 24,000 + 8,640 = Rs. 32,640

Cost of gold ring = 32,640 – 12,000 = Rs. 20, 640

Mr. B has to pay interest for 2 years more. Hence extra interest paid

= $\frac{5600 \times 2 \times 12.5}{100} = Rs. 1,400$

I = $\frac{25000 \times 18 \times 3}{100} = Rs. 13,500$

A = 25,000 + 13,500 = Rs. 38,500

Cost of ring = 38,500 – 30,000 = Rs. 8,500

Mr. B has to pay interest for 2 years more. Hence extra interest paid

= $\frac{11200 \times 10.5 \times 2}{100} = Rs. 2,352$

SI = 8

r = $\frac{100 \times 8}{1 \times 40}\% = 20\%$

Clearly profit is 4% every year

Hence $12 \times 4 = 48\%$ of amount = Rs. 32

$\Rightarrow Required\: amount\:= \frac{100}{48} \times 3,216 =Rs. 6,700$

$\frac{P \times 7 \times 5}{100}$ = P – 650

$\Rightarrow$ 7P = 20P – 13,000

$\Rightarrow$ P = Rs. 1,000

Interest for 1st year = 5% of 1,000 = Rs. 50

Amount paid = Rs. 100

Hence principal paid = Rs. 50

P for 2nd year = 1,000 – 50 = Rs. 950

Interest for second year = 5% of 950 = Rs. 47.50

Amount paid = Rs. 100

Hence principal paid = 100 – 47.5 = Rs. 52.5

P for 3rd year = 950 – 52.5 = Rs. 897.50

Interest for 3rd year = 5% of 897.50 = Rs. 44.875

Amount paid = Rs. 100

Hence principal paid = 100 – 44.875 = Rs. 55.125

P for 4th year = 897.5 – 55.125

= Rs. 842.375 $\approx$ Rs. 842.38

Interest desired = $\frac{5000 \times 6}{100} = Rs. 300$

$\Rightarrow 2,000 \times \frac{4}{100}+1,500 \times \frac{5}{100}+1500 \times \frac{r}{100} = 300$

$\Rightarrow$ 80 + 75 + 15r = 300

$\Rightarrow r=9 \frac{2}{3}$

$\frac{1}{t}= \frac{1}{60}+ \frac{1}{20}= \frac{1}{15}$

$\Rightarrow$ time = 15 days

$\frac{1}{t}= \frac{1}{20}+ \frac{1}{30}= \frac{1}{12}$

$\Rightarrow$ time = 12 days

A can type 3 pages in 1 hour

A and B can together type 5 pages in 1 hour

$\Rightarrow$ B can type 2 pages in an hour

$\Rightarrow$ B needs 21 hours to type 42 pages

$\frac{1}{a}+ \frac{1}{8}= \frac{1}{7}$

$\Rightarrow \frac{1}{a}=\frac{1}{7} - \frac{1}{8}= \frac{1}{56}$

$5 \times \frac{1}{8}+n \times \frac{1}{56}=1$

$\Rightarrow$ n = 21

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

$\Rightarrow$ time = 4 days

$\frac{1}{t}=\frac{1}{10}+ \frac{1}{12}+\frac{1}{15}= \frac{1}{4}$

$\Rightarrow$ time = 4 days

$\frac{1}{5}+ \frac{1}{4}+ \frac{1}{c}= \frac{1}{2}$

$\Rightarrow \frac{1}{C}=\frac{1}{2}- \frac{1}{5}-\frac{1}{4}= \frac{1}{20}$

$\Rightarrow$ time = 20 days

$\frac{1}{a}+ \frac{1}{b}= \frac{1}{28}$

$\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} = \frac{1}{21}$

$\Rightarrow \frac{1}{c} = \frac{1}{21}- \frac{1}{28}= \frac{1}{84}$

$\Rightarrow$ time = 84 days

$\frac{1}{a}+ \frac{1}{3}= \frac{1}{2}$

$\Rightarrow \frac{1}{a}= \frac{1}{2} -\frac{1}{3}= \frac{1}{6}$

$\Rightarrow$ time = 6 days

$\frac{1}{a}+ \frac{1}{b}= \frac{1}{24}$

$\frac{1}{b}+\frac{1}{c}= \frac{1}{18}$

$\frac{1}{c}+\frac{1}{a}=\frac{1}{36}$

Adding $2( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})= \frac{1}{24}+ \frac{1}{18}+ \frac{1}{36}= \frac{1}{8}$

