Fraction questions

Let two numbers be A and B,

A + B = 20

A - B = 10

Adding both we get 2A = 30

A = 15

From that we get B = 5

So the ratio = \(\frac{3}{1}\)

• Let the fraction be \(\frac{x}{y}\).
  Then, \(\frac{x}{y+1} = \frac{1}{2 }\)
  2x - y = 1 ..........(i) and
  \(\frac{x + 1}{ y} = 1\)
  x - y = 1 ....(ii)
• Solving (i) and (ii), we get : x = 2, y = 3.
• Hence, the required fraction is \(\frac{2}{3}\)

\( \frac{1}{4} \) \( \times \) \( \frac{1}{9} \) \( \times \)\( \frac{36}{121} \) \( \times \) \( \frac{64}{3} \) \( \times \) 2420 \( \times \) 81 = \( \frac{2420}{121} \) \( \times \) 27 \( \times \) 64 = 34560

Follow BODMAS rule to solve this question, as per the order is given below:

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first and in the bracket,

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

\(\left(\frac{1}{2} \div \frac{1}{2} \times \frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \div \frac{1}{2}\right)\)

of \(\left(\frac{1}{2}+\frac{1}{2}\right) \)

\(\Rightarrow 1 \times \frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2} \times 1 of 1\)

\(\Rightarrow \frac{1}{2}+0+\frac{1}{2}\)

\(\therefore \) answer is 1

-Total amount of sugar to be distributed \(=28 \frac{1}{2} \mathrm{~kg}=57 / 2 \mathrm{~kg}\) Amount of sugar distributed to every needy student \(=3 / 4 \mathrm{~kg}\) Number of students among which the sugar is distributed \(=\frac{\frac{57}{2}}{\frac{3}{4}}\) \[ \Rightarrow \frac{57}{2} \times \frac{4}{3} \] \[ \Rightarrow 38 \] Hence, the required number is 38 .

\(\frac{\left [ \frac{3}{7}+\left [ \frac{-5}{14} \right ] \right ]}{\frac{-1}{2}}\)<br> \(\therefore\) LCM of 7 and 2 = 14<br> \(\frac{\left [ \frac{3}{7}\times\frac{2}{2}-\frac{5}{14} \right ]}{\frac{-1}{2}}\) = \(\frac{6-5}{14}\times\frac{-2}{1}\)=\(\frac{1}{14}\times\frac{-2}{1}\)<br> =\(\frac{-1}{7}\)

\(\frac{3}{8} \times 100 \)

\(=3 \times 12.5 \)

= \(37.5\)

• Calculation :

  

Let the numerator be \(x\) and denominator be \(y\).
When 1 is added to the numerator and denominator,
\(\frac{x+1}{y+1}=\frac{4}{5}\)
\(5x+5=4y+4,\)
\( 5x-4y=-1\;\;\;\) -(i)
When 5 is subtracted from both numerator and denominator,
\(\frac{x-5}{y-5}=\frac{1}{2}\)
\(2x-10=y-5\)
\(2x-y=5\;\;\;\) -(ii)
(ii)\(\times4\) we get \(8x-4y=20\;\;\;\) - (iii)
(iii) - (i) we get,
\(3x=21,\;\;\; x=7\)
putting the value of \(x\) in (i) we get \(y=9\)
\(\therefore \) The fraction \(=\frac{x}{y}=\frac{7}{9}\)

\(\frac{m}{n}=\frac{3}{5}\)

\(\frac{6 m-3 n}{5 n+4 m}\)

\(=\frac{6 \times 3-3 \times 5}{5 \times 5+4 \times 3}=\frac{3}{37}\)

Hence, option \(B\) is correct.

Given expression is ,
\((1\frac{1}{2} + 11\frac{1}{2}+111\frac{1}{2}+1111\frac{1}{2})\)
\(=\frac{3}{2}+\frac{23}{2}+\frac{223}{2}+\frac{2223}{2}\)
\(=\frac{3+23+223+2223}{2}\)

\(=\frac{2472}{2}\)

\(=1236\)

The sum of the numerator and denominator of a fraction is 15. If 1 is added to its denominator and 1 is subtracted from the numerator, the value of the new fraction is \(\frac{2}{3}\).

