Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 7 Comparing Quantities are provided here with simple stepbystep explanations. These solutions for Comparing Quantities are extremely popular among class 7 students for Maths Comparing Quantities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 7 Maths Chapter 7 are provided here for you for free. You will also love the adfree experience on Meritnationâ€™s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 7 Maths are prepared by experts and are 100% accurate.
Page No 196:
Question 1:
In the question, there are four options, out of which one is correct. Write the correct one.
20% of 700 m is
(a) 560 m
(b) 70 m
(c) 210 m
(d) 140 m
Answer:
20% of 700 m $=\frac{20}{100}\times 700\mathrm{m}\phantom{\rule{0ex}{0ex}}=140\mathrm{m}$
Hence, the correct answer is option (d).
Page No 196:
Question 2:
In the question, there are four options, out of which one is correct. Write the correct one.
Gayatri’s income is â‚¹1,60,000 per year. She pays 15% of this as house rent and 10% of the remainder on her child’s education. The money left with her is
(a) â‚¹136000
(b) â‚¹120000
(c) â‚¹122400
(d) â‚¹14000
Answer:
Gayatri's yearly income = â‚¹1,60,000
House rent paid by her = 15% of 1,60,000
$=\frac{15}{100}\times 1,60,000\phantom{\rule{0ex}{0ex}}=\u20b924,000$
Remaining amount = â‚¹1,60,000 − â‚¹24,000
Amount paid by her on her child's education
$=10\%\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{amount}\phantom{\rule{0ex}{0ex}}=10\%\mathrm{of}\u20b91,36,000\phantom{\rule{0ex}{0ex}}=\frac{10}{100}\times 1,36,000\phantom{\rule{0ex}{0ex}}=\u20b913,600$
Money left with her = â‚¹1,36,000 − â‚¹13,600
= â‚¹1,22,400
Hence, the correct answer is option (a).
Page No 196:
Question 3:
In the question, there are four options, out of which one is correct. Write the correct one.
The ratio of Fatima’s income to her savings is 4 : 1. The percentage of money saved by her is :
(a) 20%
(b) 25%
(c) 40%
(d) 80%
Answer:
Ration of Fatima's income to her savings = 4 : 1
Sum of ratio = 4 + 1
= 5
Percentage of money saved by her $=\left(\frac{1}{5}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=20\%$
Hence, the correct answer is option (a).
Page No 196:
Question 4:
In the question, there are four options, out of which one is correct. Write the correct one.
0.07 is equal to
(a) 70%
(b) 7%
(c) 0.7%
(d) 0.07%
Answer:
$\begin{array}{rcl}0.07& =& \left(\frac{7}{100}\times 100\right)\%\\ & =& 7\%\end{array}$
Hence, the correct answer is option (b).
Page No 196:
Question 5:
In the question, there are four options, out of which one is correct. Write the correct one.
In a scout camp, 40% of the scouts were from Gujarat State and 20% of these were from Ahmedabad. The percentage of scouts in the camp from Ahmedabad is:
(a) 25
(b) 32.5
(c) 8
(d) 50
Answer:
Let the total number of scouts be x.
Percentage of scouts in the camp from Gujarat state = 40%
Number of scouts from Gujarat stat = 40% of â€‹x
$=\frac{40}{100}\times x\phantom{\rule{0ex}{0ex}}=\frac{2}{5}x$
Percentage of scouts from Ahmedabad $=20\%\mathrm{of}\frac{2}{5}x\phantom{\rule{0ex}{0ex}}=\frac{20}{100}\times \frac{2}{5}\times x\phantom{\rule{0ex}{0ex}}=\frac{4\times 2}{100}\times x\phantom{\rule{0ex}{0ex}}=\frac{8}{100}\times x\phantom{\rule{0ex}{0ex}}=8\%\mathrm{of}x$
Hence, the correct answer is option (c).
Page No 196:
Question 6:
In the question, there are four options, out of which one is correct. Write the correct one.
What percent of â‚¹ 4500 is â‚¹ 9000?
(a) 200
(b) $\frac{1}{2}$
(c) 2
(d) 50
Answer:
Let x percent of â‚¹4500 be â‚¹9000.
∴ x% of 4500 = 9000
$\Rightarrow \frac{x}{100}\times 4500=9000\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{9000}{45}\phantom{\rule{0ex}{0ex}}\Rightarrow x=200\%$
Hence, the correct answer is option (a).
Page No 197:
Question 7:
In the question, there are four options, out of which one is correct. Write the correct one.
5.2 is equal to
(a) 52%
(b) 5.2%
(c) 520%
(d) 0.52%
Answer:
$\begin{array}{rcl}5.2& =& \left(\frac{52}{10}\times 100\right)\%\\ & =& 520\%\end{array}$
Hence, the correct answer is option (c).
Page No 197:
Question 8:
In the question, there are four options, out of which one is correct. Write the correct one.
The ratio 3 : 8 is equal to
(a) 3.75%
(b) 37.5%
(c) 0.375%
(d) 267%
Answer:
$\begin{array}{rcl}\frac{3}{8}& =& \left(\frac{3}{8}\times 100\right)\%\\ & =& \left(\frac{3}{2}\times 25\right)\%\\ & =& \left(\frac{75}{2}\right)\%\\ & =& 37.5\%\end{array}$
Hence, the correct answer is option (b).
Page No 197:
Question 9:
In the question, there are four options, out of which one is correct. Write the correct one.
225% is equal to
(a) 9 : 4
(b) 4 : 9
(c) 3 : 2
(d) 2 : 3
Answer:
$\begin{array}{rcl}225\%& =& \frac{225}{100}\\ & =& \frac{9}{4}\\ & =& 9:4\end{array}$
Hence, the correct answer is option (a).
Page No 197:
Question 10:
In the question, there are four options, out of which one is correct. Write the correct one.
A bicycle is purchased for â‚¹ 1800 and is sold at a profit of 12%. Its selling price is
(a) â‚¹ 1584
(b) â‚¹ 2016
(c) â‚¹ 1788
(d) â‚¹ 1812
Answer:
Cost Price of bicycle (P) = â‚¹1800
Profit % = 12%
Selling Price of bicycle (SP) $=\mathrm{CP}\times \left(1+\frac{\mathrm{P}\%}{100}\right)\phantom{\rule{0ex}{0ex}}=1800\left(1+\frac{12}{100}\right)\phantom{\rule{0ex}{0ex}}=1800\times \frac{112}{100}\phantom{\rule{0ex}{0ex}}=\u20b92016$
Thus, the selling price of bicycle is â‚¹2016.
Hence, the correct answer is option (b).
Page No 197:
Question 11:
In the question, there are four options, out of which one is correct. Write the correct one.
A cricket bat was purchased for â‚¹ 800 and was sold for â‚¹ 1600. Then profit earned is
(a) 100%
(b) 64%
(c) 50%
(d) 60%
Answer:
Cost Price of cricket bat (CP) = â‚¹800
Selling Price of cricket bat (SP) = â‚¹1600
Profit = SP − CP
= 1600 − 800
= â‚¹800
$\begin{array}{rcl}\mathrm{Profit}\%& =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{\u20b9800}{\u20b9800}\times 100\right)\%\\ & =& 100\%\end{array}$
Thus, the profit earned is 100%.
Hence, the correct answer is option (a).
Page No 197:
Question 12:
In the question, there are four options, out of which one is correct. Write the correct one.
A farmer bought a buffalo for â‚¹44000 and a cow for â‚¹18000. He sold the buffalo at a loss of 5% but made a profit of 10% on the cow. The net result of the transaction is
(a) loss of â‚¹200
(b) profit of â‚¹400
(c) loss of â‚¹400
(d) profit of â‚¹200
Answer:
Cost Price of buffalo = â‚¹44000
loss % on buffalo = 5%
loss on buffalo = 5% of 44000
$=\frac{5}{100}\times 44000\phantom{\rule{0ex}{0ex}}=\u20b92,200$
∴ Selling Price of buffalo $=\mathrm{CP}\left(1\frac{\mathrm{Loss}\%}{100}\right)\phantom{\rule{0ex}{0ex}}=44,000\left(1\frac{5}{100}\right)\phantom{\rule{0ex}{0ex}}=44,000\times \frac{95}{100}\phantom{\rule{0ex}{0ex}}=\u20b941,800$
Cost Price of cow = â‚¹18000
Profit % on cow = 10%
∴ Selling Price of cow $=\mathrm{CP}\left(1+\frac{\mathrm{Profit}\%}{100}\right)\phantom{\rule{0ex}{0ex}}=18000\left(1+\frac{10}{100}\right)\phantom{\rule{0ex}{0ex}}=18000\times \frac{110}{100}\phantom{\rule{0ex}{0ex}}=19,800$
Total cost price = â‚¹44000 + 18000
= â‚¹62000
Total selling price = â‚¹41,800 + â‚¹19800
= â‚¹61600
Total selling price < Total cost price
∴ Net loss = Total CP − Total SP
= â‚¹62,000 − â‚¹61600
= â‚¹400
Hence, the correct answer is option (b).
Page No 197:
Question 13:
In the question, there are four options, out of which one is correct. Write the correct one.
If Mohan’s income is 25% more than Raman’s income, then Raman’s income is less than Mohan’s income by
(a) 25%
(b) 80%
(c) 20%
(d) 75%
Answer:
Let Raman's income be â‚¹x.
∴ Mohan's income = x + 25% of x
$=x\left(1+\frac{25}{100}\right)\phantom{\rule{0ex}{0ex}}=x\left(\frac{125}{100}\right)\phantom{\rule{0ex}{0ex}}=x\left(\frac{5}{4}\right)$
Difference in Raman and Mohan's income
$=\frac{5}{4}xx\phantom{\rule{0ex}{0ex}}=x\left(\frac{5}{4}1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{4}x$
Raman's income is less than Mohan's income by
$=\left(\frac{{\displaystyle \frac{1}{4}}x}{{\displaystyle \frac{5}{4}}x}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{5}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=20\%$
Hence, the correct answer is option (c).
Page No 197:
Question 14:
In the question, there are four options, out of which one is correct. Write the correct one.
The interest on â‚¹ 30000 for 3 years at the rate of 15% per annum is
(a) â‚¹ 4500
(b) â‚¹ 9000
(c) â‚¹ 18000
(d) â‚¹ 13500
Answer:
Principal (P) = â‚¹30000
Time (T) = 3 years
Rate (R) = 15%
Simple Interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{30000\times 15\times 3}{100}\phantom{\rule{0ex}{0ex}}=\u20b913,500$
Hence, the correct answer is option (d).
Page No 198:
Question 15:
In the question, there are four options, out of which one is correct. Write the correct one.
Amount received on â‚¹ 3000 for 2 years at the rate of 11% per annum is
(a) â‚¹ 2340
(b) â‚¹ 3660
(c) â‚¹ 4320
(d) â‚¹ 3330
Answer:
Principal (P) = â‚¹3000
Time (T) = 2 years
Rate (R) = 11%
Simple Interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{3000\times 11\times 2}{100}\phantom{\rule{0ex}{0ex}}=\u20b9660$
Amount (A) = P + SI
= â‚¹3000 + â‚¹660
= â‚¹3660
Hence, the correct answer is option (b).
Page No 198:
Question 16:
In the question, there are four options, out of which one is correct. Write the correct one.
