Straight Lines Test 24

Result

Straight Lines Test 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Fill in the blank: $$62, 66, 63, 66, 64,$$ __$$, 65,....$$

A
60

B
61

C
62

D
63

Solution

The series alternates the addition of $$4$$ with the subtraction of $$3$$.
So, the missing number is $$61$$.
Question 2
1 / -0

Identify the missing integer: $$9, 45,$$ ____$$, 1125, 5625...$$

A
$$220$$

B
$$223$$

C
$$224$$

D
$$225$$

Solution

This is continuous series multiplied by $$5$$.
$$9 \times 5 = 45$$
$$45 \times 5 = 225$$
$$225 \times 5 = 1,125$$
$$1,125 \times 5= 5,625.$$
So, the missing integer is $$225$$.
Question 3
1 / -0

The area of the triangle whose vertices are $$(0, 1), (1, 4)$$ and $$(1, 2)$$ is ___ square units.

A
1

B
2

C
3

D
4

Solution

Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 0$$, $$y_{1} = 1$$, $$x_{2} = 1$$, $$y_{2} = 4$$, $$x_{3} = 1$$ and $$y_{3} = 2$$
Substitute the values, we get
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[0(4 - 2) + 1(2 - 1) + 1(1 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[0 + 1 - 3]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -2\right|$$
$$=$$ $$\left|- 1 \right|$$
Area always in absolute value.
So, area of the triangle is $$ 1$$ square units.
Question 4
1 / -0

$$(9, 2), (5, -1) $$ and $$ (7, -5)$$ are the vertices of the triangle. Find its area.

A
10 square units

B
11 square units

C
12 square units

D
13 square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 9$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = -1$$, $$x_{3} = 7$$ and $$y_{3} = -5$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[9(-1 + 5) + 5(-5 - 2) + 7(2 + 1)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 35 + 21]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 22\right|$$
$$=$$ $$\left|11 \right|$$
Area always in absolute value.
So, area of the triangle $$= 11$$ square units.
Question 5
1 / -0

If n is an integer with $$0\le n \le 11$$ then the minimum value of $$n!(11-n)!$$ is attained when a value of n =

A
11

B
5

C
7

D
9

Solution

Let $$y=n!(11-n)!$$
Consider $$x= ^{11}C_n=\dfrac{11!}{n!(11-n)!}$$
For maximum value of $$x$$ we must have $$n=6$$ or $$n=5$$
Thus $$x_{max}=^{11}C_6=\dfrac{11!}{5!6!}=\dfrac{11!}{y_{min}}$$
$$\Rightarrow y_{min}=5!6!$$ i.e. $$n=6$$ or $$n=5$$
Question 6
1 / -0

What is the area of the triangle whose vertices are: $$(-3, 15), (6, -7) $$ and $$(10, 5)$$?

A
$$94$$ square units

B
$$96$$ square units

C
$$97$$ square units

D
$$98$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is $$A=\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
Here $$x_{1} = -3$$, $$y_{1} = 15$$, $$x_{2} = 6$$, $$y_{2} = -7$$, $$x_{3} = 10$$ and $$y_{3} = 5$$

Substituting the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[-3(-7 - 5) + 6(5 - 15) + 10(15 + 7)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[36 - 60 + 220]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 196\right|$$
$$=$$ $$\left|98 \right|$$

Area is always in absolute value.
So, area of the triangle $$= 98$$ square units.
Question 7
1 / -0

$$6, 10, 18, 34, 66$$
The first number in the list above is $$6$$. Determine a rule for finding each successive number in the list.

A
Add $$4$$ to the preceding number.

B
Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

C
Double the preceding number and then subtract $$2$$ from that result.

D
Subtract $$2$$ from the preceding number and then double that result.

E
Triple the preceding number and then subtract $$8$$ from that result.

Solution

The series is $$6,10,18,34,66$$
In the given series:
$$6+6=12-2=10$$
$$10+10=20-2=18$$
$$18+18=36-2=34$$
$$34+34=68-2=66$$
Hence in this each successive number is obtained by double the preceding number and then subtract $$2$$ from the result.
Question 8
1 / -0

What is the area of the triangle for the following points $$(6, 2), (5, 4)$$ and $$(3, -1)$$?

A
2.3 square units

B
4.5 square units

C
4.1 square units

D
3.6 square units

Solution

Formula for area of triangle is $$\left|\dfrac{1}{2}[x_{1}(y_{2}-y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \right|$$
where $$x_{1} = 6$$, $$y_{1} = 2$$, $$x_{2} = 5$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = -1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[6(4 + 1) + 5(-1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[30 - 15 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times 9\right|$$
$$=$$ $$\left|4.5 \right|$$
Area always in absolute value.
So, area of the triangle $$= 4.5$$ square units.
Question 9
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 10
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 $$6$$.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Straight Lines Test 24

Result

Straight Lines Test 24

Score

-

Rank

-

SHARING IS CARING

Question 1

60

61

62

63

Solution

Question 2

$$220$$

$$223$$

$$224$$

$$225$$

Solution

Question 3

1

2

3

4

Solution

Question 4

10 square units

11 square units

12 square units

13 square units

Solution

Question 5

11

5

7

9

Solution

Question 6

$$94$$ square units

$$96$$ square units

$$97$$ square units

$$98$$ square units

Solution

Question 7

Add $$4$$ to the preceding number.

Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

Double the preceding number and then subtract $$2$$ from that result.

Subtract $$2$$ from the preceding number and then double that result.

Triple the preceding number and then subtract $$8$$ from that result.

Solution

Question 8

2.3 square units

4.5 square units

4.1 square units

3.6 square units

Solution

Question 9

$$20$$

$$23$$

$$26$$

$$33$$

Solution

Question 10

$$6$$

$$7$$

$$9$$

$$10$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.