Let the fraction be \(\frac{x}{y}\)

And, x + y = 15

Now, x -1 + y + 1 = x + y = 15

So, x -1 + y + 1 = 15

Now, \(\frac{x - 1}{y + 1}\)= \(\frac{2a}{3a}\)

So, 2a + 3a = 15

5a = 15

a = 3

so fraction is, \(\frac{2 \times 3}{3 \times 3} \)= \(\frac{6}{9}\)

Actual fraction is \(\frac{6 + 1}{9 -1}\) = \(\frac{7}{8}\)

The required answer is \(\frac{7}{8}\)

\( \frac{\frac{5}{16}\times \frac{8}{25}\times \frac{10}{19}\times \frac{114}{7}}{\frac{12}{14}}\\\\ =\:\frac{\frac{1\times 10\times 114}{2\times 5\times 19\times 7}}{\frac{12}{14}}\\\\ =\:\frac{\frac{114}{19\times 7}}{\frac{12}{14}}\\\\ =\:\frac{\frac{6}{7}}{\frac{12}{14}}\\\\ =\:\frac{6}{7}\times \frac{14}{12}\\\\ =\:\frac{2}{2}\\\\ =\:1 \)

\(\text { Let } \frac{1}{2}=a, \frac{1}{4}=b \text { and } \frac{1}{5}=c\)

\(\therefore\) Expression

\(=\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a^{2}+b^{2}+c^{2}-a b-a c-b c}\)

\(=\frac{(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-a c-b c\right)}{a^{2}+b^{2}+c^{2}-a b-a c-b c}\)

\(=a+b+c\)

\(=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}\)

On cross multiplying we get \( 2x + 3 = 5x + \frac{5}{4} \)<br>

\( 3x = 3-\frac{5}{4} = \frac{7}{4} \) <br>

\( x = \frac{7}{12} \)

• Given fraction is a)\(\frac{25}{31}\), b)\(\frac{11}{14}\), c)\(\frac{15}{19}\), d)\(\frac{41}{53}\)
• a)\(\frac{25}{31}\) = 0.806
• b)\(\frac{11}{14}\) = 0.785
• c)\(\frac{15}{19}\) = 0.789
• d)\(\frac{41}{53}\) = 0.773
• Here it is clear that, the ascending order of the given fractions is 'dbca'

fraction be \( \frac{x}{y} \)

5 added to numerator,\(\frac{(x+5)}{y}= \frac{ 6}{5}\)

\( 5x+25=6y \)

\( \frac{x}{y+4}= \frac{1}{2} \)

\( 2x = y+4 \)

multiply above equation by, 6

=> \( 12 x= 6y + 24 \)

\( 6y=12x-24 \)

\( 5x+25=12x-24 \)

\( 12x-5x= 25+24=49 \)

\( x=7 \)

Then \( 2\times 7=y+4 \)\( => y =10 \)

fraction= \( \frac{x}{y}= \frac{7}{10} \)

Given:

Weight of packets in which sugar to be packed = 1/8 kg

Weight of one packet of sugar \(=2 \frac{3}{4} \mathrm{~kg}\)

Weight of the second pack of sugar \(=3 \frac{1}{4} \mathrm{~kg}\)

\( \begin{aligned} &2 \frac{3}{4} k g=\frac{11}{4} k g \\ &3 \frac{1}{4} k g=\frac{13}{4} k g \\ &\Rightarrow \text { Total weight of sugar }=\frac{11}{4}+\frac{13}{4}=\frac{24}{4} \\ &\Rightarrow \text { Total weight of sugar }=6 \mathrm{~kg} \end{aligned} \)

Number of packets \(=\) Total weight/Weight of each packet \( \begin{aligned} &\Rightarrow \text { Number of packets }=\frac{6}{\frac{1}{8}} \\ &\Rightarrow \text { Number of packets }=6 \times 8 / 1=48 \end{aligned} \)

\(\therefore\) Total 48 packets can be made.

