Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 7 - Comparing Quantities

Answer:

20% of 700 m $$\begin{gathered} =\frac{20}{100}\times 700 \mathrm{m} \\ =140 \mathrm{m} \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Gayatri's yearly income = ₹1,60,000
House rent paid by her = 15% of 1,60,000
                                     $$\begin{gathered} =\frac{15}{100}\times 1,60,000 \\ =₹24,000 \end{gathered}$$
Remaining amount = ₹1,60,000 − ₹24,000
Amount paid by her on her child's education
$$\begin{gathered} =10\% \mathrm{of} \mathrm{the} \mathrm{remaining} \mathrm{amount} \\ =10\% \mathrm{of} ₹1,36,000 \\ =\frac{10}{100}\times 1,36,000 \\ =₹13,600 \end{gathered}$$
Money left with her = ₹1,36,000 − ₹13,600
= ₹1,22,400

Hence, the correct answer is option (a).
 

Answer:

Ration of Fatima's income to her savings = 4 : 1
Sum of ratio = 4 + 1
= 5
Percentage of money saved by her $$\begin{gathered} =\left(\frac{1}{5}\times 100\right)\% \\ =20\% \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

$\begin{array}{rcl}0.07& =& \left(\frac{7}{100}\times 100\right)\%\\ & =& 7\%\end{array}$

Hence, the correct answer is option (b).

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
                                                      $$\begin{gathered} =\frac{40}{100}\times x \\ =\frac{2}{5}x \end{gathered}$$
Percentage of scouts from Ahmedabad $$\begin{gathered} =20\% \mathrm{of} \frac{2}{5}x \\ =\frac{20}{100}\times \frac{2}{5}\times x \\ =\frac{4\times 2}{100}\times x \\ =\frac{8}{100}\times x \\ =8\% \mathrm{of} x \end{gathered}$$

Hence, the correct answer is option (c).
 

Answer:

Let x percent of ₹4500 be ₹9000.
∴ x% of 4500 = 9000
$$\begin{gathered} \Rightarrow \frac{x}{100}\times 4500=9000 \\ \Rightarrow x=\frac{9000}{45} \\ \Rightarrow x=200\% \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

$\begin{array}{rcl}5.2& =& \left(\frac{52}{10}\times 100\right)\%\\ & =& 520\%\end{array}$

Hence, the correct answer is option (c).

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).

Answer:

$\begin{array}{rcl}225\%& =& \frac{225}{100}\\ & =& \frac{9}{4}\\ & =& 9 : 4\end{array}$

Hence, the correct answer is option (a).

Answer:

Cost Price of bicycle (P) = ₹1800
Profit % = 12%
Selling Price of bicycle (SP) $$\begin{gathered} =\mathrm{CP}\times \left(1+\frac{\mathrm{P}\%}{100}\right) \\ =1800\left(1+\frac{12}{100}\right) \\ =1800\times \frac{112}{100} \\ =₹2016 \end{gathered}$$

Thus, the selling price of bicycle is ₹2016.

Hence, the correct answer is option (b).

Answer:

Cost Price of cricket bat (CP) = ₹800
Selling Price of cricket bat (SP) = ₹1600
Profit = SP − CP
= 1600 − 800
= ₹800
$\begin{array}{rcl}\mathrm{Profit}\%& =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{₹800}{₹800}\times 100\right)\%\\ & =& 100\%\end{array}$
Thus, the profit earned is 100%.

Hence, the correct answer is option (a).

Answer:

Cost Price of buffalo = ₹44000
loss % on buffalo = 5%
loss on buffalo = 5% of 44000
                        $$\begin{gathered} =\frac{5}{100}\times 44000 \\ =₹2,200 \end{gathered}$$
∴ Selling Price of buffalo $$\begin{gathered} =\mathrm{CP}\left(1-\frac{\mathrm{Loss}\%}{100}\right) \\ =44,000\left(1-\frac{5}{100}\right) \\ =44,000\times \frac{95}{100} \\ =₹41,800 \end{gathered}$$
Cost Price of cow = ₹18000
Profit % on cow = 10%
∴ Selling Price of cow $$\begin{gathered} =\mathrm{CP}\left(1+\frac{\mathrm{Profit}\%}{100}\right) \\ =18000\left(1+\frac{10}{100}\right) \\ =18000\times \frac{110}{100} \\ =19,800 \end{gathered}$$
Total cost price = ₹44000 + 18000
= ₹62000
Total selling price = ₹41,800 + ₹19800
= ₹61600
Total selling price < Total cost price
∴ Net loss = Total CP − Total SP
= ₹62,000 − ₹61600
= ₹400

Hence, the correct answer is option (b).

Answer:

Let Raman's income be ₹x.
∴ Mohan's income = x + 25% of x
                            $$\begin{gathered} =x\left(1+\frac{25}{100}\right) \\ =x\left(\frac{125}{100}\right) \\ =x\left(\frac{5}{4}\right) \end{gathered}$$
Difference in Raman and Mohan's income
$$\begin{gathered} =\frac{5}{4}x-x \\ =x\left(\frac{5}{4}-1\right) \\ =\frac{1}{4}x \end{gathered}$$
Raman's income is less than Mohan's income by
$$\begin{gathered} =\left(\frac{{\displaystyle \frac{1}{4}}x}{{\displaystyle \frac{5}{4}}x}\times 100\right)\% \\ =\left(\frac{1}{5}\times 100\right)\% \\ =20\% \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

Principal (P) = ₹30000
Time (T) = 3 years
Rate (R) = 15%
Simple Interest (SI) $$\begin{gathered} =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{30000\times 15\times 3}{100} \\ =₹13,500 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Principal (P) = ₹3000
Time (T) = 2 years
Rate (R) = 11%
Simple Interest (SI) $$\begin{gathered} =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{3000\times 11\times 2}{100} \\ =₹660 \end{gathered}$$
Amount (A) = P + SI
= ₹3000 + ₹660
= ₹3660

Hence, the correct answer is option (b).

