Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 2 - Fractions & Decimals

Answer:

$\begin{array}{rcl}\frac{2}{5}\times 5\frac{1}{5} & =& \frac{2}{5}\times \frac{26}{5}\\ & =& \frac{2\times 26}{5\times 5}\\ & =& \frac{52}{25}\end{array}$

Hence, the correct answer is option (b).

Answer:

$$\begin{gathered} 3\frac{3}{4}÷\frac{3}{4}=\frac{15}{4}\times \frac{4}{3} \\ =5 \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

Total length of ribbon = $5\frac{1}{4}$m
Length of each small piece = $\frac{3}{4}\mathrm{m}$
∴ Number of pieces $=5\frac{1}{4}÷\frac{3}{4}$
$$\begin{gathered} =\frac{21}{4}\times \frac{4}{3} \\ =7 \end{gathered}$$
Thus, number of pieces will be 7.

Hence, the correct answer is option (c).

Answer:

LCM of 3, 7 and 21 is 21.
$$\begin{gathered} \frac{2}{3}=\frac{2\times 7}{3\times 7}=\frac{14}{21}, \\ \frac{6}{7}=\frac{6\times 3}{7\times 3}=\frac{18}{21}, \\ \frac{13}{21} \end{gathered}$$

Now, 
$$\begin{gathered} \because \frac{13}{21}<\frac{14}{21}<\frac{18}{21} \\ \Rightarrow \frac{13}{21}<\frac{2}{3}<\frac{6}{7} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

$\begin{array}{rcl}\mathrm{Reciprocal} \mathrm{of} \frac{2}{3}& =& \frac{1}{\frac{2}{3}}\\ & =& \frac{3}{2}\end{array}$
Hence, the correct answer is option (d).

Answer:

$\begin{array}{rcl}\frac{11}{13}\times 4& =& \frac{11}{13}\times \frac{4}{1}\\ & =& \frac{44}{13}\\ & =& 3\frac{5}{13}\end{array}$
Hence, the correct answer is option (d).

Answer:

$\begin{array}{rcl}3\times 4\frac{2}{5}& =& \frac{3}{1}\times \frac{22}{5}\\ & =& \frac{66}{5}\\ & =& 13\frac{1}{5}\end{array}$
Hence, the correct answer is option (c).

Answer:

$3\times \frac{2}{3} =2$
Thus, the Pictorial representation of 3×$\frac{2}{3} \mathrm{is}$

Hence, the correct answer is option (b).

Answer:

$\begin{array}{rcl}\frac{1}{5}÷\frac{4}{5}& =& \frac{1}{5}\times \frac{5}{4}\\ & =& \frac{1}{4}\end{array}$

Hence, the correct answer is option (d).

Answer:

$\begin{array}{rcl}0.03 \times 0.9& =& \frac{3}{100}\times \frac{9}{10}\\ & =& \frac{27}{1000}\\ & =& 0.027\end{array}$
Hence, the correct answer is option (c).

Answer:

$\begin{array}{rcl}\frac{5}{7}÷6& =& \frac{5}{7}\times \frac{1}{6}\\ & =& \frac{5}{42}\end{array}$
Hence, the correct answer is option (b).

Answer:

$\begin{array}{rcl}5\frac{1}{6}÷\frac{9}{2} & =& 5\frac{1}{6}\times \frac{2}{9}\\ & =& \frac{31}{6}\times \frac{2}{9}\\ & =& \frac{31}{27}\end{array}$

Hence, the correct answer is option (d).

Answer:

$\frac{1}{3}\mathrm{of} \frac{1}{6}=\frac{1}{3}\times \frac{1}{6}$
Hence, the correct answer is option (c).

Answer:

Hence, the correct answer is option (d).

Answer:

Flour required in 1 packet = $2\frac{1}{2}$ cups
Sugar required in 1 packet = â€‹$1\frac{2}{3}$ cups
Total quantity of both ingredients used in 1 packet = quantity of flour + quantity of sugar
                                     
  $$\begin{gathered} =2\frac{1}{2}+1\frac{2}{3} \\ =\frac{5}{2}+\frac{5}{3} \\ =\frac{15+10}{6} \\ =\frac{25}{6} \\ =4\frac{1}{6} \mathrm{cups} \end{gathered}$$

Total quantity of both ingredients used in 10 packet = 10 × Total quantity of both ingredients used in 1 packet 
$$\begin{gathered} =10\times 4\frac{1}{6} \mathrm{cups} \\ =10\times \frac{25}{6} \mathrm{cups} \\ =\frac{125}{3} \mathrm{cups} \\ =41\frac{2}{3} \mathrm{cups} \end{gathered}$$

∵ 40 < $41\frac{2}{3}$< 50
Thus, the estimated total quantity of both ingredients used in 10 such packets of biscuits will be between 40 cups and 50 cups.

Hence, the correct answer is option (c).

Answer:

$\begin{array}{rcl}7\times 6\frac{3}{4}& =& \frac{7}{1}\times \frac{27}{4}\\ & =& \frac{189}{4}\\ & =& 47\frac{1}{4}\end{array}$
Hence, the correct answer is option (b).

Answer:

$\begin{array}{rcl}7÷\frac{2}{5}& =& 7\times \frac{5}{2}\\ & =& \frac{7}{1}\times \frac{5}{2}\\ & =& \frac{35}{2}\\ & & \end{array}$
Hence, the correct answer is option (d).

Answer:

$\begin{array}{rcl}2\frac{2}{3}÷5& =& \frac{8}{3}÷5\\ & =& \frac{8}{3}\times \frac{1}{5}\\ & =& \frac{8}{15}\end{array}$
Hence, the correct answer is option (a).

Answer:

The quantity of apples used on Monday = $\frac{4}{5}$ of 5 kg
                                                                $$\begin{gathered} =\frac{4}{5}\times 5 \mathrm{kg} \\ =4 \mathrm{kg} \end{gathered}$$
The quantity of apples left after Monday = 5 kg − 4 kg = 1 kg
The quantity of apples used on the next day, i.e., Tuesday = $\frac{1}{3}$ of what was left was used
                                                                                              $\begin{array}{rcl}& =& \frac{1}{3}\times 1 \mathrm{kg}\\ & =& \frac{1}{3} \mathrm{kg}\\ & & \end{array}$

The quantity of apples left after Tuesday =$1 \mathrm{kg}-\frac{1}{3} \mathrm{kg}$
$$\begin{gathered} =\frac{3-1}{3}\mathrm{kg} \\ =\frac{2}{3}\mathrm{kg} \end{gathered}$$

Thus, weight (in kg) of apples left now is $\frac{2}{3} \mathrm{kg}$.
Hence, the correct answer is option (c).
 

