Mathematics Test

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Mathematics Test - 25

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Question 1
1 / -0

If HCF (p, q, r) = 2, LCM (p, q, r) = 120, HCF (p, q) = 4, HCF (q, r) = 2, HCF (p, r) = 2, p = 12, q = 8 then find the no ‘r’.

A
6

B
10

C
8

D
None of these

Solution

$\begin{array}{l} LCM (p, q, r) = \frac{p \cdot q \cdot r HCF (p, q, r)}{H C P (p, q) HCF (q, r) H C F (p, r)} \\ 120 = \frac{r \times 12 \times 8 \times 2}{4 \times 2 \times 2} \\ \frac{120}{12} = r \\ 10 = r \end{array}$
Question 2
1 / -0

An aeroplane at an altitude of 1200 meters finds that two ships are sailing towards it in the same direction. The angle of depression of the ships as observed from the aeroplane are 60° and 30° respectively. Find the distance between the two ships.

A
800√2

B
800√3

C
600√3

D
600√2

Solution

Let the aeroplane be at $\mathrm{B}$ and let the two ships be at C $\&$ D such that their angles of depression from B are
$30^{\circ}$ and $60^{\circ}$ respectively.
Question 3
1 / -0

If $\tan 2 \theta=\cot \left(\theta+6^{\circ}\right),$ where $2 \theta$ and $\theta+6^{\circ}$ are acute angles, find the value of $\theta$

A
$27 °$

B
$28 °$

C
$10 °$

D
$36 °$

Solution

We have

$\begin{array}{l} \tan 2 θ = \cot (θ + 6^{\circ}) \\ \Rightarrow \cot (90^{\circ} - 2 θ) = \cot (θ + 6^{\circ}) \\ \Rightarrow 90^{\circ} - 2 θ = θ + 6^{\circ} \\ \Rightarrow 3 θ = 84^{\circ} \Rightarrow θ = 28^{\circ} \end{array}$
Question 4
1 / -0

A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all a round to a width of 5m to form an embankment. Find the height of embankment.

A
4.66m

B
5.66m

C
3.66m

D
6.66m

Solution

We have, volume of the earth dugout $=\left(\pi r^{2} h\right) m^{3}$ Volume of the earth dugout $=\frac{22}{7} \times 5 \times 14$ $=1100 \mathrm{m}^{3}$
Area of the embankment (shaded region) $=\pi\left(R^{2}-r^{2}\right)$ $=\pi\left(10^{2}-5^{2}\right)=\frac{22}{7} \times 75 m^{2}$
$\therefore$ Height of the embankment $=\frac{\text {Volume of the earth dugout}}{\text {Area of the embankment}}$
$=\frac{1100}{\frac{32}{7} \times 75}=\frac{7 \times 1100}{22 \times 75}$
$=4.66 \mathrm{m}$

We have, volume of the earth dugout $=\left(\pi r^{2} h\right) m^{3}$ Volume of the earth dugout $=\frac{22}{7} \times 5 \times 14$ $=1100 \mathrm{m}^{3}$
Area of the embankment (shaded region) $=\pi\left(R^{2}-r^{2}\right)$ $=\pi\left(10^{2}-5^{2}\right)=\frac{22}{7} \times 75 m^{2}$
$\therefore$ Height of the embankment $=\frac{\text {Volume of the earth dugout}}{\text {Area of the embankment}}$
$=\frac{1100}{\frac{32}{7} \times 75}=\frac{7 \times 1100}{22 \times 75}$
$=4.66 \mathrm{m}$
Question 5
1 / -0

Consider the function $f (x) = \{\begin{matrix} \frac{\tan k x}{x}, & x < 0 \\ 3 x + 2 k^{2}, & x \geq 0 \end{matrix}$ what is the non-zero value of k for which thefunction is continuous at $x=0 ?$

A
1/4

B
1/2

C
1

D
2

Solution

We have, $f(x)=\left\{\begin{array}{c}\frac{\tan k x}{x} x<0 \\ 3 x+2 k^{2} \quad x \geq 0\end{array}\right.$
$\because f(x)$ is continuous at $x=0$ $\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan k x}{x}=\lim _{x \rightarrow 0^{+}} 3 x+2 k^{2}=3(0)+2 k^{2}$
$\Rightarrow \lim _{x \rightarrow 0-h} \frac{\tan k(0-h)}{0-h}=\lim _{x \rightarrow 0+h} 3(0+h)+2 k^{2}=2 k^{2}$
$\Rightarrow \lim _{h \rightarrow 0} \frac{\tan k h}{h}=2 k^{2}$
$\Rightarrow \lim _{h \rightarrow 0} k \frac{\tan k h}{k h}=2 k^{2}$
$\Rightarrow k=2 k^{2} \quad\left[\because \lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]$
$\therefore \mathrm{k}=1 / 2$
Question 6
1 / -0

The ratio of the sums of $\mathrm{m} \& \mathrm{n}$ terms of an $\mathrm{A}$. $\mathrm{P}$ is $\mathrm{m}^{2}: \mathrm{n}^{2}$. The ratio of the $\mathrm{m}^{\text {th }} \& \mathrm{n}^{\text {th }}$ terms is?