Answer:

Principal (P) = ₹12000
Time (T) = 1 month $=\frac{1}{12}\mathrm{year}$
Rate (R) = 10%
Simple Interest (SI) $$\begin{gathered} =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{12000\times 10\times {\displaystyle \frac{1}{12}}}{100} \\ =₹100 \end{gathered}$$

Hence, the correct answer is option (c).
 

Answer:

For Rajni
Principal (P) = ₹3000
Time (T) = 3 years
Rate (R) = 10%
Simple Interest (SI) $$\begin{gathered} =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{3000\times 10\times 3}{100} \\ =₹900 \end{gathered}$$
Amount received by Rajni = P + SI
= 3000 + 900
= ₹3900
For Mohini
Principal (P) = ₹4000
Time (T) = $2\frac{1}{2}\mathrm{years}=\frac{5}{2}\mathrm{years}$
Rate (R) = 10%
Simple Interest (SI) $$\begin{gathered} =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{4000\times 10\times {\displaystyle \frac{5}{2}}}{100} \\ =₹1000 \end{gathered}$$
Amount received by Rajni = P + SI
= 4000 + 1000
= ₹5000
Difference of the amount received by them
= ₹5000 − ₹3900
= ₹1100

Hence, the correct answer is option (d).

Answer:

90% of x = 315 km
$$\begin{gathered} \Rightarrow \frac{90}{100}\times x=315 \mathrm{km} \\ \Rightarrow x=\frac{100}{90}\times 315 \mathrm{km} \\ \Rightarrow x=350 \mathrm{km} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Selling Price of article (SP) = ₹329
loss% = 6%
Let Cost Price of article (CP) be x.
$$\begin{gathered} \therefore \mathrm{SP}=\mathrm{CP}\left(1-\frac{\mathrm{Loss}\%}{100}\right) \\ \Rightarrow 329=x\left(1-\frac{6}{100}\right) \\ \Rightarrow 329=x\left(\frac{94}{100}\right) \\ \Rightarrow \frac{329\times 100}{94}=x \\ \Rightarrow ₹350=x \end{gathered}$$
Thus, the cost price of the article is ₹350.

Hence, the correct answer is option (d).

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).

Answer:

Let the Principal be P.
Simple Interest (SI) = ₹126
Time (T) = 2 years
Rate (R) = 14%
Since, Simple interest (SI) $=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}$
$$\begin{gathered} \Rightarrow 126=\frac{\mathrm{P}\times 14\times 2}{100} \\ \Rightarrow \mathrm{P}=\frac{126\times 100}{14\times 2} \\ \Rightarrow \mathrm{P}=₹450 \end{gathered}$$

Hence, the correct answer is option (c).

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 $$\begin{gathered} =\left(\frac{40}{100}\times 100\right)\% \\ =40\% \end{gathered}$$

Hence, the correct answer is option (c).

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 $$\begin{gathered} =\left(\frac{36}{100}\times 100\right)\% \\ =36\% \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

2 : 3 $$\begin{gathered} =\left(\frac{2}{3}\times 100\right)\% \\ =\left(\frac{200}{3}\right)\% \\ =66.67\% \end{gathered}$$

Hence, 2 : 3 = 66.67%.

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, $18\frac{3}{4}\%=\underline{3 : 16}$

Answer:

30% of ₹360 $$\begin{gathered} =\frac{30}{100}\times 360 \\ =₹102 \end{gathered}$$

Hence, 30% of ₹360 = ₹102.

Answer:

120 % of 50 km $$\begin{gathered} =\left(\frac{120}{100}\times 50 \mathrm{km}\right) \\ =60 \mathrm{km} \end{gathered}$$

Hence, 120% of 50 km = 60 km.

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%.

Answer:

$\begin{array}{rcl}\frac{8}{5}& =& \left(\frac{8}{5}\times 100\right)\%\\ & =& 160\%\end{array}$

Hence, $\frac{8}{5}=\underline{160\%}.$

Answer:

A fraction with its denominator 100 is called a percent.

Answer:

Let the required percentage be x.
∴ x% of 50 kg = 15 kg
$$\begin{gathered} \Rightarrow \frac{x}{100}\times 50 \mathrm{kg}=15 \mathrm{kg} \\ \Rightarrow x=15\times 2 \\ \Rightarrow x=30\% \end{gathered}$$

Hence, 15 kg is 30% of 50 kg.

Answer:

Initial weight = 60 kg
Final weight = 66 kg
Increase in weight = 66 kg − 60 kg
= 6 kg
Percentage increase in weight $$\begin{gathered} =\left(\frac{6}{60}\times 100\right)\% \\ =10\% \end{gathered}$$

Hence, weight of Nikhil increased from 60 kg to 66 kg.
Then the increase in weight is 10%.

Answer:

Number of students = 50
Percentage of absent students = 8%
Number of absent students $$\begin{gathered} =\frac{8}{100}\times 50 \\ =4 \end{gathered}$$
Number of present students = 50 − 4
= 46

Hence, the number of students present on that day was 46.

Answer:

Marks obtained = 440
Total Marks = 500
Percentage $$\begin{gathered} =\left(\frac{440}{500}\times 100\right)\% \\ =88\% \end{gathered}$$

Hence, she secured 88% marks in the examination.

Answer:

Total deposit in bank account = ₹1500
Withdraw percent = 40%
Balance = ₹1500 − (40% of ₹1500)
= 1500 − $\left(\frac{40}{100}\times 1500\right)$
= 1500 − 600
= ₹1100
Now, the balance in her account is ₹1100.

