Mathematics Test

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Mathematics Test - 12

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Question 1
1 / -0

In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.

The number of students who had taken only physics is:

A
2

B
3

C
5

D
6

Solution

Only Physics $=12-(1+3+6)=2$
Question 2
1 / -0

In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.

The number of students who had taken only two subjects is:

A
7

B
8

C
9

D
10

Solution

Only two subjects = 6 + 2 + 1 = 9
Question 3
1 / -0

In a survey of 25 students, it was found that 15 had taken Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.

Consider the following statements:

1. The number of students who had taken only one subject is equal to the number of students who had taken only two subject.

2. The number of students who had taken at least two subjects is four times the number of students who had taken all the three subjects.

Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Statement 1:
Students, who had taken only one subject =
2 +5+ 4 = 11
Students, who had taken only two subjects =
6 + 2 + 1 = 9
1 ≠ 9
Statement 2:
Students who had taken at least two subject =
1 + 2 + 6 + 3 = 12
Students who had taken all three subjects =
4 × 3 = 12
Question 4
1 / -0

In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{\text {n }}$ where $n$ is a positive integer, the sum of the coefficients of $x^{5}$ and $x^{10}$ is 0 . What is n equal to?

A
5

B
10

C
15

D
None of these

Solution

$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$

General term, $T_{r+1}=^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(\frac{1}{x^{2}}\right)^{r}$

$=^{n} C_{r .} 3^{(3 n-3 r)} \cdot(-1)^{r} \cdot r^{-2 r}$

$=^{n} C_{r \cdot}(-1)^{r} \cdot x^{(3 n-5 r)}$

For the coefficient $x^{5}$ Put $3 n-5 r=5$

$5 r=3 n-5$

$r=\frac{3 n}{5}-1$

$\therefore$ Coefficient of $x^{5}=^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}$

For the coefficient of $x^{10}$ Put $3 n-5 r=10$

$5 r=3 n-10$

$r=\frac{3 n}{5}-2$

$\therefore$ Coefficent of $x^{10}=^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}$

The sum of the coefficient of $x^{5}$ and $x^{10}=0$ $\Rightarrow^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}+^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}=0$

$\Rightarrow(-1)^{\frac{3 n}{5}}\left[n C\left(\frac{3 n}{5}-1\right) \cdot(-1)^{-1}+^{n} C\left(\frac{3 n}{5}-2\right)(-1)^{(-2)}\right]=0$

$\Rightarrow-^{n} C_{\left(\frac{3 n}{5}-1\right)}+^{n} C_{\left(\frac{3 n}{5}-2\right)}=0 \ldots \ldots . .(\mathrm{ii})$

From equation (ii) $^{n} C_{\left(\frac{3 n}{5}-2\right)}=^{n} C_{\left(\frac{3 n}{5}-1\right)}$

$\Rightarrow \mathrm{n}=\left(\frac{3 n}{5}-2\right)+\left(\frac{3 n}{5}-1\right) \quad\left[\because^{n} C_{x}=^{n} C_{y} \Rightarrow n=x+y\right]$

$\Rightarrow n=\frac{6 n}{5}-3 \Rightarrow \frac{6 n}{5}-n=3$

$\Rightarrow \frac{n}{5}=3 \quad \therefore \mathrm{n}=15$
Question 5
1 / -0

In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where $n$ is a positive integer, the sum of the coefficients of $x^{5}$ and $x^{10}$ is 0 .

What is the value of the independent term?