Interest on â‚¹ 12000 for 1 month at the rate of 10 % per annum is
(a) â‚¹ 1200
(b) â‚¹ 600
(c) â‚¹ 100
(d) â‚¹ 12100
Answer:
Principal (P) = â‚¹12000
Time (T) = 1 month $=\frac{1}{12}\mathrm{year}$
Rate (R) = 10%
Simple Interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{12000\times 10\times {\displaystyle \frac{1}{12}}}{100}\phantom{\rule{0ex}{0ex}}=\u20b9100$
Hence, the correct answer is option (c).
Page No 198:
Question 17:
In the question, there are four options, out of which one is correct. Write the correct one.
Rajni and Mohini deposited â‚¹ 3000 and â‚¹ 4000 in a company at the rate of 10% per annum for 3 years and $2\frac{1}{2}$ years respectively. The difference of the amounts received by them will be
(a) â‚¹ 100
(b) â‚¹ 1000
(c) â‚¹ 900
(d) â‚¹ 1100
Answer:
For Rajni
Principal (P) = â‚¹3000
Time (T) = 3 years
Rate (R) = 10%
Simple Interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{3000\times 10\times 3}{100}\phantom{\rule{0ex}{0ex}}=\u20b9900$
Amount received by Rajni = P + SI
= 3000 + 900
= â‚¹3900
For Mohini
Principal (P) = â‚¹4000
Time (T) = $2\frac{1}{2}\mathrm{years}=\frac{5}{2}\mathrm{years}$
Rate (R) = 10%
Simple Interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{4000\times 10\times {\displaystyle \frac{5}{2}}}{100}\phantom{\rule{0ex}{0ex}}=\u20b91000$
Amount received by Rajni = P + SI
= 4000 + 1000
= â‚¹5000
Difference of the amount received by them
= â‚¹5000 − â‚¹3900
= â‚¹1100
Hence, the correct answer is option (d).
Page No 198:
Question 18:
In the question, there are four options, out of which one is correct. Write the correct one.
If 90% of x is 315 km, then the value of x is
(a) 325 km
(b) 350 km
(c) 405 km
(d) 340 km
Answer:
90% of x = 315 km
$\Rightarrow \frac{90}{100}\times x=315\mathrm{km}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{100}{90}\times 315\mathrm{km}\phantom{\rule{0ex}{0ex}}\Rightarrow x=350\mathrm{km}$
Hence, the correct answer is option (b).
Page No 198:
Question 19:
In the question, there are four options, out of which one is correct. Write the correct one.
On selling an article for â‚¹ 329, a dealer lost 6%. The cost price of the article is
(a) â‚¹ 310.37
(b) â‚¹ 348.74
(c) â‚¹ 335
(d) â‚¹ 350
Answer:
Selling Price of article (SP) = â‚¹329
loss% = 6%
Let Cost Price of article (CP) be x.
$\therefore \mathrm{SP}=\mathrm{CP}\left(1\frac{\mathrm{Loss}\%}{100}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 329=x\left(1\frac{6}{100}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 329=x\left(\frac{94}{100}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{329\times 100}{94}=x\phantom{\rule{0ex}{0ex}}\Rightarrow \u20b9350=x\phantom{\rule{0ex}{0ex}}$
Thus, the cost price of the article is â‚¹350.
Hence, the correct answer is option (d).
Page No 198:
Question 20:
In the question, there are four options, out of which one is correct. Write the correct one.
$\frac{25\%\mathrm{of}50\%\mathrm{of}100\%}{25\times 50}$is equal to
(a) 1.1%
(b) 0.1%
(c) 0.01%
(d) 1%
Answer:
$\begin{array}{rcl}\frac{25\%\mathrm{of}50\%\mathrm{of}100\%}{25\times 50}& =& \frac{{\displaystyle \frac{25}{100}}\times \left({\displaystyle \frac{50}{100}}\times \left({\displaystyle \frac{100}{100}}\right)\right)}{25\times 50}\\ & =& \frac{1}{100\times 100}\\ & =& \left(\frac{\left({\displaystyle \frac{1}{100}}\right)}{100}\right)\\ & =& \frac{0.01}{100}\\ & =& \left(\frac{0.01}{100}\times 100\right)\\ & =& 0.01\%\end{array}$
Hence, the correct answer is option (c).
Page No 198:
Question 21:
In the question, there are four options, out of which one is correct. Write the correct one.
The sum which will earn a simple interest of â‚¹ 126 in 2 years at 14% per annum is
(a) â‚¹ 394
(b) â‚¹ 395
(c) â‚¹ 450
(d) â‚¹ 540
Answer:
Let the Principal be P.
Simple Interest (SI) = â‚¹126
Time (T) = 2 years
Rate (R) = 14%
Since, Simple interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$
$\Rightarrow 126=\frac{\mathrm{P}\times 14\times 2}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\frac{126\times 100}{14\times 2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\u20b9450$
Hence, the correct answer is option (c).
Page No 199:
Question 22:
In the question, there are four options, out of which one is correct. Write the correct one.
The per cent that represents the unshaded region in the figure.
(a) 75%
(b) 50%
(c) 40%
(d) 60%
Answer:
Number of rows = 10
Number of columns = 10
∴ Total number of squares = 10 $\times $ 10
= 100
Number of shaded squares = 36 + 20 + 4
= 60
Number of unshaded squares = 100 − 60
= 40
Percentage of unshaded region $=\left(\frac{40}{100}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=40\%$
Hence, the correct answer is option (c).
Page No 199:
Question 23:
In the question, there are four options, out of which one is correct. Write the correct one.
The per cent that represents the shaded region in the figure is
(a) 36%
(b) 64%
(c) 27%
(d) 48%
Answer:
Number of rows = 10
Number of columns = 10
∴ Total number of squares = 10 $\times $ 10 = 100
Number of shaded squares = 12 + 12 + 12
= 36
Percentage of shaded region $=\left(\frac{36}{100}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=36\%$
Hence, the correct answer is option (a).
Page No 199:
Question 24:
Fill in the blanks to make the statements true.
2 : 3 = ________ %
Answer:
2 : 3 $=\left(\frac{2}{3}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=\left(\frac{200}{3}\right)\%\phantom{\rule{0ex}{0ex}}=66.67\%$
Hence, 2 : 3 = 66.67%.
Page No 200:
Question 25:
Fill in the blanks to make the statements true.
$18\frac{3}{4}\%$= _______ : _______
Answer:
$\begin{array}{rcl}18\frac{3}{4}\%& =& \left(\frac{75}{4}\right)\%\\ & =& \frac{75}{4\times 100}\\ & =& \frac{3}{4\times 4}\\ & =& \frac{3}{16}\\ & =& 3:16\end{array}$
Hence, $18\frac{3}{4}\%=\overline{)3:16}$
Page No 200:
Question 26:
Fill in the blanks to make the statements true.
30% of â‚¹ 360 = ________.
Answer:
30% of â‚¹360 $=\frac{30}{100}\times 360\phantom{\rule{0ex}{0ex}}=\u20b9102$
Hence, 30% of â‚¹360 = â‚¹102.
Page No 200:
Question 27:
Fill in the blanks to make the statements true.
120 % of 50 km = ________.
Answer:
120 % of 50 km $=\left(\frac{120}{100}\times 50\mathrm{km}\right)\phantom{\rule{0ex}{0ex}}=60\mathrm{km}$
Hence, 120% of 50 km = 60 km.
Page No 200:
Question 28:
Fill in the blanks to make the statements true.
2.5 = ________%
Answer:
$\begin{array}{rcl}2.5& =& \left(2.5\times 100\right)\%\\ & =& \left(\frac{25}{10}\times 100\right)\%\\ & =& 250\%\end{array}$
Hence, 2.5 = 250%.
Page No 200:
Question 29:
Fill in the blanks to make the statements true.
$\frac{8}{5}$= _______ %
Answer:
$\begin{array}{rcl}\frac{8}{5}& =& \left(\frac{8}{5}\times 100\right)\%\\ & =& 160\%\end{array}$
Hence, $\frac{8}{5}=\overline{)160\%}.$
Page No 200:
Question 30:
Fill in the blanks to make the statements true.
A _______ with its denominator 100 is called a per cent.
Answer:
A fraction with its denominator 100 is called a percent.
Page No 200:
Question 31:
Fill in the blanks to make the statements true.
15 kg is _______ % of 50 kg.
Answer:
Let the required percentage be x.
∴ x% of 50 kg = 15 kg
$\Rightarrow \frac{x}{100}\times 50\mathrm{kg}=15\mathrm{kg}\phantom{\rule{0ex}{0ex}}\Rightarrow x=15\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow x=30\%$
Hence, 15 kg is 30% of 50 kg.
Page No 200:
Question 32:
Fill in the blanks to make the statements true.
Weight of Nikhil increased from 60 kg to 66 kg. Then, the increase in weight is _______ %.
Answer:
Initial weight = 60 kg
Final weight = 66 kg
Increase in weight = 66 kg − 60 kg
= 6 kg
Percentage increase in weight $=\left(\frac{6}{60}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=10\%$
Hence, weight of Nikhil increased from 60 kg to 66 kg.
Then the increase in weight is 10%.
Page No 200:
Question 33:
Fill in the blanks to make the statements true.
In a class of 50 students, 8 % were absent on one day. The number of students present on that day was ________.
Answer:
Number of students = 50
Percentage of absent students = 8%
Number of absent students $=\frac{8}{100}\times 50\phantom{\rule{0ex}{0ex}}=4$
Number of present students = 50 − 4
= 46
Hence, the number of students present on that day was 46.
Page No 200:
Question 34:
Fill in the blanks to make the statements true.
Savitri obtained 440 marks out of 500 in an examination. She secured _______ % marks in the examination.
Answer:
Marks obtained = 440
Total Marks = 500
Percentage $=\left(\frac{440}{500}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=88\%$
Hence, she secured 88% marks in the examination.
Page No 200:
Question 35:
Fill in the blanks to make the statements true.
Out of a total deposit of â‚¹ 1500 in her bank account, Abida withdrew 40% of the deposit. Now the balance in her account is ______.
Answer:
Total deposit in bank account = â‚¹1500
Withdraw percent = 40%
Balance = â‚¹1500 − (40% of â‚¹1500)
= 1500 − $\left(\frac{40}{100}\times 1500\right)$
= 1500 − 600
= â‚¹1100
Now, the balance in her account is â‚¹1100.
Page No 200:
Question 36:
Fill in the blanks to make the statements true.
________ is 50% more than 60
Answer:
Let the required number be x.
So, x = 50% of 60 + 60
$x=\left(\frac{50}{100}\times 60\right)+60\phantom{\rule{0ex}{0ex}}x=30+60\phantom{\rule{0ex}{0ex}}x=90$
Hence, 90 is 50% more than 60.
Page No 200:
Question 37:
Fill in the blanks to make the statements true.
John sells a bat for â‚¹ 75 and suffers a loss of â‚¹ 8. The cost price of the bat is ________.
Answer:
Selling Price of bat (SP) = â‚¹75
Loss (L) = â‚¹8
Cost Price of bat (CP) = SP + L
= â‚¹75 + â‚¹8
= â‚¹83
Hence, the cost price of the bat is â‚¹83.
Page No 200:
Question 38:
Fill in the blanks to make the statements true.
