Sequences and Series Test

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Sequences and Series Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Complete the addition square by finding the missing numbers.
Find $$(A-B)-(C-D)$$
$$+$$ $$10,923$$ $$8,473$$
$$18,732$$ $$A$$ $$B$$
$$9,018$$ $$C$$ $$D$$

A
$$0$$

B
$$450$$

C
$$1000$$

D
$$300$$

Solution

From the given table, we have
$$A=18,732+10,923=29,655$$
$$B=18,732+8,473=27,205$$
$$C=10,923+9,018=19,941$$
and $$D=8473+9,018=17,491$$
Now $$A-B=29,655-27,205=2,450$$
$$C-D=19,941-17,491=2,450$$
$$\therefore$$ $$(A-B)-(C-D)=2,450-2,450=0$$
Question 2
1 / -0

Fathom is a unit once used by sailors to measure the depth of water. If a sunken ship was located underwater at $$240$$ feet, which expression would describe the location of the ship in fathoms?
$$ { 1\ fathom=6\ feet } $$

A
$$240\times 6$$

B
$$240\div 6$$

C
$$240-6$$

D
$$240+6$$

Solution

We have $$1fathom=6feet$$
$$1feet=\cfrac{1}{6}fathom$$
Ship was located underwater at $$240$$ feet
$$\therefore$$ location of ship underwater in fathoms $$=240\div6$$
Question 3
1 / -0

Choose the correct answer from the alternatives given :
Given 1 + 2 + 3 4 + + 10 = 55 then the sum 6 + 12 + 18 + 24 + + 60 is equal to

A
300

B
655

C
330

D
455

Solution

$$1 + 2 + 3 + 4 + .. + 10 = 55 (given)$$
$$6 + 12 + 18 + + 60 = ?$$
$$6(1 + 2 + 3 + 4 + + 10)$$
$$6\times 55$$ = $$330$$
Question 4
1 / -0

Select the missing number from the given alternatives.
$$14$$ $$25$$ $$42$$
$$2$$ $$4$$ $$6$$
$$3$$ $$3$$ ?
$$4$$ $$7$$ $$9$$

A
$$3$$

B
$$6$$

C
$$7$$

D
$$4$$

Solution

$$4\times 3 + 2 = 14$$
$$7\times 3 + 4 = 25$$
$$9\times (4) + 6 = 42$$
Hence, $$? = 4$$.
Question 5
1 / -0

A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as in the two matrices given below. The columns and rows of Matrix I are numbered from $$0$$ to $$4$$ and that of Matrix $$I$$ are numbered from $$5$$ to $$9$$. A letter from these matrices can be represented first by its row and next by its column, e.g. $$'A'$$ can be represented by $$00, 12, 23$$ etc., and $$'P'$$ can be represented by $$58, 69, 75$$ etc. Similarly, you have to identify the set for the word 'POET'.
Matrix I
$$0$$ $$1$$ $$2$$ $$3$$ $$4$$
$$0$$ $$A$$ $$R$$ $$S$$ $$N$$ $$C$$
$$1$$ $$N$$ $$C$$ $$A$$ $$R$$ $$S$$
$$2$$ $$S$$ $$N$$ $$C$$ $$A$$ $$R$$
$$3$$ $$R$$ $$S$$ $$N$$ $$C$$ $$A$$
$$4$$ $$C$$ $$A$$ $$R$$ $$S$$ $$N$$
Matrix II
$$5$$ $$6$$ $$7$$ $$8$$ $$9$$
$$5$$ $$O$$ $$E$$ $$L$$ $$P$$ $$T$$
$$6$$ $$T$$ $$O$$ $$E$$ $$L$$ $$P$$
$$7$$ $$P$$ $$T$$ $$O$$ $$E$$ $$L$$
$$8$$ $$L$$ $$P$$ $$T$$ $$O$$ $$E$$
$$9$$ $$E$$ $$L$$ $$P$$ $$T$$ $$O$$

A
$$69, 88, 67, 65$$

B
$$75, 56, 65, 67$$

C
$$77, 88, 98, 78$$

D
$$75, 66, 76, 78$$

Solution

From matrix,
$$P\Rightarrow 58, 69, 75, 86, 97$$
$$O\Rightarrow 55, 66, 77, 88, 99$$
$$E\Rightarrow 56, 67, 78, 87, 98$$
$$T\Rightarrow 69, 88, 67, 65$$
$$\therefore POET\Rightarrow 69, 88, 67, 65$$.
Question 6
1 / -0

