Tag: maths

Questions Related to maths

Trishika bought some strawberries. $\cfrac{1}{8}$ of the strawberries were rotten and had to be thrown away. Trishika then used $25$ strawberries to bake a cake. She then had $\cfrac{1}{4}$ of the total strawberries left.
(a) How many strawberries did she have at first?
(b) If Trishika shared the remaining strawberries between her two sons equally, what fraction of the strawberries did each of them receive?

vedic methods of multiplication history of mathematics maths

  1. $(a) 70; (b) \dfrac{1}{8}$

  2. $(a) 50; (b) \dfrac{4}{8}$

  3. $(a) 40; (b) \dfrac{1}{8}$

  4. $(a) 40; (b) \dfrac{1}{4}$

Show answer
Correct Option: C
Explanation:
Let, the total no of strawberries Trishika had $x$
No of rotten strawberries $=$ $\dfrac{1}{8}x$
No of strawberries used to bake a cake$=$25
$\therefore$ No of strawberries left$=$$x-(\dfrac{1}{8}x+25)$
A.T.Q,
$x-\dfrac{1}{8}x-25$=$\dfrac{1}{4}x$
$\implies x-\dfrac{1}{8}x-\dfrac{1}{4}x$=$25$
$\implies \dfrac{8x-x-2x}{8}$=$25$
$\implies \dfrac{5x}{8}$=$25$
$ \implies x$=$40$
b)  No of remaining strawberries $=$ $\dfrac{1}{4}x$
Shared among two sons equally$=$$\dfrac{1}{4}x \div 2$
Share of strawberries recieveed by each son$=$$\dfrac{1}{8}x$

$\cfrac{1}{10}$ of a rod is coloured red, $\cfrac{1}{20}$ orange, $\cfrac{1}{30}$ yellow, $\cfrac{1}{40}$ green, $\cfrac{1}{50}$ blue, $\cfrac{1}{60}$ black and the rest violet. If the length of the violet portion is $12.06m$, then what is the length of the rod?

vedic methods of multiplication history of mathematics maths

  1. $16m$

  2. $18m$

  3. $20m$

  4. $30m$

Show answer
Correct Option: A
Explanation:

Let the length of the rod be $x$ $m$
$\therefore$ Length of the rod coloured be red$=\cfrac{1}{10}$ $x$ $m$
$\therefore$ Length of the rod coloured be orange$=\cfrac{1}{20}$ $x$ $m$
Length of the rod coloured be yellow$=\cfrac{1}{30}$ $x$ $m$
Length of the rod coloured be green$=\cfrac{1}{40}$ $x$ $m$
Length of the rod coloured be blue$=\cfrac{1}{50}$ $x$ $m$
Length of the rod coloured be black$=\cfrac{1}{60}$ $x$ $m$
Length of the rod coloured be violet$=12.08m$
$\therefore$ According to question
$\cfrac { 1 }{ 10 } x+\cfrac { 1 }{ 20 } x+\cfrac { 1 }{ 30 } x+\cfrac { 1 }{ 40 } x+\cfrac { 1 }{ 50 } x+12.08=x$
$\Rightarrow x-\cfrac { 49 }{ 200 } x=12.08\Rightarrow \cfrac { 151 }{ 200 } x=12.08\Rightarrow x=16$
$\therefore$ Length of the rod $=16m$

If $'+'$ means $'\times', '-'$ means $'\div', '\times'$ means $'-'$ and $'\div'$ means $'+'$, then $2 + 15 \div 15 - 3 \times 8$ is equal to

vedic methods of multiplication history of mathematics maths

  1. $43$

  2. $27$

  3. $35$

  4. $28$

Show answer
Correct Option: B
Explanation:

After interchanging the signs,
$2\times 15 + 15\div 3 - 8 = ?$
$\Rightarrow 2\times 15 + 5 - 8 = ?\Rightarrow 30 + 5 - 8 =?$
$? = 35 - 8 = 27$.