$\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=\frac{1}{16}$

$\Rightarrow$ time = 16 days

$\frac{1}{a}+ \frac{1}{b}=\frac{1}{12}$

$\frac{1}{b}+ \frac{1}{c}=\frac{1}{15}$

$\frac{1}{c}+ \frac{1}{a}=\frac{1}{20}$

$2(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}) =\frac{1}{5}$

$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c} =\frac{1}{10}$

$\frac{1}{c} =\frac{1}{10}-(\frac{1}{a}+\frac{1}{b})= \frac{1}{10}- \frac{1}{12}= \frac{1}{60}$

$\Rightarrow$ time = 60 days

B needs half the time

$\Rightarrow B\;needs\:24 \times \frac{1}{2} =12 days$

$\frac{1}{x}+ \frac{1}{2x}= \frac{1}{24}$

$\Rightarrow \frac{3}{2x}= \frac{1}{24}$

$\Rightarrow$ x = 36

Hence A requires 36 days

Work done by 1 man in 1 day= $\frac{1}{3 \times 43}= \frac{1}{129}$

Work done by 1 woman in 1 day = $\frac{1}{5 \times 43}= \frac{1}{215}$

Work done by 5 men and 6 women in 1 day = $\frac{5}{129}+ \frac{6}{215}= \frac{1}{15}$

Hence work will be over in 15 days.

Work done by 1 man in 1 day = $\frac{1}{3 \times 43}= \frac{1}{129}$

Work done by 1 woman in 1 day= $\frac{1}{4 \times 43}= \frac{1}{172}$

Work done by 7 women and 5 women in 1 day

= $\frac{7}{129}+ \frac{5}{172}= \frac{1}{43} \big( \frac{7}{3}+ \frac{5}{4} \big)= \frac{1}{12}$

Hence work will be over in 12 days.

$\frac{30}{m}+ \frac{40}{b} =\frac{1}{21}$ …… (i)

$3 \times \frac{1}{m} = 5 \times \frac{1}{b}$

$\Rightarrow \frac{1}{m}= \frac{5}{3b}$

Substituting in (i)

We get $\frac{1}{b}= \frac{1}{1050}\:and\: \frac{1}{m}= \frac{1}{630}$

Work done by 20 men and 4 boys in 1 day

= $\frac{20}{630}+\frac{4}{1050}= \frac{13}{315}$

Hence work will be over in $\frac{315}{13}=24 \frac{3}{13}\:days$

$\frac{1}{c}= \frac{2}{x}+ \frac{1}{x}= \frac{3}{x}$

$\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}= \frac{2}{x}+ \frac{1}{x}+\frac{3}{x}= \frac{6}{x}= \frac{1}{7}$

$\Rightarrow$ x = 42

$\Rightarrow$ time = 42 days

$\frac{3}{a}= \frac{4}{c}\:and\: \frac{5}{b}= \frac{6}{c}$

$\Rightarrow \frac{3}{4a}= \frac{1}{c}\:and\: \frac{5}{6b}= \frac{1}{c}$

$\Rightarrow \frac{3}{4a}= \frac{5}{6b}$

But $\frac{1}{a}= \frac{1}{18}$

$\Rightarrow \frac{1}{b}= \frac{3}{4} \times \frac{1}{18}\times \frac{6}{5}= \frac{1}{20}$

$\Rightarrow$ b needs 20 days to finish the work

A : B = $\frac{1}{15}: \frac{1}{10}=10:15$

A’ share = $\frac{10}{10+15} \times 2250=900$

A:B = $\frac{1}{8}: \frac{1}{12}=12:8$

B’s share = $\frac{8}{8+12} \times 200 = 80$

A : B = $\frac{1}{12}: \frac{1}{18} =18 : 12$

A’ share = $\frac{18}{12+18} \times 29850 = 17910$

A : B : C = $\frac{1}{5}: \frac{1}{6}: \frac{1}{7}=42:35:30$

C’s share = $\frac{30}{42+35+30} \times 5350 = 1500$

Wages of $45 \times 48$ women days = 93150

Wages of 1 woman day = $\frac{93150}{45 \times 48} = Rs. 43.125$

Wage of 1 man day = $2 \times 43.125 = Rs. 86.25$

$n \times 16 \times 86.25 =34500$

n = 25

$\frac{1}{w}= \frac{1}{10 \times 4}= \frac{1}{40}$

$\frac{1}{b}= \frac{1}{6 \times 10}= \frac{1}{60}$

$\frac{1}{m}= \frac{1}{2 \times 22}= \frac{1}{24}$

$\frac{1}{m}+ \frac{2}{w}+ \frac{3}{b} = \frac{1}{24}+ \frac{1}{20}+ \frac{1}{20}$