If the price of sugar is decreased by 20%, then the new price of 3kg sugar originally costing â‚¹ 120 will be ________.
Answer:
Initial Cost Price of 3 kg of Sugar = â‚¹120
Initial Cost Price of 1 kg of Sugar $=\frac{\u20b9120}{3}\phantom{\rule{0ex}{0ex}}=\u20b940$
Percentage decrease in cost = 20%
Decrease in cost $=20\%\mathrm{of}\u20b940\phantom{\rule{0ex}{0ex}}=\frac{20}{100}\times 40\phantom{\rule{0ex}{0ex}}=\u20b98$
Decreased cost of 1 kg Sugar = â‚¹40 − â‚¹8
= â‚¹32
Decreased cost of 3 kg Sugar = â‚¹32 $\times $ 3
= â‚¹96
Hence, the price of 3 kg sugar originally costing â‚¹120 will be â‚¹96.
Page No 200:
Question 39:
Fill in the blanks to make the statements true.
Mohini bought a cow for â‚¹ 9000 and sold it at a loss of â‚¹ 900. The selling price of the cow is ________.
Answer:
Cost Price of cow (CP) = â‚¹9000
Loss (L) = â‚¹900
Selling Price of cow (SP) = CP − L
= â‚¹9000 − â‚¹900
= â‚¹8100
Hence, the selling price of the cow is â‚¹8100.
Page No 201:
Question 40:
Fill in the blanks to make the statements true.
Devangi buys a chair for â‚¹ 700 and sells it for â‚¹ 750. She earns a profit of ________ % in the transaction.
Answer:
Cost Price of Chair (CP) = â‚¹700
Selling Price of Chair (SP) = â‚¹750
Profit (P) = SP − CP
= â‚¹750 − â‚¹700
= â‚¹50
Profit % $=\left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=\left(\frac{50}{700}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=7.14\%$
Hence, she earns a profit of 7.14% in the transaction.
Page No 201:
Question 41:
Fill in the blanks to make the statements true.
Sonal bought a bed sheet for â‚¹ 400 and sold it for â‚¹ 440. Her ________% is ________.
Answer:
Cost Price of bed sheet (CP) = â‚¹400
Selling Price of bed sheet (SP) = â‚¹440
Profit (P) = SP − CP
= â‚¹440 − â‚¹400
= â‚¹40
Profit % $=\left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=\left(\frac{40}{400}\times 100\right)\%\phantom{\rule{0ex}{0ex}}=10\%$
Hence, Sonal bought a bed sheet for â‚¹400 and sold it for â‚¹400.
Her Profit 40% is 10%.
Page No 201:
Question 42:
Fill in the blanks to make the statements true.
Nasim bought a pen for â‚¹ 60 and sold it for â‚¹ 54. His ________% is ________.
Answer:
Cost Price of pen (CP) = â‚¹60
Selling Price of pen (SP) = â‚¹54
Loss (L) $=\mathrm{CP}\mathrm{SP}\phantom{\rule{0ex}{0ex}}=\u20b960\u20b954\phantom{\rule{0ex}{0ex}}=\u20b96$
$\begin{array}{rcl}\mathrm{Loss}\%& =& \left(\frac{\mathrm{Loss}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{6}{60}\times 100\right)\%\\ & =& 10\%\end{array}$
Hence, his loss % is 10%.
Page No 201:
Question 43:
Fill in the blanks to make the statements true.
Aahuti purchased a house for â‚¹ 50,59,700 and spent â‚¹ 40300 on its repairs. To make a profit of 5%, she should sell the house for â‚¹ ________.
Answer:
Purchasing amount of house = â‚¹50,59,700
Amount spent on repairing the house = â‚¹40,300
Cost price of house (CP) = â‚¹50,59,700 + â‚¹40,300
= â‚¹51,00,000
Profit % = 5%
Selling Price of house (SP) $=\mathrm{CP}\left(1+\frac{\mathrm{P}\%}{100}\right)$
$=\u20b951,00,000\left(1+\frac{5}{100}\right)\phantom{\rule{0ex}{0ex}}=\u20b951,00,000\times \frac{105}{100}\phantom{\rule{0ex}{0ex}}=\u20b953,55,000$
Hence, to make a profit of 5%, she should sell the house for â‚¹53,55,000.
Page No 201:
Question 44:
Fill in the blanks to make the statements true.
If 20 lemons are bought for â‚¹ 10 and sold at 5 for three rupees, then ________ in the transaction is ________%.
Answer:
Cost Price of 20 lemons = â‚¹10
Cost Price of 1 Lemon $=\frac{\u20b910}{20}\phantom{\rule{0ex}{0ex}}=\u20b90.5$
Selling Price of 5 lemons = â‚¹3
Selling Price of 1 lemon $=\u20b9\frac{3}{5}\phantom{\rule{0ex}{0ex}}=\u20b90.6$
Profit = SP − CP
= â‚¹0.6 − â‚¹0.5
= â‚¹0.1
$\begin{array}{rcl}\mathrm{Profit}\%& =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{0.1}{0.5}\times 100\right)\%\\ & =& 20\%\end{array}$
Hence, if 20 lemons are brought for â‚¹10 and sold at 5 for three rupee, then profit in the transaction is 20%.
Page No 201:
Question 45:
Fill in the blanks to make the statements true.
Narain bought 120 oranges at â‚¹ 4 each. He sold 60 % of the oranges at â‚¹ 5 each and the remaining at â‚¹ 3.50 each. His ________ is ________%.
Answer:
Lost of 1 orange = â‚¹4
Lost of 120 oranges = â‚¹4 × 120
= â‚¹480
$\begin{array}{rcl}60\%\mathrm{of}120& =& \frac{60}{100}\times 120\\ & =& 72\end{array}$
Selling price of 72 oranges = â‚¹5 × 72
= â‚¹360
Selling price of 48 (120 − 72) orange = â‚¹3.50 × 48
= â‚¹168
Total selling price of 120 oranges = â‚¹360 + â‚¹168
= â‚¹528
Profit = SP − CP
= â‚¹528 − â‚¹480
= â‚¹48
$\begin{array}{rcl}\mathrm{Profit}\%& =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times \mathrm{CP}\right)\%\\ & =& \left(\frac{48}{480}\times 100\right)\%\\ & =& 10\%\end{array}$
Hence, his profit is 10%.
Page No 201:
Question 46:
Fill in the blanks to make the statements true.
A fruit seller purchased 20 kg of apples at â‚¹ 50 per kg. Out of these, 5% of the apples were found to be rotten. If he sells the remaining apples at â‚¹ 60 per kg, then his _________is _________%.
Answer:
Weight of apples bought = 20 kg
Cost price of 1 kg apples = â‚¹50
Cost price of 20 kg apples = â‚¹50 × 20
= â‚¹1000
Percentage of rotten apples = 5%
Quantity of rotten apples = 5% of 20 kg
$=\frac{5}{100}\times 20\mathrm{kg}\phantom{\rule{0ex}{0ex}}=1\mathrm{kg}$
Quantity of good apples = (20 − 1) kg
= 19 kg
Selling price of 1 kg of apples = â‚¹60
Selling Price of 1 kg of apples = â‚¹60 × 19
= â‚¹1140
Profit = SP − CP
= â‚¹(1140 − 1000)
= â‚¹140
$\begin{array}{rcl}\mathrm{Profit}\%& =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \left(\frac{140}{1000}\times 100\right)\%\\ & =& 14\%\end{array}$
Hence, his Profit is 14%
Page No 201:
Question 47:
Fill in the blanks to make the statements true.
Interest on â‚¹ 3000 at 10% per annum for a period of 3 years is ________.
Answer:
Principle (P) = â‚¹3000
Rate (R) = 10%
Time (T) = 3 years
$\begin{array}{rcl}\mathrm{Interest}& =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{3000\times 10\times 3}{100}\\ & =& \u20b9900\end{array}$
Thus, interest on â‚¹3000 at 10% per annum for a period of 3 years is â‚¹900.
Page No 201:
Question 48:
Fill in the blanks to make the statements true.
Amount obtained by depositing â‚¹ 20,000 at 8 % per annum for six months is ________.
Answer:
Principle (P) = â‚¹20,000
Rate (R) = 8%
Time = 6 months
$=\frac{6}{12}=\frac{1}{2}\mathrm{year}$
$\mathrm{Interest}\left(\mathrm{I}\right)=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}=\frac{20000\times 8\times {\displaystyle \frac{1}{2}}}{100}\phantom{\rule{0ex}{0ex}}\mathrm{I}=\u20b9800$
Amount (A) = P + I
= â‚¹(20000 + 800)
= â‚¹20800
Page No 201:
Question 49:
Fill in the blanks to make the statements true.
Interest on â‚¹ 12500 at 18% per annum for a period of 2 years and 4 months is ________.
Answer:
Principle (P) = â‚¹125000
Rate (R) = 18%
Time (T) = 2 years 4 month
$=2\mathrm{years}+\frac{4}{12}\mathrm{year}\phantom{\rule{0ex}{0ex}}=\left(2+\frac{1}{3}\right)\mathrm{year}\phantom{\rule{0ex}{0ex}}=\frac{7}{3}\mathrm{year}$
$\begin{array}{rcl}\mathrm{Interest}\left(\mathrm{I}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{125000\times 18\times {\displaystyle \frac{7}{3}}}{100}\\ & =& \u20b952,500\end{array}$
Thus, interest on â‚¹12500 at 18% per annum for a period of 2 years and 4 months is â‚¹52,500.
Page No 201:
Question 50:
Fill in the blanks to make the statements true.
25 ml is _________ per cent of 5 litres.
Answer:
Let x% of 5 litres be 25 ml.
5 litres = 5000 ml
∴ x% of 5000 ml = 25 ml
$\Rightarrow \frac{x}{100}\times 5000\mathrm{ml}=25\mathrm{ml}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{25}{50}\phantom{\rule{0ex}{0ex}}\Rightarrow x=0.5$
Thus, 25 ml is 0.5 percent of 5 litres.
Page No 201:
Question 51:
Fill in the blanks to make the statements true.
If A is increased by 20%, it equals B. If B is decreased by 50%, it equals C. Then __________ % of A is equal to C.
Answer:
A + 20% of A = B
$\Rightarrow \mathrm{A}+\frac{20}{100}\times \mathrm{A}=\mathrm{B}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{120\mathrm{A}}{100}=\mathrm{B}\phantom{\rule{0ex}{0ex}}\Rightarrow 12\mathrm{A}=10\mathrm{B}\phantom{\rule{0ex}{0ex}}\Rightarrow 6\mathrm{A}=5\mathrm{B}.....\left(1\right)$
Also, B − 50% of B = C
$\Rightarrow \mathrm{B}\frac{50}{100}\times \mathrm{B}=\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}\frac{1}{2}\mathrm{B}=\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\mathrm{B}=\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}=2\mathrm{C}.....\left(2\right)$
Substituting (2) in (1), we get
⇒ 6A = 5(2C)
⇒ 6A = 10C
$\Rightarrow \frac{6\mathrm{A}}{10}=\frac{\mathrm{C}}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{C}=\frac{6\mathrm{A}\times 10}{10\times 10}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{C}=\frac{60}{100}\mathrm{A}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{C}=60\%\mathrm{of}\mathrm{A}$
Then, 60% of A is equal to C.
Page No 202:
Question 52:
Fill in the blanks to make the statements true.