Answer:

Let the required number be x.
So, x = 50% of 60 + 60
$$\begin{gathered} x=\left(\frac{50}{100}\times 60\right)+60 \\ x=30+60 \\ x=90 \end{gathered}$$

Hence, 90 is 50% more than 60.

Answer:

Selling Price of bat (SP) = ₹75
Loss (L) = ₹8
Cost Price of bat (CP) = SP + L
= ₹75 + ₹8
= ₹83

Hence, the cost price of the bat is ₹83.

Answer:

Initial Cost Price of 3 kg of Sugar = ₹120
Initial Cost Price of 1 kg of Sugar $$\begin{gathered} =\frac{₹120}{3} \\ =₹40 \end{gathered}$$
Percentage decrease in cost = 20%
Decrease in cost $$\begin{gathered} =20\% \mathrm{of} ₹40 \\ =\frac{20}{100}\times 40 \\ =₹8 \end{gathered}$$
Decreased cost of 1 kg Sugar = ₹40 − ₹8
= ₹32
Decreased cost of 3 kg Sugar = ₹32 $\times$ 3
= ₹96

Hence, the price of 3 kg sugar originally costing ₹120 will be ₹96.

Answer:

Cost Price of cow (CP) = ₹9000
Loss (L) = ₹900
Selling Price of cow (SP) = CP − L
= ₹9000 − ₹900
= ₹8100

Hence, the selling price of the cow is ₹8100.

Answer:

Cost Price of Chair (CP) = ₹700
Selling Price of Chair (SP) = ₹750
Profit (P) = SP − CP
= ₹750 − ₹700
= ₹50
Profit % $$\begin{gathered} =\left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\% \\ =\left(\frac{50}{700}\times 100\right)\% \\ =7.14\% \end{gathered}$$

Hence, she earns a profit of 7.14% in the transaction.

Answer:

Cost Price of bed sheet (CP) = ₹400
Selling Price of bed sheet (SP) = ₹440
Profit (P) = SP − CP
= ₹440 − ₹400
= ₹40
Profit % $$\begin{gathered} =\left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\% \\ =\left(\frac{40}{400}\times 100\right)\% \\ =10\% \end{gathered}$$

Hence, Sonal bought a bed sheet for ₹400 and sold it for ₹400.
Her Profit 40% is 10%.

Answer:

Cost Price of pen (CP) = ₹60
Selling Price of pen (SP) = ₹54
Loss (L) $$\begin{gathered} =\mathrm{CP}-\mathrm{SP} \\ =₹60-₹54 \\ =₹6 \end{gathered}$$
$\begin{array}{rcl}\mathrm{Loss}\%& =& \left(\frac{\mathrm{Loss}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{6}{60}\times 100\right)\%\\ & =& 10\%\end{array}$

Hence, his loss % is 10%.

Answer:

Purchasing amount of house = ₹50,59,700
Amount spent on repairing the house = ₹40,300
Cost price of house (CP) = ₹50,59,700 + ₹40,300
= ₹51,00,000
Profit % = 5%
Selling Price of house (SP) $=\mathrm{CP}\left(1+\frac{\mathrm{P}\%}{100}\right)$
                                             $$\begin{gathered} =₹51,00,000\left(1+\frac{5}{100}\right) \\ =₹51,00,000\times \frac{105}{100} \\ =₹53,55,000 \end{gathered}$$

Hence, to make a profit of 5%, she should sell the house for ₹53,55,000.

Answer:

Cost Price of 20 lemons = ₹10
Cost Price of 1 Lemon $$\begin{gathered} =\frac{₹10}{20} \\ =₹0.5 \end{gathered}$$
Selling Price of 5 lemons = ₹3
Selling Price of 1 lemon $$\begin{gathered} =₹\frac{3}{5} \\ =₹0.6 \end{gathered}$$
Profit = SP − CP
= ₹0.6 − ₹0.5
= ₹0.1
$\begin{array}{rcl}\mathrm{Profit}\%& =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\right)\%\\ & =& \left(\frac{0.1}{0.5}\times 100\right)\%\\ & =& 20\%\end{array}$

Hence, if 20 lemons are brought for ₹10 and sold at 5 for three rupee, then profit in the transaction is 20%.

Answer:

Lost of 1 orange = ₹4
Lost of 120 oranges = ₹4 × 120
= ₹480
$\begin{array}{rcl}60\% \mathrm{of} 120 & =& \frac{60}{100}\times 120\\ & =& 72\end{array}$
Selling price of 72 oranges = ₹5 × 72
= ₹360
Selling price of 48 (120 − 72) orange = ₹3.50 × 48
= ₹168
Total selling price of 120 oranges = ₹360 + ₹168
= ₹528
Profit = SP − CP
= ₹528 − ₹480
= ₹48
$\begin{array}{rcl}\mathrm{Profit} \% & =& \left(\frac{\mathrm{Profit}}{\mathrm{CP}}\times \mathrm{CP}\right)\%\\ & =& \left(\frac{48}{480}\times 100\right)\%\\ & =& 10\%\end{array}$
Hence, his profit is 10%. 


 

Answer:

Weight of apples bought = 20 kg
Cost price of 1 kg apples = ₹50
Cost price of 20 kg apples = ₹50 × 20
= ₹1000
Percentage of rotten apples = 5%
Quantity of rotten apples = 5% of 20 kg
$$\begin{gathered} =\frac{5}{100}\times 20 \mathrm{kg} \\ =1 \mathrm{kg} \end{gathered}$$
Quantity of good apples = (20 − 1) kg
= 19 kg
Selling price of 1 kg of apples = ₹60
Selling Price of 1 kg  of apples = ₹60 × 19
= ₹1140
Profit = SP − CP
= ₹(1140 − 1000)
= ₹140
$\begin{array}{rcl}\mathrm{Profit}\% & =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \left(\frac{140}{1000}\times 100\right)\%\\ & =& 14\%\end{array}$
Hence, his Profit is 14% 




 

Answer:

Principle (P) = ₹3000
Rate (R) = 10%
Time (T) = 3 years
$\begin{array}{rcl}\mathrm{Interest} & =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{3000\times 10\times 3}{100}\\ & =& ₹900\end{array}$
Thus, interest on ₹3000 at 10% per annum for a period of 3 years is ₹900.