= $\frac{5+6+6}{120}= \frac{17}{120}$

Work will be over in $\frac{120}{17}\:days$

Wages = $\frac{120}{17}(153+2 \times 93+3 \times 57)$

= Rs.3600

Relative speed = 40 + 50 90 km/hr

Distance = 300 km

Time = $\frac{300}{90}$ = 3 hrs 20 minutes.

Let the normal speed by x km/hr

Then, $\frac{300}{x}- \frac{300}{x+5}=2$ Solving x = 25

Stoppage time is the time taken to cover a distance of (54-45) km at the speed of 54 km/h

54 Km : 60 min = 9 km : x min

Solving x = 10

For express train : time to run 600 km = 6hrs. But this also includes 7 stoppages of 3 minutes each. Hence total time = 6 hrs 21 minutes or (381 minutes)

The local train covers 25 km in every 31 minutes.

$\frac{381}{31} \approx 12.29.....$

This mean train has covered $25 \times 12 km\:(300 km)\:in\:12 \times 31\:minutes\:(372 minutes)$

still 9 minutes are left.

The train covers another $\frac{9}{60} \times 50 km =7.5 km$

Hence total distance travelled = 307.5 km.

Had the bus not stopped it would have covered 18 km more in one hour. So stoppage time is the time needed to cover 18 km at the speed of 54 km/hr

54 km : 60 min = 18 km : x min

X= 20

The insect ascends (3-2) m or 1 min ever two minutes.
To ascend 7 m the insect needs 14 minutes.
Now in the fifteenth minute he ascends 3 m and reaches the top of the pole.

Let the distance be x km and the required time be t hrs

Then, $\frac{x}{30}=t+ \frac{1}{6}$

And $\frac{x}{42}=t- \frac{1}{6}$

$\frac{x}{30} - \frac{x}{42}= \big\{t+ \frac{1}{6} \big\}- \big\{t- \frac{1}{6} \big\}$

$1500 \frac{m}{min}= \frac{1500\:m}{60\:sec}=25\:m/sec$

$90\frac{km}{hr}= 90 \times \frac{5}{18}\:m/sec=25\:m/sec$

All options (a), (b) and (c) have equal value. Hence option (d) is correct.

$33 \frac{1}{3}\;m/sec = \frac{100}{3}\:m/sec$

= $\frac{100}{3} \times \frac{18}{5}\:km/hr$

= 120 km/hr.

Let speeds of A and B be x km/hr and y km/hr respectively.

Then, x +y = $\frac{25}{1}=25$

And x – y = $\frac{25}{5}=5$

Solving: x = 15

$\therefore$ Speed of A is 15 km/hr

$\frac{x}{20}=t+ \frac{1}{4}$

$\frac{x}{40}=t- \frac{1}{4}$

$\frac{x}{20} - \frac{x}{40}= (t+ \frac{1}{4}) - (t- \frac{1}{4})$

$\frac{x}{40}= \frac{1}{2}$

X = 20

Time for C = 42 min

Time for B = $\frac{42}{3}$ = 14 min

Time for A = $\frac{142}{2}$ = 7 min

Relative speed = (90 – 70) km/hr = 20 km/hr

= $20 \times \frac{5}{18}\:m/sec$

= $\frac{50}{9}\:m/sec$

Time = 36 seconds

$\Rightarrow length\:of\:faster\:train= \frac{50}{9} \times 36\:m$

= 200 m

Distance = (120 + 120) m

= 240 m

Relative speed = (40 + 20) km/hr = 60 km/hr

= $\frac{50}{3}\:m/sec$

Time = $\frac{240}{ \big( \frac{50}{3} \big) }$ sec = 14.4 sec

Let the length of the train be x m and its speed be s m/sec.