Interest = $\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$, where
T is ____________
R% is ____________ and
P is ____________.
Answer:
$\mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100},\mathrm{where}$
T is Time,
R% is Rate of interest and,
P is principle.
Page No 202:
Question 53:
Fill in the blanks to make the statements true.
The difference of interest for 2 years and 3 years on a sum of â‚¹ 2100 at 8% per annum is _________.
Answer:
Principal (P) = â‚¹21,00.
Rate (R) = 8%
Case I: Time (T_{1}) 2 years.
$\begin{array}{rcl}{\mathrm{SI}}_{1}& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_{1}}{100}\\ & =& \frac{2100\times 8\times 2}{100}\\ & =& \u20b9336\end{array}$
Case II: Time (T_{2}) = 3 years
$\begin{array}{rcl}{\mathrm{SI}}_{2}& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_{2}}{100}\\ & =& \frac{2100\times 8\times 3}{100}\\ & =& \u20b9504\end{array}$
Difference of interest for 2 years and 3 years = â‚¹504 − 336
= â‚¹168
Thus, the difference of interest for 2 years and 3 years on a sum of â‚¹2100 at 8% per annum is â‚¹168.
Page No 202:
Question 54:
Fill in the blanks to make the statements true.
To convert a fraction into a per cent, we _________ it by 100.
Answer:
To convert a fraction into a percent, we multiply it by 100.
Page No 202:
Question 55:
Fill in the blanks to make the statements true.
To convert a decimal into a percent, we shift the decimal point two places to the _________.
Answer:
To convert a decimal into a percent, we shift the decimal point two places to the right.
Page No 202:
Question 56:
Fill in the blanks to make the statements true.
The _________ of interest on a sum of â‚¹ 2000 at the rate of 6% per annum for $1\frac{1}{2}$ years and 2 years is â‚¹ 420.
Answer:
Principal (P) = â‚¹2000
Rate (R) = 6%
Case I: Time (T_{1}) = $1\frac{1}{2}\mathrm{years}$
$\begin{array}{rcl}\mathrm{Interest}\left({\mathrm{I}}_{1}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_{1}}{100}\\ & =& \frac{2000\times 6\times {\displaystyle \frac{3}{2}}}{100}\\ & =& \u20b9180\end{array}$
Case II: Time (T_{2}) = 2 years
$\begin{array}{rcl}\mathrm{Interest}\left({\mathrm{I}}_{2}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_{2}}{100}\\ & =& \frac{2000\times 6\times {\displaystyle 2}}{100}\\ & =& \u20b9240\end{array}$
Now, I_{1} + I_{2} = 180 + 240 = â‚¹420
Thus, the sum of interest on a sum of â‚¹2000 at the rate of 6% per annum for year and $1\frac{1}{2}$ 2 years is â‚¹420.
Page No 202:
Question 57:
Fill in the blanks to make the statements true.
When converted into percentage, the value of 6.5 is _________ than 100%.
Answer:
6.5 = (6.5 × 100)%
= 650%
650% > 100%
∴ 6.5 > 100%
Thus, when converted into percentage, the value of 6.5 is more than 100%.
Page No 202:
Question 58:
In the given question, copy the number line. Fill in the blanks so that each mark on the number line is labelled with a per cent, a fraction
and a decimal. Write all fractions in lowest terms.
Answer:
Page No 203:
Question 59:
In the given question, copy the number line. Fill in the blanks so that each mark on the number line is labelled with a per cent, a fraction
and a decimal. Write all fractions in lowest terms.
Answer:
Per cent  Fraction  Decimal 
0%  0  0 
12.5%  $\frac{1}{8}$  0.125 
25%  $\frac{1}{4}$  0.25 
37.5%  $\frac{3}{8}$  0.375 
50%  $\frac{1}{2}$  0.5 
62.5%  $\frac{5}{8}$  0.625 
75%  $\frac{3}{4}$  0.75 
87.5%  $\frac{7}{8}$  0.875 
100%  1  1 
Graphics to be created
Page No 203:
Question 60:
State whether the statements are True or False.
$\frac{2}{3}=66\frac{2}{3}\%$
Answer:
$\begin{array}{rcl}\frac{2}{3}& =& \left(\frac{2}{3}\times \frac{100}{1}\right)\%\\ & =& \left(\frac{200}{3}\right)\%\\ & =& 66\frac{2}{3}\%\end{array}$
Hence, the statement is true.
Page No 203:
Question 61:
State whether the statements are True or False.
When an improper fraction is converted into percentage then the answer can also be less than 100.
Answer:
When an improper fraction is converted into a percentage, the answer is always greater than 100.
Hence, the statement is False.
Page No 203:
Question 62:
State whether the statements are True or False.
8 hours is 50% of 4 days.
Answer:
4 days = 4 × 24 hours = 96 hours
50% of 4 days = 50% × 4 days
$=\frac{50}{100}\times 96\mathrm{hours}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 96\mathrm{hours}\phantom{\rule{0ex}{0ex}}=48\mathrm{hours}$
Hence, the statement is False.
Page No 203:
Question 63:
State whether the statements are True or False.
The interest on â‚¹ 350 at 5% per annum for 73 days is â‚¹ 35.
Answer:
Principal (P) = â‚¹350
Rate of interest (R) = 5%
Time (T) = 73 days $=\frac{73}{365}=0.2\mathrm{years}$
$\begin{array}{rcl}\mathrm{Simple}\mathrm{Interest}\left(SI\right)& =& \frac{P\times R\times T}{100}\\ & =& \frac{\u20b9350\times 5\times 0.2}{100}\\ & =& \u20b93.5\end{array}$
Hence, the statement is False.
Page No 203:
Question 64:
State whether the statements are True or False.
The simple interest on a sum of â‚¹ P for T years at R% per annum is given by the formula: Simple Interest = $\frac{\mathrm{T}\times \mathrm{P}\times \mathrm{R}}{100}$ .
Answer:
$\begin{array}{rcl}\mathrm{Simple}\mathrm{Interest}\left(\mathrm{SI}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{\mathrm{T}\times \mathrm{P}\times \mathrm{R}}{100}\end{array}$
Hence, the statement is True.
Page No 203:
Question 65:
State whether the statements are True or False.
75% = $\frac{4}{3}.$
Answer:
$\begin{array}{rcl}75\%& =& \frac{75}{100}\\ & =& \frac{75\xf725}{100\xf725}\\ & =& \frac{4}{3}\end{array}$
Hence, the statement is True.
Page No 203:
Question 66:
State whether the statements are True or False.
12% of 120 is 100.
Answer:
$\begin{array}{rcl}12\%\mathrm{of}120& =& \frac{12}{100}\times 120\\ & =& \frac{1440}{100}\\ & =& \frac{144}{10}\\ & =& 14.4\end{array}$
Hence, the statement is False.
Page No 204:
Question 67:
State whether the statements are True or False.
If Ankita obtains 336 marks out of 600, then percentage of marks obtained by her is 33.6.
Answer:
Total marks = 600
Marks obtained by Ankita = 336
$\begin{array}{rcl}\mathrm{Percentage}\mathrm{of}\mathrm{her}\mathrm{marks}\mathrm{obtained}& =& \frac{336}{600}\times 100\\ & =& 56\%\end{array}$
Hence, the statement is False.
Page No 204:
Question 68:
State whether the statements are True or False.
0.018 is equivalent to 8%.
Answer:
$\begin{array}{rcl}0.018& =& \frac{18}{1000}\\ & =& \left(\frac{18}{1000}\times 100\right)\%\\ & =& 1.8\%\end{array}$
Hence, the statement is False.
Page No 204:
Question 69:
State whether the statements are True or False.
50% of â‚¹ 50 is â‚¹ 25.
Answer:
$\begin{array}{rcl}50\%\mathrm{of}\u20b950& =& \frac{50}{100}\times \u20b950\\ & =& \u20b925\end{array}$
Hence, the statement is True.
Page No 204:
Question 70:
State whether the statements are True or False.
250 cm is 4% of 1 km.
Answer:
1 km = 1000 m
$\begin{array}{rcl}4\%\mathrm{of}1\mathrm{km}& =& \frac{4}{100}\times 1000\mathrm{m}\\ & =& 40\mathrm{m}\end{array}$
Hence, the statement is False.
Page No 204:
Question 71:
State whether the statements are True or False.
Out of 600 students of a school, 126 go for a picnic. The percentage of students that did not go for the picnic is 75.
Answer:
Total number of students = 600
Number of students who went on a picnic = 126
∴ Number of students who did not go for the picnic = 600 – 126 = 474
Percentage of students who did not go to the picnic =
Hence, the statement is False.
Page No 204:
Question 72:
State whether the statements are True or False.
By selling a book for â‚¹ 50, a shopkeeper suffers a loss of 10%. The cost price of the book is â‚¹ 60.
Answer:
Selling price of a book = â‚¹50
Loss % = â‚¹10
We know that
$\begin{array}{rcl}\mathrm{CP}& =& \frac{\mathrm{SP}\times 100}{\left(110\mathrm{Loss}\right)\%}\\ & =& \frac{50\times 100}{\left(10010\right)}\\ & =& \frac{5000}{90}\\ & =& \u20b955.55\end{array}$
So, the cost price of the book is â‚¹55.5.
Hence, the statement is False.
Page No 204:
Question 73:
State whether the statements are True or False.
If a chair is bought for â‚¹ 2000 and is sold at a gain of 10%, then selling price of the chair is â‚¹ 2010.
Answer:
CP of a chair = â‚¹2000
Gain % = 10
We know that,
$\begin{array}{rcl}\mathrm{SP}& =& \mathrm{CP}\left(\frac{100+\mathrm{G}\%}{100}\right)\\ & =& 2000\left(\frac{100+10}{100}\right)\\ & =& \frac{2000\times 110}{100}\\ & =& \u20b92200\end{array}$
Hence, the statement is False.
Page No 204:
Question 74:
State whether the statements are True or False.
If a bicycle was bought for â‚¹ 650 and sold for â‚¹ 585, then the percentage of profit is 10.
Answer:
CP of a bicycle = â‚¹ 650
SP of the bicycle = â‚¹ 585
Therefore, there is a loss of â‚¹ (650 – 585) = â‚¹65
$\begin{array}{rcl}\mathrm{Loss}\%& =& \frac{65}{650}\times 100\%\\ & =& 10\%\end{array}$
Hence, the statement is False.
Page No 204:
Question 75:
State whether the statements are True or False.
Sushma sold her watch for â‚¹ 3320 at a gain of â‚¹ 320. For earning a gain of 10% she should have sold the watch for â‚¹ 3300.
Answer:
SP of watch = â‚¹3320
Gain = â‚¹320
∴ CP of the watch = â‚¹(3320 – 320) = â‚¹3000
Required gain % = 10
$\begin{array}{rcl}\mathrm{SP}& =& \mathrm{CP}\left(\frac{100+\mathrm{G}\%}{100}\right)\\ & =& 3000\left(\frac{100+10}{100}\right)\\ & =& \frac{3000\times 110}{100}\\ & =& \u20b93300\end{array}$
Hence, the statement is False.
Page No 204:
Question 76:
State whether the statements are True or False.
Interest on â‚¹ 1200 for $1\frac{1}{2}$ years at the rate of 15% per annum is â‚¹ 180.
