RD Sharma Solutions Class 7 Percentage Exercise 11.5

RD Sharma Solutions Class 7 Chapter 11 Exercise 11.5

RD Sharma Class 7 Solutions Chapter 11 Ex 11.5 PDF Free Download

Exercise 11.5

Q1. What per cent of

(i) 24 is 6?

\(Required\;percentage=\frac{6}{24}\times100\\=\frac{100}{4}\\=25\%\\ Hence\;6\;is\;25\%\;of\;24\)

(ii) Rs.125 is Rs.10?

\(Required\;percentage=Rs.\left(\frac{10}{125}\times100 \right )\\=\frac{1000}{125}\\=8\%\\ Hence\;Rs.10\;is\;8\%\;of\;Rs.125\)

(iii) 4km is 160 metres?

\(Required\;percentage=km\left(\frac{160}{4}\times100 \right )\\1km=1000metres\;\;\\4km=4000metres\\km\left(\frac{160}{4000}\times100 \right )\\ \frac{16000}{4000}=4\%\\ Hence\;160metres\;of\;4km\;is\;4\%\)

(iv) Rs.8 is 25 paise?

\(We\; know\; that, Rs.1=100 paise\\ Therefore Rs.8=800 paise\\ Required\;percentage=paise\left(\frac{25}{800}\times100 \right )\\ =\frac{25}{8}\%\\ =3.125\%\\ Hence\;25paise\;is\;3.125\%\;of\;Rs.8\)

(v) 2 days is 8 hours?

We know that,

1day is 24 hours

\(hour=\frac{1}{24}day\\ 8\;hours=\frac{8}{24}day=\frac{1}{3}day\\ Therefore\; required\;percentage=\frac{\frac{1}{3}}{2}\times100\\ =\frac{100}{6}\%\\ Hence\;8\;hours\;is\;16\frac{2}{3}\%\;of\;2\;days\)

(vi) 1 lire is 175ml

We know that


\(Therefore\; required\;percentage=\frac{175ml}{1litre}\\ =\frac{175ml}{1000ml}\times100\\ =17.5\%\\ Hence\;175ml\;is\;17.5\%\;of\;1\;litre\)

Q2. What per cent is equivalent to \(\frac{3}{8}\)?

\(\frac{3}{8}\times100\\ =\frac{300}{8}\\ =37.5\)

Q3. Find the following:

(i) 8 is 4% of which number

\(let\;x\;be\;the\;required\;number.\;Then,\\ 4\%\;of\;x=8\\ \left(\frac{4}{100}\times x \right )=8\\ x=\frac{800}{4}\\ x=200\)

(ii) 6 is 60% of which number

\(let\;x\;be\;the\;required\;number.\;Then,\\ 60\%\;of\;x=6\\ \left(\frac{60}{100}\times x \right )=6\\ x=\frac{600}{6}\\ x=10\)

(iii) 6 is 30% of which number

\(let\;x\;be\;the\;required\;number.\;Then,\\ 30\%\;of\;x=6\\ \left(\frac{30}{100}\times x \right )=6\\ x=\frac{6\times100}{30}\\ x=20\)

(iv) 12 is 25% of which number

\(let\;x\;be\;the\;required\;number.\;Then,\\ 25\%\;of\;x=12\\ \left(\frac{25}{100}\times x \right )=12\\ x=\frac{12\times100}{25}\\ x=48\)26

Q4. Convert each of the following pairs into percentages and find out which is more?

(i) 25 marks out of 30, 35 marks out of 40

\(25\;marks\;out\;of\;30=\frac{25}{30}\times100\\ =\frac{250}{3}\%\\ =83.33\%\\ 35\;marks\;out\;of\;40=\frac{35}{40}\times100\\ =\frac{7}{8}\times100\%\\ =87.5\%\)

Therefore 35 marks out of 40(87.5%) is more than 25 marks out of 30

(ii) 100 runs scored off 110 balls, 50 runs scored off 55 balls

\(100\;runs\;scored\;off\;110\;balls=\frac{100}{110}\times100\\ =90.91\%\\ 50\;runs\;scored\;off\;55\;balls=\frac{50}{55}\times100\\ =90.91\%\\ Both\;are\;same\;(90.91\%)\)


Q5. Find 20% more than Rs.200.

\(We\;have\;\\ 20\%of\;Rs.200=\frac{20}{100}\times200=Rs.40\\ Therefore\;20\%\;more\;than\;Rs.200=Rs.200+Rs.40\\ =Rs.240\)


Q6. Find 10% less than Rs.150

\(We\;have\;\\ 10\%of\;Rs.150=\frac{10}{100}\times150=Rs.15\\ Therefore\;10\%\;lss\;than\;Rs.150=Rs.150+Rs.15\\ =Rs.135\)

Practise This Question

Sum of any two sides of a triangle is always __________ third side in a triangle.