Permutations and Combinations Test 31

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Permutations and Combinations Test 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$15! =2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}\cdot 7^{\delta}\cdot 11^{\theta}\cdot 13^{\Phi}$$, then the value of expression $$\alpha -\beta +\gamma -\delta +\theta -\Phi$$ is

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

$$15!=1\times 2\times 3\times 4 \times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13\times 14\times 15$$

$$15!=2\times 3\times 2^2 \times 5\times 3\times 2 \times 7\times 2^3\times 3^2\times 2\times 5\times 11\times 3\times 2^2 \times 13\times 2\times 7 \times 3\times 5$$

$$15!=2^{11}\times 3^6\times 5^3\times 7^ 2\times 11\times 13 $$

$$\implies \alpha =11$$

$$\implies \beta =6$$

$$\implies \gamma=3$$

$$\implies \delta =2$$

$$\implies \theta =1$$

$$\implies \Phi=1$$

$$\alpha-\beta+\gamma-\delta+\theta-\Phi=11-6+3-2+1-1=6$$
Question 2
1 / -0

Geometric mean of $$7,\ { 7 }^{ 2 },\ { 7 }^{ 3 }....{ 7 }^{ n }$$ is

A
$$\displaystyle { 7 }^{ \dfrac { \left( n+1 \right) }{ 2 } }$$

B
$${ 7 }^{ { n }/{ 2 } }$$

C
$${ 7 }^{ \dfrac { \left( n-1 \right) }{ 2 } }$$

D
$${ 7 }^{ \left( \dfrac { n+1 }{ 3 } \right) }$$

Solution

$$Gm$$ of two no $$a$$ & $$b=\sqrt {ab}$$
For $$aGP$$
$$GM=\sqrt {a_1. an}$$
$$\Rightarrow \ GM $$ of $$7, 7^2, ..... 7^n$$
$$=\sqrt {7.7^n}$$
$$=7\dfrac {(n+1)}{2}$$
Question 3
1 / -0

A telephone number $$d_1d_2d_3d_4d_5d_6d_7$$ is called memorable if the prefix sequence $$d_1d_2d_3$$ is exactly the same as either of the sequence $$d_4d_5d_6$$ or $$d_5d_6d_7$$(or possibly both). If each $$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$$, then number of distinct memorable telephone number is(are).

A
$$19810$$

B
$$19,910$$

C
$$19,990$$

D
$$20,000$$

Solution
Question 4
1 / -0

The number of ways in which $$6$$ rings can be worn on the four fingers of one hand is

A
$$4^{6}$$

B
$$^{6}C_{4}$$

C
$$6^{4}$$

D
None of these

Solution

Each ring can be worn on $$4$$ fingers, means each have $$4$$ possibilities.
$$\therefore$$ Total ways$$=4\times4\times4\times4\times4\times4=4^{6}$$
Hence, $$(A)$$
Question 5
1 / -0

Choose the correct answer from the alternatives given.
A loss of $$20%$$ is incurred when $$6$$ articles are sold for a rupee. To gain $$20% $$ how many articles should be sold for a rupee?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

: Given that, SP = $$1$$. Loss$$ % = 20%$$ We know that, $$SP = 0.8$$ $$\times$$ $$CP$$
$$CP \, = \, \dfrac{5}{4}$$
To gain$$ 20%$$,SP=1.2 $$\displaystyle \times \, CP \, = \, \dfrac{120}{100} \, \times \, \dfrac{5}{4} \, = \, \dfrac{3}{2} \, \Rightarrow \, = \, \dfrac{3}{2}$$
For$$ Rs. 3/2$$ number of articles sold
For $$Rs. 1$$ number of articles to be sold $$= 6 $$ $$\times$$ $$ \dfrac{2}{3}$$ = $$ 4 $$ articles.
Question 6
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 7
1 / -0

Sum of the three digit numbers (no digit being zero) having the property that all digits are perfect squares, is

A
$$3108$$

B
$$6216$$

C
$$13986$$

D
$$None\ of\ these$$

Solution

Perfect square digits from $$1 $$ to $$9 $$ are - $$1,4,9$$

The 3 digit nos. having the property that all digits are perfect squares are,

$$149, 194, 491, 419, 914, 941$$

The sum is,

$$149+194+491+419+914+941=3108$$
Question 8
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 9
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 10
1 / -0

Select the missing number from the given alternatives.
$$3$$ $$4$$ $$2$$ $$14$$
$$6$$ $$5$$ $$4$$ $$44$$
$$5$$ $$2$$ $$7$$ ?

A
$$58$$

B
$$14$$

C
$$49$$

D
$$4$$

Solution

$$3\times 2 + 4\times 2 = 6 + 8 = 14$$
$$6\times 4 + 5\times 4 = 24 + 20 = 44$$
Hence, $$5\times 7 + 2 \times 7 = 35 + 14 = 49$$.

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Re-Attempt Weekly Quiz Competition

5	2	7
?	3	1
4	5	2
-15	7	13

$$3$$	$$4$$	$$2$$	$$14$$
$$6$$	$$5$$	$$4$$	$$44$$
$$5$$	$$2$$	$$7$$	?

Permutations and Combinations Test 31

Result

Permutations and Combinations Test 31

Score

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Rank

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SHARING IS CARING

Question 1

$$4$$

$$6$$

$$8$$

$$10$$

Solution

Question 2

$$\displaystyle { 7 }^{ \dfrac { \left( n+1 \right) }{ 2 } }$$

$${ 7 }^{ { n }/{ 2 } }$$

$${ 7 }^{ \dfrac { \left( n-1 \right) }{ 2 } }$$

$${ 7 }^{ \left( \dfrac { n+1 }{ 3 } \right) }$$

Solution

Question 3

$$19810$$

$$19,910$$

$$19,990$$

$$20,000$$

Solution

Question 4

$$4^{6}$$

$$^{6}C_{4}$$

$$6^{4}$$

None of these

Solution

Question 5

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 6

$$1$$

$$5$$

$$9$$

$$7$$

Solution

Question 7

$$3108$$

$$6216$$

$$13986$$

$$None\ of\ these$$

Solution

Question 8

$$7590$$

$$5060$$

$$24200$$

$$12100$$

Solution

Question 9

$$26$$

$$1$$

$$30$$

$$8$$

Solution

Question 10

$$58$$

$$14$$

$$49$$

$$4$$

Solution

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