Straight Lines Test 19

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Straight Lines Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If the area of a triangle formed by the points (k, 2k) (-2, 6) and (3, 1) is 20 square units. Find the value of k.

A
$$5$$

B
$$4$$

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

D
$$\displaystyle \frac{2}{3}$$

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, area of the triangle with given vertices $$ \left| \frac { k(6-1)-2(1-2k)+3(2k-6) }{ 2 } \right| = 20 $$

$$ \Longrightarrow \left| \frac { 6k-k-2+4k+6k-18 }{ 2 } \right| = 20 $$

$$ \left| \frac { 15k-20 }{ 2 } \right| = 20 $$

$$ \left| 15k-20 \right| = 40 $$

$$ 15k - 20 = 40 $$

$$ 15k = 60 $$

$$ k = 4$$
Question 2
1 / -0

If the coordinates of two points A and B are $$(3, 4)$$ and $$(5, -2)$$, respectively, then the coordinates of any point P if $$PA = PB$$ and area of $$\displaystyle \Delta PAB=10$$ is

A
$$(7, 2)$$ or $$(1, 0)$$

B
$$(-7, 2)$$ or $$(3, 0)$$

C
$$(7, -2)$$ or $$(5, 0)$$

D
$$(7, -2)$$ or $$(-1, 0)$$

Solution

Let P $$ = (x, y) $$
Given, $$ PA = PB $$
$$=> {PA}^{2} = {PB}^{2} $$
$$ => {(x-3)}^{2} + {(y-4)}^{2} = {(x-5)}^{2} + {(y+2)}^{2} $$
$$=> {x} ^{2} -6x + 9 + {y} ^{2} -8y + 16 = {x}^{2} -10x + 25 + {y}^{2} + 4y + 4 $$
$$=> 4x - 12y = 4 $$
$$=> x - 3y = 1 $$ .......(i)

Also, Area of $$ \Delta PAB=10 $$
$$\left| \cfrac {x (4+2)+3(-2-y)+5(y-4)}{ 2 } \right|=10 $$
$$ \left| \cfrac { 6x -6-2y +5y -20 }{ 2 } \right| = 10 $$
$$ \cfrac {6x+3y -26}{2} = \pm 10 $$
$$ 6x+2y-26= \pm 20 $$
$$ 6x+2y = 46 $$ ........(ii)
or $$ 6x + 2y = 6 $$ .......(iii)
Solving equation (i) and (ii), we get $$ x = 7, y = 2 $$
Solving equation (i) and (iii), we get $$ x = 1, y = 0 $$.
So, the co-ordinates of P are $$ (7, 2)$$ or $$(1, 0) $$.
Question 3
1 / -0

- nmmn - mmnm - mnnm -

A
nmmn

B
mnnm

C
nnmm

D
nmnm

Solution

The series is n n m m / n n m m / n n m m / n n m m
Question 4
1 / -0

If the area of a triangle is $$68 $$ sq. units and the vertices are $$(6, 7), (-4, 1)$$ and $$(a, -9) $$ then the value of $$a$$ is

A
1

B
2

C
3

D
4

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is:
$$A= \left| \dfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (6,7) $$ ; $$({ x }_{ 2

},{ y }_{ 2 }) = (-4,1) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (a,-9)$$ in the formula for area, we get:

Area of triangle $$ = \left| \dfrac { (6)(1+9)+(-4)(-9-7) + a(7-1) }{ 2 }

\right| = 68 $$
$$ \left| \dfrac { 60 + 64 + 6a }{ 2 } \right| = 68 $$
$$ \dfrac{124 + 6a}{2} = 68 $$
$$ 124 + 6a = 136 $$
$$ 6a = 12 $$
$$ \implies a = 2 $$
Question 5
1 / -0

Find pair of consecutive odd natural numbers, both of which are larger than 13 such that their sum is less than 40

A
(15, 17)

B
(13, 17)

C
(19, 21)

D
(21, 17)

Solution

Let the consecutive odd numbers be $$ x, x + 2 $$

Given, $$ x > 13 $$ -- (1)

Also, $$ x + 2 > 13 $$
$$ x > 11 $$ -- (2)

And $$ x + x + 2 < 40 $$
$$ 2x + 2 < 40 $$
$$ 2x < 38 $$
$$ x < 19 $$ -- (3)

From, $$ 1, 2, 3 $$
$$ 13 < x < 19 $$
This means, the numbers are $$ 15, 17 $$
Question 6
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

A
$$18$$

B
$$40$$

C
$$21$$

D
$$24$$

Solution
Question 7
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

A
$$86$$

B
$$138$$

C
$$76$$

D
$$120$$

Solution

Sum of outer digits $$\times3$$ is the answer
$$\Rightarrow (4+6+3+5)\times3=18\times3=54$$
$$\Rightarrow (2+3+6+5)\times4=16\times4=64$$
$$\Rightarrow (3+5+7+9)\times5=120$$
Question 8
1 / -0

YEB, CFI, DHL, ....

A
QOL

B
QGL

C
TOL

D
QNL

E
none of these

Solution

First letter of each term is -2 steps; Second of each term +1, +2, +3, +4 steps forward Third letter is alternatively moved +2, +3 steps
Question 9
1 / -0

There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to Delhi. In how many ways a Person can travel from Kota to Delhi via Jaipur by bus?

A
80

B
8

C
10

D
160

Solution

Let $$E_1$$ be the event of travelling from Kota to Jaipur & $$E_2$$ be the event of travelling from Jaipur to Delhi by the person.
$$E_1$$ can happen in 8 ways and $$E_2$$ can happen in 10 ways.
Since both the events $$E_1$$ and $$E_2$$ are to be happened 12 in order, simultaneously,the number of ways $$=8\times10=80$$
Question 10
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle. Choose the correct answer for the term marked by the symbol (?)

A
$$22$$

B
$$88$$

C
$$69$$

D
$$153$$

Solution

Multiply opp digits & substract:
$$\;\rightarrow (8\times9)-(2\times3)=72-6=66$$
$$\;\;\;\;\rightarrow (7\times8)-(3\times4)=56-12=44$$
$$\;\;\;\;\rightarrow (9\times11)-(6\times5)=99-30=69$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 19

Result

Straight Lines Test 19

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SHARING IS CARING

Question 1

$$5$$

$$4$$

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

$$\displaystyle \frac{2}{3}$$

Solution

Question 2

$$(7, 2)$$ or $$(1, 0)$$

$$(-7, 2)$$ or $$(3, 0)$$

$$(7, -2)$$ or $$(5, 0)$$

$$(7, -2)$$ or $$(-1, 0)$$

Solution

Question 3

nmmn

mnnm

nnmm

nmnm

Solution

Question 4

1

2

3

4

Solution

Question 5

(15, 17)

(13, 17)

(19, 21)

(21, 17)

Solution

Question 6

$$18$$

$$40$$

$$21$$

$$24$$

Solution

Question 7

$$86$$

$$138$$

$$76$$

$$120$$

Solution

Question 8

QOL

QGL

TOL

QNL

none of these

Solution

Question 9

80

8

10

160

Solution

Question 10

$$22$$

$$88$$

$$69$$

$$153$$

Solution

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