Tag: maths

Questions Related to maths

Which of the following is an example of indirect proportion?

maths direct proportion and inverse proportion inverse proportion rule of three types of proportions

  1. Number of mangoes in a bag and weight of the bag.

  2. Earning and hours worked.

  3. More the speed, lesser the time taken.

  4. None of these.

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Correct Option: C
Explanation:

Indirect proportion: When one quantity increases another quantity decreases.
In option C, if the speed of the vehicle increases then the time taken by it to cover a distance, decreases.

A fort is provisioned for $42$ days. After $10$ days, a reinforcement of $200$ men arrives and the food will now last only $24$ days. How many men were there in the fort?

maths direct proportion and inverse proportion inverse proportion rule of three types of proportions

  1. $200$

  2. $600$

  3. $500$

  4. $300$

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Correct Option: B
Explanation:

Let $x$ soldiers had provision for $42$ days in a fort.

After $10$ days: $x$ soldiers will remain with $32$ days of provision
therefore for $(x+200)$ soldiers provision will decrease to $24$ days
therefore By inverse proportion,
$x\times32= (200+x)\times 24$
$\therefore \dfrac{32}{24} = \dfrac{200 + x}{x}$
$\therefore \dfrac{32}{24} = \dfrac{200}{x} + 1$
$\therefore \dfrac{32}{24} - 1 = \dfrac{200}{x}$
$\therefore \dfrac{8}{24} = \dfrac{200}{x}$
$\therefore x = \dfrac{200\times 24}{8}$
$\therefore x = 600$
Answer is $600$ days

 In any triangle are  the circumcentre, the centroid, the nine point centre and the orthocentro are all collinear ?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. True

  2. False

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Correct Option: B

Mid-point theorem states that:

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. The line segment joining the mid-points of two sides of a triangle is not parallel to the third side and equal to half the length of the third side.

  2. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to one-third the length of the third side.

  3. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to the length of the third side.

  4. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

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Correct Option: D
Explanation:

Mid-point theorem states that

The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Hence, option D is correct.

In $\Delta ABC$, AB$ =5$cm, $BC=8$cm and $CA=7$cm. If D and E are respectively, the mid-points of AB and BC, then determine the length of DE.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $3.5$ cm

  2. $2.5$ cm

  3. $2.8$ cm

  4. $2.0$ cm

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Correct Option: A
Explanation:

Mid-Point Theorem : 
The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

On applying the midpoint theorem, we get
 $DE = \cfrac {AC}{2} = \cfrac 72 = 3.5$cm

Suppose $ABCD$ is a rhombus. A straight line passing through $C$ meet $AD$ which is produced at $P$ and meet $AB$ produced at $Q$. Therefore if $DP=\dfrac {1}{2}AB$, then find the ratio between $BQ$ and $AB$?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $2:1$

  2. $1:1$

  3. $1:3$

  4. $3:1$

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Correct Option: A
Explanation:

$ ABCD $ is a rhombus. $ AB =BC = CD = DA $


$ \displaystyle \frac{DP}{AB} = \frac{1}{2} \Rightarrow DP =1 AB =2 $

In rhombus  $ \angle \theta _1 = \angle \theta _2 $

$ \angle Q $ is common for the $ \triangle BCQ \, and \, \triangle APQ $

$ \because \angle APQ = \angle BCQ $

$ \because \triangle BCQ $ is similar to $ \triangle APQ $ by $ AAA $ property. 

$ \displaystyle \frac{AP}{BC} = \frac{AQ}{BQ} = \frac{AD + DP}{BC} = \frac{3}{2} $

$ \displaystyle \frac{AQ}{BQ} = \frac{3}{2} \Rightarrow \frac{AB + BQ}{BQ} = \frac{3}{2} $

$ \displaystyle \frac{AB}{BQ} = \frac{3}{2} - 1 = \frac{1}{2} $

$ \displaystyle \frac{BQ}{AB} = \frac{2}{1} $

The median $AD$ of the triangle $ABC$ is bisected at $E$, $BE$ meets $AC$ in $F$, then $AF:AC$ is equal to ?

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $3:4$

  2. $1:3$

  3. $1:2$

  4. $none\ of\ these$

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Correct Option: A

The incentre of the triangle formed by $(0, 0, 0), (3, 0, 0), (0, 3, 0)$.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $(\dfrac{3}{4}, \dfrac{3}{4}, 0)$

  2. $(1, \dfrac{3}{4}, 0)$

  3. $(0, 1, 1)$

  4. $(1, 1, 1)$

Show answer
Correct Option: B
Explanation:

using formula incenter $x= \dfrac{ax _{1}+bx _{2}+cx _{3}}{a+b+c}$ 


    $y=\dfrac{ay _{1}+by _{2}+cy _{3}}{a+b+c}$ 
    $z=\dfrac{az _{1}+bz _{2}+cx _{3}}{a+b+c}$
    $a=5$    $x=\dfrac{5(0)+4(3)+3(0)}{5+4+3}$       ;     $x=1$


    $b=4$      $y=\dfrac{5(0)+4(0)+3(3)}{5+4+3}$       ; $y=\dfrac{3}{4}$


    $c=3$      $z=\dfrac{5(0)+4(0)+3(0)}{5+4+3}$        ; $z=0$

 incentre =$(1,\dfrac{3}{4},0)$ 

The mid-points of the sides of a triangle are $D(6,1),E(3,5)$ and $F(-1,-2)$ then vertex opposite to D is 

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. $(-4,2)$

  2. $(-4,5)$

  3. $(2,2)$

  4. $(10,8)$

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Correct Option: A

ABC is an isosceles triangle with AB=AC. D,E, F are mid point of sides BC,AB and AC respectively then line segment $A D \perp E F$ and is bisected by it.

maths mid-point and its converse application of the mid-point theorem mid point theorem mid-point theorem and its converse

  1. True

  2. False

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Correct Option: A