Permutations and Combinations Test 11

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Permutations and Combinations Test 11

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Weekly Quiz Competition

Question 1
1 / -0

_______ is a series of successive events.

A
Series

B
Preceding sequence

C
Progression

D
Geometric progression

Solution

Progression is a series of successive events.
Question 2
1 / -0

The average IQ of $$4$$ people is $$110$$. If three of these people each have an IQ of $$105$$, what is the IQ of the fourth person?

A
110

B
115

C
120

D
125

Solution

Average IQ of $$4$$ people $$=110$$
$$\therefore$$ total IQ of $$4$$ people $$=$$ $$110\times 4=440$$
Average IQ of $$3$$ people $$=105$$
$$\therefore $$ total IQ of $$3$$ people $$=$$ $$105\times 3=315$$
Then IQ of forth person $$=$$ $$440-315=125$$
Question 3
1 / -0

Find the next number in the series.
$$3, 6, 9, 12, 15,....$$

A
$$17$$

B
$$18$$

C
$$19$$

D
$$20$$

Solution

The next number is $$18$$.
Since the numbers are multiple of $$3$$.
$$3, 6, 9, 12, 15, \underline {18 }$$.
Question 4
1 / -0

Based on this information answer the questions given below.
(i) $$\displaystyle ^{ n }{ C }_{ p }= r!^{ n }{ C }_{ r }$$
(ii) $$\displaystyle ^{ n }{ C }_{ r }+^{ n }{ C }_{ r-1 }=^{ n+1 }{ C }_{ r }$$
What is the value of $$\displaystyle ^{ 8 }{ C }_{ 4 }+^{ 8 }{ C }_{ 3 }$$?

A
$$\displaystyle ^{ 8 }{ c }_{ 5 }$$

B
$$63$$

C
$$35$$

D
$$\displaystyle ^{ 9 }{ c }_{ 4 }$$

Solution

$$ ^{n}C_{r}+^{n}C_{r-1} = ^{n+1}C_{r} $$
$$ \Rightarrow ^{8}C_{4}+^{8}C_{3} = \boxed{^{9}C_{4}} $$
Question 5
1 / -0

A bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise coins in the ratio $$3 : 8 : 10$$. What is the number of $$50$$-paise coins?

A
$$112$$

B
$$128$$

C
$$96$$

D
$$24$$

Solution

Here bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise.
Coin ratio is $$ 3 : 8 : 10$$.
Value ratio $$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$$
$$= 3 : 4 : 1$$
Number of $$50$$ paise coins $$= \dfrac {4}{8} \times$$ Rs. $$112 =$$ Rs. $$56$$
Therefore, number of $$50$$ paise coins $$= 2\times 56 = 112$$.
Question 6
1 / -0

The area of the triangle formed by three vertices $$O(0, 0), A(1, 0), B(0, 1)$$ is _____ sq. units.

A
$$0$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$2$$

Solution

Points are given as $$O(0,0), A(1,0), B(0,1)$$.
When plotted on the cartesian plane, these points make a right angled triangle $$OAB$$ with
$$OA=1$$, $$OB=1$$
Area of this right angled triangle $$=\dfrac12\times OA\times OB$$
$$=\cfrac12\times1\times1\ sq.unit$$
$$=\cfrac12\ sq.unit$$
Question 7
1 / -0

Find the number of ways in which $$5$$ persons $$A,B,C,D,E$$ can be seated round a table such that A and B always sit together?

A
$$24$$

B
$$12$$

C
$$6$$

D
$$36$$

Solution

Consider $$AB$$ as single unit .
There are total 4 units and arrange them in a circular table.
It will be $$(4-1)!=3!$$
and rest $$"AB"$$ can also arrange themselves.
Total number of permutations$$=3! \times 2 $$
$$=12$$
Question 8
1 / -0

A garrison of '$$n$$' men had enough food to last for $$30$$ days. After $$10$$ days, $$50$$ more men joined them. If the food now lasted for $$16$$ days, what is the value of $$n$$?

A
$$200$$

B
$$240$$

C
$$280$$

D
$$320$$

Solution

After $$10$$ days, the food for n men is there for $$20$$ days.
This food can be eaten by $$(n + 50)$$ men in $$16$$ days.
$$\therefore 20n = 16(n + 50)$$
$$\therefore n = 200$$
Question 9
1 / -0

The sum of two numbers is $$80$$. If the larger number exceeds four times the smaller by $$5$$, what is the smaller number?

A
$$5$$

B
$$15$$

C
$$20$$

D
$$25$$

Solution

Given, sum of two numbers is $$80$$.
Let the smaller number be $$x$$
Thus the larger will be $$80 - x$$.
Also given, larger number exceeds four times the smaller number by $$5$$.
Therefore, $$ 80 - x = 4x + 5$$
$$\Rightarrow 5x = 75$$
$$\Rightarrow x = 15$$
Thus the smaller number is $$15$$.
Question 10
1 / -0

A man has 9 friends, 4 boys and 5 girls. In how many ways can be invite them, if there have to be exactly three girls in the invites?

A
$$320$$

B
$$160$$

C
$$80$$

D
$$200$$

Solution

He can select three girls in $$^5C_3$$ ways = 10 ways
He can select boys in $$ [^4C_0+^4C_1 +^4C_2 +^4C_3 + ^4C_4]$$ ways = 16 ways
Total number of ways $$= 10\times16$$ $$ = 160$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 11

Result

Permutations and Combinations Test 11

Score

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Rank

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SHARING IS CARING

Question 1

Series

Preceding sequence

Progression

Geometric progression

Solution

Question 2

110

115

120

125

Solution

Question 3

$$17$$

$$18$$

$$19$$

$$20$$

Solution

Question 4

$$\displaystyle ^{ 8 }{ c }_{ 5 }$$

$$63$$

$$35$$

$$\displaystyle ^{ 9 }{ c }_{ 4 }$$

Solution

Question 5

$$112$$

$$128$$

$$96$$

$$24$$

Solution

Question 6

$$0$$

$$\dfrac{1}{2}$$

$$1$$

$$2$$

Solution

Question 7

$$24$$

$$12$$

$$6$$

$$36$$

Solution

Question 8

$$200$$

$$240$$

$$280$$

$$320$$

Solution

Question 9

$$5$$

$$15$$

$$20$$

$$25$$

Solution

Question 10

$$320$$

$$160$$

$$80$$

$$200$$

Solution

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