Which of the following fraction equals $\displaystyle101\frac{3}{5}\%$

vedic methods of multiplication history of mathematics maths

  1. $\displaystyle\frac{508}{5}$

  2. $\displaystyle\frac{254}{5}$

  3. $\displaystyle\frac{51}{25}$

  4. $\displaystyle\frac{127}{125}$

Show answer
Correct Option: A
Explanation:

$ 101 \dfrac {3}{5} = \dfrac {101 \times 5 + 3}{5} = \dfrac {505+3}{5} = \dfrac {508}{5}$

Two candles are of different lengths and thickness. The short and the long ones can burn respectively for $3.5$ hour and $5$ hour. After burning for 2 hour, the lengths of the candles become equal in length. What fraction of the long candle's height was the short candle initially ?

vedic methods of multiplication history of mathematics maths

  1. $\displaystyle\frac{2}{7}$

  2. $\displaystyle\frac{5}{7}$

  3. $\displaystyle\frac{3}{5}$

  4. $\displaystyle\frac{4}{5}$

Show answer
Correct Option: B
Explanation:

Shorter candle ray ${L} _{1}$ burns in 3.5 hours.
So in 2 hours $\displaystyle\frac{{L} _{1}}{3.5} \times 2$ of the candle will burn
Remaining length $= {L} _{1} - \displaystyle\frac{2{L} _{1}}{3.5} = \displaystyle\frac{1.5}{3.5}{L} _{1}$
The longer candle, ${L} _{2}$ burns in 5 hours. So in 2 hours $\displaystyle\frac{{L} _{2}}{5} \times 2$ will burn
Remaining length $= {L} _{2} - \displaystyle\frac{2{L} _{2}}{5} = \displaystyle\frac{3}{5} {L} _{2}$
Equating $\displaystyle\frac{1.5}{3.5} {L} _{1} = \displaystyle\frac{3}{5}{L} _{2}$
$\Rightarrow   \displaystyle\frac{{L} _{1}}{{L} _{2}} = \displaystyle\frac{3}{5} \times \displaystyle\frac{3.5}{1.5} = \displaystyle\frac{7}{5}$
Note : There is a mistake in question. It should read the short and the long ones can burn respectively for 5 hours and 3.56 hours. The answer $\left(\displaystyle\frac{{L} _{1}}{{L} _{2}}\right)$ will come out to be $\displaystyle\frac{5}{7}$ in that case.

Fraction  $\displaystyle \frac { 2 }{ 5 }, \frac {3} {10} , \frac {9} {10}, \frac {16} {35} $ in ascending order are:

vedic methods of multiplication history of mathematics maths

  1. $\displaystyle \frac { 2 }{ 5 }, \frac {3} {10} , \frac {9} {10}, \frac {16} {35} $

  2. $\displaystyle \frac { 3 }{ 10 }, \frac {2} {5}, \frac {16} {35}, \frac {9} {14} $

  3. $\displaystyle \frac { 3 }{ 10 } , \frac {9} {14} , \frac {16} {35}, \frac {2} {5} $

  4. $\displaystyle \frac { 16 }{ 35 } , \frac {2} {5} , \frac {3} {10}, \frac {9} {14} $

Show answer
Correct Option: B
Explanation:

The given Fraction are $\displaystyle \frac { 2 }{ 5 }, \frac {3} {10} , \frac {9} {10}, \frac {16} {35} $
LCM of 5, 10, 14, 35 = $\displaystyle (5\times 2\times 7\times)= 70 $
Now change each of the following into an equivalent fraction having 70 as its denominator.
Now, $\displaystyle \frac { 2 }{ 5 } =\frac { 2\times 14 }{ 5\times 14 } =\frac { 28 }{ 70 } \ \frac { 3 }{ 10 } =\frac { 3\times 7 }{ 10\times 7 } =\frac { 21 }{ 70 } \ \frac { 9 }{ 14 } =\frac { 9\times 5 }{ 14\times 5 } =\frac { 45 }{ 70 } \ and\quad \frac { 16 }{ 35 } =\frac { 16\times 2 }{ 35\times 2 } =\frac { 32 }{ 70 } \ Clearly,\quad \frac { 28 }{ 70 } >\frac { 21 }{ 70 } <\frac { 45 }{ 70 } >\frac { 32 }{ 70 } \ Hence,\quad \frac { 3 }{ 10 } <\frac { 2 }{ 5 } <\frac { 16 }{ 35 } <\frac { 9 }{ 14 } $