Answer:

Principle (P) = ₹20,000
Rate (R) = 8%
Time = 6 months
$=\frac{6}{12}=\frac{1}{2} \mathrm{year}$
$$\begin{gathered} \mathrm{Interest} \left(\mathrm{I}\right) =\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ =\frac{20000\times 8\times {\displaystyle \frac{1}{2}}}{100} \\ \mathrm{I}=₹800 \end{gathered}$$
Amount (A) = P + I
= ₹(20000 + 800)
= ₹20800

 

Answer:

Principle (P) = ₹125000
Rate (R) = 18%
Time (T) = 2 years 4 month
$$\begin{gathered} =2 \mathrm{years} + \frac{4}{12} \mathrm{year} \\ = \left(2+\frac{1}{3}\right) \mathrm{year} \\ = \frac{7}{3} \mathrm{year} \end{gathered}$$
$\begin{array}{rcl}\mathrm{Interest} \left(\mathrm{I}\right) & =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{125000\times 18\times {\displaystyle \frac{7}{3}}}{100}\\ & =& ₹52,500\end{array}$

Thus, interest on ₹12500 at 18% per annum for a period of 2 years and 4 months is ₹52,500.

Answer:

Let x% of 5 litres be 25 ml.
5 litres = 5000 ml
∴ x% of 5000 ml = 25 ml
$$\begin{gathered} \Rightarrow \frac{x}{100}\times 5000 \mathrm{ml}=25 \mathrm{ml} \\ \Rightarrow x=\frac{25}{50} \\ \Rightarrow x=0.5 \end{gathered}$$
Thus, 25 ml is 0.5 percent of 5 litres.

Answer:

A + 20% of A = B
$$\begin{gathered} \Rightarrow \mathrm{A}+\frac{20}{100}\times \mathrm{A}=\mathrm{B} \\ \Rightarrow \frac{120\mathrm{A}}{100}=\mathrm{B} \\ \Rightarrow 12\mathrm{A}=10\mathrm{B} \\ \Rightarrow 6\mathrm{A}=5\mathrm{B} .....\left(1\right) \end{gathered}$$
Also, B − 50% of B = C
$$\begin{gathered} \Rightarrow \mathrm{B}-\frac{50}{100}\times \mathrm{B}=\mathrm{C} \\ \Rightarrow \mathrm{B}-\frac{1}{2}\mathrm{B}=\mathrm{C} \\ \Rightarrow \frac{1}{2}\mathrm{B}=\mathrm{C} \\ \Rightarrow \mathrm{B}=2\mathrm{C} .....\left(2\right) \end{gathered}$$
Substituting (2) in (1), we get
⇒ 6A = 5(2C)
⇒ 6A = 10C
$$\begin{gathered} \Rightarrow \frac{6\mathrm{A}}{10}=\frac{\mathrm{C}}{8} \\ \Rightarrow \mathrm{C}=\frac{6\mathrm{A}\times 10}{10\times 10} \\ \Rightarrow \mathrm{C}=\frac{60}{100}\mathrm{A} \\ \Rightarrow \mathrm{C}=60\% \mathrm{of} \mathrm{A} \end{gathered}$$
Then, 60% of A is equal to C.


 

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.

Answer:

Principal (P) = ₹21,00.
Rate (R) = 8%
Case I: Time (T₁) 2 years.
$\begin{array}{rcl}{\mathrm{SI}}_1& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_1}{100}\\ & =& \frac{2100\times 8\times 2}{100}\\ & =& ₹336\end{array}$
Case II: Time (T₂) = 3 years
$\begin{array}{rcl}{\mathrm{SI}}_2& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_2}{100}\\ & =& \frac{2100\times 8\times 3}{100}\\ & =& ₹504\end{array}$
Difference of interest for 2 years and 3 years = ₹504 − 336
= ₹168
Thus, the difference of interest for 2 years and 3 years on a sum of ₹2100 at 8% per annum is ₹168.

Answer:

To convert a fraction into a percent, we multiply it by 100.

Answer:

To convert a decimal into a percent, we shift the decimal point two places to the right.

Answer:

Principal (P) = ₹2000
Rate (R) = 6%
Case I: Time (T₁) = $1\frac{1}{2} \mathrm{years}$
$\begin{array}{rcl}\mathrm{Interest} \left({\mathrm{I}}_1\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_1}{100}\\ & =& \frac{2000\times 6\times {\displaystyle \frac{3}{2}}}{100}\\ & =& ₹180\end{array}$
Case II: Time (T₂) = 2 years
$\begin{array}{rcl}\mathrm{Interest} \left({\mathrm{I}}_2\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times {\mathrm{T}}_2}{100}\\ & =& \frac{2000\times 6\times {\displaystyle 2}}{100}\\ & =& ₹240\end{array}$
Now, I₁ + I₂ = 180 + 240 = ₹420
Thus, the sum of interest on a sum of ₹2000 at the rate of 6% per annum for year and $1\frac{1}{2}$ 2 years is ₹420.

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%.

Answer:

Answer:

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.

Answer:

When an improper fraction is converted into a percentage, the answer is always greater than 100.
Hence, the statement is False.

