Permutations and Combinations Test 26

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Permutations and Combinations Test 26

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Weekly Quiz Competition

Question 1
1 / -0

$$6, 10, 18, 34, 66$$
The first number in the list above is $$6$$. Determine a rule for finding each successive number in the list.

A
Add $$4$$ to the preceding number.

B
Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

C
Double the preceding number and then subtract $$2$$ from that result.

D
Subtract $$2$$ from the preceding number and then double that result.

E
Triple the preceding number and then subtract $$8$$ from that result.

Solution

The series is $$6,10,18,34,66$$
In the given series:
$$6+6=12-2=10$$
$$10+10=20-2=18$$
$$18+18=36-2=34$$
$$34+34=68-2=66$$
Hence in this each successive number is obtained by double the preceding number and then subtract $$2$$ from the result.
Question 2
1 / -0

For all numbers a and b, let $$\displaystyle a\bigodot b$$ be defined by $$\displaystyle a\bigodot b=ab+a+b$$. Then for the numbers $$x$$, $$y$$ and $$z$$, which of the following is/are true?
I. $$\displaystyle x\bigodot y=y\bigodot x$$
II. $$\displaystyle \left( x-1 \right) \bigodot \left( x+1 \right) =\left( x\bigodot x \right) -1$$
III. $$\displaystyle x\bigodot \left( y+z \right) =\left( x\bigodot y \right) +\left( x\bigodot z \right) $$

A
I only

B
II only

C
III only

D
I and II only

E
I, II, and III

Solution

$$For(I)$$
$$x\odot y=y\odot x$$
$$x\odot y=xy+x+y\Rightarrow x\odot y=y\odot x$$
$$y\odot x=yx+y+x$$

$$For(II)$$
$$\left( x-1 \right) \odot \left( x+1 \right)$$
$$=(x-1)(x+1)+x-1+x+1$$
$$={ x }^{ 2 }-1+2x$$
$$\left( x\odot x \right) -1=x\times x+x+x-1$$
$${ x }^{ 2 }+2x-1$$
$$\therefore \left( x-1 \right) \odot \left( x+1 \right) =\left( x\odot x \right) -1$$

$$For(III)$$
$$x\odot \left( y+z \right) $$
$$=x\left( y+z \right) +x+y+z$$
$$=xy+xz+x+y+z$$
$$x\odot y=xy+x+y$$
$$x\odot z=xz+x+z$$
$$x\odot y+x\odot z=xy+xz+x+y+x+z$$
$$=xy+xz+2x+y+z$$
$$\Rightarrow x\odot \left( y+z \right) \neq x\odot y+x\odot z$$

$$\therefore (I)\& (II)$$ are true.
Question 3
1 / -0

$$m, 2m, 4m, . . . $$
The first term in the sequence above is $$m$$, and each term thereafter is equal to twice the previous term. If $$m$$ is an integer, which of the following could NOT be the sum of the first four terms of this sequence?

A
$$-26$$

B
$$-15$$

C
$$45$$

D
$$75$$

E
$$120$$

Solution

The first term in given Geometric series, $$a_1=m$$
The common ratio $$r$$ $$=\dfrac{4m}{2m}=2$$
No. of terms, $$n$$ $$=4$$
Applying sum of GP formula,
$$S_n=\dfrac{a(1-r^n)}{1-r}$$
$$=\dfrac{m(1-2^4)}{1-2}$$
$$=\dfrac{m(1-16)}{-1}=15m$$
If $$m$$ is an integer the sum of the first four terms of this sequence should be in multiple of $$15$$.
Hence, option A is correct which is not a multiple of $$15$$.
Question 4
1 / -0

$$N=a^2 + b^2$$ is a three-digit number which is divisible by 5. a = 10x + y and b = 10x + z, where z is a prime number, and x and y are natural numbers. If a + b = 31, find the value of N.

A
565

B
485

C
505

D
485 or 505

Solution
Question 5
1 / -0

If $$m * n = m+(m-1)+(m-2)+ ...... +(m-n)$$, evaluate $$7 * 5$$.

A
$$25$$

B
$$20$$

C
$$27$$

D
$$18$$

Solution

$$\Rightarrow$$ $$m * n=m+(m-1)+(m-2)+....+(m-n)$$
$$\Rightarrow$$ In $$7 * 5$$, $$m=7$$ and $$n=5$$
$$\Rightarrow$$ $$7 * 5=7+(7-1)+(7-2)+(7-3)+(7-4)+(7-5)$$
$$\Rightarrow$$ $$7 * 5=7+6+5+4+3+2$$
$$\therefore$$ $$7 * 5=27$$
Question 6
1 / -0

The value of $$\dfrac {(n + 2)! - (n + 1)!}{n!} $$ is:

A
$$(n + 2)!$$

B
$$(n + 1)!$$

C
$$(n + 2)^{2}$$

D
$$(n + 1)^{2}$$

E
$$n$$

Solution

The value of $$\dfrac { (n+2)!-(n+1)! }{ n! } $$
$$=\dfrac { (n+2)(n+1)n!-(n+1)n! }{ n! } $$
$$=\dfrac { (n+1)(n!)(n+2-1) }{ n! } $$
$$=\dfrac { (n+1)(n+1) }{ 1 } $$
$$={ (n+1) }^{ 2 }$$
Question 7
1 / -0

