Differentiation Test 11

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Differentiation Test 11

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Question 1
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 $$= 10\times16$$ $$ = 160$$
Question 2
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 3
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 4
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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 5
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Differentiate
$$2x^{3/2} + 2x^{5/2} +C$$

A
$$\cfrac { dy }{ dx } =\sqrt { x } \left( 3+5x \right) $$

B
$$ \cfrac { dy }{ dx } =\sqrt { x } \left( 3-5x \right) $$

C
$$ \cfrac { dy }{ dx } =-\sqrt { x } \left( 3+5x \right) $$

D
None of these

Solution

Let $$y=2{ x }^{ 3/2 }+2{ x }^{ 5/2 }+C$$

As we know, $$\cfrac { d }{ dx } \left( { x }^{ n } \right) =n{ x }^{ n-1 }$$

$$\therefore \cfrac { dy }{ dx } =\cfrac { d }{ dx } \left( 2{ x }^{ 3/2 } \right) +\cfrac { d }{ dx } \left( 2{ x }^{ 5/2 } \right) $$

$$\quad \quad =2.\cfrac { 3 }{ 2 } { x }^{ 1/2 }+2.\cfrac { 5 }{ 2 } { x }^{ 3/2 }$$

$$\therefore \cfrac { dy }{ dx } =\sqrt { x } \left( 3+5x \right) $$
Question 6
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The number of ways in which ten candidates $$A_1, A_2,......A_{10}$$ can be ranked such that $$A_1$$ is always above $$A_{10}$$ is

A
$$5!$$

B
$$2(5!)$$

C
$$10!$$

D
$$\dfrac{1}{2}(10!)$$

Solution

Total number of ways to arrange $$A_1,A_2 .. A_{10} = 10!$$
In $$\dfrac{10! }{ 2}$$ combinations $$A_1$$ will be above $$A_2$$ and in the other half $$A_2$$ will be above $$A_1$$.

Ans : $$\dfrac{10! }{2}$$
Question 7
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 8
1 / -0

Find $$\dfrac{{dy}}{{dx}}$$ of the following $$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$$

A
$$x+2x^2+6x^3+(n-1)(n-2)x^{n-3}$$

B
$$2+6x+(n-1)x^{n-2}$$

C
$$2+6x+(n-1)(n-2)x^{n-3}$$

D
None

Solution

$$y = 1 + 2x + 3{x^2} + \left( {n - 1} \right){x^{n - 2}}$$
$$\dfrac{{dy}}{{dx}} = 2 + 6x + \left( {n - 1} \right)\left( {n - 2} \right){x^{n - 3}}$$
Question 9
1 / -0

Select the missing number from the given matrix:
5 2 4
4 4 7
2 5 3
18 30 ?

A
$$43$$

B
$$42$$

C
$$33$$

D
$$32$$

Solution

In first column $$(5+4)\times 2=18$$ In second column $$(2+4)\times 5=30$$

so in third column ans

$$=(4+7)\times3$$

$$=11\times 3=33.$$
Question 10
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From a well shuffled pack of $$52$$ playing cards two cards drawn at random. The probability that either both are red or both are kings is:

A
$$\dfrac{{\left( {^{26}{C_2} + \,{\,^4}{C_2}} \right)}}{{^{52}{C_2}}}$$

B
$$\dfrac{{\left( {^{26}{C_2} + {\,^4}{C_2} - {\,^2}{C_2}} \right)}}{{^{52}{C_2}}}$$

C
$$\dfrac{{^{30}{C_2}}}{{^{52}{C_2}}}$$

D
$$\dfrac{{^{39}{C_2}}}{{^{52}{C_2}}}$$

Solution

Assuming, cards are drawn without replacement:

Total possible events $$=^{52}C_2$$

P(both red) $$=\dfrac{ ^{26}C_2}{^{52}C_2}$$

P(both king) $$=\dfrac{^{4}C_2}{^{52}C_2}$$

P(both red & king) $$=\dfrac {^2C_2}{{^{52}C_2}}$$

We use the basic addition rule,

P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)

Required probability $$=\dfrac{^{26}C_2+^{4}C_2-^2C_2}{^{52}C_2}$$