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(c) 87

3a = 4b = 6c \(\Rightarrow\) 4b = 6c \(\Rightarrow\) b = \(\frac{3}{2}c\)

and 3a = 4b \(\Rightarrow\) a = \(\frac{4}{3}b\) = \(\frac{4}{3}\times \frac{3}{2}c\) = 2c

\(\therefore\) a + b + c = \(27\sqrt{29}\)

\(\Rightarrow\) 2c + \(\frac{3}{2}c\) + c = \(27\sqrt{29}\)

\(\Rightarrow\) \(\frac{9}{2}c\) = \(27\sqrt{29}\) \(\Rightarrow\) c = \(6\sqrt{29}\)

Now, \(\sqrt{a^2+b^2+c^2}\)

= \(\sqrt{(a+b+c)^2-2(ab+bc+ca)}\)

= \(\sqrt{(27\sqrt{29})^2-2(2c\times\frac{3}{2}c+\frac{3}{2}c\times c+c\times 2c)}\)

= \(\sqrt{729 \times29 - 2(3c^2+\frac{3}{2}c^2+2c^2})\)

= \(\sqrt{729 \times29 - 2\times \frac{13c^2}{2}}\)

= \(\sqrt{729 \times29 - 13\times(6\sqrt{29)^2}}\)

= \(\sqrt{29(729-468)}\) = \(\sqrt{29\times261}\) = \(\sqrt{29\times29\times9}\)

=29 x 3 = 87

By resolving given number into prime factors, we get

9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Therefore,

\(\sqrt{9216}\) = 2 × 2 × 2 × 2 × 2 × 3 = 96

Given √(16/81)

We can write it as √(16/81) = √16/√81

But we know that √16=4 and √81=9

By substituting the values we get √(16/81) = 4/9

By using prime factorization method, we get

225 = 3 × 3 × 5 × 5

\(\sqrt{225}\) =  3 × 5 = 15

We find the square roots of 121 and 169 by the method of repeated subtraction

121 – 1 = 120

120 – 3 = 117

117 – 5 = 112

112 – 7 = 115

115 – 9 = 106

106 – 11 = 95

95 – 13 = 82

82 – 15 = 67

67 – 17 = 50

50 – 19 = 31

31 – 21 = 10

Hence, we have performed operation 11 times.

Now,

∴ √121 = 11

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Clearly, we have performed subtraction 13 times

∴ √169 = 13

Given √(64/225)

We can write it as √(64/225) = √64/√225

But we know that √64=8 and √225=15

By substituting the values we get √(64/225) =8/15

Given √(121/256)

We can write it as √(121/256) = √121/√226

But we know that √121=11 and √256=16

By substituting the values we get √ (121/256) =11/16

(i) \(23\frac{394}{729}\)

√(23394/729) 

= √(17161/729) 

= \(4\frac{23}{27}\)

(ii) \(21\frac{51}{169}\)

= \(\sqrt{21\frac{51}{169}}\)

= √(3600/169)  

= 60/13

= \(4\frac{8}{13}\)

(iii) \(10\frac{151}{225}\)

= \(\sqrt{10\frac{151}{225}}\)

= √(2401/225) 

= 49/15 

= \(3\frac{4}{15}\)

 The given area = \(80\frac{244}{729}\) m²

= 58564/729 m²

If L is length of each side

L² = 58564/729

L = √ (58564/729) 

= √58564/√729

= 242/27

= \(8\frac{26}{27}\)

Hence,  Length is \(8\frac{26}{27}\)

Given √(3(13/36))

Converting the given fraction into improper we get √(121/36)

We can write it as √(121/36) = √121/√36

But we know that √36=6 and √121=11

By substituting the values we get √(121/36) =11/6

Given √(625/729)

