Straight Lines Test 26

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Straight Lines Test 26

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Question 1
1 / -0

If $$a\odot b = 6\times a - 3\times b$$, evaluate $$(5\odot 3) \odot 20$$

A
$$2$$

B
$$41$$

C
$$66$$

D
$$1$$

Solution

$$\Rightarrow$$ $$a\odot b=6\times a-3\times b$$ [ Given ]
$$\Rightarrow$$ $$(5\odot 3)\odot 20$$
$$\Rightarrow$$ Let $$a=5$$ and $$b=3$$
$$\Rightarrow$$ $$(6\times 5-3\times 3)\odot 20$$
$$\Rightarrow$$ $$(30-9)\odot 20$$
$$\Rightarrow$$ $$21\odot 20$$
$$\Rightarrow$$ Let $$a=21$$ and $$b=20$$
$$\Rightarrow$$ $$6\times 21-3\times 20$$
$$\Rightarrow$$ $$126-60$$
$$\Rightarrow$$ $$66$$
$$\therefore$$ $$(5\odot 3)\odot 20=66$$
Question 2
1 / -0

Two vertices of a triangle are (2, 1) and (3, -2). Its third vertex is (x, y) such that $$\displaystyle y=x+3$$. If its area is 5 sq. units, what are the co-ordinates of the third vertex?

A
(3.5, -6.5)

B
(3.5, 6.5)

C
(-1.5, -1.5)

D
(1.5, -1.5)

Solution

Given third vertex such that,
$$y=(x+3)$$
since, Area of triangle $$ABC=55q$$ units
$$\pm \dfrac{1}{2}{x(1+2)+2(-2-y)+3(y-1)}=\xi$$
$$\Rightarrow \pm \dfrac{1}{2}{x+2x-4-2y+3y-3}=\xi$$
$$\Rightarrow {3x+y-7}=\pm10$$
$$\Rightarrow 3x+y-17=0$$ ------------(1)
and $$3x+y+3=0$$ ------------(2)
Given that $$A(x,y)$$ lies any $$=(x+3)$$ ------------(3)
from equation (L) and (3),
$$x=\dfrac{7}{2},y=\dfrac{13}{2}$$
$$\Rightarrow \boxed{x=3.\xi\,and \, y=6.\xi}$$
from equation (2) and (3) we get,
$$x=\dfrac{-3}{2}\,and\, y=1.\xi $$
Question 3
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 $$\dfrac{504}{60} = 8.4$$.
Question 4
1 / -0

Area of the triangle formed by the points $$\left( 0,0 \right) ,\left( 2,0 \right) $$ and $$\left( 0,2 \right) $$ is

A
$$1$$ sq. unit

B
$$2$$ sq. units

C
$$4$$ sq. units

D
$$8$$ sq. units

Solution

Given: $$A=(x_1,y_1)=(0,0)$$

$$B=(x_2,y_2)=(2,0)$$ and

  $$C=(x_3,y_3)=(0,2)$$

Area of triangle $$=\dfrac {1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)++x_3(y_1-y_2)]$$

$$=\dfrac {1}{2}[0(0-2)+2(2-0)+0(0-0)]$$

         $$=\dfrac {1}{2}[2(2)]$$

         $$=2$$ sq. units.
Question 5
1 / -0

With the help of match-sticks, Zalak prepared a pattern as shown below. When $$97$$ matchsticks are used, the serial number of the figure will be ...........

A
Figure 32

B
Figure 95

C
Figure 49

D
Figure 48

Solution

Pattern associated with the number of matchsticks:
Figure 1: No. of sticks $$=3=3+0$$
Figure 2: No. of sticks $$=5=3+2$$
Figure 3: No. of sticks $$=7=3+2+2$$
Figure 4: No. of sticks $$=9=3+2+2+2$$
----------------------------------------
The rule associated with it can be given as no. of sticks $$=3+2(n-1)$$ where $$n$$ is the number of the figure.
Now there are $$97$$ matchsticks
$$\implies 3+2(n-1)=97$$
$$\implies 3+2n-2=97$$
$$\implies 2n=96$$
$$\implies n=48$$
Hence, serial number of the figure having $$97$$ matchsticks is $$48$$.
Question 6
1 / -0

Area of the triangle formed by the points $$(0,0),(2,0)$$ and $$(0,2)$$ is

A
$$1$$ sq.units

B
$$2$$ sq.units

C
$$4$$ sq.units

D
$$8$$ sq.units

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Therefore,
$$=\dfrac{1}{2}\left|0(0-2)+2(2-0)+0(0-0)\right|$$

$$=\dfrac{1}{2}\left|0+4+0\right|$$

$$=\dfrac{1}{2}\times 4=2\ sq.\ units$$

Hence, this is the answer.
Question 7
1 / -0

Find the value of $$a$$ if area of the triangle is $$17$$ square units whose vertices are $$(0,0), (4,a), (6,4)$$.

