Sequences and Series Test 17

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Sequences and Series Test 17

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Question 1
1 / -0

The sum of the value of the digits at the tens place of all the numbers formed with the help of $$3, 4, 5, 6$$ taken all at a time is

A
$$1080$$

B
$$4320$$

C
$$360$$

D
$$180$$

Solution

Total number of numbers formed with the digits $$3,4,5,6$$ is $$4!$$ $$=$$ $$24$$
If a digit is fixed in tens place, number of numbers formed will be $$6$$
i.e for every digit in tens place there will be $$6$$ numbers formed
Now, sum of the digits $$ = (6\times3)+(6\times4)+(6\times5)+(6\times6) $$ $$ = 108 $$.
Now its value in tens place $$= 108\times10=1080 $$.
Question 2
1 / -0

The points $$A (2, 9), B (a, 5)$$ and $$C (5, 5) $$ are the vertices of a triangle $$ABC$$ right angled at $$B$$. Find the values of $$a$$ and hence the area of $$\Delta $$ $$ABC$$.

A
$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

B
$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

C
$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

D
$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$.
So, $$AC$$ is the hypotenuse.
$$ \therefore$$ by Pythagoras theorem, we have
$$ { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$ .........(i)
Now, by distance formula
$$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } $$
So, $$AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-a \right) }^{ 2 }+{ \left( 9-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-4a+20 } $$,
$$BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-a \right) }^{ 2 }+{ \left( 5-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-10a+25 } $$ and
$$AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 5-9 \right) }^{ 2 } } $$ units $$=\sqrt { 25 } $$ units.
$$ \therefore$$ by (i), we get
$${ a }^{ 2 }-4a+20{ +a }^{ 2 }-10a+25=25$$
$$ \Rightarrow { a }^{ 2 }-7a+10=0$$
$$ \Rightarrow a=5,2$$
We reject $$a=5$$, since $$B$$ and $$C$$ will coincide in that case and $$\Delta ABC$$ will collapse.
So, $$a=2. i.e. AB=\sqrt { { a }^{ 2 }-4a+20 } =\sqrt { 4-8+20 } 3$$ units $$=43$$ units
$$ BC=\sqrt { { a }^{ 2 }-10a+25 } =\sqrt { 4-20+25 } 3$$ units $$=3$$ units.
Now,
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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Therefore, required area is $$6$$ square units.
Question 3
1 / -0

The area of triangle ABC (in sq. units) is :

A
$$15$$ sq. units

B
$$10$$ sq. units

C
$$7.5$$ sq. units

D
$$2.5$$ sq. units

Solution

The area of a triangle is given as:
Area $$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$
the given points are $$ A(1,3),~B(-1,0) $$and $$C(4,0)$$
by Substituting ,
$$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$$
$$\therefore$$ Area of the given triangle ABC is $$7.5$$ sq. units
Question 4
1 / -0

Given: $$\dfrac {20!}{18!}=380$$

A
True

B
False

C
Either

D
Neither

Solution

$$\cfrac { 20! }{ 18! } =?\\ 20!=20\times 19\times 18\times ........3\times 2\times 1\\ =20\times 19\times (18!)\\ \cfrac { 20! }{ 18! } =\cfrac { 20\times 19\times (18!) }{ 18! } =20\times 19=380$$
Hence the value matches so the equation is true.
Question 5
1 / -0

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

A
True

B
False

C
Ambiguous

D
Data insufficient

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
$$= 0$$
Question 6
1 / -0

The numbers 1, 2, ..., 100 are arranged in the squares of an table in the following way: the numbers 1, ... , 10 are in the bottom row in increasing order, numbers 11, ... ,20 are in the next row in increasing order, and so on. One can choose any number and two of its neighbors in two opposite directions (horizontal, vertical, or diagonal). Then either the number is increased by 2 and its neighbors are decreased by 1, or the number is decreased by 2 and its neighbors are increased by 1. After several such operations the table again contains all the numbers 1, 2, ... , 100. Prove that they are in the original order.