We can write it as √(625/729) = √625/√729

But we know that √625=25 and √729=27

By substituting the values we get √ (625/729) =25/27

(i) False 

(ii) False 

(iii) False 

(iv) True 

(v) False 

(vi) True 

(vii) False

(viii) False 

(ix) False 

(x) False

Let the tank be full in x min. Then,

\(\frac{X}{6}+\frac{X}{8}\) + \(\frac{X-4}{12}\) = 1 \(\Rightarrow\) \(\frac{4X+3X+2(X-4)}{24}\) = 1

\(\Rightarrow\) 9x – 8 = 24

\(\Rightarrow\) 9x = 32

\(\Rightarrow\) x = \(3\frac{5}{9}\) min

(c) 14

Part of the tank filled by (A + B + C) in 2 hours = 2 x \(\frac{1}{6}\) = \(\frac{1}{3}\) 

Remaining part = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)

\(\therefore\) \(\frac{2}{3}\)rd of the tank is filled by (A + B) in 7 hours.

\(\therefore\) The whole tank is filled by (A + B) in \(\big(7\times\frac{3}{2}\big)\) hours = \(\frac{21}{2}\) hours

\(\therefore\) Part of the tank filled by C in 1 hour = Part of the tank filled by (A + B + C) in 1 hour - Part of the tank filled by (A + B) in 1 hour

= \(\frac{1}{6}\) - \(\frac{2}{21}\) = \(\frac{3}{42}\) = \(\frac{1}{14}\)

\(\therefore\) C can fill the whole tank in 14 hours.

(a) 167 min

Work done in 3 min. = \(\big(\frac{1}{20}+\frac{1}{30}-\frac{1}{15}\big)\) = \(\frac{1}{60}\)

\(\therefore\) Work done in 3 × 55 = 165 min = \(\frac{55}{60}\)

\(\therefore\) Remaining tank = \(\big(1-\frac{55}{60}\big)\) = \(\frac{5}{60}\) = \(\frac{1}{12}\)

Now its A’s turn, \(\frac{1}{20}\) part of the tank is filled by A in 1 min.

Since there is still \(\big(\frac{1}{12}-\frac{1}{20}\big)\) = \(\frac{1}{30}\) tank to be filled, which will be filled by B in 1 min. Therefore, required time = (165 + 2) = 167 min.

(c) 40 min

Let one pipe take x minutes to fill the tank, then the other will take (x + 20) minutes.

Given, \(\frac{1}{X}+\frac{1}{X+20}\) = \(\frac{1}{24}\)

\(\Rightarrow\) \(\frac{X+20+X}{X^2+20X}\)= \(\frac{1}{24}\)  \(\Rightarrow\) (2x + 20)24 = x² + 20x

\(\Rightarrow\) x² – 28x – 480 = 0 \(\Rightarrow\) x 2 – 40x + 12x – 480 = 0

\(\Rightarrow\) x(x – 40) + 12(x – 40) = 0

\(\Rightarrow\) (x – 40) (x +12) = 0

\(\Rightarrow\) x = 40 or –12

Neglecting negative values x = 40 min

Part of the tank filled when all the three pipes are opened in 1 minute = \(\frac{1}{15}\)

Part of the tank filled when (A + B) are opened in 1 minute = \(\frac{1}{20}\)+\(\frac{1}{24}\)= \(\frac{6+5}{120}\) = \(\frac{11}{120}\)

\(\therefore\) Part of the tank emptied by C in 1 min. = \(\frac{11}{120}\) - \(\frac{1}{15}\) = \(\frac{1}{40}\)

\(\therefore\) Capacity of \(\frac{1}{40}\) part of the tank = 3 gallons

\(\therefore\) Capacity of the tank = (3 × 40) gallons = 120 gallons.