The correct option is 2 i.e. 2

Required difference = \( \frac{83}{26} - \frac{26}{83} \)
\( = \frac{83^2 - 26^2}{2158} \)
\( = \frac{6889 - 676}{2158} \)
\( = \frac{6213}{2158} \)

Question 1

108 \(\div\)36 of \(\frac{1}{4}\)+\(\frac{2}{5}\)\(\times\)3\(\frac{1}{4}\)

=108 \(\div\)36 \(\times\) \(\frac{1}{4}\)+\(\frac{2}{5}\)\(\times\)3\(\frac{1}{4}\)

=108 \(\div\)9+\(\frac{2}{5}\)\(\times\)3\(\frac{1}{4}\)

=12+\(\frac{13}{10}\)

=\(\frac{120+13}{10}\)

=\(\frac{133}{10}\)

=13\(\frac{3}{10}\)

Question 2

\[\begin{aligned} &\frac{\frac{11}{4}}{\frac{11}{6}} \div \frac{7}{8}\left(\frac{4+3}{12}\right)+\frac{5}{7} \div \frac{3}{4} \text { of } \frac{3}{7} \\ &=\left(\frac{11}{4} \times \frac{6}{11}\right) \div \frac{7}{8} \times \frac{7}{12}+\frac{5}{7} \div\left(\frac{3}{4} \times \frac{3}{7}\right) \\ &=\frac{3}{2} \div \frac{7}{8} \times \frac{7}{12}+\frac{5}{7} \div \frac{9}{28} \\ &=\frac{3}{2} \times \frac{8}{7} \times \frac{7}{12}+\frac{5}{7} \times \frac{28}{9} \\ &=1+\frac{20}{9}=\frac{9+20}{9}=\frac{29}{9}=3 \frac{2}{9} \end{aligned}\]

\(\frac{\frac{1 }{2 }+\frac{1 }{2 }}{\frac{1 }{2 }\times\frac{1 }{2 }}\)\(\div\frac{1 }{2 }\)

= \(\frac{1 }{\frac{1 }{4 }}\div\frac{1 }{2 }\)

= \(4\div\frac{1 }{2 }\)

= \(4\times\frac{2 }{1 }\) = 8

\(A=2B,B=\frac{1}{2}A\),

\(AB=\frac{2}{25}\) ,

\(\frac{1}{2}A^2=\frac{2}{25}\) ,\( \Rightarrow \)

\(A=\frac{2}{5}\)

\(A=\frac{2}{5},B=\frac{1}{5}\)

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

\(\frac{3}{5}\)\(\times\)\(\frac{2}{3}\)\(\times\)x - \(\frac{2}{5}\)\(\times\)\(\frac{1}{4}\)\(\times\)x = 288

\(\frac{2}{5}\)x-\(\frac{2}{20}\)x = 288

\(\frac{8x-2x}{20}\) = 288

x = \(\frac{288\times20}{6}\) = 960

• To add mixed fraction ,change into improper fraction .
• \(4\frac{1}{2}=\frac{9}{2},3\frac{3}{4}=\frac{15}{4}\)
• \(\frac{9}{2}+\frac{15}{4}=\frac{33}{4}=8\frac{1}{4}.\)

\( \frac{1}{2}+ \frac{1}{6}+ \frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132} \) can be wriiten as,

\( \frac{1}{1\times 2}+ \frac{1}{2 \times 3}+ \frac{1}{3 \times 4}+\frac{1}{4 \times 5}+\frac{1}{5 \times 6}+\frac{1}{6 \times 7}+\frac{1}{7 \times 8}+\frac{1}{9 \times 8}+\frac{1}{9 \times 10}+\frac{1}{10 \times 11}+\frac{1}{11 \times 12} = 1-\frac{1}{12} = \frac{11}{12} \)

Given, \(\frac{a}{b}\) +\( 7\times\)\(\frac{b}{a}\) = \(\frac{11}{2}\)
Simplifying, \(2a^2+14b^2-11ab\) = 0
= \((2a-7b)(a-2b)\) = 0
So, \((2a-7b)\) = 0
\(\frac{a}{b}\) = \(\frac{7}{2}\)

\(1\frac{5}{16}=1+\frac{5}{16}=\frac{21}{16}\)

its reciprocal is \(\frac{16}{21}\)

Now the difference =\(\frac{21}{16}-\frac{16}{21}=\frac{21\times 21-16\times 16}{16\times 21} \) =\(\frac{441-256}{336}\) =\(\frac{185}{336}\)

\(\frac{144}{0.144}= \frac{14.4}{x}\)
\(\frac{144 \times 1000}{144} = \frac{14.4}{x}\)
\(x= \frac{14.4}{1000}=0.0144\)

\(\left(5 \frac{1}{4} \div \frac{3}{7} o f \frac{1}{2}\right) \times\left(5 \frac{1}{4} \times \frac{3}{7} \div \frac{1}{2}\right) \div\left(5 \frac{1}{4} \div \frac{3}{7} \times \frac{1}{2}\right)\)