Answer:

4 days = 4 × 24 hours = 96 hours

50% of 4 days = 50% × 4 days
                       $$\begin{gathered} =\frac{50}{100}\times 96 \mathrm{hours} \\ =\frac{1}{2}\times 96 \mathrm{hours} \\ =48 \mathrm{hours} \end{gathered}$$

Hence, the statement is False.
 

Answer:

Principal (P) = â‚¹350
Rate of interest (R) = 5%
Time (T) = 73 days $=\frac{73}{365}=0.2 \mathrm{years}$

 $\begin{array}{rcl}\mathrm{Simple} \mathrm{Interest} \left(SI\right) & =& \frac{P\times R\times T}{100}\\ & =& \frac{₹350\times 5\times 0.2}{100}\\ & =& ₹3.5\end{array}$

Hence, the statement is False.

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.

Answer:

$\begin{array}{rcl}75\%& =& \frac{75}{100}\\ & =& \frac{75÷25}{100÷25}\\ & =& \frac{4}{3}\end{array}$
Hence, the statement is True.

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.

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.

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.

Answer:

$\begin{array}{rcl}50\% \mathrm{of} ₹50& =& \frac{50}{100}\times ₹50\\ & =& ₹25\end{array}$
Hence, the statement is True.

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.

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.

Answer:

Selling price of a book = ₹50
Loss % = ₹10
We know that
$\begin{array}{rcl}\mathrm{CP}& =& \frac{\mathrm{SP}\times 100}{\left(110-\mathrm{Loss}\right)\%}\\ & =& \frac{50\times 100}{\left(100-10\right)}\\ & =& \frac{5000}{90}\\ & =& ₹55.55\end{array}$
So, the cost price of the book is ₹55.5.
Hence, the statement is False.

Answer:

CP of a chair = ₹2000
Gain % = 10
We know that,
$\begin{array}{rcl}\mathrm{SP}& =& \mathrm{CP}\left(\frac{100+\mathrm{G}\%}{100}\right)\\ & =& 2000\left(\frac{100+10}{100}\right)\\ & =& \frac{2000\times 110}{100}\\ & =& ₹2200\end{array}$
Hence, the statement is False.

 

Answer:

CP of a bicycle = ₹ 650
SP of the bicycle = ₹ 585
Therefore, there is a loss of ₹ (650 – 585) = ₹65
$\begin{array}{rcl}\mathrm{Loss}\%& =& \frac{65}{650}\times 100\%\\ & =& 10\%\end{array}$
Hence, the statement is False.

Answer:

SP of watch = ₹3320
Gain = ₹320
∴ CP of the watch = ₹(3320 – 320) = ₹3000
Required gain % = 10
$\begin{array}{rcl}\mathrm{SP}& =& \mathrm{CP}\left(\frac{100+\mathrm{G}\%}{100}\right)\\ & =& 3000\left(\frac{100+10}{100}\right)\\ & =& \frac{3000\times 110}{100}\\ & =& ₹3300\end{array}$
Hence, the statement is False.

Answer:

Principal (P) = ₹1200,
Time (T)= $1\frac{1}{2}$years =$\frac{3}{2}$ years 
Rate (R)= 15% pa

Hence, the statement is False.

Answer:

Principal (P) = ₹ 800
Time (T)= 3 years
Rate (R)= 12% pa
$\begin{array}{rcl}\mathrm{Interest} \left(\mathrm{I}\right)& =& \frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100}\\ & =& \frac{₹800\times 12\times {\displaystyle 3}}{100}\\ & =& ₹288\end{array}$


Now, amount = Principal + Interest
= ₹ (800 + 288)
= ₹1088

Hence, the statement is False.

Answer:

For Feroz: Principal = ₹6400
Rate = 15% pa


For Rashmi : Principal = ₹6400,
Rate = 15% pa
Time = 5 years

Required difference in interests = ₹(4800 – 3360) = ₹1440
Hence, the statement is False.

Answer:

Total number of lemons = 720
CP of 100 lemons = ₹120
$$\begin{gathered} \therefore \mathrm{CP} \mathrm{of} 1 \mathrm{lemon}=\frac{₹120}{100} \\ \mathrm{CP} \mathrm{of} 720 \mathrm{lemons}=\frac{₹120}{100}\times 720=₹864 \end{gathered}$$
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 = ₹(36 + 810) = ₹ 846
Here, CP > SP
Therefore, there is a loss of ₹(864 – 846) = ₹18
Hence, the statement is False.

Answer:

(i) 8% of â‚¹ x is 100     
$$\begin{gathered} \Rightarrow \frac{8}{100}\times x=100 \\ \Rightarrow x=\frac{100\times 100}{8} \\ \Rightarrow x=\frac{100\times 100}{8} \\ \Rightarrow x=1250 \end{gathered}$$

(ii) 32% of x kg is 400 kg
$$\begin{gathered} \Rightarrow \frac{32}{100}\times x=400 \\ \Rightarrow x=\frac{400\times 100}{32} \\ \Rightarrow x=\frac{400\times 100}{32} \\ \Rightarrow x=1250 \end{gathered}$$

(iii) 35% of â‚¹ x is â‚¹280 
$$\begin{gathered} \Rightarrow \frac{35}{100}\times x=280 \\ \Rightarrow x=\frac{280\times 100}{35} \\ \Rightarrow x=800 \end{gathered}$$             

(iv) 45% of marks x is 405
$$\begin{gathered} \Rightarrow \frac{45}{100}\times x=405 \\ \Rightarrow x=\frac{405\times 100}{45} \\ \Rightarrow x=900 \end{gathered}$$

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
∴ 110 of 100=$\frac{110}{100}\times 100$=10
Each small square is worth 3
∴ 10 small square worth is 10 × 3 = 30

Answer:

$\frac{1}{6}=\left(\frac{1}{6}\times 100\right)\%=16.67\%$

Answer:

$\frac{9}{40}=\left(\frac{9}{40}\times 100\right)\%=22.5\%$

Answer:

$\frac{1}{100}=\left(\frac{1}{100}\times 100\right)\%=1\%$

Answer:

$80\%=\frac{80}{100}=\frac{4}{5}$

Answer:

$\begin{array}{rcl}33\frac{1}{3}\%& =& \frac{100}{3}\%\\ & =& \frac{100}{3}\times \frac{1}{100}\\ & =& \frac{1}{3}\\ & =& 1:3\end{array}$

Answer:

$\begin{array}{rcl}16\frac{2}{3}\%& =& \frac{50}{3}\%\\ & =& \frac{50}{3}\times \frac{1}{100}\\ & =& \frac{1}{6}\\ & =& 1:6\end{array}$

Answer:

$\begin{array}{rcl}150\%& =& \frac{150}{100}\\ & =& \frac{3}{2}\\ & =& 3:2\end{array}$

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.

Answer:

$\begin{array}{rcl}0.089& =& \frac{89}{1000}\\ & =& \left(\frac{89}{1000}\times 100\right)\%\\ & =& 8.9\%\end{array}$

Answer:

$\begin{array}{rcl}1.56& =& \frac{156}{100}\\ & =& \left(\frac{156}{100}\times 100\right)\%\\ & =& 156\%\end{array}$

Answer:

$\begin{array}{rcl}15\%& =& \frac{15}{100}\\ & =& 0.15\end{array}$
Thus, 15% of 20 is 0.15.

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.
 

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.

Answer:

Let x% of 1 hour is 30 minutes.
Now, 1 hour = 60 minutes
$$\begin{gathered} \therefore \frac{x}{100}\times 60=30 \\ \Rightarrow x=\frac{30\times 100}{60}=50 \end{gathered}$$
Thus, Per cent of 1 hour is 30 minutes is 50%.

Answer:

Let x% of 1 day is 1 minute,
Now, 1 day =  24 hours = (24 × 60) minutes
$$\begin{gathered} \therefore \frac{x}{100}\times 1440=1 \\ \Rightarrow x=\frac{100}{1440}=0.069 \end{gathered}$$
Thus, per cent of 1 day is 1 minute is 0.069.

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%.

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%.

Answer:

Let x% of ₹ 80 is ₹ 100

$$\begin{gathered} \therefore \frac{x}{100}\times 80=100 \\ \Rightarrow x=\frac{100\times 100}{80} \\ \Rightarrow x=125 \end{gathered}$$

Answer:

Total percentage = 100
Percentage of men = 45
Percentage of women = 40
∴ Percentage of children = [100 – (45 + 40)]% = [100 – 85]% = 15%
 

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

Answer:

Total amount of chalk$=2\frac{1}{2}\mathrm{kg}=\frac{5}{2}\mathrm{kg}$
Amount of calcium = 10% of $\frac{5}{2}\mathrm{kg}$
$$\begin{gathered} =\left(\frac{1}{10}\times \frac{5}{2}\right)\mathrm{kg} \\ =0.25 \mathrm{kg} \\ =\left(0.25\times 1000\right) \mathrm{g} \\ =250 \mathrm{g} \end{gathered}$$

Amount of carbon = 3% of $\frac{5}{2}\mathrm{kg}$
$$\begin{gathered} =\left(\frac{3}{100}\times \frac{5}{2}\right)\mathrm{kg} \\ =0.075 \mathrm{kg} \\ =\left(0.075\times 1000\right) \mathrm{g} \\ =75 \mathrm{g} \end{gathered}$$
Thus, the amount of carbon and calcium (in grams) in $2\frac{1}{2}$kg of chalk are 250g and 75g.

 

Answer:

Total mass of mortar = 800 kg
Mass of sand in mortar = 55% of 800 kg
$$\begin{gathered} =\left(\frac{55}{100}\times 800\right)\mathrm{kg} \\ =440 \mathrm{kg} \end{gathered}$$
Mass of cement in mortar = 33% of 800 kg
$$\begin{gathered} =\left(\frac{33}{100}\times 800\right)\mathrm{kg} \\ =264 \mathrm{kg} \end{gathered}$$
∴ Mass of lime in mortar = [800 – (440 + 264)] kg = 96 kg

Answer:

CP of 1 table = ₹450
∴ CP of 24 tables = ₹ (24 × 450) = ₹10800
SP of 16 tables = ₹ {(16 × 600) = ₹9600
SP of remaining tables i.e., 8 tables
= ₹(8 × 400) = 3200
SP of 24 tables = ₹(9600 + 3200) = ₹12800
Here, SP > CP
Therefore, there is a gain of ₹(12800 – 10800) = ₹2000
$\begin{array}{rcl}\mathrm{Gain}\%& =& \frac{2000}{10800}\times 100\\ & =& 18.5\%\end{array}$

Answer:

Let the amount of money Medha had ₹ 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 = $20\% \mathrm{of}\frac{4}{5}x$
$$\begin{gathered} =\frac{20}{100}\times \frac{4}{5}x \\ =\frac{4}{25}x \end{gathered}$$
Amount of money left with her =$=\frac{4}{5}x-\frac{4}{25}x$
$=\frac{16}{25}x$
According to question,
$\frac{16}{25}x=4800$
$\Rightarrow x=\frac{4800\times 25}{16}=7500$
 

Answer:

Let the original cost of a flower vase = ₹ x
The cost of the flower vase increased by

$\begin{array}{rcl}12\%& =& x+12\% of x\\ & =& x+\frac{12}{100}\times x\end{array}$

According to the question,
$$\begin{gathered} \frac{28}{25}x=896 \\ \Rightarrow x=\frac{896\times 25}{28}=800 \end{gathered}$$