Let it take x minutes to fill the cistern, then work done by (Pipe A in 5 min. + Pipe B in x min.) = 1

\(\Rightarrow\) \(\frac{5}{20}\)+ \(\frac{X}{25}\)= 1 \(\Rightarrow\) \(\frac{25+4X}{100}\) = 1 \(\Rightarrow\) 25 + 4x = 100 \(\Rightarrow\) 4x = 75 \(\Rightarrow\)x = 18.75 min

(d) 30 minutes

Let the total time taken to fill the tank be x minutes.Then,

work by B in half hour + work done by (A + B) in half hour = 1

\(\big(\frac{1}{40}\times\frac{X}{2}\big)\) + \(\big(\frac{1}{60}+\frac{1}{40}\big)\) x \(\frac{X}{2}\) = 1

\(\Rightarrow\) \(\frac{X}{80}\) + \(\frac{X}{48}\) = 1  \(\Rightarrow\) \(\frac{3X+5X}{240}\) = 1

\(\Rightarrow\) 8x = 240 \(\Rightarrow\) x = 30 min.

(d) 4.5 min

Let the obstruction remain for x minutes only

\(\therefore\) Part of cistern filled in x minutes + Part of cistern filled in 3 minutes = cistern filled 

\(\Big[\big(\frac{7}{8}\times\frac{X}{12}\big)+\big(\frac{5}{6}\times\frac{X}{16}\big)\Big]\) + \(\Big[\frac{3}{12}+\frac{3}{16}\Big]\) = 1

\(\Rightarrow\) \(\frac{12X}{96}\) + \(\frac{7}{16}\) = 1 \(\Rightarrow\) \(\frac{12X}{96}\) = \(\frac{9}{16}\)

\(\Rightarrow\) x = \(\frac{9\times96}{16\times12}\) = 4.5 min

If an outlet empties a full tank in m hours, then it will empty \(\frac{1}{m}\)th part of the tank in 1 hour, i.e., work done by it is \(-\frac{1}{m}\).

Number of minutes Pipe A requires to fill an empty tank : 20 minutes 

Number of minutes Pipe B requires to fill an empty tank : 30 minutes

Amount of water filled by Pipe A in empty tank in one minute: \(\frac{1}{20}\)

Amount of water filled by Pipe B in one minute: \(\frac{1}{30}\)

Amount of water filled by Pipe A and Pipe B together in one minute: 

\(\frac{1}{20}+\frac{1}{30}=\frac{5}{60}=\frac{1}{12}\)

∴ They can fill the tank together in 12 minutes.

In one minute, \(\frac{1}{16}\) - \(\frac{1}{8}\) = - \(\frac{1}{16}\) of the tank can be filled, i.e.,\(\frac{1}{16}\) of the whole tank can be emptied.

\(\therefore\) The whole tank can be emptied in 16 minutes.

Since the tank is half full, therefore it will be emptied in 8 minutes.

(a) 1800 litres

Part of the tank that can be filled by the two filler taps = \(\big(\frac{1}{30}+\frac{1}{20}\big)\)

= \(\frac{2+3}{60}\) = \(\frac{5}{60}\) = \(\frac{1}{12}\)

\(\therefore\) The two taps can fill the tank in 12 minutes.

\(\Rightarrow\) Half of the tank will be filled in 6 minutes

\(\Rightarrow\) The time taken by the remaining half of the tank to be filled when the outlet pipe is open = (24 – 6) min = 18 min.

\(\Rightarrow\) Part of that half of the tank that can be emptied in 1 hour 

= \(\frac{1}{6}\) - \(\frac{1}{18}\) = \(\frac{1}{9}\)

\(\therefore\) Time taken to empty half of the cistern = 9 minutes

\(\therefore\) Capacity of the tank = 100 × 9 × 2 = 1800 litres.

Number of hours Pipe A requires to fill an empty tank : 10 hours 

Number of hours Pipe B requires to fill an empty tank : 15 hours

Amount of water filled by Pipe A in empty tank in one hour: \(\frac{1}{10}\)

Amount of water filled by Pipe B in one hour: \(\frac{1}{15}\)

Amount of water filled by Pipe A and Pipe B together in one hour: \(\frac{1}{10}+\frac{1}{15}=\frac{5}{30}=\frac{1}{6}\)

∴ They can fill the tank together in 6 hours.