Answer:

$\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=3\times \frac{1}{4}=\frac{3}{4}$

Hence, the correct answer is option (b).

Answer:

Let the whole part of the cake is 1.

Part of the cake Rani ate = $\frac{2}{7}\times 1=\frac{2}{7}$

∴ Part of the cake left = $1-\frac{2}{7}=\frac{7-2}{7}=\frac{5}{7}$

Now, part of the cake Ravi ate = $\frac{4}{5}\times \frac{5}{7}=\frac{4}{7}$

Thus, part of the cake left is $\frac{4}{7}$.
 

Answer:

Reciprocal of $\frac{3}{7}=\frac{1}{\frac{3}{7}}=\frac{7}{3}$

Answer:

$\begin{array}{rcl}\frac{2}{3} \mathrm{of} 27 & =& \frac{2}{3}\times 27\\ & =& 2\times 9\\ & =& 18\end{array}$
Thus, $\frac{2}{3} \mathrm{of} 27 \mathrm{is} \underline{\mathbf{ }\mathbf{18}\mathbf{ }}$.

Answer:

$\begin{array}{rcl}\frac{4}{5} \mathrm{of} 45 & =& \frac{4}{5} \times 45 \\ & =& 4\times 9\\ & =& 36\end{array}$
Thus, $\frac{4}{5} \mathrm{of} 45 \mathrm{is} \underline{ 36 }.$

Answer:

$\begin{array}{rcl}4\times 6\frac{1}{3} & =& 4\times \frac{19}{3}\\ & =& \frac{76}{3}\\ & =& 25\frac{1}{3}\end{array}$
Thus, $4\times 6\frac{1}{3} \mathrm{is} \mathrm{equal} \mathrm{to} \underline{25\frac{1}{3} }.$

Answer:

$\begin{array}{rcl}\frac{1}{2} \mathrm{of} 4\frac{2}{7}& =& \frac{1}{2}\times \frac{30}{7}\\ & =& \frac{15}{7}\\ & =& 2\frac{1}{7}\end{array}$

Thus, $\frac{1}{2} \mathrm{of} 4\frac{2}{7} \mathrm{is} \underline{ 2\frac{1}{7} }.$

Answer:

$\begin{array}{rcl}\frac{1}{9} \mathrm{of} \frac{6}{5}& =& \frac{1}{9}\times \frac{6}{5}\\ & =& \frac{2}{15}\end{array}$

Thus, $\frac{1}{9} \mathrm{of} \frac{6}{5} \mathrm{is} \underline{ \frac{2}{15} }.$

Answer:

$\begin{array}{rcl}2\frac{3}{7}\times \frac{7}{9}& =& \frac{17}{7}\times \frac{7}{9}\\ & =& \frac{17}{9}\\ & =& 1\frac{8}{9}\end{array}$

Thus, $2\frac{3}{7}\times \frac{7}{9} \mathrm{is} \underline{ 1\frac{8}{9}} .$

Answer:

$\begin{array}{rcl}\frac{4}{5}÷4& =& \frac{4}{5}\times \frac{1}{4}\\ & =& \frac{1}{5}\end{array}$

Thus, $\frac{4}{5}÷4 \mathrm{is} \mathrm{equal} \mathrm{to} \underline{ \frac{1}{ 5 } }.$

Answer:

$\begin{array}{rcl}\frac{2}{5} \mathrm{of} 25 & =& \frac{2}{5}\times 25\\ & =& 10\end{array}$

Thus, $\frac{2}{5} \mathrm{of} 25 \mathrm{is} \mathrm{equals} \mathrm{to} \underline{ 10 }.$

Answer:

$\frac{1}{5}÷\frac{5}{6}=\frac{1}{5} \underline{\mathbf{ }\mathbf{\times }\mathbf{ }}\frac{6}{5}$

Answer:

$3.2\times 10=\frac{32}{10}\times 10=\underline{\mathbf{ }\mathbf{32}\mathbf{ }}$

Answer:

$\begin{array}{rcl}25.4 \times 1000 & =& \frac{254}{10}\times 1000\\ & =& 25400\end{array}$

Thus, 25.4 × 1000 = 25400.
 

Answer:

$93.5 \times 100 =\frac{935}{10}\times 100=9350$

Thus, 93.5 × 100 = 9350

Answer:

$\begin{array}{rcl}4.7 ÷ 10 & =& \frac{47}{10}\times \frac{1}{10}\\ & =& \frac{47}{100}\\ & =& 0.47\end{array}$

Thus, 4.7 ÷ 10 =  0.47  .
 

Answer:

$\begin{array}{rcl}4.7÷100& =& \frac{47}{10}\times \frac{1}{100}\\ & =& \frac{47}{1000}\\ & =& 0.047\end{array}$

Thus, 4.7 ÷ 100 =  0.047 .

Answer:

$\begin{array}{rcl}4.7÷1000& =& \frac{47}{10}\times \frac{1}{1000}\\ & =& \frac{47}{10000}\\ & =& 0.0047\end{array}$

Thus, 4.7 ÷ 100 =  0.0047 .

Answer:

The product of two proper fractions is smaller than each of the fractions that are multiplied.

Answer:

While dividing a fraction by another fraction, we multiply the first fraction by the reciprocal of the other fraction.