Find the sum of $$1^3-2^3+3^3-4^3+....+9^3$$

A
$$400$$

B
$$425$$

C
$$450$$

D
$$475$$

Solution

$$1^3-2^3+3^3-4^3+...+9^3$$

$$=1^3+2^3+3^3+...+9^3-2(2^3+4^3+6^3+8^3)$$

$$=[\dfrac{9 \times 10}{2}]^2-2.2^3[1^3+2^3+3^3+4^3]$$

$$=(45)^2-16[\dfrac{4 \times 5}{2}]^2$$

$$=2025-1600=425$$
Question 7
1 / -0

The models are shaded to show which of the following?

A
$$\cfrac { 1 }{ 3 } =\cfrac { 2 }{ 4 } $$

B
$$\cfrac { 1 }{ 4 } >\cfrac { 1 }{ 3 } $$

C
$$\cfrac { 2 }{ 3 } <\cfrac { 2 }{ 4 } $$

D
$$\cfrac { 2 }{ 4 } <\cfrac { 2 }{ 3 } $$

Solution

Fraction of shaded part in upper rectangle $$=2\times\dfrac{1}{3}=\dfrac{2}{3}$$
Fraction of shaded part in lower rectangle $$=2\times\dfrac{1}{4}=\dfrac{2}{4}$$
In upper rectangle shaded part is greater
So, $$\dfrac { 2 }{ 4 } <\dfrac { 2 }{ 3 } $$
Option D is correct.
Question 8
1 / -0

In this question, the sets of numbers given in the alternatives are represented by two classes of alphabets as in two matrices given below. The columns and rows of Matrix I are numbered from $$0$$ to $$4$$ and that of Matrix II are numbered from $$5$$ to $$9$$. A letter from these matrices can be represented first by its row and next by its column, e.g., $$'B'$$ can be represented by $$00, 13,$$ etc., and $$'A'$$ can be represented by $$55, 69$$, etc. Similarly you have to identify the set for the word 'GIRL'.
Matrix I
$$0$$ $$1$$ $$2$$ $$3$$ $$4$$
$$0$$ $$B$$ $$N$$ $$G$$ $$L$$ $$D$$
$$1$$ $$G$$ $$L$$ $$D$$ $$B$$ $$N$$
$$2$$ $$D$$ $$B$$ $$N$$ $$G$$ $$L$$
$$3$$ $$N$$ $$G$$ $$L$$ $$D$$ $$B$$
$$4$$ $$L$$ $$D$$ $$B$$ $$N$$ $$G$$
Matrix II
$$5$$ $$6$$ $$7$$ $$8$$ $$9$$
$$5$$ $$A$$ $$I$$ $$K$$ $$O$$ $$R$$
$$6$$ $$I$$ $$K$$ $$O$$ $$R$$ $$A$$
$$7$$ $$K$$ $$O$$ $$R$$ $$A$$ $$I$$
$$8$$ $$O$$ $$R$$ $$A$$ $$I$$ $$K$$
$$9$$ $$R$$ $$A$$ $$I$$ $$K$$ $$O$$

A
$$02, 56, 97, 24$$

B
$$31, 79, 68, 42$$

C
$$23, 97, 77, 11$$

D
$$11, 88, 95, 23$$

Solution

From matrix,
$$G\Rightarrow 02, 10, 23, 31, 44$$
$$I\Rightarrow 56, 65, 79, 88, 97$$
$$R\Rightarrow 59, 68, 77, 86, 95$$
$$L\Rightarrow 03, 11, 24, 32, 40$$
$$\therefore GIRL\Rightarrow 23, 97, 77, 11$$.
Question 9
1 / -0

Select a figure from the options which will replace the question mark to complete the given series.

