Self Studies
Self Studies
Self Studies
Self Studies

Straight Lines Test 15

Result Self Studies

Straight Lines Test 15
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The fourth vertex $$D$$ of a parallelogram $$ABCD$$ whose three vertices are $$A (2, 3), B (6, 7)$$ and $$C (8, 3)$$ is
    Solution
    Let $$A(2,3), B(6,7)$$ and $$C(8,3)$$ be the vertices of parallelogra

    And let $$D(x,y)$$ be the fourth vertex.

    Midpoint of $$AC$$ $$=\left (\dfrac {2+8}{2},\dfrac {3+3}{2}\right)=(5,3)$$

    Also, midpoint of $$BD$$ $$=\left (\dfrac {x+6}{2}, \dfrac {y+7}{2}\right)$$

    Since the diagonals of parallelogram bisect eachother at $$O$$

    Therefore, midpoint of $$AC$$ $$=$$ midpoint of $$BD$$

    $$\Rightarrow \left (\dfrac {x+6}{2},\dfrac {y+7}{2}\right)=(5,3)$$
    $$\Rightarrow \dfrac  {x+6}{2}=5, \dfrac {y+7}{2}=3$$

    $$\Rightarrow x+6=10, y+7=6$$

    $$\Rightarrow x=4, y=-1$$

    Therefore, the coordinates of the fourth vertex will be $$(x,y)=(4,-1)$$.
  • Question 2
    1 / -0
    The value of $$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{ 3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }  $$ is,
    Solution
    $$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) }  $$

    $$\quad =\displaystyle \frac{\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}}{\displaystyle \frac{20\cdot 21\cdot 22.............36}{1\cdot 2\cdot 3\cdot 4...............17}}$$


    $$ \ \ \ =\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{20\cdot 21\cdot 22.............36}$$


    $$\quad = \displaystyle \frac{18\cdot 19\cdot 20.............36}{20\cdot 21\cdot 22.............36}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{1\cdot 2\cdot 3\cdot 4...............19}$$


    $$\displaystyle  \ \ \  =\frac{18\cdot 19}{1}\times \frac{1}{18\cdot 19}=1$$
  • Question 3
    1 / -0
    The sum of the value of the digits at the tens place of all the numbers formed with the help of $$3, 4, 5, 6$$ taken all at a time is
    Solution
    Total number of numbers formed with the digits $$3,4,5,6$$ is $$4!$$ $$=$$ $$24$$
    If a digit is fixed in tens place, number of numbers formed will be $$6$$
    i.e for every digit in tens place there will be $$6$$ numbers formed
    Now, sum of the digits $$ = (6\times3)+(6\times4)+(6\times5)+(6\times6) $$ $$ = 108 $$.
    Now its value in tens place $$= 108\times10=1080 $$.
  • Question 4
    1 / -0
    Three Men have $$4$$ coats $$5$$ waist Coats, and $$6$$ caps. The number of ways they can wear them is
    Solution
    Coats $$\rightarrow  ^4P_3$$
    4 3 2
    Waist Coats $$\rightarrow   ^5P_3$$
    5 4 3
    Caps $$\rightarrow   ^6P_3$$
    6 5 4
    Total no. of ways of wearing them= $$ ^4P_3$$ x $$ ^5P_3$$ x $$ ^6P_3$$
  • Question 5
    1 / -0
    Plot $$(3, 0), (5, 0)$$ and $$(0, 4) $$ on cartesian plane. Name the figure formed by joining these points and find its area.
    Solution
    The figure is a triangle.
    We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
    Therefore, 
    Area $$= \dfrac{1}{2} [3(0-4)+5(4-0)+0] = 4$$ square units.

  • Question 6
    1 / -0
    The area of a triangle with vertices $$A (3, 0), B (7, 0)$$ and $$C (8, 4)$$ is
    Solution
    The area of the  $$ \Delta ABC$$ with vertices 
    $$ A\left( { x }_{ 1 },{ y }_{ 1 } \right)=(3,0), B\left( { x }_{ 2 },{ y }_{ 2 } \right)=(7,0)\ \&\ C\left( { x }_{ 3 },{ y }_{ 3 } \right)=(8,4)$$ is 
    $$ Ar.\Delta ABC=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right)  \right]$$
    $$=\dfrac { 1 }{ 2 } \left\{ 3\left( 0-4 \right) +7\left( 4-0 \right) +8\left( 0-0 \right)  \right\}$$ sq.units $$=8$$ sq. units.
    Hence, option C.
  • Question 7
    1 / -0
    A point $$(a, b)$$ is called a good point if both $$a$$ and $$b$$ are integers. Number of good points on the curve $$xy$$ $$=$$ $$225$$ are
    Solution
    The order pair $$(x, y)$$ satisfying $$xy=225$$ are $$(1, 225), (3, 75) (5, 45), (9, 25), (15, 15)$$. Order can be changed in the first four pairs and both $$x$$ and $$y$$ can be negative also, so the no. of pairs $$=2(2\times 4+1)=18$$
  • Question 8
    1 / -0
     In a class there are $$10$$ boys and $$8$$ girls. The teacher wants to select either a boy or a girl to represent the class in a function. The number of ways the teacher can make this selection.
    Solution
    There are $$10$$ boys and $$8$$ girls in a class. 
    The teacher wants to select just one to represent the class. 
    One boy out of $$10$$ can be selected in $$10$$ ways. 
    Similarly, one girl out of $$8$$ can be selected in $$8$$ ways. 
    So, if he has to select either a boy OR a girl, he can do it in total $$10+8$$ number of ways. i.e $$18$$
  • Question 9
    1 / -0
    The area of a triangle with vertices $$(a, b + c), (b, c + a)$$ and $$(c, a + b)$$ is
    Solution
    The area of $$\triangle ABC$$ with vertices $$A\equiv(x_1, y_1)$$, $$B\equiv(x_2, y_2)$$ and $$C\equiv(x_3, y_3)$$ is given as,
    $$A(\triangle ABC) = \left|\dfrac12[x_1(y_3-y_2) +x_2(y_1 - y_3) + x_3(y_2-y_1) ]\right|$$
    In this problem,
    $$A(\triangle ABC) = \left|\dfrac12[a(a+b-(c+a)) +b(b+c - (a+b)) + c(c+a-(b+c)) ]\right|$$
    $$\therefore A(\triangle ABC) = \left|\dfrac12[ab-ac + bc-ba + ca-cb]\right|$$
    $$\therefore A(\triangle ABC) = 0$$
    Hence, the correct Option is D.
  • Question 10
    1 / -0
    The area of triangle ABC (in sq. units) is :

    Solution
    The area of a triangle is given as:
    Area $$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$
    the given points are $$ A(1,3),~B(-1,0) $$and $$C(4,0)$$
    by Substituting ,
    $$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$$ 
    $$\therefore$$ Area of the given triangle ABC is $$7.5$$ sq. units
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now