Straight Lines Test 36

Result

Straight Lines Test 36

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

The midpoints of the sides of a triangle are $$\left(1,1\right),\left(4,3\right)$$ and $$\left(3,5\right)$$. The area of the triangle is ___ square units.

A
$$14$$

B
$$16$$

C
$$18$$

D
$$20$$

Solution
Question 2
1 / -0

Ten persons, amongst whom are $$A$$,$$B$$ and $$C$$ to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$ AND $$B$$ wants to speak before $$C$$ is

A
$$\dfrac{{10!}}{6}$$

B
$$3!$$ $$7!$$

C
$$^{10}{P_3}.7!$$

D
none of these

Solution
Question 3
1 / -0

The number of ways in which a mixed doubles tennis game can be arranged between 10 players consisting of 6 men and 4 women is .

A
180

B
90

C
48

D
12

Solution

Number of ways to select two men out of six men
$$ = ^{6}C_{2} $$
Number of ways to select two women out of four
women $$ = ^{4}C_{2} $$
Number of ways of selecting the players for the
mixed double $$ = ^{6}C_{2}\times ^{4}C_{2} = 15\times 16 = 90 $$
Let $$ M_{1},M_{2},W_{1}\& W_{2} $$ are selected players for we
mixed double tennis game
if $$ M_{1} $$ chooses $$W_{1}$$ then $$ M_{2}$$ has $$W_{2}$$ as the partner or if
$$ M_{1}$$ chooses $$ W_{2}$$, lean $$M_{2}$$ has $$ W_{1}$$ as the partner
$$ \therefore $$ There are $$2$$ choice for the teams.
Thus, Number of ways in which mixed double
tennis game be arranged $$ 90\times 2 = 180 $$
Question 4
1 / -0

solve that :-

A
$$14$$

B
$$18$$

C
$$11$$

D
$$26$$

Solution

Given,

Image $$A$$

$$4+2+5+3=14 \times 2=28$$

Image $$B$$

$$7+5+4+3=19 \times 2=38$$

$$\therefore$$ Image $$C$$

$$2+1+3+7=13 \times 2=26$$
Question 5
1 / -0

The maximum number of points of intersection of $$7$$ staright lines and $$5$$ circles when $$3$$ straight lines are parallel and $$2$$ circles are concentric,is /are

A
$$106$$

B
$$96$$

C
$$90$$

D
$$None\ of\ these$$

Solution
Question 6
1 / -0

Choose the odd one out .

A
KWIJ

B
TPRN

C
CHAF

D
OZMX

Solution
Question 7
1 / -0

Let the eleven letters, $$A, B, ....K$$ denote an artbitrary permutation of the integers $$(1,2,....11)$$, then $$(A-1)(B-2)(C-3)...(K-11)$$ is

A
Necessarily zero

B
Always odd

C
Always evem

D
None of these

Solution

Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
Question 8
1 / -0

Consider all permutations of the letters of the word MORADABAD.
The number of permutations which contain the word BAD is:

A
$$21 \times 5!$$

B
$$7 \times 5!$$

C
$$6 \times 5!$$

D
$$2 \times 5!$$

Solution

Total number of letters = $$9$$

Consider BAD as a group, then the number of total letters = $$6+1=7$$

$$\therefore$$ Permutation of 7 things = $$7!$$

but letters remaining after removing BAD will have two A's

$$\therefore$$ Total permutation = $$\dfrac{7!}{2}$$ = $$21\cdot 5!$$
Question 9
1 / -0

If $$x$$ and $$y$$ are the number of possibilities that $$A$$ can assume such that the unit digit of A and $$A^3$$ are same and the unit digit of $$A^2$$ and $$A^3$$ are same respectively ,then the value of $$x-y$$ is (where $$A$$ is a single digit number)

A
$$4$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

Here, according to the question
if $$A=1$$, then $${ A }^{ 3 }={ 1 }$$,
We see that unit digit number of $$A$$ and $${ A }^{ 3 }$$ are same.
for, $$A=2\Rightarrow { A }^{ 3 }=8,$$, unit digits are not same
$$A=3\Rightarrow { A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 3 }=64,$$ unit digits are same
$$A=5\Rightarrow { A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 3 }=216,$$ unit digits are not same
$$A=7\Rightarrow { A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are same
So, there are five possible solutions where unit digits of $$A$$ and $${ A }^{ 3 }$$ are same.
$$\therefore x=5$$
Again, if we take $$A=1,{ A }^{ 2 }=1,{ A }^{ 3 }={ 1 }$$, unit digits are same
$$A=2\Rightarrow { A }^{ 2 }=4,{ A }^{ 3 }=8,$$ unit digits are not same
$$A=3\Rightarrow { A }^{ 2 }=9,{ A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 2 }=16,{ A }^{ 3 }=64,$$ unit digits are not same
$$A=5\Rightarrow { A }^{ 2 }=25,{ A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 2 }=36,{ A }^{ 3 }=216,$$ unit digits are same
$$A=7\Rightarrow { A }^{ 2 }=49,{ A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 2 }=64,{ A }^{ 3 }=512,$$ unit digits are not same
$$A=9\Rightarrow { A }^{ 2 }=81,{ A }^{ 3 }=729$$ unit digits are not same
Here, we can see that there are three possible solutions where unit digits of $${ A }^{ 2 },{ A }^{ 3 }$$ are same.
So, $$y=3$$
Hence, $$x-y=5-3=2.$$
Question 10
1 / -0

The number of permutation of the letters of the word $$HINDUSTAN$$ such that neither the pattern $$'HIN'$$ nor $$'DUS'$$ nor $$'TAN'$$ appears, are :

A
$$166674$$

B
$$169194$$

C
$$166680$$

D
$$181434$$

Solution

Total number of letters $$=9$$ in which N is repeated twice

Therefore total number of permutation $$=\dfrac{9!}{2}$$

The number of permutation in which HIN comes as block $$=9-3+1=7!$$

The number of permutation in which DUS comes as block $$=7!$$

The number of permutation in which TAN comes as block $$=\dfrac{7!}{2}$$

The number of permutation in which HIN and DUS comes as block $$=5!$$

The number of permutation in which DUS and TAN comes as block $$=5!$$

The number of permutation in which HIN and TAN comes as block $$=5!$$

Therefore the required number of permutations $$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$$

$$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$$

$$=181440-\left(10080+2520-360+6\right)$$

$$=181440-12246$$

$$=169194$$