Mathematical Skills Test - 2

Given:

\(\frac{\sin 7 x+6 \sin 5 x+17 \sin 3 x+12 \sin x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=\frac{\sin 7 x+\sin 5 x+5 \sin 5 x+5 \sin 3 x+12 \sin 3 x+12 \sin x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

As we know, 

\(\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)\)

\(=\frac{2 \sin 6 x \cos x+10 \sin 4 x \cos x+24 \sin 2 x \cos x}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=\frac{2 \cos x(\sin 6 x+5 \sin 4 x+12 \sin 2 x)}{\sin 6 x+5 \sin 4 x+12 \sin 2 x}\)

\(=2 \cos x\)

Water in original mixture \(=70 \%\) of \(50=35 \mathrm{gm}\)

Milk in original mixture \(=(100-70 \%)=30 \%\)

Milk in original mixture \(=50-35=15 \mathrm{gm}\)

Milk in new mixture \(=\frac{15}{50+10}=\frac{15}{60}=\frac{5}{20} \times 100=25 \%\)

Water in new mixture \(=(100-25) \%=75 \%\)

Thus, If \(10 \mathrm{gm}\) of water is added then in new mixture contains \(75 \%\) of water in it.

For 1st Six months A, B, C all three have equal share of Rs. 10000

⇒ As after 6 months A & C Withdraw Rs. 5000/- each from the business hence for next 6 months the Share in business ⇒ A = 5000, B = 10000, C = 5000 Respectively

So, the ratio Becomes:

⇒ A : B : C = (10000 × 6 + 5000 × 6) : (10000 × 12) : (10000 × 6 + 5000 × 6)

⇒ A : B : C = 90000 : 120000 : 90000 = 3 : 4 : 3

⇒ A's Share in Profit $=\frac{3}{3+4+3}\times 100000$

$=\frac{3}{10}\times 100000$

$=30,000$

 ⇒ B's Share in Profit $=\frac{4}{3+4+3}\times 100000$

$=\frac{4}{10}\times 100000$

$=40,000$

⇒ C's Share in Profit $=\frac{3}{3+4+3}\times 100000$

$=\frac{3}{10}\times 100000$

$=30,000$

So, profit of A and C = 60000

We know that there are \(366\) days in a leap year.

\(\Rightarrow 52\) weeks \(+2\) days

These \(2\) days can \(\mathrm{b}\):

(a) Monday, Tuesday

(b) Tuesday, Wednesday,

(c) Wednesday, Thursday

(d) Thursday, Friday

(e) Friday, Saturday

(f) Saturday, Sunday

(g) Sunday, Monday.

\(\therefore\) Total sample space \(\mathrm{n}(\mathrm{s})=7\)

\(\Rightarrow\) Let \(\mathrm{A}\) be the event that a leap year contains \(53\) Fridays.

\(\mathrm{n}(\mathrm{A})=\{\) Thursday, Friday \(\},\{\) Friday, Saturday\(\}\)

\(=2 \)

Required probability \(\mathrm {p(B)=\frac{n(B)} {n(S)}}\)

\(=\frac{2}{7}\)

Given:

Total employees \(=50\)

Time is taken by 50 employees to complete work \(=23\) days

After every 5 days, 5 employees more join them.

Total work is always equal to the product of time and number of people.

Total employees \(=50\)

Time is taken by 50 employees to complete work \(=23\) days

Then total work \(=50 \times 23=1150\) units

So, work that done in the first 5 days by 50 employees \(=5 \times 50=250\) units

After every 5 days, 5 employees more join them.

After 5 days total employees \(=50+5=55\)

So, work that done in the next 5 days by 45 employees \(=5 \times 55=275\) units

Again after 5 days total employees \(=55+5=60\)

So, work that done in next 5 days by 60 employees \(=5 \times 60=300\) units

Again after 5 days total employees \(=60+5=65\)

So, work that done in the next 5 days by 65 employees \(=5 \times 65=325\) units

Total work that done \(=250+275+300+325=1150\) units

Thus, the total work is done.

Now, time taken by employees to complete work \(=5+5+5+5=20\) days

\(\therefore\) In 20 days the work will be completed.

If the tank is filled in 12 hours and emptied in 15 hours.

LCM of 12 and 15 = 60

So, we let the tank capacity is 60 ltr.

So, tap A can fill it in 12 hours

$\therefore$ Tap A can fill per hours = $\frac{60}{12}$ = 5 ltr

Tap B can emptied it in 15 hours-

$\therefore$ Tap B can empty per hours = $\frac{60}{15}$ = 4 ltr

Tank filled after 2 hour = 5 - 4 = 1 ltr

(If taps open alternatively)

In last hour tap A filled tank 5 ltr

So, rest part of tank need to be filled is = 60 - 5 = 55 ltr

In 2 hours tank filled = 1 ltr

So, 55 ltr to be filled in = 55 × 2 = 110

Therefore, total time = 110 + 1 = 111 hours

Given: 

\(A(0,0), B(0,10), C(8,2)\) and \(D(8,7)\) are the vertices of a quadilateral \(A B C D\).

Here, we have to find the area of quadilateral \(\mathrm{ABCD}\)

Area of quadilateral \(\mathrm{ABCD}=\) Area of \(\triangle \mathrm{ABC}+\) Area of \(\triangle \mathrm{ACD}\)

Let's find out the area of \(\triangle \mathrm{ABC}\).

\(\because \mathrm{A}(0,0), \mathrm{B}(0,10), \mathrm{C}(8,2)\) are the vertices of \(\triangle \mathrm{ABC}\)

As we know that, if \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) and \(\mathrm{C}\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)\) be the vertices of a \(\triangle \mathrm{ABC}\), then area of \(\triangle\)

\(\mathrm{ABC}=\frac{1}{2}\cdot \left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|\)

Area of \(\triangle \mathrm{ABC}=\frac{1}{2} \cdot\left|\begin{array}{ccc}0 & 0 & 1 \\0 & 10 & 1 \\8 & 2 & 1\end{array}\right|\)

Area of \(\triangle \mathrm{ABC}=40\) sq. units

Similarly, let's find out the area of \(\triangle \mathrm{ACD}\)

\(\because \mathrm{A}(0,0), \mathrm{C}(8,2)\) and \(\mathrm{D}(8,7)\)

Area of \(\triangle \mathrm{ACD}=\frac{1}{2} \cdot\left|\begin{array}{lll}0 & 0 & 1 \\8 & 2 & 1 \\8 & 7 & 1\end{array}\right|\)

Area of \(\triangle \mathrm{ACD}=20\) sq. units

Area of quadilateral \(\mathrm{ABCD}=\) Area of \(\triangle \mathrm{ABC}+\) Area of \(\triangle \mathrm{ACD}=[40+20]\) sq. units \(=60 \) sq. units

Given:

Pipe A alone can fill the tank = 8 hrs

Pipe B alone can fill the tank = 12 hrs

Time ratio of Pipe A and B = 8 ∶ 12 = 2 ∶ 3

Concept used:

Efficiency is inversely proportional to time.

Calculation:

Efficiency ratio of pipe A and B = 3 ∶ 2

Total work = 3 × 8 = 24

Both pipes opened alternately for 1 hr.

Total work in 2 hrs = 3 + 2 = 5

Work in 8 hrs = 5 × 4 = 20

Work in next 1 hr (9th) = 20 + 3 = 23

Remaining work = 24 - 23 = 1

B alone can work in 1 hr = 2

B alone can work in \(\frac{1}{2}\) hr = 1

So, the total work complete in \(9 \frac{1}{2}\) hrs.