Permutations and Combinations Test 51

Result

Permutations and Combinations Test 51

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
1 / -0

If $$ '-'$$ denotes $$'\div ',\ '\div '$$ denotes $$'\times ',\ '+'$$ denotes' - 'and $$'\times '$$ denotes $$'+'$$, then find the value of $$116+9\div 52-4\times 5$$.

A
$$16$$

B
$$8$$

C
$$9$$

D
$$4$$

Solution

$$-$$ denoted $$\div$$
$$\div$$ denoted $$\times $$
$$+$$ denoted $$-$$
$$\times $$ denoted $$+$$ $$\boxed{BODMAS}$$
$$116+9\div 52-4\times 5$$
$$116-9\times 52\div 4+5$$
$$116-9\times 13+5$$
$$116-117+5$$
$$116-112=4$$
Question 2
1 / -0

If $$n\ \in\ N$$ & $$n$$ is even, then $$\dfrac {1}{1\ .\ (n-1)\ !}+\dfrac {1}{3\ !(n-3)\ !}+\dfrac {1}{5\ !\ (n-5)\ !}+....+\dfrac {1}{(n-1)\ !\ 1\ !}=$$

A
$$2^{n}$$

B
$$\dfrac {2^{n-1}}{n\ !}$$

C
$$2^{n}n\ !$$

D
$$none\ of\ these$$

Solution
Question 3
1 / -0

$$40280625, 732375, 16275, 465, 18.6, 1.24,?$$

A
$$0.248$$

B
$$0.336$$

C
$$0.424$$

D
$$0.512$$

E
$$0.639$$

Solution

$$40280625, 732375,16275,465,18.6,1.24$$
$$\cfrac{40280625}{732375}=55$$
$$\cfrac{732375}{16275}=45$$
$$\cfrac{16275}{465}=35$$
$$\cfrac{465}{18.6}=25$$
$$\cfrac{18.6}{1.24}=15$$
$$\cfrac{1.24}{x}=5$$
$$x=0.248$$
Question 4
1 / -0

$$435, 354, 282, 219, 165, ?$$

A
$$103$$

B
$$112$$

C
$$120$$

D
$$130$$

E
$$none\ of\ these$$

Solution

$$\therefore 165-45=x$$
$$x=120$$.
Question 5
1 / -0

Find the number .

A
$$473$$

B
$$623$$

C
$$389$$

D
$$584$$

Solution

$$ \begin{array}{l} 4^{3}+7^{3}+3^{3}=434 \\ \text { similarly, } \\ 6^{3}+5^{3}+2^{3}=349 \\ \Rightarrow \text { ? }=8^{3}+4^{3}+2^{3} \\ \Rightarrow \text { ? }=584 \\ \text { option } D \text { is correct. } \end{array} $$
Question 6
1 / -0

There is a specific relationship between the numbers that are given in the following figures. On the basis of the relationship choose the correct alternative to replace the question mark.

A
$$210$$

B
$$266$$

C
$$288$$

D
$$318$$
Question 7
1 / -0

if the points A(z),B(-z) and C(z+1) are vertices of an equilateral triangle then ---- test question

A
area of triangle is $$\left( \sqrt { 3 } /4 \right) $$sq. units

B
Re(z)=1/2

C
perimeter of the triangle is 3 units

D
Re(z)=1/4

Solution
Question 8
1 / -0

If three lines $$x-3y=p,ax+2y=q$$ and $$ax+y=r$$ form a right angled triangle, then

A
$${ a }^{ 2 }-9a+18=0$$

B
$${ a }^{ 2 }-6a-12=0$$

C
$${ a }^{ 2 }-6a+9=0$$

D
$${ a }^{ 2 }-9a+12=0$$

Solution
Question 9
1 / -0

If the tangent at $$\theta =\frac { \pi }{ 4 } $$ to the curve $$x=a\cos ^{ 3 }{ \theta } ,y=a\sin ^{ 3 }{ \theta } $$ meets the x and y axes in A and B then the area of the triangle OAB is

A
$$\frac { { a }^{ 2 } }{ 4 } sq.units$$

B
$$\frac { { a }^{ 2 } }{ 2 } sq.units$$

C
$$\frac { { 3a }^{ 2 } }{ 4 } sq.units$$

D
$$\frac { { 5a }^{ 2 } }{ 4 } sq.units$$

Solution
Question 10
1 / -0

Area of a triangle whose vertices are $$ (a \cos \theta, b \sin \theta),(-a \sin \theta, b \cos \theta) $$ and$$ (-a \cos \theta,-b \sin \theta) $$ is $$ - $$

A
a b sin $$ \theta \cos \theta $$

B
$$ a \cos \theta \sin \theta $$

C
$$ \frac{1}{2} a b $$

D
$$ a b $$

Solution