Answer:

8.4 ÷ ______ = 2.1
Let x be the required number.

$$\begin{gathered} 8.4 ÷ x= 2.1 \\ \Rightarrow \frac{84}{10}\times \frac{1}{x} = \frac{21}{10} \\ \Rightarrow \frac{84}{10}\times \frac{10}{21}= x \\ \Rightarrow 4= x \end{gathered}$$

Thus, 8.4 ÷ 4 = 2.1 

 

Answer:

52.7 ÷ _______ = 0.527

Let x be the required number.

$$\begin{gathered} \Rightarrow 52.7 ÷x=0.527 \\ \Rightarrow 52.7 \times \frac{1}{x}=0.527 \\ \Rightarrow \frac{527 }{10}÷\frac{527}{1000}=x \\ \Rightarrow \frac{527 }{10}\times \frac{1000}{527}=x \\ \Rightarrow 100=x \end{gathered}$$

Thus, 52.7 ÷ 100 = 0.527

Answer:

$\begin{array}{rcl}0.5 \times 0.7 & =& \frac{5}{10}\times \frac{7}{10}\\ & =& \frac{35}{100}\\ & =& 0.35\end{array}$

Thus, 0.5 × 0.7 = 0.35

Answer:

$\begin{array}{rcl}2\times \frac{5}{3}& =& \frac{2}{1}\times \frac{5}{3}\\ & =& \frac{10}{3}\end{array}$

Thus, 2 × $\frac{5}{3}=\frac{10}{3}$

Answer:

$\begin{array}{rcl}2.001 ÷ 0.003 & =& \frac{2001}{1000}÷\frac{3}{1000}\\ & =& \frac{2001}{1000}\times \frac{1000}{3}\\ & =& \frac{2001}{3}\\ & =& 667\end{array}$

Thus, 2.001 ÷ 0.003 = 667

Answer:

Let $\frac{a}{b} \mathrm{be} \mathrm{a} \mathrm{proper} \mathrm{fraction} \mathrm{such} \mathrm{that} a<b.$
$\begin{array}{rcl}\mathrm{Reciprocal} \mathrm{of} \frac{a}{b}& =& \frac{1}{\frac{a}{b}}\\ & =& \frac{b}{a}\end{array}$
Since, 
$$\begin{gathered} b>a\Rightarrow \frac{b}{a}>1 \\ \Rightarrow \frac{b}{a} \mathrm{is} \mathrm{an} \mathrm{improper} \mathrm{fraction} \end{gathered}$$

Thus, the reciprocal of a proper fraction is a improper fraction.
So, the statement is False.


 

Answer:

Let $\frac{a}{b} \mathrm{be} \mathrm{an} \mathrm{improper} \mathrm{fraction} \mathrm{such} \mathrm{that} a>b.$
$\begin{array}{rcl}\mathrm{Reciprocal} \mathrm{of} \frac{a}{b}& =& \frac{1}{\frac{a}{b}}\\ & =& \frac{b}{a}\end{array}$
Since, 
$$\begin{gathered} b<a\Rightarrow \frac{b}{a}<1 \\ \Rightarrow \frac{b}{a} \mathrm{is} a \mathrm{proper} \mathrm{fraction} \end{gathered}$$

Thus, the reciprocal of an improper fraction is a proper fraction.
So, the statement is False.

 

Answer:

Let the two fractions be $\frac{a}{b} \mathrm{and}\frac{c}{d}$.
So, $\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}$

Thus, Product of two fractions = $\frac{\mathrm{Product} \mathrm{of} \mathrm{their} \mathrm{numerators}}{\mathrm{Product} \mathrm{of} \mathrm{their} \mathrm{denominators}}$
So, the statement is False.

Answer:

Let the two improper fractions be $\frac{a}{b} \mathrm{and}\frac{c}{d}$ where a < b and c < d.
Since, a < b and c < d
⇒ a × c < b × d
$\Rightarrow \frac{ac}{bd}<1$

As improper fractions are greater than 1.
Therefore, $\frac{ac}{bd}<1<\frac{a}{b} \mathrm{and} \frac{ac}{bd}<1<\frac{c}{d}$.

Thus, the product of two improper fractions is less than both fractions.

So, the statement is False.

Answer:

Let a fraction be $\frac{a}{b}$.
The reciprocal of the fraction $\begin{array}{rcl}\frac{a}{b}& =& \frac{1}{\frac{a}{b}}=\frac{b}{a}\end{array}$.
Thus, the reciprocal of a fraction is obtained by inverting it upside down.
So, the statement is True.
 

Answer:

For example, consider the decimal number 8.9765.
On multiplying 8.9765 by 1000
8.9765 × 1000 = 8976.5
Thus, to multiply a decimal number by 1000, we have moved the decimal point in the number to the right by three places.

So, the statement is True.


 

Answer:

For example, consider the decimal number 456.7.
On dividing 456.7 by 100
456.7 ÷ 100 = 4.567
Thus, to divide a decimal number by 100, we move the decimal point in the number to the left by two places.
So, the statement is True.

Answer:

True, 1 is the only number which is it's own reciprocal.

Answer:

$\begin{array}{rcl}\frac{2}{3}\mathrm{of} 8& =& \frac{2}{3}\times \frac{8}{1}\\ & =& \frac{16}{3}\end{array}$

$\begin{array}{rcl}\frac{2}{3}÷8& =& \frac{2}{3}\times \frac{1}{8}\\ & =& \frac{1}{4}\end{array}$
$\frac{16}{3}\ne \frac{1}{4}$
∴ $\frac{2}{3}$ of 8 is not same as $\frac{2}{3}÷8.$

So, the statement is False.

Answer:

The reciprocal of $\frac{4}{7 }=\frac{1}{{\displaystyle \frac{4}{7}}}=\frac{7}{4}.$

Thus, $\frac{4}{7 }$ is not reciprocal of $\frac{4}{7 }$.

Answer:

Adding 5 to the numerator and denominator of the fraction $\frac{5}{9}$.
$\frac{5+5}{9+5}=\frac{10}{14}=\frac{5}{7}$
Yes, the value of the fraction changes.
$\frac{5}{9}<\frac{5}{7}$
Thus, the value increases.

Answer:

The value of a fraction will increase if the denominator of the fraction is decreased while the numerator is kept unchanged.

Answer:

Letter among A and J are A, B, C, D, E, F, G, H, I, J.
Ttotal letters among A and J =10
$\frac{2}{5}\times 10=4$
∴ 4th letter among A and J = D

Answer:

Let the number be x.
$$\begin{gathered} \frac{2}{3} \mathrm{of} x =10 \\ \Rightarrow x=10\times \frac{3}{2} \\ \Rightarrow x=15 \end{gathered}$$
1.75 times of the number 15 =1.75 × 15
= 26.25

Thus, 1.75 times of that number is 26.25.