The value of the expression $\dfrac { 1 }{ \sqrt { 11-2\sqrt { 30 }  }  } -\dfrac { 3 }{ \sqrt { 7-2\sqrt { 10 }  }  } -\dfrac { 4 }{ \sqrt { 8+4\sqrt { 3 }  }  } $ after simplification is

vedic methods of multiplication history of mathematics maths

  1. $\sqrt { 30 } $

  2. $2\sqrt { 10 } $

  3. $1$

  4. $0$

Show answer
Correct Option: D
Explanation:

$\dfrac { 1 }{ \sqrt { 11-2\sqrt { 30 }  }  } -\dfrac { 3 }{ \sqrt { 7-2\sqrt { 10 }  }  } -\dfrac { 4 }{ \sqrt { 8+4\sqrt { 3 }  }  } $


$=\left (\dfrac { 1 }{ { 11-2\sqrt { 30 }  }  }\right)^{\frac{1}{2}} -\left( \dfrac { 3 }{ { 7-2\sqrt { 10 }  }  }\right)^{\frac{1}{2}} -\left( \dfrac { 4 }{ { 8+4\sqrt { 3 }  }  }\right)^{\frac{1}{2}} $


$=\left (\dfrac { 1 }{ { 5+6-2\sqrt {5}\sqrt{6}}}\right)^{\frac{1}{2}} -\left( \dfrac { 3 }{ { 5+2-2\sqrt {5}\sqrt{2}}}\right)^{\frac{1}{2}} -\left( \dfrac { 4 }{ {6+2+2\sqrt {6}\sqrt{2}  }  }\right)^{\frac{1}{2}} $

$={ \left( \dfrac { 1 }{ (\sqrt { 6 } -\sqrt { 5 } )^2 }  \right)  }^{ \frac { 1 }{ 2 }  }-3{ \left( \dfrac { 1 }{( \sqrt { 5 } -\sqrt { 2 } )^2 }  \right)  }^{ \frac { 1 }{ 2 }  }-4{ \left( \dfrac { 1 }{ (\sqrt { 6 } +\sqrt { 2 } )^2 }  \right)  }^{ \frac { 1 }{ 2 }  }$

$={ \left( \dfrac { 1 }{ \sqrt { 6 } -\sqrt { 5 }  }  \right)  }^{ \frac { 2 }{ 2 }  }-3{ \left( \dfrac { 1 }{ \sqrt { 5 } -\sqrt { 2 }  }  \right)  }^{ \frac { 2 }{ 2 }  }-4{ \left( \dfrac { 1 }{ \sqrt { 6 } +\sqrt { 2 }  }  \right)  }^{ \frac { 2 }{ 2 }  }$

$={ \left( \dfrac { 1 }{ \sqrt { 6 } -\sqrt { 5 }  } \times {\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}} \right)  }^{ \frac { 2 }{ 2 }  }-3{ \left( \dfrac { 1 }{ \sqrt { 5 } -\sqrt { 2 }  } \times{\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}} \right)  }^{ \frac { 2 }{ 2 }  }-4{ \left( \dfrac { 1 }{ \sqrt { 6 } +\sqrt { 2 }  } \times{\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}} \right)  }^{ \frac { 2 }{ 2 }  }$

$=\left( \sqrt { 6 } +\sqrt { 5 }  \right) -\left( \sqrt { 5 } +\sqrt { 2 }  \right) -\left( \sqrt { 6 } -\sqrt { 2 }  \right) =0$
Hence, option D is correct.

