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,

$\frac{1}{3}< \frac{6}{11} <\frac{5}{6}$