Answer:

Total number of students in the class = 40
Fraction of students who like to eat rice only = $\frac{1}{5}$
Number of students who like to eat rice only = $\frac{1}{5}\times 40=8$
Fraction of students who like to eat chapati only = $\frac{2}{5}$
Number of students who like to eat chapati only = $\frac{2}{5}\times 40=16$
The total number of students like to eat both = 40 − (8 + 16)
= 40 − 24
= 16

Thus, the total number of students who like to eat both are 16. 
 

Answer:

Part of her homework in 2 hours = $\frac{2}{3}$
Part of her homework in 1 hours = $\frac{2}{3}÷2=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3}$
Part of her homework in $1\frac{1}{4}$ hours  = $\frac{1}{3}\times 1\frac{1}{4}=\frac{1}{3}\times \frac{5}{4}=\frac{5}{12}$

Thus, the part of her homework she had completed in $1\frac{1}{4}$ hours is $\frac{5}{12}$.

Answer:

Let total number of pages be x.
Number of pages read by Reemu =15x
According to question,
$$\begin{gathered} \frac{1}{5}x+40=\frac{7}{10}x \\ \Rightarrow \frac{7}{10}x-\frac{1}{5}x=40 \\ \Rightarrow \left(\frac{7-2}{10}\right)x=40 \\ \Rightarrow \left(\frac{5}{10}\right)x=40 \\ \Rightarrow \frac{1}{2}x=40 \\ \Rightarrow x=80 \end{gathered}$$
Number of pages read by Reemu = $\frac{7}{10}\times 80=56$

Number of pages left to be read by Reemu = $80-56=24$

Answer:

Let the number in the box ☐ be x.
$$\begin{gathered} \frac{3}{7}\times x=\frac{15}{98} \\ \Rightarrow x=\frac{15}{98}\times \frac{7}{3} \\ \Rightarrow x=\frac{5}{14} \end{gathered}$$

Answer:

$$\begin{gathered} 7\frac{1}{6}÷3\frac{2}{3} \\ =\frac{43}{6}÷\frac{11}{3} \\ =\frac{43}{6}\times \frac{3}{11} \\ =\frac{43}{22} \\ =1.95 \end{gathered}$$

1.95 > 1.5

Thus, the quotient $7\frac{1}{6}÷3\frac{2}{3}$ of the given fractions is greater than 1.5.

 

Answer:

Method I: Conversion of both numbers $\frac{13}{17}$ and 0.82 into decimals.
$\frac{13}{17}$ = 0.76 and 0.82 into decimals.

Method II: Conversion of both numbers $\frac{13}{17}$ and 0.82 into fractions.
$0.82=\frac{82}{100}=\frac{41}{50}$ and $\frac{13}{17}$
 
∴ Method of converting numbers into decimals is easier because it can easily be compared in this method but in other method it is also required to convert the fractions into like fractions to compare the numbers.

Answer:

(a) Each tablet weighs $\frac{4}{25}$ gram
∴Weight of 4 tablets = $\frac{4}{25}\times 4 \mathrm{gram}=\frac{16}{25} \mathrm{gram}$

(b) An adult weighing 46 kg takes only $2\frac{1}{2}$ tablets.
∴ Weight of $2\frac{1}{2}$ tablets =  $\frac{4}{25}\times \frac{5}{2}=\frac{2}{5} \mathrm{grams}$
=425×52gram=25gram

Answer:

(a) Total dosage given to a 18 kg adult dog = $\left(\frac{1}{2}+\frac{1}{2}\right)$ tsp = 1tsp
(b) Total dosage given to a 6 kg cat$=6\times \frac{1}{4}=\frac{3}{2}=1\frac{1}{2}$ tsp

(c) Total dosage given to a 18 kg pregnant dog $=\frac{1}{2}\times \frac{1}{4.5}\times 18=\frac{1}{9}\times 18=2$ tsp
 

Answer:

$\frac{1}{16}$ kg chocolates to be filled in 1 box.
∴ 1 kg chocolates to be filled in $\frac{1}{16}=1÷\frac{1}{16}=1\times \frac{16}{1}=16 \mathrm{boxes}$ 

Now, $1\frac{1}{2}$ kg chocolates to be filled in $16\times 1\frac{1}{2}=16\times \frac{3}{2}=24$ boxes
 

Answer:

Length of one bookmarker = $10\frac{1}{2} \mathrm{cm}=\frac{21}{2}\mathrm{cm}$

∴ Number of bookmarkers Anvi can make from 15 m ix., 15 × 100 cm long ribbon
$$\begin{gathered} =\left(\frac{1}{{\displaystyle \frac{21}{2}}}\times 15\times 100\right) \\ =\left(\frac{2}{21}\times 15\times 100\right) \\ =\frac{1000}{7} \\ =142.85 \end{gathered}$$
=142
Thus, Anvi can make 142 bookmaker from a 15 m long ribbon

Answer:

(a) Length of a side of the square = 83 cm
∴ Length of the diagonal of the square = 1.414 × 8.3 cm
= 11.7362 cm
≈ 11.74 cm

(b) Length of a side of the square = 7.875 cm
Length of the diagonal of the square = 1.414 × 7.875 cm
= 11.13525 cm
≈ 11.14 cm

Answer:

(a) Diameter of the circle = 14.35 cm 
∴ Length of the side of the square = 0.707 × 14.35 cm
= 10.14545
= 10.15 cm

(b) Diameter of the circle = 8.63 cm
∴ Length of the side of the square = 0.707 × 8.63 cm
= 6.10141 cm
= 6.10 cm

Answer:

(a) Diameter of the disc = 18.7 cm
∴ The distance around the disc = 3.14 × 18.7 cm
= 58.718 cm

(b) Radius of the disc = 6.45 cm
∴ Diameter of the disc = 2 × 6.45 cm = 12.9 cm
∴ The distance around the disc = 3.14 × 12.9 cm
= 40.506 cm

Answer:

Cost of 1 metre of cloth = ₹53.50
∴ Cost of 27.5 metres of cloth = ₹ (53.50 × 27.5) = ₹1471.25

Answer:

(a) When Nidhi is $\frac{4}{6}$of the way through the race, she will be at hurdle D.
(b) When Nidhi is $\frac{5}{6}$of the way through the race, she will be at hurdle E.
(c) When Nidhi is over hurdle C, she would finish the $\frac{3}{6}=\frac{1}{2}$ or middle part of the race.