Answer:
Principal (P) = â‚¹1200,
Time (T)
Rate (R)= 15% pa
Hence, the statement is False.
Page No 204:
Question 77:
State whether the statements are True or False.
Amount received after depositing â‚¹ 800 for a period of 3 years at the rate of 12% per annum is â‚¹896.
Answer:
Principal (P) = â‚¹ 800
Time (T)= 3 years
Rate (R)= 12% pa
$\begin{array}{rcl}\mathrm{Interest}\left(\mathrm{I}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{\u20b9800\times 12\times {\displaystyle 3}}{100}\\ & =& \u20b9288\end{array}$
Now, amount = Principal + Interest
= â‚¹ (800 + 288)
= â‚¹1088
Hence, the statement is False.
Page No 204:
Question 78:
State whether the statements are True or False.
â‚¹ 6400 were lent to Feroz and Rashmi at 15% per annum for $3\frac{1}{2}$ and 5 years respectively. The difference in the interest paid by them is â‚¹ 150.
Answer:
For Feroz: Principal = â‚¹6400
Rate = 15% pa
For Rashmi : Principal = â‚¹6400,
Rate = 15% pa
Time = 5 years
Required difference in interests = â‚¹(4800 – 3360) = â‚¹1440
Hence, the statement is False.
Page No 204:
Question 79:
State whether the statements are True or False.
A vendor purchased 720 lemons at â‚¹ 120 per hundred.10% of the lemons were found rotten which he sold at â‚¹ 50 per hundred. If he sells the remaining lemons at â‚¹ 125 per hundred, then his profit will be 16%.
Answer:
Total number of lemons = 720
CP of 100 lemons = â‚¹120
$\therefore \mathrm{CP}\mathrm{of}1\mathrm{lemon}=\frac{\u20b9120}{100}\phantom{\rule{0ex}{0ex}}\mathrm{CP}\mathrm{of}720\mathrm{lemons}=\frac{\u20b9120}{100}\times 720=\u20b9864$
Rotten lemons = 10%
$\mathrm{Number}\mathrm{of}\mathrm{rotten}\mathrm{lemons}=\frac{10}{100}\times 720=72$
Number of remaining lemons which are not rotten = 720 – 72 = 648
SP of 720 lemons = â‚¹(36 + 810) = â‚¹ 846
Here, CP > SP
Therefore, there is a loss of â‚¹(864 – 846) = â‚¹18
Hence, the statement is False.
Page No 205:
Question 80:
Find the value of x if
(i) 8% of â‚¹ x is â‚¹ 100 (ii) 32% of x kg is 400 kg
(iii) 35% of â‚¹ x is â‚¹ 280 (iv) 45% of marks x is 405.
Answer:
(i) 8% of â‚¹ x is 100
$\Rightarrow \frac{8}{100}\times x=100\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{100\times 100}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{100\times 100}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1250$
(ii) 32% of x kg is 400 kg
$\Rightarrow \frac{32}{100}\times x=400\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{400\times 100}{32}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{400\times 100}{32}\phantom{\rule{0ex}{0ex}}\Rightarrow x=1250$
(iii) 35% of â‚¹ x is â‚¹280
$\Rightarrow \frac{35}{100}\times x=280\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{280\times 100}{35}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow x=800$
(iv) 45% of marks x is 405
$\Rightarrow \frac{45}{100}\times x=405\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{405\times 100}{45}\phantom{\rule{0ex}{0ex}}\Rightarrow x=900$
Page No 205:
Question 81:
Imagine that a 10 × 10 grid has value 300 and that this value is divided evenly among the small squares. In other words, each small square is worth 3. Use a new grid for each part of this problem, and label each grid “Value : 300.”
(a) Shade 25% of the grid. What is 25% of 300? Compare the two answers.
(b) What is the value of 25 squares?
(c) Shade 17% of the grid. What is 17% of 300? Compare the two answers.
(d) What is the value of $\frac{1}{10}$ of the grid?
Answer:
(a) $25\%\mathrm{of}300=\frac{25}{100}\times 300=75$
Now, each small square is worth 3.
∴ 25 small square worth is 25 × 3 = 75
Both the values are same.
(b) From part (a), value of 25 small squares is 75.
$\left(\mathrm{c}\right)17\%\mathrm{of}300=\frac{17}{100}\times 300=51$
Now, each small square is worth 3
∴ 17 small square worth is 17 × 3 = 51
Both the values are same.
(d) Total numbers in grid = 10 × 10 = 100
∴
Each small square is worth 3
∴ 10 small square worth is 10 × 3 = 30
Page No 205:
Question 82:
Express $\frac{1}{6}$ as a per cent.
Answer:
$\frac{1}{6}=\left(\frac{1}{6}\times 100\right)\%=16.67\%$
Page No 206:
Question 83:
Express $\frac{9}{40}$as a per cent.
Answer:
$\frac{9}{40}=\left(\frac{9}{40}\times 100\right)\%=22.5\%$
Page No 206:
Question 84:
Express $\frac{1}{100}$ as a per cent.
Answer:
$\frac{1}{100}=\left(\frac{1}{100}\times 100\right)\%=1\%$
Page No 206:
Question 85:
Express 80% as fraction in its lowest term.
Answer:
$80\%=\frac{80}{100}=\frac{4}{5}$
Page No 206:
Question 86:
Express $33\frac{1}{3}\%$ as a ratio in the lowest term.
Answer:
$\begin{array}{rcl}33\frac{1}{3}\%& =& \frac{100}{3}\%\\ & =& \frac{100}{3}\times \frac{1}{100}\\ & =& \frac{1}{3}\\ & =& 1:3\end{array}$
Page No 206:
Question 87:
Express $16\frac{2}{3}\%$ as a ratio in the lowest form.
Answer:
$\begin{array}{rcl}16\frac{2}{3}\%& =& \frac{50}{3}\%\\ & =& \frac{50}{3}\times \frac{1}{100}\\ & =& \frac{1}{6}\\ & =& 1:6\end{array}$
Page No 206:
Question 88:
Express 150% as a ratio in the lowest form.
Answer:
$\begin{array}{rcl}150\%& =& \frac{150}{100}\\ & =& \frac{3}{2}\\ & =& 3:2\end{array}$
Page No 206:
Question 89:
Sachin and Sanjana are calculating 23% of 800.
Answer:
I. 52% of 700 = (1% of 700) × 52
= (0.01 × 700) × 52
= 7 × 52
= 364
II.
$\begin{array}{rcl}52\%\mathrm{of}700& =& \frac{52}{100}\mathrm{of}700\\ & =& \frac{52}{100}\times 700\\ & =& 52\times 7\\ & =& 364\end{array}$
We find that the way II is easier.
Page No 207:
Question 90:
Write 0.089 as a per cent.
Answer:
$\begin{array}{rcl}0.089& =& \frac{89}{1000}\\ & =& \left(\frac{89}{1000}\times 100\right)\%\\ & =& 8.9\%\end{array}$
Page No 207:
Question 91:
Write 1.56 as a per cent.
Answer:
$\begin{array}{rcl}1.56& =& \frac{156}{100}\\ & =& \left(\frac{156}{100}\times 100\right)\%\\ & =& 156\%\end{array}$
Page No 207:
Question 92:
What is 15% of 20?
Answer:
$\begin{array}{rcl}15\%& =& \frac{15}{100}\\ & =& 0.15\end{array}$
Thus, 15% of 20 is 0.15.
Page No 207:
Question 93:
What is 800% of 800?
Answer:
$\begin{array}{rcl}800\%\mathrm{of}800& =& \left(\frac{800}{100}\times 800\right)\\ & =& 8\times 800\\ & =& 6400\end{array}$
Thus, 800% of 800 is 6400.
Page No 207:
Question 94:
What is 100% of 500?
Answer:
100% of 500
$\begin{array}{rcl}100\%\mathrm{of}500& =& \left(\frac{100}{100}\times 500\right)\\ & =& 500\end{array}$
Thus, 100% of 500 is 500.
Page No 207:
Question 95:
What per cent of 1 hour is 30 minutes?
Answer:
Let x% of 1 hour is 30 minutes.
Now, 1 hour = 60 minutes
$\therefore \frac{x}{100}\times 60=30\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{30\times 100}{60}=50$
Thus, Per cent of 1 hour is 30 minutes is 50%.
Page No 207:
Question 96:
What per cent of 1 day is 1 minute?
Answer:
Let x% of 1 day is 1 minute,
Now, 1 day = 24 hours = (24 × 60) minutes
$\therefore \frac{x}{100}\times 1440=1\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{100}{1440}=0.069$
Thus, per cent of 1 day is 1 minute is 0.069.
Page No 207:
Question 97:
What per cent of 1 km is 1000 metres?
Answer:
Let x% of 1 km is 1000 metres.
Now, 1 km = 1000 metres
$\therefore \frac{x}{100}\times 1000=1000$
$\begin{array}{rcl}& \Rightarrow & x=\frac{1000\times 100}{1000}\\ & =& 100\end{array}$
Thus, per cent of 1 km is 1000 metres is 100%.
Page No 207:
Question 98:
Find out 8% of 25 kg.
Answer:
$\begin{array}{rcl}8\%\mathrm{of}25\mathrm{kg}& =& \left(\frac{8}{100}\times 25\right)\%\\ & =& 2\%\end{array}$
Thus, 8% of 25 kg is 2%.
Page No 207:
Question 99:
What percent of â‚¹ 80 is â‚¹ 100?
Answer:
Let x% of â‚¹ 80 is â‚¹ 100
$\therefore \frac{x}{100}\times 80=100\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{100\times 100}{80}\phantom{\rule{0ex}{0ex}}\Rightarrow x=125$
Page No 207:
Question 100:
45% of the population of a town are men and 40% are women. What is the percentage of children?
Answer:
Total percentage = 100
Percentage of men = 45
Percentage of women = 40
∴ Percentage of children = [100 – (45 + 40)]% = [100 – 85]% = 15%
Page No 207:
Question 101:
The strength of a school is 2000. If 40 % of the students are girls then how many boys are there in the school?
Answer:
Total number of students = 2000
$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{girls}& =& 40\%\mathrm{of}2000\\ & =& \frac{40}{100}\times 2000\\ & =& 800\end{array}$
∴ Number of boys = 2000 – 800 = 1200
Page No 207:
Question 102:
Chalk contains 10% calcium, 3% carbon and 12% oxygen. Find the amount of carbon and calcium (in grams) in $2\frac{1}{2}$kg of chalk.
Answer:
Total amount of chalk$=2\frac{1}{2}\mathrm{kg}=\frac{5}{2}\mathrm{kg}$
Amount of calcium = 10% of $\frac{5}{2}\mathrm{kg}$
$=\left(\frac{1}{10}\times \frac{5}{2}\right)\mathrm{kg}\phantom{\rule{0ex}{0ex}}=0.25\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\left(0.25\times 1000\right)\mathrm{g}\phantom{\rule{0ex}{0ex}}=250\mathrm{g}$
Amount of carbon = 3% of $\frac{5}{2}\mathrm{kg}$
$=\left(\frac{3}{100}\times \frac{5}{2}\right)\mathrm{kg}\phantom{\rule{0ex}{0ex}}=0.075\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\left(0.075\times 1000\right)\mathrm{g}\phantom{\rule{0ex}{0ex}}=75\mathrm{g}$
Thus, the amount of carbon and calcium (in grams) in $2\frac{1}{2}$kg of chalk are 250g and 75g.
