Sequences and Series Test 19

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Sequences and Series Test 19

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Weekly Quiz Competition

Question 1
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 2
1 / -0

The triangle with vertices A(4, 4), B(-2, -6) and C(4, -1) is shown in the diagram. The area of $$\Delta$$ ABC is _______

A
5 sq. units

B
12 sq. units

C
15 sq. units

D
20 sq. units

Solution

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 $$= \cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$

$$=\dfrac{1}{2}|4(-6-(-1))-2(-1-4)+4(4-(-6))|$$

$$=\dfrac{1}{2}|4(-5)-2(-5)+4(10)|$$

$$=\dfrac{1}{2}|30|$$

$$=15 $$ sq. units
Question 3
1 / -0

The coordinates of $$A$$ for which area of triangle, whose vertices are $$A(a, 2a),\ B(-2, 6)$$ and $$C(3, 1)$$ is $$10$$ square units, are:

A
$$(0,\:3)$$

B
$$(5,\:8)$$

C
$$(\displaystyle 3, \frac{8}{3})$$

D
None of these

Solution

Given: Vertices of the triangle are $$A(a, 2a),\ B(-2, 6),\ C(3,1)$$

The area of the triangle is $$10$$ square units.

$$\Rightarrow \dfrac{1}{2}\bigg | a(6-1)+(-2)(1-2a)+3(2a-6) \bigg |=10$$

$$\Rightarrow 5a-2+4a+6a-18=20$$

$$\Rightarrow 15a=40$$

$$\Rightarrow a=\dfrac{40}{15}=\dfrac{8}{3}$$

$$\therefore $$ The coordinates of vertex $$A $$ are $$\displaystyle \left ( \dfrac{8}{3}, \frac{16}{3} \right )$$
Question 4
1 / -0

The number of positive fractions $$\dfrac{m}{n}$$ such that $$\dfrac{1}{3}< \dfrac{m}{n}<1 $$ and having the property that the fraction remains the same by adding some positive integer to the numerator and multiplying the denominator by the same positive integer is:

A
$$1$$

B
$$3$$

C
$$6$$

D
infinite

Solution

Given $$m,n$$ are integers $$\dfrac{1}{3}<\dfrac{m}{n}<1$$
$$\Rightarrow \dfrac{m+x}{nx}=\dfrac{m}{n}$$
$$\Rightarrow m+x=mx$$
$$\Rightarrow x=\dfrac{m}{m-1},$$ $$x$$ must be an integer.
So only possible value of $$m$$ satisfying above equation is $$2.$$
$$\Rightarrow m=2$$
$$\Rightarrow \dfrac{m}{n}>\dfrac{1}{3}\Rightarrow 3m>n\Rightarrow 6>n$$

$$\Rightarrow \dfrac{m}{n}<1\Rightarrow m<n\Rightarrow 2<n$$
$$\Rightarrow n=3,4,5$$
$$\therefore 3$$ Possible fractions.
Hence, the answer is $$3.$$
Question 5
1 / -0

Three points A, B and C have coordinates $$(a, b + c), \ (b, c + a)$$ and $$(c, a + b)$$, respectively. The area of the triangle ABC will be:

A
$$\displaystyle a^{2}+b^{2}+c^{2}$$

B
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}2$$

C
$$\displaystyle \dfrac{a^{2}+b^{2}+c^{2}}4$$

D
$$0$$

Solution

Area of triangle $$=\dfrac{ 1 }{ 2 }\left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$

Given the points: $$A(a, b+c) , B(b, c+a) , C(c, a+b)$$
Therefore, area:

$$A =\dfrac{ 1 }{ 2 }\left[ a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a) \right]$$

$$A =\dfrac{ 1 }{ 2}{\left[ a(c-b)+b(a-c)+c(b-a) \right] }$$

$$A=\dfrac{ 1 }{ 2}{\left[ ac-ab+ba-bc+cb-ac \right] }=0$$
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 (?)
$$0,\,3,\,9,\,18,\,30,\,?$$

A
$$48$$

B
$$42$$

C
$$36$$

D
$$45$$

Solution

The difference between the consecutive numbers increases by $$3$$.
$$3-0 = 3$$
$$9 - 3 = 6$$
$$18-9 = 9$$
$$30-18 = 12$$.
Pattern is adding multiples of 3
Next term is $$30+15=45$$
Question 7
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 8
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 9
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 10
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 $$ 15, 17 $$