We know that number of days taken to work is reciprocal of the rate of the work.

∴ Work done by A: \(\frac{1}{3}\)

Work done by B: \(\frac{1}{4}\)

∴ Ratio of work done is \(\frac{1}{3}:\frac{1}{4}=4:3\)

(b) 25 : 18 : 10

Given, A + B + C = 53          … (i) 

Also, A = B + 7 and B = C + 8

∴ From (i), we get (B + 7) + B + (B – 8) = 53 

⇒ 3B = 54 ⇒ B = 18 

⇒ A = 25 and C = 10 

∴ A : B : C = 25 : 18 : 10

Given,

 Inlet A can fill the cistern in = 12 hours

 Inlet B can fill it in = 15 hours

 Outlet pipe can empty it in = 10 hours

 ∵ Work done by pipe A in 1 hour \(=\frac{1}{12}\)

∵ Work done by pipe B in 1 hour =\(\frac{1}{15}\)

 ∵ work done by outlet pipe in 1 hour \(=\frac{1}{10}\)

∴ Net work done by 3 pipe in 1 hour \(=[{\frac{1}{12}+\frac{1}{15}}]-\frac{1}{10}=\frac{9}{60}-\frac{1}{10}=\frac{9-6}{60}=\frac{3}{60}=\frac{1}{20}\)part

Hence, time taken by 3 pipes to fill the tank \(=\frac{1}{\frac{1}{20}}\)= 20 hours

Given,

 Tap A can fill a tank in= 10 hours 

Tap B can fill the tank in = 15 hours 

Work done by Tap A in 1 hour \(\frac{1}{10}\)

Work done by tap B in 1 hour \(=\frac{1}{15}\)

∴ Work done by both tap in 1 hour \(=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\)part

Work done by both tap in 4 hours \(={4}\times\frac{1}{6}=\frac{2}{3}\)part

Remaining part \(={1}-\frac{2}{3}=\frac{1}{3}\)part

Hence, time taken by A to fill remaining part \(=\frac{\frac{1}{3}}{\frac{1}{10}}=\frac{10}{3}={3}\frac{1}{3}\)hours

∵Working 8 hours a day Ashu can complete a work in = 18 days

 ∴ Working 1 hours a day he can complete it in = 18 × 8 = 144 days

 ∴ Number of hours he should work to complete the work in 12 days \(=\frac{144}{12}\)=12 hours/day

(c) Rs 240

Suppose the daily wages of a man, a woman and a boy are 4k, 3k and 2k respectively.

∴ 15 x 4k + 18 x 3k + 12 x 2k = 2070 

⇒ 60k + 54k + 24k = 2070 

⇒ 138k = 2070 ⇒ k = 15. 

∴ Daily wages of a man, a woman and a boy are Rs 60, Rs 45 

∴ Daily wages of 1 man + 2 women + 3 boys 

= Rs 60 + 2 × Rs 45 + 3 × Rs 30 

= Rs 60 + Rs 90 + Rs 90 = Rs 240

Given,

 A pipe can fill a cistern in = 10 hours 

Due to leakage it get filled in = 12 hours 

∵ Without leakage work done by pipe in 1 hour \(=\frac{1}{10}\)part

∵ With leakage work done by pipe in 1 hour \(=\frac{1}{12}\)part

∴ Work done by leakage in 1 hour =\(\frac{1}{10}-\frac{1}{12}=\frac{12-10}{120}=\frac{2}{120}=\frac{1}{60}\)part

Hence, the cistern will be emptied by leakage in \(=\frac{1}{\frac{1}{60}}\)=60 hours.

6 persons working 8 hours a day earn Rs 8400 in a week

6 persons working 1 hour a day earn Rs \(\frac{8400}{8}\) in a week.

\(\therefore\) 1 person working 1 hour a day earns Rs \(\frac{8400}{6\times8}\) in a week. (less persons, less days, less earnings)

1 person working 6 hours a day earns Rs \(\frac{8400\times9\times6}{6\times8}\) in a week. (more persons, more hours, more earning) 

= Rs 9450.