Then $\frac{x+0}{15}=s$ (we assume the horizontal length of the pole to be o m)

And $\frac{x+100}{30}=s$

$\Rightarrow \frac{x}{15}= \frac{x+100}{30}$

Solving x =100

$\Rightarrow$ Length of train is 100 m

54 km/hr = $54 \times \frac{5}{8}\:m/s=15\:m/s$

Time = 20 sec

Let length of the train be x, then 100 + x = $15 \times 20=300$

So length of the train is = 200 m

Speed = $\frac{120+55}{10}\:m/s$ = 17.5 m/s = 63 km/hr

Relative speed 40 + 32 = 72 km/hr = $72 \times \frac{5}{18}\:m/s=20\:m/s$

Sum of lengths = 200 m so time = $\frac{200}{20}\:sec=10\:sec$

Relative speed = 50 + 40 = 90 km/hr = $90 \times \frac{5}{18}=25\:m/s,time = 8 sec.$

Let length of the first train be x m, then (125 + x) = $25 \times 8=200$

Hence length of the first train = 75 m

Sum of lengths = 280 m

Time = 8 seconds

Relative speed = $\frac{280}{8}\:m/s$ = 35 m/s = 126 km/hr

70 + x = 136

Or x = 56 km/hr

Distance downstream = distance upstream

$(x+40) \times 20= (x-4) \times 25$

$\Rightarrow$ x = 36

Distance = 800 km

Speed of downstream = $\frac{33}{5}$ km/min = 36 km/hr

Speed upstream = 33 km/hr

Speed of boat in still water = $\frac{36+33}{2}$ km/hr = 34.5 km/hr

Let speed of boat in still water b x km/hr speed of stream by y km/hr.

Then x + y = 11

And x – y = 5

Solving x = 8

Hence speed of boat in still water = 8 km/hr

Let speed of boat in still water be x km/hr and speed of stream by y km/hr

Then x + y = 8

And x – y =2

Solving y = 3

Hence speed of stream = 3 km/hr.

Let speed of boat in still water = 6 km/hr

Let speed of stream be x km/hr

Time upstream = $3 \times time\:downstream$

$\Rightarrow Speed\:upstream= \frac{1}{3} \times speed \:downstream$

6 – x = $\frac{1}{3}\times (6+x) \Rightarrow 6-x=2+ \frac{x}{3}$

$\Rightarrow 4= \frac{4x}{3}$

$\Rightarrow$ x = 3

Hence speed of current is 3 km/hr.

Let $x+ \sqrt{x+ \sqrt{x \sqrt{x...}}}$ = y

$2+ \sqrt{2+ \sqrt{2 \sqrt{2+ \sqrt{2...} } } }\: = y$

$2+ \sqrt{y}=y$

$( \sqrt{y} )^{2}=(y-2)^{2}$

$y^{2}-4y+4=y$

$y^{2}-5y+4=0$

(y – 4) (y – 1) = 0

Y = 1, 4

Use $(a^{3}+b^{3})=(a+b)(a^{2}+b^{2}-ab)$

So, $\big( \frac{1}{x^{3} }+ \frac{1}{y^{3} } \big)= \big( \frac{1}{x}+ \frac{1}{y} \big) ( \frac{1}{x^{2} }+ \frac{1}{y^{2} }- \frac{1}{xy} )$

= $\big( \frac{x+y}{xy} \big) \big( \frac{ x^{2}+y^{2} }{(xy)^{2} }- \frac{1}{xy} \big)$