A

B

C

D

Solution

Each element moves on step forward anticlockwise direction.
Option B.
Question 10
1 / -0

If $$n\in N>1$$, then the sum of real part of roots of $$z^n=(z+1)^n$$ is equal to

A
$$\dfrac {n}{2}$$

B
$$\dfrac {(n-1)}{2}$$

C
$$-\dfrac {n}{2}$$

D
$$\dfrac {(1-n)}{2}$$

Solution

$${ z }^{ n }={ \left( z+1 \right) }^{ n }\\ { \left( \cfrac { z+1 }{ z } \right) }^{ n }=1\\ \therefore \cfrac { z+1 }{ z } ={ e }^{ i\cfrac { 2\pi m }{ n } },m=0,1,2,........,n-1\\ \cfrac { 1 }{ z } =-2\sin ^{ 2 }{ \left( \cfrac { \pi m }{ n } \right) } +i2\sin { \left( \cfrac { \pi m }{ n } \right) } \cos { \left( \cfrac { \pi m }{ n } \right) } =2i\sin { \left( \cfrac { \pi m }{ n } \right) } \left[ \cos { \left( \cfrac { \pi m }{ n } \right) } +i\sin { \left( \cfrac { \pi m }{ n } \right) } \right] \\ z=\cfrac { \cos { \left( \cfrac { \pi m }{ n } \right) } -i\sin { \left( \cfrac { \pi m }{ n } \right) } }{ 2i\sin { \left( \cfrac { \pi m }{ n } \right) } } =\cfrac { 1 }{ 2 } \left( -1-i\cot { \left( \cfrac { \pi m }{ n } \right) } \right) $$
$$\therefore$$ Sum of real parts of roots = $$n\left( -\cfrac { 1 }{ 2 } \right) =-\cfrac { n }{ 2 } $$ ($$\because$$ n roots are there)

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$$+$$	$$10,923$$	$$8,473$$
$$18,732$$	$$A$$	$$B$$
$$9,018$$	$$C$$	$$D$$

		Matrix I
	$$0$$	$$1$$	$$2$$	$$3$$	$$4$$
$$0$$	$$A$$	$$R$$	$$S$$	$$N$$	$$C$$
$$1$$	$$N$$	$$C$$	$$A$$	$$R$$	$$S$$
$$2$$	$$S$$	$$N$$	$$C$$	$$A$$	$$R$$
$$3$$	$$R$$	$$S$$	$$N$$	$$C$$	$$A$$
$$4$$	$$C$$	$$A$$	$$R$$	$$S$$	$$N$$

		Matrix II
	$$5$$	$$6$$	$$7$$	$$8$$	$$9$$
$$5$$	$$O$$	$$E$$	$$L$$	$$P$$	$$T$$
$$6$$	$$T$$	$$O$$	$$E$$	$$L$$	$$P$$
$$7$$	$$P$$	$$T$$	$$O$$	$$E$$	$$L$$
$$8$$	$$L$$	$$P$$	$$T$$	$$O$$	$$E$$
$$9$$	$$E$$	$$L$$	$$P$$	$$T$$	$$O$$

Sequences and Series Test - 14

Result

Sequences and Series Test - 14

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Rank

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SHARING IS CARING

Question 1

$$0$$

$$450$$

$$1000$$

$$300$$

Solution

Question 2

$$240\times 6$$

$$240\div 6$$

$$240-6$$

$$240+6$$

Solution

Question 3

300

655

330

455

Solution

Question 4

$$3$$

$$6$$

$$7$$

$$4$$

Solution

Question 5

$$69, 88, 67, 65$$

$$75, 56, 65, 67$$

$$77, 88, 98, 78$$

$$75, 66, 76, 78$$

Solution

Question 6

$$400$$

$$425$$

$$450$$

$$475$$

Solution

Question 7

$$\cfrac { 1 }{ 3 } =\cfrac { 2 }{ 4 } $$

$$\cfrac { 1 }{ 4 } >\cfrac { 1 }{ 3 } $$

$$\cfrac { 2 }{ 3 } <\cfrac { 2 }{ 4 } $$

$$\cfrac { 2 }{ 4 } <\cfrac { 2 }{ 3 } $$

Solution

Question 8

$$02, 56, 97, 24$$

$$31, 79, 68, 42$$

$$23, 97, 77, 11$$

$$11, 88, 95, 23$$

Solution

Question 9

Solution

Question 10

$$\dfrac {n}{2}$$

$$\dfrac {(n-1)}{2}$$

$$-\dfrac {n}{2}$$

$$\dfrac {(1-n)}{2}$$

Solution

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