Mathematical Skills Test - 6

Given: Center is (0, 0)

So standard equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)

Since (4, 3) lies on ellipse satisfy the equation of ellipse

\(\frac{16}{{a}^{2}}+\frac{9}{{b}^{2}}=1\) .....(i)

Similarly for (2,5)

\(\frac{4}{a^{2}}+\frac{25}{b^{2}}=1\) ....(ii)

Substracting the equations (i) from 4 × (ii)

\(\frac{4 \times 25}{b^{2}}-\frac{9}{b^{2}}=4-1\)

\(\Rightarrow \frac{91}{b^{2}}=3\)

\(\Rightarrow b^{2}=\frac{91}{3}\)

Putting it back in equation (i)

\(\frac{16}{a^{2}}+\frac{9 \times 3}{91}=1\)

\(\Rightarrow \frac{16}{a^{2}}=1-\frac{27}{91}\)

\(\Rightarrow \frac{16}{a^{2}}=\frac{64}{91}\)

\(\Rightarrow a^{2}=\frac{91}{4}\)

Equation of ellipse

\(\frac{4 x^{2}}{91}+\frac{3 y^{2}}{91}=1\)

\(\Rightarrow 4 x^{2}+3 y^{2}=91\)

Let the litres of water which the tank hold be ‘w’ litres.

Given, leak can completely empty the tank in 10 hours.

∴ In 1 hour, volume of tank emptied = \(\frac{w}{10}\) litres ----(1)

Given, tap admits 4 litres water in a minute.

Thus in 1 hr water filled by the tap = 4 × 60 = 240 litres ----(2)

Now, the leak takes 15 hours to empty the tank when tap is simultaneously opened

⇒ In 1 hr, volume of tank emptied = \(\frac{w}{15}\)litres ----(3)

Thus,

Combining 1, 2 and 3 we have

240 = \(\frac{w}{10}\)–\(\frac{w}{15}\)

⇒ 240 =\(\frac{w}{30}\)

⇒ w = 7200 litres

Given:

A family gets 1 kg more sugar for Rs. 80 due to reduction of price of sugar by Rs. 4 per kg.

Formula:

Consumption =$\frac{\mathrm{Expenditure}}{\mathrm{Price}}$

Percentage =$\frac{\mathrm{Actual}\mathrm{value}}{\mathrm{Original}\mathrm{Value}}\times 100$

Let the original price of sugar be Rs. x/kg.

According to the question,

$\Rightarrow \frac{80}{\mathrm{x}-4}-\frac{80}{\mathrm{x}}=1$

$\Rightarrow \frac{80\left(\mathrm{x}-\mathrm{x}+4\right)}{\left(\mathrm{x}-4\right)\left(\mathrm{x}\right)}=1$

⇒ 320 = (x – 4) × (x)

⇒ 320 = x² – 4x

Solving the equation, x = 20

Original price of sugar = Rs. 20/kg

Reduced price = Rs. 16/kg

∴ Required percentage =$\left(\frac{4}{20}\right)$ × 100 = 20%

∴ Reduction in the price of sugar is 20%.

The equation of parabola is x = y² - 6y + c

⇒ x = y² - 6y + 9 - 9 + c

⇒ x = (y - 3)² - 9 + c

⇒ (y - 3)² = x + 9 - c

⇒ (y - 3)² = [x - (c - 9)]

Comparing it with (y - k)² = 4a(x - h)

Thus vertex is (c - 9, 3)

∵ The vertex lies on y -axis.

⇒ c - 9 = 0 ⇒ c = 9

Given that:

Speed of a boat in still water = 5 km/hr

Speed of stream = 2 km/hr

Distance to be traveled downstream = 70 km

We know that,

Downstream speed (S_D) = Speed of boat + Speed of stream

T_D=$\frac{D_D}{S_D}$

WhereS_D, D_D, and T_Dis speed, distance, and time respectivelyduring downstream

S_D= Speed of boat + Speed of stream

⇒ S_D= 5 + 2

⇒ S_D= 7 km/hr

T_D=$\frac{D_D}{S_D}$

⇒ Time =$\frac{154}{7}$

∴Time taken to cover 154 km in 22 hours.