A
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

B
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

C
$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

D
$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Label the table entry in the i th row and j th column by $$a_{ ij }$$
where the bottom-left corner is in the first row and first column.
Let $$\displaystyle b_{ij}= 10(i-1)+j$$ be the number originally in the i th row and j th column. Observe that $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
is invariant. Indeed, every time entries
$$\displaystyle a_{mn}, a_{pq}, a_{rs}$$ are changed (with m+r= 2p and n+s= 2q), P increases or decreases by $$\displaystyle b_{mn}-2b_{pq}+b_{rs},$$
But this equals $$\displaystyle 10\left ( \left ( m-1 \right )+ \left ( r-1 \right )-2\left ( p-1 \right )+\left ( n+s-2q \right )\right )= 0.$$
In the beginning $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$ at the end, the entires $$a_{ij}$$ equal the $$b_{ij}$$
In some order,we now have $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
By the rearrangement inequality, this is at least $$\displaystyle P= \sum_{i,j=1}^{10}a_{ij}b_{ij}$$
with equality only when each $$a_{ij}=b_{ij}$$
The equality does occur since P is invariant. Therefore the $$a_{ij}$$ do indeed equal the $$b_{ij}$$
in the same order, and thus the entries $$1, 2, ... , 100$$ appear in their original order.
Question 7
1 / -0

Unit digit in the number $$(12357)^{655}$$ is

A
$$1$$

B
$$3$$

C
$$7$$

D
$$9$$

Solution

unit digit of $$(12357)^{655}$$

$$=$$ unit digit of $$(7^{655})=7^{4\times 163+3}=7^{4\times 143}+7^3$$

$$=$$ unit digit is $$(1\times 3)=$$ unit digit is 3.
Question 8
1 / -0

In a class of 80 students it is found that 40 students like Tajmahal and 50 students like Charminar and 18 like both. Then the number of students who do not like neither are.

A
18

B
72

C
50

D
8

Solution

Given that
Total students$$=80$$
Students who like Tajmahal $$=40 \quad n(T)$$
Students who like charminar$$=50 \quad n(C)$$
Students who like both $$=18 \quad n(T\cap C)$$
Say $$'x'$$ like neither then,
$$80=n(T)+n(C)-n(T\cap C)+x\\ \Rightarrow 80=40+50-18+x\\ \Rightarrow x=8$$
Question 9
1 / -0

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is agraph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any other point through a sequence of edges. The number of edges "e" in the graph must satisfy the condition

A
$$11 \leq e \leq 66$$

B
$$10 \leq e \leq 66$$

C
$$11 \leq e \leq 65$$

D
$$0 \leq e \leq 11$$

Solution

(A) Since every edge connects a pair of points, the given 12 points have to be joined using lines. We may have minimum number of edges if all the 12 points are collinear.
No. of edges in this particular case
$$=12-1=11$$
Maximum number of edges are possible when all the 12 points are non-collinear. In this particular case number of different straight lines that can be formed using 12 points which is equal to $$^12C_{2}$$
$$=\frac{12\times 11}{2}=66$$
Therefore, following inequality holds for "e"
$$11 \leq e \leq 66$$
Question 10
1 / -0

Out of nine cells of a square, one cell is left blank and in the rest of the cells, numbers are written follow some rule. Get the rule and find out the proper option for the blank cell
2
72
56
?
0
42
12
20
30

A
4

B
6

C
8

D
10

Solution

The pattern of numbers are as followed
$$1^2-1=0$$
$$2^2-1=3$$
$$3^2-3=6$$
$$4^2-4=12$$
$$5^2-5=20$$
$$6^2-6=30$$
$$7^2-7=42$$
$$8^2-8=56$$
$$9^2-9=72$$

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Re-Attempt Weekly Quiz Competition

Sequences and Series Test 17

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Sequences and Series Test 17

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SHARING IS CARING

Question 1

$$1080$$

$$4320$$

$$360$$

$$180$$

Solution

Question 2

$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Question 3

$$15$$ sq. units

$$10$$ sq. units

$$7.5$$ sq. units

$$2.5$$ sq. units

Solution

Question 4

True

False

Either

Neither

Solution

Question 5

True

False

Ambiguous

Data insufficient

Solution

Question 6

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 appear in their original order.

$$\displaystyle b_{ij}$$ is in the same order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

$$\displaystyle b_{ij}$$ is in the different order, and thus the entries 1, 2, ... , 100 do not appear in their original order.

Solution

Question 7

$$1$$

$$3$$

$$7$$

$$9$$

Solution

Question 8

18

72

50

8

Solution

Question 9

$$11 \leq e \leq 66$$

$$10 \leq e \leq 66$$

$$11 \leq e \leq 65$$

$$0 \leq e \leq 11$$

Solution

Question 10

4

6

8

10

Solution

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