Mathematics Test - 10

In an Linear programming problem, the restrictions or limitations under which the objective function is to be optimised are called Constraints.

Constraints:

• These are the restrictions on the variables of an linear programming problem are called as linear constraints.
• The final solution of the objective function must satisfy these constraints.

The number of way to choose a sample of relements from a set of n distinct objects where order does not matter is \(={ }^{n} C_{r}=\frac{n !}{(n-r) ! r !}\).

Given: 5 Indian couples and 5 American couples.

There are 10 males and 10 females present in the party.

Total handshakes \(={ }^{20} C_{2}\)

\(\Rightarrow \frac{20 !}{(20-2) ! 2 !}\)

\(\Rightarrow \frac{20 !}{18 ! 2 !}=\frac{20 \times 19 \times 18 !}{18 ! \times 2}=190\)

Indian women handshakes with male \(={ }^{5} C_{1} \times{ }^{10} C_{1}\)

\(\Rightarrow \frac{5 !}{(5-1) ! 1 !} \times \frac{10 !}{(10-1) ! 1 !}\)

\(\Rightarrow \frac{5 !}{4 !} \times \frac{10 !}{9 !}=\frac{5 \times 4 !}{4 !} \times \frac{10 \times 9 !}{9 !}=5 \times 10=50\)

It includes the handshakes with her husband too.

American women handshakes with their own husband= 5 ways

So, required handshakes \(=190-50-5=135\) ways

Since 135 is divided by 5 and 3.

The distance between the parallel lines ax + by + c₁ and ax + by + c₂ is:

\(D=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

Calculation:

The 2 given lines are:

3y + 4x - 12 = 0

3y + 4x - 7 = 0

a = 4, b = 3, c₁ = -12 and c₂ = -7

∴ The distance between the lines

\(D=\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\)

\(\Rightarrow D=\left|\frac{-12-(-7)}{\sqrt{4^{2}+3^{2}}}\right|\)

\(\Rightarrow D=\left|\frac{-5}{5}\right|=1\)

Consider \(P (n)=7^{n}-3^{n}\)

\(P(1)= 7^{1}-3^{1}=4\)

\(4\) is divisible by \(4\).

\(P(2)= 7^{2}-3^{2}\)

\(=49-9=40\)

\(40\)is divisible by \(4\).

\(P(3)= 7^{3}-3^{3}\)

\(=343-27=316\)

\(316\)is divisible by \(4\).

So, \(P(n)= 7^{n}-3^{n}\) is divisible by \(4\).

Given:

\(\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{20^{\circ}+100^{\circ}}{2} \cos \frac{100^{\circ}-20^{\circ}}{2}+\cos 140^{\circ}\)

\(=2 \cos 60^{\circ} \cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \times \frac{1}{2} \cos 40^{\circ}+\cos 140^{\circ}\)

\(=\cos 40^{\circ}+\cos 140^{\circ}\)

\(=2 \cos \frac{140^{\circ}+40^{\circ}}{2} \cos \frac{140^{\circ}-40^{\circ}}{2}\)

\(=2 \cos 90^{\circ} \cos 50^{\circ}\)

\(=0 \quad\left(\because \cos 90^{\circ}=0\right)\)

Concept:

In mathematical logic, for two statements p and q: ~(p ∧ q) = (~p) ∨ (~q), where the logical operator '~' stands for negation. It is read as "not (p and q) is equal to (not p) or (not q)".

Calculation:

The logical equivalent of "72 is divisible by 2 and 3" is "72 is divisible by 2 ∧ 72 is divisible by 3".

Since, ~(p ∧ q) = (~p) ∨ (~q), we can say that:

~(72 is divisible by 2 ∧ 72 is divisible y 3)

= ~(72 is divisible by 2) ∨ ~(72 is divisible y 3)

= 72 is not divisible by 2 or 72 is not divisible by 3.

As \( \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \)

\(\Rightarrow \frac{1+\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}+\frac{1+\cos 2 \gamma}{2}=1 \)

\( \Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=-1\)

When two dice are throw, then total outcome = 36

A doubled: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Favourable outcome = 6

Sum is 10: {(4, 6), (5, 5), (6, 4)}

Favourable outcome = 3

Again, A doubled and sum is 10: (5, 5)

Favourable outcome = 1

Now, P(either doubled or a sum of 10 appears) = P(A doubled appear) + P(sum is 10) – P(A doubled appear and sum is 10)

⇒ P(either doubled or a sum of 10 appears) = \(\frac{6}{36} + \frac{3}{36} – \frac{1}{36}\)

= \(\frac{6 + 3 – 1}{36}\)

= \(\frac{8}{36}\)

= \(\frac{2}{9}\)

So, P(neither doubled nor a sum of 10 appears) = 1 – \(\frac{2}{9}\)

= \(\frac{7}{9}\)