Answer:

Diameter of Earth = 12756000 m
Diameter of a new planet = $\frac{5}{86}\times 12756000 \mathrm{m}$
= 741627.90 m

= $\frac{741627.90 }{1000}\mathrm{km}$ 

= 741.6 km

Thus, the diameter of this planet is 741.6 km.

Answer:

Reciprocal of $\frac{5}{129}=\frac{1}{\frac{5}{129}}=\frac{129}{5}$
The product of $\frac{5}{129}$ and its reciprocal = $\frac{5}{129}\times \frac{129}{5}=1$
Thus, the product of $\frac{5}{129}$ and its reciprocal is 1.
 

Answer:

$\begin{array}{rcl}\frac{2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}}{2\frac{1}{2}÷\frac{1}{5}}& =& \frac{{\displaystyle \frac{5}{2}}+{\displaystyle \frac{1}{5}}}{\frac{5}{2}÷\frac{1}{5}}\\ & =& \frac{{\displaystyle \frac{25+2}{10}}}{\frac{5}{2}\times \frac{5}{1}}\\ & =& \frac{{\displaystyle \frac{27}{10}}}{\frac{25}{2}}\\ & =& \frac{27}{10}\times \frac{2}{25}\\ & =& \frac{27}{125}\end{array}$

Thus, $\frac{2{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}}{2\frac{1}{2}÷\frac{1}{5}}=\begin{array}{rcl}& & \frac{27}{125}\end{array}$ 

Answer:

$\begin{array}{rcl}\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{3}{8}}\times {\displaystyle \frac{3}{5}}}& =& \frac{{\displaystyle \frac{5}{20}+\frac{4}{20}}}{1-{\displaystyle \frac{9}{40}}}\\ & =& \frac{{\displaystyle \frac{9}{20}}}{\frac{40-9}{40}}\\ & =& \frac{{\displaystyle \frac{9}{20}}}{\frac{31}{40}}\\ & =& \frac{9}{20}\times \frac{40}{31}\\ & =& \frac{18}{31}\end{array}$
Thus, $\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{3}{8}}\times {\displaystyle \frac{3}{5}}}=\frac{18}{31}$

Answer:

$\begin{array}{rcl}\frac{3}{10} \mathrm{by} \left(\frac{1}{4} \mathrm{of} \frac{3}{5}\right)& =& \frac{3}{10}÷ \left(\frac{1}{4} \times \frac{3}{5}\right)\\ & =& \frac{3}{10}÷ \left(\frac{3}{20}\right)\\ & =& \frac{3}{10}\times \frac{20}{3}\\ & =& 2\end{array}$
Thus, $\frac{3}{10} \mathrm{by} \left(\frac{1}{4} \mathrm{of} \frac{3}{5}\right)=2$

Answer:

Let the number be x.
According to the question,
$$\begin{gathered} \frac{1}{8} \text{o}\mathrm{f} x=\frac{2}{5}÷\frac{1}{20} \\ \Rightarrow \frac{1}{8} \times x=\frac{2}{5}\times \frac{20}{1} \\ \Rightarrow x=\frac{2}{5}\times \frac{20}{1}\times \frac{8}{1} \\ \Rightarrow x=2\times 4\times 8 \\ x=64 \end{gathered}$$
Thus, the number is x.

Answer:

The amount of money Heena’s father has = ₹500
He paid an electric bill of ₹385.70
∴ The amount of money he should have received = ₹(500 − 385.70) = ₹114.30
Thus, the change he should have received is â‚¹114.30.
 

Answer:

The normal body temperature = 98.6°F
Savitri’s temperature, when she was ill = 103.1°F
∴ Savitri’s temperature above normal = 103.1°F – 98.6°F
= 4.5°F

Answer:

(a) –0.17°C < –0.10°C < 0.02°C < 0.10°C < 0.54°C
–0.17°C < –0.10°C < ${\left(\frac{1}{50}\right)}^{0^{}}\mathrm{C}$ < 0.10°C < 0.54°C
The order of five years from coldest to warmest is 1964, 1965, 1978, 1958, 2002.

(b) In 1946, the average temperature varied by –0.03°C.
Since, –0.10°C< –0.03°C < ${\left(\frac{1}{50}\right)}^{0^{}}\mathrm{C}$

∴ 1946 should fall between 1965 and 1978.

Answer:

The atomic weight of Helium = 4.0030
The atomic weight of Hydrogen = 1.0080
The atomic weight of Oxygen – 16.0000
(a) Difference between the atomic weights of Oxygen and Hydrogen = 16.0000 – 1.0080
= 14.9920
(b) Difference between atomic weights of Oxygen and Helium = 16.0000 – 4.0030 = 11.9970
(c) Difference between atomic weights of Helium and Hydrogen = 4.0030 – 1.0080 = 2.9950

Answer:

The correct length of the iron rod = 19.33 cm
The length of the rod measured by Ravi = 19.34 cm
∴ The error made by Ravi = (19.34 − 19.33) cm = +0.01 cm
The length of the rod measured by Kamal = 19.25 cm
∴ The error made by Kamal = (19.25 − 19.33) cm = −0.08 cm
The length of the rod measured by Tabish = 19.27 cm
∴ The error made by Tabish = (19.27 – 19.33) cm = −0.06 cm

Answer:

$\begin{array}{rcl}0.02964÷0.004& =& \frac{2964}{100000}÷\frac{4}{1000}\\ & =& \frac{2964}{100000}\times \frac{1000}{4}\\ & =& \frac{741}{100}\\ & =& 7.41\end{array}$
Thus, when 0.02964 is divided by 0.004, the quotient will be 7.41.

Answer:

Let the number be x.
x divided by 520 = 85 divided by 0.625
⇒ x ÷ 520 = 85 ÷ 0.625
$$\begin{gathered} \Rightarrow \frac{x}{520}= \frac{85}{0.625} \\ \Rightarrow \frac{x}{520}=136 \\ \Rightarrow x=136\times 520 \\ \Rightarrow x=70720 \end{gathered}$$

Thus, the required number is 70720.

