Permutations and Combinations Test 25

Result

Permutations and Combinations Test 25

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

Which pair of numbers comes next?
$$10, 18, 12, 21, 15 ,25, ,....$$

A
$$19, 30$$

B
$$19, 27$$

C
$$30, 19$$

D
$$27, 19$$

Solution

This is an alternating addition series
First series = $$10, 10+2=12, 12+3=15$$. $$\therefore$$ next number= $$15+4=19$$
Second series = $$18, 18+3=21, 21+4=25$$ $$\therefore$$ next number = $$25+5=30$$
The next pair of numbers are $$19, 30$$.
Question 2
1 / -0

The area of the triangle formed from points $$(1, 2), (2, 4)$$ and $$(3, 1)$$ is ____ square units.

A
$$\dfrac{5}{2}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{2}$$

D
$$2$$

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 $$x_{1} = 1$$, $$y_{1} = 2$$, $$x_{2} = 2$$, $$y_{2} = 4$$, $$x_{3} = 3$$ and $$y_{3} = 1$$
Substitute the values, we get,
Area of triangle $$=$$ $$\left|\dfrac{1}{2}\times[1(4 - 1) + 2(1 - 2) + 3(2 - 4)]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times[3 - 2 - 6]\right|$$
$$=$$ $$\left|\dfrac{1}{2}\times -5\right|$$
$$=$$ $$\left|-\frac{5}{2} \right|$$
Area always in absolute value.
So, area of the triangle $$= \dfrac{5}{2}$$ square units.
Question 3
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 4
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 5
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 6
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 7
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 8
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 9
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 10
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 $$k=11$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 25

Result

Permutations and Combinations Test 25

Score

-

Rank

-

SHARING IS CARING

Question 1

$$19, 30$$

$$19, 27$$

$$30, 19$$

$$27, 19$$

Solution

Question 2

$$\dfrac{5}{2}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{2}$$

$$2$$

Solution

Question 3

10 square units

11 square units

12 square units

13 square units

Solution

Question 4

$$94$$ square units

$$96$$ square units

$$97$$ square units

$$98$$ square units

Solution

Question 5

2.3 square units

4.5 square units

4.1 square units

3.6 square units

Solution

Question 6

1

2

3

4

Solution

Question 7

$$20$$

$$23$$

$$26$$

$$33$$

Solution

Question 8

$$6$$

$$7$$

$$9$$

$$10$$

Solution

Question 9

11

5

7

9

Solution

Question 10

$$-10$$

$$-9$$

$$-5$$

$$5$$

$$6$$

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.