Given: \(\cos A=\frac{3}{4}\)

As we know that, sin² x + cos² x = 1

\(\Rightarrow \sin ^{2} A=1-\cos ^{2} A=1-\frac{9}{16}=\frac{7}{16}\)

\(\Rightarrow \sin A=\frac{\sqrt{7}}{4}\)

\(\Rightarrow \sin \left(\frac{A}{2}\right) \sin \left(\frac{3 A}{2}\right)=\frac{1}{2} \times\left[2 \times \sin \left(\frac{A}{2}\right) \sin \left(\frac{3 A}{2}\right)\right]\)

As we know that, 2 sin A sin B = cos (A - B) – cos (A + B)

\(\Rightarrow \sin \left(\frac{A}{2}\right) \sin \left(\frac{3 A}{2}\right)=\frac{1}{2} \times(\cos A-\cos 2 A)\) .....(i)

As we know that, cos 2A = cos² A – sin² A

\(\Rightarrow \cos 2 A=\frac{9}{16}-\frac{7}{16}=\frac{1}{8}\)

By substituting the value of cos A and cos 2A in equation (i), we get

\(\Rightarrow \sin \left(\frac{A}{2}\right) \sin \left(\frac{3 A}{2}\right)=\frac{1}{2} \times\left(\frac{3}{4}-\frac{1}{8}\right)=\frac{5}{16}\)

Given:

The price of the article in 1999 is 7200.

Profit and loss:

⇒ Price of the article in 1999 = 7200

⇒ Price of the article in 1998$=\mathrm{Price}\times \left(\frac{100}{100-\mathrm{loss}}\right)$

= (7200) × $\left(\frac{100}{90}\right)$ = 8000

⇒ Price of the article in 1997 = (8000) × $\left(\frac{100}{125}\right)$ = 6400

⇒ Price of article in 1996 = (6400) × $\left(\frac{3}{2}\right)$ = 9600

∴ The required result will be 9600.

Given:

The speed of a faster train doubles the speed of a slower train.

As we know,

Speed = \(\frac{Distance} {Time}\)

Let be assume the speed of the faster train is 2x and the slower train is x.

Therefore,

\(\frac{(100+100)}{(2 x+x)}=12\)

⇒ x = \(\frac{200}{36}\)m/s

= 20 kmph

Speed of faster train = 2x

= 2 × 20

= 40

∴ The required result will be 40 kmph.

Given

Total students = 550

Ratio of boys and girls = 6 : 5

Number of boys =$550\times \left(\frac{6}{11}\right)=300$

Number of girls = 550 ×$\left(\frac{5}{11}\right)$ = 250

Let the number of girls joining the class be x.

Now,

$\frac{300}{(250+x)}=\frac{5}{6}$

⇒ 250 + x = 300 ×$\left(\frac{6}{5}\right)$

⇒ 250 + x = 360

⇒ x = 360 - 250 = 110

∴ The required girls is 110.

Given:

Total sum in the bank = Rs. 25,000

Simple interest on one part for 5 years @ 6% p.a. = Simple interest on second part for 4 years @ 5% p.a.

Formula:

Simple interest =$\frac{\mathrm{Principal}\times \mathrm{Time}\times \mathrm{Rate}\mathrm{of}\mathrm{interest}}{100}$

Let assume that one part of sum = x

And second part = y

According to the formula,

Interest on one part = x × 5 × 6% =$\frac{30\mathrm{x}}{100}$

Interest on second part = y × 4 × 5% =$\frac{20\mathrm{y}}{100}$

According to question,

Interest on one part = Interest on second part

$\Rightarrow \frac{30\mathrm{x}}{100}=\frac{20\mathrm{y}}{100}$

$\Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{2}{3}$

⇒ x ∶ y = 2 ∶ 3

⇒ Rs. 25,000 is divided in 2 ∶ 3.

Second part of the sum lent = Rs. 25,000 × $\frac{3}{5}$= Rs. 15,000

∴ The second part of the sum lent is Rs. 15,000.