A
5005

B
7200

C
-5005

D
-7200

Solution

$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$

General term, $T_{r+1}=^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(\frac{1}{x^{2}}\right)^{r}$

$=^{n} C_{r_{r}} 3^{(3 n-3 r)} \cdot(-1)^{r} \cdot r^{-2 r}$

$=^{n} C_{r \cdot}(-1)^{r} \cdot x^{(3 n-5 r)}$

For the coefficient $x^{5}$ Put $3 n-5 r=5$

$5 r=3 n-5$

$r=\frac{3 n}{5}-1$

$\therefore$ Coefficient of $x^{5}=^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}$

For the coefficient of $x^{10}$ Put $3 n-5 r=10$

$5 r=3 n-10$

$r=\frac{3 n}{5}-2$

$\therefore$ Coefficent of $x^{10}=^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}$

The sum of the coefficient of $x^{5}$ and $x^{10}=0$ $\Rightarrow^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}+^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}=0$

$\Rightarrow(-1)^{\frac{3 n}{5}}\left[^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{(-2)}\right]=0$

$\Rightarrow-^{n} C_{\left(\frac{3 n}{5}-1\right)}+^{n} C_{(3 n-2)}^{3 n}=0 \ldots \ldots(\mathrm{ii})$

For the independent term, Put $3 n-5 r=0 \quad[\text { from eq. }(i)]$

$\Rightarrow 5 r=3 n=3 \times 15$

$5 r=3 \times 3 \times 5$

Putting the value of $r$ in eq. (i) we get $T_{9+1}=^{15} C_{9 .}(-1)^{9} \cdot x^{(3 \times 15-5 \times 9)}$

$\Rightarrow T_{10}=-^{15} C_{9} \cdot x^{0}=-^{15} C_{9}$

$\Rightarrow T_{10}=-^{15} C_{6}\left[\because^{n} C_{r}=^{n} C_{n-r}\right]$

$=\frac{-15 !}{6 ! 9 !}\left[\because n C_{r} \frac{n !}{r !(n-r)}\right]$

$=-5005$
Question 6
1 / -0

In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{\text {n }}$ where $n$ is a positive integer, the sum of the coefficients of $x^{5}$ and $x^{10}$ is 0 .

What is the sum of the coefficients of the two middle terms?

A
0

B
1

C
-1

D
None of these

Solution

$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$

General term, $T_{r+1}=^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(\frac{1}{x^{2}}\right)^{r}$

$=^{n} C_{r} 3^{(3 n-3 r)} \cdot(-1)^{r} \cdot r^{-2 r}$

$=^{n} C_{r \cdot}(-1)^{r} \cdot x^{(3 n-5 r)}$

For the coefficient $x^{5}$ Put $3 n-5 r=5$

$5 r=3 n-5$

$r=\frac{3 n}{5}-1$

$\therefore$ Coefficient of $x^{5}=^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}$

For the coefficient of $x^{10}$ Put $3 n-5 r=10$

$5 r=3 n-10$

$\therefore r=\frac{3 n}{5}-2$

$\therefore$ Coefficent of $x^{10}=^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}$

The sum of the coefficient of $x^{5}$ and $x^{10}=0$ $\Rightarrow^{n} C_{\left(\frac{3 n}{5}-1\right)}(-1)^{\left(\frac{3 n}{5}-1\right)}+^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{\left(\frac{3 n}{5}-2\right)}=0$

$\Rightarrow(-1)^{\frac{3 n}{5}}\left[^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+^{n} C_{\left(\frac{3 n}{5}-2\right)}(-1)^{(-2)}\right]=0$

$\Rightarrow-^{n} C_{\left(\frac{3 n}{5}-1\right)}+^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$

$\mathrm{n}=15$

Total term in the expansion of $\left(x^{3}-\frac{1}{2}\right)^{15}$ is 16

$\therefore$ middle term $=8^{\text {th }}$ term and $9^{\text {th }}$ term $T_{8}=T_{(7+1)}=^{15} C_{7} \cdot(-1)^{7} \cdot x^{(3 \times 15-5 \times 7)}$

$=-^{15} C_{7} \cdot x^{10} \quad(\text { from eq. }(\mathrm{i}))$

$T_{9}=T_{(8+1)}=^{15} C_{8} \cdot(-1)^{8} \cdot x^{(3 \times 15-5 \times 8)}$

$=-15 C_{8} \cdot x^{5} \quad(\text { from eq. }(\text { ii }))$

The sum of the coefficients of the two middle terms $=-^{15} C_{7}+^{15} C_{8}=-^{15} C_{7}+^{15} C_{7} \cdot\left[\because^{n} C_{r}=^{n} C_{n-r}\right]$
Question 7
1 / -0

Given that C(n, r) : C(n, r + 1) = 1 : 2 and C(n, r + 1) : C(n, r + 2) = 2 : 3.What is n equal to?