Answer:

Total amount borrowed by Radhika = ₹12000
I: Principal = ₹4000, Rate = 18% p.a.,
Time = 3 years
$\mathrm{Interest}=\frac{4000\times 18\times 3}{100}=2160$
II: Principal = ₹ (12000 – 4000) = ₹8000,
Rate = 15%, Time = 3 years
$\mathrm{Interest}=\frac{8000\times 15\times 3}{100}=3600$
Total interest = ₹ (2160 + 3600) = ₹5760

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$=\left(\frac{240}{300}\times 100\right)\%=80\%$

Answer:

S.P. of a chair = ₹1440
Loss% = 10
$\begin{array}{rcl}\mathrm{CP}& =& \frac{\mathrm{SP}\times 100}{100-\mathrm{Loss}\%}\\ & =& \frac{1440\times 100}{100-10}\\ & =& \frac{14400}{90}\\ & =& 1600\end{array}$
So, the shopkeeper bought the chair at ₹1600.

Answer:

Let the principal be ₹P.
Rate (R) = 12% p.a.,
Interest (T) = ₹ 1620
Time = 2 years
$$\begin{gathered} \mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \Rightarrow 1620=\frac{\mathrm{P}\times 12\times 2}{100} \\ \Rightarrow \mathrm{P}=\frac{1620\times 100}{24} \\ \Rightarrow \mathrm{P}=6750 \end{gathered}$$
So, amount of money invested by Dhruvika is ₹6750.

Answer:

Let the CP of a scooter = ₹x
SP of the scooter for a loss of 25%$=\left(\frac{100-25}{100}\right)\times x=\frac{3}{4}x$
SP of the scooter for a loss of 20%$=\left(\frac{100-20}{100}\right)\times x=\frac{4}{5}x$

The difference in both SPs of the scooter$=\frac{4}{5}x-\frac{3}{4}x=\frac{1}{20}x$
According to question,
$$\begin{gathered} \frac{1}{20}x=4000 \\ \Rightarrow x=4000\times 20 \\ \Rightarrow x=80000 \end{gathered}$$

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.

Answer:

Cost of 250 tickets of ₹ 400 per ticket = ₹ (250 × 400) = ₹ 100000
Cost of 500 tickets of ₹ 100 per ticket = ₹ (500 × 100) = ₹ 50000
Tax on tickets of ₹ 400$=\frac{40}{100}\times 100000=₹40000$
Tax on tickets of 100₹$=\frac{20}{100}\times 50000=₹10000$

So, total tax collected from the programme = ₹(40000 + 10000) = ₹50000

Answer:

Monthly income of Bhavya = ₹50000
Spending amount = 80% of ₹50000$=\frac{80}{100}\times 50000=₹40000$
New income – 20% of 50000 + 50000$=\frac{20}{100}\times 50000+50000=₹10000+₹50000$ = ₹ 60000
New spending amount = 40000 + 20% of 40000$=40000+\frac{20}{100}\times 40000$

= 40000 + 8000 = ₹ 48000
∴ New savings of Bhavya = ₹(60000 – 48000) = ₹12000

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,
$$\begin{gathered} \frac{\left(128+x\right)}{300}\times 100=70 \\ \Rightarrow 128+x=3\times 70 \\ \Rightarrow x=210-128 \\ \Rightarrow x=82 \end{gathered}$$
Thus, the candidate must obtain 82 marks in the third paper.

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$
 

Answer:

(a) Let the length of the largest reticulated python – x cm
According to the question,60% of x = 558
$$\begin{gathered} \Rightarrow \frac{60}{100}\times x=558 \\ \Rightarrow x=\frac{558\times 100}{60} \end{gathered}$$
⇒ x = 930

(b) We have,
$$\begin{gathered} \frac{\mathrm{mass} 1}{\mathrm{length} 2}=\frac{\mathrm{mass} 2}{\mathrm{length} 1} \\ \therefore \frac{24}{2}=\frac{\mathrm{mass} 2}{3} \\ \Rightarrow \mathrm{mass} 2=\frac{24\times 3}{2} \\ \Rightarrow \mathrm{mass} 2=36 \mathrm{kg} \end{gathered}$$

(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$=\frac{0.0000001}{2}\times 17=0.00000085 \mathrm{mm}$
 

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’$=\frac{9}{14}\times 100=64.28\%$

Answer:

$0.06\%=\frac{0.06}{100}=\frac{6}{10000}$
Therefore 6 out of every 10000 are defective T-shifts.

Answer:

34=3×44×4=1216≠916
$\frac{3}{4}=\frac{3\times 4}{4\times 4}=\frac{12}{16}\ne \frac{9}{16}$   
Therefore, $\frac{3}{4}$ and $\frac{9}{16}$  are not proportional.

Answer:

Cost price for 4 T-shirts = ₹1024 .
$\therefore \mathrm{Cost} \mathrm{price} \mathrm{of} 1 \mathrm{T}‐\mathrm{shirt}=\frac{₹1024}{4}=₹256$
Therefore, the correct unit price is ₹256.

Answer:

Let quantity of first variety and second variety of tea be 5x kg and 4x kg respectively.
Cost of first variety of tea = ₹(200 × 5x) = ₹1000 x
Cost of second variety of tea = ₹(300 × 4x) = ₹1200x
Total cost price of tea = ₹(1000x + 1200x) = ₹2200x
Selling price of tea = ₹(9x × 275) = ₹2475x
Here, SP > CP
Therefore, there is a profit of = ₹(2475x – 2200x) = ₹275x
$\begin{array}{rcl}\mathrm{Profit}\%& =& \frac{\mathrm{Profit}}{\mathrm{CP}}\times 100\\ & =& \frac{275}{2200\mathrm{x}}\times 100\\ & =& 12.5\%\end{array}$

Answer:

Length of cloth = 5 m
Shrinking percentage after washing = 10%
∴ Length of cloth, after washing
= 5 – 10% of 5
$$\begin{gathered} =5-\frac{10}{100}\times 5 \\ =5-\frac{1}{2} \\ =4.5 \end{gathered}$$

Answer:

Percentage of marks obtained by Nancy $=\frac{426}{600}\times 100=71\%$
Percentage of marks obtained by Rohit$=\frac{560}{800}\times 100=70\%$
So, Nancy’s performance is better.

