Differentiation Test 28

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Differentiation Test 28

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Weekly Quiz Competition

Question 1
1 / -0

In figure, if the midpoints of segments $$\overline{GH}, \overline{JK}$$, and $$\overline{LM}$$ are connected, calculate the area of the resulting triangle.

A
$$20$$

B
$$23$$

C
$$26$$

D
$$33$$

Solution

Midpoint of $$GH$$ is $$\left (0, \dfrac {15}{4}\right)$$
Midpoint of $$JK$$ is $$(-5,-2)$$
Midpoint of $$ML$$ is $$(3,-2)$$
If the coordinates of triangle is $$(a,b) , (c,d) , (e,f)$$ then the area formed by the coordinates of triangle is $$\dfrac{1}{2} \times |a(d-f) + c(f-b) + e(b-d)|$$
Therefore, the area enclosed by those midpoints will be
$$ \dfrac12 \times \left |0(-2+2) -5\left (-2-\dfrac {15}{4}\right) +3\left (\dfrac {15}{4}+2\right)\right| $$
$$= \dfrac {1}{2} \times |0+46| $$
$$= 23$$
Question 2
1 / -0

Calculate the area of a triangle with vertices $$(1, 1), (3, 1)$$ and $$(5, 7)$$.

A
$$6$$

B
$$7$$

C
$$9$$

D
$$10$$

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Area of triangle $$=\dfrac 12 \left|1(21-5)-1(7-5)+1(1-3) \right|$$
$$=\dfrac 12 \left| 16-2-2\right| = 6$$
Hence, area of the triangle with the given coordinates is $$6$$.
Question 3
1 / -0

The value of $$\dfrac {(n + 2)! - (n + 1)!}{n!} $$ is:

A
$$(n + 2)!$$

B
$$(n + 1)!$$

C
$$(n + 2)^{2}$$

D
$$(n + 1)^{2}$$

E
$$n$$

Solution

The value of $$\dfrac { (n+2)!-(n+1)! }{ n! } $$
$$=\dfrac { (n+2)(n+1)n!-(n+1)n! }{ n! } $$
$$=\dfrac { (n+1)(n!)(n+2-1) }{ n! } $$
$$=\dfrac { (n+1)(n+1) }{ 1 } $$
$$={ (n+1) }^{ 2 }$$
Question 4
1 / -0

Find the area of a triangle whose vertices are $$(0, 6\sqrt {3}), (\sqrt {35}, 7)$$, and $$(0, 3)$$.

A
$$15.37$$

B
$$17.75$$

C
$$21.87$$

D
$$25.61$$

E
$$39.61$$

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 whose vertices are $$(0,6\sqrt{3}),(\sqrt{35},7) ,(0,3)$$
$$=\dfrac{1}{2}\left [ 0(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7)) \right ]$$
$$=\dfrac{1}{2}\left [ \sqrt{35}(3-6\sqrt{3}) \right ]$$
$$=\dfrac{1}{2}\left [ 5.91(3-6\times 1.73) \right ]$$
$$=\dfrac{1}{2}\times 5.91\times 7.40$$
$$=\dfrac{1}{2}\times 43.74$$
$$=21.87$$
Question 5
1 / -0

The area of the triangle with coordinates $$(1, 2), (5, 5)$$ and $$(k, 2)$$ is $$15$$ square units. Calculate a possible value for $$k$$.

A
$$-10$$

B
$$-9$$

C
$$-5$$

D
$$5$$

E
$$6$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
Area of triangle is $$15 = \dfrac {1}{2} \times |1(5-2) + 5(2-2) + k(2-5)| $$
$$ \Rightarrow \dfrac {1}{2} \times |3+0-3k| = \dfrac {1}{2} \times |3-3k| = 15$$
$$\Rightarrow |3-3k|=30$$ => $$|1-k| = 10$$
$$\Rightarrow 1-k = 10$$ and $$1-k=-10$$
$$\Rightarrow k=-9$$ and $$k=11$$
Question 6
1 / -0

$$m, 2m, 4m, . . . $$
The first term in the sequence above is $$m$$, and each term thereafter is equal to twice the previous term. If $$m$$ is an integer, which of the following could NOT be the sum of the first four terms of this sequence?