A
11

B
12

C
13

D
14

Solution

$\frac{n_{C_{r}}}{n_{C_{r+1}}}=\frac{1}{2}$

$\frac{n !(r+1) !(n-r-1) !}{r !(n-r) ! n !}=\frac{1}{2}$

$\frac{r+1}{n-r}=\frac{1}{2} \Rightarrow 3 r-n+2=0$

$\frac{n C_{r+1}}{n_{C r+2}}=\frac{2}{3}$

$\frac{n !(r+2) !(n-r-2) !}{(r+1) !(n-r-1) ! n !}=\frac{2}{3}$

$\frac{r+2}{n-r-1}=\frac{2}{3} \Rightarrow 5 r-2 n+8=0$

Solving equations (i) and (ii), we get $\mathrm{n}=14, \mathrm{r}=4$
Question 8
1 / -0

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?

A
$\frac{2}{3}$

B
$\frac{8}{21}$

C
$\frac{3}{7}$

D
$\frac{9}{22}$

Solution

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green
=event that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
Question 9
1 / -0

Given that C(n, r) : C(n, r + 1) = 1 : 2 and C(n, r + 1) : C(n, r + 2) = 2 : 3.What is P(n, r) : C(n, r) equal to?

A
6

B
24

C
120

D
720

Solution

$\frac{n_{C_{r}}}{n_{C_{r+1}}}=\frac{1}{2}$

$\frac{n !(r+1) !(n-r-1) !}{r !(n-r) ! \cdot n !}=\frac{1}{2}$

$\frac{r+1}{n-r}=\frac{1}{2} \Rightarrow 3 r-n+2=0$

$\frac{n_{C r+1}}{n_{C r+2}}=\frac{2}{3}$

$\frac{n !(r+2) !(n-r-2) !}{(r+1) !(n-r-1) ! n !}=\frac{2}{3}$

$\frac{r+2}{n-r-1}=\frac{2}{3} \Rightarrow 5 r-2 n+8=0$

Solving equations (i) and (ii), we get $\mathrm{n}=14, \mathrm{r}=4$

$\mathrm{P}(\mathrm{n}, \mathrm{r}): \mathrm{C}(\mathrm{n}, \mathrm{r})=| r=24$
Question 10
1 / -0

The complete solution 3 tan² x = 1 is given by :

A
$x=n \pi \pm \frac{\pi}{3}$

B
$x=n \pi+\frac{\pi}{3}$ only

C
$x=n \pi \pm \frac{\pi}{6}$

D
$x=n \pi+\frac{\pi}{6}$ only

Solution

$3 \tan ^{2} x=1$

$\tan x=\pm \frac{1}{\sqrt{3}}$

$\tan x=\tan \left(\pm \frac{\pi}{6}\right)$

$x=n \pi \pm \frac{\pi}{6}$

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 12

Result

Mathematics Test - 12

Score

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Rank

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SHARING IS CARING

Question 1

2

3

5

6

Solution

Question 2

7

8

9

10

Solution

Question 3

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 4

5

10

15

None of these

Solution

Question 5

5005

7200

-5005

-7200

Solution

Question 6

0

1

-1

None of these

Solution

Question 7

11

12

13

14

Solution

Question 8

23

821

37

922

Solution

Question 9

6

24

120

720

Solution

Question 10

\(x=n \pi \pm \frac{\pi}{3}\)

\(x=n \pi+\frac{\pi}{3}\) only

\(x=n \pi \pm \frac{\pi}{6}\)

\(x=n \pi+\frac{\pi}{6}\) only

Solution

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$\frac{2}{3}$

$\frac{8}{21}$

$\frac{3}{7}$

$\frac{9}{22}$