Answer:

Total donation amount = ₹500000
Interest on donation amount
$=500000\times \frac{12}{100}\times 1=₹60000$
Value of the second and third scholarships = ₹(20000 + 15000) = ₹35000
Now, value of the first scholarship = ₹(60000 – 35000) = ₹25000

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$=\frac{415}{500}\times 100=83\%$
=415500×100=83%

Answer:

SI = ₹ 9600,
Time (T) = 2 years,
Rate (R) = 16% p.a.
We know that,
$$\begin{gathered} \mathrm{SI}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \Rightarrow 9600=\frac{\mathrm{P}\times 16\times 2}{100} \\ \Rightarrow \frac{9600\times 100}{16\times 2}=\mathrm{P} \\ \Rightarrow \mathrm{P}=30000 \end{gathered}$$

Answer:

For Harish: CP = ₹900, Profit % = 5
$\mathrm{SP}=\left(\frac{100+5}{100}\right)\times 900=105\times 9= ₹945$
Now, S.P. for Harish would be the C.P. for Archana.
For Archana : C.P. = ₹ 945, Loss % = 20
$\mathrm{SP}=\left(\frac{100-20}{100}\right)\times 945=\frac{4}{5}\times 945= ₹756$
Again, SP for Archana would be the CP for Babita.
So, Babita got the gas-chullah for ₹756.

Answer:

ans

Answer:

Let the total marks be x,
Marks for accent and presentation $=20\% \mathrm{of} x$
$$\begin{gathered} =\frac{20}{100}\times x \\ =\frac{x}{5} \end{gathered}$$
Remaining marks $=x-\frac{x}{5}=\frac{4x}{5}$
Marks reserved for subject matter $60\% \mathrm{of} \frac{4x}{5}$.
$$\begin{gathered} =\frac{60}{100}\times \frac{4x}{5} \\ =\frac{12}{25}x \end{gathered}$$
 $$\begin{gathered} \mathrm{Marks} \mathrm{for} \mathrm{rebuttal}=\frac{4x}{5}-\frac{12}{25}x \\ =\frac{8}{25}x \end{gathered}$$
So, marks for rebuttal = 8
$$\begin{gathered} \therefore \frac{8}{25}x=8 \\ \Rightarrow x=25 \end{gathered}$$

Answer:

The total amount which has to be divided = ₹10000
Let the first part be = ₹x
Then, second part = ₹(10000 – x)
Interest in first part $=\frac{x\times 12\times 4}{100}=₹\frac{12}{25}x$
Interest in the second part$=\left(10000-x\right)\times \frac{16}{100}\times 4.5$
$$\begin{gathered} =\left(10000-x\right)\times \frac{16}{100}\times 4.5 \\ =₹\left(7200-\frac{18}{25}x\right) \end{gathered}$$

According to the question,

$$\begin{gathered} \frac{12}{25}x=7200-\frac{18}{25}x \\ \Rightarrow \frac{12}{25}x+\frac{18}{25}x=7200 \\ \Rightarrow \frac{30}{25}x=7200 \\ \Rightarrow x=6000 \end{gathered}$$

So, the first part = ₹6000
And second apart = ₹(10000 – 6000) = ₹4000

Answer:

Principal (P) = ₹ 9000,
Amount (A)  = ₹ 18000,
Time (T) = 8 years
∴ Interest (I) = ₹(18000 – 9000) = ₹9000
$$\begin{gathered} \mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \therefore 9000=\frac{9000\times \mathrm{R}\times 8}{100} \\ \Rightarrow \mathrm{R}=\frac{100}{8} \\ \Rightarrow \mathrm{R}=12.5\% \end{gathered}$$
 

Answer:

Let the principal be ₹ P.
Then, simple interest (I) = ₹ 4.05 × P
Rate (R) = 13.5%
$$\begin{gathered} \mathrm{Interest}=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \therefore 4.05\times \mathrm{P}=\frac{\mathrm{P}\times 13.5\times \mathrm{T}}{100} \\ \Rightarrow \mathrm{Time}=\frac{4.05\times 100}{13.5} \\ \Rightarrow \mathrm{Time}=30 \mathrm{years} \end{gathered}$$
 

Answer:

Let the sum be ₹P.
Interest for 8 years at 12% per annum
$$\begin{gathered} \mathrm{Interest} \left(\mathrm{I}\right)=\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{T}}{100} \\ \Rightarrow \mathrm{I}=\frac{\mathrm{P}\times 14\times 5}{100} \\ \Rightarrow \mathrm{I}=\frac{7\mathrm{P}}{10} \end{gathered}$$

According to the question,
$$\begin{gathered} \frac{24}{25}\mathrm{P}=3120+\frac{7}{10}\mathrm{P} \\ \Rightarrow \left(\frac{24}{25}-\frac{7}{10}\right)\mathrm{P}=3120 \\ \Rightarrow \left(\frac{96-70}{100}\right)\mathrm{P}=3120 \\ \Rightarrow \left(\frac{26}{100}\right)\mathrm{P}=3120 \\ \Rightarrow \mathrm{P}=\frac{3120\times 100}{26} \\ \Rightarrow \mathrm{P}=12000 \end{gathered}$$

Thus, the sum is â‚¹12000.