Straight Lines Test 17

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Straight Lines Test 17

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Question 1
1 / -0

Directions For Questions

Directions (27-31) : 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 (?)

...view full instructions

4
20
25
27
81
9
11
44
?

A
4

B
16

C
30

D
55

Solution

$$20 \div 4 =5 \Rightarrow 5^2=25$$
$$81 \div 27 =3 \Rightarrow 3^2 = 9$$
$$44 \div 11 =4 \Rightarrow 4^2=16$$
Question 2
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 3
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 4
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 5
1 / -0

Rajdhani Express going from Bombay to Delhi stops at five inter-mediate stations, $$10$$ passengers enter the train during the journey with $$10$$ different ticket of two classes. The number of different sets of tickets they may have is

A
$$^{15}C_{10}$$

B
$$^{20}C_{10}$$

C
$$^{30}C_{10}$$

D
None of these

Solution

For a particular class, the total number of different tickets from first intermediate station is $$5.$$
Similarly, number of different tickets from second intermediate station is $$4.$$
So the total number of different tickets is $$5+4+3+2+1=15$$.
And same number of tickets for another class is equal to total number of different tickets,
which is equal to $$30$$ and number of selection is $$^{30}C_{10}$$.
Question 6
1 / -0

Find the area of the triangle whose vertices are $$(3,2), \ (-2, -3)$$ and $$(2,3)$$.

A
$$8$$ sq.unit

B
$$7$$ sq.unit

C
$$6$$ sq.unit

D
$$5$$ sq.unit

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 });$$ $$({ x }_{ 2 },y_2)$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$
$$ = \left |\dfrac{ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2) } {2} \right |$$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (3,2) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (-2,-3) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,3)$$
in the area formula, we get

Area = $$\left | \dfrac {3(-3-3) +(-2)(3-2) + 2 (2-(-3)) }{2}\right | = 5 $$ sq. unit
Question 7
1 / -0

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

A
$$2ac$$

B
$$2bc$$

C
$$b(a + c)$$

D
$$c (a - b)$$

Solution

Area of a triangle $$ (A) =\left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (a,b+c) $$ , $$({ x }_{ 2

},{ y }_{ 2 }) = (a,b-c) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (-a,c) $$

$$ A=\left| \cfrac { a(b-c-c)+a(c-b-c)-a(b+c-b+c) }{ 2 } \right| $$
$$=\left| \cfrac { a(b-2c)+a(-b)-a(2c) }{ 2 } \right|$$
$$ =\left| \cfrac { ab-2ac-ab-2ac }{ 2 } \right| $$
$$=\left| \cfrac { -4ac }{ 2 } \right| $$
$$=2ac$$ square units
Question 8
1 / -0

Find the area of the right-angled triangle whose vertices are $$(2, -2)$$ , $$(-2, 1)$$ and $$(5, 2).$$

A
$$\displaystyle 5\sqrt{2}$$ sq. units

B
$$\displaystyle \dfrac{25}{2}$$ sq. units

C
$$\displaystyle 15\sqrt{2}$$ sq. units

D
$$10$$ sq. units

Solution

Let the points be $$ A(2,-2), B(-2,1), C(5,2) $$.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Hence, Length of side AB $$ = \sqrt { \left(-2-2 \right) ^{ 2 }+\left(1 + 2\right) ^{ 2 } } = \sqrt { 16+ 9 } = \sqrt { 25 } = 5 $$
Length of side BC $$ = \sqrt {\left(5 + 2\right) ^{ 2 }+\left(2-1\right) ^{ 2 } } = \sqrt { 49 + 1 } = \sqrt { 50} $$
Length of side AC $$ = \sqrt { \left(5-2 \right) ^{ 2 }+\left(2+ 2\right) ^{ 2 } } = \sqrt { 9+16 } = \sqrt { 25 } = 5 $$
Since, $$ {(\sqrt { 50 }) }^{2} = { 5 }^{2} + { 5 }^{2} $$,
$$ \Rightarrow {BC}^{2} ={AB}^{2} + {AC}^{2} $$
Hence, the triangle has a right angle at $$A$$, with $$AB$$ and $$AC$$ as base and height.
So, area of the triangle $$ABC = \cfrac {1}{2} \times base \times height = \cfrac {1}{2} \times 5 \times 5= \cfrac {25}{2}$$ sq units.
Question 9
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Observe the given multiples of 37.
$${37\times3=111}$$
$${37\times 6 =222}$$
$${37\times9=333}$$
$${37\times12=444}$$-------------------------------
Find the product of $${37\times27}$$

A
$$999$$

B
Greatest $$3$$ digit number

C
Either A or B

D
Smallest $$3$$ digit number

Solution

$${37\times3=37\times(3\times1)=111}$$
$${37\times 6=37\times(3\times2)=222}$$
$${37\times9=37\times(3\times3)=333}$$

$${37\times27=37\times(3\times9)=999}$$
Question 10
1 / -0

The area of the triangle whose vertices are $$A(1,1), B(7, 3)$$ and $$C(12, 2)$$ is

A
$$25$$ square units

B
$$8$$ square units

C
$$16$$ square units

D
$$12$$ square units

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
$$\therefore$$
Area $$= \dfrac{1}{2} [1(1)+ 7(1)+ 12(-2)] = -8$$, but area can not be negative
Hence, area = 8 square units.

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 17

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Straight Lines Test 17

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SHARING IS CARING

Question 1

4

16

30

55

Solution

Question 2

$$\displaystyle b_{ij}$$ 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 3

$$1$$

$$3$$

$$7$$

$$9$$

Solution

Question 4

$$11 \leq e \leq 66$$

$$10 \leq e \leq 66$$

$$11 \leq e \leq 65$$

$$0 \leq e \leq 11$$

Solution

Question 5

$$^{15}C_{10}$$

$$^{20}C_{10}$$

$$^{30}C_{10}$$

None of these

Solution

Question 6

$$8$$ sq.unit

$$7$$ sq.unit

$$6$$ sq.unit

$$5$$ sq.unit

Solution

Question 7

$$2ac$$

$$2bc$$

$$b(a + c)$$

$$c (a - b)$$

Solution

Question 8

$$\displaystyle 5\sqrt{2}$$ sq. units

$$\displaystyle \dfrac{25}{2}$$ sq. units

$$\displaystyle 15\sqrt{2}$$ sq. units

$$10$$ sq. units

Solution

Question 9

$$999$$

Greatest $$3$$ digit number

Either A or B

Smallest $$3$$ digit number

Solution

Question 10

$$25$$ square units

$$8$$ square units

$$16$$ square units

$$12$$ square units

Solution

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