\(\Rightarrow\left(\frac{21}{4} \div \frac{3}{7} \times \frac{1}{2}\right) \times\left(\frac{21}{4} \times \frac{3}{7} \div \frac{1}{2}\right) \div\left(\frac{21}{4} \div \frac{3}{7} \times \frac{1}{2}\right)\)

\(\Rightarrow\left(\frac{21}{4} \div \frac{3}{14}\right) \times\left(\frac{21}{4} \times \frac{3}{7} \times \frac{2}{1}\right) \div\left(\frac{21}{4} \times \frac{7}{3} \times \frac{1}{2}\right)\)

\(\Rightarrow\left(\frac{21}{4} \times \frac{14}{3}\right) \times\left(\frac{9}{2}\right) \div\left(\frac{49}{8}\right)\)

\(\Rightarrow \frac{49}{2} \times \frac{9}{2} \times \frac{8}{49}\)

\(\Rightarrow 18\)

Total number of toffees, \(M+N+P=1000\)</p><p>Number of tofees P gets, P\( =\frac{(M+ N)}{4}\) ------- (i)

Number of tofees N gets, N \(=\frac{5M}{3}\)

3N = 5M------- (ii) Equation (i)\( \times 3 \)

From (i) and (ii), \(P= \frac{8M}{12}\)

So, \(M+ \frac{5M}{3}+\frac{8M}{12} = 1000\)

\(\frac{12M+20M +8M}{12}=1000\)

\(\therefore M=300\). Hence, option 1.

Taking the LCM of the denominator: LCM of 6, 4, 2, 3 and 5 = 60 \begin{aligned} &\frac{5}{6}=\frac{5 \times 10}{6 \times 10}=\frac{50}{60} \\ &\frac{3}{4}=\frac{3 \times 15}{4 \times 15}=\frac{45}{60} \\ &\frac{1}{2}=\frac{1 \times 30}{2 \times 30}=\frac{30}{60} \\ &\frac{2}{3}=\frac{2 \times 20}{3 \times 20}=\frac{40}{60} \\ &\frac{3}{5}=\frac{3 \times 12}{5 \times 12}=\frac{36}{60} \end{aligned}

So, the largest fraction = 5/6 And second largest fraction = 3/4

Let the fraction 'F' be \(x / y\)

According to the question,

\(\begin{aligned} &\Rightarrow \frac{x }{(y+4)}=\frac{1 }{ 2} \\ &\Rightarrow 2 x=y+4--(i) \\ &\Rightarrow \frac{(x-5)}{y}= \frac{1 }{ 3} \\ &\Rightarrow 3 x-15=y \\ &\Rightarrow 3 x=y+15---(ii) \end{aligned}\)

Equation (i) and (ii), multiply the equation (i) by 3 and equation (ii) by 2

\(\begin{aligned} &\Rightarrow 3 y+12=2 y+30 \\ &\Rightarrow y=18 \end{aligned}\)

Put the value of \(y\) in equation (i), we get \(x=11\)

So, fraction ' \(F^{\prime}=\frac{11 }{18}\)

7\( \frac{ 2}{3} \) = \( \frac{ 21+2}{3} \) = \( \frac{ 23}{3} \)
12 \( \frac{ 3}{4} \) = \( \frac{ 48+3}{4} \) = \( \frac{ 51}{4} \)
cost per metre = \( \frac{ 51}{4} \) / \( \frac{ 23}{3} \) = \( \frac{ 153}{92} \)

LCM of 32, 60 , 24 and 15

\( = 2\times2\times2\times2\times2\times3\times5 =480 \)

\(\frac{ 15}{32} = \frac{15\times15 }{ 32\times15} =\frac{225}{480} \)

\( \frac{21}{60} =\frac{21\times8}{ 60\times8} = \frac{168}{480} \)

\(\frac{ 11}{24} = \frac{11\times20}{ 24\times20} = \frac{220}{480} \)

\( \frac{7}{15} = \frac{7\times32} {15\times32} = \frac{224}{480} \)

\(\frac{ 225}{480} > \frac{224}{480} > \frac{220}{480} > \frac{168}{480} \)

\( \frac{15}{32} > \frac{7}{15} >\frac{ 11}{24} > \frac{21}{60} \)