A
$$a=-2$$

B
$$a=-3$$

C
$$a=3$$

D
none of the above

Solution

vertices of the triangle are $$A(0,0), B(4,a), C(6,4)$$.

Then area of triangle$$ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

$$\Rightarrow 17=\dfrac{1}{2}[0(a-4)+4(4-0)+6(0-a)$$

$$\Rightarrow 34=16-6a$$

$$\Rightarrow a=-\dfrac{18}{6}=-3$$
Question 8
1 / -0

Out of $$7$$ consonants and $$4$$ vowels, words are formed each having $$3$$ consonants and $$2$$ vowels. The number of such words that can be formed is

A
$$210$$

B
$$25200$$

C
$$2520$$

D
$$302400$$

Solution

Total no. of consonants is $$7$$ and of vowels is $$4$$
Out of $$7$$ consonants, $$3$$ are chosen to form a word. So, this can be done in $${ }^{7}C_{3}$$ ways
Out of $$4$$ vowels, $$2$$ are chosen to form a word. So, that can be done in $${ }^{4}C_{2}$$ ways
So, the no. of select $$3$$ consonants and $$2$$ vowels is $$^{7}C_{3} \times ^{4}C_{2}$$
Now, we can arrange the word containing $$5$$ letters in $$5!$$ ways.
Thus, the total no. of words formed each having $$3$$ consonants and $$2$$ vowels is
$$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$$
Hence, the answer is $$25200$$.
Question 9
1 / -0

Consider an incomplete pyramid of balls on a square base having $$18$$ layers; and having $$13$$ balls on each side of the top layer. Then the total number $$N$$ of balls in that pyramid satisfies

A
$$9000 < N < 10000$$

B
$$8000 < N < 9000$$

C
$$7000 < N < 8000$$

D
$$10000 < N < 12000$$

Solution

The top layer has $$(13\times 13)$$ balls the layer below it will have $$(14\times 14)$$ balls

We have $$18$$ layers

So the total number of balls

$$N=(13\times 13)+(14\times 14)+.......(30\times 30)\\ N={ 13 }^{ 2 }+{ 14 }^{ 2 }+.....{ 30 }^{ 2 }$$

Sum of squares of first $$n$$ natural numbers is $$\dfrac { n(n+1)(2n+1) }{ 6 } \\ $$

$$\therefore N=$$ sum of first $$30$$ $$-$$ sum of first $$12$$

$$N=\dfrac { 30\times 31\times 61 }{ 6 } -\dfrac { 12\times 13\times 25 }{ 6 } \\ N=8805\\ $$

$$\Rightarrow 8000<N<9000$$
Hence, option B is correct.
Question 10
1 / -0

Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points are dividing the plane into two regions each containing four of the remaining points is

A
1

B
5

C
10

D
Dependent on the configuration of points.

Solution

You can clearly see from the attached figure
Five such lines are possible joining $$AF,BG,CH,DI,EJ$$

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Straight Lines Test 26

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Straight Lines Test 26

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Question 1

$$2$$

$$41$$

$$66$$

$$1$$

Solution

Question 2

(3.5, -6.5)

(3.5, 6.5)

(-1.5, -1.5)

(1.5, -1.5)

Solution

Question 3

4.4

8.4

12.4

16.4

20.4

Solution

Question 4

$$1$$ sq. unit

$$2$$ sq. units

$$4$$ sq. units

$$8$$ sq. units

Solution

Question 5

Figure 32

Figure 95

Figure 49

Figure 48

Solution

Question 6

$$1$$ sq.units

$$2$$ sq.units

$$4$$ sq.units

$$8$$ sq.units

Solution

Question 7

$$a=-2$$

$$a=-3$$

$$a=3$$

none of the above

Solution

Question 8

$$210$$

$$25200$$

$$2520$$

$$302400$$

Solution

Question 9

$$9000 < N < 10000$$

$$8000 < N < 9000$$

$$7000 < N < 8000$$

$$10000 < N < 12000$$

Solution

Question 10

1

5

10

Dependent on the configuration of points.

Solution

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