Mix Test 3

Formula of finding volume of cuboid = \(l\times b\times h\)

The dimensions of the cuboid are \(100\,cm \times 80\,cm\times 64\,cm\)

Therefore, the volume of the cuboid is V = \(l\times b\times h\) = 100 \(\times\) 80 \(\times\) 64

= \(512000\,cm^3\)

The dimensions of the cuboid are \(100\,cm\times 80\,cm \times 64\,cm\)

Therefore, the volume of the cuboid is \(V=l\times b\times h\) = \(100\times 80\times 64\) = \(512000\,cm^3\)

Given that the cuboid is melted and recast into a cube.

\(\therefore\) The volume of the cube = the volume of the cuboid.

\(\therefore\) The volume of the cube = \(512000\,cm^3\)

The dimensions of the cuboid are \(100\,cm\times 80\,cm\times 64\,cm\)

Therefore, the volume of the cuboid is \(V=l\times b\times h\) = \(100\times80\times 64\) = \(512000\,cm^3\).

Given that the cuboid is melted and recast into a cube.

\(\therefore\) The volume of the cube = the volume of the cuboid.

\(\therefore\) The volume of the cube = \(512000\,cm^3\)

Since, volume of the cube whose side length is a is given by \(a^3\)

\(\therefore\) \(a^3=512000\) = \(8^3\times 10^3\) = \(80^3\)

\(\Rightarrow\) a = 80 cm.

Hence, the side length of the cube is a = 80 cm.

\(\therefore\) Surface area of the cube = \(6a^2\) 

= \(6\times80^2\) = \(38400\,cm^2\)

Let first term of AP is a and common difference of AP is d.

Given that \(p^{th}\) term of AP is q.

\(\therefore\) \(a+(p-1)d=q\) ............(1)  (\(\because\) \(n^{th}\) term of AP is given by \(a_n\) = a + (n - 1)d)

And given that \(q^{th}\) term of AP is p.

Therefore, a+(q - 1)d = p.  ..........(2)  (\(\because\) \(n^{th}\) term of AP is given by a + (n - 1)d)

Now, \((p+q)^{th}\) term of given AP is

a + (p + q – 1) d = 2a+ (p + q – 1 – 1 + 1) d – a

= a + (p - 1)d + a + (q - 1)d + d – a

= p + q + d – a.   (From equations (1) and (2) )

Hence, \((p+q)^{th}\) term of given AP is = p + q + d – a.  .........(3)

Now subtracting equation (2) from equation (1), we get

a + (p –1)d – (a + (q − 1)d) = q – p

\(\Rightarrow\) a + (p – 1) d – a – (q – 1) d = q – p

\(\Rightarrow\) (p – 1 – q + 1)d = q – p

\(\Rightarrow\) (p– q)d = –(p – q).

\(\Rightarrow\) d = – 1.

Now, by equation (1), a + (q – 1) (–1) = p

\(\Rightarrow\) a = p + q – 1.

Now, putting the value of a & d in equation (3), we get

\((p+q)^{th}\) term is p + q + (-1) – (p + q – 1)

= p + q – 1 – p – q + 1 = 0.

Hence \((p+q)^{th}\) term of given AP is 0.

Given that cos θ = 0.6

Therefore sin θ = \(\sqrt{1-\cos^2\theta}\) 

= \(\sqrt{1-(0.6)^2}\) 

= \(\sqrt{1-0.36}\) 

= \(\sqrt{0.64}\) = 0.8

And tan θ = \(\frac{\sin\theta}{\cos\theta}\) 

= \(\frac{0.8}{0.6}\) = \(\frac{8}{6}\) = \(\frac{4}{3}\)

Now, 5 sinθ - 3tanθ = \(5\times 0.8-3\times \frac{4}{3}\) 

= 4 - 4 = 0

Given that string makes an angle θ with the ground level such that tan θ = \(\frac{15}{8}\)

\(\therefore\) cot θ = \(\frac{1}{\tan\theta}=\frac{1}{\frac{15}{8}}=\frac{8}{15}\)