Answer:

Length of the floor = 4.5 m
= (4.5 × 100) cm
= 450 cm

Width of the floor = 3.6 m = (3.6 × 100) cm = 360 cm
∴ Area of the floor = (450 × 360) cm² = 162000 cm²

Side of a square tile = 6 cm
∴ Area of one square tile = (6 × 6) cm² = 36 cm²
Number of tiles required to cover the floor = Area of floor ÷  Area of one tile =162000 cm² ÷ 36 cm²
= 4500

Cost of 1 tile = ₹23.25
∴Cost of 4500 tiles = ₹(23.25 × 4500)
= ₹104625

Answer:

Length of cloth Sunita has = $\frac{3}{4}$ m
∴ Length of cloth Rehana has $\frac{1}{3} \mathrm{of} \frac{3}{4}$ m
$$\begin{gathered} =\frac{1}{3}\times \frac{3}{4} \\ =\frac{1}{4}\mathrm{m} \end{gathered}$$

Thus, Rehana has $\frac{1}{4}$ m of cloth.

Answer:

Length of the garden = 22.50 m
Length of one brick = 0.25 m
∴ Number of bricks required to make the border =$\frac{ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{garden} }{\mathrm{Length} \mathrm{of} \mathrm{one} \mathrm{brick}}$
$$\begin{gathered} =\frac{22.50 \mathrm{m}}{0.25 \mathrm{m}} \\ =\frac{2250}{25} \\ =90 \end{gathered}$$

Thus, 90 bricks will be needed.

Answer:

Length of cloth required for one shirt = $2\frac{1}{4}$m + $\frac{1}{8}$m
$$\begin{gathered} =\frac{9}{4}+\frac{1}{8} \\ =\frac{18}{8}+\frac{1}{8} \\ =\frac{19}{8} \\ =2\frac{3}{8}\mathrm{m} \end{gathered}$$

∴ Length of cloth required for 6 shirts$= 6\times 2\frac{3}{8}\mathrm{m}$

$$\begin{gathered} =\frac{6}{1}\times \frac{19}{8} \\ =\frac{57}{4} \\ =14\frac{1}{4}\mathrm{m} \end{gathered}$$

Thus, cloth used in making 6 shirts will be $14\frac{1}{4}\mathrm{m}$.

Answer:

Total number of seats = 820
Sale of total tickets = 648

Number of sold tickets guessed by the first usher = $\frac{3}{4}$ of 820
$$\begin{gathered} =\frac{3}{4}\times 820 \\ =615 \end{gathered}$$

Number of sold tickets guessed by the second usher = $\frac{2}{3}$ of 820
$$\begin{gathered} =\frac{2}{3}\times 820 \\ =546.67 \\ \approx 547 \end{gathered}$$

Since, 615 is more close to 648 than 547.
Thus, the first usher made the better guess.

Answer:

Total spending amount = ₹740.25 + ₹70
= ₹810.25
Total number of students who contributed money = ₹35

Contribution of each student = $=\frac{₹810.25}{35}$
= ₹23.14

Thus, the equal amount contributed by each student is â‚¹23.14.

Answer:

Total time taken by Rohan in five races = (3.20 + 3.37 + 3.29 + 3.17 + 3.32) minutes
= 16.35 minutes

Total number of races = 5

∴ The average time taken by Rohan = $\frac{16.35}{5}$ minutes = 3.27 minutes

Answer:

Total length of a sewer line = $80\frac{1}{4}$m
$=\frac{321}{4}$m
7.5 m long sewer line completed in 1 day.

$80\frac{1}{4}$m long sewer line completed in $\frac{1}{7.5}\times \frac{321}{4}$ days
$$\begin{gathered} =\frac{10}{75}\times \frac{321}{4} \\ =\frac{1}{5}\times \frac{107}{2} \\ =\frac{107}{10} \\ =10.7 \\ \approx 11 \mathrm{days} \end{gathered}$$

Thus, the project will get complete in 3 days.

Answer:

Weight of the object on Earth = $5\frac{3}{5}$ kg
 $=\frac{28}{5}\mathrm{kg}$
Weight of an object on moon = $\frac{1}{6}$ × weight on Earth
$$\begin{gathered} =\frac{1}{6}\times \frac{28}{5}\mathrm{kg} \\ =\frac{14}{15}\mathrm{kg} \end{gathered}$$=

Answer:

Total number of students = 200

(a) Number of students who were influenced by radio$=\frac{9}{20}\times 200=90$

(b) Number of students who were influenced by music video channel$=\frac{2}{25}\times 200=16$

Number of students who were influenced by radio = 90

∴ Number of students that were influenced more by radio than by a music video channel = 90 − 16 = 74

(c) Numberofstudents who were influenced by friend or relative $=\frac{3}{20}\times 200=30$

Number of students who were influenced by hearing or seeing the CD in a shop$=\frac{1}{10}\times 200=20$

∴ Total number of students who were influenced by both friend or relative and hearing the CD in a shop = 30 + 20 = 50

Answer:

The can of milk filled by the milkman in the morning = $5\frac{1}{2}\mathrm{L}=\frac{11}{2}\mathrm{L}$ 
Milk sold by him to Renu = $\frac{3}{4}$ L
Milk sold by him to Kamla = $\frac{3}{4}$ L
Milk sold by him to Renuka = $\frac{3}{4}$ L
Milk sold by him to Shadma =  $\frac{7}{8}$ L
Milk sold by him to Jassi = $1\frac{1}{2}\mathrm{L}=\frac{3}{2}\mathrm{L}$
Total milk sold by him = $\left(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{7}{8}+\frac{3}{2}\right)\mathrm{L}$

                                     $$\begin{gathered} =\left(\frac{9}{4}+\frac{7}{8}+\frac{3}{2}\right)\mathrm{L} \\ =\left(\frac{18}{8}+\frac{7}{8}+\frac{12}{8}\right)\mathrm{L} \\ =\left(\frac{18+7+12}{8}\right)\mathrm{L} \\ =\frac{37}{8}\mathrm{L} \end{gathered}$$ 

Milk left in the can = The can of milk filled by the milkman in the morning − Total milk sold by him
                               $$\begin{gathered} =\left(\frac{11}{2}-\frac{37}{8}\right)\mathrm{L} \\ =\left(\frac{44}{8}-\frac{37}{8}\right)\mathrm{L} \\ =\left(\frac{44-37}{8}\right)\mathrm{L} \\ =\left(\frac{7}{8}\right)\mathrm{L} \end{gathered}$$ 

Thus, milk left in the can is $\frac{7}{8}\mathrm{L}$
  

Answer:

Let the whole work be represented by x.
The part of work done by Anuradha in 6 hours = x
∴ The part of work done by her in 1 hour = $\frac{x}{6}$
The part of work done by her in 5 hours = $\frac{x}{6}$× 5 = $\frac{5x}{6}$
The part of work done by her in 6 hours = $\frac{x}{6}\times 6=x$
Thus, the work can she do  ${\frac{1}{6}}^{\mathrm{th}}$ part of the work in 1 hour, ${\frac{5}{6}}^{\mathrm{th}}$ part of the work in 5 hours and complete work in 6 hours.