If $$a\odot b = 6\times a - 3\times b$$, evaluate $$(5\odot 3) \odot 20$$

A
$$2$$

B
$$41$$

C
$$66$$

D
$$1$$

Solution

$$\Rightarrow$$ $$a\odot b=6\times a-3\times b$$ [ Given ]
$$\Rightarrow$$ $$(5\odot 3)\odot 20$$
$$\Rightarrow$$ Let $$a=5$$ and $$b=3$$
$$\Rightarrow$$ $$(6\times 5-3\times 3)\odot 20$$
$$\Rightarrow$$ $$(30-9)\odot 20$$
$$\Rightarrow$$ $$21\odot 20$$
$$\Rightarrow$$ Let $$a=21$$ and $$b=20$$
$$\Rightarrow$$ $$6\times 21-3\times 20$$
$$\Rightarrow$$ $$126-60$$
$$\Rightarrow$$ $$66$$
$$\therefore$$ $$(5\odot 3)\odot 20=66$$
Question 8
1 / -0

Find the area of a triangle whose vertices are $$(0, 6\sqrt {3}), (\sqrt {35}, 7)$$, and $$(0, 3)$$.

A
$$15.37$$

B
$$17.75$$

C
$$21.87$$

D
$$25.61$$

E
$$39.61$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle whose vertices are $$(0,6\sqrt{3}),(\sqrt{35},7) ,(0,3)$$
$$=\dfrac{1}{2}\left [ 0(7-3)+\sqrt{35}(3-6\sqrt{3})+0(6\sqrt{3}-7)) \right ]$$
$$=\dfrac{1}{2}\left [ \sqrt{35}(3-6\sqrt{3}) \right ]$$
$$=\dfrac{1}{2}\left [ 5.91(3-6\times 1.73) \right ]$$
$$=\dfrac{1}{2}\times 5.91\times 7.40$$
$$=\dfrac{1}{2}\times 43.74$$
$$=21.87$$
Question 9
1 / -0

In fig., the area of triangle ABC (in sq. units) is:

A
$$15$$

B
$$10$$

C
$$7.5$$

D
$$2.5$$

Solution

Given: Coordinates of Point $$A (1,3) ,B (-1,0)$$ and $$C (4,0)$$
Construction: Drop a perpendicular from $$A$$ on $$x-$$ axis, which meets x-axis at $$D\equiv(1,0)$$
Now in $$\Delta ADC, AD = 3, DC = 3$$
Area of $$\Delta ADC = \dfrac12\times DC\times AD$$
$$= \dfrac12\times3\times3 = \dfrac92 \ cm^2$$

Now in $$\Delta ADB, AD = 3, DB = 2$$
Area of $$\Delta ADB = \dfrac12\times DB\times AD$$
$$= \dfrac12\times2\times3 = 3 \ cm^2$$

Area of $$\Delta ABC =$$ Ara of $$\Delta ADC + $$ Area of $$\Delta ABD$$
$$ = \dfrac92 + 3 = \dfrac{15}2 = 7.5\ cm^2$$
Question 10
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The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, -2). The third vertex lies on $$\displaystyle y=x+3$$. What is the third vertex?

A
$$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) $$

B
$$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) $$

C
$$\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right) $$

D
$$\displaystyle (0,0)$$

Solution

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Permutations and Combinations Test 26

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Permutations and Combinations Test 26

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Question 1

Add $$4$$ to the preceding number.

Take $$\displaystyle \frac { 1 }{ 2 } $$ of the preceding number and then add $$7$$ to that result.

Double the preceding number and then subtract $$2$$ from that result.

Subtract $$2$$ from the preceding number and then double that result.

Triple the preceding number and then subtract $$8$$ from that result.

Solution

Question 2

I only

II only

III only

I and II only

I, II, and III

Solution

Question 3

$$-26$$

$$-15$$

$$45$$

$$75$$

$$120$$

Solution

Question 4

565

485

505

485 or 505

Solution

Question 5

$$25$$

$$20$$

$$27$$

$$18$$

Solution

Question 6

$$(n + 2)!$$

$$(n + 1)!$$

$$(n + 2)^{2}$$

$$(n + 1)^{2}$$

$$n$$

Solution

Question 7

$$2$$

$$41$$

$$66$$

$$1$$

Solution

Question 8

$$15.37$$

$$17.75$$

$$21.87$$

$$25.61$$

$$39.61$$

Solution

Question 9

$$15$$

$$10$$

$$7.5$$

$$2.5$$

Solution

Question 10

$$\displaystyle \left( \frac { 7 }{ 2 } ,\frac { 13 }{ 2 } \right) $$

$$\displaystyle \left( \frac { 5 }{ 2 } ,\frac { 5 }{ 2 } \right) $$

$$\displaystyle \left( -\frac { 3 }{ 2 } ,-\frac { 3 }{ 2 } \right) $$

$$\displaystyle (0,0)$$

Solution

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