Which of the following statement is CORRECT?
Statement-1: Mrs. Soni bought 7$\dfrac{1}{2}$ litres of milk. Out of this, 5$\dfrac{3}{4}$ liters was consumed. 1$\dfrac{1}{3}$ liters of milk is left with her.
Statement-2: Amit reads $\dfrac{3}{5}$ of a book. He finds that there are still 80 pages left to be read. The total number of pages in the book is $200.$

vedic methods of multiplication history of mathematics maths

  1. Only statement-1

  2. Only Statement-2

  3. Both Statement-1 and Statement-2

  4. Neither Statement-1 nor Statement-2

Show answer
Correct Option: B
Explanation:

Statement-1: Total quantity of milk bought = 7$\dfrac{1}{2}$litres = $\dfrac{15}{2}$litres
Quantity of milk consumed = 5$\dfrac{3}{4}$litres
                                            = $\dfrac{23}{4}$ litres
$\therefore$ Quantity of milk left = $\big(\dfrac{15}{2}$ - $\dfrac{23}{4}$$\big)$ litres
= $\big(\dfrac{30-23}{4}\big)$ litres = $\dfrac{7}{4}$ = 1$\dfrac{3}{4}$ litres
Statement-2: Let total number of pages in the book be x.
According to question,
$\dfrac{3}{5}$x + 80 = x
$\Rightarrow$ x - $\dfrac{3}{5}$x = 80 $\Rightarrow$ x = $\dfrac{80\times5}{2}$
$\Rightarrow$ x = 200
$\therefore$ Total number of pages = 200
Hence, only statements-2 is true.

Two pipes $A$ and $B$ can fill a cistern in $37\dfrac {1}{2}$ minutes and $45$ minutes respectively. Both pipes are opened, the cistern will be filled just in half an hour, if the pipe $B$ is turned off after.

vedic methods of multiplication history of mathematics maths

  1. $15\ minutes$

  2. $10\ minutes$

  3. $5\ minutes$

  4. $9\ minutes$

Show answer
Correct Option: D
Explanation:

Let the capacity of cistern be $225$ units $\left (LCM\ of \dfrac {75}{2}\ and\ 45\right )$
$A$ does $= \dfrac {225}{75}\times 2 = 6\ units/ min$
$B$ does $= \dfrac {225}{45} = 5\ units/ min$.
Let pipe is turned off after $x$ minutes.
According to the question,
$6\times 30 + 5\times x = 225$
$5x = 225 - 180 = 45$
$x = 9$
After $9$ minutes, pipe $B$ is turned off.

Two trains starts from stations $A$ and $B$ and travel towards each other at speed of $50\ km/hr$ and $60\ km/hr$ respectively. At the time of their meeting, the second train has travelled $120\ km$ more than the first. The distance between $A$ and $B$ is

vedic methods of multiplication history of mathematics maths

  1. $990\ km$

  2. $1200\ km$

  3. $1320\ km$

  4. $1440\ km$

Show answer
Correct Option: C
Explanation:

Speed of train $A = 50\ kmph$
Speed of train $B = 60\ kmph$
Since, time is constant
$Speed \propto$ Distance covered
$\dfrac {S _{A}}{S _{B}} = \dfrac {D _{A}}{D _{B}}$
$\dfrac {50}{60} = \dfrac {D _{A}}{D _{B}}\Rightarrow \dfrac {5}{6} = \dfrac {D _{A}}{D _{B}}$
Given that train $B$ has travelled $120\ km$ extra.
$6x - 5x = 120$
$x = 120$
The distance between $A$ and $B = 6x + 5x = 11x$
$= 11\times 120 = 1320$
Alternate Method:
Let train $A$ start form station $A$ and $B$ from station $B$.
Let the trains $A$ and $B$ meet after/ hours.
$\therefore$ Distance covered by train $A$ in $t$ hours $= 50t$
Distance covered by train $B$ in $t$ hours $= 60t\ km$.
According to the question,
$60t - 50t = 120$
$\Rightarrow t = \dfrac {120}{10} = 12\ hours$.
$\therefore$ Distance $AB = 50\times 12 + 60 \times 12$
$= 600 + 720 = 1320\ km$.