Page No 207:
Question 103:
800 kg of mortar consists of 55% sand, 33% cement and rest lime. What is the mass of lime in mortar?
Answer:
Total mass of mortar = 800 kg
Mass of sand in mortar = 55% of 800 kg
$=\left(\frac{55}{100}\times 800\right)\mathrm{kg}\phantom{\rule{0ex}{0ex}}=440\mathrm{kg}$
Mass of cement in mortar = 33% of 800 kg
$=\left(\frac{33}{100}\times 800\right)\mathrm{kg}\phantom{\rule{0ex}{0ex}}=264\mathrm{kg}$
∴ Mass of lime in mortar = [800 – (440 + 264)] kg = 96 kg
Page No 207:
Question 104:
In a furniture shop, 24 tables were bought at the rate of â‚¹ 450 per table. The shopkeeper sold 16 of them at the rate of â‚¹ 600 per table and the remaining at the rate of 400 per table. Find her gain or loss percent.
Answer:
CP of 1 table = â‚¹450
∴ CP of 24 tables = â‚¹ (24 × 450) = â‚¹10800
SP of 16 tables = â‚¹ {(16 × 600) = â‚¹9600
SP of remaining tables i.e., 8 tables
= â‚¹(8 × 400) = 3200
SP of 24 tables = â‚¹(9600 + 3200) = â‚¹12800
Here, SP > CP
Therefore, there is a gain of â‚¹(12800 – 10800) = â‚¹2000
$\begin{array}{rcl}\mathrm{Gain}\%& =& \frac{2000}{10800}\times 100\\ & =& 18.5\%\end{array}$
Page No 207:
Question 105:
Medha deposited 20% of her money in a bank. After spending 20% of the remainder, she has â‚¹ 4800 left with her. How much did she originally have?
Answer:
Let the amount of money Medha had â‚¹ x.
Amount of money deposited in bank = 20% of x
$=\frac{20}{100}\times x=\frac{x}{5}$
Remaining amount$=x\frac{x}{5}=\frac{4}{5}x$
Amount of money spent by her = $20\%\mathrm{of}\frac{4}{5}x$
$=\frac{20}{100}\times \frac{4}{5}x\phantom{\rule{0ex}{0ex}}=\frac{4}{25}x$
Amount of money left with her =$=\frac{4}{5}x\frac{4}{25}x$
$=\frac{16}{25}x$
According to question,
$\frac{16}{25}x=4800$
$\Rightarrow x=\frac{4800\times 25}{16}=7500$
Page No 207:
Question 106:
The cost of a flower vase got increased by 12%. If the current cost is â‚¹ 896, what was its original cost?
Answer:
Let the original cost of a flower vase = â‚¹ x
The cost of the flower vase increased by
$\begin{array}{rcl}12\%& =& x+12\%ofx\\ & =& x+\frac{12}{100}\times x\end{array}$
According to the question,
$\frac{28}{25}x=896\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{896\times 25}{28}=800$
Page No 207:
Question 107:
Radhika borrowed â‚¹ 12000 from her friends. Out of which â‚¹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years?
Answer:
Total amount borrowed by Radhika = â‚¹12000
I: Principal = â‚¹4000, Rate = 18% p.a.,
Time = 3 years
$\mathrm{Interest}=\frac{4000\times 18\times 3}{100}=2160$
II: Principal = â‚¹ (12000 – 4000) = â‚¹8000,
Rate = 15%, Time = 3 years
$\mathrm{Interest}=\frac{8000\times 15\times 3}{100}=3600$
Total interest = â‚¹ (2160 + 3600) = â‚¹5760
Page No 208:
Question 108:
A man travelled 60 km by car and 240 km by train. Find what percent of total journey did he travel by car and what percent by train?
Answer:
Distance travelled by car = 60 km
Distance travelled by train = 240 km
Total distance travelled = (60+240) km = 300 km
Percentage of journey travelled by car $=\left(\frac{60}{300}\times 100\right)\%=20\%$$=\left(\frac{60}{300}\times 100\right)\%$
Percentage of journey travelled by train$=\left(\frac{240}{300}\times 100\right)\%=80\%$
Page No 208:
Question 109:
By selling a chair for â‚¹ 1440, a shopkeeper loses 10%. At what price did he buy it?
Answer:
S.P. of a chair = â‚¹1440
Loss% = 10
$\begin{array}{rcl}\mathrm{CP}& =& \frac{\mathrm{SP}\times 100}{100\mathrm{Loss}\%}\\ & =& \frac{1440\times 100}{10010}\\ & =& \frac{14400}{90}\\ & =& 1600\end{array}$
So, the shopkeeper bought the chair at â‚¹1600.
Page No 208:
Question 110:
Dhruvika invested money for a period from May 2006 to April 2008 at rate of 12% per annum. If interest received by her is â‚¹ 1620, find
the money invested.
Answer:
Let the principal be â‚¹P.
Rate (R) = 12% p.a.,
Interest (T) = â‚¹ 1620
Time = 2 years
$\mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow 1620=\frac{\mathrm{P}\times 12\times 2}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\frac{1620\times 100}{24}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=6750$
So, amount of money invested by Dhruvika is â‚¹6750.
Page No 208:
Question 111:
A person wanted to sell a scooter at a loss of 25%. But at the last moment he changed his mind and sold the scooter at a loss of 20%. If the difference in the two SP’s is â‚¹ 4000, then find the CP of the scooter.
Answer:
Let the CP of a scooter = â‚¹x
SP of the scooter for a loss of 25%
SP of the scooter for a loss of 20%$=\left(\frac{10020}{100}\right)\times x=\frac{4}{5}x$
The difference in both SPs of the scooter
According to question,
$\frac{1}{20}x=4000\phantom{\rule{0ex}{0ex}}\Rightarrow x=4000\times 20\phantom{\rule{0ex}{0ex}}\Rightarrow x=80000$
Page No 208:
Question 112:
The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. Find the ratio of the number of literate women to the total population.
Answer:
Total population of a village = 8000
Number of literate people = 80% of 8000$=\frac{80}{100}\times 8000=6400$
Number of literate women = 40% of 6400$=\frac{40}{100}\times 6400=2560$
$\mathrm{Required}\mathrm{ratio}=\frac{2560}{8000}=\frac{8}{25}=8:25$
Thus, the ratio of the number of literate women to the total population is 8:25.
Page No 208:
Question 113:
In an entertainment programme, 250 tickets of â‚¹ 400 and 500 tickets of â‚¹ 100 were sold. If the entertainment tax is 40% on ticket of â‚¹ 400
and 20% on ticket of â‚¹ 100, find how much entertainment tax was collected from the programme.
Answer:
Cost of 250 tickets of â‚¹ 400 per ticket = â‚¹ (250 × 400) = â‚¹ 100000
Cost of 500 tickets of â‚¹ 100 per ticket = â‚¹ (500 × 100) = â‚¹ 50000
Tax on tickets of â‚¹ 400$=\frac{40}{100}\times 100000=\u20b940000$
Tax on tickets of 100â‚¹$=\frac{20}{100}\times 50000=\u20b910000$
So, total tax collected from the programme = â‚¹(40000 + 10000) = â‚¹50000
Page No 208:
Question 114:
Bhavya earns â‚¹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings.
Answer:
Monthly income of Bhavya = â‚¹50000
Spending amount = 80% of â‚¹50000$=\frac{80}{100}\times 50000=\u20b940000$
New income – 20% of 50000 + 50000$=\frac{20}{100}\times 50000+50000=\u20b910000+\u20b950000$ = â‚¹ 60000
New spending amount = 40000 + 20% of 40000$=40000+\frac{20}{100}\times 40000$
= 40000 + 8000 = â‚¹ 48000
∴ New savings of Bhavya = â‚¹(60000 – 48000) = â‚¹12000
Page No 209:
Question 115:
In an examination, there are three papers each of 100 marks. A candidate obtained 53 marks in the first and 75 marks in the second paper. How many marks must the candidate obtain in the third paper to get an overall of 70 per cent marks?
Answer:
Total marks = 100 + 100 + 100 = 300
Marks obtained by the candidate in first paper = 53
Marks obtained by the candidate in second paper = 75
Let the marks obtained by the candidate in third paper = x
Total marks obtained by the candidate = 53 + 75 + x – 128 + x
According to question,
$\frac{\left(128+x\right)}{300}\times 100=70\phantom{\rule{0ex}{0ex}}\Rightarrow 128+x=3\times 70\phantom{\rule{0ex}{0ex}}\Rightarrow x=210128\phantom{\rule{0ex}{0ex}}\Rightarrow x=82$
Thus, the candidate must obtain 82 marks in the third paper.
Page No 209:
Question 116:
Health Application
A doctor reports blood pressure in millimetres of mercury (mm Hg) as a ratio of systolic blood pressure to diastolic blood pressure (such as 140 over 80). Systolic pressure is measured when the heart beats, and diastolic pressure is measured when it rests. Refer to the table of blood pressure ranges for adults.
Blood Pressure Ranges  
Normal  Prehypertension  Hypertension (very High) 

Systolic Diastolic 
Under 120 mm Hg Under 80 mm Hg 
120139 mm Hg 8089 mm Hg 
140 mm Hg and above 90 mm Hg and above 
Manohar is a healthy 37 years old man whose blood pressure is in the normal category.
(a) Calculate an approximate ratio of systolic to diastolic blood pressures in the normal range.
(b) If Manohar’s systolic blood pressure is 102 mm Hg, use the ratio from part (a) to predict his diastolic blood pressure.
(c) Calculate ratio of average systolic to average diastolic blood pressure in the prehypertension category.
Answer:
(a) Systolic blood pressure in the normal range = 120 mm Hg
Diastolic blood pressure in the normal range = 80 mm Hg
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{120}{80}=\frac{3}{2}=3:2$
(b) Systolic blood pressure of Manohar = 102 mm Hg$=\frac{80+89}{2}=\frac{169}{2}$
$\therefore \frac{102}{\mathrm{Diastolic}\mathrm{blood}\mathrm{pressure}}=\frac{3}{2}$
⇒ Diastolic blood pressure$=102\times \frac{2}{3}=68\mathrm{mm}\mathrm{Hg}$
(c) Average systolic blood pressure in the prehypertension category $=\frac{120+139}{2}=\frac{259}{2}$
Average diastolic blood pressure in the prehypertension category $=\frac{120+139}{2}=\frac{259}{2}$
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{{\displaystyle \raisebox{1ex}{$259$}\!\left/ \!\raisebox{1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$169$}\!\left/ \!\raisebox{1ex}{$2$}\right.}}=\frac{259}{169}=259:169$
Page No 210:
Question 117:
(a) Science Application:
The king cobra can reach a length of 558 cm. This is only about 60 per cent of the length of the largest reticulated python. Find the length of the largest reticulated python.
(b) Physical Science Application:
Unequal masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down. Unequal masses will balance when the following proportion is true:
$\frac{\mathrm{mass}1}{\mathrm{length}2}=\frac{\mathrm{mass}2}{\mathrm{length}1}$
Two children can be balanced on a seesaw when
$\frac{\mathrm{mass}1}{\mathrm{length}2}=\frac{\mathrm{mass}2}{\mathrm{length}1}$The child on the left and child on the right are balanced. What is the mass of the child on the right?