(c) 8640 litres

Suppose the cistern holds x litres of water. The tap admits 6 litres of water in 1 minute, i.e., it admits 360 litres of water in 1hour.

\(\therefore\) x litres of water can be filled in \(\frac{X}{360}\) hours.

The leak would empty the tank in 8 hours.

\(\therefore\) In one hour, \(\frac{1}{8}\) of the tank becomes empty. 

If the tap and the leak both are allowed simultaneously, then the tank becomes empty in 12 hours, i.e.,

\(\frac{1}{8}-\frac{360}{X}\) = \(\frac{1}{12}\) \(\Rightarrow\) \(\frac{360}{X}\) = \(\frac{1}{8}\) - \(\frac{1}{12}\) = \(\frac{3-2}{24}\) = \(\frac{1}{24}\)

\(\Rightarrow\) x = (360 × 24) litres = 8640 litres.

(b) 100 minutes

Let the third pipe empty the whole cistern in x minutes.

\(\therefore\) \(\frac{1}{60}+\frac{1}{75}\) - \(\frac{1}{X}\) = \(\frac{1}{50}\)

\(\Rightarrow\) \(\frac{1}{X}\) = \(\frac{1}{60}+\frac{1}{75}-\frac{1}{50}\) = \(\frac{5+4-6}{300}\) = \(\frac{3}{300}\) = \(\frac{1}{100}\)

The third pipe can empty the cistern in 100 minutes

Number of hours Tap A requires to fill an empty tank : 8 hours 

Number of hours Tap B requires to empty the full tank : 12 hours

Amount of water filled by Tap A in empty tank in one hour: \(\frac{1}{8}\)

Amount of water Tap B empties in one hour: \(\frac{1}{12}\)

Amount of water filled by Tap A and Tap B together in one hour: \(\frac{1}{8}-\frac{1}{12}=\frac{1}{24}\)

∴ They can fill the tank together in 24 hours.

24 men can complete a work in 16 days

\(\therefore\) 1 man’s 1 days’ work = \(\frac{1}{24\times16}\) 

\(\therefore\) 12 men’s 1 day’s work = \(\frac{12}{24\times16}\) = \(\frac{1}{32}\)

18 women can complete the work in 32 days

\(\therefore\) 1 woman’s 1 days’ work = \(\frac{1}{18\times32}\)

\(\therefore\) 6 women’s 1 day’s work = \(\frac{6}{18\times32}\) = \(\frac{1}{96}\)

\(\therefore\) (12 men’s + 6 women’s) 16 days’ work = 16 x \(\big(\frac{1}{32}+\frac{1}{96}\big)\) = 16 x \(\big(\frac{3+1}{96}\big)\)

= 16 x \(\frac{4}{96}\) = \(\frac{2}{3}\)

\(\therefore\) Remaining work = 1 - \(\frac{2}{3}\) = \(\frac{1}{3}\)

(12 men’s + 6 women’s) 2 days’ work = 2 x \(\big(\frac{1}{32}+\frac{1}{96}\big)\) = 2 x \(\big(\frac{4}{96}\big)\) = \(\frac{1}{12}\)

Remaining work = \(\frac{1}{3}\) - \(\frac{1}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)

\(\therefore\) \(\frac{1}{384}\) work is done in 1 day by 1 man

\(\therefore\) \(\frac{1}{12}\) work will be done in 2 days by \(\big(384\times\frac{1}{12}\times\frac{1}{2}\big)\)men = 16 men.

(a) Rs 7371

Given, Man : Woman : Child = 3 : 2 :1

\(\Rightarrow\) M = 3K , W = 2K, C = K, where K is a constant

\(\Rightarrow\) M = 3C , W = 2C

\(\therefore\) 20M + 30 W + 36C = 60C + 60C + 36C = 156C

and 15M + 21W + 30C = 45C + 42C + 30C = 117C

It is given that (20 men + 30 women + 36 children) = 156 children earn Rs 78 per day

\(\Rightarrow\) 1 child earns Rs \(\frac{78}{156}\) per day = Rs \(\frac{1}{2}\) per day.