Answer:

Let the total portion of the saree be represented by 1.
The portion of saree painted by Rehana in 5 hours = 1
The portion of saree painted by her in 1 hour = 

Since, $4\frac{3}{5} \mathrm{hours}=\frac{23}{5} \mathrm{hours}$
The portion of saree painted by her in $4\frac{3}{5}$ hours$=\frac{1}{5}\times 4\frac{3}{5}$
                                                                             $$\begin{gathered} =\frac{1}{5}\times \frac{23}{5} \\ =\frac{23}{25} \end{gathered}$$
Also, $3\frac{1}{2}\mathrm{hours}=\frac{7}{2}\mathrm{hours}$
The portion of saree painted by her in $3\frac{1}{2}$ hours$=\frac{1}{5}\times 3\frac{1}{2}$
                                                                              $$\begin{gathered} =\frac{1}{5}\times \frac{7}{2} \\ =\frac{7}{10} \end{gathered}$$





 

Answer:

Quantity of cotton wool Rama has = $6\frac{1}{4}\mathrm{kg}=\frac{25}{4}\mathrm{kg}$
Quantity of cotton wool one pillow takes = $1\frac{1}{4}\mathrm{kg}=\frac{5}{4}\mathrm{kg}$
Number of pillows can she make $=6\frac{1}{4}\mathrm{kg}÷1\frac{1}{4}\mathrm{kg}$
$$\begin{gathered} =\frac{25}{4}\mathrm{kg}÷\frac{5}{4}\mathrm{kg} \\ =\frac{25}{4}\times \frac{4}{5} \\ =5 \end{gathered}$$
Thus, she cam make 5 such pillows.
 

Answer:

Length of cloth required to make a shirt = $2\frac{1}{3}\mathrm{m}=\frac{7}{3}\mathrm{m}$
Length of piece of cloth Radhika has = $9\frac{1}{3}\mathrm{m}=\frac{28}{3}\mathrm{m}$
Number of shirts she can make$=9\frac{1}{3}\mathrm{m}÷2\frac{1}{3}\mathrm{m}$
                                                  $$\begin{gathered} =\left(\frac{28}{3}÷\frac{7}{3}\right) \\ =\left(\frac{28}{3}\times \frac{3}{7}\right) \\ =4 \end{gathered}$$

Thus, she can make 4 such shirts.

Answer:

Time taken by Ravi to walk $3\frac{1}{3}$km = 1 hour 
Time taken by Ravi to walk 1 km = $1÷3\frac{1}{3}$
                                                      $$\begin{gathered} =1÷\frac{10}{3} \\ =1\times \frac{3}{10} \\ =\frac{3}{10}\mathrm{hours} \end{gathered}$$
Time taken by Ravi to walk 10 km = $\frac{3}{10}\times 10 \mathrm{hours}$
                                                         = 3 hours

Thus, time taken by him to walk to his office which is 10 km from his home is 3 hours.

 

Answer:

Distance travelled by Raj with ${\frac{3}{5}}^{\mathrm{th}}$petrol tank = 360 km
Distance travelled by him with a full petrol tank = 360 km ÷ $\frac{3}{5}$
                                                                             $$\begin{gathered} =360\times \frac{5}{3} \\ =120\times 5 \\ =600 \mathrm{km} \end{gathered}$$

Thus, at the same rate with a full tank of petrol he can travel 600 km.

Answer:

Let the amount earned by Kajol be â‚¹x. 
Amount she has = â‚¹75
According to the question
$$\begin{gathered} \frac{3}{8}\mathrm{of} x=75 \\ \Rightarrow \frac{3}{8}\times x=75 \\ \Rightarrow x=75\times \frac{8}{3} \\ \Rightarrow x=25\times 8 \\ \Rightarrow x=200 \end{gathered}$$

Thus, she earns â‚¹200.

Answer:

(i) Number of trees required to make one tonne of paper = 17
Total number of trees = 221
∴ Fraction of forest will be used to make one tonne of paper =  = 
(a) Fraction of forest will be used to make 5 tonnes of paper =$\frac{1}{13}\times 5=\frac{5}{13}$
(b) Fraction of forest will be used to make 10 tonnes of paper = $\frac{1}{13}\times 10=\frac{10}{13}$
(ii) Let we have to save x tonnes of paper to save $\frac{7}{13}$ part of the forest
Fraction of forest will be used to make x tonnes of paper = $\frac{1}{13}\times x=\frac{x}{13}$

Here, $$\begin{gathered} \frac{x}{13}=\frac{7}{13} \\ \Rightarrow x=7 \end{gathered}$$
 

Answer:

$$\begin{gathered} \left(1÷\frac{2}{9}\right)+\left(1÷3\frac{1}{5}\right)+\left(1÷2\frac{2}{3}\right)=\left(1\times \frac{9}{2}\right)+\left(1÷\frac{16}{5}\right)+\left(1÷\frac{8}{3}\right) \\ =\left(1\times \frac{9}{2}\right)+\left(1\times \frac{5}{16}\right)+\left(1\times \frac{3}{8}\right) \\ =\left(\frac{9}{2}\right)+\left(\frac{5}{16}\right)+\left(\frac{3}{8}\right) \\ =\left(\frac{72}{16}\right)+\left(\frac{5}{16}\right)+\left(\frac{6}{16}\right) \\ =\frac{72+5+6}{16} \\ =\frac{83}{16} \\ =5.1875 \end{gathered}$$

Answer:

$\left(1\right) 2\times \frac{1}{4}$ represents the addition of 2 figures, each representing 1 shaded part out of 4 equal parts. Thus, (1) is represented by (d)
$\left(2\right) 2\times \frac{3}{7}$represents the addition of 2 figures, each representing 3 shaded parts out of 7 equal parts. Thus, (2) is represented by (f).$\left(3\right) 2\times \frac{1}{3}$ represents the addition of 2 figures, each representing 1 shaded part out of 3 equal parts. Thus, (3) is represented by (c).
$\left(4\right) \frac{1}{4}\times 4$represents the addition of 4 figures, each representing 1 shaded part out of 4 equal parts. Thus, (4) is represented by (b).
$\left(5\right) 3\times \frac{2}{9}$ represents the addition of 3 figures, each representing 2 shaded parts out of 9 equal parts. Thus, (5) is represented by (a).
 $\left(5\right)\frac{1}{4}\times 3$represents the addition of 3 figures, each representing 1 shaded part out of 3 equal parts. Thus, (6) is represented by (e).