A
$(2 m - 1) : (2 n - 1)$

B
$(1 - 2 m) : (2 n - 1)$

C
$(2 + m) : (2 + n)$

D
$(2 m + 1) : (2 n + 1)$

Solution

$\begin{array}{l} S_{m} = \frac{m}{2} {2 a + (m - 1) d} & S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ \frac{S m}{S n} = \frac{m^{2}}{n^{2}} \\ \frac{\frac{m}{2} (2 a + (m - 1) d)}{\frac{a}{2} (2 a + (n - 1) d)} = \frac{m^{2}}{n^{2}} \\ \frac{2 a + (m - 1) d}{2 a + (n - 1) d} = \frac{m}{n} \\ {2 a + (m - 1) d} n = {2 a + (n - 1) d} m \\ 2 a (n - m) = d {(n - 1) m - (m - 1) n} \\ 2 a (n - m) = d (n - m) \\ d = 2 a \\ ∴ \frac{T m}{T n} = \frac{a + (m - 1) d}{a + (n - 1) d} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a} = \frac{2 m - 1}{2 n - 1} \end{array}$
Question 7
1 / -0

4 chairs & 3 tables cost Rs. 2100 and 5 chairs and 2 tables cost Rs. 1750. Find the cost of a chair & a table separately.

A
500, 150

B
150, 500

C
$300, 350$

D
350, 300

Solution

Let the cost of a chair be Rs. $x$ and that of a table be Rs. $y$. Then, $4 x+3 y=2100$
Put the value of $x \text { in ( } 1)$
$
\begin{array}{l}
4 x+3 y=2100 \\
600+3 y=2100 \\
3 y=2100-600 \\
3 y=1500 \\
y=500
\end{array}
$
i.e. $x =150, y =500$
Question 8
1 / -0

A sum of money amounts to Rs. 4818 after 3 years and Rs. 7227 after 6 years on compound interest.The sum is

A
Rs. 3122

B
Rs. 3212

C
Rs. 2409

D
Rs. 2490

Solution
Question 9
1 / -0

Find the value of $(1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 - x)$ ?

A
$1 + x^{16}$

B
$1 - x^{16}$

C
$1 + x^{8}$

D
$1 - x^{8}$

Solution

We have,

$\begin{array}{l} (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 - x) \\ = (1 - x) (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ = (1 - x^{2}) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) \\ = (1 - x^{4}) (1 + x^{4}) (1 + x^{8}) \\ = (1 - x^{8}) (1 + x^{8}) \\ = (1 - x^{16}) \end{array}$
Question 10
1 / -0

The solution of the equation $\log \left(\frac{d y}{d x}\right)=a x+b y$ is

A
$\frac{e^{b y}}{b} = \frac{e^{a x}}{a} + c$

B
$\frac{e^{- b y}}{- b} = \frac{e^{a x}}{a} + c$

C
$\frac{e^{- b y}}{a} = \frac{e^{a x}}{b} + c$

D

None of these

Solution

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 25

Result

Mathematics Test - 25

Score

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Rank

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SHARING IS CARING

Question 1

6

10

8

None of these

Solution

Question 2

800√2

800√3

600√3

600√2

Solution

Question 3

27°

28°

10°

36°

Solution

Question 4

4.66m

5.66m

3.66m

6.66m

Solution

Question 5

1/4

1/2

1

2

Solution

Question 6

(2m-1):(2n-1)

(1-2m):(2n-1)

(2+m):(2+n)

(2m+1):(2n+1)

Solution

Question 7

500, 150

150, 500

300,350

350, 300

Solution

Question 8

Rs. 3122

Rs. 3212

Rs. 2409

Rs. 2490

Solution

Question 9

1+x16

1-x16

1+x8

1-x8

Solution

Question 10

ebyb=eaxa+c

e-by-b=eaxa+c

e-bya=eaxb+c

None of these

Solution

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$27 °$

$28 °$

$10 °$

$36 °$

$(2 m - 1) : (2 n - 1)$

$(1 - 2 m) : (2 n - 1)$

$(2 + m) : (2 + n)$

$(2 m + 1) : (2 n + 1)$

$300, 350$

$1 + x^{16}$

$1 - x^{16}$

$1 + x^{8}$

$1 - x^{8}$

$\frac{e^{b y}}{b} = \frac{e^{a x}}{a} + c$

$\frac{e^{- b y}}{- b} = \frac{e^{a x}}{a} + c$

$\frac{e^{- b y}}{a} = \frac{e^{a x}}{b} + c$