Differentiation Test 23

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Differentiation Test 23

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
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 5
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 6
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 7
1 / -0

Find the missing letters
A, B, D, G, .....

A
L

B
M

C
J

D
K

Solution

Pattern is $$\displaystyle ^{1}A,^{1}A+1,^{2}B+2,^{4}D+3,^{7}G+4,^{11}K$$
$$\displaystyle \therefore $$ Missing letter = K
Question 8
1 / -0

Find the missing letters
M, N, O, L, R, I, V, ........

A
H

B
K

C
E

D
S

Solution

Pattern is combination of two patterns M, O, R, V, A, .... and N, L, I, H, .......
$$\displaystyle ^{13}M,^{14}N,^{15}O,^{11}L,^{18}R,^{9}I,^{22}V,^{8}H,$$ .....
$$\displaystyle \therefore $$ Missing letter = H
Question 9
1 / -0

The area of triangle formed by $$(0, 0), (0, a)$$ and $$(b, 0)$$ is .......... .

A
$$ab$$

B
$$\displaystyle \frac{ab}{2}$$

C
$$\left | \displaystyle \frac{ab}{2} \right |$$

D
$$\left | ab \right |$$

Solution

The area of a triangle formed by joining the points $$(x_1, y_1)$$, $$(x_2, y_2)$$ and $$(x_3, y_3)$$ is

$$A=\dfrac { 1 }{ 2 } \bigg| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \bigg|$$

Therefore, the area of a triangle formed by joining the points $$(0, 0)$$, $$(0, a)$$ and $$(b, 0)$$ is:

$$A=\dfrac { 1 }{ 2 } \bigg| 0(0-b)+a(b-0)+0(0-0) \bigg| $$

$$=\dfrac { 1 }{ 2 } \bigg| 0+ab \bigg| $$

$$=\bigg| \dfrac { ab }{ 2 } \bigg|$$

Hence, the area of the triangle is $$\left| \dfrac { ab }{ 2 } \right|$$
Question 10
1 / -0

The midpoints of the sides of triangle $$ABC$$ are $$ (-1,-2), (6,1)$$ and $$(3,5) $$. The area of $$\displaystyle \triangle ABC$$ is ____ square units.

A
$$72$$

B
$$73$$

C
$$74$$

D
$$75$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle formed with midpoints
$$\displaystyle =\frac{1}{2}\left [ -1\left ( 1-5 \right )+6\left ( 5+2 \right )+3\left ( -2-1 \right ) \right ]$$
$$\displaystyle =\frac{37}{2}$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta ABC=4\times \frac{37}{2}=74$$ square units.
$$[\because$$ Area of triangle formed by joining mid-points of the sides of given triangle is $$\left(\dfrac{1}{4}\right)^{th}$$ of the area of original triangle $$]$$

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Differentiation Test 23

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Differentiation Test 23

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Question 1

1

2

3

4

Solution

Question 2

$$(0,\:3)$$

$$(5,\:8)$$

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

None of these

Solution

Question 3

$$48$$

$$42$$

$$36$$

$$45$$

Solution

Question 4

$$5$$

$$4$$

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

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

Solution

Question 5

QOL

QGL

TOL

QNL

none of these

Solution

Question 6

80

8

10

160

Solution

Question 7

L

M

J

K

Solution

Question 8

H

K

E

S

Solution

Question 9

$$ab$$

$$\displaystyle \frac{ab}{2}$$

$$\left | \displaystyle \frac{ab}{2} \right |$$

$$\left | ab \right |$$

Solution

Question 10

$$72$$

$$73$$

$$74$$

$$75$$

Solution

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