= $\frac{x+y}{(xy)^{3} }((x+y)^{2}-3xy)$

= $\frac{p}{(2)^{3} }(p^{2}-3q)$

= $\frac{p^{3}-3pq }{q^{3} }$

$\frac{1}{x^{2} }+ \frac{1}{y^{2} }+= \frac{y^{2}+ x^{2} }{x^{2}y^{2} }$

= $\frac{(x+y)^{2}-2xy }{(xy)^{2} }$

= $\frac{25-12}{36}= \frac{13}{36}$

Let $x- \frac{1}{x- \frac{1}{x....} }=y$

$x- \frac{1}{y}=y \Rightarrow 2y-1=y^{2}$

$\Rightarrow y^{2}-2y+1=0$

$\Rightarrow (y+1)^{2} =0 \Rightarrow y=1$

$a^{x}=b^{3} \Rightarrow (a^{x})^{y} =(b^{3})^{y} \Rightarrow a^{xy}=b^{3y}$

$a^{xyz}=((c^{3})^{3})^{3}$

$a^{xyz}=(c^{3})^{9}=a^{27}$

$\Rightarrow xyz=27$

$2^{x+y}=2^{z},2^{x+2y}=(2)^{-3}$

X + y = 2

X – 2y = -3

Put y = 2 – x

x – 2(2 – x) = -3

$\Rightarrow 3x=1 \Rightarrow x= \frac{1}{3}$

Difference = $\frac{1}{x^{2}-y^{2} }- \frac{1}{x^{2}+y^{2} }$

= $\frac{(x^{2}+y^{2})-(x^{2}-y^{2}) }{(x^{2}-y^{2} )(x^{2}+y{2} )}$

= $\frac{2y^{2} }{x^{4}-y^{4} }$

$7x- \frac{1}{ x^{2}-4 }=21- \frac{1}{ x^{2}-4 }$

$\Rightarrow 7x=21 \Rightarrow x=3$

$f( \frac{1}{3} )=( \frac{1}{3} )^{3}-2(- \frac{1}{3} )+1$

= $\frac{1}{27}- \frac{2}{3}+1$

$f( \frac{1}{3} )= \frac{10}{27}$

$\frac{1}{3}f( \frac{1}{3} )= \frac{10}{81}$

$4^{x+2y}=2^{4x+10y}=4^{2x+5y}$

X + 2y = 2x + 5y

$\Rightarrow$ x = -3

Y = $- \frac{x}{3}$

$2y^{2}-9y+4$

= $2y^{2}-8y-y+4$

= 2y(y – 4) – 1(y – 4)

= (2y – 1) (y – 4)

$2^{x+y}=2^{5},2^{x-y}=2^{4}$

X + y = 5

X - y = 4

$\Rightarrow$ 2x = 9

$\Rightarrow x= \frac{9}{2},y= \frac{1}{2}$

So $(x^{2}+y^{2} )= \frac{81}{4}+ \frac{1}{4}= \frac{41}{2}$

$\left ( y+\frac{1}{y} \right )^{3} =y^{3}+\frac{1}{y^{3} }+3(y+ \frac{1}{y} )$

$\Rightarrow y^{3}+\frac{1}{y^{3} }=(12)^{3}-3\:(12)=1692$

x + y = 12, xy = 32

$x+ \frac{32}{x}=12 \Rightarrow x^{2}-12x+32=0$

$\Rightarrow$ (x – 8) (x – 4) = 0

$\Rightarrow$ x = 4, 8

$\Rightarrow$ y = 8, 4

$\Rightarrow \frac{1}{x}+ \frac{1}{y}= \frac{1}{4}+ \frac{1}{8}$

= $\frac{3}{8}$

$3^{ \frac{n}{2} }=3^{4}$

N = 8

$\Rightarrow \frac{n^{2}+8 }{8}=9$

$(x+8) \times (x-5)$

= $x^{2}+3x-40$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

Where $\alpha=0$

$\beta \gamma = \frac{c}{a}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

$0+0+0= \frac{c}{a}$

$\Rightarrow$ c = 0

Zeroes are equal

$\Rightarrow$ polynomial is perfect square

$\Rightarrow$ c and a have the same sign

Put x =-1,

$2(-1)^{2}+k(-1)=0$

$\Rightarrow k=2$

2x + 5 = 0

$\Rightarrow x=- \frac{5}{2}$

Put x = -1

$(-1)^{51}+51$

= 0

Put x = -1

We get the value of polynomial $x^{3}+x^{2}+x+1$ is 0 so (x + 1) is a factor of (b)

$p-27pq^{3}$

= $p(1-27q^{3})$

= $p \begin{bmatrix}(1)^{3}-(3q)^{3} \end{bmatrix}$

= $p(1-3q)(1+3q+9q^{2})$

$(25 x^{2}-1)+(5x+1)^{2}$

= (5x + 1) (5x + 1) + $(5x+1)^{2}$

= (5x + 1) (5x -1 + 5x + 1)

= (5x + 1) (10x)

$(249)^{2}-(248)^{2}$

= (249 – 248) (249 + 248)

= 1 (497)