(c) Life Science Application
A DNA model was built using the scale 2 cm : 0.0000001 mm. If the model of the DNA chain is 17 cm long, what is the length of the actual chain?
Answer:
(a) Let the length of the largest reticulated python – x cm
According to the question,60% of x = 558
$\Rightarrow \frac{60}{100}\times x=558\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{558\times 100}{60}$
⇒ x = 930
(b) We have,
$\frac{\mathrm{mass}1}{\mathrm{length}2}=\frac{\mathrm{mass}2}{\mathrm{length}1}\phantom{\rule{0ex}{0ex}}\therefore \frac{24}{2}=\frac{\mathrm{mass}2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{mass}2=\frac{24\times 3}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{mass}2=36\mathrm{kg}\phantom{\rule{0ex}{0ex}}$
(c) Length of actual chain 0.0000001 mm is measured as 2 cm in the model.
∴ Length of actual chain when length of chain is 17 cm in the model$=\frac{0.0000001}{2}\times 17=0.00000085\mathrm{mm}$
Page No 211:
Question 118:
Language Application
Given below are few Mathematical terms.
Find
(a) The ratio of consonants to vowels in each of the terms.
(b) The percentage of consonants in each of the terms.
Answer:
a) (i) Number of consonants in Hypotenuse’ = 6
Number of vowels in ‘Hypotenuse’ = 4
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{6}{4}=\frac{3}{2}=3:2$
(ii) Number of consonants in ‘Congruence’ = 6
Number of vowels in ‘Congruence’ = 4
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{6}{4}=\frac{3}{2}=3:2$
(iii) Number of consonants in ‘Perpendicular’ = 8
Number of vowels in ‘Perpendicular = 5
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{8}{5}=8:5$
(iv) Number of consonants in ‘Transversal’ = 8
Number of vowels in ‘Transversal’ = 3
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{8}{3}=8:3$
(v) Number of consonants in ‘Correspondence’ = 9
Number of vowels in ‘Correspondence’ = 5
$\therefore \mathrm{Required}\mathrm{ratio}=\frac{9}{5}=9:5$
(b)(i) Percentage of consonants in ‘Hypotenuse’$=\frac{6}{10}\times 100=60\%$
(ii) Percentage of ‘consonants in Congruence’$=\frac{6}{10}\times 100=60\%$
(iii) Percentage of consonants in ‘Perpendicular’$=\frac{8}{13}\times 100=61.53\%$
(iv) Percentage of consonants in ‘Transversal’$=\frac{8}{11}\times 100=72.72\%$
(v) Percentage of consonants in ‘Correspondence’$=\frac{9}{14}\times 100=64.28\%$
Page No 212:
Question 119:
What’s the Error? An analysis showed that 0.06 per cent of the Tshirts made by one company were defective. A student says this is 6 out of every 100. What is the student’s error?
Answer:
$0.06\%=\frac{0.06}{100}=\frac{6}{10000}$
Therefore 6 out of every 10000 are defective Tshifts.
Page No 212:
Question 120:
What’s the Error? A student said that the ratios $\frac{3}{4}$ and $\frac{9}{16}$ were proportional. What error did the student make?
Answer:
$\frac{3}{4}=\frac{3\times 4}{4\times 4}=\frac{12}{16}\ne \frac{9}{16}$
Therefore, $\frac{3}{4}$ and $\frac{9}{16}$
Page No 212:
Question 121:
What’s the Error? A clothing store charges â‚¹ 1024 for 4 Tshirts. A student says that the unit price is â‚¹ 25.6 per Tshirt. What is the
error? What is the correct unit price?
Answer:
Cost price for 4 Tshirts = â‚¹1024 .
$\therefore \mathrm{Cost}\mathrm{price}\mathrm{of}1\mathrm{T}\u2010\mathrm{shirt}=\frac{\u20b91024}{4}=\u20b9256$
Therefore, the correct unit price is â‚¹256.
Page No 212:
Question 122:
A tea merchant blends two varieties of tea in the ratio of 5 : 4. The cost of first variety is â‚¹ 200 per kg and that of second variety is â‚¹ 300 per kg. If he sells the blended tea at the rate of â‚¹ 275 per kg, find out the percentage of her profit or loss.
Answer:
Let quantity of first variety and second variety of tea be 5x kg and 4x kg respectively.
Cost of first variety of tea = â‚¹(200 × 5x) = â‚¹1000 x
Cost of second variety of tea = â‚¹(300 × 4x) = â‚¹1200x
Total cost price of tea = â‚¹(1000x + 1200x) = â‚¹2200x
Selling price of tea = â‚¹(9x × 275) = â‚¹2475x
Here, SP > CP
Therefore, there is a profit of = â‚¹(2475x – 2200x) = â‚¹275x
$\begin{array}{rcl}\mathrm{Profit}\%& =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \frac{275}{2200\mathrm{x}}\times 100\\ & =& 12.5\%\end{array}$
Page No 212:
Question 123:
A piece of cloth 5 m long shrinks 10 per cent on washing. How long will the cloth be after washing?
Answer:
Length of cloth = 5 m
Shrinking percentage after washing = 10%
∴ Length of cloth, after washing
= 5 – 10% of 5
$=5\frac{10}{100}\times 5\phantom{\rule{0ex}{0ex}}=5\frac{1}{2}\phantom{\rule{0ex}{0ex}}=4.5$
Page No 212:
Question 124:
Nancy obtained 426 marks out of 600 and the marks obtained by Rohit are 560 out of 800. Whose performance is better?
Answer:
Percentage of marks obtained by Nancy $=\frac{426}{600}\times 100=71\%\phantom{\rule{0ex}{0ex}}$
Percentage of marks obtained by Rohit$=\frac{560}{800}\times 100=70\%\phantom{\rule{0ex}{0ex}}$
So, Nancy’s performance is better.
Page No 212:
Question 125:
A memorial trust donates â‚¹ 5,00,000 to a school, the interest on which is to be used for awarding 3 scholarships to students obtaining first three positions in the school examination every year. If the donation earns an interest of 12 per cent per annum and the values of the second and third scholarships are â‚¹ 20,000 and â‚¹ 15,000 respectively, find out the value of the first scholarship.
Answer:
Total donation amount = â‚¹500000
Interest on donation amount
$=500000\times \frac{12}{100}\times 1=\u20b960000$
Value of the second and third scholarships = â‚¹(20000 + 15000) = â‚¹35000
Now, value of the first scholarship = â‚¹(60000 – 35000) = â‚¹25000
Page No 213:
Question 126:
Ambika got 99 per cent marks in Mathematics, 76 per cent marks in Hindi, 61 per cent in English, 84 per cent in Science, and 95% in Social Science. If each subject carries 100 marks, then find the percentage of marks obtained by Ambika in the aggregate of all the subjects.
Answer:
Total maximum marks = 5 × 100 = 500
Total marks obtained by Ambika in all subjects = 99 + 76 + 61 + 84 + 95 = 415
Percentage of marks obtained by Ambika$=\frac{415}{500}\times 100=83\%$
Page No 213:
Question 127:
What sum of money lent out at 16 per cent per annum simple interest would produce â‚¹ 9600 as interest in 2 years?
Answer:
SI = â‚¹ 9600,
Time (T) = 2 years,
Rate (R) = 16% p.a.
We know that,
$\mathrm{SI}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow 9600=\frac{\mathrm{P}\times 16\times 2}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9600\times 100}{16\times 2}=\mathrm{P}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=30000$
Page No 213:
Question 128:
Harish bought a gaschullah for â‚¹ 900 and later sold it to Archana at a profit of 5 per cent. Archana used it for a period of two years and later sold it to Babita at a loss of 20 per cent. For how much did Babita get it?
Answer:
For Harish: CP = â‚¹900, Profit % = 5
$\mathrm{SP}=\left(\frac{100+5}{100}\right)\times 900=105\times 9=\u20b9945$
Now, S.P. for Harish would be the C.P. for Archana.
For Archana : C.P. = â‚¹ 945, Loss % = 20
$\mathrm{SP}=\left(\frac{10020}{100}\right)\times 945=\frac{4}{5}\times 945=\u20b9756$
Again, SP for Archana would be the CP for Babita.
So, Babita got the gaschullah for â‚¹756.
Page No 213:
Question 129:
Match each of the entries in Column I with the appropriate entries in Column II:
Column I  Column II  
(i)  3 : 5  (A)  â‚¹ 54 
(ii)  2.5  (B)  â‚¹ 47 
(iii)  100%  (C)  â‚¹ 53 
(iv)  $\frac{2}{3}$  (D)  â‚¹ 160 
(v)  $6\frac{1}{4}\%$  (E)  60 % 
(vi)  12.5 %  (F)  25 % 
(vii)  SP when CP = â‚¹ 50 and loss = 6 %  (G)  $\frac{1}{16}$ 
(viii)  SP when CP = â‚¹ 50 and profit = â‚¹ 4  (H)  250 % 
(ix)  Profit% when CP = â‚¹ 40 and SP = â‚¹ 50  (I)  â‚¹ 159 
(x)  Profit% when CP = â‚¹ 50 and SP = â‚¹ 60  (J)  $66\frac{2}{3}\%$ 
(xi)  Interest when principal = â‚¹ 800, Rate of interest = 10% per annum and period = 2 years  (K)  20 % 
(xii)  Amount when principal = â‚¹ 150, Rate of interest = 6% per annum and period = 1 year  (L)  0. 125 
(M)  3 : 2  
(N)  â‚¹ 164  
(O)  3 : 3 
Answer:
ans
Page No 214:
Question 130:
In a debate competition, the judges decide that 20 per cent of the total marks would be given for accent and presentation. 60 per cent of the rest are reserved for the subject matter and the rest are for rebuttal. If this means 8 marks for rebuttal, then find the total marks.
Answer:
Let the total marks be x,
Marks for accent and presentation $=20\%\mathrm{of}x$
$=\frac{20}{100}\times x\phantom{\rule{0ex}{0ex}}=\frac{x}{5}$
Remaining marks $=x\frac{x}{5}=\frac{4x}{5}$
Marks reserved for subject matter $60\%\mathrm{of}\frac{4x}{5}$.
$=\frac{60}{100}\times \frac{4x}{5}\phantom{\rule{0ex}{0ex}}=\frac{12}{25}x$
$\mathrm{Marks}\mathrm{for}\mathrm{rebuttal}=\frac{4x}{5}\frac{12}{25}x\phantom{\rule{0ex}{0ex}}=\frac{8}{25}x$
So, marks for rebuttal = 8
$\therefore \frac{8}{25}x=8\phantom{\rule{0ex}{0ex}}\Rightarrow x=25$
Page No 214:
Question 131:
Divide â‚¹ 10000 in two parts so that the simple interest on the first part for 4 years at 12 per cent per annum may be equal to the simple interest on the second part for 4.5 years at 16 per cent per annum.