\(\therefore\) (15 men + 21 women + 30 children) = 117

children should get \(\big(Rs\, \frac{1}{2}\times117\big)\) per day,

i.e., Rs \(\big(\frac{1}{2}\times117\times18\times7\big)\) = Rs 7371 in 18 weeks.

In ‘If I was a Tree’, the friendship of the tree with the cool breeze and the leaves would be sweet.

(b) 24

1 man’s 1 days’ work = \(\frac{1}{24\times16}\)

\(\therefore\) 16 men’s 1days’ work = \(\frac{16}{24\times16}\) = \(\frac{1}{24}\)

1 woman’s 1 days’ work = \(\frac{1}{32\times24}\)

16 women’s 1 days’ work = \(\frac{16}{32\times24}\) = \(\frac{1}{48}\)

\(\therefore\) 12 days’ work of (16 men + 16 women) = 12\(\big(\frac{1}{24}+\frac{1}{48}\big)\)

= 12\(\big(\frac{2+1}{48}\big)\) = 12 x \(\frac{3}{48}\) = \(\frac{3}{4}\)

Remaining work = 1 - \(\frac{3}{4}\) = \(\frac{1}{4}\)

Now (16 men’s + 16 women’s) 2 days’ work = 2\(\big(\frac{1}{24}+\frac{1}{48}\big)\) = 2 x \(\frac{1}{16}\) = \(\frac{1}{8}\)

\(\therefore\) Remaining work = \(\frac{1}{4}\) - \(\frac{1}{8}\) = \(\frac{1}{8}\)

\(\frac{1}{384}\) work is done in 1 day by 1 man

\(\therefore\) \(\frac{1}{8}\) work will be done in 2 days by \(\big(384\times\frac{1}{8}\times\frac{1}{2}\big)\) = 24 men.

Number of days required by 3 men to finish the work : 12 days 

Number of days required by 5 women to finish the work : 12 days 

Number of days required by 1 man to finish the work : 3 x 12 = 36 days 

Number of days required by 1 woman to finish the work : 5 x 12 = 60 days

Work done by a man in one day: \(\frac{1}{36}\)

Work done by a woman in one day: \(\frac{1}{60}\)

Work done by 6 men in one day: \(6\times\frac{1}{36}=\frac{1}{6}\)

Work done by 5 women in one day: \(5\times\frac{1}{60}=\frac{1}{12}\)

Work done by 6 men and 5 women together in one day: \(\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

∴ 6 men and 5 women together take 4 days to complete the work.

Number of days required by 2 men to finish the work : 16 days 

Number of days required by 3 women to finish the work : 16 days 

Number of days required by 1 man to finish the work : 2 x 16 = 32 days 

Number of days required by 1 woman to finish the work : 3 x 16 = 48 days

Work done by a man in one day: \(\frac{1}{32}\)

Work done by a woman in one day: \(\frac{1}{48}\)

Work done by 4 men in one day: \(4\times\frac{1}{32}=\frac{1}{8}\)

Work done by 6 women in one day: \(6\times\frac{1}{48}=\frac{1}{8}\)

Work done by 4 men and 6 women together in one day: \(\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\)

∴ 4 men and 6 women together take 4 days to complete the work.

(a) 7

Let 1 woman’s 1 days’ work = x. Then

1 man’s 1 days’ work = \(\frac{X}{2}\) and 1 child’s 1 days’ work = \(\frac{X}{4}\)

Given, (3 men’s + 4 women’s + 6 children’s) 1 days’ work = \(\frac{1}{7}\)

\(\Rightarrow\) \(\frac{3X}{2}\)+ 4x + \(\frac{6X}{4}\) = \(\frac{1}{7}\) \(\Rightarrow\) \(\frac{6X+16X+6X}{4}\) = \(\frac{1}{7}\)

\(\Rightarrow\) \(\frac{28X}{4}\) = \(\frac{1}{7}\)  \(\Rightarrow\) x = \(\frac{1}{49}\)

\(\therefore\) 1 woman can alone complete the work in 49 days To complete the work in 7 days, number of women required = \(\frac{49}{7}\) = 7.

