Mathematical Skills Test - 9

Given:

\(\frac{\tan \mathrm{a}}{1-\cot \mathrm{a}}+\frac{\cot \mathrm{a}}{1-\tan \mathrm{a}}\)

\(=\frac{\tan \mathrm{a}}{1-\frac{1}{\tan \mathrm{a}}}+\frac{\operatorname{cota}}{1-\tan \mathrm{a}}\)

\(=\frac{\tan ^{2} \mathrm{a}}{\tan \mathrm{a}-1}-\frac{\operatorname{cota}}{\tan \mathrm{a}-1}\)

\(=\frac{\tan ^{2} \mathrm{a}-\operatorname{cota}}{\tan \mathrm{a}-1}\)

\(=\frac{\tan ^{3} \mathrm{a}-1}{\tan \mathrm{a}(\tan \mathrm{a}-1)}\)

\(=\frac{(\tan \mathrm{a}-1) \times\left(\tan ^{2} \mathrm{a}+1+\tan \mathrm{a}\right)}{\tan \mathrm{a} \times(\tan \mathrm{a}-1)}\)

\(=\frac{\left(\tan ^{2} \mathrm{a}+1+\tan \mathrm{a}\right)}{\tan \mathrm{a}}\)

\(=\frac{\left(\sec ^{2} \mathrm{a}+\tan \mathrm{a}\right)}{\tan \mathrm{a}}\)

\(=\frac{\left(\frac{1}{\cos ^{2} \mathrm{a}}+\frac{\operatorname{sina}}{\operatorname{cosa}}\right)}{\frac{\sin 3}{\operatorname{cosa}}}\)

\(=\frac{1}{\cos \mathrm{a}} \cdot \frac{1}{\sin \mathrm{a}}+1\)

= sec a × cosec a + 1

Given:

350 is increased by 20% and then decreased by 20%.

Formula:

Percentage increase/decrease =$\left(\frac{\mathrm{Increased}\mathrm{or}\mathrm{decreased}\mathrm{value}}{\mathrm{Original}\mathrm{value}}\right)\times 100$

Calculation:

Original number = 350

On increasing it by 20%, we get

The new value = 350 ×$\left(\frac{100+20}{100}\right)$

⇒ 350 ×$\frac{120}{100}$ = 420

Now, on decreasing it by 20%, we get

The new value = 420 ×$\left(\frac{100-20}{100}\right)$

⇒ 420 ×$\frac{80}{100}$ = 336

⇒ Decreased value = 350 - 336 = 14

⇒ Decreased % =$\frac{14}{350}$ × 100 = 4%

So, the percentage decrease is 4%.

Here, we have to find the distance of the point (2, 3, 4) from the plane 3x - 6y + 2z + 11 = 0

As we know that, the perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x₁, y₁, z₁) is given by:

$\mathrm{D}=\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1+\mathrm{d}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$

Here, x₁ = 2, y₁ = 3, z₁ = 4, a = 3, b = - 6, c = 2and d = 11

So, the required distance between the given point and plane is$\mathrm{D}=\frac{\left|3\times 2-6\times 3+2\times 4+11\right|}{\sqrt{3^2+{\left(-6\right)}^2+2^2}}$

⇒ D = 1

Let the capacity of the cistern be 120 litres (LCM of 12, 8 and 10)

Then, in one-hour A fill 10 litres, B fill 15 litres and C empties 12 litres.

Till 1 pm, A filled for four hours and B filled for two hours.

⇒ A filled (4 × 10) = 40 litres

⇒ B filled (2 × 15) = 30 litres

i.e. (40 + 30) = 70 litres filled by 1 pm.

⇒ C should have emptied 10 litres by 1 pm so that the cistern remains half full.

⇒ C will empty 10 litres in 50 minutes

∴ C should be started at 12:10 pm.

Given:

The lines \(\frac{x-3}{1}=\frac{y+2}{-4}=\frac{z}{5}\) and \(\frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-a}{5}\) are coplanar

Here, we have to find the value of a

As we know that, if two lines \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) and \(\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}\) are coplanar then \(\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0\)

Here, \(x_{1}=3, y_{1}=-2, z_{1}=0, a_{1}=1, b_{1}=-4, c_{1}=5\)

Similarly, \(x_{2}=4, y_{2}=3, z_{2}=a, a_{2}=1, b_{2}=-4\) and \(c_{2}=5\)

\(\Rightarrow\left|\begin{array}{ccc}1 & 5 & a \\ 1 & -4 & 5 \\ 1 & -4 & 5\end{array}\right|=0\)

\(\Rightarrow 1 \times(-20+20)-5 \times(5-5)+a \times(-4+4)=0\)

\(\Rightarrow\) a can be any real number.

Given, Ratio of speed of trains \(=6: 7\)

Second train covers \(364\) kms in 4 hours

Then, its speed \(=\frac{364}{4}=91\)km/hr

In the question it is given that speed of the second train \(=7\) units

But actual speed \(=91\)km/hr

i.e. 7 units \(\rightarrow 91\) km

1 unit \(\rightarrow 13\) km

Therefore,

Speed of the first train is:

\(=6\)

\(=6 \times 13\)

\(=78\)km/hr

After reduction of Rs. 25, let the shares of A, B and C be x, 3x and 2x, respectively.

Total amount = 6x

Initially, A + B + C = 735 …(i)

After reduction of Rs. 25,

A – 25 + B – 25 + C – 25 = 6x

⇒ A + B + C – 75 = 6x …(ii)

From (i) and (ii), we get

735 – 75 = 6x

⇒ x = 110

After deduction, share of C = 2 × Rs. 110 = Rs. 220

∴ Before deduction, share of C = Rs. 220 + Rs. 25 = Rs. 245

Given:

Time = 6 years

Amount = 4 × Principal

Formula:

Simple interest(S.I) =$\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{t}}{100}$

Amount = P + S.I

Where P = Principal,R = Rate of Interest,t = Time

Let the sum be Rs. P.

The amount will be Rs. 4P.

Let the rate of interest be R%.

And the time to become 9 times of itself is t years.

So, Simple interest will become Rs. 3P

Simple interest(S.I) =$\frac{\mathrm{P}\times \mathrm{R}\times \mathrm{t}}{100}$

⇒ 3P =$\frac{\mathrm{P}\times \mathrm{R}\times 6}{100}$

⇒ R =$\frac{300}{6}$ = 50%

In the question asked 9 times the principal

So, Interest will 8P in T years

⇒ 8P =$\frac{\mathrm{P}\times 50\times \mathrm{T}}{100}$

⇒ T =$\frac{800}{50}$ = 16 years.

∴ The sum will take 16 years to become 9 times itself.