Answer:
The total amount which has to be divided = â‚¹10000
Let the first part be = â‚¹x
Then, second part = â‚¹(10000 – x)
Interest in first part $=\frac{x\times 12\times 4}{100}=\u20b9\frac{12}{25}x$
Interest in the second part$=\left(10000x\right)\times \frac{16}{100}\times 4.5$
$=\left(10000x\right)\times \frac{16}{100}\times 4.5\phantom{\rule{0ex}{0ex}}=\u20b9\left(7200\frac{18}{25}x\right)$
According to the question,
$\frac{12}{25}x=7200\frac{18}{25}x\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{12}{25}x+\frac{18}{25}x=7200\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{30}{25}x=7200\phantom{\rule{0ex}{0ex}}\Rightarrow x=6000$
So, the first part = â‚¹6000
And second apart = â‚¹(10000 – 6000) = â‚¹4000
Page No 214:
Question 132:
â‚¹ 9000 becomes â‚¹ 18000 at simple interest in 8 years. Find the rate per cent per annum.
Answer:
Principal (P) = â‚¹ 9000,
Amount (A) = â‚¹ 18000,
Time (T) = 8 years
∴ Interest (I) = â‚¹(18000 – 9000) = â‚¹9000
$\mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\therefore 9000=\frac{9000\times \mathrm{R}\times 8}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{R}=\frac{100}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{R}=12.5\%$
Page No 214:
Question 133:
In how many years will the simple interest on a certain sum be 4.05 times the principal at 13.5 per cent per annum?
Answer:
Let the principal be â‚¹ P.
Then, simple interest (I) = â‚¹ 4.05 × P
Rate (R) = 13.5%
$\mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\therefore 4.05\times \mathrm{P}=\frac{\mathrm{P}\times 13.5\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Time}=\frac{4.05\times 100}{13.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Time}=30\mathrm{years}$
Page No 214:
Question 134:
The simple interest on a certain sum for 8 years at 12 per cent per annum is â‚¹ 3120 more than the simple interest on the same sum for 5 years at 14 per cent per annum. Find the sum.
Answer:
Let the sum be â‚¹P.
Interest for 8 years at 12% per annum
$\mathrm{Interest}\left(\mathrm{I}\right)=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{I}=\frac{\mathrm{P}\times 14\times 5}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{I}=\frac{7\mathrm{P}}{10}$
According to the question,
$\frac{24}{25}\mathrm{P}=3120+\frac{7}{10}\mathrm{P}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{24}{25}\frac{7}{10}\right)\mathrm{P}=3120\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{9670}{100}\right)\mathrm{P}=3120\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{26}{100}\right)\mathrm{P}=3120\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=\frac{3120\times 100}{26}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=12000\phantom{\rule{0ex}{0ex}}$
Thus, the sum is â‚¹12000.
Page No 215:
Question 135:
The simple interest on a certain sum for 2.5 years at 12 per cent per annum is â‚¹ 300 less than the simple interest on the same sum for 4.5 years at 8 per cent per annum. Find the sum.
Answer:
Let the sum be â‚¹ P.
Interest for 2.5 years at 12% per annum$=\frac{\mathrm{P}\times 12\times 2.5}{100}$
$=310\times \mathrm{P}$
Interest for 4.5 years at 8% per annum$=\frac{\mathrm{P}\times 8\times 4.5}{100}$
$=925\times \mathrm{P}$
According to the question,
$\frac{3\mathrm{P}}{10}=\frac{9}{25}\mathrm{P}300\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{9}{25}\frac{3}{10}\right)\mathrm{P}=300\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{3630}{100}\right)\mathrm{P}=300\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{6}{100}\right)\mathrm{P}=300\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{P}=5000\phantom{\rule{0ex}{0ex}}$
Thus, the sum is â‚¹5000.
Page No 215:
Question 136:
Designing a Healthy Diet
When you design your healthy diet, you want to make sure that you meet the dietary requirements to help you grow into a healthy adult. As you plan your menu, follow the following guidelines
1. Calculate your ideal weight as per your height from the table given at the end of this question.
2. An active child should eat around 55.11 calories for each kilogram desired weight.
3. 55 per cent of calories should come from carbohydrates. There are 4 calories in each gram of carbohydrates.
4. 15 per cent of your calories should come from proteins. There are 4 calories in each gram of proteins.
5. 30 per cent of your calories may come from fats. There are 9 calories in each gram of fat.
Following is an example to design your own healthy diet.
Example
1. Ideal weight = 40 kg.
2. The number of calories needed = 40 × 55.11 = 2204.4
3. Calories that should come from carbohydrates
= 2204.4 × 0.55 = 1212.42 calories.
Therefore, required quantity of carbohydrates
= $\frac{1212.42}{4}$= 300 g. (approx).
4. Calories that should come from proteins
= 2204.4 × 0.15 = 330.66 calories.
Therefore, required quantity of protein
$=\frac{330.66}{4}$ g = 82.66 g.
5. Calories that may come from fat = 2204.4 × 0.3
= 661.3 calories.
Therefore, required quantity of fat
$=\frac{661.3}{9}\mathrm{g}$ = 73.47 g.
Answer the Given Questions
1. Your ideal desired weight is __________ kg.
2. The quantity of calories you need to eat is _______.
3. The quantity of protein needed is ________ g.
4. The quantity of fat required is ___________ g.
5. The quantity of carbohydrates required is __________ g.
Answer:
(1) Ideal weight – 45 kg
(2) Quantity of calories needed = 45 × 55.11 = 2479.95 calories
(3) Calories that should come from carbohydrates$=\frac{2479.95\times 55}{100}$
$=2479.95\times 0.55\phantom{\rule{0ex}{0ex}}=1363.97\mathrm{calories}$
$\begin{array}{rcl}\therefore \mathrm{Required}\mathrm{quantity}\mathrm{of}\mathrm{carbohydrates}& =& \frac{1363.97}{4}\mathrm{g}\\ & =& 340.99\mathrm{g}\end{array}$
(4) Calories that should come from proteins$=\frac{2479.95\times 15}{100}$
$=2479.95\times 0.15\phantom{\rule{0ex}{0ex}}=1371.99\mathrm{calories}$
$\begin{array}{rcl}\therefore \mathrm{Required}\mathrm{quantity}\mathrm{of}\mathrm{protien}& =& \frac{371.99}{4}\mathrm{g}\\ & =& 392.99\mathrm{g}\end{array}$
(5) Calories that may come from fat$=\frac{2479.95\times 30}{100}$
$=2479.95\times 0.3\phantom{\rule{0ex}{0ex}}=743.985\mathrm{calories}$
$\begin{array}{rcl}\therefore \mathrm{Required}\mathrm{quantity}\mathrm{of}\mathrm{carbohydrates}& =& \frac{743.985}{4}\mathrm{g}\\ & =& 185.99\mathrm{g}\end{array}$
Page No 218:
Question 137:
150 students are studying English, Maths or both. 62 per cent of students study English and 68 per cent are studying Maths. How many students are studying both?
Answer:
Total number of students = 150
Number of students studying English $=62\%\mathrm{of}150$
$=\frac{62}{100}\times 150\phantom{\rule{0ex}{0ex}}=93$
Number of students studying Maths$=68\%\mathrm{of}150$
$=\frac{68}{100}\times 150\phantom{\rule{0ex}{0ex}}=102$
∴ Number of students studying both subjects = (93 + 102) – 150 = 45
Page No 218:
Question 138:
Earth Science: The table lists the world’s 10 largest deserts.
Largest Deserts in the World
Desert  Area (km^{2}) 
Sahara (Africa)  8,800,000 
Gobi (Asia)  1,300,000 
Australian Desert (Australia)  1,250,000 
Arabian Desert (Asia)  850,000 
Kalahari Desert (Africa)  580,000 
Chihuahuan Desert (North America)  370,000 
Takla Makan Desert (Asia)  320,000 
Kara Kum (Asia)  310,000 
Namib Desert (Africa)  310,000 
Thar Desert (Asia)  260,000 
(a) What are the mean, median and mode of the areas listed?
(b) How many times the size of the Gobi Desert is the Namib Desert?
(c) What percentage of the deserts listed are in Asia?
(d) What percentage of the total area of the deserts listed is in Asia?
Answer:
(a) Total area of 10 Deserts = 8800000 + 1300000 + 1250000 + 850000 + 580000 + 370000 + 320000 + 310000 + 310000 + 260000 = 14350000 km^{2}
$\begin{array}{rcl}\therefore \mathrm{Mean}& =& \frac{14350000}{10}{\mathrm{km}}^{2}\\ & =& 1435000{\mathrm{km}}^{2}\end{array}$
Since, n = 10 which is even
$\begin{array}{rcl}\therefore \mathrm{Median}& =& \frac{\left({5}^{\mathrm{th}}\mathrm{observation}+{6}^{\mathrm{th}}\mathrm{observation}\right)}{2}\\ & =& \frac{580000+370000}{2}\\ & =& \frac{950000}{2}\\ & =& 475000{\mathrm{km}}^{2}\end{array}$
Since, area = 310000 km^{2} occurs two times.
∴ Mode = 310000 km^{2}
(b) Area of Gobi Desert = 1300000 km^{2}
Area of Namib Desert = 310000 km^{2}
Now, 1300000 = x × 310000
So, size of the Gobi Desert is 4.19 times the Namib Desert.
(c) Total number of Deserts = 10
Deserts listed in Asia = 5
$\therefore \mathrm{Required}\mathrm{percentage}=\frac{5}{10}\times 100\%=50\%$
(d) Total area of Deserts listed in Asia = 1300000 + 850000 + 320000+ 310000 + 260000 = 3040000 km^{2}
Total area of all Deserts = 14350000 km^{2}
Percentage of Deserts in Asia
Page No 218:
Question 139:
Geography Application: Earth’s total land area is about 148428950 km^{2}. The land area of Asia is about 30 per cent of this total. What is the approximate land area of Asia to the nearest square km?
Answer:
Earth’s total land area = 148128950 km^{2}
Land area of Asia = 30% of 148128950 km^{2}
$=\frac{30}{100}\times 148428950\phantom{\rule{0ex}{0ex}}=44528685{\mathrm{km}}^{2}$
Thus, the approximate land area of Asia to the nearest square km is 44528685 km^{2}.
Page No 218:
Question 140:
The pieces of Tangrams have been rearranged to make the given shape.
By observing the given shape, answer the following questions:
• What percentage of total has been coloured?
(i) Red (R) = _________
(ii) Blue (B) = ________
(iii) Green (G) = _______
• Check that the sum of all the percentages calculated above should be 100.
• If we rearrange the same pieces to form some other shape, will the percentatge of colours change?
Answer:
(i) Total traction for Red (R)
$=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$
$\begin{array}{rcl}\frac{3}{8}& =& \frac{3}{8}\times \frac{100}{100}\\ & =& \frac{300}{8}\%\\ & =& 37.5\%\end{array}$
∴ Percentage of red colour = 37.5%
(ii) Total fraction for Blue (B)
$=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
$\begin{array}{rcl}\frac{1}{2}& =& \left(\frac{1}{2}\times 100\right)\%\\ & =& \frac{100}{2}\%\\ & =& 50\%\end{array}$
∴ Percentage of blue colour = 50%
(iii) Total fraction for Green (G)
$=\frac{1}{16}+\frac{1}{16}=\frac{2}{16}=\frac{1}{8}$
$\begin{array}{rcl}\frac{1}{8}& =& \left(\frac{1}{8}\times 100\right)\%\\ & =& \frac{100}{8}\%\\ & =& 12.5\%\end{array}$
∴ Percentage of green colour = 12.5%
Now, sum of all percentage
= 37.5% + 50% +12.5% = 100%
If we rearrange the same pieces to form some other shape, then percentage of colours will not change.
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