Straight Lines Test 15

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Straight Lines Test 15

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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

A
$(0, 1)$

B
$(0, -1)$

C
$(4,-1)$

D
$(-1,4)$

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,

A
$1$

B
$^{36}C_{17}$

C
$\dfrac {2}{19}$

D
$^{36}C_{18}$

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

A
$1080$

B
$4320$

C
$360$

D
$180$

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

A
${^1}{^5}P_3$

B
$4^3 5^3 6^3$

C
${^4}P_3 {^5}P_3 {^6}P_3$

D
$180$

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.

A
Figure formed is a Rectangle and Area $= 52$ sq. unit

B
Figure formed is a triangle and Area $= 4$ sq. unit

C
Figure formed is a square and Area $= 40$ sq. unit

D
None of these

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

A
$14$

B
$28$

C
$8$

D
$6$

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

A
20

B
18

C
16

D
14

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.

A
$18$

B
$80$

C
${^1}{^0}P_8$

D
${^1}{^0}C_8$

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

A
$a$

B
$b$

C
$c$

D
$0$

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 :

A
$15$ sq. units

B
$10$ sq. units

C
$7.5$ sq. units

D
$2.5$ sq. units

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

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Straight Lines Test 15

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Straight Lines Test 15

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Question 1

(0,1)(0, 1)(0,1)

(0,−1)(0, -1)(0,−1)

(4,−1)(4,-1)(4,−1)

(−1,4)(-1,4)(−1,4)

Solution

Question 2

111

36C17 ^{36}C_{17} 36C17​

219\dfrac {2}{19}192​

36C18 ^{36}C_{18} 36C18​

Solution

Question 3

108010801080

432043204320

360360360

180180180

Solution

Question 4

15P3{^1}{^5}P_315P3​

4353634^3 5^3 6^3435363

4P35P36P3{^4}P_3 {^5}P_3 {^6}P_34P3​5P3​6P3​

180180180

Solution

Question 5

Figure formed is a Rectangle and Area =52= 52=52 sq. unit

Figure formed is a triangle and Area =4= 4=4 sq. unit

Figure formed is a square and Area =40= 40=40 sq. unit

None of these

Solution

Question 6

141414

282828

888

666

Solution

Question 7

20

18

16

14

Solution

Question 8

181818

808080

10P8{^1}{^0}P_810P8​

10C8{^1}{^0}C_810C8​

Solution

Question 9

aaa

bbb

ccc

000

Solution

Question 10

151515 sq. units

101010 sq. units

7.57.57.5 sq. units

2.52.52.5 sq. units

Solution

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$(0, 1)$

$(0, -1)$

$(4,-1)$

$(-1,4)$

$1$

$^{36}C_{17}$

$\dfrac {2}{19}$

$^{36}C_{18}$

$1080$

$4320$

$360$

$180$

${^1}{^5}P_3$

$4^3 5^3 6^3$

${^4}P_3 {^5}P_3 {^6}P_3$

$180$

Figure formed is a Rectangle and Area $= 52$ sq. unit

Figure formed is a triangle and Area $= 4$ sq. unit

Figure formed is a square and Area $= 40$ sq. unit

$14$

$28$

$8$

$6$

$18$

$80$

${^1}{^0}P_8$

${^1}{^0}C_8$

$a$

$b$

$c$

$0$

$15$ sq. units

$10$ sq. units

$7.5$ sq. units

$2.5$ sq. units