Sequences and Series Test

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Sequences and Series Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

If $$9@ 3 = 12, 15 @ 4 = 22, 16 @ 14 = 4$$, then what is the value of $$6 @ 2 = ?$$

A
$$26$$

B
$$1$$

C
$$30$$

D
$$8$$

Solution

$$(9 - 3) \times 2 = 12, (15 - 4) \times 2$$
$$= 22, (16 - 14) \times 2 = 4$$
Hence, $$(6 - 2) \times 2 = 8$$.
Question 2
1 / -0

5 2 7
? 3 1
4 5 2
-15 7 13
Select the missing number from the given alternatives.

A
$$1$$

B
$$5$$

C
$$9$$

D
$$7$$

Solution

Columnwise
First Number x Third Number - Second Number = Lowermost Number

First Column
$$5 \times 4 - ? = 15 \Rightarrow 20 - ? = 15$$
$$\therefore ? = 20 - 15 = 5$$

Second Column
$$ 2 \times 5 - 3 = 10 - 3 = 7$$

Third Column
$$ 7 \times 2 - 1 = 14 - 1 = 13$$
Question 3
1 / -0

Directions for questions 1 to 3: Find the related word/ letters/numbers from given alternatives.
12:72::8:?

A
36

B
32

C
38

D
40

Solution

12 $$\times \, \frac{12}{2}$$ = 72 similarly, 8 $$\times \, \frac{8}{2}$$ = 32
Question 4
1 / -0

Select the missing number from the given alternatives.

A
110

B
115

C
121

D
54

Solution

48 $$\div$$ 2 = 24;
24 $$\times$$ 3 = 72: 72 $$\div$$ 2 = 36;
36 $$\div$$ 3 = 108; 108 $$\div$$ 2 = 54.
Question 5
1 / -0

Choose the correct answer from the alternatives given.
If $$1^2+2^2$$ +.... + $$x^2$$ = $$\frac{x(x + 1)(2x + 1)}{6}$$then $$1^2$$ + $$3^2$$+ $$5^2$$ + .... + $$19^2$$ is equal to

A
$$1330$$

B
$$2100$$

C
$$2485$$

D
$$2470$$

Solution

$$(1^2+3^2+5^2+...........+19^2)$$
= $$\displaystyle\, $$(1^2$$ \, + \, $$2^2$$ \, + \, $$3^2$$ \, + \, ........... \, + \, 20^2) \, - \, (2^2 \, + \, 4^2 \, + \, 6^2 \, + \, ........... \, + \, 20^2)$$
= $$\displaystyle\, (1^2 \, + \, 2^2 \, + \, 3^2 \, + \, .......... \, + \, 20^2) - 4(1 \, + \, 2^2 \, + \, 3^2 \, + \, .......... \, + \, 10^2)$$
Using the formula,
$$\displaystyle\, = \frac{20 \times 21 \times 41}{6} - 4 \left ( \frac{10 \times 11 \times 21}{6} \right ) = 2870 - 1540 = 1330$$
Question 6
1 / -0

What is the value of $$\dfrac {m^{2} + mn}{m^{2} + n^{2}} + \dfrac {m^{2} - mn}{m^{2} + n^{2}} + \dfrac {n^{2} + m}{m^{2} + n^{2}} + \dfrac {n^{2} - mn}{m^{2} + n^{2}} $$

A
$$1$$

B
$$0$$

C
$$2$$

D
$$4$$

Solution

Given that,
$$\dfrac {m^{2} + mn}{m^{2} + n^{2}} + \dfrac {m^{2} - mn}{m^{2} + n^{2}} + \dfrac {n^{2} + m}{m^{2} + n^{2}} + \dfrac {n^{2} - mn}{m^{2} + n^{2}}$$
$$= \dfrac {m^{2} + mn + m^{2} - mn + n^{2} + mn + n^{2} - mn}{m^{2} + n^{2}}$$
$$= \dfrac {2(m^{2} + n^{2})}{(m^{2} + n^{2})} = 2$$.
Question 7
1 / -0

Choose the correct answer from the alternatives given :
The sum of $$\dfrac{1}{\sqrt 2 + 1} \, + \, \dfrac{1}{\sqrt 3 + \sqrt 2} \, + \, \dfrac{1}{\sqrt 4 + \sqrt 3} \, + \, ..... \, + \dfrac{1}{\sqrt {100} + \sqrt {99}}$$ is

A
$$9$$

B
$$10$$

C
$$11$$

D
None of these

Solution

$$\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + .......+\frac{1}{\sqrt{100} + \sqrt{99}}$$
Rationalising the denominator,
$$\displaystyle\, = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} + \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$
$$\displaystyle\, + \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} + ......+ \frac{1}{\sqrt{100} + \sqrt{99}} \times \frac{\sqrt{100} - \sqrt{99}}{\sqrt{100} - \sqrt{99}}$$
$$\displaystyle\, \sqrt{2} 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + .+ \sqrt{100} - \sqrt{99}$$
= $$-1 + 10 = 9$$
Question 8
1 / -0

Choose the correct answer alternatives given.
Select the missing number from the given alternatives.

A
18

B
21

C
5

D
17

Solution

80 = 8 $$\times$$ (8 + 2)
143 = 11 $$\times$$ (11 + 2)
323 = ? $$\times$$ (? + 2)
$$\therefore$$ ? = 17
Question 9
1 / -0

If $$1^{3} + 2^{3} + .... + 10^{3} = 3025$$, then the value of $$2^{3} + 4^{3} + ..... + 20^{3}$$ is

A
$$7590$$

B
$$5060$$

C
$$24200$$

D
$$12100$$

Solution

Given that, $$1^{3} + 2^{3} + .... + 10^{3} = 3025$$
Now, $$2^{3} + 4^{3} + .... + 20^{3}$$
$$= 2^{3} (1 + 2^{3} + .... + 10^{3}) = 8\times 3025 = 24200$$.
Question 10
1 / -0

In our number system the base is ten. If the base were changed to four you would count as follows: $$1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30,....$$ The twentieth number would be:

A
$$20$$

B
$$38$$

C
$$44$$

D
$$104$$

E
$$110$$

Solution

If we change the base from 10 to 4 then the number 20 will be written as
$$20$$= $$1\times 4^2$$+$$1\times4^1$$+$$0\times4^0$$.
so the required number is 110.
we can also find by dividing the number by 4
$$\dfrac{20}{4}$$ =$$5$$. now $$\frac{5}{4}$$=1 as dividend, now $$\frac{1}{4}$$=0 as dividend. so the required number is 110.
so option $$E$$ is correct option.

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Sequences and Series Test - 33

Result

Sequences and Series Test - 33

Score

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Rank

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SHARING IS CARING

Question 1

$$26$$

$$1$$

$$30$$

$$8$$

Solution

Question 2

$$1$$

$$5$$

$$9$$

$$7$$

Solution

Question 3

36

32

38

40

Solution

Question 4

110

115

121

54

Solution

Question 5

$$1330$$

$$2100$$

$$2485$$

$$2470$$

Solution

Question 6

$$1$$

$$0$$

$$2$$

$$4$$

Solution

Question 7

$$9$$

$$10$$

$$11$$

None of these

Solution

Question 8

18

21

5

17

Solution

Question 9

$$7590$$

$$5060$$

$$24200$$

$$12100$$

Solution

Question 10

$$20$$

$$38$$

$$44$$

$$104$$

$$110$$

Solution

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