Permutations and Combinations Test 30

Result

Permutations and Combinations Test 30

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

Directions For Questions

In the questions, numbers are placed in figures on the basis of some rules. One place in the figure is indicated by the interrogation sign (?). Find out the correct alternative to replace the question mark and indicate your answer by filling the circle of the corresponding letter of alternatives in the OMR Answer sheet.

...view full instructions

The figure is given

A
$$P$$

B
$$H$$

C
$$M$$

D
$$L$$

Solution

We can observe the figure as :
Position of $$N$$ is $$14$$
Position of $$T$$ is $$20$$
Position of $$B$$ is $$2$$
Position of $$P$$ is $$16$$
Question 2
1 / -0

Find the missing number in above figure :

A
$$110$$

B
$$1$$

C
$$55$$

D
$$441$$

Solution

$$\dfrac {11\times 10}{2} = 55$$
Question 3
1 / -0

Find the missing number in above figure :

A
$$38$$

B
$$66$$

C
$$68$$

D
$$70$$

Solution

Sum of all digits is $$7$$.
example $$142$$ $$i.e$$ $$1+4+2 = 7$$
$$\therefore$$ The solution is $$70$$.
Question 4
1 / -0

$$4C$$ $$2B$$ $$3A$$
$$28A$$ ? $$45B$$
$$7C$$ $$8A$$ $$15B$$

A
$$16C$$

B
$$12C$$

C
$$13C$$

D
$$7C$$

Solution
Question 5
1 / -0

Select the INCORRECT match

A
$$3249-MMMCCXLIX$$

B
$$1667-MDCLXVII$$

C
$$207-CCXVII$$

D
$$499-CDXCIX$$

Solution

(a) $$MMMCCXLIX=3000+200+40+9=3249$$
(b) $$MDCLXVII=1000+500+100+60+7=1667$$
(c) $$CCXVII=217=200+17$$
(d) $$CDXCIX=400+90+9=499$$
Hence the correct match is option C.
Question 6
1 / -0

Figures $$1$$ and $$2$$ are related in a particular manner. Establish the same relationship between figures $$3$$ and $$4$$ by choosing a figure from amongst the options.

A

B

C

D

Solution

Hexagon is rotating in anti-clockwise direction two times to get the next figure. (1 to 2).
Same for getting the figure (3 to 4) rotate the figure in an anti-clockwise direction (two times) we will get the desired figure.
Question 7
1 / -0

Find the missing number in above figure :

A
$$100$$

B
$$10$$

C
$$200$$

D
$$9$$

Solution

$$5 \times 3 = 15, 8 \times 3 = 24$$
$$14 \times 2 = 28, 16 \times 2= 32$$
$$25 \times 2 = 50, 10 \times 2 = 20$$.

$$\therefore$$ The solution is $$10$$.
Question 8
1 / -0

Find the area (in square units) of the triangle whose vertices are $$(a, b+c), (a, b-c) $$ and $$(-a, c). $$

A
$$2ac$$

B
$$2bc$$

C
$$b(a+c)$$

D
$$c(a-h)$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times |[ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] |$$
Given the vertices of triangle,
$$A (a,b+c) ,B(a,b-c) ,C(-a,c)$$
Therefore, area is given by
$$=\dfrac { 1 }{ 2 } |\left[ a\left[ b-c-c \right] +a\left[ c-b-c \right] +(-a)\left[ b+c-b+c \right] \right]| \\ =\dfrac { 1 }{ 2 } |\left[ a(b-2c)+a(-b)-a(2c) \right]| \\ =\dfrac { 1 }{ 2 } |\left[ ab-2ac-ab-2ac \right]| =\left| \dfrac { -4ac }{ 2 } \right| =2ac$$
Question 9
1 / -0

13, 74, 290, 650,.......

A
$$1248$$

B
$$1370$$

C
$$1346$$

D
$$1452$$

E
$$1625$$

Solution

Here we observe the first term 13 can be written as: $$2^2+3^2$$
Similarly other numbers can be written as:
$$74: 5^2+7^2$$
$$290: 11^2+13^2$$
$$650: 17^2+19^2$$
Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.
Like: $$(2,3);(5,7);(11,13);(17,19)$$ So, other pair of prime number will be $$ (23,29).$$
Hence next term of given series will be:
$$\Rightarrow 23^2+29^2=1370$$
So, correct answer is $$1370$$.
Question 10
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

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

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.