= 497

$\frac{x}{y}+ \frac{y}{x}=-1$

$\Rightarrow x^{2}+ y^{2}+xy=0$

$\Rightarrow x^{3} y^{3}=(x-y)( x^{3}+xy+ y^{2})$

= 0

$49 x^{2}-b= \begin{bmatrix}7x+ \frac{1}{2} \end{bmatrix} \begin{bmatrix}7x- \frac{1}{2} \end{bmatrix}$

$49 x^{2}-b= 49 x^{2}- \frac{1}{4}$

$\Rightarrow b= \frac{1}{4}$

$(x+y)^{3}-( x^{3}+ y^{3})$

= $x^{3}+ y^{3}+3xy(x+y)= x^{3}- y^{3}$

= 3xy(x + y)

$a^{3} +b^{3} + c^{3}-3abc$

= (a + b + c) $(a^{2} +b^{2} + c^{2}-ab -bc- ca)$

= $a^{3} +b^{3} + c^{3}-3 abc =0$

= $\Rightarrow a^{3} +b^{3} + c^{3}=3 abc$

Putting x = 1

We get $(1)^{3}-2(1)^{2}+1+1=1$

Put x = 1

$p(1)^{2}+2(1)-2=0$

P + 2 – 2 = 0

P = 0

p(- 1) = $2+ \frac{(-1)}{2}+(-1)^{2}- \frac{(-1)^{3} }{3}$

= $2- \frac{1}{2}+1+ \frac{1}{3}$

= $\frac{10}{3}- \frac{1}{2}= \frac{17}{6}$

$12 x^{2}-7x+1$

= $12 x^{2}-4x-3x+1$

= 4x (3x – 1) -1(3x – 1)

= (4x – 1) (3x – 1)

$(2x+3y)^{2}=64$

$4 x^{2}+9 y^{2}+12xy=64$

$4 x^{2}+9 y^{2}=64-12(4)$

$4 x^{2}+9 y^{2}=64-48$

$4 x^{2}+9 y^{2}=16$

$(3x+y)^{2}=9 x^{2}+ y^{2}+6xy$

= 31 + 18

$(3x+y)^{2}=49$

3x + y =7

$\begin{bmatrix} \frac{x-1}{x} \end{bmatrix}^{2}= x^{2}+ \frac{1}{ x^{2} }-2$

= 38 -2

$\begin{bmatrix}x- \frac{1}{x} \end{bmatrix}^{2}=36$

$x- \frac{1}{x}=6$

(3x + 2y) (3x – 2y) $(9 x^{2}+4 y^{2})$

= $(9 x^{2}-4 y^{2} ) (9 x^{2}+4 y^{2})$

= $81 x^{4}-16 y^{4}$

$\begin{bmatrix}x- \frac{1}{x} \end{bmatrix} \begin{bmatrix}x+ \frac{1}{x} \end{bmatrix} \begin{bmatrix} x^{2}+ \frac{1}{ x^{2} } \end{bmatrix} \begin{bmatrix} x^{4}+ \frac{1}{ x^{4} } \end{bmatrix}$

= $\begin{bmatrix} x^{2}- \frac{1}{ x^{2} } \end{bmatrix} \begin{bmatrix} x^{2}+ \frac{1}{ x^{2} } \end{bmatrix} \begin{bmatrix} x^{4}+ \frac{1}{ x^{4} } \end{bmatrix}$

= $\begin{bmatrix} x^{4}- \frac{1}{ x^{4} } \end{bmatrix} \begin{bmatrix} x^{4}+ \frac{1}{ x^{4} } \end{bmatrix}= x^{8}- \frac{1}{ x^{8} }$

$3 x^{2}-x-y=3 x^{2}-3x+4x-4$

= 3x (x – 1) + 4 (x -1)

= 3 (x – 1) (3x + 4)

p(x) = $27 x^{3}-9x^{2}-x+3$

$p(- \frac{1}{3} )=0$

$p(- \frac{1}{3} )=27(- \frac{1}{3} )^{3}-a(- \frac{1}{3} )^{2}$

$-(- \frac{1}{3} )+2=0$

= $27(- \frac{1}{27} )-a (\frac{1}{9})+ \frac{1}{3}+2=0$

$- \frac{a}{9}=- \frac{7}{3}+1$

$- \frac{a}{9}=- \frac{4}{3}$

a = 12

$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$

$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

$(10)^{2}$ = 30 + 2 (ab + bc + ca)

$\frac{100-30}{2}$ = ab + bc + ca = 35