Answer:

(0.3) × (0.3) – (0.2) × (0.2)
$$\begin{gathered} =\frac{3}{10}\times \frac{3}{10}-\frac{2}{10}\times \frac{2}{10} \\ =\frac{9}{100}-\frac{4}{100} \\ =\frac{5}{100} \\ =0.05 \end{gathered}$$

Thus, (0.3) × (0.3) – (0.2) × (0.2) = 0.05

Answer:

$$\begin{gathered} \frac{0.6}{0.3}+\frac{0.16}{0.4}=\frac{6}{3}+\frac{16}{40} \\ =\frac{2}{1}+\frac{2}{5} \\ =\frac{10+2}{5} \\ =\frac{12}{5} \\ =2.4 \end{gathered}$$
Thus, $\frac{0.6}{0.3}+\frac{0.16}{0.4}$ = 2.4
 

Answer:

$\begin{array}{rcl}\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}& =& \frac{\left({\displaystyle \frac{2}{10}}\times {\displaystyle \frac{14}{100}}\right)+\left({\displaystyle \frac{5}{10}}+{\displaystyle \frac{91}{100}}\right)}{\left(\frac{1}{10}\times \frac{2}{10}\right)}\\ & =& \frac{\left(\frac{28}{1000}\right)+\left({\displaystyle \frac{455}{1000}}\right)}{{\displaystyle \frac{2}{100}}}\\ & =& \frac{\frac{483}{1000}}{\frac{2}{100}}\\ & =& \frac{483}{1000}\times \frac{100}{2}\\ & =& \frac{241.5}{10}\\ & =& 24.15\\ & & \end{array}$

Thus, $\frac{\left(0.2\times 0.14\right)+\left(0.5\times 0.91\right)}{\left(0.1\times 0.2\right)}$= 24.15

Answer:

Side of the equilateral triangle = $\frac{4}{3}$cm
∴ Side of the square = $\frac{4}{3}$cm

∴ Perimeter of the figure formed = $\left(\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}+\frac{4}{3}\right)$ cm
$$\begin{gathered} =\left(\frac{20}{3}\right) \mathrm{cm} \\ =6\frac{2}{3}\mathrm{cm} \end{gathered}$$

Thus,  the perimeter of figure formed is $6\frac{2}{3}\mathrm{cm}$.
 

Answer:

 Area of the carpet bought by Rita = 4 m × $6\frac{2}{3}$m
                                                      $$\begin{gathered} =\frac{4}{1}\times \frac{20}{3}{\mathrm{m}}^2 \\ =\frac{80}{3}{\mathrm{m}}^2 \end{gathered}$$
Area of the room = $3\frac{1}{3}$m ×  $5\frac{1}{3}$m
                           $$\begin{gathered} =\frac{10}{3}\times \frac{16}{3}{\mathrm{m}}^2 \\ =\frac{160}{9}{\mathrm{m}}^2 \end{gathered}$$

Area to be cut off to fit wall to wall carpet into the room = $\left(\frac{80}{3}-\frac{160}{9}\right){\mathrm{m}}^2$
                                                                                          $$\begin{gathered} =\left(\frac{240}{9}-\frac{160}{9}\right){\mathrm{m}}^2 \\ =\left(\frac{240-160}{9}\right){\mathrm{m}}^2 \\ =\frac{80}{9}{\mathrm{m}}^2 \\ =8\frac{8}{9}{\mathrm{m}}^2 \end{gathered}$$

Thus, area to be cut off to fit wall to wall carpet into the room is $8\frac{8}{9}{\mathrm{m}}^2$.

Answer:

Length of the photograph = $14\frac{2}{5}$cm $=\frac{72}{5}\mathrm{cm}$
Breadth of the photograph = $10\frac{2}{5}$cm$=\frac{52}{5}\mathrm{cm}$
Width of the boarded = $2\frac{3}{5}$ cm = $\frac{13}{5}\mathrm{cm}$
Length of the framed photograph $=\left(\frac{72}{5}+\frac{13}{5}+\frac{13}{5}\right)\mathrm{cm}=\left(\frac{98}{5}\right)\mathrm{cm}$
Breadth of the framed photograph $=\left(\frac{52}{5}+\frac{13}{5}+\frac{13}{5}\right)\mathrm{cm}=\left(\frac{78}{5}\right)\mathrm{cm}$

Area of the framed photograph $=\frac{98}{5}\times \frac{78}{5}{\mathrm{cm}}^2$
                                                $$\begin{gathered} =\frac{7644}{25}{\mathrm{cm}}^2 \\ =305\frac{19}{25}{\mathrm{cm}}^2 \end{gathered}$$

Answer:

Cost of 1 burger = ₹$20\frac{3}{4}$ 
Cost of 4 burger = ₹$20\frac{3}{4}$ × 4
                          $$\begin{gathered} =₹\frac{83}{4}\times 4 \\ =₹83 \end{gathered}$$

Cost of 1 Macpuff = ₹ $15\frac{1}{2}$
Cost of 14 Macpuff = ₹ $15\frac{1}{2}$ × 14
                               $$\begin{gathered} =₹\frac{31}{2}\times 14 \\ =₹217 \end{gathered}$$

So, the cost of 4 burgers and 14 macpuffs = â‚¹83 + â‚¹217
= â‚¹300

Thus, the cost of 4 burgers and 14 macpuffs is â‚¹300.