A
$$-26$$

B
$$-15$$

C
$$45$$

D
$$75$$

E
$$120$$

Solution

The first term in given Geometric series, $$a_1=m$$
The common ratio $$r$$ $$=\dfrac{4m}{2m}=2$$
No. of terms, $$n$$ $$=4$$
Applying sum of GP formula,
$$S_n=\dfrac{a(1-r^n)}{1-r}$$
$$=\dfrac{m(1-2^4)}{1-2}$$
$$=\dfrac{m(1-16)}{-1}=15m$$
If $$m$$ is an integer the sum of the first four terms of this sequence should be in multiple of $$15$$.
Hence, option A is correct which is not a multiple of $$15$$.
Question 7
1 / -0

In fig., the area of triangle ABC (in sq. units) is:

A
$$15$$

B
$$10$$

C
$$7.5$$

D
$$2.5$$

Solution

Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$

Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$

Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$
Question 8
1 / -0

For all numbers a and b, let $$\displaystyle a\bigodot b$$ be defined by $$\displaystyle a\bigodot b=ab+a+b$$. Then for the numbers $$x$$, $$y$$ and $$z$$, which of the following is/are true?
I. $$\displaystyle x\bigodot y=y\bigodot x$$
II. $$\displaystyle \left( x-1 \right) \bigodot \left( x+1 \right) =\left( x\bigodot x \right) -1$$
III. $$\displaystyle x\bigodot \left( y+z \right) =\left( x\bigodot y \right) +\left( x\bigodot z \right) $$

A
I only

B
II only

C
III only

D
I and II only

E
I, II, and III

Solution

$$For(I)$$
$$x\odot y=y\odot x$$
$$x\odot y=xy+x+y\Rightarrow x\odot y=y\odot x$$
$$y\odot x=yx+y+x$$

$$For(II)$$
$$\left( x-1 \right) \odot \left( x+1 \right)$$
$$=(x-1)(x+1)+x-1+x+1$$
$$={ x }^{ 2 }-1+2x$$
$$\left( x\odot x \right) -1=x\times x+x+x-1$$
$${ x }^{ 2 }+2x-1$$
$$\therefore \left( x-1 \right) \odot \left( x+1 \right) =\left( x\odot x \right) -1$$

$$For(III)$$
$$x\odot \left( y+z \right) $$
$$=x\left( y+z \right) +x+y+z$$
$$=xy+xz+x+y+z$$
$$x\odot y=xy+x+y$$
$$x\odot z=xz+x+z$$
$$x\odot y+x\odot z=xy+xz+x+y+x+z$$
$$=xy+xz+2x+y+z$$
$$\Rightarrow x\odot \left( y+z \right) \neq x\odot y+x\odot z$$

$$\therefore (I)\& (II)$$ are true.
Question 9
1 / -0

$$N=a^2 + b^2$$ is a three-digit number which is divisible by 5. a = 10x + y and b = 10x + z, where z is a prime number, and x and y are natural numbers. If a + b = 31, find the value of N.

A
565

B
485

C
505

D
485 or 505

Solution
Question 10
1 / -0

Let $$\boxed { n }$$ be defined as $$\frac{(n+2)!}{(n-1)!}$$, what is the value of $$\frac{\boxed{7}}{\boxed {3}}$$ ?

A
4.4

B
8.4

C
12.4

D
16.4

E
20.4

Solution

Given, box $$n$$ is equal to $$\dfrac{(n+2)!}{(n-1)!}=(n+2)(n+1)n$$
We get box $$7$$ is equal to $$9 \times 8 \times 7 =504$$
We get box $$3$$ is equal to $$5 \times 4 \times 3 = 60$$
Therefore, If we divide those two, we get $$\dfrac{504}{60} = 8.4$$.

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Re-Attempt Weekly Quiz Competition

Differentiation Test 28

Result

Differentiation Test 28

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SHARING IS CARING

Question 1

$$20$$

$$23$$

$$26$$

$$33$$

Solution

Question 2

$$6$$

$$7$$

$$9$$

$$10$$

Solution

Question 3

$$(n + 2)!$$

$$(n + 1)!$$

$$(n + 2)^{2}$$

$$(n + 1)^{2}$$

$$n$$

Solution

Question 4

$$15.37$$

$$17.75$$

$$21.87$$

$$25.61$$

$$39.61$$

Solution

Question 5

$$-10$$

$$-9$$

$$-5$$

$$5$$

$$6$$

Solution

Question 6

$$-26$$

$$-15$$

$$45$$

$$75$$

$$120$$

Solution

Question 7

$$15$$

$$10$$

$$7.5$$

$$2.5$$

Solution

Question 8

I only

II only

III only

I and II only

I, II, and III

Solution

Question 9

565

485

505

485 or 505

Solution

Question 10

4.4

8.4

12.4

16.4

20.4

Solution

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