Mix Test 2

Given that a and b are prime numbers.

Therefore, HCF (a, b) = 1.

We know that for any two positive integers a and b, HCF (a, b) \(\times\) LCM (a, b) = a \(\times\) b.

Therefore, LCM (a, b) = \(\frac{ab}{HCF(a,b)}\) = \(\frac{ab}{1}\) = ab.

Hence, if a and b are prime numbers then LCM (a, b) = a \(\times\) b.

Given system of equations are

2x + 5y = 17,

5x + 3y = 14.

Comparing given system of equations with

\(a_1\mathrm x+b_1y=c_1\)

\(a_2\mathrm x+b_2y=c_2\)

We get \(a_1=2,\) \(b_1=5,\) \(c_1=17\)

And \(a_2=5,b_2=3,c_2=14\)

Now, \(\frac{a_1}{a_2}=\frac{2}{5}\) and \(\frac{b_1}{b_2}=\frac{5}{3}\ne \frac{2}{5}=\frac{a_1}{a_2}\)

Since, \(\frac{a_1}{a_2}\ne \frac{b_1}{b_2}\)

Therefore, given system of equations has a unique solution.

Let the radius of the circle is r cm.

Given that the circumference of the circle is \(2\pi r\) = 8 cm.

\(\therefore\) r = \(\frac{8}{2\pi}\) = \(\frac{4}{\frac{22}{7}}\) = \(\frac{14}{11}cm.\)

\(\because\) Area of the circle is \(\pi r^2\)

\(\therefore\) Area of the sector whose central angle is 360° = \(\pi r^2\,cm^2\).

\(\therefore\) Area of the sector whose central angle is 1° = \(\frac{\pi r^2}{360^o}cm^2\)

\(\therefore\) Area of the sector whose central angle is 72° = \(\frac{\pi r^2}{360^o}\times 72^o cm^2\)

= \(\frac{22}{7}\times \frac{14}{11}\times \frac{14}{11}\times \frac{72}{360}\) = \(2\times 2\times \frac{14}{11}\times \frac{1}{5}\) = \(\frac{56}{55}\) = \(1.02cm^2\)

Hence, the area of the sector whose central angel is 72° is \(1.02cm^2\).

A die is thrown once.

Therefore, possible outcomes are {1, 2, 3, 4, 5, 6}.

Hence, total possible outcomes = n(S) = 6.

Let the event E be the event of getting an even prime number.

Since, only even prime number is 2.

Therefore, number of outcomes favorable to event E is n(E) = 1.

Hence, the probability of getting an even prime number = \(\frac{n(E)}{n(S)}=\frac{1}{6}\)

If the edge of the cube is a then, the volume of the cube is given by = \(a^3\).