Mathematical Skills Test - 5

Given: The dice is thrice as likely to show an odd number than an even number.

Let the odd number be event A and the even number be event B.

P(A) = 3 P(B)

let P(B) = p ⇒ P(A) = 3p

Now p + 3p = 1

p =$\frac{1}{4}$

When thrown twice, the sum of two numbers thrown is even when both are odd or both are even, i.e.

Possible chances are:

1) Even in 1st throw and Even in 2nd throw =$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

2) Odd in 1st throw and Odd in 2nd throw =$\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

∴ The total probability will be:

$\mathrm{P}=\frac{1}{16}+\frac{9}{16}$

P = 1/16 + 9/16 = 5/8

We have,\(\frac{\cos 7 \mathrm{x}+\cos 6 \mathrm{x}+\cos 5 \mathrm{x}}{\sin 7 \mathrm{x}+\sin 6 \mathrm{x}+\sin 5 \mathrm{x}}\)

We know that,

\(\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

\(\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

Using the above formula, we get

\(=\frac{(\cos 7 x+\cos 5 x)+\cos 6 x}{(\sin 7 x+\sin 5 x)+\sin 6 x}\)

\(=\frac{2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)+\cos 6 x}{2 \sin \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)+\sin 6 x}\)

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

\(=\frac{\cos 6 x(2 \cos x+1)}{\sin 6 x(2 \cos x+1)}\)

= cot 6x

Given:

A shopkeeper sells his items using 20% less than the true weight

Profit earns by shopkeeper = 30%

Let the true weight = 1000 gm

Sells by shopkeeper = 20% less than the true weight

= (100 – 20)% of 1000

= 80% × 1000

= 800 gm

Let the cost price per gram = Rs. 1

∴ The cost price of 800 gm = Rs. 800

Profit earns by shopkeeper = 30%

∴ Selling price = 30% profit on 1000 = (100 + 30)% of 1000

⇒ 130% of 1000 = 1300

So, the profit of the shopkeeper = 1300 – 800 = 500

∴ Profit% =$\frac{500}{800}$ × 100 = 62.5%

∴ Profit earns by the shopkeeper = 62.5%

There are 5 white (W) socks, 4 red (R) socks and 3 green (G) socks

Total socks = 5 + 4 + 3 = 12

Probability of 1st sock white =$\frac{5}{12}$

Probability of 2nd sock white =$\frac{4}{11}$

∴ Probability of two white socks =$\frac{5}{12}\times \frac{4}{11}=\frac{5}{33}$

Similarly, Probability of two red =$\frac{4}{12}\times \frac{3}{11}=\frac{1}{11}$

Probability of two green =$\frac{3}{12}\times \frac{2}{11}=\frac{1}{22}$

We want socks of same color

i.e., 2 white or 2 red or 2 green

∴ Total probability =$\frac{5}{33}+\frac{1}{11}+\frac{1}{22}=\frac{1}{11}\left(\frac{5}{3}+1+\frac{1}{2}\right)$

$=111\left(\frac{19}{6}\right)=\frac{19}{66}$

A polygon of 12 sides has 12 vertices.

By joining any two of the vertices, we obtain either a side or a diagonal of the polygon.

Number of all straight lines obtained by joining 2 vertices at a time\(={ }^{12} \mathrm{C}_{2}=\frac{12 \times 11}{2 \times 1}\)

= 66

These straight lines include 12 sides of the polygon.

∴ The number of diagonals of the polygon = 66 - 12

= 54

Given:

Time taken by pipe A to fill the tank = 40 minute.

Time taken by pipe B to fill the tank = 80 minute.

Total time to fill the tank = 30 min.

Total work is LCM of 40 minute and 80 minute,

Total work = 80 unit

⇒ Work done by A in 1 minute = \(\frac{80}{40}\) = 2 unit,

⇒ Work done by B in 1 minute = \(\frac{80}{80}\)= 1 unit.

Given that tank is filled in 30 minute,

According to the question,

A filled the tank till 30 minute.

Since in 1 minute, A does 2 unit of work,

⇒ In 30 minutes A can do 60 units of work.

Remaining work = 80 unit – 60 unit = 20 unit.

Since in 1 minute, B does 1 unit of work,

⇒ For 20 unit work, time taken by B = 20 × 1 minute = 20 minute

∴ Tank B was open for 20 minutes.

Given circle:x² + y² + 2x + 2y - 14 = 0

⇒ (x² + 2x +1) + (y² + 2y +1) -16 = 0

⇒ (x + 1)² + (y + 1)² = 16 = 42

∴ center = (-1, -1) and radius = 4

Intercept on x-axis is when y = 0

⇒ (x + 1)² + (0 + 1)² = 16

⇒ (x + 1) = \(\sqrt{15}\)

⇒ x = \(\sqrt{15}\) - 1

Intercept on y-axis is when x = 0

⇒ (0 + 1)² + (y + 1)² = 16

⇒ (y + 1) = \(\sqrt{15}\)

⇒ y = \(\sqrt{15}\) - 1

Rate of flow of water in each pipe \(=2^{2}:\left(\frac{4}{3}\right)^{2}: 3^{2}=36: 16: 81\)

Part of the cistern filled by pipe A in 1 min = 36k

Part of the cistern filled by pipe B in 1 min = 16k

And that filled by C in 1 min = 81k

The largest pipe fills the tank in 133 min.

Hence, cistern filled by C in 133 min = 133 \(\times\) 81k = 1

\(k=\frac{1}{(133 \times 81)}\)

Now part of the cistern filled by all the three in 1 min = 36k + 16k + 81k = 133k \(= \frac{133}{(133 \times 81)} = \frac{1}{81}\)

So, all the three can fill the cistern in 81 min.

Given:

Selling price = cost price – 25% of cost price

Let the cost price for 1000 gram = Rs. 1000

The selling price of 1000 gram = Rs. 750

But, due to false weighing,

The selling price of 600 gram = Rs. 750

Cost price of 600 gram = Rs. 600

So, there is a profit

Profit = Rs. 750 – Rs. 600

= Rs. 150

Profit% =$\left[\frac{\mathrm{Profit}}{\mathrm{Cost}\mathrm{price}}\times 100\right]\%$

=$\left[\left(\frac{150}{600}\right)\times 100\right]$%

= 25%

Given:

Original income = Rs. 50,000; original expenditure = Rs. 25,000

Percentage increase in income = 20%; percentage increase in expenditure = 10%

Formula:

Percentage of increase in his savings =$\frac{\mathrm{Change}\mathrm{in}\mathrm{saving}}{\mathrm{Original}\mathrm{saving}}\times 100$

Calculation:

Original saving = Rs. (50,000 – 25,000) ⇒ Rs. 25,000

New income = Original income + Increase income

= Rs. 50,000 + 20 % of 50,000

= Rs. 60,000

New expenditure = Original expenditure + Increase expenditure

= Rs. 25,000 + 10% of Rs. 25,000

= Rs. 27,500

New saving = New income – New expenditure

= Rs. (60,000 – 27,500)

= Rs. 32,500

Change in saving = New saving – Original saving

= Rs. (32,500 – 25,000)

= Rs. 7,500

Percentage of increase in his saving =$\frac{7,500}{25,000}$ × 100

= 30%