1) \(1 \mathrm{~kg}\) Watermelon \(=\) Rs. \(22.50\)

\(3 \frac{1}{2} \mathrm{~kg}\) Watermelon \(=3.5 \mathrm{~kg}\) Watermelon

\(\Rightarrow\) The cost of \(3.5 \mathrm{~kg}\) Watermelon \(=3.5 \times 22.50\)

\(\Rightarrow\) The cost of \(3.5 \mathrm{~kg}\) Watermelon \(=\) Rs. \(78.75\)

2) \(1 \mathrm{~kg}\) Mango = Rs. \(80.60\)

\(\Rightarrow\) The cost of \(2 \mathrm{~kg}\) mangoes \(=2 \times 80.60\)

\(\Rightarrow\) The cost of \(2 \mathrm{~kg}\) mangoes \(=\) Rs. \(161.2\)

3) \(1 \mathrm{~kg}\) Apple \(=\) Rs. \(120.50\)

\(1 \frac{1}{2} \mathrm{~kg}\) apples \(=1.5 \mathrm{~kg}\) apples

\(\Rightarrow\) The cost of \(1.5 \mathrm{~kg}\) apples \(=1.5 \times 120.50\)

\(\Rightarrow\) The cost of \(1.5 \mathrm{~kg}\) apples \(=\) Rs. \(180.75\)

4) \(1 \mathrm{~kg}\) grapes = Rs. \(110.40\)

\(\frac{3}{4} \mathrm{~kg}\) grapes \(=0.75 \mathrm{~kg}\) grapes

\(\Rightarrow\) The cost of \(0.75 \mathrm{~kg}\) grapes \(=0.75 \times 110.40\)

\(\Rightarrow\) The cost of \(0.75 \mathrm{~kg}\) grapes \(=\) Rs. \(82.8\)

The total cost of all the fruits \(=\) The cost of \(3.5 \mathrm{~kg}\) Watermelon \(+\) The cost of \(2 \mathrm{~kg}\) mangoes \(+\) The cost of \(1.5 \mathrm{~kg}\) apples \(+\) The cost of \(0.75 \mathrm{~kg}\) grapes \(\Rightarrow\) The total cost of all the fruits \(=\) Rs. \(78.75+\) Rs. \(161.2+\) Rs. \(180.75+\) Rs. \(82.8\)

\(\Rightarrow\) The total cost of all the fruits = Rs. \(503.5\)

Money given to the shopkeeper by Sneha = Rs. 2000

The money that Sneha get back = Money given to the shopkeeper - The total cost of all the fruits

\(\Rightarrow\) The money that Sneha gets back = Rs. 2000 - Rs. \(503.5\)

\(\Rightarrow\) The money that Sneha gets back \(=\) Rs. \(1496.5\)

Hence, the correct answer is Rs. \(1496.5\)

on cancelling with numerators and denominators,

we get, 11\(\times\)23\(\times\)13\(\div\)(7\(\times\)9) = \(\frac{3289}{63}\)

• 0.66666.... is a non terminating repeating decimal.
• 0.666... =0.\( \overline{6} \) = \( \frac{6}{9} \)

\begin{aligned}
&\text { } \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{1}{2}\\
&\text { LCM of } 3,4,5,2=60\\
&\begin{array}{c}
\frac{40,45,48,30}{60} \\
48,45,40,30
\end{array}\\
&\frac{4}{5}>\frac{3}{4}>\frac{2}{3}>\frac{1}{2}
\end{aligned}

\( \frac{1}{4}\times{2}{5}x – \frac{3}{5} \times \frac{2}{6} x = 6 \)<br>

\( \frac{1}{10}x – \frac{1}{5}x = 6 \)<br>

\( \frac{-x}{10}=6 \)<br>

\( X= -60 \)

\( 66\times \frac{1}{100} \) \( =\frac{33}{50} \)

LCM of \(11, 6, 3 = 66 \)

\(\frac{6}{11}=\frac{6}{11} \times\frac{6}{6}=\frac{36}{66}\)

\( \frac{5}{6}=\frac{5}{6} \times\frac{11}{11}=\frac{55}{66}\)

\( \frac{1}{3}=\frac{1}{3} \times\frac{22}{22}=\frac{22}{66}\)

Now the denominators are same,the fraction with smallest numerator is the smallest.Ascending order of the fractions,

\(\frac{1}{3}< \frac{6}{11} <\frac{5}{6} \)