Straight Lines Test 25

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Straight Lines Test 25

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
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 5
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 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

The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, -2). The third vertex lies on $\displaystyle y=x+3$ . What is the third vertex?

A
$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right)$

B
$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right)$

C
$\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right)$

D
$\displaystyle (0,0)$

Solution
Question 8
1 / -0

In the $xy$ -plane, the vertices of a triangle are $(-1,3), (6,3)$ and $(-1,-4)$ . The area of the triangle is ___ square units.

A
$10$

B
$17.5$

C
$24.5$

D
$35$

Solution

If $(x_1,y_1)$ , $(x_2,y_2)$ and $(x_3,y_3)$ are the vertices of a triangle then its area is given by,
$A=\left| \dfrac { 1 }{ 2 } (x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 })) \right|$
Therefore, with the vertices $(-1,3)$ , $(6,3)$ and $(-1,-4)$ .
Area of triangle is given by,
$A=\left| \dfrac { 1 }{ 2 } (-1(3-(-4))+6(-4-3)+(-1)(3-3)) \right|$
$=\left| \dfrac { 1 }{ 2 } (-7-42) \right|$
$=\left| \dfrac { 1 }{ 2 } (-49) \right|$
$=\left| -24.5 \right|$
$=24.5$ square units.
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

If $m * n = m+(m-1)+(m-2)+ ...... +(m-n)$ , evaluate $7 * 5$ .

A
$25$

B
$20$

C
$27$

D
$18$

Solution

$\Rightarrow$ $m * n=m+(m-1)+(m-2)+....+(m-n)$
$\Rightarrow$ In $7 * 5$ , $m=7$ and $n=5$
$\Rightarrow$ $7 * 5=7+(7-1)+(7-2)+(7-3)+(7-4)+(7-5)$
$\Rightarrow$ $7 * 5=7+6+5+4+3+2$
$\therefore$ $7 * 5=27$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 25

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Straight Lines Test 25

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Question 1

(n+2)!(n + 2)!(n+2)!

(n+1)!(n + 1)!(n+1)!

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

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

nnn

Solution

Question 2

15.3715.3715.37

17.7517.7517.75

21.8721.8721.87

25.6125.6125.61

39.6139.6139.61

Solution

Question 3

−10-10−10

−9-9−9

−5-5−5

555

666

Solution

Question 4

I only

II only

III only

I and II only

I, II, and III

Solution

Question 5

151515

101010

7.57.57.5

2.52.52.5

Solution

Question 6

−26-26−26

−15-15−15

454545

757575

120120120

Solution

Question 7

(72,132)\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) (27​,213​)

(52,52)\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) (25​,25​)

(−32,−32)\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right) (−23​,−23​)

(0,0)\displaystyle (0,0)(0,0)

Solution

Question 8

101010

17.517.517.5

24.524.524.5

353535

Solution

Question 9

565

485

505

485 or 505

Solution

Question 10

252525

202020

272727

181818

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

$(n + 2)!$

$(n + 1)!$

$(n + 2)^{2}$

$(n + 1)^{2}$

$n$

$15.37$

$17.75$

$21.87$

$25.61$

$39.61$

$-10$

$-9$

$-5$

$5$

$6$

$15$

$10$

$7.5$

$2.5$

$-26$

$-15$

$45$

$75$

$120$

$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right)$

$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right)$

$\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right)$

$\displaystyle (0,0)$

$10$

$17.5$

$24.5$

$35$

$25$

$20$

$27$

$18$