(b) 7.5 days

10 women can complete a piece of work in 15 days

\(\therefore\) 1 woman can complete in 1 day, \(\frac{1}{10\times15}\) = \(\frac{1}{150}\) part of work

6 men can complete the same work in 10 days,

\(\therefore\) 1 man can complete in 1 day, \(\frac{1}{6\times10}\) = \(\frac{1}{60}\) part of the work

\(\therefore\) (6 men’s + 5 women’s) 1 days’ work

= \(\frac{6}{60}\)+ \(\frac{5}{150}\) = \(\frac{1}{10}\) + \(\frac{1}{30}\) = \(\frac{3+1}{30}\) = \(\frac{4}{30}\) = \(\frac{2}{15}\)

\(\therefore\) 6 men and 5 women can complete the whole work in \(\frac{15}{2}\) = 7.5 days

The sunlight would embrace the tree in the poem ‘If I was a Tree’

The poem ‘If I was a tree’ is a satire in which the poet makes an attempt to expose the subtle ways in which the upper caste society has been discriminating against untouchables for centuries. The speaker juxtaposes the world of nature with the human world so as to accuse human beings of practising untouchability and being meaner than the world of nature.

The speaker speaks in the persona of an untouchable but asks the reader to imagine that he is a tree. Next, he presents a few everyday instances of untouchability. In the poem, the tree is a metaphor for the world of nature and, using the tree as the point of contact between nature and the people, he points out how elements of nature like the ‘cool breeze’ and the ‘sunlight’ do not discriminate between people and treat everyone equally. Thus he highlights nature’s idea of equality.

The speaker argues that if he was a tree, and when sunlight falls on him, his shadow would not feel defiled. On the contrary, if he were an untouchable and when sunlight falls on him and his shadow falls on a person of the upper caste he or she would feel that he or she has been defiled by the shadow of an untouchable. Similarly, since he is a tree and not an untouchable, the friendship between the cool breeze and the leaves of the tree would be sweet. On the contrary, if he were an untouchable there would be no such sweet friendship. Thus, the speaker wants to argue that the elements of nature enforce equality whereas some sections of human beings enforce inequality.

(a) 15

According to question, 10 women = 7 men

\(\therefore\) 20 women = 14 men

14 men + 20 women = 14 men + 14 men = 28 men.

A wall of 100 m is built by 7 men in 10 days

A wall of 100 m is built by 1 man in (10 × 7) days

A wall of 1 m is built by 1 man in \(\frac{10\times7}{100}\)days

(less length, less days)

A wall of 600 m is built by 1 man in \(\frac{10\times7\times600}{100}\) days (more length,more days)

A wall of 600 m is built by 28 men in \(\frac{10\times7\times600}{100\times28}\) days = 15 days.

In the poem, ‘If I was a Tree’, the shadow of the tree would not feel defiled when sunlight embraces it.

If he were to be a tree.

(b) 30

Let the number of extra men employed be x.

30 men in 60 days complete 120 m long road.

1 man in 1 day will complete \(\frac{120}{30\times60}\)m long road.

Also, (30 + x) men in 60 days complete 240 m long road.

\(\therefore\) 1 man in 1 day will complete \(\frac{240}{(30+X)\times60}\)m long road.

\(\Rightarrow\) \(\frac{120}{30\times60}\) = \(\frac{240}{(30+X)\times60}\)m \(\Rightarrow\) \(\frac{1}{15}\) = \(\frac{4}{(30+X)}\)

\(\Rightarrow\) 30 + x = 60 \(\Rightarrow\) x = 30

\(\therefore\) 30 more men have to be employed